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Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II

Students can Download Accountancy Chapter 13 Final Accounts of Sole Proprietors – II Questions and Answers, Notes Pdf, Samacheer Kalvi 11th Accountancy Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II

Samacheer Kalvi 11th Accountancy Final Accounts of Sole Proprietors – II Text Book Back Questions and Answers

I. Multiple Choice Questions
Choose the Correct Answer

Question 1.
A prepayment of insurance premium will appear in ………………
(a) The trading account on the debit side
(b) The profit and loss account on the credit side
(c) The balance sheet on the assets side
(d) The balance sheet on the liabilities side
Answer:
(c) The balance sheet on the assets side

Question 2.
Net profit is ………………
(a) Debited to capital account
(b) Credited to capital account
(c) Debited to drawings account
(d) Credited to drawings account
Answer:
(b) Credited to capital account

Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II

Question 3.
Closing stock is valued at ………………
(a) Cost price
(b) Market price
(c) Cost price or market price whichever is higher
(d) Cost price or net realisable value whichever is lower
Answer:
(d) Cost price or net realisable value whichever is lower

Question 4.
Accrued interest on investment will be shown ………………
(a) On the credit side of profit and loss account
(b) On the assets side of balance sheet
(c) Both (a) and (b)
(d) None of these
Answer:
(c) Both (a) and (b)

Question 5.
If there is no existing provision for doubtful debts, provision created for doubtful debts is ………………
(a) Debited to bad debts account
(b) Debited to sundry debtors account
(c) Credited to bad debts account
(d) Debited to profit and loss account
Answer:
(d) Debited to profit and loss account

II. Very Short Answer Questions

Question 1.
What are adjusting entries?
Answer:
Adjustment entries are the journal entries made at the end of the accounting period to account for items which are omitted in trial balance and to make adjustments for outstanding and prepaid expenses and revenues accrued and received in advance.

Question 2.
What is outstanding expense?
Answer:
Expenses which have been incurred in the accounting period but not paid till the end of the accounting period are called outstanding expenses.

Question 3.
What is prepaid expense?
Answer:
Prepaid expenses refer to any expense or portion of expense paid in the current accounting year but the benefit or services of which will be received in the next accounting period. They are also called as unexpired expenses.

Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II

Question 4.
What are accrued incomes?
Answer:
Accrued income is an income or portion of income which has been earned during the current accounting year but not received till the end of that accounting year.

Question 5.
What is provision for discount on debtors?
Answer:
Cash discount is allowed by the suppliers to customers for prompt payment of amount due either on or before the due date. A provision created on sundry debtors for allowing such discount is called provision for discount on debtors.

III. Short Answer Questions

Question 1.
What is the need for preparing final accounts?
Answer:

  1. To record omissions in trial balance such as closing stock, interest on captial, interest on drawings, etc.
  2. To bring into account outstanding and prepaid expenses.
  3. To bring into account income accrued and received in advance.
  4. To create reserves and provisions.

Question 2.
What is meant by provision for doubtful debts? Why is it created?
Answer:
Provision for bad and doubtful debts refers to amount set aside as a charge against profit to meet any loss arising due to bad debt in future. The amount of doubtful debts is calculated on the basis of some percentage on debtors at the end of the accounting period after deducting further bad debts (if any). A provision for doubtful debts is created and is charged to profit and loss account.

Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II

Question 3.
Explain how closing stock is treated in final accounts?
Answer:
The unsold goods in the business at the end of the accounting period are termed as closing stock. As per As-2 (Revised), the stock is valued at cost price or net realisable value, whichever is lower.

Presentation in final accounts:

  1. In the trading account: Shown on the credit side.
  2. In the balance sheet: Shown on the assets side under current assets.

Question 4.
Give the adjusting entries for interest on capital and interest on drawings.
Answer:
Adjusting Entry: Interest on Capital
Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II
Adjusting Entry: Interest on Drawings
Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II

Question 5.
Explain the accounting treatment of bad debts, provision for doubtful debts and provision for discount on debtors.
Answer:

  1. Bad Debts: When it is definitely known that amount due from a customer (debtor) to whom goods were sold on credit, cannot be realised at all, it is treated as bad debts.
  2. Provision for bad and doubtful debts refers to amount set aside as a charge against profit to meet any loss arising due to bad debt in future.
  3. Cash discount is allowed by the suppliers to customers for prompt payment of amount due either on or before the due date.

IV. Exercises

Question 1.
Pass adjusting entries for the following:
(a) The closing stock was valued at ₹ 5,000
(b) Outstanding salaries ₹ 150
(c) Insurance prepaid ₹ 450
(d) ₹ 20,000 was received in advance for commission.
(e) Accrued interest on investments is ₹ 1,000.
Answer:
Adjusting Entries
Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II

Question 2.
For the fol owing adjustments, pass adjusting entries:
(a) Outstanding wages ₹ 5,000.
(b) Depreciate machinery by ₹ 1,000.
(c) Interest on capital @ 5% (Capital: ₹ 20,000)
(d) Interest on drawings ₹ 50
(e) Write off bad debts ₹ 500
Answer:
Adjusting Entries
Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II

Question 3.
On preparing final accounts of Suresh, bad debt account has a balance of ₹ 800 and sundry debtors account has a balance of ₹ 16,000 of which ₹ 1,200 is to be written off as further bad debts. Pass adjusting entry for bad debts. And also show how it would appear in profit and loss account and balance sheet.
Answer:
Adjusting Entry
Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II
Profit and Loss Account
Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II
Balance Sheet
Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II

Question 4.
The trial balance on March 31, 2016 shows the following:
Sundry debtors ₹ 30,000; Bad debts ₹ 1,200
It is found that 3% of sundry debtors is doubtful of recovery and is to be provided for. Pass journal entry for the amount of provision and also show how it would appear in the profit and loss account and balance sheet.
Answer:
Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II
Profit and Loss Account for the year ended 31.03.2016
Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II
Balance Sheet as on 31.03.2016
Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II

Question 5.
The trial balance of a trader on 31st December, 2016 shows debtors as ₹ 50,000.
Adjustments:
(a) Write off ₹ 1,000 as bad debts
(b) Provide 5% for doubtful debts
(c) Provide 2% for discount on debtors
Show how these items will appear in the profit and loss A/c and balance sheet of the trader.
Answer:
Profit and Loss Account for the year ended 31st December, 2016
Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II
Balance Sheet as on 31st December, 2016
Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II

Question 6.
On 1st January, 2016, provision for doubtful debts account had a balance of ₹ 3,000. On December 31, 2016, sundry debtors amounted to ₹ 80,000. During the year, bad debts to be written off were ₹ 2,000. A provision for 5% was required for next year. Pass journal entries and show how these items would appear in the final accounts.
Answer:
Adjusting Entries
Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II
Profit and Loss Account for the year ended 31st December, 2016
Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II
Balance Sheet as on 31.12.2016
Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II

Question 7.
The following are the extracts from the trial balance.
Sundry debtors ₹ 30,000; Bad debts ₹ 5,000 Additional information:
(a) Write off further bad debts ₹ 3,000.
(b) Create 10% provision for bad and doubtful debts.
You are required to pass necessary adjusting entries and show how these items will appear in profit and loss account and balance sheet.
Answer:
Adjusting Entries
Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II
Profit and Loss Account
Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II
Balance Sheet
Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II

Question 8.
The following are the extracts from the trial balance.
Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II
Additional information:
(a) Additional bad debts ₹ 3,000.
(b) Keep a provision for bad and doubtful debts @ 10% on sundry debtors.
You are required to pass necessary adjusting entries and show how these items will appear in profit and loss account and balance sheet.
Answer:
Adjusting Entries
Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II
Profit and Loss Account
Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II
Balance Sheet
Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II

Question 9.
The accounts of Lakshmi traders showed the following balance on 31st March, 2016.
Sundry debtors ₹ 60,000; Bad debts ₹ 2,000
Provision for doubtful debts ₹ 4,200
At the time of preparation of final accounts on 31st March, it was found that out of sundry debtors, ₹ 1,000 will be irrecoverable. It was decided to create a provision of 2% on debtors to meet any future possible bad debts.
Pass necessary journal entries and show how these items would appear in the final accounts.
Answer:
Adjusting Entries
Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II
Profit and Loss Account for the year ended 31.03.2016
Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II
Balance Sheet as on 31.03.2016
Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II

Question 10.
Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II
The following are the extracts from the trial balance.
Additional information:
(a) Create a provision for doubtful debts @ 10% on sundry debtors.
(b) Create a provision for discount on debtors @ 5% on sundry debtors.
You are required to pass necessary adjusting entries and show how these items will appear in the final accounts.
Answer:
Adjusting Entries
Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II
Profit and Loss Account
Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II
Balance Sheet
Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II

Question 11.
Following are the extracts from the trial balance.
Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II
Additional information:
(a) Additional bad debts 1,000
(b) Create a provision for doubtful debts @ 5% on sundry debtors.
(c) Create a provision for discount on debtors @ 2% on sundry debtors.
You are required to pass necessary journal entries and show how these items will appear in the final accounts.
Answer:
Adjusting Entries
Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II
Profit and Loss Account
Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II
Balance Sheet
Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II

Question 12.
The following are the extracts from the trial balance.
Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II
Answer:
Profit and Loss Account
Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II
Balance Sheet
Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II

Question 13.
Prepare trading account of Archana for the year ending 31st December, 2106 from the following information.
Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II
Adjustments:
(a) Closing stock ₹ 1,00,000
(b) Wages outstanding ₹ 12,000
(c) Freight inwards paid in advance ₹ 5,000
Answer:
Trading A/c of Archana for the year ended 31.12.2016
Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II

Question 14.
Prepare profit and loss account of Manoj for the year ending on 31st March, 2016
Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II
Adjustments:
(a) Salary outstanding ₹ 400
(b) Rent paid in advance ₹ 50
(c) Commission receivable ₹ 100
Answer:
Profit and Loss A/c of Manoj for the year ended 31.03.2016
Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II

Question 15.
From the trial balance of Sumathi and the adjustments prepare the trading and profit and loss account for the year ended 31st March, 2016, and a balance sheet as on that date.
Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II
Adjustments
(a) Six months interest on loan is outstanding.
(b) Two months rent is due from tenant, the monthly rent being ₹ 25.
(c) Salary for the month of March 2016, ₹ 75 is unpaid.
(d) Stock in hand on March 31, 2016 was valued at ₹ 1,030.
Answer:
Trading and Profit & Loss A/c of Sumathi
for the year ended 31st March, 2016
Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II
Balance Sheet of Sumathi as on 31.03.2016
Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II

Question 16.
The following trial balance was extracted from the books of Arun Traders as on 31st March, 2018.
Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II
Answer:
Prepare trading and profit and loss account for the year ending 31st March, 2018 and balance sheet as on that date after considering the following:
(a) Depreciate Plant and machinery @ 20%
(b) Wages outstanding amounts to ₹ 750.
(c) Half of repairs and maintenance paid is for the next year.
(d) Closing stock was valued at ₹ 15,000.
Answer:
Trading and Profit & Loss A/c of Arun Traders for the year ended 31.03.2018
Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II
Balance Sheet of Arun Traders as on 31.03.2018
Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II

Question 17.
The following is the trial balance of Babu as on 31st December, 2016.
Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II
Prepare trading and profit and loss account for the year ended 31st December, 2016 and a balance sheet as on that date after the following adjustments.
(a) Salaries outstanding ₹ 500
(b) Interest on investments receivable at 10%.
(c) Provision required for bad debts is 5%.
(d) Closing stock is valued at ₹ 9,900.
Answer:
Trading and Profit & Loss A/c for the year ended 31.12.2016
Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II
Balance Sheet as on 31.12.2016
Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II

Question 18.
From the following trial balance of Ramesh as on 31st March, 2017, prepare the trading and profit and loss account and the balance sheet as on that date.
Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II
Answer:
Adjustments:
Closing stock was valued at ₹ 35,000
(b) Unexpired advertising ₹ 250
(c) Provision for bad and doubtful debts is to be increased to ₹ 3,000
(d) Provide 2% for discount on debtors.
Answer:
Trading and Profit & Loss A/c of Ramesh for the year ended 31.03.2017
Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II
Balance Sheet of Ramesh as on 31.03.2017
Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II

Question 19.
Following are the ledger balances of Devi as on 31st December, 2016.
Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II
Prepare trading and profit and loss account for the year ended 31st December, 2016 and balance sheet as on that date.
(a) Stock on 31st December, 2016 ₹ 5,800.
(b) Write off bad debts ₹ 500.
(c) Make a provision for bad debts @ 5%.
(d) Provide for discount on debtors @ 2%.
Answer:
Trading and Profit & Loss A/c of Devi for the year ended 31.12.2016
Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II
Balance Sheet of Devi as on 31.12.2016
Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II

Question 20.
From the following trial balance of Mohan for the year ended 31st March, 2017 and additional information, prepare trading and profit and loss account and balance sheet.
Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II
Additional information:
(a) Closing stock is valued at ₹ 15,500
(b) Write off ₹ 500 as bad debts and create a provision for bad debts @ 10% on debtors.
(c) Depreciation @ 10% required
Answer:
Trading and Profit and Loss A/c of Mohan for the year ended 31.03.2017
Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II
Balance Sheet of Mohan as on 31.03.2017
Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II

Question 21.
From the following trial balance of Subramaniam, prepare his trading and profit and loss account and balance sheet as on 31st December, 2016.
Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II
Take into account the following adjustments:
(a) Charge interest on drawings at 8%.
(b) Outstanding salaries ₹ 3,000
(c) Closing stock was valued at ₹ 48,000
(d) Provide for 5% interest on capital.
Answer:
Trading and Profit & Loss A/c of Subramaniam for the year ended 31.12.2016
Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II
Balance Sheet of Subramaniam as on 31.12.2016
Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II

Question 22.
Prepare trading and profit and loss account and balance sheet from the following trial balance of Madan as on 31st March, 2018.
Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II
Adjustments:
(a) The closing stock was ₹ 80,000
(b) Provide depreciation on plant and machinery @ 20%
(c) Write off ₹ 800 as further bad debts
(d) Provide the doubtful debts @ 5% on sundry debtors
Answer:
Trading and Profit & Loss A/c for the year ended 31.03.2018
Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II
Balance Sheet as on 31.03.2018
Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II

Question 23.
From the following information prepare trading and profit and loss account and balance sheet of Kumar for the year ending 31st December, 2017.
Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II
Adjustments:
(a) The closing stock on 31st December, 2017 was valued at ₹ 3,900.
(b) Carriage inwards prepaid ₹ 250
(c) Rent received in advance ₹ 100
(d) Manager is entitled to receive commission @ 5% of net profit after providing such commission.
Answer:
Trading and Profit & Loss Account of Kumar for the year ended 31.12.2017
Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II
Balance Sheet of Kumar as on 31.12.2017
Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II

Question 24.
From the following information, prepare trading and profit and loss account and balance sheet in the books of Sangeetha for the year ending 31st March, 2018.
Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II
Adjustments:
(a) Stock on 31st March, 2018 ₹ 14,200
(b) Income tax of Sangeetha paid ₹ 800
(c) Charge interest on drawings @ 12% p.a.
(d) Provide managerial remuneration @ 10% of net profit before charging such commission.
Answer:
Trading and Profit & Loss Account of Sangeetha for the year ended 31.03.2018
Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II
Balance Sheet of Sangeetha as on 31.03.2018
Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II

Textbook Case Study Solved

Question 1.
James is a trader who sells washing machines on credit. But, he does not remember the due date to collect the money from his debtors. Some of his customers do not pay on time. His cash inflow is becoming worse. As a result, he could not pay his telephone bill and rent at the end of the accounting period. Hence, he showed only the amount paid as expense. He has many washing machines unsold at the year end. He is worried about the performance of his business. So, he is planning to appoint a manager to take care of his business. The new manager insists James to apply the accounting principle of prudence and matching and also to allow cash discount.
Now, discuss on the following points:

Question 1.
Why does James sell on credit?
Answer:
James sells goods on credit to increase the sales volume and reduce the stock.

Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II

Question 2.
Are there any ways to encourage his debtors to make the payment on time?
Answer:
Yes, there are many ways to encourage his debtors to make the payment on time by way of cash discount and trade discount.

Question 3.
What might happen if the debtors do not pay?
Answer:
If the debtors do not pay, the bad debts will be increased in the business.

Question 4.
In what ways prudence and matching principles can be applied for the business of James?
Answer:
Prudence principle can be applied for the business here closing stock was valued on cost price or market price whichever is lower under the prudence principle. Matching principle can be applied here for revenue and expense.

Question 5.
What will be the impact on income statement and the balance sheet, if the outstanding expenses are not adjusted?
Answer:
Outstanding expenses to be added with the concerned expenditure in the income Statement and the outstanding expenses will be recorded in liabilities side.

Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II

Question 6.
On what basis the unsold washing machines should be valued?
Answer:
The unsold washing machine should be valued at cost price or market price, whichever is lower under prudence principle. Managerial commission can be given to motivate the new manager to retain him in the business of James.

Samacheer Kalvi 11th Accountancy Final Accounts of Sole Proprietors – II Additional Questions and Answers

I. Multiple Choice Questions (Other important questions)
Choose the correct answer

Question 1.
If closing stock is already adjusted, adjusted purchase account and ………………. stock will appear in trial balance.
(a) Opening
(b) Closing
(c) Average
(d) None of these
Answer:
(b) Closing

Question 2.
Outstanding expense account is a ………………. account.
(a) Nominal
(b) Real
(c) Representative personal
(d) Personal
Answer:
(c) Representative personal

Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II

Question 3.
When bad debts already appears in the trial balance, it is taken only to debit side of ………………. account.
(a) Profit and Loss
(b) Balance sheet
(c) Asset side
(d) None of these
Answer:
(a) Profit and Loss

Question 4.
Income tax paid by the business for the proprietor is treated as ……………….
(a) Expense
(b) Profit and Loss A/c
(c) Drawings
(d) None of these
Answer:
(c) Drawings

Question 5.
Commission on net profit after charging such commission:
Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II
Answer:
Samacheer Kalvi 11th Accountancy Solutions Chapter 13 Final Accounts of Sole Proprietors – II

Samacheer Kalvi 12th Bio Zoology Solutions Chapter 7 Human Health and Diseases

Students can Download Bio Zoology Chapter 7 Human Health and Diseases Questions and Answers, Notes Pdf, Samacheer Kalvi 12th Bio Zoology Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Bio Zoology Solutions Chapter 7 Human Health and Diseases

Samacheer Kalvi 12th Bio Zoology Human Health and Diseases Text Book Back Questions and Answers

Question 1.
A 30 year old woman has bleedy diarrhoea for the past 14 hours, which one of the following organisms is likely to cause this illness?
(a) Streptococcus pyogens
(b) Clostridium difficile
(c) Shigella dysenteriae
(d) Salmonella enteritidis
Answer:
(c) Shigella dysenteriae

Question 2.
Exo-erythrocytic schizogony of Plasmodium takes place in __________
(a) RBC
(b) Leucocytes
(c) Stomach
(d) Liver
Answer:
(d) Liver

Question 3.
The sporozoites of Plasmodium vivax are formed from __________
(a) Gametocytes
(b) Sporoblasts
(c) Oocysts
(d) Spores
Answer:
(c) Oocysts

Samacheer Kalvi 12th Bio Zoology Solutions Chapter 7 Human Health and Diseases

Question 4.
Amphetamines are stimulants of the CNS, whereas barbiturates are __________
(a) CNS stimulant
(b) both a and b
(c) hallucinogenic
(d) CNS depressants
Answer:
(d) CNS depressants

Question 5.
Choose the correctly match pair.
(a) Amphetamines – Stimulant
(b) LSD-Narcotic
(c) Heroin – Psychotropic
(d) Benzodiazepine – Pain killer
Answer:
(a) Amphetamines – Stimulant

Question 6.
The Athlete’s foot disease in human is caused by __________
(a) Bacteria
(b) Fungi
(c) Virus
(d) Protozoan
Answer:
(b) Fungi

Question 7.
Cirrhosis of liver is caused by chronic intake of __________
(a) Opium
(b) Alcohol
(c) Tobacco
(d) Cocaine
Answer:
(b) Alcohol

Question 8.
The sporozoite of the malarial parasite is present in __________
(a) saliva of infected female Anopheles mosquito.
(b) RBC of human suffering from malaria.
(c) Spleen of infected humans.
(d) Gut of female Anopheles mosquito
Answer:
(a) saliva of infected female Anopheles mosquito.

Question 9.
Where do the following events in the life cycle of Plasmodium takes place?
(a) Fertilization – __________
(b) Development of gametocytes – __________
(c) Release of sporozoites – __________
(d) Schizogony – __________
Answer:
(a) Fertilization – Gut of mosquito
(b) Development of gametocytes – Human RBC’s
(c) Release of sporozoites – From Mosquito to the human blood
(d) Schizogony – Human liver cells

Question 10.
Paratope is an __________
(a) Antibody binding site on variable regions
(b) Antibody binding site on heavy regions
(c) Antigen binding site on variable regions
(d) Antigen binding site on heavy regions
Answer:
(c) Antigen binding site on variable regions

Question 11.
Allergy involves __________
(a) IgE
(b) IgG
(c) IgA
(d) IgM
Answer:
(a) IgE

Question 12.
Spread of cancerous cells to distant sites is termed as __________
(a) Metastasis
(b) Oncogenes
(c) Proto-oncogenes
(d) Malignant neoplasm
Answer:
(a) Metastasis

Question 13.
AIDS virus has __________
(a) Single stranded RNA
(b) Double stranded RNA
(c) Single stranded DNA
(d) Double stranded DNA
Answer:
(a) Single stranded RNA

Question 14.
B cells that produce and release large amounts of antibody are called __________
(a) Memory cells
(b) Basophils
(c) Plasma cells
(d) killer cells
Answer:
(c) Plasma cells

Question 15.
Given below are some human organs.
Answer:
Primary lymphoid organ – thymus
Secondary lymphoid organ – tonsils
Immunocompetency of T cells is achieved in Thymus
Whereas tonsils help to fight viral and bacterial infections.

Question 16.
Name and explain the type of barriers which involve macrophages.
Answer:
Phagocytic barriers involves macrophages (monocytes, neutrophils) which phagocytose and . digest whole micro organisms.

Question 17.
What are interferons? Mention their role.
Answer:
Interferons are the antiviral proteins which induce antiviral state in the infected cells.

Question 18.
List out chemical alarm signals produced during inflammation.
Answer:
Serotonin, histamine and prostaglandins.

Question 19.
Explain the process of replication of retrovirus after it gains entry into the human body.
Answer:
After getting into the body of the person, the retrovirus enters into macrophages where RNA genome of the virus replicates to form viral DNA with the help of the enzyme reverse transcriptase. This viral DNA gets incorporated into the DNA of host cells and directs the infected cells to produce viral particles. The macrophages continue to produce virus and in this way acts like an HIV factory.

Question 20.
Explain the structure of immunoglobulin with suitable diagram.
Answer:
An antibody molecule is a Y-shaped structure that comprises of four polypeptide chains, two identical light chains (L) of molecular weight 25,000 Da (approximately 214 amino acids) and two identical heavy chains (H) of molecular weight 50,000 Da (approximately 450 amino acids). The polypeptide chains are linked together by di-sulphide (S-S) bonds. One light chain is attached to each heavy chain and two heavy chains are attached to each other to form a Y shaped structure. Hence, an antibody is represented by H2 L2 The heavy chains have a flexible hinge region at their approximate middles Antigen binding
Samacheer Kalvi 12th Bio Zoology Solutions Chapter 7 Human Health and Diseases
Each chain (L and H) has two terminal They are C – terminal (Carboxyl) and amino or N-terminal. Each chain (L and H) has two regions. They have variable (V) region at one end and a much larger constant (C) region at the other end. Antibodies responding to different antigens have very different (V) regions but their (C) regions are the same in all antibodies. In each arm of the monomer antibody, the (V) regions of the heavy and light chains combines to form an antigen-binding

site shaped to ‘fit’ a specific antigenic determinant. Consequently, each antibody monomer has two such antigen-binding regions. The (C) regions that form the stem of the antibody monomer determine the antibody class and serve common functions in all antibodies.

Question 21.
What are the cells involved innate immune system?
Answer:
Cells involved in innate immunity are monocytes (macrophages), neutrophils, helper T-cells, B-cells, dendritic cells.

Question 22.
What is vaccine? What are its types?
Answer:
A vaccine is a biological preparation that provides active acquired immunity to a particular disease and resembles a disease causing microorganism and is often made from weakened or attenuated or killed forms of the microbes, their toxins, or one of its surface proteins. The vaccines are classified as first, second and third-generation vaccines.

Question 23.
A person is infected by HIV. How will you diagnose with AIDS?
Answer:
A simple blood test is available that can determine whether the person has been infected with HIV. The ELISA test (Enzyme-Linked Immunosorbent Assay) detects the presence of HIV antibodies. It is a preliminary test. Western blot test is more reliable and a confirmatory test. It detects the viral core proteins. If both tests detect the presence of the antibodies, the person is considered to be HIV positive.

Question 24.
Autoimmunity is a misdirected immune response. Justify.
Answer:
Autoimmune diseases: Autoimmunity is due to an abnormal immune response in which the immune system fails to properly distinguish between self and non-self and attacks its own body. Our body produces antibodies (autoantibodies) and cytotoxic T cells that destroy our own tissues. If a disease-state results, it is referred to as an auto-immune disease. Thus, autoimmunity is a misdirected immune response.

Question 25.
List the causative agent, mode of transmission and symptoms for Diphtheria and Typhoid
Answer:
Samacheer Kalvi 12th Bio Zoology Solutions Chapter 7 Human Health and Diseases

Question 26.
A patient was hospitalized with fever and chills. Merozoites were observed in her blood. What is your diagnosis?
Answer:
Appearance of merozoites in a patient’s blood may be an indication of malarial parasite – plasmodium.

Samacheer Kalvi 12th Bio Zoology Solutions Chapter 7 Human Health and Diseases

Question 27.

  1. Write the scientific name of the filarial worm that causes filariasis.
  2. Write the symptoms of filariasis.
  3. How is this disease transmitted?

Answer:

  1. The scientific name of the filarial worm – Wuchereria bancrofti
  2. Symptoms of filariasis – Inflammation of lymph nodes, obstruction of lymph vessels causes elephantiasis of limbs, genital etc.
  3. Mode of disease transmission: Vector transmission (female Culex mosquito)

Question 28.
List the common withdrawal symptoms of drugs and alcohol abuse.
Answer:
The withdrawal symptoms may range from mild tremors or convulsions, severe agitation and fits, depressed mood, anxiety, nervousness, restlessness, irritability, insomnia, dryness of throat, etc, depending on the type of drug abuse.

Question 29.
Why do you think it is not possible to produce a vaccine against common cold’?
Answer:
Vaccines target specific pathogens and elicit immunity. In case of the common cold, more than 200 strains of viruses are responsible for causing the disease. Hence it is practically difficult to develop a vaccine against it.

Samacheer Kalvi 12th Bio Zoology Human Health and Diseases Additional Questions and Answers

1 – Mark Questions

Question 1.
Pick out the disease which is caused by a virus.
(a) Candidiasis
(b) Ascariasis
(c) Poliomyelitis
(d) Dysentery
Answer:
(c) Poliomyelitis

Question 2.
__________ test is done to confirm typhoid.
(a) ELISA
(b) Western blot
(c) Widal test
(d) Southern blot
Answer:
(c) Widal test

Question 3.
Match the following. Disease
Samacheer Kalvi 12th Bio Zoology Solutions Chapter 7 Human Health and Diseases
Answer:
(a) a – iv, b – iii, c – i, d- ii

Question 4.
Identify the mismatched pair.
Samacheer Kalvi 12th Bio Zoology Solutions Chapter 7 Human Health and Diseases
Answer:
(c) Measles – Intestine

Question 5.
Identify the wrong statement regarding polio disease.
(a) Polio is caused by an RNA virus.
(b) One of the infection site of polio is the intestine.
(c) Culex mosquito acts as a vector for polio.
(d) Paralysis and respiratory failure are the symptoms
Answer:
(c) Culex mosquito acts as a vector for polio.

Question 6.
Yellow fever is a __________ type of disease.
(a) Neurotropic
(b) Viscerotropic
(c) Pneumotropic
(d) Dermotropic
Answer:
(b) Viscerotropic

Question 7.
Which one of the following pairs is wrong.
(a) Amoebiasis – Home fly
(b) African sleeping sickness – Tsetse flies
(c) Kala-azar – Sandfly
(d) Malaria – female Anopheles mosquito
Answer:
(d) Malaria – female Anopheles mosquito

Question 8.
__________ is the malignant form of malaria.
Answer:
P. falciparum

Question 9.
Schizogony of plasmodium parasite in human liver cells returns in __________
(a) sporozoites
(b) merozoites
(c) trophozoites
(d) schizont
Answer:
(b) merozoites

Samacheer Kalvi 12th Bio Zoology Solutions Chapter 7 Human Health and Diseases

Question 10.
Incubation period of malaria is about __________
Answer:
12 days

Question 11.
Samacheer Kalvi 12th Bio Zoology Solutions Chapter 7 Human Health and Diseases
Answer:
(a) a – iii, b – i, c – iv, d – ii

Question 12.
Assertion (A): Plasmodium vivax is a digenic parasite
Reason (R): The primary host of P. vivax is man.
(a) Both (A) and (R) are true. (R) explains (A)
(b) (A) is true (R) is false
(c) Both (A) and (R) are false
(d) (A) is false (R) is true
Answer:
(b) (A) is true (R) is false

Question 13.
Assertion (A): Dermatomycosis is a cutaneous infection.
Reason (R): Fungus belongs to the order Trichophyton
(a) Both (A) and (R) are true. (R) explains (A)
(b) (A) is true (R) is false
(c) Both (A) and (R) are false.
(d) (A) is false (R) is true
Answer:
(a) Both (A) and (R) are true. (R) explains (A)

Question 14.
Assertion (A): Spleen is a primary lymphoid organ
Reason (R): Primary lymphoid organs trap antigen and destroy them.
(a) Both (A) and (R) are true. (R) explains (A)
(b) (A) is true (R) is false
(c) Both (A) and (R) are false.
(d) (A) is false (R) is true
Answer:
(c) Both (A) and (R) are false.

Question 15.
Assertion (A): Paratope is the antigen-binding site.
Reason (R): It is a part of antibody
(a) Both (A) and (R) are true. (R) explains (A)
(b) (A) is true (R) is false
(c) Both (A) and (R) are false.
(d) (A) is false (R) is true
Answer:
(b) (A) is true (R) is false

Question 16.
Assertion (A): HIV is a DNA virus.
Reason (R): HIV belongs to genus Lentivirus
(a) Both (A) and (R) are true. (R) explains (A)
(b) (A) is true (R) is false
(c) Both (A) and (R) are false.
(d) (A) is false (R) is true
Answer:
(d) (A) is false (R) is true

Question 17.
Secretion of HCL in stomach is an example for __________
(a) Anatomical barriers
(b) Phagocytic barriers
(c) Physiological barriers
(d) Inflammatory barriers
Answer:
(c) Physiological barriers

Question 18.
Identify the incorrect statement.
(a) Antibody Mediated Immunity was elicited by T cells.
(b) It is a character of vertebrates only
(c) Immunoglobulins act against pathogens and kill them.
(d) It is also called humoral immunity
Answer:
(a) Antibody-Mediated Immunity was elicited by T cells.

Samacheer Kalvi 12th Bio Zoology Solutions Chapter 7 Human Health and Diseases

Question 19.
Production process of blood cells in bone marrow is called __________
Answer:
hematopoiesis

Question 20.
Which of the following is not a feature of passive immunity?
(a) It is transient and less effective
(b) Immunological memory is present
(c) Immunity develops immunity
(d) Antibodies are obtained from outside
Answer:
(b) Immunological memory is present

Question 21.
__________ is a primary lymphoid organ of birds.
Answer:
Bursa of Fabricius

Question 22.
Match the following.
(a) Peyer’s patches (i) trachea
(b) BALT (ii) intestine
(c) Adenoid (iii) heart
(d) Thymus (iv) roof of the mouth
Answer:
a – ii b – i c – iv d – iii

Question 23.
Which is not a granulocyte?
(a) Lymphocytes
(b) Neutrophils
(c) Basophils
(d) Eosinophils
Answer:
(a) Lymphocytes

Question 24.
The L and H chains of immunoglobulin are joined by
(a) Hydrogen bonds
(b) disulphide bonds
(c) phosphodiester bonds
(d) ionic bond
Answer:
(b) disulphide bonds

Question 25.
__________ type of Immunoglobulin is involved in allergic reactions.
Answer:
IgE

Question 26.
Identify the wrong statement.
(a) Vaccine provide passively acquired immunity
(b) It is made from attenuated or killed microbes.
(c) Vaccines teach our body how to defend from microbes.
(d) MMR is a first generation vaccine.
Answer:
(a) Vaccine provide passively acquired immunity

Question 27
__________ developed first vaccine for smallpox.
Answer:
Edward Jenner

Question 28.
The enzyme attached to RNA of HIV is __________
(a) RNA polymerase
(b) reverse transcriptase
(c) primase
(d) endonuclease
Answer:
(b) reverse transcriptase

Samacheer Kalvi 12th Bio Zoology Solutions Chapter 7 Human Health and Diseases

Question 29.
Infection of Ascariasis occur due to __________
(a) Sand fly
(b) contaminated food
(c) mosquito bite
(d) stagnant water
Answer:
(b) contaminated food

Question 30.
Which of the following statement(s) is true regarding cancer cells?
(a) Neoplasm or tumor cells show uncontrolled growth
(b) They are metastatic
(c) They lack contact inhibition
(d) They may be benign or malignant.
(a) (a) only
(b) (b) and (c)
(c) (d) only
(d) All the above
Answer:
(d) All the above

Question 31.
Study dealing with body’s defence mechanism against disease is called __________
(a) Pathology
(b) Immunology
(c) Microbiology
(d) Dermatology
Answer:
(b) Immunology

Question 32.
AIDS is characterized by sharp reduction in number of __________
(a) helper T cells
(b) killer T cells
(c) superior T cells
(d) B-cells
Answer:
(a) helper T cells

Question 33.
Plague and malaria are caused by __________ and __________ respectively.
(a) bacteria and virus
(b) fungi and protozoa
(c) bacteria and protozoan
(d) fungi and bacteria
Answer:
(c) bacteria and protozoan

Question 34.
A pair of fungal disease __________
(a) Amoebiasis, Kala-azar
(b) Candidiasis, Athlete’s foot
(c) Ascariasis, Filariasis
(d) Poliomyelitis, Amoebiasis
Answer:
(b) Candidiasis, Athlete’s foot

Question 35.
Plant source of Heroin is __________
(a) Poppy plants
(b) Cannabis plants
(c) Datura species
(d) Atropa species
Answer:
(a) Poppy plants

Samacheer Kalvi 12th Bio Zoology Solutions Chapter 7 Human Health and Diseases

Question 36.
The test that confirms HIV positive is __________
(a) Western blot
(b) Northern blot
(c) Southern blot
(d) all the above
Answer:
(a) Western blot

Question 37.
Bacillary dysentery is caused due to __________
(a) Salmonella
(b) Shigella
(c) Clostridium
(d) Yersinia
Answer:
(b) Shigellosis

Question 38.
Cocaine is a __________ potent.
(a) Sedative
(b) Hallucinogen
(c) pain reliever
(d) neurotransmitter
Answer:
(b) Hallucinogen

Question 39.
Alkaloid found in tobacco is __________
(a) Atropine
(b) cocaine
(c) heroin
(d) nicotine
Answer:
(d) nicotine

Question 40.
__________ is a chronic memory disorder due to alcohol misuse.
(a) Cushing’s syndrome
(b) Turners’ syndrome
(c) Klinefelters’ syndrome
(d) Korsakoff syndrome
Answer:
(d) Korsakoff syndrome

2 – Mark Questions

Question 1.
What steps should be taken to maintain good health?
Answer:
Personal hygiene, regular exercise and balanced diet are very important to maintain good health.

Question 2.
Define disease.
Answer:
Disease can be defined as a disorder or malfunction of the mind or body. It involves morphological, physiological and psychological disturbances which may be due to environmental factors or pathogens or genetic anomalies or life style changes.

Question 3.
According to the WHO, what is health?
Answer:
The World Health Organization [WHO] defines health as ‘a state of complete physical, mental and social wellbeing and not merely absence of diseases’.

Question 4.
Differentiate between infectious and non-infections disease.
Answer:

  1. Infectious disease: Disease which are transmitted form one person to another are called infectious disease, e.g. Common cold
  2. Non-infectious disease: Disease which do not transmitted form one person to another are called non-infectious disease, e.g. Anaemia.

Question 5.
Name any two fungal disease and helminthic disease.
Answer:

  1. Fungal disease : Candidiasis, Athlete’s foot
  2. Helminthic disease : Ascariasis, Filariasis

Question 6.
Mention the causative organism of the following.

  1. Tetanus
  2. Bubonic plague
  3. Pneumonia
  4. Cholera

Answer:

  1. Tetanus : Clostridium tetani
  2. Bubonic plague : Yersinia pestis
  3. Pneumonia : Streptococcus pneumoniae
  4. Cholera : Vibrio cholerae

Question 7.
Classify viral disease based on the organ of infection.
Answer:

  1. Pneumotropic diseases
  2. Dermotropic diseases
  3. Viscerotropic diseases
  4. Neurotropic diseases

Question 8.
Write the symptoms of viral hepatitis.
Answer:

  1. Liver damage
  2. jaundice
  3. nausea
  4. yellowish eyes
  5. fever and pain in the abdomen.

Question 9.
Name the different species of malarial parasites and also mention which is the fatal one?
Answer:

  1. Plasmodium vivax
  2. Plasmodium ovale
  3. Plasmodium malariae
  4. Plasmodium falciparum .
  5. Plasmodium falciparum causes fatal malaria.

Question 10.
Name the causative agent and confirmatory test for Typhoid.
Answer:
Typhoid is caused by Salmonella typhi. Widal test can be done to confirm typhoid.

Question 11.
Mention the three phases in the life of plasmodium parasite with their respective host.
Answer:

  1. Schizogony occurs in man.
  2. Gamogony occurs in man.
  3. Sporogony occurs in gut of mosquito.

Question 12.
What causes shivering in malarial patient?
Answer:
After infection of RBC with sporozoite, when the RBC ruptures a toxic substance haemozoin is released which causes shivering chills and high fever followed by sweating.

Question 13.
Malaria Eradication programme launched by WHO is a failure. Why?
Answer:
In the 1950’s the World Health Organisation (WHO) introduced the Malaria eradication programme. This programme was not successful due to the resistance of Plasmodium to the drugs used to treat it and resistance of mosquito’s to DDT and other insecticides.

Question 14.
Filarial worm and Plasmodium both are digenic parasites. Man is a common host for both parasites. Name the other host.
Answer:
The other host of filarial worm is female Culex mosquito The other host of plasmodium is female Anopheles mosquito

Question 15.
Name the causative organism of filariasis and mention the site of infection of parasite.
Answer:
Causative organisms of filariasis is Wuchereria bancrofti (filarial worm). The site of infection is lymph vessels and lymph nodes.

Question 16.
What do you mean by the term personal hygiene?
Answer:
Personal hygiene refers to maintaining one’s body clean by bathing, washing hands, trimming fingernails, wearing clean clothes and also includes attention to keeping surfaces in the home and workplace, including toilets, bathroom facilities, clean and pathogen-free.

Question 17.
Define Immunity and Susceptibility.
Answer:
The overall ability of body to fight against the disease causing pathogen is called immunity. It is also called disease resistance and the lack of immunity is known as susceptibility.

Question 18.
How skin and mucus membrane act as barriers for infections?
Answer:
Skin prevents the entry of microbes. Its acidic environment (pH 3-5) retards the growth of microbes.

Samacheer Kalvi 12th Bio Zoology Solutions Chapter 7 Human Health and Diseases

Question 19.
Mucus membrane entraps foreign microorganisms and competes with microbes for attachment. What is diapedesis?
Answer:
Tissue damage and infection induce leakage of vascular fluid, containing chemotactic signals like serotonin, histamine and prostaglandins. They influx the phagocytic cells into the affected area. This phenomenon is called diapedesis.

Question 20.
Write any two differences between CMI and AMI.
Answer:
Cell Mediated Immunity (CMI):

  1. In CMI, pathogens are destroyed by cells without producing antibodies.
  2. It is carried out by T cells, Macrophages, NK cells

Antibody Mediated Immunity (AMI):

  1. In AMI, pathogens are destroyed by antibodies.
  2. It is carried out by B cells, T helper cells, APC cells.

Question 21.
Define haematopoiesis.
Answer:
The process of production of blood cells in the bone marrow is called haematopoiesis.

Question 22.
Which is the primary lymphoid organs of birds? Mention its location and role.
Answer:
Bursa of Fabricius is a primary lymphoid organ of birds. It is attached to the dorsal side of the cloaca. B lymphocytes mature in the bursa and bring about humoral immunity.

Question 23.
What are the primary lymphoid organs of mammals?
Answer:
Bone marrow and thymus gland.

Question 24.
What are Peyer’s patches?
Answer:
Peyer’s patches are oval-shaped areas of thickened tissue that are embedded in the mucus- secreting lining of the small intestine of humans and other vertebrate animals. Peyer’s patches contain a variety of immune cells, including macrophages, dendritic cells, T cells, and B cells.

Question 25.
Point out any four peripheral lymphoid organs.
Answer:
Lymph nodes, spleen, tonsils, MALT.

Question 26.
Write a brief note on GALT.
Answer:
Gut-associated lymphoid tissue (GALT) is a component of the mucosa associated lymphoid tissue (MALT) which works in the immune system to protect the body from invasion in the gut.

Question 27.
Why secondary immune response is more effective than primary immune response?
Answer:
Due to the establishment of immunological memory of the first encounter, the secondary immune response highly intense and effective.

Question 28.
Name the Agranulocytes involved in immune response.
Answer:

  1. Lymphocyte
  2. Monocytes

Question 29.
Why dendritic cells are called so?
Answer:
Dendritic cells are called so because its covered with long, thin membrane extensions that resemble dendrites of nerve cells. These cells present the antigen to T-helper cells.

Question 30.
How many dendritic cells are identified? Name them.
Answer:
Four types of dendritic cells are known. They are langerhans, interstitial cells, myeloid and lymphoid cells.

Question 31.
What are Haptens?
Answer:
Haptens are substance that are non-immunogenic but can react with the products of a specific immune response.

Question 32.
Distinguish between Epitope and Paratope.
Answer:

  1. Epitope: Epitope is an antigenic determinant and is ‘ the active part of an antigen.
  2. Paratope: A paratope is the antigen – binding site and is a part of an antibody which recognizes and binds to an antigen.

Question 33.
Draw a simplified diagram of immunoglobulin.
Answer:
Samacheer Kalvi 12th Bio Zoology Solutions Chapter 7 Human Health and Diseases

Question 34.
On which basis, the antibodies are classified? Name them.
Answer:
The antibodies are classified into five major categories, based on their physiological and biochemical properties. They are IgG (gamma), IgM(mu), IgA (alpha), IgD(delta) and IgE (epsilon).

Question 35.
Name any four functions of antibodies.
Answer:
The functions of immunoglobulin are agglutination, precipitation, opsonisation, neutralization.

Samacheer Kalvi 12th Bio Zoology Solutions Chapter 7 Human Health and Diseases

Question 36.
Which type of bonds are developed between an antigen and antibody? Name them.
Answer:
The bonds that hold the antigen to the antibody combining site are all non-covalent in nature. These include hydrogen bonds, electrostatic bonds, Van der Waals forces and hydrophobic bonds.

Question 37.
What is antibody affinity?
Answer:
Antibody affinity is the strength of the reaction between a single antigenic determinant and a single combining site on the antibody.

Question 38
What do you mean by third generation vaccine?
Answer:
Third generation vaccine contains the purest and the highest potency vaccines which are synthetic in generation. The latest revolution in vaccine is DNA vaccine or recombinant vaccine.

Question 39
Define vaccination.
Answer:
Vaccination is the process of administrating a vaccine into the body or the act of introducing a vaccine into the body to produce immunity to a specific disease.

Question 40.
Define allergy and allergen.
Answer:
The exaggerated response of the immune system to certain antigens present in the environment is called allergy (allo-altered, erg-reaction). The substances to which such an immune response is produced are called allergens.

Question 41.
What is Anaphylaxis?
Answer:
Anaphylaxis is the classical immediate hypersensitivity reaction. It is a sudden, systematic, severe and immediate hypersensitivity reaction occurring as a result of rapid generalized mast-cell degranulation.

Question 42.
Expand

  1. MALT
  2. NACO

Answer:

  1. MALT – Mucosal-Associated Lymphoid Tissue
  2. NACO – National AIDS Control Organisation

Question 43.
How will you define Autoimmunity?
Answer:
Autoimmunity is due to an abnormal immune response in which the immune system fails to properly distinguish between self and non-self and attacks its own body.

Question 44.
What do you mean by drug abuse?
Answer:
The intake of certain drugs for a purpose other than their normal clinical use in an amount and frequency that impair one’s physical, physiological and psychological functions is called drug abuse.

Question 45.
Name any 4 natural cannabinoids.
Answer:
Marijuana, Ganja, Hashish and Charas.

Question 46.
Mention any two drugs to treat insomnia patient.
Answer:

  1. methamphetamine
  2. Lysergic acid diethylamide (LSD)

Question 47.
Name the antibody responsible for allergic reaction. Also mention two chemicals released during allergic response.
Answer:
IgE (epsilon) antibody is responsible for allergic reaction. Histamine and serotonin are the chemicals released during allergic reaction.

Question 48.
Name an opioid drug and its source plant. How it is useful in medical field?
Answer:

  1. Morphine is an opioid drug which is extracted from flowers of the poppy plant.
  2. It is used as a strong pain killer during surgery.

Question 49.
What is withdrawal symptom? Name any two symptoms.
Answer:
If the intake of the drug or alcohol is abruptly stopped, he or she would develop withdrawal symptoms. In a sense, the body becomes confused and protests against the absence of the drug. The withdrawal symptoms are nervousness and insomnia.

Question 50.
Name the plant source of Cocaine and Heroin.
Answer:

  1. Cocaine is obtained from Erythroxylum coca.
  2. Heroin is obtained from poppy plant.

Question 51.
What is Liver cirrhosis?
Answer:
Alcohol interferes with the ability of the liver to break down fat. Over time fat accumulation and high levels of alcohol destroy the liver cells and a scar tissue grows in the place of dead cells. This scarring of the liver is called “Liver cirrhosis”.

Question 52.
Define Korsakoff syndrome.
Answer:
Korsakoff syndrome, a chronic memory disorder is most commonly caused by alcohol misuse.

Question 53.
What are the benefits of exercising our body?
Answer:
Participating in an exercise programme can:

  1. Increase self-esteem
  2. Boost self-confidence
  3. Create a sense of empowerment
  4. Enhance social connections and relationships

3 – Mark Questions

Question 54.
Write the name of causative agent, infection site, mode of transmission and any two symptoms of Chikungunya.
Answer:

  1. Causative agent – Alpha virus
  2. Infection site – Nervous system
  3. Mode of transmission – Aedes aegypti (Mosquito)
  4. Symptoms – Fever, headache, joint pain and swelling.

Question 55.
Draw and label the parts of Entamoeba histolytica.
Answer:
Samacheer Kalvi 12th Bio Zoology Solutions Chapter 7 Human Health and Diseases

Question 56.
Give a brief account of Kala-azar.
Answer:
Kala – azar or visceral leishmaniasis is caused by Leishmania donovani, which is transmitted by the vector Phlebotomus (sand fly). Infection may occur in the endothelial cells, bone marrow, liver, lymph glands and blood vessels of the spleen. Symptoms of Kala azar are weight loss, anaemia, fever, enlargement of spleen and liver.

Samacheer Kalvi 12th Bio Zoology Solutions Chapter 7 Human Health and Diseases

Question 57.
Comment on Malaria vaccine.
Answer:
Malaria vaccine is used to prevent malaria. The only approved vaccine as of 2015 is RTS,S(Mosquirix). It requires four injections and has relatively low efficacy (26-50%). Due to this low efficacy, WHO does not recommend the use of RTS,S vaccine in babies between 6 lr and 12 weeks of age.

Question 58.
What is innate immunity?
Answer:
Innate immunity is the natural phenomenon of resistance to infection which an individual possesses right from the birth. The innate defense mechanisms are non-specific in the sense that they are effective against a wide range of potentially infectious agents. It is otherwise known as non-specific immunity or natural immunity

Question 59.
Explain

  1. Cell Mediated Immunity
  2. Antibody Mediated Immunity

Answer:

  1. Cell-mediated immunity: When pathogens are destroyed by cells without producing antibodies, then it is known as cell mediated immune response or cell mediated immunity. This is brought about by T cells, macrophages and natural killer cells.
  2. Antibody mediated immunity or humoral immunity: When pathogens are destroyed by the production of antibodies, then it is known as antibody mediated or humoral immunity. This is brought about by B cells with the aid of antigen presenting cells and T helper cells. Antibody production is the characteristic feature of vertebrates only.

Question 60.
Secondary response is a booster response – Explain.
Answer:
The secondary immune response occurs when a person is exposed to the same antigen again. During this time, immunological memory has been established and the immune system can start producing antibodies immediately. Within hours after recognition of the antigen, a new army of plasma cells are generated. Within 2 to 3 days, the antibody concentration in the blood rises steeply to reach much higher level than primary response. This is also called as “booster response”.

Question 61.
What are lymphoid organs? Mention its types.
Answer:
The organs involved in the origin, maturation and proliferation of lymphocytes are called lymphoid organs. Based on their functions, they are classified into primary or central lymphoid organs and secondary or peripheral lymphoid organs.

Question 62.
Classify antigens based on origin.
Answer:
On the basis of origin, antigens are classified into exogenous antigens and endogenous antigens. The antigens which enter the host from the outside in the form of microorganisms, pollens, drugs, or pollutants are called exogenous antigens. The antigens which are formed within the individual are endogenous antigens. The best examples are blood group antigens.

Question 63.
Point out the factors that determine the binding force between antigen – antibody reactions.
Answer:
The binding force between antigen and antibody is due to three factors. They are closeness between antigen and antibody, noncovalent bonds or intermolecular forces and affinity of antibody.

Question 64.
Why it is important to study antigen-antibody interaction?
Answer:
The chief application of antigen-antibody reactions is to determine blood groups for transfusion, to study serological ascertainment of exposure to infectious agents, to develop immunoassays for the quantification of various substances, to detect the presence or absence, of protein in serum and to determine the characteristics of certain immunodeficiency diseases.

Question 65.
Explain the opsonization property of antibodies.
Answer:
Opsonisation or enhanced attachment is the process by which a pathogen is marked of ingestion and destruction by a phagocyte. Opsonization involves the binding of an opsonin i.e antibody, to a receptor on the pathogen’s cell membrane. After opsonin binds to the membrane, phagocytes are attracted to the pathogen. So, opsonization is a process in which pathogens are coated with a substance called an opsonin, marking the pathogen out for destruction by the immune system. This results in much more efficient phagocytosis.

Samacheer Kalvi 12th Bio Zoology Solutions Chapter 7 Human Health and Diseases

Question 66.
Give an example for

  1. First generation vaccine
  2. Second generation vaccine
  3. Third generation vaccine

Answer:

  1. First-generation vaccine – MMR vaccine
  2. Second generation vaccine – Hepatitis-B vaccine
  3. Third generation vaccine – DNA Vaccine

Question 67.
Name the diseases for which vaccines were developed by Louis Pasteur.
Answer:
Cholera, Anthrax and Rabies

Question 68.
How AIDS patient fail to develop immunity?
Answer:
AIDS is caused by Human Immuno Deficiency Virus (HIV). It selectively infects helper T cells. The infected helper T cells will not stimulate antibody production by B-cells resulting in loss of natural defence against viral infection.

Question 69.
Suggest few methods to prevent AIDS.
Answer:
Prevention of AIDS is the best option. Advocating safe sex and promoting regular check-up, safe blood for transfusion, use of disposable needles, use of condoms during sexual contact, prevention of drug abuse, AIDS awareness programme by NACO (National AIDS Control Organisation), NGOs (Non-Governmental Organisations) and WHO are to prevent the spreading of AIDS.

Question 70.
State immunological surveillance theory.
Answer:
The concept of immunological surveillance postulates that the primary function of the immune system is to “seek and destroy” malignant cells that arise by somatic mutation. The efficiency of the surveillance mechanism reduces either as a result of ageing or due to congenital or acquired immunodeficiencies, leads to increased incidence of cancer. Thus, if immunological surveillance is effective, cancer should not occur. The development of tumour represents a lapse in surveillance.

Question 71.
What is contact inhibition? How it is related to tumours growth?
Answer:
Normal cells show a property called contact inhibition, which inhibits uncontrolled growth. Cancer cells do not have this property. As a result, cancerous cells divide continuously giving rise to mass of tissues called tumours.

Question 72.
Differentiate between normal cells and cancer cells.
Answer:
Normal Cells:

  1. Small uniformly shaped nuclei Relatively large cytoplasmic volume
  2. Conformity in cell size and shape Cells arranged into discrete tissue
  3. May possess differentiated cell structures Normal presentation of cell surface markers
  4. Lower levels of dividing cells Cell tissues clearly demarcated

Cancer Cells:

  1. Large, variable shaped nuclei Relatively small cytoplasmic volume
  2. Variation in cell size and shape Disorganised arrangement of cells
  3. Loss of normal specialised features Elevated expression of certain cell markers
  4. Large number of dividing cells Poorly defined tumor boundaries

Question 73.
Write a note on Heroin.
Answer:
Heroin (smack) is chemically diacetylmorphine, which is white, odourless and bitter crystalline compound. It is obtained by acetylation of morphine, which is extracted from flowers of the poppy plant.

Samacheer Kalvi 12th Bio Zoology Solutions Chapter 7 Human Health and Diseases

Question 74.
“Smoking and Tobacco chewing is injurious to health” – Comment on the statement.
Answer:
Tohacco is smoked, chewed and used as snuff. It increases the carbon monoxide content of blood and reduces the concentration of haem bound oxygen, thus causing oxygen deficiency in the body. Tobacco contains nicotine, carbon monoxide and tars, which cause problems in the heart, lung and nervous system. Adrenal glands are stimulated by nicotine to release adrenaline and noradrenaline which increases blood pressure and heartbeat.

Question 75.
Point out the symptoms of mental depression.
Answer:
Signs and symptoms of mental depression:

  1. Loss of self-confidence and self-esteem
  2. Anxiety
  3. Not being able to enjoy things that are usually pleasurable or interesting.

5-Mark Questions 

Question 76.
Name any five viral diseases, their causative agents, infection site, mode of transmission and their symptoms.
Answer:
Samacheer Kalvi 12th Bio Zoology Solutions Chapter 7 Human Health and Diseases

Question 77.
Describe the life cycle of the plasmodium parasite.
Answer:
Plasmodium vivax is a digenic parasite, involving two hosts, man as the secondary host and female Anopheles mosquito as the primary host. The life cycle of Plasmodium involves three phases namely schizogony, gamogony and sporogony.

The parasite first enters the human bloodstream through the bite of an infected female Anopheles mosquito. As it feeds, the mosquito injects the saliva containing the sporozoites. ‘The sporozoite within the bloodstream immediately enters the hepatic cells of the liver. Further in the liver they undergo multiple asexual fission (schizogony) and produce merozoites. After being released from liver cells, the merozoites penetrate the RBC’s.

Inside the RBC, the merozoite begins to develop as unicellular trophozoites. The trophozoite grows in size and a central vacuole develops pushing them to one side of cytoplasm and becomes the signet ring stage. The trophozoite nucleus then divides asexually to produce the schizont. The large schizont shows yellowish-brown pigmented granules called Schuffners granules. The schizont divides and produces mononucleated merozoites. Eventually, the erythrocyte lyses, releasing the merozoites and haemozoin toxin into the bloodstream to infect other erythrocytes.

Lysis of red blood cells results in cycles of fever and other symptoms. This erythrocytic stage is cyclic and repeats itself approximately every 48 to 72 hours or longer depending on the species of Plasmodium involved. The sudden release of merozoites triggers an attack on the RBCs.

Occasionally, merozoites differentiate into macrogametocytes and microgametocytes. When these are ingested by a mosquitoe, they develop into male and female gametes respectively. In the mosquito’s gut, the infected erythrocytes lyse and male and female gametes fertilize to form a diploid zygote called ookinete. The ookinete migrates to the mosquito’s gut wall and develop into an oocyte.

The oocyte undergoes meiosis by a process called sporogony to form sporozoites. These sporozoites migrate to the salivary glands of the mosquito. The cycle is now completed and when the mosquito bites another human host, the sporozoites are injected and the cycle begins anew.

The pathological changes caused by malaria affects not only the erythrocytes but also the spleen and other visceral organs. The incubation period of malaria is about 12 days. The early symptoms of malaria are headache, nausea and muscular pain. The classic symptoms first develop with the synchronized release of merozoites, haemozoin toxin and erythrocyte debris into the bloodstream resulting in malarial paroxysms – shivering chills, high fever followed by sweating. Fever and chills are caused partly by malarial toxins that induce macrophages to release tumour necrosis factor (TNF-a) and interleukin.
Samacheer Kalvi 12th Bio Zoology Solutions Chapter 7 Human Health and Diseases

Question 78.
Give an account of helminthic disease.
Answer:
Helminthes are mostly endoparasitic in the gut and blood of human beings and cause diseases called helminthiasis. The two most prevalent helminthic diseases are Ascariasis and Filariasis. Ascaris is a monogenic parasite and exhibits sexual dimorphism. Ascariasis is a disease caused by the intestinal endoparasite Ascaris lumbricoides commonly called the roundworms. It is transmitted through ingestion of embryonated eggs through contaminated food and water.

Children playing in contaminated soils are also prone to have a chance of transfer of eggs from hand to mouth. The symptoms of the disease are abdominal pain, vomiting, headache, anaemia, irritability and diarrhoea. A heavy infection can cause nutritional deficiency and – severe abdominal pain and causes stunted growth in children. It may also cause enteritis, hepatitis and bronchitis.

Filariasis is caused by Wuchereria bancrofti, commonly called a filarial worm. It is found in the lymph vessels and lymph nodes of man. Wuchereria bancrofti is sexually dimorphic, viviparous and digenic. The life cycle is completed in two hosts, man and the female Culex mosquitoes. The female filarial worm gives rise to juveniles called microfilariae larvae. In the lymph glands, the juveniles develop into adults. The accumulation of the worms blocks the lymphatic system resulting in inflammation of the lymph nodes. In some cases, the obstruction of lymph vessels causes elephantiasis or filariasis of the limbs, scrotum and mammary glands.

Question 79.
Tabulate the various types of innate immunity and their action mechanism.
Answer:
Samacheer Kalvi 12th Bio Zoology Solutions Chapter 7 Human Health and Diseases

Question 80.
Point out the differences between active and passive immunity.
Answer:
Active Immunity:

  1. Active immunity is produced actively by the host’s immune system.
  2. It is produced due to contact with a pathogen or by its antigen.
  3. It is durable and effective in protection.
  4. Immunological memory is present.
  5. Booster effect on subsequent dose is possible.
  6. Immunity is effective only after a short period.

Passive Immunity:

  1. Passive immunity is received passively and there is no active host participation.
  2. It is produced due to antibodies obtained from the outside.
  3. It is transient and less effective.
  4. No memory.
  5. A subsequent dose is less effective.
  6. Immunity develops immediately.

Samacheer Kalvi 12th Bio Zoology Solutions Chapter 7 Human Health and Diseases

Question 81.
How primary immune response differ from the secondary immune response?
Answer:
Primary Immune Response:

  1. It occurs as a result of primary contact with an antigen.
  2. Antibody level reaches peak in 7 to 10 days.
  3. Prolonged period is required to establish immunity.
  4. There is rapid decline in antibody level.
  5. It appears mainly in the lymph nodes and spleen.

Secondary Immune Response:

  1. It occurs as a result of second and subsequent contacts with the same antigen.
  2. Antibody level reaches peak in 3 to 5 days.
  3. It establishes immunity in a short time.
  4. Antibody level remains high for longer period.
  5. It appears mainly in the bone marrow, followed by the spleen and lymph nodes.

Question 82.
Explain the structure and role of thymus in primary lymphoid organ.
Answer:
Thymus is most active during the neonatal and pre-adolescent periods. The thymus is a flat and bilobed organ located behind the stemun, above the heart. Each lobe of the thymus contains numerous lobules, separated from each other by connective tissue called septa. Each lobule is differentiated into an outer cortex and an inner medulla. Within each lobule, the developing T cells called thymocytes are arranged into outer cortex and an inner medulla. The inner medulla contains immature thymocytes and outer cortex has matured thymocytes. One of its main secretions is the hormone thymosin. It stimulates the cell to become mature and immunocompetent. By the early teens, the thymus begins to atrophy and is replaced by adipose tissue.

Question 83.
Describe the structure of lymph node with a labelled diagram.
Answer:
Lymph node has three zones. They are the cortex, paracortex and medulla. The outermost layer of the lymph node is called cortex, which consists of B-lymphocytes, macrophages, and follicular dendritic cells. The paracortex zone is beneath the cortex, which is richly populated by lymphocytes and interdigitating dendritic cells. The innermost zone is called the medulla

which is sparsely populated by lymphocytes, but many of Afferentymiatic them are plasma cells, which actively secrete antibody molecules. As the lymph enters, it slowly percolates through the cortex, paracortex and medulla, giving sufficient chance for the phagocytic cells and dendritic cells to trap the antigen brought by the lymph.
Samacheer Kalvi 12th Bio Zoology Solutions Chapter 7 Human Health and Diseases
The lymph leaving a node carries enriched antibodies secreted by the medullary plasma cells against the antigens that enter the lymph node. Sometimes visible swelling of lymph nodes occurs due to active immune response and increased concentration of lymphocytes. Thus swollen lymph nodes may signal an infection. There are several groups of lymph nodes. The most frequently enlarged lymph nodes are found in the neck, under the chin, in the armpits and in the groin.

Samacheer Kalvi 12th Bio Zoology Solutions Chapter 7 Human Health and Diseases

Question 84.
Explain the types of cells involved in the immune system.
Answer:
Lymphocytes: About 20-30% of the white blood cells are lymphocytes. They have a large nucleus filling most of the cell, surrounded by a little cytoplasm. The two main types of lymphocytes are B and T lymphocytes. Both these are produced in the bone marrow. B lymphocytes (B cells) stay in the bone marrow until they are mature. Then they circulate around the body. Some remain in the blood, while others accumulate in the lymph nodes and spleen. T lymphocytes leave the bone marrow and mature in the thymus gland.

Once mature, T cells also accumulate in the same areas of the body as B cells. Lymphocytes have receptor proteins on their surface. When receptors on a B cell bind with an antigen, the B cell becomes activated and divides rapidly to produce plasma cells. The plasma cells produce antibodies. Some B cells do not produce antibodies but become memory cells.

These cells are responsible for secondary immune response. T lymphocytes do not produce antibodies. They recognize antigen-presenting cells and destroy them. The two important types of T cells are Helper T cells and Killer T cells. Helper T cells release a chemical called cytokine which activates B cells. Killer cells more around the body and destroy cells which are damaged or infected. Apart from these cells neutrophils and monocytes destroy foreign cells by phagocytosis. Monocytes when they mature into large cells, they are called macrophages which performs phagocytosis on any foreign organism.

Dendritic cells are called so because its covered with long, thin membrane extensions that resemble dendrites of nerve cells. These cells present the antigen to T-helper cells. Four types of dendritic cells are known. They are langerhans, interstitial cells, myeloid and lymphoid cells.

Question 85.
Write in detail about various types of antigen-antibody reactions.
Different types of antigen and antibody reactions:
Answer:
The reaction between soluble antigen and antibody leads to visible precipitate formation, which is Called precipitin reaction. Antibodies that bring about precipitate formation on reacting with antigens are called as precipitins. Whenever a particulate antigen interacts with its antibody, it would result in clumping or agglutination of the particulate antigen, which is called agglutination reaction. The antibody involved in bringing about agglutination reaction is called agglutinin.

Opsonisation or enhanced attachment is the process by which a pathogen is marked of ingestion and destruction by a phagocyte. Opsonisation involves the binding of an opsonin antibody, to a receptor on the pathogen’s cell membrane. After opsonin binds to the membrane, phagocytes are attracted to the pathogen. So, opsonisation is a process in which pathogens are coated with a substance called an opsonin, marking the pathogen out for destruction by the immune system. This results in a much more efficient phagocytosis.

The neutralization reactions are the reactions of antigen-antibody that involve the elimination of harmful effects of bacterial exotoxins or a virus by specific antibodies. These neutralizing substances i.e. antibodies are known as antitoxins. This specific antibody is produced by a host cell in response to a bacterial exotoxin or corresponding toxoid (inactivated toxin).

Question 86.
Describe the structure of HIV with a diagram.
Answer:
The human immunodeficiency virus belongs to the genus Lentivirus. When observed under the electron microscope, HIV is seen as a spherical virus, 100-120 nm in diameter, containing a dense .core surrounded by a lipoprotein envelope.The envelope has glycoprotein (gp) spikes termed gp 41 and gp 120. At the core, there are two large single stranded RNA. Attached to the RNA are molecules of reverse transcriptase.

Samacheer Kalvi 12th Bio Zoology Solutions Chapter 7 Human Health and Diseases
It also contains enzymes like protease and ribonuclease. The core is covered by a capsid made of proteins. This is followed by another layer of matrix proteins as shown above.

Question 95.
Suggest some of the ways to prevent drug and alcohol abuse.
Answer:

  1. Effectively dealing with peer pressure: The biggest reason for teens to start on drugs is due to their friends/peer groups imposing pressure on them. Hence, it is important to have a better group of friends to avoid such harmful drugs and alcohol.
  2. Seeking help from parents and peers: Help from parents and peer group should be sought immediately so that they can be guided appropriately. Help may even be sought from close and trusted friends. Getting proper advice to sort out their problems would help the young to vent their feelings of anxiety and guilt.
  3. Education and counselling: Education and counselling create positive attitude to deal with many problems and to accept disappointments in life.
  4. Looking for danger signs: Teachers and parents need to look for sign that indicates a tendency to go in for addiction.
  5. Seeking professional and medical assistance: Assistance is available in the form of highly qualified psychologists, psychiatrists and de-addiction and rehabilitation programmes to help individuals to overcome their problems.

Higher Order Thinking Skills (HOTs) Questions

Question 1.
Identify the mismatched pair and give a reason.
(a) Plague – Yersinia pestis
(b) Filariasis – Wuchereria bancrofti
(c) Dermatomycosis – Trypanosoma gambiense
(c) Common cold – Rhinovirus
Answer:
(c) Dermatomycosis – Trypanosoma gambiense is the mismatched pair. Dermatomycosis is a cutaneous infection caused by fungus.

Samacheer Kalvi 12th Bio Zoology Solutions Chapter 7 Human Health and Diseases

Question 2.
In which form does the malarial parasite enter the human body through mosquito? What is the target site of the parasite immediately after entering the host body?
Answer:

  1. Sporozoites.
  2. After entering the bloodstream of the host it finally reaches the liver cell.

Question 3.
A boy of ten years had suffered from chicken-pox. His grandmother consoled him that he is not expected to have it for the rest of his life. Whether his grandmother is right? If so how it happens?
Answer:
The infection produces not only antibodies but also generates memory in the lymphocytes. These cells recognize the pathogen on the subsequent attacks and generate antibodies to neutralize the pathogen, so his grandmother is right and the boy will not get the disease for the rest of his life.

Samacheer Kalvi 11th Bio Botany Solutions Chapter 6 Cell: The Unit of Life

Students can Download Bio Botany Chapter 6 Cell: The Unit of Life Questions and Answers, Notes Pdf, Samacheer Kalvi 11th Bio Botany Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Bio Botany Solutions Chapter 6 Cell: The Unit of Life

Samacheer Kalvi 11th Bio Botany Cell: The Unit of Life Text Book Back Questions and Answers

I. Multiple Choice Questions
Choose the correct answer
Question 1.
The two subunits of ribosomes remain united at critical ion level of …………… .
(a) Magnesium
(b) Calcium
(c) Sodium
(d) Ferrous
Answer:
(a) Magnesium

Question 2.
Sequences of which of the following is used to know the phylogeny?
(a) mRNA
(b) rRNA
(c) tRNA
(d) Hn RNA
Answer:
(b) rRNA

Samacheer Kalvi 11th Bio Botany Solutions 6 Cell: The Unit of Life

Question 3.
Many cells function properly and divide mitotically even though they do not have …………… .
(a) Plasma membrane
(b) Cytoskeleton
(c) Mitochondria
(d) Plastids
Answer:
(d) Plastids

Question 4.
Keeping in view the fluid mosaic model for the structure of cell membrane, which one of the following statements is correct with respect to the movement of lipids and proteins from one lipid monolayer to the other…………… .
(a) Neither lipid nor proteins can flip – flop
(b) Both lipid and proteins can flip – flop
(c) While lipids can rarely flip – flop proteins cannot
(d) While proteins can flip – flop lipids cannot
Answer:
(c) While lipids can rarely flip – flop proteins cannot

Question 5.
Match the columns and identify the correct option:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 6 Cell The Unit of Life 1
Answer:
(c) iii, iv, i, ii.

Question 6.
Bring out the significance of phase contrast microscopy.
Answer:
Phase contrast microscope is used to observe living cells, tissues and the cells cultured invitro during mitosis.

Question 7.
State the protoplasm theory.
Answer:
Protoplasm theory was proposed by Max Schultze which states that the protoplasm is the living content of cell and is a complex colloidal system.

Samacheer Kalvi 11th Bio Botany Solutions 6 Cell: The Unit of Life

Question 8.
Distinguish between prokaryotes and eukaryotes.
Answer:
Between prokaryotes and eukaryotes:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 6 Cell The Unit of Life 2

Question 9.
Difference between plant and animal ceil.
Answer:

Plant Cell

Animal Cell

1. Usually they are larger than animal cells 1. Usually smaller than plant cells
2. Cell wall present in addition to plasma membrane and consists of middle lamellae, primary and secondary walls 2. Cell wall absent
3. Plasmodesmata present 3. Plasmodesmata absent
4. Chloroplast present 4. Chloroplast absent
5. Vacuole large and permanent 5. Vacuole small and temporary
6. Tonoplast present around vacuole 6. Tonoplast absent
7. Centrioles absent except motile cells of lower plants 7. Centrioles present
8. Nucleus present along the periphery of the cell 8. Nucleus at the centre of the cell
9. Lysosomes are rare 9. Lysosomes present
10. Synthesizes amino acids, coenzymes and vitamins required by them 10. Cannot synthesize aminoacids, coenzymes and vitamins required by them
11. Storage material is starch grains 11. Storage material is a glycogen granules

Question 10.
Draw the ultra structure of Plant Cell.
Answer:
The ultra structure of Plant Cell:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 6 Cell The Unit of Life 3

Samacheer Kalvi 11th Bio Botany Cell: The Unit of Life Additional Questions and Answers

I. Multiple Choice Questions
Choose the correct answer:
Question 1.
Scientist who named the unicellular particles as ‘animalcules’ …………… .
(a) Aristotle
(b) Robert Brown
(c) Antonie von Leeuwenhoek
(d) Robert Hooke
Answer:
(c) Antonie van Leeuwenhoek

Question 2.
Cell theory was modified by …………… .
(a) Schwann
(b) Schleiden
(c) Virchow
(d) Dutrochet
Answer:
(c) Virchow

Samacheer Kalvi 11th Bio Botany Solutions 6 Cell: The Unit of Life

Question 3.
Which of the following microscope produce 3D – image?
(a) Phase contrast
(b) TEM
(c) SEM
(d) Dark field
Answer:
(c) SEM

Question 4.
Which of the following electron opaque chemical is used in Electron microscope?
(a) Strontium
(b) Deuterium
(c) Palladium
(d) Uranium
Answer:
(c) Palladium

Question 5.
Medium for electron movement in TEM is …………… .
(a) Air
(b) Oil
(c) Water
(d) Vacuum
Answer:
(d) Vacuum

Samacheer Kalvi 11th Bio Botany Solutions 6 Cell: The Unit of Life

Question 6.
Resolving power of SEM is …………… .
(a) 5 – 10 Å
(b) 2 – 10 Å
(c) 5 – 20 nm
(d) 5 – 20 m
Answer:
(c) 5 – 20 nm

Question 7.
Which among the following is NOT an exception to cell theory?
(a) Viruses
(b) Viroids
(c) Prions
(d) Fungi
Answer:
(d) Fungi

Question 8.
Scientist who named the cytoplasm as “Sarcode” is …………… .
(a) Dujardin
(b) Corti
(c) Purkinje
(d) Hugo Van Mohl
Answer:
(a) Dujardin

Samacheer Kalvi 11th Bio Botany Solutions 6 Cell: The Unit of Life

Question 9.
The pH of protoplasm is around …………… .
(a) 6.6
(b) 6.7
(c) 6.8
(d) 6.9
Answer:
(c) 6.8

Question 10.
The refractive index of protoplasm is …………… .
(a) 1.4
(b) 2.4
(c) 3.4
(d) 0.4
Answer:
(a) 1.4

Question 11.
Histone proteins are seen in the DNA of …………… .
(a) Pseudokaryotes
(b) Prokaryotes
(c) Mesokaryotes
(d) Eukaryotes
Answer:
(d) Eukaryotes

Samacheer Kalvi 11th Bio Botany Solutions 6 Cell: The Unit of Life

Question 12.
Which of the following organelle is believed to be an endosymbiont?
(a) Ribosomes
(b) Mitochondrion
(c) Golgi bodies
(d) Nucleus
Answer:
(b) Mitochondrion

Question 13.
is the living content of the cell …………… .
(a) Cytoplasm
(b) Protoplasm
(c) Nucleoplasm
(d) Nucleus
Answer:
(b) Protoplasm

Question 14.
Fungal cell wall is made of …………… .
(a) Cutin
(b) Chitin
(c) Hemicellulose
(d) Pectin
Answer:
(b) Chitin

Question 15.
…………… acts as a channel between the protoplasm of adjacent cells.
(a) Middle lamellae
(b) Pits
(c) Plasmodesmata
(d) Primary cell wall
Answer:
(c) Plasmodesmata

Question 16.
Fluid mosaic model was proposed by …………… .
(a) Schleiden and Schwann
(b) Singer and Nicolson
(c) Binning and Roher
(d) G. Palade
Answer:
(b) Singer and Nicolson

Samacheer Kalvi 11th Bio Botany Solutions 6 Cell: The Unit of Life

Question 17.
Which is the largest of the internal membranes?
(a) Golgi bodies
(b) Endoplasmic reticulum
(c) Tonoplast
(d) Nuclear membrane
Answer:
(b) Endoplasmic reticulum

Question 18.
In plant cells, golgi bodies are found as small vesicles called …………… .
(a) Polysomes
(b) Cytosomes
(c) Cytosol
(d) Dictyosomes
Answer:
(d) Dictyosomes

Question 19.
organelle plays a major role in post translation process of protein …………… .
(a) Golgi bodies
(b) Nucleolus
(c) Ribosomes
(d) ER
Answer:
(a) Golgi bodies

Question 20.
Zymogen granules are synthesized in …………… .
(a) Lysosomes
(b) Golgi bodies
(c) Mitochondria
(d) Chloroplast
Answer:
(b) Golgi bodies

Samacheer Kalvi 11th Bio Botany Solutions 6 Cell: The Unit of Life

Question 21.
Altmann named mitochondrion as …………… .
(a) Apoplast
(b) Elaioplast
(c) Symplast
(d) Bioplast
Answer:
(d) Bioplast

Question 22.
DNA of mitochondrion is …………… .
(a) Helical
(b) Dumb – bell
(c) Circular
(d) Spiral
Answer:
(c) Circular

Question 23.
Mitochondria are inherited from parent …………… .
(a) Male
(b) Female
(c) Both
(d) None
Answer:
(b) Female

Question 24.
F1 particles are also called as …………… .
(a) Polysomes
(b) Glyoxysomes
(c) Peroxisomes
(d) Oxysomes
Answer:
(d) Oxysomes

Question 25.
Elaioplasts store …………… .
(a) Starch
(b) Lipid
(c) Protein
(d) Chlorophyll
Answer:
(b) Lipid

Question 26.
The photosynthetic units are called as …………… .
(a) Oxysomes
(b) Quantosomes
(c) Thylakoids
(d) Chloroplasts
Answer:
(b) Quantosomes

Question 27.
Which organelle is not membrane bound?
(a) Mitochondrion
(b) Golgi bodies
(c) Chloroplast
(d) Ribosomes
Answer:
(d) Ribosomes

Samacheer Kalvi 11th Bio Botany Solutions 6 Cell: The Unit of Life

Question 28.
Ribosomes of mitochondrion are …………… .
(a) 16 S
(b) 80 S
(c) 70 S
(d) 50 S
Answer:
(c) 70 S

Question 29.
…………… mineral is required for structural cohesion of ribosomes.
(a) Ca2+
(b) H+
(c) Mg2+
(d) Cl
Answer:
(c) Mg2+

Question 30.
Lysosomes originate from …………… .
(a) Mitochondrion
(b) Nucleus
(c) ER
(d) Golgi bodies
Answer:
(d) Golgi bodies

Question 31.
In mammals, peroxisomes are seen in …………… cells.
(a) Brain
(b) Lung
(c) Liver
(d) Heart
Answer:
(c) Liver

Question 32.
Which organelle has a single unit membrane?
(a) Ribosomes
(b) Glyoxysomes
(c) Polysomes
(d) Nucleus
Answer:
(b) Glyoxysomes

Samacheer Kalvi 11th Bio Botany Solutions 6 Cell: The Unit of Life

Question 33.
The single unit membrane of vacuoles is called as …………… .
(a) Tonoplast
(b) Symplast
(c) Apoplast
(d) Amyloplast
Answer:
(a) Tonoplast

Question 34.
Vacuoles of Apple cells store …………… .
(a) Sucrose
(b) Malic acid
(c) Citrate
(d) Flavanoid
Answer:
(b) Malic acid

Question 35.
Ribosomal biogenesis occur at …………… .
(a) Mitochondrion
(b) Polysomes
(c) Nucleolus
(d) Chromosomes
Answer:
(c) Nucleolus

Question 36.
The term chromosome was introduced by …………… .
(a) Bridges
(b) Strasburger
(c) Waldeyer
(d) Poster
Answer:
(c) Waldeyer

Question 37.
Stability to chromosome is offered by …………… .
(a) Satellite
(b) Telomere
(c) Kinetochore
(d) Nucleolus
Answer:
(b) Telomere

Question 38.
Life span of the cells is determined by …………… .
(a) Kinetochore
(b) Satellite
(c) Chromatin
(d) Telomere
Answer:
(d) Telomere

Samacheer Kalvi 11th Bio Botany Solutions 6 Cell: The Unit of Life

Question 39.
The metacentric chromosomes are …………… shaped.
(a) L
(b) V
(c) J
(d) I
Answer:
(b) V

Question 40.
Polytene chromosomes are observed in …………… of Drosophila.
(a) Endocrine gland
(b) Gall bladder
(c) Salivary gland
(d) Exocrine gland
Answer:
(c) Salivary gland

Question 41.
Lampbrush chromosomes occur at …………… stage of meiotic Prophase I.
(a) Leptotene
(b) Diplotene
(c) Zygotene
(d) Pachytene
Answer:
(b) Diplotene

Samacheer Kalvi 11th Bio Botany Solutions 6 Cell: The Unit of Life

Question 42.
Number of basal rings in gram positive bacteria …………… .
(a) 2
(b) 4
(c) 6
(d) 8
Answer:
(a) 2

Question 43.
Microtubules are made of …………… .
(a) Dyenin
(b) Tubin
(c) Tubulin
(d) Nexin
Answer:
(c) Tubulin

Question 44.
Cytoplasm is stained …………… by eosin.
(a) Pink
(b) Blue
(c) Greenish blue
(d) Green
Answer:
(a) Pink

Question 45.
Key difference between plant cell & animal cell is …………… .
(a) Ribosomes
(b) Vacuoles
(c) Cell wall
(d) Centrioles
Answer:
(c) Cell wall

II. Very Short Answer Type Questions (2 Marks)

Question 1.
Name the scientist who proposed the cell theory.
Answer:
Matthias Schleiden and Theodor Schwann.

Question 2.
Define resolving power of a microscope.
Answer:
Resolving power or resolution refers to the ability of the lenses to show the details of object lying between two points. It is the finest detail available from an object. It can be calculated using the following formula.
⇒ Resolution = \(\frac {0.61λ}{NA}\),
where λ, = wavelength of the light and
NA is the numerical aperture.

Samacheer Kalvi 11th Bio Botany Solutions 6 Cell: The Unit of Life

Question 3.
Define magnification. How will you calculate it?
Answer:
The optical increase in the size of an image is called magnification. It is calculated by the following formula
Samacheer Kalvi 11th Bio Botany Solutions Chapter 6 Cell The Unit of Life 4

Question 4.
List out the types of objective lenses used in Bright field microscope.
Answer:
The types of objective lenses used in Bright field microscope:

  1. 5x
  2. 10x
  3. 45x
  4. 100x

Question 5.
In Bright field microscope, where does the primary & secondary magnification occurs?
Answer:
Primary magnification is obtained through, objective lens and secondary magnification is obtained through eye piece lens.

Question 6.
List out the components of Electron microscope.
Answer:
The components of the microscope are as follows:

  1. Electron Generating System
  2. Electron Condensor
  3. Specimen Objective
  4. Tube Lens
  5. Projector

Question 7.
Name the organisms that are exceptions to cell theory.
Answer:
Viruses, Virions and Prions.

Samacheer Kalvi 11th Bio Botany Solutions 6 Cell: The Unit of Life

Question 8.
Name the types of cells based on nuclear characteristics.
Answer:
The types of cells based on nuclear characteristics:

  1. Prokaryotes
  2. Mesokaryotes and
  3. Eukaryotes.

Question 9.
Define nucleoid.
Answer:
Nucleoid refers to the primitive nucleus which is not bound by nuclear membrane and the DNA is without histone protein.

Question 10.
Name the organelles which are believed to be endosymbiont in Eukaryote cell.
Answer:
Chloroplast and Mitochondrion.

Question 11.
Write a note on endosymbiont theory.
Answer:
Two eukaryotic organelles believed to be the descendants of the endosymbiotic prokaryotes. The ancestors of the eukaryotic cell engulfed a bacterium and the bacteria continued to function inside the host cell.

Question 12.
Point out any four prokaryotes.
Answer:Four prokaryotes:

  1. Bacteria
  2. Blue Green algae
  3. Mycoplasma and
  4. Rickettsiae.

Question 13.
Why spirochaetae is said to be a prokaryote?
Answer:
Spirochaetae is a prokaryote since its DNA lies in nucleoid and not bound by nuclear membrane and also devoid of histone protein.

Samacheer Kalvi 11th Bio Botany Solutions 6 Cell: The Unit of Life

Question 14.
Name any two unique structures / organelles of a plant cell.
Answer:
Cell wall and Chloroplast.

Question 15.
What are the components of protoplasm?
Answer:
Protoplasm is composed of a mixture of small particles, such as ions, amino acids, monosaccharides, water, macro-molecules like nucleic acids, proteins, lipids and polysaccharides.

Question 16.
What is the cell wall composition of the following organism?
(a) Fungi
(b) Bacteria
(c) Algae
Answer:
(a) Fungi – Chitin and fungal cellulose.
(b) Bacteria – Peptidoglycan
(c) Algae – Cellulose, mannan and galactan.

Question 17.
Which cell wall layer of plant cell is laid during maturation? Mention its role.
Answer:
Secondary wall is laid during maturation. It plays a key role in determining the shape of a cell.

Question 18.
How pits are formed?
Answer:
In plant cell, at few regions the secondary wall layer is laid unevenly, whereas the primary wall and middle lamellae are laid continuously such regions are called pits.

Samacheer Kalvi 11th Bio Botany Solutions 6 Cell: The Unit of Life

Question 19.
What is the role of plasmodesmata in a plant cell?
Answer:
Plasmodesmata act as a channel between the protoplasm of adjacent cells through which many substances pass through.

Question 20.
Name the two types of Pits.
Answer:
The two types of Pits:

  1. Simple pit and
  2. Bordered pit.

Question 21.
Define flip – flop movement.
Answer:
The movement of membrane lipids from one side of the membrane to the other side by vertical movement is called flip – flopping or flip – flop movement.

Question 22.
Name the two types of protein seen in cell membrane.
Answer:
Two types of protein seen in cell membrane:

  1. Integral proteins and
  2. peripheral proteins.

Question 23.
Define cytoplasmic streaming.
Answer:
Cytoplasmic streaming refers to the movement of the cytoplasm along with the cellular materials inside the cell.

Samacheer Kalvi 11th Bio Botany Solutions 6 Cell: The Unit of Life

Question 24.
Name any four endomembrane structures.
Answer:
Four endomembrane structures:

  1. Nuclear membrane
  2. endoplasmic reticulum
  3. vascular membrane
  4. golgi bodies.

Question 25.
What are dictyosomes?
Answer:
In plant cells, the golgi bodies are found as small vesicles which are called as dictyosomes.

Question 26.
What are porins?
Answer:
Porins are the proteins in the outer membrane of mitochondrion which forms the channel that allows the free diffusion of molecules smaller than about 1000 daltons.

Question 27.
What are oxysomes?
Answer:
The inner membrane of mitochondrion consists of stalked particles called elementary particles or Fernandez Moran particles. F1 particles or oxysomes.

Samacheer Kalvi 11th Bio Botany Solutions 6 Cell: The Unit of Life

Question 28.
How mitochondria helps in evolutionary studies?
Answer:
Mitochondrial DNA is used to track and date recent evolutionary time because it mutates 5 to 10 time faster than DNA in the nucleus.

Question 29.
Why mitochondrion is called as semi – autonomous organelle?
Answer:
Since mitochondria has it own DNA, it is called as Organelle.

Question 30.
Why mitochondria are called “Power houses of cell?
Answer:
Mitochondria are called Power house of a cell, as they produce energy rich (Adenosine Triphosphate) ATP.

Question 31.
Classify plastids based on colour.
Answer:
Chloroplast, Phaeoplast and Rhodoplast.

Samacheer Kalvi 11th Bio Botany Solutions 6 Cell: The Unit of Life

Question 32.
Classify plastids based on storage & mention their storage component.
Answer:

Plastids

Storage Content

1. Amyloplast 1. Starch
2. Elaioplast 2. Lipids / Oils
3. Aleuroplast or Proteoplast 3. Protein

Question 33.
What are Quantosomes?
Answer:
Quantosomes are small, rounded photosynthetic units present in thylakoids.

Question 34.
List out the functions of chloroplast.
Answer:
Photosynthesis, photo respiration.

Question 35.
What is Svedberg unit?
Answer:
Svedberg unit is a measure of a particle size dependent on the speed with which particle sediment in the ultracentrifuge.

Question 36.
Where the biogenesis of ribosomes occur?
Answer:
Biogenesis of ribosome are denova formation, auto replication and nucleolar origin.

Question 37.
What are Polysomes? State its function.
Answer:
During protein synthesis many ribosomes are attached to the single mRNA called polysomes or polyribosomes. The function of polysomes is the formation of several copies of a particular polypeptide during protein synthesis.

Samacheer Kalvi 11th Bio Botany Solutions 6 Cell: The Unit of Life

Question 38.
Define Autolysis.
Answer:
Autolysis is the process, where the lysosome causes self destruction of cell on insight of disease they destroy the cells.

Question 39.
List out the enzymes of lysosomes.
Answer:
Acid Hydrolases, Nuclease, Proteases, Glycosidases, Lipases, Phosphatases, Sulphatases and Phospholipidases.

Question 40.
Name any two organelles involved in photorespiration.
Answer:
Peroxisomes and Chloroplast.

Question 41.
Name few single unit membrane bound organelles.
Answer:
Peroxisomes, Glyoxysomes and Sphaerosomes.

Question 42.
What are Tonoplast?
Answer:
In plant cells vacuoles are large, bounded by a single unit membrane called Tonoplast.

Question 43.
How vacuoles helps to maintain the structure of a plant cell?
Answer:
The major function of plant vacuole is to maintain water pressure known as turgor pressure, which maintains the plant structure.

Samacheer Kalvi 11th Bio Botany Solutions 6 Cell: The Unit of Life

Question 44.
What are metachromatic granules?
Answer:
Inorganic inclusions in bacteria are polyphosphate granules (volutin granules) and sulphur granules. These granules are also known as metachromatic granules.

Question 45.
Which is the largest organelle in a cell? State its function.
Answer:
Nucleus. It controls all the cellular activities.

Question 46.
What is a pore complex?
Answer:
The pores are enclosed by circular structures called annuli. The pore and annuli forms the pore complex.

Samacheer Kalvi 11th Bio Botany Solutions 6 Cell: The Unit of Life

Question 47.
What are nucleolar organizers region?
Answer:
Secondary constrictions contains the genes for ribosomal RNA which induce the formation of nucleoli and are called nucleolar organizer regions.

Question 48.
Draw the types of chromosomes based on centromere position.
Answer:
The types of chromosomes based on centromere position:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 6 Cell The Unit of Life 5

Question 49.
What is a telomere?
Answer:
Telomere is the terminal part of chromosome. It offers stability to the chromosome. DNA of the telomere has specific sequence of nucleotides.

Question 50.
Classify chromosomes based on function.
Answer:
Autosomes and sex chromosomes.

Question 51.
Name any two giant chromosomes.
Answer:
Polytene chromosomes and lamp brush chromosomes.

Question 52.
Define proton motive force.
Answer:
The proton motive force is the force derived from the electrical potential and the hydrogen ion gradient across the cytoplasmic membrane.

Samacheer Kalvi 11th Bio Botany Solutions 6 Cell: The Unit of Life

Question 53.
Name the basal rings found in flagella of gram negative bacteria.
Answer:
The basal rings found in flagella of gram negative bacteria:

  • L – Lipopolysacchride ring
  • P – Peptidoglycan ring
  • S – Super membrane ring and
  • M – membrane ring.

Question 54.
(a) Define histochemistry.
(b) What is Somatic pairing?
Answer:
(a) The technique of staining the cells and tissue is called histochemical staining or histo chemistry.
(b) Maternal and paternal homologues remain associated side by side is called somatic pairing.

III. Short Answer Type Questions (3 Marks)

Question 1.
Explain the principle involved in Scanning Electron Microscope (SEM).
Answer:
In Scanning Electron Microscope (SEM) electrons are focused by means of lenses into a very fine point. The interaction of electrons with the specimen results in the release of different forms of radiation (such as auger electrons, secondary electrons, back scattered electrons) from the surface of the specimen. These radiations are then captured by an appropriate detector, amplified and then imaged on fluorescent screen.

Question 2.
Write a note on solation gelation property of protoplasm.
Answer:
The protoplasm exist either in semisolid (jelly – like) state called ‘gel’ due to suspended particles and various chemical bonds or may be liquid state called ‘sol’. The colloidal protoplasm which is in gel form can change into sol form by solation and the sol can change into gel by gelation. These gel – sol conditions of colloidal system are prime basis for mechanical behaviour of cytoplasm.

Question 3.
Explain the nuclear characters of Mesokaryotes.
Answer:
Mesokaryotes contains well organized nucleus with nuclear membrane and the DNA is organized into chromosomes but without histone protein components divides through amitosis similar with prokaryotes.

Samacheer Kalvi 11th Bio Botany Solutions 6 Cell: The Unit of Life

Question 4.
List out the unique characters of a Eukaryotic cell.
Answer:
Eukaryotes have true nucleus. The DNA is associated with protein bound histones forming the chromosomes. Membrane bound organelles are present.

Question 5.
Name the chemicals seen in the cell wall of plant cells.
Answer:
Cellulose, hemicellulose, pectin, lignin, cutin, suberin and silica.

Question 6.
Name the 3 distinct regions of plant cell wall.
Answer:
In plant, cell wall shows three distinct regions

  1. Primary wall
  2. Secondary wall and
  3. Middle lamellae.

Question 7.
Explain the role of hemicellulose, pectin & glycoprotein in primary cell wall.
Answer:
Hemicellulose binds the microfibrils with matrix and glycoproteins control the orientation of microfibrils while pectin serves as filling material of the matrix.

Samacheer Kalvi 11th Bio Botany Solutions 6 Cell: The Unit of Life

Question 8.
In cell membrane, phospholipids undergo flip – flop movement but not the protein. Why?
Answer:
The phospholipids can have flip – flop movement because the phospholipids have smaller polar regions, whereas the proteins cannot flip – flop because the polar region is extensive.

Question 9.
Define signal transduction.
Answer:
The process by which the cells receive information from outside and respond is called signal transduction.

Question 10.
Distinguish between rough endoplasmic reticulum and smooth endoplasmic reticulum.
Answer:
Between rough endoplasmic reticulum and smooth endoplasmic reticulum:

RER

SER

1. Ribosomes are present in the outer surface of membrane 1. Ribosomes are absent on the membrane
2. It is involved in protein synthesis 2. It is involved in lipid synthesis

Question 11.
Write the major roles of Golgi bodies.
Answer:
Golgi complex plays a major role in post translational modification of proteins and glycosidation of lipids.

Question 12.
Which is the most abundant protein on Earth? Where it is encoded?
Answer:
Rubisco is the abundant protein on Earth. It is encoded by the chloroplast DNA.

Samacheer Kalvi 11th Bio Botany Solutions 6 Cell: The Unit of Life

Question 13.
Classify ribosomes with an example.
Answer:
Types of Ribosomes
1. 70S Ribosomes (subunit 30S and 50S): 3 RNA molecule

  • 16SrRNA in 30S subunit
  • 23S and 5S in 50S large subunit
    (Prokaryotic cells of Blue green algae, Bacteria, Mitochondria and Chloroplast of many Algae and higher plants)

2. 80S Ribosomes (subunits 40S and 60S): 4 RNA molecule

  • 18Sr RNA in 40S small sub unit
  • 28S, 5.8S and 5S in larger 60S subunit.
    (Eukaryotic cells of plants and animals).

Question 14.
Write a note on Glyoxysomes.
Answer:
Glyoxysome is a single membrane bound organelle. It is a sub cellular organelle and contains enzymes of glyoxylate pathway – oxidation of fatty acid occurs in glyoxysomes of germinating seeds.

Question 15.
What are cell inclusions? Give example.
Answer:
The cell inclusions are the non – living materials present in the cytoplasm. They are organic and inorganic compounds.  Example: Phosphate granules.

Samacheer Kalvi 11th Bio Botany Solutions 6 Cell: The Unit of Life

Question 16.
Draw and label the structure of Nucleus.
Answer:
The structure of Nucleus:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 6 Cell The Unit of Life 6Samacheer Kalvi 11th Bio Botany Solutions Chapter 6 Cell The Unit of Life 7

Question 17.
Distinguish between Euchromatin & Heterochromatin.
Answer:
The portion of Eukaryotic chromosome which is transcribed into mRNA contains active genes that are not tightly condensed during interphase is called Euchromatin. The portion of a Eukaryotic chromosome that is not transcribed into mRNA which remains condensed during interphase and stains intensely is called Heterochromatin.

Question 18.
Define Kinetochore.
Answer:
The centromere contains a complex system of protein fibres called kinetochore. Kinetochore is the region of chromosome which is attached to the spindle fibre during mitosis.

Question 19.
Write a note on SAT – chromosome.
Answer:
A satellite or SAT – chromosome are short chromosomal segment or rounded body separated from main chromosome by a relatively elongated secondary constriction. It is a morphological entity in certain chromosomes.

Samacheer Kalvi 11th Bio Botany Solutions 6 Cell: The Unit of Life

Question 20.
How telomeres helps in cancer studies?
Answer:
Maintenance of telomeres appears to be an important factor in determining the life span and reproductive capacity of cells so studies of telomeres and telomerase have the promise of providing new insights into conditions such as ageing and cancer.

Question 21.
Name the three types of centromere in eukaryotes.
Answer:
The three types of centromere in eukaryotes:

  1. Point centromere
  2. Regional centromere and
  3. Holocentromere.

Question 22.
What are giant chromosomes?
Answer:
These chromosomes are larger in size and are called giant chromosomes. In certain plants they are found in the suspensors of the embryo. The polytene chromosome and lamp brush chromosome occur in animals and are also called as giant chromosomes.

Samacheer Kalvi 11th Bio Botany Solutions 6 Cell: The Unit of Life

Question 23.
How polyteny condition is achieved?
Answer:
Polyteny is achieved by repeated replication of chromosomal DNA several times without nuclear division and the daughter chromatids aligned side by side and do not separate (endomitosis).

Question 24.
What is Microphotography?
Answer:
Images of structures observed through microscopes can be further magnified, projected and saved by attaching a camera to the microscope by a microscope coupler or eyepiece adaptor. Picture taken using a inbuilt camera in a microscope is called microphotography or microphotograph.

Question 25.
Draw the Structure of Peroxisomes.
Answer:
The Structure of Peroxisomes:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 6 Cell The Unit of Life 7

IV. Long Answer Type Questions (5 Marks)

Question 1.
Write a detailed account on Dark field microscope.
Answer:
The dark field microscope was discovered by Z. Sigmondy (1905). Here the field will be dark but object will be glistening so the appearance will be bright. A special effect in an ordinary microscope is brought about by means of a special component called “Patch Stop Carrier”. It is fixed in metal ring of the condenser component.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 6 Cell The Unit of Life 8
Patch stop is a small glass device which has a dark patch at centre of the disc leaving a small area along the margin through which the light passes. The light passing through the margin will travel oblique like a hollow cone and strikes the object in the periphery, therefore the specimen appears glistening in a dark background.

Question 2.
Compare Transmission Electron Microscope with Scanning Electron Microscope.
Answer:
Transmission Electron Microscope with Scanning Electron Microscope:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 6 Cell The Unit of Life 9

Question 3.
List out the features of Cell Doctrine.
Answer:
The features of cell doctrine are as follows:

  1. All organisms are made up of cells.
  2. New cells are produced from the pre – existing cells.
  3. Cell is a structural and functional unit of all living organisms.
  4. A cell contains hereditary information which is passed on from cell to cell during cell division.
  5. All the cells are basically the same in chemical composition and metabolic activities.
  6. The structure and function of cell is controlled by DNA.
  7. Sometimes the dead cells may remain functional as tracheids and vessels in plants and horny cells in animals.

Question 4.
Enumerate the properties of protoplasm.
Answer:
The properties of protoplasm:

  1. Protoplasm is translucent, odourless and polyphasic fluid.
  2. It is a crystal colloid solution which is a mixture of chemical substances forming crystalloid and colloidal solution.
  3. Protoplasm exhibits three Brownian movement, amoeboid movement and cytoplasmic streaming or cyclosis. Viscosity of protoplasm is 2 – 20 centipoises. The Refractive index of the protoplasm is 1.4.
  4. The pH of the protoplasm is around 6.8, contain 90% water (10% in dormant seeds).
  5. Approximately 34 elements are present in protoplasm but only 13 elements are main or universal elements i.e. C, H, O, N, Cl, Ca, P, Na, K, S, Mg, I and Fe. Carbon, Hydrogen, Oxygen and Nitrogen form the 96% of protoplasm.
  6. Protoplasm is neither a good nor a bad conductor of electricity.
  7. Cohesiveness: Particles or molecules of protoplasm are adhered with each other by forces, such as Van der Waal’s bonds, that hold long chains of molecules together. This property varies with the strength of these forces.
  8. Contractility: The contractility of protoplasm is important for the absorption and removal of water especially stomatal operations.
  9. Surface Tension: The proteins and lipids of the protoplasm have less surface tension, hence they are found at the surface forming the membrane. On the other hand the chemical substances (NaCl) have high surface tension, so they occur in deeper parts of the cell protoplasm.

Samacheer Kalvi 11th Bio Botany Solutions 6 Cell: The Unit of Life

Question 5.
List out the functions of the Cell Wall.
Answer:
The cell wall plays a vital role in holding several important functions given below.

  1. Offers definite shape and rigidity to the cell.
  2. Serves as barrier for several molecules to enter the cells.
  3. Provides protection to the internal protoplasm against mechanical injury.
  4. Prevents the bursting of cells by maintaining the osmotic pressure.
  5. Plays a major role by acting as a mechanism of defense for the cells.

Question 6.
Explain in detail about Fluid mosaic model.
Answer:
Jonathan Singer and Garth Nicolson (1972) proposed fluid model: It is made up of lipids and proteins together with a little amount of carbohydrate. The lipid membrane is made up of phospholipid. The phospholipid molecule has a hydrophobic tail and hydrophilic head. The hydrophobic tail repels water and water loving polar molecule are called hydrophilic molecule. They have polar phosphate group responsible for attracting water. Water hating non – polar molecule are called as hydrophobic molecule. They have fatty acid which is non – polar which cannot attract water.

Hydrophilic head attracts water. The proteins of the membrane are globular proteins which are found intermingled between the lipid bilayer most of which are projecting beyond the lipid bilayer. These proteins are called as integral proteins. Few are superficially attached on either surface of the lipid bilayer which are called as peripheral proteins. The proteins are involved in transport of molecules across the membranes and also act as enzymes, receptors or antigens.

Question 7.
Describe Endocytosis and Exocytosis.
Answer:
Cell surface membrane are able to transport individual molecules and ions. There are processes in which a cell can transport a large quantity of solids and liquids into cell endocytosis or out of cell exocytosis.

1. Endocytosis: During endocytosis the cell wraps the cell surface membrane around the material and brings it into cytoplasm inside a vesicle. There are two types of endocytosis:

  • Phagocytosis – Particle is engulfed by membrane, which folds around it and forms a vesicle. The enzymes digest the material and products are absorbed by cytoplasm.
  • Pinocytosis – Fluid droplets are engulfed by membrane, which forms vesicles around them.

2. Exocytosis: Vesicles fuse with plasma membrane and eject contents. This passage of material out the cell is known as exocytosis. This material may be a secretion in the case of digestive enzymes, hormones or mucus.

Question 8.
List out the functions of Golgi bodies.
Answer:
Functions of Golgi bodies:

  1. Glycoproteins and glycolipids are produced.
  2. Transporting and storing lipids.
  3. Formation of lysosomes.
  4. Production of digestive enzymes.
  5. Cell plate and cell wall formation
  6. Secretion of carbohydrates for the formation of plant cell walls and insect cuticles.
  7. Zymogen granules (proenzyme / pre – cursor of all enzyme) are synthesized.

Samacheer Kalvi 11th Bio Botany Solutions 6 Cell: The Unit of Life

Question 9.
Draw & describe the structure of mitochondrion.
Answer:
Mitochondria are ovoid, rounded, rod shape and pleomorphic structures. Mitochondrion consists of double membrane, the outer and inner membrane. The outer membrane is smooth, highly permeable to small molecules and it contains proteins called Porins, which form channels that allows free diffusion of molecules smaller than about 1000 daltons and the inner membrane divides the mitochondrion into two compartments, outer chamber between two membranes and the inner chamber filled with matrix.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 6 Cell The Unit of Life 10
The inner membrane is convoluted (infoldings), called crista (plural: cristae). Cristae contain most of the enzymes for electron transport system. Inner chamber of the mitochondrion is filled with proteinaceous material called mitochondrial matrix. The inner membrane consists of stalked particles called elementary particles or Fernandez Moran particles. F1 particles or Oxysomes. Each particle consists of a base, stem and a round head. In the head ATP synthase is present for oxidative phosphorylation. Inner membrane is impermeable to most ions, small molecules and maintains the proton gradient that drives oxidative phosphorylation.

Question 10.
Draw & describe the structure of Nucleus.
Answer:
Nucleus is an important unit of cell which control all activities of the cell. Nucleus holds the hereditary information. It is the largest among all cell organelles. It may be spherical, cuboidal, ellipsoidal or discoidal. It is surrounded by a double membrane structure called nuclear envelope, which has the inner and outer membrane. The inner membrane is smooth without ribosomes and the outer membrane is rough by the presence of ribosomes and is continues with irregular and infrequent intervals with the endoplasmic reticulum.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 6 Cell The Unit of Life 6
The membrane is perforated by pores known as nuclear pores which allows materials such as mRNA, ribosomal units, proteins and other macromolecules to pass in and out of the nucleus. The pores enclosed by circular structures called annuli. The pore and annuli forms the pore complex. The space between two membranes is called perinuclear space. Nuclear space is filled with nucleoplasm, a gelatinous matrix has uncondensed chromatin network and a conspicuous nucleoli. The chromatin network is the uncoiled, indistinct and remain thread like during the interphase. It has little amount of RNA and DNA bound to histone proteins in eukaryotic cells.

Question 11.
Write in detail about the 3 types of centromere in eukaryotes.
Answer:
There are three types of centromere in Eukaryotes. They are as follows:

  1. Point Centromere: The type of centromere in which the kinetochore is assembled as a result of protein recognition of specific DNA sequences. Kinetochores assembled on point centromere bind a single microtubule. It is also called as localized centromere. It occurs in budding yeasts.
  2. Regional Centromere: In regional centromere where the kinetochore is assembled on a variable array of repeated DNA sequences. Kinetochore assembled on regional centromeres bind multiple microtubules. It occurs in fission yeast cell, humans and so on.
  3. Holocentromere: The microtubules bind all the along the mitotic chromosome. Example: Caenorbabditis elegans (transparent nematode) and many insects.

Question 12.
List the functions of Nucleus.
Answer:
Functions of the Nucleus:

  1. Controlling all the cellular activities
  2. Storing the genetic or hereditary information
  3. Coding the information in the DNA for the production of enzymes and proteins.
  4. DNA duplication and transcription takes place in the nucleus.
  5. In nucleolus ribosomal biogenesis takes place.

Question 13.
Explain the structure and movement of Eukaryotic flagella.
Answer:
Structure: Eukaryotic Flagella are enclosed by unit membrane and it arises from a basal body. Flagella is composed of outer nine pairs of microtubules with two microtubules in its centre (9 + 2 arrangement). Flagella are microtubule projection of the plasma membrane. Flagellum is longer than cilium (as long as 200 m). The structure of flagellum has an axoneme made up microtubules and protein tubulin.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 6 Cell The Unit of Life 11
Movement: Outer microtubule doublet is associated with axonemal dynein which generates force for movement. The movement is ATP driven. The interaction between tubulin and dynein is the mechanism for the contraction of cilia and flagella. Dynein molecules uses energy from ATP to shift the adjacent microtubules. This movement bends the cilium or flagellum.

Question 14.
Describe the steps involved in cytologieal techniques.
Answer:
There are different types of mounting based on the portion of a specimen to be observed.

  1. Whole mount: The whole organism or smaller structure is mounted over a slide and observed.
  2. Squash: Is a preparation where the material to be observed is crushed/squashed onto a slide so as to reveal their contents. Example: Pollen grains, mitosis and meiosis in root tips and flower buds to observe chromosomes.
  3. Smears: Here the specimen is in the fluid (blood and microbial cultures etc) are scraped, brushed or aspirated from surface of organ. Example: Epithelial cells.
  4. Sections: Free hand sections from a specimen and thin sections are selected, stained and mounted on a slide. Example: Leaf and stem of plants.

Question 15.
Name any 5 common stains their colour & their affinity used in cytologieal studies.
Answer:
Common Stains used in Histochemistry
Samacheer Kalvi 11th Bio Botany Solutions Chapter 6 Cell The Unit of Life 12

V. Higher Order Thinking Skills (HOTs)

Question 1.
What makes the plant cell more rigid than animal cells?
Answer:
Plants cells posses cell wall which provides sufficient rigidity and proper shape to them whereas in case of animal cells, cell wall is totally absent.

Question 2.
Cleaning organelle in the cell – Explain.
Answer:
Lysosomes contains a variety of hydrolytic enzymes, which can digest the materials within the cell. Thus lysosomes act as cleaning organelle of the cell.

Samacheer Kalvi 11th Bio Botany Solutions 6 Cell: The Unit of Life

Question 3.
Ribosomes are single membrane organelles present in both prokaryotes & eukaryotes. List out the sites where ribosomes are present in plant cell.
Answer:
The sites where ribosomes are present in plant cell:

  1. Cytoplasm
  2. On rough endoplasmic reticulum
  3. Inside mitochondira & chloroplast.

Question 4.
What does ‘S’ refer in a 70 S and an 80 S ribosomes?
Answer:
‘S’ refers to Svedberg units named after Theoder Svedberg. The size of ribosome and their subunits are represented by Svedberg unit.

Question 5.
Briefly give the contributions of the following scientists in the field of cytology.
(a) Schleiden and Schwann
(b) Singer and Nicolson
Answer:
(a) Schleiden and Schwann proposed the cell theory.
(b) Singer & Nicolson proposed the fluid mosaic model of plasmomembrane.

Samacheer Kalvi 11th Bio Botany Solutions 6 Cell: The Unit of Life

Question 6.
Is extra genomic DNA is present in prokaryotes & Eukaryotes? If yes, locate them in both the types of organisms.
Answer:
Locate them in both the types of organisms:

  1. In prokaryotes like bacteria, plasmids are the extra genomic DNA present in cytoplasm.
  2. In Eukaryotes, the circular DNA present in matrix of mitochondria & chloroplast are extragenomic DNA.

Question 7.
Give possible reasons to call mitochondria & chloroplast as semi – autonomous organelles.
Answer:
Mitochondria & chloroplasts are considered as semi – autonomoius organelle due to following
facts.

  1. Both mitochondria & chloroplasts have their own DNA, which can replicate independently.
  2. They have their own ribosomes by which they self-synthesize same of their proteins without depending on cellular DNA.

Samacheer Kalvi 11th Computer Science Solutions Chapter 1 Introduction to Computers

Students can Download Computer Science Chapter 1 Introduction to Computers Questions and Answers, Notes Pdf, Samacheer Kalvi 11th Computer Science Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Computer Science Solutions Chapter 1 Introduction to Computers

Samacheer Kalvi 11th Computer Science Introduction to Computers Text Book Back Questions and Answers

PART – 1
I. Choose The Correct Answer

Question 1.
First generation computers used ………………
(a) Vacuum tubes
(b) Transistors
(c) Integrated circuits
(d) Microprocessors
Answer:
(a) Vacuum tubes

Question 2.
Name the volatile memory.
(a) ROM
(b) PROM
(c) RAM
(d) EPROM
Answer:
(c) RAM

Question 3.
Identify the output device ………………
(a) Keyboard
(b) Memory
(c) Monitor
(d) Mouse
Answer:
(c) Monitor

Samacheer Kalvi 11th Computer Science Solutions Chapter 1 Introduction to Computers

Question 4.
Identify the input device ………………
(a) Printer
(b) Mouse
(c) Plotter
(d) Projector
Answer:
(b) Mouse

Question 5.
……………… Output device is used for printing building plan, flex board, etc.
(a) Thermal printer
(b) Plotter
(c) Dot matrix
(d) inkjet printer
Answer:
(b) Plotter

Question 6.
Which one of the following is used in ATM machine?
(a) Touch Screen
(b) speaker
(c) Monitor
(d) Printer
Answer:
(a) Touch Screen

Samacheer Kalvi 11th Computer Science Solutions Chapter 1 Introduction to Computers

Question 7.
When a system restarts, which type of booting is used.
(a) Warm booting
(b) Cold booting
(c) Touch boot
(d) Real boot
Answer:
(a) Warm booting

Question 8.
Expand POST ……………….
(a) Post on self Test
(b) Power on Software Test
(c) Power on Self Test
(d) Power on Self Text
Answer:
(c) Power on Self Test

Question 9.
Which one of the following is the main memory?
(a) ROM
(b) RAM
(c) Flash drive
(d) Hard disk
Answer:
(b) RAM

Question 10.
Which generation of computer used IC’s?
(a) First
(b) Second
(c) Third
(d) Fourth
Answer:
(c) Third

PART – 2
II. Short Answers

Question 1.
What is a computer?
Answer:
Computer is an Electronic Machine, capable of performing basic operations like addition, multiplication, division etc. Computer accepts data as input, process it, produce output and stores it for future.

Question 2.
Distinguish between data and information.
Answer:
Data:

  • Data is a collection of facts from which the information may be derived. It is an unprocessed collection of facts in a manner suitable for communication, interpretation or processing.
  • Example: 134, 16, Kavitha Does not give meaning.

Information:

  • Information is a collection of facts from the conclusions can be drawn. Information is a processed facts, active, Business based and transformed from data.
  • Example: Kavitha is 16 years old. Conveys meaning.

Question 3.
What are the components of a CPU?
Answer:
The Components of CPU are:

  1. Control Unit
  2. Arithmetic and Logic Unit (ALU)
  3. Memory Unit.

Question 4.
What is the function of an ALU?
Answer:
The ALU is a part of CPU where various computing functions are performed on data. ALU performs arithmetic operations such as addition, subtraction, multiplication, division and logical operations.

Samacheer Kalvi 11th Computer Science Solutions Chapter 1 Introduction to Computers

Question 5.
Write the functions of control unit.
Answer:
The control unit controls the flow of data between the CPU, memory I/O devices. It also controls the entire operations of a computer.

Question 6.
What is the function of memory?
Answer:
Memory enables the computer to store the program. The memory unit is of two types:

  1. Primary memory
  2. Secondary memory

The primary memory is used to store data temporarily and secondary memory stores the data permanently.

Question 7.
Differentiate Input and output unit.
Answer:
Input unit:

  • An input device feeds information to a computer system for processing.
  • Input devices only allow for input of data to a computer.
  • It is a peripheral device used to provide data and control signals to an information processing system such as computer.
  • Example: keyboard, mouse.

Output unit:

  • Output device reproduces or displays the results of that processing.
  • Output devices only receive the output of data from another device.
  • It is a unit which sends data from the computer to another device.
  • Example: printer, monitor.

Question 8.
Distinguish Primary and Secondary memory.
Answer:
Primary Memory:

  • It is directly accessed by CPU.
  • The access of memory is done by making use of addresses or data buses.
  • The Primary memory is embedded with 2 types of technologies. They are RAM and ROM.
  • Volatile in nature.
  • Much faster than the secondary Memory.
  • Costlier than secondary memory.
  • It is temporary memory.

Secondary Memory:

  • It is not directly accessed by CPU.
  • Here the Memory access is done through the input and output channels.
  • The secondary memory are stored in mass storage units like CD ROM, CD, DVD, Floppy disk storage.
  • Non – Volatile in nature.
  • Slower than the Primary Memory.
  • Cheaper than the Primary Memory.
  • Secondary memory is permanent.

PART – 3
III. Explain in Brief

Question 1.
What are the characteristics of a computer?
Answer:
The Basic characteristics of the computers are:

  1. High speed
  2. Stores Huge Data
  3. They do millions of different tasks
  4. Very versatile
  5. Accurate
  6. Diligence

Question 2.
Write the applications of computer.
Answer:
Applications of Computer: Banking, Insurance, Education, Marketing, Health care, Engineering design, Communication, Government, Weather forecasting.

Question 3.
What is an input device? Give two examples.
Answer:
Input Device: It is a peripheral device used to provide data and control signals to an information processing system.
Example: Keyboard, Mouse, Scanners, Digital cameras, Joysticks.

Question 4.
Name any three output devices.
Answer:
The three output devices are (a) Plotter (b) Speakers (c) Multimedia Projector.

Samacheer Kalvi 11th Computer Science Solutions Chapter 1 Introduction to Computers

Question 5.
Differentiate optical and Laser mouse
Answer:
Optical Mouse:

  • It is less sensitive towards the surface
  • Tracking power is less
  • Old technology
  • This uses the light source instead of ball to judge the motion of the pointer
  • Optical mouse has 3 buttons

Laser Mouse:

  • Highly sensitive
  • Tracking power is more
  • Latest technology
  • This uses Laser light
  • No. of buttons will vary from 3 to many

Question 6.
Write short note on impact printer.
Answer:

  1. Impact printer refers to the class of printers that work by banging a head or needle against an ink ribbon to make a mark on the paper.
  2. Impact printers are very noisy.
  3. These printers can print on multi-part.
  4. Example: Dot Matrix, Line printers.

Question 7.
Write the characteristics of sixth generation.
Answer:

  1. Natural Language Processing
  2. Development of Robotics
  3. Parallel and Distributed computing
  4. Artificial Intelligence
  5. Development of Robotics

Question 8.
Write the significant features of monitor.
Answer:

  1. A computer monitor is an output device which displays information in pictorial form.
  2. Monitors can be of monochrome which displays text or images in Black and white (or) in color.
  3. Different types of monitors are CRT, LCD and LED.

PART – IV

Question 1.
Explain the basic components of a computer with a neat diagram.
Answer:
(i) Basic components of Computer:
Samacheer Kalvi 11th Computer Science Solutions Chapter 1 Introduction to Computers
The Basic components are I/O Unit, CPU, ALU, Control Unit, Storage Unit.

Input unit:
Input unit is used to feed any form of data to the computer. It can be stored in the memory unit for further processing.
Ex. Keyboard, Mouse, Scanner, Retinal scanner, Track ball.

Central processing unit:
CPU is the major component and it controls the operation of all other components such as memory, input and output units. The CPU has three components namely control unit, Arithmetic and logic unit (ALU) and memory unit.

Arithmetic and logic unit:
The ALU is a part of the CPU where arithmetic operations like addition, subtraction, multiplication, divisions and logical operations will takes place.

Control unit:
The control unit controls the flow of data between the CPU, memory and I/O devices. It also f controls the entire operation of a computer.

Output unit:
An output device is a hardware component that conveys information to the user in an understandable form.
Example: Monitor, printer, plotter

Memory unit:
The memory stores everything that computer works with. The memory unit is of two types namely primary memory and secondary memory. The primary memory stores the data and instructions temporarily. Whereas the secondary memory stores the data permanently. The primary memory is volatile and secondary memory is non – volatile.

Question 2.
Discuss the various generations of computers.
Answer:
Samacheer Kalvi 11th Computer Science Solutions Chapter 1 Introduction to Computers

Question 3.
Explain the following
(a) Inkjet Printer
(b) Multimedia projector
(c) Bar code / QR code Reader
Answer:
(a) Inkjet Printer:

  1. Inkjet printer use colour cartridges which combined magenta, yellow and cyan inks to create colour tones.
  2. A black cartridge is used for monochrome output. Inkjet printer works by spraying ionised ink at a sheet of paper.
  3. The speed of inkjet printers generally range from 1 – 20 ppm (page per minute).
  4. An inkjet printer can spread millions of dots of ink at the paper every single second.
  5. They use the technology of firing ink by heating it so that it explodes towards the paper in bubbles or by using piezo electricity in which tiny electric currents are controlled by electronic circuits.

(b) Multimedia Projector:

  1. Multimedia projectors are used to produce computer output on a big screen.
  2. These are used to display presentations in meeting halls or in class rooms.

(c) Barcode/ QR code Reader:

  1. A Bar code is a pattern printed in lines of different thickness.
  2. The Bar code reader scans the information on the bar codes transmits to the computer for further processing.
  3. The system gives fast and error free entry of information into the computer.

Quick response (QR) code: The QR code is the two dimension bar code which can be read by a camera and processed to interpret the image.

Samacheer kalvi 11th Computer Science Introduction to Computers Additional Questions and Answers

PART – 1
I. Choose the correct answer

Question 1.
Natural language processing is a component of ………………
(a) ULSI
(b) AI
(c) ENIAC
(d) OCR
Answer:
(b) AI

Question 2.
Computer is an ……………… device.
(a) Electrical
(b) Electronic
(c) Digital
(d) Memory
Answer:
(b) Electronic

Samacheer Kalvi 11th Computer Science Solutions Chapter 1 Introduction to Computers

Question 3.
Main component of second generation ………………
(a) vacuum tubes
(b) microprocessor
(c) integrated circuits
(d) transistor
Answer:
(d) transistor

Question 4.
Microprocessor is the component of ……………… generation.
(a) first
(b) fourth
(c) third
(d) all the above
Answer:
(b) fourth

Question 5.
Period of fourth generation ………………
(a) 1975 – 1980
(b) 1964 – 1975
(c) 1955 – 1964
(d) 1942 – 1955
Answer:
(a) 1975 – 1980

Question 6.
Hardware is the ……………… component of a computer.
(a) physical
(b) electrical
(c) electronic
(d) user
Answer:
(a) physical

Question 7.
Which word can be related to the inkjet printer ………………
(a) Airlines
(b) Piezo – electricity
(c) matrix
(d) plotter
Answer:
(b) Piezo – electricity

Question 8.
The printing speed of Impact printers varies from ………………
(a) 40 to 1540 CPS
(b) 50 to 150 CPS
(c) 90 to 1500 CPS
(d) 30 to 1550 CPS
Answer:
(d) 30 to 1550 CPS

Question 9.
Which printer using the carbon papers ………………
(a) laser printer
(b) non – impact printers
(c) Impact printer
(d) all the above
Answer:
(c) Impact printer

Question 10.
The speed of inkjet printers generally range from ………………
(a) 1 – 20 PPM
(b) 1 – 22 PPM
(c) 10 – 20 PPM
(d) 11 – 20 PPM
Answer:
(a) 1 – 20 PPM

Samacheer Kalvi 11th Computer Science Solutions Chapter 1 Introduction to Computers

Question 11.
Retinal scanner uses the technique of ………………
(a) GUI
(b) UI
(c) Biometric
(d) None of these
Answer:
(c) Biometric

Question 12.
Biometric technique followed by ………………
(a) printer
(b) plotter
(c) finger print scanner
(d) OCR
Answer:
(c) finger print scanner

Question 13.
The components of CPU ………………
(a) control unit
(b) ALU
(c) Memory unit
(d) all the above
Answer:
(d) all the above

Question 14.
The primary memory is embedded with ……………… types of technologies.
(a) 2
(b) 3
(c) 5
(d) 7
Answer:
(a) 2

Question 15.
Computer monitor displays the information in the form of ……………….
(a) vertical
(b) pictorial
(c) horizontal
(d) numeric
Answer:
(b) pictorial

Question 16.
Processed data is ………………
(a) information
(b) primary data
(c) data
(d) message
Answer:
(a) information

Question 17.
Who invented analytical engine?
(a) Charles Babbage
(b) John von Newman
(c) Blaise pascal
(d) Dennis Richard
Answer:
(a) Charles Babbage

Question 18.
Assembly language was introduced in which computer generation?
(a) First
(b) Second
(c) Third
(d) Fourth
Answer:
(b) Second

Question 19.
In which generation UNIVACI was used?
(a) First
(b) Second
(c) Third
(d) Fourth
Answer:
(a) First

Question 20.
1 BM 1401 belongs to which computer generation?
(a) First
(b) Second
(c) Third
(d) Fourth
Answer:
(b) Second

Question 21.
IBM 1620 belongs to which computer generation of computers?
(a) I
(b) II
(c) III
(d) IV
Answer:
(b) II

Question 22.
UNIVAC 1108 belongs to which generation?
(a) First
(b) Third
(c) Second
(d) Fourth
Answer:
(c) Second

Question 23.
Honeywell 6000 series belongs to ……………… generation.
(a) First
(b) Second
(c) Third
(d) Fourth
Answer:
(c) Third

Question 24.
Which is the first fully functional electronic computer?
(a) EBSAC
(b) ENIAC
(c) EDSAC
(d) EDIAC
Answer:
(b) ENIAC

Question 25.
NLP is a component of ………………
(a) AI
(b) Hardware
(c) Circuit
(d) Electronics
Answer:
(a) AI

Question 26.
Which is a raw fact about an entity?
(a) Information
(b) Processed data
(c) data
(d) record
Answer:
(c) data

Question 27.
Which input device is a pointing device?
(a) Keyboard
(b) Monitor
(c) Mouse
(d) Scanner
Answer:
(c) Mouse

Question 28.
Which controls the entire operation of a computer?
(a) ALU
(b) CU
(c) BUS
(d) I/O unit
Answer:
(b) CU

Question 29.
Arithmetic and logical computation are done by ………………
(a) CU
(b) ALU
(c) BUS
(d) memory
Answer:
(b) ALU

Question 30.
Which of the following stores the instructions and data?
(a) ALU
(b) CU
(c) BUS
(d) memory
Answer:
(d) memory

Question 31.
Which conveys information to the user in an understandable form?
(a) Input unit
(b) CU
(c) Output unit
(d) Bus
Answer:
(c) Output unit

Question 32.
Which is a volatile memory?
(a) Primary memory
(b) Secondary memory
(c) ROM
(d) EPROM
Answer:
(a) Primary memory

Samacheer Kalvi 11th Computer Science Solutions Chapter 1 Introduction to Computers

Question 33.
CD – ROM, DVD – ROM, DVD – ROM are examples of which of the following memory.
(a) read / write
(b) volatile
(c) primary
(d) non – volatile
Answer:
(d) non – volatile

Question 34.
Hard disk. CD – ROM, DVD – ROM are examples of which of the following memory.
(a) read only
(b) primary
(c) secondary
(d) volatile
Answer:
(c) secondary

Question 35.
Which one of the following is true?
(a) The no. of keys in any keyboard is always the same.
(b) All keyboard have strictly has the same layout.
(c) All keyboards are wired keyboards.
(d) All keyboards have alphabet and numeric keys.
Answer:
(d) All keyboards have alphabet and numeric keys.

Question 36.
Caps lock key, Num lock key are ………………
(a) Functional keys
(b) Lock keys
(c) GUI keys
(d) Direction keys
Answer:
(b) Lock keys

Question 37.
Mechanical, optical and laser are types of which input device.
(a) Keyboard
(b) Mouse
(c) Scanner
(d) Printer
Answer:
(b) Mouse

Question 38.
Who invented mouse?
(a) Douglas Engelbart
(b) Blaise paseal
(c) Bill gates
(d) Eckert
Answer:
(a) Douglas Engelbart

Question 39.
Which one of the following mouse type has more than 3 bottons and can be programmed?
(a) Mechanical
(b) Optical
(c) Laser
(d) 3D
Answer:
(c) Laser

Question 40.
The non – impact printer using similar technology used by photo copier is ……………… printer.
(a) Inkjet
(b) dot matrix
(c) laser
(d) line matrix
Answer:
(c) laser

Question 41.
The device that reads the information directly into the computer’s memory and works like a Xerox machine is ………………
(a) plotter
(b) scanner
(c) touch screen
(d) track ball
Answer:
(b) scanner

Question 42.
The oinput device used to display computer output on big screen is ………………
(a) line matrix printer
(b) dot matrix printer
(c) multimedia projector
(d) monitor
Answer:
(c) multimedia projector

Question 43.
The output device similar to upside-down design of a mouse ………………
(a) laser mouse
(b) optical mouse
(c) mechanical mouse
(d) track ball
Answer:
(d) track ball

Question 44.
Which is false?
(a) Light pen is very easy to use and very accurate input device
(b) Mouse and light pen are pointing devices.
(c) Monochrome monitors do not display multiple colours.
(d) Retinal scanner and finger print scanner are used for security.
Answer:
(a) Light pen is very easy to use and very accurate input device

Question 45.
Which of the following uses biometrics and unique pattern of retinal blood vessels?
(a) Retinal track
(b) Finger print scanner
(c) Optical scanner
(d) Retinal scanner
Answer:
(d) Retinal scanner

Question 46.
Which input device is a pointing device?
(a) Keyboard
(b) Monitor
(c) light pen
(d) Scanner
Answer:
(c) light pen

Question 47.
The input device that detects characters printed or written on paper is ………………
(a) Voice input system
(b) Track ball
(c) Optical character reader
(d) 3D mouse
Answer:
(c) Optical character reader

Question 48.
……………… converts spoken words to machine-readable form.
(a) Voice input system
(b) Speaker
(c) Optical character reader
(d) Scanner
Answer:
(a) Voice input system

Samacheer Kalvi 11th Computer Science Solutions Chapter 1 Introduction to Computers

Question 49.
CCD stands for ………………
(a) Code Converting Device
(b) Code Change Device
(c) Change Code Device
(d) Charge Coupled Device
Answer:
(d) Charge Coupled Device

Question 50.
The input device in which 4 to 50 keys are arranged in the cluster.
(a) Keyboard
(b) Mouse keys
(c) Keyer
(d) Scanner
Answer:
(c) Keyer

Question 51.
……………… are picture elements.
(a) Picture Point
(b) Monitor
(c) Routers
(d) Pixels
Answer:
(d) Pixels

Question 52.
Match the following
Samacheer Kalvi 11th Computer Science Solutions Chapter 1 Introduction to Computers
(a) (4) (3) (1) (2)
(b) (1) (2) (3) (4)
(c) (2) (3) (1) (2)
(d) (1) (2) (4) (3)
Answer:
(a) (4) (3) (1) (2)

Question 53.
The ……………… printer use the same technology used by photo copier.
(a) Inkjet
(b) dot matrix
(c) line
(d) laser
Answer:
(d) laser

Question 54.
Which is the first step when you on the computer?
(a) Default application is executed
(b) BIOS starts
(c) Printer drivers are loaded
(d) Checks FAT
Answer:
(b) BIOS starts

Question 55.
Printer are of types.
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(a) 2

Question 56.
The two types of booting are ………………
(a) soft and hard
(b) warm and cold
(c) heavy and light
(d) standard and default
Answer:
(b) warm and cold

Question 57.
Who is considered to be the father of computers?
(a) Charles Babbage
(b) John Von Nuemann
(c) John Napier
(d) Dennis Ritchie
Answer:
(a) Charles Babbage

Question 58.
Analytify engine was developed in the year.
(a) 1827
(b) 1837
(c) 1847
(d) 1857
Answer:
(b) 1837

Question 59.
Who invented ENIAC?
(a) J. Presper Eckert
(b) J. Napier
(c) J. Van Nueman
(d) J. Mauchaley
Answer:
(a) J. Presper Eckert

Question 60.
Identify the computer which belongs to third generation?
(a) EDVAC
(b) ENIAC
(c) IBM 1620
(d) IBM 360
Answer:
(d) IBM 360

Question 61.
Which is used as a component of second generation computers?
(a) Vacuum Tubes
(b) Transistor
(c) IC
(d) VLSI
Answer:
(b) Transistor

Question 62.
Expand BIOS?
(a) Basic Input Output System
(b) Biased Input Output System
(c) Battery Input Output System
(d) Booting Input Output System
Answer:
(a) Basic Input Output System

Samacheer Kalvi 11th Computer Science Solutions Chapter 1 Introduction to Computers

Question 63.
Expand CPS.
(a) Correction Per Second
(b) Characters Per Second
(c) Calculations Per Second
(d) Cording Per Second
Answer:
(b) Characters Per Second

Question 64.
Expand ENIAC.
(a) Electronic Number Integrated Algebra Calculation
(b) Electronic Numerical Integrator and Calculator
(c) Electronic Null Interpreter and Compiler
(d) Electronic Null Interpreter and Compiler
Answer:
(b) Electronic Numerical Integrator and Calculator

Question 65.
Which is the first known calculating device?
(a) Slide rule
(b) Rotating wheel calculator
(c) Abacus
(d) Daisywheel
Answer:
(c) Abacus

Question 66.
Which component is used in second generation computers?
(a) Vacuum tubes
(b) Transistor
(c) IC
(d) VLT
Answer:
(b) Transistor

Question 67.
Artificial Intelligence was introduced in which generation of computers?
(a) V
(b) II
(c) III
(d) IV
Answer:
(a) V

Question 68.
Match the following.
Samacheer Kalvi 11th Computer Science Solutions Chapter 1 Introduction to Computers
(a) (4) (3) (2) (1)
(b) (1) (2) (3) (4)
(c) (2) (3) (1) (2)
(d) (1) (2) (4) (3)
Answer:
(a) (4) (3) (2) (1)

Question 69.
In which generation of computers, NLP was developed?
(a) First
(b) Second
(c) Fifth
(d) Sixth
Answer:
(d) Sixth

Question 70.
Identify which is true?
(a) Portable computers were introduced in fourth generation.
(b) ULSI was used in the fifth generation.
(c) High level languages were used in HI generation.
(d) All the above statements are true.
Answer:
(d) All the above statements are true.

Question 71.
Expand NLP?
(a) Natural Language Processing
(b) Netural Language Processing
(c) New Laptop Processor
(d) New Language Processor
Answer:
(a) Natural Language Processing

Question 72.
OCR stands for ………………
(a) Optimal Compiler Recorder
(b) Optimal Character Recorder
(c) Optimum Charge Recorder
(d) Optimal Character Resolution
Answer:
(b) Optimal Character Recorder

Question 73.
Which is the meaning for the term computer?
(a) To estimate
(b) To calculate
(c) To connect
(d) To think
Answer:
(b) To calculate

Question 74.
Which is not a hardware component?
(a) Information
(b) Monitor
(c) Motherboard
(d) Keyboard
Answer:
(a) Information

Question 75.
What is the expansion of IPO?
(a) Input Process Output
(b) Internal Process Outsourcing
(c) Integrated program Output
(d) Integral project Output
Answer:
(a) Input Process Output

Question 76.
Identify the statement which is wrong?
(a) ALU performs addition, subtraction
(b) ALU controls flow of data between the CPU, memory and I/O devices
(c) Control unit controls the entire operation of computer
(d) The logical operation of ALU promotes the decision making
Answer:
(b) ALU controls flow of data between the CPU, memory and I/O devices

Samacheer Kalvi 11th Computer Science Solutions Chapter 1 Introduction to Computers

Question 77.
How many major classifications of memory are there?
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(a) 2

Question 78.
Which of the following input device scan the book?
(a) OMR
(b) OCR
(c) ECR
(d) OVR
Answer:
(b) OCR

Question 79.
Which of the folliwng is the two dimensional bar code?
(a) QR
(b) OCR
(c) OMR
(d) MICR
Answer:
(a) QR

Question 80.
Match the following.
Samacheer Kalvi 11th Computer Science Solutions Chapter 1 Introduction to Computers
(a) (4) (3) (1) (2)
(b) (1) (2) (3) (4)
(c) (1) (2) (4) (3)
(d) (4) (3) (2) (1)
Answer:
(a) (4) (3) (1) (2)

Question 81.
Which of the following input devices are classified as tactile, ergonomic, gaming?
(a) Keyboard
(b) Printer
(c) Monitor
(d) Mouse
Answer:
(d) Mouse

Question 82.
The main advantage of using the light pen is ……………….
(a) easy to use
(b) accurate
(c) easy to detect the characters
(d) drawing directly onto the screen
Answer:
(d) drawing directly onto the screen

Question 83.
When was the first computer monitor released?
(a) March 1, 1973
(b) March 1, 1972
(c) March 1, 1974
(d) March 1, 1970
Answer:
(a) March 1, 1973

Question 84.
The individual keys for letters, numbers and special characters are collectively called ………………. keys.
(a) character
(b) functional
(c) lock
(d) special
Answer:
(a) character

Question 85.
Which of the following device converts photographs into digital format?
(a) Digital camera
(b) Mouse
(c) Scanner
(d) Light pen
Answer:
(c) Scanner

Samacheer Kalvi 11th Computer Science Solutions Chapter 1 Introduction to Computers

Question 86.
Wired, wireless and virtual are the categories of ……………….
(a) mouse
(b) keyboard
(c) printer
(d) monitor
Answer:
(b) keyboard

Question 87.
Line printers can print how many lines per minute?
(a) 1500
(b) 1520
(c) 1000
(d) 1020
Answer:
(c) 1000

Question 88.
Which one of the following is the main characteristics of laser printer?
(a) Speed
(b) Resolution
(c) Reliability
(d) Durability
Answer:
(b) Resolution

Question 89.
Each dot in dot matrix printers produced by a tiny metal rod is called ……………….
(a) binary
(b) pixel
(c) resolution
(d) wire or pin
Answer:
(d) wire or pin

Question 90.
Expand DPI ……………….
(a) Dots Per Inch
(b) Dark Pen Ink
(c) Dark Page Ink
(d) Double Part ink
Answer:
(a) Dots Per Inch

PART – 2
II. Short Answers

Question 1.
Expand ANN, OCR
Answer:
ANN – Artificial Neural networks; OCR – Optical Character Recognition.

Question 2.
Name the different types of mouse available.
Answer:
Mechanical mouse, Optical mouse,-Laser mouse, Air mouse, 3D mouse, Tactile mouse, Ergonomic mouse and gaming mouse.

Question 3.
What are the advantages and disadvantages of light pen?
Answer:

  1. Advantage: Drawing directly on to the screen.
  2. Disadvantage: Hard to use and not accurate.

Question 4.
What is CCD?
Answer:
CCD is charge coupled device. Digital camera uses CCD electric chip. When light falls on the chip through the lens, it converts light rays into digital format.

Question 5.
Name some mouse actions.
Answer:
Move, click, double click, right click, drag and drop are some of the mouse actions.

Question 6.
Name the different types of keys available in the keyboard.
Answer:
Character keys, modifier keys, system and GUI keys, enter and editing keys, function keys, navigation keys, numeric keypad and lock keys.

Samacheer Kalvi 11th Computer Science Solutions Chapter 1 Introduction to Computers

Question 7.
What is NLP?
Answer:
NLP stands for Natural Language Processing which is a component of Artificial Intelligence (AI). It provides the ability to develop the computer program to understand human language.

Question 8.
Name the different types of monitors available.
Answer:
CRT (Cathode Ray Tube), LCD(Liquid Crystal Display), LED (Light Emitting Diode).

PART -2
III. Explain in Brief

Question 1.
Differentiate warm booting and cold booting?
Answer:
Cold Booting:
When the system starts from initial state i.e. it is switched on, we call it cold booting or Hard Booting. When the user presses the Power button, the instructions are read from the ROM to initiate the booting process.

Warm Booting:
When the system restarts or when Reset button is pressed, we call it Warm Booting or Soft Booting. The system does not start from initial state and so all diagnostic tests need not be carried out in this case. There are chances of data loss and system damage as the data might not have been stored properly.

Question 2.
Define non – impact printer.
Answer:
These printers do not use striking mechanism for printing. They use electrostatic or laser technology. Quality and speed of these printers are better than Impact printers. For example, Laser printers and Inkjet printers are non – impact printers.

Question 3.
Define Keyer?
Answer:
A Keyer is a device for signaling by hand, by way of pressing one or more switches. Modem keyers have a large number of switches but not as many as a full size keyboard. Typically, this number is between 4 and 50. A keyer differs from a keyboard, which has “no board”, but the keys are arranged in a cluster.

Samacheer Kalvi 11th Computer Science Solutions Chapter 1 Introduction to Computers

Question 4.
Write notes on Analytical Engine.
Answer:

  1. Analytical engine was developed by Charles Babbage in 1837.
  2. It has (a) Arithmetic Logic Unit (b) Basic Flow Control (c) Integrated memory.
  3. The concept of Analytical engine led to the development of modem computers.

Question 5.
List out the input and output devices of a computer?
Answer:
Input Devices:

  1. Keyboard
  2. Mouse
  3. Scanner
  4. Fingerprint Scanner
  5. Track Ball
  6. Retinal Scanner
  7. Light Pen
  8. Optical Character Reader
  9. Bar Code / QR Code Reader
  10. Voice Input Systems
  11. Digital Camera
  12. Touch Screen
  13. Keyer

Output Devices:

  1. Monitor
  2. Plotter
  3. Printers

Question 6.
Explain the mechanism of laser mouse?
Answer:

  1. Measures the motion and acceleration of pointer.
  2. Laser Mouse uses Laser Light
  3. Laser Mouse is highly sensitive and able to work on any hard surface.

Question 7.
Explain fingerprint scanner and retinal scanner?
Answer:
Fingerprint Scanner:
Finger print Scanner is a fingerprint recognition device used for computer security, equipped with the fingerprint recognition feature that uses biometric technology. Fingerprint Reader / Scanner is a very safe and convenient device for security instead of using passwords, which is vulnerable to fraud and is hard to remember.

Retinal Scanner:
This performs a retinal scan which is a biometric technique that uses unique patterns on a person’s retinal blood vessels.

Question 8.
Define touch screen?
Answer:
Touch Screen:
A touch screen is a display device that allows the user to interact with a computer by using the finger. It can be quite useful as an alternative to a mouse or keyboard for navigating a Graphical User Interface (GUI). Touch screens are used on a wide variety of devices such as computers, laptops, monitors, smart phones, tablets, cash registers and information kiosks. Some touch screens use a grid of infrared beams to sense the presence of a finger instead of utilizing touch-sensitive input.

PART – 4
IV. Explain in Detail

Question 1.
What is mouse? And explain its types?
Answer:
Mouse:
Mouse (wired/wireless) is a pointing device used to control the movement of the cursor on the display screen.
Samacheer Kalvi 11th Computer Science Solutions Chapter 1 Introduction to Computers

Question 2.
Explain input devices of a computer?
Input Devices:
Keyboard : Keyboard (wired / wireless, virtual) is the most common input device used today. The individual keys for letters, numbers and special characters are collectively known as character keys. This keyboard layout is derived from the keyboard of original typewriter.

Mouse:
Mouse (wired/wireless) is a pointing device used to control the movement of the cursor on the display screen.

Types of Mouses:

  1. Mechanical Mouse
  2. Optical Mouse.
  3. Laser Mouse

Scanner : Scanners are used to enter the information directly into the computer’s memory. This device works like a Xerox machine. The scanner converts any type of printed or written information including photographs into a digital format, which can be manipulated by the computer.

Track Ball : Track ball is similar to the upside – down design of the mouse. The user moves the ball directly, while the device itself remains stationary. The user spins the ball in various directions to navigate the screen movements.

Retinal Scanner : This performs a retinal scan which is a biometric technique that uses unique patterns on a person’s retinal blood vessels.

Light Pen : A light pen is a pointing device shaped like a pen and is connected to a monitor. The tip of the light pen contains a light-sensitive element which detects the light from the screen enabling the computer to identify the location of the pen on the screen. Light pens have the advantage of ‘drawing’ directly onto the screen, but this becomes hard to use, and is also not accurate.

Optical Character Reader : It is a device which detects characters printed or written on a paper with OCR, a user can scan a page from a book. The Computer will recognize the characters in the page as letters and punctuation marks and stores. The Scanned document can be edited using a word processor.

Bar Code / QR Code Reader : A Bar code is a pattern printed in lines of different thickness. The Bar code reader scans the information on the bar codes transmits to the Computer for further processing. The system gives fast and error free entry of information into the computer.

QR (Quick response) Code : The QR code is the two dimension bar code which can be read by a camera and processed to interpert the image

Voice Input Systems : Microphone serves as a voice Input device. It captures the voice data and send it to the Computer. Using the microphone along with speech recognition software can offer a completely new approach to input information into the Computer.

Digital Camera : It captures images / videos directly in the digital form. It uses a CCD (Charge Coupled Device) electronic chip. When light falls on the chip through the lens, it converts light rays into digital format.

Touch Screen : A touch screen is a display device that allows the user to interact with a computer by using the finger. It can be quite useful as an alternative to a mouse or keyboard for navigating a Graphical User Interface (GUI). Touch screens are used on a wide variety of devices such as computers, laptops, monitors, smart phones, tablets, cash registers and information kiosks. Some touch screens use a grid of infrared beams to sense the presence of a finger instead of utilizing touch – sensitive input.

Keyer : A Keyer is a device for signaling by hand, by way of pressing one or more switches. Modem keyers have a large number of switches but not as many as a full size keyboard. Typically, this number is between 4 and 50.

Samacheer Kalvi 12th Bio Zoology Solutions Chapter 5 Molecular Genetics

Students can Download Bio Zoology Chapter 5 Molecular Genetics Questions and Answers, Notes Pdf, Samacheer Kalvi 12th Bio Zoology Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Bio Zoology Solutions Chapter 5 Molecular Genetics

Samacheer Kalvi 12th Bio Zoology Molecular Genetics Text Book Back Questions and Answers

Question 1.
Hershey and Chase experiment with bacteriophage showed that
(a) Protein gets into the bacterial cells
(b) DNA is the genetic material
(c) DNA contains radioactive sulphur
(d) Viruses undergo transformation
Answer:
(b) DNA is the genetic material

Question 2.
DNA and RNA are similar with respect to
(a) Thymine as a nitrogen base
(b) A single-stranded helix shape
(c) Nucleotide containing sugars, nitrogen bases and phosphates
(d) The same sequence of nucleotides for the amino acid phenyl alanine
Answer:
(c) Nucleotide containing sugars, nitrogen bases and phosphates

Question 3.
A mRNA molecule is produced by
(a) Replication
(b) Transcription
(c) Duplication
(d) Translation
Answer:
(b) Transcription

Samacheer Kalvi 12th Bio Zoology Solutions Chapter 5 Molecular Genetics

Question 4.
The total number of nitrogenous bases in human genome is estimated to be about
(a) 3.5 million
(b) 35000
(c) 35 million
(d) 3.1 billion
Answer:
(d) 3.1 billion

Question 5.
E. coli cell grown on 15N medium are transferred to 14N medium and allowed to grow for two generations. DNA extracted from these cells is ultracentrifuged in a cesium chloride density gradient. What density distribution of DNA would you expect in this experiment?
(a) One high and one low density band
(b) One intermediate density band
(c) One high and one intermediate density band
(d) One low and one intermediate density band
Answer:
(d) One low and one intermediate density band

Question 6.
What is the basis for the difference in the synthesis of the leading and lagging strand of DNA molecules?
(a) Origin of replication occurs only at the 5’ end of the molecules
(b) DNA ligase works only in the 3’ → 5’ direction
(c) DNA polymerase can join new nucleotides only to the 3 ’ end of the growing stand
(d) Helicases and single-strand binding proteins that work at the 5’ end
Answer:
(d) Helicases and single-strand binding proteins that work at the 5’ end

Question 7.
Which of the following is the correct sequence of event with reference to the central dogma?
(a) Transcription, Translation, Replication
(b) Transcription, Replication, Translation
(c) Duplication, Translation, Transcription
(d) Replication, Transcription, Translation
Answer:
(d) Replication, Transcription, Translation

Question 8.
Which of the following statements about DNA replication is not correct?
(a) Unwinding of DNA molecule occurs as hydrogen bonds break
(b) Replication occurs as each base is paired with another exactly like it
(c) Process is known as semi – conservative replication because one old strand is conserved in the new molecule
(d) Complementary base pairs are held together with hydrogen bonds
Answer:
(b) Replication occurs as each base is paired with another exactly like it

Question 9.
Which of the following statements is not true about DNA replication in eukaryotes?
(a)) Replication begins at a single origin of replication.
(b) Replication is bidirectional from the origins.
(c) Replication occurs at about 1 million base pairs per minute.
(d) There are numerous different bacterial chromosomes, with replication occurring in each at the same time.
Answer:
(d) There are numerous different bacterial chromosomes, with replication occurring in each at the same time.

Question 10.
The first codon to be deciphered was which codes for
(a) AAA, proline
(b) GGG, alanine
(c) UUU, Phenylalanine
(d) TTT, arginine
Answer:
(c) UUU, Phenylalanine

Question 11.
Meselson and Stahl’s experiment proved __________
(a) Transduction
(b) Transformation
(c) DNA is the genetic material
(d) Semi-conservative nature of DNA replication
Answer:
(d) Semi-conservative nature of DNA replication

Question 12.
Ribosomes are composed of two subunits; the smaller subunit of a ribosome has a binding site for and the larger subunit has two binding sites for two
Answer:
mRNA, tRNA

Question 13.
Anoperonisa:
(a) Protein that suppresses gene expression
(b) Protein that accelerates gene expression
(c) Cluster of structural genes with related function
(d) Gene that switched other genes on or off
Answer:
(d) Gene that switched other genes on or off

Question 14.
When lactose is present in the culture medium:
(a) Transcription of lacy, lac z, lac a genes occurs
(b) Repressor is unable to bind to the operator
(c) Repressor is able to bind to the operator
(d) Both (a) and (b) are correct
Answer:
(d) Both (a) and (b) are correct

Question 15.
Give reasons: Genetic code is ‘universal’.
Answer:
The genetic code is universal. It means that all known living systems use nucleic acids and the same three base codons (triplet codon) direct the synthesis of protein from amino acids. . For example, the mRNA (UUU) codon codes for phenylalanine in all cells of all organisms. Some exceptions are reported in prokaryotic, mitochondrial and chloroplast genomes. However, similarities are more common than differences.

Question 16.
Name the parts marked ‘A’ and ‘B’ in the given transcription unit:
Answer:
Samacheer Kalvi 12th Bio Zoology Solutions Chapter 5 Molecular Genetics img 1

Question 17.
Differentiate – Leading strand and lagging strand
Answer:

  1. DNA polymerase I Involved DNA repair mechanism
  2. DNA polymerase II Involved DNA repair mechanism
  3. DNA polymerase III Involved in DNA replication

Question 18.
Differentiate – Template strand and coding strand.
Answer:
Template Strand: During replication, DNA strand having the polarity 3’ → 5’ act as template strand.

Coding Strand: During replication, DNA strand having the polarity 5’ → 3’ act as coding strand.

Question 19.
Mention any two ways in which single nucleotide polymorphism (SNPs) identified in the human genome can bring a revolutionary change in biological and medical science.
Answer:
Scientists have identified about 1.4 million locations, where single-base DNA differences (SNPs – Single nucleotide polymorphism – pronounced as ‘snips’) occur in humans. Identification of ‘SNIPS’ is helpful in finding chromosomal locations for disease-associated sequences and tracing human history.

Samacheer Kalvi 12th Bio Zoology Solutions Chapter 5 Molecular Genetics

Question 20.
State any three goals of the human genome project.
Answer:

  • Identify all the genes (approximately 30000) in human DNA.
  • Determine the sequence of the three billion chemical base pairs that make up the human DNA.
  • To store this information in databases.

Question 21.
In E.coli, three enzymes 0- galactosidase, permease, and transacetylase are produced in the presence of lactose. Explain why the enzymes are not synthesized in the absence of lactose.
Answer:
In the absence of lactose, the repressor protein binds to the operator and prevents the transcription of the structural gene by RNA polymerase, hence the enzymes are not produced. However, there will always be a minimal level of lac operon expression even in absence of lactose.

Question 22.
Distinguish between structural gene, a regulatory gene, and operator gene.
Answer:
Structure of the operon: Each operon is a unit of gene expression and regulation and consists of one or more structural genes and an adjacent operator gene that controls the transcriptional, activity of the structural gene.

  • The structural gene codes for proteins, rRNA, and tRNA required by the cell.
  • Promoters are the signal sequences in DNA that initiate RNA synthesis. RNA polymerase binds to the promoter prior to the initiation of transcription.
  • The operators are present between the promoters and structural genes. The repressor protein binds to the operator region of the operon.

Question 23.
A low level of expression of lac operon occurs at all the time in E-coli. Justify the statement.
Answer:
One of the enzyme synthesized by lac operon is permease which is involved in the transport of lactose into the cells. If the lac operon gets inactivated, permease is not synthesized hence lactose cannot enter the cell. Lactose acts as an inducer, binding to the repressor protein and switch on the operator to initiate gene expression.

Question 24.
Why the human genome project is called a megaproject?
Answer:
The international human genome project was launched in the year 1990. It was a mega project and took 13 years to complete. The human genome is about 25 times larger than the genome of any organism sequenced to date and is the first vertebrate genome to be completed. Human genome is said to have approximately 3 x 109 bp. HGP was closely associated with the rapid development of a new area in biology called bioinformatics.

Question 25.
From their examination of the structure of DNA, What did Watson and Crick infer about the probable mechanism of DNA replication, coding capability and mutation?
Answer:
Inference of Watson and Crick on DNA replication: They concluded that each of the DNA strand in a helix act as template during DNA replication leading to formation of new daughter DNA molecules, which are complementary to parental strand, (i.e., Semi-conservative method of replication) Inference on coding capability: During transcription, the genetic information in the DNA strand is coded to mRNA as complementary bases, (except for uracil in place of thymine in RNA) Inference on mutation: Any changes in the nucleotide sequence of DNA leads to the corresponding alteration in amino acid sequence of specific protein thus confirming the validity of genetic code.

Question 26.
Why tRNA is called an adapter molecule?
Answer:
The transfer RNA, (tRNA) molecule of a cell acts as a vehicle that picks up the amino acids scattered through the cytoplasm and also reads specific codes of mRNA molecules. Hence it is called an adapter molecule. This term was postulated by Francis Crick.

Question 27.
What are the three structural differences between RNA and DNA?
Answer:
DNA:

  1. Sugar is deoxyribose sugar.
  2. Double-stranded structure.
  3. Nitrogen bases are Adenine, Guanine, Cytosine and Thymine.

RNA:

  1. Sugar is ribose sugar.
  2. Single-stranded molecule.
  3. Nitrogen bases are Adenine, Guanine, Cytosine and Uracil.

Question 28.
Name the anticodon required to recognize the following codons:
AAU, CGA, UAU, and GCA.
Answer:
UUA, GCU, AUA and CGU.

Question 29.
(a) Identify the figure given below
(b) Redraw the structure as a replicating fork and label the parts
(c) Write the source of energy for this replication and name the enzyme involved in this process.
(d) Mention the differences in the synthesis of protein, based on the polarity of the two template strands.
Answer:
(a) Replication fork
(b)
Samacheer Kalvi 12th Bio Zoology Solutions Chapter 5 Molecular Genetics img 2
(c) Deoxy nucleotide, triphosphate acts as an energy source for replication. DNA polymerase is used for replication
(d) mRNA contacting information for protein synthesis will develop from DNA strand having polarity 5’ → 3’

Question 30.
If the coding sequence in a transcription unit is written as follows:
5’ TGCATGCATGCATGCATGCATGCATGC 3’
Write down the sequence of mRNA.
Answer:
mRNA sequence is 3’ACGUACGUACGUUCGUACGUACGUACG5’

Question 31.
How is the two-stage process of protein synthesis advantageous?
Answer:
The split gene feature of eukaryotic genes is almost entirely absent in prokaryotes. Originally each exon may have coded for a single polypeptide chain with a specific function. Since exon arrangement and intron removal are flexible, the exon coding for these polypeptide subunits act as domains combining in various ways to form new genes. Single genes can produce different functional proteins by arranging their exons in several different ways through alternate splicing patterns, a mechanism known to play an important role in generating both protein and functional diversity in animals. Introns would have arosen before or after the evolution of eukaryotic gene.

If introns arose late how did they enter eukaryotic gene? Introns are mobile DNA sequences that can splice themselves out of, as well as into, specific ‘target sites’ acting like mobile transposon-like elements (that mediate transfer of genes between organisms – Horizontal Gene Transfer – HGT). HGT occurs between lineages of prokaryotic cells, or from prokaryotic to eukaryotic cells and between eukaryotic cells. HGT is now hypothesized to have played a major role in the evolution of life on Earth.

Question 32.
Why did Hershey and Chase use radioactively labelled phosphorous and sulphur only? Would they have got the same result if they use radiolabelled carbon and nitrogen?
Answer:
Generally proteins contain sulphur but not phosphorous and nucleic acid (DNA) contains , phosphorous but not sulphur. Hence Hershey – Chase used radioactive isotopes of sulphur (35S) and phosphorus (32P) to keep separate track of viral protein and nucleic acid in culture medium. The expected result cannot be achieved, if radioactive carbon and nitrogen is used, since these molecules are present in both DNA and proteins.

Samacheer Kalvi 12th Bio Zoology Solutions Chapter 5 Molecular Genetics

Question 33.
Explain the formation of a nucleosome.
Answer:
Komberg proposed a model for the nucleosome, in which 2 molecules of the four histone proteins H2A, H2B, H3 and H4 are organized to form a unit of eight molecules called histone octamere.
The negatively charged DNA is wrapped around the positively charged histone octamere to form a structure called nucleosome. A typical nucleosome contains 200 bp of DNA helix. The histone octameres are in close contact and DNA is coiled on the outside of nucleosome.
Samacheer Kalvi 12th Bio Zoology Solutions Chapter 5 Molecular Genetics img 12

Question 34.
It is established that RNA is the first genetic material. Justify giving reasons.
Answer:
Three molecular biologists in the early 1980’s (Leslie Orgel, Francis Brick and Carl Woese) independently proposed the ‘RNA world’ as the first stage in the evolution of life, a stage when RNA catalysed all molecules necessary for survival and replication. The term ‘RNA world’ first used by Walter Gilbert in 1986, hypothesizes RNA as the first genetic on Earth. There is now enough evidence to suggest that essential life processes (such as metabolism, translation and splicing etc.,) evolved around RNA. RNA has the ability to act as both genetic material and catalyst. There are several biochemical reactions in living systems that are catalyzed by RNA. This catalytic RNA is known as a ribozyme. But, RNA being a catalyst was reactive and hence unstable.

This led to the evolution of a more stable form of DNA, with certain chemical modifications. Since DNA is a double-stranded molecule having a complementary strand, it has resisted changes by evolving a process of repair. Some RNA molecules function as gene regulators by binding to DNA and affect gene expression. Some viruses use RNA as genetic material. Andrew Fire and Craig Mellow (recipients of the Nobel Prize in 2006) were of the opinion that RNA is an active ingredient in the chemistry of life.

Samacheer Kalvi 12th Bio Zoology Molecular Genetics Additional Questions and Answers

1 – Mark Questions

Question 1.
The term‘gene’was coined by ___________
Answer:
Wilhelm Johannsen

Question 2.
Whose experiment finally provided convincing evidence that DNA is the genetic material?
(a) Griffith experiment
(b) Avery, Macleod, and McCarty’s experiment
(c) Hershey-Chase experiment
(d) Urey-Miller’s experiment
Answer:
(c) Hershey-Chase experiment

Question 3.
In Hershey – Chase experiment, the DNA of T2 phase was made radioactive by using ___________
(a) 32P
(b) 32S
(c) 35P
(d) 32S
Answer:
(a) 32P

Question 4.
A nucleoside is composed of ___________
(a) Sugar and Phosphate
(b) Nitrogen base and Phosphate
(c) Sugar and a Nitrogen base
(d) Sugar, Phosphate, and Nitrogenous base
Answer:
(c) Sugar and a Nitrogen base

Question 5.
Identify the incorrect statement
(a) a base is a substance that accepts H+ ion
(b) Both DNA and RNA have four bases
(c) Purines have a single carbon-nitrogen ring
(d) Thymine is unique for DNA
Answer:
(c) Purines have a single carbon-nitrogen ring

Samacheer Kalvi 12th Bio Zoology Solutions Chapter 5 Molecular Genetics

Question 6.
Watson and Crick proposed their double helical DNA model based on the X-ray diffraction analysis of ___________
(a) Erwin Chargaff
(b) Meselson and Stahl
(c) Wilkins and Franklin
(d) Griffith
Answer:
(c) Wilkins and Franklin

Question 7.
The term ‘RNA world’ was first used by ___________
Answer:
Walter Gilbert

Question 8.
The distance between two consecutive base pairs in DNA is ___________
(a) 0.34 nm
(b) 3.4 nm
(c) 0.034 nm
(d) 34 nm
Answer:
(a) 0.34 nm

Question 9.
If the length of E. coli DNA is 1.36 mm, the number of base pairs is ___________
(a) 0.36 × 106m
(b) 4 × 106m
(c) 0.34 × 10-9nm
(d) 4 × 10-9m
Answer:
(b) 4 × 106m

Question 10.
Identify the proper sequence in the organisation of the eukaryotic chromosome.
(a) Nucleosome – Solenoid – Chromatid
(b) Chromatid – Nucleosome – Solenoid
(c) Solenoid – chromatin – DNA
(d) Nucleosome – solenoid – genophore
Answer:
(a) Nucleosome – Solenoid – Chromatid

Question 11.
Assertion (A) : Genophore is noticed in prokaryotes.
Reason (R) : Bacteria possess circular DNA without chromatin organization.
(a) Both A and R are correct
(b) A is correct R is incorrect
(c) R explains A
(d) A is incorrect R is correct
Answer:
(c) R explains A

Question 12.
Assertion (A): Heterochromatin is transcriptionally active.
Reason (R): Tightly packed chromatin which stains dark.
(a) Both A and R are correct
(b) A is correct R is incorrect
(c) R explains A
(d) A is incorrect R is correct
Answer:
(d) A is incorrect R is correct

Question 13.
Assertion (A) : semi-conservative model was proposed by Hershey and Chase.
Reason (R) : The daughter DNA contains only new strands.
(a) Both A and R are incorrect
(b) A is correct R is incorrect
(c) R explains A
(d) A is incorrect R is correct
Answer:
(a) Both A and R are incorrect

Question 14.
Kornberg enzyme is called as _____
Answer:
DNA polymerase I

Samacheer Kalvi 12th Bio Zoology Solutions Chapter 5 Molecular Genetics

Question 15.
Replication of DNA occurs at __________ phase of cell cycle.
(a) M
(b) S
(c) G1
(d) G2
Answer:
(b) S

Question 16.
Semi-conservative model of replication was proved by __________
(a) Hershey and Chase
(b) Griffith
(c) Meselson and Stahl
(d) Macleod and McCarty
Answer:
(c) Meselson and Stahl

Question 17.
How many types of DNA polymerases does a eukaryotic cell possess?
(a) two
(b) three
(c) four
(d) five
Answer:
(d) Five

Question 18.
Identify the incorrect statement
(a) Replication occurs at ori – site of DNA
(b) Deoxy nucleotide triphosphate acts as a substrate
(c) Unwinding of DNA strand is carried out by topoisomerase
(d) DNA polymerase catalyses the polymerization at 3-OH
Answer:
(c) Unwinding of DNA strand is carried out by topoisomerase

Question 19.
The discontinuously synthesized fragments of lagging strand are called ________
Answer:
Okazaki fragments

Question 20.
Retroviruses possess ________ as genetic material.
Answer:
RNA

Question 21.
Which is NOT a part of the transcription unit?
(a) Promoter
(b) Operator
(c) Structural gene
(d) Terminator
Answer:
(b) Operator

Question 22.
Goldberg – Hogness box of eukaryotes is equivalent to ________ of prokaryotes.
Answer:
Pribnow box

Question 23.
Okazaki fragments are joined by the enzyme ________ during DNA replication.
Answer:
DNA ligase

Question 24.
Samacheer Kalvi 12th Bio Zoology Solutions Chapter 5 Molecular Genetics img 3
Answer:
(a) A – iv, B – i, C – ii, D – iii

Question 25.
The RNA polymerase of prokaryotes binds with the factor to initiate polymerization.
(a) rho
(b) theta
(c) sigma
(d) psi
Answer:
(c) sigma

Question 26.
Samacheer Kalvi 12th Bio Zoology Solutions Chapter 5 Molecular Genetics
(a) Capping
(b) Tailing
(c) Splicing
(d) Transcribing
Answer:
(c) Splicing

Samacheer Kalvi 12th Bio Zoology Solutions Chapter 5 Molecular Genetics

Question 27.
Which of the following feature is absent in prokaryotes?
(a) Prokaryotes possess three major types of RNAs
(b) Structural genes are polycistronic
(c) Initiation process of transcription requires ‘P’ factor
(d) Split gene feature
Answer:
(d) Split gene feature

Question 28.
Which of the following sequence has completely translated?
(i) AGA, UUU, UGU, AGU, UAG
(ii) AUG, UUU, AGA, UAC, UAA
(iii) AAA, UUU, UUG, UGU, UGA
(iv) AUG,AAU,AAC,UAU,UAG
(a) i and ii
(b) ii only
(c) i and iii
(d) ii and iv
Answer:
(d) ii and iv

Question 29.
Capping of mRNA occurs using __________
(a) Poly A residues
(b) Methyl guanosine triphosphate
(c) Deoxy ribonucleotide triphosphate
(d) Ribonucleotide triphosphate
Answer:
(b) Methyl guanosine triphosphate

Question 30.
One of the aspects is not a feature of genetic code?
(a) Specific
(b) Degenerate
(c) Universal
(d) Ambiguous
Answer:
(d) Ambiguous

Question 31.
Which of the triplet codon is not a code of proline?
(i) CCU
(ii) CAU
(iii) CCG
(iv) CAA
(a) i only
(b) ii and iv
(c) iii only
(d) all the above
Answer:
(b) ii and iv

Question 32.
Coding sequences found in split genes are called.
(a) Operons
(b) Introns
(c) Exons
(d) Cistron
Answer:
(c) Exons

Question 33.
Which of the following mRNA yields 6 aminoacids after translation?
(i) UCU UAU AGU CGA UGC AGU UGA AAA UUU
(ii) UGA AGA UAG GAG CAU CCC UAC UAU GAU
(iii) GUC UGC UGG GCU GAU UAA AGG AGC AUU
(iv) AUG UAC CAU UGC UGA UGC AGG AGC CCG
Answer:
(i) UCU UAU AGU CGA UGC AGU UGA AAA UUU

Question 34.
The transcription termination factor associated with RNA polymerase in prokaryotes is
(a) θ
(b) σ
(c) ρ
(d) ∑
Answer:
(c) ρ

Samacheer Kalvi 12th Bio Zoology Solutions Chapter 5 Molecular Genetics

Question 35.
In a DNA double strand, if guanine is of 30%, what will be the percentage of thymine?
(a) 100%
(b) 20%
(c) 10%
(d) 70%
Answer:
(b) 20%

Question 36.
Identify the triplet pairs that code for Tyrosine
(a) UUU, UUC
(b) UAU, UAU
(c) UGC, UGU
(d) CAU, CAC
Answer:
(b) UAU, UAU

Question 37.
Samacheer Kalvi 12th Bio Zoology Solutions Chapter 5 Molecular Genetics img 4
Answer:
A – ii B – i C – iv D – iii

Question 38.
AUG code is for __________
(a) Arginine
(b) Tyrosine
(c) Tryptophan
(d) Methionine
Answer:
(d) Methionine

Question 39.
The sequence of bases in the coding strand of DNA is G A G T  T A G C A G G C, then the sequence of codons in the primary transcript is
(a) C U C A U A C G C C C G
(b) C U C A A U C G U C C G
(c) U C A G A U C U G C G C
(d) U U C A A U C G U G C G
Answer:
(b) C U C A A U C G U C C G

Question 40.
The promoter region of eukaryote is __________
(a) TATAA
(b) AUGUT
(c) UUUGA
(d) AAAAU
Answer:
(a) TATAA

Question 41.
Match the following:
(A) AUG – (i) Tyrosine
(B) UGA – (ii) Glycine
(C) UUU – (iii) Methionine
(D) GGG – (iv) Phenylalanine
(a) A – iii B – i C – iv D – ii
(b) A – iii B – ii C – i D – iv
(c) A – iv B – i C – iii D – ii
(d) A – ii B – i C – iv D – iii
Answer:
(d) A – ii B – i C – iv D – iii

Question 42.
__________ number of codons, codes for cystine.
Answer:
Two

Question 43.
In sickle cell anaemia, the __________ codon of β – globin gene is modified.
(a) Eighth
(b) Seventh
(c) Sixth
(d) Nineth
Answer:
(c) Sixth

Question 44.
Pick out the incorrect statement.
(a) tRNA acts as a adapter molecule
(b) Stop codons donot have tRNA’s
(c) Addition of aminoacid leads to hydrolysis of tRNA
(d) tRNA has four major loops
Answer:
(c) Addition of aminoacid leads to hydrolysis of tRNA

Question 45.
Which of the following antibiotic inhibits the interaction between tRNA and mRNA?
(a) Neomycin
(b) Streptomycin
(c) Tetracycline
(d) Chloramphenicol
Answer:
(a) Neomycin

Question 47.
The cluster of genes with related function is called _________
(a) Cistron
(b) Operon
(c) Muton
(d) Recon
Answer:
(b) Operon

Question 48.
Repressor protein of Lac operon binds to __________ of operon.
(a) Promoter region
(b) Operator region
(c) terminator region
(d) inducer region
Answer:
(b) Operator region

Question 49.
Lac Z gene codes for __________
(a) Permease
(b) transacetylase
(c) β -galactosidase
(d) Aminoacyl transferase
Answer:
(c) β -galactosidase

Samacheer Kalvi 12th Bio Zoology Solutions Chapter 5 Molecular Genetics

Question 50.
Lac operon model was proposed by __________
Answer:
Jacob and Monod

Question 51.
Approximate count of base pair in human genome is __________
Answer:
3 × 109 bp

Question 52.
Automated DNA sequences are developed by.
Answer:
Frederick Sanger

Question 53.
Which of the chromosome has higher gene density?
(a) Chromosome 20
(b) Chromosome 19
(c) Chromosome 13
(d) Chromosome Y
Answer:
(b) Chromosome 19

Question 54.
Number of genes located in chromosome Y is __________
(a) 2968
(b) 213
(c) 2869
(d) 231
Answer:
(d) 231

Question 55.
How many structural genes are located in lac operon of E.Coli?
(a) 4
(b) 3
(c) 2
(d) 1
Answer:
(b) 3

Question 56.
DNA finger printing technique was developed by
(a) Jacob and Monod
(b) Alec Jeffreys
(c) Frederick Sanger
Answer:
(b) Alec Jeffreys

Question 57.
In DNA fingerprinting, separation of DNA fragments is done by __________
(a) Centrifugation
(b) Electrophoresis
(c) X-ray diffraction
(d) denaturation
Answer:
(b) Electrophoresis

Question 58.
SNP stands for
(a) Single nucleotide Polymorphism
(b) Single Nucleoside Polypeptide
(c) Single nucleotide Polymorphism
(d) Single nucleotide polymer
Answer:
(a) Single nucleotide Polymorphism

Question 59.
Specific sequences of mRNA that are not translated are __________
Answer:
UnTranslated Regions (UTR)

Question 60.
Non-coding or intervening DNA sequence is called __________

Question 61.
_______ Intron is the monomer of DNA.
Answer:
Nucleotide

Question 62.
Which one of the following is wrongly matched?
(a) Transcription – Copying information from DNA to RNA
(b) Translation – Decoding information from mRNA to protein
(c) Replication – Making of DNA copies
(d) Splicing – Joining of exons with introns
Answer:
(d) Splicing – Joining of exons with introns

2- Mark Questions

Question 1.
Who proposed One gene – One enzyme hypothesis? Define it.
Answer:
George Beadle and Edward Tatum proposed One gene – One enzyme hypothesis which states that one gene controls the production of one enzyme.

Question 2.
Differentiate nucleoside from nucleotide.
Answer:

  1. Nucleoside: Nucleoside subunit is composed of nitrogenous bases linked to a pentose sugar molecule.
  2. Nucleotide: Nucleotide subunit is composed of nitrogenous bases, a pentose sugar and a phosphate group.

Question 3.
State the key differences between DNA and RNA.
Answer:
DNA:

  1. DNA is made of deoxyribose sugar.
  2. Nitrogenous bases of DNA are Adenine, Guanine, Cytosine and Thymine.

RNA:

  1. RNA is made of ribose sugar.
  2. Nitrogenous bases of RNA are Adenine, Guanine, Cytosine and Uracil.

Question 4.
Point out the nitrogenous bases of RNA.
Answer:
Adenine, Guanine, Cytosine and Uracil.

Question 5.
What makes the DNA and RNA as acidic molecules?
Answer:
The phosphate functional group (PO4) gives DNA and RNA the property of an acid at physiological pH, hence the name nucleic acid.

Question 6.
Which type of bond is formed

  1. between a purine and pyrimidine base?
  2. between the pentose sugar and adjacent nucleotide?

Answer:

  1. Purine and pyrimidine bases are linked by hydrogen bonds.
  2. Pentose sugar is linked to adjacent nucleotide by phosphodiester bonds.

Question 7.
DNA acts as genetic material for majority of living organisms and not the RNA. Give reasons to support the statement.
Answer:

  1. RNA was reactive and hence highly unstable.
  2. Some RNA molecules acts as gene regulators by binding to DNA and affect gene expression.
  3. Uracil of RNA is less stable than thymine of DNA.

Question 8.
Name any two viruses whose genetic material is RNA.
Answer:

  1. Tobacco Mosaic Virus (TMV)
  2. Bacteriophage 0B

Question 9.
What are the properties that a molecule must possess to act as genetic material?
Answer:

  1. Self replication
  2. Information storage
  3. Stability
  4. Variation through mutation

Question 10.
How many base pairs are present in one complete turn of DNA helix? What is the distance between two consecutive base pairs?
Answer:
There are ten base pairs in each turn with a distance of 0.34 x 109m between two adjacent base pairs.

Question 11.
What is a genophore?
Answer:
In prokaryotes such as E. coli though they do not have defined nucleus, the DNA is not scattered throughout the cell. DNA (being negatively charged) is held with some proteins (that have positive charges) in a region called the nucleoid. The DNA as a nucleoid is organized into large loops held by protein. DNA of prokaryotes is almost circular and lacks chromatin organization, hence termed genophore.

Question 12.
Whqt is nucleosome? How many base pairs are there in a typical nucleosome?
Answer:
The negatively charged DNA is wrapped around the positively charged histone octamere to form a structure called nucleosome. A typical nucleosome contains 200 bp of DNA helix.

Question 13.
Expand and define NHC
Answer:

  1. NHC : Non-histone Chromosomal protein.
  2. In eukaryotes, apart from histone proteins, additional set of proteins are required for packing of chromatin at higher level and are referred as non – histone chromosomal proteins.

Question 14.
Differentiate between Heterochromatin and Euchromatin.
Answer:
Heterochromatin:

  1. Region of nucleus where the chromatin are loosely packed and stains light are called Heterochromatin.
  2. Transcriptionally inactive.

Euchromatin:

  1. Region of nucleus where the chromatin are tightly packed and stains dark are called Euchromatin.
  2. Transcriptionally active.

Question 15.
Which is the widely accepted model of DNA replication? Who has proved it?
Answer:
Semi-conservative replication model. It was proved by Meselson and Stahl in 1958.

Question 16.
Name the chemical substance which is called by the name

  1. Kornberg Enzyme
  2. Ochoa’s enzyme

Answer:

  1. DNA polymerase I is also known as Komberg enzyme.
  2. Polynucleotide phosphorylase is also known as Ochoa’s enzyme.

Question 17.
Name the various types of prokaryotic DNA polymerase. State their role in replication process.
Answer:

  1. DNA Polymerase I Involver in DNA repair mechanism
  2. DNA Polymerase II Involver in DNA repair mechanism
  3. DNA Polymerase III Involver in DNA replication

Question 18.
What is the function of Deoxy nucleotide triphosphate in replication?
Answer:
Deoxy nucleotide triphosphate acts as substrate and also provides energy for polymerization reaction.

Question 19.
Given below are some events of eukaryotic replication. Name the enzymes involved in the process.

  1. Unwinding of DNA
  2. Joining of Okazaki fragments
  3. Addition of nucleotides to new strand
  4. Correcting the repair

Answer:

  1. Helicase
  2. DNA ligase
  3. DNA polymerase
  4. Nuclease

Samacheer Kalvi 12th Bio Zoology Solutions Chapter 5 Molecular Genetics

Question 20.
Differentiate leading strand from lagging strand
Answer:
Leading strand:

  1. Leading strand has the polarity 3’ → 5’.
  2. Replication is continuous.

Lagging strand:

  1. Lagging strand has the polarity 5’ → 3’.
  2. Replication is discontinuous.

Question 21.
What are Okazaki fragments?
Answer:
The discontinuously synthesized fragments of the lagging strand are called the Okazaki fragments are joined by the enzyme DNA ligase.

Question 22.
What is a replication fork?
Answer:
At the point of origin of replication, the helicases and topoisomerases (DNA gyrase) unwind and pull apart the strands, forming a Y-Shaped structure called the replication fork. There are two replication forks at each origin.

Question 23.
Apart from DNA polymerase, name any other four enzymes which were involved in DNA replication of eukaryotic cell.
Answer:
DNA ligase, Topoisomerase (DNA gyrase), Helicase and Nuclease.

Question 24.
Who proposed the central dogma? Write its concept.
Answer:
Francis Crick proposed the Central dogma in molecular biology which states that genetic information flows as follows:
Samacheer Kalvi 12th Bio Zoology Solutions Chapter 5 Molecular Genetics img 5

Question 25.
Define transcription and name the enzyme involved in this process.
Answer:
The process of copying genetic information from one strand of DNA into RNA is termed transcription. This process takes place in presence of DNA-dependent RNA polymerase.

Question 26.
What is TATA box? State its function.
Answer:
In eukaryotes, the promoter has AT-rich regions called TATA box or Goldberg-Hogness box. It acts as a binding site for RNA polymerase.

Question 27.
Structural gene of eukaryotes differ from prokaryotes. How?
Answer:
In eukaryotes, the structural gene is monocistronic coding for only one protein whereas in prokaryotes the structural gene is polycistronic coding for many proteins.

Question 28.
What are the two major components of prokaryotic RNA polymerase? How do they act?
Answer:
Bacterial (prokaryotic) RNA polymerase consists of two major components, the core enzyme and the sigma subunit. The core enzyme (β1, β, and α) is responsible for RNA synthesis ” whereas a sigma subunit is responsible for recognition of the promoter.

Question 29.
Distinguish between exons and introns.
Answer:

  1. Exons: Expressed sequences (Coding sequences) of an eukaryotic gene
  2. Introns: Interveining sequences (non-coding sequences) of an eukaryotic gene

Question 30.
Define splicing.
Answer:
The process of removing introns from hnRNA is called splicing.

Question 31.
What is capping and tailing?
Answer:
In capping an unusual nucleotide, methyl guanosine triphosphate is added at the 5’ end of hnRNA, whereas adenylate residues (200-300) (Poly A) are added at the 3’ end in tailing.

Question 32.
If a double stranded DNA has 20% of cytosine, calculate the percentage of adenine in DNA.
Answer:
Cytdsine = 20, hence Guanine = 20
As per ChargafFs rule (A+T) = (G+C) =100
Percent of Thymine + Adenine = 20 + 20 = 100
(T + A) = (20 + 20) =100
(T + A)=100-(20 + 20)
T +A = 100 – 40
T + A = 60
Therefore the percent of Adenine will be 60/2 = 30%.

Question 33.
Mention the dual functions of AUG.
Answer:
AUG has dual functions. It acts as a initiator codon and also codes for the amino acid methionine.

Question 34.
How many codons are involved in termination of translation. Name them.
Answer:
Three codons terminate translation process. They are UAA, UAG and UGA.

Question 35.
Degeneracy of codon – comment.
Answer:
A degenerate code means that more than one triplet codon could code for a specific amino acid. For example, codons GUU, GUC, GUA and GUG code for valine.

Samacheer Kalvi 12th Bio Zoology Solutions Chapter 5 Molecular Genetics

Question 36.
Point out the exceptional categories to universality of genetic code.
Answer:
Exceptions to universal nature of genetic code is noticed in prokaryotic mitochondrial and chloroplast genomes.

Question 37.
What are non-sense codons?
Answer:
UGA, UAA and UAG are the non-sense codons, which terminates translation.

Question 38.
Name the triplet codons that code for

  1. Tyrosine
  2. Histidine

Answer:

  1. Tyrosine – UAU, UAC
  2. Histidine – CAU, CAC

Question 39.
Why hnRNA has to undergo splicing?
Answer:
Since hnRNA contains both coding sequences (exons) and non-coding sequences (introns) it has to undergo splicing to remove introns.

Question 40.
State the role of following codons in translation process

  1. AUG
  2. UAA

Answer:

  1. AUG is the initiator codon and also codes for methionine.
  2. UAA is a terminator codon.

Question 41.
Given below is mRNA sequence. Mention the aminoacids sequence that is formed after its translation.
Answer:
3’AUGAAAGAUGGGUAA5’
Methionine – Lysine – Aspartic acid – Glycine

Question 42.
Name the four codons that codes valine.
Answer:
GUU, GUC, GUA and GUG.

Question 43.
The base sequence in one of the DNA strand is TAGC ATGAT. Mention the base sequence in its complementary strand.
Answer:
The complementary strand has ATCGTACTA.

Question 44.
Why t-RNA is called as adapter molecule?
Answer:
The transfer RNA (tRNA) molecule of a cell acts as a vehicle that picks up the amino acids scattered through the cytoplasm and also reads specific codes of mRNA molecules. Hence it is called as adapter molecule.

Question 45.
What do you mean by charging of tRNA? Name the enzyme involved in this process.
Answer:
The process of addition of amino acid to tRNA is known as aminoacylation or charging and the resultant product is called aminoacyl- tRNA (charged tRNA). Aminoacylation is catalyzed by an enzyme aminoacyl – tRNA synthetase.

Question 46.
What are UTR’s?
Answer:
mRNA also have some additional sequences that are not translated and are referred to as Untranslated Regions (UTR). UTRs are present at both 5’ end (before start codon) and at 3’ end (after stop codon).

Question 47.
What is S – D sequence?
Answer:
The 5’ end of the mRNA of prokaryotes has a special sequence which precedes the initial AUG start codon of mRNA. This ribosome binding site is called the Shine – Dalgamo sequence or S-D sequence. This sequences base-pairs with a region of the 16Sr RNA of the small ribosomal subunit facilitating initiation.

Question 48.
Define translation unit.
Answer:
A translation unit in mRNA is the sequence of RNA that is flanked by the start codon on 5’ end and stop codon on 3’ end and codes of polypeptide.

Question 49.
Mention the inhibitory role of tetracycline and streptomycin in bacterial translation.
Answer:
Tetracycline inhibits binding between aminoacyl tRNA and mRNA.Streptomycin inhibits initiation of translation and causes misreading.

Question 50.
At what stage, does the gene expression is regulated?
Answer:
Gene expression can be controlled or regulated at transcriptional or translational levels.

Question 51.
What is an operon? Give example.
Answer:
The cluster of genes with related functions is called an operon.
E.g: lac operon in E.coli.

Question 52.
Considering the lac operon of E.coli, name the products of the following genes.

  1. i gene
  2. lac Z gene
  3. lac Y gene
  4. lac a gene

Answer:

  1. i gene – repressor protein
  2. lac Z gene – fS-galactosidase
  3. Lac Y gene – Permease
  4. lac a gene – transacetylase

Question 53.
Expand

  1. ETS
  2. YAC.

Answer:

  1. ETS : Expressed Sequence Tags
  2. YAC : Yeast Artificial Chromosomes

Question 54.
Name the human chromosome that has

  1. most number of genes
  2. least number of genes

Answer:

  1. Chromosome 1 has maximum number of genes (2968 genes)
  2. Chromosome Y has least genes (231 genes)

Question 55.
What are SNPs? Mention its uses.
Answer:
SNPs : Single nucleotide polymorphism. It helps to find chromosomal locations for disease associated sequences and tracing human history.

Samacheer Kalvi 12th Bio Zoology Solutions Chapter 5 Molecular Genetics

Question 56.
Mention any four areas where DNA fingerprinting can be used.
Answer:

  1. Forensic analysis
  2. Pedigree analysis
  3. Conservation of wild life
  4. Anthropological studies

3 – Mark Questions

Question 57.
Classify nucleic acid based on sugar molecules.
Answer:
There are two types of nucleic acids depending on the type of pentose sugar. Those containing deoxyribose sugar are called Deoxyribo Nucleic Acid (DNA) and those with ribose sugar are known as Ribonucleic Acid (RNA). The only difference between these two sugars is that there is one oxygen atom less in deoxyribose.

Question 58.
Both purines and pyrimidines are nitrogen bases yet they differ. How?
Answer:
Both purines and pyrimidines are nitrogen bases. The purine bases Adenine and Guanine have double carbon – nitrogen ring, whereas cytosine and thymine bases have single carbon nitrogen ring.

Question 59.
How 5’ of DNA differ from its 3’?
Answer:
The 5’ of DNA refers to the carbon in the sugar to which phosphate (P04V) functional group is attached. The 3’ of DNA refers to the carbon in the sugar to which a hydroxyl (OH) group is attached.

Question 60.
State Chargaff’s rule.
Answer:
According to Erwin Chargaff,

  1. Adenine pairs with Thymine with two hydrogen bonds.
  2. Guanine pairs with Cytosine with three hydrogen bonds.

Question 61.
Chemically DNA is more stable than RNA – Justify.
Answer:
In DNA, the two strands being complementary, if separated (denatured) by heating can come together (renaturation) when appropriate condition is provided. Further 2 OH group present at every nucleotide in RNA is a reactive group that makes RNA liable and easily degradable. RNA is also known to be catalytic and reactive. Hence, DNA is chemically more stable and chemically less reactive when compared to RNA. Presence of thymine instead of uracil in DNA confers additional stability to DNA.

Question 62.
Write in simple about semi-conservative mode of DNA replication.
Answer:
Semi-conservative replication was proposed by Watson and Crick in 1953. This mechanism of replication is based on the DNA model. They suggested that the two polynucleotide strands of DNA molecule unwind and start separating at one end. During this process, covalent hydrogen bonds are broken. The separated single strand then acts as template for the synthesis of a new strand. Subsequently, each daughter double helix carries one polynucleotide strand from the parent molecule that acts as a template and the other strand is newly synthesised and complementary to the parent strand.

Question 63.
Draw a simplified diagram of nucleosome and label it.
Answer:
Samacheer Kalvi 12th Bio Zoology Solutions Chapter 5 Molecular Genetics img 6

Question 64.
What is a primer?
Answer:
A primer is a short stretch of RNA. It initiates the formation of new strand. The primer produces 3’-OH end on the sequence of ribonucleotides, to which deoxyribonucleotides are added to form a new strand.

Question 65.
Both strands of DNA are not copied during transcription. Give reason.
Answer:
Both the strands of DNA are not copied during transcription for two reasons.

  1. If both the strands act as a template, they would code for RNA with different sequences. This in turn would code for proteins with different amino acid sequences. This would result in one segment of DNA coding for two different proteins, hence complicate the genetic information transfer machinery.
  2. If two RNA molecules were produced simultaneously, double stranded RNA complementary to each other would be formed. This would prevent RNA from being translated into proteins.

Samacheer Kalvi 12th Bio Zoology Solutions Chapter 5 Molecular Genetics

Question 66.
What do you mean by a template strand and coding strand?
Answer:
DNA dependent RNA polymerase catalyses the polymerization in only one direction, the strand that has the polarity 3’→ 5’ acts as a template, and is called the template strand. The other strand which has the polarity 5’→ 3’ has a sequence same as RNA (except thymine instead of uracil) and is displaced during transcription. This strand is called coding strand.

Question 67.
Name the factors that are responsible for initiation and termination of transcription in prokaryotes.
Answer:

  1. Sigma factor is responsible for initiation of transcription.
  2. Rho factor is responsible for termination of transcription.

Question 68.
Name the major RNA types of prokaryotes and mention their role.
Answer:
In prokaryotes, there are three major types of RNAs: mRNA, tRNA, and rRNA. All three \ RNAs are needed to synthesize a protein in a cell. The mRNA provides the template, tRNA brings amino acids and reads the genetic code, and rRNAs play structural and catalytic role
during translation.

Question 69.
Define genetic code.
Answer:
The order of base pairs along DNA molecule controls the kind and order of amino acids found in the proteins of an organism. This specific order of base pairs is called genetic code.

Question 70.
Explain Wobble hypothesis.
Answer:
Wobble Hypothesis is proposed by Crick (1966) which states that tRNA anticodon has the ability to wobble at its 5’ end by pairing with even non-complementary base of mRNA codon.’ According to this hypothesis, in codon-anticodon pairing the third base may not be complementary.

The third base of the codon is called wobble base and this position is called wobble position. The actual base pairing occurs at first two positions only. The importance of Wobbling hypothesis is that it reduces the number of tRNAs required for polypeptide synthesis and it overcomes the effect of code degeneracy.

Question 71.
Explain the nature of eukaryotic ribosome.
Answer:
The ribosomes of eukaryotes (80 S) are larger, consisting of 60 S and 40 S sub units. ‘S’ denotes the sedimentation efficient which is expressed as Svedberg unit (S). The larger subunit in eukaryotes consist of a 23 S RNA and 5Sr RNA molecule and 31 ribosomal proteins. The smaller eukaryotic subunit consist of 18Sr RNA component and about 33 proteins.

Question 72.
Expand and define ORF.
Answer:
Any sequence of DNA or RNA, beginning with a start codon and which can be translated into a protein is known as an Open Reading Frame (ORF).

Question 73.
What are the components of initiation complex of prokaryotic translation?
Answer:
Initiation of translation in E. coli begins with the formation of an initiation complex, consisting of the 30S subunits of the ribosome, a messenger RNA and the charged N-formyl methionine tRNA (fmet -1 RNA fmet), three proteinaceous initiation factors (IF 1, IF2, IF3), GTP (Guaniner Tri Phosphate) and Mg 2+.

Question 74.
Explain the components of operon.
Answer:
Structure of the operon: Each operon is a unit of gene expression and regulation and consists
of one or more structural genes and an adjacent operator gene that controls transcriptional
activity of the structural gene.

  1. The structural gene codes for proteins, rRNA and tRNA required by the cell.
  2. Promoters are the signal sequences in DNA that initiate RNA synthesis. RNA polymerase I binds to the promoter prior to the initiation of transcription.
  3. The operators are present between the promoters and structural genes. The repressor protein binds to the operator region of the operon.

5 – Mark Question

Question 75.
Describe Hershey and Chase experiment. What is concluded by their experiment?
Answer:
Alfred Hershey and Martha Chase (1952) conducted experiments on bacteriophages that infect bacteria. Phage T2 is a virus that infects the bacterium Escherichia coli. When phages (virus) are added to bacteria, they adsorb to the outer surface, some material enters the bacterium, and then later each bacterium lyses to release a large number of progeny phage. Hershey and Chase wanted to observe whether it was DNA or protein that entered the bacteria. All nucleic acids contain phosphorus, and contain sulphur (in the amino acid cysteine and methionine). Hershey and Chase designed an experiment using radioactive isotopes of Sulphur (35S) and phosphorus (32P) to keep separate track of the viral protein and nucleic acids during the infection process.

The phages were allowed to infect bacteria in culture medium which containing the radioactive isotopes 35S or 32P. The bacteriophage that grew in the presence of 35S had labelled proteins and bacteriophages grown in the presence of 32P had labelled DNA. The differential labelling thus enabled them to identity DNA and proteins of the phage. Hershey and Chase mixed the labelled phages with unlabeled E. coli and allowed bacteriophages to attack and inject their genetic material. Soon after infection (before lysis of bacteria), the bacterial cells were gently agitated in a blender to loosen the adhering phase particles.

It was observed that only 32P was found associated with bacterial cells and 35S was in the surrounding medium and not in the bacterial cells. When phage progeny was studied for radioactivity, it was found that it carried only 32P and not 35 S. These results clearly indicate that only DNA and not protein coat entered the bacterial cells. Hershey and Chase thus conclusively proved that it was DNA, not protein, which carries the hereditary information from virus to bacteria.

Samacheer Kalvi 12th Bio Zoology Solutions Chapter 5 Molecular Genetics

Question 76.
Explain the properties of DNA that makes it an ideal genetic material.
Answer:
1. Self Replication: It should be able to replicate. According to the rule of base pairing and complementarity, both nucleic acids (DNA and RNA) have the ability to direct duplications. Proteins fail to fulfill this criterion.

2. Stability: It should he stable structurally and chemically. The genetic material should be stable enough not to change with different stages of life cycle, age or with change in physiology of the organism. Stability as one of the properties of genetic material was clearly evident in Griffith’s transforming principle. Heat which killed the bacteria did not destroy some of the properties of genetic material. In DNA the two strands being complementary.

if separated (denatured) by heating can come together (renaturation) when appropriate condition is provided. Further 2’ OH group present at every nucleotide in RNA is a reactive group that makes RNA liable and easily degradable. RNA is also known to be catalytic and reactive. Hence, DNA is chemically more stable and chemically less reactive when compared to RNA. Presence of thymine instead of uracil in DNA confers additional stability to DNA.

3. Information storage: It should be able to express itself in the form of ‘Mendelian characters’. RNA can directly code for protein synthesis and can easily express the characters. DNA, however depends on RNA for synthesis of proteins. Both DNA and RNA can act as a genetic material, but DNA being more stable stores the genetic information and RNA transfers the genetic information.

4. Variation through mutation: It should be able to mutate. Both DNA and RNA are able to mutate. RNA being unstable mutates at a faster rate. Thus viruses having RNA genome with shorter life span can mutate and evolve faster. The above discussion indicates that both RNA and DNA can function as a genetic material. DNA is more stable, and is preferred for storage of genetic information.

Question 77.
How the DNA is packed in a eukaryotic cell? ft
Answer:
In eukaryotes, organization is more complex. Chromatin is formed by a series of repeating units called nucleosomes. Komberg proposed a model for the nucleosome, in which 2 molecules of the four histone proteins H2A, H2B, H3 and H4 are organized to form a unit of eight molecules called histone octamere. The negatively charged DNA is wrapped around the positively charged histone octamere to form a structure called a nucleosome. A typical nucleosome contains 200 bp of DNA helix. The histone octameres are in close contact and DNA is coiled on the outside of the nucleosome. Neighbouring nucleosomes are connected by linker DNA (HI) that is exposed to enzymes.

The DNA makes two complete turns around the histone octameres and the two turns are sealed off by an HI molecule. Chromatin lacking HI has a beads-on-a-string appearance in which DNA inters and leaves the nucleosomes at random places. HI of one nucleosome can interact with 33l of the neighbouring nucleosomes resulting in the further folding of the fibre.

The chrof&atin fiber in interphase nuclei and mitotic chromosomes have a diameter that vary between 200-300 nm and represents inactive chromatin. 30 nm fibre arises from the folding of nucfeosbme, chains into a solenoid structure having six nucleosomes per turn. This structure is stabilized by interaction between different HI molecules. DNA is a solenoid and packed about,%)_folds. The hierarchical nature of chromosome structure is illustrated.

Additional set of proteins are required for packing of chromatin at higher level and are referred to as non-histone chromosomal proteins (NHC). In*,a typical nucleus, some regions of chromatin are Ibosely packed (lightly stained) and are referred to as euchromatin. The chromatin that is,-tightly packed (stained darkly) is called heterochromatin. Euchromatin is transcriptionally active and heterochromatin is transcriptionally inactive.

Question 78.
Meselson and Stahl’s experiment proved the semi-coflBptervation mode of DNA replication. Explain.
Answer:
The mode of DNA replication was determined in 1958 by Meselson and Stahl. They designed an experiment to distinguish between semi-conservative, conservative and dispersive replications. In their experiment, they grew two cultures of E.coli for many generations in separate media. The ‘heavy’ culture was grown in a medium in which the nitrogen source (NH4CI) contained the heavy isotope 15N and the ‘ light’ culture was grown in a medium in which the nitrogen source contained light isotope 14H for many generations. At the end of growth, they observed that the bacterial DNA in the heavy culture contained only 15N and in the light culture only 14N. The heavy DNA could be distinguished from light DNA (15N from 14N) with a technique called Cesium Chloride (CsCl) density gradient centrifugation. In this process, heavy and light DNA extracted from cells in thtytwo cultures settled into two distinct and separate bands (hybrid DNA).

The heavy culture (15N) was then transferred into a medium that had only NH4CI, and took samples at various definite time intervals (20 minutes duration). After the first replication, they extracted DNA and subjected it to density gradient centrifugation. The DNA settled into a band that was intermediate in position between the previously determined heavy and light bands. After the second replication (40 minutes duration), they again extracted DNA samples,and this time found the DNA settling into two bands, one at the light band position and one at intermediate position. These results confirm Watson and Crick’s semi-conservative replication hypothesis.

Question 79.
Give a detailed account of a transcription unit.
Answer:
A transcriptional unit in DNA is defined by three regions, a promoter, the structural gene and a terminator. The promoter is located towards the 5 ’ end. It is a DNA sequence that provides binding site for RNA polymerase. The presence of promoter in a transcription unit, defines the template and coding strands. The terminator region located towards the 3’ end of the coding strand contains a DNA sequence that causes the RNA polymerase to stop transcribing. In eukaryotes the promoter has AT rich regions called TATA box (Goldberg- Hogness box) ‘ and in prokaryotes this region is called Pribnow box.

Besides promoter, eukaryotes also require an enhancer. The two strands of the DNA in the structural gene of a transcription unit have opposite polarity. DNA dependent RNA polymerase catalyses the polymerization in only one direction, the strand that has the polarity 3’→5’ acts as a template, and is called the template strand. The other strand which has the polarity 5’→ 3’ has a sequence same as RNA (except thymine instead of uracil) and is displaced during transcription. This strand is called coding strand

The structural gene may be monocistronic (eukaryotes) or polycistronic (prokaryotes). In eukaryotes, each mRNA carries only a single gene and encodes information for only a single protein and is called monocistronic mRNA. In prokaryotes, clusters of related genes, known as operon, often found next to each other on the chromosome are transcribed together to give a single mRNA and hence are polycistronic.

Question 80.
Explain the transcription process in prokaryotes with the needed diagram.
Answer:
In prokaryotes, there are three major types of RNAs:
mRNA, tRNA, and rRNA. All three RNAs are needed to synthesize a protein in a cell. The mRNA provides the template, tRNA brings amino acids and reads the genetic code, and rRNAs play a structural and catalytic role during translation. There is a single DNA-dependent RNA polymerase that catalyses the transcription of all types of RNA. It binds to the promoter and initiates transcription (Initiation).
Samacheer Kalvi 12th Bio Zoology Solutions Chapter 5 Molecular Genetics img 7
The polymerase binding sites are called promoters. It uses nucleoside triphosphate as substrate and polymerases in a template dependent fashion following the rule of complementarity. After the initiation of transcription, the polymerase continues to elongate the RNA, adding one nucleotide after another to the growing RNA chain. Only a short stretch of RNA remains bound to the enzyme, when the polymerase reaches a terminator at the end of a gene, the, nascent RNA falls off, so also the RNA polymerase. The RNA polymerase is only capable of catalyzing the process of elongation. The RNA polymerase associates transiently with initiation factor sigma (a) and termination factor rho (p) to initiate and terminate the transcription, respectively.

Association of RNA with these factors instructs the RNA polymerase either to initiate or terminate the process of transcription. In bacteria, since the mRNA does not require any processing to become active and also since transcription and translation take place simultaneously in the same compartment since there is no separation of cytosol and nucleus in bacteria), many times the translation can begin much before the mRNA is fully transcribed. This is because the genetic material is not separated from other cell organelles by a nuclear membrane consequently; transcription and translation can be coupled in bacteria.

Samacheer Kalvi 12th Bio Zoology Solutions Chapter 5 Molecular Genetics

Question 81.
Write the salient features of genetic code.
Answer:
The salient features of genetic code are as follows:

  1. The genetic codon is a triplet code and 61 codons code for amino acids and 3 codons do not code for any amino acid and function as a stop codon (Termination).
  2. The genetic code is universal. It means that all known living systems use nucleic acids and the same three base codons (triplet codon) direct the synthesis of protein from amino acids. For example, the mRNA (UUU) codon codes for phenylalanine in all cells of all organisms. Some exceptions are reported in prokaryotic, mitochondrial and chloroplast genomes. However, similarities are more common than differences.
  3. A non-overlapping codon means that the same letter is not used for two different codons. For instance, the nucleotide sequence GUTJ and GUC represents only two codons.
  4. It is comma less, which means that the message would be read directly from one end to the other i.e., no punctuation are needed between two codes.
  5. A degenerate code means that more than one triplet codon could code for a specific amino acid. For example, codons GUU, GUC, GUA, and GUG code for valine.
  6. Non-ambiguous code means that one codon will code for one amino acid.
  7. The code is always read in a fixed direction i.e. from 5’→3’ direction called polarity.
  8. AUG has dual functions. It acts as an initiator codon and also codes for the amino acid methionine.
  9. UAA, UAG (tyrosine) and UGA (tryptophan) codons are designated as termination (stop) codons and also are known as “non-sense” codons.

Question 82.
Mutations in genetic code affect the phenotype. Describe with an example.
Answer:
The simplest type of mutation at the molecular level is a change in nucleotide that substitutes one base for another. Such changes are known as base substitutions which may occur spontaneously or due to the action of mutagens. A well-studied example is sickle cell anaemia in humans which results from a point mutation of an allele of β-hemoglobin gene (βHb).

A haemoglobin molecule consists of four polypeptide chains of two types, two a chains and two P-chains. Each chain has a heme group on its surface. The heme groups are involved in the binding of oxygen. The jruman blood disease, sickle cell anaemia is due to abnormal haemoglobin. This abnormality in hemoglobin is due to a single base substitution at the sixth codon of the beta globingene from GAG to GTG in p -chain of haemoglobin.

It results in a change of amino acid glufeniic acid to valine at the 6th position of the p -chain. This is the classical example of point mutation that results in the change of amino acids residue glutamic acid to valine. The mutant haemoglobin undergoes polymerisation under oxygen tension causing the change in the shape of the RBC from biconcave to a sickle shaped structure.
Samacheer Kalvi 12th Bio Zoology Solutions Chapter 5 Molecular Genetics img 8

Question 83.
Explain the mechanism of AteArperon of the E-coli.
Answer:
The Lac (Lactose) operon: The metabolism of lactose in E.coli requires three enzymes – permease, P-galactosidase (P-gat) and transacetylase. The enzyme permease is needed for entry of lactose into the cell, Pjgglactosidase brings about hydrolysis of lactose to glucose and galactose, while transacety transfers acetyl group from acetyl Co A to P-galactosidase. The lac operon consists of one-regulator gene (T gene refers to inhibitor) promoter sites (p), and operator site (o). Besides these, it has three structural genes namely lac z, y and lac a. The lac ‘z’ gene codes for P-gaiaqtttsidase, lac ‘y’ gene codes for permease and ‘a’ gene codes for transacetylase.

Jacob and Monod proposed the classical model of Lac operon to explain gene expression and regulation in E.coli. In lac a polycistronic structural gene is regulated by a common promoter and regulatory genfc When the cell is using its normal energy source as glucose, the ‘i’ gene transcribes a repressor mRNA and after its translation, a repressor protein is produced. It binds to the operator region of the operon and prevents translation, as a result, P-galactosidase is not produced. In the absence of preferred carbon source such as glucose, if lactose is available as an energy source for the bacteria then lactose enters the cell as a result of permease enzyme. Lactose acts as an inducer and interacts with the repressor to inactivate it.

The repressor protein binds to the operator of the operon and prevents RNA polymerase from transcribing the operon. In the presence of inducer, such as lactose or allolactose, the repressor is inactivated by interaction with the inducer. This allows RNA polymerase to bind to the promotor site and transcribe the operon to produce lac mRNA which enables formation of all the required enzymes needed for lactose metabolism. This regulation of lac operon by the repressor is an example of negative control of transcription initiation.
Samacheer Kalvi 12th Bio Zoology Solutions Chapter 5 Molecular Genetics img 9

Question 84.
What are the objectives of Human Genome project?
Answer:
The main goals of Human Genome Project are as follows:

  1. Identify all the genes (approximately 30000) in human DNA.
  2. Determine the sequence of the three billion chemical base pairs that makeup the human DNA.
  3. To store this information in databases.
  4. Improve tools for data analysis.
  5. Transfer related technologies to other sectors, such as industries.
  6. Address the ethical, legal and social issues (ELSI) that may arise from the project.

Samacheer Kalvi 12th Bio Zoology Solutions Chapter 5 Molecular Genetics

Question 85.
Write the salient features of Human Genome Project.
Answer:

  1. Although human genome contains 3 billion nucleotide bases, the DNA sequences that encode proteins make up only about 5% of the genome.
  2. An average gene consists of 3000 bases, the largest known human gene being dystrophin with 2.4 million bases.
  3. The function of 50% of the genome is derived from transposable elements such as LINE and ALU sequence.
  4. Genes are distributed over 24 chromosomes. Chromosome 19 has the highest gene density. Chromosome 13 and Y chromosome have lowest gene densities.
  5. The chromosomal organization of human genes shows diversity.
  6. There may be 35000-40000 genes in the genome and almost 99.9 nucleotide bases are exactly the same in all people.
  7. Functions for over 50 percent of the discovered genes are unknown.
  8. Less than 2 percent of the genome codes for proteins.
  9. Repeated sequences make up very large portion of the human genome. Repetitive sequences have no direct coding functions but they shed light on chromosome structure, dynamics and evolution (genetic diversity).
  10. Chromosome 1 has 2968 genes, whereas chromosome ’Y’ has 231 genes.
  11. Scientists have identified about 1.4 million locations, where single-base DNA differences (SNPs – Single nucleotide polymorphism – pronounce as ‘snips’) occur in humans. Identification of ‘SNIPS’ is helpful in finding chromosomal locations for disease-associated sequences and tracing human history.

Question 86.
Describe the principle involved in DNA fingerprinting technique.
Answer:
The DNA fingerprinting technique was first developed by Alec Jeffreys in 1985. The DNA of a person and finger prints are unique. There are 23 pairs of human chromosomes with 1.5 million pairs of genes. It is a well known fact that genes are segments of DNA that differ in the sequence of their nucleotides. Not all segments of DNA code for proteins, some DNA segments have a regulatory function, while others are intervening sequences (introns) and still others are repeated DNA sequences. In DNA fingerprinting, short repetitive nucleotide sequences are specific for a person. These nucleotide sequences are called as variable number tandem repeats (VNTR). The VNTRs of two persons generally show variations and are useful as genetic markers.

DNA fingerprinting involves identifying differences in some specific regions in the DNA sequence called repetitive DNA because in these sequences, a small stretch of DNA is repeated many times. These repetitive DNA are separated from bulk genomic DNA as different peaks during density gradient centrifugation. The bulk DNA forms a major peak and the other small peaks are referred to as satellite DNA. Depending on base composition (A: T rich or G : C rich), length of segment and number of repetitive units, the satellite DNA is classified into many subcategories such as micro-satellites and mini satellites, etc.

These sequences do not code for any proteins, but they form a large portion of human genome. These sequences show high degree of polymorphism and form the basis of DNA fingerprinting. DNA isolated from blood, hair, skin cells, or other genetic evidences left at the scene of a crime can be compared through VNTR patterns, with the DNA of a criminal suspect to determine guilt or innocence. VNTR patterns are also useful in establishing the identity of a homicide victim, either from DNA found as evidence or from the body itself.

Samacheer Kalvi 12th Bio Zoology Solutions Chapter 5 Molecular Genetics

Question 87.
Draw a flow chart depicting the steps of DNA finger printings technique
Answer:
Samacheer Kalvi 12th Bio Zoology Solutions Chapter 5 Molecular Genetics img 10

Higher Order Thinking Skills (HOTs) Questions

Question 1.
An mRNA strand has a series of triplet codons of which the first three codons are given below
(a) AUG
(b) UUU
(c) UGC
(i) Name the amino acid encoded by these triplet codons.
(ii) Mention the DNA sequence from which these triplet codons would have transcribed?
Answer:
(i) AUG codes for Methionine
UUU codes for Phenylalanine
UGC codes for Cysteine
(ii) TAC sequence of DNA is transcribed to AUG
AAA sequence of DNA is transcribed to UUU
ACG sequence of DNA is transcribed to UGC

Question 2.
Given below are the structures of tRNA molecules which are involved in the translation process. In one tRNA, the codon is mentioned but not the amino acid. In another tRNA molecule, the amino acid is named and not the codon. Complete the figure by mentioning the respective amino acids and codons.
Answer:
Samacheer Kalvi 12th Bio Zoology Solutions Chapter 5 Molecular Genetics img 11

Question 3.
A DNA fragment possesses 32 adenine bases and 32 cytosine bases. How many total numbers of nucleotides does that DNA fragment contain? Explain.
Answer:
128 nucleotides. Adenine always pairs Thymine base. If there are 32 adenine bases then there must be 32 Thymine bases. Similarly cytosine pairs with guanine. If cytosine bases are 32 in number the guanine bases will be equal to cytosine. So it makes a total of 128 nucleotides.

Samacheer Kalvi 12th Bio Zoology Solutions Chapter 5 Molecular Genetics

Question 4.
Following is a DNA sequence representing a part of the gene TAC TCG CCC TAT UAA CCC AAA ACC TCT using this derive A.

  1. The RNA transcript
  2. The spliced mRNA (consider all the codons with two Aderine bases are introns)
  3. The total number of amino acids coded by the mRNA

Answer:

  1. RNA transcript: AUG UGC GGG AUA GGG UUU UGG AGA
  2. Spliced mRNA: AUG UGC GGG GGG UUU UGG
  3. 6 amino acids are coded by mRNA

Question 5.
Complete the molecular processes by naming them

  1. DNA → DNA
  2. mRNA → Protein
  3. RNA transcript → mRNA

Answer:

  1. Replication
  2. Translation
  3. Splicing

Samacheer Kalvi 11th Accountancy Solutions Chapter 8 Bank Reconciliation Statement

Students can Download Accountancy Chapter 8 Bank Reconciliation Statement Questions and Answers, Notes Pdf, Samacheer Kalvi 11th Accountancy Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Accountancy Solutions Chapter 8 Bank Reconciliation Statement

Samacheer Kalvi 11th Accountancy Bank Reconciliation Statement Text Book Back Questions and Answers

I. Multiple Choice Questions
Choose the Correct Answer

Question 1.
A bank reconciliation statement is prepared by ………………
(a) Bank
(b) Business
(c) Debtor to the business
(d) Creditor to the business
Answer:
(b) Business

Question 2.
A bank reconciliation statement is prepared with the help of ………………
(a) Bank statement
(b) Cash book
(c) Bank statement and bank column of the cash book
(d) Petty cash book
Answer:
(c) Bank statement and bank column of the cash book

Samacheer Kalvi 11th Accountancy Solutions Chapter 8 Bank Reconciliation Statement

Question 3.
Debit balance in the bank column of the cash book means ………………
(a) Credit balance as per bank statement
(b) Debit balance as per bank statement
(c) Overdraft as per cash book
(d) None of the above
Answer:
(a) Credit balance as per bank statement

Question 4.
A bank statement is a copy of ………………
(a) Cash column of the cash book
(b) Bank column of the cash book
(c) A customer’s account in the bank’s book
(d) Cheques issued by the business
Answer:
(c) A customer’s account in the bank’s book

Question 5.
A bank reconciliation statement is prepared to know the causes for the difference between:
(a) The balance as per the cash column of the cash book and bank column of the cash book
(b) The balance as per the cash column of the cash book and bank statement
(c) The balance as per the bank column of the cash book and the bank statement
(d) The balance as per petty cash book and the cash book
Answer:
(c) The balance as per the bank column of the cash book and the bank statement

Question 6.
When money is withdrawn from bank, the bank ………………
(a) Credits customer’s account
(b) Debits customer’s account
(c) Debits and credits customer’s account
(d) None of these
Answer:
(b) Debits customer’s account

Question 7.
Which of the following is not the salient feature of bank reconciliation statement?
(a) Any undue delay in the clearance of cheques will be shown up by the reconciliation
(b) Reconciliation statement will discourage the accountant of the bank from embezzlement
(c) It helps in finding the actual position of the bank balance
(d) Reconciliation statement is prepared only at the end of the accounting period
Answer:
(d) Reconciliation statement is prepared only at the end of the accounting period

Question 8.
Balance as per cash book is ₹ 2,000. Bank charge of ₹ 50 debited by the bank is not yet shown in the cash book. What is the bank statement balance now?
(a) ₹ 1,950 credit balance
(a) ₹ 1,950 credit balance
(b) ₹ 1,950 debit balance
(c) ₹ 2,050 debit balance
(d) ₹ 2,050 credit balance
Answer:
(a) ₹ 1,950 credit balance

Question 9.
Balance as per bank statement is ₹ 1,000. Cheque deposited, but not yet credited by the bank is ₹ 2, 000. What is the balance as per bank column of the cash book?
(a) ₹ 3,000 overdraft
(b) ₹ 3,000 favourable
(c) ₹ 1,000 overdraft
(d) ₹ 1,000 favourable
Answer:
(b) ₹ 3,000 favourable

Samacheer Kalvi 11th Accountancy Solutions Chapter 8 Bank Reconciliation Statement

Question 10.
Which one of the following is not a timing difference?
(a) Cheque deposited but not yet credited
(b) Cheque issued but not yet presented for payment
(c) Amount directly paid into the bank
(d) Wrong debit in the cash book
Answer:
(d) Wrong debit in the cash book

II. Very Short Answer Questions

Question 1.
What is meant by bank overdraft?
Answer:
It is not possible to have unfavourable cash balance in the cash book. But, it is possible to have unfavourable balance in the bank account. When the business
is not having sufficient money in its bank account, it can borrow money from the bank. As a result of this, amount is overdrawn from bank.

Question 2.
What is bank reconciliation statement?
Answer:
If every entry in the cash book matches with the bank statement, then bank balance will be the same in both the records. But, practically it may not be possible. When the balances do not agree with each other, the need for preparing a statement to explain the causes arises. This statement is called bank reconciliation statement (BRS).

Question 3.
State any two causes of disagreement between the balance as per bank column of cash book and bank statement.
Answer:
(a) Cheques issued but not yet presented for payment.
(b) Cheques deposited into bank but not yet credited.

Samacheer Kalvi 11th Accountancy Solutions Chapter 8 Bank Reconciliation Statement

Question 4.
Give any two expenses which may be paid by the banker as per standing instruction.
Answer:
Insurance premium, loan instalment, etc., paid as per standing instructions.

Question 5.
Substitute the following statements with one word/phrase
(a) A copy of customer’s account issued by the bank.
(b) Debit balance as per bank statement.
(c) Statement showing the causes of disagreement between the balance as per cash book and balance as per bank statement.
Answer:
(a) Pass book
(b) Pass book favourable
(c) (1) Timing difference, (2) Errors in recording

Question 6.
Do you agree on the following statements? Write “yes” if you agree, and write “no” if you disagree.
(a) Bank reconciliation statement is prepared by the banker.
(b) Adjusting the cash book before preparing the bank reconciliation statement is compulsory.
(c) Credit balance as per bank statement is an overdraft.
(d) Bank charges debited by the bank increases the balance as per bank statement.
(e) Bank reconciliation statement is prepared to identify the causes of differences between balance as per bank column of the cash book and balance as per cash column of the cash book.
Answer:
(a) No
(b) No
(c) No
(d) No
(e) Yes

III. Short Answer Questions

Question 1.
Give any three reasons for preparing bank reconciliation statement.
Answer:
The main reasons for preparing bank reconciliation statement are:

  1. To identify the reasons for the difference between the bank balance as per the cash book and bank balance as per bank statement.
  2. To identify the delay in the clearance of cheques.
  3. To ascertain the correct balance of bank column of cash book.

Question 2.
What is meant by the term “cheque not yet presented?”
Answer:
When the cheques are issued by the business, it is immediately entered on the credit side of the cash book by the business. But, this may not be entered in the bank statement on the same day. It will be entered in the bank statement only after it is presented with the bank.

Question 3.
Explain why does money deposited into bank appear on the debit side of the cash book, but on the credit side of the bank statement?
Answer:
When the cheques are deposited into bank, the amount is debited in the cash book on the same day. But, these may not be shown in the bank pass book on the same day because these will be entered in the bank statement only after the collection of the cheques.

Samacheer Kalvi 11th Accountancy Solutions Chapter 8 Bank Reconciliation Statement

Question 4.
What will be the effect of interest charged by the bank, if the balance is an overdraft?
Answer:
The bank has to cover the cost of running the customer’s account. So debit is given to the account of the business towards bank charges. Also, if the business had taken any loan or overdrawn, interest has to be paid by the business. These entries for bank charges and interest are made in the bank statement. But, the entry is made in the cash book only when the bank statement is received by the business. Till then, the Cash book shows more balance than bank statement.

Question 5.
State the timing differences in BRS with examples.
Answer:
The timing differences in BRS are:
(a) cheques issued but not yet presented for payment
(b) cheques deposited into bank but not yet credited
(c) bank charges and interest on loan and overdraft
(d) interest and dividends collected by the bank
(e) dishonour of cheques and bills
(f) amount paid by parties directly into the bank
(g) payment made directly by the bank to others
(h) bills collected by the bank on behalf of its customer

IV. Exercises

Question 1.
From the following particulars prepare a bank reconciliation statement of Jayakumar as on 31st December, 2016. (3 Marks)
(a) Balance as per cash book ₹ 7,130
(b) Cheque deposited but not cleared ₹ 1,000
(c) A customer has deposited ₹ 800 into the bank directly
Answer:
Bank reconciliation statement of Jayakumar as on 31st December, 2016
Samacheer Kalvi 11th Accountancy Solutions Chapter 8 Bank Reconciliation Statement

Question 2.
From the following particulars of Kamakshi traders, prepare a bank reconciliation statement as on 31st March, 2018. (3 Marks)
(a) Debit balance as per cash book ₹ 10,500
(b) Cheque deposited into bank amounting to ₹ 5,500 credited by bank, but entered twice in the cash book
(c) Cheques issued and presented for payment amounting to ₹ 7,000 omitted in the cash book
(d) Cheque book charges debited by the bank ₹ 200 not recorded in the cash book.
(e) Cash of ₹ 1,000 deposited by a customer of the business in cash deposit machine not recorded in the cash book.
Answer:
Bank reconciliation statement of Kamakshi as on 31st March, 2018
Samacheer Kalvi 11th Accountancy Solutions Chapter 8 Bank Reconciliation Statement

Question 3.
From the following information, prepare bank reconciliation statement to find out the bank statement balance as on 31st December, 2017. (5 Marks)
Samacheer Kalvi 11th Accountancy Solutions Chapter 8 Bank Reconciliation Statement
Answer:
Bank reconciliation statement as on 31st December, 2017
Samacheer Kalvi 11th Accountancy Solutions Chapter 8 Bank Reconciliation Statement

Question 4.
On 31st March, 2017, Anand’s cash book showed a balance of ₹ 1,12,500. Prepare bank reconciliation statement. (5 Marks)
(a) He had issued cheques amounting to ₹ 23,000 on 28.3.2017, of which cheques amounting to ₹ 9,000 have so far been presented for payment.
(b) A cheque for ₹ 6,300 deposited into bank on 27.3.2017, but the bank credited the same only on 5th April 2017.
(c) He had also received a cheque for ₹ 12,000 which, although entered by him in the cash book, was not deposited in the bank.
(d) Wrong credit given by the bank on 30th March 2017 for ₹ 2,000.
(e) On 30th March 2017, a bill already discounted with the bank for ₹ 3,000 was dishonoured, but no entry was made in the cash book.
(f) Interest on debentures of ₹ 700 was received by the bank directly.
(g) Cash sales of ₹ 4,000 wrongly entered in the bank column of the cash book.
Answer:
Bank reconciliation statement of Anand as on 31st March, 2017
Samacheer Kalvi 11th Accountancy Solutions Chapter 8 Bank Reconciliation Statement

Question 5.
From the following particulars of Siva and Company, prepare a bank reconciliation statement as on 31st December, 2017. (2 Marks)
(a) Credit balance as per cash book ₹ 12,000.
(b) A cheque of ₹ 1,200 issued and presented for payment to the bank, wrongly credited in the cash book as ₹ 2,100.
(c) Debit side of bank statement was undercast by ₹ 100.
Answer:
Bank reconciliation statement of Siva and Company as on 31st December, 2017
Samacheer Kalvi 11th Accountancy Solutions Chapter 8 Bank Reconciliation Statement

Question 6.
From the following particulars of Raheem traders, prepare a bank reconciliation statement as on 31st March, 2018. (3 Marks)
(a) Overdraft as per cash book ₹ 2,500
(b) Debit side of cash book was undercast by ₹ 700
(c) Amount received by bank through RTGS amounting to ₹ 2,00,000, omitted in the cash book.
(d) Two cheques issued for ₹ 1,800 and ₹ 2,000 on 29th March 2018. Only the second cheque is presented for payment.
(e) Insurance premium on car for ₹ 1,000 paid by the bank as per standing instruction not recorded in the cash book.
Answer:
Bank reconciliation statement of Raheem as on 31st March, 2018
Samacheer Kalvi 11th Accountancy Solutions Chapter 8 Bank Reconciliation Statement

Question 7.
From the following information, prepare bank reconciliation statement as on 31st December, 2017 to find out the balance as per bank statement. (5 Marks)
Samacheer Kalvi 11th Accountancy Solutions Chapter 8 Bank Reconciliation Statement
Answer:
Bank reconciliation statement as on 31st December, 2017
Samacheer Kalvi 11th Accountancy Solutions Chapter 8 Bank Reconciliation Statement

Question 8.
Prepare bank reconciliation statement from the following data. (5 Marks)
Samacheer Kalvi 11th Accountancy Solutions Chapter 8 Bank Reconciliation Statement
Answer:
Samacheer Kalvi 11th Accountancy Solutions Chapter 8 Bank Reconciliation Statement

Question 9.
From the following particulars of Veera traders, prepare a bank reconciliation statement as on 31st December, 2017. (2 Marks)
(a) Credit balance as per bank statement ₹ 6,000
(b) Amount received by bank through NEFT for ₹ 3,500, entered twice in the cash book.
(c) Cheque dishonoured amounting to ₹ 2,500, not entered in cash book.
Answer:
Bank reconciliation statement of Veera Traders as on 31st December, 2017
Samacheer Kalvi 11th Accountancy Solutions Chapter 8 Bank Reconciliation Statement

Question 10.
Prepare bank reconciliation statement from the following data and find out the balance as per cash book as on 31st March, 2018. (3 Marks)
Samacheer Kalvi 11th Accountancy Solutions Chapter 8 Bank Reconciliation Statement
Answer:
Bank reconciliation statement as on 31st March, 2018
Samacheer Kalvi 11th Accountancy Solutions Chapter 8 Bank Reconciliation Statement

Question 11.
Ascertain the cash book balance from the following particulars as on 31st December, 2017: (5 Marks)

  1. Credit balance as per bank statement ₹ 2,500
  2. Bank charges of ₹ 60 have not been entered in the cash book
  3. Cheque deposited on 28th December 2017 for ₹ 1,000 was not yet credited by the bank
  4. Cheque issued on 24th December 2017 for ₹ 700, not yet presented for payment
  5. A dividend of ₹ 400 collected by the bank directly but not entered in the cash book
  6. A cheque of ₹ 600 had been dishonoured, but no entry was made in the cash book
  7. Interest on term loan ₹ 1,200 debited by bank but not accounted in cash book
  8. No entry had been made in the cash book for a trade subscription of ₹ 500 paid vide banker’s order on 23rd December 2014

Answer:
Bank reconciliation statement as on 31st December, 2017
Samacheer Kalvi 11th Accountancy Solutions Chapter 8 Bank Reconciliation Statement

Question 12.
From the following particulars of Raja traders, prepare a bank reconciliation statement as on 31st January, 2018. (5 Marks)
(a) Balance as per bank statement ₹ 5,000
(b) Cheques amounting to ₹ 800 had been recorded in the cash book as having been deposited into the bank on 25th January 2018, but were entered in the bank statement on 2nd February 2018.
(c) Amount received by bank through NEFT amounting to ₹ 3,000, omitted in the cash book.
(d) Two cheques issued for ₹ 3,000 and ₹ 2,000 on 29th March 2018. Only the first cheque is presented for payment.
(e) Insurance premium on motor vehicles for ₹ 1,000 paid by the bank as per standing instruction not recorded in the cash book.
(f) Credit side of cash book was undercast by ₹ 700
(g) Subsidy received directly by the bank from the state government amounting to ₹ 10,000, not entered in cash book.
Answer:
Bank reconciliation statement of Raja Traders as on 31st January, 2018
Samacheer Kalvi 11th Accountancy Solutions Chapter 8 Bank Reconciliation Statement

Question 13.
From the following particulars of Simon traders, prepare a bank reconciliation statement as on 31st March, 2018. (2 Marks)
(a) Debit balance as per bank statement ₹ 2,500
(b) Cheques deposited amounting to ₹ 10,000, not yet credited by bank.
(c) Payment through net banking for ₹ 2,000, omitted in the cash book
Answer:
Bank reconciliation statement of Simon Traders as on 31st March, 2018
Samacheer Kalvi 11th Accountancy Solutions Chapter 8 Bank Reconciliation Statement

Question 14.
From the following particulars, ascertain the cash book balance as on 31st December, 2016.

  1. Overdraft balance as per bank statement ₹ 1,26,640 (3 Marks)
  2. Interest on overdraft entered in the bank statement, but not yet recorded in cash book ₹ 3,200
  3. Bank charges entered in bank statement, but not found in cash book ₹ 600
  4. Cheques issued, but not yet presented for payment ₹ 23,360
  5. Cheques deposited into the bank but not yet credited ₹ 43,400
  6. Interest on investment collected by the bank ₹ 24,000

Answer:
Bank reconciliation statement as on 31st December, 2016
Samacheer Kalvi 11th Accountancy Solutions Chapter 8 Bank Reconciliation Statement

Question 15.
From the following particulars of John traders, prepare a bank reconciliation statement as on 31st March, 2018. (5 Marks)
(a) Bank overdraft as per bank statement ₹ 4,000
(b) Cheques amounting to ₹ 2,000 had been recorded in the cash book as having been deposited into the bank on 26th March 2018, but were entered in the bank statement on 4th April 2018.
(c) Amount received by bank through cash deposit machine amounting to ₹ 5,000, omitted in the cash book.
(d) Amount of ₹ 3,000 wrongly debited to John traders account by the bank, for which no details are available.
(e) Bills for collection credited by the bank till 29th March 2017 amounting to ₹ 4,000, but no advice received by John traders.
(f) Electricity charges made through net banking for ₹ 900 was wrongly entered in cash column of the cash book instead of bank column.
(g) Cash sales wrongly recorded in the bank column of the cash book for ₹ 4,000.
Answer:
Bank reconciliation statement of John Traders as on 31st March, 2018
Samacheer Kalvi 11th Accountancy Solutions Chapter 8 Bank Reconciliation Statement

Question 16.
Prepare bank reconciliation statement from the following data. (3 Marks)
Samacheer Kalvi 11th Accountancy Solutions Chapter 8 Bank Reconciliation Statement
Answer:
Bank reconciliation statement
Samacheer Kalvi 11th Accountancy Solutions Chapter 8 Bank Reconciliation Statement

Question 17.
Prepare bank reconciliation statement as on 31st March, 2017 from the following extracts of cash book and bank statement.
Samacheer Kalvi 11th Accountancy Solutions Chapter 8 Bank Reconciliation Statement
Bank Statement
Samacheer Kalvi 11th Accountancy Solutions Chapter 8 Bank Reconciliation Statement
Answer:
Samacheer Kalvi 11th Accountancy Solutions Chapter 8 Bank Reconciliation Statement

Question 18.
A trader received his bank statement on 31st December, 2017 which showed an overdraft balance of ₹ 12,000. On the same day, his cash book showed a debit balance of ₹ 2,000.
Analyse the following transactions. Choose the possible causes and prepare a bank reconciliation statement to show the causes of differences.
(a) Cheque deposited for ₹ 2,000 on 21st December, 2017. Bank credited the same on 26th December, 2017.
(b) Cheque issued for payment on 26th December, 2017 amounting to ₹ 2,500, not yet presented until 31st, December, 2017.
(c) Bank charges amounting to ₹ 200 not yet entered in the cash book.
(d) Online payment for ₹ 1,500 entered twice in the cash book.
(e) Cheque deposited amounting to ₹ 1,000, but omitted in the cash book. The same cheque was dishonoured by bank, but not yet entered in cash book.
(f) Cheque deposited, not yet credited by bank amounting to ₹ 17,800.
Answer:
Bank reconciliation statement as on 31st December, 2017
Samacheer Kalvi 11th Accountancy Solutions Chapter 8 Bank Reconciliation Statement
Note: Transactions (a) and (c) have been entered in both cash book and bank statement also. So we need not false in to accounts.

Textbook Case Study Solved

Magesh, an enthusiastic young entrepreneur, started a business on 1st  December, 2017. He opened a current account with a nationalised bank for his business transaction. In the same bank, he maintains his personal savings bank account too. He did not find time to maintain his cash book. So he appointed a person called Dinesh to take care of bank transactions. But that person was inexperienced.

Question 1.
On 1st December, 2017, the opening balance as per cash book and bank record was the same. On 2nd December, Magesh issued a cheque for ₹ 2,000 to a supplier, but the same was entered in the credit side of the cash book as ₹ 200.
Answer:
Credit balance as per bank statement is ₹ 19,700.

Question 2.
On 3rd, December, Magesh issued his savings bank account cheque for his personal expenses amounting to ₹ 2,500, but Dinesh assumed this as current account cheque and the same was entered in the cash book as drawings.
Answer:
Over casting of debit side of bank column of the cash book is ₹ 1800.

Question 3.
Dinesh was asked to deposit cash of ₹ 1,000 in cash deposit machine in order to make a payment to one of the business’ supplier. He credited the same in the bank column of the cash book.
Answer:
Wrong debit in cash book is ₹ 2,500.

Samacheer Kalvi 11th Accountancy Solutions Chapter 8 Bank Reconciliation Statement

Question 4.
On 15th December, one of his customers made online payment to Magesh’s current account, amounting to ₹ 1,000. There was no entry in the cash book for this.
Answer:
Wrong credit in cash book is ₹ 1,000.
Instead of personal bank account he can open business bank account (i.e.) current account.

Question 5.
Dinesh received his salary in cash for ₹ 5,000. He credited this amount in the bank column of cash book.
Answer:
Online payment no recorded in cash book is ₹ 1,000.

Question 6.
Bank made payment on 23rd December, amounting to ₹ 2,500, as per standing instruction. But, there is no entry in the cash book for the same.
Answer:
Wrong credit in the cash book of bank column is ₹ 5,000.

Question 7.
On 31st, December 2017, Magesh received a bank statement from his bank, which showed a credit balance of ₹ 19,700. He instructed Dinesh to check the statement with the cash book. On comparing both, Dinesh found that the cash book showed a balance of ₹ 14,500. He was puzzled. He needs your help to reconcile the balances.
Answer:
Insurance premium paid but not entered in cash book is ₹ 2,500.
Bank reconciliation cash book as on 31st December 2017
Samacheer Kalvi 11th Accountancy Solutions Chapter 8 Bank Reconciliation Statement

Samacheer Kalvi 12th Bio Zoology Solutions Chapter 4 Principles of Inheritance and Variation

Students can Download Bio Zoology Chapter 4 Principles of Inheritance and Variation Questions and Answers, Notes Pdf, Samacheer Kalvi 12th Bio Zoology Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Bio Zoology Solutions Chapter 4 Principles of Inheritance and Variation

Samacheer Kalvi 12th Bio Zoology Principles of Inheritance and Variation Text Book Back Questions and Answers

Question 1.
Haemophilia is more common in males because it is a ____________
(a) Recessive character carried by Y-chromosome
(b) Dominant character carried by Y-chromosome
(c) Dominant trait carried by X-chromosome
(d) Recessive trait carried by X-chromosome
Answer:
(d) Recessive trait carried by X-chromosome

Question 2.
ABO blood group in man is controlled by ____________
(a) Multiple alleles
(b) Lethal genes
(c) Sex linked genes
(d) Y-linked genes
Answer:
(a) Multiple alleles

Question 3.
Three children of a family have blood groups A, AB and B. What could be the geno types of their parents?
(a) IA IB and ii
(b) IA 1O and IB IO
(c) IB IB and IA IA
(d) IA IA and ii
Answer:
(b) IA IO and IB IO

Samacheer Kalvi 12th Bio Zoology Solutions Chapter 4 Principles of Inheritance and Variation

Question 4.
Which of the following is not correct?
(a) Three or more alleles of a trait in the population are called multiple alleles.
(b) A normal gene undergoes mutations to form many alleles
(c) Multiple alleles map at different loci of a chromosome
(d) A diploid organism has only two alleles out of many in the population
Answer:
(c) Multiple alleles map at different loci of a chromosome

Question 5.
Which of the following phenotypes in the progeny are ____________
(a) A and B only
(b) A,B only and AB
(c) AB only
(d) A,B,AB and O
Answer:
(d) A,B,AB and O

Question 6.
Which of the following phenotypes is not possible in the progeny of the parental genotypic combination IAIO × lAlB ?
(a) AB
(b) O
(c) A
(d) B
Answer:
(b) O

Question 7.
Which of the following is true about Rh factor in the offspring of a parental combination DdXDd (both Rh positive)?
(a) All will be Rh positive
(b) Half will be Rh positive
(c) About 3/4 will be Rh negative
(d) About one fourth will be Rh negative
Answer:
(d) About one fourth will be Rh negative

Question 8.
What can be the blood group of offspring when both parents have AB blood group?
(a) AB only
(b) A, B and AB
(c) A, B, AB and O
(d) A and B only
Answer:
(b) A, B and AB

Question 9.
If the childs blood group is ‘O’ and fathers blood group is ‘A’ and mother’s blood group is ‘B’ the genotype of the parents will be _________
(a) IAIA and IAIO
(b) IAIO and IBIO
(c) IAIO and IOIO
(d) IOIO and IBIB
Answer:
(b) IAIO and IBIO

Question 10.
XO type of sex determination and XY type of sex determination are examples of _________
(a) Male heterogamety
(b) Female heterogamety
(c) Male homogamety
(d) Both (b) and (c)
Answer:
(a) Male heterogamety

Question 11.
In an accident there is great loss of blood and there is no time to analyse the blood group which blood can be safely transferred?
(a) ‘O’ and Rh negative
(b) ‘O’ and Rh positive
(c) ‘B’ and Rh negative
(d) ‘AB’ and Rh positive
Answer:
(a) ‘O’ and Rh negative

Question 12.
Father of a child is colour blind and mother is carrier for colour blindness, the probability of the child being colour blind is _________
(a) 25%
(b) 50%
(c) 100%
(d) 75%
Answer:
(b) 50%

Question 13.
A marriage between a colour blind man and a normal woman produces _________
(a) All carrier daughters and normal sons
(b) 50% carrier daughters, 50% normal daughters
(c) 50% colour blind sons, 50% normal sons
(d) All carrier off springs
Answer:
(a) All carrier daughters and normal sons

Question 14.
Mangolism is a genetic disorder which is caused by the presence of an extra chromosome number.
(a) 20
(b) 21
(C) 4
(d) 23
Answer:
(b) 21

Question 15.
Klinefelters’ syndrome is characterized by a karyotype of _________
(a) XYY
(b) XO
(c) XXX
(d) XXY
Answer:
(d) XXY

Question 16.
Females with Turners’syndrome have _________
(a) Small uterus
(b) Rudimentary ovaries
(c) Underdeveloped breasts
(d) All of these
Answer:
(d) All of these

Samacheer Kalvi 12th Bio Zoology Solutions Chapter 4 Principles of Inheritance and Variation

Question 17.
Pataus’ syndrome is also referred to as _________
(a) 13-Trisomy
(b) 18-Trisormy
(c) 21-Trisormy
(d) None of these
Answer:
(a) 13-Trisomy

Question 18.
Who is the founder of Modem Eugenics movement?
(a) Mendel
(b) Darwin
(c) Fransis Galton
(d) Karl pearson
Answer:
(c) Fransis Galton

Question 19.
Improvement of human race by encouraging the healthy persons to marry early and produce large number of children is called _________
(a) Positive eugenics
(b) Negative eugenics
(c) Positive euthenics
(d) Positive euphenics
Answer:
(a) Positive eugenics

Question 20.
The _________ deals with the control of several inherited human diseases especially inborn errors of metabolism.
(a) Euphenics
(b) Eugenics
(c) Euthenics
(d) All of these
Answer:
(a) Euphenics

Question 21.
“Universal Donor” and “Universal Recipients” blood group are _________ and _________ respectively.
(a) AB, O
(b) O, AB
(c) A, B
(d) B, A
Answer:
(b) O, AB

Question 22.
ZW-ZZ system of sex determination occurs in _________
(a) Fishes
(b) Reptiles
(c) Birds
(d) All of these
Answer:
(d) All of these

Question 23.
Co-dominant blood group is _________
(a) A
(b) AB
(c) B
(d) O
Answer:
(b) AB

Question 24.
Which of the following is incorrect regarding ZW-ZZ type of sex determination?
(a) It occurs in birds and some reptiles
(b) Females are homogametic and males are heterogametic
(c) Male produce two types of gametes
(d) It occurs in gypsy moth
Answer:
(b) Females are homogametic and males are heterogametic

Question 25.
What is haplodiploidy?
Answer:
In haplodiploidy, the sex of the offspring is determined by the number of sets of chromosomes it receives. Fertilized eggs develop into females (Queen or Worker) and unfertilized eggs develop into males (drones) by parthenogenesis. It means that the males have half the number of chromosomes (haploid) and the females have double the number (diploid).

Question 26.
Distinguish between heterogametic and homogametic sex determination systems
Answer:
Heterogametic Sex:

  1. Organisms producing two different types of gametes.
  2. Example: Human male.
  3. Sperm with X chromosome Sperm with a Y chromosome

Heterogametic Sex:

  1. Organisms producing two different types of gametes.
  2. Example: Human female.
  3. Every egg produced contain X chromosomes.

Question 27.
What is Lyonisation?
Answer:
Lyonisation is a process of inactivation of one of the X chromosomes in some females.

Question 28.
What is criss-cross inheritance?
Answer:

  1. Inheritance of genes from a male parent to female child and then to male grandchild or female parent to male child and then to female grandchild.
  2. E.g., X-linked gene inheritance.

Question 29.
Why are sex-linked recessive characters more common in male human beings?
Answer:
Sex-linked inherited traits are more common in males than females because males are hemizygous and therefore express the trait when they inherit one mutant allele.

Question 30.
What are holandric genes?
Answer:
The genes present in the differential region of Y chromosome are called Y- linked or holandric genes. The Y linked genes have no corresponding allele in X chromosome.

Question 31.
Mention the symptoms of Phenylketonuria.
Answer:
Severe mental retardation, light pigmentation of skin and hair. Phenylpyruvic acid is excreted in the urine.

Question 32.
Mention the symptoms of Down’s syndrome.
Answer:
Severe mental retardation, defective development of the central nervous system, increased separation between the eyes, flattened nose, ears are malformed, mouth is constantly open and the tongue protrudes.

Samacheer Kalvi 12th Bio Zoology Solutions Chapter 4 Principles of Inheritance and Variation

Question 33.
Differentiate Intersexes from Supersexes.
Answer:
Intersexes :
Intersexes refers to the individuals having the characteristics of both female and male sexes and their sexual anatomy does not seem to fit the typical definition of male or female.
Example: Super males in humans human beings have 44+XYY chromosomes.

Supersexes:
Supersexes are formed as a result of an abnormal combination of sex chromosomes.
Example: Super males in humans human beings have 44+XYY chromosomes.

Question 34.
Explain the genetic basis of ABO blood grouping in man.
Multiple allele inheritance of ABO blood groups
Answer:
Blood differs chemically from person to person. When two different incompatible blood types are mixed, agglutination (clumping together) of erythrocytes (RBC) occurs. The basis of these chemical differences is due to the presence of antigens (surface antigens) on the membrane of RBC and epithelial cells. Karl Landsteiner discovered two kinds of antigens called antigen ‘A’ and antigen ‘B’ on the surface of RBC’s of human blood. Based on the presence or absence of these antigens three kinds of blood groups, type ‘A’, type ‘B’, and type ‘O’ (universal donor) were recognized. The fourth and the rarest blood group ‘AB’ (universal recipient) was discovered in 1902 by two of Landsteiner’s students Von De Castelle and Sturli.

Bernstein in 1925 discovered that the inheritance of different blood groups in human beings is determined by a number of multiple allelic series. The three autosomal alleles located on chromosome 9 are concerned with the determination of blood group in any person. The gene controlling blood type has been labeled as ‘L’ (after the name of the discoverer, Landsteiner) or I (from isoagglutination). The I gene exists in three allelic forms, IA, IB and 1°. IA specifies A antigen. IB allele determines B antigen and 1° allele specifies no antigen. Individuals who possess these antigens in their fluids such as the saliva are called secretors.

Each allele (IA and IB) produces a transferase enzyme. IA allele produces N-acetyl galactose transferase and can add N-acetyl galactosamine (NAG) and IB allele encodes for the enzyme galactose transferase that adds galactose to the precursor (i.e. H substances). In the case of IO/IO allele no terminal transferase enzyme is produced and therefore called “null” allele and hence cannot add NAG or galactose to the precursor.

From the phenotypic combinations it is evident that the alleles IA and IB are dominant to 1°, but co-dominant to each other (IA = IB). Their dominance hierarchy can be given as (IA=IB> 1O). A child receives one of three alleles from each parent, giving rise to six possible genotypes and four possible blood types (phenotypes). The genotypes are IAIA, IAIO, IBIB, IBIO, IAIB and IO IO

Question 35.
How is sex determined in human
Answer:
Samacheer Kalvi 12th Bio Zoology Solutions Chapter 4 Principles of Inheritance and Variation img 1
Genes determining sex in human beings are located on two sex chromosomes, called allosomes.
In mammals, sex determination is associated with chromosomal differences between the two sexes, typically XX females and XY males. 23 pairs of human chromosomes include 22 pairs of autosomes (44A) and one pair of sex chromosomes (XX or XY). Females are homogametic producing only one type of gametes (egg), each containing one X chromosome while the males are heterogametic producing two types of sperms with X and Y chromosomes. An independently evolved XX: XY system of sex chromosomes also exist in Drosophila.

Question 36.
Explain male heterogamety.
Answer:
Male heterogamety (XY males) is a type of sex determination in which males produce two different types of gametes. For example, human males produce two kinds of sperms that is sperm with X-chromosome and sperms with Y-chromosome.

Question 37.
Brief about female heterogamety.
Answer:
Female heterogamety (ZO females) refers to the condition, where the female produces two types of egg cells. Some with Z chromosome and some without the Z chromosome.

Question 38.
Give an account of genetic control of Rh factor?
Genetic control of Rh factor
Answer:
Samacheer Kalvi 12th Bio Zoology Solutions Chapter 4 Principles of Inheritance and Variation img 2
Fisher and Race hypothesis: Rh factor involves three different pairs of alleles located on three different closely linked loci on the chromosome pair. This system is more commonly in use today, and uses the ‘Cde’ nomenclature. In the given figure, three pairs of Rh alleles (Cc, Dd and Ee) occur at 3 different loci on homologous chromosome pair-1.

The possible genotypes will be one C or c, one D or d, one E or e from each chromosome. For e.g. CDE/cde; CdE/cDe; cde/cde; CDe/CdE etc. All genotypes carrying a dominant ‘D’ allele will produce Rh+positive phenotype and double recessive genotype ‘dd’ will give rise to Rh negative phenotype.

Wiener Hypothesis:
Wiener proposed the existence of eight alleles (R1, R2, R0, RZ, r, r1, r11, ry) at a single Rh locus. All genotypes carrying a dominant ‘R allele’ (R1, R2 ,R0 ,Rz) will produce ‘Rh-positive’ ^phenotype and double recessive genotypes (rr, rr1, rr11, rry) will give rise to Rh-negative phenotype.

Question 39.
Explain the mode of sex determination in honeybees.
Haplodiploidv in Honeybees:
Answer:
Haplodiploidv in Honeybees:
In hymenopteran insects such as honeybees, ants and wasps, a mechanism of sex determination called haplodiploidy mechanism of sex determination is common. In this system, the sex of the offspring is determined by the number of sets of chromosomes it receives. Fertilized eggs develop into females (Queen or Worker) and unfertilized eggs develop into males (drones) by parthenogenesis. It means that the males have half the number of chromosomes (haploid) and the females have double the number (diploid), hence the name haplodiplody for this system of sex determination.

This mode of sex determination facilitates the evolution of sociality in which only one diploid female becomes a queen and lays the eggs for the colony. All other females which are diploid having developed from fertilized eggs help to raise the queen’s eggs and so contribute to the queen’s reproductive success and indirectly to their own, a phenomenon known as Kin Selection. The queen constructs their social environment by releasing a hormone that suppresses fertility of the workers.

Samacheer Kalvi 12th Bio Zoology Solutions Chapter 4 Principles of Inheritance and Variation

Question 40.
Discuss the genic balance mechanism of sex determination with reference to Drosophila?
Answer:
XX-XY type (Lygaeus Type) sex determination is seen in Drosophila. The females are homogametic with XX chromosome, while the males are heterogametic with X and Y
Samacheer Kalvi 12th Bio Zoology Solutions Chapter 4 Principles of Inheritance and Variation img 3
chromosome. Homogametic females produce only one kind of egg, each with one X chromosome, while the heterogametic males produce two kinds of sperms some with X chromosome and some with Y chromosome.

The sex of the embryo depends on the fertilizing sperm. An egg fertilized by an ‘X’ bearing sperm produces a female, if fertilized by a ‘Y’ bearing sperm, a male is produced.

Question 41.
What are the applications of Karyotyping?
Answer:

  1. Karyotyping helps in gender identification.
  2. It is used to detect chromosomal aberrations like deletion, duplication, translocation, non-disjunction of chromosomes.
  3. It helps to identify the abnormalities of chromosomes like aneuploidy.
  4. It is also used in predicting the evolutionary relationships between species.
  5. Genetic diseases in human beings can be detected by this technique.

Question 42.
Explain the inheritance of sex-linked characters in human beings.
Answer:
Haemophilia is commonly known as bleeder’s disease, which is more common in men than women. This hereditary disease was first reported by John Cotto in 1803. Haemophilia is caused by a recessive X-linked gene. A person with a recessive gene for haemophilia lacks a normal clotting substance (thromboplastin) in blood, hence minor injuries cause continuous ’bleeding, leading to death. The females are carriers of the disease and would transmit the disease to 50% of their sons even if the male parent is normal. Haemophilia follows the characteristic criss-cross pattern of inheritaitce.

Question 43.
What is extrachromosomal inheritance? Explain with an example.
Answer:
The cytoplasmic extranuclear genes have a characteristic pattern of inheritance which does not resemble genes of nuclear chromosomes and are known as Extrachromosomal/ Cytoplasmic inheritance.

Question 44.
Comment on the methods of Eugenics.
Answer:
Eugenics refers to the study of the possibility of improving the qualities of human population. Methods of Eugenics:

  1. Sex-education in school and public forums.
  2. Promoting the uses of contraception.
  3. Compulsory sterilization for mentally retarded and criminals.
  4. Egg donation.
  5. Artificial insemination by donors.
  6. Prenatal diagnosis of genetic disorders and performing MTP
  7. Gene therapy
  8. Cloning
  9. Egg/sperm donation of healthy individuals.

Samacheer Kalvi 12th Bio Zoology Principles of Inheritance and Variation Additional Questions and Answers

1 – Mark Questions

Question 1.
If a colorblind female marries a normal male, their sons will be _________
(a) All normal visioned
(b) All color blinded
(c) One-half normal visioned another half colorblind
(d) Three fourth colorblind one fourth normal
Answer:
(c) One-half normal visioned another half colorblind

Question 2.
Excess hair growth on pinna is a feature noticed only in males because _________
(a) Males produce more testosterone
(b) a gene responsible for the character is located in Y-chromosome
(c) Estrogen suppresses the character in females
(d) females act only as a carrier for this character
Answer:
(b) gene responsible for the character is located in Y-chromosome

Question 3.
ABO blood group is a classical example for _________
(a) Multiple allelism
(b) Pleotropism
(c) Incomplete dominance
(d) Polygenic mechanism
Answer:
(a) Multiple allelism

Question 4.
Unit of heredity is _________
(a) allele
(b) allelomorph
(d) gene
Answer:
(d) gene

Question 5.
Identify the proper dominance hierarchy.
(a) IB = 1° > IB
(b) IA = IB > 0
(c) IO = IB > 
(d) IB = IA > O
Answer:
(b) IA = IB > 0

Question 6.
Haemophilia is more common in human males than human females. The reason is due to ______
(а) X-linked dominant gene
(b) X-linked recessive gene
(c) Y-linked recessive gene
(d) Allosomal abnormality
Answer:
(b) X-linked recessive gene

Question 7.
Identify the correct statement.
(a) Homozygous sex chromosome (XX) produce males in Drosophila
(b) Homozygous sex chromosome (ZZ) determine female sex in birds
(c) Heterozygous sex chromosome (XO) determine male sex in grasshopper
(d) Heterozygous sex chromosome (ZW) determine male sex in gypsy moth
Answer:
(c) Heterozygous sex chromosome (XO) determine male sex in grasshopper

Question 8.
Which blood group does not possess antibodies?
(a) IAIB
(b) IOIO
(c) IAO
(d) IBIB
Answer:
(a) IAIB

Question 9.
Assertion (A): On diagnosis, Ramu is reported to have underdeveloped testis and gynecomastia.
Reason (R): His karyotype reveals XXY condition.
(а) A is right but R is wrong
(b) R explains A
(c) Both A and R are wrong
(d) Both and R are right but R is not the correct explanation of A
Answer:
(b) R explains A

Question 10.
Pick out the odd man.
(a) Klinefelter’s syndrome
(b) Turner’s syndrome
(c) Huntington’s chorea
(d) 13-Trisomy
Answer:
(c) Huntington’s chorea

Question 11.
Pick the odd one out regarding Mendelian disorder.
(a) Thalassemia
(b) phenylketonuria
(c) Albinism
(d) Huntington’s chorea
Answer:
(d) Huntington’s chorea

Question 12.
Samacheer Kalvi 12th Bio Zoology Solutions Chapter 4 Principles of Inheritance and Variation
Answer:
(a) A – iv, B – ii, D – i, D – iii

Question 13.
Identify the proper ratio of normal visioned individuals against colorblind individuals, if colorblind carrier female marries a normal male.
(a) 1 : 1
(b) 3 : 1
(c) 1 : 3
(d) All four are normal visioned
Answer:
(c) 1 : 3

Samacheer Kalvi 12th Bio Zoology Solutions Chapter 4 Principles of Inheritance and Variation

Question 14.
Pick out the correct statement.
(i) Karyotyping helps in gender identification
(ii) Holandric genes are located on X-chromosome
(iii) Trisomy-21 is an allosomal abnormality
(iv) Cooley’s anaemia is an autosomal recessive disorder
(a) i, iii, iv are correct
(c) i and iv are correct
Answer:
(c) i and iv are correct

Question 15.
DOPA stands for _________
(a) 3,4 – dihydroxy phenylacetate
(b) 3,4 – dihydroxy phenyle alanine
(c) 3,4 – dihydroxy phenyl asparate
(d) 3,4 – dihydroxy phenyle aldehyde
Answer:
(b) 3,4 – dihydroxy phenyle alanine

Question 16.
The type of antibody generated against Rh antigen is _________
(a) IgE
(b) IgG
(c) IgA
Answer:
(b) IgG

Question 17.
Which of the following symbol is used in pedigree analysis to represent unspecified sex?
Samacheer Kalvi 12th Bio Zoology Solutions Chapter 4 Principles of Inheritance and Variation img 4
Answer:
Samacheer Kalvi 12th Bio Zoology Solutions Chapter 4 Principles of Inheritance and Variation img 5

Question 18.
A colorblind man marries a woman with normal sight who has no history of color blindness in her family. What is the probability of their grandson being colorblind?
(a) 1/4
(b) 3/4
(c) 2/4
(d) 4/4
Answer:
(a) 1/4

Question 19.
Multiple alleles are located _________
(a) at different loci on homologous chromosome
(b) at same locus on homologous chromosome
(c) at different loci on non-homologous chromosome
(d) at different chromosomes
Answer:
(b) at same locus on homologous chromosome

Question 20.
Identify the incorrect statement regarding haplodiploidy.
(g) Haplodiploidy is noticed in honeybees and drosophila
(b) Unfertilized eggs develop into drones
(c) Fertilized eggs develop into queen and worker bees
(d) Males have half the total chromosomal number
Answer:
(a) Haplodiploidy is noticed in honeybees and drosophila

Question 21.
IA and IB genes of ABO blood group are _________
(a) Co-dominant
(b) Pleotropic
(c) Dominant and recessive
(d) Epistatic
Answer:
(a) Co-dominant

Question 22.
Which one of the following crosses show 3 : 1 ratio of normal visioned versus carrier blind?
(a) XC XC × X+Y
(b) X+ XC × XC Y
(c) X+XC × X+Y
(d) X+X+× XC Y
Answer:
(c) X+XC × X+Y

2 – Mark Questions

Question 1.
Define multiple allelism.
Answer:
When three or more alleles of a gene that control a particular trait occupy the same locus on the homologous chromosome of an organism, they are called multiple alleles and their inheritance is called multiple allelism.

Question 2.
Name the discoverers of antigen A, B and AB.
Answer:
Antigens A and Antigen B was discovered by Karl Landsteiner. Antigen AB was discovered by Von De Castelle and Sturli.

Question 3.
What happens if type A blood is injected to a person having B blood group? Explain the reason.
Answer:
When two different incompatible blood types are mixed, agglutination (clumping together) of erythrocytes (RBC) occurs. The basis of these chemical differences is due to the presence of antigens (surface antigens) on the membrane of RBC and epithelial cells.

Question 4.
State the allelic forms of I gene and mention its chromosomal location.
Answer:
The I gene exists in three forms: IA, IB and I°. The alleles are located on chromosome 9.

Question 5.
Write the possible genotypes for a person having B-blood group.
Answer:
The possible genotypes of a B-blood group person are IBIB or IBI°.

Question 6.
State Wiener Hypothesis on Rh-factor.
Answer:
Wiener proposed the existence of eight alleles (R1, R2, R0, Rz, r, r1, r11, ry) at a single Rh locus. All genotypes carrying a dominant ‘R allele’ (R1, R2 ,R0 ,Rz) will produce ‘Rh-positive’ phenotype and double recessive genotypes (rr, rr1, rr11, rry) will give rise to Rh-negative phenotype.

Question 7.
Distinguish between homogametic and heterogametic condition with example.
Answer:
Homogametic organism:

  1. Organism producing only one type of gametes.
  2. e.g. Human female (Only X)

Heterogametic organism:

  1. Organism producing two different types of gametes.
  2. e.g. Human Male (X and Y)

Question 8.
Name any four organism expressing ZW-ZZ type of sex determination.
Answer:
Gypsy moth, fishes, reptiles and birds.

Question 9.
Expand

  1. SRY
  2. TDF

Answer:

  1. SRY – Sex Determining region
  2. TDF – Testes Determining Factor

Question 10.
Define Barr body.
Answer:
In 1949, Barr and Bertram first observed a condensed body in the nerve cells of female cat which was absent in the male. This condensed body was called sex chromatin by them and was later referred as Barr body.

Question 11.
Based on Lyon’s hypothesis, mention the number of Barr bodies in XXY males, XO females.
Answer:

  1. XXY males – One Barr body.
  2. XO females – No Barr body.

Question 12.
State Lyon’s hypothesis.
Answer:
Lyon’s hypothesis states that in mammals the necessary dosage compensation is accomplished by the inactivation of one of the X chromosome in females so that both males and females have only one functional X chromosome per cell.

Mary Lyon suggested that Barr bodies represented an inactive chromosome, which in females becomes tightly coiled into a heterochromatin, a condensed and visible form of chromatin (Lyon’s hypothesis). The number of Barr bodies observed in cell was one less than the number of X-Chromosome. XO females have no Barr body, whereas XXY males have one Barr body.

Samacheer Kalvi 12th Bio Zoology Solutions Chapter 4 Principles of Inheritance and Variation

Question 13.
Mention few X-linked inherited diseases.
Answer:
Red-green colour blindness or daltonism, haemophilia and Duchenne’s muscular dystrophy.

Question 14.
Define Karyotyping.
Answer:
Karyotyping is a technique through which a complete set of chromosomes is separated from a cell and the chromosomes are arranged in pairs. An idiogram refers to a diagrammatic representation of chromosomes.

Question 15.
Explain the inheritance pattern of Y-linked genes with example.
Answer:
Genes in the non-homologous region of the Y-chromosome are inherited directly from male to male. In humans, the Y-linked or holandric genes for hypertrichosis (excessive development of hairs on pinna of the ear) are transmitted directly from father to son, because males inherit the Y chromosome from the father. Female inherits only X chromosome from the father and are not affected.

Question 16.
Observe the symbol used in pedigree analysis and give the proper terms they represent.
Samacheer Kalvi 12th Bio Zoology Solutions Chapter 4 Principles of Inheritance and Variation img 6
Answer:
Samacheer Kalvi 12th Bio Zoology Solutions Chapter 4 Principles of Inheritance and Variation img 7

Question 17.
Write a brief note on pedigree analysis.
Answer:
Pedigree is a “family tree”, drawn with standard genetic symbols, showing the inheritance pathway for specific phenotypic characters. Pedigree analysis is the study of traits as they have appeared in a given family line for several past generations.

Question 18.
What do you mean by ‘Mendelian disorder’.
Answer:
Alteration or mutation in a single gene causes Mendelian disorders. These disorders are transmitted to the off springs on the same line as the Mendelian pattern of inheritance.
E.g., Thalassemia.

Question 19.
Name any four Mendelian disorders.
Answer:

  1. Thalassemia
  2. Albinism
  3. sickle cell anaemia
  4. Huntington’s chorea

Question 20.
What is the phenotype of

  1. IAIO
  2. IOIO

Answer:

  1. IAIO – A blood group person
  2. IOIO – O blood group person

Question 21.
On which chromosomes does HBA1 gene and HBB genes are located?
Answer:

  1. HBA1 gene is located on chromosome 16.
  2. HBB gene is located on chromosome 11.

Question 22.
Complete the equation.
Samacheer Kalvi 12th Bio Zoology Solutions Chapter 4 Principles of Inheritance and Variation img 8
Answer:
(a) A = Phenylalanine hydroxylase
(b) B = Tyrosinase

Question 23.
Write a note on Huntington’s chorea.
Answer:
Huntington’s chorea is inherited as an autosomal dominant lethal gene in man. It is characterized by involuntary jerking of the body and progressive degeneration of the nervous system, accompanied by gradual mental and physical deterioration. The patients with this disease usually die between the age of 35 and 40.

Question 24.
Comment on Trisomy-21.
Answer:
Trisomic condition of chromosome – 21 results in Down’s syndrome. It is characterized by severe mental retardation, defective development of the central nervous system, increased separation between the eyes, flattened nose, ears are malformed, mouth is constantly open and the tongue protrudes.

Question 25.
Mention the genetic makeup of Turner’s syndrome person and Klinefelter’s syndrome , person.
Answer:
Klinefelter’s syndrome – 44AA+XXY Turner’s syndrome – 44AA+XO

Question 26.
List out any four clinical symptoms of Klinefelter’s syndrome.
Answer:
Gynaecomastia, high pitched voice, under developed genetalia and tall with long limbs.

3 – Mark Questions

Question 27.
Write the types of sex-determination mechanisms does the following crosses as shown. Give an example for each.

  1. Female XX with Male XO
  2. Female ZW with Male ZZ

Answer:

  1. Male heterogamety. e.g., Human beings.
  2. Female heterogamety. e.g., Birds.

Question 28.
What are the enzymes encoded by the alleles IA, IB and I°?
Answer:
IA allele produces N-acetyl galactose transferase and can add N-acetyl galactosamine (NAG) and IB allele encodes for the enzyme galactose transferase that adds galactose to the precursor (i.e. H substances). In the case of IO/IO allele no terminal transferase enzyme is produced and therefore called “null” allele and hence cannot add NAG or galactose to the precursor.

Samacheer Kalvi 12th Bio Zoology Solutions Chapter 4 Principles of Inheritance and Variation

Question 29.
Draw a tabular column representing various types of blood group in human beings, their phenotypes, genotypes, antigens and respective antibodies.
Answer:
Genetic basis of the human ABO blood groups:
Samacheer Kalvi 12th Bio Zoology Solutions Chapter 4 Principles of Inheritance and Variation img 9

Question 30.
Give an account on Rhesus factor.
Answer:
Rhesus or Rh – Factor: The Rh factor or Rh antigen is found on the surface of erythrocytes. It was discovered in 1940 by Karl Landsteiner and Alexander Wiener in the blood of rhesus monkey, Macaca rhesus and later in human beings. The term ‘Rh factor’ refers to “immunogenic D antigen of the Rh blood group system. An individual having D antigen are Rh D positive (Rh+) and those without D antigen are Rh D negative (Rh”)”. Rhesus factor in the blood is inherited as a dominant trait. Naturally occurring Anti D antibodies are absent in the plasma of any normal individual. However if an Rh” (Rh negative) person is exposed to Rh+ (Rh positive) blood cells (erythrocytes) for the first time, anti D antibodies are formed in the blood of that individual. On the other hand, when an Rh positive person receives Rh negative blood no effect is seen.

Question 31.
How Erythroblastosis foetalis can be prevented?
Answer:
If the mother is Rh negative and foetus is Rh positive, anti D antibodies should be administered to the mother at 28th and 34th week of gestation as a prophylactic measure. If the Rh negative mother delivers Rh positive child then anti D antibodies should be administered to the mother soon after delivery. This develops passive immunity and prevents the formation of anti D antibodies in the mothers blood by destroying the Rh foetal RBC before the mother’s immune system is sensitized. This has to be done whenever the woman attains pregnancy.

Question 32.
Explain XX-XO type of sex determination.
Answer:
Samacheer Kalvi 12th Bio Zoology Solutions Chapter 4 Principles of Inheritance and Variation img 10
XX-XO method of sex determination is seen in bugs, some insects such as cockroaches and grasshoppers. Pi The female with two X chromosomes are homogametic Gametes (XX) while the males with only one X chromosome are heterogametic (XO). The presence of an unpaired X chromosomes determines the male sex. The males PI Generation with unpaired ‘X’ chromosome produce two types of sperms, one half with X chromosome and other half without X chromosome. The sex of the offspring depends upon the sperm that fertilizes the egg.

Question 33.
Name the type of sex-determination mechanism of the following organisms.

  1. Gypsy moth
  2. Human beings
  3. Butterflies

Answer:

  1. Gypsy moth -ZW – ZZ type (ZW-females, ZZ – males)
  2. Human beings – XX – XY type (XX-females, XY – males)
  3. Butterflies – ZO – ZZ type (ZO-females, ZZ – males)

Question 34.
Complete the following cross.
AAZW × AAZZ (female) (male)
Answer:
Samacheer Kalvi 12th Bio Zoology Solutions Chapter 4 Principles of Inheritance and Variation img 11

Question 35.
Role of Y- chromosome is crucial for maleness – Justify.
Answer:
Current analysis of Y chromosomes has revealed numerous genes and regions with potential genetic function; some genes with or without homologous counterparts are seen on the X. Present at both ends of the Y chromosome are the pseudoautosomal regions (PARs) that are similar with regions on the X chromosome which synapse and recombine during meiosis. The remaining 95% of the Y chromosome is referred as the Non-combining Region of the Y (NRY).

The NRY is divided equally into functional genes (euchromatic) and non-functional genes (heterochromatic). Within the euchromatin regions, is a gene called Sex determining region Y (SRY). In humans, absence of Y chromosome inevitably leads to female development and this SRY gene is absent in X chromosome. The gene product of SRY is the testes determining factor (TDF) present in the adult male testis.

Question 36.
Color blindness is a perfect example for criss-cross of inheritance – Justify the statement.
Answer:
A marriage between a colour blind man and a normal visioned woman will produce normal visioned male and female individuals in F1 generation but the females are carriers. The marriage between a F1 normal visioned carrier woman and a normal visioned male will produce one normal visioned female, one carrier female, one normal visioned male and one colour blind male. Colour blind trait is inherited from the male parent to his grandson through carrier daughter, which is an example of criss-cross pattern of inheritance.
image 12

Samacheer Kalvi 12th Bio Zoology Solutions Chapter 4 Principles of Inheritance and Variation

Question 37.
How the Karyotype of lymphocytes was prepared by Tjio and Levan?
Answer:
Preparation of Karyotype Tjio and Levan (1960) described a simple method of culturing lymphocytes from the human blood. Mitosis is induced followed by addition of colchicine to arrest cell division at metaphase stage and the suitable spread of metaphase chromosomes is photographed. The individual chromosomes are cut from the photograph and are arranged in an orderly fashion in homologous pairs. This arrangement is called a karyotype. Chromosome banding permits structural definitions and differentiation of chromosomes.

Question 38.
What is a genetic disorder? Mention its types?
Answer:
A genetic disorder is a disease or syndrome that is caused by an abnormality in an individual -DNA. Abnormalities can range from a small mutation in a single gene to the addition or subtraction of an entire chromosome or even a set of chromosomes. Genetic disorders are of two types namely, Mendelian disorders and chromosomal disorders.

Question 39.
Explain the genetic basis of Phenylketonuria.
Answer:
Phenylketonuria is an inborn error of Phenylalanine metabolism caused due to a pair of autosomal recessive genes. It is caused due to mutation in the gene PAH (phenylalanine hydroxylase gene) located on chromosome 12 for the hepatic enzyme “phenylalanine hydroxylase”.

This enzyme is essential for the conversion of phenylalanine to tyrosine. Affected individual lacks this enzyme, so phenylalanine accumulates and gets converted to phenylpyruvic acid and other derivatives. It is characterized by severe mental retardation, light pigmentation of skin and hair. Phenylpyruvic acid is excreted in the urine.
Samacheer Kalvi 12th Bio Zoology Solutions Chapter 4 Principles of Inheritance and Variation img 13

Question 40.
Give an account on Patau’s syndrome.
Answer:
Trisomic condition of chromosome 13 results in Patau’s syndrome. Meiotic non-disjunction is thought to be the cause for this chromosomal abnormality. It is characterized by multiple and severe body malformations as well as profound mental deficiency. Small head with small eyes, cleft palate, malformation of the brain and internal organs are some of the symptoms of this syndrome.

Question 41.
Define aneuploidy.
Answer:
Failure of chromatids to segregate during cell division resulting in the gain or loss of one or more chromosomes is called aneuploidy. It is caused by non-disjunction of chromosomes.

Question 42.
What do you mean by “syndrome”? Give two examples.
Answer:
Group of signs and symptoms that occur together and characterize a particular abnormality is called a syndrome, e.g., Down’s syndrome and Turner’s syndrome.

5 – Mark Questions

Question 42.
Explain in detail about Erythroblastosis foetalis.
Answer:
Rh incompatability has great significance in child birth. If a woman is Rh negative and the man is Rh positive, the foetus may be Rh positive having inherited the factor from its father. The Rh negative mother becomes sensitized by carrying Rh positive foetus within her body. Due to damage of blood vessels, during child birth, the mother’s immune system recognizes the Rh antigens and gets sensitized. The sensitized mother produces Rh antibodies. The antibodies are IgG type which are small and can cross placenta and enter the foetal circulation. By the time the mother gets sensitized and produce anti ‘D’ antibodies, the child is delivered.

Usually no effects are associated with exposure of the mother to Rh positive antigen during the first child birth, subsequent Rh positive children carried by the same mother, may be exposed to antibodies produced by the mother against Rh antigen, which are carried across the placenta into the foetal blood circulation. This causes haemolysis of foetal RBCs resulting in haemolytic jaundice and anaemia. This condition is known as Erythroblastosis foetalis or Haemolytic disease of the new bom (HDN).

Question 43.
Decribe female heterogamy and its types.
Answer:
Heterogametic Females:
In this method of sex determination, the homogametic male possesses two ‘X’ chromosomes as in certain insects and certain vertebrates like fishes, reptiles and birds producing a single type of gamete; while females produce dissimilar gametes. The female sex consists of a single ‘X’ chromosome or one ‘X’ and one ‘Y’ chromosome. Thus the females are heterogametic and produce two types of eggs. Heterogametic females are of two types, ZO-ZZ type and ZW-ZZ type.
Samacheer Kalvi 12th Bio Zoology Solutions Chapter 4 Principles of Inheritance and Variation img 14
ZO-ZZ Type:
This method of sex determination is seen in certain moths, butterflies and domestic chickens. In this type, the female possesses
single ‘Z’ chromosome in its body cells and is heterogametic (ZO) producing two kinds of eggs some with ‘Z’ chromosome and some without ‘Z’ chromosome, while the male possesses two ‘Z’ chromosomes and is homogametic (ZZ).
Samacheer Kalvi 12th Bio Zoology Solutions Chapter 4 Principles of Inheritance and Variation img 15
ZW-ZZ type:
This method of sex determination occurs in certain insects (gypsy moth) and in vertebrates such as fishes, reptiles and birds. In this method the female has one ‘Z’ and one ‘ W’ chromosome (ZW) producing two types of eggs, some carrying the Z chromosomes and some carry the W chromosome. The male sex has two ‘Z’ chromosomes and is homogametic (ZZ) producing a single type of sperm.

Question 44.
Write elaborately about the following Mendelian disorders.
Answer:
(a) Thalassemia
(b) Albinism
Answer:
(a) Thalassemia: Thalassemia is an autosomal recessive disorder. It is caused by gene mutation resulting in excessive destruction of RBC’s due to the formation of abnormal haemoglobin molecules. Normally haemoglobin is composed of four polypeptide chains, two alpha and two beta globin chains. Thalassemia patients have defects in either the alpha or beta globin chain causing the production of abnormal haemoglobin molecules resulting in anaemia.

Thalassemia is classified into alpha and beta based on which chain of haemoglobin molecule _ is affected. It is controlled by two closely linked genes HBA1 and HBA2 on chromosome 16. Mutation or deletion of one or more of the four alpha gene alleles causes Alpha Thalassemia. In Beta Thalassemia, production of beta globin chain is affected. It is controlled by a single gene (HBB) on chromosome 11. It is the most common type of Thalassemia and is also known as Cooley’s anaemia. In this disorder, the alpha chain production is increased and damages the membranes of RBC.

(b) Albinism: Albinism is an inborn error of metabolism, caused due to an autosomal recessive gene. Melanin pigment is responsible for skin colour. Absence of melanin results in a condition called albinism. A person with the recessive allele lacks the tyrosinase enzyme system, which is required for the conversion of dihydroxyphenyl alanine (DOPA) into melanin pigment inside the melanocytes. In an albino, melanocytes are present in normal numbers in their skin, hair, iris, etc., but lack melanin pigment.
Samacheer Kalvi 12th Bio Zoology Solutions Chapter 4 Principles of Inheritance and Variation img 16

Question 45.
Discuss any two Allosomal anomalies in human.
Allosomal abnormalities in human beings
Answer:
Mitotic or meiotic non-disjunction of sex chromosomes causes allosomal abnormalities. Several sex chromosomal abnormalities have been detected.
E.g. Klinefelter’s syndrome and Turner’s syndrome.

1. Klinefelter’s Syndrome (XXY Males): This genetic disorder is due to the presence of an additional copy of the X chromosome resulting in a karyotype of 47,XXY. Persons with this syndrome have 47 chromosomes (44AA+XXY). They are usually sterile males, tall, obese, with long limbs, high pitched voice, under developed genetalia and have feeble breast (gynaecomastia) development.

2. Turner’s Syndrome (XO Females): This genetic disorder is due to the loss of a X chromosome resulting in a karyotype of 45,X. Persons with this syndrome have 45 chromosomes (44 autosomes and one X chromosome) (44AA+XO) and are sterile females. Low stature, webbed neck, under developed breast, rudimentary gonads lack of menstrual cycle during puberty, are the main symptoms of this syndrome.

Higher Order Thinking Skills (HOTs) Questions

Question 1.
On analysis, a person’s karyotype reveals an extra one chromosome of twenty first pair. What does this condition represents? which type of symptoms can be noticed in the person?
Answer:

  1. Trisomy-21 or Down’s syndrome.
  2. Symptoms – Mental retardation, malformed ears, protruded tongue, mouth is constantly open etc.

Samacheer Kalvi 12th Bio Zoology Solutions Chapter 4 Principles of Inheritance and Variation

Question 2.
A female whose blood group is AB- got conceived and later it is diagnoised that her – foetus possess B+. What measures would be taken to prevent the foetus from Haemolytic
disease of Newborn (HDN)
Answer:
If the mother is Rh negative and foetus is Rh positive, anti D antibodies should be administered to the mother at 28th and 34th week of gestation as a prophylactic measure. If the Rh negative mother delivers Rh positive child then anti D antibodies should be administered to the mother soon after delivery. This develops passive immunity and prevents the formation of anti D antibodies in the mothers blood by destroying the Rh foetal RBC before the mother’s immune system is sensitized. This has to be done whenever the woman attains pregnancy.

Question 3.
The following table shows the genotypes for ABO blood grouping and this phenotypes. Complete the table by filling the gaps.
Answer:
Samacheer Kalvi 12th Bio Zoology Solutions Chapter 4 Principles of Inheritance and Variation img 17

Question 4.
Give one example for each of the following group of drugs,

  1. Stimulants
  2. Analgesic
  3. Hallucinogens

Answer:

  1. Stimulants – Eg : Nicotine
  2. Analgesic – Eg : Opium
  3. Hallucinogens – Phencyclidine

Samacheer Kalvi 12th Bio Zoology Solutions Chapter 2 Human Reproduction

Students can Download Bio Zoology Chapter 2 Human Reproduction Questions and Answers, Notes Pdf, Samacheer Kalvi 12th Bio Zoology Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Bio Zoology Solutions Chapter 2 Human Reproduction

Samacheer Kalvi 12th Bio Zoology Human Reproduction Text Book Back Questions and Answers

Question 1.
The mature sperms are stored in the ________
(a) Seminiferous tubules
(b) Vas deferens
(c) Epididymis
(d) Seminal vesicle
Answer:
(c) Epididymis

Question 2.
The male sex hormone testosterone is secreted from ________
(a) Sertoli cells
(b) Leydig cell
(c) Epididymis
(d) Prostate gland
Answer:
(b) Leydig cell

Samacheer Kalvi 12th Bio Zoology Solutions Chapter 2 Human Reproduction

Question 3.
The glandular accessory organ which produces the largest proportion of semen is ________
(a) Seminal vesicle
(b) Bulbourethral gland
(c) Prostate gland
(d) Mucous gland
Answer:
(a) Seminal vesicle

Question 4.
The male homologue of the female clitoris is ________
(a) Scrotum
(b) Penis
(c) Urethra
(d) Testis
Answer:
(b) Penis

Question 5.
The site of embryo implantation is the ________
(a) Uterus
(b) Peritoneal cavity
(c) Vagina
(d) Fallopian tube
Answer:
(a) Uterus

Question 6.
The foetal membrane that forms the basis of the umbilical cord is ________
(a) Allantois
(b) Amnion
(c) Chorion
(d) Yolk sac
Answer:
(a) Allantois

Question 7.
The most important hormone in initiating and maintaining lactation after birth is ________
(a) Oestrogen
(b) FSH
(c) Prolactin
(d) Oxytocin
Answer:
(c) Prolactin

Question 8.
Mammalian egg is ________
(a) Mesolecithal and non-cleidoic
(b) Microlecithal and non-cleidoic
(c) Alecithal and non-cleidoic
(d) Alecithal and cleidoic
Answer:
(c) Alecithal and non-cleidoic

Question 9.
The process which the sperm undergoes before penetrating the ovum is ________
(a) Spermiation
(b) Cortical reaction
(c) Spermiogenesis
(d) Capacitation
Answer:
(d) Capacitation

Question 10.
The milk secreted by the mammary glands soon after child birth is called ________
(a) Mucous
(b) Colostrum
(c) Lactose
(d) Lactose
Answer:
(b) Colostrum

Question 11.
Colostrum is rich in ________
(a) IgE
(b) IgA
(c) IgD
(d) Ig M
Answer:
(b) IgA

Samacheer Kalvi 12th Bio Zoology Solutions Chapter 2 Human Reproduction

Question 12.
The Androgen Binding Protein (ABP) is produced by ________
(a) Leydig cells
(b) Hypothalamus
(c) Sertoli cells
(d) Pituitary gland
Answer:
(c) Sertoli cells

Question 13.
Which one of the following menstrual irregularities is correctly matched?
(a) Menorrhagia – excessive menstruation
(b) Amenorrhoea – absence of menstruation
(c) Dysmenorrhoea – irregularity of menstruation
(d) Oligomenorrhoea – painful menstruation
Answer:
(b) Amenorrhoea – absence of menstruation

Question 14.
Find the wrongly matched pair:
(a) Bleeding phase – fall in oestrogen and progesterone
(b) Follicular phase – rise in oestrogen
(c) Luteal phase – rise in FSH level
(d) Ovulatory phase – LH surge
Answer:
(c) Luteal phase – rise in FSH level

Question 15.
A – In human male, testes are extra abdominal and lie in scrotal sacs.
R – Scrotum acts as thermoregulator and keeps temperature lower by 2°C for normal sperm production.
Assertion (A) and Reason (R)
(a) A and R are true, R is the correct explanation of A
(b) A and R are true, R is not the correct explanation of A
(c) A is true, R is false
(d) Both A and R are false
Answer:
(a) A and R are true, R is the correct explanation of A

Question 16.
A – Ovulation is the release of ovum from the Graafian follicle.
R – It occurs during the follicular phase of the menstrual cycle.
Assertion (A) and Reason (R)
(a) A and R are true, R is the correct explanation of A
(b) A and R are true, R is not the correct explanation of A
(c) A is true, R is false
(d) Both A and R are false
Answer:
(c) A is true, R is false

Question 17.
A – Head of the sperm consists of acrosome and mitochondria.
R – Acrosome contains spiral rows of mitochondria.
(d) Both A and R are false
Assertion (A) and Reason (R)
(a) A and R are true, R is the correct explanation of A
(b) A and R are true, R is not the correct explanation of A
(c) A is true, R is false
(d) Both A and R are false
Answer:
(d) Both A and R are false

Question 18.
Mention the differences between spermiogenesis and spermatogenesis
Answer:
Spermiogenesis: Transformation of spermatids into mature sperm.

Spermatogenesis: Spermatogenesis is the sequence of events in the seminiferous tubules of testes that produces male gametes, the sperms.

Question 19.
At what stage of development are the gametes formed in newborn male and females?
Answer:
In males, at puberty, the spermatogonia (sperm mother cells) begin to undergo meiotic division and produces sperms throughout life, whereas in females during the stage of foetal development, the germinal epithelial cells undergo mitosis and produce oogonia (egg mother cells) and they further enter prophase-I of meiosis-I forming primary oocytes and get arrested. No more oogonia is formed further. At puberty, out of million eggs (prime oocytes) produced at birth only 300-400 will ovulate till menopause.

Question 20.
Expand the acronyms

  1. FSH
  2. LH
  3. HCG
  4. HPL

Answer:

  1. FSH – Follicle Stimulating Hormone
  2. LH – Luteinizing Hormone
  3. hCG – human Chorionic Gonadotropin
  4. hPL – human Placental Lactogen

Question 21.
How is polyspermy avoided in humans?
Answer:
Once fertilization is accomplished, cortical granules from the cytoplasm of the ovum form a barrier called the fertilization membrane around the ovum preventing further penetration of other sperms. Thus polyspermy is prevented.

Question 22.
What is colostrum? Write its significance.
Answer:
The mammary glands secrete a yellowish fluid called colostrum during the initial few days after parturition. It has less lactose than milk and almost no fat, but it contains more proteins, vitamin A and minerals. Colostrum is also rich in IgA antibodies. This helps to protect the infant’s digestive tract against bacterial infection.

Question 23.
Placenta is an endocrine tissue. Justify.
Answer:
During pregnancy, the placenta acts as a temporary endocrine gland and produces large quantities of human Chorionic Gonadotropin (hCG), human Chorionic Somatomammotropin (hCS) or human Placental Lactogen (hPL), oestrogens and progesterone which are essential For a normal pregnancy. A hormone called relaxin is also secreted during the later phase of pregnancy which helps in relaxation of the pelvic ligaments at the time of parturition.

Question 24.
Draw a labeled sketch of a spermatozoan.
Answer:
Samacheer Kalvi 12th Bio Zoology Solutions Chapter 2 Human Reproduction img 1
Question 25.
What is inhibin? State its functions.
Answer:
inhibin is a hormone secreted by Sertoli cells of the testes which is involved in the negative feedback control of sperm production.

Question 26.
Mention the importance of the position of the testes in humans.
Answer:
The testes are positioned in such a way hanging out from the body in the scrotal sac that provides optimal temperature 2°C to 3°C lower than internal body temperature for effective sperm production.

Question 27.
What is the composition of semen?
Answer:
Semen or seminal fluid is a milky white fluid which contains sperms and seminal plasma, which is secreted from the seminal vesicles, prostate gland and bulbourethral glands.

Samacheer Kalvi 12th Bio Zoology Solutions Chapter 2 Human Reproduction

Question 28.
Name the hormones produced from the placenta during pregnancy, human Chorionic Gonadotropin (hCG)
Answer:

  1. human Placental Lactogen (hPL)
  2. Relaxin.

Question 29.
Define gametogenesis.
Answer:
Gametogenesis is the process of formation of gametes i.e., sperms and ovaries from the primary sex organs in all sexually reproducing organisms. Meiosis plays the most significant role in the process of gametogenesis.

Question 30.
Describe the structure of the human ovum with a neat labelled diagram.
Answer:
Samacheer Kalvi 12th Bio Zoology Solutions Chapter 2 Human Reproduction img 2
Human ovum is non-cleidoic, alecithal and microscopic in nature. Its cytoplasm called ooplasm contains a large nucleus called the germinal vesicle. The ovum is surrounded by three coverings namely an inner thin transparent vitelline membrane, middle thick zona pellucida and outer thick coat of follicular cells called corona radiata. Between the vitelline membrane and zona pellucida is a narrow perivitelline space.

Question 31.
Give a schematic representation of spermatogenesis and oogenesis in humans
Answer:
Samacheer Kalvi 12th Bio Zoology Solutions Chapter 2 Human Reproduction img 3

Question 32.
Explain the various phases of the menstrual cycle.
Answer:
Menstrual cycle: The menstrual or ovarian cycle occurs approximately once in every 28/29 days during the reproductive life of the female from menarche (puberty) to menopause except during pregnancy. The cycle of events starting from one menstrual period till the next one is called the menstrual cycle during which cyclic changes occurs in the endometrium every month. Cyclic menstruation is an indicator of normal reproductive phase.

Menstrual cycle comprises of the following phases

  1. Menstrual phase
  2. Follicular or proliferative phase
  3. Ovulatory phase
  4. Luteal or secretory phase

1. Menstrual phase: The cycle starts with the menstrual phase when menstrual flow occurs and lasts for 3-5 days. Menstrual flow is due to the breakdown of endometrial lining of the uterus, and its blood vessels due to decline in the level of progesterone and oestrogen. Menstruation occurs only if the released ovum is not fertilized. Absence of menstruation may be an indicator of pregnancy. However it could also be due to stress, hormonal disorder and anaemia.

2. Follicular or proliferative phase: The follicular phase extends from the 5th day of the cycle until the time of ovulation. During this phase, the primary follicle in the ovary grows to become a fully mature Graafian follicle and simultaneously, the endometrium regenerates through proliferation. These changes in the ovary and the uterus are induced by the secretion of gonadotropins like FSH and LH, which increase gradually during the follicular phase. It stimulates follicular development and secretion of oestrogen by the follicle cells.

3. Ovulatory phase: Both LH and FSH attain peak level in the middle of the cycle (about the 14th day). Maximum secretion of LH during the mid cycle called LH surge induces the rupture of the Graafian follicle and the release of the ovum (secondary oocyte) from the ovary wall into the peritoneal cavity. This process is called as ovulation.

4. Luteal or secretory phase: During luteal phase, the remaining part of the Graafian follicle is transformed into a transitory endocrine gland called corpus luteum. The corpus luteum secretes large amount of progesterone which is essential for the maintenance of the endometrium. If fertilization takes place, it paves way for the implantation of the fertilized ovum. The uterine wall secretes nutritious fluid in the uterus for the foetus. So, this phase is also called as a secretory phase. During pregnancy, all events of the menstrual cycle stop and there is no menstruation.
In the absence of fertilization, the corpus luteum degenerates completely and leaves a scar tissue called corpus Albicans. It also initiates the disintegration of the endometrium leading to menstruation, marking the next cycle.

Question 33.
Explain the role of oxytocin and relaxin in parturition and lactation.
Answer:
Relaxin is the hormone secreted by the placenta that causes the contraction of pelvic joints and promotes parturition (childbirth).
Oxytocin causes the Let-down reflex – the actual ejection of milk from the alveoli of mammary glands. Oxytocin also stimulates the uterus to regain its pre-pregnancy size after childbirth.

Question 34.
Identify the given image and label its parts marked as a, b, c, and d.
Samacheer Kalvi 12th Bio Zoology Solutions Chapter 2 Human Reproduction img 4
Answer:
The given image is the diagram of the human egg cell or ovum
a – vitelline membrane
b – Nucleus
c – Zona pellucida
d – Corona Radiata

Question 35.
The following is the illustration of the sequence of ovarian events (a-i) in a human female.
Samacheer Kalvi 12th Bio Zoology Solutions Chapter 2 Human Reproduction img 5
(a) Identify the figure that illustrates ovulation and mention the stage of oogenesis it represents.
(b) Name the ovarian hormone and the pituitary hormone that have caused the above- mentioned events.
(c) Explain the changes that occurs in the uterus simultaneously in anticipation.
(d) Write the difference between C and H.
Answer:
(a) name of types:
A- Primordial follicle
B- Primary follicle
C- Secondary follicle
D-Tertiary follicle
E- Mature graafian follicle
F- Ovulation (release of egg)
G- Empty Graafian follicle
H- Corpus luteum
I – Corpus Albicans.

(b) Pituitary hormones: Follicle Stimulating Hormones (FSH) and Lutenizing Hormone (LH). Ovarian hormones: Estrogen and Progesterone.

(c) At the start of menstrual cycle, the endometrium of uterus starts regenerating through proliferation of cells induced by FSH and CH. After ovulation, the progesterone secreted by corpus luteum prepares the endometrium (uterine wall) to receive the egg if it is fertilized.

(d) C- Secondary follicle H – Corpus luteum During development of ovum, the primary follicle gets surrounded by many layers of granular Cells and forms a new layer called secondary follicle. Corpus luteum is the empty graafian follicle that remains after ovulation. It acts as a transitory endocrine gland secreting progesterone to maintain pregnancy.

Samacheer Kalvi 12th Bio Zoology Human Reproduction Additional Questions and Answers

1 – Mark Questions

Question 1.
The developing spermatozoa are nourished by _________
(a) Leydig cells
(b) Sertoli cells
(c) Follicular cells
(d) Epididymis
Answer:
(b) Sertoli cells

Samacheer Kalvi 12th Bio Zoology Solutions Chapter 2 Human Reproduction

Question 2.
Identify the correct sequence of reproductive events in human beings.
(a) Insemination, Implantation, Fertilization, Parturition and Placentation.
(b) Implantation, Fertilization, Insemination, Placentation and Parturition.
(c) Implantation, Insemination, Fertilization, Parturition and Placentation.
(d) Insemination, Fertilization, Implantation, Placentation and Parturition.
Answer:
(d) Insemination, Fertilization, Implantation, Placentation and Parturition.

Question 3.
Expulsion of baby from the mother’s womb is referred as _________
Answer:
(a) Ejection
(b) Relaxation
(c) Parturition
(d) Implantation
Answer:
(c) Parturition

Question 4.
Match the Column I with Column II
Samacheer Kalvi 12th Bio Zoology Solutions Chapter 2 Human Reproduction img 6
Answer:
(a) a – ii, b – i, c – iv, d – iii

Question 5.
Which of the following statement is not correct?
(i) Interstitial cells are seen surrounding the seminiferous tubule.
(ii) Nurse cells secrete inhibin.
(iii)Males have single prostate gland which encircles the urethra.
(iv) Insemination, Fertilization, Implantation, Placentation and Parturition.
(a) i and ii
(b) iii only
(c) iii and iv
(d) iv only
(d) iv only
Answer:
(d) iv only

Question 6.
Assertion (A): In scrotum, the temperature is maintained 2 – 3°C lower than body temperature.
Reason (R): Reduced temperature results in efficient sperm production.
(a) R explains A.
(b) A is right R is wrong.
(c) A and R are right. R does not explain A.
(d) Both A and R are wrong.
Answer:
(a) R explains A.

Question 7.
Assertion (A): The acrosome of the sperm cell contains sperm lysin.
Reason (R): Sperm lysin destroys the deformed sperm cells.
(a) R explains A.
(b) A is right, R is wrong.
(c) A and R are right. R does not explains A.
(d) Both A and R are wrong.
Answer:
(A) A is right, R is wrong.

Question 8.
Assertion (A): Human ovum is non – cieidoic
Reason (R): Human does not contain yolk.
(a) R explains A.
(b) A is right, R is wrong.
(c) A and R are right. R does not explains A.
(d) Both A and R are wrong.
Answer:
(c) A and R are right. R does not explains A.

Question 9.
Assertion (A): Menopause refers to the absence of menstruation during pregnancy.
Reason (R): Ovulation occurs during menstrual phase.
(a) R explains A.
(b) A is right, R is wrong.
(c) A and R are right. R does not explains A.
(d) Both A and R are wrong.
Answer:
(d) Both A and R are wrong.

Question 10.
Assertion (A): Cervix is common site of ectopic pregnancies
Reason (R): Implantation of fertilized ovum outside uterus.
(a) A is wrong, R is right.
(b) A is right, R is wrong.
(c) A and R are right. R does not explains A.
(d) Both A and R are wrong.
Answer:
(a) A is wrong, R is right.

Question 11.
Which of the following contributes to the seminal plasma?
(i) Cowper’s gland
(ii) Seminal vesicles
(iii) Prostate gland
(iv) Bulbourethral gland
(a) ii, iii and ii
(b) i, ii, and iii
(c) i, iii and iv
(d) all the above
Answer:
(d) all the above

Question 12.
Organ of copulation in human female is _________
(a) Cevix
(b) Fundus
(c) Vagina
(d) Uterus
Answer:
(c) Vagina

Question 13.
Identify the gland which is homologous to the Cowper’s glands of male.
(a) Bartholin’s gland
(b) Bulbourethral gland
(c) Prostate gland
(d) Skene’s gland
Answer:
(a) Bartholin’s gland

Samacheer Kalvi 12th Bio Zoology Solutions Chapter 2 Human Reproduction

Question 14.
Find out the proper sequence representing the parts of female reproductive system.
(a) Vagina → Ovary → Uterus → Cervix → Infundibulum → Oviduct
(b) Vagina → Ovary → Oviduct → Infundibulum → Cervix → Uterus
(c) Ovary → Infundibulum → Oviduct → Uterus → Cervix → Vagina
(d) Oviduct → Ovary → Uterus → Infundibulum Vagina → Cervix
Answer:
(c) Ovary- Infundibulum → Oviduct → Uterus → Cervix → Vagina

Question 15.
Spermatid → spermatozoa. What does ‘A’ stands for?
(a) Spermatogenesis
(b) Spermiation
(c) Spermiogenesis
(d) Gametogenesis
Answer:
(c) Spermiogenesis

Question 16.
An adult male produces an average of…………….. sperms per day
(a) 200 million
(b) 300 million
(c) 300 billion
(d) 120 million
Answer:
(a) 200 million

Question 17.
Statement (1): During spermiation, the sperms are released into the cavity of I seminiferous tubule.
Statement (2): During spermiogenesis, the spermatids get mature into sperms.
(a) Statement 1 is correct; statement 2 is incorrect.
(b) Statement 1 is incorrect; statement 2 is correct.
(c) Both statements 1 and 2 are correct.
(d) Both statements 1 and 2 are incorrect.
Answer:
(c) Both statements 1 and 2 are correct.

Question 18.
Statement (1): Siamese twins are conjoined twins who are joined during birth.
Statement (2): Dizygotic twins will be of same-sex.
(a) Statement 1 is correct; statement 2 is incorrect.
(b) Statement 1 is incorrect; statement 2 is correct.
(c) Both statements 1 and 2 are correct.
(d) Both statements 1 and 2 are incorrect.
Answer:
(a) Statment 1 is correct; statement 2 is incorrect.

Question 19.
Statement (1): The endometrium acts as a transitory endocrine gland secreting
progesterone
Statement (2): Progesterone maintains pregnancy
(а) Statement 1 is correct; statement 2 is incorrect.
(b) Statement 1 is incorrect; statement 2 is correct.
(c) Both statements 1 and 2 are correct.
(d) Both statements 1 and 2 are incorrect.
Answer:
(b) Statement 1 is incorrect; statement 2 is correct

Question 20.
Statement (1): Human pregnancy lasts for 35 weeks.
Statement (2): During gestation, the embryo’s heat develops during the 12th week.
(a) Statement 1 is correct; statement 2 is incorrect.
(b) Statement 1 is incorrect; statement 2 is correct.
(c) Both statements 1 and 2 are correct.
(d) Both statements 1 and 2 are incorrect.
Answer:
(d) Both statements 1 and 2 are incorrect.

Question 21.
Statement (1): Menstrual cycle occurs once in every 29 days.
Statement (2): The average age of menopause is 45-50 years.
(a) Statement 1 is correct; statement 2 is incorrect.
(b) Statement 1 is incorrect; statement 2 is correct.
(c) Both statements 1 and 2 are correct.
(d) Both statements 1 and 2 are incorrect.
Answer:
(c) Both statements 1 and 2 are correct.

Question 22.
The first ejaculation of the semen in male is called as ___________
Answer:
Spermarche

Question 23.
Identify the mismatched pair.
(a) Castration – Orchidectomy
(b) Spermiogenesis – Release of sperms into the cavity of seminiferous tubule
(c) Ovulation – Release of egg from ovary
(d) Capacitation – Process enabling the sperm to penetrate the egg.
Answer:
(b) Spermiogenesis – Release of sperms into the cavity of seminiferous tuble.

Question 24.
Given below are the extraembryonic membranes of which identify the outermost membrane
(a) Amnion
(b) Chorion
(c) Yolk sac
(d) Allantois
Answer:
(b) Chorion

Question 25.
Identify the given figure and select the correct option representing X, Y and Z.
Samacheer Kalvi 12th Bio Zoology Solutions Chapter 2 Human Reproduction img 7

Samacheer Kalvi 12th Bio Zoology Solutions Chapter 2 Human Reproduction img 8
Answer:
(b) Acrosome, Nucleus and Mitochondria

Question 26.
The entire process of spermatogenesis takes about ________ days
(a) 60 days
(b) 44 days
(c) 64 days
(d) 50 days
Answer:
(c) 64 days

Samacheer Kalvi 12th Bio Zoology Solutions Chapter 2 Human Reproduction

Question 27.
Observe the diagram and select the correct option denoting the proper sequence of parts.
Samacheer Kalvi 12th Bio Zoology Solutions Chapter 2 Human Reproduction img 9

Samacheer Kalvi 12th Bio Zoology Solutions Chapter 2 Human Reproduction img 10
Answer:
(b) Fimbriae, Infundibulum, Uterus and Isthmus

Question 28.
Pick out the incorrect statements.
(a) The upper rounded portion of uterus is fundus.
(b) Uterus open into vagina through narrow cervix.
(c) Cervix is the organ of copulation in female.
(d) Vagina extends from the cervix and opens to exterior.
Answer:
(c) Cervix is the organ of copulation in female.

Question 29.
What is the role of fimbriae?
(a) Secretion of oestrogen and prolactin.
(b) Helps in the collection of the ovum after ovulation.
(c) Attaches the ovary to the abdominal cavity.
(d) Connects oviduct with ovary.
Answer:
(b) Helps in the collection of the ovum after ovulation.

Question 30.
Name the enzyme found in the acrosomal tip of sperm cell.
Answer:
Hyaluronidase

Question 31.
Which is not a correct statement regarding Oogenesis?
Answer:
(i) During foetal development, cells in germinal epithelium of foetal ovary undergo, mitosis and produce oogonia.
(ii) Oogonial cell divide and enter into prophase I of meiosis I and from primary oocytes.
(iii)Primary oocytes later develop into primary follicles.
(iv) No oogonia is formed or added after the foetal birth.
(a) Only i
(b) ii and iii
(c) iv only
(d) None of the above
Answer:
(d) None of the above

Question 32.
In embryo development of human beings, how long does it takes for a zygote to convert into morula?
(a) 24hrs
(b) 36hrs
(c)48hrs
(d) 72hrs
Answer:
(d) 72 hrs

Question 33.
Identify the hormone which is produced only during the time of pregnancy
(a) Relaxin
(b) Oxytocin
(c) Progesterone
(d) Cortisol
Answer:
(a) Relaxin

Question 34.
The type of antibodies present in colostrum.
(a) IgE
(b) IgM
(c) IgA
(d) IgB
Answer:
(c) IgA

2 – Mark Questions

Question 1.
Enumerate the functions of the reproductive system.
Answer:
The reproductive system has four main functions namely,

  1. to produce the gametes namely sperms and ova
  2. to transport and sustain these gametes
  3. to nurture the developing offspring
  4. to produce hormones

Question 2.
Define the terms

  1. Insemination
  2. Fertilization.

Answer:

  1. Insemination: Transfer of sperms by the male into the female genital tract.
  2. Fertilization: Fusion of male and female gametes to form zygote, called fertilization.

Samacheer Kalvi 12th Bio Zoology Solutions Chapter 2 Human Reproduction

Question 3.
What are seminiferous tubules? Mention its role.
Answer:
Seminiferous tubules are highly coiled tubules seen in the lobules of testis. They occupy 80% – of testicular substance. They are the site for sperm production.

Question 4.
Name the cells noticed in the epithelial layer of seminiferous tubule.
Answer:

  1. Sertoli cells or Nurse cells
  2. Spermatogonic cells or male germ cells.

Question 5.
Mention the role epididymis.
Answer:

  1. Epididymis is a temporary store house for sperms.
  2. Sperms undergo physiological maturation, increased motility and fertilizing capacity inside epididymis.

Question 6.
Seminal plasma is acidic or alkaline. Write its composition.
Answer:

  1. Seminal plasma is alkaline in nature.
  2. It contains fructose, ascorbic acid, prostaglandins and a coagulating enzyme called vesiculase.

Question 7.
Define Semen.
Answer:
Semen or seminal fluid is a milky white fluid which contains sperms and seminal plasma, which is secreted from the seminal vesicles, prostate gland, and bulbourethral glands.

Question 8.
Why do males have Penis?
Answer:

  1. Penis is the male external genitalia.
  2. It functions as both excretory and copulatory organ.
  3. It is made of special tissue that erects the penis to facilitate insemination.

Question 9.
Point out the female accessory organs.
Answer:
Fallopian tubes, Uterus and Vagina.

Question 10.
Define the nature of uterus.
Answer:
The uterus or womb is a hollow, thick-walled, muscular, highly vascular and inverted pear shaped structure lying in the pelvic cavity between the urinary bladder and rectum.

Question 11.
What are the components that make up external genitalia female?
Answer:
Labia Majora, Labia Minora, Hymen and Clitoris.

Question 12.
Name the accessory reproductive glands in female which are homologous to (a) Cowper’s gland and (b) Prostate gland.
Answer:
In female, Bartholin’s gland is homologous to Cowper’s gland and Skene’s gland is homologous to prostate gland.

Question 13.
Define Gametogenesis.
Answer:
Gametogenesis is the process of formation of gametes i.e., sperms and ovary from the primary ,sex organs in all sexually reproducing organisms. Meiosis plays the most significant role in the process of gametogenesis.

Question 14.
Define the terms

  1. Spermiogenesis
  2. Spermination

Answer:

  1. Spermiogenesis: Transformation of spermatids into mature sperm.
  2. Spermiation: Release of mature sperm into the lumen of the seminiferous tubule.

Question 15.
What do you mean by ‘Sperm lysin’? Mention its function.
Answer:

  1. Sperm lysin is a proteolytic enzyme secreted in the acrosome of sperm.
  2. It helps to penetrate the ovum during fertilization.
  3. It is also called hyaluronidase.

Samacheer Kalvi 12th Bio Zoology Solutions Chapter 2 Human Reproduction

Question 16.
Name the four phases of the menstrual cycle.
Answer:

  1. Menstrual phase
  2. Follicular or proliferative phase
  3. Ovulatory phase
  4. Luteal or secretory phase

Question 17.
What is corpus Albicans?
Answer:
In the absence of fertilization, the corpus luteum degenerates completely and leaves a scar tissue called corpus Albicans. It also initiates the disintegration of the endometrium leading to menstruation, marking the next cycle.

Question 18.
Define menopause.
Answer:
Menopause is the phase in a women’s life when ovulation and menstruation stops. The average age of menopause is 45-50 years. It indicates the permanent cessation of the primary functions of the ovaries.

Question 19.
When does capacitation occur? Define it.
Answer:
The sperms deposited in the female reproductive tract undergo capacitation. It is a biochemical event that makes the sperm to penetrate and fertilize the egg.

Question 20.
Write a brief note on ectopic pregnancy?
Answer:
If the fertilized ovum is implanted outside the uterus it results in ectopic pregnancy. About 95 percent of ectopic pregnancies occur in the fallopian tube. The growth of the embryo may cause internal bleeding, infection, and in some cases even death due to rupture of the fallopian tube.

Question 21.
Point out the extraembryonic membranes of the human embryo.
Answer:

  1. amnion
  2. Chorion
  3. allantois
  4. Yolk nac

Question 22.
What is the placenta?
Answer:
Placenta is a temporary endocrine organ formed during pregnancy and it connects the foetus to the uterine wall through the umbilical cord. It is the organ by which the nutritive, respiratory and excretory functions are fulfilled.

Question 23.
Name the organs developed from embryonic ectoderm.
Answer:
Brain and spinal cord (CNS), peripheral nervous system (PNS), epidermis and its derivatives, and mammary glands.

Question 24.
Mention the hormones secreted by the placenta during pregnancy.
Answer:

  1. human Chorionic Gonadotropin (hCG)
  2. human Chorionic Somatomammotropin (hCS)
  3. human Placental Lactogen (hPL)
  4. Oestrogen and progesterone and relaxin.

Question 25.
Name the hormones that are secreted in human only during pregnancy.
Answer:

  1. human Chorionic Gonadotropin (hCG)
  2. human Chorionic Somatomammotropin (hCS)
  3. relaxin

Question 26.
State the role of relaxin.
Answer:
Relaxin is an hormone secreted by the placenta during the later phase of pregnancy. It helps in the relaxation of the pelvis during childbirth.

Question 27.
Define parturition and labour.
Answer:
Parturition is the completion of pregnancy and giving birth to the baby. The series of events that expels the infant from the uterus is collectively called “labour”.

Question 28.
What do you mean by ‘false labour’?
Answer:
Throughout pregnancy the uterus undergoes periodic episodes of weak and strong contractions, fhese contractions called Braxter-Hick’s contractions lead to false labour.

Question 29.
Explain the term C-section.
Answer:
When normal vaginal delivery is not possible due to factors like position of the baby and nature of the placenta, the baby is delivered through a surgical incision in the woman’s abdomen and uterus. It is also termed as abdominal delivery or Caesarean Section or ‘C’ Section.

3 – Mark Questions

Question 1.
Compare gametogenesis with organogenesis.
Answer:
Gametogenesis:

  1. Formation of gametes i.e., Sperm and Ova.
  2. It takes place in gonads.

Organogenesis:

  1. Formation of tissues, organs and organ system.
  2. It takes place in the embryonic germ layers.

Samacheer Kalvi 12th Bio Zoology Solutions Chapter 2 Human Reproduction

Question 2.
What are primary reproductive organs? What role does they play in organisms?
Answer:
The primary reproductive organs namely the ovary and testis are responsible for producing the ova and sperms respectively. Hormones secreted by the pituitary gland and the gonads help in the development of the secondary sexual characteristics, maturation of the reproductive system and regulation of normal functioning of the reproductive system.

Question 3.
Scrotum acts as a thermoregulator – Justify.
Answer:
The scrotum is a sac of skin that hangs outside the abdominal cavity. Since viable sperms cannot be produced at normal body temperature, the scrotum is placed outside the abdominal cavity to provide a temperature 2-3°C lower than the normal internal body temperature. Thus, the scrotum acts as a thermoregulator for spermatogenesis.

Question 4.
Write any three statements on Sertoli cells.
Answer:

  1. Sertoli cells are elongated and pyramidal cells.
  2. They provide nourishment to sperm till maturation.
  3. They secrete a hormone called inhibin which is involved in negative feedback control of sperm production.

Question 5.
Give a brief account on leydig cells.
Answer:

  1. Interstitial cells of Leydig are seen embedded in the soft connective tissue surrounding the seminiferous tubules.
  2. These cells are endocrine natured and produce testosterone (androgen).
  3. These cells are characteristic to mammalian testes.

Question 6.
Name the accessory glands of male reproductive system.
Answer:

  1. A pair of seminal vesicles.
  2. A pair of bulbourethral gland (Cowper’s gland).
  3. A single prostate gland.

Question 7.
State the location and secretion of prostate gland.
Answer:
The prostate encircles the urethra and is just below the urinary bladder and secretes a slightly acidic fluid that contains citrate, several enzymes and prostate specific antigens.

Question 8.
Write a note on hymen.
Answer:
The external opening of the vagina is partially closed by a thin ring of tissue called the hymen. The hymen is often tom during the first coitus (physical union). However in some women it remains intact. It can be stretched or tom due to a sudden fall or jolt and also during strenuous physical activities such as cycling and horseback riding, etc., and therefore cannot be considered as an indicator of a woman’s virginity.

Question 9.
“Role of hormones in spermatogenesis” – comment on the statement.
Answer:
Spermatogenesis starts at the age of puberty and is initiated due to the increase in the release of Gonadotropin Releasing Hormone (GnRH) by the hypothalamus. GnRH acts on the anterior pituitary gland and stimulates the secretion of two gonadotropins namely Follicle Stimulating Hormone (FSH) and Lutenizing Hormone (LH). FSH stimulates testicular growth and enhances the production of Androgen Binding Protein (ABP) by the Sertoli cells and helps in the process of spermiogenesis. LH acts on the Leydig cells and stimulates the synthesis of testosterone which in turn stimulates the process of spermatogenesis.

Question 10.
Define menstrual cycle.
Answer:
The menstrual or ovarian cycle occurs approximately once in every 28/29 days during the reproductive life of the female from menarche (puberty) to menopause except during pregnancy. The cycle of events starting from one menstrual period till the next one is called the menstrual cycle during which cyclic changes occurs in the endometrium every month. Cyclic menstruation is an indicator of normal reproductive phase.

Question 11.
Luteal phase of menstrual cycle is also called as secretory phase. Why?
Answer:
After ovulation, the graafian follicle turns into corpus luteum (a transistory endocrine gland) which secrets progesterone. Progesterone maintain endometrium for implantation of fertilized ovum, the endometrium of uterus also secretes nutritious fluid for the foetus. Hence this phase is also referred as secretory phase.

Question 12.
Menstrual hygiene is essential for women. Why?
Answer:
Menstrual hygiene is vital for good health, well-being, dignity, empowerment and productivity of women. The impact of poor menstrual hygiene on girls is increased stress levels, fear and embarrassment during menstruation. This can keep girls inactive during such periods leading to absenteeism from school.

Question 13.
Name the absorbents or materials used to manage menstruation.
Answer:
Clean and safe absorbable clothing materials, sanitary napkins, pads, tampons and menstrual cups have been identified as materials used to manage menstruation.

Question 14.
Explain acrosomal reaction.
Answer:
The follicular cells of egg are held together by an adhesive cementing substance called hyaluronic acid. The acrosomal membrane disintegrates releasing the proteolytic enzyme, hyaluronidase during sperm entry through the corona radiata and zona pellucida. This is called
acrosomal reaction.

Question 15.
Differentiate between monozygotic and Dizygotic twins.
Answer:
Monozygotic twins:

  1. Monozygotic (Identical) twins are produced when a single fertilized egg splits into two during the first cleavage.
  2. They are of the same sex, look alike and share the same genes.

Dizygotic twins:

  1. Dizygotic (Fraternal) twins are produced when two separate eggs are fertilized by two separate sperms.
  2. The twins may be of the same sex or different sex and are non-identical.

Question 16.
What is morula?
Answer:
The first cleavage in zygote produces two identical cells called blastomeres. These produce 4 cells, then 8 and so on. After 72 hours of fertilization, a loose collection of cells forms a berry shaped cluster of 16 or more cells called the morula.

Question 17.
Explain gastrulation.
Answer:
The inner cell mass in the blastula is differentiated into epiblast and hypoblast immediately after implantation. The hypoblast is the embryonic endoderm and the epiblast is the ectoderm. The cells remaining in between the epiblast and the endoderm form the mesoderm. Thus the transformation of the blastocyst into a gastrula with the primary germ layers by the movement of the blastomeres is called gastrulation.

Question 18.
Name the three primary germ layers of embryo. Also mention any three organs or ortgan systems developing from each layer.
Answer:
Samacheer Kalvi 12th Bio Zoology Solutions Chapter 2 Human Reproduction img 11

Question 19.
Define lactation and explain its hormonal background.
Answer:
Lactation is the production of milk by mammary glands. The mammary glands show changes during every menstrual cycle, during pregnancy and lactation. Increased level of oestrogens, progesterone and human Placental Lactogen (hPL) towards the end of pregnancy stimulate the hypothalamus towards prolactin-releasing factors. The anterior pituitary responds by secreting prolactin which plays a major role in lactogenesis.

Question 20.
What is “let-down reflex”?
Answer:
Oxytocin causes the “Let-Down” reflex the actual ejection of milk from the alveoli of the mammary glands. During lactation, oxytocin also stimulates the recently emptied uterus to contract, helping it to return to pre-pregnancy size.

Question 21.
Explain Foetal-ejection reflex.
Answer:
As the pregnancy progresses, increase in the oestrogen concentration promotes uterine contractions. These uterine contractions facilitate moulding of the foetus and downward movement of the foetus. The descent of the foetus causes dilation of cervix of the uterus and vaginal canal resulting in a neurohumoral reflex called Foetal ejection reflex or Ferguson reflex. This initiates the secretion of oxytocin from the neurohypophysis which in turn brings about the powerful contraction of the uterine muscles and leads to the expulsion of the baby through the birth canal.

Samacheer Kalvi 12th Bio Zoology Solutions Chapter 2 Human Reproduction

Question 22.
Give an account on uterine wall layers.
Answer:
The wall of the uterus has three layers of tissues. The outermost thin membranous serous layer called the perimetrium, the middle thick muscular layer called myometrium and the inner glandular layer called endometrium. The endometrium undergoes cyclic changes during the menstrual cycle while myometrium exhibits strong contractions during parturition

5 – Mark Questions

Question 1.
Describe the structure of human ovary.
Answer:

  1. Ovaries are the primary female sex organ producing eggs (ovum).
  2. They are located one on each side of lower abdomen (pelvis).
  3. The ovary is attached in pelvic wall uterus by an ovarian ligament called mesovarium.
  4. It is an elliptical structure of 2-4 cm long
  5. Each ovary is covered by thin cuboidal germinal epithelium encloses ovarian stroma.
  6. Below germinal epithelium is a dense connective tissue called tunica albuginea.
  7. The stroma is differentiated into outer cortex and inner medulla.
  8. The cortex is dense and granular due to follicular cells at varying development stages.
  9. The medulla is a loose connective tissue with blood vessels, lymph vessels and nerve

Question 2.
Explain the structure and function of mammary glands.
Answer:
The mammary glands are modified sweat glands present in both sexes. It is rudimentary in the males and functional in the females. A pair of mammary glands is located in the thoracic region. It contains glandular tissue and variable quantities of fat with a median nipple surrounded by a pigmented area called the areola. Several sebaceous glands called the areolar glands are found on the surface and they reduce cracking of the skin of the nipple. Internally each mammary gland consists of 2-25 lobes, separated by fat and connective tissues.

Each lobe is made up of lobules which contain acini or alveoli lined by epithelial cells. Cells of the alveoli secrete milk. The alveoli open into mammary tubules. The tubules of each lobe join to form a mammary duct. Several mammary ducts join to form a wider mammary ampulla which is connected to the lactiferous duct in the nipple. Under the nipple, each lactiferous duct expands to form the lactiferous sinus which serves as a reservoir of milk. Each lactiferous duct opens separately by a minute pore on the surface of the nipple.

Normal development of the breast begins at puberty and progresses with changes during each menstrual cycle. In non-pregnant women, the glandular structure is largely underdeveloped and the breast size is largely due to the amount of fat deposits. The size of the breast does not have an influence on the efficiency of lactation.

Question 3.
Describe the spermatogenesis with a diagram.
Answer:
Spermatogenesis is the sequence of events in the seminiferous tubules of the testes that produce the male gametes, the sperms.
During development, the primordial germ cells migrate into the testes and become immature germ cells called sperm mother cells or spermatogonia in the inner surfaces of the seminiferous tubules. The spermatogonia begin to undergo mitotic division at puberty and continues throughout life.

In the first stage of spermatogenesis, the spermatogonia migrate among Sertoli cells towards the central lumen of the seminiferous tubule and become modified and enlarged to form primary spermatocytes which are diploid with 23 pairs i.e., 46 chromosomes. Some of the primary spermatocytes undergo the first meiotic division to form two secondary spermatocytes which are haploid with 23 chromosomes each.
Samacheer Kalvi 12th Bio Zoology Solutions Chapter 2 Human Reproduction img 12
The secondary spermatocytes undergo second meiotic division to produce four haploid spermatids. The spermatids are transformed into mature spermatozoa (sperms) by the process called spermiogenesis. Sperms are finally released into the cavity of seminiferous tubules by a process called spermiation The whole process of spermatogenesis takes about 64 days. At any given time, different regions of the seminiferous tubules contain spermatocytes in different stages of development.

The sperm production remains nearly constant at a rate of about 200 million sperms per day. Spermatogenesis starts at the age of puberty and is initiated due to the increase in the – release of Gonadotropin-Releasing Hormone (GnRH) by the hypothalamus. GnRH acts on the anterior pituitary gland and stimulates the secretion of two gonadotropins namely Follicle Stimulating Hormone (FSH) and Lutenizing Hormone (LH). FSH stimulates testicular growth and enhances the production of Androgen Binding Protein (ABP) by the Sertoli cells and helps in the process of spermiogenesis. LH acts on the Leydig cells and stimulates the synthesis of testosterone which in turn stimulates the process of spermatogenesis.

Question 4.
Describe the structure of the human spermatozoan.
Answer:
The human sperm is a microscopic, flagellated and motile gamete. The whole body of the sperm is enveloped by P plasma membrane and is composed of a head, neck and tail. The head comprises of two parts namely the acrosome and nucleus. Acrosome is a small cap like pointed structure present at the tip of the nucleus and is formed mainly from the Golgi body of the spermatid. It contains hyaluronidase, a proteolytic enzyme, popularly known as sperm lysin which I helps to penetrate the ovum during fertilization. The nucleus is flat and oval. The neck is very short and is present between the head and the middle piece.
Samacheer Kalvi 12th Bio Zoology Solutions Chapter 2 Human Reproduction img 13
It contains the proximal centriole towards the nucleus which plays a role in the first division of the zygote and the distal centriole gives rise to I the axial filament of the sperm. The middle piece possesses mitochondria spirally twisted around the axial filament called mitochondrial spiral or nebenkem. It produces energy in the form of ATP molecules for the movement of sperms. The tail is the longest part of the sperm and is slender and tapering. It is formed of a central axial filament or axoneme and an outer protoplasmic sheath. The lashing movements of the tail, push the sperm forward.

Question 5.
Explain the process of oogenesis.
Answer:
Oogenesis is the process of development of the female gamete or ovum or egg in the ovaries. During foetal development, certain cells in the germinal epithelium of the foetal ovary divide by mitosis and produce millions of egg mother cells or oogonia. No more oogonia are formed or added after birth.

The oogonial cells start dividing and enter into Prophase-I of meiotic division-I to form the primary oocytes which are temporarily arrested at this stage. The primary oocytes then get surrounded by a single layer of granulosa cells to form the primordial or primary follicles. A large number of follicles degenerate during the period from birth to puberty, so at puberty, only 60,000 to 80,000 follicles are left in each ovary.
Samacheer Kalvi 12th Bio Zoology Solutions Chapter 2 Human Reproduction img 14
The primary follicle gets surrounded by many layers of granulosa cells and a new theca layer to form the secondary follicle. A fluid filled Oogenesis space, the antrum develops in the follicle and gets transformed into a tertiary follicle. The theca layer gets organized into an inner theca interna and an outer theca externa. At this time, the primary oocyte within the tertiary follicle grows in size and completes its first meiotic division and forms the secondary oocyte.

It is an unequal division resulting in the formation of a large haploid secondary oocyte and a first polar body. The first polar body disintegrates. During fertilization, the secondary oocyte undergoes second meiotic division and produces a large cell, the ovum and a second polar body. The second polar body also degenerates. The tertiary follicle eventually becomes a mature follicle or Graafian follicle. If fertilisation does not take place, second meiotic division is never completed and the egg disintegrates. At the end of gametogenesis in females, each primary oocyte gives rise to only one haploid ovum.

Question 6.
Write a note on embryonic membranes
Answer:
The extra embryonic membranes include amnion, yolk sac, allantois and chorion. They protect the embryo from dessication, mechanical shock, absorption of nutrients, gaseous exchange and placental formatio
Answer:
Samacheer Kalvi 12th Bio Zoology Solutions Chapter 2 Human Reproduction img 15

Question 7.
Name the three primary germ layers of embryo. Also mention any three organs or organ systems developing from each layer.
Answer:
Samacheer Kalvi 12th Bio Zoology Solutions Chapter 2 Human Reproduction img 16

Higher Order Thinking Skills (HOTs) Questions

Questio 1.
Give the alternate terminologies for

  1. Spermatogonia
  2. Embryonic ectoderm

Answer:

  1. Spermatogonia = Sperm mother cell
  2. Embryonic ectoderm = hypoblast

Samacheer Kalvi 12th Bio Zoology Solutions Chapter 2 Human Reproduction

Question 2.
Mention the production site and active site of the following hormones.
(a) GnRH
(b) Relaxin
Answer:
Samacheer Kalvi 12th Bio Zoology Solutions Chapter 2 Human Reproduction img 17

Question 3.
May 28th is celebrated as the annual Menstrual Hygiene Day (MHD). State its importance.
Answer:
MHgD aims to create awareness of the importance for women and girls to hygienically manage their menstruation. Menstrual hygiene is vital for good health, well-being, dignity, empowerment, and productivity of women.

Question 4.
Suggest a few hygiene tips for face healthy and happy menses.
Answer:

  1. Change the napkins periodically for 4 to 6 hours.
  2. Wash your genitals properly using clean lukewarm water.
  3. Avoid using soaps and vaginal hygiene products.
  4. Discard the sanitary napkins by incinerating them.
  5. Wear clean and comfortable underwear. Avoid tight clothing.
  6. Have a healthy diet rich in Iron content and vitamins.

Samacheer Kalvi 11th Accountancy Solutions Chapter 9 Rectification of Errors

Students can Download Accountancy Chapter 9 Rectification of Errors Questions and Answers, Notes Pdf, Samacheer Kalvi 11th Accountancy Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Accountancy Solutions Chapter 9 Rectification of Errors

Samacheer Kalvi 11th Accountancy Rectification of Errors Text Book Back Questions and Answers

I. Multiple Choice Questions
Choose the Correct Answer

Question 1.
Error of principle arises when ………………
(a) There is complete omission of a transaction
(b) There is partial omission of a transaction
(c) Distinction is not made between capital and revenue items
(d) There are wrong postings and wrong castings
Answer:
(c) Distinction is not made between capital and revenue items

Question 2.
Errors not affecting the agreement of trial balance are ………………
(a) Errors of principle
(b) Errors of overcasting
(c) Errors of undercasting
(d) Errors of partial omission
Answer:
(a) Errors of principle

Question 3.
The difference in trial balance is taken to ………………
(a) The capital account
(b) The trading account
(c) The suspense account
(d) The profit and loss account
Answer:
(c) The suspense account

Samacheer Kalvi 11th Accountancy Solutions Chapter 9 Rectification of Errors

Question 4.
A transaction not recorded at all is known as an error of ………………
(a) Principle
(b) Complete omission
(c) Partial omission
(d) Duplication
Answer:
(b) Complete omission

Question 5.
Wages paid for installation of machinery wrongly debited to wages account is an errs of ………………
(a) Partial omission
(b) Principle
(c) Complete omission
(d) Duplication
Answer:
(b) Principle

Question 6.
Which of the following errors will not affect the trial balance?
(a) Wrong balancing of an account
(b) Posting an amount in the wrong account but on the correct side
(c) Wrong totalling of an account
(d) Carried forward wrong amount in a ledger account
Answer:
(b) Posting an amount in the wrong account but on the correct side

Question 7.
Goods returned by Senguttuvan were taken into stock, but no entry was passed in the books. While rectifying this error, which of the following accounts should be debited?
(a) Senguttuvan account
(b) Sales returns account
(c) Returns outward account
(d) Purchases returns account
Answer:
(b) Sales returns account

Question 8.
A credit purchase of furniture from Athiyaman was debited to purchases account. Which of the following accounts should be debited while rectifying this error?
(a) Purchases account
(b) Athiyaman account
(c) Furniture account
(d) None of these
Answer:
(c) Furniture account

Question 9.
The total of purchases book was overcast. Which of the following accounts should be debited in the rectifying journal entry?
(a) Purchases account
(b) Suspense account
(c) Creditor account
(d) None of the above
Answer:
(b) Suspense account

Samacheer Kalvi 11th Accountancy Solutions Chapter 9 Rectification of Errors

Question 10.
Which of the following errors will be rectified using suspense account?
(a) Purchases returns book was undercast by ₹ 100
(b) Goods returned by Narendran was not recorded in the books
(c) Goods returned by Akila ₹ 900 was recorded in the sales returns book as ₹ 90
(d) A credit sale of goods to Ravivarman was not entered in the sales book
Answer:
(a) Purchases returns book was undercast by ₹ 100

II. Very Short Answer Questions

Question 1.
What is meant by rectification of errors?
Answer:
Correction of errors in the books of accounts is not done by erasing, rewriting or striking the figures which are incorrect. Correcting the errors that has occured is called Rectification.

Question 2.
What is meant by error of principle?
Answer:
It means the mistake committed in the application of fundamental accounting principles in recording a transaction in the books of accounts.

Question 3.
What is meant by error of partial omission?
Answer:
When the accountant has failed to record a part of the transaction, it is known as error of partial omission. This error usually occurs in posting. This error affects only one account.

Samacheer Kalvi 11th Accountancy Solutions Chapter 9 Rectification of Errors

Question 4.
What is meant by error of complete omission?
Answer:
It means the failure to record a transaction in the journal or subsidiary book or failure to post both the aspects in ledger. This error affects two or more accounts.

Question 5.
What are compensating errors?
Answer:
The errors that make up for each other or neutralize each other are known as compensating errors. These errors may occur in related or unrelated accounts. Thus, excess debit or credit in one account may be compensated by excess credit or debit in some other account. These are also known as offsetting errors.

III. Short Answer Questions

Question 1.
Write a note on error of principle by giving an example.
Answer:
It means the mistake committed in the application of fundamental accounting principles in recording a transaction in the books of accounts.
Example:
Entering the purchase of an asset in the purchases book. Machinery purchased on credit for ₹ 10,000 by M/s. Anbarasi garments manufacturing company entered in the purchases book.

Question 2.
Write a note on suspense account.
Answer:
When the trial balance does not tally, the amount of difference is placed to the debit (when the total of the credit column is higher than the debit column) or credit (when the total of the debit column is higher than the credit column) to a temporary account is known as ‘suspense account’.

Question 3.
What are the errors not disclosed by a trial balance?
Answer:
Certain errors will not affect the agreement of trial balance. Though such errors occur in the books of accounts, the total of debit and credit balance will be the same. The trial balance will tally. Errors of complete omission, error of principle, compensating error, wrong entry in the subsidiary books are not disclosed by the trial balance.

Samacheer Kalvi 11th Accountancy Solutions Chapter 9 Rectification of Errors

Question 4.
What are the errors disclosed by a trial balance?
Answer:
Certain errors affect the agreement of trial balance. If such errors have occurred in the books of accounts, the total of debit and credit balances will not be the same. The trial balance will not tally. Error of partial omission and error of commission affect the agreement of trial balance.

Question 5.
Write a note on one – sided errors and two – sided errors.
Answer:

  1. One – sided errors: When preparing the trial’balance, if the total of debit balances and credit balances are not the same, there is disagreement of trial balance.
  2. Two – sided errors: Rectification of two – sided errors at the time of preparing the trial balance is just similar to that of their rectification before preparation of trial balance.

IV. Exercises

Question 1.
State the account/s affected in each of the following errors: (2 marks)
(a) Goods purchased on credit from Saranya for ₹ 150 was posted to the debit side of her account.
(b) The total of purchases book ₹ 4,500 was posted twice.
Answer:
(a) Purchases from Saranya should have been posted to the credit of Saranya’s A/c, but it has been debited. Hence, credit Saranya’s A/c with double the amount i.e., Rs. 300.
(b) Credit the Purchases A/c.

Question 2.
State the account/s affected in each of the following errors: (2 marks)
(a) Goods sold to Vasu on credit for ₹ 1,000 was not recorded in the sales book.
(b) The total of sales book ₹ 2,500 was posted twice.
Answer:
(a)
Samacheer Kalvi 11th Accountancy Solutions Chapter 9 Rectification of Errors
(b) Debit the Sales A/c

Question 3.
Rectify the following errors discovered before the preparation of the trial balance: (2 marks)
(a) Sales book was undercast by ₹ 100
(b) Purchases returns book was overcast by ₹ 200
Answer:
(a) Sales account should be credited with ₹ 100
(b) Purchases returns account should be debited with ₹ 200

Question 4.
Rectify the following errors before the preparation of trial balance: (3 marks)
(a) Returns outward book was undercast by ₹ 2,000
(b) Returns inward book total was taken as ₹ 15,000 instead of ₹ 14,000
(c) The total of the purchases account was carried forward ₹ 100 less.
Answer:
(a) Returns outward Account should be credited with ₹ 2,000
(b) Sales returns account should be credited with ₹ 1,000
(c) Purchases account should be debited with ₹ 100

Question 5.
Rectify the following errors assuming that the trial balance is yet to be prepared: (5 marks)
(a) Sales book was undercast by ₹ 400
(b) Sales returns book was overcast by ₹ 500
(c) Purchases book was undercast by ₹ 600
(d) Purchases returns book was overcast by ₹ 700
(e) Bills receivable book was undercast by ₹ 800
Answer:
(a) Sales account should be credited with ₹ 400
(b) Sales returns account should be credited with ₹ 500
(c) Purchases account should be debited with ₹ 600
(d) Purchases returns account should be debited with ₹ 700
(e) Bills receivable account should be debited with ₹ 800

Samacheer Kalvi 11th Accountancy Solutions Chapter 9 Rectification of Errors

Question 6.
Rectify the following errors before preparing trial balance: (5 marks)
(a) The total of purchases book was carried forward ₹ 90 less.
(b) The total of purchases book was carried forward ₹ 180 more
(c) The total of sales book was carried forward ₹ 270 less.
(d) The total of sales returns book was carried forward ₹ 360 more.
(e) The total of purchases returns book was carried forward ₹ 450 less.
Answer:
(a) Purchases account should be debited ₹ 90
(b) Purchases account should be credited ₹ 180
(c) Sales account should be credited ₹ 270
(d) Sales returns account should be credited with ₹ 360
(e) Purchases returns account should be credited with ₹ 450

Question 7.
The following errors were located by the accountant before preparation of trial balance. Rectify them. (5 marks)
(a) The total of the discount column of ₹ 1,100 on the debit side of the cash book was not yet posted.
(b) The total of the discount column on the credit side of the cash book was undercast by ₹ 500.
(c) Purchased goods from Anbuchelvan on credit for ₹ 700 was posted to the debit side of his account.
(d) Sale of goods to Ponmukil on credit for ₹ 78 was posted to her account as ₹ 87.
(e) The total of sales returns book of ₹ 550 was posted twice.
Answer:
Samacheer Kalvi 11th Accountancy Solutions Chapter 9 Rectification of Errors

Question 8.
The accountant of a firm located the following errors before preparing the trial balance. Rectify them. (5 marks)
(a) Machinery purchased for ₹ 3,000 was debited to purchases account.
(b) Interest received ₹ 200 was credited to commission account.
(c) An amount of ₹ 1,000 paid to Tamilselvan as salary was debited to his personal account.
(d) Old furniture sold for ₹ 300 was credited to sales account.
(e) Goods worth ₹ 800 purchased from Soundarapandian on credit was not recorded in the books of accounts.
Answer:
Rectifying Journal
Samacheer Kalvi 11th Accountancy Solutions Chapter 9 Rectification of Errors

Question 9.
Rectify the following errors which were located before preparing the trial balance. (5 marks)
(a) Wages paid ₹ 2,000 for the erection of machinery was debited to wages account.
(b) Sales returns book was short totalled by ₹ 1,000.
(c) Goods purchased for ₹ 200 was posted as ₹ 2,000 to purchases account.
(d) The sales book was overcast by ₹ 1,500.
(e) Cash paid to Mukil ₹ 2,800 which was debited to Akhil’s account as ? 2,000.
Answer:
Rectifying Journal
Samacheer Kalvi 11th Accountancy Solutions Chapter 9 Rectification of Errors

Question 10.
Rectify the following errors which were located at the time of preparing the trial balance: (5 marks)
(a) The total of the discount column on the debit side of the cash book of ₹ 225 was posted twice.
(b) Goods of the value of ₹ 75 returned by Ponnarasan was not posted to his account.
(c) Cash received from Yazhini ₹ 1,000 was not posted.
(d) Interest received ₹ 300 has not been posted.
(e) Rent paid ₹ 100 was posted to rent account as ₹ 10.
Answer:
Samacheer Kalvi 11th Accountancy Solutions Chapter 9 Rectification of Errors

Question 11.
The following errors were located at the time of preparing trial balance. Rectify them. (5 marks)
(a) A personal expense of the proprietor ₹ 200 was debited to travelling expenses account.
(b) Goods of ₹ 400 purchased from Ramesh on credit was wrongly credited to Ganesh’s account.
(c) An amount of ₹ 500 paid as salaries to Mathi was debited to his personal account.
(d) An amount of ₹ 2,700 paid for extension of the building was debited to repairs account.
(e) A credit sale of goods of ₹ 700 on credit to Mekala was posted to Krishnan’s account.
Answer:
Samacheer Kalvi 11th Accountancy Solutions Chapter 9 Rectification of Errors

Question 12.
Rectify the following journal entries. (5 marks)
Samacheer Kalvi 11th Accountancy Solutions Chapter 9 Rectification of Errors
Answer:
Samacheer Kalvi 11th Accountancy Solutions Chapter 9 Rectification of Errors

Question 13.
Rectify the following errors discovered after the preparation of the trial balance: (5 marks)
(a) Rent paid was carried forward to the next page ₹ 500 short.
(b) Wages paid was carried forward ₹ 250 excess.
Answer:
(a) Rent account is to be debited with ₹ 500.
(b) Wages account is to be credited with ₹ 250.

Question 14.
Rectify the following errors after preparation of trial balance: (5 marks)
(a) Salary paid to Ram ₹ 1,000 was wrongly debited to his personal account.
(b) A credit sale of goods to Balu for ₹ 450 was debited to Balan.
Answer:
Samacheer Kalvi 11th Accountancy Solutions Chapter 9 Rectification of Errors

Question 15.
Pass necessary journal entries to rectify the following errors located after the preparation of trial balance: (5 marks)
(a) Sales book was undercast by ₹ 1,000.
(b) An amount of ₹ 500 paid for wages was wrongly posted to machinery Account.
Answer:
(a) Sales account should be credited ₹ 1,000.
(b)
Samacheer Kalvi 11th Accountancy Solutions Chapter 9 Rectification of Errors

Question 16.
Give journal entries to rectify the following errors discovered after the preparation of trial balance: (5 marks)
(a) Purchases book was overcast by ₹ 10,000.
(b) Repairs to furniture of ₹ 500 was debited to furniture account.
(c) A credit sale of goods to Akilnilavan for ₹ 456 was credited to his account as ₹ 654.
Answer:
(a) Purchases account should be credited ₹ 10,000.
Samacheer Kalvi 11th Accountancy Solutions Chapter 9 Rectification of Errors

Question 17.
Rectify the following errors located after the preparation of trial balance: (5 marks)
(a) Purchases book was undercast by ₹ 900.
(b) Sale of old furniture for ₹ 1,000 was credited to sales account.
(c) Purchase of goods from Arul for ₹ 1,500 on credit was not recorded in the books.
Answer:
(a) Purchases account should be debited with ₹ 900.
Samacheer Kalvi 11th Accountancy Solutions Chapter 9 Rectification of Errors

Question 18.
The following errors were located after the preparation of trial balance. Pass journal entries to rectify them. Assume that there exists a suspense account. (5 marks)
(a) The total of sales book was undercast by ₹ 350.
(b) The total of the discount column on the debit side of cash book ₹ 420 was not posted.
(c) The total of one page of the purchases book of ₹ 5,353 was carried forward to the next page as ₹ 5,533.
(d) Salaries ₹ 2,400 was posted as ₹ 24,000.
(e) Purchase of goods from Sembiyanmadevi on credit for ₹ 180 was posted to her account as ₹ 1,800.
Answer:
Rectifying Journals
Samacheer Kalvi 11th Accountancy Solutions Chapter 9 Rectification of Errors

Question 19.
Rectify the following errors assuming that, the trial balance is already prepared and the difference was placed to suspense account: (5 marks)
(a) Sales book was undercast by ₹ 250.
(b) Purchases book was undercast by ₹ 120.
(c) Sales book was overcast by ₹ 130.
(d) Bills receivable book was undercast by ₹ 75.
(e) Purchases book was overcast by ₹ 35.
Answer:
(a) Sales account should be credited ₹ 250.
(b) Purchases account should be debited ₹ 120.
(c) Sales account should be debited ₹ 130.
(d) Bills receivable account should be debited ₹ 75.
(e) Purchases account should be credited ₹ 35.

Question 20.
The following errors were located after the preparation of trial balance. The difference in trial balance has been taken to suspense account. Rectify them. (5 marks)
(a) The total of purchases book was carried forward ₹ 70 less.
(b) The total of sales book was carried forward ₹ 340 more.
(c) The total of purchases book was carried forward ₹ 150 more.
(d) The total of sales book was carried forward ₹ 200 less.
(e) The total of purchases returns book was carried forward ₹ 350 less.
Answer:
Rectifying Journals
Samacheer Kalvi 11th Accountancy Solutions Chapter 9 Rectification of Errors

Question 21.
The following errors were located by the accountant after the preparation of trial balance. There exists a suspense account. Rectify them. (5 marks)
(a) The total of the discount column of ₹ 1,180 on the debit side of the cash book was not posted.
(b) Purchase of goods from Arivuchelvan on credit for ₹ 600 was posted to the debit side of his account.
(c) The total of the discount column on the credit side of the cash book was undercast by ₹ 400.
(d) The total of sales returns book of ₹ 570 was posted twice.
(e) Sold goods to Mukil on credit for ₹ 87 was posted to her account as ₹ 78.
Answer:
Samacheer Kalvi 11th Accountancy Solutions Chapter 9 Rectification of Errors

Question 22.
The accountant of a firm located the following errors after preparing the trial balance. Rectify them assuming that there is a suspense account. (5 marks)
(a) Machinery purchased for ₹ 3,500 was debited to purchases account.
(b) ₹ 1,800 paid to Raina as salary was debited to his personal account.
(c) Interest received ₹ 200 was credited to commission account.
(d) Goods worth ₹ 1,800 purchased from Amudhanila on credit was not recorded in the books of accounts.
(e) Used furniture sold for ₹ 350 was credited to sales account.
Answer:
Samacheer Kalvi 11th Accountancy Solutions Chapter 9 Rectification of Errors

Question 23.
The book – keeper of a firm found that the trial balance was out by ₹ 922 (excess credit). He placed the amount in the suspense account and subsequently found the following errors: (5 marks)
(a) The total of discount column on the credit side of the cash book ₹ 78 was not posted in the ledger.
(b) The total of purchases book was short by ₹ 1,000.
(c) A credit sale of goods to Natarajan for ₹ 375 was entered in the sales book as ₹ 735.
(d) A credit sale of goods to Mekala for ₹ 700 was entered in the purchases book. You are required to give rectification entries and prepare suspense account.
Answer:
Samacheer Kalvi 11th Accountancy Solutions Chapter 9 Rectification of Errors
Suspense Account
Samacheer Kalvi 11th Accountancy Solutions Chapter 9 Rectification of Errors

Question 24.
The books of Raman did not agree. The accountant placed the difference of ₹ 1,270 to the debit of suspense account. Rectify the following errors and prepare the suspense account:
(a) Goods taken by the proprietor for his personal use ₹ 75 was not entered in the books.
(b) A credit sale of goods to Shanmugam for ₹ 430 was credited to his account as ₹ 340.
(c) A purchase of goods on credit for ₹ 400 from Vivek was entered in the sales book. However, Vivek’s account was correctly credited.
(d) The total of the purchases returns book ₹ 300 was not posted.
Answer:
Samacheer Kalvi 11th Accountancy Solutions Chapter 9 Rectification of Errors
Suspense Account
Samacheer Kalvi 11th Accountancy Solutions Chapter 9 Rectification of Errors

Textbook Case Study Solved

Question 1.
Rameela, a class 11 student, visited one of her relative’s furniture shop. She met the accountant of the shop. He was busy with preparing final accounts. At that time, one of the staff approached the accountant with a list of errors found in ledger postings. Rameela asked the accountant, in a surprised tone, “Is it possible to rectify the errors before preparing the final accounts?” The accountant replied, “Yes, it is!” final accounts?” The accountant replied,
“Yes, it is!”
Rameela was curious to analyse the errors. She found the following:

  1. Furniture sold on credit to Siva and company for ₹ 12,000 was debited to Sam and company;
  2. Rent paid ₹ 2,500, was debited to rent account as ₹ 250.
  3. The total of purchase journal was undercast by ₹ 1,000.
  4. A sales invoice for ₹ 2,000, completely omitted from the books.
  5. Stationery bought for ₹ 250, was posted to purchases account.

Can you help Rameela to identify and rectify the errors?
Answer:
Samacheer Kalvi 11th Accountancy Solutions Chapter 9 Rectification of Errors

Samacheer Kalvi 11th Accountancy Rectification of Errors Additional Questions and Answers

I. Multiple Choice Questions
Choose the correct answer

Question 1.
The errors can be classified into ……………… types.
(a) One
(b) Two
(c) Three
(d) Four
Answer:
(d) Four

Question 2.
When the accountant has failed to record a part of the transaction is known as ………………
(a) Error of partial omission
(b) Error of commission
(c) Compensating errors
(d) Error of principle
Answer:
(a) Error of partial omission

Samacheer Kalvi 11th Accountancy Solutions Chapter 9 Rectification of Errors

Question 3.
The errors that make up for each other or neutralise each other are known as ………………
(a) Errors of commission
(b) Errors of principle
(c) Errors of omission
(d) Compensating errors
Answer:
(d) Compensating errors

Question 4.
The total of salary account is carried forward ₹ 1200 excess ……………….
(a) Errors in carry forward
(b) Errors in posting
(c) Errors in casting
(d) Errors of commission
Answer:
(a) Errors in carry forward

Question 5.
Sales book is undercast by ₹ 100, classify the error
(a) Errors of principle
(b) Errors of commission
(c) Errors in casting
(d) Errors of omission
Answer:
(c) Errors in casting

II. Very Short Answer Questions

Question 1.
What is error of omission?
Answer:
The failure of the accountant to record a transaction or an item in the books of accounts is known as an error of omission. It can be complete omission or partial omission.

Question 2.
What is error of commission?
Answer:
When a transaction is incorrectly recorded, it is known as error of commission. It usually occurs due to lack of concentration or carelessness of the accountant.

Question 3.
What do you mean by errors?
Answer:
Errors means recording or classifying or summarising the accounting transactions wrongly or omissions to record them by a clerk or an accountant unintentionally.

III. Short Answer Questions

Question 1.
What are the types of errors at the stage of journalising?
Answer:

  1. Error of omission
  2. Error of commission
  3. Error of principle

Samacheer Kalvi 11th Accountancy Solutions Chapter 9 Rectification of Errors

Question 2.
What are the types of errors at the stage of posting?
Answer:
(i) Errors of Omission:
(a) Error of complete omission
(b) Error of partial omission

(ii) Errors of Commission:
(a) Posting to wrong account
(b) Posting of wrong account
(c) Posting to the wrong side

Question 3.
What are the types of errors at the stage of preparing trial balance?
Answer:
(i) Error of Omission
(ii) Error of Commission:

(a) Entering to wrong account
(b) Entering wrong amount
(c) Entering to the wrong side of trial balance, etc.

Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry

Students can Download Chemistry Chapter 10 Surface Chemistry Questions and Answers, Notes Pdf, Samacheer Kalvi 12th Chemistry Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry

Samacheer Kalvi 12th Chemistry Chapter 10 Surface Chemistry Textual Evaluation Solved

Samacheer Kalvi 12th Chemistry Surface Chemistry Multiple Choice Questions

Question 1.
For freudlich isotherm a graph of log \(\frac{x}{m}\) is plotted against log P. The slope of the line and its y – axis intercept respectively corresponds to
(a) \(1 / n\), k
(b) log \(1 / n\), k
(c) \(1 / n\), log k
(d) log \(1 / n\), log k
Answer:
(c) \(1 / n\), log k
\(\frac{x}{m}\) = \(\mathrm{k} \cdot \mathrm{p}^{1 / \mathrm{n}}\)
log\((\frac{x}{m})\) = log k + \(\frac { 1 }{ n }\)log p
y = c + mx
m = \(\frac { 1 }{ n }\) and c = log k

Question 2.
Which of the following is incorrect for physisorption?
(a) reversible
(b) increases with increase in temperature
(c) low heat of adsorption
(d) increases with increase in surface area
Answer:
(b) increases with increase in temperature
Physisorption is an exothermic process. Hence increase in temperature decreases the physisorption.

Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry

Question 3.
Which one of the following characteristics are associated with adsorption?
(a) ∆G and ∆H are negative but ∆S is positive
(b) ∆G and ∆S are negative but ∆H is positive
(c) ∆G is negative but ∆H and ∆S are positive
(d) ∆G. AH and ∆S all are negative.
Answer:
(d) ∆G, ∆H and ∆S all are negative.
Adsorption leads to decrease in randomness (entropy).i.e. ∆S < 0 for the adsorption to occur, ∆G should be – ve. We know that ∆G = ∆H – T∆S if ∆S is – ve, T∆S is + ve. It means that ∆G will become negative only when ∆H is – ve and ∆H > T∆S

Question 4.
Fog is colloidal solution of ……………..
(a) solid in gas
(b) gas in gas
(c) liquid in gas
(d) gas in liquid
Answer:
(c) liquid in gas
dispersion medium-gas, dispersed phase-liquid

Question 5.
Assertion: Coagulation power of Al3+ is more than Na.
Reason: greater the valency of the flocculating ion added, greater is its power to cause precipitation
(a) if both assertion and reason are true and reason is the correct explanation of assertion.
(b) if both assertion and reason are true but reason is not the correct explanation of assertion.
(c) assertion is true but reason is false
(d) both assertion and reason are false
Answer:
(a) if both assertion and reason are true and reason is the correct explanation of assertion. (Hardy-Sechuize nile)

Question 6.
Statement: To stop bleeding from an injury, ferric chloride can be applied. Which comment about the statement is justified?
(a) It is not true, ferric chloride is a poison.
(b) It is true, Fe3+ ions coagulate blood which is a negatively charged sol
(c) It is not true; ferric chloride is ionic and gets into the blood stream.
(d) It is true, coagulation takes place because of formation of negatively charged sol with Cl.
Answer:
(b) It is true, Fe3+ ions coagulate blood which is a negatively charged sol

Question 7.
Hair cream is …………..
(a) gel
(b) emulsion
(c) solid sol
(d) sol.
Answer:
(b) emulsion
Emulsion dispersed phase, Dispersion medium -liquid

Question 8.
Which one of the following is correctly matched?
(a) Emulsion – Smoke
(b) Gel – butter
(c) foam – Mist
(d) whipped cream – sol
Answer:
(b) Gel – butter

Question 9.
The most effective electrolyte for the coagulation of As2S3 Soils
(a) NaCI
(b) Ba(NO3)2
(c) K3[Fe(CN)6]
(d) AI3+
Answer:
(d) AI3+
As2S3 is a – vely charged colloid. It will be most effectively coagulated by the cation with greater valency. i.e., Al3+.

Question 10.
Which one of the is  not a surfactant?
(a) CH3 – (CH2)15 – N – (CH3)2CH2Br
(b) CH3 – (CH2)15 – NH2
(c) CH3 – (CH2)16 – CH2OSO2 – Na+
(d) OHC – (CH2)14 – CH2 – COONa+
Answer:
(b) CH3 – (CH2)15 – NH2

Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry

Question 11.
The phenomenon observed when a beam of light is passed through a colloidal solution is ………….
(a) Cataphoresis
(b) Electrophoresis
(c) Coagulation
(d) Tyndall effect
Answer:
(d) Tyndall effect-scattering of light

Question 12.
In an electrical field, the particles of a colloidal system move towards cathode. The coagulation of the same sol is studied using K2SO4
(i). Na3PO4
(ii). K4[Fe(CN)6]
(iii). and NaCI
(iv). Their coagulating power should be …………..
(a) II > I >IV > III
(b) III > II > I > IV
(c) I > II > III > IV
(d) none of these
Answer:
(b) III > II > I > IV

Question 13.
Collodion is a 4% solution of which one of the following compounds in alcohol – ether mixture?
(a) Nitroglycerine
(b) Cellulose acetate
(c) Glycoldinitrate
(d) Nitrocellulose
Answer:
(a) Nitrocellulose
pyroxylin (nitro cellulose)

Question 14.
Which one of the following is an example for homogeneous catalysis?
(a) manufacture of ammonia by Haber’s process
(b) manufacture of sulphuric acid by contact process
(c) hydrogenation of oil
(a) Hydrolysis of sucrose in presence of all HCI
Answer:
(a) Hydrolysis of sucrose in presence of all HCl
Both reactant and catalyst are in same phase. i.e. (1)

Question 15.
Match the following.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-1
Answer:
(a) (iv) (i) (ii) (iii)

Question 16.
The coagulation values in millimoles per litre of the electrolytes used for the coagulation of AS2S3 are given below
(I) (NaCI) = 52
(II) (BaCl) = 0.69
(III) (MgSO4) = 0.22
The correct order of their coagulating power is ……….
(a) III > II > I
(b) I > II > III
(c) I >III > II
(d) II > III > I
Answer:
(a) III > II > I
coagulating power ± \(\frac{1}{\text { coagulation value }}\)

Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry

Question 17.
Adsorption of a gas on solid metal surface is spontaneous and exothermic, then ……………
(a) ∆H increases
(b) ∆S increases
(c) ∆G increases
(d) ∆S decreases
Answer:
(a) ∆S decreases – ∆S is -ve

Question 18.
If x is the amount of adsorbate and m is the amount of adsorbent, which of the following relations is not related to adsorption process?
(a) x/m = f(P) at constant T
(b) x/m = f(T) at constant P
(c) P = f(T) at constant x/m
(d) x/m = PT
Answer:
(d) x/m = mPT

Question 19.
On which of the following properties does the coagulating power of an ion depend?
(a) Both magnitude and sign of the charge on the ion.
(b) Size of the ion alone
(c) the magnitude of the charge on the ion alone
(d) the sign of charge on the ion alone.
Answer:
(a) Both magnitude and sign of the charge on the ion.

Question 20.
Match the following.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-2
Answer:
(d) (iv) (iii) (ii) (i)

Samacheer Kalvi 12th Chemistry Surface Chemistry Short Answer Questions

Question 1.
Give two important characteristics of physisorption.
Answer:

Important characteristics of physisorption:

  1. It is reversible
  2. It has low heat of adsorption
  3. It has weak van der Waals forces of attraction with adsorbent.
  4. It increases with increase in pressure.
  5. It forms multi molecular layer.

Question 2.
Differentiate physisorption and chemisorption.
Answer:
Chemical adsorption or Chemisorption or Activated adsorption

  1. It is very slow
  2. It is very specific depends on nature of adsorbent and adsorbate.
  3. chemical adsorption is fast with increase pressure, it can not alter the amount.
  4. When temperature is raised chemisorption first increases and then decreases.
  5. Chemisorption involves transfer of electrons between the adsorbent and adsorbate,
    Heat of adsorption is high i.e., from 40 – 400kJ/mole.
  6. Monolayer of the adsorbate is formed.
  7. Adsorption occurs at fixed sites called active centres. It depends on surface area.
  8. Chemisorption involves the formation of activated complex with appreciable activation energy.
  9. Physical adsorption or van der Waals adsorption or Physisorptlon
  10. It is irreversible.

Physical adsorption or van der Waals adsorption or physisorption.

  1. It is instantaneous
  2. It is non-specific
  3. In Physisorption. when pressure increases the amount of adsorption increases.
  4. Physisorption decreases with an increase in temperature.
  5. No transfer of electrons
  6. Heat of adsorption is low in the order of 40 kJ/mole.
  7. Multilayer of the adsorbate is formed on the adsorbent.
  8. It occurs on all sides.
  9. Activation energy is insignificant.
  10. It is reversible.

Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry

Question 3.
In the case of chemisorption, why adsorption first increases and then decrease with temperature?
Answer:
1. Chemisorption involves high activation energy so it is also referred to as activated adsorption.

2. It is found in chemisorption that it first increases and then decreases with increase in temperature. When adsorption is plotted, the graph first increases and then decreases with temperature.

3. The initial increase illustrates the requirement of activation of the surface for adsorption is due to fact that the formation of activated complex requires certain energy. But later it decreases at high temperature is due to desorption as the kinetic energy of the adsorbate increases (exothermic nature)

Question 4.
Which will be adsorbed more readily on the surface of charcoal and why; NH3 or CO2?
Answer:
1. The gases having low critical temperature are adsorbed slowly, while gases with high critical temperature are adsorbed readily.

2. Among CO2, and NH3, NH3 will be more readily adsorbed on the surface of the charcoal. This is because the critical temperature of ammonia gas is quite high than the CO2. Hence, it easily combines with the materials than the CO2 whether it is solid, liquid or any gases.

Question 5.
Heat of adsorption is greater for chemisorptions than physisorption. Why?
Answer:
Chemisorption has higher heat of adsorption. because in chemisorption the chemical bonds are much stronger. In adsorbed state the adsorbate is hold on the surface of adsorbent by attractive forces (bond). And chemisorption is irreversible one. Therefore, heat of adsorption is greater for chenil sorptions than physisorption. Chemisorption, heat of adsorption range 40 – 400kJ/mole.

Question 6.
In a coagulation experiment 10 mL of a colloid (X) is mixed with distilled water and 0.1M solution of an electrolyte AB so that the volume is 20 mL. It was found that all solutions containing more than 6.6 mL of AB coagulate with in 5 minutes. What is the flocculation values of AB for sol (X)?
Answer:
A minimum of 6.6mL of AB is required to coagulate the sol. The moles of AB in the sol is
\(\frac{6.6 \times 0.01}{20}\) = 0.033 moles
This means that a minimum of 0.033 moles or 0.0033 x 1000 = 3.3 milli moles are required for coagulating one litre of sol. Flocculation value of AB for X = 3.3

Question 7.
Peptising agent is added to convert precipitate into colloidal solution. Explain with an example.
Answer:
1. Ions either positive or negative of peptizing agent (electrolyte) are adsorbed on the particles of precipitate. They repel and hit each other and break the particles of the precipitate into colloidal size.

2. For example, when we add a small volume of very dilute hydrochloric acid solution peptising agent to a fresh precipitate of a silver chloride, it leads to formation of silver chloride colloidal solution,
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-3

Question 8.
What happens when a colloidal sol of Fe(OH)3 and As2S3 are mixed?
Answer:
On mixing Fe(OH)3 positive sol and As2S3 negative sol, mutual coagulation occurs which causes precipitation. When these sol got mixed with each other, due to Fe3+ and S2- ions neutralisation of charges will happen and precipitate will be formed.
Fe(OH)3 + As2S3 → Fe2S3 + As(OH)3

Question 9.
What is the difference between sol and a gel?
Answer:
Sol

  1. The liquid state of a colloidal solution is called sol.
  2. The sol does not have a definite structure.
  3. The dispersion medium of the sol may be water.
  4. The sol can be converted to gel by cooling The sol can be easily dehydrated.
  5. The viscosity of the sol is very low.
  6. Sol is categorized into lyophobic and lyophilic sols.
  7. Example: Blood

Gel

  1. The solid or semi-solid state of a colloidal solution is called gel.
  2. The gel possesses honey comb-like structure.
  3. The dispersion medium of gel will be hydrated colloid particles.
  4. The gel can be converted sol by heating.
  5. The gel cannot be dehydrated.
  6. The viscosity of the gel is very high.
  7. There is no such classification of gel.
  8. Example: Fruit jelly, cooked gelatin jelly.

Question 10.
Why are lyophillic colloidal sols are more stable than lyophobic colloidal sol?
Answer:
1. A lyophilic colloidal sols are stable due to the charge and the hydration of sol particles.

2. Lyophilic sols are more stable than lyophobilc sols because they are highly hydrated in the solution. And since more is the hydration more will be its stability.

3. Lyophilic sols are stabilized by electrostatic charge and hydration where as lyophobile sols are only stabilized by charge, so they easily gets coagulated and requires a stabilising agent. Hence, lyophilic sols are more stable than lyophobilc sols.

Question 11.
Addition of Alum purifies water. Why?
Answer:
Purification of drinking water is activated by coagulation of suspended impurities in water using alums containing Al3+. That is why we are adding to purify water.

Question 12.
What are the factors which influence the adsorption of a gas on a solid?
Answer:
Factors which influence the adsorption of a gas on a solid is as follows:
1. Nature of the gas:
Easily liquefiable gases such as NH3, HCl etc are adsorbed to a great extent in comparison to gases such as H2, O2, etc. This is because van der Waal’s forces are stronger is easily liquifiable gases.

2. Surface area of the solid:
The greater the surface area of the adsorbent, the greater is the adsorption of gas on the solid surface.

3. Effect of pressure:
Adsorption is a reversible process and is accompanied by a decrease in pressure. Therefore, adsorption increases with an increase in pressure.

4. Effect of temperature:
Adsorption is an exothermic process. Thus in accordance with Le – Chatelier’s principle, the magnitude of adsorption decreases with an increase in temperature.

Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry

Question 13.
What are enzymes? Write a brief note on the mechanism of enzyme catalysis.
Answer:
Enzymes are complex protein molecules with three-dimensional structures. They catalyse the chemical reaction in a living organism. They are often present in colloidal state and extremely specific in catalytic action.

Each enzyme produced in a particular living cell can catalyse a particular reaction in the cell. Mechanism of enzyme catalysis: Mechanism of enzyme catalysed reaction is known as lock and key mechanism.
1. Enzymes arc highly specific in their action.

2. This specificity is due to the pressure of active sites. The shape of active site of any given enzyme is like cavity such that only a specific substrate can fit into it.

In the same way a key fits into lock. The specific binding needs the formation of an enzyme-substrate complex which accounts for high specificity of enzyme-catalyzed reactions.

3. Once the proper orientation is attained the substrate molecules react to form the product in two steps.

4. Since the product molecule does not have an affinity for the enzyme they leave the enzyme surface making room for the fresh substrate.

Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-4

step 1: Formation of the enzyme-substrate complex
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-5
Step 2: Dissociation of the enzyme-substrate complex to form product
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-6
The rate of the formation of the product depends upon the concentration of ES.

Question 14.
What do you mean by activity and selectivity of catalyst?
Answer:
1. Activity of Catalyst:
The activity of a catalyst is its ability to increase the rate of a particular reaction, Chemisorption is the main factor in deciding the activity of a catalyst. The adsorption of reactants in the catalyst surface should be neither too strong nor too weak. It should just be strong enough to make the catalyst active.

2. Selectivity of the catalyst:
The ability of the catalyst to direct a reaction to yield a particular product is referred to as the selectivity of the catalyst. For example, by using different catalysts, we can get different products for the reaction between H2 and CO.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-7

Question 15.
Describe some features of catalysis b Zeolites.
Answer:

  1. Zeolites are microporous, crystalline, hydrated aluminosilicates, made of silicon and aluminium tetrahedra.
  2. There are about 50 natural zeolites and 1 50 synthetic zeolites. As silicon is tetravalent and aluminium is trivalent, the zeolite matrix carries an extra negative charge. To balance the negative charge, there are extra framework cations, for example, H+ or Na+ ions.
  3. Zeolites earring protons are used as solid acids, catalysis and they are extensively used in the petrochemical industry for cracking heavy hydrocarbon fractions into gasoline, diesel, etc.
  4. Zeolites earring Na ions are used as basic catalysis.
  5. One of the most important applications of zeolites is their shape selectivity. In zeolites, the active sites namely protons are lying inside their pores. So, reactions occur only inside the pores of zeolites.

Question 16.
Give three uses of emulsions.
Answer:

  1. The cleansing action of soap is due to emulsions.
  2. It is used in the preparation of vanishing cream.
  3. It is used in the preparation of cold liver oil.

Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry

Question 17.
Why does the bleeding stop by rubbing moist alum?
Answer:
Blood is a colloidal sol. When we nib the injured part with moist alum then coagulation of blood takes place. Hence main reason is coagulation, which stops the bleeding. Therefore bleeding stop by rubbing moist alum.

Question 18.
Why is desorption important for a substance to act as good catalyst?
Answer:
Desorption is important for a substance to act as a good catalyst, so that after the reaction, the products found on the surface separate out (desorbed) to create free surface again for other reactant molecules to approach the surface and react. If desorption does not occur then other reactants are left with no space on the surface of the catalyst for adsorption and the reaction will stop.

Question 19.
Comment on the statement: Colloid is not a substance but it is a state of substance.
Answer:
The statement is true. Because the same substance may exist as a colloid under certain conditions and as a crystalloid under certain other conditions. For example. NaCl in water behaves as a crystalloid while in benzene, it behaves as a colloid. Similarly, dilute soap solution behaves

like a crystalloid while concentrated solutions behaves as a colloid. It is the size of the particles which matters. That is the state in which the substance exists. If the size of the particles lies in the range 1 nm to 1oo nm, it is in the colloidal state.

Question 20.
Explain any one method for coagulation
Answer:
The flocculation and setting down of the sol particles is called coagulation. Various method of coagulation are given below:

  1. Addition of electrolytes
  2. Electrophoresis
  3. Mining oppositely charged sols
  4. Boiling.

Addition of electrolytes
A negative ion causes the precipitation of positively charged sol and vice versa. When the valency of ion is high, the precipitation power is increased. For example, the precipitation power of some cations and anions varies in the following order
Al3+ > Ba2+ > Na+, Similarly [Fe(CN)6]-3 > SO4-2 > Cl
The precipitation power of electrolyte is determined by finding the minimum concentration (millimoles/lit) required to cause precipitation of a sol in 2hours. This value is called the flocculation value. The smaller the flocculation value greater will be precipitation.

Question 21.
Write a note on electroosmosis.
Answer:
Electro osmosis:
A sol is electrically neutral. Hence the medium carries an equal but opposite charge to that of dispersed particles. When sol particles are prevented from moving, under the influence of electric field the medium moves in a direction opposite to that of the soil particles. This movement of dispersion medium under the influence of electric potential is called electro-osmosis.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-8

Question 22.
Write a note on catalytic poison
Answer:
Catalytic poison:
Certain substances when added to a catalysed reaction, decreases or completely destroys the activity of catalyst and they are often known as catalytic poisons. For example, In the reaction,
2SO2 + O2 → 2SO3 with a Pt catalyst, the poison is AS2O3.
i.e., AS2O3 destroys the activity of pt. AS2O3 blocks the activity of the catalyst. So, the activity is lost.

Question 23.
Explain the intermediate compound formation theory of catalysis with an example.
Answer:
The intermediate compound formation theory:
A catalyst acts by providing a new path with low energy of activation. in homogeneously catalysed reactions a catalyst may combine with one or more reactant to form an intermediate which reacts with other reactant or decompose to give products and the catalyst is regenerated.

Consider the reactions:
A + B → AB ……………(1)
A + C → AC (intermediate) ………….(2)
C is the catalyst
AC + B → AB + C …………(3)
Activation energies for the reactions (2) and (3) are lowered compared to that of (1). Hence the formation and decomposition of the intermediate accelerate the rate of the reaction.
Example:
The mechanIsm of Fridel crafts reaction is given below
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-9
The action of catalyst is explained as follows .
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-10

This theory describes,

  1. The specificity of a catalyst.
  2. The increase in the rate of the reaction with increase in the concentration of a catalyst.

Limitations

  1. The intermediate compound theory fails to explain the action of catalytic poison and activators (promoters).
  2. This theory is unable to explain the mechanism of heterogeneous catalysed reactions.

Question 24.
What is the difference between homogenous and hetrogenous catalysis?
Answer:
Hornogenous Catalysis:

  1. In a catalysed reaction the reactants, products and catalyst are present in the same phase.
  2. For example.
    Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-11
    Hence NO act as catalyst.
  3. Homogeneous catalysis explained by intermediate compound formation theory.

Heterogeneous Catalysis:

  1. In a reaction, the catalyst is present in a different phase. i.e., catalyst is not present in the same phase as that of reactants and products.
  2. For example.
    Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-12
    Hence Pt(s) act as catalyst.
  3. Hetenogeneous catalysis explained by adsorption theory.

Question 25.
Describe adsorption theory of catalysis.
Answer:
Adsorption theory:
Langmuir explained the action of catalyst in heterogeneous catalysed reactions based on adsorption. The reactant molecules are adsorbed on the catalyst surfaces, so this can also he called as contaçt catalysis.

According to this theory, the reactants arc adsorbed on the catalyst surface to form an activated complex which subsequently decomposes and gives the product. The various steps involved in a heterogeneous catalysed rcacton arc given as follows:

  1. Reactant molecules diffuse from bulk to the catalyst surface.
  2. The reactant molecules are adsorbed on the surface of the catalyst.
  3. The adsorbed reactant molecules are activated and form activated complex which is decomposed to form the products.
  4. The product molecules are desorbed.
  5. The product diffuse away from the surface of the catalyst.

Advantages of adsorption theory:
The adsorption theory explains the following .
1. Increase in the activity of a catalyst by increasing the surface area. Increase in the surface area of metals and metal oxides by reducing the particle size increases the rate of the reaction.

2. The action of catalytic poison occurs when the poison blocks the active centres of the catalyst.

3. A promoter or activator increases the number of active centres on the surfaces.

Samacheer Kalvi 12th Chemistry Surface Chemistry Additional Questions

Samacheer Kalvi 12th Chemistry Surface Chemistry 1 mark Questions and Answers

I. Choose the correct answer and write it. Answers are in bold it.

Question 1.
Which one of the following is used to absorb colourants from sugar?
(a) Silica gel
(b) Magnesia
(c) Charcoal
(d) Alumina
Answer:
(c) Charcoal

Question 2.
Silica gel is usually adsorbed………….
(a) Colourants
(b) Hydrogen
(c) Liquid Helium
(d) Water
Answer:
(d) Water

Question 3.
Which one of the following is called adsorbate?
(a) Charcoal
(b) Silica gel
(c) Ammonia
(d) Magnesia
Answer:
(c) Ammonia

Question 4.
Which of the following can act as adsorbent?
(a) Silica gel
(b) Ammonia
(c) Colourants
(d) Water
Answer:
(a) Silica gel

Question 5.
The surface of separation of two phases where the concentration of adsorbed molecule is high is known as …………..
(a) adsorbate
(b) adsorbent
(c) interface
(d) residual phase
Answer:
(c) interface

Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry

Question 6.
Consider the following statement.
(i) High adsorption is the result of high surface area of the adsorbent.
(ii) The process of removing an adsorbed substance is called absorption.
(iii) Adsorbed substance is called an adsorbate.

Which of the above statement is / are not correct?
(a) (i) & (ii)
(b) (ii) & (iii)
(c) (ii) only
(d) (iii) only
Answer:
(c) (ii) only

Question 7.
Which metal cannot act as adsorbent?
(a) Pt
(b) Ag
(c) Pd
(d) Al
Answer:
(d) Al

Question 8.
Consider the following statements.
(i) Adsorption is spontaneous process.
(ii) Adsorption is always accompanied by an increase in free energy.
(iii) Adsorption is an endothermic reaction.

Which of the above statement is/are not correct?
(a) (i) only
(b) (ii) & (iii)
(c) (ii) & (i)
(d) (i) only
Answer:
(b) (ii) & (iii)

Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry

Question 9.
Absorption and adsorption if simultaneously occurs, it is called ……………
(a) occlusion
(b) sorption
(c) desorption
(d) dissolution
Answer:
(b) sorption

Questioin 10.
The process of sorption of gases on the metal surface is called ……………
(a) Desorption
(b) Dissolution
(c) Occlusion
(d) Condensation
Answer:
(c) Occlusion

Question 11.
When gas molecules are held to the surface by the formation of the chemical bond the heat energy released is nearly equal to
(a) 40 kJ/mole
(b) 800 kJ/mole
(c) 400 kJ/mole
(d) 4 kJ/mole
Answer:
(c) 400 kJ/mole

Question 12.
Which of the following is physical adsorption?
(a) Adsorption of H2 on nickel
(b) Friedel crafts reaction
(c) Synthesis of SO3 in the presence of NO
(d) Corrosion of iron
Answer:
(a) Adsorption of H2 on nickel

Question 13.
Which one of the following is chemical adsorption?
(a) Adsorption of O2 on tungsten
(b) Adsorption of ethyl alcohol vapours on nickel
(c) Adsorption of N2 on mica
(d) Rusting of iron
Answer:
(d) Rusting of iron

Question 14.
Which of the following occurs at low temperatures?
(a) Adsorption of O2 on tun gsen
(b) Adsorption of N2 on mica
(c) Adsorption of ethyl alcohol vapours on nickel
(d) Adsorption of H2 on nickel
Answer:
(b) Adsorption of N2 on mica

Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry

Question 15.
Consider the following statements:
(i) Chemical adsorption is an instantaneous process
(ii) Multilayer of the adsorbate is formed on the adsorbent
(iii) Chemisorption involves the formation of the activated complex.

Which of the above statement is/are not correct?
(a) (i) & (ii)
(b) (iii) only
(c) (i) only
(d) (ii) only
Answer:
(a) (i) & (ii)

Question 16.
Consider the following statements:
(i) In chemisorption, heat of adsorption is high
(ii) Monolayer of the adsorbate is formed during chemisorption
(iii) Physisorption increases with an increase in temperature.

Which of the above statement is/are not correct’?
(a) (i) & (ii)
(b) (iii) only
(c) (ii) only
(d) (i) only
Answer:
(b) (iii) only

Question 17.
The extent of surface adsorption does not depend on
(a) Nature of the adsorbent
(b) Pressure
(c) Temperature
(d) Density
Answer:
(d) Density

Question 18.
Which of the following gases is not a permanent gas?
(a) NH3
(b) H2
(c) N2
(d) O2
Answer:
(a) NH3

Question 19.
Which of the following is liquefiable gas?
(a) SO2
(b) H2
(c) N2
(d) O2
Answer:
(a) SO2

Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry

Question 20.
Which one of the following is a permanent gas?
(a) NH3
(b) SO3
(c) N2
(d) CO2
Answer:
(c) N2

Question 21.
Consider the following statements.
(i) When pressure increases, the amount of physisorplion also increases.
(ii) Permanent gases like H2, N2 and O2 cannot be liquefied easily.
(iii) Lesser is the surface area, higher is the amount adsorbed.

Which of the above statement is/are correct?
(a) (i) & (ii)
(b) (iii) only
(c) (ii) only
(d) (i) only
Answer:
(a) (i) & (ii)

Question 22.
Which one of the following is used in blast furnace for drying air?
(a) Activated charcoal
(b) Silica gel
(c) Alumina
(d) Permutit
Answer:
(b) Silica gel

Question 23.
Which is employed in the softening of hardwater to absorb Ca2+ and Mg2+ ions?
(a) Alumina
(b) Silica gel
(c) Permutit
(d) Charcoal
Answer:
(c) Permutit

Question 24.
The formula for permit is …………
(a) Ca Al2 Si4 O12
(b) CaAl3SiO2. xH2O
(c) Na2 Al2 Si4 O12
(d) Na2 SiO3
Answer:
(c) Na2 Al2 Si4 O12

Question 25.
Which one of the following is used to regenerate permutit in the softening of hard water?
(a) Common salt
(b) Baking soda
(c) Washing soda
(d) Quick lime
Answer:
(a) Common salt

Question 26.
Which of the following is used to demineralise water?
(a) Permutit
(b) Common salt
(c) Ion exchange resin
(d) Charcoal

Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry

Question 27.
Which of the following is used during world war as gas masks?
(a) Permutit
(b) Silica gel
(c) Fuller’s earth
(d) Charcoal
Answer:
(d) Charcoal

Question 28.
Which of the following is used in petroleum refining and refining of vegetable oils?
(a) Charcoal
(b) Silica gel
(c) Pcrmutit
(d) Nickel
Answer:
(b) Silica gel

Question 29.
The catalyst used in the hydrogenation of oils to obtain vanaspati is …………
(a) Iron
(b) Molybdenum
(c) Nickel
(d) Copper
Answer:
(c) Nickel

Question 30.
The catalyst and promoter used in Haber’s process are respectively ………..
(a) Mo, Fe
(b) Fe, Mo
(c) Pt, H2S
(d) Pt, V2O5
Answer:
(b) Fe, Mo

Question 31.
Which method is used for identification, detection, and estimation of many substances even if they are in micro quantities?
(a) Lassaigne’s test
(b) Canus method
(c) Kjeldhals method
(d) Chromatography
Answer:
(d) Chromatography

Question 32.
Which one of the following is used in the identification of Al3+ ion in Al(OH)3?
(a) Red litmus
(b) Blue litmus
(c) Phenol red
(d) Sodium hydroxide
Answer:
(b) Blue litmus

Question 33.
Which ores are concentrated by the froth floatation process?
(a) Oxide ore
(b) Carbonate ore
(c) Sulphate ores
(d) Suiphide ores
Answer:
(d) Suiphide ores

Question 34.
In froth floatation process, the lighter ore particles are wetted by …………..
(a) Olive oil
(b) Pine oil
(c) Soap oil
(d) Neem oil
Answer:
(b) Pine oil

Question 35.
Which one of the following is an example of homogeneous catalysis?
(a) Decomposition of acetaldehyde by 12 catalysts
(b) Hydrolysis of cane sugar with a mineral acid
(c) Ester hydrolysis with alkali
(d) All the above
Answer:
(d) All the above

Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry

Question 36.
Which one of the following is an example for homogeneous catalysis?
(a) Manufacture of sulphuric acid by contact process
(b) Manufacture of ammonia by Haber’s process
(c) Oxidation of ammonia carried out in the presence of platinum gauze
(d) Hydrolysis of cane sugar with mineral acid
Answer:
(d) Hydrolysis of cane sugar with mineral acid

Question 37.
Which one of the following is an example for heterogeneous catalysis?
(a) Decomposition of acetaldehyde by I2 catalyst
(b) Decomposition of H2O2 in the presence of Pt catalyst
(c) Acid hydrolysis of ester
(d) Hydrolysis of cane sugar with mineral acid
Answer:
(b) Decomposition of H2O2 in the presence of Pt catalyst

Question 38.
Which one of the following is not an example for homogeneous catalysis?
(a) Contact process of manufacture of H2SO4
(b) Haber’s process of manufacture of NH3
(c) Acid hydrolysis of ester
(d) Freidel crafts reaction
Answer:
(c) Acid hydrolysis of ester

Question 39.
Consider the following statements:
(i) A catalyst needed in very small quantity
(ii) A catalyst can initiate a reaction
(iii) Catalyst are highly specific in nature

Which of the above statement is/are not correct?
(a) (i) & (iii)
(b) (ii) & (iii)
(c) (iii) only
(d) (ii) only
Answer:
(d) (ii) only

Question 40.
Consider the following statements.
(i) A solid catalyst will be more effective if it is taken in a finely divided form
(ii) A catalyst cannot initiate a reaction
(iii) For a chemical reaction4 catalyst is needed in very large quantity

Which of the above statement is / are not correct?
(a) (i) & (ii)
(b) (ii) & (iii)
(c) (iii) only
(d) (i) & (iii)
Answer:
(a) (i) & (ii)

Question 41.
The catalyst poison in the contact process of manufacture of SO3 is …………
(a) As2O3
(b) H2S
(c) CO
(d) As2S3
Answer:
(a) As2O3

Question 42.
In Haber’s process of manufacture of ammonia, the Fe catalyst is poisoned by the pressure of …………….
(a) Mo
(b) Co
(c) H2S
(d) As2O3
Answer:
(c) H2S

Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry

Question 43.
In the reaction 2H2 + O2 → 2H2O acts as a catalytic poison for Pt catalyst.
(a) Co
(b) Mo
(c) As2O3
(d) H2S
Answer:
(a) Co

Question 44.
The negative catalyst in the decomposition of H2O2 is …………..
(a) Ethanol
(b) Acetic acid
(c) Ethanoic acid
(d) Methanol
Answer:
(a) Ethanol

Question 45.
Which one of the following is an example tar an autocatalysis?
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-13

Question 46.
In the decomposition of hydrogen peroxide which acts as a negative catalyst?
(a) Dilute acid
(b) Glycerol
(c) a (or) b
(d) Ethanol
Answer:
(c) a (or) b

Question 47.
The energy required for the reactants to reach the activated complex is called …………
(a) threshold energy
(b) activation energy
(c) internal energy
(d) Gibbs free energy
Answer:
(b) activation energy

Question 48.
Which of the following is explained by intermediate compound formation theory?
(a) Mechanism of Friedel crafts reaction
(b) Thermal decomposition of KClO3 in the presence of MnO2
(c) Oxidation of HCl by air in the presence of CuCl2
(d) Manufacture of NH3 by Haber’s process
Answer:
(d) Manufacture of NH3 by Haber’s process

Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry

Question 49.
Consider the following statements.
(i) Intermediate compound theory describes the specificity of a catalyst.
(ii) Intermediate compound theory explains the action of catalytic poison and activators.
(iii) Intermediate compound theory is unable to explain the mechanism of heterogeneously catalysed reactions.

Which of the above statement is/are not correct?
(a) (ii) only
(b) (i) & (iii)
(c) (iii) only
(d) (i) & (ii)
Answer:
(a) (ii) only

Question 50.
Who explained the action of catalyst in adsorption theory?
(a) Berzellius
(b) Langmuir
(c) Thomas Graham
(d) Dalton
Answer:
(b) Langmuir

Question 51.
Consider the following statements.
(i) The action of catalytic poison occurs when the poison blocks the active centres of the catalyst.
(ii) A promoter decreases the number of active centres on the surfaces.
(iii) Increase in the activity of a catalyst by increasing the surface area.

Which of the above statement is/are correct?
(a) (ii) only
(b) (iii) only
(c) (i) & (iii)
(d) (ii) & (iii)
Answer:
(c) (i) & (iii)

Question 52.
Which of the following catalyse the chemical reaction in a living organisms?
(a) enzymes
(b) protein
(c) lipids
(d) serum
Answer:
(c) lipids

Question 53.
Which of the following enzyme catalyse the hydrolysis of starch into maltose?
(a) maltase
(b) irivertase
(c) diastase
(d) zymase
Answer:
(c) diastase

Question 54.
Which enzyme catalyses the conversion of glucose into ethanol?
(a) maltase
(b) invertase
(c) diastase
(d) zymase
Answer:
(c) diastase

Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry

Question 55.
Which of the following act as catalyst in the oxidation of alcohol into acetic acid?
(a) pepsin
(b) diastase
(c) micro derma
(d) urease
Answer:
(c) micro derma

Question 56.
Which catalyst is used in the hydrolysis of urea?
(a) micro derma
(b) zymase
(c) pepsin
(d) urease
Answer:
(d) urease

Question 57.
Which of the following enzyme is present in soya beans?
(a) urease
(b) zymase
(c) pepsin
(d) lactase
Answer:
(a) urease

Question 58.
Consider the following statements.
(i) Enzymes are complex protein molecules with three-dimensional structures.
(ii) Enzymes catalyse the chemical reaction in living organism.
(iii) Enzymes arc not specific in catalytic action.

Which of the above statement is J are correct?
(a) (iii) only
(b) (ii) & (iii)
(c) (i) & (ii)
(d) (i) & (iii)
Answer:
(c) (i) & (ii)

Question 59.
Consider the following statements.
(i) Enzyme catalysed reaction has maximum rate at optimum temperature
(ii) Enzyme catalysis is highly specific in nature
(iii) Catalytic activity of enzyme is decreased by coenzyrnes or activators.

Which of the above statement is / are not correct?
(a) (iii) only
(b) (i) only
(c) (ii) only
(d) (i) & (ii)

Question 60.
The temperature at which enzyme activity is high (or) maximum is called ………….
(a) critical temperature
(b) optimum temperature
(c) low temperature
(d) high temperature
Answer:
(b) optimum temperature

Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry

Question 61.
Enzymes can be active in human body at a temperature of ………….
(a) 98°F
(b) 105°F
(c) 37°F
(d) 50°F
Answer:
(a) 98°F

Question 62.
Consider the following statements.
(i) Zeolites are alumino silicates made of silicon and aluminium tetrahedra.
(ii) Zeolites carrying Na ions are used as basic catalyst.
(iii) As silicon is tetravalent and aluminium is trivalent, the zeolite matrix carries extra positive charge.

Which of the above statement is / are correct?
(a) (i) & (ii)
(b) (i), (ii) & (iii)
(c) (iii) only
(d) (ii) only
Answer:
(a) (i) & (ii)

Question 63.
Which one of the following is used in the petrochemical industry for cracking heavy hydrocarbon fractions into gasoline, diesel, etc.?
(a) permutit
(b) zeolite
(c) pepsin
(d) protein
Answer:
(b) zeolite

Question 64.
Which one of the following is used as a catalyst in the conversion of Lindane to cyclohexane?
(a) Fe°/Pd°
(b) Ni
(c) Zn + HCl
(d) LiAIH4
Answer:
(a) Fe°/Pd°

Question 65.
Which one of the following is used as catalyst in homogeneous and heterogeneous catalysis?
(a) enzymes
(b) zeolite
(c) nanocatalyst
(d) coenzyme
Answer:
(c) nanocatalyst

Question 66.
Who studied and analysed about colloids?
(a) Berzelius
(b) Thomas Graham
(c) Langmuir
(d) Robert Brown
Answer:
(b) Thomas Graham

Question 67.
Which one of the following is lyophillic colloid?
(a) Protein sol
(b) Gold sol
(c) Silver sol
(d) Platinum sol
Answer:
(a) Protein sol

Question 68.
Which one of the following is lyophobic colloid?
(a) Protein sol
(b) Starch sol
(c) Gel
(d) Gold sol
Answer:
(d) Gold sol

Question 69.
An example of liquid aerosol is ………..
(a) Soda water
(b) Milk
(c) Fog
(d) Inks
Answer:
(c) Fog

Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry

Question 70.
Which of the following is an example of Emulsion?
(a) mayonnaise
(b) shaving cream
(c) fumes
(d) paint
Answer:
(a) mayonnaise

Question 71.
The dispersed phase and dispersion medium in smoke, fumes and dust are …………..
(a) gas, solid
(b) solid, gas
(c) gas, liquid
(d) solid, liquid
Answer:
(b) solid, gas

Question 72.
Inks, paints arc considered as …………
(a) liquid in solid
(b) solid in liquid
(c) gas in gas
(d) solid in solid
Answer:
(b) solid in liquid

Question 73.
Which of the following is an example for gel?
(a) Pumice stone
(b) Pearls
(c) Coloured glass
(d) Butter
Answer:
(c) Coloured glass

Question 74.
Which one of the following is an example for solid sol?
(a) Butter
(b) Cheese
(c) Pearls
(d) Pumice stone
Answer:
(c) Pearls

Question 75.
Soda water is an example for ………..
(a) gel
(b) emulsion
(c) foam
(d) sol
Answer:
(c) foam

Question 76.
Colloidal ink and graphite are prepared by …………
(a) colloid mill
(b) Bredig’s arc
(c) ultrasonic homogenizer
(d) peptisation
Answer:
(a) colloid mill

Question 77.
Which method is used to prepare metal sols?
(a) ultrasonic dispersion
(b) mechanical dispersion
(c) Bredigs arc method
(d) peptisation
Answer:
(c) Bredigs arc method

Question 78.
Who prepared non-aqueous inflammable liquids like Benzene and ether by Bredig’s arc method?
(a) George Bredig
(b) Sved berg
(c) Thomas Graham
(d) Robert Brown
Answer:
(b) Sved berg

Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry

Question 79.
Which method is used to prepare mercury colloid?
(a) peptisation
(b) mechanical dispersion
(c) ultrasonic dispersion
(d) Bredig’s arc method
Answer:
(c) ultrasonic dispersion

Question 80.
Mercury sol is obtained by subjecting it to sound waves of frequency more than ………..
(a) 20 Hz
(b) 20 kHz
(c) 200 kHz
(d) 2000 kHz
Answer:
(b) 20 kHz

Question 81.
The conversion of a precipitate into colloid is called …………..
(a) coagulation
(b) hydrolysis
(c) condensation
(d) peptisation
Answer:
(d) peptisation

Question 82.
Gold sol is prepared by reduction of auric chloride using …………..
(a) water
(b) HCHO
(c) CH3CHO
(d) CH3COOH
Answer:
(b) HCHO

Question 83.
Which method is suitable to prepare I2 sol and Se sol?
(a) Reduction
(b) Hydrolysis
(c) oxidation
(d) peptisation
Answer:
(c) oxidation

Question 84.
Which condensation method is used to prepare sulphur sol?
(a) Hydrolysis
(b) Decomposition
(c) Reduction
(d) Peptisation
Answer:
(b) Decomposition

Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry

Question 85.
Arsenic sulphide colloid is prepared by ………..
(a) hydrolysis
(b) reduction
(c) double decomposition
(d) decomposition
Answer:
(c) double decomposition

Question 86.
By which method phosphorous colloid can be prepared’?
(a) Decomposition
(b) Exchange of solvent
(c) Hydrolysis
(d) Reduction
Answer:
(b) Exchange of solvent

Question 87.
Which one of the following is not used to purify colloids?
(a) Dialysis
(b) Peptisation
(c) Electrodialysis
(d) Uhrafilteration
Answer:
(b) Peptisation

Question 88.
The process of conversion of colloidal solution into precipitate is known as …………..
(a) peptisation
(b) dispersion
(c) coagulation
(d) decomposition
Answer:
(c) coagulation

Question 89.
Which one of the following is named collodion?
(a) 4% solution of nitro cellulose in a mixture of alcohol and water
(b) 40% solution of cellulose acetate in acetic acid.
(c) agar-agar along with gel
(d) semipermeable membrane
Answer:
(a) 4% solution of nitro cellulose in a mixture of alcohol and water

Question 90.
Which of the following is the size of the colloidal particle?
(a) 100 μm diameter – 1000 μm diameter
(b) 1 mμ to 1 μm diameter
(c) 1 mμ to 100 μm diameter
(d) 1 μm to 1 μm diameter
Answer:
(b) 1 mμ to 1 μm diameter

Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry

Question 91.
Consider the following statements.
(i) Colloidal solutions are quite stable and are not affected by gravity
(ii) Colloids diffuse more readily through membranes
(iii) Colloidal solutions show colligative properties

Which of the above statement is / are correct?
(a) (i) & (iii)
(b) (ii) & (iii)
(c) (ii) only
(d) (iii) only
Answer:
(a) (i) & (iii)

Question 92.
The shape of tungstic acid W3O5 sol is ………….
(a) spherical
(b) disc
(c) plate-like
(d) rod like
Answer:
(d) rod like

Question 93.
Which one of the following colloids has spherical shape?
(a) AS2S3
(b) Fe(OH)3
(c) W3O5
(d) dust
Answer:
(a) AS2S3

Question 94.
Tyndall effect is possible in colloid due to ……………
(a) absorption of light
(b) adsorption of light
(c) scattering of light
(d) reflection of light
Answer:
(c) scattering of light

Question 95.
Which one of the following does not show Tyndali effect and Brownian movement?
(a) Milk
(b) common salt solution
(c) smoke
(d) tungstic acid sol
Answer:
(b) common salt solution

Question 96.
The migration of sol particles under the influence of electric field is called ……………
(a) electro osmosis
(b) electro dialysis
(c) electrophoresis
(d) dialysis
Answer:
(c) electrophoresis

Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry

Question 97.
Which one of the following is used for detection of pressure of charge on sol particles?
(a) Cataphoresis
(b) Electrodialysis
(c) Dialysis
(d) Ultrafiltration
Answer:
(a) Cataphoresis

Question 98.
Which of the following is positively charged colloid?
(a) haemoglobin
(b) starch
(c) clay
(d) AS2S3
Answer:
(a) haemoglobin

Question 99.
Which one of the following is a positively charged colloid?
(a) Ag
(b) AU
(c) Basic dyes
(d) Clay
Answer:
(c) Basic dyes

Question 100.
Which one of the following is a negatively charged colloid?
(a) Pt
(b) Al(OH)3
(c) Fe (OH)3
(d) Basic dyes
Answer:
(a) Pt

Question 101.
Which one of the following is a negatively charged colloid?
(a) Ferric hydroxide
(b) Clay
(c) Basic dyes
(d) Haemoglobin
Answer:
(b) Clay

Question 102.
The movement of dispersion medium under the influence of electric potential is called ………….
(a) Electrophoresis
(b) Cataphoresis
(c) Electro osmosis
(d) Electro dialysis
Answer:
(c) Electro osmosis

Question 103.
Which one of the following is added to gold sol to protect it?
(a) Gelatine sol
(b) Gum
(c) Starch
(d) Basic dye
Answer:
(a) Gelatine sol

Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry

Question 104.
Consider the following statements.
(i) Smaller the gold number, greater the protective power
(ii) Greater the gold number, greater the protective power
(iii) Colloidal sols with opposite charges are mixed, mutual coagulation takes place.

Which of the above statement is / are not correct?
(a) (i) only
(b) (i) & (iii)
(c) (ii) only
(d) (ii) & (iii)
Answer:
(c) (ii) only

Question 105.
Which one of the following can act as emulsifier?
(a) glue
(b) dye
(c) water
(d) starch
Answer:
(a) glue

Question 106.
Which one of the following is not used to identify the types of emulsion?
(a) dye test
(b) viscosity test
(c) conductivity test
(d) Tollen’s test
Answer:
(d) Tollen’s test

Question 107.
By adding which one of the following oil in water emulsion containing potassium soap can be converted into water in oil emulsion?
(a) MCl3
(b) NaCI
(c) KCI
(d) C6H5Cl
Answer:
(a) MCl3

Question 108.
Which of the following colloid is used as a medicine for stomach troubles?
(a) colloidal Au
(b) colloidal Ca
(c) milk of magnesia
(d) silver sol
Answer:
(c) milk of magnesia

Question 109.
Which one of the following is used in the purification of drinking water?
(a) silver sol protected by gelatine
(b) milk of magnesia
(c) Alum containing Al3+
(d) Argyrol
Answer:
(c) Alum containing Al3+

Question 110.
Which of the following is used as tonics?
(a) milk of magnesia
(b) Argyrol
(c) colloidal Au & colloidal Ca
(d) Alum
Answer:
(c) colloidal Au & colloidal Ca

Question 111.
Which one of the following is used in tanning of leather?
(a) chromium salt
(b) colloidal Au
(c) Argyrol
(d) Fe (OH)3
Answer:
(a) chromium salt

Question 112.
Carbon dust in air is solidified by ………..
(a) cottrell’s precipitator
(b) colloidal mill
(c) Bredig’s arc
(d) peptisation
Answer:
(a) cottrell’s precipitator

Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry

Question 113.
Which of the following voltage is used in cottrell’s precipitator?
(a) 5000 V
(b) 50,000 V
(c) 1,000V
(d) 10,000V
Answer:
(b) 50,000 V

Question 114.
The blue colour of the sky is due to …………
(a) coagulation
(b) peptisation
(c) Tyndall effect
(d) Brownian movement
Answer:
(c) Tyndall effect

Question 115.
Which one of the following is used to distinguish Natural honey and artificial honey?
(a) Ammoniacal AgNO3
(b) Fehling’s solution
(c) Arsenic suiphide sol
(d) gelatin
Answer:
(a) Ammoniacal AgNO3

Question 116.
Which one of the following is the catalyst poison in Haber’s process?
(a) AS2S3
(b) AS2O3
(c) Co
(d) H2S
Answer:
(d) H2S

Question 117.
Which one of the following is an example for water in oil emulsion?
(a) Milk
(b) Vanishing cream
(c) Butter
(d) Soap
Answer:
(c) Butter

Question 118.
Which of the following is contributed towards the extra stability of lyophillic colloid?
(a) Hydration
(b) Charge
(c) Colour
(d) Tyndall effect
Answer:
(a) Hydration

Question 119.
A catalyst is a substance which
(a) increases the equilibrium concentration of the product
(b) changes the equilibrium constant of the reaction
(c) shortens the time to reach equilibrium
(d) supplies energy to the reaction
Answer:
(c) shortens the time to reach equilibrium

Question 120.
The ability of an ion to bring about coagulation of a given colloid depends upon …………
(a) its size
(b) magnitude of its charge
(c) the sign of its charge
(d) both the magnitude and sign of the charge
Answer:
(d) both the magnitude and sign of the charge

Question 121.
Which one of the following is an incorrect statement for physisorption?
(a) It is a reversible process
(b) It requires less heat of adsorption
(c) It requires activation energy
(d) It take place at low temperature
Answer:
(c) It requires activation energy

Question 122.
Which is not a colloid?
(a) Chlorophyll
(b) Egg
(c) Ruby glass
(d) Milk
Answer:
(a) Chlorophyll

Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry

Question 123.
Which of the following electrolytes is most effective in the coagulation of gold sol?
(a) NaNO3
(b) K4[Fe(CN)6]
(c) Na3PO4
(d) MgCl2
Answer:
(b) K4[Fe(CN)6]

Question 124.
Gold number gives ………………..
(a) the amount of gold present in the colloid
(b) the amount of gold required to break the colloid
(c) the amount of gold required to protect the colloid
(d) the measure of protective power of a lyophillic colloid
Answer:
(d) the measure of protective power of a lyophillic colloid

Question 125.
Identify the gas which is readily adsorbed by activated charcoal?
(a) N2
(b) SO2
(c) H2
(d) O2
Answer:
(b) SO2

Question 126.
Starch dispersed in hot water is an example of …………..
(a) emulsion
(b) hydrophobic sol
(c) lyophilic sol
(d) associated colloid
Answer:
(c) lyophilic sol

Question 127.
Which one is an example of gel?
(a) soap
(b) cheese
(c) milk
(d) fog

Question 128.
The random, zig-zag motion of colloidal particles in the dispersion medium is referred to as …………..
(a) Electrophoresis
(b) Brownian movement
(c) Tyndall effect
(d) Electro osmosis
Answer:
(b) Brownian movement

Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry

Question 129.
Which of the following electrolytes is least effective in causing flocculation of ferric hydroxide sol?
(a) K4 [Fe(CN)6]
(b) K2CrO4
(c) KBr
(d) K2SO4
Answer:
(c) KBr

Question 130.
Gelatin is mostly used in making icecream in order to …………..
(a) prevent making of colloid
(b) to stabilize the colloid and to prevent the crystallization
(c) to stabilise the mixture
(d) to enrich the aroma
Answer:
(b) to stabilize the colloid and to prevent the crystallization

Question 131.
Which one of the following is not a colloidal solution?
(a) smoke
(b) ink
(c) air
(d) coffee
Answer:
(c) air

Question 132.
Milk can be preserved by adding a few drops of …………
(a) HCOOH
(b) HCHO
(c) CHCOOH
(d) CH3CHO
Answer:
(b) HCHO

Question 133.
Bleeding is stopped by the application of ferric chloride. This is because …………
(a) ferric chloride seal the blood cells
(b) blood starts flowing in ohter direction
(c) blood is coagulated and blood vessel is sealed
(d) blood is peptised
Answer:
(c) blood is coagulated and blood vessel is sealed

Question 134.
Delta at the rivers are formed due to …………
(a) peptisation
(b) coagulation
(c) hydrolysis
(d) precipitation
Answer:
(b) coagulation

Question 135.
Alum purifies muddy water by …………..
(a) dialysis
(b) adsorption
(c) coagulation
(d) forming a true solution
Answer:
(c) coagulation

Question 136.
Reactions of zeolite catalysts depend on ………….
(a) pores
(b) apertures
(c) size of cavily
(d) all of these
Answer:
(d) all of these

Question 137.
chemisorption ……………
(a) increases with increase in temperature
(b) decreases with increase in temperature
(c) remains unaffected by the, change of temperature
(d) first increases and then decreases.
Answer:
(d) first increases and then decreases.

Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry

Question 138.
Adsorption is always …………
(a) endothermic
(b) exothermic
(c) iso thermic
(d) either a (or) b
Answer:
(a) endothermic

Question 139.
Which one of the following can be explained by the adsorption theory?
(a) Homogeneous catalysis
(b) Acid-base catalysis
(c) Heterogeneous catalysis
(d) Enzyme catalysis
Answer:
(c) Heterogeneous catalysis

Question 140.
Physical adsorption is inversly proportional to ……………..
(a) volume
(b) concentration
(c) temperature
(d) all of these
Answer:
(c) temperature

Question 141.
Noble gases are adsorbed by ……………
(a) anhydrous CaCl2
(b) Fe(OH)3
(c) Conc. H2SO4
(d) activated charcoal
Answer:
(d) activated charcoal

Question 142.
Animal charcoal is used in decolourising agent in the manufacture of sugar because it is a good …………..
(a) adsorbate
(b) adsorbent
(c) oxidising agent
(d) dehydrating agent
Answer:
(a) adsorbate

Question 143.
Gold number is associated only with …………
(a) lyophobic colloids
(b) lyophilic colloids
(c) both lyophobic and lyophilic colloids
(d) Au in water
Answer:
(b) lyophilic colloids

Question 144.
Which of the following forms a colloidal solution with water?
(a) NaCl
(b) Glucose
(c) Starch
(d) Barium sulphate
Answer:
(c) Starch

Question 145.
Which one of the following is an example for homogeneous catalysis?
(a) Hydrogenation of oil
(b) manufacture of NH3 by Haber’s process
(c) manufacture of sulphuric acid by contact process
(d) hydrolysis of sucrose in the presence of dilute hydrochloric acid
Answer:
(d) hydrolysis of sucrose in the presence of dilute hydrochloric acid

Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry

Question 146.
Which of the following does not involve coagulation?
(a) peptisation
(b) formation of delta regions
(c) treatment of drinking water by potash alum
(d) clotting of blood by the use of ferric chloride
Answer:
(a) peptisation

Question 147.
Among the electrolytes Na2SO4, CaCI2, Al2(SO4)3 and NH4Cl, the most effective coagulating agent for Sb2S3 Sol is ……………
(a) Na,SO4
(b) CaCl2
(c) Al2(SO4)3
(d) NH4Cl
Answer:
(c) Al2(SO4)3

Question 148.
Which of the following statement is incorrect regarding physisorption?
(a) It occurs because of Van der Waals forces
(b) more easily liquefiable gases are adsorbed easily
(c) under high pressure, it resuLts into multimolecular layer on adsorbent surface
(d) enthalpy of adsorption is low and positive
Answer:
(d) enthalpy of adsorption is low and positive

Question 149.
Rate of physical adsorption increase with ………..
(a) increase in temperature
(b) decrease in pressure
(c) decrease in temperature
(d) decrease in surface are
Answer:
(c) decrease in temperature

Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry

Question 150.
Gold numbers of protective eolloids, A, B, C and D are respectively 0.50, 0.01, 0.10 and 0.005. The correct order of the stability of colloids is ……..
(a) B < D < A < C
(b) D < A < C < B
(c) C < B < D < A
(d) A < C < B < D
Answer:
(d) A < C < B < D

II. Fill in the blanks

  1. The surface of separation of the two phases where the concentration of adsorbed molecule is high is known as …………..
  2. In adsorption, if the concentration of a substance in the interface is high, it is called ……………
  3. The process of removing a adsorbed substance from the surface is called ……………
  4. Adsorption is always accompanied by decrease in ……………
  5. A term …………… is used for sorption of gases on metal surfaces
  6. M.C. Bain introduced a term …………… to represent the simultaneous adsorption and absorption.
  7. In chemical adsorption, gas molecules are held to the surface by formation of chemical bond and nearly …………… is given out as heat of adsorption
  8. In physical adsorption …………… exist between adsorbent and adsorbate
  9. Heat of adsorption is low hence physical adsorption occurs at ……………
  10. …………… involves the formation of activated complex with appreciable activation energy.
  11. Adsorption occurs at fixed sites called ……………
  12. Gases like NH3, SO3 and CO2 are …………… as have greater Van der Waals force of attraction.
  13. gases like H2, N2, O2 have low critical temperature and ………….. slowly
  14. A plot between the amount of adsorbate adsorbed and pressure or concentration of adsorbate at constant temperature is called …………..
  15. Freundlich adsorption isothermal equation is applicable fo adsorption of …………..
  16. During World War I ………….. gas mask was employed
  17. ………….. is used to create high vaccum in vessels.
  18. In blast furnace ………….. is used for drying air
  19. For dehydration and also purification of gases like CO2, N2, CI2, O2 and He ………….. and are employed.
  20. …………..is employed in the softening of hard water to adsorb Ca2+ and Mg2+ ions
  21. In the process of softening of hard water, exhausted permutit is regenerated by adding a solution of …………..
  22. ………….. are used to demineralise water.
  23. ………….. and ………….. are used in petroleum refining and refining of vegetable oil.
  24. ………….. is used to decolourising agent in manufacture of sugar from molasses.
  25. In forth floation process, the suiphide are particles are wetted by …………..
  26. ………….. is defined as a substance which alters the rate of a chemical reaction without itself undergoing chemical change.
  27. The decomposition of acetaldehyde by I2 catalyst is an example of ………….. catalysis.
  28. Manufacture of sulphuric acid by contact process is an example of ………….. catalysis.
  29. Friedel crafts reaction is an example of ………….. catalysis.
  30. The substances increases the activity of a catalyst are called …………..
  31. ………….. or ………….. used as promoter for iron in Haber’s process
  32. ………….. destroys the activity of platinum in contact process
  33. In the reaction CH3COOC2H5 + H2O → CH3COOH + C2H5OH ………….. is act as auto calyst.
  34. The decomposition of H2O2 rate is decreased by …………..
  35. As …………..is lowered in the presence of catalyst, more molecules take part in the reaction and hence the rate of the reaction increases.
  36. The mechanism of friedel crafts reaction is explained by ………….. theory.
  37. The catalyst used for the oxidation of HCl by air is …………..
  38. Thermal decomposition of KClO3 in the presence of ………….. follows ………….. theory.
  39. Intermediate compound formation theory is unable to explain the mechanism of …………..
  40. Hydrogenation of ethylene in the presence of nickel catalyst follows ………….. theory.
  41. ………….. are complex protein molecules and catalyse the chemical reaction in living organism.
  42. The enzyme ………….. hydrolyses starch into maltose.
  43. The enzyme ………….. oxidises alcohol into acetic acid.
  44. Enzyme catalysed reaction has ………….. rate at optimum temperature.
  45. ………….. in hibits the action of bacteria and used for curing pneumonia.
  46. ………….. are microporous, hydrated alumino silicates.
  47. Zeolites carrying ………….. are used as basic catalysts
  48. Like heterogeneous catalyst ………….. can be recovered and recycled.
  49. Sols of gold, silver, platinum and copper are examples of …………..
  50. So is of protein and starch are examples of …………..
  51. A liquid in liquid colloid is called …………..
  52. Pearls, opals and Ruby red glass are belong the colloid named …………..
  53. Rubber forns colloidal solution with …………..
  54. In colloid mill, the two metal plates rotating in opposite directions of nearly ………….. revolution per minute.
  55. Colloidal solutions of ink and graphite are prepared by ………….. method.
  56. A brown colloidal solution of platinum was prepared by ………….. in 1898.
  57. ………….. is added as an stabilising agent for making platinum colloid.
  58. Metal hydroxide is added as an ………….. for making noble metal sols.
  59. Claus obtained ………….. by subjecting ………….. to high frequency ultrasonic vibrations,
  60. In the preparation of AgCI colloid from AgCl precipitate, the peptising or dispersing agent used is …………..
  61. Gold sol is prepared by reduction of auric chloride using …………..
  62. Arsenic sulphide colloid can be prepared by ………….. method.
  63. 12 sol is obtained from HIO3 by ………….. method.
  64. The process of conversion of colloidal solution into precipitate is called …………..
  65. In the dialysis of kidney. recycling of patient’s blood is done through semipermeable tube in an ………….. solution
  66. Collodion is 4% solution of ………….. in a mixture of …………..
  67. The size of colloidal particles ranges from …………..
  68. The shape of blue gold sol (or) Fe(OH)3 sol is …………..
  69. Pollen grains suspended in water showed …………..
  70. The flocculation and setting down of sol particles is called …………..
  71. When the valency of ion is high ………….. power is increased in colloids.
  72. The smaller the ………….. value, greater will be precipitation of colloids.
  73. ………….. is added to gold sol to protect it.
  74. ………….. introduced the term gold number as a measure of protecting power of a colloid.
  75. An oil in watrer emulsion containing potassium soap as emulsifying agent can be converted into water in oild emulsion by adding ………….. or …………..
  76. Synthetic polymers like polystyrene, silicones and PVC, are …………..
  77. ………….. colloid is used as eye lotion.
  78. ………….. protected by gelatin is known as Argyrol.
  79. ………….. salts arc used in tanning of leather.
  80. Natural honey is distinguished artificial honey by adding …………..

Answer:

  1. interface
  2. positive adsorption
  3. desorption
  4. free energy
  5. occlusion
  6. sorption
  7. 400kJ/rnole
  8. Van der Waals forces of attraction
  9. low temperature
  10. Chemisorption
  11. active centres
  12. easily liquefiable
  13. Permanent, adsorbed
  14. Adsorption isotherm
  15. gases on solid surface
  16. charcoal
  17. Activated charcoal
  18. silica gel
  19. alumina, silica
  20. Permutit
  21. common salt
  22. Con exchange resins
  23. Fuller’s earth, silica gel
  24. Animal charcoal
  25. pine oil
  26. catalyst
  27. homogeneous
  28. heterogeneous
  29. heterogeneous
  30. promoters
  31. Mo (or) Al2O3
  32. AS2O3 catalyst poison
  33. CH3COOH
  34. ethanol (or) Glycerol, negative catalyst
  35. Activation energy
  36. intermediate compound formation
  37. CuCI2
  38. MnO2, intermediate compound formation theory
  39. heterogeneous catalysis
  40. adsorption
  41. enzymes
  42. diastase
  43. micoderma aceti
  44. maximum
  45. penicillin
  46. zeolites
  47. Na ions
  48. nano catalyst
  49. lyophobic sols (or) irreversible sols
  50. lyophilic sols (or) reversible sols
  51. emulsion
  52. solid sol
  53. benzene
  54. 7000
  55. mechanical dispersion
  56. George Bredig
  57. alkali hydroxide
  58. stabilising agent
  59. mercury sol, mercury
  60. HCl
  61. formaldehyde
  62. double decomposition
  63. oxidation
  64. coagulation
  65. isotonic saline
  66. nitro cellulose, alcohol and water
  67. 1 mi to llim diameter
  68. disc or plate like
  69. Brownian movement
  70. coagulation
  71. precipitation
  72. flocculation
  73. geLatine sol
  74. Zsigmondy
  75. CaCl2, AlCl3
  76. colloids
  77. Argyrol
  78. Silver sol
  79. chromium
  80. Tollens reagent (or) Ammoniacal AgNO3

III. Match the column I & II using the code given below the coulum.

Question 1.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-14
Answer:
(a) 4 3 1 2

Question 2.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-15
Answer:
(c) 3 4 1 2

Question 3.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-16
Answer:
(a) 4 3 1 2

Question 4.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-17
Answer:
(a) 2 3 4 1

Question 5.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-19
Answer:
(b) 3 4 2 1

Question 6.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-20
Answer:
(a) 3 1 4 2

Question 7.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-21
Answer:
(a) 4 1 2 3

Question 8.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-23
Answer:
(a) 4 3 2 1

Question 9.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-25
Answer:
(a) 2 4 1 3

Question 10.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-26
Answer:
(a) 4 3 1 2

Question 11.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-27
Answer:
(c) 3 1 4 2

Question 12.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-28
Answer:
(a) 4 3 1 2

Question 13.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-29
Answer:
(d) 3 1 4 2

Question 14.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-30
Answer:
(b) 4 3 2 1

Question 15.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-31
Answer:
(a) 2 4 1 3

Question 16.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-32
Answer:
(b) 3 4 2 1

Question 17.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-33
Answer:
(a) 4 1 2 3

Question 18.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-34
Answer:
(a) 3 1 4 2

IV. Assertion and Reason

Question 1.
Assertion (A) : Absorption is a bulk phenomenon.
Reason (R) : The absorbed molecules are distributed throughout the absorbent.
(a) Both A and R are correct and R is the correct explanation of A.
(b) A is correct but R is wrong.
(c) A is wrong but R is correct.
(d) Both A and R are wrong.
Answer:
(a) Both A and R are correct and R is the correct explanation of A.

Question 2.
Assertion (A) : Adsorption is a spontaneous process.
Reason (R) : Adsorption is always accompanied by decrease in free energy. When molecules are adsorbed, there is always a decrease in randomness of the molecules.
(a) Both A and R are correct and R is the correct explanation of A.
(b) Both A and R are correct but R is not correct explanation of A.
(c) Both A and R are wrong.
(d) A is correct but R is wrong.
Answer:
(a) Both A and R are correct and R is the correct explanation of A.

Question 3.
Assertion (A) : Chemical adsorption is an exothermic process.
Reason (R) : In chemical adsorption, gas molecules are held to the surface by formation of chemical bonds. Since strong bond is formed, nearly 400kJ/mole is given out as heat adsorption.
(a) Both A and R are correct and R is the correct explanation of A.
(b) Both A and R are correct but R is not correct explanation of A.
(c) Both A and R are wrong.
(d) A is correct but R is wrong.
Answer:
(a) Both A and R are correct and R is the correct explanation of A.

Question 4.
Assertion (A) : Physical adsorption occurs at low temperature.
Reason (R) : ¡n physical adsorption, weak Van der Waals force of attraction exist. Other weak forces exist in physical adsorption are dipole-dipole interaction and dispersion forces. As these forces are weak, heat of adsorption is low.
(a) Both A and R are correct but R is not correct explanation of A.
(b) Both A and R are correct and R is the correct explanation of A.
(c) Both A and R are wrong.
(d) A is correct but R is wrong.
Answer:
(b) Both A and R are correct and R is the correct explanation of A.

Question 5.
Assertion (A) : Platinised asbestos is a better adsorbent than platinum block.
Reason (R) : Higher the surface area, higher is the amount adsorbed. In platinum coated asbestos the surface area is more and so it act as a better adsorbent.
(a) Both A and R are correct and R is the correct explanation of A.
(b) Both A and R are wrong.
(c) A is correct but R is wrong.
(d) A is wrong but R is correct.
Answer:
(a) Both A and R are correct and R is the correct explanation of A.

Question 6.
Assertion (A) : Gases like SO2, NH3 and CO2 are readily adsorbed.
Reason (R) : SO2, NH3 and CO2 are easily liquefiable as have greater van der WaaI’s forces of attraction and adsorbed readily.
(a) Both A and R are correct and R is the correct explanation of A.
(b) Both A and Rare wrong.
(c) A is correct but R is wrong.
(d) A is wrong but R is correct.
Answer:
(a) Both A and R are correct and R is the correct explanation of A.

Question 7.
Assertion (A) : Permanent gases like H2, N2 and O2 cannot be adsorbed readily.
Reason (R) : Permanent gases having low critical temperature and adsorbed slowly.
(a) Both A and R are wrong.
(b) A is correct and R is the correct explanation of A.
(c) A is wrong but R is correct.
(d) A is correct but R is wrong.
Answer:
(b) A is correct and R is the correct explanation of A

Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry

Question 8.
Assertion (A) : Chromatography is a very effective method and used for identification, detection and estimation of micro quantities of many substances.
Reason (R) : Chromatography technique is applied for separation and detection of components in a mixture it is mainly based on adsorption of components on the surface of adsorbents.
(a) Both A and R are correct and R is the correct explanation of A.
(b) Both A and R are correct.
(c) A is correct but R is wrong.
(d) A is wrong but R is correct.
Answer:
(a) Both A and R are correct and R is the correct explanation of A.

Question 9.
Assertion (A) : Ester hydrolysis of acid (or) alkali catalyst is an example of homogeneous catalysis.
Reason (R) : Ester, H2O acid (or) alkali and the products are in liquid form.
(a) Both A and R are correct but R is the correct explanation of A.
(b) Both A and R are wrong.
(c) A is correct but R is wrong.
(d) A is wrong but R is correct,
Answer:
(a) Both A and R are correct but R is the correct explanation of A.

Question 10.
Assertion (A) : The manufacture of sulphuric acid by contact process is an example of heterogeneous catalysis.
Reason (R) : The catalyst Pt (or) VO5, reactants and products are in different phases in contact process.
(a) Both A and R are correct and R is the correct explanation of A.
(b) Both A and R are correct but R is not correct explanation of A.
(c) Both A and R are wrong.
(d) A is wrong but R is correct.
Answer:
(a) Both A and R are correct and R is the correct explanation of A.

Question 11.
Assertion (A) : Acid hydrolysis of ethylacetate by water to produce acetic acid and ethanol is an example of auto catalysis.
Reason (R) : In acid hydrolysis of ester, the product acetic acid act as catalyst and this process is called autocatalysis.
(a) Both A and R are correct and R is the correct explanation of A.
(b) Both A and R are wrong.
(c) A is correct hut R is wrong.
(d) A is wrong but R is correct.
Answer:
(a) Both A and R are correct and R is the correct explanation of A.

Question 12.
Assertion (A) : Effective and efficient conversion is the special characteristic of enzyme catalysed reactions.
Reason (R) : An enzyme may transform a million molecules of reactants to products in a minute.
(a) Both A and R are correct and R is the correct explanation of A.
(b) Both A and R are correct but R is not correct explanation of A.
(c) A is correct but R is wrong.
(d) A is wrong but R is correct.
Answer:
(a) Both A and R are correct and R is the correct explanation of A.

Question 13.
Assertion (A) : lyophillic colloids will not get precipitated easily.
Reason (R) : In lyophillic colloids, definite attractive forces exists between dispersion medium and dispersed phase and they are more stable.
(a) Both A and R are correct and R is the correct explanation of A.
(b) Both A and R are correct but R is not correct explanation of A.
(c) Both A and R are wrong.
(d) A is correct but R is wrong.
Answer:
(a) Both A and R are correct and R is the correct explanation of A.

Question 14.
Assertion (A) : Lyophobic colloids like sols of gold will precipitate readily.
Reason (R) : In lyophobic colloids, no attractive force exists between the dispersed phase and dispersion medium and are less stable.
(a) Both A and R are correct and R is the correct explanation of A.
(b) Both A and R are correct but R is not correct explanation of A.
(c) Both A and R are correct.
(d) A is correct but R is wrong.
Answer:
(a) Both A and R are correct and R is the correct explanation of A.

Question 15.
AssertIon (A) : Iron colloid cannot be prepared by Bredig’s arc method.
Reason (R) : iron cannot react with alkali hydroxide stabilising agent added in water.
(a) Both A and R are correct and R is the correct explanation of A.
(b) A is correct but R is wrong.
(c) A is wrong but R is correct.
(d) Both A and R are wrong.
Answer:
(a) Both A and R are correct and R is the correct explanation of A.

V. Find the odd one out.

Question 1.
He, Ne, O2, N2, Pt
Answer:
Pt. It is adsorbent and all others are adsorbates.

Question 2.
SO2, NH3, NaCI, Silica gel
Answer:
Silica gel. It is an adsorbent whereas others are adsorbates.

Question 3.
Silica gel, Pt, Ag, Pd, NH3
Answer:
NH3 It is an adsorbate whereas others are adsorbents.

Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry

Question 4.
Coconut charcoal, silica gel, mica, SO2, Animal charcoal
Answer:
SO2 It is an adsorbate whereas others are adsorbents.

Question 5.
Mica, Nickel, Charcoal, Tungsten, Ethyl alcohol vapours
Answer:
Ethyl alcohol vapours. It is an adsorbate whereas others are adsorbents.

Question 6.
Pt, Glycerol, MnO2, Ni, I2
Answer:
Glycerol. It is a negative catalyst whereas others are positive catalyst.

Question 7.
Fe, Anhydrous AlCl3, V2O5, Pt, Ethyl alcohol
Answer:
Ethyl alcohol is a negative catalyst whereas others are positive catalyst.

Question 8.
Decomposion of acetaldehyde by I2 Decomposition of H2O2 by Pt, Ester hydrolysis with acid, Hydrolysis of cane sugar.
Answer:
Decomposition of H2O2 by Pt. It is heterogeneous catalysis whereas others are homogeneous catalysis.

Question 9.
Friedel crafts reactopm, Haner’s process, Hydrolysis of cane sugar, contact process.
Answer:
Hydrolysis of cane sugar. It is homogeneous catalysis whereas others are heterogeneous catalysis.

Question 10.
Pepsin, Zymase, Maltase, Diastase, Maltose, Urease.
Answer:
Maltose. It is a carbohydrate whereas others are enzyme catalysts.

Question 11.
Pt, Ni, Fe°/Pd°, Fe/Mo, Sn/HCI
Answer:
Fe°/Pd°. It is a nano catalyst where as others ordinary catalyst.

Question 12.
Milk, coffee, smoke, common salt solution, dust.
Answer:
Common salt solution. It is a true solution whereas others are colloids.

Question 13.
Soda water, Butter, Starch solution, Cheese, Cream.
Answer:
Starch solution. It is a suspension whereas others are colloids.

Question 14.
Ink, Milk, Cream, Mayonnaise.
Answer:
Ink. It is a sol whereas others are Emulsion.

Question 15.
Pearls, Opals, Coloured glass, Alloys, Pumice stone.
Answer:
Pumice stone. It is a solid foam whereas others are solid sol.

Question 16.
Smoke, Froth, Fumes, Dust, Air pollutants.
Answer:
Froth. It is a foam whereas others are solid aerosol.

Question 17.
Pumice stone, Foam, Milk, Rubber band.
Answer:
Milk. It is an emulsion whereas others are solid foam.

Question 18.
Mechanical dispersion, Bredigts arc method, Peptisation, Double decomposition, Ultrasonic dispersion.
Answer:
Double decomposition. It is a condensation method of preparation of colloids whereas others are dispersion methods of preparation of colloids.

Question 19.
Oxidation, Peptisation, Reduction, Decomposition, Hydrolysis.
Answer:
Peptisation. It is a dispersion method of preparation ofcoiloid whereas others are condensation methods of preparation of colloids.

Question 20.
Dialysis, electrophoresis, Ultrafi iteration. Electrodialysis.
Answer:
Electrophoresis is an electrical property of colloids but others are purification methods of colloids.

Samacheer Kalvi 12th Chemistry Surface Chemistry 2 Marks Questions and Answers

Question 1.
Define

  1. Adsorbent
  2. Adsorbate with an example.

Answer:

  1. Adsorbent is a material on which adsorption takes place. eg., silica gel and metals like Ni, Cu, Pt.
  2. Adsorbate is a substance which is adsorbed on the adsorbent. e.g., Gaseous molecules like He, Ne, O2, N2 and solutions of NaCI (or) KCI.

Question 2.
Define

  1. Interface
  2. Desorption.

Answer:
1. Interface:
The surface of separation of the two phases where the concentration of adsorbed molecule is high is known as interface.

2. Desorption:
The process of removing an adsorbed substance from the substance is called desorption.

Question 3.
What is adsorption? What is meant by positive and negative adsorption?
Answer:
Adsorption is the odhesion of atoms, ions or molecules from a gas, liquid or dissolved solid to a surface. In adsorption, if the concentration of a substance in the interface is high, then it is called positive adsorption. 1f it is less, then it is called negative adsorption.

Question 4.
Define chemical adsorption? Give example.
Answer:
Chemical adsorption is a process in which gas molecules are held to the surface by formation of chemical bonds. e.g., Adsorption of O2 on tungsten, Adsorption of H2 on nickel.

Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry

Question 5.
Chemical adsorption is an exothermic process. Justify this statement.
Answer:
In chemical adsorption, gas molecules are held to the surface by the formation of strong bond. Due to this 400kJ/mole is given out as heat of adsorption. Since heat is evolved, it is an exothermic process.

Question 6.
Why physical adsorption take place at low temperature?
Answer:
In physical adsorption, the forces between the adsorbent and adsorbate are very weak, heat of adsorption is low and hence physical adsorption occurs at low temperature.

Question 7.
What are the forces exist in physical adsorption?
Answer:
In physical adsorption, physical forces like van der Waals force of attraction exist between adsorbent and adsorbate.
The other forces that can cause physical adsorption are

  1. dipole-dipole interaction
  2. dispersion forces.

Question 8.
What is physical adsorption? Give example.
Answer:
Physcial adsorption is a process in which adsorbate are attached to adsorbent by means of weak van der Waals forces of attraction. e.g.,

  1. Adsorption of N2 on mica
  2. Adsorption of gases on charcoal.

Question 9.
Finely divided Nickel is a better adsorbent than Nickel crystal.
Answer:
As the adsorption is a surface phenomenon it depends on the surface area of adsorbent. i.e., higher the surface area, higher is the amount adsorbed. Finely divided Nickel has longer surface area and it is a better adsorbent than Nickel crystal.

Question 10.
NH3, CO2 are readily adsorbed where as H2, N2 are slowly adsorbed. Give reason.
Answer:
1. The nature of adsorbate can influence the adsorption. Gases like NH3 CO2 are easily liquefiable as have greater van der Waals forces of attraction and hence readily adsorbed due to high critical temperature.

2. But permanent gases like H2, N2 can not be easily liquefied and having low critical temperature and adsorbed slowly.

Question 11.
What is meant by adsorption isotherm?
Answer:
A plot between the amount of adsorbate adsorbed and pressure or concentration of adsorbate at constant temperature is called adsorption isotherm.

Question 12.
Mention Freundlinch adsorption isothermal equation.
Answer:
Freundlinch adsorption isotherm
\(\frac{x}{m}=k p^{1 / n}\)
where x is the amount of adsorbate or adsorbed on ‘m’ gm of adsorbent at a pressure of p. k and n arc constants.

Question 13.
What are the limitations of Freundlich adsorption isotherm?
Answer:

  1. The Freundlich adsorption isotherm is purely empirical and valid over a limited pressure range.
  2. The values of constants ‘k’ and ‘n’ also found vary with temperature. No theoritical explanations were given.

Question 14.
How is adsorption applied in the decolourisation of sugar?
Answer:
Sugar prepared from molasses is decolourised to remove coloured impurities. Animcal charcoal is applied to adsorb colouring impurities and sugar is purified.

Question 15.
What is chromatography?
Answer:
1. The chromatographic technique is a process of separation of components in a mixture which is mainly based on adsorption of components on the surface of adsorbents.

2. This method is very effective and used for identification, detection and estimation of many substances even if they are contained in micro quantities.

Question 16.
Explain the application of adsorption ¡n qualitative analysis with an example.
Answer:
1. In the identification of Al3+ ion in Al(OH)3 is obtained as blue lake. Due to adsorption of blue litmus solution added to the aluminium ion solution.

2. A red colouration is seen due to the acidic nature of the solution. Ammonium hydroxide is added to it until the bkie colour develops.

3. During this addition, Al(OH)3 precipitate is formed with blue colour due to the adsorption of litmus compound.

Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry

Question 17.
Define catalyst. Give example.
Answer:
A catalyst is defined as a substance which alter the rate of chemical reaction without itself undergoing chemical change. The phenomenon which involves the action of a catalyst is called catalysis.
e.g.,Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-35

Question 18.
What is homogeneous catalysis? Give example.
Answer:
Homogeneous catalysis is a process in which the reaction, products and catalyst are present in the same phase. NO
e.g. Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-36
In the above reaction, the catalyst NO, the reactants SO2, O2 and product SO3 are present in the gaseous form

Question 19.
What is heterogeneous catalysis? Give example.
Answer:
Heterogeneous catalysis is a process in which the reactants, products and catalyst are present in different phases.
e.g. Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-37

Question 20.
What are promoters? Explain with example.
Answer:
1. In a catalysed reaction, the presence of a certain substance increases the activity of a catalyst. Such substance is called a promoter.

2. For example, in Haber’s process of manufacture of ammonia, the activity of the iron catalyst is increased by the presence of molybdenum. Hence molybdenum is called a promoter.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-38

Question 21.
What is meant by catalyst poison?
Answer:
1. The substances when added to a catalysed reaction decreases or completely destroys the activity of a catalyst are often known as catalytic poisons.

2. In the reaction 2SO2 + O2 → 2SO3 with Pt catalyst, the catalyst poison is AS2O3.

Question 22.
Which is the catalyst and catalyst poison In Haber’s process?
Answer:
In the Haber’s process of the manufacture of ammonia, the Fe is catalyst and it is poisoned by the presence of H2S.
3H2 + N2 → 2NH3 Fe – catalyst, H2S – Catalyst poison

23. In the reaction 2H2 + O2 → 2H2O, which ¡s the catalyst and catalyst poison?
Answer:

  1. Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-39 Pt – catalyst
  2. CO act as catalytic poison for Pt catalyst.

Question 24.
Explain the relation between activation energy and the rate of the reaction using catalyst.
Answer:
1. For a chemical reaction to occur, the reactants are to be activated to form a activated complex. The energy required for the reactants to reach the activated complex is called the activation energy. The activation energy can be decreased by increasing the reaction temperature.

2. In the presence of catalyst, the reactants are activated at reduced temperatures. i.e., the activation energy is lowered. The catalyst adsorbs the reactants activates them by weakening the bonds and allow them to react to form products.

3. As activation energy is lowered in the presence of a catalyst, more molecules take part in the reaction and hence the rate of the reaction increases.

Question 25.
What are the merits and limitations of the intermediate compound theory?
Answer:
Merits:

  1. The specificity of a catalyst
  2. The mercase in the rate of the reaction with increase in the concentration of catalyst.

Limitations:

  1. This theory fails to explain the action of catalytic poison and promoters.
  2. This theory is unable to explain the mechanism of heterogeneous catalysed reactions.

Question 26.
What are active centres?
Answer:

  1. The surfiice of a catalyst is not smooth. It bears steps, cracks and corners. Hence the atoms on such locations of the surface are coordinatively unsaturated.
  2. The have residual force of attraction. Such sites are called active centres. So the surface cames high surface free energy.

Question 27.
Enzyme catalysis are more effective and efficient than ordinary catalysis. Prove this statement.
Answer:
1. Effective and efficient conversion ¡s the special chareteristic of enzyme catalysed reactions. An enzyme may transform a million molecules of reactant to product in a minute.
For e.g. 2H2O2 → 2H2O + O2

2. For this reaction colloidal platinum as a catalyst the activation energy is 11 .7 k.cal/mole. But with the enzyme catalyst the activation energy of this reaction is less than 2k.cal/mole.

Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry

Question 28.
Enzyme catalysis is highly specific in nature. Justify this statement.
Answer:
1. Urea can be hydrolysed to urea by the enzyme urease.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-40
The catalyst mease (enzyme) catalyses urea but it does not hydrolyse methyl urea.

Question 29.
Enzyme catalysed reaction has maximum rate at optimum temperature. Prove it.
Answer:
1. At first rate of reaction increases with the increase of temperature, but above a particular temperature, the activity of enzyme is destroyed. The rate may even drop to zero. The temperature at which enzymic activity in high or maximum is called optimum temperature.

2. For e.g., enzymes involved in human body have an optimum temperature 37°C/98°F.

Question 30.
What are the types of colloids based on dispersion medium?
Answer:

1. If the dispersion medium considered as water, then the colloids are referred as hydrosols (or) aquasols.
2. If the dispersion medium is an alcohol, the colloid is termed as alcosol. and if benzene is the dispersion medium, it is called benzosol.

Question 31.
Explain about

  1. Liquid aerosol
  2. solid aerosol with example.

Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-41

Question 32.
Explain oxidation method of preparation of colloids with two examples.
Answer:
Oxidation:
1. When hydrolodic acid is treated with iodic acid 12 sol is obtained.
HIO3 + 5HI → 3H2O + 3I2 (sol)

2. When O2 is passed through H2Se, a sol of selenium is obtained.
H2Se + O2 → 2H2O + Se (sol)

Question 33.
Explain the method of preparation of gold sol by reduction method.
Answer:
Gold sol is prepared by the reduction of auric chloride using formaldehyde.
2AuCl3 + 3HCHO + 3H2O → 2Au (sol) + 6HCI + 3HCOOH

Question 34.
How would you prepare ferric hydroxide sol by hydrolysis method?
Answer:
Fe(OH)3 sol can be preparcd by the hydrolysis of FeCl3
FeCI3 + 3H2O → Fe(OH)3 (soL) + 3 HCI

Question 35.
How would you prepare colloid by the exchange of solvent method?
Answer:
1. Colloidal solution of few substances like phosphorous or sulphur is obtained by the solutions in alcohol and pouring them into water.
2. As they are insoluble in water, they form colloidal solution.
P in alcohol + water → Psol.

Question 36.
Why colloids are to be purified? If not what will happen?
Answer:

  1. The colloidal solutions due to their different methods ofpreparation may contain impurities.
  2. If they are not removed, they may destabilise and precipitate the colloidal solution. This is called coagulation.
  3. Hence the impurities mainly electrolytes should be removed to increase the stabilisation of colloid.

Question 37.
How is human kidney dialysis take place?
Answer:

  1. Kidney malfunction results in the building up of electrolyte concentration within the blood to toxic levels.
  2. In the dialysis. recycling of patient’s blood is done through considerable length of semipermeable tube in an isotomic saline solution.

Question 38.
Write a note about Helmoholtz double laver.
Answer:
The surface of a colloidal particle adsorbs one type of ion due to preferential adsorption. This Stem layer
layer attracts the oppositely charged ions in the. medium and hence the boundary separating the Slpping plane two electrical double layers are set up. This is called Helmholtz electrical double layer. As the particles nearby are having similar charges. they cannot come close and condense. Hence this helps to explain the stability of the colloid.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-42

Question 39.
What is meant by gold number?
Answer:

  1. Zsigmondy introduced the term “Gold number” as a measure of protective power of a colloid.
  2. Gold number is defined as the number of milligrams of hydrophilic colloid that will just prevent the precipitation of 10 ml of gold sol on the addition of 1 ml of 10% NaCI solution.

Question 40.
Potato starch is less stable than gelatin. Why?
Answer:

  1. The gold number of gelatin is 0.005 – 1 and the gold number of potato starch is 25.
  2. Smaller the gold number, greater the protective power.

Question 41.
Write a note about Cortrell’s precipitator.
Answer:
Carbon dust in air is solidified by Cortrell’s precipitator. In it. a high potential difference of about 50.000V is used. The change on carbon is neutralised and solidified. Thus the air is free from carbon particles.

Question 42.
Explain about

  1. blue colour of the sky
  2. formation of delta.

Answer:

  1. The blue colour of the sky in nature is due to Tyndall effect of air particles.
  2. The electrolytes in sea and river water coagulate the solid particles in river water at their intersection. By this way delta is formed.

Question 43.
Distinguish between the meaning of the terms adsorption and absorption. Give one example each.
Answer:
Adsorption

  1. Substance is concentrated only at the surface and does notpcnctrate from the surface to the bulkofthe adsorbent.
  2. It is a surface phenomenon.
    Example: Silical gel absorbs water vapour on its surface.

Absorption

  1. Substnce is uniformly distributed throughout the bulk of the solid.
  2. It is a bulk phenomenon.
    Example: Anhydrous CaCl, absorbs water vapours in it.

Question 44.
Explain the following term giving a suitable example, emulsification.
Answer:
The process of making emulsion using a mixture of two immiscible or partially miscible liquids is called emulsification. For example. Cod liver oil is an emulsion made up of water in oil.

Question 45.
What is the reason for the stability of colloidal sols?
Answer:
Reason for the stability of colloidal sols are:

  1. Coagulation of the colloidal sol is prevented because of the presence of equal and similar charges on the colloidal particles.
  2. Colloidal particles are covered by a sheath of liquid in which they are extensively solvated because of which they acquire stability.

Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry

Question 46.
Dialysis is a method of purification of sols. But prolonged dialysis of the sol makes it unstable. Why?
Answer:
Traces of electrolytes in the sol, impart charge to the dispersed phase particles making it stable. Prolonged dialysis removes all the electrolytes, thus making the sol unstable.

Question 47.
Why the sun looks red at the time of setting? Explain on the basis of colloidal properties.
Answer:
At the time of setting, the sun is at the horizon. The light emitted by the sun has to travel a longer distance. As a result the blue part of the light is scattered away by the dust particles in the atmosphere. Hence the red part is visible.

Question 48.
What are emulsions? What are their different types? Give one example of each type.
Answer:
An emulsion is a colloidal dispersion in which both the dispersed phase and the dispersion medium are liquids and the two liquids involved are otherwise immiscible.
1. Oil in water type, in which oil is a dispersed phase and water is the dispersion medium, For example: milk is as emulsion of liquid fat dispersed in water.

2. Water in oil type. in which water is the dispersed phase and oil is thedispersion medium. For example: cod liver oil is an emulsion of water in oil in which water is the dispersed phase and oil is the dispersion medium.

Question 49.
How does chemical adsorption of a gas on a solid vary with temperature?
Answer:
Chemical adsorption first increases with increase in temperature and then it decreases, if the pressure remains constant.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-43

Question 50.
What are lyophilic and lyophobic sols? Give one example of each type. Why are hydrophobic sols easily coagulated?
Answer:
Lyophilic Sols:
Colloidal sols directly formed by rnixing substances like gums, gelatin, starch, rubber. etc. with a suitable liquid (The dispersion medium) are lyophilic sols. An important characteristic of these sols is that if the dispersion medium is separated from the dispersed phase

(say by evaporation) the sol can be reconstituted by simply remixing with the dispersion medium. That is why these sols are also called reversible sols. These sols are quite stable and cannot be easily coagulated.

Lyophobic sols:
These colloidal sols can only be prepared by some special methods. These sols are readily precipitated on the addition of small amount of electrolytes, by heating or by shaking and hence are not stable.

Hydrophobic sols are water hating. They are formed by indirect method. These sols are irreversible sols. These sols are readily precipitated by the addition of small amount of electrolytes, by heating or by shaking and hence are not stable.

Samacheer Kalvi 12th Chemistry Surface Chemistry 3 Marks Questions and Answers

Question 1.
What are the characteristics of adsorption?
Answer:
1. Adsorption can occur in all interfacial faces i.e., the adsorption can occur in between gas – solid, liquid – solid, liquid – liquid, solid – solid and gas – liquid.

2. Adsorption is always accompanied by decrease in free energy. When ∆G reaches zero, the equilibrium is attained.

3. Adsorption is a spontaneous process.

4. When molecules are getting adsorbed, there is always decrease in randomness of the molecules.
∆G = ∆H – T∆S where ∆G = change in free energy
∆H change in enthalpy
∆S change in entropy
∆H = ∆G + T∆S

5. Adsorption is exothermic and it is a quick process.

6. If simultaneous adsorption and absorption take place, it is termed as ‘sorption and sorption of gases on metal surface is called occlusion.

Question 2.
Explain graphical representation of chemical adsorption and physical adsorption.
Answer:
1. Adsorption isotherms represents the variation of adsorption at constant temperature.

2. When amount ?f adsorption is plotted versus temperature at constant pressure is called adsorption iso bar
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-44
3. In physical adsorption x/m decreases with increase in T. But in chemical adsorption x/m increases with rise in temperature and then decreases. The increase illustrate the requirement of activation of the surface for adsorption is due to the fact that formation of activated complex require certain energy. The decrease at high temperature is due to desorption, as the kinetic energy of the adsorbate increases.

Question 3.
Write any 3 applications of adsorption.
Answer:

  1. Gas masks: During World War I. charcoal gas mask was employed.
  2. To create high vaccum in vessels, activated charcoal is used. For dehydration and purification of gases like CO2, N2, O2 and He, alumina and silica gel are employed.
  3. In blast furnace, silica gel is used for drying air.

Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry

Question 4.
Explain the function of permutit in the softening of hard water.
Answer:
1. Permutit is employed for softening 6 hard water. It adsorbs Ca2 and Mg2 ions in its surface, there is an ion exchange as shown below it occurs on the surface.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-45

2. Exhausted permutit is generated by adding a solution of commonsalt.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-46

Question 5.
Explain about the application of ion exchange resins In adsorption.
Answer:

  1. Ion exchange resins are working only based on the process of adsorption.
  2. Ion exchange resins are used to demmeralise water. This process is carried out by passing water through two columns of cation and anion exchange resins.

Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-47

Question 6.
What is catalysis? Explain with two examples.
Answer:
Catalysis is a process in which a chemical reaction takes place in a faster rate with the help of catalyst.
Haber’s process:
N2 + 3H2 → 2NH3 In this process. Fe is the catalyst and Mo is the promoter.
The surface of Fe catalyses the reaction.

Vanaspathi preparation:
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-48
Nickel surface catalyses the reaction.

Question 7.
In the following fields, how adsorption ¡s applied?

  1. Medicine
  2. Metallurgy
  3. Mordant & Dyes
  4. indicators

Answer:
1. Medicine:
Drugs cure diseases by adsorption on body tissues.

2. Metallurgy:
Sulphide ores are concentrated by a process called froth floation in which lighter ore particles are adsorbed by pine oil. .

3. Mordants and Dyes:
Most of the dyes are adsorbed on the surface of the fabric. Mordants are the substances used for fixing dyes onto the fabric.

4. In the precipitation titrations, the end point is indicated by an external indicator which changes its colour aller getting absorbed on precipitate. It is used to indicate the end point of filtration.

Question 8.
Differentiate homogeneous and heterogeneous catalysis.
Answer:
Homogeneous catalysis

  1. The reaction in which, the reactants, products and catalyst are present in the same phase, is called homogeneous catalysis.
  2. Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-49
  3. Homogeneous catalysis mechanism is ex plained by intermediate compound formation theory.

Heterogeneous catalysis

  1. The reaction in which the reactants, products and catalyst are present in different phases is called heterogeneous calalysis.
  2. Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-50
  3. Heterogeneous catalysis mechanism is explained by adsorption theory.

Question 9.
Give three examples for homogeneous catalysis.
Answer:
1. Decomposition of acetaldehyde by I2 catalyst.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-51

2. Hydrolysis of cane sugar with mineral acid
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-52

3. Acid hydrolysis of an ester
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-53

Question 10.
Give three examples for heterogeneous catalysis
Answer:
1. hydrogenation of ethylene in the presence of Nickel
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-54

2. Decomposition of H2O2 with Pt catalyst
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-55

3. Friedel crafts reaction
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-56

Question 11.
What is auto ctalysis? Give two examples
Answer:
The process in which one of the product formed act as a catalyst is termed auto catalysis
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-57
Acetic acid acts as auto catalyst.
Decomposition of Arsenic hydride
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-58
Arsenic act as auto catalyst.

Question 12.
What is negative catalysis? Explain with example.
Answer:

1. In certain reactions, presence of certain substances decreases the rate of the reaction. Such substances are called negative catalyst and the process is called negative catalysis.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-59

2. In oxidation of chloroform, ethanol decreases the rate of the reaction and ethanol act as negative catalyst.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-60

Question 13.
Explain the formation of water with copper catalyst by Intermediate compound formation theory.
Answer:
1. Formation of water due o the reaction of H2 and O2 in the presence of Cu can be given as

2. Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-61

3. Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-62

Question 14.
Explain the mechanism of oxidation of HCl by air in the presence of CuCl2
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-63

Question 15.
Explain the thermal decomposition of potssium chlorate by intermediate compound formation theory.
Answer:
1. Thermal decomposition of KClO3 in the presence of MnO2 proceeds as follows.

2. Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-64

Question 16.
Describe the action of active centres present in the catalyst.
Answer:
1. Active centres increases the rate of the reaction by adsorbing and activating the reactants.

2. Increase in the activity of a catalyst by increasing the surface area. Increase in the surface area of metals and metal oxides by reducing the particle size increases the rate of the reaction.

3. The action of catalytic poison occurs when the poison blocks the active centres of the catalyst.

4. A promoter (or) activator increases the number of active centres on the surface.

Question 17.
Write a note about nano catalyst.
Answer:

  1. Nano materials such as metallic nano particles, metal oxide are used as catalyst in many chemical transformation.
  2. Nano catalysts carry the advantages of both homogeneous and heterogeneous catalysis.
  3. Like homogeneous catalyst the nino catalyst give 100% selective transformations and excellent yield and show extremely high reactivity.
  4. Like the heterogeneous catalyst, nano catalysts can be recovered and recycled.
  5. Nano catalysts are actually soluble heterogeneous catalyst.Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-65

Question 18.
Differentiate lyophillic and lyophobic colloids
Answer:
Lyophiltic colloids

  1. In lyophillic colloids (or) sols definite attractive force (or) affinity exists between dispersion medium and dispersed phase.
  2. e.g., Sols of protein, starch
  3. They are more stable and will not get precipitated readily.
  4. They can be brought back to colloidal solution even after the precipitation by addition of dispersion medium
  5. They arc called reversible sols

Lyophobic colloids

  1. In lyophobic colloid no attractive force exists between dispersed phase and dispersed phase and dispersion medium.
  2. e.g., sols of gold, silver
  3. They are less stable and get precipitated readily
  4. They cannot be produced again by adding the dispersion medium.
  5. They are called irreversible sols.

Question 19.
Explain about dispersion mediu, dispersed phase and example of

  1. foam,
  2. emulsion
  3. sol

Answer:

Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-66

Question 20.
Explain about dispersion medium, dispersed phase and example of

  1. solid foam
  2. Gel
  3. Solid sol.

Answer:

Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-67

Question 21.
How would you prepare colloids of ink and graphite? (OR)
Answer:
Explain about mechanical ‘dispersion method.
1. Using a colloid mill, the solid is ground to colloidal dimension. Coarse dispersion The colloid mill consists of two metal plates rotating in opposite directions at very high speed of nearly 7000 revolution per minute.

2. The colloidal particles of required colloidal size is obtained by adjusting the distance between the two plates.

3. By this method, colloidal solutions of ink and graphite and prepared.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-68

Question 22.
Explain about Bredic’s arc method (or) Electro dispersion method (or) How would you prepare colloids of noble metals?
Answer:
1. An electrical arc is struck between electrodes dispersed in water surrounded by ice. When a current of I amp! 100V is passed, an arc produced forms vapours of metal which immediately condense to from colloidal solution.

2. By this method, colloidal solution of many metals like copper, silver, gold. platinum can be prepared.

3. Alkali hydroxide is an added an stabilising agent for the colloid solution.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-69

Question 23.
Explain about ultrasonic dispersion. (or) How would you prepare mercury colloid?
Answer:
1. Sound waves of frequency more than 20kHz (audible limit) could cause transformation of coarse suspension to colloidal solution.

2. Claus obtained mercury sol by subjecting mercury to sufficiently high frequency ultrasonic vibrations.

3. The ultrasonic vibrations produced by generator spread the oil and transfer the vibration to the vessel with mercury in water.

Question 24.
Explain the methods of preparation of colloids of

  1. AS2S3
  2. S.

Answer:
1. Double decomposition:
When hydrogen suiphide gas is passed through a solution of arsenic oxide, a yellow coloured arsenic sulphide is obtained as a colloidal solution.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-70

2. Decomposition:
When few drops of an acid is added to a dilute solution of sodium thio sulphate, the insoluble free sulphur produced by the decomposition of sodium thiosuiphate accumulates into clusters which impart various colours blue, yellow and even red to the system depending on their growth within the size of colloidal dimensions.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-71

Question 25.
Describe about

  1. Dialysis
  2. Electro dialysis.

Answer:
1. Dialysis:
Thomas Graham separated the electrolyte from a colloid using a semipermeable membrane. In this method, the colloidal solution is taken in a bag made up of semipermeable membrane. It is suspended in a trough of flowing water, the electrolytes diffuse out of the membrane and they are carried away by water. membrane and they are carried away by water.

2. Electro dialysis:
The presence of electric field increases the speed of removal of electrolytes from colloidal solution. The colloidal solution containing an electrolyte as impurity is placed between two dialysing membranes enclosed into two compartments filled with water.

When current is passed, the impurities pass into water compartment and get removed periodically. This process is faster than dialysis, as the rate of diffusion of electrolytes is increased by The application of electricity.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-72

Question 26.
Explain about ultrafiltration.
Answer:
1. The pores of ordinary filter papers permit the passage of colloidal solutions. In ultrafilteration, the membranes are made by using collodion, cellophane or visiking.

2. Vhen a colloidal solution is filtered using such a filter, colloidal particles are separated on the filter and the impurities are removed as washings.

3. This process is quickened by application of pressure. The separation of sol particles from electrolyte by fikration through an ultrafilter is called ultrafiltration.

4. Collodian is 4% solution of nitrocellulose in a mixmre of alcohol and water.

Question 27.
Write a note about shape of colloidal particles.
Answer:
Colloidal particles possess various shapes.
Collidal particle

  1. AS2S3
  2. Fe(OH)3 (blue gold sol)
  3. W3O5 sol(tungstic acid sol)

Shape

  1. Spherical
  2. Disc (or) plate like
  3. Rod like

Question 28.
What is meant by Tyndall effect? (or) Explain about the optical property of colloid.
Answer:
Answer:

  1. Colloids have optical property. When a homogeneous solution is seen in the direction of light, it appears clear but it appears dark in a perpendicular direction.
  2. When light passes through colloidal solution, it is scattered in all direction and it is known as Tyndall effect.
  3. The colloidal particles absorbs a portion of light and remaining portion is scattered from the surface of the colloid. Hence the path of light is made clear.

Question 29.
What is meant by Brownian movement?
Answer:
1. Robert Brown observed that when the pollen grains suspended in water were viewed through an ultra microscope, they showed a random, zigzag, ceaseless motion. This is called Brownian movement of colloidal particles.

2. The colloidal sol particles are continuously bombard with the molecules of dispersion mediuri and hence they follow a zigzag. random, continuous movement.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-73

Question 30.
Mention the uses of Brow nian movement.
Answer:
1. Brownian movement enables us to calculate Avogadro Number.

2. It is used to confirm kinetic theory which considers the ceaseles rapid movement of molecules that increases with increase in temperature.

3. It is used to understand the stability of colloids. As the particles are in continuous rapid movement, they do not come close and hence not get condensed. That is Brownian movement does not allow the particles to be acted on by force of gravity.

Question 31.
What is coagulation? Mention the method used to coagulate a colloid.
Answer:

  1. The flocculation and setting down of the sol particles is called coagulation.
  2. Various methods of coagulation are
  3. addition of an electrolyte
  4. Electrophoresis
  5. mixing oppositely changed sols
  6. boiling

Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry

Question 32.
The precipitation power of ions are in the order Al3+ > Ba2+ > Na+.
Similarly [Fe(CN)6]4-SO42- Cl. Give the reason behind this.
Answer:
1. A negative ion causes the precipitation of positively changed sol and vice versa.

2. When the valency of ion is high, the precipitation power increased. So in cations the precipitation power order is Al3+ > Ba2+ > Na and in anions the precipitation power order is [Fe(CN)6]4- > SO42- > Cl.

3. The precipitation power of electrolyte is determined by finding the minimum concentration required to cause precipitation of sol in 2hrs. This value is called flocculation value. The smaller the flocculation value, greater will be precipitation.

Question 33.
Explain how coagulation of colloid is carried out by

  1. Flectrophoresis
  2. By mixing two oppositely changed sols
  3. By boiling.

Answer:
1. Electrophoresis:
In eleetrophoresis, charged particles migrate to the electrodes of opposite sign. It is due to neutralization of the charge of the colloids. The particles are discharged and sothey get precipitated.

2. By mixing two oppositely charged sols: When colloidal sols with opposite charges are mixed, mutual coagulation takes place. It is due to migration of ions from the surface of the particles.

3. By boiling – When colloidal sol is boiled, due to increased collisions, the sol particles combine and settle down.

Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry

Question 34.
Explain about protective action of colloid.
Answer:
1. Lyophobic sols are precipitated readily even with small amount of electrolytes. But they are stabilised by the addition of small amount of lyophillic colloid.

2. A small amount of gelatine sol is added to gold sol to protect gold sol.

3. Gold number is a measure of protecting power of a colloid, Gold number is defined as the number of milligrams of hydrophillic colloid that will just prevent the precipitation of 10ml of gold sol on the addition of imi of 10% NaG solution. Smaller the gold number, greater the protective power.

Question 35.
What are emulsions? Give its types. Explain with examples.
Answer:

  1. Emulsions are colloidal solution in which a liquid is dispersed in another liquid.
  2. There are two types of emulsions.
    • Oil in water (0/W)
    • Water in Oil (W/O)
  3. Oil in Water : Oil dispersed in water. e.g., mayonnaise. icecrearn.
    Water in Oil: Water dispersed in Oil. e.g., stiff grease, butter, cold cream.

Question 36.
Write 3 examples for emulsifiers.
Answer:

  1. Most of the lyophillic colloids act as emulsifiers. Example, glue. gelatin.
  2. Long chain compounds with polar groups like soap and suiphonic acid.
  3. Insoluble powders like clay and lamp black also act as emulsifier,

Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry

Question 37.
What is meant by inversion of phase? Explain with example.
Answer:

  1. The change of W/O emulsion into O/W emulsion is called inversion of phase.
  2. Example: An oil in water emulsion containing potassium soap as emulsifying agent can be converted into water in oil emulsion by adding CaCl2 (or) AlCl3. This is called inversion of phase.

Question 38.
Write a note about medicinal applications of colloids.
Answer:

  1. Antibodies such as penicillin and streptomycin are produced in colloidal form for suitable injections. They cure pneumonia.
  2. Colloidal gold and colloidal calcium are used as tonics.
  3. Milk of magnesia is used for stomach troubles.
  4. Silver sol protected by gelatine known as Argyrol is used as eye lotion.

Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry

Question 39.
How colloids are used in

  1. Tanning of leather
  2. Rubber industry
  3. Sewage disposal.

Answer:
1. Tanning of leather:
Skin and hides are protein containing positively charged particles which are coagulated by adding tannin to give hardened leather for further application. Chromium salts are used for tanning of leather. Chrome tanning can produce soft and polishable leather.

2. Rubber industry:
Latex is the emulsion of natural rubber with negative particles. By heating rubber with sulphur, vulcanized rubbers are produced for tyres, tubes etc.

3. Sewage disposal:
Sewage containing dirt, mud and wastes dispersed in water. Th e passage of electnic current deposits the wastes materials which can be used as a manure.

Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry

Question 40.
How would you distinguish natural honey from artificial honey?
Answer:

  1. Natural honey is a colloidal sol. It is distinguished from artificial one by adding ammonia cal AgNO3.
  2. In the case of natural honey, a metallic silver is produced assumes a reddish yellow colour due to the traces of albumin or ethereal oil which acts as a protective colloid,
  3. In case of artificial honey, a dark yellow (or) greenish yellow precipitate is formed.

Question 41.
Give four uses of emulsions.
Answer:

  1. The cleansing action of soap is due to emulsions.
  2. It is used in the preparation of vanishing cream.
  3. It is used in the preparation of cod liver oiL.
  4. It is used in the preparation of butter, cream etc.

Question 42.

  1. Adsorption of a gas on the surface of solid is generally accompanied by a decrease in entropy. Still ¡t is a spontaneous process. Explain.
  2. How does an increase in temperature affect both physical as well as chemical adsorption?

Answer:
1. According to the equation ∆G = ∆H – T∆S For a process to be spontaneous, ∆G should be negative. ∆H of adsorption is always negative. For a gas. ∆S is also negative. Thus, in an adsorption process, which is spontaneous, a. combination of these two factors always makes ∆G negative.

2. On increasing the temperature desorption occurs in case of physical adsorption. Chemical adsorption first increases and then decreases with increase in temperature.

Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry

Question 43.

  1. What is the difference between a colloidal solution and an emulsion? Give one example of each.
  2. What are emulsifiers?

Answer:
1. In a colloidal solution, the dispersed phase is a solid and the dispersion medium is liquid. In an emulsion, both the dispersed phase and the dispersion medium are liquid. Example – Colloidal sol — cell fluids, muddy water. Emulsion – milk, cold cream

2. Emulsifiers – The substance which are added to stabilise the emulsions are called emulsifiers. Example – Soaps of various kinds and lyophilic colloids (proteins, gum etc.)

Question 44.
Explain what is observed when.

  1. KCl, an electrolyte, is added to an hydrated ferric hydroxide sol.
  2. An electric current is passed through a colloidal solution.
  3. A beam of light is passed through a colloidal solution.

Answer:

  1. Feme hydroxide Fe(OH)3 is a positively charged sol, so it gets coagulated by the chloride ions released from KCI solution.
  2. When an electric current is passed through a colloidal solution due to charge on the colloidal particles they migrate towards the oppositely charged electrode.
  3. When a beam of light is passed through a colloidal solution the path of light becomes visible.

Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry

Question 45.
Write three distinct differences between physical adsorption and chemisorption.
Answer:
Physical adsorption

  1. Forces of attraction between the adsorbent and adsorbate molecules are weak Van der Waals’ forces,
  2. Heat of adsorption is low (20-40 kJ mol1)
  3. It is temporary and reversible.

Chemical adsorption

  1. Forces between the adsorbent and the adsorbate are strong chemical bond.
  2. Heat of adsorption is high (80-24OkJ mol1)
  3. It is permanent and irreversible.

Question 46.
Explain the following observations.
Answer:

  1. Lyophilic colloid is more stable than lyophobic colloid.
  2. Coagulation takes place when sodium chloride solution added to a colloidal solution of ferric hydroxide.
  3. Sky appears blue in colour.

Answer:
1. A lyophilic sol is stable duc to the charge and the hydration of the sol particles. Such a sol can only be coagulatd by removing the water and adding solvents like alcohol, acetone, etc. and then an electrolyte. On the other hand, a lyophobic soils stable due to charge only and hence it can he easily coagulated by adding small amount of an electrolyte.

2. The colloidal particles get precipitated. i.e., ferric hydroxide is precipitated.

3. The atmospheric particles of colloidal range scatter blue component of the white sunlight preferentially. That is why the sky appears blue.

Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry

Question 47.

  1. Heat of adsorption is greater for chemisorption than physisorption. Why?
  2. What is colloldion?
  3. Differentiate between peptization and coagulation.

Answer:

  1. Due to the formation of chemical bond between adsorbate and adsorbent.
  2. 4% solution of nitro cellulose in a mixture of alcohol and ether.
  3. Peptisation is the process of converting a precipitate into colloidal sol by adding an electrolyte. Rut coagulation is the settling of colloidal particles.

Question 48.
Give reasons for the following.

  1. Enzyme catalysts are highly specific in their action.
  2. The path of light beconies visible wheti it is passed through As2S3 sol ¡n water.
  3. The enthalpy in case of chemisorption is usually higher than that of physisorption.

Answer:

1. This is because each enzyme has a specific active site on which only a particular substrate can bind. ,

2. This is because of Tyndall effect caused due to the scattering of light by colloidal particles of As2S3 sol.

3. Chemisorption involves the formation ofa chemical bond between adsorbent and adsorbate molecule which involves high energy changes while in physisorption, the molecules of adsorbate and adsorbent arc held by weak van der Waals’ interactions.

Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry

Question 49.
What is adsorption? How does adsorption ola gas on a solid surface vary with pressure? Illustrate with the help of an appropriate graph.
Answer:
Adsorption is the phenomenon of attracting and retaining the molecules of a substance on the surface of a liquid or a solid, resulting in higher concentration of the molecules on the surface.

Effect of pressure:
At constant temperature, the adsorption of gas increases with increase of pressure of the gas. At low pressure. it increases rapidly. At an equilibrium pressure. the extent of adsorption (x/m) reaches its maximum value after which adsorption is independent of pressure.
\(\frac{\mathrm{x}}{\mathrm{m}}=\mathrm{kP}^{1 / \mathrm{n}}\)
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-74

Question 50.
How do size of particles of adsorbent, pressure of a gas and prevailing temperature influence of extent of adsorption of a gas on a solid?
Answer:
1. Effect of size of the particles of adsorbent: Greater the specific area of the solid available for adsorption of adsorbent, greater would be its adsorbing power. That is why porous or finely divided forms of adsorbents more strongly. However, the size of pores should be large enough to allow the diffusion of gas molecules.

2. Effect of pressure:
Increase in pressure initially increases the adsorption which later attains equilibrium at high pressure.
\(\frac{\mathrm{x}}{\mathrm{m}}=\mathrm{kP}^{1 / \mathrm{n}}\) (n>1)

Samacheer Kalvi 12th Chemistry Surface Chemistry 5 Mark Questions and Answers

Question 1.
What is adsorption isotherm? Explain about Freundlich adsorption isotherm.
Answer:
1. Adsorption isotherms represents the variation of adsorption at constant temperature. Adsorption isoterm can be studied quantitatively.

2. A plot between the amount of adsorbate adsorbed and pressure or concentration of adsorbate at constant temperature is called adsorption isotherms.

3. Freundlich adsorption isotherm. According to Freundlich
\(\frac{\mathrm{x}}{\mathrm{m}}=\mathrm{kP}^{1 / \mathrm{n}}\)
Where x is the amount of adsorbate (or) adsorbed on ‘m’ gm of adsorbent at a pressure of P. k and n are constants. Value of’n’ is always less than unity.

4. This equation is applicable for adsorption of gases on solid surfaces. The Same equation becomes \(\frac{\mathrm{x}}{\mathrm{m}}=\mathrm{kc}^{1 / \mathrm{n}}\) when used for adsorption in solutions with ‘c’ as concentration.

5. These equation quantitatively predict the effect of pressure (or concentration) on the adsorption of gases (or adsorbates) at constant temperature.

6. Taking log on both sides of equation
\(\frac{\mathrm{x}}{\mathrm{m}}=\mathrm{kP}^{1 / \mathrm{n}}\)
log \(\frac { x }{ m }\) = log k + \(\frac { 1 }{ n }\) log P
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-75

7. Hence the intercept represents the value of log k and the slope \(\frac { b }{ q }\) gives \(\frac { 1 }{ n }\)

8. This equation explains the increase of with increase in pressure. But experimental values shows the deviation at low pressure.

9. Limitations:

  1. This equation is purely empirical and valid over a limited pressure range.
  2. The value of k and n also found vary with temperatures. No theoretical explanations were given.

Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry

Question 2.
Define catalyst. What are the characteristics of catalysts?
Answer:
A catalyst is defined as a substance which alters the rate of chemical reaction without itself undergoing chemical change.
Characteristics of catalyst:
1. For chemical reaction, catalyst is needed in very small quantity. Generally, a pinch of catalyst is enough for a reaction in bulk.

2. There may be some physical changes, but the catalyst remains unchanged in mass and chemical composition in a chemical reaction.

3. A catalyst itself cannot initiate a reaction. It means it cannot start a reaction which is not taking place. Rut, lithe reaction is taking place in a slow rate it can increase its rate.

4. A solid catalyst will be more effective if it is taken in a finely divided form.

5. A catalyst can catalyse a particular type of reaction, hence they are said to be specific in nature.

6. In an equilibrium reaction, presence of catalyst reduces the time for attainment of equilibrium and hence it does not affect the position
of equilibrium and the value of equilibrium constant.

7. A catalyst is highly effective at a particular temperature called as optimum temperature.

8. Presence of a catalyst generally does not change the nature of products.

9. For example 2SO2 + O2 → 2SO3
This reaction is slow in the absence of a catalyst, but fast in the presence of Pt catalyst.

Question 3.
What is enzyme catalysis? Give the characteristics of enzyme catalysed reaction?
Answer:
Enzymes are complex protein molecules with three dimensional structures. They catalyse the chemical reaction in living organism. They are often present in colloidal state and extremely specific in catalytic action. This process is called enzyme catalysis.
Special characteristics of enzyme catalysis:
1. Effective and efficient conversion is the special characteristic of enzyme catalysed reactions. An enzyme may transform a million molecules of reactant in a minute.
For e.g., 2H2O2 → 2H2O + O2
For this reaction, activation energy is 1 8k.cal/mole without a catalyst. With colloidal platinum as a catalyst, the activation energy is 11.7 k.cal/mole. But with the enzyme catalyst. the activation energy of this reaction is less than 2 k.cal / mole.

2. Enzyme catalysis is highly specific in nature.
H2N – CO – NH2 + H2O → 2NH3 + H2O
The enzyme urcase which catalyses the reaction of urea does not catalyse the hydrolysis of methyl urea (NH2 – CO – NHCH3)

3. Enzyme catalysed reaction has maximum rate at optimum temperature. At first rate of the reaction increases with increase of temperature, but above a particular temperature, the activity of enzyme is destroyed. The rate may even drop to zero. The temperature at which enzyrnic activity is high (or) maximum is called as optimum temperature. e.g., enzyme involved in human body have an optimum temperature 37°C / 98°F.

4. The rate of enzyme catalysed reactions varies with the pH of the system. The rate is maximum at a pH called optimum pH.

5. Enzymes can be inhibited i.e., poisoned. Activity of an enzyme is decreased and destroyed by a poison. The physiological action of drugs is related to their inhibiting action. e.g.. sulpha drugs, penicillin inhibits the action of bacteria and used for curing diseases like
pneumonia, dysentery, cholera.

6. Catalytic activity of enzymes is increased by coenzymes or activators. A small non protein (vitamin) called a coenzyme promotes the catalytic activity of enzyme.

Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry

Question 4.
Explain about phase transfer catalysis.
Answer:
1. Consider the reactant of a reaction is present in one solvent and the other reactant is present in an another solvent. The reaction between them is very slow, if the solvents are immisible.

2. As the solvents form separate phases, the reactants have to migrate across the boundary to react. But migration of reactants across the boundary is not easy. For such situation, a third solvent is added which is miscible with both. So, the phase boundary is eliminated the reactants freely mix and react fast.

3. But for large scale preparation of any product, use of a third solvent is not convenient as it may be expensive. For such problems, phase transfer catalysis provides a simple solution, which avoids the use of solvents.

4. it directs the use of a phase transfer catalyst (a phase transfer reagent) to facilitate transport of a reactant in one solvent to other solvent where the second reactant is present. As the reactants are now brought together, they rapidly react and form the product.

5. Example – Substitution of C1 and CN in the following reaction.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-76
R Cl = 1 – chloro octane
R CN = 1 – cyano octane

6. By direct heating of two phase mixture of organic 1 – chioro Octane with aqueous sodium cyanide for several days, 1 – cyano octane is not obtained. However, if a small amount of quarternary ammonium salt like tetra alkyl ammonium cation which has hydrophobic and hydrophilic ends, transports CN – from the aqueous phase to the organic phase using its hydrophilic end and facilitates the reactions with 1 – chloro octane as shown below
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-77

7. So phase transfer catalyst, speeds up the reaction by transporting one reactant from one phase to another.

Question 5.
Explain about the classification of colloids based on the physical state of dispersed phase and dispersion medium with example.
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-78

Question 6.
Describe about condensation methods of preparation of colloids. (OR) Describe chemical methods of preparation of colloids. When the substance for colloidal particle is present as smalL sized particle, molecule or ion, they are brought to the colloidal dimension by condensation methods.

Answer:
1. Oxidation method:
When hydroiodic acid is treated with iodle acid J, sol is obtained.
HIO3 + 5HI → 3H2O + 3I2(sol)

2. Reduction method:
Gold sol is prepared by reduction of aune chloride using formaldehyde.
2 AuCl3 + 3HCHO + 3H2O → 2 Au(sol) + 6HCI + 3HCOOH

3. Hydrolysis:
Ferric chloride is hydrolysed to get ferric hydroxide colloid
FeCl3 + 3H2O → Fe(OH)3(sol) + 3HCI

4. Double decomposition:
When hydrogen suiphide gas is passed through a solution of arsenic oxide, a yellow coloured arsenic sulphide is obtained as a colloidal solution.
As2O3 – 3H2S → As2S3 + 3H2O

5. Decomposition:
When few drops of an acid is added to a dilute solution of sodium thiosulphate, sulphur colloid is produced by the decomposition of sodium thio sulphate.
S2O32- + 2H+ → S(sol) + H2O + SO2

Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry

7. Describe about the properties of colloids.
Answer:
1. Colour:
The colour of a sol is not always the same as the colour of the substance in the bulk. For example, bluish tinge is given by diluted milk in reflected light and reddish tinge in transmitted light.

2. Size:
The size of colloidal particles ranges from Imp to 1pm diameter.

3. Colloidal solutions are heterogeneous in nature:
They have two distinct phases. Experiments like dialysis. ultrafiltration show the heterogeneous in nature but they are considered as borderline cases.

4. Filterability:
As the size of pores in ordinary filter paper are large, the colloidal particles easily pass through the ordinary filter papers.

5. Non setting nature:
Colloidal solutions are quite stable i.e., they are not affected by gravtiy.

6. Concentration and density:
When the colloidal solution is dilute, it is stable. When the volume of medium is decreased, coagulation occurs. Density of sol decreases with
decrease in the concentration.

7. Diffusability:
Unlike true solution, colloids diffuse less readily through membranes.

8. Colligative properties:
The colloidal solutions show colligative properties such as elevation of boiling point, depression in freezing point and osmotic pressure. These. properties are used to determine molecular weight of colloidal particles.

9. Shape of colloidal particles:
Colloidal particles have various shapes

Collidal particle

  1. AS2S3
  2. Fe(OH)3 (blue gold sol)
  3. W3O5 sol(tungstic acid sol)

Shape

  1. Spherical
  2. Disc (or) plate like
  3. Rod like

10. Optical property:
The path of the light is visible when it is passes through a colloidal solution due to the scattering of light by colloidal particles. This is known as Tyndall effect.

11. Kinetic property:
When colloidal solution is viewed through an ultra microscope, they showed a random, zigzag ceaseless motion which is called Brownian movement.

12. Electrical property:
Helmholtz double layer. electrophoresis and electro osmosis are electrical properties of colloids.

Question 8.
Explain about Electrophoresis (or) Cataphoresis (or) How would you detect the presence of charges on sol particles? (or) Explain about the method used to detect the presence of charge on sol particles. Electrophoresis (or) cataphoresis:

Answer:
1. When electric potential is applied across two platinum electrodes dipped in a hydrophilic sol, the dispersed particles move toward one or other electrode. This migration of sol particles under the influence of electric field is called electrophoresis.

2. If the sol particles migrate to the cathode, then they possess positive charges and if the sol particles migrate to the anode, they have negative charges. Thus from the direction Afiet of migration of sol particles, we can determine the charge of the sol particles. Hence electrophoresis is used for the detection of presence of change on the sol particles.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-79

3.
Positively charged coiloids

Fe(OH)3, Al(OH)3
Basic dyes, Haemoglobin

Negatively charged colloids

Ag, Au & Pt, AS2S3,
Clay, Starch

Question 9.
What are emulsion? Mention its type with example. What is emulsification? How the types of emulsions are identified?
Answer:
1. Emulsions arc colloidal solution in which a liquid dispersed in another liquid.

2. Two types of emulsions

  • Oil in Water (O/W) e.g., mayonnaise
  • Water in Oil (W/O) e.g., Butter

3. The process of preparation of emulsion by the dispersal of one liquid in another liquid is called emulsification.

4. Identification of types of emulsions. The two types of emulsions can be identified by the following tests

  • Dye test – A small amount of dye soluble in oil is added to emulsion. The emulsion is shaken well. The aqueous emulsion will not take the colour whereas oily emulsion will take up the colour of the dye.
  • Viscosity test – Viscosity of the emulsion is determined by experiments. Oily emulsions will have higher value than aqueous emulsions.
  • Conductivity test – Conductivity of aqueous solutions are always higher than oily emulsions.
  • Spreading test – Oily emulsions spread readily than aqueous emulsion when spread on an oily surface.

Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry

Question 10.
What is deeniulsification? Explain about the various techniques of deemulsification.
Answer:
Emulsion can be separated into two separate layers. This process is called Deemulsification. Various deemulsilication techniques are given below.

  1. Distilling of one component
  2. Adding an electrolyte to destroy the charge.
  3. Destroying the emulsifier using chemical methods.
  4. Using solvent extraction to remove one component.
  5. By freezing one of the components.
  6. By applying centrifugal force.
  7. Adding dehydrating agents for water in oil type.
  8. Using ultrasonic waves.
  9. Heating at high pressures.

Question 11.

(a) How can a colloidal solution and a true solution of the same colour be distinguished from each other?
(b) List four applications of adsorption.

Answer:

  1. The path of light becomes visible when passed through a colloidal solution while it is not visible in case of a true solution. This is because of Tyndall effect caused by the scattering of light by colloidal particles.
  2. Applications of adsorption:
    • Activated charcoal is used in gas masks to remove poisonous gases such as CH4, CO, etc.
    • Animal charcoal is used as decoloriser in the manufacture of sugar.
    • Silica is used for removing moisture.
    • The ion exchange resins are used for removing hardness of water.

Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry

Question 12.
Illustrate with examples.

  1. Lyophilic and Lyophobic sols
  2. Homogeneous and Heterogeneous catalysis.

Answer:
1. The substances such as proteins, starch, rubber, etc. directly passes into the colloidal state when brought in contact with the solvent. Such colloids are known as lyophillic sols, The substances like metals, their suiphides. hydroxides, etc. do not form colloidal sol readily when mixed with dispersion medium. The colloidal sols can only be prepared by some special methods. Such sols are called lyophobic sols.

2. Homogeneous catalysis:
Here the reactants and catalyst are present in the same phase. For example, lead chamber process for the manufacture of H2SO4.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-80

Heterogeneous catalysis:
Here the reactants and catalyst are present in different phases. For example, contact process for manufacture of H2SO4.
Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry-81
Common Errors

  1. Students may get confused with the words absorption and adsorption
  2. There may be a confusion between physical adsorption and chemical adsorption
  3. Catalyst poison and negative catalyst are different.
  4. Colloids : There may be a doubt between dispersed phase and dispersion medium.
  5. Electro dialysis, Electrophoresis, Electro osmosis may get confused.

Rectifications

1. In surface chemistry, we should use only adsorption. Adsorption mainly explains surface phenomenon.
Physical adsorption – Weak Vander walls forces, multilayer formation, Heterogenous catalysis.

2. Chemical adsorption:
Bond formation, mono-layer formation, Homogeneous catalysis.
Catalyst poison: decreases the catalytic activity of catalyst. Negative catalyst: reduces the rate of a chemical reaction.

3. Dispersion medium:
Larger in amount. Dispersed phase – Smaller in amount. Colloid is dispersed phase in dispersion

4. medium. For e.g., smoke is colloid in which carbon solid particles dispersed in air dispersion medium.
Electrodialysis – Purification of colloid by electric current.

5. Electrophoresis:
Movement of colloidal particles (dispersed phase) towards the oppositely charged electrode under the influence of electric current. Electro osmosis: Movement of dispersion medium towards the oppositely charged electrode under the influence of electric current.