Class 12

Students can Download Computer Science Chapter 4 Algorithmic Strategies Questions and Answers, Notes Pdf, Samacheer Kalvi 12th 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 12th Computer Science Solutions Chapter 4 Algorithmic Strategies

Samacheer Kalvi 12th Computer Science Algorithmic Strategies Text Book Back Questions and Answers

PART – I
I. Choose The Best Answer

Question 1.
The word comes from the name of a Persian mathematician Abu Jafar Mohammed ibn – i Musa al Khowarizmi is called?
(a) Flow chart
(b) Flow
(c) Algorithm
(d) Syntax
Answer:
(c) Algorithm

Question 2.
From the following sorting algorithms which algorithm needs the minimum number of swaps?
(a) Bubble sort
(b) Quick sort
(c) Merge sort
(d) Selection sort
Answer:
(d) Selection sort

Question 3.
Two main measures for the efficiency of an algorithm are ……………………………
(a) Processor and memory
(b) Complexity and capacity
(c) Time and space
(d) Data and space
Answer:
(c) Time and space

Question 4.
The complexity of linear search algorithm is ……………………………
(a) O(n)
(b) O(log n)
(c) O(n2)
(d) O(n log n)
Answer:
(a) O(n)

Question 5.
From the following sorting algorithms which has the lowest worst case complexity?
(a) Bubble sort
(b) Quick sort
(c) Merge sort
(d) Selection sort
Answer:
(c) Merge sort

Question 6.
Which of the following is not a stable sorting algorithm?
(a) Insertion sort
(b) Selection sort
(c) Bubble sort
(d) Merge sort
Answer:
(b) Selection sort

Question 7.
Time complexity of bubble sort in best case is ……………………………
(a) θ(n)
(b) θ(n log n)
(c) θ(n2)
(d) θ(n(logn) 2)
Answer:
(a) θ(n)

Question 8.
The \(\Theta\) notation in asymptotic evaluation represents
(a) Base case
(b) Average case
(c) Worst case
(d) NULL case
Answer:
(b) Average case

Question 9.
If a problem can be broken into subproblems which are reused several times, the problem possesses which property?
Answer:
(a) Overlapping subproblems
(b) Optimal substructure
(c) Memoization
(d) Greedy
Answer:
(a) Overlapping subproblems

Question 10.
In dynamic programming, the technique of storing the previously calculated values is called?
(a) Saving value property
(b) Storing value property
(c) Memoization
(d) Mapping
Answer:
(c) Memoization

PART – II
II. Answer The Following Questions

Question 1.
What is an Algorithm?
Answer:
An algorithm is a finite set of instructions to accomplish a particular task. It is a step-by-step procedure for solving a given problem. An algorithm can be implemented in any suitable programming language.

Question 2.
Define Pseudo code?
Answer:

1. Pseudo code is a mix of programming – language – like constructs and Plain English.
2. Pseudo code is a notation similar to programming languages. Algorithms expressed in pseudo code are not intended to be executed by computers, but for communication among people.

Question 3.
Who is an Algorist?
Answer:

1. A person skilled in the design of algorithms are called as Algorist.
2. An algorithmic artist.

Question 4.
What is Sorting?
Answer:
Sorting is a method of arranging group of items in an ascending or descending order. Various sorting techniques in algorithms are Bubble Sort, Quick Sort, Heap Sort, Selection Sort, Insertion Sort.

Question 5.
What is searching? Write its types?
Answer:
A search algorithm is the step-by-step procedure used to locate specific data among a collection of data. Types of searching algorithms are

1. Linear search
2. Binary search
3. Hash search
4. Binary Tree search

PART – III
III. Answer The Following Questions

Question 1.
List the characteristics of an algorithm?
Answer:
Input, Output, Finiteness, Definiteness, Effectiveness, Correctness, Simplicity, Unambiguous, Feasibility, Portable and Independent.

Question 2.
Discuss about Algorithmic complexity and its types?
Answer:
The complexity of an algorithm f(n) gives the running time and/or the storage space required by the algorithm in terms of n as the size of input data.

Time Complexity:
The Time complexity of an algorithm is given by the number of steps taken by the algorithm to complete the process.

Space Complexity:
Space complexity of an algorithm is the amount of memory required to run to its completion.

Question 3.
What are the factors that influence time and space complexity?
Answer:
Time Complexity:
The Time complexity of an algorithm is given by the number of steps taken by the algorithm to complete the process.

Space Complexity:
Space complexity of an algorithm is the amount of memory required to run to its completion. The space required by an algorithm is equal to the sum of the following two components:

A fixed part is defined as the total space required to store certain data and variables for an algorithm. For example, simple variables and constants used in an algorithm. A variable part is defined as the total space required by variables, which sizes depends on the problem and its iteration. For example: recursion used to calculate factorial of a given value n.

Question 4.
Write a note on Asymptotic notation?
Answer:
Asymptotic Notations:
Asymptotic Notations are languages that uses meaningful statements about time and space complexity.

(I) Big O
Big O is often used to describe the worst – case of an algorithm.

(II) Big Ω
Big Omega is the reverse Big O, if Bi O is used to describe the upper bound (worst – case) of a asymptotic function, Big Omega is used to describe the lower bound (best-case).

(III) Big \(\Theta\)
When an algorithm has a complexity with lower bound = upper bound, say that an algorithm has a complexity O (n log n) and Ω (n log n), it’s actually has the complexity \(\Theta\) (n log n), which means the running time of that algorithm always falls in n log n in the best – case and worst – case.

Question 5.
What do you understand by Dynamic programming?
Answer:
Dynamic programming is an algorithmic design method that can be used when the solution to a problem can be viewed as the result of a sequence of decisions. Dynamic programming approach is similar to divide and conquer. The given problem is divided into smaller and yet smaller possible sub – problems.

PART – IV
IV. Answer The Following Questions

Question 1.
Explain the characteristics of an algorithm?
Answer:

1. Input – Zero or more quantities to be supplied.
2. Output – At least one quantity is produced.
3. Finiteness – Algorithms must terminate after finite number of steps.
4. Definiteness – All operations should be well defined. For example operations involving division by zero or taking square root for negative number are unacceptable.
5. Effectiveness – Every instruction must be carried out effectively.
6. Correctness – The algorithms should be error free.
7. Simplicity – Easy to implement.
8. Unambiguous – Algorithm should be clear and unambiguous. Each of its steps and their inputs/outputs should be clear and must lead to only one meaning.
9. Feasibility – Should be feasible with the available resources.
10. Portable – An algorithm should be generic, independent of any programming language or an operating system able to handle all range of inputs.
11. Independent – An algorithm should have step-by-step directions, which should be independent of any programming code.

Question 2.
Discuss about Linear search algorithm.?
Answer:
Linear Search:
Linear search also called sequential search is a sequential method for finding a particular value in a list. This method checks the search element with each element in sequence until the desired element is found or the list is exhausted. In this searching algorithm, list need not be ordered.

Pseudo code:
(I) Traverse the array using for loop
(II) In every iteration, compare the target search key value with the current value of the list.

1. If the values match, display the current index and value of the array
2. If the values do not match, move on to the next array element.

(III) If no match is found, display the search element not found.
To search the number 25 in the array given below, linear search will go step by step in a sequential order starting from the first element in the given array if the search element is found that index is returned otherwise the search is continued till the last index of the array. In this example number 25 is found at index number 3.

Example 1:
Input: values[ ] = {5, 34, 65, 12, 77, 35}
target = 77
Output: 4
Example 2:
Input: values[ ] = {101, 392, 1, 54, 32, 22, 90, 93}
target = 200
Output: -1 (not found)

Question 3.
What is Binary search? Discuss with example?
Answer:
Binary Search:
Binary search also called half – interval search algorithm. It finds the position of a search element within a sorted array. The binary search algorithm can be done as divide-and-conquer search algorithm and executes in logarithmic time.

Pseudo code for Binary search:
(I) Start with the middle element:

• If the search element is equal to the middle element of the array i.e., the middle value = number of elements in array/2, then return the index of the middle element.
• If not, then compare the middle element with the search value,
• If the search element is greater than the number in the middle index, then select the elements to the right side of the middle index, and go to Step-1.
• If the search element is less than the number in the middle index, then select the elements to the left side f the middle index, and start with Step-1.

(II) When a match is found, display success message with the index of the element matched.
(III) If no match is found for all comparisons, then display unsuccessful message.

Binary Search Working principles:
List of elements in an array must be sorted first for Binary search. The following example describes the step by step operation of binary search. Consider the following array of elements, the array is being sorted so it enables to do the binary search algorithm. Let us assume that the search element is 60 and we need to search the location or index of search element 60 using binary search.

First, we find index of middle element of the array by using this formula:
mid = low + (high – low) / 2
Here it is, 0 + (9 – 0 ) / 2 = 4 (fractional part ignored). So, 4 is the mid value of the array.

Now compare the search element with the value stored at mid value location 4. The value stored at location or index 4 is 50, which is not match with search element. As the search value 60 is greater than 50.

Now we change our low to mid + 1 and find the new mid value again using the formula, low to mid – 1
mid = low + (high – low) / 2
Our new mid is 7 now. We compare the value stored at location 7 with our target value 31.

The value stored at location or index 7 is not a match with search element, rather it is more than what we are looking for. So, the search element must be in the lower part from the current mid value location

The search element still not found. Hence, we calculated the mid again by using the formula.
high = mid – 1
mid = low + (high – low) / 2
Now the mid value is 5.

Now we compare the value stored at location 5 with our search element. We found that it is a match.

We can conclude that the search element 60 is found at location or index 5. For example if we take the search element as 95, For this value this binary search algorithm return unsuccessful result.

Question 4.
Explain the Bubble sort algorithm with example?
Answer:
Bubble sort algorithm:
Bubble sort is a simple sorting algorithm. The algorithm starts at the beginning of the list of values stored in an array. It compares each pair of adjacent elements and swaps them if they are in the unsorted order. This comparison and passed to be continued until no swaps are needed, which indicates that the list of values stored in an array is sorted. The algorithm is a comparison sort, is named for the way smaller elements “bubble” to the top of the list.

Although the algorithm is simple, it is too slow and less efficient when compared to insertion sort and other sorting methods. Assume list is an array of n elements. The swap function swaps the values of the given array elements.

Pseudo code:

1. Start with the fist element i.e., index = 0, compare the current element with the next element of the array.
2. If the current element is greater than the next element of the array, swap them.
3. If the current element is less than the next or right side of the element, move to the next element. Go to Step 1 and repeat until end of the index is reached.

Let’s consider an array with values {15, 11, 16, 12, 14, 13} Below, we have a pictorial representation of how bubble sort will sort the given array.

The above pictorial example is for iteration – 1. Similarly, remaining iteration can be done. The final iteration will give the sorted array.
At the end of all the iterations we will get the sorted values in an array as given below:

Question 5.
Explain the concept of Dynamic programming with suitable example?
Answer:
Dynamic programming:
Dynamic programming is an algorithmic design method that can be used when the solution to a problem can be viewed as the result of a sequence of decisions. Dynamic programming approach is similar to divide and conquer. The given problem is divided into smaller and yet smaller possible sub – problems.

Dynamic programming is used whenever problems can be divided into similar sub-problems, so that their results can be re-used to complete the process. Dynamic programming approaches are used to find the solution in optimized way. For every inner sub problem, dynamic algorithm will try to check the results of the previously solved sub-problems. The solutions of overlapped sub – problems are combined in order to get the better solution.

Steps to do Dynamic programming:

1. The given problem will be divided into smaller overlapping sub-problems.
2. An optimum solution for the given problem can be achieved by using result of smaller sub – problem.
3. Dynamic algorithms uses Memoization.

Fibonacci Series – An example:
Fibonacci series generates the subsequent number by adding two previous numbers. Fibonacci series starts from two numbers – Fib 0 & Fib 1. The initial values of fib 0 & fib 1 can be taken as 0 and 1.
Fibonacci series satisfies he following conditions:
Fibn = Fib_n-1 + Fib_n-2
Hence, a Fibonacci series for the n value 8 can look like this
Fib₈ = 0 1 1 2 3 5 8 13

Fibonacci Iterative Algorithm with Dynamic programming approach:
The following example shows a simple Dynamic programming approach for the generation of Fibonacci series.
Initialize f0 = 0, f1 = 1
step – 1: Print the initial values of Fibonacci f0 and f1
step – 2: Calculate fibanocci fib ← f0 + f1
step – 3: Assign f0 ← f1, f1 ← fib
step – 4: Print the next consecutive value of Fibonacci fib
step – 5: Go to step – 2 and repeat until the specified number of terms generated
For example if we generate fibonacci series up to 10 digits, the algorithm will generate the series as shown below:
The Fibonacci series is: 0 1 1 2 3 5 8 1 3 2 1 3 4 5 5

Samacheer kalvi 12th Computer Science Algorithmic Strategies Additional Questions and Answers

PART – I
I. Choose The Best Answer

Question 1.
Which one of the following is not a data structure?
(a) Array
(b) Structures
(c) List, tuples
(d) Database
Answer:
(d) Database

Question 2.
The word Algorithm has come to refer to a method ……………………………
(a) Solve a problem
(b) Insert a data
(c) Delete data
(d) Update data
Answer:
(a) Solve a problem

Question 3.
Which is wrong fact about the algorithm?
(a) It should be feasible
(b) Easy to implement
(c) It should be independent of any programming languages
(d) It should be generic
Answer:
(c) It should be independent of any programming languages

Question 4.
Complete the diagram

Answer:
Process

Question 5.
An algorithm that yields expected output for a valid input is called as ……………………………
Answer:
Algorithmic solution.

Question 6.
Program should be written for the selected language with specific ……………………………
Answer:
Syntax

Question 7.
…………………………… is an expression of algorithm in a programming language.
Answer:
Program

Question 8.
How many different phases are there in the analysis of algorithms and performance evaluations?
(a) 1
(b) 2
(c) 3
(d) Many
Answer:
(b) 2

Question 9.
Which one of the following is a theoretical performance analysis of an algorithm?
(a) A posteriori testing
(b) A priori estimates
(c) A preposition
(d) A post preori
Answer:
(b) A priori estimates

Question 10.
…………………………… is called performance measurement.
(a) A posteriori testing
(b) A priori estimates
(c) A preposition
(d) A post preori
Answer:
(a) A posteriori testing

Question 11.
Time is measured by counting the number of key operations like comparisons in the sorting algorithm. This is called as ……………………………
(a) Space Factor
(b) Key Factor
(c) Priori Factor
(d) Time Factor
Answer:
(d) Time Factor

Question 12.
Which of the following statement is true?
(a) Space Factor is the maximum memory space required by an algorithm
(b) Space Factor is the minimum memory spaces required by an algorithm
Answer:
(a) Space Factor is the maximum memory space required by an algorithm

Question 13.
In space complexity, the space required by an algorithm is equal to the sum of …………………………… part and …………………………… part.
Answer:
Fixed, Variable

Question 14.
…………………………… is an example for variable part of space complexity.
Answer:
Recursion

Question 15.
A …………………………… or …………………………… trade off is a way of solving in less time by using more storage space or by solving a given algorithm in very little space by spending more time.
Answer:
Space – timw, time – memory

Question 16.
Which is true related to the efficiency of an algorithm?
(I) Less time, more storage space
(II) More time, very little space
(a) I is correct
(b) II is correct
(c) Both are correct
(d) Both are wrong
Answer:
(c) Both are correct

Question 17.
How many asymptotic notations are used to represent time complexity of an algorithms?
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(c) 3

Question 18.
Which one of the following is not an Asymptotic notations?
(a) Big
(b) Big \(\Theta\)
(c) Big Ω
(d) Big ⊗
Answer:
(d) Big ⊗

Question 19.
………………………… is the reverse of Big O
(a) Big Ω
(b) Big \(\Theta\)
(c) Big C
(d) Big ⊗
Answer:
(a) Big Ω

Question 20.
………………………… describes the worst case of an algorithm.
(a) Big Q
(b) Big \(\Theta\)
(c) Big O
(d) Big C
Answer:
(c) Big O

Question 21.
…………………….. describes the lower bound of an algorithm.
(a) Big Ω
(b) Big \(\Theta\)
(c) Big O
(d) Big ⊗
Answer:
(a) Big Ω

Question 22.
Which search technique is also called sequential search techniques?
(a) Binary
(b) Binary Tree
(c) Hash
(d) Linear
Answer:
(d) Linear

Question 23.
What value will be returned by the linear search technique if value is not found?
(a) 0
(b) 1
(c) -1
(d) +1
Answer:
(c) -1

Question 24.
Which search algorithm is called as Half – Interval search algorithm?
(a) Binary
(b) Binary Tree
(c) Hash
(d) Linear
Answer:
(a) Binary

Question 25.
Which technique is followed by Binary Search algorithm?
(a) Subroutines
(b) Mapping
(c) Divide and conquer
(d) Namespaces
Answer:
(c) Divide and conquer

Question 26.
In Binary Search, if the search element is …………………….. to the middle element of the array, then index of the middle element is returned.
(a) >
(b) <
(c) =
(d) < >
Answer:
(c) =

Question 27.
In Binary search, if the search element is greater than the number in the middle index, then select the elements to the side of the middle index.
(a) Right
(b) Left
(c) Middle
(d) Bottom
Answer:
(a) Right

Question 28.
Fill in the box [Formula for Binary Search]
mid = low + (high – low) /
Answer:
2

Question 29.
……………………… is a simple sorting algorithm.
(a) Binary
(b) Bubble
(c) Selection
(d) Insertion
Answer:
(b) Bubble

Question 30.
Which one of the following is not a characteristics of Bubble Sort?
(a) Simple
(b) Too slow
(c) Too fast
(d) Less efficient
Answer:
(c) Too fast

Question 31.
In selection sort, there will be ……………………….. exchange for every pass through the list.
(a) 0
(b) 1
(c) 2
(d) 3
Answer:
(b) 1

Question 32.
How many number of passes are used in the Insertion Sort to get the final sorted list?
(a) 0
(b) 1
(c) n
(d) n -1
Answer:
(d) n – 1

Question 33.
………………………….. approach is similar to divide and conquer.
Answer:
Dynamic programming

Question 34.
………………………… is an example for dynamic programming approach.
(a) Fibonacci
(b) Prime
(c) Factorial
(d) Odd or Even
Answer:
(a) Fibonacci

Question 35.
Match the following.
(1) Linear search – (i) o(n²)
(2) Binary – (ii) o(n)
(3) Bubble Sort – (iii) o(log n)
(4) Merge Sort – (iv) o(n log n)
(a) 1 – (ii), 2 – (iii), 3 – (i), 4 – (iv)
(b) 1 – (i), 2 – (ii), 3 – (iii), 4 – (iv)
(c) 1 – (iv), 2 – (iii), 3 – (ii), 4 – (i)
(d) 1 – (iv), 2 – (ii), 3 – (i), 4 – (iii)
Answer:
(a) 1 – (ii), 2 – (iii), 3 – (i), 4 – (iv)

Question 36.
Time complexity of bubble sort in worst case is …………………………..
(a) θ(n)
(b) θ(n log n)
(c) θ(n2)
(d) θ(n(log n)2)
Answer:
(c) θ(n2)

Question 37.
The complexity of Merge Sort is …………………………
Answer:
o (n log n)

Question 38.
The complexity of Bubble Sort is …………………………
Answer:
o (n2)

Question 39.
The complexity of Binary search is ……………………….
Answer:
o (log n)

Question 40.
Pick the odd one out.
Merge Sort, Bubble, Binary, Insertion.
Answer:
Binary

PART – II
II. Answer The Following Questions

Question 1.
Define fixed part in the space complexity?
Answer:
A fixed part is defined as the total space required to store certain data and variables for an algorithm. For example, simple variables and constants used in an algorithm.

Question 2.
What is space – Time Trade off?
Answer:
A space – time or time – memory trade off is a way of solving in less time by using more storage space or by solving a given algorithm in very little space by spending more time.

PART – III
III. Answer The Following Questions

Question 1.
Design an algorithm to find square of the given number and display the result?
Answer:
Problem: Design an algorithm to find square of the given number and display the result. The algorithm can be written as:
Step 1 – start the process
Step 2 – get the input x
Step 3 – calculate the square by multiplying the input value ie., square ← x* x
Step 4 – display the result square
Step 5 – stop
Algorithm could be designed to get a solution of a given problem. A problem can be solved in many ways. Among many algorithms the optimistic one can be taken for implementation.

Question 2.
Differentiate Algorithm and Program?
Answer:
Algorithm:

1. Algorithm helps to solve a given problem logically and it can be contrasted with the program.
2. Algorithm can be categorized based on their implementation methods, design techniques etc.
3. There is no specific rules for algorithm writing but some guidelines should be followed.
4. Algorithm resembles a pseudo code which can be implemented in any language

Program:

1. Program is an expression of algorithm in a programming language.
2. Algorithm can be implemented by structured or object oriented programming approach.
3. Program should be written for the selected language with specific syntax
4. Program is more specific o a programming language

Question 3.
What are the two phases in the Analysis of algorithms and performance evaluation?
Answer:
Analysis of algorithms and performance evaluation can be divided into two different phases:
(a) A Priori estimates: This is a theoretical performance analysis of an algorithm. Efficiency of an algorithm is measured by assuming the external factors.
(b) A Posteriori testing: This is called performance measurement. In this analysis, actual statistics like running time and required for the algorithm executions are collected.

Question 4.
Name the factors where the program execution time depends on?
The program execution time depends on:

1. Speed of the machine
2. Compiler and other system Software tools
3. Operating System
4. Programming language used
5. Volume of data required

Question 5.
Write note on Big \(\Theta\)?
Answer:
Big \(\Theta\)
When an algorithm has a complexity with lower bound = upper bound, say that an algorithm has a complexity O (n log n) and Ω . (n log n), it’s actually has the complexity \(\Theta\) (n log n), which means the running time of that algorithm always falls in n log n in the best – case and worst – case.

PART – IV
IV. Answer The Following Questions

Question 1.
Explain Selection Sort?
Answer:
Selection sort
The selection sort is a simple sorting algorithm that improves on the performance of bubble sort by making only one exchange for every pass through the list. This algorithm will first find the smallest elements in array and swap it with the element in the first position of an array, then it will find the second smallest element and swap that element with the element in the second position, and it will continue until the entire array is sorted in respective order. This algorithm repeatedly selects the next-smallest element and swaps in into the right place for every pass. Hence it is called selection sort.

Pseudo code:
(I) Start from the first element (i.e.), index – 0, we search the smallest element in the array, and replace it with the element in the first position.

(II) Now we move on to the second element position, and look for smallest element present in the sub-array, from starting index to till the last index of sub – array.

(III) Now replace the second smallest identified in step-2 at the second position in the or original array, or also called first position in the sub array.

(IV) This is repeated, until the array is completely sorted.
Let’s consider an array with values {13, 16, 11, 18, 14, 15}
Below, we have a pictorial representation of how selection sort will sort the given array

In the first pass, the smallest element will be 11, so it will be placed at the first position. After that, next smallest element will be searched from an array. Now we will get 13 as the smallest, so it will be then placed at the second position.

Then leaving the first element, next smallest element will be searched, from the remaining elements. We will get 13 as the smallest, so it will be then placed at the second position. Then leaving 11 and 13 because they are at the correct position, we will search for the next smallest element from the rest of the elements and put it at third position and keep doing this until array is sorted.
Finally we will get the sorted array end of the pass as shown above diagram.

Question 2.
Explain Insertion Sort?
Answer:
Insertion Sort
Insertion sort is a simple sorting algorithm. It works by taking elements from the list one by one and inserting then in their correct position in to a new sorted list. This algorithm builds the final sorted array at the end. This algorithm uses n-1 number of passes to get the final sorted list as per the previous algorithm as we have discussed.
Pseudo for Insertion sort
Step 1 – If it is the first element, it is already sorted.
Step 2 – Pick next element
Step 3 – Compare with all elements in the sorted sub-list
Step 4 – Shift all the elements in the sorted sub-list that is greater than the value to be sorted
Step 5 – Insert the value
Step 6 – Repeat until list is sorted

At the end of the pass the insertion sort algorithm gives the sorted output in ascending order as shown below:

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Tamilnadu Samacheer Kalvi 12th Bio Botany Solutions Chapter 1 Asexual and Sexual Reproduction in Plants

Samacheer Kalvi 12th Bio Botany Asexual and Sexual Reproduction in Plants Text Book Back Questions and Answers

Question 1.
Choose the correct statement from the following
(a) Gametes are involved in asexual reproduction
(b) Bacteria reproduce asexually by budding
(c) Conidia formation is a method of sexual reproduction
(d) Yeast reproduce by budding
Answer:
(c) Conidia formation is a method of sexual reproduction

Question 2.
An eminent Indian embryologist is ____________
(a) S.R. Kashyap
(b) P. Maheswari
(c) M.S. Swaminathan
(d) K.C. Mehta
Answer:
(b) P. Maheswari

Question 3.
Identity the correctly matched pair ____________
(a) Tuber – Allium cepa
(b) Sucker – Pistia
(c) Rhizome – Musa
(d) Stolon – Zingiber
Answer:
(c) Rhizome – Musa

Question 4.
Pollen tube was discovered by ____________
(a) J.G. Kolreuter
(b) G.B. Amici
(c) E. Strasburger
(d) E. Hanning
Answer:
(b) G.B. Amici

Question 5.
Size of pollen grain in Myosotis ____________
(a) 10 micrometer
(b) 20 micrometer
(c) 200 micrometer
(d) 2000 micrometer
Answer:
(a) 10 micrometer

Question 6.
First cell of male gametophyte in angiosperm is ____________
(a) Microspore
(b) Megaspore
(c) Nucleus
(d) Primary Endosperm Nucleus
Answer:
(a) Microspore

Question 7.
Match the following ____________
(I) External fertilization – (i) pollen grain
(II) Androecium – (ii) anther wall
(III) Male gametophyte – (iii) algae
(IV) Primary parietal layer – (iv) stamens
(a) I-(iv); II-(i); III-(ii); IV-(iii)
(b) I-(iii); II-(iv); III-(ii); IV-(i)
(c) l-(iii); II-(iv); III-(ii); IV-(i)
(d) I-(iii); II-(i); III-(iv); IV-(ii)
Answer:
(b) I-(iii); II-(iv); III-(ii); IV-(i)

Question 8.
Arrange the layers of anther wall from locus to periphery
(a) Epidermis,middle layers, tapetum, endothecium
(b) Tapetum, middle layers, epidermis, endothecium
(c) Endothecium, epidermis, middle layers, tapetum
(d) Tapetum, middle layers endothecium epidermis
Answer:
(d) Tapetum, middle layers endothecium epidermis

Question 9.
Identify the incorrect pair
(a) sporopollenin – exine of pollen grain
(b) tapetum – nutritive tissue for developing microspores
(c) Nucellus – nutritive tissue for developing embryo
(d) obturator – directs the pollen tube into micropyle
Answer:
(c) Nucellus – nutritive tissue for developing embryo

Question 10.
Assertion : Sporopollenin preserves pollen in fossil deposits
Reason : Sporopollenin is resistant to physical and biological decomposition
(a) assertion is true; reason is false
(b) assertion is false; reason is true
(c) Both Assertion and reason are not true
(d) Both Assertion and reason are true.
Answer:
(b) assertion is false; reason is true

Question 11.
Choose the correct statement(s) about tenuinucellate ovule
(a) Sporogenous cell is hypodermal
(b) Ovules have fairly large nucellus
(c) sporogenous cell is epidermal
(d) ovules have single layer of nucellus tissue
Answer:
(a) Sporogenous cell is hypodermal

Question 12.
Which of the following represent megagametophyte ____________
(a) Ovule
(b) Embryo sac
(c) Nucellus
(d) Endosperm
Answer:
(b) Embryo sac

Question 13.
In Haplopappus gracilis, number of chromosomes in cells of nucellus is 4. What will be the chromosome number in Primary endosperm cell?
(a) 8
(b) 12
(c) 6
(d) 2
Answer:
(b) 12

Question 14.
Transmitting tissue is found in
(a) Micropylar region of ovule
(b) Pollen tube wall
(c) Stylar region of gynoecium
(d) Integument
Answer:
(c) Stylar region of gynoecium

Question 15.
The scar left by funiculus in the seed is ____________
(a) tegmen
(b) radicle
(c) epicotyl
(d) hilum
Answer:
(d) hilum

Question 16.
A Plant called X possesses small flower with reduced perianth and versatile anther. The probable agent for pollination would be ____________
(a) water
(b) air
(c) butterflies
(d) beetles
Answer:
(b) air

Question 17.
Consider the following statement(s)

1. In Protandrous flowers pistil matures earlier
2. In Protogynous flowers pistil matures earlier
3. Herkogamy is noticed in unisexual flowers
4. Distyly is present in Primula

(a) i and ii are correct
(b) ii and iv are correct
(c) ii and Hi are correct
(d) i and iv are correct
Answer:
(b) ii and iv are correct

Question 18.
Coelorhiza is found in ____________
(a) Paddy
(b) Bean
(c) Pea
(d) Tridax
Answer:
(a) Paddy

Question 19.
Parthenocarpic fruits lack ____________
(a) Endocarp
(b) Epicarp
(c) Mesocarp
(d) seed
Answer:
(d) seed

Question 20.
In majority of plants pollen is liberated at ____________
(a) 1 celled stage
(b) 2 celled stage
(c) 3 celled stage
(d) 4 celled stage
Answer:
(b) 2 celled stage

Question 21.
What is reproduction?
Answer:

• It is a vital process for the existence of a species.
• It brings suitable changes through variation in offsprings.
• Plant reproduction is important for the existence of all other organisms.

Question 22.
Mention the contribution of Hofmeister towards Embryology.
Answer:
Hofmeister described the structure of pollen tetrad

Question 23.
List out two sub-aerial stem modifications with example.
Answer:
Subaerial stem modifications.
The stem is partly aerial and partly underground.

a) Runner. (Ex. oxalis, Centella Asiatica)

• It is running horizontally on the soil surface.
• Nodes have axillary buds, scale leaves, and adventitious roots.
• Runner arises from the axillary bud.
• Mother plant produces many runners in all directions.
• They break off and grow into individual plants.

b) Sucker. (Ex. Musa (banana), chrysanthemum)
Grows horizontally for a distance under the soil. Then it emerges obliquely upwards.

c) Stolon (Ex. Strawberry, Vallisneria)
Develop from underground stems.
They grow horizontally outwards.

d) Offset (condensed runners)
Unlike runners, they produce tilt of leaves above and duster of roots below Ex. Pistia, Eichhornia.

Question 24.
What is layering?
Answer:
Layering is a conventional propagation method, where the stem of a parent plant is allowed to develop roots while still intact. When the root develops, the rooted part is cut and planted to grow as a raw individual. E.g.: Jasminum.

Question 25.
What are clones?
Answer:
The individuals (Ex. Bacteria) formed by this method are morphologically and genetically identical. They are called clones.

Question 26.
A detached leaf of Bryophyllum produces new plants. How?
Answer:
In Bryophyllum, the leaf is succulent and notched on its margin. Adventious buds develop at these notches and are called epiphyllous buds. They develop into new plants forming a root system and become independent plants when the leaf gets decayed.

Question 27.
Differentiate Grafting and Layering.
Answer:
Grafting

1. Two different plants are involved.
2. Two different plants are joined.
3. They continue to grow as one plant.
4. The plant in the soil is called stock.
5. The plant used for grafting is called the scion.
6. Ex. Citrus, Mango, Apple

Layering:

1. Only parent plant is involved.
2. Stem of the parent plant is allowed to develop roots.
3. The rooted part is cut and grown as a new plant.
4. Ex. Ixora, Jasminum

Question 28.
Tissue culture is the best method for propagating rare and endangered plant species”- Discuss.
Answer:
Micropropagation of plants in-vitro through tissue culturing is a modem and alternative tool to conserve and safeguard rare plant species. Since the basic principle behind PTC is totipotency. With the help of a single explant, it is possible to generate a huge population of plantlets within a short span of time. Conservation through micropropagation offers the possibility to rescue endangered and endemic species.

Question 29.
Distinguish mound layering and air layering
Answer:
Mound layering

1. Flexible branch is buried in the soil.
2. Roots emerge from the buried stem.
3. Buried part after cutting from for parent, grows into a new plant.

Air layering

1. Nodal region is girdled.
2. Hormones are applied.
3. Rooting is promoted.
4. This area is covered by moist soil.
5. Roots emerge in 2-4 months.
6. These branches removed from parent and grown separately.

Air Layering:
In air layering, the stem is girdled at the nodal part and hormones are applied and covered with moist soil using a polythene sheet. Roots emerge in these branches after 2-4 months. Such branches are removed from parent plant and grown separately.

Question 30.
Explain the conventional methods adopted in vegetative propagation of higher plants.
Answer:
The common methods of conventional propagation are cutting, grafting and layering.
a. Cutting: It is the method of producing a new plant by cutting the plant parts such as root, stem and leaf from the parent plant. The cut part is placed in a suitable medium for growth. It produces root and grows into a new plant. Depending upon the part used it is called as root cutting (Malus), stem cutting (Hibiscus, Bougainvillea and Moringa) and leaf cutting (Begonia, Bryophyllum). Stem cutting is widely used for propagation.

b. Grafting: In this, parts of two different plants are joined so that they continue to grow as one plant. Of the two plants, the plant which is in contact with the soil is called stock and the plant used for grafting is called scion. Examples are Citrus, Mango and Apple. There are different types of grafting based on the method of uniting the scion and stock. They are bud grafting, approach grafting, tongue grafting, crown grafting and wedge grafting.

(i) Bud grafting: AT- shaped incision is made in the stock and the bark is lifted. The scion bud with little wood is placed in the incision beneath the bark and properly bandaged with a tape.

(ii) Approach grafting: In this method both the scion and stock remain rooted. The stock is grown in a pot and it is brought close to the scion. Both of them should have the same thickness. A small slice is cut from both and the cut surfaces are brought near and tied together and held by a tape. After 1-4 weeks the tip of the stock and base of the scion are cut off and detached and grown in a separate pot.

(iii) Tongue grafting: A scion and stock having the same thickness is cut obliquely and the scion is fit into the stock and bound with tape.

(iv) Crown grafting: When the stock is large in size scions are cut into wedge shape and are
inserted on the slits or clefts of the stock and fixed in position using graft wax.

(v) Wedge grafting: In this method, a slit is made in the stock or the bark is cut. A twig of the scion is inserted and tightly bound so that the cambium of the two is joined.

c. Layering: In this method, the stem of a parent plant is allowed to develop roots while still intact. When the root develops, the rooted part is cut and planted to grow as a new plant.
Examples: Ixora mdJasminum Mound layering and Air layering are few types of layering.

(i) Mound layering: This method is applied for the plants having flexible branches. The lower branch with leaves is bent to the ground and part of the stem is buried in the soil and tip of ( the branch is exposed above the soil. After the roots emerge from the part of the stem buried in the soil, a cut is made in parent plant so that the buried part grow into a new plant.

(ii) Air layering: In this method the stem is girdled at nodal region and hormones are applied to this region which promotes rooting. This portion is covered with damp or moist soil using a polythene sheet. Roots emerge in these branches after 2-4 months. Such branches are removed-from the parent plant and grown in a separate pot or ground.

Question 31.
Highlight the milestones from the history of plant embryology.
Answer:
Milestones in Plant Embryology

1. 1682 – Nehemiah Grew mentioned stamens as the male organ of a flower.
2. 1694 – R.J.Camerarius described the structure of a flower, anther, pollen, and ovule
3. 1761 – J.G. Kolreuter gave a detailed account of the importance of insects in pollination
4. 1824 – G.B.Amici discovered the pollen tube.
5. 1848 – Hofmeister described the structure of pollen tetrad
6. 1870 – Hanstein described the development of the embryo in Capsella and Alisma
7. 1878 – E.Strasburger reported polyembryony
8. 1884 – E.Strasburger discovered the process of Syngamy.
9. 1898 -99 S.G.Nawaschin and L. Guignard independently discovered Double fertilization
10. 1904 – E.Hanning initiated embryo cultures.

Question 32.
Discuss the importance of Modern methods in the reproduction of plants.
Answer:
Advantages of modern methods

1. Plants with desired characteristics can be multiplied rapidly in a short duration.
2. Plants produced are genetically identical.
3. Tissue culture can be carried out in any season to produce plants.
4. Plants which do not produce viable seeds and seeds that are difficult to germinate can be propagated by tissue culture.
5. Rare and endangered plants can be propagated.
6. Disease free plants can be produced by meristem culture.
7. Cells can be genetically modified and transformed using tissue culture.

Question 33.
What is Cantharophily?
Answer:

• It is the cross-pollination of flowers by beetles. They feed on pollen or juicy tissues of their flower.
• The plants using this mode of pollination
• Er. Nymphaea species of plants – Rhinoceros beetle.
• Giant Water lily – Scarab beetle
• Illicium plant – Diptera files.

Question 34.
List any two strategies adopted by bisexual flowers to prevent self-pollination.
Answer:

• Protandry or protogyny
• Herkogamy

Question 35.
What is endothelium?
Answer:

• Some ovules are unitegmic (with one integument) tenuinucellate type, (with a single layer of micellar tissue).
• In these types of ovules, the inner layer of the integument is specialized for nutritive function for embryosac. It is called endothelium (Integumentary tapetum) Ex. Asteraceae.

Question 36.
“The endosperm of angiosperm is different from gymnosperm”. Do you agree. Justify your answer.
Answer:
The endosperm of Angiosperm:

1. Develops as a result of double fertilization.
2. Endosperm is generally triploid (polyploid).

Endosperm of Gymnosperm:

1. Develops before the fertilization process.
2. Endosperm is haploid.

Question 37.
Define the term Diplospory.
Answer:

• A diploid embryo sac is formed from a megaspore mother cell without a regular meiotic division.
• Examples: Eupatorium and Aerva.

Question 38.
What is polyembryony? How it can commercially exploited.
Answer:
The occurrence of more than one embryo in a seed is called polyembryony.

1. Embryos developed through polyembryony are found virus free.
2. The seedlings formed from nuclear tissue in citrus are found on better clones for orchards.

Question 39.
Why does the zygote divide only after the division of Primary endosperm cell?
Answer:

• The Zygote needs nourishment during its development.
• Fertilized embryo sac offers little nourishment to the Zygote.
• The primary endosperm cell divides and generates endosperm tissue.
• This nourishes the Zygote. So, the Zygote divides after the primary Endosperm cell.

Question 40.
What is Mellitophily?
Answer:
Pollination carried out by Bees is said to be mellitophily.

Question 41.
“Endothecium is associated with dehiscence of anther” Justify the statement.
Answer:

• Endothecium is a layer in the anther wall.
• It has a single layer of radially elongated cells. It is below the epidermis.
• The tangential wall or radial wall has lignified thickenings.
• These cells are hygroscopic. This nature helps in the dehiscence of anther at maturity.

Question 42.
List out the functions of the tapetum.
Answer:

1. It supplies nutrition to the developing microspores.
2. It contributes sporopollenin through ubisch bodies thus plays an important role in pollen wall formation.
3. The pollenkitt material is contributed by tapetal cells and is later transferred to the pollen surface.
4. Exine proteins responsible for the ‘rejection reaction’ of the stigma are present in the cavities of the exine. These proteins are derived from tapetal cells.

Question 43.
Write a short note on Pollenkitt.
Answer:

• Pollen kitt is an oily layer on the pollen surface. It is a viscous coating.
• It is contributed by tapetum.
• It is coloured yellow or orange.
• It is made of carotenoids, flavonoids.
• It attracts insects.
• It protects from damage by Uv radiation.

Question 44.
Distinguish tenuinucellate and crassinucellate ovules.
Answer:
Tenuinucellate Ovule:

1.  Ovules with hypodermal sporogerous cell with unilayerd nucellus tissue is called tenuinucellate type.
2. They have very small nucellus

Crassinucellate Ovule:

1. Ovule with subhypodermal sporogenous cell is called crassinucellate type.
2. They have large nucellus

Question 45.
‘Pollination in Gymnosperms is different from Angiosperms’ – Give reasons.
Answer:
Gymnosperms:

1. Pollination in gymnosperms is direct.
2. The pollens are deposited directly on the exposed ovules.

Angiosperms:

1. In Angiosperms it is indirect.
2. The pollens are deposited on the stigma of the pistil.

Question 46.
Write short note on Heterostyly.
Answer:
Heterostyly: Some plants produce two or three different forms of flowers that are different in their length of stamens and style. Pollination will take place only between organs of the same length.
E.g: Primula.

Question 47.
Enumerate the characteristic features of Entomophilous flowers.
Answer:

• Generally large. It is small aggregated in the inflorescence. ex: Asteraceae flowers.
• Brightly coloured to attract insects.
  ex: poinsettia and Bougainvillea the bracts become coloured.
• Scented with nectar.
• Pollen and nectar are floral rewards. Pollen is used for consumption.
• Foul odour also attracts flies and beetles
• Juicy cells of flowers are pierced and sucked by insects.

Question 48.
Discuss the steps involved in Microsporogenesis.
Answer:
Microsporogenesis: The stages involved in the formation of haploid microspores from diploid microspore mother cell through meiosis is called Microsporogenesis. The primary sporogenous cells directly, or may undergo a few mitotic divisions to form sporogenous tissue. The last generation of sporogenous tissue functions as microspore mother cells.

Each microspore mother cell divides meiotically to form a tetrad of four haploid microspores (microspore tetrad). Microspores soon separate from one another and remain free in the anther locule and develop into pollen grains.

Question 49.
With a suitable diagram explain the structure of an ovule.
Answer:
Structure of ovule(Megasporangium):

Ovule is also called megasporangium and is protected by one or two covering called integuments. A mature ovule consists of a Raphe stalk and a body. The stalk or the funiculus (also called funicle) is present at the base and it attaches the ovule to the placenta. The point of attachment of funicle to the body of the ovule is known as hilum. It represents the junction ovule and funicle. In an inverted ovule, the funicle is adnate to the body of the ovule forming a ridge called raphe. The body of the ovule is made up of a central mass of parenchymatous tissue called nucellus which has large reserve food materials. The nucellus is enveloped by one or two protective coverings called integuments. Integument encloses the nucellus completely except at the top where it is free and forms a pore called micropyle.

The ovule with one or two integuments are said to be unitegmic or bitegmic ovules respectively. The basal region of the body of the ovule where the nucellus, the integument and the funicle meet or merge is called chalaza. There is a large, oval, sac-like structure in the nucellus toward the micropylar end called embryo sac or female gametophyte. It develops from the functional megaspore formed within the nucellus. In some species(unitegmic tenuinucellate) the inner layer of the integument may become specialized to perform the nutritive function for the embryo sac and is called as endothelium or integumentary tapetum (Example: Asteraceae).

Question 50.
Give a concise account on steps involved in the fertilization of an angiosperm plant.
Answer:
Steps involved in fertilization of angiosperms plant:

1. Germination of pollen grain on stigma.
2. Formation of pollen tube in stigma.
3. Growth of pollen tube inside the style.
4. Direction of pollen tube towards the micropyle of ovule.
5. Entry of pollen tube into the synergid of embryo sac.
6. Discharge of male gametes from the pollen tube.
7. Fusion of male gamete with egg cell (syngany)
8. Fusion of second male gamete with polar nuclei (triple fusion/double

Question 51.
What is endosperm? Explain the types.
Answer:
1. Endosperm: The primary endosperm nucleus (PEN) divides immediately after fertilization but before the zygote starts to divide, into an endosperm. The primary endosperm nucleus is the result of triple fusion (two polar nuclei and one sperm nucleus) and thus has 3n number of chromosomes. It is a nutritive tissue and regulatory structure that nourishes the developing embryo. Depending upon the mode of development three types of endosperm are recognized in angiosperms. They are nuclear endosperm, cellular endosperm and helobial endosperm.

2. Nuclear endosperm: Primary Endosperm Nucleus undergoes several mitotic divisions without cell As, wall formation thus a free nuclear condition exists in the endosperm.
Examples: Coccinia, Capsella and Arachis.

3. Cellular endosperm: Primary endosperm nucleus divides into 2 nuclei and it is immediately followed by wall formation. Subsequent divisions also follow cell wall formation.
Examples: Adoxa, Helianthus and Scoparia.

4. Helobial endosperm: Primary Endosperm Nucleus moves towards base of embryo sac and divides into two nuclei. Cell wall formation takes place leading to the formation of a large micropylar and small chalazal chamber. The nucleus of the micropylar chamber undergoes several free nuclear division whereas that of chalazal chamber may or may not divide.
Examples : Hydrilla and Vallisneria.

5. Ruminate endosperm: The endosperm with irregularity and unevenness in its surface forms ruminate endosperm. Examples : Areca catechu, Passiflora and Myristica

Question 52.
Differentiate the structure of Dicot and Monocot seed
Answer:
Dicot Seed:

1. Possess two cotyledons
2. Absence of coleoptile and coleorhiza
3. Endosperm is scarce or absent

Monocot Seed:

1. Possess only one cotyledon
2. Presence of coleoptile and colerhiza surrounding plumule and radicle respectively.
3. Endosperm from the major storage tissue.

Question 53.
Give a detailed account of parthenocarpy. Add a note on its significance.
Answer:
1. Parthenocarpy
Development of fruit like structures from the ovary without fertilization. These fruits are parthenocarpic fruits. They have no true seeds. Commercially they are seedless fruits.

Genetic Parthenocarpy (Ex. Citrus)
Due to hybridization, Mutation.
Ex: Citrus, Cucurbita

Environmental Parthenocarpy
Environmental condition induces parthenocarpy. Ex) Low temperature for 3-19 hours.

Chemically Induced Parthenocarpy.
Growth promoting Auxins, Gibberellins induce parthenocarpy.

Significance

• Significance of seedless fruits in horticulture.
• Commercial Importance
• To prepare jam, jelly, sauce, fruit drinks.
• A high proportion of edible part due to absence of seed.

Samacheer Kalvi 12th Bio Botany Asexual and Sexual Reproduction in Plants Additional Questions and Answers

1 – Mark Questions

Question 1.
Match the following:
(1) Conidia – (i) Yeast
(2) Budding – (ii) Bacteria
(3) Gemma cups – (iii) Aspergilus
(4) Binary fission – (iv) Marchantia
(a) 1 – (iii), 2 – (i), 3 – (iv), 4 – (ii)
(b) 1 – (ii), 2 – (iv), 3 – (iii), 4 – (i)
(c) 1 – (iv), 2 – (ii), 3 – (i), 4 – (iii)
(d) 1 – (i), 2 – (iii), 3 – (ii), 4 – (iv)
Answer:
(a) 1 – (iii), 2 – (i), 3 – (iv), 4 – (ii)

Question 2.
The unit of reproductive structure used in vegetative propagation is called as
(a) Diplospores
(b) Aplanospores
(c) Diaspores
(d) Condiospores
Answer:
(c) Diaspores

Question 3.
Which of the following aquatic plant is popularly known as the “Terror of Bengal”?
(a) Eichornia crassipes
(b) Vallisneria spiralis
(c) Pistia stratiotes
(d) Zostera marina
Answer:
(a) Eichornia crassipes

Question 4.
Identify the incorrect statement regarding vegetative reproduction.
Answer:
(a) Only one parent is required for propagation.
(b) New individuals are genetically dissimilar.
(c) Easy mode of reproduction.
(d) Variation does not exist.
Answer:
(b) New individuals are genetically dissimilar.

Question 5.
The genetic ability of a plant cell to produce the entire plant is said to be
(a) Multipotency
(b) Totipotency
(c) Pleuripotency
(d) Differentiation
Answer:
(b) Totipotency

Question 6.
A typical anther is
(a) Bisporangiate
(b) Tetrasporangiate
(c) Unisporangiate
(d) Multisporangiate
Answer:
(b) Tetrasporangiate

Question 7.
Match the following:
Vegetative Reproductive structures

(a) 1 – (ii), 2 – (i), 3 – (iv), 4 – (iii)
(b) 1 – (ii), 2 – (iv), 3 – (iii), 4 – (i)
(c) 1 – (iv), 2 – (ii), 3 – (i), 4 – (iii)
(d) 1 – (i), 2 – (iii), 3 – (ii), 4 – (iv)
Answer:
(a) 1 – (ii), 2 – (i), 3 – (iv), 4 – (iii)

Question 8.
Innermost layer of anther wall is
(a) Endothecium
(b) Endothecum
(c) Endothelium
(d) Tapetum
Answer:
(d) Tapetum

Question 9.
Identify the mismatched pair:
(a) Epidermal layer – Protective infunction
(b) Endothecium layer – Helps in dehiscence of anther
(c) Middle layer – Persistent layer
(d) Tapetum – Nutritive in function
Answer:
(c) Middle layer – Persistent layer

Question 10.
Name the person who discovered the pollen tube?
(a) E.Strasburger
(b) Hofineister
(c) Nehemiah Grew
(d) G.B.Amici
Answer:
(d) G.B.Amici

Question 11.
Identify the mismatched pair
(i) Sucker – Chrysanthemum
(ii) Bulbils – Agave
(iii) Stolon – Fragaria
(iv) Runner – Lilium
(a) i only (b) ii only (c) iii only (d) iv only
Answer:
(d) iv only

Question 12.
Assertion (A): Epidermis is protective in function.
Reason (R): Epidermis is outermost unilayer of anther wall.
(a) A is correct R is incorrect.
(b) R explains A.
(c) Both A and R are incorrect.
(d) Both A and R are correct. R does not explain A.
(b) R explains A.
Answer:
(b) R explains A.

Question 13.
Assertion (A) : Microspores are the first cell of male gametophyte.
Reason (R) : Microspores undergo development and forms pollen grains.
(a) A is correct R is incorrect.
(b) R explains A.
(c) Both A and R are incorrect.
(d) Both A and R are correct. R does not explain A.
Answer:
(b) R explains A.

Question 14.
Assertion (A) : Carica papaya is a dioecious plant.
Reason (R): Both male and female flowers are borne on same plant.
(a) A is correct R is incorrect.
(b) R explains A.
(c) Both A and R are incorrect.
(d) Both A and R are correct. R does not explain A.
Answer:
(a) A is correct R is incorrect.

Question 15.
Assertion (A) : Anemophilous pollination occurs by animals.
Reason (R) : Pollen grains are sticky for easy attachment on animals.
(a) A is correct R is incorrect.
(b) R explains A.
(c) Both A and R are incorrect.
(d) Both A and R are correct. R does not explain A.
Answer:
(c) Both A and R are incorrect.

Question 16.
Assertion (A) : Fusion of male and female gametes results in zygote.
Reason (R): Product of triple fusion is PEN.
(a) A is correct R is incorrect.
(b) R explains A.
(c) Both A and R are incorrect.
(d) Both A and R are correct. R does not explain A.
Answer:
(d) Both A and R are correct. R does not explain A.

Question 17.
Assertion (A) : Zea mays is a monocotyledonous plant.
Reason (R) : Shield shaped cotyledon is called scutellum.
(a) A is correct R is incorrect.
(b) R explains A.
(c) Both A and R are incorrect.
(d) Both A and R are correct. R does not explain A.
Answer:
(b) R explain A.

Question 18.
Assertion (A) : In Bryophyllum, vegetative propagation occurs through leaf.
Reason (R) : Epiphyllous buds are noticed in Bryophyllum.
(a) A is correct R is incorrect.
(b) R explains A.
(c) Both A and R are incorrect.
(d) Both A and R are correct. R does not explain A.
Answer:
(b) R explain A.

Question 19.
Assertion (A): Androecium and Gynoecium are essential whorls of flower
Reason (R) : Androecium and Gynoecium assist the reproduction.
(a) A is correct R is incorrect
(b) R explains A
(c) Both A and R are incorrect
(d) Both A and R are correct. R does not explain A
Answer:
(a) A is correct R is incorrect

Question 20.
Identify the correct statement.
(a) Grafting is a modem method of artificial propagation.
(b) The plant which is used for graft is scion.
(c) In tongue grafting, the scion bud is placed inside the incision beneath bark.
(d) Grafting is usually carried out in monocot plants.
Answer:
(b) The plant which is used for graft is scion.

Question 21.
Statement 1: Flower is a highly condensed shoot for reproductive purpose.
Statement 2: A complete flower possess four whorls.
(a) Both the statements are incorrect.
(b) Statement 1 is correct and Statement 2 is incorrect.
(c) Both the statements are correct.
(d) Statement 1 is incorrect and statement 2 is correct.
Answer:
(c) Both the statements are correct.

Question 22.
Identify the incorrect statement.
(a) One seeded fruit of paddy is caryopsis.
(b) Primitive root is called coleorhiza.
(c) Sctellum is a part of monocot seed.
(d) Embryonic axis above the cotyledon is epicotyl.
Answer:
(b) Primitive root is called coleorhiza.

Question 23.
Cleavage polyembryony is noticed in
(a) Orchids
(b) Casuarina
(c) Balanophora
(d) Syzygium
Answer:
(a) Orchids

Question 24.
Pick out the non-spermous seed
(a) Wheat
(b) Sunflower
(c) Bean
(d) Orchids
Answer:
(c) Bean

Question 25.
The type of endosperm noticed in Hydrilla seed
(a) Ruminate endosperm
(b) Nuclear endosperm
(c) Cellular endosperm
(d) Helobial endosperm
Answer:
(b) Nuclear endosperm

Question 26.
Which is note a part of mature seed?
(a) Funiculus
(b) Testa and tegma
(c) hilm
(d) Chalaza
Answer:
(d) Chalaza

Question 27.
Select the wrong statement(s) regarding cross pollination.
(a) Pollination depends on external agent and so it is certain.
(b) New varieties are produced.
(c) Continuous cross pollination leads to weaker progeny.
(d) Germination capacity is highly declined.
(i) a and d
(ii) b and c
(iii) a, b and d
(iv) a, c and d
(iv) a, c and d
Answer:
(iv) a, c and d

Question 28.
Which of the following characters does not exist in Omithophilous flowers?
(a) Huge sized flowers
(b) Bright coloured
(c) Scented flowers
(d) Nectar is secreted in large
Answer:
(c) Scented flowers

Question 29.
Which of the following plant was introduced as a contaminant into India along with wheat?
(a) Parthenium hysterophorus
(b) Zea mays
(c) Rosaindica
(d) Mangifera indica
Answer:
(a) Parthenium hysterophorus

Question 30.
_____ is a carotenoid derivative of exine layer which provides resistance to pollen grains.
Answer:
Sporopollenin

Question 31.
The most common type of ovule noticed in dicots and monocots is ______
(a) Orthotropus
(b) Anatropous
(c) Campylotropus
(d) Amphitropous
Answer:
(b) Anatropous

Question 32.
Identity the incorrect statement.
(а) The stalk of the ovule is funiculus.
(b) Nucellus is composed of sclerenchymatous tissue.
(c) Basal region of the ovule is chalaza end.
(d) Micropyle is always oriented opposite to chalaza.
Answer:
(b) Nucellus is composed of sclerenchymatous tissue.

Question 33.
Generally the pollen grains are liberated from anther at
(a) 2-celled stage
(b) 4-celled stage
(c) 6-celled stage
(d) 8-celled stage
Answer:
(a) 2-celled stage

Question 34.
Assertion (A) : Self – pollination is certain in cleistogamous flowers.
Reason (R) : Flowers never open and do not expose reproductive organs.
(a) Both A and R are incorrect.
(b) A is correct R is incorrect.
(c) R explains A.
(d) Both A and R are correct. R is not correct explanation for A.
Answer:
(c) R explains A.

Question 35.
Assertion (A): Entamophily is the most common type of pollination.
Reason (R) : Birds and animals brings out effective pollination.
(a) Both A and R are incorrect.
(b) A is correct R is incorrect.
(c) R explains A.
(d) Both A and R are correct. R is not a correct explanation for A.
Answer:
(d) Both A and R are correct. R is not a correct explanation for A.

Question 36.
Statement 1: Primary sporogenous cell functions as megaspore mother cell.
Statement 2: Megaspore mother cell undergoes mitotic division producing megaspores.
(a) Statement 1 is correct and statement 2 is incorrect.
(b) Statement 1 is incorrect and statement 2 is correct.
(c) Both the statements 1 and 2 are correct.
(d) Both the statements 1 and 2 are incorrect.
Answer:
(a) Statement 1 is correct and statement 2 is incorrect.

Question 37.
Statement 1: Apomixis does not involve meiosis and syngamy.
Statement 2: The term Apomixis was introduced by Winkler.
(a) Statement 1 is correct and statement 2 is incorrect.
(b) Statement 1 is incorrect and statement 2 is correct.
(c) Both the statements 1 and 2 are correct.
(d) Both the statements 1 and 2 are incorrect.
Answer:
(c) Both the statements 1 and 2 are correct.

Question 38.
Statement 1: The pollen grains are deposited on the receptive surface of style.
Statement 2: After landing, the first visible change in pollen is hydration.
(a) Statement 1 is correct and statement 2 is incorrect.
(b) Statement 1 is incorrect and statement 2 is correct.
(c) Both the statements 1 and 2 are correct.
(d) Both the statements 1 and 2 are incorrect.
Answer:
(b) Statement 1 is incorrect and statement 2 is correct.

Question 39.
Which of the following post fertilization change is incorrectly matched?
(a) Secondary nucellus – Endosperms
(b) Antipodals – Degenerates
(c) Nucellus – Testa and tegma
(d) Funicle – Seed stalk
Answer:
(c) Nucellus – Testa and tegma Identify the parthenocarpic fruit

Question 40.
Identify the parthenocarpic
(a) Banana
(b) Pear
(c) Papaya
(d) More than one option is correct
Answer:
(d) More than one option is correct

Question 41.
A mature angiospermic embryo sac is
(a) 8 celled and 8 nucleated
(b) 7 celled and 8 nucleated
(c) 8 celled and 7 nucleated
(d) 7 celled and 8 nucleated
Answer:
(A) 7 celled and 8 nucleated

Question 42.
Identify the type of ovule, where the nucellus acquires a horse-shoe shaped structure.
(a) Anatropus
(b) Hemianatropus
(c) Campylotropus
(d) Amphitropus
Answer:
(d) Amphitropus

Question 43
(a) 1 egg cell and 2 anti
(b) 1 egg cell and 2 polar nuclei
(c) 1 egg cell and 1 secondary nuycleus
(d) 1 egg cell and 2 synergids
Answer:
(d) 1 egg cell and 2 synergids

Question 44.
Match the following :
(1) Hemianatropous
(2) Circinotropus
(3) Campylotropus
(4) Anatropous

Question 45.
Product of triple fusion is
(a) PEN
(b) PEG
(c) PVC
(d) PPT
Answer:
(a) PEN

Question 46.
Ex-albuminous seeds are
(a) Pea, castor, paddy
(b) Paddy, Coconut, Groundnut
(c) Beans, coconut, castor
(d) Groundnut, pea, beans
Answer:
(d) Groundnut, pea, beans

Question 47.
The white edible part of coconut is…
(a) Epicarp
(b) Endosperm
(c) Embryo
(d) Mesocarp
Answer:
(b) Endosperm

Question 48.
Observe the diagram and select the correct option mentioning the parts A,B,C and D.


Answer:
(b) (A) Plumule, (B) Cotyledon, (C) Testa, (D) Radicle

Question 49.


Answer:
(d) (A) Hemiamphitropus, (B) Campylotropus, (c) Amphitropus, (d) Circinotropus

Question 50.
Attractants and rewards are required for.
(а) Anemophily
(b) Entamophily
(c) Malacophily
(d) Cheiropterophily
Answer:
(b) Entamophily

Question 51.
Filiform apparatus is a special cellular thickening which is seen in
(a) Antipodals
(b) Polar nuclei
(c) Nucellus
(d) Synergids
Answer:
(d) Synergids

Question 52.
In anatropous ovule, the micropyle faces
(a) Right side
(b) Leftside
(c) Upward
(d) Downward
Answer:
(d) Downward

Question 53.
Observe the diagram and select the correct option mentioning the parts A,B,C and D.

Answer:
(A) Synergid, (B) Egg, (C) polar nuclei, (D) Antipodals

Question 54.
Identify the correct adaptation that checks autogamy
(a) Homogamy
(b) Cleistogamy
(c) Herkogamy
(d) None of the above
Answer:
(c) Herkogamy

Question 55.
In monoecious plants,
(a) Both autogamy and geitonogamy are prevented
(b) Both autogamy and geitonogamy are takes place
(c) Autogamy takes place preventing geitonogamy
(d) Autogamy is prevented whereas geitonogamy takes place
Answer:
(d) Autogamy is prevented whereas geitonogamy takes place

Question 56.
Antipodals are located at of embryo sac.
Answer:
Chalazal end

Question 57.
Identify the correct sequence of anther wall layers from periphery towards core part.
(a) Epidermis → endothelium → stomium → tapetum
(b) Epidermis → middle layer → endothecium → endothelium
(c) Epidermis → endothelium → middle layers → tapetum
(d) Epidermis → endothelium → endothecium → tapetum
Answer:
(c) Epidermis → endothelium → middle layers → tapetum

Question 58.
The proteins responsible for rejection reaction present in exine cavities of pollen is a- derivative of.
(a) Stomium
(b) Endothecium
(c) Tapetum
(d) Ubisch bodies
Answer:
(c) Tapetum

Question 59.
Pick out the mismatched pair:
(a) Entamophily – Insects
(b) Malacophily – Mammals
(c) Cheiropterophily – Bats
(d) Omithophily – Birds
Answer:
(b) Malacophily – Mammals

Question 60.
Which is the most common type of style seen in monocots?
(a) Open type
(b) Closed type
(c) Solid type
(d) Half closed type
Answer:
(a) Open type

2 – Mark Questions

Question 1.
Write the names of organisms that undergo the following types of asexual reproduction.
(a) Budding
(b) fragmentation
(c) Regeneration
(d) Gemma cup formation.
Answer:
(a) Budding – Hydra
(b) Fragmentation – Spirogyra
(c) Regeneration – Planaria
(d) Gemma cup formation – Marchantia

Question 2.
What are diaspores?
Answer:
The unit of reproductive structures used in vegetative propagation is called diaspore or reproductive propagules.

Question 3.
Name the vegetative propagules of the following plants.
(a) Allium cepa
(b) Zingiber officinalis
(c) Agave
(d) Colocasia
Answer:
(a) Allium cepa – Bulb
(b) Zingiber officinalis – Rhizome
(c) Agave – Bulbils
(d) Colocasia – Corm

Question 4.
Point out the advantages of natural vegetative reproduction.
Answer:
(a) Only one parent is required.
(b) New individuals are genetically similar
(c) Rapid spreading
(d) Large scale production

Question 5.
Mention any two conventional propagation techniques.
Answer:
(a) Cutting
(b) Grafting.

Question 6.
What do you mean by terms‘stock’and‘scion’in grafting technique?
Answer:
In Grafting, parts of two different plants are joined so that they continue to grow as one plant. Of the two plants, the plant which is in contact with the soil is called stock and the plant used for grafting is called scion.

Question 7.
Name any four types of grafting.
Answer:
(a) Bud grafting
(b) Tongue grafting
(c) Crown grafting
(d) Wedge grafting

Question 8.
Define totipotency.
Answer:
The genetic ability of a plant cell to produce the entire plant under suitable conditions is said to be totipotency.

Question 9.
What are the term micropropogation refers to?
Answer:
The regeneration of a whole plant from single cell, tissue or small pieces of vegetative structures through tissue culture is called micropropagation. It is one of the modem methods used to propagate plants.

Question 10.
Name the four whorls of a typical flower.
Answer:
(a) Calyx
(b) Corolla
(c) Androecium
(d) Gynoecium

Question 11.
Write any four valid points on Androecium
Answer:
(a) Androecium is the male part of a flower
(b) It is made up of stamens
(c) Each stamen possess anther and a filament
(d) Anthers bear pollen grains (male gametophyte)

Question 12.
What is poilinium? Give example.
Answer:
In some plants, all the microspores in a microsporangium remain held together called poilinium.
Example: Calotropis.

Question 13.
Name the four anther wall layers.
Answer:
(a) Epidermis
(b) Endothecium
(c) Middle layers
(d) Tapetum

Question 14.
Tapetum is dual in origin – Justify.
Answer:
Tapetum is derived partly from peripheral anther wall layer and partly from the connective tissue of anther lining the anther locule. Hence it is said to dual in origin.

Question 15.
Name the two types of tapetum. Mention any one function of tapetum.
Answer:
Tapetum are of two types. Secretory tapetum and Invasive tapetum. Tapetum nourishes the pollen grains.

Question 16.
Differentiate between Exine and Intine layers of pollen grain.
Answer:
Exine:

1. Thick outer layers
2. It is made of cellulose, sporopollenin and pollenkitt.
3. Ununiform layer

Intine:

1. Thin inner layer
2. It is made up of pectin, hemicellulose, cellulose and callose
3. Uniform layer

Question 17.
What are the chemical components that make up the wall layers of pollen grains?
Answer:
Pectin, cellulose, hemicellulose, callose, sporopollenin, pollenkitt and other proteins.

Question 18.
Draw and label the structure of a typical pollen grain
Answer:

Question 19.
At which cellular stage, does the pollen grains are usually liberated from anther? What happens to the generative cell if the pollen reaches the stigma?
Answer:
In general, the pollen grains are liberated at 2 – celled stage. On reaching the stigma at this stage, the generative cell divides meiotically and form two male gametes (male cells)

Question 20.
Write the equivalent botanical terms for the following words / sentences.
(a) Landing platform of pollen
(b) Ovarian cavity
(c) Megasporangium
(d) Basal swollen part of pistil
Answer:
(a) Stigma
(b) Locule
(c) Ovule
(d) Ovary

Question 21.
What is Nucellus?
Answer:
The body of the ovule is made up of a central mass of parenchymatous tissue called nucellus which has large reserve food materials.

Question 22.
What is integumentary tapetum?
Answer:
In Asteraceae species, the inner layer of integument get specialized for nourishing the embryosac and this is called integumentary tapetum or endothelium.

Question 23.
Mention the types of ovule seen in the members of
Answer:
(a) Cactaceae
(b) Leguminosae
(c) Polygonaceae
(d) Primulaceae.
Answer:
(a) Cactaceae – Circinotropous ovule
(b) Leguminosae – Campylotropous ovule
(c) Polygonaceae – Orthotropous ovule
(d) Primulaceae. – Hemianatropous ovule

Question 24.
What does the term ‘Bisporic development of embryo sac’ refers to? Give example.
Answer:
In gynoecium, out of four megaspores formed, if two are involved in embryo sac formation then it is said to be bisporic embryo sac. E.g. Peperomia

Question 25.
State the role of filiform apparatus found in embryo sac of angiosperm.
Answer:
(a) Filiform apparatus helps in absorption conduction of nutrients from nucellus to embryo sac
(b) It guides the pollen tube into the egg.

Question 26.
What is pollination? Mention its types.
Answer:
The process of transfer of pollen grains from the anther to a stigma of a flower is called pollination. The pollination is classified into two kinds, namely, self-pollination (Autogamy) and cross pollination(Allogamy).

Question 27.
Why pollination in gymnosperm is said to be direct?
Answer:
Pollination in gymnosperms is said to be direct as the pollens are deposited directly on the exposed ovules.

Question 28.
Define Homogamy with an example
Answer:
When the stamens and stigma of a flower mature at the same time it is said to be homogamy. It favours self-pollination to occur. Example: Mirabilis jalapa.

Question 29.
What is cross – pollination? What are its types?
Answer:
Cross – pollination refers to the transfer of pollens on the stigma of another flower.
The cross-pollination is of two types:

1. Geitonogamy
2. Xenogamy.

Question 30.
Distinguish between monoecious and dioecious plants.
Answer:
Monoecious:
Male & Female flowers develop on the same plant.
E.g: Maize

Dioecious:
Male and Female flowers develop on different plants.
E.g: Papaya

Question 31.
Define the following terms.
(a) Protandry
(b) Protogyny
Answer:
(a) Protandry: The stamens mature earlier than stigma. E.g: Helianthus
(b) Protogyny: The stigma mature earlier than stamens. E.g: Aristolochia bracteata

Question 32.
What is Heterostyly? Give example.
Answer:
Heterostyly: Some plants produce two or three different forms of flowers that are different in their length of stamens and style. Pollination will take place only between organs of the same length. E.g: Primula.

Question 33.
How self-pollination is avoided in Abutilon?
Answer:
In Abutilon, the self-pollination is avoided by self sterility or self-incompatibility, in which if the pollen grain reaches the stigma of the same flower, it will be prevented from germination. It is a genetic mechanism.

Question 34.
Name the agents of the following types of pollination
(a) Anemophily
(b) Ornithophily
(c) Cheiropterophily
(d) Malacophily
Answer:
(a) Winds
(b) Birds
(c) Bats
(d) Snails and slugs

Question 35.
Give any four unique characters exhibited by anemophilous flowers.
Answer:
(a) Flowers are small and inconspicuous
(b) Colourless
(c) Non-scented
(d) No nectar secretion

Question 36.
How the pollen grains of Vallisneria protect themselves?
Answer:
Vallisneria is an aquatic plant. Pollen grains of vallisneria are covered by mucilage coating which protects them from wetting.

Question 37.
Point out the differences between anemophilous flowers and ornithophilous flowers.
Answer:

Anemophilous Flowers	Ornithophilous Flowers
(a) Small sized flowers	Large sized flowers
(b) Colourless	Brightly coloured
(c) Donor produce nectar	Produce large quantity of nectar
E.g: Grasses	Bombax

Question 38.
Mention any two disadvantages of self-pollination.
Answer:

• Continuous self-pollination, generation after generation results in weaker progeny.
• Chances of producing new species and varieties are meager.

Question 39.
What is pollen-pistil interaction?
Answer:
The events from pollen deposition on the stigma to the entry of pollen tube in to the ovule is called pollen – pistil interaction. It is a dynamic process which involves recognition of pollen and to promote or inhibit its germination and growth.

Question 40.
What are the major post fertilization events in a flower?
Answer:
Endosperm development, embryo development, seed formation and fruit formation.

Question 41.
What is perisperm? Give example.
Answer:
In some plants, the nucellar tissue is not utilized by the embryo completely, a small portion will remain as a storage tissue in the seed, which is called as perisperm.
E.g: Black Pepper.

Question 42.
What happens to the following floral parts, after the fertilization process?
(a) Ovary
(b) Secondary nucleus
(c) Outer integument of ovule
(d) Funicle
Answer:
(a) Ovary → Fruit
(b) Secondary nucleus → Endosperm
(c) Outer integument of ovule → Outer seed coat (Testa)
(d) Funicle → Stalk of the seed

Question 43.
What is the product of triple fusion? Mention its ploidy.
Answer:
The product of triple fusion is Primary Endosperm Nucleus (PEN). It is triploid (3n) in condition.

Question 44.
Coconut is an albmunious seed. Why?
Answer:
Since coconut seed possess endosperm, it is called as albmunious seed. Endosperm nourishes the embryo during seed germination.

Question 45.
Name the various types of endosperms.
Answer:
(a) Nuclear Endosperm
(b) Helobial Endosperm
(c) Cellular Endosperm
(d) Ruminate Endosperm

Question 46.
What is scutellum?
Answer:
In monocot seeds, the embryo is small and consists of single shield-shaped cotyledon known as scutellum present towards lateral side of embryonal axis.

Question 47.
What is coleoptile and coleorhiza?
Answer:
In monocot embryo, the plumule is covered by a protective sheath called coleoptile and the radicle along with root cap is covered by a protective sheath called coleorhiza.

Question 48.
Who coined the term Apomixis? Define it.
Answer:
The term Apomixis was introduced by Winkler in the year 1908. It is defined as the substitution of the usual sexual system (Amphimixis) by a form of reproduction which does not involve meiosis and syngamy.

Question 49.
What are parthenocarpic fruits?
Answer:
In some plants, fruits develop from ovary without fertilization process. Such fruits are called parthenocarpic fruits. They lack true seeds.
E.g: Grapes.

3 – Mark Questions

Question 50.
How tongue grafting differs from wedge grafting?
Answer:
Tongue grafting:
In tongue grafting A scion and stock having the same thickness is cut obliquely and the scion is fit into the stock and bound with a tape.

Wedge grafting:
In wedge grafting a slit is made in the stock or the bark is cut. A twig of scion is inserted and tightly bound so that the cambium of the two is joined.

Question 51.
List any three advantages of micropropagation.
Answer:

1. Tissue culture can be carried out in any season to produce plants.
2. Plants which do not produce viable seeds and seeds that are difficult to germinate can be propagated by tissue culture.
3. Cells can be genetically modified and transformed using tissue culture.

Question 52.
Where the stomium is located? What is its role?
Answer:
In a mature anther, the cells along with junction or the two sporangia of an anther lobe lack cellulose and lignin thickening. This region is called stomium. Stomium along with hygroncopic nature of Endothecium helps in the dehiscence of anther at maturity.

Question 53.
Briefly explain about the types of tapetum.
Answer:
There are two types of tapetum based on its behaviour.
They are:
Secretory tapetum (parietal/glandular/ cellular): The tapetum retains the original position hnd cellular integrity and nourishes the developing microspores.

Invasive tapetum (periplasmodial): The cells loose their inner tangential and radial walls and the protoplast of all tapetal cells coalesces to form a periplasmodium.

Question 54.
Enumerate the functions of tapetum.
Answer:

1. It supplies nutrition to the developing microspores.
2. It contributes sporopollenin through ubisch bodies thus plays an important role in pollen wall formation.
3. The pollenkitt material is contributed by tapetal cells and is later transferred to the pollen surface.
4. Exine proteins responsible for ‘rejection reaction’ of the stigma are present in the cavities of the exine. These proteins are derived from tapetal cells.

Question 55.
State the significance of sporopollenin.
Answer:
The wall material sporopollenin is contributed by both pollen cytoplasm and tapetum. It is derived from carotenoids. It is resistant to physical and biological decomposition. It helps to withstand high temperature and is resistant to strong acid, alkali and enzyme action. Hence, it preserves the pollen for long periods in fossil deposits, and it also protects pollen during its journey from anther to stigma.

Question 56.
What do you know about bee pollen?
Answer:

1. Bee pollen is a natural substance with high proteins, carbohydrate and trace amount of vitamins and minerals.
2. It is used as dietary supplement as tablets.
3. It increases performance of athletes, race horses and also heals bum wounds.

Question 57.
Write a note on pollenkitt.
Answer:
Pollenkitt is contributed by the tapetum and coloured yellow or orange and is chiefly made of carotenoids or flavonoids. It is an oily layer forming a thick viscous coating over pollen surface. It attracts insects and protects damage from UV radiation.

Question 58.
Draw and label the structure of a mature embryo sac of angiosperm.
Answer:

Question 59.
How many synergid cells are there in an mature embryo sac. Mention any two major role of synergids.
Answer:
The egg apparatus possess two synergids.
Synergids secrete chemotropic substances that attracts the pollen tube.
Synergids also guides the pollen tube into the egg.

Question 60.
A mature female gametophyte (embryo sac / egg apparatus) of angiosperms is 7 celled with 8 nucleus. Name the individual cells and mention their count.
Answer:
(a) Egg – 1
(b) Secondary nucleus – 1
(c) Synergids – 2
(d) Antipodalcells – 3

Question 61.
Differentiate between chasmogamy and cleistogamy
Answer:
Chasmogamy:

1. In chasmogamy, the flowers opens and express its mature anthers and stigma for pollination.
2. Self-pollination is uncertain
3. Depend on pollinating agents
4. Example: Hibiscus

Cleistogamy:

1. In cleistogamy the pollination occurs without opening of flowers and exposing their sex organs.
2. Self pollination is certain.
3. No need of pollinating agents.
4. Example: Commelina

Question 62.
Geitonogamy is similar to autogamy. Justify the statement.
Answer:
When the pollen deposits on another flower of the same individual plant, it is said to be geitonogamy. It usually occurs in plants which show monoecious condition. It is functionally cross-pollination but is similar to autogamy because the pollen comes from same plant.

Question 63.
Explain the Herkogamy mechanism with suitable examples.
Answer:
In bisexual flowers the essential organs, the stamens and stigmas, are arranged in such a way that self-pollination becomes impossible. For example in Gloriosa superba, the style is reflexed away from the stamens and in Hibiscus the stigmas project far above the stamens.

Question 64.
Give a brief account on pollination process in Zea mays.
Answer:
The maize is monoecious and unisexual. The male inflorescence (tassel) is borne terminally and female inflorescence (cob) laterally at lower levels. Maize pollens are large and heavy and cannot be carried by light breeze. However, the mild wind shakes the male inflorescence to release the pollen which falls vertically below. The female inflorescence has long stigma (silk) measuring upto 23 cm in length, which projects beyond leaves. The pollens drop from the tassel is caught by the stigma.

Question 65.
Explain the role of water as a pollinating agent in Vallisneria spiralis.
Answer:
Pollination in Vallisneria spiralis: It is a dioecious, submerged and rooted hydrophyte. The female plant bears solitary flowers which rise to the surface of water level using a long coiled stalk at the time of pollination. A small cup shaped depression is formed around the female flower on the surface of the water.

The male plant produces male flowers which get detached and float on the surface of the water. As soon as a male flower comes in closer to a female flower, it gets settled in the depression and contacts with the stigma thus bringing out pollination. Later the stalk of the female flower coils and brings back the flower from surface to under water where fruits are produced.

Question 66.
Enumerate the characters of ornithophilous flowers.
Answer:
The ornithophilous flowers have the following characteristic features:

1. The flowers are usually large in size.
2. The flowers are tubular, cup shaped or um- shaped.
3. The flowers are brightly coloured, red, scarlet, pink, orange, blue and yellow which attracts the birds.
4. The flowers are scentless and produce nectar in large quantities. Pollen and nectar form the floral rewards for the birds visiting the flowers.
5. The floral parts are tough and leathery to withstand the powerful impact of the visitors.

Question 67.
How the flowers of salvia are adopted for mellitophily?
Answer:
Pollination in Salvia (Lever mechanism): The flower of Salvia is adapted for Bee pollination. The flower is protandrous and the corolla is bilabiate with 2 stamens. A lever mechanism helps in pollination. Each anther has an upper fertile lobe and lower sterile lobe which is separated by a long connective which helps the anthers to swing freely. When a bee visits a flower, it sits on the lower lip which acts as a platform. It enters the flower to suck the nectar by pushing its head into the corolla.

During the entry of the bee into the flower the body strikes against the sterile end of the connective. This makes the fertile part of the stamen to descend and strike at the back of the bee. The pollen gets deposited on the back of the bee. When it visits another flower, the pollen gets rubbed against the stigma and completes the act of pollination in Salvia.

Question 68.
Mention any three advantages of cross pollination.
Answer:

1. It always results in bringing out much healthier off springs.
2. Germination capacity is much better.
3. New varieties may be produced.
4. The adaptability of the plants to their environment is better.

Question 69.
Why pollination has to occur?
Answer:

1. Pollination is a pre-requisite for the process of fertilisation. Fertilisation helps in the formation of fruits and seeds.
2. It brings the male and female gametes closer for the process of fertilisation.
3. Cross-pollination introduces variations in plants due to the mixing up of different genes. These variations help the plants to adapt to the environment and results in speciation.

Question 70.
How the pollen germination and compatability is regulated by stigma of Gynoecium?
Answer:
The receptive surface of the stigma receives the pollen. If the pollen is compatible with the stigma it germinates to form a tube. This is facilitated by the stigmatic fluid in wet stigma and pellicle in dry stigma. These two also decide the incompatibility and compatibility of the pollen through recognition-rejection protein reaction between the pollen and stigma surface.

Question 71.
Give a brief account on solid style.
Answer:
It is common among dicots. It is characterized by the presence of central core of elongated, highly specialised cells called transmitting tissue.This is equivalent to the lining cells of hollow style and does the same function. Its contents are also similar to the content of those cells. The pollen tube grows through the intercellular spaces of the transmitting tissue.

Question 72.
What are the ways through which the pollen tube enters into ovule? Explain.
Answer:
Entry of pollen tube into the ovule: There are three types of pollen tube entry into the ovule.
Porogamy: when the pollen tube enters through the micropyle.
Chalazogamy: when the pollen tube enters through the chalaza.
Mesogamy: when the pollen tube enters through the integument.

Question 73.
What is the fate of pollen tube after reaching the embryo sac?
Answer:
After reaching the embryo sac, a pore is formed in pollen tube wall at its apex or just behind the apex. The content of the pollen tube (two male gametes, vegetative nucleus and cytoplasm) are discharged into the synergids into which pollen tube enters. The pollen tube does not grow beyond it, in the embryo sac. The tube nucleus disorganizes.

Question 74.
Double fertilization and triple fusion are correlated terms. Comment.
Answer:
The two male gametes released from a male gametophyte are involved in the fertilization. They fertilize two different components of the embryo sac. Since both the male gametes are involved in fertilization, the phenomenon is called double fertilization and is unique to angiosperms. One of the male gametes fuses with the egg nucleus (syngamy) to form , Zygote.

The second gamete migrates to the central cell where it fuses with the polar nuclei or their fusion product, the secondary nucleus and forms the primary endosperm nucleus (PEN). Since this involves the fusion of three nuclei, this phenomenon is called triple fusion. This
act results in endosperm formation which forms the nutritive tissue for the embryo.

Question 75.
Write a short note on endosperm.
Answer:
The primary endosperm nucleus (PEN) divides immediately after fertilization but before 1 the zygote starts to divide, into an endosperm. The primary endosperm nucleus is the result of triple fusion (two polar nuclei and one sperm nucleus) and thus has 3n number of chromosomes. It is a nutritive tissue and regulatory structure that nourishes the developing embryo. Depending upon the mode of development three types of endosperm are recognized in angiosperms. They are nuclear endosperm, cellular endosperm and helobial endosperm.

Question 76.
How nuclear endosperm is different from cellular endosperm.
Answer:
Nuclear endosperm:
Nuclear endosperm: Primary Endosperm Nucleus undergoes several mitotic divisions without cell wall formation thus a free nuclear condition exists in the endosperm.
Examples: Coccinia, Capsella and Arachis.

Cellular endosperm:
Cellular endosperm: Primary endosperm nucleus divides into 2 nuclei and it is immediately followed by wall formation. Subsequent divisions also follow cell wall formation.
Examples: Adoxa, Helianthus and Scoparia.

Question 77.
Give an account on Helobial endosperm.
Answer:
Helobial endosperm: Primary Endosperm Nucleus moves towards base of embryo sac and divides into two nuclei. Cell wall formation takes place leading to the formation of a large micropylar and small chalazal chamber. The nucleus of the micropylar chamber undergoes several free nuclear division whereas that of chalazal chamber may or may not divide.
Examples : Hydrilla and Vallisneria.

Question 78.
Differentiate between albuminous seed and ex-albuminous seed.
Answer:
Albuminous Seed:
The seeds with endosperm are called albuminous seed of endospermous seeds.
E.g: Coconut

Ex-albuminous Seed:
The seeds without endosperm are called ex-albuminous seeds or non-endospermous seeds.
E.g: Beans

Question 79.
Draw and label the structure of nuclear endosperm and Helobial endosperm
Answer:

Question 80.
Point out the function of endosperm.
Answer:
Functions of endosperm:

1. It is the nutritive tissue for the developing embryo.
2. In majority of angiosperms, the zygote divides only after the development of endosperm.
3. Endosperm regulates the precise mode of embryo development.

Question 81.
What are the components of mature dicot embryo.
Answer:
The mature dicot embryo has a radicle, two cotyledons and a plumule.

Question 82.
What is apospory?
Answer:
Megaspore mother cell undergoes the normal meiosis and four megaspores formed gradually disappear. A nucellar cell becomes activated and develops into a diploid embryo sac. This type of apospory is also called somatic apospory. Examples Hieracium and Parthenium.

5 – Mark Questions

Question 83.
Give a comperative account on Anther wall layers.
Answer:
Anther wall : The mature anther wall consists of the following layers
a. Epidermis
b. Endothecium
c. Middle layers
d. Tapetum.
a. Epidermis: It is single layered and protective in function. The cells undergo repeated anticlinal divisions to cope up with the rapidly enlarging internal tissues.

b. Endothecium: It is generally a single layer of radially elongated cells found below the epidermis. The inner tangential wall develops bands (sometimes radial walls also) of a cellulose (sometimes also slightly lignified). The cells are hygroscopic. In the anthers of aquatic plants, saprophytes, cleistogamous flowers and extreme parasites endothecial differentiation is absent. The cells along the junction of the two sporangia of an anther lobe lack these thickenings. This region is called stomium. This region along with the hygroscopic nature of endothecium helps in the dehiscence of anther at maturity.

c. Middle layers: Two to three layers of cells next to endothecium constitute middle layers. They are generally ephemeral. They disintegrate or get crushed during maturity.

d. Tapetum: It is the innermost layer of anther wall and attains its maximum development at the tetrad stage of microsporogenesis. It is derived partly from the peripheral wall layer and partly from the connective tissue of the anther lining the anther locule. Thus, the tapetum is dual in origin. It nourishes the developing sporogenous tissue, microspore mother cells andmicrospores.

The cells of the tapetum may remain uninucleate or may contain more than one nucleus or the nucleus may become polyploid. It also contributes to the wall materials, sporopollenin, pollenkitt, tryphine and number of proteins that control incompatibility reaction .Tapetum also controls the fertility of sterility of the microspores or pollen grains.

Question 84.
Explain the development process of male gametophyte.
Answer:

The microspore is the first cell of the male gametophyte and is haploid. The development of male gametophyte takes place while they are still in the microsporangium.

The nucleus of the microspore divides to form a vegetative and a generative nucleus. A wall is laid around the generative nucleus resulting in the formation of two unequal cells, a large irregular nucleus bearing with abundant food reserve called vegetative cell and a small generative cell. At this 2 celled stage, the pollens are liberated from the anther. In some plants the generative cell again undergoes a division to form two male gametes. In these plants,

the pollen is liberated at 3 celled stage. In 60% of the angiosperms pollen is liberated in 2 celled stage. Further, the growth of the male gametophyte occurs only if the pollen reaches the right stigma. The pollen on reaching the stigma absorbs moisture and swells.

The inline grows as pollen tube through the germ pore. Incase the pollen is liberated at 2 celled stage the generative cell divides in the pollen into 2 male cells (sperms) after reaching the stigma or in the pollen tube before reaching the embryo sac.

Question 85.
Explain any five types of angiospermic ovules.
Answer:
Orthotropous: In this type of ovule, the micropyle is at the distal end and the micropyle, the funicle and the chalaza lie in one straight vertical line. Examples: Piperaceae, Polygonaceae

Anatropous: The body of the ovule becomes completely inverted so that the micropyle and funiculus come to lie very close to each other. This is the common type of ovules found in dicots and monocots.

Hemianatropous: In this, the body of the ovule is placed transversely and at right angles to the funicle. Example: Primulaceae.

Campylotropous: The body of the ovule at the micropylar end is curved and more or less bean shaped. The embryo sac is slightly curved. All the three:, hilum, micropyle and chalaza are adjacent to one another, with the micropyle oriented towards the placenta.
Example: Leguminosae.

Amphitropous: The distance between hilum and chalaza is less. The curvature of the ovule leads to horse-shoe shaped nucellus.
Example: some Alismataceae.

Question 86.
Describe the development of monrosporic embryo sac.
Answer:
The functional megaspore is the first cell of the embryo sac or female gametophyte. The megaspore elongates along micropylar- chalazal axis. The nucleus undergoes a mitotic division. Wall formation does not follow the nuclear division. A large central vacuole now appears between the two daughter nuclei.

The vacuole expands and pushes the nuclei towards the opposite poles of the embryo sac. Both the nuclei divide twice mitotically, forming four nuclei at each pole. At this stage all the eight nuclei are present in a common cytoplasm (free nuclear division). After the last nuclear division the cell undergoes appreciable elongation, assuming a sac-like appearance. This is followed by cellular organization of the embryo sac. Of the four nuclei at the micropylar end of the embryo sac, three organize into an egg apparatus, the fourth one is left free in the cytoplasm of the central cell as the upper polar nucleus.

Three nuclei of the chalazal end form three antipodal cells whereas the fourth one functions atk the lower polar nucleus. Depending on the plant the 2 polar nuclei may remain free or iftiay fuse to form a secondary nucleus , (central cell). The egg apparatus is made up of a central egg celkand two synergids, one on each side of the egg cell. Synergids secrete chemotropic substances that help to attract the pollen tube. The special cellular thickening called filiform apparatus of synergids help in the absorption, conduction of nutrients from the nucellus to embryo sac. It also guides the pollen tube into the egg. Thus, a 7 celled with 8 nucleated embryo sac is formed.

Question 87.
Enumerate the characters of anemophilous flowers Anemophilous plants have the following characteristic features:
Answer:

1. The flowers are produced in pendulous, catkin-like or spike inflorescence.
2. The axis of inflorescence elongates so that the flowers are brought well above the leaves.
3. The perianth is absent or highly reduced.
4. The flowers are small, inconspicuous, colourless, not scented, do not secrete nectar.
5. The stamens are numerous, filaments are long, exerted and versatile.
6. Anthers produce enormous quantities of pollen grains compared to number of ovules available for pollination. They are minute, light and dry so that they can be carried to long distances by wind.
7. In some plants anthers burst violently and release the pollen into the air. Example: Urtica.
8. Stigmas are comparatively large, protruding, sometimes branched and feathery, adapted to catch the pollen grains. Generally single ovule is present.
9. Plant produces flowers before the new leaves appear, so the pollen can be carried without hindrance of leaves.

Question 88.
Describe the structure of a dicot seed.
Answer:
Structure of a Cicer seed as an example for Dicot seed The mature seeds are attached to the fruit wall by a stalk called funiculus. The funiculus disappears leaving a scar called hilum. Below the hilum a small pore called micropyle is present. It facilitates entry of oxygen and water into the seeds during germination. Each seed has a thick outer covering called seed coat. The seed coat is developed from integuments of the ovule. The outer coat is called testa and is hard whereas the inner coat is thin, membranous and is called tegmen.

In Pea plant the tegmen and testa are fused. Two cotyledons laterally attached to the embryonic axis are present. It stores the food materials in pea whereas in other seeds like castor the endosperm contains reserve food and the cotyledons are thin. The portion of embryonal axis projecting beyond the cotyledons is called radicle or embryonic root. The other end of the axis called embryonic shoot is the plumule.

Embryonal axis above the level of cotyledon is called epicotyl whereas the cylindrical region between the level of cotyledon is called hypocotyl. The epicotyl terminates in plumule whereas the hypocotyl ends in radicle.

Higher Order Thinking Skills (HOTs) Questions

Question 1.
Cleistogamous flower and chasmogamous flower. In which type does the autogamy is certain? Why?
Answer:
Autogamy is certain in cleistogamous flowers since they never open and expose the reproductive organs.

Question 2.
Position of essential whorls and inhibition of autogamy in Gloriosa superba – comment.
Answer:
In the bisexual flowers of Gloriosa superba, the style of the gynoecium is reflexed away from the stamens assuring that self-pollination (autogamy) is impossible.

Question 3.
Anemophilous flowers produce abundant pollen grains. Give reason.
Answer:
Anemophily is a chance event. The liberated pollen may or may not reach the target flower and are wasted during the transmission from one flower to another. Hence enormous pollen grains are produced to assure pollination.

Question 4.
Observe the picture and answer the questions.

(а) Label the part – A
(b) Name the types of vegetative propagule
(c) Give one example for such type of vegetative propagule.
Answer:
(a) A – Bud from eye
(b) Tuber
(c) Solanum tuberosum

Question 5.
Arrange the following events in a proper sequence.
Embryogenesis, Zygote formation, Syngamy, Gametogenesis
Answer:
Gametogenesis → Syngamy → Zygote formation → Embryogenesis

Question 6.
Name the process through which microspore tetrads are formed. What would be the ploidy of the cells of terad?
Answer:
Microspores are formed by the process of microsporogenesis. The cells of microspore tetrad are haploid(n).

Question 7.
Anemophilous flowers are colourless and non-scented. What may be the reason?
Answer:
Production of coloured and scented flowers are to attract the pollinating agents. Where as wind acts as a pollinating agent for anemophilous flowers. Hence it is unnecessory to produce coloured and scented flowers.

Question 8.
If you break open the coconut fruit, we can observe a fluid part and the white kernel.What does those parts represent?
Answer:
The fluid part of the coconut represents free-nuclear endosperm and the white kernel represents cellular endosperm.

Question 9.
Cite one common feature and one contrast feature shared between apomixis and parthenocarpy
Answer:
Common features : Fertilization (Syngamy) is absent in both apomixis and parthenocarpy.
Contrast features : Parthenocarpic fruits does not develop seeds, whereas in apomixis seeds are developed.

Students can Download Bio Botany Chapter 2 Classical Genetics Questions and Answers, Notes Pdf, Samacheer Kalvi 12th 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 12th Bio Botany Solutions Chapter 2 Classical Genetics

Samacheer Kalvi 12th Bio Botany Classical Genetics Text Book Back Questions and Answers

Question 1.
Extra nuclear inheritance is a consequence of presence of genes in __________
(a) Mitrochondria and chloroplasts
(b) Endoplasmic reticulum and mitrochondria
(c) Ribosomes and chloroplast
(d) Lysosomes and ribosomes
Answer:
(a) Mitrochondria and chloroplasts

Question 2.
In order to find out the different types of gametes produced by a pea plant having the genotype AaBb, it should be crossed to a plant with the genotype __________
(a) aaBB
(b) AaBB
(c) AABB
(d) aabb
Answer:
(d) aabb

Question 3.
How many different kinds of gametes will be produced by a plant having die genotype AABbCC?
(a) Three
(b) Four
(c) Nine
(d) Two
Answer:
(d) Two

Question 4.
Which one of the following is an example of polygenic inheritance?
(a) Flower colour in MirabilisJalapa
(b) Production of male honey bee
(c) Pod shape in garden pea
(d) Skin Colour in humans
Answer:
(d) Skin Colour in humans

Question 5.
In Mendel’s experiments with garden pea, round seed shape (RR) was dominant over wrinkled seeds (rr), yellow cotyledon (YY) was dominant over green cotyledon (yy). What are the expected phenotypes in the F₂ generation of the cross RRYY xrryy?
(a) Only round seeds with green cotyledons
(b) Only wrinkled seeds with yellow cotyledons
(c) Only wrinkled seeds with green cotyledons
(d) Round seeds with yellow cotyledons an wrinkled seeds with yellow cotyledons
Answer:
(d) Round seeds with yellow cotyledons an wrinkled seeds with yellow cotyledons

Question 6.
Test cross involves __________
(a) Crossing between two genotypes with recessive trait
(b) Crossing between two F₁ hybrids
(c) Crossing the F₁ hybrid with a double recessive genotype
(d) Crossing between two genotypes with dominant trait
Answer:
(c) Crossing the F₁ hybrid with a double recessive genotype

Question 7.
In pea plants, yellow seeds are dominant to green. If a heterozygous yellow seed pant is crossed with a green seeded plant, what ratio of yellow and green seeded plants would you expect in generation?
(a) 9:3
(b) 1:3
(c) 3:1
(d) 50:50
Answer:
(d) 50:50

Question 8.
The genotype of a plant showing the dominant phenotype can be determined by __________
(a) Back cross
(b) Test cross
(c) Dihybrid cross
(d) Pedigree analysis
Answer:
(b) Test cross

Question 9.
Select the correct statement from the ones given below with respect to dihybrid cross
(a) Tightly linked genes on the same chromosomes show very few combinations
(b) Tightly linked genes on the same chromosomes show higher combinations
(c) Genes far apart on the same chromosomes show very few recombinations
(d) Genes loosely linked on the same chromosomes show similar recombinations as the tightly I linked ones
Answer:
(a) Tightly linked genes on the same chromosomes show very few combinations

Question 10.
Which Mendelian idea is depicted by a cross in which the F₁ generation resembles both the parents
(a) Incomplete dominance
(b) Law of dominance
(c) Inheritance of one gene
(d) Co-dominance
Answer:
(d) Co-dominance

Question 11.
Fruit colour in squash is an example of __________
(a) Recessive epistasis
(b) Dominant epistasis
(c) Complementary genes
(d) Inhibitory genes
Answer:
(b) Dominant epistasis

Question 12.
In his classic experiments on Pea plants, Mendel did not use __________
(a) Flowering position
(b) Seed colour
(c) Pod length
(d) Seed shape
Answer:
(c) Pod length

Question 13.
The epistatic effect, in which the hybrid cross 9:3:3:1 between AaBb Aabb is modified as
(a) Dominance of one allele on another allele of both loci
(b) Interaction between two alleles of different loci
(c) Dominance of one allele to another alleles of same loci
(d) Interaction between two alleles of some loci
Answer:
(b) Interaction between two alleles of different loci

Question 14.
In a test cross involving F₁ dihybrid flies, more parental type offspring were produced than the recombination type offspring. This indicates __________
(a) The two genes are located on two different chromosomes
(b) Chromosomes failed to separate during meiosis
(c) The two genes are linked and present on the some chromosome
(d) Both of the characters are controlled by more than one gene
Answer:
(c) The two genes are linked and present on the some chromosome

Question 15.
The genes controlling the seven pea characters studied by Mendel are known to be located on h6w many different chromosomes?
(a) Seven
(b) Six
(c) Five
(d) Four
Answer:
(a) Seven

Question 16.
Which of the following explains how progeny can posses the combinations of traits that none
of the parent possessed?
(a) Law of segregation
(b) Chromosome theory
(c) Law of independent assortment
(d) Polygenic inheritance
Answer:
(d) Polygenic inheritance

Question 17.
“Gametes are never hybrid”. This is a statement of __________
(a) Law of dominance
(b) Law of independent assortment
(c) Law of segregation
(d) Law of random fertilization
Answer:
(c) Law of segregation

Question 18.
Gene which suppresses other genes activity but does not lie on the same locus is called as __________
(a) Epistatic
(b) Supplement only
(c) Hypostatic
(d) Codominant
Answer:
(c) Hypostatic

Question 19.
Pure tall plants are crossed with pure dwarf plants. In the F₁ generation, all plants were tall. These tall plants of generation were selfed and the ratio of tall to dwarf plants obtained was 3:1. This is called __________
(a) Dominance
(b) Inheritance
(c) Codominance
(d) Heredity
Answer:
(a) Dominance

Question 20.
The dominant epistatis ratio is _________
(a) 9:3:3:1
(b) 12:3:1
(c) 9:3:4
(d) 9:6:1
Answer:
(b) 12:3:1

Question 21.
Select the period for Mendel’s hybridization experiments.
(a) 1856 – 1863
(b) 1850 – 1870
(c) 1857-1869
(d) 1870 – 1877
Answer:
(a) 1856 – 1863

Question 22.
Among the following characters which one was not considered by Mendel in his experimentation pea?
(a) Stem – Tall or dwarf
(b) Trichomal glandular or non-glandular
(c) Seed – Green or yellow
(d) Pod – Inflated or constricted
Answer:
(b) Trichomal glandular or non-glandular

Question 23.
Name the seven contrasting traits of Mendel.
Answer:

1. Tall x Dwarf
2. yellow x Green
3. Purple x white
4. Inflated x constricted
5. axial x terminal
6. Round x wrinkled
7. Green x Yellow

Question 24.
What is meant by true-breeding or pure breeding lines/strain?
Answer:
True-breeding lines (Pure-breeding strains) means it has undergone continuous self-pollination having stable trait inheritance from parent to offspring. Matings within pure breeding lines produce offsprings having specific parental traits that are constant in inheritance and expression for many generations. Pure line breed refers to homozygosity only.

Question 25.
Give the names of the scientists who rediscovered Mendelism.
Answer:

• Hugo Devries
• Carl Correns
• Erich Von Tschermak

Question 26.
What is back cross?
Answer:
Back cross is a cross of F₁ hybrid with any one of the parental genotypes. The back cross is of two types; they are dominant back cross and recessive back cross. It involves the cross between the F₁ offspring with either of the two parents.

Question 27.
Define Genetics.
Answer:
It is the branch of biological science which deals with the mechanism of transmission of characters from parent to off-springs.

Question 28.
What are multiple alleles?
Answer:
Three or more alternative forms of a gene that occupy the same locus and control the expression of a single trait.
E.g: ABO blood group

Question 29.
What are the reasons for Mendel’s successes in his breeding experiment?
Answer:

• He concentrated in one or few characters at a time
• Factor are considered as character
• Pisum sativum has large bi sexual flower so emasculation is very easy for the hybridization process or technique.
• By naturally it is a self-pollinated crop
• This crop has short duration so three or four-generation can be raised in a year

Question 30.
Explain the law of dominance in monohybrid cross.
Answer:
Law of dominance states that the offsprings of an individual with contrasting (dissimilar) traits will only express the dominant trait in F₁ generation and both the characters are expressed in F₂ generation. This law also explains the proportion of 3 : 1 ratio in F₂ generation.

Question 31.
Differentiate incomplete dominance and codominance.
Answer:
Incomplete dominance:

1. Effect of one of the two alleles is more conspicuous.
2. it produces a fine mixture of the expression of two alleles.
3. The effect of hybrid is intermediate expression of the two alleles
4. It produces new phenotype.
5. The expressed new phenotype has no allele of its own
6. It has a quantitative effect.

Co – dominance :

1. The effect of both alleles is equally conspicuous.
2. No mixing effect of the two alleles
3. Both the alleles produces their effect independently.
4. Does not produce new phenotype.
5. The phenotype is combination of two phenotype and their alleles
6. A quantitative effect is absent

Question 32.
What is meant by cytoplasmic inheritance
Answer:
DNA is a universal genetic material. Genes located in nuclear chromosomes follow Mendelian inheritance. But certain traits are governed either by the chloroplast or mitochondrial genes. This phenomenon is known as extra nuclear inheritance. It is a kind of Non-Mendelian inheritance. Since it involves cytoplasmic organelles such as chloroplast and mitochondrion that act as inheritance vectors, it is also called Cytoplasmic inheritance.

Question 33.
Describe dominant epistasis with an example.
Answer:
Dominant Epistasis – It is a gene interaction in which two alleles of a gene at one locus interfere and suppress or mask the phenotypic expression of a different pair of alleles of another gene at another locus. The gene that suppresses or masks the phenotypic expression of a gene at another locus is known as epistatic.

The gene whose expression is interfered by non-allelic genes and prevents from exhibiting its character is known as hypostatic. When both the genes are present together, the phenotype is determined by the epistatic gene and not by the hypostatic gene.

In the summer squash the fruit colour locus has a dominant allele ‘W’ for white colour and a recessive allele ‘w’ for coloured fruit. ‘W’ allele is dominant that masks the expression of any colour.

Dominant epistasis in summer squash

In another locus hypostatic allele ‘G’ is for yellow fruit and its recessive allele ‘g’ for green fruit. In the first locus the white is dominant to colour where as in the second locus yellow is dominant to green. When the white fruit with genotype WWgg is crossed with yellow fruit with genotype wwGG, the F₁ plants have white fruit and are heterozygous (WwGg). When F₁ heterozygous plants are crossed.

they give rise to F₂ with the phenotypic ratio of 12 white : 3 yellow : 1 green.Since W is epistatic to the alleles ‘G’ and ‘g’, the white which is dominant, masks the effect of yellow or green. Homozygous recessive ww genotypes only can give the coloured fruits (4/16). Double recessive ‘wwgg’ will give green fruit (1/16). The Plants having only ‘G’ in its genotype (wwGg or wwGG) will give the yellow fruit(3/l 6).

Question 34.
Explain polygenic inheritance with an example.
Answer:
Polygenic inheritance – Several genes combine to affect a single trait. A group of genes that together Dark Red determine (contribute) a characteristic of an organism is called polygenic inheritance. It gives explanations to the inheritance of continuous traits which are compatible with Mendel’s Law. The first experiment on polygenic inheritance was demonstrated by Swedish Geneticist H.

Nilsson-Ehle (1909) in wheat kernels. Kernel colour is controlled by two genes each with two alleles, one with red kernel colour was dominant to white. He crossed the two pure breeding wheat varieties dark red and a white. Dark red genotypes F₁ generation R₁R₁R₂R₂ and white genotypes are r₁r₁r₂r₂ – F₁ generation medium red were obtained with the genotype R₁r₁R₂r₂. F₁ wheat plant produces

four types of gametes R₁R₂, R₁r₂, r,₁r₂. The intensity of the red colour is determined by the number of R genes in the F₂ generation. Four R genes: A dark red kernel colour is obtained. Three R genes: Medium – dark red kernel colour is obtained. Two R genes: Medium-red kernel colour is obtained. One R gene: Light red kernel colour is obtained. Absence of R gene: Results in White kernel colour.

The R gene in an additive manner produces the red kernel colour. The number of each phenotype is plotted against the intensity of red kernel colour which produces a bell shaped curve. This represents the distribution of phenotype.

Conclusion: Finally the loci that was studied by Nilsson – Ehle were not linked and the genes assorted independently. Later, researchers discovered the third gene that also affect the kernel colour of wheat. The three independent pairs of alleles were involved in wheat kernel colour. Nilsson – Ehle found the ratio of 63 red : 1 white in F₂ generation – 1 : 6 : 15 : 20 : 15 : 6 : 1 in F₂ generation.

Question 35.
Differentiate continuous variation with discontinuous variation.
Answer:
1. Discontinuous Variation: Within a population there are some characteristics which show a limited form of variation.
Example: Style length in Primula, plant height of garden pea. In discontinuous variation, the characteristics are controlled by one or two major genes which may have two or more allelic forms.

These variations are genetically determined by inheritance factors. Individuals produced by this variation show differences without any intermediate form between them and there is no overlapping between the two phenotypes. The phenotypic expression is unaffected by environmental conditions. This is also called as qualitative inheritance

2. Continuous Variation: This variation may be due to the combining effects of environmental and genetic factors. In a population most of the characteristics exhibit a complete gradation, from one extreme to the other without any break. Inheritance of phenotype is determined by the combined effects of many genes, (polygenes) and environmental factors. This is also known as quantitative inheritance.
Example: Human height and skin color.

Question 36.
Explain with an example how single genes affect multiple traits and alleles the phenotype of an organism.
Answer:
In Pleiotropy, the single gene affects multiple traits and alter the phenotype of the organism. The Pleiotropic gene influences a number of characters simultaneously and such genes are called a pleiotropic gene. Mendel noticed pleiotropy while performing a breeding experiment with peas (Pisum sativum).

Peas with purple flowers, brown seeds, and dark spot on the axils of the leaves were crossed with a variety of peas having white flowers, light-colored seeds, and no spot on the axils of leaves, the three traits for flower colour, seed colour, and leaf axil spot all were inherited together as a single unit. This is due to the pattern of inheritance where the three traits were controlled by a single gene with dominant and recessive alleles. Example: sickle cell anemia.

Question 37.
Bring out the inheritance of the chloroplast gene with an example.
Answer:
Chloroplast Inheritance

It is found in the 4 ‘O’Clock plant (Mirabilis jalapa). In this, there are two types of variegated leaves namely dark green-leaved plants and pale green-leaved plants. When the pollen of dark green leaved plant (male) is transferred to the stigma of pale green leaved plant (female) and pollen of pale green leaved plant is transferred to the stigma of dark green leaved plant, the F₁ generation of both the crosses must be identical as per Mendelian inheritance. But in the reciprocal cross, the F₁ plant differs from each other.

In each cross, the F plant reveals the character of the plant which is used as a female plant. This inheritance is not through the nuclear gene. It is due to the chloroplast gene found in the ovum of the female plant which contributes to the cytoplasm during fertilization since the male gamete contributes only to the nucleus but not the cytoplasm.

Samacheer Kalvi 12th Bio Botany Classical Genetics Additional Questions and Answers

1 – Mark Questions

Question 1.
The term ‘Genetics’ was introduced by __________
(a) Gregor Mendel
(b) Bateson
(c) Hugo de vries
(d) Carl Correns
Answer:
(b) Bateson

Question 2.
Which is not a correct statement?
(A) Variations are the raw materials for evolution
(B) Variations provide genetic material for natural selection
(C) It helps the individual to adapt to the changing environment
(D) Variations allow breeders to improve the crop field
(a) A and D
(b) B only
(c) C and D
(d) nono of the above
Answer:
(d) nono of the above

Question 3.
The process of removal of anthers from the flower is called __________
Answer:
Emasculation

Question 4.
An allele is __________
(a) another word for a gene
(b) alternate forms of a gene
(c) morphological expression of a gene
(d) genetic
Answer:
(b) alternate forms of a gene

Question 5.
Gregor Mendel __________
(i) was born in Czechoslovakia
(ii) did his experiments in Pisum fulvum
(iii) was the first systemic researcher in genetics
(iv) Published his results in the paper “Experiments on Plant Hybrids”
(a) All are correct
(b) (ii),(iii), (iv) are correct
(c) (i), (iii),(iv) are correct
(d) (i), (iii),(iv) are correct
Answer:
(c) (i), (iii),(iv) are correct

Question 6.

Answer:
A – (ii) B – (iv) C – (iii) D – (i)

Question 7.
How many characters studied by Mendel in pisum sativum
(a) Three
(b) Five
(c) Seven
(d) Nine
Answer:
(c) Seven

Question 8.
Mendel’s work were rediscovered by __________
(a) Hugo de Vries
(b) Tschermak
(c) Carl Correns
(d) All the above
Answer:
(d) All the above

Question 9.
Crossing of F₁, to any one of the parent refers to __________
(a) selling
(b) back cross
(c) test cross
(d) all of the above
Answer:
(b) back cross

Question 10.
Match the following

Answer:
A – (ii)
B – (iv)
C – (i)
D – (iii)

Question 11.
In an intergenic interaction, the gene that suppresses the pherotype of a gene is said to Crossing of F, to any one of the parent refers to __________
(a) Dominant
(b) Inhibitory
(c) Epistatic
(d) Hypostatic
Answer:
(c) Epistatic

Question 12.
Assertion (A) : Test cross is done between F₂ hybrid with F₁ recessive
Reason (R) : It helps to identify the homozygosity of hybrids
(a) A and R are correct R explains A
(b) A and R are incorrect
(c) A is correct R is incorrect
(d) A is incorrect R is correct
Answer:
(b) A and R are incorrect

Question 13.
Assertion (A) : Codominance is an example for intragenic interaction
Reason (R) : Interaction take place between the alleles of same gene
(a) A and R are correct R explains A
(b) A and R are incorrect
(c) A is correct R is incorrect
(d) A is incorrect R is correct
Answer:
(a) A and R are correct R explains A

Question 14.
Assertion (A) : Pleiotropic gene affects multiple traits
Reason (R) : ABO blood group is an example for Pleiotropism
(a) A and R are correct R explains A
(b) A and R are incorrect
(c) A is correct R is incorrect
(d) A is incorrect R is correct
Answer:
(c) A is correct R is incorrect

Question 15.
Assertion (A) : Cytoplasmic male sterility is a Mendelian inheritance
Reason (R) : The genes for cytoplasmic male sterility in peal maize is located at mitochondrial DNA
(a) A and R are correct R explains A
(b) A and R are incorrect
(c) A is correct R is incorrect
(d) A is incorrect R is correct
Answer:
(d) A is incorrect R is correct

Question 16.
What is the phenotypic ratio in case of incomplete dominance
(a) 9 : 7
(b) 3 : 1
(c) 1 : 2 : 1
(d) 1 : 1 : 1 : 1
Answer:
(c) 1 : 2 : 1

Question 17.
Identify the mismatched pair
(a) Chloroplast inheritance – Gregor Mendel
(b) Polygenic inheritance – H. Nilsson
(c) Lethal genes – E. Baur
(d) Incomplete dominance – Carl Correns
Answer:
(a) Chloroplast inheritance – Gregor Mendel

Question 18.
Statement 1 : Intergenic gene interaction occurs between alleles at same locus
Statement 2 : Co-dominance is an example for intergenic gene interaction
(a) Statement 1 is correct & Statement 2 is incorrect
(b) Statement 1 is incorrect & Statement 2 is correct
(c) Both Statements 1 & 2 are correct
(d) Both Statements 1 & 2 are incorrect
Answer:
(c) Both Statements 1 & 2 are correct

Question 19.
Statement 1 : Test cross is done between F₁ individual with homozygous recessive
Statement 2 : If F₁ individual is homozygous, the rate of a monohybrid cross will be 1:1
(a) Statement 1 is correct & Statement 2 is incorrect
(b) Statement 1 is incorrect & Statement 2 is correct
(c) Both Statements 1 & 2 are correct
(d) Both Statements 1 & 2 are incorrect
Answer:
(a) Statement 1 is correct & Statement 2 is incorrect

Question 20.
Identify the incorrect statement
Answer:
(a) In incomplete dominance, the traits are blended not the genes
(b) Incomplete dominance is noticed in Mirabilis jalapa by Carl Correns
(c) It is a type of Intragenic gene interaction
(d) Incomplete dominance F₂ ratio is 1 : 3 : 1
Answer:
(d) Incomplete dominance F₁ ratio is 1 : 3 :1

Question 21.
In case of co-dominance, monohybrid F_{1 __________} is 1 : 2 : 1
(a) Genotype ratio
(b) Phenotype ratio
(c) Both genotype & Phenotype ratio
(d) Ratio is wrong
Answer:
(c) Both genotype & Phenotype ratio

Question 22.
Identify the wrong statement (s)
(i) Monohybrid cross involve the inhertance of teo alleles of a gene
(ii) The dwarf traits reappeared in F₂
(iii) Law of dominance was proved by monohybrid cross
(iv) F₁ monohybrid was an hererozygous
(a) i and ii
(b) iii and iv
(c) i only
(d) none of the above
Answer:
(d) none of the above

Question 23.
Result of incomplete dominance is __________
(а) Intermediate genotype
(b) Intermediate phenotype
(c) Recessive phenotype
(d) Epistasis
Answer:
(b) Intermediate phenotype

Question 24.
Heterozygous Tall mono hybrid is cross with homozygous dwarf. What will be characteristic of offspring?
(a) 25 % recessive 75% dominant
(b) 75 % recessive 25% dominant
(c) 50 % recessive 50% dominant
(d) All are dominance
Answer:
(c) 50 % recessive 50% dominant

Question 25.
ABO blood group is a classical example for __________
(a) Polygenic inheritance
(b) Incomplete Dominance
(c) Epistasis
(d) Dominance
Answer:
(d) Dominance

Question 26.
RR (Red) flower of Mirabilis is crossed with White (WW) flowers. Resultant offspring are pink RW. This is an example of __________
(a) Epistasis
(b) Co-dominance
(c) Incomplete dominance
(d) Pleiotropism
Answer:
(c) Incomplete dominance

Question 27.
How many genetically different gametes are produced by a plant have genotype TtYyRr?
(a) 2
(b) 4
(c) 6
(d) 8
Answer:
(d) 8

Question 28.
When a single gene influences multiple traits then the phenomenon is called __________
(a) Pleiotropy
(b) Polygenic inheritance
(c) Epistasis
(d) Atavism
Answer:
(a) Pleiotropy

Question 29.
According to Mendel which character shown dominance.
(a) Yellow flower color
(b) Yellow cotyledon color
(c) Wrinkled seeds
(d) Inflated pod
Answer:
(d) Inflated pod

Question 30.
Ratio of recessive epistasis is __________
(a) 12 : 3 : 1
(b) 9 : 7
(c) 9 : 3 : 4
(d) 9 : 6 : 1
Answer:
(c) 9 : 3 : 4

Question 31.
According to Mendel, which is not a dominant trait?
(a) Wrinkled seeds
(b) Purple flower
(c) Inflated pod form
(d) Axial flower portion
Answer:
(a) Wrinkled seeds

Question 32.
Identify the allelic interaction.
(a) Domination epistasis
(b) Co – dominance
(c) Recessive epistasis
(d) Duplicate genes
Answer:
(b) Co – dominance

Question 33.
Gametes are never hybrid’ is concluded by __________
(a) Law of dominance
(b) Law of segregation
(c) Law of independent environment
(d) Law of lethality
Answer:
(b) Law of segregation

Question 34.
Factor hypothesis was proposed by __________
(a) Reginald Punnett
(b) W. Bateson
(c) Gregor Mende
(d) Carl Correns
Answer:
(b) W. Bateson

Question 35.
The 1:2:1 ratio of co-dominance process Mendel’s __________
(a) Law of dominance
(b) Law of recessiveness
(c) Law of segregation
(d) Law of independent assortment
Answer:
(b) Law of recessiveness

Question 36.
Match the following:
Epistatic interaction Example
(A) Complementary genes (i) Seed capsule in xxxxx
(B) Supplementary genes (ii) Leaf color in rice plant
(C) Inhibitory genes (iii) Grain color in maize
(D) Duplicate genes (iv) Flower color in sweet peas
Answer:
A – (iv)
B – (iii)
C – (ii)
D – (i)

2 – Mark Questions

Question 1.
Who coined the term genetics? Also define it.
Answer:
“Genetics” is the branch of biological science which deals with the mechanism of transmission of characters from parents to off springs. The term Genetics was introduced by W. Bateson in 1906.

Question 2.
Name the four major subdisciplines of genetics.
Answer:
(a) Classical genetics
(b) Molecular genetics
(c) Population genetics
(d) Quantitative genetics

Question 3.
Define Heredity and variations.
Answer:
Heredity : Heredity is the transmission of characters from parents to off springs.
Variations : The organisms belonging to the same natural population or species that shows a difference in the characteristics is called variation.

Question 4.
Mendel’s theory is a particulate theory – justify.
Answer:
Mendel’s theory of inheritance, known as the Particulate theory, establishes the existence of minute particles or hereditary units or factors, which are now called as genes.

Question 5.
Which organism was studied by Gregor Mendel? How many traits does he considered on his experiments?
Answer:
Gregor Mendel selected seven pairs of characters in Pisum sativum (garden pea)

Question 6.
Name any four characters of pisum sativum that was studied by Mendel.
Answer:
Seed shape, flower color, flower position & pod color.

Question 7.
Define the terms

1. Emasculation
2. Alleles.

Answer:

1. Emasculation : Removal of anthers from the flower
2. Alleles : Alternate forms of a gene

Question 8.
Name the first and second law of Mendel.
Answer:

1. The Law of Dominance
2. The Law of Segregation

Question 9.
What is genotype & phenotype?
Answer:
genotype & phenotype

1. The term genotype is the genetic constitution of an individual.
2. The term phenotype refers to the observable characteristic of an organism.

Question 10.
Write the phenotypic and genotypic ratio of monohybrid cross.
Answer:
(a) Phenotypic ratio = 3:1.
(b) Genotypic ratio =1 : 2 : 1

Question 11.
What is test cross? Why it is done?
Answer:

1. Test cross is crossing an individual of unknown genotype with a homozygous recessive.
2. Test cross is used to identify whether an individual is homozygous or heterozygous for dominant character.

Question 12.
State the law of independent assortment.
Answer:
When two pairs of traits are combined in a hybrid, segregation of one pair of characters is independent to the other pair of characters. Genes that are located in different chromosomes assort independently during meiosis.

Question 13.
Give the phenotypic ratio of
(a) Dihybrid cross
(b) Dihybrid test cross
Answer:
(a) Dihybrid cross ratio = 9 : 3 : 3 : 1
(b) Dihybrid test cross ratio = 1 : 1 : 1 : 1

Question 14.
RrYyf (F₁ hybrid)  rryy (recessive parent). Name the type of cross. Mention its ratio.
Answer:
Dihybrid test cross and the ratio is 1 : 1 : 1 : 1

Question 15.
How many types of gametes are produced by a dihybrid plant. If the same plant is self fertilized, how many second generation offsprings are developed?
Answer:
Four different gametes are produced by a dihybrid plant and on selfing, it yield 16 off springs.

Question 16.
Write the phenotypic ratio of trihybrid cross.
Answer:
27 : 9 : 9 : 9 : 3 : 3 : 3 : 1

Question 17.
Define gene interaction.
Answer:
A single phenotype is controlled by more than one set of genes, each of which has two or more alleles. This phenomenon is called Gene Interaction.

Question 18.
Classify gene interactions with an example.
The gene interactions may be
(a) Intragenic gene interaction. E.g.: Codominance
(b) Intergenic gene interaction. E.g.: Epistasis

Question 19.
Provide any four intergenic gene interactions.
Answer:
(a) Incomplete dominance
(b) Codominance
(c) Multiple alleles
(d) Pleiotropic genes are common examples for intragenic interaction.

Question 20.
Define intragenic interaction
Answer:
Interactions take place between the alleles of the same gene i.e., alleles at the same locus is called intragenic or intralocus gene interaction.

Question 21.
In which plant does the incomplete dominance was studied by Carl Correns? Write the ratio of the cross.
Answer:
Mirabilis Jalapa (4 o’ clock plant). Incomplete dominance ratio is 1 : 2 : 1

Question 22.
What are lethal alleles? Give example.
Answer:
An allele which has the potential to cause the death of an organism is called a Lethal Allele.
E.g : Recessive lethality in Antirrhinum species.

Question 23.
Give the proper terminologies for the following statement
(a) Single gene affecting multiple traits
(b) Single trait affected by many genes.
Answer:
(a) Pleiotropism
(b) Poly genic inheritance

Question 24.
What is intergenic gene interactions? Give example
Answer:
Interlocus interactions take place between the alleles at different loci i.e. between alleles of different genes.
Eg: Dominant Epistasis

Question 25.
Name any two extranuclear inheritance.
Answer:
(a) Chloroplast inheritance
(b) Mitrochondrial inheritance

Question 26.
What are plasmogenes?
Answer:
Plasmogenes are independent, self-replicating, extra-chromosomal units located in cytoplasmic organelles, chloroplast and mitochondrion

Question 27.
What are extra nuclear inheritance?
Answer:
Certain characters/traits are governed and inherited by genes located in cytoplasmic organelles (chloroplast or mitochondrion) other than nucleus. This is called extra nuclear inheritance.

Question 28.
Why extranuclear inheritance is called as cytoplasmic inheritance.
Answer:
Extra nuclear inheritance is due to genes located on the cytoplasmic organelles such as chloroplast and mitochondrion hence it is called cytoplasmic inheritance.

Question 29.
What is cytoplasmic male sterility?
Answer:
In Sorghum vulgare (Pearl maize), the gene located for the sterility pollens are located in the mitochondrial DNA. This phenomenon is called as cytoplasmic male sterility.

3 – Mark Questions

Question 30.
Point out any three importance of variations.
Answer:

1. They help the individuals to adapt themselves to the changing environment.
2. Variations allow breeders to improve better yield, quicker growth, increased resistance and lesser input.
3. They constitute the raw materials for evolution.

Question 31.
Why Mendel selected pea plants for his experiments.
Answer:
He choose pea plant because,

1. It is an annual plant and has clear contrasting characters that are controlled by a single gene separately.
2. Self-fertilization occurred under normal conditions in garden pea plants. Mendel used both self-fertilization and cross-fertilization.
3. The flowers are large hence emasculation and pollination are very easy for hybridization.

Question 32.
State the law of segregation.
Answer:
The Law of Segregation (Law of Purity of gametes): Alleles do not show any blending. During the formation of gametes, the factors or alleles of a pair separate and segregate from each other such that each gamete receives only one of the two factors. A homozygous parent produces similar gametes and a heterozygous parent produces two kinds of gametes each having one allele with equal proportion. Gametes are never hybrid.

Question 33.
How many types of gametes are produced by heterozygous dihybrid plant with a genotypeRrYy? Write them.
Answer:
Four gametes – RY, Ry, rY, ry

Question 34.
Define trihybrid cross. Mention its F₂ phenotypic ratio.
Answer:
A cross between homozygous parents that differ in three gene pairs (i.e. producing trihybrids) is called trihybrid cross, F₂ Phenotypic ratio -27 : 9 : 9 : 9 : 3 : 3 : 3 : 1

Question 35.
Define co-dominance. How it is proved by using Gossypium species?
Answer:
The phenomenon in which two alleles are both expressed in the heterozygous individual is known as codominance. The codominance was demonstrated in plants with the help of electrophoresis or chromatography for protein or flavonoid substance.

Example: Gossypium hirsutum and Gossypium sturtianum, their F₁ hybrid (amphiploid) was tested for seed proteins i by electrophoresis. Both the parents have different banding patterns for their seed proteins. In hybrids, additive banding pattern was noticed. Their hybrid shows the presence of both the types of proteins similar to their parents.

Question 36.
Give an account on cytoplasmic male sterility.
Answer:
Male sterility found in pearl maize (Sorgum vulgare) is the best example for mitochondrial cytoplasmic inheritance. So it is called cytoplasmic male sterility. In this, male sterility is inherited maternally. The gene for cytoplasmic male sterility is found in the mitochondrial DNA.

Question 37.
Write a short note on Atavism.
Answer:
Atavism is a modification of a biological structure whereby an ancestral trait reappears after having been lost through evolutionary changes in the previous generations. Evolutionary traits that have disappeared phenotypically do not necessarily disappear from an organism’s DNA. The gene sequence often remains, but is inactive.

Such an unused gene may remain in the genome for many generations. As long as the gene remains intact, a fault in the genetic control suppressing the gene can lead to the reappearance of that character again. Reemergence of sexual reproduction in the flowering plant Hieracium pilosella is the best example for Atavism in plants.

5 – Mark Questions

Question 38.
Explain Dihybrid cross in pea plant.
Answer:
The crossing of two plants differing in two pairs of contrasting traits is called dihybrid cross. In dihybrid cross, two characters (colour and shape) are considered at a time. Mendel considered the seed shape (round and wrinkled) and cotyledon colour (yellow & green) as the two characters. In seed shape round (R) is dominant over wrinkled (r); in cotyledon colour yellow (Y) is dominant over green (y).

Hence the pure breeding round yellow parent is represented by the genotype RRYY and the pure breeding green wrinkled parent is represented by the genotype rryy. During gamete formation the paired genes of a character assort out ‘ independently of the other pair. During the F₁ x F, fertilization each zygote with an equal probability receives one of the four combinations from each parent. The resultant gametes thus will be genetically different and they are of the following four types:

(1) Yellow round (YR) – 9/16
(2) Yellow wrinkled (Yr) – 3/16
(3) Green round (yR) – 3/16
(4) Green wrinkled (yr) -1/16

These four types of gametes of F₁ dihybrids unite randomly in the process of fertilization and produce sixteen types of individuals in F₂ in the ratio of 9:3:3:1 as shown in the figure. Mendel’s 9:3:3:1 dihybrid ratio is an ideal ratio based on the probability including segregation, independent assortment and random fertilization. In sexually reproducing organism / plants from the garden peas to human beings, Mendel’s findings laid the foundation for understanding inheritance and revolutionized the field of biology. The dihybrid cross and its result led Mendel to propose a second set of generalisations that we called Mendel’s Law of independent assortment.

Question 39.
How does the wrinkled gene make Mendel’s peas wrinkled? Find out the molecular explanation.
Answer:
The protein called starch branching enzyme (SBEI) is encoded by the wild-type allele of the gene (RR) which is dominant. When the seed matures, this enzyme SBEI catalyzes the formation of highly branched starch molecules. Normal gene (R) has become interrupted by the insertion of extra piece of DNA (0.8 kb) into the gene, resulting in allele. In the homozygous mutant form of the gene (R) which is recessive, the activity of the enzyme SBEI is lost resulting in wrinkled peas.

The wrinkled seed accumulates more sucrose and high water content. Hence Ore osmotic pressure inside the seed rises. As a result, the seed absorbs more water and when it matures it loses water as it dries. So it becomes wrinkled at maturation. When the seed has at least one copy of normal dominant gene heterozygous, the dominant allele helps to synthesize starch, amylopectin an insoluble carbohydrate, with the osmotic balance which minimises the loss of water resulting in smooth structured round seed.

Question 40.
Describe incomplete dominance exhibited by Mirabilis jalapa.
Answer:
The German Botanist Carl Correns’s (1905) Experiment – In 4 O’ clock plant, Mirabilis jalapa when the pure breeding homozygous red (R¹R¹) parent is crossed with homozygous white (R²R²), the phenotype of the F₁ hybrid is heterozygous pink (R¹R²). The F₁ heterozygous phenotype differs from both the parental homozygous phenotype. This cross did not exhibit the character of the dominant parent but an intermediate colour pink. When one allele is not completely dominant to another allele it shows incomplete dominance. Such allelic interaction is known as incomplete dominance. F₁ generation produces intermediate phenotype pink coloured flower.

When pink coloured plants of F₁ generation were interbred in F₂ both phenotypic and genotypic ratios were found to be identical as 1 : 2 :1(1 red: 2 pink: 1 white). Genotypic ratio is 1 R¹ R¹ : 2 R¹R² : 1 R²R². From this we conclude that the alleles themselves remain discrete and unaltered proving the Mendel’s Law of Segregation. The phenotypic and genotypic ratios are the same. There is no blending of genes. In the F¹ generation R¹ and R² genes segregate and recombine to produce red, pink and white in the ratio of 1 : 2 : 1. R¹ allele codes for an enzyme responsible for the formation of red pigment. R² allele codes for defective enzyme.

R¹ and R² genotypes produce only enough red pigments to make the flower pink. Two R¹ R² are needed for producing red flowers. Two R²R² genes are needed for white flowers. If blending had taken place, the original pure traits would not have appeared and all F₂ plants would have pink flowers. It is very clear that Mendel’s particulate inheritance takes place in this cross which is confirmed by the reappearance of original phenotype in F₂.

Higher Order Thinking Skills (HOTs) Questions

Question 1.
A yellow colour flower plant indicated by YY is crossed with white color flower plant denoted by yy.
(a) following the Mendelian inheritance pattern, what would be the flower color is first filial generation?
(b) Which Mendelian principle is illustrated in this cross?
(c) Derive the cross and state the phenotypic ratio of yellow flowers to white flowers in F₂ generation?
Answer:
(a) F₁ plants produce yellow colour flower plants.
(b) Law of dominance and Law of segregation
(c)

Question 2.
Mala is a genetic research student. She was given a plant to identify whether it is a homozygous or heterozygous for a particular trait. How will she proceed further?
Answer:
To identify the plant genotype whether homozygous or heterozygous Mala can perform test cross, where the individual is crossed with homozygous recessive for the trait. If the plant is heterozygous then the resultant progenies would be in the ratio 50:50

Question 3.
In the chart given below, ‘AA’ are the genes located in a chromosome of Pisum sativum.
Answer:

Observe the chart and mention the genetic phenomenon does it indicates.
Pleitrophy – A single gene affecting many traits. Here the single gene AA controls the traits – for flower colour, seed colour and leaf axil spot.

Question 4.
Give the F₂ phenotypic ratio of
(a) Supplementary genes
(b) Complementary genes
(c) Dominant epistasis
Answer:
(a) Supplementary genes – 9 : 3 : 4
(b) Complementary genes – 9 : 7
(c) Dominant epistasis -12 : 3 : 1

Question 5.
Name the respective pattern of inheritance where F₁ phenotype
(a) resembles any one of the two parents
(b) is an intermediate between two parental traits.
Answer:
(a) Dominance
(b) Incomplete dominance

Students can Download Computer Science Chapter 7 Python Functions Questions and Answers, Notes Pdf, Samacheer Kalvi 12th 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 12th Computer Science Solutions Chapter 7 Python Functions

Samacheer Kalvi 12th Computer Science Python Functions Text Book Back Questions and Answers

PART – I
I. Choose The Best Answer

Question 1.
A named blocks of code that are designed to do one specific job is called as …………………………
(a) Loop
(b) Branching
(c) Function
(d) Block
Answer:
(c) Function

Question 2.
A Function which calls itself is called as …………………………
(a) Built – in
(b) Recursion
(c) Lambda
(d) Return
Answer:
(b) Recursion

Question 3.
Which function is called anonymous un – named function?
(a) Lambda
(b) Recursion
(c) Function
(d) Define
Answer:
(a) Lambda

Question 4.
Which of the following keyword is used to begin the function block?
(a) Define
(b) For
(c) Finally
(d) Def
Answer:
(d) Def

Question 5.
Which of the following keyword is used to exit a function block?
(a) Define
(b) Return
(c) Finally
(d) Def
Answer:
(b) Return

Question 6.
While defining a function which of the following symbol is used.
(a) ; (semicolon)
(b) . (dot)
(c) : (colon)
(d) $ (dollar)
Answer:
(c) : (colon)

Question 7.
In which arguments the correct positional order is passed to a function?
(a) Required
(b) Keyword
(c) Default
(d) Variable – length
Answer:
(a) Required

Question 8.
Read the following statement and choose the correct statement(s).
(I) In Python, you don’t have to mention the specific data types while defining function.
(II) Python keywords can be used as function name.

(a) (I) is correct and (II) is wrong
(b) Both are correct
(c) (I) is wrong and (II) is correct
(d) Both are wrong
Answer:
(a) (I) is correct and (II) is wrong

Question 9.
Pick the correct one to execute the given statement successfully.
if_: print (x,” is a leap year”)
(a) x % 2 = 0
(b) x % 4 = = 0
(c) x / 4 = 0
(d) x % 4 = 0
Answer:
(b) x % 4 = = 0

Question 10.
Which of the following keyword is used to define the function testpython():?
(a) Define
(b) Pass
(c) Def
(d) While
Answer:
(c) Def

PART – II
II. Answer The Following Questions

Question 1.
What is function?
Answer:
Functions are named blocks of code that are designed to do specific job. If you need to perform that task multiple times throughout your program, you just call the function dedicated to handling that task.

Question 2.
Write the different types of function?
Answer:

1. User – defined Functions
2. Built – in Functions
3. Lambda Functions
4. Recursion Functions

Question 3.
What are the main advantages of function?
Answer:
Main advantages of functions are:

1. It avoids repetition and makes high degree of code reusing.
2. It provides better modularity for your application.

Question 4.
What is meant by scope of variable? Mention its types?
Answer:
Scope of variable refers to the part of the program, where it is accessible, i.e., area where you can refer (use) it. We can say that scope holds the current set of variables and their values. The two types of scopes are: local scope and global scope.

Question 5.
Define global scope?
Answer:
A variable, with global scope can be used anywhere in the program. It can be created by defining a variable outside the scope of any function/block.

Question 6.
What is base condition in recursive function?
Answer:
The condition that is applied in any recursive function is known as base condition. A base condition is must in every recursive function otherwise it will continue to execute like an infinite loop.

Question 7.
How to set the limit for recursive function? Give an example?
Answer:
Python also allows you to change the limit using sys.setrecursionlimit (limit value).
Example:
import sys
sys.setrecursionlimit(3000)
def fact (n):
if n = = 0:
return 1
else:
return n * fact (n – 1)
print (fact (2000))

PART – III
III. Answer The Following Questions

Question 1.
Write the rules of local variable?
Answer:
Rules of local variable:

1. A variable with local scope can be accessed only within the function/block that it is created in.
2. When a variable is created inside the function/block, the variable becomes local to it.
3. A local variable only exists while the function is executing.
4. The formate arguments are also local to function.

Question 2.
Write the basic rules for global keyword in python?
Answer:
The basic rules for global keyword in Python are:

1. When we define a variable outside a function, it’s global by default. You don’t have to use global keyword.
2. We use global keyword to read and write a global variable inside a function.
3. Use of global keyword outside a function has no effect

Question 3.
What happens when we modify global variable inside the function?
Answer:
It will change the global variable value outside the function also.

Question 4.
Differentiate ceil ( ) and floor ( ) function?
Answer:

Question 5.
Write a Python code to check whether a given year is leap year or not?
Leap year or not:
Program code:
Answer:
n = int (input(“Enter any year”))
if (n % 4 = = 0):
print “Leap year”
else:
print “Not a Leap year”
Output:
Enter any year 2001
Not a Leap year

Question 6.
What is composition in functions?
Answer:
The value returned by a function may be used as an argument for another function in a nested manner. This is called composition. For example, if we wish to take a numeric value or an expression as a input from the user, we take the input string from the user using the function input ( ) and apply eval ( ) function to evaluate its value.

Question 7.
How recursive function works?
Answer:

1. Recursive function is called by some external code.
2. If the base condition is met then the program gives meaningful output and exits.
3. Otherwise, function does some required processing and then calls itself to continue recursion.

Question 8.
What are the points to be noted while defining a function?
Answer:

1. Function blocks begin with the keyword “def ” followed by function name and parenthesis ( ).
2. Any input parameters or arguments should be placed within these parentheses when you define a function.
3. The code block always comes after a colon (:) and is indented.
4. The statement “return [expression]” exits a function, optionally passing back an expression to the caller. A “return” with no arguments is the same as return None

PART – IV
IV. Answer The Following Questions

Question 1.
Explain the different types of function with an example?
Answer:
Types of Functions:
Basically, we can divide functions into the following types:

1. User – defined Functions
2. Built – in Functions
3. Lambda Functions
4. Recursion Functions

Functions:
User – defined functions
Built – in functions
Lambda functions
Recursion functions

Description:
Functions defined by the users themselves.
Functions that are inbuilt with in Python.
Functions that are anonymous un-named function.
Functions that calls itself is known as recursive.

(I) Syntax for User defined function
def <function_name ( [parameter1, parameter!…] ) > :
<Block of Statements> return <expression /None>
Example:
def hello ( ):
print (“hello – Python”)
return

Advantages of User – defined Functions:

1. Functions help us to divide a program into modules. This makes the code easier to manage.
2. It implements code reuse. Every time you need to execute a sequence of statements, all you need to do is to call the function.
3. Functions, allows us to change functionality easily, and different programmers can work on different functions.

(II) Anonymous Functions:
In Python, anonymous function is a function that is defined without a name. While normal functions are defined using the def keyword, in Python anonymous functions are defined using the lambda keyword. Hence, anonymous functions are also called as lambda functions.

The use of lambda or anonymous function:

1. Lambda function is mostly used for creating small and one-time anonymous function.
2. Lambda functions are mainly used in combination with the functions like filter ( ), map ( ) and reduce ( ).

Syntax of Anonymous Functions
The syntax for anonymous functions is as follows:
lambda [argument(s)] expression
Example:
sum = lambda argl, arg2: argl + arg2
print (‘The Sum is sum (30, 40))
print (‘The Sum is :’, sum (-30, 40))
Output:
The Sum is: 70
The Sum is: 10
The above lambda function that adds argument argl with argument arg2 and stores the result in the variable sum. The result is displayed using the print ( ).

(III) Functions using libraries:
Built – in and Mathematical functions

(IV) Recursive functions:
When a function calls itself is known as recursion. Recursion works like loop but sometimes 1 ’ it makes more sense to use recursion than loop. You can convert any loop to recursion.
Example:
def fact(n):
if n = = 0:
return 1
else:
return n * fact (n – 1)
print (fact (0))
print (fact (5))
Output:
1
120

Question 2.
Explain the scope of variables with an example?
Answer:
Scope of Variables:
Scope of variable refers to the part of the program, where it is accessible, i.e., area where you can refer (use) it. We can say that scope holds the current set of variables and their values.
The two types of scopes are local scope and global scope.

(I) Local scope:
A variable declared inside the function’s body or in the local scope is called as local variable.

Rules of local variable:

1. A variable with local scope can be accessed only within the function/block that it is created in.
2. When a variable is created inside the function/block; the variable becomes local to it.
3. A local variable only exists while the function is executing.
4. The formate arguments are also local to function.

Example: Create a Local Variable
def loc ( ):
y = 0 # local scope
print (y)
loc ( )
Output:
0
(II) Global Scope:
A variable, with global scope can be used anywhere in the program. It can be created by defining a variable outside the scope of any function/block.

Rules of global Keyword:
The basic rules for global keyword in Python are:

1. When we define a variable outside a function, it’s global by default. You don’t have to useglobal keyword.
2. We use global keyword to read and write a global variable inside a function.
3. Use of global keyword outside a function has no effect

Example: Global variable and Local variable with same name
x = 5 def loc ( ):
x = 10
print (“local x:”, x)
loc ( )
print (“global x:”, x)
Output:
local x: 10
global x: 5
In above code, we used same name ‘x’ for both global variable and local variable. We get different result when we print same variable because the variable is declared in both scopes, i.e. the local scope inside the function loc() and global scope outside the function loc ( ).
The output:- local x: 10, is called local scope of variable.
The output: – global x: 5, is called global scope of variable.

Question 3.
Explain the following built-in functions?
(a) id ( )
(b) chr ( )
(c) round ( )
(d) type ( )
(e) pow ( )
Answer:

Question 4.
Write a Python code to find the L.C.M. of two numbers?
Answer:
Method I using functions:
def lcm (x, y):
if x > y:
greater = x
else:
greater = y while (true):
if ((greater % x = = 0) and (greater % y = = 0)):
lcm = greater break
greater + = 1
return lcm
num 1 = int (input(“Enter first number : “))
num 2 = int (input(“Enter second number : “))
print (“The L.C.M of”, numl, “and”, num, “is”, lcm(num1, num2))

Method II
(without using functions)
a = int (input (“Enter the first number :”))
b = int (input (“Enter the second number :”))
if a > b:
mini = a
else:
min 1 = b
while(1):
if (min 1 % a = = 0 and mini 1 % b = = 0):
print (“LCM is:”, mini)
break
mini = min 1 + 1
Output:
Enter the first number: 15
Enter the second number: 20
LCM is: 60

Question 5.
Explain recursive function with an example?
Answer:
Python recursive functions
When a function calls itself is known as recursion. Recursion works like loop but sometimes it makes more sense to use recursion than loop. You can convert any loop to recursion.
A recursive function calls itself. Imagine a process would iterate indefinitely if not stopped by some condition! Such a process is known as infinite iteration. The condition that is applied in any recursive function is known as base condition. A base condition is must in every recursive function otherwise it will continue to execute like an infinite loop.

Working Principle:

1. Recursive function is called by some external code.
2. If the base condition is met then the program gives meaningful output and exits.
3. Otherwise, function does some required processing and then calls itself to continue recursion. Here is an example of recursive function used to calculate factorial.

Example:
def fact (n):
if n = = 0:
return 1
else:
return n * fact (n – 1)
print (fact (0))
print (fact (5))
Output:
1
120

Practice Programs

Question 1.
Try the following code in the above program?
Answer:

Output:
1. Error

2. Name: Sri
Salary: 3500
Salary: 3500

3. Name: Balu
Salary: 3500

4. Name: Jose
Salary: 1234

5. Name:
Salary: 1234

Question 2.
Evaluate the following functions and write the output?
Answer:

Output:

1. 30
2. 9
3. 8
4. 9

Question 3.
Evaluate the following functions and write the output?
Answer:


Output:
(i) 1. 13
2. 3.2

(ii) 1. 50
2. 36

(iii) <class ‘str’>

(iv) 0b10000

(v). 1. CR (carriage return)
2. It moves the cursor to the beginning of same line

(vi) 1. 8.2
2. 18.0
3. 0.510
4. 0.512

(vii) 1. B
2. a
3. A
4. 6
5. 10

(viii) 1. 0.125
2. 8.0
3. 1

Samacheer kalvi 12th Computer Science Python Functions Additional Questions and Answers

PART – 1
I. Choose The Best Answer

Question 1.
Name of the function is followed by ………………………….
(a) ( )
(b) [ ]
(c) <>
(d) { }
Answer:
(a) ( )

Question 2.
A …………………………. is one or more lines of code, grouped together.
(a) Code –
(b) Block
(c) Function
(d) Arguments
Answer:
(b) Block

Question 3.
A block of code begins when a line is indented by ……………………. spaces usually.
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(c) 4

Question 4.
A block within a block is called …………………………… block.
Answer:
Nested

Question 5.
If the return has no argument, …………………………….. will be displayed as the last statement of the output.
(a) No
(b) None
(c) Nothing
(d) No value
Answer:
(b) None

Question 6.
How many types of functions arguments are there?
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(c) 4

Question 7.
The arguments can be given in improper order in ………………………. arguments.
(a) Required
(b) Keyword
(c) Default
(d) Variable – length
Answer:
(b) Keyword

Question 8.
What is the symbol used to denote variable – length arguments?
(a) +
(b) *
(c) &
(d) ++
Answer:
(b) *

Question 9.
How many methods of arguments passing are there in variable length method.
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(a) 2

Question 10.
Non – keyword variable arguments are called ……………………………….
Answer:
Tuples

Question 11.
In python’s ……………………….. function supports variable length arguments.
(a) Input
(b) Write
(c) Output
(d) Print
Answer:
(d) Print

Question 12.
Lambda functions cannot be used in combination with ………………………….
(a) Filter
(b) Map
(c) Print
(d) Reduce
Answer:
(c) Print

Question 13.
Lambda function can only access ……………………….. variables.
(a) Local
(b) Function
(c) Global
(d) Nested
Answer:
(c) Global

Question 14.
How many types of scopes are there?
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(a) 2

Question 15.
Find the correct one:
(a) Global keyword outside the function has no effect
(b) Global keyword outside the function has effect
Answer:
(a) Global keyword outside the function has no effect

Question 16.
Read the following statement and choose the wrong statements
(a) Without using the global keyword, we cannot modify the global variable
(b) Using global keyword we can modify the global variable
(c) Without global keyword, we can modify the global variable
Answer:
(c) Without global keyword, we can modify the global variable

Question 17.
Find the correct statement.
(a) Local and global variables cannot be used in the same code
(b) Local and global variables can be used in the same code
Answer:
(b) Local and global variables can be used in the same code

Question 18.
The ……………………… function is the inverse of chr ( ) function.
(a) Ord ( )
(b) Abs ( )
(c) Chr ( )
(d) Bin ( )
Answer:
(a) Ord ( )

Question 19.
…………………………. function is the alternative for bin ( ) function.
(a) Ord ( )
(b) Format ( )
(c) Binary ( )
(d) Ord ( )
Answer:
(b) Format ( )

Question 20.
bin ( ) returns the binary string prefixed with ………………………… for the given integer number
(a) b
(b) ob
(c) obin
(d) bin
Answer:
(b) ob

Question 21.
Find the output.
d = 43
print(‘A =ord(d))
(a) 67
(b) 95
(c) 97
(d) 65
Answer:
(d) 65

Question 22.
Find the output,
d = 43
print (chr(d))
(a) –
(b) +
(c) *
(d) /
Answer:
(b) +

Question 23.
The default precision for fixed point number is ………………………….
(a) 2
(b) 4
(c) 6
(d) 8
Answer:
(c) 6

Question 24.
How many formats are there for the format ( ) functions?
(a) 12
(b) 5
(c) 3
(d) 1
Answer:
(c) 3

Question 25.
………………………. function returns the smallest integer greater than or equal to x
(a) Sqrt
(b) Flow
(c) Floor
(d) Cell
Answer:
(d) Cell

Question 26.
………………………. function is used to evaluate the input value.
(a) Input
(b) Valuate
(c) Eval
(d) Val
Answer:
(c) Eval

Question 27.
Any Loop can be converted to recursive functions.
True / False
Answer:
True

Question 28.
Find true statement
(a) Recursive function call itself
(b) Recursive function have to be called externally
Answer:
(a) Recursive function call itself

PART – II
II. Answer The Following Questions.

Question 1.
Define nested blocks?
Answer:
Nested Block:
A block within a block is called nested block. When the first block statement is indented by a single tab space, the second block of statement is indented by double tab spaces.

Question 2.
Give the syntax for passing parameters in functions?
Answer:
Parameters or arguments can be passed to functions
def function _ name (parameter (s) separated by comma):

Question 3.
Differentiate parameters and arguments?
Answer:
Parameters are the variables used in the function definition whereas arguments are the values we pass to the function parameters.

Question 4.
Classify Function Arguments?
Function Arguments:

1. Required arguments
2. Keyword arguments
3. Default arguments
4. Variable – length arguments

Question 5.
Define default arguments?
Answer:
Default Arguments
In Python the default argument is an argument that takes a default value if no value is provided in the function call. The following example uses default arguments, that prints default salary when no argument is passed, def printinfo(sal=3500):

Question 6.
What are the two methods of passing arguments in variable length arguments?
Answer:
In Variable Length arguments we can pass the arguments using two methods.

1. Non keyword variable arguments
2. Keyword variable arguments

PART – III
III. Answer The Following Questions.

Question 1.
Write the output for the program given below?
Answer:
Program:
def printdata (name, age):
print (“Example – 3 Keyword arguments”)
print (“Name :”,name)
print (“Age age)
return
# Now you can call printdata ( ) function
printdata (age = 25, name = “Gshan”)
Output:
Example – 3 Keyword arguments
Name: Gshan
Age: 25

Question 2.
Give the syntax for defining variable – length arguments.?
Syntax – Variable – Length Arguments:
Answer:
def function _name (*args):
function_body
return_statement

Question 3.
Write note on return statement?
Answer:
The return Statement
1. The return statement causes your function to exit and returns a value to its caller. The point of functions in general is to take inputs and return something.

2. The return statement is used when a function is ready to return a value to its caller. So, only one return statement is executed at run time even though the function contains multiple return statements.

3. Any number of ‘return’ statements are allowed in a function definition but only one of them is executed at run time.

Question 4.
Find the output?
Answer:
Program:
Answer:
x = 0 # global variable
def add ( ):
global x
x = x + 5 # increment by 2
print (“Inside add ( ) function x value is:”, x)
add ( )
print (“In main x value is x)
Output:
Inside add ( ) function x value is: 5.
In main x value is: 5

Question 5.
Write note on format function?
Answer:

PART – IV
IV. Answer The Following Questions.

Question 1.
Write any 5 built in functions?
Answer:

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Tamilnadu Samacheer Kalvi 12th Computer Science Solutions Chapter 5 Python – Variables and Operators

Samacheer Kalvi 12th Computer Science Python – Variables and Operators Text Book Back Questions and Answers

PART – 1
I. Choose The Best Answer

Question 1.
Who developed Python?
(a) Ritche
(b) Guido Van Rossum
(c) Bill Gates
(d) Sunder Pitchai
Answer:
(b) Guido Van Rossum

Question 2.
The Python prompt indicates that Interpreter is ready to accept instruction?
(a) >>>
(b) <<<
(c) #
(d) <<
Answer:
(a) >>>

Question 3.
Which of the following shortcut is used to create new Python Program?
(a) Ctrl + C
(b) Ctrl + F
(c) Ctrl + B
(d) Ctrl + N
Answer:
(d) Ctrl + N

Question 4.
Which of the following character is used to give comments in Python Program?
(a) #
(b) &
(c) @
(d) $
Answer:
(a) #

Question 5.
This symbol is used to print more than one item on a single line?
(a) Semicolon
(b) Dollor($)
(c) Comma(,)
(d) Colon(;)
Answer:
(c) Comma(,)

Question 6.
Which of the following is not a token?
(a) Interpreter
(b) Identifiers
(c) Keyword
(d) Operators
Answer:
(a) Interpreter

Question 7.
Which of the following is not a Keyword in Python?
(a) Break
(b) While
(c) Continue
(d) Operators
Answer:
(d) Operators

Question 8.
Which operator is also called as Comparative operator?
(a) Arithmetic
(b) Relational
(c) Logical
(d) Assignment
Answer:
(b) Relational

Question 9.
Which of the following is not Logical operator?
(a) And
(b) Or
(c) Not
(d) Assignment
Answer:
(d) Assignment

Question 10.
Which operator is also called as Conditional operator?
(a) Ternary
(b) Relational
(c) Logical
(d) Assignment
Answer:
(a) Ternary

PART – II
II. Answer The Following Questions

Question 1.
What are the different modes that can be used to test Python Program?
Answer:
In Python, programs can be written in two ways namely Interactive mode and Script mode. The Interactive mode allows us to write codes in Python command prompt (>>>) whereas in script mode programs can be written and stored as separate file with the extension .py and executed. Script mode is used to create and edit python source file.

Question 2.
Write short notes on Tokens?
Answer:
Python breaks each logical line into a sequence of elementary lexical components known as Tokens. The normal token types are;

1. Identifiers
2. Keywords
3. Operators
4. Delimiters
5. Literals

Whitespace separation is necessary between tokens, identifiers or keywords.

Question 3.
What are the different operators that can be used in Python?
Answer:
In computer programming languages operators are special symbols which represent computations, conditional matching etc. The value of an operator used is called operands. Operators are categorized as Arithmetic, Relational, Logical, Assignment etc.

Question 4.
What is a literal? Explain the types of literals?
Answer:
Literal is a raw data given in a variable or constant. In Python, there are various types of literals.

1. Numeric
2. String
3. Boolean

Question 5.
Write short notes on Exponent data?
Answer:
An Exponent data contains decimal digit part, decimal point, exponent part followed by one or more digits.
12.E04, 24.e04 # Exponent data

PART – III
III. Answer The Following Questions

Question 1.
Write short notes on Arithmetic operator with examples?
Answer:
Arithmetic operators:
An arithmetic operator is a mathematical operator that takes two operands and performs a calculation on them. They are used for simple arithmetic. Most computer languages contain a set of such operators that can be used within equations to perform different types of sequential calculations.
Python supports the following Arithmetic operators.

Question 2.
What are the assignment operators that can be used in Python?
Answer:
Assignment operators:
In Python, = is a simple assignment operator to assign values to variable. Let a = 5 and b = 10 assigns the value 5 to a and 10 to b these two assignment statement can also be given as a, b = 5, 10 that assigns the value 5 and 10 on the right to the variables a and b respectively. There are various compound operators in Python like + =, – =,* =, / = % =,** = and //= are also available.

Question 3.
Explain Ternary operator with examples?
Answer:
Conditional operator:
Ternary operator is also known as conditional operator that evaluate something based on a condition being true or false. It simply allows testing a condition in a single line replacing the multiline if – else making the code compact.
The Syntax conditional operator is,
Variable Name = [on – true] if [Test expression] else [on – false]
Example:
min = 50 if 49 < 50 else 70 # min = 50 min = 50 if 49 > 50 else 70 # min = 70

Question 4.
Write short notes on Escape sequences with examples?
Answer:
Escape Sequences:
In Python strings, the backslash “\” is a special character, also called the “escape” character. It is used in representing certain whitespace characters: “\t” is a tab, “\n” is a newline, and “\r” is a carriage return. For example to print the message “It’s raining”, the Python command is >>> print (“It\’s rainning”)
It’s rainning
Python supports the following escape sequence characters.

Question 5.
What are string literals? Explain?
Answer:
String Literals:
In Python a string literal is a sequence of characters surrounded by quotes. Python supports single, double and triple quotes for a string. A character literal is a single character surrounded by single or double quotes. The value with triple – quote is used to give multi – line string literal.
Strings = “This is Python”
char = “C”
multiline _ str = “This is a multiline string with more than one line code”.

PART – IV
IV. Answer The Following Questions

Question 1.
Describe in detail the procedure Script mode programming?
Answer:
Script mode Programming:
Basically, a script is a text file containing the Python statements. Python Scripts are reusable code. Once the script is created, it can be executed again and again without retyping. The Scripts are editable.
Creating Scripts in Python:
(I) Choose File → New File or press Ctrl + N in Python shell window.

(II) An untitled blank script text editor will be displayed on screen

(III) Type the following code in Script editor
a = 100
b = 350
c = a + b
print (“The Sum = “, c)

Saving Python Script
(I) Choose File → Save or Press Ctrl + S

(II) Now, Save As dialog box appears on the screen

(III) In the Save As dialog box, select the location where you want to save your Python code, and type the file name in File Name box. Python files are by default saved with extension .py. Thus, while creating Python scripts using Python Script editor, no need to specify the file extension.

(IV) Finally, click Save button to save your Python script.
Executing Python Script

(I) Choose Run → Run Module or Press F5

(II) If your code has any error, it will be shown in red color in the IDLE window, and Python describes the type of error occurred. To correct the errors, go back to Script editor, make corrections, save the file using Ctrl + S or File → Save and execute it again.

(III) For all error free code, the output will appear in the IDLE window of Python:

Question 2.
Explain input ( ) and print ( ) functions with examples?
Answer:
Input and Output Functions:
A program needs to interact with the user to accomplish the desired task; this can be achieved using Input – Output functions. The input ( ) function helps to enter data at run time by the user and the output function print ( ) is used to display the result of the program on the screen after execution.
The print ( ) function
In Python, the print Q function is used to display result on the screen. The syntax for print Q is as follows:
Example
print (“string to be displayed as output ”)
print (variable)
print (“String to be displayed as output ”, variable)
print (“String1 ”, variable, “String 2”, variable, “String 3”)
Example
>>> print (“Welcome to Python Programming”)
Welcome to Python Programming
>>> x = 5
>>> y = 6
>>> z = x + y
>>> print (z)
11
>>> print (“The sum = ”, z)
The sum = 11
>>> print (“The sum of”, x, “and”, y, “is”, z)
The sum of 5 and 6 is 11
Th print ( ) evaluates the expression before printing it on the monitor. The print ( ) displays an entire statement which is specified within print ( ). Comma ( , ) is used as a separator in print ( ) to print more than one item.
input ( ) function
In Python, input ( ) function is used to accept data as input at run time. The syntax for input ( ) function is,
Variable = input (“prompt string”)
Where, prompt string in the syntax is a statement or message to the user, to know what input can be given.
If a prompt string is used, it is displayed on the monitor; the user can provide expected data from the input device. The input ( ) takes whatever is typed from the keyboard and stores the entered data in the given variable. If prompt string is not given in input ( ) no message is displayed on the screen, thus, the user will not know what is to be typed as input.

Example 1:
input ( ) with prompt string
>>> city = input (“Enter Your City: ”)
Enter Your City: Madurai
>>> print (“I am from “, city)
I am from Madurai

Example 2:
input ( ) without prompt string
>>> city = input ( )
Rajarajan
>>> print (I am from”, city)
I am from Rajarajan
Note that in example – 2, the input ( ) is not having any prompt string, thus the user will not know what is to be typed as input. If the user inputs irrelevant data as given in the above example, then the output will be unexpected. So, to make your program more interactive, provide prompt string with input ( ).
The input ( ) accepts all data as string or characters but not as numbers. If a numerical value is entered, the input values should be explicitly converted into numeric data type. The int ( ) function is used to convert string data as integer data explicitly. We will leam about more such functions in later chapters.

Example 3:
x = int (input(“Enter Number 1: ”))
y = int (input(“Enter Number 2: ”))
print (“The sum =”, x + y)
Output:
Enter Number 1:34
Enter Number 2:56
The sum = 90

Example 4:
Alternate method for the above program
x, y = int (input(“Enter Number 1 :”)), int(input(“Enter Number 2:”))
print (”X = “,x,”Y = “,y)
Output:
Enter Number 1:30
Enter Number 2:50
X = 30 Y= 50

Question 3.
Discuss in detail about Tokens in Python?
Answer:
Tokens:
Python breaks each logical line into a sequence of elementary lexical components known as Tokens. The normal token types are

1. Identifiers
2. Keywords
3. Operators
4. Delimiters
5. Literals.

Whitespace separation is necessary between tokens, identifiers or keywords.

(I) Identifiers:
An Identifier is a name used to identify a variable, function, class, module or object.

1. An identifier must start with an alphabet (A..Z or a..z) or underscore ( _ ).
2. Identifiers may contain digits (0 .. 9)
3. Python identifiers are case sensitive i.e. uppercase and lowercase letters are distinct.
4. Identifiers must not be a python keyword.
5. Python does not allow punctuation character such as %, $, @ etc., within identifiers.

Example of valid identifiers
Sum, total _ marks, regno, num 1

Example of invalid identifiers
12 Name, name$, total – mark, continue

(II) Keywords
Keywords are special words used by Python interpreter to recognize the structure of program. As these words have specific meaning for interpreter, they cannot be used for any other purpose.

(III) Operators
In computer programming languages operators are special symbols which represent computations, conditional matching etc. The value of an operator used is called operands. Operators are categorized as Arithmetic, Relational, Logical, Assignment etc. Value and variables when used with operator are known as operands.

Arithmetic operators:
An arithmetic operator is a mathematical operator that takes two operands and performs a calculation on them. They are used for simple arithmetic. Most computer languages contain a set of such operators that can be used within equations to perform different types of sequential calculations.
Python supports the following Arithmetic operators.

Relational or Comparative operators:
A Relational operator is also called as Comparative operator which checks the relationship between two operands. If the relation is true, it returns True; otherwise it returns False.

Logical operators:
In python, Logical operators are used to perform logical operations on the given relational expressions. There are three logical operators they are and, or and not.

Assignment operators:
In Python, = is a simple assignment operator to assign values to variable. Let a = 5 and b = 10 assigns the value 5 to a and 10 to b these two assignment statement can also be given as a, b = 5, 10 that assigns the value 5 and 10 on the right to the variables a and b respectively. There are various compound operators in Python like + =, – =, * =, / =, % =, ** = and //= are also available.

Conditional operator:
Ternary operator is also known as conditional operator that evaluate something based on a condition being true or false. It simply allows testing a condition in a single line replacing the multiline if – else making the code compact.
The Syntax conditional operator is,
Variable Name = [on _ true] if [Test expression] else [on _ false]
Example:
min = 50 if 49 < 50 else 70 # min = 50 min = 50 if 49 > 50 else 70 # min = 70

(IV) Delimiters
Python uses the symbols and symbol combinations as delimiters in expressions, lists, dictionaries and strings. Following are the delimiters.

(V) Literals
Literal is a raw data given in a variable or constant. In Python, there are various types of literals.

1. Numeric
2. String
3. Boolean

1. Numeric Literals:
Numeric Literals consists of digits and are immutable (unchangeable). Numeric literals can belong to 3 different numerical types Integer, Float and Complex.

2. String Literals:
In Python a string literal is a sequence of characters surrounded by quotes. Python supports single, double and triple quotes for a string. A character literal is a single character surrounded by single or double quotes. The value with triple-quote is used to give multi-line string literal.

3. Boolean Literals:
A Boolean literal can have any of the two values: True or False.

Escape Sequences:
In Python strings, the backslash “\” is a special character, also called the “escape” character. It is used in representing certain whitespace characters: “\t” is a tab, “\n” is a newline, and “\r” is a carriage return. For example to print the message “It’s raining”, the Python command is
>>>print (“It\’s rainning”)
It’s rainning
Python supports the following escape sequence characters.

Samacheer kalvi 12th Computer Science Python – Variables and Operators Additional Questions and Answers

PART – 1
I. Choose The Best Answer:

Question 1.
Python language was released in the year …………………………….
(a) 1991
(b) 1993
(c) 1995
(d) 1997
Answer:
(a) 1991

Question 2.
CWI means ……………………………
Answer:
Centrum Wiskunde & Information

Question 3.
Python got its name from ……………………………
Answer:
Monthly Python’s Flying Circus

Question 4.
Find the wrong statement from the following.
(a) Python supports procedural approaches
(b) Python supports object oriented approaches
(c) Python is DBMS tool
Answer:
(c) Python is DBMS tool

Question 5.
IDLE means ……………………………
Answer:
Integrated Development Learning Environment

Question 6.
How many modes of programming are there in python?
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(a) 2

Question 7.
The extension for python file is ……………………………
(a) .pyt
(b) .pt
(c) .py
(d) .pyth
Answer:
(c) .py

Question 8.
…………………………… mode is used to create and edit python source file.
(a) Line
(b) Script
(c) Interactive
(d) Interface
Answer:
(b) Script

Question 9.
Which mode can be used as a simple calculator?
(a) Line
(b) Script
(c) Interactive
(d) Interface
Answer:
(c) Interactive

Question 10.
What does prompt (>>>) indicator?
(a) Compiler is ready to debug
(b) Results are ready
(c) Waiting for the Input data
(d) Interpreter is ready to accept Instructions
Answer:
(d) Interpreter is ready to accept Instructions

Question 11.
Which command is used to get output ……………………………
(a) Cout
(b) Print
(c) Print f
(d) Write
Answer:
(b) Print

Question 12.
Find the correct one from the following.
(a) Scripts are reusable and not editable
(b) Scripts are not reusable and they are editable
(c) Scripts are not reusuable, not editable
(d) Scripts are both reusable and editable
Answer:
(d) Scripts are both reusable and editable

Question 13.
Which command is selected from file menu create new script text editor?
(a) New
(b) New file
(c) New editor
(d) New Script file
Answer:
(b) New file

Question 14.
What is the default name for blank script text editor?
(a) Untitled
(b) Untitled 1
(c) Document 1
(d) Editor 1
Answer:
(b) Untitled 1

Question 15.
What is the keyboard shortcut to Run?
(a) F5
(b) Alt + F5
(c) Shift + F5
(d) Ctrl + F5
Answer:
(a) F5

Question 16.
…………………………… command is used to execute python script?
(a) Run
(b) Compile
(c) Run → Run Module
(d) Compile → Compile Run
Answer:
(c) Run → Run Module

Question 17.
Errors in the python script appears in …………………………… color.
(a) Yellow
(b) Red
(c) Blue
(d) Black
Answer:
(b) Red

Question 18.
…………………………… function helps to enter data at run time by the user.
Answer:
input ( )

Question 19.
If …………………………… is not given in input ( ), no message is displayed on the screen.
Answer:
Prompt sting

Question 20.
Identify the wrong statement from the following. The input ( ) accepts all data as
(a) Strings
(b) Characters
(c) Numbers
(d) All the above
Answer:
(c) Numbers

Question 21.
The …………………………… function is used to convert string data as integer data explicitly.
Answer:
int ( )

Question 22.
…………………………… are ignored by the python interpreter.
(a) Keywords
(b) Tokens
(c) Delimiters
(d) Comments
Answer:
(d) Comments

Question 23.
Comments are of …………………………… or …………………………… lines.
Answer:
Single, multi

Question 24.
…………………………… are used to indicate blocks of codes in python.
(a) Whitespaces
(b) { }
(c) [ ]
(d) < >
Answer:
(a) Whitespaces

Question 25.
Python breaks each logical line into a sequence of elementary lexical components called ……………………………
Answer:
Tokens

Question 26.
How many types of tokens are there?
(a) 1
(b) 2
(c) 5
(d) 10
Answer:
(c) 5

Question 27.
Pick the odd one out.
Identifiers, Keywords, Delimiters, Comments, Literals
Answer:
Comments

Question 28.
An identifier is a name used to identify a ……………………………
(a) Variable
(b) Function
(c) Class
(d) All of these
Answer:
(d) All of these

Question 29.
Pick the odd one out.
(a) Sum, regno, numl, 12Name, – Marks
(b) False, class, is, as, if, end
(c) Relational, while, logical, Assignment,
(d) + * % ** = =
Answer:
(a) 12 Name, (b) end, (c) while, (d) =

Question 30.
Find the correct statement from the following
(a) Continue is an identifier
(b) Sum is a keyword
(c) ** = is a delimiter
(d) = = is an assignment operator
Answer:
(c) ** = is a delimiter

Question 31.
How many types of operators are there?
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(d) 5

Question 32.
Match the following
1. // – (i) Modulus
2. # – (ii) Floor division
3. % – (iii) Strings
4. ||| ||| – (iv) Comments
(a) 1 – (ii), 2 – (iv), 3 – (i), 4 – (iii)
(b) 1 – (i), 2 – (ii), 3 – (iii), 4 – (iv)
(c) 1 – (iv), 2 – (ii), 3 – (i), 4 – (iii)
(d) 1 – (iv), 2 – (i), 3 – (iii), 4 – (ii)
Answer:
(a) 1 – (ii), 2 – (iv), 3 – (i), 4 – (iii)

Question 33.
…………………………… are special words used by Python Interpreter
Answer:
Keywords

Question 34.
The value of an operator used is ……………………………
(a) 0
(b) 1
(c) Operands
(d) NULL
Answer:
(c) Operands

Question 37.
Match the following expressions with their equivalent output a = 100, b = 10
(1) >>> a % 30 – (i) 100
(2) >>> a // 30 – (ii) 10.0
(3) >>> a/b – (iii) 3
(4) >>> a * b – (iv) 10
(a) 1 – (iv), 2 – (iii), 3 – (ii), 4 – (i)
(b) 1 – (i), 2 – (ii), 3 – (iii), 4 – (iv)
(c) 1 – (i), 2 – (ii), 3 – (iv), 4 – (iii)
(d) 1 – (iv), 2 – (ii), 3 – (iii), 4 – (i)
Answer:
(a) 1 – (iv), 2 – (iii), 3 – (ii), 4 – (i)

Question 38.
Which operator checks the relationship between two operands.
(a) Relational
(b) Comparative
(c) Both
(d) None of the these
Answer:
(c) Both

Question 39.
Assume a = 100 and b = 35. Find the true statements.
(i) >>> a > b
(ii) >>> a = = b
(iii) >>> a ! = b
(a) (i), (iii) are true
(b) (ii), (iii) are true
(c) (i), (ii) are true
Answer:
(a) (i), (iii) are true

Question 40.
Identify Not equal to operator in python?
(a) < >
(b) ==
(c) NOT EQUAL
(d) ! =
Answer:
(d) ! =

Question 41.
How many logical operators are there?
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(b) 3

Question 42.
How many comparative operators are there?
(a) 4
(b) 5
(c) 6
(d) 7
Answer:
(c) 6

Question 43.
…………………………… is the simple assignment operator.
(a) ! =
(b) >
(c) >>
(d) =
Answer:
(d) =

Question 44.
Compound operators comes under the category of …………………………… operators.
Answer:
Assingnment

Question 45.
Find the wrongly matched pair.
(a) ! = – relational operator
(b) not a > b – Comparative Operator
(c) **= – delimitors operator
(d) Assingnment Operator
Answer:
(b) not a > b – Comparative Operator

Question 46.
Identify which is not a delimiter
(a) & =
(b) :
(c) ;
(d) ::
Answer:
(d) ::

Question 47.
How many types of literals are there?
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(b) 3

Question 48.
Numeric literals are ……………………………
(a) Integer, Float, Complex
(b) Int, Float, Void
(c) Int, Float, Char
(d) Int, Float, Boolean
Answer:
(a) Integer, Float, Complex

Question 49.
Strings in python are represented using
(a) ”
(b) “”
(c) “‘ “‘
(d) All of these
Answer:
(d) All of these

Question 50.
Multiline string literal is given by ……………………………
Answer:
“‘ “‘ triple quotes

Question 51.
The two values accepted by Boolean literals are …………………………… or ……………………………
Answer:
True, False

Question 52.
…………………………… is the escape character.
(a) +
(b) t
(c) \
(d) %
Answer:
(c) \

Question 53.
Which one of the following is the newline character?
(a) \t
(b) \v
(c) \r
(d) \n
Answer:
(d) \n

Question 54.
…………………………… is the escape character for carriage return.
Answer:
\r

Question 55.
Pick the odd one out.
(a) Tuples, for, list, dictionaries, Number
(b) \n, \”, V, \r, \k
(c) and, or, not, true
(d) >, > =, <, < =, < >
Answer:
(a) for, (b) \k, (c) true, (d) <>

Question 56.
How many Interger data are there?
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(b) 3

Question 57.
OX represents …………………………… integer.
Answer:
Hexadecimal

Question 58.
Octal integer uses …………………………… to denote octal digits
(a) OX
(b) O
(c) OC
(d) Od
Answer:
(b) O

Question 59.
Find the hexadecimal Integer.
(a) 0102
(b) 0876
(c) 0432
(d) 0X102
Answer:
(d) 0X102

Question 60.
How many floating point values are there in a complex number?
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(b) 2

Question 61.
What is the another name for fundamental data type?
(a) Class
(b) Built – in
(c) Typedef
(d) User defined
Answer:
(b) Built – in

Question 62.
Which one of the following statement is wrong?
(a) Octal Integer uses upper and lower case O
(b) Hexadecimal Integer uses upper and lower case OX
(c) Long integer uses upper and lower case 1.
Answer:
(c) Long integer uses upper and lower case 1.

PART – II
II. Answer The Following Questions.

Question 1.
How will you create scripts in python?
Answer:
(I) Choose File → New File or press Ctrl + N in Python shell window.

(II) An untitled blank script text editor will be displayed on screen.

Question 2.
Write note on keywords. Give examples?
Answer:
Keywords are special words used by Python interpreter to recognize the structure of program. As these words have specific meaning for interpreter, they cannot be used for any other purpose. Eg, While, if.

PART – III
III. Answer The Following Questions.

Question 1.
What are the key features of python?
Answer:
Key features of Python:
It is a general purpose programming language which can be used for both scientific and non – scientific programming
It is a platform independent programming language.
The programs written in Python are easily readable and understandable

Question 2.
How will you execute python script?
Executing Python Script
(I) Choose Run → Run Module or Press F5

(II) If your code has any error, it will be shown in red color in the IDLE window, and Python describes the type of error occurred. To correct the errors, go back to Script editor, make corrections, save the file using Ctrl + S or File → Save and execute it again.

(III) For all error free code, the output will appear in the IDLE window of Python.

Question 3.
Give the syntax for print ( ) function?
Answer:
The syntax for print ( ) is as follows:
Example:
print (“string to be displayed as output ” )
print (variable)
print (“String to be displayed as output ”, variable)
print (“String 1 ”, variable, “String 2”, variable, “String 3”)

Question 4.
Explain the Syntax of input ( ) function?
Answer:
The syntax for inputQ function is,
Variable = input (“prompt string”)
Where, prompt string in the syntax is a statement or message to the user, to know what input can be given.

If a prompt string is used, it is displayed on the monitor; the user can provide expected data from the input device. The input) ) takes whatever is typed from the keyboard and stores the entered data in the given variable. If prompt string is not given in input() no message is displayed on the screen

Question 5.
Give an example for input with and without prompt string?
Example 1: input ( ) with prompt string
>>> city = input (“Enter Your City: ”)
Enter Your City: Madurai

Example 2: input ( ) without prompt string
>>> city = input ( )
Rajarajan

Question 6.
Write a short note on comments?
Answer:
Comments in Python
In Python, comments begin with hash symbol (#). The lines that begins with # are considered as comments and ignored by the Python interpreter. Comments may be single line or no multi – lines. The multiline comments should be enclosed within a set of # as given below.
# It is Single line Comment
# It is multiline comment
which contains more than one line #

Question 7.
Mention some rules for Identifiers?
Answer:

1. An identifier must start with an alphabet (A..Z or a..z) or underscore (_).
2. Identifiers may contain digits (0 .. 9)
3. Python identifiers are case sensitive i.e. uppercase and lowercase letters are distinct.
4. Identifiers must not be a python keyword.
5. Python does not allow punctuation character such as %,$, @ etc., within identifiers.

Question 8.
Give any 6 keywords in Python?
Answer:
false, class, none, continue, finally, return, is

Question 9.
Give an example program for Ternary operator?
To test Conditional (Ternary) Operator:
# Program to demonstrate conditional operator
a, b = 30, 20
# Copy value of a in min if a < b else copy b
min = a if a < b else b print (“The Minimum of A and B is “,min) # End of the Program Output: The Minimum of A and B is 20.

Question 10.
Write any 6 delimiters in python?
Answer:

PART – IV
IV. Answer The Following Questions

Question 1.
Explain operators in detail?
Answer:
Operators: In computer programming languages operators are special symbols which represent computations, conditional matching etc. The value of an operator used is called operands. Operators are categorized as Arithmetic, Relational, Logical, Assignment etc. Value and variables when used with operator are known as operands.

(I) Arithematic operators An arithmetic operator is a mathematical operator that takes two operands and performs a calculation on them. They are used for simple arithmetic. Most computer languages contain a set of such operators that can be used within equations to perform different types of sequential calculations.

Python supports the following Arithmetic operators.

To test Arithmetic Operators: #Demo Program to test Arithmetic Operators a = 100 b = 10 print (“The Sum = “,a + b) print (“The Difference = “,a – b) print (“The Product = “,a * b) print (“The Quotient = “,a / b) print (“The Remainder = “,a % 30) print (“The Exponent = “, a ** 2) print (“The Floor Division =”, a // 30) #Program End Output: The Sum = 110 The Difference = 90 The Product = 1000 The Quotient = 10.0 The Remainder = 10 The Exponent = 10000 The Floor Division = 3

(II) Relational or Comparative operators A Relational operator is also called as Comparative operator which checks the relationship between two operands. If the relation is true, it returns True; otherwise it returns False.

To test Relational Operators: #Demo Program to test Relational Operators a = int (input(“Enter a Value for A:”)) b = int (input(“Enter a Value for B:”)) print (“A = “,a,” and B = “,b) print (“The a = = b = “,a = = b) print (“The a > b = “,a > b)
print (“The a < b = “,a < b) print (“The a > = b = “, a > = b)
print (“The a! = b = “,a! = 0)
print (“The a! = b = “,a! = b)
#Program End

Output:
Enter a Value for A: 35
Enter a Value for B: 56
A = 35 and B = 56
The a = =b = False
The a > b = False
The a < b = True The a > = b = False
The a < = b = False
The a ! = b = True

(III) Logical operators
In python, Logical operators are used to perform logical operations on the given relational expressions. There are three logical operators they are and, or and not.

To test Logical Operators:

(IV) Assignment operators
In Python, = is a simple assignment operator to assign values to variable. Let a = 5 and b = 10 assigns the value 5 to a and 10 to b these two assignment statement can also be given as a, b = 5, 10 that assigns the value 5 and 10 on the right to the variables a and b respectively. There are various compound operators in Python like + =, – =, * =, /=, % =, ** = and //= are also available.

To test Assingnment Operators:

(V) Conditional operator
Ternary operator is also known as conditional operator that evaluate something based on a condition being true or false. It simply allows testing a condition in a single line replacing the multiline if – else making the code compact.
The Syntax conditional operator is,
Variable Name = [on _ true] if [Test expression] else [on _ false]
Example:
min = 50 if 49 < 50 else 70 # min = 50 min = 50 if 49 > 50 else 70 # min = 70
To test Conditional (Ternary) Operator:
# Program to demonstrate conditional operator
a, b = 30, 20
# Copy value of a in min if a < b else copy b
min = a if a < b else b print (“The Minimum of A and B is “,min) # End of the Program Output: The Minimum of A and B is 20.

Question 2.
Write the Output for the given program?
Answer:

Question 3.
Explain literals and its types?
Answer: Literal is a raw data given in a variable or constant. In Python, there are various types of literals.

(I) Numeric Literals Numeric Literals consists of digits and are immutable (unchangeable). Numeric literals can belong to 3 different numerical types Integer, Float and Complex. To demonstrate Numeric literals # Program to demonstrate Numeric Literals a = Ob 1010 #Binary Literals b = 100 #Decimal Literal c = 0o310 #Octal Literal d = Ox 12c #Hexadecimal Literal print (“Integer Literals :”,a, b, c, d) #Float Literal float_1 = 10.5 float_2 = 1.5e2 print (“Float Literals :”,float_1, float_2) #Complex Literal x = 1 + 3.14 j print (“Complex Literals:”, x) Print (“x = “, x , “Imaginary part of x = “, x.imag, “Real part of x = “, x.real) #End of the Program Output: Integer Literals: 10 100 200 300 Float Literals: 10.5 150.0 Complex Literals: x = (1.3.14) Imaginary part of x = 3.14 Real part of 9 x = 1.0

(II) String Literals: In Python a string literal is a sequence of characters surrounded by quotes. Python supports single, double and triple quotes for a string. A character literal is a single character surrounded by single or double quotes. The value with triple-quote is used to give multi – line string literal. To test String Literals # Demo Program to test String Literals strings = “This is Python” char = “C” multiline_str = ‘”This is a multiline string with more than one line code.'” print (strings) print (char) print (multiline_str) # End of the Program Output: This is Python C C This is a multiline string with more than one line code.

(III) Boolean Literals A Boolean literal can have any of the two values: True or False. # Demo Program to test String Literals boolean _ 1 = True boolean _ 2 = False print (“Demo Program for Boolean Literals”) print (“Boolean Value 1 :”,boolean_1) print (“Boolean Value2 :”,boolean_2) # End of the Program Output: Demo Program for Boolean Literals Boolean Value 1: True Boolean Value2: False (iv) Escape Sequences In Python strings, the backslash “\” is a special character, also called the “escape” character. It is used in representing certain whitespace characters: “\t” is a tab, “\n” is a newline, and “\r” is a carriage return. For example to print the message “It’s raining”, the Python command is >>> print (“ItVs rainning”)
It’s rainning
Python supports the following escape sequence characters.

Question 4.
Explain data types in python?
Answer:
Python Data types:
All data values in Python are objects and each object or value has type. Python has Built-in or Fundamental data types such as Number, String, Boolean, tuples, lists and dictionaries.

Number Data type:
The built – in number objects in Python supports integers, floating point numbers and complex numbers.
Integer Data can be decimal, octal or hexadecimal. Octal integer use O (both upper and lower case) to denote octal digits and hexadecimal integer use OX (both upper and lower case) and L (only upper case) to denote long integer.
Example:
102, 4567, 567 # Decimal integers
0102, o876, 0432 # Octal integers
0X102, oX876, 0X432 # Hexadecimal integers
34L, 523L # Long decimal integers
A floating point data is represented by a sequence of decimal digits that includes a decimal point. An Exponent data contains decimal digit part, decimal point, exponent part followed by one or more digits.
Example :
123.34, 456.23, 156.23 # Floating point data
12.E04, 24.e04 # Exponent data
Complex number is made up of two floating point values, one each for the real and imaginary parts.

Boolean Data type:
A Boolean data can have any of the two values: True or False.

Example:
Bool_varl = True
Bool_var2 = False

String Data type:
String data can be enclosed with single quote or double quote or triple quote.

Example:
Char_data = ‘A’
String_data = “Computer Science”
Multiline_data= “““String data can be enclosed with single quote or double quote or triple quote.”””

Students can Download Accountancy Chapter 4 Goodwill in Partnership Accounts Questions and Answers, Notes Pdf, Samacheer Kalvi 12th 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 12th Accountancy Solutions Chapter 4 Goodwill in Partnership Accounts

Samacheer Kalvi 12th Accountancy Goodwill in Partnership Accounts Text Book Back Questions and Answers

I. Choose the Correct Answer

Question 1.
Which of the following statements is true?
(a) Goodwill is an intangible asset
(b) Goodwill is a current asset
(c) Goodwill is a fictitious asset
(d) Goodwill
Answer:
(a) Goodwill is an intangible asset

Question 2.
Super profit is the difference between ………………..
(a) Capital employed and average profit
(b) Assets and liabilities
(c) Average profit and normal profit
(d) Current year’s profit and average profit
Answer:
(c) Average profit and normal profit

Question 3.
The average rate of return of similar concerns is considered as ………………..
(a) Average profit
(b) Normal rate of return
(c) Expected rate of return
(d) None of these
Answer:
(b) Normal rate of return

Question 4.
Which of the following is true?
(a) Super profit = Total profit / number of years
(b) Super profit = Weighted profit / number of years
(c) Super profit = Average profit – Normal profit
(d) Super profit = Average profit x Years of purchase
Answer:
(c) Super profit = Average profit – Normal profit

Question 5.
Identify the incorrect pair ………………..
(a) Goodwill under Average profit method – Average profit x Number of years of purchase
(b) Goodwill under Super profit method – Super profit x Number of years of purchase
(c) Goodwill under Annuity method – Average profit x Present value of annuity factor
(d) Goodwill under Weighted average profit method – Weighted average profit x Number of years of purchase
Answer:
(c) Goodwill under Annuity method – Average profit x Present value of annuity factor

Question 6.
When the average profit is ₹ 25,000 and the normal profit is ₹ 15,000, super profit is ………………..
(a) ₹ 25,000
(b) ₹ 5,000
(c) ₹ 10,000
(d) ₹ 15,000
Answer:
(c) ₹ 10,000

Question 7.
Book profit of 2017 is ₹ 35,000; non – recurring income included in the profit is ₹ 1,000 and abnormal loss charged in the year 2017 was ₹ 2,000, then the adjusted profit is ………………..
(a) ₹ 36,000
(b) ₹ 35,000
(c) ₹ 38,000
(d) ₹ 34,000
Answer:
(a) ₹ 36,000

Question 8.
The total capitalised value of a business is ₹ 1,00,000; assets are ₹ 1,50,000 and liabilities are ₹ 80,000. The value of goodwill as per the capitalisation method will be ………………..
(a) ₹ 40,000
(b) ₹ 70,000
(c) ₹ 1,00,000
(d) ₹ 30,000
Answer:
(d) ₹ 30,000

II. Very Short Answer Questions

Question 1.
What is goodwill?
Answer:
Goodwill is the good name or reputation of the business which brings benefit to the business. It enables the business to earn more profit. It is the present value of a firm’s future excess earnings. It is an intangible asset as it has no physical existence.

Question 2.
What has acquired goodwill?
Answer:
Goodwill acquired by making payment in cash or kind is called acquired or purchased goodwill. When a firm purchases an existing business, the price paid for purchase of such business may exceed the net assets (Assets – Liabilities) of the business acquired.

Question 3.
What is a super profit?
Answer:
Super profit is the excess of average profit over the normal profit of a business.
Super profit = Average profit – Normal profit.
Average profit is calculated by dividing the total of adjusted actual profit of certain number of years by the total number of such years. Normal profit is the profit earned by the similar business firms under normal conditions.
Normal profit = Capital employed x Normal rate of return Capital employed = Fixed assets + Current assets – Current liabilities

Question 4.
What is the normal rate of return?
Answer:
Normal rate of return = It is the rate at which profit is earned by similar business entities in the industry under normal circumstances.

Question 5.
State any two circumstances under which goodwill of a partnership firm is valued?
Answer:

1. When there is a change in the profit-sharing ratio.
2. When a new partner is admitted into a firm.
3. When an existing partner retires from the firm or when a partner dies.
4. When a partnership firm is dissolved.

III. Short Answer Questions

Question 1.
State any six factors determining goodwill.
Answer:

1. The profitability of the firm
2. Favourable location of the business enterprises
3. good quality of goods or services offered
4. Tenure of the business enterprises
5. Efficiency of management.
6. Degree of Competition

Question 2.
How is goodwill calculated under the super-profits method?
Answer:
1. Purchase of super profit method: Goodwill is calculated by multiplying the super profit by a certain number of years of purchase.
Goodwill = super profit x No. of years of purchase

2. Annuity method: The value of goodwill is calculated by multiplying the super profit with the present value of the annuity.
Goodwill = Super profit x Present value annuity factor

3. Capitalisation of super profit method: Goodwill = x 100

Question 3.
How is the value of goodwill calculated under the capitalisation method?
Answer:
Capitalisation method:
Under Capitalisation method, goodwill is the excess of capitalised value of average profit of the business over the actual capital employed in the business.
Goodwill = Total capitalised value of the business – Actual capital employed
The total capitalised value of the business is calculated by capitalising the average profits on the basis of the normal rate of return.
Capitalised value of the business = x 100
Actual capital employed = Fixed assets (excluding goodwill) + Current assets – Current liabilities

Question 4.
Compute average profit from the following information.
Answer:
Calculation of Average profit:
2016 – ₹ 8,000; 2017 – ₹ 10,000; 2018 – ₹ 9,000

Valuation of goodwill = ₹ 9,000

Question 5.
Calculate the value of goodwill at 2 years purchase of average profit when the average profit is ₹ 15,000.
Answer:
Goodwill: ₹ 30,000

IV. Exercises

Simple average profit method:
Question 1.
The following are the profits of a firm in the last five years:

Answer:

Valuation of goodwill = Average profit x No. of years purchase
= ₹ 12,000 x 2 years
= ₹ 24,000

Question 2.
From the following information, calculate the value of goodwill on the basis of 3 years purchase of average profits of last four years.

Answer:
Calculation of goodwill:

Valuation of goodwill = Average profit x No. of years purchase = ₹ 4,000 x 3
= ₹ 12,000

Question 3.
From the following information relating to a partnership firm, find out the value of its goodwill based on 3 years purchase of average profits of the last 4 years:

1. Profits of the years 2015, 2016, 2017 and 2018 are ₹ 10, 000, ₹ 12, 500, ₹ 12, 000 and ₹ 11, 500, respectively.
2. The business was looked after by a partner and his fair remuneration amounts to ₹ 1, 500 per year. This amount was not considered in the calculation of the above profits.

Answer:
Valuation of goodwill
Calculation of average profit Year Profit

(-) Remuneration = ₹ 1,500
Net average profit = ₹ 10,000
Valuation of goodwill = Average profit x No. of years purchase
= ₹ 10,000 x 3
= ₹ 30,000

Question 4.
From the following information relating to Sridevi enterprises, calculate the value of goodwill on the basis of 4 years purchase of the average profits of 3 years.

1. Profits for the years ending 31^st December 2016, 2017 and 2018 were ₹ 1,75,000, ₹ 1,50,000 and ₹ 2,00,000, respectively.
2. A non – recurring income of ₹ 45,000 is included in the profits of the year 2016.
3. The closing stock of the year 2017 was overvalued by ₹ 30,000.

Answer:
Calculation of adjusted profit

Average profit = <
Average profit = \(\frac { 4, 80, 000 }{ 3 }\) = ₹ 1,60,000
Goodwill = Average profit x No. of year purchase = ₹ 1,60,000 x 4
= ₹ 64,00,000

Question 5.
The following particulars are available in respect of the business carried on by a partnership firm:

1. Profits earned: 2016: ₹ 25,000; 2017: ₹ 23,000 and 2018: ₹ 26,000.
2. Profit of 2016 includes a non – recurring income of ₹ 2,500.
3. Profit of 2017 is reduced by ₹ 3,500 due to stock destroyed by fire.
4. The stock was not insured. But, it is decided to insure the stock in future. The insurance premium is estimated to be ₹ 250 per annum.

You are required to calculate the value of goodwill of the firm on the basis of 2 years purchase of average profits of the last three years.
Answer:
Calculation of adjusted profit

Total profit = 22,250 + 26, 250 + 25, 750 = ₹ 74, 250

Valuation of goodwill = Average profit x No.of year purchase
= ₹ 24, 750 x 2 years = ₹ 49, 500

Weighted average profit method:
Question 6.
Find out the value of goodwill at three years purchase of weighted average profit of last four years.

Answer:
Calculation of weighted average profit

Goodwill = Weighted average profit x No. of years purchase
= ₹ 15,400 x 3 = ₹ 46,200

Purchase of super profit method:
Question 7.
From the following details, calculate the value of goodwill at 2 years purchase of super profit:

1. Total assets of a firm are ₹ 5,00,000
2. The liabilities of the firm are ₹ 2,00,000
3. Normal rate of return in this class of business is 12.5 %.
4. Average profit of the firm is ₹ 60,000.

Answer:
Capital employed = fixed assets + current assets – current liabilities
= 5, 00, 000 – 2, 00, 000 = 3, 00, 000
Normal profit = Capital employed x Normal rate of return
= 3,00,000 x \(\frac { 12.5 }{ 100 }\) = 3, 75, 000
Super profit = Average profit – Normal profit
= 60, 000 – 37, 500 = 22, 500
Goodwill = Super profit x Number of years of purchase
= ₹ 22,500 x 2
= ₹ 45,000

Question 8.
A partnership firm earned net profits during the last three years as follows:
2016: ₹ 20,000; 2017: ₹ 17,000 and 2018: ₹ 23,000
The capital investment of the firm throughout the above mentioned period has been ₹ 80,000. Having regard to the risk involved, 15% is considered to be a fair return on capital employed in the business. Calculate the value of goodwill on the basis of 2 years purchase of super profit.
Answer:
Calculation of average profit Year Profit

Normal profit = Capital employed x Normal rate of return
= 80,000 x \(\frac { 15 }{ 100 }\) = 12,000
Super profit = 8,000
Valuation of goodwill = Super profit x No. of years purchase
= ₹ 8,000 x 2
= ₹ 16,000

Annuity method:
Question 9.
From the following information, calculate the value of goodwill under annuity method:

1. Average profit – ₹ 14,000
2. Normal profit – ₹ 4,000
3. Normal rate of return – 15%
4. Years of purchase of goodwill – 5

Present value of ₹ 1 for 5 years at 15% per annum as per the annuity table is 3.352
Answer:
Super profit = Average profit – Normal profit
= 14, 000 – ₹ 4, 000 = ₹ 10, 000
Goodwill = Super profit x Present value of annuity factor
= ₹ 10,000 x 3.352 = ₹ 33, 520

Capitalisation of super profit method:
Question 10.
Find out the value of goodwill by capitalising super profits:

1. Normal Rate of Return 10%
2. Profits for the last four years are ₹ 30,000, ₹ 40,000, ₹ 50,000 and ₹ 45,000.
3. Anon – recurring income of ₹ 3,000 is included in the above mentioned profit of ₹ 30,000.
4. Average capital employed is ₹ 3,00,000.

Answer:

Average profit = = \(\frac { 1, 62, 000 }{ 4 }\)
= ₹ 40,500
Normal profit = Capital employed x Normal rate of return
= 3,00,00 x \(\frac { 10 }{ 100 }\) = 30,000
Super profit = 10,500
Capitalisation super profit method = x 100
= \(\frac { 10, 500 }{ 10 }\) = ₹ 1, 05, 000

Capitalisation method:
Question 11.
From the following information, find out the value of goodwill by capitalisation method :

1. Average profit ₹ 20, 000
2. Normal rate of return 10%
3. Tangible assets of the firm ₹ 2, 20, 000
4. Liabilities of the firm ₹ 70, 000

Answer:
Capital Assets – Liabilities
= 2, 20, 000 – 70, 000 = ₹ 1, 50, 000
Capitalised value of business = x 100
= \(\frac { 20, 000 }{ 10 }\) x 100 = ₹ 2, 00, 000
Value of goodwill = Total capitalised average profit – Capital employed
= 2, 00, 000 – 1, 50, 000
= ₹ 50,000

Samacheer Kalvi 12th Accountancy Goodwill in Partnership Accounts Additional Questions and Answers

I. Choose the correct answer

Question 1.
Goodwill is valued under ………………
(a) Average profit method
(b) Super profit method
(c) Capitalisation method
(d) All of these
Answer:
(d) All of these

Question 2.
Average profit method can be further divided ………………
(a) Simple average profit
(b) Weighted average profit
(c) Both (a) and (b)
(d) None of these
Answer:
(c) Both (a) and (b)

Question 3.
Super profit is
(a) Average profit – Normal profit
(b) Normal profit – Average profit
(c) Both (a) and (b)
(d) None of these
Answer:
(a) Average profit – Normal profit

Question 4.
Kinds of goodwill ………………
(a) Purchased goodwill
(b) Self – generated goodwill
(c) None of these
(d) Both (a) and (b)
Answer:
(d) Both (a) and (b)

Question 5.
Nature of goodwill is ………………
(a) Reputation of business firm
(b) Profit of firm
(c) total asset of firm
(d) None of these
Answer:
(a) Reputation of business firm

Question 6.
Factor of goodwill is ………………
(a) Location of the business
(b) Efficient management
(c) Nature of the goods
(d) All of these
Answer:
(b) Efficient management

Question 7.
Arrange the following in ascending or descending order:
1. Super profit = Weighted average profit x No. of years purchase
2. Capitalistion method = x 100
3. Weighted Average method = Average profit – Normal profit
4. Capitalisation super profit = Total profit/No. of years purchase
(a) 1, 2, 3, 4
(b) 2, 4, 1, 2
(c) 3, 4, 1, 2
(d) 4, 3, 2, 1
Answer:
(c) 3, 4, 1, 2

Question 8.
Goodwill of the firm on the basis of 2 years purchase of average profit of the last 3 years is ₹ 25,000. Find average profit ………………
(a) ₹ 50,000
(b) ₹ 25,000
(c) ₹ 10,000
(d) ₹ 12,500
Answer:
(a) ₹ 50,000

Question 9.
Calculate the value of goodwill at 3 years purchase when capital employed is ₹ 2,50,000. Average profit ₹ 30,000 and normal rate of return is 10% ………………
(a) ₹ 3,000
(b) ₹ 25,000
(c) ₹ 30,000
(d) ₹ 15,000
Answer:
(d) ₹ 15,000

Question 10.
Adjusted profit = Actual profit
(a) + past expenses not required in the future
(b) – past revenue not likely to be eached in the future
(c) + additional income expected in the future
(d) All of these
Answer:
(d) All of these

II. Fill in the blanks:

Question 11.
Goodwill brought in cash by new partner is divided among the old partner by debiting is ________ A/c and crediting ________ A/c.
Answer:
Goodwill; capital.

Question 12.
Calculation of goodwill under simple average profit method is ________
Answer:
Total profit =

Question 13.
Super profit method normal profit is ________
Answer:
Normal profit = Capital employed x Normal rate of return

Question 14.
Under average profit method ________
Answer:
x 100

Question 15.
Super profit = ________
Answer:
Average profit – Normal profit

III. Short answer questions

Question 1.
What are the nature of goodwill?
Answer:
The nature of the goodwill can be described as follows:

1. Goodwill is an tangible fixed asset. It is tangible because it has no physical existence. It cannot be seen or touched.
2. It has a definite value depending on the profitability of the business enterprise.
3. It cannot be separated from the business.
4. It helps in earning more profit and attracts more customers.
5. It can be purchased or sold only when the business is purchased or sold in full or in part.

Question 2.
What is average profit method?
Answer:
Under this method, goodwill is calculated as certain years of purchase of average profits of the past years. The number of years of purchase is generally determined on the basis of the average period a new business will take in order to bring it to the current state of the existing business.

Question 3.
What is meant by simple average profit method?
Answer:
Goodwill is calculated by multiplying the average profit by a certain number of years of purchase. Simple average profit is calculated by adding the adjusted profits of certain number of years by dividing the total number of such years.
Goodwill = Average profit x Number of years purchase Total profit.
Average profit =

Question 4.
Weighted average profit method.
Answer:
Goodwill is calculated by multiplying the weighted average profit by a certain number of years of purchase.
Goodwill = Weighted average profit x Number of years purchase
In this method, weights are assigned to each year’s profit. Weighted profit is ascertained by multiplying the weights assigned with the respective year’s profit.
Weighted average profit =

Question 5.
What is meant by annuity method of valuation of goodwill?
Answer:
Value of goodwill is calculated by multiplying the super profit with the present value of annuity.
Goodwill = Super profit x Present value annuity factor
Present value annuity factor is the present value of annuity of rupee one at a given time. It can be found out from annuity table or by using formula.

Question 6.
What is annuity factor?
Answer:
Annuity refers to series of uniform cash flows at regular intervals. The table value gives the present value of annuity of rupee one received at the end of every year for a specified number of years.

where, i = interest rate ; n = estimated number of years

Question 7.
What is capitalisation of super profit method?
Answer:
Under this method, value of goodwill is calculated by capitalising the super profit at normal rate of return, that is, goodwill is the capitalised value of super profit.
Goodwill = x 100

IV. Exercise

Question 1.
Goodwill is to be valued at three years purchase of five years average profits. The profit for the last five years were 2010 – ₹ 4,200; 2011 — ₹ 4,500; 2012 – ₹ 4,700; 2013 – ₹ 4,600; and 2014 – ₹ 5,000.
Solution:
Calculation of average profit:

Calculation of goodwill = Average profit x No. of years purchase
= ₹ 4,600 x 3 = ₹ 13,800

Question 2.
A firm’s profit for the last 5 years were ₹ 20,000, ₹ 30,000, ₹ 40,000. ₹ 50,000, and ₹ 60,000. Calculated the value of firm’s goodwill on the basis of three years purchase of weighted average profit after using weight of 1, 2, 3, 4, 5 respectively.
Solution:

Goodwill = Weighted avg. profit x Number of years purchase = 46,667 x 3 = ₹ 1,40,000 (appr.)

Question 3.
A business has earned average profit ₹ 1,00,000 during the last few years and the normal rate of return in similar business is 10%. Find out the value of goodwill.

1. Capitalisation of super profit method.
2. Super profit method of the goodwill is valued at 3 years purchase of super profit. The assets of the business were ₹ 10,00,000 and liabilities of ₹ 1,80,000.

Solution:
1. Capital employed = Assets – libilities
= 10,00,000 – 1,80,000
= ₹ 8,20,000
Normal profit = Capital employed x
8,20,000 = 1 x \(\frac { 10 }{ 100 }\) = 82,000
Super profit = Average profit – Normal profit
= ₹ 1,00,000 – ₹ 82,000 = ₹ 18,000

2. Capitalisation method Super profit
Goodwill = x 100
= \(\frac { 18, 000 }{ 10 }\) = ₹ 1, 80,000
as per super profit method Goodwill = super profit x No. of years purchase
= ₹ 18000 x 3 = ₹ 54,000

Question 4.
The average profit earned by the firm is Z 80,000 which includes under valuation of stock an average basis the capital invested in the business and normal rate of return is 8%. Calculate the goodwill of the firm on the basis of times the super profit.
Solution:
Average profit – ₹ 80,000
(A) Under value of stock – ₹ 8,000
Actual Average profit – ₹ 88,000
Normal profit = Capital investment x Normal rate of return
= ₹ 88,000 – ₹ 64,000
= ₹ 24,000
Goodwill = Super profit x 7
= ₹ 24,000 x 7
= ₹ 1,68,000

Question 5.
Capital investment is ₹ 5,00,000; firms profit ₹ 1,50,000 Assuming that normal rate of return is 20%. Calculate the goodwill

1. Capitalisation method.
2. Super profit method if the goodwill is valued @ 2 years purchase

Solution:
1. Capitalisation of average profit method.
Goodwill = x 100
= \(\frac { 1,50,000 }{ 20 }\) = ₹ 7,50,000
Goodwill = Total capital employed – capital employed
= ₹ 7,50,000 – ₹ 5,00,000 = ₹ 2,50,000

2. Super profit method
Normal profit = Capital employed x
Normal profit = ₹ 5,00,000 x \(\frac { 20 }{ 100 }\) = ₹ 1,00,000
Super profit = Average profit – Normal profit
= ₹ 1,50,000 – 1,00,000
= ₹ 50,000
Value of Goodwill = Super profit x No. of years purchase
= ₹ 50,000 x 2
= ₹ 1,00,000

Question 6.
Calculate the value of goodwill of the firm of 2 partners
(a) At the 3 years purchase of average profits
(b) At 3 years purchase of super profits
(c) On the basis of capitalisation of super profits
(d) On the basis of capitalisation of average profits

(i) Average capital employed ₹ 7,00,000
(ii) Net trading results of the firm 2014 – ₹ 1,47,600 2015 – Loss ₹ 1,48,100. Profit for 2016 – ₹ 4,48,700
(iii) Rate of interest on capital @ 18%
(iv) Remunuration ₹ 500/- per month
Solution:
Calculation of Average profit and super profit
Total profit = ₹ 1,47,600 – ₹ 1,48,100 + ₹ 4,48,700 = ₹ 4,48,200
Average profit = = ₹ 1,49,400
Less: Remunration of 2 partner (2 x 500 x 1/2) \(\frac { 12,000 }{ 1,37,400 }\)
Less: Normal profit = Capital employed x
= 7,00,000 x \(\frac { 18 }{ 100 }\) = ₹ 1,26,000
Super profit = ₹ 11,400

(a) Average profit – Average profit x No. of year purchase
= ₹ 1,37,400 x 3 = ₹ 4,12,200

(b) Super profit – Super profit x No. of year purchase
= ₹ 11,400 x 3 = ₹ 34,200

(c) Capitalisation of super profit – x 100
= \(\frac { 11,400 }{ 18 }\) x 100
= ₹ 63,333

(d) Capitalisation of average profit = x 100
= \(\frac { 1,37,400 }{ 100 }\) x 100
= ₹ 7,63,333
(-) Actual capital employed = ₹ 7,00,000
Goodwill = ₹ 63,333

Students can Download Accountancy Chapter 5 Admission of a Partner Questions and Answers, Notes Pdf, Samacheer Kalvi 12th 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 12th Accountancy Solutions Chapter 5 Admission of a Partner

Samacheer Kalvi 12th Accountancy Admission of a Partner Text Book Back Questions and Answers

I. Choose the Correct Answer

Question 1.
Revaluation A/c is a ……………..
(a) Real A/c
(b) Nominal A/c
(c) Personal A/c
(d) Impersonal A/c
Answer:
(b) Nominal A/c

Question 2.
On revaluation, the increase in the value of assets leads to ……………..
(a) Gain
(b) Loss
(c) Expense
(d) None of these
Answer:
(a) Gain

Question 3.
The profit or loss on revaluation of assets and liabilities is transferred to the capital account of ……………..
(a) The old partners
(b) The new partner
(c) All the partners
(d) The Sacrificing partners
Answer:
(a) The old partners

Question 4.
If the old profit sharing ratio is more than the new profit sharing ratio of a partner, the difference is called ……………..
(a) Capital ratio
(b) Sacrificing ratio
(c) Gaining ratio
(d) None of these
Answer:
(b) Sacrificing ratio

Question 5.
At the time of admission, the goodwill brought by the new partner may be credited to the capital accounts of ……………..
(a) all the partners
(b) the old partners
(c) the new partner
(d) the sacrificing partners
Answer:
(d) the sacrificing partners

Question 6.
Which of the following statements is not true in relation to admission of a partner?
(a) Generally mutual rights of the partners change
(b) The profits and losses of the previous years are distributed to the old partners
(c) The firm is reconstituted under a new agreement
(d) The existing agreement does not come to an end
Answer:
(d) The existing agreement does not come to an end

Question 7.
Match List I with List II and select the correct answer using the codes given below:

Answer:
(b) 3 2 4 1

Question 8.
Select the odd one out:
(a) Revaluation profit
(b) Accumulated loss
(c) Goodwill brought by new partner
(d) Investment fluctuation fund
Answer:
(c) Goodwill brought by new partner

Question 9.
James and Kamal are sharing profits and losses in the ratio of 5:3. They admit Sunil as a partner giving him 1/5 share of profits. Find out the sacrificing ratio.
(a) 1:3
(b) 3:1
(c) 5:3
(d) 3:5
Answer:
(c) 5:3

Question 10.
Balaji and Kamalesh are partners sharing profits and losses in the ratio of 2:1. They admit Yogesh into partnership. The new profit sharing ratio between Balaji, Kamalesh and Yogesh is agreed to 3:1:1. Find the sacrificing ratio between Balaji and Kamalesh.
(a) 1:3
(b) 3:1
(c) 2:1
(d) 1:2
Answer:
(d) 1:2

II. Very Short Answer Questions

Question 1.
What is meant by the revaluation of assets and liabilities?
Answer:
When a partner is admitted into the partnership, the assets and liabilities are revealed as the current value may differ from the book value. Determination of current values of assets and liabilities is called revaluation of assets and liabilities.

Question 2.
How are accumulated profits and losses distributed among the partners at the time of admission of a new partner?
Answer:
Profits and losses of previous years which are not distributed to the partners are called accumulated profit and losses. This belongs to the old partners and hence these should be distributed to the old partners in the old profit sharing ratio.

Question 3.
What is sacrificing ratio?
Answer:
The sacrificing ratio is the proportion of the profit which is sacrificed or foregone by the old partners in favour of the new partner.
The share sacrificed is calculated by deducting the new share from the old share.
Share sacrificed = Old share – New share

Question 4.
Give the journal entry for writing off existing goodwill at the time of admission of a new partner.
Answer:
Journal Entry

Question 5.
State whether the following will be debited or credited in the revaluation account.

1. Depreciation on assets
2. Unrecorded liability
3. Provision for outstanding expenses
4. Appreciation of assets

Answer:

1. Depreciation on assets – Debited
2. Unrecorded liability – Debited
3. Provision for outstanding expenses – Debited
4. Appreciation of assets – Debited

III. Short Answer Questions

Question 1.
What are the adjustments required at the time of admission of a partner?
Answer:
The following adjustment is necessary at the time of admission of a partner:

• Distribution of accumulated profits, reserves, and losses
• Revaluation of assets and liabilities
• Determination of new profit-sharing ration and sacrificing ratio
• Adjustment for goodwill
• Adjustment of capital on the basis of new profit sharing ratio (if so agreed)

Question 2.
What are the journal entries to be passed on revaluation of assets and liabilities?
Answer:

Question 3.
Write a short note on the accounting treatment of goodwill.
Answer:
1. For the goodwill brought in cash credited to old partner’s capital account

2. For the goodwill brought in kind (in the form of the asset) credited to the old partner’s capital account

3. For withdrawal of cash received for goodwill by the old partners

III. Exercises

Question 1.
Arul and Anitha are partners sharing profits and losses in the ratio of 4 : 3. On 31.3.2018, Ajay was admitted as a partner. On the date of admission, the book of the firm showed a general reserve of ₹ 42,000. Pass the journal entry to distribute the general reserve.
Answer:

Question 2.
Anjali and Nithya are partners of firm sharing profits and losses in the ratio of 5 : 3. They admit Pramila on 1.1.2018. On that date, their balance sheet showed an accumulated loss of ₹ 40,000 on the asset side of the balance sheet. Give the journal entry to transfer the accumulated loss on admission.
Answer:

Question 3.
Oviya and Kavya are partners in firm sharing profits and losses in the ratio of 5:3. They admit Agalya into the partnership. Their balance sheet as on 31^st March, 2019 is as follows:
Balance Sheet as on 31^st March 2019

Pass journal entry to transfer the accumulated profits and reserve on admission.

Question 4.
Hari, Madhavan and Kesavan are partners, sharing profits and losses in the ratio of 5:3:2. As of 1^st April 2017, Vanmathi is admitted into the partnership and the new profit sharing ratio is decided as 4:3:2:1. The following adjustments are to be made.

1. Increase the value of premises by ₹ 60,000.
2. Depreciate stock by ₹ 5,000, furniture by ₹ 2,000 and machinery by ₹ 2,500.
3. Provide for an outstanding liability of ₹ 500.

Pass journal entries and prepare revaluation account.
Journal Entries

Revaluation Account

Question 5.
Seenu and Siva are partners sharing profits and losses in the ratio of 5:3. In view of Kowsalya admission, they decided

1. To increase the value of the building by ₹ 40,000.
2. To bring into record investments at ₹ 10,000, which have not so far been brought into account.
3. To decrease the value of machinery by ₹ 14,000 and furniture by ₹ 12,000.
4. To write off sundry creditors by ₹ 16,000.

Pass journal entries and prepare revaluation account.

Revaluation Account

Question 6.
Sai and Shankar are partners, sharing profits and losses in the ratio of 5 : 3. The firm’s balance sheet as on 31^st December 2017, was as follows:

On 31^st December 2017 Shanmugam was admitted into the partnership for 1/4 share of profit with ₹ 12,000 as capital subject to the following adjustments.

1. Furniture is to be revalued at ₹ 5,000 and building is to be revalued at ₹ 50,000
2. Provision for doubtful debts is to be increased to ₹ 5,500
3. An unrecorded investment of ₹ 6,000 is to be brought into account
4. An unrecorded liability ₹ 2,500 has to be recorded now

Pass journal entries and prepare Revaluation Account and capital account of partners after admission.
Journal Entries

Revaluation Account

Capital Account

Question 7.
Amal and Vimal are partners in firm sharing profits and losses in the ratio of 7 : 5. Their balance sheet as on 31^st March, 2019, is as follows:

Nirmal is admitted as a new partner on 1.4.2018 by introducing a capital of 30,000 for 1/3 share in the future profit subject to the following adjustments.

1. Stock to be depreciated by ₹ 5,000
2. Provision for doubtful debts to be created for ₹ 3,000
3. Land to be appreciated by ₹ 20,000

Prepare revaluation account and capital account of partners after admission.
Answer:
Revaluation Account

Capital Account

Question 8.
Praveena and Dhanya are partners sharing profits in the ratio of 7 : 3. They admit Malini into the firm. The new ratio among Praveena, Dhanya and Malini is 5 : 2 : 3. Calculate the sacrificing ratio.
Answer:
Sacrificing Ratio = Old Ratio – New Ratio
Praveena = \(\frac { 7 }{ 10 }\) – \(\frac { 5 }{ 10 }\) = \(\frac { 2 }{ 10 }\)
Dhanya = \(\frac { 3 }{ 10 }\) – \(\frac { 2 }{ 10 }\) = \(\frac { 1 }{ 10 }\)
Sacrificing Ratio = 2 : 1

Question 9.
Ananth and Suman are partners sharing profits and losses in the ratio of 3 : 2. They admit Saran for 1/5 share, which he acquires entirely from Ananth. Find out the new profit sharing ratio and sacrificing ratio.
Answer:
New Profit Sharing Ratio:
Ananth = \(\frac { 3 }{ 5 }\) – \(\frac { 1 }{ 5 }\) = \(\frac { 2 }{ 5 }\)
Suman = \(\frac { 2 }{ 5 }\) – = \(\frac { 2 }{ 5 }\)
Ananth = \(\frac { 1}{ 5 }\) – = \(\frac { 1 }{ 5 }\)
New Profit sharing Ratio = 2 : 2 : 1
Sacrificing Ratio = 1 : 0

Question 10.
Raja and Ravi are partners, sharing profits in the ratio of 3 : 2. They admit Ram for 1/4 share of the profit. He takes 1/20 share from Raja and 4/20 from Ravi. Calculate the new profit sharing ratio and sacrificing ratio.
Answer:
New Profit Sharing Ratio = Old Ratio – Sacrificing Ratio
Raja = \(\frac { 3 }{ 5 }\) – \(\frac { 1 }{ 20 }\) = \(\frac { 12 – 1 }{ 20 }\) = \(\frac { 11 }{ 20 }\)
Ravi = \(\frac { 2 }{ 5 }\) – \(\frac { 4 }{ 20 }\) = \(\frac { 8 – 4 }{ 20 }\) = \(\frac { 4 }{ 20 }\)
Ram = \(\frac { 1 }{ 20 }\) + \(\frac { 4 }{ 20 }\) = \(\frac { 5 }{ 20 }\)
New Profit sharing Ratio = 11 : 4 : 5
Sacrificing Ratio = \(\frac { 1 }{ 20 }\) : \(\frac { 4 }{ 20 }\)

Question 11.
Vimala and Kamala are partners, sharing profits and losses in the ratio of 4:3. Vinitha enters into the partnership and she acquires 1/14 from Vimala and 1/14 from Kamala. Find out the new profit sharing ratio and sacrificing ratio.
Answer:
New Profit Sharing Ratio = Old Ratio – Sacrificing Ratio
Vimala = \(\frac { 4 }{ 7 }\) – \(\frac { 1 }{ 14 }\) = \(\frac { 8 – 1 }{ 14 }\) = \(\frac { 7 }{ 14 }\)
Kamala = \(\frac { 3 }{ 7 }\) – \(\frac { 1 }{ 14 }\) = \(\frac { 6 – 1 }{ 14 }\) = \(\frac { 5 }{ 14 }\)
Vinitha = \(\frac { 1 }{ 14 }\) + \(\frac { 1 }{ 14 }\) = \(\frac { 2 }{ 14 }\)
New Profit Sharing Ratio = 7 : 5 : 2
Sacrificing Ratio = 1 : 1

Question 12.
Govind and Gopal are partners in a firm sharing profits in the ratio of 5 : 4. They admit Rahim as a partner. Govind surrenders 2/9 of his share in favour of Rahim. Gopal surrenders 1/9 of his share in favour of Rahim. Calculate the new profit sharing ratio and sacrificing ratio.
Answer:
New Profit Sharing Ratio:
Govind = \(\frac { 5 }{ 9 }\) x \(\frac { 2 }{ 9 }\) = \(\frac { 10 }{ 81 }\) ; \(\frac { 5 }{ 9 }\) – \(\frac { 10 }{ 81 }\) = \(\frac { 45 – 10 }{ 81 }\) = \(\frac { 35 }{ 81 }\)
Gopal = \(\frac { 4 }{ 9 }\) x \(\frac { 1 }{ 9 }\) = \(\frac { 4 }{ 81 }\) ; \(\frac { 3 }{ 8 }\) – \(\frac { 3 }{ 64 }\) = \(\frac { 24 – 3 }{ 64 }\) = \(\frac { 21 }{ 64 }\)
Rahim = \(\frac { 2 }{ 9 }\) + \(\frac { 1 }{ 9 }\) = \(\frac { 3 }{ 9 }\) ; \(\frac { 10 }{ 81 }\) + \(\frac { 4 }{ 81 }\) = \(\frac { 14 }{ 81 }\)
New Profit Sharing Ratio = 35 : 32 : 14
Sacrificing Ratio = \(\frac { 10 }{ 81 }\) : \(\frac { 4 }{ 81 }\) = 5 : 2

Question 13.
Prema and Chandra share profits in the ratio of 5:3. Hema is admitted as a partner. Prema surrendered 1/8 of her share and Chandra surrendered 1/8 of her share in favour of Hema. Calculate the new profit sharing ratio and sacrificing ratio.
Answer:
New Profit Sharing Ratio:
Prema = \(\frac { 5 }{ 8 }\) x \(\frac { 1 }{ 8 }\) = \(\frac { 5 }{ 64 }\) ; \(\frac { 5 }{ 8 }\) – \(\frac { 5 }{ 64 }\) = \(\frac { 40 – 5 }{ 64 }\) = \(\frac { 35 }{ 64 }\)
Chandra = \(\frac { 3 }{ 8 }\) x \(\frac { 1 }{ 8 }\) = \(\frac { 3 }{ 64 }\) ; \(\frac { 3 }{ 8 }\) – \(\frac { 3 }{ 64 }\) = \(\frac { 24 – 3 }{ 64 }\) = \(\frac { 21 }{ 64 }\)
Hema = \(\frac { 5 }{ 64 }\) + \(\frac { 3 }{ 64 }\) = \(\frac { 8 }{ 64 }\)
New Profit Sharing Ratio = 35 : 21 : 8
Sacrificing Ratio = 5 : 3

Question 14.
Karthik and Kannan are equal partners. They admit Kailash with 1/4 share of the profit. Kailash acquired his share from old partners in the ratio of 7:3. Calculate the new profit sharing ratio and sacrificing ratio.
New Profit Sharing Ratio:
Karthik = \(\frac { 1 }{ 4 }\) x \(\frac { 7 }{ 10 }\) = \(\frac { 7 }{ 40 }\) ; \(\frac { 1 }{ 2 }\) – \(\frac { 7 }{ 40 }\) = \(\frac { 20 – 7 }{ 40 }\) = \(\frac { 13 }{ 40 }\)
Kannan = \(\frac { 1 }{ 4 }\) x \(\frac { 3 }{ 10 }\) = \(\frac { 3 }{ 40 }\) ; \(\frac { 1 }{ 2 }\) – \(\frac { 3 }{ 40 }\) = \(\frac { 20 – 3 }{ 40 }\) = \(\frac { 17 }{ 40 }\)
Kailash = \(\frac { 1 }{ 4 }\) x 10 = \(\frac { 10 }{ 40 }\)
New Profit Sharing Ratio = 13 : 17 : 10
Sacrificing Ratio = 7 : 3

Question 15.
Selvam and Senthil are partners sharing profit in the ratio of 2:3. Siva is admitted into the firm with 1/5 share of profit. Siva acquires equally from Selvam and Senthil. Calculate the new profit sharing ratio and sacrificing ratio.
Answer:
New Profit Sharing Ratio:
Siva’s share = \(\frac { 1 }{ 5 }\) x \(\frac { 1 }{ 2 }\) = \(\frac { 1 }{ 10 }\)
Selvam’s share = \(\frac { 2 }{ 5 }\) – \(\frac { 1 }{ 10 }\) = \(\frac { 4 – 1 }{ 10 }\) = \(\frac { 3 }{ 10 }\)
Senthil’s share = \(\frac { 3 }{ 5 }\) – \(\frac { 1 }{ 10 }\) = \(\frac { 6 – 1 }{ 10 }\) = \(\frac { 5 }{ 10 }\)
Siva’s share = \(\frac { 1 }{ 10 }\) + \(\frac { 1 }{ 10 }\) = \(\frac { 2 }{ 10 }\)
New Profit sharing Ratio = 3 : 5 : 2
Sacrificing Ratio = 1 : 1

Question 16.
Mala and Anitha are partners, sharing profits and losses in the ratio of 3 : 2. Mercy is admitted into the partnership with 1/5 share in the profits. Calculate new profit sharing ratio and sacrificing ratio.
Answer:
Calculate New Profit Sharing Ratio = 1 – \(\frac { 1 }{ 5 }\) = \(\frac { 4 }{ 5 }\)
Mala = \(\frac { 3 }{ 5 }\) x \(\frac { 4 }{ 5 }\) = \(\frac { 12 }{ 25 }\)
Anitha = \(\frac { 2 }{ 5 }\) x \(\frac { 4 }{ 5 }\) = \(\frac { 8 }{ 25 }\)
Mercy = \(\frac { 1 }{ 5 }\) x 5 = \(\frac { 5 }{ 25 }\)
New Profit Sharing Ratio = 12 : 8 : 5
Sacrificing Ratio = Old share – New share
Mala = \(\frac { 3 }{ 5 }\) – \(\frac { 12 }{ 25 }\) = \(\frac { 15 – 12}{ 25 }\) = \(\frac { 3 }{ 25 }\)
Anitha = \(\frac { 2 }{ 5 }\) – \(\frac { 8 }{ 25 }\) = \(\frac { 2 }{ 25 }\)
Sacrificing Ratio = 3 : 2

Question 17.
Ambika, Dharani, and Padma are partners in firm sharing profits in the ratio of 5:3:2. They admit Ramya for 25% profit. Calculate the new profit sharing ratio and sacrificing ratio.
Answer:
Ramya share = 25% = 100% – 25% = 75% = \(\frac { 3 }{ 4 }\)
Ambika’s share = \(\frac { 5 }{ 10 }\) x \(\frac { 3 }{ 4 }\) = \(\frac { 15 }{ 40 }\)
Dharani’s share = \(\frac { 3 }{ 10 }\) x \(\frac { 3 }{ 4 }\) = \(\frac { 9 }{ 40 }\)
Padma’s share = \(\frac { 2 }{ 10 }\) x \(\frac { 3 }{ 4 }\) = \(\frac { 6 }{ 40 }\)
Mercy’s share
New Profit Sharing Ratio
Sacrificing Ratio = Old share – New share
Ambika’s share = \(\frac { 5 }{ 10 }\) – \(\frac { 15 }{ 40 }\) = \(\frac { 20 – 15 }{ 40 }\) = \(\frac { 5 }{ 40 }\)
Dharani’s share = \(\frac { 3 }{ 10 }\) – \(\frac { 9 }{ 40 }\) = \(\frac { 12 – 9 }{ 40 }\) = \(\frac { 3 }{ 40 }\)
Padma’s share = \(\frac { 2 }{ 10 }\) – \(\frac { 6 }{ 40 }\) = \(\frac { 8 – 6 }{ 40 }\) = \(\frac { 2 }{ 40 }\)
Sacrificing Ratio = 5 : 3 : 2

Question 18.
Aparna and Priya are partners who share profits and losses in the ratio of 3 : 2. Brindha joins the firm for 1/5 share of profits and brings in cash for her share of the goodwill of ₹ 10,000. Pass necessary journal entry for adjusting goodwill on the assumption that the fluctuating capital method is followed and the partners withdraw the entire amount of their share of goodwill.
Adjustment of goodwill
Journal Entries

Question 19.
Deepak, Senthil, and Santhosh are partners sharing profits and losses equally. They admit Jerald into a partnership for 1/3 share in future profits. The goodwill of the firm is valued at ₹ 45,000 and Jerald brought cash for his share of goodwill. The existing partners withdraw half of the amount of their share of goodwill. Pass necessary journal entries for adjusting goodwill on the assumption that the fluctuating capital method is followed.
Journal Entries

Jerald’s Share of goodwill = ₹ 45000 x 1/3 = ₹ 1,500
As the sacrifice made by the existing partners is not mentioned. It is assumed that they sacrifice in their old share profit ratio = 1 : 1 : 1.
Therefore, Sacrificing ratio = 1 : 1 : 1.

Question 20.
Malathi and Shobana are partners sharing profits and losses in the ratio of 5 : 4. They admit Jayasri into a partnership for 1/3 share of profit. Jayasri pays cash ₹ 6,000 towards her share of goodwill. The new ratio is 3 : 2 : 1. Pass necessary journal entry for adjusting goodwill on the assumption that the fixed capital method is followed.
Adjusting Goodwill: Old share – New Share
Malathi’s share = \(\frac { 5 }{ 9 }\) – \(\frac { 3 }{ 6 }\) = \(\frac { 10 – 9 }{ 18 }\) = \(\frac { 1 }{ 18 }\)
Shobana’s share = \(\frac { 4 }{ 9 }\) – \(\frac { 2 }{ 6 }\) = \(\frac { 8 – 6 }{ 18 }\) = \(\frac { 2 }{ 18 }\)

Question 21.
Anu and Arul were partners in firm sharing profits and losses in the ratio of 4 : 1. They have decided to admit Mano into the firm for 2/5 share of profits. The goodwill of the firm on the date of admission was valued at ₹ 25,000. Mano is not able to bring in cash for his share of goodwill. Pass necessary journal entry for goodwill on the assumption that the fluctuating capital method is followed.
Answer:
Mano’s shares: 25000 x \(\frac { 2 }{ 5 }\) = Rs 10000

Question 22.
Varun and Barath are partners sharing profits and losses 5 : 4. They admit Dhamu into partnership. The new profit sharing ratio is agreed at 1 : 1 : 1. Dhamu’s share of goodwill is valued at ₹ 15,000 of which he pays ₹ 10,000 in cash. Pass necessary journal entries for adjustment of goodwill on the assumption that the fluctuating capital method is followed.
Answer:
Dhamu’s share of goodwill against sacrificing ratio:
Sacrificing Ratio = Old share – New share
Varun = \(\frac { 5 }{ 9 }\) – \(\frac { 1 }{ 3 }\) = \(\frac { 5 – 3 }{ 9 }\) = \(\frac { 2 }{ 9 }\)
Barath = \(\frac { 4 }{ 9 }\) – \(\frac { 1 }{ 3 }\) = \(\frac { 4 – 3 }{ 9 }\) = \(\frac { 1 }{ 9 }\)
Goodwill value = \(\frac { 15000 }{ 3 }\) = ₹ 5000.

Question 23.
Sam and Jose are partners in a firm sharing profits and losses in the ratio of 3 : 2. On 1^st April 2018, they admitted Joel as a partner. On the date of Joel’s admission, goodwill appeared in the books of the firm at ₹ 30,000. By assuming fluctuating capital method, pass the necessary journal entry if the partners decide to

1. write off the entire amount of existing goodwill
2. write off ₹ 20,000 of the existing goodwill.

Answer:
Journal Entries
(a) Write off the entire amount of existing goodwill

(b) Write off ₹ 20,000 of the existing goodwill

Question 24.
Raj an and Selva are partners sharing profits and losses in the ratio of 3:1. Their balance sheet as on 31^st March 2017 is as under:

On 1.4.2017, they admit Ganesan as a new partner on the following arrangements:

1. Ganesan brings ₹ 10,000 as capital for 1/5 share of profit.
2. Stock and furniture are to be reduced by 10%, a reserve of 5% on debtors for doubtful debts is to be created.
3. Appreciate buildings by 20%.

Prepare revaluation account, partners’ capital account, and the balance sheet of the firm after admission.
Revaluation Account

Capital Account

Balance Sheet

Question 25.
Sundar and Suresh are partners sharing profits in the ratio of 3:2. Their balance sheet as on 1^st January, 2017 was as follows:

They decided to admit Sugumar into a partnership for 1/4 share in the profits on the following terms:

1. Sugumar has to bring in ₹ 30,000 as capital. His share of goodwill is valued at ₹ 5,000. He could not bring cash towards goodwill.
2. That the stock is valued at ₹ 20, 000.
3. That the furniture is depreciated by ₹ 2,000.
4. That the value of the building is depreciated by 20%.

Prepare necessary ledger accounts and the balance sheet after admission.
Answer:
Revaluation Account

Capital Account

Balance Sheet as on 31.12.17

Question 26.
The following is the balance sheet of Janies and Justina as of 1.1.2017. They share profits and losses equally.

On the above date, Balan is admitted as a partner with a 1/5 share in future profits. Following are the terms for his admission:

1. Balan brings ₹ 25,000 as capital.
2. His share of goodwill is ₹ 10,000 and he brings cash for it.
3. The assets are to be valued as under:

Building ₹ 80,000; Debtors ₹ 18,000; Stock ₹ 33,000 Prepare necessary ledger accounts and the balance sheet after admission.
Answer:
Revaluation Account

Capital Account

Cash Account

Balance Sheet as on 01.01.2017

Question 27.
Anbu and Shankar are partners in a business sharing profits and losses in the ratio of 7 : 5. The balance sheet of the partners on 31.03.2018 is as follows:

Rajesh is admitted for 1/5 share on the following terms:

1. Goodwill of the firm is valued at ₹ 80,000 and Rajesh brought cash ₹ 6,000 for his share of goodwill.
2. Rajesh is to bring ₹ 1,50,000 as his capital.
3. Motor car is valued at ₹ 2,00,000; stock at ₹ 3,80,000 and debtors at ₹ 3,50,000.
4. Anticipated claim on workmen compensation fund is ₹ 10,000
5. Unrecorded investment of ₹ 5,000 has to be brought into account.

Prepare revaluation account, capital accounts, and balance sheet after Rajesh’s admission.
Revaluation Account

Capital Account

Balance Sheet as on 31.03.2018

Samacheer Kalvi 12th Accountancy Admission of a Partner Additional Questions and Answers

I. Choose the correct answer

Question 1.
At the time of admission of a partner calculation of net profit ratio is ………………
(a) not necessary
(b) necessary
(c) optional
Answer:
(b) necessary

Question 2.
In admission, undistributed profit or loss transferred to ………………
(a) New Partners only
(b) Old Partners only
(c) All the Partners
Answer:
(b) Old Partners only

Question 3.
To get sacrificing ratio should be deducted from old share ………………
(a) Gaining share
(b) New share
(c) Neither of the two
Answer:
(b) new share

Question 4.
A person who is admitted to the firm is known as ………………
(a) Outgoing partners
(b) Incoming partner
(c) Both
Answer:
(b) incoming partner

Question 5.
At the time of admission of a new partner, the following are revalued ………………
(a) Assets
(b) Liabilities
(c) Both
Answer:
(c) Both

Question 6.
New profit ratio is calculated at the time of admission, by deducting ………………
(a) Sacrifice from the old ratio
(b) Old ratio from the sacrifice
(c) Sacrifice from the new ratio
Answer:
(a) Sacrifice from the old ratio

Question 7.
On the admission of a new partner ………………
(a) Old firm has to be dissolved
(b) Old partnership has to be dissolved
(c) Both old firm and partnership have to be dissolved
(d) Neither partnership nor firm has to be dissolved
Answer:
(b) Old partnership has to be dissolved

Question 8.
When a new partner brings his share of goodwill in cash, the amount is debited to ………………
(a) Premium A/c
(b) Cash A/c
(c) Capital A/c of old partner
(d) Capital A/c of new partner
Answer:
(b) Cash A/c

Question 9.
Goodwill already appearing in the Balance sheet at the time of admission of a partner is transferred to ………………
(a) New Partners’ Capital A/c
(b) Old Partners’ Capital A/c
(c) Revaluation A/c
(d) None of the above
Answer:
(b) Old Partners’ Capital A/c