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Samacheer Kalvi 11th Bio Botany Solutions Chapter 12 Mineral Nutrition

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Tamilnadu Samacheer Kalvi 11th Bio Botany Solutions Chapter 12 Mineral Nutrition

Samacheer Kalvi 11th Bio Botany Mineral Nutrition Text Book Back Questions and Answers

Question 1.
Identify correct match.

1. Die back disease of citrus (i) Mo
2. Whip tail disease (ii) Zn
3. Brown heart of turnip (iii) Cu
4. Little leaf (iv) B

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

Question 2.
If a plant is provided with all mineral nutrients but, Mn concentration is increased, what will be the deficiency?
(a) Mn prevent the uptake of Fe, Mg but not Ca
(b) Mn increase the uptake of Fe, Mg and Ca
(c) Only increase the uptake of Ca
(d) Prevent the uptake Fe, Mg, and Ca
Answer:
(a) Mn prevent the uptake of Fe, Mg but not Ca

Samacheer Kalvi 11th Bio Botany Solutions 12 Mineral Nutrition

Question 3.
The element which is not remobilized?
(a) Phosphorus
(b) Potassium
(c) Calcium
(d) Nitrogen
Answer:
(c) Calcium

Question 4.
Match the correct combination.

Minerals

Role

(a) Molybdenum 1. Chlorophyll
(b) Zinc 2. Methionine
(c) Magnesium 3. Auxin
(d) Sulphur 4. Nitrogenase

(a) A – 1, B – 3, C – 4, D – 2
(b) A – 2, B – 1, C – 3, D – 4
(c) A – 4, B – 3, C – 1, D – 2
(d) A – 4, B – 2, C – 1, D – 3
Answer:
(c) A – 4, B – 3, C – 1, D – 2

Question 5.
Identify the correct statement:
(i) Sulphur is essential for amino acids Cystine and Methionine
(ii) Low level of N, K, S and Mo affect the cell division
(iii) Non – leguminous plant Alnus which contain bacterium Frankia
(iv) Denitrification carried out by nitrosomonas and nitrobacter.

(a) (i), (ii) are correct
(b) (i), (ii), (iii) are correct
(c) I only correct
(d) all are correct
Answer:
(b) (i), (ii), (iii) are correct

Question 6.
The nitrogen is present in the atmosphere in huge amount but higher plants fail to utilize it. Why?
Answer:
The higher plants do not have the association of bacteria or fungi, which are able to fix atmospheric nitrogen.

Question 7.
Why is that in certain plants deficiency symptoms appear first in younger parts of the plants while in others, they do so in mature organs?
Answer:
In certain plants, the deficiency symptom appears first in the younger part of the plant, due to the immobile nature of certain minerals like calcium, sulphur, iron, boron and copper.

Question 8.
Plant A in a nutrient medium shows whiptail disease plant B in a nutrient medium shows a little leaf disease. Identify mineral deficiency of plant A and B?
Answer:
Mineral deficiency of plant A and B:

  1. Plant A is deficient of the mineral molybdenum (Mo).
  2. Plant B is deficient of the mineral zinc (Zn).

Question 9.
Write the role of nitrogenase enzyme in nitrogen fixation?
Answer:
The role of nitrogenase enzyme in nitrogen fixation:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 12 Mineral Nutrition 2

Question 10.
Explain the insectivorous mode of nutrition in angiosperms?
Answer:
Plants which are growing in nitrogen deficient areas develop insectivorous habit to resolve nitrogen deficiency.
(i) Nepenthes (Pitcher plant): Pitcher is a modified leaf and contains digestive enzymes. Rim of the pitcher is provided with nectar glands and acts as an attractive lid. When insect is trapped, proteolytic enzymes will digest the insect.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 12 Mineral Nutrition 3

(ii) Drosera (Sundew): It consists of long club shaped tentacles which secrete sticky digestive fluid which looks like a sundew.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 12 Mineral Nutrition 4

(iii) Utricularia (Bladder wort): Submerged plant in which leaf is modified into a bladder to collect insect in water.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 12 Mineral Nutrition 5

(iv) Dionaea (Venus fly trap): Leaf of this plant modified into a colourful trap. Two folds of lamina consist of sensitive trigger hairs and when insects touch the hairs it will close.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 12 Mineral Nutrition 6

Samacheer Kalvi 11th Bio Botany Mineral Nutrition Additional Questions & Answers

I. Choose the correct answer (1 Mark)
Question 1.
Plants naturally obtain nutrients from:
(a) atmosphere
(b) water
(c) soil
(d) all of these
Answer:
(d) all of these

Question 2.
Which of the following are included under micro nutrients:
(a) sodium, carbon and hydrogen
(b) magnesium, nitrogen and silicon
(c) sodium, cobalt and selenium
(d) calcium, sulphur and potassium
Answer:
(c) sodium, cobalt and selenium

Question 3.
Who coined the term ‘Hydroponics’:
(a) Julius Von Sachs
(b) William Frederick Goerick
(c) Liebig
(d) Wood word
Answer:
(b) William Frederick Goerick

Question 4.
Selenium is essential for plants:
(a) to prevent water lodging
(b) to enhance growth
(c) to resist drought
(d) to prevent transpiration
Answer:
(a) to prevent water lodging

Question 5.
Actively mobile minerals are:
(a) nitrogen and phosphorus
(b) iron and manganese
(c) sodium and cobalt
(d) silicon and selenium
Answer:
(a) nitrogen and phosphorus

Question 6.
Copper shows deficiency symptoms first that appear in young leaves due to:
(a) less active movement of minerals to younger leaves
(b) active movement of minerals
(c) the immobile nature of mineral
(d) none of the above
Answer:
(c) the immobile nature of mineral

Question 7.
Molybdenum is essential for the reaction of:
(a) hydrolase enzyme
(b) nitrogenase enzyme
(c) carboxylase enzyme
(d) dehydrogenase enzyme
Answer:
(b) nitrogenase enzyme

Question 8.
Match the following:

A. Magnesium (i) dehydrogenase
B. Nickel (ii) ion exchange
C. Zinc (iii) chlorophyll
D. Potassium (iv) urease

(a) A – (ii); B – (i); C – (iv); D – (iii)
(b) A – (iii); B – (ii); C – (i); D – (iv)
(c) A – (ii); B – (iv); C – (i); D – (iii)
(d) A – (iii); B – (iv); C – (i); D – (ii)
Answer:
(d) A – (iii); B – (iv); C – (i); D – (ii)

Question 9.
Nitrogen is the essential component of:
(a) carbohydrate
(b) fatty acids
(c) protein
(d) none of these
Answer:
(c) protein

Question 10.
Which of the element is involved in the synthesis of DNA and RNA:
(a) calcium
(b) magnesium
(c) sulphuric
(d) potassium
Answer:
(b) magnesium

Question 11.
The deficiency of magnesium is the plant, causes:
(a) necrosis
(b) interveinal chlorosis
(c) sand drown of tobacco
(d) all the above
Answer:
(d) all the above

Question 12.
Sulphur is an essential components of amino acids like:
(a) histidine, leucine and aspartic acid
(b) valene, alkaline and glycine
(c) cystine, cysteine and methionine
(d) none of the above
Answer:
(c) cystine, cysteine and methionine

Question 13.
Indicate the correct statements:
(i) Iron is the essential element for the synthesis of chlorophyll and carotenoid
(ii) Iron is the activator of carboxylene enzyme
(iii) Iton is the component of cytochrome
(iv) lvon is the component of plastocyanin

(a) (i) and (ii)
(b) (ii) and (iv)
(c) (ii) and (iii)
(d) (i) and (iii)
Answer:
(d) (i) and (iii)

Question 14.
Khaira disease of rice is caused by:
(a) deficiency of boron
(b) deficiency of zinc
(c) deficiency of iron
(d) deficiency of all the three
Answer:
(b) deficiency of zinc

Question 15.
Match the following:

A. Marginal chlorosis (i) nitrogen
B. Anthocyanin formation (ii) zinc
C. Hooked leaf tip (iii) potassium
D. Little leaf (iv) calcium

(a) A – (ii); B – (iii); C – (i); D – (iv)
(b) A – (iii), B – (ii); C – (iv); D – (i)
(c) A – (iii); B – (i); C – (iv); D – (ii)
(d) A – (iv); B – (iii); C – (i); D – (ii)
Answer:
(c) A – (iii); B – (i); C – (iv); D – (ii)

Question 16.
Increased concentration of manganese in plants will prevent the uptake of:
(a) calcium and potassium
(b) sodium and potassium
(c) boron and silicon
(d) iron and magnesium
Answer:
(d) iron and magnesium

Question 17.
Which of the statement is not correct?
(a) Aluminium toxicity causes the appearance of brown spots in the leaves.
(b) Aluminium toxicity causes the precipitation of nucleic acid.
(c) Aluminium toxicity inhibits ATPase activity
(d) Aluminium toxicity inhibits cell division.
Answer:
(a) Aluminium toxicity causes the appearance of brown spots in the leaves.

Question 18.
The techniques of Aeroponics was developed by:
(a) Goerick
(b) Amon and Hoagland
(c) Soifer Hillel and David Durger
(d) Von Sachs
Answer:
(c) Soifer Hillel and David Durger

Question 19.
Nitrogen occurs in atmosphere in the form of N2, two nitrogen atoms joined together by strong:
(a) di – covalent bond
(b) triple covalent bond
(c) non – valent bond
(d) none of these
Answer:
(b) triple covalent bond

Question 20.
The process of converting atmospheric nitrogen (N2) into ammonia is termed as:
(a) nitrogen cycle
(b) nitrification
(c) nitrogen fixation
(d) ammonification
Answer:
(c) nitrogen fixation

Question 21.
Find out the odd organism:
(a) Rhizobium
(b) Cyanobacteria
(c) Azolla
(d) Pistia
Answer:
(d) Pistia

Question 22.
The legume plants secretes phenolics to attract:
(a) Azolla
(b) Rhizobium
(c) Nitrosomonas
(d) Streptococcus
Answer:
(b) Rhizobium

Question 23.
Which are the organisms help in nitrogen fixation of lichens:
(a) Anabaena and Nostoc
(b) Anabaena alone
(c) Nostoc alone
(d) Anabaena azollae
Answer:
(a) Anabaena and Nostoc

Question 24.
Nitrogenase enzyme is active:
(a) only in aerobic condition
(b) only in anaerobic condition
(c) both in aerobic and anaerobic condition
(d) only in toxic condition
Answer:
(b) only in anaerobic condition

Question 25.
Ammonia (NH3+) is converted into nitrite (NO2) by a bacterium called:
(a) Nitrobacter bacterium
(b) Rhizobium
(c) Anabaena azollae
(d) Nitrosomonas
Answer:
(d) Nitrosomonas

Question 26.
Decomposition of organic nitrogen (proteins and amino acids) from dead plants and animals into ammonia is called:
(a) nitrification
(b) ammonification
(c) nitrogen fixation
(d) denitrification
Answer:
(b) ammonification

Question 27.
The bacteria involved in the denitrification process are:
(a) E.coli and Anabaena
(b) Streptococcus and Bacillus vulgaris
(c) Pseudomonas and Thiobacillus
(d) none of the above
Answer:
(c) Pseudomonas and Thiobacillus

Question 28.
In the process of ammonium assimilation:
(a) Ammonia is converted into nitrites
(b) Ammonia is converted into atmospheric nitrogen
(c) Ammonia is converted into ammonium ions
(d) Ammonia is converted into amino acids
Answer:
(d) Ammonia is converted into amino acids

Question 29.
The transfer of amino group (NH2) from glutamic acid to keto group of keto acid is termed as:
(a) Transamination
(b) Hydrogenation
(c) Nitrification
(d) Denitrification
Answer:
(a) Transamination

Question 30.
Monotrapa (Indian pipe) absorbs nutrients through:
(a) Rhizobium association
(b) mycorrhizal association
(c) microbial association
(d) animal association
Answer:
(b) mycorrhizal association

Question 31.
Cuscuta is a:
(a) partial parasite
(b) total root parasite
(c) obligate stem parasite
(d) partial stem parasite
Answer:
(c) obligate stem parasite

Question 32.
Indicate the correct statement:
(a) Loranthus grows on banana and coconut
(b) Loranthus grows on fig and mango trees
(c) Balanophora is a stem parasite
(d) Viscum is a root parasite
Answer:
(b) Loranthus grows on fig and mango trees

Question 33.
The association of mycorrhizae with higher plants is termed as:
(a) Parasitism
(b) Mutualism
(c) Symbiosis
(d) Saprophytic
Answer:
(c) Symbiosis

Question 34.
In Utricularia, the bladder is a modified form of:
(a) leaf
(b) stem
(c) tentacle
(d) lamina
Answer:
(a) leaf

Question 35.
Lichens are the indicators of:
(a) carbon monoxide
(b) nitrogen oxide
(c) sulphur di oxide
(d) hydrogen sulphide
Answer:
(c) sulphur di oxide

II. Answer the following (2 Marks)

Question 1.
Define micro nutrients of plants.
Answer:
Essential minerals which are required in less concentration called are as Micro nutrients.

Question 2.
Mention any two actively mobile minerals.
Answer:
Nitrogen and Phosphorus.

Question 3.
What is the role of molybdenum in the conversion of nitrogen into ammonia?
Answer:
Molybdenum (Mo) is essential for nitrogenase enzyme during reduction of atmospheric nitrogen into ammonia.

Question 4.
What is the role of potassium on osmotic potential of the cell?
Answer:
Potassium (K) plays a key role in maintaining osmotic potential of the cell. The absorption of water, movement of stomata and turgidity are due to osmotic potential.

Question 5.
What are the deficiency symptoms of nitrogen?
Answer:
Chlorosis, stunted growth, anthocyanin formation.

Question 6.
Explain the role of sulphur in plant biochemistry.
Answer:
Essential component of amino acids like cystine, cysteine and methionine, constituent of coenzyme A, Vitamins like biotin and thiamine, constituent of proteins and ferredoxin plants utilise sulphur as sulphate (SO4) ions.

Question 7.
Define the term Siderophores.
Answer:
Siderophores (iron carriers) are iron chelating agents produced by bacteria. They are used to chelate ferric iron (Fe3+) from environment and host.

Question 8.
List out any two iron deficiency symptoms in plants.
Answer:
Interveinal chlorosis, formation of short and slender stalk and inhibition of chlorophyll formation.

Question 9.
What is the role of Boron in plant physiology.
Answer:
Translocation of carbohydrates, uptake and utilisation of Ca++, pollen germination, nitrogen metabolism, fat metabolism, cell elongation and differentiation. It is absorbed as borate BO3-  ions.

Question 10.
Write down the deficiency symptoms of molybdenum in plants.
Answer:
Chlorosis, necrosis, delayed flowering, retarded growth and whip tail disease of cauliflower.

Question 11.
Explain briefly about aluminium toxicity on plants.
Answer:
Aluminium toxicity causes precipitation of nucleic acid, inhibition of ATPase, inhibition of cell division and binding of plasma membrane with Calmodulin.

Question 12.
Define Aeroponics.
Answer:
It is a system where roots are suspended in air and nutrients are sprayed over the roots by a motor driven rotor.

Question 13.
Define nitrogen fixation.
Answer:
The process of converting atmospheric nitrogen (N2) into ammonia is termed as nitrogen fixation. Nitrogen fixation can occur by two methods:

  1. Biological
  2. Non – Biological.

Question 14.
Mention any two ways of non – biological nitrogen fixation.
Answer:
Two ways of non – biological nitrogen fixation:

  1. Nitrogen fixation by chemical process in industry.
  2. Natural electrical discharge during lightening fixes atmospheric nitrogen.

Question 15.
Match the following.

A. Lichens (i) Anabaena Azolla
B. Anthoceros (ii) Frankia
C. Azolla (iii) Anabaena and Nostoc
D. Casuarina (iv) Nostoc

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

Question 16.
Define the term Nitrate assimilation.
Answer:
The process by which nitrate is reduced to – ammonia is called nitrate assimilation and occurs during nitrogen cycle.

Question 17.
Explain.the term Transamination.
Answer:
Transfer of amino group (NH3+) from glutamic acid glutamate to keto group of keto acid. Glutamic acid is the main amino acid from which other amino acids are synthesised by transamination.

Question 18.
Explain briefly about total stem parasite.
Answer:
The leafless stem twine around the host and produce haustoria. eg: Cuscuta (Dodder), a rootless plant growing on Zizyphus, Citrus and so on.

Question 19.
Give two examples of symbiotic mode of nutrition.
Answer:
Two examples of symbiotic mode of nutrition:

  1. Lichens: It is a mutual association of Algae and Fungi. Algae prepares food and fungi absorbs water and provides thallus structure.
  2. Mycorrhizae: Fungi associated with roots of higher plants including Gymriosperms. eg: Pinus.

Question 20.
Explain briefly about insectivorous mode of nutrition.
Answer:
Plants which are growing in nitrogen deficient areas develop insectivorous habit to resolve nitrogen deficiency.

III. Answer the following (3 Marks)

Question 1.
What are the criteria required for essential minerals in plants?
Answer:
The criteria required for essential minerals in plants:

  1. Elements necessary for growth and development.
  2. They should have direct role in the metabolism of the plant.
  3. It cannot be replaced by other elements.
  4. Deficiency makes the plants impossible to complete their vegetative and reproductive phase.

Question 2.
Explain the unclassified minerals required for plants.
Answer:
Minerals like Sodium,Silicon, Cobalt and Selenium are not included in the list of essential nutrients but are required by some plants, these minerals are placed in the list of unclassified minerals. These minerals play specific roles for example, Silicon is essential for pest resistance, prevent water lodging and aids cell wall formation in Equisetaceae (Equisetum), Cyperaceae and Gramineae.

Question 3.
Distinguish between macro and micro nutrients?
Answer:
Macro nutrients:

  • Excess than 10 mmole Kg-1 in tissue concentration or 0.1 to 10 mg per gram of dry weight.
  • eg: C, H, O, N, P, K, Ca, Mg and S.

Micro nutrients:

  • Less than 10 mmole Kg-1 in tissue concentration or equal or less than 0.1 mg per gram of dry weight.
  •  eg: Fe, Mn, Cu, Mo, Zn, B, Cl and Ni.

Question 4.
Explain briefly the functions and deficiency symptoms of potassium.
Answer:
Functions: Maintains turgidity and osmotic potential of the cell, opening and closure of stomata, phloem translocation, stimulate activity of enzymes, anion and cation balance by ion – exchange. It is absorbed as K+ ions. Deficiency symptoms: Marginal chlorosis, necrosis, low cambial activity, loss of apical dominance, lodging in cereals and curled leaf margin.

Question 5.
What is meant by Chelating agents? Explain the role of EDTA as chemical chelating agent.
Answer:
Plants which are growing in alkaline soil when supplied with all nutrients including iron will show iron deficiency. To rectify this, we have to make iron into a soluble complex by adding a chelating agent like EDTA (Ethylene Diamine Tetra Acetic acid) to form Fe – EDTA.

Question 6.
Explain the term critical concentration of minerals.
Answer:
To increase the productivity and also to avoid mineral toxicity knowledge of critical concentration is essential. Mineral nutrients lesser than . critical concentration cause deficiency symptoms. Increase of mineral nutrients more than the normal concentration causes toxicity. A concentration, at which 10% of the dry weight of tissue is reduced, is considered as toxic critical concentration.

Question 7.
Describe the competitive behaviour of iron and manganese.
Answer:
Iron and Manganese exhibit competitive behaviour. Deficiency of Fe and Mn shows similar symptoms. Iron toxicity will affect absorption of manganese. The possible reason for iron toxicity is excess usage of chelated iron in addition with increased acidity of soil (pH less than 5.8) Iron and manganese toxicity will be solved by using fertilizer with balanced ratio of Fe and Mn.

Question 8.
Who are people responsible for developing hydroponics?
Answer:
Hydroponics or Soil less culture: Von Sachs developed a method of growing plants in nutrient solution. The commonly used nutrient solutions are Knop solution (1865) and Amon and Hoagland Solution (1940). Later the term Hydroponics was coined by Goerick (1940) and he also introduced commercial techniques for hydroponics. In hydroponics roots are immersed in the solution containing nutrients and air is supplied with help of tube.

Question 9.
List out the free living bacteria and fungi responsible for non-symbiotic nitrogen fixation.
Answer:
Free living bacteria and fungi also fix atmospheric nitrogen.

Aerobic Azotobacter, Beijerneckia and Derxia
Anaerobic Clostridium
Photosynthetic Chlorobium and Rhodospirillum
Chemosynthetic Disulfovibrio
Free living fungi Yeast and Pullularia
Cyanobacteria Nostoc, Anabaena and Oscillatoria.

Question 10.
Define the term Ammonification.
Answer:
Decomposition of organic nitrogen (proteins and amino acids) from dead plants and animals into ammonia is called ammonification. Organisim involved in this process are Bacillus ramosus and Bacillus vulgaris.

Question 11.
Explain briefly Catalytic amination.
Answer:
Glutamate amino acid combines with ammonia to form the amide glutamine.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 12 Mineral Nutrition 8
Glutamine reacts with a ketoglutaric acid to form two molecules of glutamate.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 12 Mineral Nutrition 9
(GOGAT – Glutamine – 2 – Oxoglutarate aminotransferase)

Question 12.
Compare the partial stem parasite and partial root parasite.
Answer:
The partial stem parasite and partial root parasite:

  1. Partial Stem Parasite: eg: Loranthus and Viscum (Mistletoe) Loranthus grows on fig and mango trees and absorb water and minerals from xylem.
  2. Partial root parasite: eg: Santalum album (Sandal wood tree) in its juvenile stage produces haustoria which grows on roots of many plants.

Question 13.
Explain the mode of nutrition in pitcher plant.
Answer:
Pitcher is a modified leaf and contains digestive enzymes. Rim of the pitcher is provided with nectar glands and acts as an attractive lid. When insect is trapped proteolytic enzymes will digest the insect.

Question 14.
What is meant by saprophytic mode of nutrition?
Answer:
Saprophytes derive nutrients from dead and decaying matter. Bacteria and fungus are main saprophytic organisms. Some angiosperms also follow saprophytic mode of nutrition. eg: Neottia. Roots of Neottia (Bird’s Nest Orchid) associate with mycorrhizae and absorb nutrients as a saprophyte. Monotropa (Indian Pipe) grow on humus rich soil found in thick forests. It absorbs nutrient through mycorrhizal association.

Question 15.
Describe briefly the method of nitrogen fixation in leguminous plants.
Answer:
Rhizobium bacterium is found in leguminous plants and fix atmospheric nitrogen. This kind of symbiotic association is beneficial for both the bacterium and plant. Root nodules are formed due to bacterial infection. Rhizobium enters into the-host cell and proliferates, it remains separated from the host cytoplasm by a membrane.

IV. Answer the following (5 Marks)

Question 1.
Write an essay on the functions and deficiency symptoms of macro nutrients.
Answer:
Macronutrients, their functions, their mode of absorption, deficiency symptoms and deficiency diseases are discussed here:
(i) Nitrogen (N): It is required by the plants in greatest amount. It is an essential component of proteins, nucleic acids, amino acids, vitamins, hormones, alkaloids, chlorophyll and cytochrome. It is absorbed by the plants as nitrates (NO3).

Deficiency symptoms: Chlorosis, stunted growth, anthocyanin formation.

(ii) Phosphorus (P): Constituent of cell membrane, proteins, nucleic acids, ATP, NADP, phytin and sugar phosphate. It is absorbed as H2PO4+ and HPO4 ions.

Deficiency symptoms: Stunted growth, anthocyanin formation, necrosis, inhibition of cambial activity, affect root growth and fruit ripening.

(iii) Potassium (K): Maintains turgidity and osmotic potential of the cell, opening and closure of stomata, phloem translocation, stimulate activity of enzymes, anion and cation balance by ion – exchange. It is absorbed as K+ ions.

Deficiency symptoms: Marginal chlorosis, necrosis, low cambial activity, loss of apical dominance, lodging in cereals and curled leaf margin.

(iv) Calcium (Ca): It is involved in synthesis of calcium pectate in middle lamella, mitotic spindle formation, mitotic cell division, permeability of cell membrane, lipid metabolism, activation of phospholipase, ATPase, amylase and activator of adenyl kinase. It is absorbed as Ca2+ exchangeable ions.

Deficiency symptoms: Chlorosis, necrosis, stunted growth, premature fall of leaves and flowers, inhibit seed formation, Black heart of Celery, Hooked leaf tip in Sugar beet, Musa and Tomato.

(v) Magnesium (Mg): It is a constituent of chlorophyll, activator of enzymes of carbohydrate metabolism (RUBP Carboxylase and PEP Carboxylase) and involved in the synthesis of DNA and RNA. It is essential for binding of ribosomal sub units. It is absorbed as Mg2+ ions.

Deficiency symptoms: litter veinal chlorosis, necrosis, anthocyanin (purple) formation and Sand drown of tobacco.

(vi) Sulphur (S): Essential component of amino acids like cystine, cysteine and methionine, constituent of coenzyme A, Vitamins like biotin and thiamine, constituent of proteins and ferredoxin. plants utilise sulphur as sulphate (SO4) ions.

Deficiency symptoms: Chlorosis, anthocyanin formation, stunted growth, rolling of leaf tip and reduced nodulation in legumes.

Question 2.
Describe the role of micro nutrients on plant health and function.
Answer:
Micronutrients even though required in trace amounts are essential for the metabolism of plants. They play key roles in many plants. eg: Boron is essential for translocation of sugars, molybdenum is involved in nitrogen metabolism and zinc is needed for biosynthesis of auxin. Here, we will study about the role of micro nutrients, their functions, their mode of absorption, deficiency symptoms and deficiency diseases.

(i) Iron (Fe): Iron is required lesser than macronutrient and larger than micronutrients, hence, it can be placed in any one of the groups. Iron is an essential element for the synthesis of chlorophyll and carotenoids. It is the component of cytochrome, ferredoxin, flavoprotein, formation of chlorophyll, porphyrin, activation of catalase, peroxidase enzymes. It is absorbed as ferrous (Fe2+) and ferric (Fe3+) ions. Mostly fruit trees are sensitive to iron.

Deficiency: Interveinal Chlorosis, formation of short and slender stalk and inhibition of chlorophyll formation.

(ii) Manganese (Mn): Activator of Carboxylases, oxidases, dehydrogenases and kinases, involved in splitting of water to liberate oxygen (photolysis). It is absorbed as manganous (Mn2+) ions.

Deficiency: Interveinal chlorosis, grey spot on oats leaves and poor root system.

(iii) Copper (Cu): Constituent of plastocyanin, component of phenolases, tyrosinase, enzymes involved in redox reactions, synthesis of ascorbic acid, maintains carbohydrate and nitrogen balance, part of oxidase and cytochrome oxidase. It is absorbed as cupric (Cu2+) ions,

Deficiency: Die back of citrus, Reclamation disease of cereals and legumes, chlorosis, necrosis and Exanthema in Citrus.

(iv) Zinc (Zn): Essential for the synthesis of Indole acetic acid (Auxin) activator of carboxylases, alcohol dehydrogenase, lactic dehydrogenase, glutamic acid dehydrogenase, carboxy peptidases and tryptophan synthetase. It is absorbed as Zn2+ ions.

Deficiency: Little leaf and mottle leaf due to deficiency of auxin, Inter veinal chlorosis, stunted growth, necrosis and Khaira disease of rice.

(v) Boron (B): Translocation of carbohydrates, uptake and utilisation of Ca++, pollen germination, nitrogen metabolism, fat metabolism, cell elongation and differentiation. It is absorbed as borate BO3- ions.

Deficiency: Death of root and shoot tips, premature fall of flowers and fruits, brown heart of beet root, internal cork of apple and fruit cracks.

(vi) Molybdenum (Mo): Component of nitrogenase, nitrate reductase, involved in nitrogen metabolism, and nitrogen fixation. It is absorbed as molybdate (Mo2+) ions.

Deficiency: Chlorosis, necrosis, delayed flowering, retarded growth and whip tail disease of cauliflower.

(vii) Chlorine (Cl): It is involved in Anion – Cation balance, cell division, photolysis of water. It is absorbed as Cl ions.
Deficiency: Wilting of leaf tips.

(viii) Nickel (Ni): Cofactor for enzyme urease and hydrogenase.

Deficiency: Necrosis of leaf tips.

Question 3.
Give the details of minerals and their deficiency symptoms.
Answer:
Name of the deficiency disease and symptoms:

  1. Chlorosis (Overall)
    • (a) Interveinal chlorosis
    • (b) Marginal chlorosis
  2. Necrosis (Death of the tissue)
  3. Stunted growth
  4. Anthocyanin formation
  5. Delayed flowering
  6. Die back of shoot, Reclamation disease, Exanthema in citrus (gums on bark)
  7. Hooked leaf tip
  8. Little Leaf
  9. Brown heart of turnip and Internal cork of apple
  10. Whiptail of cauliflower and cabbage
  11. Curled leaf margin

Deficiency minerals:

  1. Nitrogen, Potassium, Magnesium, Sulphur, Iron, Manganese, Zinc and Molybdenum. Magnesium, Iron, Manganese and Zinc Potassium
  2. Magnesium, Potassium, Calcium, Zinc, Molybdenum and Copper.
  3. Nitrogen, Phosphorus, Calcium, Potassium and Sulphur.
  4. Nitrogen, Phosphorus, Magnesium and Sulphur
  5. Nitrogen, Sulphur and Molybdenum
  6. Copper
  7. Calcium
  8. Zinc
  9. Boron
  10. Molybdenum
  11. Potassium

Question 4.
Give the schematic diagram of nitrogen cycle.
Answer:
The schematic diagram of nitrogen cycle:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 12 Mineral Nutrition 7

Question 5.
Describe the modes of biological nitrogen fixation.
Answer:
Symbiotic bacterium like Rhizobium fixes atmospheric nitrogen. Cyanobacteria found in Lichens, Anthoceros, Azolla and coralloid roots of Cycas also fix nitrogen. Non – symbiotic (free living bacteria) like Clostridium also fix nitrogen. Symbiotic nitrogen fixation:
1. Nitrogen fixation with nodulation: Rhizobium bacterium is found in leguminous plants and fix atmospheric nitrogen. This kind of symbiotic association is beneficial for both the bacterium and plant. Root nodules are formed due to bacterial infection. Rhizobium enters into the host cell and proliferates, it remains separated from the host cytoplasm by a membrane.

2. Stages of Root nodule formation:

  • Legume plants secretes phenolics which attracts Rhizobium.
  • Rhizobium reaches the rhizosphere and enters into the root hair, infects the root hair and leads to curling of root hairs.
  • Infection thread grows inwards and separates the infected tissue from normal tissue.
  • A membrane bound bacterium is formed inside the nodule and is called bacteroid.
  • Cytokinin from bacteria and auxin from host plant promotes cell division and leads to nodule formation

3. Non – Legume: Alnus and Casuarina contain the bacterium Frankia Psychotria contains the bacterium Klebsiella.
Nitrogen fixation without nodulation. The following plants and prokaryotes are involved in nitrogen fixation:

  • Lichens – Anabaena and Nostoc
  • Anthoceros – Nostoc
  • Azolla – Anabaena azollae
  • Cycas – Anabaena and Nostoc.

Solution To Activity
Textbook Page No: 95

Question 1.
Collect leaves showing mineral deficiency. Tabulate the symptoms like Marginal Chlorosis, Interveinal Chlorosis, Necrotic leaves, Anthocyanin formation in leaf, Little leaf and Hooked leaf. (Discuss with your teacher about the deficiency of minerals)
Answer:
Symptoms:

  1. Marginal Chlorosis
  2. interveinal Chlorosis
  3. Necrotic leaves
  4. Anthocyanin formation in leaves
  5. Little leaf
  6. Hooked leaf

Minerals:

  1. Potassium (K)
  2. Magnesium (Mg)
  3. Nickel (Ni)
  4. Phosphorus (P)
  5. Zinc (Zn)
  6. Calcium (Ca)

Textbook Page No: 98

Question 1.
Preparation of Solution Culture to find out Mineral Deficiency
1. Take a glass jar or polythene bottle and cover with black paper (to prevent algal growth and roots reacting with light).
2. Add nutrient solution.
3. Fix a plant with the help of split cork.
4. Fix a tube for aeration.
5. Observe the growth by adding specific minerals.
Answer:
The deficiency of minerals like nitrogen, phosphorus, calcium, potassium and sulphur cause stunted growth in plants.

Textbook Page No: 99

Question 1.
Collect roots of legumes with root nodules.
• Take cross section of the root nodule.
• Observe under microscope. Discuss your observations with your teacher.
Answer:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 12 Mineral Nutrition 1

Samacheer Kalvi 11th Bio Botany Solutions Chapter 10 Secondary Growth

Students can Download Bio Botany Chapter 10 Secondary Growth Questions and Answers, Notes Pdf, Samacheer Kalvi 11th Bio Botany Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Bio Botany Solutions Chapter 10 Secondary Growth

Samacheer Kalvi 11th Bio Botany Secondary Growth Text Book Back Questions and Answers

Question 1.
Consider the following statements In spring season vascular cambium:
(i) is less active
(ii) produces a large number of xylary elements
(iii) forms vessels with wide cavities of these

(a) (i) is correct but (ii) and (iii) are not correct
(b) (i) is not correct but (ii) and (iii) are correct
(c) (i) and (ii) are correct but (iii) is not correct
(d) (i) and (ii) are not correct but (iii) is correct
Answer:
(b) (i) is not correct but (ii) and (iii) are correct

Samacheer Kalvi 11th Bio Botany Solutions 10 Secondary Growth

Question 2.
Usually, the monocotyledons do not increase their girth, because:
(a) They possess actively dividing cambium
(b) They do not possess actively dividing cambium
(c) Ceases activity of cambium
(d) All are correct
Answer:
(b) They do not possess actively dividing cambium

Question 3.
In the diagram of lenticel identify the parts marked as A,B,C,D.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 10 Secondary Growth 1
(a) A. Phellem, B. Complementary tissue, C. Phelloderm, D. Phellogen.
(b) A. Complementary tissue, B. Phellem, C. Phellogen, D. Phelloderm.
(c) A. Phellogen, B. Phellem, C. Pheiloderm, D. complementary tissue
(d) A. Phelloderm, B. Phellem, C. Complementary tissue, D. Phellogen
Answer:
(a) A. Phellem, B. Complementary tissue, C. Phelloderm, D. Phellogen.

Question 4.
The common bottle cork is a product of:
(a) Dermatogen
(b) Phellogen
(c) Xylem
(d) Vascular cambium
Answer:
(b) Phellogen

Question 5.
What is the fate of primary xylem in a dicot root showing extensive secondary growth?
(a) It is retained in the center of the axis
(b) It gets crushed
(c) May or may not get crushed
(d) It gets surrounded by primary phloem
Answer:
(b) It gets crushed

Samacheer Kalvi 11th Bio Botany Solutions 10 Secondary Growth

Question 6.
In a forest, if the bark of a tree is damaged by the horn of a deer, How will the plant overcome the damage?
Answer:
When the bark is damaged, the phellogem forms a complete cylinder around the stem and it gives rise to ring barks.

Question 7.
In which season the vessels of angiosperms are larger in size, why?
Answer:
In spring season the vessels are larger in size, because the cambium cells are very active during spring season.

Question 8.
Continuous state of dividing tissue is called meristem. In connection to this, what is the role of lateral meristem?
Answer:
Apical meristems produce the primary plant body. In some plants, the lateral meristem increase the girth of a plant. This type of growth is secondary because the lateral meristem are not directly produced by apical meristems. Woody plants have two types of lateral meristems: a vascular cambium that produces xylem, phloem tissues and cork cambium that produces the bark of a tree.

Question 9.
A timber merchant bought 2 logs of wood from a forest & named them A & B, The log A was 50 year old & B was 20 years old. Which log of wood will last longer for the merchant? Why?
Answer:
The wood of 50 years old will last longer than 20 years old wood, because timber from hard wood is more durable and more resistant to the attack of micro organisms and insect than the timber from sap wood.

Samacheer Kalvi 11th Bio Botany Solutions 10 Secondary Growth

Question 10.
A transverse section of the trunk of a tree shows concentric rings which are known as growth rings. How are these rings formed? What are the significance of these rings?
Answer:
The annual ring denotes the combination of early wood and late wood and the ring becomes evident to our eye due to the high density of late wood. Sometimes annual rings are called growth rings but it should be remembered all the growth rings are not annual. In some trees more than one growth ring is formed with in a year due to climatic changes. Additional growth rings are developed within a year due to adverse natural calamities like drought, frost, defoliation, flood, mechanical injury and biotic factors during the middle of a growing season, which results in the formation of more than one annual ring.

Such rings are called pseudo – or false – annual rings. Each annual ring corresponds to one year’s growth and on the basis of these rings, the age of a particular plant can easily be calculated. The determination of the age of a tree by counting the annual rings is called dendrochronology.

Samacheer Kalvi 11th Bio Botany Secondary Growth Other Important Questions & Answers

I. Choose the correct answer. (I Marks)
Question 1.
The roots and stems grow in length with the help of:
(a) cambium
(b) secondary growth
(c) apical meristem
(d) vascular parenchyma
Answer:
(c) apical meristem

Question 2.
The increase in the girth of plant is called:
(a) primary growth
(b) tertiary growth
(c) longitudinal growth
(d) secondary growth
Answer:
(d) secondary growth

Samacheer Kalvi 11th Bio Botany Solutions 10 Secondary Growth

Question 3.
The secondary vascular tissues include:
(a) secondary xylem and secondary phloem
(b) secondary xylem, cambium strip and secondary phloem
(c) secondary phloem and fascicular cambium
(d) secondary xylem and primary phloem
Answer:
(a) secondary xylem and secondary phloem

Question 4.
Choose the correct statements.
(i) A strip of vascular cambium is present between xylem and phloem of the vascular bundle.
(ii) Vascular cambium is believed originate from fusiform initials.
(iii) The vascular cambium is originated from procambium of vascular bundle
(iv) Vascular cambium is present between fusiform initials and ray initials

(a) (i) and (iv)
(b) (i) and (iii)
(c) (ii) and (iii)
(d) (ii) and (iv)
Answer:
(b) (i) and (iii)

Question 5.
Match the following:

A. Xylem (i) Treachery elements
B. Secondary xylem (ii) Water transport
C. Phloem (iii) Sieve elements
D. Secondary phloem (iv) Food transport

(a) B – (i); A – (ii); C – (iii); D – (iv)
(b) B – (ii); A – (iii); C – (i); D – (iv)
(c) A – (ii); B – (i); C – (iv); D – (iii)
(d) A – (i); B – (ii); C – (iii); D – (iv)
Answer:
(c) A – (ii); B – (i); C – (iv); D – (iii)

Samacheer Kalvi 11th Bio Botany Solutions 10 Secondary Growth

Question 6.
The axial system of the secondary xylem includes:
(a) treachery elements, sieve elements, fibers and axial parenchyma
(b) treachery elements, fibers and axial parenchyma
(c) treachery elements and fibers
(d) sieve elements and axial parenchyma
Answer:
(b) treachery elements, fibers and axial parenchyma

Question 7.
The study of wood by preparing sections for microscopic observation is termed as:
(a) histology
(b) xylotomy
(c) phoemtomy
(d) anatomy
Answer:
(b) xylotomy

Question 8.
Ray cells are present between:
(a) primary xylem and phloem
(b) primary xylem and secondary xylem
(c) secondary xylem and phloem
(d) secondary phloem and cambium
Answer:
(c) secondary xylem and phloem

Samacheer Kalvi 11th Bio Botany Solutions 10 Secondary Growth

Question 9.
The axial system Consists of vertical files of:
(a) treachery elements and sieve elements
(b) treachery elements and apical parenchyma
(c) sieve elements are fibers
(d) treachery elements, fibers and wood parenchyma
Answer:
(d) treachery elements, fibers and wood parenchyma

Question 10.
Morus rubra has:
(a) porous wood
(b) soft wood
(c) spring wood
(d) sap wood
Answer:
(a) porous wood

Question 11.
Which of the statement is not correct?
(a) In temperate regions, the cambium is very active in winter season.
(b) In temperate regions, the cambium is very active in spring season.
(c) In temperate regions, cambium is less active in winter season.
(d) In temperate regions early wood is formed in spring season.
Answer:
(a) In temperate regions, the cambium is very active in winter season.

Question 12.
Usually more distinct annual rings are formed:
(a) in tropical plants
(b) in seashore plants
(c) in temperate plants
(d) in desert plants
Answer:
(c) in temperate plants

Samacheer Kalvi 11th Bio Botany Solutions 10 Secondary Growth

Question 13.
False annual rings are formed due to:
(a) rain
(b) adverse natural calamities
(c) severe cold
(d) none of the above
Answer:
(b) adverse natural calamities

Question 14.
determination of the age of a tree by counting the annual rings is called:
(a) chronology
(b) dendrochronology
(c) palaeology
(d) histology
Answer:
(c) palaeology

Question 15.
The age of American sequoiadendron tree is about:
(a) 350 years
(b) 3,000 years
(c) 3400 years
(d) 3500 years
Answer:
(d) 3500 years

Question 16.
The wood of Acer plant has:
(a) ring porous
(b) diffuse porous
(c) central porous
(d) none of the above
Answer:
(b) diffuse porous

Samacheer Kalvi 11th Bio Botany Solutions 10 Secondary Growth

Question 17.
In fully developed tyloses:
(a) only starchy crystals are present
(b) resin and gums only are present
(c) oil and tannins are present
(d) starchy crystals, resins, gums, oils, tannins or colored substances are present
Answer:
(d) starchy crystals, resins, gums, oils, tannins or colored substances are present

Question 18.
In bombax:
(a) the sieve tubes are blocked by tylose like outgrowths
(b) the resin ducts are blocked by tylose like outgrowths
(c) the phloem tube is blocked by tylose like out growths
(d) none of the above
Answer:
(a) the sieve tubes are blocked by tylose like outgrowths

Question 19.
Which of the statement is not correct?
(a) Sap wood and heart wood can be distinguished in the secondary xylem
(b) Sap wood is paler in colour
(c) Heart wood is darker in colour
(d) The sap wood conducts minerals, while the heart wood conduct water
Answer:
(d) The sap wood conducts minerals, while the heart wood conduct water

Question 20.
Timber from heart wood is:
(a) more fragile and resistant to the attack of insects
(b) more durable and more resistant to the attack of micro organism and insects
(c) more hard and less resistant to the attack of micro organism
(d) less durable and more resistant to the attack of micro organism and insects
Answer:
(b) more durable and more resistant to the attack of micro organism and insects

Samacheer Kalvi 11th Bio Botany Solutions 10 Secondary Growth

Question 21.
The dye, haematoxylin is obtained from:
(a) the heart wood of haematoxylum campechianum
(b) the sap wood of haematoxylum campechianum
(c) cambium cells of haematoxylum campechianum
(d) the seeds of haematoxylum campechianum
Answer:
(a) the heart wood of haematoxylum campechianum

Question 22.
Canada balsam is produced from:
(a) Pisum sativum
(b) resin of Arjuna plant
(c) Abies balsamea
(d) the root of Vinca rosea
Answer:
(c) Abies balsamea

Question 23.
Some commercially important phloem or bast fibres are obtained from:
(a) banana
(b) bamboo
(c) vinca rosea
(d) cannabis sativa
Answer:
(d) cannabis sativa

Question 24.
Phellogen comprises:
(a) homogeneous sclerenchyma cells
(b)homogeneous meristamatic cells
(c) homogeneous collenchyma cells
(d) none of the above cells
Answer:
(b)homogeneous meristamatic cells

Samacheer Kalvi 11th Bio Botany Solutions 10 Secondary Growth

Question 25.
Phelloderm is otherwise called as:
(a) primary cortex
(b) cork wood
(c) secondary cortex
(d) rhytidome
Answer:
(c) secondary cortex

Question 26.
Lenticel is helpful in:
(a) transportation of food
(b) photosynthesis
(c) exchanges of gases and transpiration
(d) transportation of water
Answer:
(c) exchanges of gases and transpiration

Question 27.
The antimalarial compound quinine is, extracted from:
(a) seeds of cinchona
(b) bark of cinchona
(c) leaves of cinchona
(d) flowers of cinchona
Answer:
(b) bark of cinchona

Question 28.
Gum Arabic is obtained from:
(a) Hevea brasiliensis
(b) Acacia Senegal
(c) Pinus
(d) Dilonix regia
Answer:
(b) Acacia Senegal

Samacheer Kalvi 11th Bio Botany Solutions 10 Secondary Growth

Question 29.
Turpentine used as thinner of paints is obtained from:
(a) Acacia Senegal
(b) Vinca rosea
(c) Hevea brasiliensis
(d) Pinus
Answer:
(d) Pinus

Question 30.
Rubber is obtained from:
(a) Bombax mori
(b) Hevea brasiliensis
(c) Quercus suber
(d) Morus rubra
Answer:
(b) Hevea brasiliensis

II. Answer the following. (2 Marks)

Question 1.
Define primary growth?
Answer:
The roots and stems grow in length with the help of apical meristems. This is called primary growth or longitudinal growth.

Question 2.
Mention the two lateral meristem responsible for secondary growth.
Answer:
The secondary growth in dicots and gymnosperms is brought about by two lateral meristems.

  1. Vascular Cambium and
  2. Cork Cambium

Question 3.
What is meant by vascular cambium?
Answer:
The vascular cambium is the lateral meristem that produces the secondary vascular tissues. i.e., secondary xylem and secondary phloem.

Question 4.
Define intrafascicular or fascicular cambium?
Answer:
A strip of vascular cambium that is believed to originate from the procambium is present between xylem and phloem of the vascular bundle. This cambial strip is known as intrafascicular or fascicular cambium.

Question 5.
Define interfascicular cambium?
Answer:
In between the vascular bundles, a few parenchymatous cells of the medullary rays that are in line with the fascicular cambium become meristematic and form strips of vascular cambium. It is called interfascicular cambium.

Samacheer Kalvi 11th Bio Botany Solutions 10 Secondary Growth

Question 6.
What is vascular cambial ring?
Answer:
This interfascicular cambium joins with the intrafascicular cambium on both sides to form a continuous ring. It is called a vascular cambial ring.

Question 7.
What is meant by stratified cambium?
Answer:
If the fusiform initials are arranged in horizontal tiers, with the end of the cells of one tier appearing at approximately the same level, as seen in Tangential Longitudinal Section (TLS), it is called storied (stratified) cambium.

Question 8.
Explain non – stratified cambium.
Answer:
In plants with long fusiform initials, they strongly overlap at the ends, and this type of cambium is called non – storied (non – startified) cambium.

Samacheer Kalvi 11th Bio Botany Solutions 10 Secondary Growth

Question 9.
Give a brief note on ray initials.
Answer:
These are horizontally elongated cells. They give rise to the ray cells and form the elements of the radial system of secondary xylem and phloem.

Question 10.
How does secondary xylem or wood form?
Answer:
The secondary xylem, also called wood, is formed by a relatively complex meristem, the vascular cambium, consisting of vertically (axial) elongated fusiform initials and horizontally (radially) elongated ray initials.

Question 11.
What is meant by spring wood?
Answer:
In the spring season, cambium is very active and produces a large number of xylary elements having vessels / tracheids with wide lumen. The wood formed during this season is called spring wood or early wood.

Samacheer Kalvi 11th Bio Botany Solutions 10 Secondary Growth

Question 12.
How does the autumn wood form?
Answer:
In winter, the cambium is less active and forms fewer xylary elements that have narrow vessels /  tracheids and this wood is called autumn wood or late wood.

Question 13.
Define growth rings?
Answer:
The annual ring denotes the combination of early wood and late wood and the ring becomes evident to our eye due to the high density of late wood. Sometimes annual rings are called growth rings.

Question 14.
Define dendroclimatology?
Answer:
It is a branch of dendrochronology concerned with constructing records of past climates and climatic events by analysis of tree growth characteristics, especially growth rings.

Question 15.
Explain diffuse porous woods with an example.
Answer:
Diffuse porous woods are woods in which the vessels or pores are rather uniform in size and distribution throughout an annual ring. eg: Acer

Samacheer Kalvi 11th Bio Botany Solutions 10 Secondary Growth

Question 16.
What is meant by ring porous woods?
Answer:
The pores of the early wood are distinctly larger than those of the late wood. Thus rings of wide and narrow vessels occur.

Question 17.
Define tyloses?
Answer:
In many dicot plants, the lumen of the xylem vessels is blocked by many balloon like ingrowths from the neighbouring parenchymatous cells. These balloons like structure are called tyloses.

Question 18.
Mention two plants from which bast fibres are obtained.
Answer:
Two plants from which bast fibres are obtained:

  1. Flax – Linum ustitaissimum
  2. Hemp – Cannabis sativa

Question 19.
Define Rhytidome?
Answer:
Rhytidome is a technical term used for the outer dead bark which consists of periderm and isolated cortical or phloem tissues ? formed during successive secondary growth, eg: Quercus.

Samacheer Kalvi 11th Bio Botany Solutions 10 Secondary Growth

Question 20.
What is polyderm? Explain briefly.
Answer:
Polyderm is found in the roots and underground stems. eg: Rosaceae. It refers to a special type of protective tissues consisting of uniseriate suberized layer alternating with multiseriate nonsuberized cells in periderm.

Question 21.
Define’bark’?
Answer:
The term ‘bark’ is commonly applied to all the tissues outside the vascular cambium of stem (i.e., periderm, cortex, primary phloem and secondary phloem).

Question 22.
What are the functions of lenticel?
Answer:
Lenticel is helpful in exchange of gases and transpiration called lenticular transpiration.

Question 23.
Explain briefly phelloderm.
Answer:
It is a tissue resembling cortical living parenchyma produced centripetally (inward) from the phellogen as a part of the periderm of stems and roots in seed plants.

Question 24.
What is the function of secondary phloem?
Answer:
Secondary phloem is a living tissue that transports soluble organic compounds made during photosynthesis to various parts of plant.

Samacheer Kalvi 11th Bio Botany Solutions 10 Secondary Growth

Question 25.
what is periderm?
Answer:
Whenever stems and roots increase in thickness by secondary growth, the periderm, a protective tissue of secondary origin replaces the epidermis and Often primary cortex. The periderm consists of phellem, phellogen, and phelloderm.

III. Answer the following. (3 Marks)

Question 1.
Distinguish between primary and secondary growth.
Answer:
1. Primary growth: The plant organs originating from the apical meristems pass through a period of expansion in length and width. The roots and stems grow in length with the help of apical meristems. This is tailed primary growth or longitudinal growth.

2. Secondary growth: The gymnosperms and most angiosperms, including some monocots, show an increase in thickness of stems and roots by means of secondary growth or latitudinal growth.

Question 2.
Explain fusiform initials.
Answer:
These are vertically elongated cells. They give rise to the longitudinal or axial system of the secondary xylem (treachery elements, fibers, and axial parenchyma) and phloem (sieve elements, fibers, and axial parenchyma).

Question 3.
Explain briefly about false annual rings.
Answer:
Additional growth rings are developed within a year due to adverse natural calamities like drought, frost, defoliation, flood, mechanical injury and biotic factors during the middle of a growing season, which results in the formation of more than one annual ring. Such rings are called pseudo – or false – annual rings.

Samacheer Kalvi 11th Bio Botany Solutions 10 Secondary Growth

Question 4.
Write down the differences between spring wood and autumn wood.
Answer:
The differences between spring wood and autumn wood:

Spring wood or Early wood

Autumn wood or Late wood

1. The activity of cambium is faster. 1. Activity of cambium is slower.
2. Produces large number of xylem elements. 2. Produces fewer xylem elements.
3. Xylem vessels /  trachieds have wider lumen. 3. Xylem vessels / trachieds have narrow lumen.
4. Wood is lighter in colour and has lower density. 4. Wood is darker in colour and has a higher density.

Question 5.
How do you distinguish between sap wood and heart wood?
Answer:

Sap wood (Alburnum)

Heart wood (Duramen)

1. Living part of the wood. 1. Dead part of the wood.
2. It is situated on the outer side of wood. 2.It is situated in the certre part of wood.
3. It is less in coloured. 3. It is dark in coloured.
4. Very soft in nature. 4. Hard in nature.
Tyloses are absent.  Tyloses are present.
5. It is not durable and not resistant to microorganisms. 5. It is more durable and resists microorganisms.

Question 6.
What are fossil resins? Explain with an example.
Answer:
Plants secrete resins for their protective benefits. Amber is a fossilized tree resinespecially from the wood, which has been appreciated for its colour and natural beauty since neolithic times. Much valued from antiquity to the present as a gemstone, amber is made into a variety of decorative objects. Amber is used in jewellery. It has also been used as a healing agent in folk medicine.

Samacheer Kalvi 11th Bio Botany Solutions 10 Secondary Growth

Question 7.
Write briefly about Cork cambium.
Answer:
It is a secondary lateral meristem. It comprises homogenous meristematic cells unlike vascular cambium. It arises from epidermis, cortex, phloem or pericycle (extrastelar in origin). Its cells divide periclinally and produce radially arranged files of cells. The cells towards the outer side differentiate into phellem (cork) and those towards the inside as phelloderm (secondary cortex).

Question 8.
Explain the term lenticel.
Answer:
Lenticel is raised opening or pore on the epidermis or bark of stems and roots. It is formed during secondary growth in stems. When phellogen is more active in the region of lenticels, a mass of loosely arranged thin – walled parenchyma cells are formed. It is called complementary tissue or filling tissue. Lenticel is helpful in exchange of gases and transpiration called lenticular transpiration.

Question 9.
Mention the benefits of bark in a tree.
Answer:
Bark protects the plant from parasitic fungi and insects, prevents water loss by evaporation and guards against variations of external temperature. It is an insect repellent, decay proof, fireproof and is used in obtaining drugs or spices. The phloem cells of the bark are involved in conduction of food while secondary cortical cells involved in storage.

Samacheer Kalvi 11th Bio Botany Solutions 10 Secondary Growth

Question 10.
Distinguish between Intrafascicular Interfascicular cambium.
Answer:
Between Intrafascicular Interfascicular cambium:

Intrafascicular cambium

Interfascicular cambium

1. Present inside the vascular bundles 1. Present in between the vascular bundles.
2. Originates from the procambium. 2. Originates from the medullary rays.
3. Initially it forms a part of the primary meristem. 3. From the beginning it forms a part of the secondary meristem.

IV. Answer In detail
Question 1.
Describe the activity of vascular with the help of diagram.
Answer:
Activity of Vascular Cambium:
The vascular cambial ring, when active, cuts off new cells both towards the inner and outer side. The cells which are produced outward form secondary phloem and inward secondary xylem. At places, cambium forms some narrow horizontal bands of parenchyma which passes through secondary phloem and xylem. These are the rays. Due to the continued formation of secondary xylem and phloem through vascular cambial activity, both the primary xylem and phloem get gradually crushed.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 10 Secondary Growth 6

Question 2.
Describe the formation of sap wood and heart wood with suitabie diagram.
Answer:
Sap wood and heart wood can be distinguished in the secondary xylem. In any tree the outer part of the wood, which is paler in colour, is called sap wood are alburnum. The centre part of the wood, which is darker in colour is called heart wood or duramen. The sap wood conducts water while the heart wood stops conducting water. As vessels of the heart wood are blocked by tyloses, water is not conducted through them.

Due to the presence of tyloses and their contents the heart wood becomes coloured, dead and the hardest part of the wood. From the economic point of view, generally the heartwood is more useful than the sapwood. The timber form the heartwood is more durable and more resistant to the attack of microorganisms and insects than the timber from sapwood.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 10 Secondary Growth 4

Question 3.
Draw and label the transverse section of dicot stem showing the secondary growth.

Answer:
The transverse section of dicot stem showing the secondary growth:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 10 Secondary Growth 5

Question 4.
Distinguish between Phellem and Phelloderm.
Answer:
Phellem (Cork):

  1. It is formed on the outer side of phellogen.
  2. Cells are compactly arranged in regular tires and rows without intercellular spaces.
  3. Protective in function.
  4. Consists of nonliving cells with suberized walls.
  5. Lenticels are present.

Phelloderm (Secondary cortex):

  1. It is formed on the inner side of phellogen.
  2. Cells are loosely arranged with intercellular spaces.
  3. As it contains chloroplast, it synthesises and stores food.
  4. Consists of living cells, parenchymatous in nature and does not have suberin.
  5. Lenticels are absent.

Question 5.
Write down the economic importance of tree bark.
Answer:
The economic importance of tree bark:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 10 Secondary Growth 3

Question 5.
Draw the different stages of secondary growth in a dicot root and label the parts.
Answer:
Stages of secondary growth in a dicot root and label the parts:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 10 Secondary Growth 2

Solution To Activity
Textbook Page No: 38
Question 1.
Generally monocots do not have secondary growth, but palms and bamboos have woody stems. Find the reason.
Answer:
Some of the monocots like palm and bamboos show an increase in thickness of stems by means of secondary growth or latitudinal growth.

Textbook Page No: 48
Question 2.
Be friendly with your environment (Eco friendly) Why should not we use the natural products which are made by plant fibres like rope, fancy bags, mobile pouch, mat and gunny bags etc., instead of using plastics or nylon?
Answer:
We should not use the natural products, which are made by plants fibres, because, if we use more of plant products the greedy people will exploit the plant resources for making plant products and thereby depleting the tree cover, which in turn causes reduction in rain fall.

Samacheer Kalvi 11th Bio Botany Solutions 10 Secondary Growth

Samacheer Kalvi 11th Bio Botany Solutions Chapter 7 Cell Cycle

Students can Download Bio Botany Chapter 7 Cell Cycle Questions and Answers, Notes Pdf, Samacheer Kalvi 11th Bio Botany Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Bio Botany Solutions Chapter 7 Cell Cycle

Samacheer Kalvi 11th Bio Botany Cell Cycle Text Book Back Questions and Answers

I. Multiple Choice Questions
Choose the correct answer

Question 1.
The correct sequence in cell cycle is …………… .
(a) S-M-G1-G2
(b) S-G1-G2-M
(c) G1-S-G2-M
(d) M-G-G2-S
Answer:
(c) G1-S-G2-M

Question 2.
If mitotic division is restricted in G1 phase of the cell cycle then the condition is known as …………… .
(a) S Phase
(b) G2 Phase
(c) M Phase
(d) G0 Phase
Answer:
(d) G0 Phase

Samacheer Kalvi 11th Bio Botany Solutions 7 Cell Cycle

Question 3.
Anaphase promoting complex APC is a protein degradation machinery necessary for proper mitosis of animal cells. If APC is defective in human cell, which of the following is expected to occur?
(a) Chromosomes will be fragmented
(b) Chromosomes will not condense
(c) Chromosomes will not segregate
(d) Recombination of chromosomes will occur
Answer:
(b) Chromosomes will not condense

Question 4.
In S phase of the cell cycle …………… .
(a) Amount of DNA doubles in each cell
(b) Amount of DNA remains same in each cell
(c) Chromosome number is increased
(d) Amount of DNA is reduced to half in each cell
Answer:
(a) Amount of DNA doubles in each cell

Question 5.
Centromere is required for …………… .
(a) Transcription
(b) Crossing over
(c) Cytoplasmic cleavage
(d) Movement of chromosome towards pole
Answer:
(d) Movement of chromosome towards pole

Question 6.
Synapsis occur between …………… .
(a) mRNA and ribosomes
(b) Spindle fibres and centromeres
(c) Two homologous chromosomes
(d) A male and a female gamete
Answer:
(c) Two homologous chromosomes

Samacheer Kalvi 11th Bio Botany Solutions 7 Cell Cycle

Question 7.
In meiosis crossing over is initiated at …………… .
(a) Diplotene
(b) Pachytene
(c) Leptotene
(d) Zygotene
Answer:
(b) Pachytene

Question 8.
Colchicine prevents the mitosis of the cells at which of the following stage …………… .
(a) Anaphase
(b) Metaphase
(c) Prophase
(d) Interphase
Answer:
(b) Metaphase

Question 9.
The paring of homologous chromosomes on meiosis is known as …………… .
(a) Bivalent
(b) Synapsis
(c) Disjunction
(d) Synergids
Answer:
(b) Synapsis

Question 10.
Anastral mitosis is the characteristic feature of …………… .
(a) Lower animals
(b) Higher animals
(c) Higher plants
(d) All living organisms
Answer:
(c) Higher plants

Samacheer Kalvi 11th Bio Botany Solutions 7 Cell Cycle

Question 11.
Write any three significance of mitosis.
Answer:
Exact copy of the parent cell is produced by mitosis (genetically identical).

  1. Genetic stability – daughter cells are genetically identical to parent cells.
  2. Repair of tissues – damaged cells must be replaced by identical new cells by mitosis.
  3. Regeneration – Arms of star fish.

Question 12.
Differentiate between Mitosis and Meiosis.
Answer:
Difference Between Mitosis and Meiosis:

Difference Between Mitosis and Meiosis

Mitosis

Meiosis

1. One division 1. Two divisions
2. Number of chromosomes remains the same 2. Number of chromosomes is halved
3. Homologous chromosomes line up separately on the metaphase plate 3. Homologous chromosomes line up in pairs at the metaphase plate
4. Homologous chromosome do not pair up 4. Homologous chromosome pairup to form bivalent
5. Chiasmata do not form and crossing over never occurs 5. Chiasmata form and crossingover occurs
6. Daughter cells are genetically identical 6. Daughter cells are genetically different from the parent cells
7. Two daughter cells are formed 7. Four daughter cells are formed

Question 13.
Given an account of G0 phase.
Answer:
Some cells exit G1 and enters a quiescent stage called G0, where the cells remain metabolically active without proliferation. Cells can exist for long periods in G0 phase. In G0 cells cease growth with reduced rate of RNA and protein synthesis. The G0 phase is not permanent. Mature neuron and skeletal muscle cell remain permanently in G0 .Many cells in animals remains in G0 unless called onto proliferate by appropriate growth factors or other extracellular signals. G0 cells are not dormant.

Question 14.
Differentiate Cytokinesis in plant cells and animal cells.
Answer:
1. Cytokinesis in Plant Cells:
Division of the cytoplasm often starts during telophase. In plants, cytokinesis cell plate grows from centre towards lateral walls centrifugal manner of cell plate formation. Phragmoplast contains microtubules, actin filaments and vesicles from golgi apparatus and ER. The golgi vesicles contains carbohydrates such as pectin, hemicellulose which move along the microtubule of the pharagmoplast to the equator fuse, forming a new plasma membrane and the materials which are placed their becomes new cell wall.

The first stage of cell wall construction is a line dividing the newly forming cells called a cell plate. The cell plate eventually stretches right across the cell forming the middle lamella. Cellulose builds up on each side of the middle lamella to form the cell walls of two new plant cells.

2. Cytokinesis in Animal Cells:
It is a contractile process. The contractile mechanism contained in contractile ring located inside the plasma membrane. The ring consists of a bundle of microfilaments assembled from actin and myosin. This fibril helps for the generation of a contractile force. This force draws the contractile ring inward forming a cleavage furrow in the cell surface dividing the cell into two.

Samacheer Kalvi 11th Bio Botany Solutions 7 Cell Cycle

Question 15.
Write about Pachytene and Diplotene of Prophase I.
Answer:
1. Pachytene: At this stage bivalent chromosomes are clearly visible as tetrads. Bivalent of meiosis I consists of 4 chromatids and 2 centromeres. Synapsis is completed and recombination nodules appear at a site where crossing over takes place between non – sister chromatids of homologous chromosome. Recombination of homologous chromosomes is completed by the end of the stage but the chromosomes are linked at the sites of crossing over. This is mediated by the enzyme recombinase.

2. Diplotene: Synaptonemal complex disassembled and dissolves. The homologous chromosomes remain attached at one or more points where crossing over has taken place. These points of attachment where ‘X’ shaped structures occur at the sites of crossing over is called Chiasmata. Chiasmata are chromatin structures at sites where recombination has been taken place. They are specialised chromosomal structures that hold the homologous chromosomes together.

Sister chromatids remain closely associated whereas the homologous chromosomes tend to separate from each other but are held together by chiasmata. This substage may last for days or years depending on the sex and organism. The chromosomes are very actively transcribed in females as the egg stores up materials for use during embryonic development. In animals, the chromosomes have prominent loops called lampbrush chromosome.

Samacheer Kalvi 11th Bio Botany Cell Cycle Additional Questions and Answers

I. Multiple Choice Questions
Choose the correct answer:
Question 1.
Most of the neurons in the brain are in …………… stage.
(a) G1
(b) S
(c) G2
(d) G0
Answer:
(d) G0

Question 2.
Short, constricted region in the chromosome is …………… .
(a) Kinetochore
(b) Centromere
(c) Satellite
(d) Telomere
Answer:
(b) Centromere

Samacheer Kalvi 11th Bio Botany Solutions 7 Cell Cycle

Question 3.
Robert Brown discovered the nucleus in the cells of …………… roots.
(a) Mirabilas
(b) Orchid
(c) Moringa
(d) Oryza
Answer:
(b) Orchid

Question 4.
Scientist who described chromosomes for the first time is …………… .
(a) Robert Brown
(b) Anton van Leeuwenhoek
(c) Boveri
(d) Anton Schneider
Answer:
(d) Anton Schneider

Question 5.
Number of chromosomes in onion cell is …………… .
(a) 8
(b) 16
(c) 32
(d) 64
Answer:
(a) 16

Samacheer Kalvi 11th Bio Botany Solutions 7 Cell Cycle

Question 6.
Longest part of the cell cycle is …………… .
(a) Prophase
(b) G1 Phase
(c) Interphase
(d) Sphase
Answer:
(c) Interphase

Question 7.
Eukaryotic cells divides every …………… .
(a) 12
(b) 24
(c) 1
(d) 6
Answer:
(b) 24

Question 8.
Cell cycle was discovered by …………… .
(a) Singer & Nicolson
(b) Prevost & Dumans
(c) Schleider & Schwann
(d) Boveri
Answer:
(b) Prevost & Dumans

Question 9.
G0 stage is called as …………… stage.
(a) Quiescent
(b) Metabolically active
(c) Synthesis of DNA
(d) Replication
Answer:
(a) Quiescent

Samacheer Kalvi 11th Bio Botany Solutions 7 Cell Cycle

Question 10.
…………… protein acts as major check point in phase.
(a) Porins
(b) Kinases
(c) Cyclins
(d) Ligases
Answer:
(c) Cyclins

Question 11.
Replication of DNA occurs at …………… phase.
(a) G0
(b) G1
(c) S
(d) G2
Answer:
(c) S

Question 12.
Condensation of interphase chromosomes into mitotic forms is done by …………… proteins.
(a) MPF
(b) APF
(c) AMF
(d) MAF
Answer:
(a) MPF

Samacheer Kalvi 11th Bio Botany Solutions 7 Cell Cycle

Question 13.
Which of the following is also called as direct division?
(a) Amitosis
(b) Meiosis
(c) Mitosis
(d) Reduction division
Answer:
(a) Amitosis

Question 14.
Cells of mammalian cartilage undergoes …………… .
(a) Amitosis
(b) Meiosis
(c) Mitosis
(d) Equational division
Answer:
(a) Amitosis

Question 15.
Yeast cells undergo …………… .
(a) Open mitosis
(b) Closed mitosis
(c) Amitosis
(d) Meiosis
Answer:
(b) Closed mitosis

Samacheer Kalvi 11th Bio Botany Solutions 7 Cell Cycle

Question 16.
…………… is the longest phase in mitosis.
(a) Anaphase
(b) Telophase
(c) Prophase
(d) Interphase
Answer:
(c) Prophase

Question 17.
The DNA protein complex present in the centromere is …………… .
(a) Cyclin
(b) Kinesis
(c) MPF
(d) Kinetochore
Answer:
(d) Kinetochore

Question 18.
…………… protein induces the break down of cohesion proteins leading to chromatid separation during mitosis.
(a) APC
(b) MPF
(c) Cyclin
(d) Kinetochore
Answer:
(a) APC

Question 19.
Regeneration of arms of star fish is due to …………… .
(a) Meiosis
(b) Amitosis
(c) Mitosis
(d) Budding
Answer:
(c) Mitosis

Samacheer Kalvi 11th Bio Botany Solutions 7 Cell Cycle

Question 20
…………… is called as reduction division.
(a) Meiosis
(b) Mitosis
(c) Amitosis
(d) Budding
Answer:
(a) Meiosis

Question 21.
Bivalents occur at …………… stage.
(a) Zygotene
(b) Leptotene
(c) Pachytene
(d) Diplotene
Answer:
(a) Zygotene

Question 22.
Recombination of chromosomes occur at …………… .
(a) Zygotene
(b) Leptotene
(c) Pachytene
(d) Diplotene
Answer:
(c) Pachytene

Question 23.
Terminalisation of chiasmata occurs at …………… .
(a) Zygotene
(b) Leptotene
(c) Diakinesis
(d) Pachytene
Answer:
(c) Diakinesis

Samacheer Kalvi 11th Bio Botany Solutions 7 Cell Cycle

Question 24.
Number of daughter cells formed at the end of Meiosis I is …………… .
(a) 2
(b) 4
(c) 1
(d) 0
Answer:
(a) 2

Question 25.
…………… division leads to genetic variability.
(a) Mitotic
(b) Amitotic
(c) Meiotic
(d) Equational
Answer:
(c) Meiotic

Question 26.
Crossing over occurs at …………… stage.
(a) Leptotene
(b) Zygotene
(c) Pachytene
(d) Diplotene
Answer:
(c) Pachytene

Samacheer Kalvi 11th Bio Botany Solutions 7 Cell Cycle

Question 27.
Which of the following is not a mitogen?
(a) Giberellin
(b) Ethylene
(c) Kinetin
(d) Colchicine
Answer:
(d) Colchicine

Question 28.
In plants mitosis occurs at …………… cells.
(a) Sclerenchyma
(b) Meristem
(c) Xylem
(d) Parenchyma
Answer:
(b) Meristem

Samacheer Kalvi 11th Bio Botany Solutions 7 Cell Cycle

Question 29.
Which of the following alone is formed in the division of plant cells?
(a) Aster
(b) Centrioles
(c) Spindle
(d) Microtubules
Answer:
(c) Spindle

Question 30.
Amphiastral type cell division is seen in …………… cells.
(a) Fungal
(b) Algal
(c) Plant cells
(d) Animal
Answer:
(d) Animal

II. Very Short Answer Type Questions (2 Marks)

Question 1.
Name the two types of nuclear division.
Answer:
The two types of nuclear division:

  1. Mitosis and
  2. Meiosis.

Question 2.
Define Cell Cycle.
Answer:
A series of events leading to the formation of new cell is known as cell cycle.

Question 3.
Who discovered the Cell Cycle?
Answer:
Prevost & Dumans in 1824.

Samacheer Kalvi 11th Bio Botany Solutions 7 Cell Cycle

Question 4.
Draw a tabular column showing the duration of various phase in the cell cycle of human cell.
Answer:
A tabular column showing the duration of various phase in the cell cycle of human cell:

Cell cycle of a proliferating human cell

Phase

Time Duration (in hrs)

1. G2 1. 11
2. S 2. 8
3. G2 3. 4
4. M 4. 1

Question 5.
Define C – Value.
Answer:
C – Value is the amount in picograms of DNA contained within a haploid nucleus.

Question 6.
Which is the longest phase of cell cycle? What happens during that phase?
Answer:
Interphase is the longest phase. Cells are metabolically active and involved in protein synthesis and growth.

Question 7.
Name the phases which comprises the Interphase.
Answer:
The phases which comprises the Interphase:

  1. G1 Phase
  2. S Phase and
  3. G2 Phase.

Samacheer Kalvi 11th Bio Botany Solutions 7 Cell Cycle

Question 8.
Name the proteins involved in the activation of genes & their proteins to perform cell division.
Answer:
Kinases & Cyclins.

Question 9.
What do you mean by G0 stage?
Answer:
G0 stage is called as quiescent stage, where the cells remain metabolically active without proliferation.

Question 10.
What is the role of MPF in Cell cycle?
Answer:
Maturation Promoting Factor (MPF) brings about condensation of interphase chromosomes into the mitotic form.

Question 11.
Distinguish between Karyokinesis & Cytokinesis.
Answer:
Between Karyokinesis & Cytokinesis:

  • Karyokinesis: Karyokinesis refers to the nuclear division.
  • Cytokinesis: Cytokinesis refers to the cytoplasmic division.

Question 12.
Point out any two cell – types which remain G0 phase.
Answer:
Mature neurons and Skeletal muscle cells.

Samacheer Kalvi 11th Bio Botany Solutions 7 Cell Cycle

Question 13.
Why amitosis is called as incipient cell division?
Answer:
Amitosis is also called incipient cell division. Since there is no spindle formation and chromatin material does not condense.

Question 14.
List out the disadvantages of Amitosis.
Answer:
The disadvantages of Amitosis:

  • Causes unequal distribution of chromosomes.
  • Can lead to abnormalities in metabolism and reproduction.

Question 15.
Mitosis also called as equational division – Justify.
Answer:
At the end of mitosis the number of chromosomes in the parent and the daughter (Progeny) cells remain the same so it is also called as equational division.

Samacheer Kalvi 11th Bio Botany Solutions 7 Cell Cycle

Question 16.
Enumerate the stages of mitosis.
Answer:
Mitosis is divided into four stages prophase, metaphase, anaphase and telophase.

Question 17.
Define an aster.
Answer:
In animal cell the centrioles extend a radial array of microtubules towards the plasma membrane when they reach the poles of the cell. This arrangement of microtubules is called an aster. Plant cells do not form asters.

Question 18.
What is metaphase plate?
Answer:
The alignment of chromosome into compact group at the equator of the cell is known as metaphase plate.

Question 19.
What is Kinetochore?
Answer:
Kinetochore is a DNA – Protein complex present in the centromere DNA, where the microtubules are attached. It is a trilaminar disc like plate.

Question 20.
How will you calculate the length of the S period.
Answer:
Length of the S period = Fraction of cells in DNA replication × generation time.

Samacheer Kalvi 11th Bio Botany Solutions 7 Cell Cycle

Question 21.
Which type of cell division occurs in reproductive cells? What will be the result?
Answer:
Meiosis takes place in the reproductive organs. It results in the formation of gametes with half the normal chromosome number.

Question 22.
Define Synapsis.
Answer:
In Zygotene, pairing of homologous chromosomes takes place and it is known as synapsis.

Question 23.
What do you understand by independent assortment?
Answer:
The random distribution of homologous chromosomes in a cell in Metaphase I is called independent assortment.

Question 24.
Define Mitogen. Give an example.
Answer:
The factors which promote cell cycle proliferation is called mitogen.
Example: gibberellin. These increase mitotic rate.

Question 25.
What are mitotic poisons.
Answer:
Certain chemical components act as inhibitors of the mitotic cell division and they are called mitotic poisons.

Samacheer Kalvi 11th Bio Botany Solutions 7 Cell Cycle

Question 26.
Distinguish between Anastral & Amphiastral.
Answer:
Between Anastral & Amphiastral:
Anastral:

  1. This is present only in plant cells.
  2. No asters or centrioles are formed only spindle fibres are formed during cell division.

Amphiastral:

  1. This is found in animal cells.
  2. Aster and centrioles are formed at each pole of the spindle during cell division.

Question 27.
Draw a simple diagram to show the Amitosis.
Answer:
The Amitosis:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 7 Cell Cycle 1

III. Short Answer Type Questions (3 Marks)

Question 1.
What is the role of nucleus in the cell?
Answer:
The role of nucleus in the cell:

  • Control activities of the cell.
  • Genetic information copied from cell to cell while the cell divides.
  • Hereditary characters are passed onto new individuals when gametic cells fuse together in sexual reproduction.

Question 2.
What are restriction points? Mention its role in Cell cycle.
Answer:
The checkpoint called the restriction point at the end of G1 it determines a cells fate whether it will continue in the cell cycle and divide or enter a stage called G0 as a quiescent stage and probably as specified cell or die.

Samacheer Kalvi 11th Bio Botany Solutions 7 Cell Cycle

Question 3.
Point out the reasons responsible for the arresting of the cell in G1 phase?
Answer:
Cells are arrested in G1 due to:

  • Nutrient deprivation
  • Lack of growth factors or density dependant inhibition
  • Undergo metabolic changes and enter into G0 state.

Question 4.
Write a note on G0 phase.
Answer:
Some cells exit G1 and enters a quiescent stage called G0, where the cells remain metabolically active without proliferation. Cells can exist for long periods in G0 phase. In G0 cells cease growth with reduced rate of RN A and protein synthesis. The G0 phase is not permanent. Mature neuron and skeletal muscle cell remain permanently in G0. Many cells in animals remains in G0 unless called onto proliferate by appropriate growth factors or other extracellular signals. G0 cells are not dormant.

Question 5.
List out the events taking place in S – Phase.
Answer:
S Phase – Synthesis phase – cells with intermediate amounts of DNA Growth of the cell continues as replication of DNA occur, protein molecules called histones are synthesised and attach to the DNA. The centrioles duplicate in the cytoplasm. DNA content increases from 2C to 4C.

Samacheer Kalvi 11th Bio Botany Solutions 7 Cell Cycle

Question 6.
Distinguish between Karyokinesis & Cytokinesis.
Answer:
Karyokinesis:

  1. Involves division of nucleus.
  2. Nucleus develops a constriction at the center and becomes dumbellshaped.
  3. Constriction deepens and divides the nucleus into two.

Cytokinesis:

  1. Involves division of cytoplasm.
  2. Plasma membrane develops a constriction along nuclear constriction.
  3. It deepens centripetally and finally divides the cell into two cells.

Question 7.
Explain the differences between closed and open mitosis.
Answer:
Between closed and open mitosis:

  1. In closed mitosis, the nuclear envelope remains intact and chromosomes migrate to opposite poles of a spindle within the nucleus. Example: Yeast and slime molds.
  2. In open mitosis, the nuclear envelope breaks down and then reforms around the 2 sets of separated chromosome. Example: Most plants and animals cells.

Question 8.
What happens to plant cells at the end of Telophase in Mitosis?
Answer:
In plants, phragmoplast are formed between the daughter cells. Cell plate is formed between the two daughter cells, reconstruction of cell wall takes place. Finally the cells are separated by the distribution of organelles, macromolecules into two newly formed daughter cells.

Samacheer Kalvi 11th Bio Botany Solutions 7 Cell Cycle

Question 9.
Bring out the significance of Meiosis.
Answer:
The significance of Meiosis:

  • Meiosis maintains a definite constant number of chromosomes in organisms.
  • Crossing over takes place and exchange of genetic material leads to variations among species. These variations are the raw materials to evolution. Meiosis leads to genetic variability by partitioning different combinations of genes into gametes through independent assortment.
  • Adaptation of organisms to various environmental stress.

Question 10.
Differentiate between the mitosis of Plant Cell & Animal Cell.
Answer:
Plants:

  1. Centrioles are absent
  2. Asters are not formed
  3. Cell division involves formation of a cell plate
  4. Occurs mainly at meristem.

Animals:

  1. Centrioles are present
  2. Asters are formed
  3. Cell division involves furrowing and cleavage of cytoplasm
  4. Occurs in tissues throughout the body.

Question 11.
Explain briefly about Endomitosis.
Answer:
The replication of chromosomes in the absence of nuclear division and cytoplasmic division resulting in numerous copies within each cell is called endomitosis. Chromonema do not separate to form chromosomes, but remain closely associated with each other. Nuclear membrane does not rupture. So no spindle formation. It occurs notably in the salivary glands of Drosophila and other flies. Cells in these tissues contain giant chromosomes (polyteny), each consisting of over thousands of intimately associated, or synapsed, chromatids. Example: Polytene chromosome.

Samacheer Kalvi 11th Bio Botany Solutions 7 Cell Cycle

Question 12.
How G0 cells help in Closing Technology?
Answer:
Since the DNA of cells in G0, do not replicate. The researcher are able to fuse the donor cells from a sheep’s mammary glands into G0, state by culturing in the nutrient free state. The G0, donor nucleus synchronised with cytoplasm of the recipient egg, which developed into the clone Dolly.

IV. Long Answer Type Questions (5 Marks)

Question 1.
Draw and label the various stages of Prophase I.
Answer:
Label the various stages of Prophase I:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 7 Cell Cycle 2

Question 2.
Explain in detail about the various stages of Prophase I.
Answer:
The various stages of Prophase I:
1. Prophase I – Prophase I is of longer duration and it is divided into 5 substages – Leptotene, Zygotene, Pachytene, Diplotene and Diakinesis.

2. Leptotene – Chromosomes are visible under light microscope. Condensation of chromosomes takes place. Paired sister chromatids begin to condense.

3. Zygotene – Pairing of homologous chromosomes takes place and it is known as synapsis. Chromosome synapsis is made by the formation of synaptonemal complex. The complex formed by the homologous chromosomes are called as bivalent (tetrads).

4. Pachytene – At this stage bivalent chromosomes are clearly visible as tetrads. Bivalent of meiosis I consists of 4 chromatids and 2 centromeres. Synapsis is completed and recombination nodules appear at a site where crossing over takes place between non – sister chromatids of homologous chromosome. Recombination of homologous chromosomes is completed by the end of the stage but the chromosomes are linked at the sites of crossing over. This is mediated by the enzyme recombinase.

5. Diplotene – Synaptonemal complex disassembled and dissolves. The homologous chromosomes remain attached at one or more points where crossing over has taken place. These points of attachment where ‘X’ shaped structures occur at the sites of crossing over is called.

6. Chiasmata: Chiasmata are chromatin structures at sites where recombination has been taken place. They are specialised chromosomal structures that hold the homologous chromosomes together. Sister chromatids remain closely associated whereas the homologous chromosomes tend to separate from each other but are held together by chiasmata. This substage may last for days or years depending on the sex and organism. The chromosomes are very actively transcribed in females as the egg stores up materials for use during embryonic development. In animals, the chromosomes have prominent loops called lampbrush chromosome.

7. Diakinesis – Terminalisation of chiasmata. Spindle fibres assemble. Nuclear envelope breaks down. Homologous chromosomes become short and condensed. Nucleolus disappears.

Question 3.
Describe the process of Cytokinesis in Plant cell & Animal Cell.
Answer:
1. Cytokinesis in Plant Cell: Division of the cytoplasm often starts during telophase. In plants, cytokinesis cell plate grows from centre towards lateral walls – centrifugal manner of cell plate formation. Phragmoplast contains microtubules, actin filaments and vesicles from golgi apparatus and ER. The golgi vesicles contains carbohydrates such as pectin, hemicellulose which move along the microtubule of the pharagmoplast to the equator fuse, forming a new plasma membrane and the materials which are placed there becomes new cell wall.

The first stage of cell wall construction is a line dividing the newly forming cells called a cell plate. The cell plate eventually stretches right across the cell forming the middle lamella. Cellulose builds up on each side of the middle lamella to form the cell walls of two new plant cells.

2. Cytokinesis in Animal Cells:
It is a contractile process. The contractile mechanism contained in contractile ring located inside the plasma membrane. The ring consists of a bundle of microfilaments assembled from actin and myosin. This fibril helps for the generation of a contractile force. This force draws the contractile ring inward forming a cleavage furrow in the cell surface dividing the cell into two.

Samacheer Kalvi 11th Bio Botany Solutions 7 Cell Cycle

Question 4.
What are the significances of Mitosis.
Answer:
Exact copy of the parent cell is produced by mitosis (genetically identical).

  1. Genetic stability – Daughter cells are genetically identical to parent cells.
  2. Growth – As multicellular organisms grow, the number of cells making up their tissue increases. The new cells must be identical to the existing ones.
  3. Repair of tissues – Damaged cells must be replaced by identical new cells by mitosis.
  4. Asexual reproduction – Asexual reproduction results in offspring that are identical to the parent. Example Yeast and Amoeba.
  5. In flowering plants, structure such as bulbs, corms, tubers, rhizomes and runners are produced by mitotic division. When they separate from the parent, they form a new individual. The production of large numbers of offsprings in a short period of time, is possible only by mitosis. In genetic engineering and biotechnology, tissues are grown by mitosis (i.e. in tissue culture).
  6. Regeneration – Arms of star fish

Question 5.
Explain the various phases in Cell Cycle.
Answer:
The different phases of cell cycle are as follows:
1. Interphase: Longest part of the cell cycle, but it is of extremely variable length. At first glance the nucleus appears to be resting but this is not the case at all. The chromosomes previously visible as thread like structure, have dispersed. Now they are actively involved in protein synthesis, at least for most of the interphase. C – Value is the amount in picograms of DNA contained within a haploid nucleus.

2. G1 Phase: The first gap phase – 2C amount of DNA in cells of G1 The cells become metabolically active and grows by producing proteins, lipids, carbohydrates and cell organelles including mitochondria and endoplasmic reticulum. Many checkpoints control the cell cycle. The checkpoint called the restriction point at the end of G1 it determines a cells fate whether it will continue in the cell cycle and divide or enter a stage called G0 as a quiescent stage and probably as specified cell or die. Cells are arrested in G1 due to:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 7 Cell Cycle 3
3. Nutrient deprivation: Lack of growth factors or density dependant inhibition. Undergo metabolic changes and enter into G0 state. Biochemicals inside cells activates the cell division. The proteins called kinases and cyclins activate genes and their proteins to perform cell division. Cyclins act as major checkpoint which operates in G1 to determine whether or not a cell divides.

4. G0 Phase: Some cells exit G1 and enters a quiescent stage called G0, where the cells remain metabolically active without proliferation. Cells can exist for long periods in G0 phase. In G0 cells cease growth with reduced rate of RNA and protein synthesis. The G0 phase is not permanent. Mature neuron and skeletal muscle cell remain permanently in G0. Many cells in animals remains in G0 unless called on to proliferate by appropriate growth factors or other extracellular signals. G0 cells are not dormant.

5. S phase – Synthesis phase – cells with intermediate amounts of DNA. Growth of the cell continues as replication of DNA occur, protein molecules called histones are synthesised and attached to the DNA. The centrioles duplicate in the cytoplasm. DNA content increases from 2C to 4C.

6. G2 – The second Gap phase – 4C amount of DNA in cells of G2 and mitosis. Cell growth continues by protein and cell organelle synthesis, mitochondria and chloroplasts divide. DNA content remains as 4C. Tubulin is synthesised and microtubules are formed. Microtubles organise to form spindle fibre. The spindle begins to form and nuclear division follows.

One of the proteins synthesized only in the G2 period is known as Maturation Promoting Factor (MPF). It brings about condensation of interphase chromosomes into the mitotic form. DNA damage checkpoints operates in G1 S and G2 phases of the cell cycle.

Samacheer Kalvi 11th Bio Botany Solutions 7 Cell Cycle

Question 6.
List out the important features of Chromosomes.
Answer:
The four important features of the chromosome are:
1. The shape of the chromosome is specific: The long, thin, lengthy structured chromosome contains a short, constricted region called centromere. A centromere may occur any where along the chromosome, but it is always in the same position on any given chromosome. The number of chromosomes per species is fixed: For example the mouse has 40 chromosomes, the onion has 16 and humans have 46.

2. Chromosomes occur in pairs: The chromosomes of a cell occur in pairs, called homologous pairs. One of each pair come originally from each parent. Example, human has 46 chromosomes, 23 coming originally from each parent in the process of sexual reproduction. Chromosomes are copied: Between nuclear divisions, whilst the chromosomes are uncoiled and cannot be seen, each chromosome is copied. The two identical structures formed are called chromatids.

V. Higher Order Thinking Skills (HOTs)

Question 1.
Given that the average duplication time of E.coli is 20 minutes. How much time will two E.coli cells takes to become 32 cells?
Answer:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 7 Cell Cycle 4
One cells takes 80 minutes to form 16 cells. If 2 cells undergoes division simultaneously, it take 160 minutes (2 hours 40 minutes) to form 32 cells.

Question 2.
Complete the cell cycle by filling the gaps with respective phases.
Answer:
X= S phase or Synthesis phase
Y= M phase or Mitosis phase
Z= G0 phase
Samacheer Kalvi 11th Bio Botany Solutions Chapter 7 Cell Cycle 5

Question 3.
Telophase is reverse of prophase – Comment
Answer:
Events in Prophase:

  1. Nuclear membrane disappears
  2. Nucleolus disappear
  3. Spindle fibre begins to form
  4. Chromosomes threads condeme to form chromosomes

Events in Telophase:

  1. Nuclear membrane reappears
  2. Nucleolus reappears
  3. Spindle fibre disappears
  4. Chromosomes decondeme to form chromosomes

Question 4.
Name the pathological condition when uncontrolled cell division occurs.
Answer:
Uncontrolled cell division & abnormal growth of cells leads to the pathological condition called tumor or cancer.

Question 5.
Microspores are produced in the multiples of four, why?
Answer:
Microspores are haploid spores produced from diploid microspores mother cells. Each microspores mother cell (2n) undergoes meiosis producing four Microspores (n). Because a complete meiotic division yields 4 cells. Thus microspores are produced in multiples of four.

Samacheer Kalvi 11th Bio Botany Solutions 7 Cell Cycle

Question 6.
Between Prokaryotes & Eukaryotes, which cell has a shorter cell division time.
Answer:
Prokaryotes like bacteria undergo simple form of cell division called binary fission which will get completed with in a hour, whereas Eukaryotic cell division (mitosis) takes nearly 24 hours to get completed. Hence Prokaryotes have shorter cell division time.

Question 7.
Though Prokaryotic cell division differs from Eukaryotic cell division, both show certain common aspects during cell division. Explain.
Answer:
Whether a cell is Prokaryote or Eukaryote, while undergoing division, the following events must occur in common.

  1. Replication of DNA.
  2. Cytokinesis at the end of cell division.

Question 8.
An anther has 1204 pollen grains. How many Pollen mother cells must have been there to produce them?Explain.
Answer:
301 – Pollen mother cells: 301 Pollen mother cells undergo meiosis producing 1204 pollen grains. Because at the end of meiosis, each pollen mother cells produces 4 pollen grains.

Samacheer Kalvi 11th Bio Botany Solutions 7 Cell Cycle

Question 9.
A cell has 32 chromosomes. It undergoes mitosis. What will be the chromosome number during metaphase?
Answer:
During S phase of interphase, the genetic material of the cell is duplicated. So during metaphase, the chromosome number(chromatid number) will be doubled thus 64 chromosomes (chromatids) will be present.

Question 10.
Why sibilings show disimilarities?
Answer:
Though born to same parents, siblings show dissimilarities and variation due to the crossing over and recombination of chromosomes during meiosis.

Question 11.
Ramu’s met with an accident while riding cycle and got wounded in his leg. After few days, the wound was healed and the skin becomes normal. How?
Answer:
Ramu’s wound was healed because of the mitotic division. As a result of mitosis, new cells are produced and damaged tissues were repaired resulting the damaged skin to become normal.

Question 12.
A flower of tomato plant following the process of sexual reproduction produces 240 viable seeds. What is the minimum number of microspore mother cells involved in this process?
Answer:
60 microspore mother cells are involved in providing 240 pollen grains. Because each microspore mother cell undergoes meiosis producing four pollen grains (i.e. 60 × 4 = 240). Each pollen grain produces two male gametes of which one undergoes true fertilization of ovule producing seeds. Other male gamete participate in double fertilization.

Samacheer Kalvi 11th Bio Botany Solutions Chapter 8 Biomolecules

Students can Download Bio Botany Chapter 8 Biomolecules Questions and Answers, Notes Pdf, Samacheer Kalvi 11th Bio Botany Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Bio Botany Solutions Chapter 8 Biomolecules

Samacheer Kalvi 11th Bio Botany Biomolecules Text Book Back Questions and Answers

I. Multiple Choice Questions
Choose the correct answer
Question 1.
The most basic amino acid is …………… .
(a) Arginine
(b) Histidine
(c) Glycine
(d) Glutamine
Answer:
(a) Arginine

Question 2.
An example of feedback inhibition is  …………… .
(a) Cyanide action on cytochrome
(b) Sulpha drug on folic acid synthesiser bacteria
(c) Allosteric inhibition of hexokinase by glucose – 6 – phosphate
(d) The inhibition of succinic dehydrogenase by malonate
Answer:
(c) Allosteric inhibition of hexokinase by glucose – 6 – phosphate

Samacheer Kalvi 11th Bio Botany Solutions 8 Biomolecules

Question 3.
Enzymes that catalyse interconversion of optical, geometrical or positional isomers are …………… .
(a) Ligases
(b) Lyases
(c) Hydrolases
(d) Isomerases
Answer:
(d) Isomerases

Question 4.
Proteins perform many physiological functions. For example some functions as enzymes. One of the following represents an additional function that some proteins discharge …………… .
(a) Antibiotics
(b) Pigment conferring colour to skin
(c) Pigments making colours of flowers
(d) Hormones
Answer:
(d) Hormones

Question 5.
Given below is the diagrammatic representation of one of the categories of small molecular weight organic compounds in the living tissues. Identify the category shown & one blank component “X” in it …………… .
Samacheer Kalvi 11th Bio Botany Solutions Chapter 8 Biomolecules 1
Answer:
(a) Nucleoside
(b) Uracil.

Samacheer Kalvi 11th Bio Botany Solutions 8 Biomolecules

Question 6.
Distinguish between nitrogenous base and a base found in inorganic chemistry.
Answer:
Nitrogenous Base:

  1. Nitrogenous bases are organic molecules containing the element nitrogen & acts as a base in chemical reaction.
  2. e.g. Adenine, Thymine

Base:

  1. Bases are the substance that release hydroxide (OH ) ions in aqueous solution.
  2. e.g. NaOH and Ca(OH)2

Question 7.
What are the factors affecting the rate of enzyme reaction?
Answer:
(a) Temperature: Heating increases molecular motion. Thus the molecules of the substrate and enzyme move more quickly resulting in a greater probability of occurrence of the reaction. The temperature that promotes maximum activity is referred to as optimum temperature.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 8 Biomolecules 2
(b) pH: The optimum pH is that at which the maximum rate of reaction occurs. Thus the pH change leads to an alteration of enzyme shape, including the active site. If extremes of pH are encountered by an enzyme, then it will be denatured.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 8 Biomolecules 3
(c) Substrate Concentration: For a given enzyme concentration, the rate of an enzyme reaction increases with increasing substrate concentration.

(d) Enzyme Concentration: The rate of reaction is directly proportional to the enzyme concentration.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 8 Biomolecules 4
The Michaelis – Menton Constant (Km) and Its Significance:
When the initial rate of reaction of an enzyme is measured over a range of substrate concentrations (with a fixed amount of enzyme) and the results plotted on a graph. With increasing substrate concentration, the velocity increases – rapidly at lower substrate concentration. However the rate increases progressively, above a certain concentration of the substrate the curve flattened out. No further increase in rate occurs. This shows that the enzyme is working at maximum velocity at this point. On the graph, this point of maximum velocity is shown as VMax.

Samacheer Kalvi 11th Bio Botany Solutions 8 Biomolecules

Question 8.
Briefly outline the classification of enzymes.
Answer:
Enzymes are classified into six groups based on their mode of action.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 8 Biomolecules 5

Question 9.
Write the characteristic feature of DNA.
Answer:
The characteristic feature of DNA.

  1. If one strand runs in the 5′ – 3′ direction, the other runs in 3′ – 5′ direction and thus are antiparallel (they run in opposite direction). The 5′ end has the phosphate group and 3’end has the OH group.
  2. The angle at which the two sugars protrude from the base pairs is about 120°, for the narrow angle and 240° for the wide angle. The narrow angle between the sugars generates a minor groove and the large angle on the other edge generates major groove.
  3. Each base is 0.34 nm apart and a complete turn of the helix comprises 3.4 nm or 10 base pairs per turn in the predominant B form of DNA.
  4. DNA helical structure has a diameter of 20 Å and a pitch of about 3 Å. X – ray crystal study of DNA takes a stack of about 10 bp to go completely around the helix (360°).
  5. Thermodynamic stability of the helix and specificity of base pairing includes
    • (a) The hydrogen bonds between the complementary bases of the double helix
    • (b) stacking interaction between bases tend to stack about each other perpendicular to the direction of helical axis. Electron cloud interactions (\({ \Pi -{ \Pi } }\)) between the bases in the helical stacks contribute to the stability of the double helix.
  6. The phosphodiester linkages gives an inherent polarity to the DNA helix. They form strong covalent bonds, gives the strength and stability to the polynucleotide chain.
  7. Plectonemic coiling – the two strands of the DNA are wrapped around each other in a helix, making it impossible to simply move them apart without breaking the entire structure. Whereas in paranemic coiling the two strands simply lie alongside one another, making them easier to pull apart.
  8. Based on the helix and the distance between each turns, the DNA is of three forms – A DNA, B DNA and Z DNA.

Question 10.
Explain the structure and function of different types of RNA.
Answer:
1. mRNA (messenger RNA): Single stranded, carries a copy of instructions for assembling amino acids into proteins. It is very unstable and comprises 5% of total RNA polymer. Prokaryotic mRNA (Polycistronic) carry coding sequences for many polypeptides. Eukaryotic mRNA (Monocistronic) contains information for only one polypeptide.

2. tRNA (transfer RNA): Translates the code from mRNA and transfers amino acids to the ribosome to build proteins. It is highly folded into an elaborate 3D structure and comprises about 15% of total RNA. It is also called as soluble RNA.

3. rRNA (ribosomal RNA): Single stranded, metabolically stable, makeup the two subunits of ribosomes. It constitutes 80% of the total RNA. It is a polymer with varied length from 120 – 3000 nucleotides and gives ribosomes their shape. Genes for rRNA are highly conserved and employed for phylogenetic studies.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 8 Biomolecules 6

Entrance Examination Questions Solved
Choose the correct answer:
Question 1.
Who invented electron microscope? (2010 AIIMS, 2008 JIPMER)
(a) Janssen
(b) Edison
(c) Knoll and Ruska
(d) Landsteiner
Answer:
(c) Knoll and Ruska

Question 2.
Specific proteins responsible for the flow of materials and information into the cellare called …………… . (2009 AIIMS)
(a) Membrane receptors
(b) carrier proteins
(c) integral proteins
(d) none of these
Answer:
(b) carrier proteins

Samacheer Kalvi 11th Bio Botany Solutions 8 Biomolecules

Question 3.
Omnis – cellula – e – cellula was given by …………… . (2007 AIIMS)
(a) Virchow
(b) Hooke
(c) Leeuwenhoek
(d) Robert Brown
Answer:
(a) Virchow

Question 4.
Which of the following is responsible for the mechanical support, protein synthesis and enzyme transport? (2007 AIIMS)
(a) cell membrane
(b) mitochondria
(c) dictyosomes
(d) endoplasmic reticulum
Answer:
(d) endoplasmic reticulum

Question 5.
Genes present in the cytoplasm of eukaryotic cells are found in …………… . (2006 AIIMS)
(a) mitochondria and inherited via egg cytoplasm
(b) lysosomes and peroxisomes
(c) Golgi bodies and smooth endoplasmic reticulum
(d) Plastids inherited via male gametes
Answer:
(a) mitochondria and inherited via egg cytoplasm

Samacheer Kalvi 11th Bio Botany Solutions 8 Biomolecules

Question 6.
In which one the following would you expect to find glyoxysomes? (2005 AIIMS)
(a) Endosperm of wheat
(b) Endosperm of castor
(c) Palisade cells in leaf
(d) Root hairs
Answer:
(b) Endosperm of castor

Question 7.
A quantosome is present in …………… . (JIPMER 2012)
(a) Mitochondria
(b) Chloroplast
(c) Golgi bodies
(d) ER
Answer:
(b) Chloroplast

Question 8.
In mitochondria the enzyme cytochrome oxidase is present in …………… . (2012 JIPMER)
(a) Outer mitochondrial membrane
(b) inner mitochondrial membrane
(c) Stroma
(d) Grana
Answer:
(b) inner mitochondrial membrane

Question 9.
Which organelle is present in higher number in secretory cell? (2008 JIPMER)
(a) Mitochondria
(b) Chloroplast
(c) Nucleus
(d) Dictyosomes
Answer:
(d) Dictyosomes

Question 10.
Major site for the synthesis of lipids …………… . (2013 NEET)
(a) Rough ER
(b) smooth ER
(c) Centriole
(d) Lysosome
Answer:
(b) smooth ER

Samacheer Kalvi 11th Bio Botany Solutions 8 Biomolecules

Question 11.
Golgi complex plays a major role in …………… . (2013 NEET)
(a) post translational modification of proteins and glycosidation of lipids
(b) translation of proteins
(c) Transcription of proteins
(d) Synthesis of lipid
Answer:
(a) post translational modification of proteins and glycosidation of lipids

Question 12.
Main arena of various types of activities of a cell is …………… . (2010 AIPMT)
(a) Nucleus
(b) Mitochondria
(c) Cytoplasm
(d) Chloroplast
Answer:
(c) Cytoplasm

Question 13.
The thylakoids in chloroplast are arranged in …………… . (2005 JIPMER)
(a) regular rings
(b) linear array
(c) diagonal direction
(d) stacked discs
Answer:
(d) stacked discs

Question 14.
Sequences of which of the following is used to know the phylogeny rRNA? (20022JIPMER)
(a) mRNA
(b) rRNA
(c) tRNA
(d) Hn RNA
Answer:
(b) rRNA

Question 15.
Structures between two adjacent cells which is an effective transport pathway? (2010 AIPMT)
(a) Plasmodesmata
(b) Middle lamella
(c) Secondary wall layer
(d) Primary wall layer
Answer:
(a) Plasmodesmata

Question 16.
In active transport carrier proteins are used, which use energy in the form of ATP to …………… .
(a) transport molecules against concentration gradient of cell wall
(b) transport molecules along concentration gradient of cell membrane
(c) transport molecules against concentration gradient of cell membrane
(d) transport molecules along concentration gradient of cell wall
Answer:
(c) transport molecules against concentration gradient of cell membrane

Samacheer Kalvi 11th Bio Botany Solutions 8 Biomolecules

Question 17.
The main organelle involved in modification and routing of newly synthesised protein to their destinations is …………… . (AIPMT 2005)
(a) Mitochondria
(b) Glyoxysomes
(c) Spherosomes
(d) Endoplasmic reticulum
Answer:
(d) Endoplasmic reticulum

Question 18.
Algae have cell wall made up of …………… . (AIPMT 2010)
(a) Cellulose, galactans and mannans
(b) Cellulose, chitin and glucan
(c) Cellulose, Mannan and peptidoglycan
Answer:
(a) Cellulose, galactans and mannans

Samacheer Kalvi 11th Bio Botany Biomolecules Additional Questions and Answers

Question 1.
The percentage of water in the total cellular mass is …………… .
(a) 50%
(b) 60%
(c) 70%
(d) 80%
Answer:
(c) 70%

Question 2.
The metabolites which does not show any direct function in growth is called …………… metabolite.
(a) Primary
(b) Secondary
(c) Tertiary
(d) Quartemary
Answer:
(b) Secondary

Question 3.
Molecular formula for carbohydrates is …………… .
(a) (CH2O)2
(b) (CH6O)
(C) (C2H2O)n
(d) (CH6O)n
Answer:
(a) (CH2O)2

Question 4.
Number of carbon molecule in glucose is …………… .
(a) 4
(b) 6
(c) 8
(d) 12
Answer:
(b) 6

Samacheer Kalvi 11th Bio Botany Solutions 8 Biomolecules

Question 5.
Number of sugar units in oligo saccharides are …………… .
(a) 6 to 10
(b) 1 to 10
(c) 2 to 8
(d) 2 to 10
Answer:
(d) 2 to 10

Question 6.
Which of the following is a trisaccharide?
(a) Maltose
(b) Stachyose
(c) Ramnose
(d) Aldose
Answer:
(c) Ramnose

Question 7.
…………… are also called as Glycan.
(a) Monosaccharides
(b) Disaccharides
(c) Polysaccharides
(d) Multisaccharides
Answer:
(c) Polysaccharides

Question 8.
Sucrose is a combination of …………… and fructose.
(a) α – glucose
(b) β – glucose
(c) Ketoses
(d) Maltose
Answer:
(a) α – glucose

Samacheer Kalvi 11th Bio Botany Solutions 8 Biomolecules

Question 9.
…………… is also called as animal starch.
(a) Amylose
(b) Glycogen
(c) Glucose
(d) Glycerol
Answer:
(b) Glycogen

Question 10.
…………… reagent is used in starch test.
(a) Potassium permanganate
(b) Potassium iodide
(c) Calcium chloride
(d) Calcium iodide
Answer:
(b) Potassium iodide

Question 11.
Glycogen is not seen in …………… cells.
(a) liver
(b) skeletal
(c) muscle
(d) brain
Answer:
(d) Brain

Question 12.
Benedicts solution is nothing but …………… .
(a) Copper II sulphate
(b) Cuprous sulphate
(c) Cupric sulphate
(d) Copper I sulphate
Answer:
(a) Copper II sulphate

Question 13.
…………… is not a reducing sugar.
(a) Glucose
(b) Fructose
(c) Sucrose
(d) Ketose
Answer:
(c) Sucrose

Samacheer Kalvi 11th Bio Botany Solutions 8 Biomolecules

Question 14.
…………… form the exoskeleton of insects & arthropods.
(a) N – acetyl glucosamine
(b) N – butyl glucosamine
(c) N – phenyl glucosamine
(d) N – methyl glucosamine
Answer:
(a) N – acetyl glucosamine

Question 15.
Number of fatty acids in triglyceride is …………… .
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(c) 3

Question 16.
The major structural component of cell membrane is …………… .
(a) glucolipids
(b) phospholipids
(c) proteolipids
(d) triglycerides
Answer:
(b) phospholipids

Question 17.
There are …………… different amino acids existing naturally.
(a) about 20
(b) about 10
(c) about 25
(d) about 22
Answer:
(a) about 20

Question 18.
A zwitterion also called as …………… ion.
(a) dipolar
(b) monopolar
(c) tripolar
(d) nonpolar
Answer:
(a) dipolar

Question 19.
…………… test is used as an indicator of the presence of protein.
(a) Biuret test
(b) Iodine test
(c) Benedict’s test
(d) Starch test
Answer:
(a) Biuret test

Samacheer Kalvi 11th Bio Botany Solutions 8 Biomolecules

Question 20.
The competitive inhibitor is …………… for succinic dehydrogenase.
(a) malonate
(b) succinate
(c) oxalate
(d) citrate
Answer:
(a) malonate

Question 21.
…………… is the abundant protein in whole biosphere.
(a) RUBP
(b) NAD+
(c) NADPH
(d) RUBISCO
Answer:
(d) RUBISCO

Question 22.
…………… is an active enzyme with its non – protein component.
(a) Apoenzyme
(b) Holoenzyme
(c) Coenzymes
(d) Enzymes
Answer:
(b) Holoenzyme

Question 23.
Flavin adenine dinucleotide contains …………… which helps to accept hydrogen.
(a) ascolac acid
(b) cyanocobalamin
(c) riboflavin
(d) keratinine
Answer:
(c) riboflavin

Question 24.
…………… is a catalytic RNA.
(a) mRNA
(b) Ribozyme
(c) Ribonuclease
(d) rRNA
Answer:
(b) Ribozyme

Samacheer Kalvi 11th Bio Botany Solutions 8 Biomolecules

Question 25.
…………… protects the end of the chromosomes from damage.
(a) Satellite
(b) Kinetochore
(c) Primary constriction
(d) Telomere
Answer:
(d) Telomere

Question 26.
Which is not a pyrimidine base?
(a) Cytosine
(b) Uracil
(c) Guanine
(d) Thymine
Answer:
(c) Guanine

Question 27.
Which type of DNA was described by Watson & Crick?
(a) Z – DNA
(b) α – DNA
(c) B – DNA
(d) A – DNA
Answer:
(c) B – DNA

Question 28.
According to Chargaff’s rule, the hydrogen bonding between Adenine and Thymine is …………… .
(a) 2
(b) 3
(c) 4
(d) Nil
Answer:
(a) 2

Question 29.
The first clear crystallographic evidence for helical structure of DNA was produced by …………… .
(a) Maurice Wilkins
(b) Rosalind Franklin
(c) Francis Crick
(d) Chargaff
Answer:
(b) Rosalind Franklin

Samacheer Kalvi 11th Bio Botany Solutions 8 Biomolecules

Question 30.
According to Cargaff’s rule, A : T = G : C = …………… .
(a) 0
(b) 1
(c) >1
(d) <1
Answer:
(b) 1

Question 31.
A complete turn of the helix comprises …………… .
(a) 34 nm
(b) 3.4 nm
(c) 20 nm
(d) 2nm
Answer:
(b) 3.4 nm

Question 32.
Diameter of DNA helix is …………… .
(a) 34 Å
(b) 20 nm
(c) 34 nm
(d) 20 Å
Answer:
(d) 20 Å

Question 33.
RNA is …………… .
(a) Single stranded and stable
(b) Single stranded and unstable
(c) Double stranded and stable
(d) Double stranded and unstable
Answer:
(b) Single stranded and unstable

Question 34.
rRNA constitutes …………… of total RNA.
(a) 20%
(b) 70%
(c) 80%
(d) 15%
Answer:
(c) 80%

Question 35.
Shape to the ribosomes is provided by …………… .
(a) rRNA
(b) tRNA
(c) mRNA
(d) DNA
Answer:
(a) rRNA

Question 36.
Which RNA is also called as soluble RNA?
(a) rRNA
(b) tRNA
(c) mRNA
(d) ssRNA
Answer:
(b) tRNA

Question 37.
Which is the left – handed DNA?
(a) B – DNA
(b) A – DNA
(c) Z – DNA
(d) dsDNA
Answer:
(c) Z – DNA

Samacheer Kalvi 11th Bio Botany Solutions 8 Biomolecules

Question 38.
Which of the following does not contain cell wall?
(a) Fungi
(b) Bacteria
(c) Mycoplasma
(d) Algae
Answer:
(c) Mycoplasma

Question 39.
The amino acid which is both an acid and a base is called …………… .
(a) Amphibolic
(b) Amphoteric
(c) Amphipathetic
(d) Anabolic
Answer:
(b) Amphoteric

Question 40.
…………… leads to the loss of 3D structure of protein.
(a) Annealing
(b) Extension
(c) Denaturation
(d) Polymerisation
Answer:
(c) Denaturation

Question 41.
Which of the following polysaccharides is used as solidifying agent in culture medium?
(a) Inulin
(b) Heparin
(c) Agar
(d) Keratan sulphate
Answer:
(c) Agar

Question 42.
Which is an anticoagulant?
(a) Inulin
(b) Heparin
(c) Agar
(d) Keratan sulphate
Answer:
(b) Heparin

Question 43.
Insulin is a polymer of …………… .
(a) sucrose
(b) fructose
(c) glucose
(d) maltose
Answer:
(b) fructose

II. Very Short Answer Type Questions (2 Marks)

Question 1.
Define cell pool and mention its constituents.
Answer:
The cell components are made of collection of molecules called as cellular pool, which consists of both inorganic and organic compounds.

Question 2.
Draw the molecular structure of water.
Answer:
the molecular structure of water:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 8 Biomolecules 7

Question 3.
Point out the percentage of water in human cell & a plant cell.
Answer:
Water makes upto 70% of human cell and upto 95% of mass of a plant cell.

Question 4.
What are metabolites?
Answer:
Metabolites are the organic compounds synthesized by plants, fungi and various microbes. They are the intermediates & products of metabolism.

Samacheer Kalvi 11th Bio Botany Solutions 8 Biomolecules

Question 5.
Write the molecular formula for carbohydrates?
Answer:
(CH2O)n

Question 6.
Give an example for simple sugar with its formula.
Answer:
Glucose – C6H12O6

Question 7.
Which type of sugar does sucrose belongs to? Write its monomer units.
Answer:
Sucrose is a disaccharides composed of α – glucose & fructose.

Question 8.
Classify polysaccharides based on function.
Answer:
Depending on the function, polysaccharides are of two types:

  1. storage polysaccharide and
  2. structural polysaccharide.

Question 9.
What are Glycans?
Answer:
Polysaccharides are also called as Glycans. They are made of hundreds of monosaccharide units.

Question 10.
Which is a common storage polysaccharide? Mention its monomer units.
Answer:
Starch is a storage polysaccharide made up of repeated units of amylose and amylopectin.

Samacheer Kalvi 11th Bio Botany Solutions 8 Biomolecules

Question 11.
Which is an animal starch? Where can we see it in our body?
Answer:
Glycogen. It is found in liver cells & skeletal muscles.

Question 12.
Why oil does not get mixed if added with water?
Answer:
Oil is a lipid. Lipids are long hydrocarbon chains that are non-polar & thus hydrophobic, which avoids the oil to dissolve in water.

Question 13.
How saturated fatty acids differ from unsaturated fatty acids?
Answer:
Saturated factty acids have the hydrocarbon chain with single bond, whereas in unsaturated fatty acids the hydrocarbon chain will have double bonds.

Samacheer Kalvi 11th Bio Botany Solutions 8 Biomolecules

Question 14.
How waxes are formed?
Answer:
Waxes are esters formed between a long chain alcohol and saturated fatty acids.

Question 15.
Why amino acids are amphoteric?
Answer:
The amino acid is both an acid and a base and is called amphoteric.

Question 16.
Name the various groups attached to the 4 valencies of carbon in an amino acid.
Answer:
The 4 valencies of carbon in an amino acid:

  1. (NH2)
  2. an acidic carboxylic group (COOH) and
  3. a hydrogen atom (H)
  4. and side chain or variable R group.

Question 17.
Where the peptide bond is formed?
Answer:
A peptide bond is formed when the amino group of one amino acid reacts with carbonyl group of another amino acid.

Samacheer Kalvi 11th Bio Botany Solutions 8 Biomolecules

Question 18.
Which was the first sequenced protein? Who had done it?
Answer:
First protein is insulin and it was sequenced by Fred Sanger.

Question 19.
Why proteins undergo conformational changes after its synthesis?
Answer:
After synthesis, the protein attains conformational change into a specific 3D form for proper functioning.

Question 20.
Mention the levels of protein organisation based on folding.
Answer:
According to the mode of folding, four levels of protein organisation have been recognised namely primary, secondary, tertiary and quaternary.

Question 21.
Define enzymes.
Answer:
Enzymes are globular proteins that catalyse the thousands of metabolic reactions taking place within cells and organism.

Samacheer Kalvi 11th Bio Botany Solutions 8 Biomolecules

Question 22.
Name any four factors that affect enzyme reactions.
Answer:
Four factors that affect enzyme reactions:

  1. pH
  2. temperature
  3. enzyme concentration and
  4. substrate concentration.

Question 23.
What are inhibitors? Mention its types.
Answer:
Certain substances present in the cells may react with the enzyme and lower the rate of reaction. These substances are called inhibitors. It is of two types:

  1. Competitive and
  2. Non – competitive.

Question 24.
Differentiate Apoenzyme from Holoenzyme.
Answer:
Differ Apoenzyme from Holoenzyme:

Apoenzyme

Holoenzyme

1. Active enzyme with its non – protein component 1. Inactive enzyme without its non – protein component

Question 25.
What are Prosthetic groups? Give example.
Answer:
Prosthetic groups are organic molecules that assist in catalytic function of an enzyme. Example: Flavin adenine dinucleotide (FAD).

Question 26.
Draw a diagram showing the various components of enzymes.
Answer:
Catalytic site, Cofactor and Holoenzyme:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 8 Biomolecules 8

Question 27.
Write a note on Ribozyme.
Answer:
Ribozyme – Non – Protein Enzyme. A Ribozyme, also called as catalytic RNA; is a ribonucleic acid that acts as enzyme. It is found in ribosomes.

Samacheer Kalvi 11th Bio Botany Solutions 8 Biomolecules

Question 28.
Give an example for following enzyme groups.
Answer:
An example for following enzyme groups:

  1. Transferase – Ex: Transaminase
  2. Isomerase – Ex: Isomerase
  3. Oxidoreductase – Ex: Dehydrogenase
  4. Lyase – Ex: Decarboxylase

Question 29.
Write the composition of DNA & RNA.
Answer:
Nitrogen base, pentose sugar and phosphate.

Question 30.
What is a nucleoside?
Answer:
A purine or a pyrimidine and a ribose or deoxyribose sugar is called nucleoside. A nitrogenous base is linked to pentose sugar through n-glycosidic linkage and forms a nucleoside.

Question 31.
What is a nucleotide?
Answer:
When a phosphate group is attached to a nucleoside it is called a nucleotide.

Question 32.
Name the two types of Purines and Pyrimidines.
Answer:
The two types of Purines and Pyrimidines:

  1. Purines: Adenine and guanine
  2. Pyrimidines: Cytosine and thymine (Uracil)

Question 33.
How DNA differs from RNA?
Answer:
DNA has thymine base, whereas RNA has uracil base. DNA has deoxyribose sugar, whereas RNA has ribose sugar.

Question 34.
Draw a simple diagram showing basic components of DNA.
Answer:
Deoxyribose sugar:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 8 Biomolecules 9

Question 35.
Which is the secondary structure of DNA? Who discovered it?
Answer:
B – DNA is the secondary structure of DNA. Watson & Crick discovered B – DNA.

Question 36.
State Chargaff’s rule.
Answer:
Chargaff’s Rule:
A = T; G = C
A + G = T + C,
A : T = G : C = 1.

Question 37.
Name the three forms of DNA.
Answer:
The three forms of DNA:

  1. A – DNA
  2. B – DNA and
  3. Z – DNA.

Question 38.
Which is the soluble forms of RNA. Write its percentage composition of total RNA.
Answer:
tRNA is the soluble RNA which is about 15% of total RNA.

Samacheer Kalvi 11th Bio Botany Solutions 8 Biomolecules

Question 39.
Name the types of RNA?
Answer:
The types of RNA:

  1. mRNA
  2. tRNA and
  3. rRNA.

Question 40.
Draw the structure of transfer RNA.
Answer:
Transfer RNA (tRNA):
Samacheer Kalvi 11th Bio Botany Solutions Chapter 8 Biomolecules 10

III. Short Answer Type Questions (3 Marks)

Question 1.
Distinguish between Macronutrients & Micronutrients.
Answer:
Macronutrients:

  • Nutrients required in larger quantities for plant growth are called Macronutrients.
  • e.g. Potassium and Calcium

Micronutrients:

  • Nutrients required in trace amount for plant growth are called Micronutrients
  • e.g. Zinc and Bora

Question 2.
Tabulate the various cellular components with their percentage.
Answer:
The various cellular components with their percentage:

Component

% of the total cellular mass

1. Water 1. 70
2. Proteins 2. 15
3. Carbohydrates 3. 3
4. Lipids 4. 2
5. Nucleic acids 5. 6
6. Ions 6. 4

Question 3.
List out the properties of Water.
Answer:
Properties of Water:

  1. Adhesion and cohesion property
  2. High latent heat of vaporisation
  3. High melting and boiling point
  4. Universal solvent
  5. Specific heat capacity

Question 4.
How lattice formation occurs in water molecule?
Answer:
Two electro negative atoms of oxygen share a hydrogen bonds of two water molecule. Thus, they can stick together by cohesion and results in lattice formation.

Samacheer Kalvi 11th Bio Botany Solutions 8 Biomolecules

Question 5.
Distinguish between Primary metabolite & Secondary metabolite.
Answer:
Between Primary metabolite & Secondary metabolite:

  • Primary metabolites are those that are required for the basic metabolic processes like photosynthesis, respiration, etc Example: Lipase, a protein.
  • Secondary metabolites does not show any direct function in growth and development of organisms. Example: Ricin, gums.

Question 6.
Define Polymerization.
Answer:
Polymerization, is a process in which repeating subunits termed monomers are bound into chains of different lengths. These chains of monomers are called polymers.

Question 7.
Explain the bond formation in sucrose molecule.
Answer:
Sucrose is formed from a molecule of α – glucose and a molecule of fructose. This is a condensation reaction releasing water. The bond formed between the glucose and fructose molecule by removal of water is called glycosidic bond. This is another example of strong, covalent bond.

Samacheer Kalvi 11th Bio Botany Solutions 8 Biomolecules

Question 8.
How will you identify the presence of starch in a food sample.
Answer:
The presence of starch is identified by adding a solution of iodine in potassium iodide. Iodine molecules fit nearly into the starch helix, creating a blue – black colour.

Question 9.
Write a note on steroids.
Answer:
Steroids are complex compounds commonly found in cell membrane and animal hormones. e.g. Cholesterol which reinforces the structure of the cell membrane in animal cells and in an unusual group of cell wall deficient bacteria – Mycoplasma.

Question 10.
Draw the structure of basic amino acid.
Answer:
The structure of basic amino acid:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 8 Biomolecules 11

Question 11.
What is a Zwitterion? or What is an isoelectric point?
Answer:
A zwitterion also called as dipolar ion, is a molecule with two or more functional groups, of which at least one has a positive and other has a negative electrical charge and the net charge of the entire molecule is zero. The pH at which this happens is known as the isoelectric point.

Question 12.
Write briefly about protein denaturation.
Answer:
Denaturation is the loss of 3D structure of protein. Exposure to heat causes atoms to vibrate violently, and this disrupts the hydrogen and ionic bonds. Under these conditions, protein molecules become elongated, disorganised strands. Agents such as soap, detergents, acid, alcohol and some disinfectants disrupt the interchain bond and cause the molecule to be non – functional.

Question 13.
Why do some people have curly hair?
Answer:
Human hair is made of protein. The more the distance between the sulphur atoms, the more the proteins bend; the more the hair curls.

Samacheer Kalvi 11th Bio Botany Solutions 8 Biomolecules

Question 14.
Write a note on RUBISCO.
Answer:
Ribulose biphosphate carboxylase oxygenase (RUBISCO) is an enzyme that catalyses the reaction between CO2 and the CO2 acceptor molecule in photosynthesis. It is the most abundant protein in the whole biosphere.

Question 15.
Differentiate Anabolic reaction and Catabolic reaction.
Answer:
Anabolic reaction:

  • Anabolic reaction involves the building up of organic molecules.
  • Ex: Synthesis of protein from amino acids.

Catabolic reaction:

  • Catabolic reaction involves the breaking down of larger molecules.
  • Ex: Breaking down of sugar in respiration.

Question 16.
What are Allosteric inhibitors?
Answer:
Compounds which modify enzyme activity by causing a reversible change in the structure of the enzyme active site. This in turn affects the ability of the substrate to bind to the enzyme. Such compounds are called allosteric inhibitors, e.g. The enzyme hexokinase which catalysis glucose to glucose – 6 phosphate in glycolysis is inhibited by glucose – 6 phosphate. This is an example for feedback allosteric inhibitor.

Question 17.
Explain in brief about End – product inhibitor. (Negative Feedback Inhibition)
Answer:
When the end product of a metabolic pathway begins to accumulate, it may act as an allosteric inhibitor of the enzyme controlling the first step of the pathway. Thus the product starts to switch off its own production as it builds up. The process is self – regulatory. As the product is used up, its production is switched on once again. This is called end – product inhibition.

Samacheer Kalvi 11th Bio Botany Solutions 8 Biomolecules

Question 18.
Draw the structure of Purine & Pyrimidine.
Answer:
The structure of Pyrimidine & Purine:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 8 Biomolecules 12

Question 19.
Why the sugar in DNA is a deoxyribose?
Answer:
The sugar in DNA molecule is called 2’ – deoxyribose because there is no hydroxyl group at 2’ position.

Question 20.
How dinucleotide & polynucleotides are formed?
Answer:
Two nucleotides join to form dinucleotide that are linked through 3′ – 5′ phosphodiester linkage by condensation between phosphate groups of one with sugar of other. This is repeated many times to make polynucleotide.

Question 21.
Compare Plectonemic & Paranemic Coiling.
Answer:
Plectonemic coiling – the two strands of the DNA are wrapped around each other in a helix, making it impossible to simply move them apart without breaking the entire structure. Whereas in Paranemic coiling the two strands simply lie alongside one another, making them easier to pull apart.

Samacheer Kalvi 11th Bio Botany Solutions 8 Biomolecules

Question 22.
Differentiate between Polycistronic & Monocistronic mRNA.
Answer:
Between Polycistronic & Monocistronic mRNA:

Polycistronic mRNA

Monocistronic mRNA

1. Polycistronic mRNA carry coding sequences for many Polypeptides 1. Monocistronic mRNA carry coding sequences for only one Polypeptide
2. Prokaryotic mRNA are polycistronic 2. Eukaryotic mRNA are monocistronic

Question 23.
What are Proteins?
Answer:
Proteins are polymers of 20 different amino acids, each of which has a distinct side chain with specific chemical properties. Each protein has a unique amino acid sequence which determines its 3D structure.

Question 24.
Herbivores can digest cellulose rich food, Why can’t human beings?
Answer:
Human cannot digest cellulose but herbivores can digest them with the help of bacteria present in the gut which produces enzymes cellulase. This is an example of mutualism.

Samacheer Kalvi 11th Bio Botany Solutions 8 Biomolecules

Question 25.
How will you identify the presence of protein in food samples?
Answer:
The biuret test is used as an indicator of the presence of protein because it gives a purple colour in the presence of peptide bonds (-C- N-). To a protein solution an equal quantity of sodium hydroxide solution is added and mixed. Then a few drops of 0.5% copper (II) sulphate is added with gentle mixing. A distinct purple colour develops without heating.

Question 26.
Write a note on peptide bonds between amino acids.
Answer:
The amino group of one amino acid reacts with carboxyl group of other amino acid, forming a peptide bond. Two amino acids can react together with the loss of water to form a dipeptide. Long strings of amino acids linked by peptide bonds are called polypeptides. In 1953, Fred Sanger first sequenced the Insulin protein.

Question 27.
Which was the first alkaloid discovered? Mentions its uses.
Answer:
Morphine is the first alkaloid to be found. It comes from the plant Opium poppy (Papaver somniferum). It is used as a pain reliever in patients with severe pain levels and cough suppressant.

IV. Long Answer Type Questions (5 Marks)

Question 1.
How will you identify the presence of glucose in a given food sample?
Answer:
Aldoses and ketoses are reducing sugars. This means that, when heated with an alkaline solution of copper (II) sulphate (a blue solution called Benedict’s solution), the aldehyde or ketone group reduces Cu2+ ions to Cu+ ions forming brick red precipitate of copper (I) oxide. In the process, the aldehyde or ketone group is oxidised to a carboxyl group (-COOH).

This reaction is used as test for reducing sugar and is known as Benedict’s test. The results of Benedict’s test depends on concentration of the sugar. If there is no reducing sugar it remains blue. Sucrose is not a reducing sugar The greater the concentration of reducing sugar, the more is the precipitate formed and greater is the colour change.

Samacheer Kalvi 11th Bio Botany Solutions 8 Biomolecules

Question 2.
Write a note on various levels of protein organisation.
Answer:
The primary structure is linear arrangement of amino acids in a polypeptide chain. Secondary structure arises when various functional groups are exposed on outer surface of the molecular interaction by forming hydrogen bonds. This causes the amino acid chain to twist into coiled configuration called α – helix or to fold into a flat β – pleated sheets.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 8 Biomolecules 13
Tertiary protein structure arises when the secondary level proteins fold into globular structure called domains. Quaternary protein structure may be assumed by some complex proteins in which more than one polypeptide forms a large multiunit protein. The individual polypeptide chains of the protein are called subunits and the active protein itself is called a multimer.

Question 3.
Enumerate the properties of Enzyme.
Answer:
The properties of Enzyme:

  • Enzymes are globular proteins.
  • They act as catalysts and effective even in small quantity.
  • They remain unchanged at the end of the reaction.
  • They are highly specific.
  • They have an active site where the reaction takes place.
  • Enzymes lower activation energy of the reaction they catalyse.

Question 4.
Draw a Flow Chart depicting the Carbohydrate Classification
Answer:
Flow Chart depicting the Carbohydrate Classification:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 8 Biomolecules 19

Question 5.
Explain the various types of chemical bonding in proteins.
Answer:
1. Hydrogen Bond: It is formed between some hydrogen atoms of oxygen and nitrogen in polypeptide chain. The hydrogen atoms have a small positive charge and oxygen and nitrogen have small negative charge. Opposite charges attract to form hydrogen bonds. Though these bonds are weak, large number of them maintains the molecule in 3D shape.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 8 Biomolecules 14
2. Ionic Bond: It is formed between any charged groups that are not joined together by peptide bond. It is stronger than hydrogen bond and can be broken by changes in pH and temperature.

3. Disulfide Bond: Some amino acids like cysteine and methionine have sulphur. These form disulphide bridge between sulphur atoms and amino acids.

4. Hydrophobic Bond: This bond helps some protein to maintain structure. When globular proteins are in solution, their hydrophobic groups point inwards away from water.

Question 6.
Explain Lock & Key Mechanism of Enzymatic reaction.
Answer:
Lock and Key Mechanism of Enzyme: In a enzyme catalysed reaction, the starting substance is the substrate. It is converted to the product. The substrate binds to the specially formed pocket in the enzyme – the active site, this is called lock and key mechanism of enzyme action. As the enzyme and substrate form a ES complex, the substrate is raised in energy to a transition state and then breaks down into products plus unchanged enzyme.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 8 Biomolecules 15

Question 7.
Describe the Competitive & Non – Competitive Inhibitors of enzyme.
Answer:
1. Competitive Inhibitor: Molecules that resemble the shape of the substrate and may compete to occupy the active site of enzyme are known as competitive inhibitors. For Example: the enzyme that catalyses the reaction between carbon dioxide and the CO2 acceptor molecule in photosynthesis, known as ribulose biphosphate carboxylase oxygenase (RUBISCO) is competitively inhibited by oxygen / carbon – di – oxide in the chloroplast. The competitive inhibitor is malonate for succinic dehydrogenase.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 8 Biomolecules 16
2. Non – competitive Inhibitors: There are certain inhibitors which may be unlike the substrate molecule but still combines with the enzyme. This either blocks the attachment of the substrate to active site or change the shape so that it is unable to accept the substrate. For example the effect of the amino acids alanine on the enzyme pyruvate kinase in the final step of glycolysis.

Certain non – reversible / irreversible inhibitors bind tightly and permanently to an enzyme and destroy its catalytic properties entirely. These could also be termed as poisons. Example – cyanide ions which blocks cytochrome oxidase in terminal oxidation in cell aerobic respiration, the nerve gas sarin blocks a neurotransmitter in synapse transmission.

Question 8.
Give a detailed account on Enzyme Co – factors.
Answer:
Many enzymes require non – protein components called co – factors for their efficient activity. Co – factors may vary from simple inorganic ions to complex organic molecules.
They are of three types:

  1. Inorganic ions, prosthetic groups and coenzymes.
  2. Holoenzyme – active enzyme with its non – protein component.
  3. Apoenzyme – the inactive enzyme without its non – protein component.

Inorganic ions help to increase the rate of reaction catalysed by enzymes. Example: Salivary amylase activity is increased in the presence of chloride ions. Prosthetic groups are organic molecules that assist in catalytic function of an enzyme. Flavin adenine dinucleotide (FAD) contains riboflavin(vit B2), the function of which is to accept hydrogen. ‘Haem’ is an iron – containing prosthetic group with an iron atom at its centre. Coenzymes are organic compounds which act as cofactors but do not remain attached to the enzyme. The essential chemical components of many coenzymes are vitamins. Eg. NAD, NADP, Coenzyme A, ATP.

Question 9.
Tabulate the various features of different forms of DNA.
Answer:
The various features of different forms of DNA:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 8 Biomolecules 17

Question 10.
Compare DNA with RNA?
Answer:
Compare DNA with RNA:

DNA

RNA

1. Deoxyribose sugar is present 1. Ribose sugar is present
2. Thymine is present 2. Uracil is present
3. More stable 3. Less stable
4. Double stranded 4. Single stranded
5. Types: A – DNA, B – DNA, Z – DNA 5. Type: mRNA, tRNA, rRNA
6. Genetic material for most of living organism except few viruses 6. Genetic material for few viruses only

V. Higher Order Thinking Skills (HOTs)

Question 1.
In which form does the glucose is stored in animal cells? Specify the cells?
Answer:
Glucose is stored in the form of glycogen. Glycogen is stored in liver cells and skeletal muscles, etc.

Question 2.
State the key differences between DNA & RNA.
Answer:
The key differences between DNA & RNA:

DNA

RNA

1. Double stranded 1. Single stranded
2. Thymine is the pyrimidine base 2. Uracil is the pyrimidine base

Samacheer Kalvi 11th Bio Botany Solutions 8 Biomolecules

 

Question 3.
Aminoacids are the monomers of proteins. Similarly mention the monomers of nucleic acids along with its composition.
Answer:
The monomer unit of nucleic acids are nucleotides, which are composed of nitrogen base, pentose sugar and phosphoric acid.

Question 4.
Complete the equations.
(a) Nitrogen base + …………… . = Nucleoside.
(b) …………… + nucleoside = Nucleotide.
(c) Glucose + fructose = …………… .
Answer:
(a) sugar
(b) phosphoric acid and
(c) sucrose.

Question 5.
What happens if the sucrose is hydrolysed?
Answer:
On hydrolysis, the glycosidic bonds in sucrose gets splitted yielding glucose and fructose.

Samacheer Kalvi 11th Bio Botany Solutions 8 Biomolecules

Question 6.
Name the types of bonds.
(a) Between amino acids of protein
(b) Between carboxyl group and glycerol of fatty acids and
(c) Between glucose units of cellulose.
Answer:
(a) Peptide bond
(b) Ester bond
(c) Glycosidic bond

Question 7.
Study the following equation and name the reaction A and B.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 8 Biomolecules 18
Answer:
Reaction A is glycogenolysis. Reaction B is glycogenesis.

Question 8.
Whether waxes are found in living organisms?
Answer:
Yes. Fur, feathers, fruits, leaves, skin, and exoskeleton of insects are naturally water – proofed with a coating of wax.

Samacheer Kalvi 11th Bio Botany Solutions 8 Biomolecules

Question 9.
If dsDNA has 40% Guanine. Calculate the percentage of Adenine.
Answer:
According to Chargaff’s rule:
Guanine pairs with cytosine. It Guanine is 40%, then cytosine will also be 40%. Similarly, Adenine pairs with thymine, if guanine is 40%, the remaining 60% will be Adenine. So Thymine will also be 60%.
Thus, A : T = G : C = 1
and 60 : 60 = 40 : 40 = 1.

Question 10.
In an Eukaryotic cell, totally there are 10000 RNA molecules. Calculate the number of mRNA’s and tRNA’s if the count of rRNA is 8000.
Answer:
In a cell, rRNA contributes 80%, tRNA constitutes 15% and mRNA constitutes 5%. If rRNA is 8000 (80%), then tRNA count is 1500 (15%) and mRNA is 500 (5%).

Question 11.
Despite made of two different monomers amylose and amylopectin, starch is a homopolysaccharide – Comment.
Answer:
Starch is made up of amylose and amylopectin. Both are glucose polymers, hence starch is considered as homopolysaccharides.

Samacheer Kalvi 11th Bio Botany Solutions 8 Biomolecules

Question 12.
How do you call a fatty acid as saturated or unsaturated?
Answer:
If the hydrocarbon chain is single bonded, then the fatty acid is said to be saturated. In unsaturated fatty acids, the hydrocarbon chain is double bonded.

Question 13.
Enzymes are biocatalysts – Justify.
Answer:
Enzymes are globular proteins that catalyze thousands of metabolic reactions taking place within cells and organisms. Hence enzymes are called as biological catalysts.

Question 14.
Starch, cellulose, glycogen and chitin are polysaccharides found among the following. Choose the one appropriate and write against each.
(a) Cotton fibre – …………… .
(b) Exoskeleton of ant – …………… .
(c) Liver –  …………… .
(d) Peeled potato – …………… .
Answer:
(a) Cellulose
(b) Chitin
(c) Glycogen and
(d) Starch.

Samacheer Kalvi 11th Bio Botany Solutions 8 Biomolecules

Question 15.
Sucrose is not a reducing sugar. Why?
Answer:
Sucrose is a non – reducing sugar since it does not possess aldehyde or ketone group, which is responsible for reducing the alkaline solutions like copper (II) sulphate.

Question 16.
A DNA segment has a total of 1000 nucleotides, out of which 240 are adenine containing nucleotides. How many pyrimidine bases this DNA segment possess?
Answer:
Pyrimidine = 500.
According to Chargaff’s rule,
A = T,
A = 240, hence T = 240.
A + T = 240 + 240 = 480.
So, G + C = 1000 – 480 = 520.
G = C, Therefore, C = \(\frac {520}{2}\) = 260.
Thus, pyrimidine = C + T = 260 + 240 = 500.

Samacheer Kalvi 11th Bio Botany Solutions Chapter 11 Transport in Plants

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

Tamilnadu Samacheer Kalvi 11th Bio Botany Solutions Chapter 11 Transport in Plants

Samacheer Kalvi 11th Bio Botany Transport in Plants Text Book Back Questions and Answers

I. Choose the correct answers.
Question 1.
In a fully turgid cell:
(a) DPD = 10 atm; OP = 5 atm; TP = 10 atm
(b) DPD = 0 atm; OP =10 atm; TP = 10 atm
(c) DPD = 0 atm; OP = 5 atm; TP = 10 atm
(d) DPD = 20 atm; OP = 20 atm; TP = 10 atm
Answer:
(b) DPD = 0 atm; OP =10 atm; TP = 10 atm

Samacheer Kalvi 11th Bio Botany Solutions 11 Transport in Plants

Question 2.
Which among the following is correct?
(i) Apoplast is fastest and operate in nonliving part
(ii) Trahsmembrane route includes vacuole
(iii) Symplast interconnect the nearby cell through plasmadesmata
(iv) Symplast and transmembrane route are in living part of the cell

(a) (i) and (ii) only
(b) (ii) and (iii) only
(c) (iii) and (iv) only
(d) All of these
Answer:
(c) (iii) and (iv) only

Question 3.
What type of transpiration is possible in the xerophyte Opuntia?
(a) Stomatal
(b) Lenticular
(c) Cuticular
(d) All the above
Answer:
(b) Lenticular

Samacheer Kalvi 11th Bio Botany Solutions 11 Transport in Plants

Question 4.
Stomata of a plant open due to:
(a) Influx of K+
(b) Efflux of K+
(c) Influx of Cl
(d) Influx of OH
Answer:
(a) Influx of K+

Question 5.
Munch hypothesis is based on:
(a) translocation of food due to TP gradient and imbibition force
(b) ranslocation of food due to TP
(c) translocation of food due to imbibition force
(d) None of the above
Answer:
(b) ranslocation of food due to TP

Question 6.
If the concentration of salt in the soil is too high and the plants may wilt even if the field is thoroughly irrigated. Explain.
Answer:
The salts present in the soil dissolve in the irrigated water and form hypertonic solution outside the root hairs of the plant and the root hairs cannot absorb water from hypertonic solution, since water molecules cannot move from hypertonic solution to hypotonic solution in the cells of root hair. Hence the plants become wilt even the field is irrigated.

Samacheer Kalvi 11th Bio Botany Solutions 11 Transport in Plants

Question 7.
How phosphorylase enzyme open the stomata in starch sugar interconversion theory?
Answer:
The discovery of enzyme phosphorylase in guard cells by Hanes (1940) greatly supports the starch – sugar interconversion theory. The enzyme phosphorylase hydrolyses starch into sugar and high pH followed by endosmosis and the opening of stomata during light. The vice versa takes place during the night.

Question 8.
List out the non photosynthetic parts of a plant that need a supply of sucrose?
Answer:
The non photosynthetic parts of a plant that need a supply of sucrose:

  1. Roots
  2. Tubers
  3. Developing fruits and
  4. Immature leaves.

Question 9.
What are the parameters which control water potential?
Answer:
Water potential (Ψ) can be controlled by,

  1. Solute concentration or Solute potential (Ψs)
  2. Pressure potential (Ψp).

By correlating two factors, water potential is written as, Ψw = Ψs + Ψp.
Water Potential = Solute potential + Pressure potential.

Samacheer Kalvi 11th Bio Botany Solutions 11 Transport in Plants

Question 10.
An artificial cell made of selectively permeable membrane immersed in a beaker (in the figure). Read the values ans answer the following questions?
Samacheer Kalvi 11th Bio Botany Solutions Chapter 11 Transport in Plants 14
Ψw = 0, Ψs = 2, Ψp = 0.
(a) Draw an arrow to indicate the direction of water movement.
(b) Is the solution outside the cell isotonic, hypotonic or hypertonic?
(c) Is the cell isotonic, hypotonic or hypertonic?
(d) Will the cell become more flaccid, more turgid or stay in original size?
(e) With reference to artificial cell state, the process is endosmosis or exosmosis? Give reasons.
Answer:
(a) An arrow to indicate the direction of water movement:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 11 Transport in Plants 1
(b) Outside solution in hypotonic.
(c) The cell is hypertonic.
(d) The cell become more turgid.
(e) The process is endo – osmosis because the solvent (water) moves inside the cell.

Samacheer Kalvi 11th Bio Botany Transport in Plants Additional Questions & Answers

I. Choose the correct answer (1 Mark)
Question 1 .
In plants, cell to cell transport is aided by:
(a) diffusion alone
(b) osmosis alone
(c) imbibition alone
(d) all the three above
Answer:
(d) all the three above

Question 2.
In passive transport:
(a) no energy expenditure is required
(b) energy expenditure is required
(c) no involvement of physical forces like gravity
(d) no involvement of osmosis
Answer:
(a) no energy expenditure is required

Samacheer Kalvi 11th Bio Botany Solutions 11 Transport in Plants

Question 3.
Which of the following statements are correct?
(i) Cell membranes allow water and non polar molecules to permeate by simple diffusion.
(ii) Polar molecules like amino acids can also diffuse through membrane.
(iii) Smaller molecules diffuse faster than larger molecules.
(iv) Larger molecules diffuse faster than smaller molecules.

(a) (i) and (iv) only
(b) (i) and (iii) only
(c) (i) and (ii) only
(d) (ii) and (iv) only
Answer:
(b) (i) and (iii) only

Question 4.
In co – transport across membrane:
(a) two different molecules are transported in opposite direction.
(b) two types of molecules are transported the same direction.
(c) three types of molecules are transported in opposite direction.
(d) two types of molecules are transported in all directions.
Answer:
(b) two types of molecules are transported the same direction.

Samacheer Kalvi 11th Bio Botany Solutions 11 Transport in Plants

Question 5.
The swelling of dry seeds is due to phenomenon called:
(a) osmosis
(b) transpiration
(c) imbibition
(d) none of the above
Answer:
(c) imbibition

Question 6.
The concept of water potential was introduced by:
(a) Slatyer and Mosses
(b) Slatyer and Taylor
(c) Armusten and Taylor
(d) Mosses and Robert
Answer:
(b) Slatyer and Taylor

Question 7.
At standard temperature the water potential pure water is:
(a) 1.0
(b) -1.0
(c) 0.5
(d) zero
Answer:
(d) zero

Question 8.
Addition of solute to pure water:
(a) increases water potential
(b) does not change water potential
(c) decreases water potential
(d) does not change the gradient of water potential
Answer:
(b) does not change water potential

Samacheer Kalvi 11th Bio Botany Solutions 11 Transport in Plants

Question 9.
Osmotic pressure is increased with:
(a) decrease of dissolved solutes in the solution
(b) increase of dissolved solutes in the solution.
(c) increase of solvent in a solution
(d) isotonic condition of the solution
Answer:
(b) increase of dissolved solutes in the solution.

Question 10.
Diffusion Pressure Deficit (DPD) was termed by Meyer in:
(a) 1928
(b) 1828
(c) 1936
(d) 1938
Answer:
(d) 1938

Question 11.
The root hairs are:
(a) unicellular extensions of epidermal cells with cuticle
(b) Unicellular extensions of xylem parenchyma cells without cuticle
(c) Unicellular extensions of epidermal cells without cuticle
(d) None of the above
Answer:
(c) Unicellular extensions of epidermal cells without cuticle

Samacheer Kalvi 11th Bio Botany Solutions 11 Transport in Plants

Question 12.
Kramer (1949) recognised two distinct mechanisms, which independently operate in the absorption of water in plants are:
(a) osmosis and diffusion
(b) imbibition and diffusion
(c) diffusion and absorption
(d) active absorption and passive absorption
Answer:
(d) active absorption and passive absorption

Question 13.
Indicate the correct statements:
(i) the cell sap concentration in xylem is always high.
(ii) the cell sap concentration in xylem is not always high.
(iii) root pressure is not universal in all plants.
(iv) root pressure is universal in all plants.

(a) (i) and (iv) only
(b) (ii) and (iii) only
(c) (i) and (iii) only
(d) (ii) and (iv) only
Answer:
(b) (ii) and (iii) only

Question 14.
When respiratory inhibitors like KCN, chloroform are applied:
(a) there is a decrease in the rate of respiration and increase in the rate of absorption of water.
(b) there is an increase in the rate of respiration and decrease in the rate of absorption of water.
(c) there is a decrease in the rate of respiration and also decrease in the rate of absorption of water.
(d) there is an increase in the rate of respiration and also in the rate of absorption of water.
Answer:
(c) there is a decrease in the rate of respiration and also decrease in the rate of absorption of water.

Samacheer Kalvi 11th Bio Botany Solutions 11 Transport in Plants

Question 15.
Relay pump theory was proposed by:
(a) J.C. Bose
(b) Godlewski
(c) Stoking
(d) Strasburger
Answer:
(b) Godlewski

Question 16.
Pulsation theory was proposed by:
(a) Strasburger
(b) Godsey
(c) J.C. Bose
(d) C.V. Raman
Answer:
(c) J.C. Bose

Question 17.
The term ‘root pressure’ was coined by:
(a) Strasburger
(b) Stephen Hales
(c) Amstrong
(d) Overton
Answer:
(b) Stephen Hales

Question 18.
Indicate the correct statements:
(i) Root pressure is absent in gymnosperms.
(ii) Root pressure in totally absent in angiosperms.
(iii) There is a relationship between the ascent of sap and root pressure.
(iv) There is no relationship between the ascent of sap and root pressure.

(a) (i) and (ii)
(b) (ii) and (iii)
(c) (ii) and (iv)
(d) (i) and (iv)
Answer:
(d) (i) and (iv)

Samacheer Kalvi 11th Bio Botany Solutions 11 Transport in Plants

Question 19.
The capillary theory was suggested by:
(a) Unger
(b) J.C. Bose
(c) Boehm
(d) Sachs
Answer:
(c) Boehm

Question 20.
Cohesion and transpiration pull theory was originally proposed by:
(a) Unger and Sachs
(b) Xavier and Dixon
(c) Boehm and Jolly
(d) Dixon and Jolly
Answer:
(d) Dixon and Jolly

Question 21.
Loss of water from mesophyll cells causes:
(a) increase in water potential
(b) decrease in water potential
(c) does not change in water potential
(d) hone of the above events
Answer:
(b) decrease in water potential

Samacheer Kalvi 11th Bio Botany Solutions 11 Transport in Plants

Question 22.
The water may move through the xylem at the rate as fast as:
(a) 65 cm / min
(b) 85 cm / min
(c) 75 cm / min
(d) 45 cm / min
Answer:
(c) 75 cm / min

Question 23.
The length and breadth of stomata is:
(a) about 10 – 30μ and 2 – 10μ respectively
(b) about 10 – 14μ and 3 – 10μ respectively
(c) about 10 – 40μ and 3 – 10μ respectively
(d) about 5 – 30μ and 5 – 10μ respectively
Answer:
(c) about 10 – 40μ and 3 – 10μ respectively

Question 24.
The opening and closing of stomata depends upon the change in pH of guard cells. This is observed by:
(a) Loftfield
(b) Sayre
(c) Von Mohl
(d) Amstrong
Answer:
(b) Sayre

Samacheer Kalvi 11th Bio Botany Solutions 11 Transport in Plants

Question 25.
Who did observe that stomata open in light and close in the night:
(a) Unger
(b) Sachs
(c) Boehm
(d) Von Mohl
Answer:
(d) Von Mohl

Question 26.
The phosphorylase enzyme in guard cells supports the starch – sugar inter conversion theory. The above reaction is:
(a) oxidation reaction
(b) hydrolyses reaction
(c) reduction reaction
(d) none of the above
Answer:
(b) hydrolyses reaction

Question 27.
Low pH and a shortage of water in the guard cell activate the stress hormone namely:
(a) Ascorbic acid
(b) Malic acid
(c) Abscisic acid
(d) Salisilic acid
Answer:
(c) Abscisic acid

Question 28.
Accumulation of CO2 in plant cell during dark:
(a) increases the pH level
(b) decreases the pH level
(c) does not alter pH
(d) decreases in H+ ion concentration
Answer:
(b) decreases the pH level

Question 29.
Phenyl Mercuric Acetate (PMA), when applied as a foliar spray to plants:
(a) induces partial stomatal closure for two weeks.
(b) induces partial stomatal opening for two weeks.
(c) induces partial stomatal closure for four weeks.
(d) induces stomatal closure permanently
Answer:
(a) induces partial stomatal closure for two weeks.

Samacheer Kalvi 11th Bio Botany Solutions 11 Transport in Plants

Question 30.
The transpiration in plants is a “necessary evil” as stated by:
(a) Steward
(b) Sayre
(c) Curtis
(d) Meyer
Answer:
(c) Curtis

Question 31.
Sink in plants, which receives food from source is:
(a) tubers
(b) developing fruits
(c) roots
(d) all the three above
Answer:
(d) all the three above

Question 32.
Activated diffusion theory was first proposed by:
(a) Fenson and Spanner
(b) Mason and Masked
(c) Crafts and Munch
(d) Hanes and Robert
Answer:
(b) Mason and Masked

Question 33.
From sieve elements sucrose is translocated into sink organs such as root, tubers etc and this process is termed as:
(a) Xylem unloading
(b) Xylem uploading
(c) Phloem unloading
(d) Phloem uploading
Answer:
(c) Phloem unloading

Samacheer Kalvi 11th Bio Botany Solutions 11 Transport in Plants

Question 34.
In which plant, the petioles are flattened and widened, to become phyllode:
(a) Asparagus
(b) Acacia melanoxylon
(c) Vinca rosea
(d) Delonix regia
Answer:
(b) Acacia melanoxylon

Question 35.
Match the following:

(i) Opuntia (a) Cladode
(ii) Acacia (b) Guttation
(iii) Asparagus (c) Phyllode
(iv) Alocasia (d) Phylloclade

(a) i – b; ii – d; iii – a; iv – c
(b) i – b; ii – c; iii – d; iv – a
(c) i – d; ii – c; iii – a; iv – b
(d) i – c; ii – b; iii – d; iv – a
Answer:
(c) i – d; ii – c; iii – a; iv – b

Question 36.
Hydathodes are generally present in plants that grow in:
(a) dry places
(b) moist and shady places
(c) sunny places
(d) deserts
Answer:
(b) moist and shady places

Samacheer Kalvi 11th Bio Botany Solutions 11 Transport in Plants

Question 37.
Ganongs potometer is used to measure:
(a) the rate of photosynthesis
(b) the rate of gaseous exchange
(c) the rate of water transport
(d) the rate of transpiration
Answer:
(d) the rate of transpiration

Question 38.
Indicate the correct statement:
(a) Anti – transpirants increases the loss of water by transpiration.
(b) Anti – transpirants do not alter the rate of transpiration.
(c) Anti – transpirants do not decrease the loss water by transpiration in cross plants.
(d) Anti – transpirants reduce the enormous loss of wafer by transpiration in crop plants.
Answer:
(d) Anti – transpirants reduce the enormous loss of wafer by transpiration in crop plants.

Question 39.
The liquid coming out of hydathode of grasses is:
(a) pure water
(b) not pure water
(c) a solution containing a number of dissolved substances
(d) salt water
Answer:
(c) a solution containing a number of dissolved substances

Question 40.
A dry cobalt chloride strip, when hydrated, turns:
(a) white
(b) red
(c) green
(d) pink
Answer:
(d) pink

II. Answer the following (2 Marks)

Question 1.
What is the need for transport of materials in plants?
Answer:
Water absorbed from roots must travel up to leaves by xylem for food preparation by photosynthesis. Likewise, food prepared from leaves has to travel to all parts of the plant including roots.

Question 2.
What are the types of transport based on the distance travelled by the materials?
Answer:
Based on the distance travelled by water (sap) or food (solute) they are classified as

  1. Short distance (cell to cell transport)
  2. Long distance transport.

Question 3.
Define the term diffusion.
Answer:
The net movement of molecules from a region of their higher concentration to a region of their lower concentration along a concentration gradient until an equilibrium is attained.

Samacheer Kalvi 11th Bio Botany Solutions 11 Transport in Plants

Question 4.
Define the term semipermeable.
Answer:
Semipermeable allow diffusion of solvent molecules but do not allow the passage of solute molecule. eg: Parchment paper.

Question 5.
What is meant by Porin?
Answer:
Porin is a large transporter protein found in the outermembrane of plastids, mitochondria and bacteria which facilitates smaller molecules to pass through the membrane.

Question 6.
Define symport or co – transport?
Answer:
The term symport is used to denote an integral membrane protein that simultaneously transports two types of molecules across the membrane in the same direction.

Question 7.
Explain the term counter transport.
Answer:
An antiport is an integral membrane transport protein that simultaneously transports two different molecules, in opposite directions, across the membrane.

Question 8.
What is the difference between co – transport and counter transport?
Answer:
In co – transport, two molecules are transported together whereas, in counter transport two molecules are transported in opposite direction to each other.

Question 9.
Define the term Imbibition.
Answer:
Colloidal systems such as gum, starch, proteins, cellulose, agar, gelatin when placed in water, will absorb a large volume of water and swell up. These substances are called imbibants and the phenomenon is imbibition.

Question 10.
Give two examples for the phenomenon of Imbibition.
Answer:
two examples for the phenomenon of Imbibition:

  1. The swelling of dry seeds.
  2. The swelling of wooden windows, tables, doors due to high humidity during the rainy season.

Question 11.
Define the term osmotic potential.
Answer:
Osmotic potential is defined as the ratio between the number of solute particles and the number of solvent particles in a solution.

Samacheer Kalvi 11th Bio Botany Solutions 11 Transport in Plants

Question 12.
What is transpiration?
Answer:
The loss of excess of water in the form of vapour from various aerial parts of the plant is called transpiration.

Question 13.
What is meant by osmotic pressure?
Answer:
When a solution and its solvent (pure water) are separated by a semipermeable membrane, a pressure is developed in the solution, due to the presence of dissolved solutes. This is called osmotic pressure (OP).

Question 14.
Explain the term wall pressure exerted by the cell wall.
Answer:
The cell wall reacts to this turgor pressure with equal and opposite force, and the counter – pressure exerted by the cell wall towards cell membrane is wall pressure (WP).

Samacheer Kalvi 11th Bio Botany Solutions 11 Transport in Plants

Question 15.
Define the term osmosis.
Answer:
Osmosis (Latin: Osmos – impulse, urge) is a special type of diffusion. It represents the movement of water or solvent molecules through a selectively permeable membrane from the place of its higher concentration (high water potential) to the place of its lower concentration (low water potential).

Question 16.
What is meant by isotonic solution?
Answer:
Isotonic (Iso = identical; tonic = soute): It refers to two solutions having same concentration. In this condition the net movement of water molecule will be zero.

Question 17.
What are the three types of plasmolysis?
Answer:
Three types of plasmolysis occur in plants:

  1. Incipient plasmolysis
  2. Evident plasmolysis
  3. Final plasmolysis.

Question 18.
Explain briefly about root hairs.
Answer:
Root hairs are unicellular extensions of epidermal cells without cuticle. Root hairs are extremely thin and numerous and they provide a large surface area for absorption.

Samacheer Kalvi 11th Bio Botany Solutions 11 Transport in Plants

Question 19.
Define active absorption of water.
Answer:
The mechanism of water absorption due to forces generated in the root itself is called active absorption. Active absorption may be osmotic or non – osmotic.

Question 20.
Explain briefly the term stomatal transpiration.
Answer:
Stomata are microscopic structures present in high number on the lower epidermis of leaves. This is the most dominant form of transpiration and being responsible for most of the water loss (90 – 95%) in plants.

Question 21.
Give any two objections to starch – sugar inter conversion theory.
Answer:
Two objections to starch – sugar inter conversion theory:

  1. In monocots, guard cell does not have starch.
  2. There is no evidence to show the presence of sugar at a time when starch disappears and stomata open.

Question 22.
Briefly explain plant anti – transpirants.
Answer:
The term anti – transpirant is used to designate any Material applied to plants for the purpose of retarding transpiration. An ideal anti – transpirant checks the transpiration process without disturbing the process of gaseous exchange.

Samacheer Kalvi 11th Bio Botany Solutions 11 Transport in Plants

Question 23.
Mention any two uses of anti – transpirants.
Answer:
Two uses of anti – transpirants:

  1. Anti – transpirants reduce the enormous loss of water by transpiration in crop plants.
  2. Useful for seedling transplantations in nurseries.

Question 24.
What is meant by translocation of organic solutes.
Answer:
The phenomenon of food transportation from the site of synthesis to the site of utilization is known as translocation of organic solutes. The term solute denotes food material that moves in a solution.

Question 25.
Define the term Ion – Exchange.
Answer:
Ions of external soil solution are exchanged with same charged (anion for anion or cation for cation) ions of the root cells.

III. Answer the following (3 Marks)

Question 1.
Briefly explain the term aquaporin.
Answer:
Aquaporin Is a water channel protein embedded in the plasma membrane. It regulates the massive amount of water transport across the membrane. Plants contain a variety of aquaporins. Over 30 types of aquaporins are known from maize.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 11 Transport in Plants 2
Currently, they are also recognised to transport substrates like glycerol, urea, CO2, NH3, rhetalloids, and reactive oxygen species (ROS) in addition to water. They increase the permeabi lity of the membrane to water. They confer drought and salt stress tolerance.

Question 2.
What is carrier protein? Mention the. three types of carrier proteins?
Answer:
Carrier protein acts as a vehicle to carry molecules from outside of the membrane to inside the cell and vice versa. Due to association with molecules to be transported, the structure of carrier protein gets modified until the dissociation of the molecules.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 11 Transport in Plants 3
There are three types of carrier proteins classified on the basis of handling of molecules and direction of transport. They are:

  1. Uniport
  2. Symport
  3. Antiport.

Question 3.
Explain osmotic potential.
Answer:
Solute potential, otherwise known as osmotic potential denotes the effect of dissolved solute on water potential. In pure water, the addition of solute reduces its free energy and lowers the water potential value from zero to negative. Thus the value of solute potential is always negative. In a solution at standard atmospheric pressure, water potential is always equal to solute potential (Ψw = Ψs).

Question 4.
What are the types of osmosis based on the direction of the movement of water? Explain briefly.
Answer:
Based on the direction of movement of water or solvent in an osmotic system, two types of osmosis can occur, they are Endosmosis and Exosmosis:

  1. Endosmosis: Endosmosis is defined as the osmotic entry of solvent into a cell or a system when it is placed in a pure water or hypotonic solution. eg: dry raisins (high solute and low solvent) placed in the water, it swells up due to turgidity.
  2. Exosmosis: Exosmosis is defined as the osmotic withdrawal of water from a cell or system when it is placed in a hypertonic solution. Exosmosis in a plant cell leads to plasmolysis.

Question 5.
Describe the method of demonstration of endo – osmosis by potato Osmoseope.
Answer:
The method of demonstration of endo – osmosis by potato Osmoscope:

  1. Take a peeled potato tuber and make a cavity inside with the help of a knife.
  2. Fill the cavity with concentrated sugar solution and mark the initial level.
  3. Place this setup in a beaker of pure water.
  4. After 10 minutes observe the sugar solution level and record your findings.
  5. With the help of your teacher discuss the results.
    Samacheer Kalvi 11th Bio Botany Solutions Chapter 11 Transport in Plants 4

Instead of potato use beetroot or bottleguard and repeat the above experiment. Compare and discuss the results.

Question 6.
Explain the term reverse osmosis.
Answer:
Reverse Osmosis follows the same principles of osmosis, but in the reverse direction. In this process movement of water is reversed by applying pressure to force the water against a concentration gradient of the solution.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 11 Transport in Plants 5
In regular osmosis, the water molecules move from the higher concentration (pure water = hypotonic) to lower concentration (salt water = hypertonic). But in reverse osmosis, the water molecules move from the lower concentration (salt water = hypertonic) to higher concentration (pure water = hypotonic) through a selectively permeable membrane.

Uses:  Reverse osmosis is used for purification of drinking water and desalination of seawater.

Question 7.
Give details of symplast route of water movement.
Answer:
The symplast (Greek: sym = within; plast = cell) consists of the entire mass of cytosol of all the living cells in a plant, as well as the plasmodesmata, the cytoplasmic channel that interconnects them. In the symplastic route, water has to cross plasma membrane to enter the cytoplasm of outer root cell; then it will move within adjoining cytoplasm through plasmodesmata around the vacuoles without the necessity to cross more membrane, till it reaches xylem.

Question 8.
Describe the non – osmotic active absorption theory proposed by Bennet – Clark in 1936.
Answer:
Bennet – Clark (1936), Thimann (1951) and Kramer (1959) observed absorption of water even if the concentration of cell sap in the root hair is lower than that of the soil water. Such a movement requires an expenditure of energy released by respiration (ATP). Thus, there is a link between water absorption and respiration. It is evident from the fact that when respiratory inhibitors like KCN, Chloroform are applied there is a decrease in the rate of respiration and also the rate of absorption of water.

Question 9.
Mention the objections to vital force theory of Ascent of sap.
Answer:
The objections to vital force theory of Ascent of sap:

  1. Strasburger (1889) and Overton (1911) experimentally proved that living cells are not mandatory for the ascent of sap. For this, he selected an old oak tree trunk which when immersed in picric acid and subjected to excessive heat killed all the living cells of the trunk. The trunk when dipped in water, the ascent of sap took place.
  2. Pumping action of living cells should be in between two xylem elements (vertically) and not on lateral sides.

Question 10.
Explain the capillary theory of Boehm (1809).
Answer:
Capillary theory: Boehm (1809) suggested that the xylem vessels work like a capillary tube. This capillarity of the vessels under normal atmospheric pressure is responsible for the ascent of sap. This theory was rejected because the magnitude of capillary force can raise water level only up to a certain height. Further, the xylem vessels are broader than the tracheid which actually conducts more water and against the capillary theory.

Question 11.
Give a brief account of Lenticular transpiration.
Answer:
In stems of woody plants and trees, the epidermis is replaced by periderm because of secondary growth. In order to provide gaseous exchange between the living cells and outer atmosphere, some pores which looks like lens – shaped raised spots are present on the surface of the stem called Lenticels. The loss of water from lenticels is very insignificant as it amounts to only 0.1% of the total.

Question 12.
Explain the theory of photosynthesis in guard cells observed by Von Mohl with its demerits.
Answer:
Von Mohl (1856) observed that stomata open in light and close in the night. According to him, chloroplasts present in the guard cells photosynthesize in the presence of light resulting in the production of carbohydrate (Sugar) which increases osmotic pressure in guard cells. It leads to the entry of water from other cell and stomatal aperture opens. The above process vice versa in night leads to closure of stomata.

Demerits:

  1. Chloroplast of guard cells is poorly developed and incapable of performing photosynthesis.
  2. The guard cells already possess much amount of stored sugars.

Samacheer Kalvi 11th Bio Botany Solutions 11 Transport in Plants

Question 13.
What are the three types of wilting in plants? Explain them briefly.
Answer:
In general, there are three types of wilting as follows:

  1. Incipient wilting: Water content of plant cell decreases but the symptoms are not visible.
  2. Temporary wilting: On hot summer days, the freshness of herbaceous plants reduces turgor pressure at the day time and regains it at night.
  3. Permanent wilting: The absorption of water virtually ceases because the plant cell does not get water from any source and the plant cell passes into a state of permanent wilting.

Question 14.
Define guttation. Explain it with examples.
Answer:
During high humidity in the atmosphere, the rate of transpiration is much reduced. When plants absorb water in such a condition root pressure is developed due to excess water within the plant. Thus excess water exudates as liquid from the edges of the leaves and is called guttation. eg: Grasses, tomato, potato, brinjal and Alocasia.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 11 Transport in Plants 6
Guttation occurs through stomata like pores called hydathodes generally present in plants that grow in moist and shady places. Pores are present over a mass of loosely arranged cells with large intercellular spaces called epithem. This mass of tissue lies near vein endings (xylem and Phloem). The liquid coming out of hydathode is not pure water but a solution containing a number of dissolved substances.

Question 15.
What is the significance of transpiration in plants?
Answer:
Transpiration leads to loss of water, as 95% of absorbed water is lost in transpiration. It seems to be an evil process to plants. However, number of process like absorption of water, ascent of sap and mineral absorption – directly relay on the transpiration. Moreover plants withstand against scorching sunlight due to transpiration. Hence the transpiration is a “necessary evil” as stated by Curtis.

Question 16.
What do you understand by the source and sink organ of plant?
Answer:
The source organ: Source is defined as any organ in plants which are capable of exporting food materials to the areas of metabolism or to the areas of storage. eg: Mature leaves, germinating seeds.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 11 Transport in Plants 7
Sink organ: Sink is defined as any organ in plants which receives food from source. eg: Roots, tubers, developing fruits and immature leaves.

Question 17.
Why plants transport sugars as sucrose and not as starch or glucose or fructose?
Answer:
Glucose and Fructose are simple monosaccharides, whereas, Sucrose is a disaccharide composed of glucose and fructose. Starch is a polysaccharide of glucose. Sucrose and starch are more efficient in energy storage when compared to glucose and fructose, but starch is insoluble in water. So it cannot be transported via phloem and the next choice is sucrose, being water soluble and energy efficient, sucrose is chosen as the carrier of energy from leaves to different parts of the plant.

Sucrose has low viscosity even at high concentrations and has no reducing ends which makes it inert than glucose or fructose. During photosynthesis, starch is synthesized and stored in the chloroplast stroma and sucrose is synthesized in the leaf cytosol from which it diffuses to the rest of the plant.

Question 18.
What is meant by phloem unloading?
Answer:
From sieve elements sucrose is translocated into sink organs such as roots, tubers, flowers and fruits and this process is termed as phloem unloading. It consists of three steps:

  1. Sieve element unloading: Sucrose leave from sieve elements.
  2. Short distance transport: Movement of sucrose to sink cells.
  3. Storage and metabolism: The final step when sugars are stored or metabolized in sink cells.

Samacheer Kalvi 11th Bio Botany Solutions 11 Transport in Plants

Question 19.
Explain the term Donnam equilibrium.
Answer:
Within the cell, some of the ions never diffuse out through the membrane. They are trapped within the cell and are called fixed ions. But they must be balanced by the ions of opposite charge. Assuming that a concentration of fixed anions is present inside the membrane, more cations would be absorbed in addition to the normal exchange to maintain the equilibrium. Therefore, the cation concentration would be greater in the internal than in the external solution. This electrical balance or equilibrium controlled by electrical as well as diffusion phenomenon is known as the Donnan equilibrium.

IV. Answer the following (5 Marks)

Question 1.
Define the term osmosis. Give details of the types of osmosis in plants.
Answer:
1. Osmosis (Latin: Osmos – impulse, urge) is a special type of diffusion. It represents the movement of water or solvent molecules through a selectively permeable membrane from the place of its higher concentration (high water potential) to the place of its lower concentration (low water potential). Types of Solutions based on concentration:

  • Hypertonic (Hyper = High; tonic = solute): This is a strong solution (low solvent / high solute / low Ψ) which attracts solvent from other solutions.
  •  Hypotonic (Hypo – low; tonic = solute): This is a weak solution (high sol vent / low or zero solute/ high Ψ) and it diffuses water out to other solutions.
  • Isotonic (Iso = identical; tonic = soute): It refers to two solutions having same concentration. In this condition the net movement of water molecule will be zero.

The term hyper, hypo and isotonic are relative terms which can be used only in comparison with another solution.

2. Types of osmosis: Based on the direction of movement of water or solvent in an osmotic system, two types of osmosis can occur, they are Endosmosis and Exosmosis.

  • Endosmosis: Endosmosis is defined as the osmotic entry of solvent into a cell or a system when it is placed in a pure water or hypotonic solution. eg: dry raisins . (high solute and low solvent) placed in the water, it swells up due to turgidity.
  • Exosmosis: Exosmosis is defined as the osmotic withdrawal of water from a cell or system when it is placed in a hypertonic solution. Exosmosis in a plant cell leads to plasmolysis.

Question 2.
Give an account of active absorption theories with their demerits.
Answer:
The mechanism of water absorption due to forces generated in the root itself is called active absorption. Active absorption may be osmotic or non – osmotic.
1. Osmotic active absorption: The theory of osmotic active absorption was postulated by Atkins (1916) and Preistley (1923). According to this theory, the first step in the absorption is soil water imbibed by cell wall of the root hair followed by osmosis. The soil water is hypotonic and cell sap is hypertonic. Therefore, soil water diffuses into root hair along the concentration gradient (endosmosis).

When the root hair becomes fully turgid, it becomes hypotonic and water moves osmotically to the outer most cortical cell. In the same way, water enters into inner cortex, endodermis, pericycle and finally reaches protoxylem. As the sap reaches the protoxylem a pressure is developed known as root pressure. This theory involves the symplastic movement of water.

2. Objections to osmotic theory:

  • The cell sap concentration in xylem is not always high
  • Root pressure is not universal in all plants especially in trees.

3. Non – Osmotic active absorption: Bennet – Clark (1936), Thimann (1951) and Kramer (1959) observed absorption of water even if the concentration of cell sap in the root hair is lower than that of the soil water. Such a movement requires an expehditure of energy released by respiration (ATP). Thus, there is a link between water absorption and respiration. It is evident from the fact that when respiratory inhibitors like KCN, Chloroform are applied there is a decrease in the rate of respiration and also the rate of absorption of water.

Samacheer Kalvi 11th Bio Botany Solutions 11 Transport in Plants

Question 3.
Explain in detail about the cohesion tension theory proposed by Dixon and Jolly (1894).
Answer:
(i) Strong cohesive force or tensile strength of water: Water molecules have the strong mutual force of attraction called cohesive mutual force of attraction called cohesive force due to which they cannot be easily separated from one another. Further, the attraction between a water molecule and the wall of the xylem element is called adhesion. These cohesive and adhesive force works together to form an unbroken continuous water column in the xylem. The magnitude of the cohesive force is much high (350 atm) and is more than enough to ascent sap in the tallest trees.

(ii) Continuity of the water column in the plant: An important factor which can break the water column is the introduction of air bubbles in the xylem. Gas bubbles expanding and displacing water within the xylem element is called cavitation or embolism. However, the overall continuity of the water column remains undisturbed since water diffuses into the adjacent xylem elements for continuing ascent of sap.

(iii) Transpiration pull or Tension in the unbroken water column: The unbroken water column from leaf to root is just like a rope. If the rope is pulled from the top, the entire rope will move upward. In plants, such a pull is generated by the process of transpiration which is known as transpiration pull. Water vapour evaporates from mesophyll cells to the intercellular spaces near stomata as a result of active transpiration.

The water vapours are then transpired through the stomatal pores. Loss of water from mesophyll cells causes a decrease in water potential. So, water moves as a pull from cell to cell along the water potential gradient. This tension, generated at the top (leaf) of the unbroken water column, is transmitted downwards from petiole, stem and finally reaches the roots. The cohesion theory is the most accepted among the plant physiologists today.

Question 4.
Describe the theory of K+ transport theory of stomatal opening.
Answer:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 11 Transport in Plants 10
This theory was proposed by Levit (1974) and elaborated by Raschke (1975). According to this theory, the following steps are involved in the stomatal opening:
In light:

  1. In guard cell, starch is converted into organic acid (malic acid).
  2. Malic acid in guard cell dissociates to malate anion and proton (H+).
  3. Protons are transported through the membrane into nearby subsidiary cells with the exchange of K+ (Potassium ions) from subsidiary cells to guard cells. This process involves an electrical gradient and is called ion exchange.
  4. This ion exchange is an active process and consumes ATP for energy.
  5. Increased K+ ions in the guard cell are balanced by Cl ions. Increase in solute concentration decreases the water potential in the guard cell.
  6. Guard cell becomes hypertonic and favours the entry of water from surrounding cells.
  7. Increased turgor pressure due to the entry of water opens the stomatal pore.

In Dark:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 11 Transport in Plants 9

  1. In dark photosynthesis stops and respiration continues with accumulation of CO2 in the sub-stomatal cavity.
  2. Accumulation of CO2 in cell lowers the pH level.
  3. Low pH and a shortage of water in the guard cell activate the stress hormone Abscisic acid (ABA).
  4. ABA stops further entry of K+ ions and also induce K+ ions to leak out to subsidiary cells from guard cell.
  5. Loss of water from guard cell reduces turgor pressure and causes closure of stomata.

Question 5.
Give an account of external factors, which affect the rate of transpiration.
Answer:
External or Environmental factors:
1. Atmospheric humidity: The rate of transpiration is greatly reduced when the atmosphere is very humid. As the air becomes dry, the rate of transpiration is also increased proportionately.

2. Temperature: With the increase in atmospheric temperature, the rate of transpiration also increases. However, at very high – temperatures stomata closes because of flaccidity and transpiration stop.

3. Light: Light intensity increases the temperature. As in temperature, transpiration is increased in high light intensity and is decreased in low light intensity. Light also increases the permeability of the cell membrane, making it easy for water molecules to move out of the cell.

4. Wind velocity: In still air, the surface above the stomata get saturated with water vapours and there is no need for more water vapour to come out. If the wind is breezy, water vapour gets carried away near leaf surface and DPD is created to draw more vapour from the leaf cells enhancing transpiration. However, high wind velocity creates an extreme increase in water loss and leads to a reduced rate of transpiration and stomata remain closed.

5. Atmospheric pressure: In low atmospheric pressure, the rate of transpiration increases. Hills favour high transpiration rate due to low atmospheric pressure. However, it is neutralized by low temperature prevailing in the hills.

6. Water: Adequate amount of water in the soil is a pre – requisite for optimum plant growth. Excessive loss of water through transpiration leads to wilting.

Question 6.
Describe the method of Ganongs potometer to measure the rate of transpiration.
Answer:
Ganongs potometer is used to measure the rate of transpiration indirectly. In this, the amount of water absorbed is measured and assumed that this amount is equal to the amount of water transpired. Apparatus consists of a horizontal graduated tube which is bent in opposite directions at the ends. One bent end is wide and the other is narrow. A reservoir is fixed to the horizontal tube near the wider end. The reservoir has a stopcock to regulate water flow. The apparatus is filled with water from reservoir.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 11 Transport in Plants 8
A twig or a small plant is fixed to the wider arm through a split cock. The other bent end of the horizontal tube is dipped into a beaker containing coloured water. An air bubble is introduced into the graduated tube at the narrow end. Keep this apparatus in bright sunlight and observe. As transpiration takes place, the air bubble will move towards the twig. The loss is compensated by water absorption through the xylem portion of the twig. Thus, the rate of water absorption is equal to the rate of transpiration.

Question 7.
Explain Munch Mass Flow Hypothesis with its merits and objections.
Answer:
Mass flow theory was first proposed by Munch (1930) and elaborated by Crafts (1938). According to this hypothesis, organic substances or solutes move from the region of high osmotic pressure (from mesophyll) to the region of low osmotic pressure along the turgor pressure gradient. The principle involved in this hypothesis can be explained by a simple physical system as shown in figure.

Two chambers “A” and “B” made up of semipermeable membranes are connected by tube “T” immersed in a reservoir of water. Chamber “A” contains highly concentrated sugar solution while chamber “B” contains dilute sugar solution. The following changes were observed in the system.

  1. The high concentration sugar solution of chamber “A” is in a hypertonic state which draws water from the reservoir by endosmosis.
  2. Due to the continuous entry of water into chamber “A”, turgor pressure is increased.
  3. Increase in turgor pressure in chamber “A” force, the mass flow of sugar solution to chamber “B” through the tube “T” along turgor pressure gradient.
  4. The movement of solute will continue till the solution in both the chambers attains the state of isotonic condition and the system becomes inactive.
  5. However, if new sugar solution is added in chamber “A”, the system will start to run again.

A similar analogous system as given in the experiment exists in plants:
Chamber “A” is analogous to mesophyll cells of the leaves which contain a higher concentration of food material in soluble form. In short “A” is the production point called “source”. Chamber “B” is analogous to cells of stem and roots where the food material is utilized. In short “B” is consumption end called “sink”. Tube “T” is analogous to the sieve tube of phloem.

Mesophyll cells draw water from the xylem (reservoir of the experiment) of the leaf by endosmosis leading to increase in the turgor pressure of mesophyll cell. The turgor pressure in the cells of stem and the roots are comparatively low and hence, the soluble organic solutes begin to flow en masse from mesophyll through the phloem to the cells of stem and roots along the gradient turgor pressure.

In the cells of stem and roots, the organic solutes are either consumed or converted into insoluble form and the excess water is released into xylem (by turgor pressure gradient) through cambium.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 11 Transport in Plants 12
Merits:

  1. When a woody or herbaceous plant is girdled, the sap contains high sugar containing exudates from cut end.
  2. Positive concentration gradient disappears when plants are defoliated.

Objections:

  1. This hypothesis explains the unidirectional movement of solute only. However, bidirectional movement of solute is commonly observed in plants.
  2. Osmotic pressure of mesophyll cells and that of root hair do not confirm the requirements.
  3. This theory gives passive role to sieve tube and protoplasm, while some workers demonstrated the involvement of ATP.

Question 8.
Write an essay on Lunde – gardh’s cytochrome pump theory of mineral transport.
Answer:
Lundegardh and Burstrom (1933) observed a correlation between respiration and anion absorption. When a plant is transferred from water to a salt solution the rate of respiration increases which is called,as anion respiration or salt respiration. Based on this observation Lundegardh (1950 and 1954) proposed cytochrome pump theory which is based on the following assumptions:

  1. The mechanism of anion and cation absorption are different.
  2. Anions are absorbed through cytochrome chain by an active process, cations are absorbed passively.
  3. An oxygen gradient responsible for oxidation at the outer surface of the membrane and reduction at the inner surface.

According to this theory, the enzyme dehydrogenase on inner surface is responsible for the formation of protons (H+) and electrons (e). As electrons pass outward through electron transport chain there is a corresponding inward passage of anions.

Anions are picked up by oxidized cytochrome oxidase and are transferred to other members of chain as they transfer the electron to the next component. The theory assumes that cations (C+) move passively along the electrical gradient created by the accumulation of anions (A) at the inner surface of the membrane.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 11 Transport in Plants 13
Main defects of the above theory are:

  1. Cations also induce respiration.
  2. Fails to explain the selective uptake of ions.
  3. It explains absorption of anions only.

Solution To Activity
Textbook Page No: 63

Question 1.
Imbibition experiment: Collect 5 gm of gum from Drumstick tree or Babool tree or Almond tree. Immerse in 100 ml of water. After 24 hours observe the changes and discuss the results with your teacher.
Answer:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 11 Transport in Plants 11
The gum will absorb large amount of water and swells. The phenomenon is called imbibition.

Textbook Page No: 65

Question 1.
Find the role of turgor pressure in sudden closing of leaves when we touch the ‘touch me not’ plant.
Answer:
When touched, this sensitive leaf reacts to stimulus as there is a higher pressure at that point and water in the vacuoles of the cells of the leaf lose water to the adjacent cell. This causes the leaves to close. If the leaves are left undisturbed for a few seconds, they slowly open up again and regain turgidity.

Textbook Page No: 75

Question 1.
Select a leafy twig of fully grown plant in your school campus. Cover the twig with a transparent polythene bag and tie the mouth of the bag at the base of the twig. Observe the changes after two hours and discuss with your teacher.
Answer:
Two hours and discuss with your teacher:

  1. Select a leafy twig of a fully grown plant.
  2. Cover the twig in a transparent polythene bag.
  3. Tie the mouth of the bag.
  4. Observe the bag after two hours.
  5. Observation: Moisture will be observed inside the plastic bag because of transpiration of water from the plant twig.

Textbook Page No: 79

Question 1.
What will happen if an indoor plant is placed under fan and AC?
Answer:
When an indoor plant is placed under fan and AC, the transpiration of water from the plant may increase, because the wind from fan and the humidity from AC will increase transpiration of water from the plant.

Samacheer Kalvi 11th Bio Botany Solutions Chapter 13 Photosynthesis

Students can Download Bio Botany Chapter 13 Photosynthesis Questions and Answers, Notes Pdf, Samacheer Kalvi 11th Bio Botany Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Bio Botany Solutions Chapter 13 Photosynthesis

Samacheer Kalvi 11th Bio Botany Photosynthesis Text Book Back Questions and Answers

Questions 1.
Assertion (A): Increase in Proton gradient inside lumen responsible for ATP synthesis.
Reason (R): Oxygen evolving complex of PS I located on thylakoid membrane facing Stroma, releases H+ ions.
(a) Both Assertion and Reason are True.
(b) Assertion is True and Reason is False.
(c) Reason is True and Assertion is False.
(d) Both Assertion and Reason are False.
Answer:
(a) Both Assertion and Reason are True.

Samacheer Kalvi 11th Bio Botany Solutions 13 Photosynthesis

Question 2.
Which chlorophyll molecule does not have a phytol tail?
(a) Chl – a
(b) Chl – b
(c) Chl – c
(d) Chl – d
Answer:
(c) Chl – c

Question 3.
The correct sequence of flow of electrons in the light reaction is:
(a) PS II, plastoquinone, cytochrome, PS I, ferredoxin.
(b) PS I, plastoquinone, cytochrome, PS II ferredoxin.
(c) PS II, ferredoxin, plastoquinone, cytochrome, PS I.
(d) PS I, plastoquinone, cytochrome, PS II, ferredoxin.
Answer:
(a) PS II, plastoquinone, cytochrome, PS I, ferredoxin.

Question 4.
For every CO2 molecule entering the C3 cycle, the number of ATP & NADPH required:
(a) 2 ATP + 2 NADPH
(b) 2 ATP + 3 NADPH
(c) 3 ATP + 2 NADPH
(d) 3 ATP + 3 NADPH
Answer:
(c) 3 ATP + 2 NADPH.

Samacheer Kalvi 11th Bio Botany Solutions 13 Photosynthesis

Question 5.
Identify true statement regarding light reaction of photosynthesis?
(a) Splitting of water molecule is associate with PS I.
(b) PS I and PS II involved in the formation of NDPH + H+.
(c) The reaction center of PS I is Chlorophyll a with absorption peak at 680 nm.
(d) The reaction center of PS II is Chlorophyll a with absorption peak at 700 nm.
Answer:
(b) PS I and PS II involved in the formation of NDPH + H+.

Question 6.
Two groups (A & B) of bean plants of similar size and same leaf area were placed in identical conditions. Group A was exposed to light of wavelength 400 – 450 nm and Group B to light of wavelength of 500 – 550 nm. Compare the photosynthetic rate of the 2 groups giving reasons.
Answer:
The rate of photosynthesis in group A bean plants is more than what is found in Group B plants because the rate of absorption of light is more at the wavelength is less beyond the wavelength of 500 – 550 nm.

Question 7.
A tree is believed to be releasing oxygen during night time. Do you believe the truthfulness of this statement? Justify your answer by giving reasons?
Answer:
Yes, a tree is believed to be releasing O2 during night time because at night CAM plants fix CO2 with the help of phospho Enol Pyruvic acid and produce oxala acetic acid, which is converted into malic acid like C4 cycle.

Question 8.
Grasses have an adaptive mechanism to compensate. photorespiratory losses – Name and describe the mechanism.
Answer:
Rate of respiration is more in light than in dark. Photorespiration is the excess respiration taking place in photosynthetic cells due to absence of CO2 and increase of O2. This condition changes the carboxylase role of RUBISCO into oxygenase. C2 Cycle takes place in chloroplast, peroxisome and mitochondria. RUBP is converted into PGA and a 2C – compound phosphoglycolate by Rubisco enzyme in chloroplast. Since the first product is a 2C – compound, this cycle is known as C2 Cycle. Phosphoglycolate by loss of phosphate becomes glycolate.

Glycolate formed in chloroplast enters into peroxisome to form glyoxylate and hydrogen peroxide. Glyoxylate is converted into glycine and transferred into mitochondria. In mitochondria, two molecules of glycine combine to form serine. Serine enters into peroxisome to form hydroxy pyruvate. Hydroxy pyruvate with help of NADH + H+ becomes glyceric acid. Glyceric acid is cycled back to chloroplast utilising ATP and becomes Phosphoglyceric acid (PGA) and v enters into the Calvin cycle (PCR cycle). Photorespiration does not yield any free energy in the form of ATP. Under certain conditions 50% of the photosynthetic potential is lost because of Photorespiration

Samacheer Kalvi 11th Bio Botany Solutions 13 Photosynthesis

Question 9.
In Botany class, teacher explains, Synthesis of one glucose requires 30 ATPs in C4 plants and only 18 ATPs in C3 plants. The same teacher explains C4 plants are more advantageous than C3 plants. Can you identify the reason for this contradiction?
Answer:
C4 plants requires 30 ATPs and 12 NADPH + H+ to synthesize one glucose, but C3 plants require only 18 ATPs and 12 NADPH + H+ to synthesize one glucose molecule. If then, how can you say C4 plants are more advantageous? C4 plants are more advantageous than C3 plants because C4 photosynthesis is advantages over C3 plant, because C4 photosynthesis avoids photorespiration and is thus potentially more efficient than C3 plants. Due to the absense of photorespiration, carbon di oxide compensation point for C4 is lower than that of C3 plants.

Question 10.
When there is plenty of light and higher concentration of O2, what kind of pathway does the plant undergo? Analyse the reasons.
Answer:
The rate of photosynthesis decreases when there is an increase of oxygen concentration. This Inhibitory effect of oxygen was first discovered by Warburg (1920) using green algae, Chlorella.

Samacheer Kalvi 11th Bio Botany Photosynthesis Additional Questions & Answers

I. Choose the correct answer (1 Mark)
Question 1.
Photosynthesis is the major:
(a) endothermic reaction
(b) exothermic reaction
(c) endergonic reaction
(d) exergonic reaction
Answer:
(c) endergonic reaction

Question 2.
Who has first explained the importance chlorophyll in photosynthesis:
(a) Joseph Priestly
(b) Dutrochet
(c) Stephen Hales
(d) Lovoisier
Answer:
(b) Dutrochet

Samacheer Kalvi 11th Bio Botany Solutions 13 Photosynthesis

Question 3.
How many million tonnes of dry matter produced annually by photosynthesis?
(a) 1700 million tonnes
(b) 1900 million tonnes
(c) 1400 million tonnes
(d) 2000 million tonnes
Answer:
(a) 1700 million tonnes

Question 4.
Who received 1988 Nobel prize for his work on photosynthesis in Rhodobacter:
(a) Emerson and Arnold
(b) Ruben and Kamem
(c) Arnon, Allen and Whatley
(d) Huber, Michael and Dissenhofer
Answer:
(d) Huber, Michael and Dissenhofer

Question 5.
Thylakoid disc diameter is:
(a) 0.35 to 0.75 microns
(b) 0.25 to 0.8 microns
(c) 0.45 to 0.8 microns
(d) 0.50 to 0.9 microns
Answer:
(b) 0.25 to 0.8 microns

Samacheer Kalvi 11th Bio Botany Solutions 13 Photosynthesis

Question 6.
Indicate the correct statement:
(a) Grana lamellae have only PS I
(b) Stroma lamellae have only PS II
(c) Grana lamellae have both PS I and PS II
(d) Stroma lamellae have both PS I and PS II
Answer:
(c) Grana lamellae have both PS I and PS II

Question 7.
Match the following:

A. Cyanobacteria (i) Chlorophyll D
B. Green algae (ii) Chlorophyll C
C. Brown algae (iii) Chlorophyll A
D. Red algae (iv) Chlorophyll B

(a) A – (iii); B – (i); C – (iv); D – (ii)
(b) A – (ii); B – (iii); C – (iv); D – (i)
(c) A – (iii); B – (iv); C – (i); D – (ii)
(d) A – (iii); B – (iv); C – (ii); D – (i)
Answer:
(d) A – (iii); B – (iv); C – (ii); D – (i)

Question 8.
Each pyrrole ring comprises of:
(a) six carbons and one nitrogen atom
(b) three carbons and one nitrogen atom
(c) four carbons and one nitrogen atom
(d) four carbons and two nitrogen atom
Answer:
(c) four carbons and one nitrogen atom

Samacheer Kalvi 11th Bio Botany Solutions 13 Photosynthesis

Question 9.
Biosynthesis of chlorophyll ‘a’ requires:
(a) Mg, Fe, Cu, Zn, Mn, K and nitrogen
(b) Mg, Fe, Cu, Mo, Mn, K and nitrogen
(c) Mg, Cu, Zn, Mo, Mn, K and nitrogen
(d) Mg, Fe, Cu, Zn, Mo, K and nitrogen
Answer:
(a) Mg, Fe, Cu, Zn, Mn, K and nitrogen

Question 10.
Pheophytin resembles chlorophyll ‘a’ except that it lacks:
(a) Fe atom
(b) Mn atom
(c) Mg atom
(d) Cu atom
Answer:
(c) Mg atom

Question 11.
Almost all carotenoid pigments have:
(a) 50 carbon atoms
(b) 40 carbon atoms
(c) 30 carbon atoms
(d) 60 carbon atoms
Answer:
(b) 40 carbon atoms

Samacheer Kalvi 11th Bio Botany Solutions 13 Photosynthesis

Question 12.
Which one of the photosynthetic pigments is called shield pigment:
(a) carotenes
(b) chlorophyll ‘b’
(c) pheophytin
(d) carotenoids
Answer:
(d) carotenoids

Question 13.
The visible spectrum of light ranges between:
(a) 200 to 2800 nm
(b) 300 to 2600 nm
(c) 200 to 800 nm
(d) 300 to 2400 nm
Answer:
(b) 300 to 2600 nm

Question 14.
Photosynthetic rate of red light (650 nm) is equal to:
(a) 42.5
(b) 10.0
(c) 43.5
(d) 40.8
Answer:
(c) 43.5

Question 15.
Indicate the correct statement in respect to Hill’s reaction:
(i) During photosynthesis oxygen is evolved from water
(ii) Electrons for the reduction of CO2 are obtained from H2S.
(iii) During photosynthesis oxygen is evolved from CO2
(iv) Electrons for the reduction of CO2 are obtained from water

(a) (i) and (ii)
(b) (ii) and (iii)
(c) (i) and (iv)
(d) (ii) and (iv)
Answer:
(c) (i) and (iv)

Samacheer Kalvi 11th Bio Botany Solutions 13 Photosynthesis

Question 16.
Phosphorylation taking place during respiration is called as:
(a) Photophorylation
(b) Oxidative phosphorylation
(c) Reductive phosphorylation
(d) None of the above
Answer:
(b) Oxidative phosphorylation

Question 17.
Find out the odd one:
(a) Ferredoxin
(b) Succinate
(c) Cytochrome b6 – f
(d) Plastocyanin
Answer:
(b) Succinate

Question 18.
In bio – energetics of light reaction, to release one electron from pigment system it requires:
(a) two quanta of light
(b) four quanta of light
(c) one quanta of light
(d) eight quanta of light
Answer:
(a) two quanta of light

Samacheer Kalvi 11th Bio Botany Solutions 13 Photosynthesis

Question 19.
Chemiosmatic theory was proposed by:
(a) S. Michael
(b) R. Hill
(c) P. Mitchell
(d) G. Root
Answer:
(c) P. Mitchell

Question 20.
In C4 plants, how many ATPs and NADPH + H+ are utilised for the release of one oxygen molecule:
(a) 3 ATPs and 2 NADPH + H+
(b) 4 ATPs and 3 NADPH + H+
(c) 2 ATPs and 2 NADPH + H+
(d) 5 ATPs and 2 NADPH + H+
Answer:
(d) 5 ATPs and 2 NADPH + H+

Question 21.
The key enzyme in the carboxylation reaction is:
(a) Ribulose dehydrogenase
(b) Carboxylase
(c) Carboxylase oxygenase
(d) Carboxyl anhydrase
Answer:
(c) Carboxylase oxygenase

Question 22.
In sugarcane plant, the dicarboxylic acid pathway was first discovered by:
(a) Hatch and Slack
(b) Kortschak, Hart and Burr
(c) Calvin and Benson
(d) Mitchell and Root
Answer:
(b) Kortschak, Hart and Burr

Samacheer Kalvi 11th Bio Botany Solutions 13 Photosynthesis

Question 23.
In bundle sheath cells, malic acid undergoes dicarboxylation and produces 3 carbon compound:
(a) Glyceric acid and CO2
(b) Glyceraldehyde and CO2
(c) Pyruvic acid and CO2
(d) None of the above
Answer:
(c) Pyruvic acid and CO2

Question 24.
Indicate the correct answer:
(a) C4 plants are adapted to only rainy conditions
(b) C4 plants are partially adapted to drought condition
(c) C4 plants are exclusively adapted to desert condition
(d) C4 plants are adapted to aquatic condition
Answer:
(b) C4 plants are partially adapted to drought condition

Question 25.
Crassulacean acid metabolism or CAM cycle was first observed in:
(a) sugarcane
(b) bryophyllum
(c) mango
(d) banana
Answer:
(b) bryophyllum

Samacheer Kalvi 11th Bio Botany Solutions 13 Photosynthesis

Question 26.
Glycolate protects plant cells from:
(a) Photophosphorylation
(b) Photo reduction
(c) Photo oxidation
(d) Photolysis
Answer:
(c) Photo oxidation

Question 27.
The important external factors affecting photosynthesis are:
(a) light, chlorophyll, temperature
(b) light, stomatal opening, oxygen
(c) light, protoplasmic factor, oxygen
(d) light, CO2 and oxygen
Answer:
(d) light, CO2 and oxygen

Question 28.
Hormones like gibberellin:
(a) increases the rate of photosynthesis
(b) increase respiration
(c) decrease the rate of photosynthesis
(d) decrease the rate of transpiration
Answer:
(a) increases the rate of photosynthesis

Question 29.
Bacterial photosynthesis differs from higher plant photosynthesis in:
(a) utilizing water as electron donar
(b) releasing O2
(c) releasing sulphur instead of oxygen
(d) utilizing SO2 as electron donar
Answer:
(c) releasing sulphur instead of oxygen

Samacheer Kalvi 11th Bio Botany Solutions 13 Photosynthesis

Question 30.
Splitting of water molecule (photolysis) produces:
(a) hydrogen and oxygen
(b) electrons, protons and oxygen
(c) electrons and oxygen
(d) hydrogen, carbon di oxide and oxygen
Answer:
(b) electrons, protons and oxygen

II. Answer the following (2 Marks)

Question 1.
What is the function of plant in the universe?
Answer:
Plants are the major machinery which produce organic compounds like carbohydrates,lipids, proteins, nucleic acids and other biomolecules.

Question 2.
Define photosynthesis.
Answer:
Photosynthesis is referred as photochemical oxidation and reduction reactions carried out with help of light, converting solar energy into Chemical energy.

Question 3.
What is the site of photosynthesis?
Answer:
Chloroplasts are the main site of photosynthesis and both energy yielding process (Light reaction) and fixation of carbon dioxide (Dark reaction) that takes place in chloroplast.

Question 4.
What is thylakoid? Explain how they are arranged?
Answer:
A sac like membranous system called thylakoid or lamellae is present in stroma and they are arranged one above the other forming a stack of coin like structure called granum (plural grana).

Samacheer Kalvi 11th Bio Botany Solutions 13 Photosynthesis

Question 5.
Endosymbiotic hypothesis says that chloroplasts evolved from bacteria. Substantiate the statement.
Answer:
Presence of 70S ribosome and DNA gives them status of semi-autonomy and proves endosymbiotic hypothesis which says chloroplast evolved from bacteria.

Question 6.
Define photosynthetic pigment.
Answer:
A photosynthetic pigment is a pigment that is present in chloroplasts or photosynthetic bacteria which captures the light energy necessary for photosynthesis.

Question 7.
Match the following:

A. Xanthophyll (i) Lycopene
B. Phycocyanin (ii) Red algae
C. Carotene (iii) Brown algae
D. Phycoerythin (iv) Cyanobacteria

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

Samacheer Kalvi 11th Bio Botany Solutions 13 Photosynthesis

Question 8.
What are Xanthophylls?
Answer:
Yellow (C40H56O2) pigments are like carotenes but contain oxygen. Lutein is responsible for yellow colour change of leaves during autumn season. Examples: Lutein, Violaxanthin and Fueoxanthin.

Question 9.
Write down any two properties of light.
Answer:
Two properties of light:

  1. Light is a transverse electromagnetic wave.
  2. It consists of oscillating electric and magnetic fields that are perpendicular to each other and perpendicular to the direction of propagation of the light.

Question 10.
Define absorption spectrum.
Answer:
Pigments absorb different wavelengths of light. A curve obtained by plotting the amount of absorption of different wavelengths of light by a pigment is called its absorption spectrum.

Question 11.
Define the term fluorescence.
Answer:
The electron from first singlet state (SI) returns to ground state (SO) by releasing energy in the form of radiation energy (light) in the red region and this is known as fluorescence.

Samacheer Kalvi 11th Bio Botany Solutions 13 Photosynthesis

Question 12.
What is known as substrate level phosphorylation?
Answer:
Phosphorylation taking place during respiration is called as oxidative phosphorylation and ATP produced by the breakdown of substrate is known as substrate level phosphorylation.

Question 13.
Define photophosphorylation.
Answer:
Phosphorylation is the process of synthesis of ATP by the addition of inorganic phosphate to ADP. The addition of phosphate here takes place with the help of light generated electron and so it is called as photophosphorylation.

Question 14.
What are the phases of dark reaction?
Answer:
Dark reaction consists of three phases:

  • Carboxylation (fixation)
  • Reduction (Glycolytic Reversal)
  • Regeneration

Samacheer Kalvi 11th Bio Botany Solutions 13 Photosynthesis

Question 15.
What are significance of photo respiration?
Answer:
Significance of photo respiration:

  1. Glycine and Serine synthesised during this process are precursors of many biomolecules like chlorophyll, proteins, nucleotides.
  2. It consumes excess NADH + H+ generated.
  3. Glycolate protects cells from Photo oxidation.

Question 16.
What is meant by carbon dioxide compensation point?
Answer:
When the rate of photosynthesis equals the rate of respiration, there is no exchange of oxygen and carbon dioxide and this is called as carbon dioxide compensation point.

Question 17.
What, are the internal factors, that affect photosynthesis?
Answer:
Pigments, protoplasmic factor, accumulation of carbohydrates, anatomy of leaf and hormones.

Question 18.
What are the air pollutants, that affect rate of photosynthesis?
Answer:
Pollutants like SO2, NO2, O3 (Ozone) and Smog affect rate of photosynthesis.

Samacheer Kalvi 11th Bio Botany Solutions 13 Photosynthesis

Question 19.
How does water affect the rate of photosynthesis?
Answer:
Photolysis of water provides electrons and protons for the reduction of NADP, directly. Indirect roles are stomatal movement and hydration of protoplasm. During water stress, supply of NADPH + H+ is affected.

Question 20.
Name any three photosynthetic bacteria.
Answer:
Three photosynthetic bacteria:

  1. Chlorobacterium
  2. Thiospirillum
  3. Rodhospirillum

III. Answer the following. (3 Marks)

Question 1.
Mention any three significance of photosynthesis.
Answer:
Three significance of photosynthesis:

  1. Photosynthetic organisms provide food for all living organisms on earth either directly or indirectly.
  2. It is the only natural process that liberates oxygen in the atmosphere and balances the oxygen level.
  3. Photosynthesis balances the oxygen and carbon cycle in nature.

Samacheer Kalvi 11th Bio Botany Solutions 13 Photosynthesis

Question 2.
How is the chlorophyll synthesized?
Answer:
Chlorophyll is synthesized from intermediates of respiration and photosynthesis. Succinic acid an intermediate of Krebs cycle is activated by the addition of coenzyme A and it reacts with a simple amino acid glycine and the reaction goes on to produce chlorophyll ‘a’. Biosynthesis of chlorophyll ‘a’ requires Mg, Fe, Cu, Zn, Mn, K and nitrogen. The absence of any one of these minerals leads to chlorosis.

Question 3.
What are phycobilins?
Answer:
They are proteinaceous pigments, soluble in water, and do not contain Mg and Phytol tail. They exist in two forms such as:

  1. Phycocyanin found in cyanobacteria.
  2. Phycoerythin found in rhodophycean algae (Red algae).

Question 4.
What are the conclusions of Hill’s reaction?
Answer:
The conclusions of Hill’s reaction:

  1. During photosynthesis oxygen is evolved from water.
  2. Electrons for the reduction of CO2 are obtained from water.
  3. Reduced substance produced, later helps to reduce CO2.

Question 5.
What is meant by ground state?
Answer:
The action of photon plays a vital role in excitation of pigment molecules to release an electron. When the molecules absorb a photon, it is in excited state. When the light source turned off, the high energy electrons return to their normal low energy orbitals as the excited molecule goes back to its original stable condition known as ground state.

Samacheer Kalvi 11th Bio Botany Solutions 13 Photosynthesis

Question 6.
Explain the term phosphorescence.
Answer:
Electron from Second Singlet State (S2) may return to next higher energy level (S1) by losing some of its extra energy in the form of heat. From first singlet state (S1) electron further drops to first triplet state (T1). Triplet State is unstable having half life time of 10-3 seconds and electrons returns to ground state with emission of light in red region called as phosphorescence. Phosphorescence is the delayed emission of absorbed radiations. Pathway of electron during Phosphorescence:
S2 → S1 → T1 → S0

Question 7.
Describe the method of carboxylation.
Answer:
The acceptor molecule Ribulose 1,5 Bisphosphate (RUBP) a 5 carbon compound with the help of RUBP carboxylase oxygenase (RUBISCO) enzyme accepts one molecule of carbon dioxide to form an unstable 6 carbon compound. This 6C compound is broken down into two molecules of 3 – carbon compound phospho glyceric acid (PGA).
Samacheer Kalvi 11th Bio Botany Solutions Chapter 13 Photosynthesis 7

Question 8.
Explain the phase – 3 of dark reaction.
Answer:
Regeneration of RUBP involves the formation of several intermediate compounds of 6 – carbon, 5 – carbon, 4 – carbon and 7 – carbon skeleton. Fixation of one carbon dioxide requires 3 ATPs and 2 NADPH + H+, and for the fixation of 6 CO2 requires 18 ATPs and 12 NADPH + H+ during C3 cycle. One 6 carbon compound is the net gain to form hexose sugar.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 13 Photosynthesis 1
Overall equation for dark reaction:
6CO2 + 18 ATP + 12 NADPH + H+ → C6H12O6 + 6H2O + 18 ADP + 18 Pi + 12 NADP+

Question 9.
What is meant by dicarfioxylic acid pathway?
Answer:
C4 pathway is completed in two phases, first phase takes place in stroma of mesophyll cells, where the CO2 acceptor mblecule is 3 – Carbon compound, phospho enol pyruvate (PEP) to form 4 – carbon Oxalo acetic acid (OAA). The first product is a 4 – carbon and so it is named as C4 cycle. Oxalo acetic acid is a dicarbokylic acid and hence this cycle is also known as dicarboxylic acid pathway.

Question 10.
Mention the significances of C4 cycle.
Answer:
The significances of C4 cycle:

  1. Plants having C4 cycle are mainly of tropical and sub – tropical regions and are able to survive in environment with low CO2 concentration.
  2. C4 plants are partially adapted to drought conditions.
  3. Oxygen has no inhibitory effect on C4 cycle since PEP carboxylase is insensitive to O2.
  4. Due to absence of photorespiration, CO2 Compensation Point for C4 is lower than that of C3 plants.

Samacheer Kalvi 11th Bio Botany Solutions 13 Photosynthesis

Question 11.
What is the type of carbon pathway in xerophytic plants?
Answer:
Crassulacean Acid Metabolism or CAM cycle is one of the carbon pathways identified in succulent plants growing in semi – arid or xerophytic condition. This was first observed in crassulaceae family plants like Bryophyllum, Sedum, Kalanchoe and is the reason behind the name of this cycle. It is also noticed in plants from other families eg: Agave, Opuntia, Pineapple and Orchids.

Question 12.
what are the significance of CAM cycle?
Answer:
The significance of CAM cycle:

  1. It is advantageous for succulent plants to obtain CO2 from malic acid when stomata are closed.
  2. During day time stomata are closed and CO2 is not taken but continue their photosynthesis.
  3. Stomata are closed during the day time and help the plants to avoid transpiration and water loss.

IV. Answer the following (5 Marks)

Question 1.
Explain in detail about absorption spectrum and action spectrum of light.
Answer:
1. Absorption spectrum: The term absorption refers to complete retention of light, without reflection or transmission. Pigments absorb different Wavelengths of light. A curve obtained by plotting the amount of absorption of different wavelengths of light by a pigment is called its absorption spectrum.

  • Chlorophyll ‘a’ and chlorophyll ‘b’ absorb quanta from blue and red region.
  • Maximum absorption peak for different forms of chlorophyll ‘a’ is 670 to 673, 680 to 683 and 695 to 705 nm.
  • Chlorophyll ‘a’ 680 (P680) and Chlorophyll ‘a’ 700 (P700) function as trap centre for PS II and PS I respectively.

2. Action Spectrum: The effectiveness of different wavelength of light on photosynthesis is measured by plotting against quantum yield. The curve showing the rate of photosynthesis at different wavelengths of light is called action spectrum. From the graph showing action spectrum, it can be concluded that maximum photosynthesis takes place in blue and red region of the spectrum. This wavelength of the spectrum is the absorption maxima for Chlorophyll (a) and Chlorophyll (b). The Action Spectrum is instrumental in the discovery of the existence of two photosystems in O2 evolving photosynthesis.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 13 Photosynthesis 2

Question 2.
Distinguish between Photo system – I and photo system – II
Answer:
Photo system – I:

  1. The reaction centre is P700.
  2. PS I is involved in both cyclic and non – cyclic.
  3. Not involved in photolysis of water and evolution of oxygen.
  4. It receives electrons from PS II during non – cyclic photophosphorylation.
  5. Located in unstacked region granum facing chloroplast stroma.
  6. Chlorophyll and Carotenoid ratio is 20 to 30 : 1.

Photo system – II:

  1. Reaction centre is P680.
  2. PS II participates in Non – cyclic pathway.
  3. Photolysis of water and evolution of oxygen take place.
  4. It receives electrons by photolysis of water.
  5. Located in stacked region of thylakoid membrane facing lumen of thylakoid.
  6. Chlorophyll and Carotenoid ratio is 3 to 7 : 1.

Samacheer Kalvi 11th Bio Botany Solutions 13 Photosynthesis

Question 3.
Explain the process of photolysis of photolysis water with suitable diagram.
Answer:
The process of Photolysis is associated with Oxygen Evolving Complex (OEC) or water splitting complex in pigment system II and is catalysed by the presence of Mn++ and Cl. When the pigment system II is active it receives light and the water molecule splits into OH ions and H+ ions. The OHions unite to form water molecules again and release O2 and electrons. Photolysis of water is due to strong oxidant which is yet unknown and designated as Z or Yz.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 13 Photosynthesis 3
Widely accepted theory proposed by Kok et al., (1970) explaining photo – oxidation of water is water oxidizing clock (or) S’ State Mechanism. It consists of a series of 5 states called as S0, S1, S2, S3 and S4. Each state acquires positive charge by a photon (hv) and after the S4 state it acquires 4 positive charges, four electrons and evolution of oxygen. Two molecules of water go back to the S0. At the end of photolysis 4H+, 4e and O2 are evolved from water.

Question 4.
Describe the process of non – cyclic photophosphorylation.
Answer:
When photons are activated reaction centre of pigment system II (P680), electrons are moved to the high energy level. Electrons from high energy state passes through series of electron carriers like pheophytin, plastoquinone, cytochrome complex, plastocyanin and finally accepted by PS I (P700). During this movement of electrons from PS II to PS I ATP is generated. PS I (P700) is activated by light, electrons are moved to high energy state and accepted by electron acceptor molecule ferredoxin reducing Substance (FRS). During the downhill movement through ferredoxin, electrons are transferred to NADP+ and reduced into NADPH + H+ (H+ formed from splitting of water by light).
Samacheer Kalvi 11th Bio Botany Solutions Chapter 13 Photosynthesis 4
Electrons released from the photosystem II are not cycled back. It is used for the reduction of NADP+ in to NADPH + H+. During the electron transport it generates ATP and hence this type of photophosphorylation is called non – cyclic photophosphorylation. The electron flow looks like the appearance of letter ‘Z’ and so known as Z scheme.

When there is availability of NADP+ for reduction and when there is splitting of water molecules both PS I and PS II are activated. Non-cyclic electron transport PS I and PS II both are involved co – operatively to transport electrons from water to MADP+. In oxygenic species non – cyclic electron transport takes place in three stages.

  1. Electron transport from water to P680: Splitting of water molecule produce electrons, protons and oxygen. Electrons lost by the PS II (P680) are replaced by electrons from splitting of water molecule.
  2. Electron transport from P680 to P700: Electron flow starts from P680 through a series of electron carrier molecules like pheophytin, plastoquinone (PQ), cytochrome b6 – f complex, plastocyanin (PC) and finally reaches P700 (PS I).
  3. Electron transport from P700 to NADP: PS I (P700) is excited now and the electrons pass to high energy level. When electron travels downhill through ferredoxin, NADP+ is reduced to NADPH + H+.

Question 5.
Explain chemiosmotic theory with suitable I diagram.
Answer:
Chemiosmotic theory was proposed by P. Mitchell (1966). According to this theory electrons are transported along the membrane through PS I and PS II and connected by Cytochrome b6 – f complex. The flow of electrical current is due to difference in electrochemical potential of protons across the membrane. Splitting of water molecule takes place inside the membrane. Protons or H+ ions accumulate within the lumen of the thylakoid (H+ increase 1000 to 2000 times). As a result, proton concentration is increased inside the thylakoid lumen.

These protons move across the membrane because the primary acceptor of electron is located outside the membrane. Protons in stroma less in number and creates a proton gradient. This gradient is broken down due to the movement of proton across the membrane to the stroma through CFo of the ATP synthase enzyme. The proton motive force created inside the lumen of thylakoid or chemical gradient of H+ ion across the membrane stimulates ATP generation.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 13 Photosynthesis 5
The evolution of one oxygen molecule (4 electrons required) requires 8 quanta of light. C3 plants utilise 3 ATPs and 2 NADPH + H+ to evolve one Oxygen molecule. To evolve 6 molecules of Oxygen 18 ATPs and 12 NADPH + H+ are utilised. C4 plants utilise 5 ATPs and 2 NADPH + H+ to evolve one oxygen molecule. To evolve 6 molecules of Oxygen 30 ATPs and 12 NADPH + H+ are utilised.

Question 6.
Compare and contrast the photosynthetic processes in C3 and C4 plants.
Answer:
Contrast the photosynthetic processes in C3 and C4 plants:
C3 Plants:

  • CO2 fixation takes place in mesophyll cells only.
  • CO2 acceptor is RUBP only.
  • First product is 3C – PGA.
  • Kranz anatomy is not present.
  • Granum is present in mesophyll cells.
  • Normal Chloroplast.
  • Optimum temperature 20° to 25° C.
  • Fixation of CO2 at 50 ppm.
  • Less efficient due to higher photorespiration.
  • RUBP carboxylase enzyme used for fixation.
  • 18 ATPs used to synthesize one glucose.
  • Efficient at low CO2.
  • eg: Paddy, Wheat, Potato and so on.

C4 Plants:

  • CO2 fixation takes place mesophyll and bundle sheath.
  • PEP in mesophyll and RUBP in bundle sheath cells.
  • First product is 4C – OAA.
  • Kranz anatomy is present.
  • Granum present in mesophyll cells and absent in bundle sheath.
  • Dimorphic chloroplast.
  • Optimum temperature 30° to 45° C.
  • Fixation of CO2 even less than 10 ppm.
  • More efficient due to less photorespiration.
  • PEP carboxylase and RUBP carboxylase used.
  • 30 ATPs to produce one glucose.
  • Efficient at higher CO2.
  • eg: Sugar cane, Maize, Sorghum, Amaranthus and so on.

Question 7.
Give the schematic diagram of photorespiration.
Answer:
The schematic diagram of photorespiration:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 13 Photosynthesis 6

Question 8.
Distinguish between photorespiration and dark respiration.
Answer:
Photo respiration:

  • It takes place in photosynthetic green cells.
  • It takes place only in the presence of light.
  • It involves chloroplast, peroxisome and mitochondria.
  • It does not involve Glycolysis, Kreb’s Cycle, and ETS.
  • Substrate is glycolic acid.
  • It is not essential for survival.
  • No phosphorylation and yield of ATP.
  • NADH2 is oxidised to NAD+.
  • Hydrogen peroxide is produced.
  • End products are CO2 and PGA.

Dark respiration:

  • It takes place in all living cells.
  • It takes place all the time.
  • It involves only mitochondria.
  • It involves glycolysis, Kreb’s Cycle and ETS.
  • Substrate is carbohydrates, protein or fats.
  • Essential for survival.
  • Phosphorylation produces ATP energy.
  • NAD+ is reduced to NADH2.
  • Hydrogen peroxide is not produced.
  • End products are CO2 and water.

CHECK YOUR GRASP
Textbook Page No: 123

Question 1.
(i) Name the products produced from Non – Cyclic photophosphorylation?
(ii) Why does PS II require electrons from water?
(iii) Can you find the difference in the Pathway of electrons during PS I and PS II?
Answer:
(i) The products of non-cyclic phosphorylation are NADPH + H+ and ATP.
(ii) The electrons received from water are responsible for the production of ATP and NADPH + H+ through electron transport system in PS I and PS II.
(iii) Yes. Electron flow starts from P680 through a series of electron carrier molecules and finally reaches P700 (PSI). From PS I the electrons travels downhill through ferredoxin, NADP+ is recorded to NADPH + H+.

Samacheer Kalvi 11th Bio Botany Solutions Chapter 9 Tissue and Tissue System

Students can Download Bio Botany Chapter 9 Tissue and Tissue System Questions and Answers, Notes Pdf, Samacheer Kalvi 11th Bio Botany Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Bio Botany Solutions Chapter 9 Tissue and Tissue System

Samacheer Kalvi 11th Bio Botany Tissue and Tissue System Text Book Back Questions and Answers

Question 1.
Refer to the given figure and select the correct statement:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 9 Tissue and Tissue System 4
(i) A, B, and C are histogen of shoot apex
(ii) A Gives rise to medullary rays
(iii) B Gives rise to cortex
(iv) C Gives rise to epidermis
(a) (i) and (ii) only
(b) (ii) and (in) only
(c) (i) and (iii) only
(d) (iii) and (iv) only
Answer:
(c) (i) and (iii) only

Samacheer Kalvi 11th Bio Botany Solutions 9 Tissue and Tissue System

Question 2.
Read the following sentences and identify the correctly matched sentences.
(i) In exarch condition, the protoxylem lies outside of metaxylem.
(ii) In endarch condition, the protoxylem lie towards the centre.
(iii) In centarch condition, metaxylem lies in the middle of the protoxylem.
(iv) In mesarch condition, protoxylem lies in the middle of the metaxylem.
(a) (i), (ii) and (iii) only
(b) (ii), (iii) and (iv) only
(c) (i), (ii) and (iv) only
(d) All of these
Answer:
(c) (i), (ii) and (iv) only

Question 3.
In Gymnosperms, the activity of sieve tubes are controlled by:
(a) Nearby sieve tube members.
(b) Phloem parenchyma cells.
(c) Nucleus of companion cells.
(d) Nucleus of albuminous cells.
Answer:
(c) Nucleus of companion cells.

Samacheer Kalvi 11th Bio Botany Solutions 9 Tissue and Tissue System

Question 4.
When a leaf trace extends from a vascular bundle in a dicot stem, what would be the arrangement of vascular tissues in the veins of the leaf?
(a) Xylem would be on top and the phloem on the bottom
(b) Phloem would be on top and the xylem on the bottom
(c) Xylem would encircle the phloem
(d) Phloem would encircle the xylem
Answer:
(a) Xylem would be on top and the phloem on the bottom

Question 5.
Grafting is successful in dicots but not in monocots because the dicots have:
(a) vascular bundles arranged in a ring
(b) cambium for secondary growth
(c) vessels with elements arranged end to end
(d) cork cambium
Answer:
(b) cambium for secondary growth

Question 6.
Why the cells of sclerenchyma and tracheids become dead?
Answer:
The cells of sclerenchyma and tracheids become dead because they lack protoplasm.

Samacheer Kalvi 11th Bio Botany Solutions 9 Tissue and Tissue System

Question 7.
Explain sclereids with their types.
Answer:
Sclereids are dead cells, usually these are isodiametric but some are elongated too. The cell wall is very thick due to lignification. Lumen is very much reduced. The pits may simple or branched. Sclereids are mechanical in function. They give hard texture to the seed coats, endosperms etc., Sclereids are classified into the following types.

  1. Branchysclereids or Stone cells: Isodiametric sclereids, with hard cell wall. It is found in bark, pith cortex, hard endosperm and fleshy portion of some fruits. eg: Pulp of Pyrus.
  2. Macrosclereids: Elongated and rod shaped cells, found in the outer seed coat of leguminous plants. eg: Crotalaria and Pisum sativum.
  3. Osteosclereids (Bone cells): Rod shaped with dilated ends. They occur in leaves and seed coats. eg: seed coat of Pisum and Hakea.
  4. Astrosclereids: Star cells with lobes or arms diverging form a central body. They occur in petioles and leaves. eg: Tea, Nymphae and Trochodendron.
  5. Trichosclereids: Hair like thin walled sclereids. Numerous small angular crystals are embedded in the wall of these sclereids, present in stems and leaves of hydrophytes. eg: Nymphaea leaf and Aerial roots of Monstera

Question 8.
What are sieve tubes? Explain.
Answer:
Sieve tubes are long tube like conducting elements in the phloem. These are formed from a series of cells called sieve tube elements. The sieve tube elements are arranged one above the other and form vertical sieve tube. The end wall contains a number of pores and it looks like a sieve. So it is called as sieve plate. The sieve elements show nacreous thickenings on their lateral walls. They may possess simple or compound sieve plates.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 9 Tissue and Tissue System 1
The function of sieve tubes are believed to be controlled by campanion cells In mature sieve tube, Nucleus is absent. It contains a lining layer of cytoplasm. A special protein (P. Protein = Phloem Protein) called slime body is seen in it. In mature sieve tubes, the pores in the sieve plate are blocked by a substance called callose (callose plug). The conduction of food material takes lace through cytoplasmic strands. Sieve tubes occur only in Angiosperms.

Question 9.
Distinguish the anatomy of dicot root from monocot root.
Answer:
The anatomy of dicot root from monocot root:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 9 Tissue and Tissue System 2

Question 10.
Distinguish the anatomy of dicot stem from monocot stem.
Answer:
The anatomy of dicot stem from monocot stem:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 9 Tissue and Tissue System 3

Samacheer Kalvi 11th Bio Botany Tissue and Tissue System Other Important Questions & Answers

I. Choose the correct answer. (1 Mark)
Question 1.
Who is the father of plant anatomy?
(a) David Muller
(b) Katherine Esau
(c) Nehemiah Grew
(d) Hofmeister
Answer:
(c) Nehemiah Grew

Question 2.
The study of internal structure and organisation of plant is called:
(a) plant taxonomy
(b) plant anatomy
(c) plant physiology
(d) plant ecology
Answer:
(b) plant anatomy

Samacheer Kalvi 11th Bio Botany Solutions 9 Tissue and Tissue System

Question 3.
The book “Anatomy of seed plants” is written by:
(a) Hanstein
(b) Schmidt
(c) Nicholsen
(d) Katherine Esau
Answer:
(d) Katherine Esau

Question 4.
The term meristem is coined by:
(a) Nageli
(b) Robert
(c) Stevers
(d) Clowes
Answer:
(a) Nageli

Question 5.
Which of the statement is not correct?
(a) Meristematic cells are self perpetuating
(b) Meristematic cells are most actively dividing cells
(c) Meristematic cells have large vacuoles
(d) Meristematic cells have dense cytoplasm with prominent nucleus
Answer:
(c) Meristematic cells have large vacuoles

Question 6.
Apical cell theory is proposed by:
(a) David brown
(b) Hofmeister
(c) Land mark
(d) Clowes
Answer:
(b) Hofmeister

Question 7.
The tunica is:
(a) the peripheral zone of shoot apex, that forms cortex
(b) the inner zone of shoot apex, that forms stele
(c) the peripheral zone of shoot apex, that forms epidermis
(d) the inner zone of shoot apex, that forms cortex and stele
Answer:
(c) the peripheral zone of shoot apex, that forms epidermis

Question 8.
Which of the histogens gives rise to root cap?
(a) Plerome
(b) Periblem
(c) Dermatogen
(d) Calyptrogen
Answer:
(d) Calyptrogen

Samacheer Kalvi 11th Bio Botany Solutions 9 Tissue and Tissue System

Question 9.
Quiescent centre concept was proposed by:
(a) Lindall
(b) Clowes
(c) Holstein
(d) Sanio
Answer:
(b) Clowes

Question 10.
Parenchyma cells which stores resin, tannins, calcium carbonate and calcium oxalate are termed as:
(a) critoblast
(b) chromoblasts
(c) idioblasts
(d) astroblasts
Answer:
(c) idioblasts

Question 11.
Petioles of banana is composed of:
(a) storage parenchyma
(b) stellate parenchyma
(c) angular collenchyma
(d) prosenchyma
Answer:
(b) stellate parenchyma

Question 12.
Which of the following statement is not correct?
(a) Sclerenchyma is a dead cell
(b) It lacks protoplasm
(c) The cell walls of these cells are uniformly thickened
(d) Sclerenchyma are actively dividing cells
Answer:
(d) Sclerenchyma are actively dividing cells

Samacheer Kalvi 11th Bio Botany Solutions 9 Tissue and Tissue System

Question 13.
The seed coat of ground nut is made up of:
(a) stone cells
(b) osteosclereids
(c) macrosclereids
(d) parenchyma cells
Answer:
(b) osteosclereids

Question 14.
Plant fibers are modified:
(a) sclerenchyma cells
(b) collenchyma cells
(c) parenchyma cells
(d) none of the above
Answer:
(a) sclerenchyma cells

Question 15.
The term xylem was introduced by:
(a) Alexander
(b) Nageli
(c) Holstein
(d) Schemidt
Answer:
(b) Nageli

Samacheer Kalvi 11th Bio Botany Solutions 9 Tissue and Tissue System

Question 16.
What type of xylem arrangement is seen in Selaginella sp?
(a) Endarch
(b) Exarch
(c) Centrarch
(d) Mesarch
Answer:
(c) Centrarch

Question 17.
In cross section, the tracheids are:
(a) hexagonal in shape
(b) rectangular in shape
(c) triangular in shape
(d) polygonal in shape
Answer:
(d) polygonal in shape

Question 18.
In grasses the guard cells in stoma are:
(a) bean shaped
(b) irregular shaped
(c) dumbbell shaped
(d) bell shaped
Answer:
(c) dumbbell shaped

Samacheer Kalvi 11th Bio Botany Solutions 9 Tissue and Tissue System

Question 19.
Bulliform cells are present in:
(a) mango
(b) grasses
(c) ground nut
(d) potato
Answer:
(b) grasses

Question 20.
The sunken stomata:
(a) reduce water loss by transpiration
(b) increase water loss by transpiration
(c) increase heat loss by evaporation
(d) neither reduce nor increase water loss by transpiration
Answer:
(a) reduce water loss by transpiration

Question 21.
In Ocimum the trichomes are:
(a) non – glandular
(b) fibrous
(c) glandular
(d) none of these
Answer:
(c) glandular

Question 22.
In dicot stem, the hypodermis is generally:
(a) parenchymatous
(b) sclerenchymatous
(c) collenchymatous
(d) none of these
Answer:
(c) collenchymatous

Samacheer Kalvi 11th Bio Botany Solutions 9 Tissue and Tissue System

Question 23.
Casparian strips contain thickenings of:
(a) calcium carbonate and calcium oxalate
(b) carbohydrate, protein and lignin
(c) crystal of calcium oxalate
(d) lignin, suberin and some other carbohydrates
Answer:
(d) lignin, suberin and some other carbohydrates

Question 24.
Indicate the correct statement:
(a) Albuminous cells in gymnosperms are a nucleated parenchyma cells.
(b) Albuminous cells in gymnosperms are nucleated collenchyma cells.
(c) Albuminous cells in gymnosperms are nucleated, thin walled parenchyma cells.
(d) Albuminous cells in gymnosperms are a nucleated sclerenchyma cells.
Answer:
(c) Albuminous cells in gymnosperms are nucleated, thin walled parenchyma cells.

Question 25.
Secondary phloem is derived from:
(a) apical meritesm
(b) vascular cambium
(c) primary phloem
(d) none of the above
Answer:
(b) vascular cambium

Samacheer Kalvi 11th Bio Botany Solutions 9 Tissue and Tissue System

Question 26.
Which of the following statement is not correct?
(a) The outer most layer of the root is called piliferous layer.
(b) The chief function of piliferous layer is protection.
(c) Piliferous layer is made up of parenchyma cells with intracellular space.
(d) Piliferous layer is made up of parenchyma cells without intracellular space.
Answer:
(d) Piliferous layer is made up of parenchyma cells without intracellular space.

Question 27.
In beans, the metaxylem vessels are generally:
(a) polygonal in shape
(b) circular in shape
(d) rectangular in shape
(d) triangular in shape
Answer:
(a) polygonal in shape

Question 28.
Who discovered the Annular collenchyma?
(a) Clowes
(b) Sanio
(c) Nageli
(d) Duchaigne
Answer:
(d) Duchaigne

Question 29.
The main function of xylem is:
(a) to conduct the minerals to various parts of plants
(b) to conduct oxygen to various parts of plant body
(c) to conduct water and minerals from root to the other parts of the plant body
(d) to conduct stored food to various parts of plant body
Answer:
(c) to conduct water and minerals from root to the other parts of the plant body

Samacheer Kalvi 11th Bio Botany Solutions 9 Tissue and Tissue System

Question 30.
In maize the vascular bundles are:
(a) scattered
(b) concentric
(c) excentric
(d) radial
Answer:
(a) scattered

Question 31.
stomata in leaves of a plant are used for:
(a) transpiration
(b) transpiration and gas exchange
(c) gas exchange
(d) none of the above
Answer:
(b) transpiration and gas exchange

Question 32.
Which of the statement is not correct?
(a) Palisade parenchyma cells are seen beneath the upper epidermis
(b) Palisade parenchyma cells contain more chloroplasts
(c) Palisade parenchyma cells are irregularly shaped
(d) The function of palisade parenchyma is photosynthesis
Answer:
(c) Palisade parenchyma cells are irregularly shaped

Question 33.
Spongy parenchyma cells are:
(a) irregularly shaped
(b) elongated cylindrical cells
(c) very lightly arranged cells
(d) with more number of chloroplasts than palisade parenchyma
Answer:
(a) irregularly shaped

Question 34.
The main function of spongy parenchyma is:
(a) photosynthesis
(b) exchange of gases
(c) exchange of minerals
(d) water transport
Answer:
(b) exchange of gases

Samacheer Kalvi 11th Bio Botany Solutions 9 Tissue and Tissue System

Question 35.
All mesophyll cells in monocot leaf are nearly:
(a) isodiametric and thick walled
(b) irregular and thick walled
(c) isodiametric and thin walled
(d) irregular and thin willed
Answer:
(c) isodiametric and thin walled

Question 36.
Structurally, hydathodes are modified:
(a) cambium tissue
(b) parenchyma
(c) pith
(d) stomata
Answer:
(d) stomata

Question 37.
Hydathodes occurs in the leaves of:
(a) desert plants
(b) submerged aquatic plants
(c) floating aquatic weeds
(d) forest trees
Answer:
(b) submerged aquatic plants

Question 38.
The process of guttation is seen in:
(a) grasses
(b) dicot plants
(c) desert plants
(d) Nerium
Answer:
(a) grasses

Question 39.
Salt glands are present in:
(a) xerophytes
(b) hydrophytes
(c) halophytes
(d) merophytes
Answer:
(c) halophytes

Question 40.
The term sieve tubes is coined by:
(a) Schleiden
(b) Hanstein
(c) Tsehireh
(d) Hartig
Answer:
(d) Hartig

II. Answer the following. (2 Marks)

Question 1.
Define the tissue?
Answer:
A tissue is a group of cells that are alike in origin, structure and function. The study of tissue is called Histology.

Question 2. What are the different types of plant tissue?
Answer:
The two types of principal groups are:

  1. Meristematic tissues
  2. Permanent tissues.

Question 3.
Mention any, two characters of meriste – matic tissue.
Answer:
Two characters of meriste – matic tissue:

  1. The meristematic cells are isodiametric and they may be, oval, spherical or polygonal in shape.
  2. They have generally dense cytoplasm with prominent nucleus.

Samacheer Kalvi 11th Bio Botany Solutions 9 Tissue and Tissue System

Question 4.
Mention the function of apical meristem.
Answer:
Present in apices of root and shoot. It is responsible for increase in the length of the plant, it is called as primary growth.

Question 5.
What is mean by carpus?
Answer:
It is the inner zone of shoot apex,that forms cortex and stele of shoot.

Question 6.
Explain apical cell theory?
Answer:
Apical cell theory is proposed by Nageli. The single apical cell or apical initial composes the root meristem. The apical initial is tetrahedral in shape and produces root cap from one side. The remaining three sides produce epidermis, cortex and vascular tissues. It is found in vascular cryptogams.

Question 7.
What is meant by angular collenchyma?
Answer:
It is the most common type of collenchyma with irregular arrangement and thickening at the angles where cells meet., eg: Hypodermis of Datum and Nicotiana.

Samacheer Kalvi 11th Bio Botany Solutions 9 Tissue and Tissue System

Question 8.
Explain briefly Branchysciereids or Stone cells.
Answer:
Isodiametric sclereids, with hard cell wall. It is found in bark, pith cortex, hard endosperm and fleshy portion of some fruits. eg: Pulp of Pyrus.

Question 9.
What is Filiform Sclereids?
Answer:
The sclereids are present in the leaf lamina of Olea europaea. They are very much elongated fibre like and about 1mm length.

Question 10.
Distinguish between Libriform fibres and Fibre tracheids.
Answer:
Between Libriform fibres and Fibre tracheids:

Libriform fibres

Fibre tracheids

1. These fibres have slightly lignified secondary walls with simple pits. These fibres are long and narrow. 1. These are shorter than the libriform fibres with moderate secondary thickenings in the cell walls. Pits are simple or bordered.

Question 11.
What are bast fibres?
Answer:
These fibres are present in the phloem. Natural Bast fibres are strong and cellulosic. Fibres obtaining from the phloem or outer bark of jute, kenaf, flax and hemp plants. The so called pericyclic fibres are actually phloem fibres.

Samacheer Kalvi 11th Bio Botany Solutions 9 Tissue and Tissue System

Question 12.
What is meant by endarch type of xylem arrangements?
Answer:
Protoxylem lies towards the .centre and meta xylem towards the periphery this condition is called Endarch. It is seen in stems.

Question 13.
What are the types of cells present in phloem?
Answer:
Phloem consists of four types of Cells:

  1. Sieve elements
  2. Companion cells
  3. Phloem parenchyma
  4. Phloem fibres.

Question 14.
Define epiblema?
Answer:
It is made up of single layer of parenchyma cells which are arranged compactly without intercellular spaces. It is devoid of epidermal pores and cuticle. Root hair is always single celled, it absorbs water and mineral salts from the soil. The another important function of piliferous layer is protection.

Question 15.
Explain bulliform cells in grasses.
Answer:
Some cells of upper epidermis (eg: Grasses) are larger and thin walled. They are called bulliform cells or motor cells. These cells are helpful for the rolling and unrolling of the leaf according to the weather change.

Question 16.
What is meant by Sunken Stomata?
Answer:
In some Xerophytic plants (eg: Cycas, Nerium), stomata is sunken beneath the abaxial leaf surface within stomatal crypts. The sunken stomata reduce water loss by transpiration.

Samacheer Kalvi 11th Bio Botany Solutions 9 Tissue and Tissue System

Question 1 7.
Mention any two functions of epidermal tissue system in plants.
Answer:
Two functions of epidermal tissue system in plants:

  1. This system in the shoot checks excessive loss of water due to the presence of cuticle.
  2. Epidermis protects the underlying tissues.

Question 18.
Define chlorenchymo?
Answer:
Sometimes in young stem, chloroplasts develop in peripheral cortical cells, which is Called chlorenchyma.

Question 19.
What is meant by casparian strips?
Answer:
In true root endodermis, radial and inner tangential walls of endodermal cells possess thickenings of lignin, suberin and some other carbohydrates in the form of strips they are called casparian strips.

Question 20.
What are albuminous cells?
Answer:
The cytoplasmic nucleated parenchyma, is associated with the sieve cells of Gymnosperms. Albuminous cells in Conifers are analogous to companion cells of Angiosperms. It also called as strasburger cells.

Question 21.
Describe briefly radial types of vascular Bundles.
Answer:
Xylem and phloem are present on different radii alternating with each other. The bundles are separated by parenchymatous tissue. (Monocot and Dicot roots).

Samacheer Kalvi 11th Bio Botany Solutions 9 Tissue and Tissue System

Question 22.
Define stele?
Answer:
All the tissues inside the endodermis comprise the stele. This includes pericycle, vascular system and pith.

Question 23.
What is meant by cambium?
Answer:
Cambium consists of brick shaped and thin walled meristematic cells. It is one to four layers in thickness. These cells are capable of forming new cells during secondary growth.

Question 24.
Define silica Cells?
Answer:
Some of the epidermal cells of the grass are filled with silica. They are called silica cells.

Question 25.
Define, hydathode?
Answer:
A hydathode is a type of epidermal pore, commonly found in higher plants. Structurally, hydathodes are modified stomata, usually located at leaf tips or margins, especially at the teeth. Hydathodes occur in the leaves of submerged aquatic plants such as ranunculus fluitans as well as in many herbaceous land plants.

III. Answer the following. (3 Marks)

Question 1.
Explain apical cell theory.
Answer:
Apical cell theory is proposed by Hofmeister (1852) and supported by Nageli (1859). A single apical cell is the structural and functional unit. This apical cell governs the growth and development of whole plant body. It is applicable in Algae, Bryophytes and in some Pteridophytes.

Samacheer Kalvi 11th Bio Botany Solutions 9 Tissue and Tissue System

Question 2.
Explain histogen theory.
Answer:
Histogen theory is proposed by Hanstein (1868) and supported by Strassburgur. The shoot apex comprises three distinct zones.

  1. Dermatogen: It is a outermost layer. It gives rise to epidermis.
  2. Periblem: It is a middle layer. It gives rise to cortex.
  3. Plerome: It is innermost layer. It gives rise to stele.

Question 3.
What is meant by quiescent centre concept?
Answer:
Quiescent centre concept was proposed by Clowes (1961) to explain root apical meristem activity. These centre is located between root cap and differentiating cells of the roots. The apparently inactive region of cells in root promeristem is called quiescent centre. It is the site of hormone synthesis and also the ultimate source of all meristematic cells of the meristem.

Question 4.
Explain the term “sclereids”.
Answer:
Sclereids are dead cells, usually these are isodiametric but some are elongated too. The cell wall is very thick due to lignification. Lumen is very much reduced. The pits may simple or branched. Sclereids are mechanical in function. They give hard texture to the seed coats, endosperms etc.

Question 5.
Explain briefly about plant fibres.
Answer:
Fibres are very much elongated sclerenchyma cells with pointed tips. Fibres are dead cells and have lignified walls with narrow lumen. They have simple pits. They provide mechanical strength and protect them from the strong wind. It is also called supporting tissues. Fibres have a great commercial value in cottage and textile industries.

Question 6.
Write briefly about xylem fibres.
Answer:
The fibres of sclerenchyma associated with the xylem are known as xylem fibres. Xylem fibres are dead cells and have lignified walls with narrow lumen. They cannot conduct water but being stronger provide mechanical strength. They are present in both primary and secondary xylem. Xylem fibres are also called libriform fibres.
The fibres are abundantly found in many plants. They occur in patches, in continuous bands and sometimes singly among other cells. Between fibres and normal tracheids, there are many transitional forms which are neither typical fibres nor typical tracheids. The transitional types are designated as fibre – tracheids. The pits of fibre – tracheids are smaller than those of vessels and typical tracheids.

Samacheer Kalvi 11th Bio Botany Solutions 9 Tissue and Tissue System

Question 7.
Explain companion cells.
Answer:
The thin walled, elongated, specialized parenchyma cells, which are associated with die sieve elements, are called companion cells. These cells are living and they have cytoplasm and a prominent nucleus. They are connected to the sieve tubes through pits found in the lateral walls. Through these pits cytoplasmic connections are maintained between these elements. These cells are helpful in maintaining the pressure gradient in the sieve tubes. Usually the nuclei of the companion cells serve for the nuclei of sieve tubes as they lack them. The companion cells are present only in Angiosperms and absent in Gymnosperms and Pteridophytes. They assist the sieve tubes in the conduction of food materials.

Question 8.
Distinguish between meristematic tissue and permanent tissue.
Answer:
Between meristematic tissue and permanent tissue:

Meristematic tissue

Permanent tissue

1. Cells divide repeatedly 1. Do not divide
2. Cells are undifferentiated 2. Cells are fully differentiated
3. Cells are small and Isodiametric 3. Cells are variable in shape and size
4. Intercellular spaces are absent 4. Intercellular spaces are present
5. Vacuoles are absent 5. Vacuoles are present
6. Cell walls are thin 6. Cell walls maybe thick or thin
7. Inorganic inclusions are absent 7. Inorganic inclusions  are present

Question 9.
Write down the differences between tracheids and fibres.
Answer:
The differences between tracheids and fibres:

Tracheids

Fibres

1. Not much elongated 1. Very long cells
2. Posses oblique end walls 2. Posses tapering end walls
3. Cell walls are not as thick as fibtres 3. Cell wall are thick and lignified
4. Possess various types of thickenings 4. Possess only pitted thickenings
5. Responsible for the conduction and also mechanical support 5. Provide only mechanical support

Question 10.
Give a brief answer on subsidiary cells in plant leaves.
Answer:
Stomata are minute pores surrounded by two guard cells. The stomata occur mainly in the epidermis of leaves. In some plants addition to guard cells, specialised epidermal cells are present which are distinct from other epidermal cells. They are called Subsidiary cells. Based on the number and arrangement of subsidiary cells around the guard cells, the various types of stomata are recognised, The guard cells and subsidiary cells help in opening and closing of stomata during gaseous exchange and transpiration.

Samacheer Kalvi 11th Bio Botany Solutions 9 Tissue and Tissue System

Question 11.
Explain the term trichomes.
Answer:
There are many types of epidermal outgrowths in stems. The unicellular or multicellular appendages that originate from the epidermal cells are called trichomes. Trichomes may be branched or unbranched and one or more one celled thick. They assume many shapes and sizes. They may also be glandular (eg: Rose, Ocimum) or non – glandular.

Question 12.
What do you understand about hypodermis in plant tissue system.
Answer:
One or two layers of continuous or discontinuous tissue present below the epidermis, is called hypodermis. It is protective in function. In dicot stem, hypodermis is generally collenchymatous, whereas in monocot stem, it is generally sclerenchymatous. In many plants collenchyma form the hypodermis.

Question 13.
What is meant by pith?’
Answer:
The central part of the ground tissue is known as pith or medulla. Generally this is made up of thin walled parenchyma cells with intercellular spaces. The cells in the pith generally stores starch, fatty substances, tannins, phenols, calcium oxalate crystals, etc.

Question 14.
Explain the piliferous layer as epiblema.
Answer:
The outermost layer of the root is known as piliferous layer. It consists of single row of thin – walled parenchymatous cells without any intercellular space. Epidermal pores and cuticle are absent in the piliferous layer. Root hairs that are found in the piliferous layer are always unicellular. They absorb waer and mineral salt from the soil. Root hairs are generally short lived. The main function of piliferous layer is protection of the inner tissues.

Samacheer Kalvi 11th Bio Botany Solutions 9 Tissue and Tissue System

Question 15.
What is meant by stele in plant stem?
Answer:
The central part of the stem inner to the endodermis is known as stele. It consists of pericyle, vascular bundles and pith. In dicot stem, vascular bundles are arranged in a ring around the pith. This type of stele is called eustele.

Question 16.
Explain the nature of phloem in dicot stem.
Answer:
Primary phloem lies towards the periphery. It consists of protpphloem and metaphloem. Phloem consists of sieve tubes, companion cells and phloem parenchyma. Phloem fibres are absent in primary phloem. Phloem conduct organic foods material from the leaves to other parts of the plant body.

Question 17.
Explain the mesophyll layer of leaf.
Answer:
The ground tissue that is present between the upper and lower epidermis of the leaf is called mesophyll. Here, the mesophyll is not differentiated into palisade and spongy parenchyma. All the mesophyll cells are nearly isodiametric and thin walled. These cells are compactly arranged with limited intercellular spaces. They contain numerous chloroplasts.

Samacheer Kalvi 11th Bio Botany Solutions 9 Tissue and Tissue System

Question 18.
Mention any three differences between stomata and hydathodes.
Answer:
Stomata:

  1. Occur in epidermis of leaves, young stems.
  2. Stomatal aperture is guarded by two guard cells.
  3. The two guard cells are generally surrounded by subsidiary cell.

Hydathodes:

  1. Occur at the tip or margin of leaves that are grown in moist shady place.
  2. Aperture of hydathodes are surrounded by a ring of cuticularized cells.
  3. Subsidiary cells are absent.,

Question 19.
What are halophiles? Explain briefly.
Answer:
Halophiles:

  1. Plants that grow in salty environment are called Halophiles.
  2. Plant growth in saline habitat developed numerous adaptations to salt stress. The secretion of ions by salt glands is the best known mechanism for regulating the salt content of plant shoots.
  3. Salt glands typically are found in halophytes. (Plants that grow in saline environments)

IV. Answer in detail.

Question 1.
Explain Histogen theory, Korper Kappe Theory and Quiescent Centre Concept with diagrams.
Answer:
Histogen Theory: Histogen theory is proposed by Hanstein (1868) and supported by Strassburgur. The histogen theory as appilied to the root apical meristem speaks of four histogen in the meristem. They are respectively

  1. Dermatogen: It is a outermost layer. It gives rise to root epidermis.
  2. Periblem: it is a middle layer. It gives rise to cortex.
  3. Plerome: It is innermost layer. It gives rise to stele.
  4. Calyptrogen: it gives rise to root cap.
    Samacheer Kalvi 11th Bio Botany Solutions Chapter 9 Tissue and Tissue System 9

Korper Kappe Theory: Korper kappe theory is proposed by Schuepp. There are two zones in root apex – Korper and Kappe.

  1. Korper zone forms the body.
  2. Kappe zone forms The cap.

This theory is equivalent to tunica corpus theory of shoot apex.The two divisions are distinguished by the type of T (also called Y divisions). Korper is characterised by inverted T division and kappe by straight T divisions.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 9 Tissue and Tissue System 10

Quiescent Centre Concept: Quiescent centre concept was proposed by Clowes (1961) to explain root apical meristem activity. These centre is located between root cap and differentiating cells of the roots. The apparently inactive region of cells in root promeristem is called quiescent centre. It is the site of hormone synthesis and also the ultimate source of all meristematic cells of the meristem.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 9 Tissue and Tissue System 11

Question 2.
Describe the structure and function of different kinds of parenchyma tissues?
Answer:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 9 Tissue and Tissue System 12
Parenchyma is generally present in all organs of the plant. It forms the ground tissue in a plant. Parenchyma is a living tissue and made up of thin walled cells. The cell wall is made up of cellulose. Parenchyma cells may be oval, polyhedral, cylindrical, irregular, elongated or armed. Parenchyma tissue normally has prominent intercellular spaces. Parenchyma may store various types of materials like, water, air, ergastic substances. It is usually colourless. The turgid parenchyma cells help in giving rigidity to the plant body. Partial conduction of water is also maintained through parenchymatous cells. Occsionaliy Parenchyma cells which store resin, tannins, crystals of calcium carbonate, calcium oxalate are called idioblasts. Parenchyma is of different types and some of them are discussed as follows. Types of paranchyma:
(i) Aerenchyma: Parenchyma which contains air in its intercellular spaces. It helps in aeration and buoyancy. eg: Nymphae and Hydrilia.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 9 Tissue and Tissue System 13

(ii) Storage Parenchyma: parenchyma stores food materials. eg: Root and stem tubers.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 9 Tissue and Tissue System 14

(iii) Stellate Parenchyma: Star shaped parenchyma. eg: Petioles of Banana and Canna.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 9 Tissue and Tissue System 15

(iv) Chlorenchyma: Parenchyma cells with chlorophyll. Function is photosynthesis, eg: Mesophyll of leaves.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 9 Tissue and Tissue System 16

(v) Prosenchyma: parenchyma cells became elongated, pointed and slightly thick walled. It provides mechanical support.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 9 Tissue and Tissue System 17

Question 3.
Describe the types of tracheids with diagram.
Answer:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 9 Tissue and Tissue System 18
Types of secondary wall thickenings in tracheids and vessels:
Tracheids are dead, lignified and elongated cells with tapering ends. Its lumen is broader than that of fibres. In cross section, the tracheids are polygonal. There are different types of cell wall thickenings due to the deposition of secondary wall substances. They are annular (ring like), spiral (spring like), scalariform (ladder like) reticulate (net like) and pitted (uniformly thick except at pits). Tracheids are imperforated cells with bordered pits on their side walls. Only through this conduction takes place in Gymnosperms. They are arranged one above the other. Tracheids are chief water conducting elements in Gymnosperms and Pteridophytes. They also offer mechanical support to the plants.

Question 4.
Compare the different types of plant tissues.
Answer:
The different types of plant tissues:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 9 Tissue and Tissue System 5

Question 5.
Compare the vascular tissues of plant.
Answer:
The vascular tissues of plant:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 9 Tissue and Tissue System 6

Question 6.
Draw and label the various parts of T.S. of dicot root.
Answer:
Draw and label the various parts of T.S. of dicot root:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 9 Tissue and Tissue System 8

Question 7.
Explain in detail about the vascular bundles of monocot stem.
Answer:
1. Vascular bundles: Vascular bundles are scattered (atactostele) in the parenchyma ground tissue. Each vascular bundle is surrounded by a sheath of sclerenehymatous fibres called bundle sheath. The vascular bundles are conjoint, collateral, endarch and closed. Vascular bundles are numerous, small and closely arranged in the peripheral portion. Towards the centre, the bundles are comparatively large in size and loosely arranged. Vascular bundles are skull or oval shaped.

2. Phloem: The phloem in the monocot stem consists of sieve tubes and companion cells. Phloem parenchyma and phloem fibres are absent. It can be distinguished into an outer crushed protophloem and an inner metaphloem.

3. Xylem: Xylem vessels are arranged in the form of ‘Y’ the two metaxylem vessels at the base. In a mature bundle, the lowest protowylem disintegrates and forms a cavity known as protoxylem lacuna.

Question 8.
Draw and label the various parts of monocot stem.
Answer:
The various parts of monocot stem:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 9 Tissue and Tissue System 7

Question 9.
Explain the various parts of sunflower leaf with neat diagram.
Answer:
1. Anatomy of a Dicot Leaf – sunflower Leaf: Internal structure of dicotyledonous leaves reveal epidermis, mesophyll and vascular tissues.

2. Epidermis: This leaf is generally dorsiventral. It has upper and lower epidermis. The epidermis is usually made up of a single layer of cells that are closely packed. The cuticle on the upper epidermis is thicker than that of lower epidermis. The minute opening found on the epidermis are called stomata. Stomata are more in number on the lower epidermis than on the upper epidermis.

A stomata is surrounded by a pair of bean shaped cells are called guard cells. Each stoma internally opens into an air chamber. These guard cells contain chlotroplasts. The main function of epidermis is to give protection to the inner tissue called mesospyll. The cuticle helps to check transpiration. Stomata are used for transpiration and gas exchange.

3. Mesophyll: The entire tissue between the upper and lower epidermis is called mesophyll (GK meso = in the middle, phyllome = leaf). There are two regions in the mesophyll. They are palisade parenchyma and spongy parenchyma. Palisade parenchyma cells are seen beneath the upper epidermis. It consists of vertically elongated cylindrical cells in one or more layers. These are compactly arranged and are generally without intercellular spaces. Palisade parenchyma cells contain more chloroplasts than the spongy parenchyma cells.

The function of palisade parenchyma is photosynthesis. Spongy parenchyma lies below the palsied parenchyma. Spongy cells are irregularly shaped. These cells are very loosly arranged with numerous airspaces. As compared to palisade cells, the spongy cells contain number of chloroplasts. Spongy cells facilitate the exchange of gases with the help of air spaces. The air space that is found next to the stomata is called respiratory cavity or substomatal cavity. Å

4. Vascular tissue: Vascular tissue are present in the veins of leaf. Vascular bundles are conjoint collateral and closed. Xylem is present towards the upper epidermis, while the phloem towards the lower epidermis. Vascular bundles are surrounded by a compact layer by a parenchymatous cells called bundle sheath or border parenchyma.

Xylem consists of metaxylem and protoxylem elements. Protoxylem is present towards the upper epidermis, while the phloem consists of sieve tubes, companion cells and phloem parenchyma. Phloem fibres are absent. Xylem sonsists of vessels and xylem parenchyma. Tracheids and xylem fibres are absent.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 9 Tissue and Tissue System 19
Solution To Activity
Text Book Page No. 10
Question 1.
Cell lab: Students prepare the slide and identify the different types tissues.
Answer:
Preparing a slide of plant tissue.
Objective:

  1. Using hand cutting method to make thin slice of dicot root.
  2. To make slide and stain of plant sample.
  3. To observe the plant sample under microscope.

Materials:

  1. A young dicot root
  2. Compound microscope
  3. Slide
  4. Cover slip
  5. Eosin stain

Method:

  1. Place 2 cm of young dicot root on a glass slide or plate.
  2. Cut thin slices of the root through the region of maturation.
  3. Stain it with Eosin.
  4. Fix one or two of the sections in a slide and put a cover slip.
  5. To observe the sample under a compound microscope and record the parts of the sample.

Samacheer Kalvi 11th Bio Botany Solutions Chapter 4 Reproductive Morphology

Students can Download Bio Botany Chapter 4 Reproductive Morphology Questions and Answers, Notes Pdf, Samacheer Kalvi 11th Bio Botany Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Bio Botany Solutions Chapter 4 Reproductive Morphology

Samacheer Kalvi 11th Bio Botany Reproductive Morphology Text Book Back Questions and Answers

Choose the correct answer
Question 1.
Vexillary aestivation is characteristic of the family …………… .
(a) Fabaceae
(b) Asteraceae
(c) Solanaceae
(d) Brassicaceae
Answer:
(a) Fabaceae

Samacheer Kalvi 11th Bio Botany Solutions 4 Reproductive Morphology

Question 2.
Gynoecium with united carples is termed as …………… .
(a) apocarpous
(b) multicarpellary
(c) syncarpous
(d) none of the above
Answer:
(c) syncarpous

Question 3.
Aggregate fruit develops from …………… .
(a) multicarpellary, apocarpous ovary
(b) multicarpellary, syncarpous ovary
(c) multicarpellary ovary
(d) whole inflorescence
Answer:
(a) multicarpellary, apocarpous ovary

Question 4.
In an inflorescence where flowers are borne laterally in an acropetal succession the position of the youngest floral bud shall be …………… .
(a) proximal
(b) distal
(c) intercalary
(d) anywhere
Answer:
(b) distal

Samacheer Kalvi 11th Bio Botany Solutions 4 Reproductive Morphology

Question 5.
A true fruit is the one where …………… .
(a) only ovary of the flower develops into fruit
(b) ovary and calyx of the flower develops into fruit
(c) ovary, calyx and thalamus of the flower develops into fruit
(d) all floral whorls of the flower develops into fruit
Answer:
(a) only ovary of the flower develops into fruit

Question 6.
Find out the floral formula for a bisexual flower with bract, regular, pentamerous, distinct calyx and corolla, superior ovary without bracteole?
Answer:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 4 Reproductive Morphology 1

Question 7.
Give the technical terms for the following:
(a) A sterile stamen
(b) Stamens are united in one bunch
(c) Stamens are attached to the petals
Answer:
(a) A sterile stamen – Staminode
(b) Stamens are united in one bunch – Monadelphous
(c) Stamens are attached to the petals – Epipetalous (petalostemonous)

Samacheer Kalvi 11th Bio Botany Solutions 4 Reproductive Morphology

Question 8.
Explain the different types of placentation with example.
Answer:
The different types of placentation with example:

  1. Marginal: It is with the placentae along the margin of a unicarpellate ovary. Example: Fabaceae.
  2. Axile: The placentae arises from the column in a compound ovary with septa. Example: Hibiscus, tomato and lemon.
  3. Superficial: Ovules arise from the surface of the septa. Example: Nymphaeceae.
  4. Parietal: It is the placentae on the ovary walls or upon intruding partitions of a unilocular, compound ovary. Example: Mustard, argemone and cucumber.
  5. Free – central: It is with the placentae along the column in a compound ovary without septa. Example: Caryophyllaceae, Dianthus and primrose.
  6. Basal: It is the placenta at the base of the ovary. Example: Sunflower (Asteraceae) Marigold.

Question 9.
Differentiate between aggregate fruit with multiple fruit.
Answer:
1. Aggregate fruit:
Aggregate fruits develop from a single flower having an apocarpous pistil. Each of the free carpel is developed into a simple fruitlet. A collection of simple fruitlets makes an aggregate fruit. An individual ovary develops into a drupe, achene, follicle or berry. An aggregate of these fruits borne by a single flower is known as an etaerio. Example: Magnolia, Raspberry, Annona and Polyalthia.

2. Multiple or Composite fruit: A multiple or composite fruit develops from the whole inflorescence along with its peduncle on which they are borne.

  • Sorosis: A fleshy multiple fruit which develops from a spike or spadix. The flowers fused together by their succulent perianth and at the same time the axis bearing them become fleshy or juicy and the whole inflorescence forms a compact mass. Example: Pineapple, Jack fruit and Mulberry.
  • Syconus: A multiple fruit which develops from hypanthodium inflorescence. The receptacle develops further and converts into fleshy fruit which encloses a number of true fruit or achenes which develops from female flower of hypanthodium inflorescence. Example: Ficus.

Samacheer Kalvi 11th Bio Botany Solutions 4 Reproductive Morphology

Question 10.
Explain the different types of fleshy fruit with suitable example?
Answer:
The fleshy fruits are derived from single pistil, where the pericarp is fleshy, succulent and differentiated into epicarp, mesocarp and endocarp. It is subdivided into the following:

  1. Berry: Fruit develops from bicarpellary or multicarpellary, syncarpous ovary. Here the epicarp is thin, the mesocarp and endocarp remain undifferentiated. They form a pulp in which the seeds are embedded. Example: Tomato, date palm, grapes and brinjal.
  2. Drupe: Fruit develops from monocarpellary, superior ovary. It is usually one seeded. Pericarp is differentiated into outer skinny epicarp, fleshy and pulpy mesocarp and hard and stony endocarp around the seed. Example: Mango and coconut.
  3. Pepo: Fruit develops from tricarpellary inferior ovary. Pericarp terns leathery or woody which encloses, fleshy mesocarp and smooth endocarp. Example: Cucumber, watermelon, bottle gourd and pumpkin.
  4. Hesperidium: Fruit develops from multicarpellary, multilocular, syncarpous, superior ovary. The fruit wall is differentiated into leathery epicarp with oil glands, a middle fibrous mesocarp. The endocarp forms distinct chambers, containing juicy hairs. Example: Orange and lemon.
  5. Pome: It develops from multicarpellary, syncarpous, inferior ovary. The receptacle also develops along with the ovary and becomes fleshy, enclosing the true fruit. In pome the epicarp is thin skin like and endocarp is cartilagenous. Example: Apple and pear.
  6. Balausta: A fleshy indehiscent fruit developing from multicarpellary, multilocular inferior ovary whose pericarp is tough and leathery. Seeds are attached irregularly with testa being the edible portion. Example: Pomegranate.

Textbook Activity Solved

Prepare a diet chart to provide balanced diet to an adolescent (a school going child) which includes food items (fruits, vegetable and seeds) which are non – expensive and are commonly available.
Diet Chart for an Adolescent:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 4 Reproductive Morphology 2

Samacheer Kalvi 11th Bio Botany Reproductive Morphology Additional Questions and Answers

I. Multiple Choice Questions
Choose the correct answer:
Question 1.
Placentation in tomato and lemon is …………… .
(a) parietal
(b) marginal
(c) free – central
(d) axile
Answer:
(d) axile

Question 2.
The coconut water and the edible part of coconut are equivalent to …………… .
(a) endosperm
(b) endocarp
(c) mesocarp
(d) embryo
Answer:
(a) endosperm

Samacheer Kalvi 11th Bio Botany Solutions 4 Reproductive Morphology

Question 3.
Geocarpic fruits are seen in …………… .
(a) carrot
(b) groundnut
(c) radish
(d) turnip
Answer:
(b) groundnut

Question 4.
Keel is characteristic petal of the flowers of …………… .
(a) Gulmohar
(b) Cassia
(c) Calotropis
(d) Bean
Answer:
(d) Bean

Question 5.
When the calyx is coloured and showy, it is called …………… .
(a) petaloid
(b) sepaloid
(c) bract
(d) spathe
Answer:
(a) petaloid

Question 6.
Bracts are modified leaves which bear flowers in their axils. Identify the plant which has a large showy brightly coloured bract …………… .
(a) Jasmine
(b) Euphorbia
(c) Hibiscus
(d) Bougainvillea
Answer:
(d) Bougainvillea

Samacheer Kalvi 11th Bio Botany Solutions 4 Reproductive Morphology

Question 7.
A flower which can be divided into equal vertical halves, by more than one plane of division is …………… .
(a) zygomorphic
(b) cyclic
(c) actinomorphic
(d) heteromorphic
Answer:
(c) actinomorphic

Question 8.
In Theobroma cocoa, the inflorescence arise from …………… .
(a) terminal shoot
(b) axillary part
(c) trunk of plant
(d) leaf node
Answer:
(c) trunk of plant

Question 9.
The type of inflorescence seen in Caesalpinia is …………… .
(a) corymb
(b) compound corymb
(c) capitulum
(d) umbel
Answer:
(a) corymb

Question 10.
Head is the characteristic of …………… family.
(a) Fabaceae
(b) Malvaceae
(c) Asteraceae
(d) Solanaceae
Answer:
(c) Asteraceae

Question 11.
Thyrsus is a type of …………… inflorescence.
(a) raceme
(b) cyme
(c) mixed
(d) special
Answer:
(c) mixed

Samacheer Kalvi 11th Bio Botany Solutions 4 Reproductive Morphology

Question 12.
Number of whorls in a complete flower is …………… .
(a) one
(b) two
(c) three
(d) four
Answer:
(d) four

Question 13.
Monoclinous flower will have …………… .
(a) androecium
(b) gynoecium
(c) both androecium & gynoecium
(d) none
Answer:
(c) both androecium & gynoecium

Question 14.
If unisexual and bisexual flowers are seen in same plant then the plant is said to be …………… .
(a) polyphyllous
(b) polygamous
(c) hermaphroditic
(d) dioecious
Answer:
(b) polygamous

Question 15.
…………… is a raceme of cymes.
(a) Verticil
(b) Cyathium
(c) Umbel
(d) Thyrsus
Answer:
(d) Thyrsus

Question 16.
Unit of perianth is …………… .
(a) petal
(b) sepal
(c) tepal
(d) stamen
Answer:
(c) tepal

Question 17.
Number of floral parts per whorl is called …………… .
(a) curosity
(b) atrocity
(c) merosity
(d) porosity
Answer:
(c) merosity

Samacheer Kalvi 11th Bio Botany Solutions 4 Reproductive Morphology

Question 18.
What is the green cap – like part of brinjal fruit?
(a) Corolla
(b) Perianth
(c) Calyx
(d) Pistil
Answer:
(c) Calyx

Question 19.
Butterfly shaped corolla is seen in …………… type.
(a) rosaceous
(b) caryophyllaceous
(c) cruciform
(d) papilionaceous
Answer:
(d) papilionaceous

Question 20.
Inflorescence seen in Daucas carota is …………… .
(a) umbel
(b) corymb
(c) compound umbel
(d) compound corymb
Answer:
(c) compound umbel

Question 21.
Arrangement of sepals and petals in flower bud is called …………… .
(a) adhesion
(b) aestivation
(c) placentation
(d) cohesion
Answer:
(b) aestivation

Samacheer Kalvi 11th Bio Botany Solutions 4 Reproductive Morphology

Question 22.
Which is not a part of pistil?
(a) Style
(b) Stigma
(c) Connective tissue
(d) carpel
Answer:
(c) Connective tissue

Question 23.
The type of calyx in brinjal is …………… .
(a) caducous
(b) deciduous
(c) persistent
(d) fugacious
Answer:
(c) persistent

Question 24.
Sterile stamen is called …………… .
(a) pistillode
(b) sessile
(c) staminode
(d) apostamen
Answer:
(c) staminode

Samacheer Kalvi 11th Bio Botany Solutions 4 Reproductive Morphology

Question 25.
Other name for gynoecium is …………… .
(a) carpel
(b) pistil
(c) style
(d) overy
Answer:
(b) pistil

Question 26.
Cavity found inside the ovary is called …………… .
(a) lobule
(b) locule
(c) lacuna
(d) labium
Answer:
(b) locule

Question 27.
Which part of saffron flower is used as flavouring agent?
(a) Carpel
(b) Anther
(c) Style
(d) Stigma
Answer:
(d) Stigma

Question 28.
If the ovary is inferior, then the flower is …………… .
(a) hypogynous
(b) epigynous
(c) perigynous
(d) epihypogynous
Answer:
(b) epigynous

Samacheer Kalvi 11th Bio Botany Solutions 4 Reproductive Morphology

Question 29.
Fabaceae members show …………… placentation.
(a) basal
(b) parietal
(c) superficial
(d) marginal
Answer:
(d) marginal

Question 30.
The side of the flower facing mother axis is called as …………… side.
(a) anterior
(b) posterior
(c) lateral
(d) dorsi – ventral
Answer:
(b) posterior

Question 31.
Which of the following option represents calyx?
(a) C
(b) Ca
(c) K
(d) Ka
Answer:
(c) K

Question 32.
…………… are the products of pollination & fertilization.
(a) Seeds
(b) Ovules
(c) Fruits
(d) Vegetables
Answer:
(c) Fruits

Samacheer Kalvi 11th Bio Botany Solutions 4 Reproductive Morphology

Question 33.
Study of fruits is called as …………… .
(a) honology
(b) pomology
(c) horticulture
(d) apology
Answer:
(b) pomology

Question 34.
Fruit wall can also be called as …………… .
(a) endocarp
(b) epicarp
(c) pericarp
(d) mericorp
Answer:
(c) pericarp

Question 35.
Which part of the apple fruit does we eat?
(a) Perianth
(b) Involucre
(c) Thalamus
(d) Bracteole
Answer:
(c) Thalamus

Question 36.
The false septum seen in siliqua fruits is …………… .
(a) frenulum
(b) micropyle
(c) raphae
(d) replum
Answer:
(d) replum

Samacheer Kalvi 11th Bio Botany Solutions 4 Reproductive Morphology

Question 37.
The type of fruit in Ricinus in …………… .
(a) lomentum
(b) cremocarp
(c) regma
(d) nut
Answer:
(c) regma

Question 38.
Jack fruit is an example for …………… .
(a) syconus
(b) siliqua
(c) sorosis
(d) nut
Answer:
(c) sorosis

Question 39.
Which of the following is not a schizocarpic fruit?
(a) Cremocarp
(b) Regma
(c) Samara
(d) Carcerulus
Answer:
(c) Samara

Question 40.
After fertilization …………… modifies into seed.
(a) ovary
(b) ovule
(c) carpel
(d) stigma
Answer:
(b) ovule

Question 41.
In groundnuts, which part nourishes the embryo?
(a) Endosperm
(b) Albumin
(c) Cotyledons
(d) Carpel
Answer:
(c) Cotyledons

Samacheer Kalvi 11th Bio Botany Solutions 4 Reproductive Morphology

Question 42.
…………… are the means for perpetuation of species.
(a) Fruits
(b) Seeds
(c) Corolla
(d) Flowers
Answer:
(b) Seeds

Question 43.
…………… is the second whorl of the flower.
(a) Calyx
(b) Corolla
(c) Gynoecium
(d) Perianth
Answer:
(b) Corolla

Question 44.
…………… is a ripened ovule.
(a) Carpel
(b) Pistil
(c) Seed
(d) Fruit
Answer:
(c) Seed

Question 45.
Imperfect flowers will have …………… essential whorl(s).
(a) only 1
(b) 2
(c) none
(d) 4
Answer:
(a) only 1

II. Very Short Answer Type Questions (2 Marks)

Question 1.
How will you define inflorescence?
Answer:
An inflorescence is a group of flowers arising from a branched or unbranched axis with a definite pattern.

Question 2.
Where does the inflorescence axis arise in cauliflorous type of inflorescence?
Answer:
In cauliflorous type, inflorescence developed directly from a woody trunk. Example: Theobroma cocoa.

Samacheer Kalvi 11th Bio Botany Solutions 4 Reproductive Morphology

Question 3.
Name any two mixed inflorescences.
Answer:
Two mixed inflorescences:

  1. Thyrsus and
  2. Verticillaster.

Question 4.
When a flower is said to be complete?
Answer:
A flower is said to be complete when it has all the four whorls (calyx, corolla, androecium & gynoecium).

Question 5.
What is a sessile flower?
Answer:
A flower without a pedicel or stalk is said to be sessile flower.

Question 6.
Define merosity.
Answer:
Number of floral parts per whorl is called merosity.

Question 7.
Write the units of (a) Perianth and (b) Calyx.
Answer:
The units of (a) Perianth and (b) Calyx:

  1. (a) Perianth – tepals and
  2. (b) Calyx – sepals

Question 8.
Name the three types of petals in papilionoceous corolla.
Answer:
The three types of petals in papilionoceous corolla:

  1. Vexillum (standard)
  2. wings (alae)
  3. petals (carina)

Samacheer Kalvi 11th Bio Botany Solutions 4 Reproductive Morphology

Question 9.
What is a staminode? Give example.
Answer:
Sterile stamen is called staminode. e.g. Cassia

Question 10.
Define Pollinium.
Answer:
When the pollen grains are fused together as a single main, it is said to be pollinium.

Question 11.
List out the parts of a pistil.
Answer:
Ovary, style and stigma.

Question 12.
Define aestivation.
Answer:
Arrangement of sepals and petals in a floral bud.

Question 13.
What is mother axis?
Answer:
The branch that bears the flower is called mother axis.

Samacheer Kalvi 11th Bio Botany Solutions 4 Reproductive Morphology

Question 14.
What do you understand by the term “Pomology”?
Answer:
The branch of horticulture that deals with the study of fruits and their cultivation is called pomology.

Question 15.
How the seeds are classified based on endosperm?
Answer:
(a) Albuminous seed or Endospermous seed.
(b) Ex – Albuminous seed or non – Endospermous seed.

Question 16.
What is Spathe?
Answer:
In spadix, entire inflorescence is covered by a brightly coloured or hard bract called a spathe.

Question 17.
Differentiate Apocarpous and Syncarpous ovary.
Answer:
Apocarpous and Syncarpous ovary:

  1. Apocarpous: A pistil contains two or more distinct carpels. Example: Annona
  2. Syncarpous: A pistil contains two or more carpels which are connate. Example: Citrus and Tomato

Question 19.
Give examples for following fruit types: (a) Berry (b) Hesperidium
Answer:
(a) Berry: Tomato
(b) Hesperidium: Orange

III. Short Answer Type Questions (3 Marks)

Question 1.
Distinguish between Monoecious & Dioecious.
Answer:
Between Monoecious & Dioecious:

  1. Monoecious: Both male and female flowers are present in the same plant, e.g., Coconut
  2. Dioecious:  Male and female flowers are present on separate plants, e.g., Papaya

Question 2.
Explain Bilateral symmetry.
Answer:
In bilateral symmetry the flower can be divided into equal halves in only one plane. Zygomorphic flower can efficiently transfer pollen grains to visiting pollinators. Example: Pisum.

Samacheer Kalvi 11th Bio Botany Solutions 4 Reproductive Morphology

Question 3.
Differentiate Apopetalous from Sympetalous.
Answer:
Apopetalous from Sympetalous:

  1. Apopetalous (or) Polypetalous: Petals are distinct, e.g., Hibiscus.
  2. Sympetalous (or) Gamopetalous: Petals are fused, e.g., Datura.

Question 4.
Define Placentation & mention their types.
Answer:
The mode of distribution of placenta inside the ovary is called placentation. Types of placentation: Marginal, Axile, Superficial, Parietal, Free – central and Basal.

Question 5.
Write the floral formula for the Hibiscus rosasinensis.
Answer:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 4 Reproductive Morphology 3

Question 6.
What are Parthenocarpic fruit?
Answer:
Development of fruits without fertilization are called Parthenocarpic fruit. They are seedless fruits. Example: Banana.

Samacheer Kalvi 11th Bio Botany Solutions 4 Reproductive Morphology

Question 7.
From which type of flowers does the aggregate fruit develops?
Answer:
Aggregate fruits develop from a single flower having an apocarpous pistil. Each of the free carpel is develops into a simple fruitlet. A collection of simple fruitlets makes an aggregate fruit.

Question 8.
Classify seeds based on their cotyledons.
Answer:
Based on the number of cotyledons present, two types of seeds are recognized.

  1. Dicotyledonous seed: Seed with two cotyledons.
  2. Monocotyledonous seed: Seed with one cotyledon.

Question 9.
List out any 3 significances of seed.
Answer:
3 significances of seed:

  1. The seed encloses and protects the embryo for next generation.
  2. Seeds of various plants are used as food, both for animals and human.
  3. Seeds are the products of sexual reproduction so they provide genetic variations and recombination in a plant.

Question 10.
What is the importance of inflorescence.
Answer:
Function of inflorescence is to display the flowers for effective pollination and facilitate seed dispersal. The grouping of flowers in one place gives a better attraction to the visiting pollinators and maximize the energy of the plant.

Question 11.
Draw the line diagram for the following inflorescence.
Answer:
(a) Simple Dichasium:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 4 Reproductive Morphology 5

(b) Compound Dichasium:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 4 Reproductive Morphology 4

IV. Long Answer Type Questions (5 Marks)

Question 1.
Explain the various types of Schizocarpic fruit.
Answer:
This fruit type of intermediate between dehiscent and indehiscent fruit. The fruit instead of dehiscing rather splits into number of segments, each containing one or more seeds. They are of following types:

  1. Cremocarp: Fruit develops from bicarpellary, syncarpous, inferior ovary and splitting into two one seeded segments known as mericarps. e.g., Coriander and Carrot.
  2. Carcerulus: Fruit develops from bicarpellary, syncarpous, superior ovary and splitting into four one seeded segments known as nutlets, e.g., Leucas, Ocimum and Abutilon.
  3. Lomentum: The fruit is derived from monocarpellary, unilocular ovary. A leguminous fruit, constricted between the seeds to form a number of one seeded compartments that separate at maturity, e.g., Desmodium, Arachis and Mimosa.
  4. Regma: They develop from tricarpellary, syncarpous, superior, trilocular ovary and splits into one – seeded cocci which remain attached to carpophore, e.g., Ricinus and Geranium.

Question 2.
Explain the different types of flowers based on the position of ovary.
Answer:
Based on the position of ovary, a flower can be classified as:

  1. Hypogynous: The term is used for sepals, petals and stamens attached at the base of a superior ovary, e.g. Malvaceae.
  2. Epihypogynous: The term is used for sepals, petals and stamens attached at the middle of the ovary (half – inferior), e.g. Fabaceae and Rosaceae.
  3. Epigynous: The term is used for sepals, petals and stamens attached at the tip of an inferior ovary, e.g. Cucumber, apple and Asteraceae.
  4. Perigynous: The term is used for a hypanthium attached at the base of a superior ovary.
  5. Epiperigynous: The term is used for hypanthium attached at the apex of an inferior ovary.

Samacheer Kalvi 11th Bio Botany Solutions 4 Reproductive Morphology

Question 3.
Classify the anthers based on their mode of attachment.
Answer:
The anthers based on their mode of attachment:

  1. Basifixed: (Innate) Base of anther is attached to the tip of filament, e.g., Brassica, Datura
  2. Dorsifixed: Apex of filament is attached to the dorsal side of the anther, e.g. Citrus, Hibiscus
  3. Versatile: Filament is attached to the anther at midpoint, e.g., Grasses
  4. Adnate: Filament is continued from the base to the apex of anther, e.g. Verbena, Ranunculus, Nelumbo.

Question 4.
Define aestivation. Explain its types with example.
Answer:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 4 Reproductive Morphology 6

Question 5.
Distinguish between racemose and cymose inflorescence.
Answer:
Racemose inflorescence

  1. Main axis of unlimited growth
  2. Flowers arranged in an acropetal succession
  3. Opening of flowers is centripetal
  4. Usually the oldest flower at the base of the inflorescence axis.

Cymose inflorescence:

  1. Main axis of limited growth
  2. Flowers arranged in a basipetal succession
  3. Opening of flowers is centrifugal
  4. Usually the oldest flower at the top of the inflorescence axis.

Question 6.
Write in detail about head inflorescence.
Answer:
Head: A head is a characteristic inflorescence of Asteraceae and is also found in some members of Rubiaceae. Example: Neolamarkia cadamba and Mitragyna parvifolia; and in some members of Fabaceae – Mimosoideae, example: Acacia nilotica, Albizia lebbeck, Mimosa pudica (sensitive plant). Torus contains two types of florets:

  1. Disc floret or tubular floret.
  2. Ray floret or ligulate floret.

Heads are classified into two types:

  1. Homogamous head: This type of inflorescence exhibits single kind of florets. Inflorescence has disc florets alone. Example: Vernonia, Ageratum or Ray florets alone. example: Launaea, Sonchus.
  2. Heterogamous head: The inflorescence possesses both types of florets. Example: Helianthus, Tridax.

Disc florets at the centre of the head are tubular and bisexual whereas the ray florets found at the margin of the head which are ligulate pistilate (unisexual).
Samacheer Kalvi 11th Bio Botany Solutions Chapter 4 Reproductive Morphology 7
Samacheer Kalvi 11th Bio Botany Solutions Chapter 4 Reproductive Morphology 8

Question 7.
List out the significance of fruits.
Answer:
The significance of fruits:

  1. Edible part of the fruit is a source of food, energy for animals.
  2. They are source of many chemicals like sugar, pectin, organic acids, vitamins and minerals.
  3. The fruit protects the seeds from unfavourable climatic conditions and animals.
  4. Both fleshy and dry fruits help in the dispersal of seeds to distant places.
  5. In certain cases, fruit may provide nutrition to the developing seedling.
  6. Fruits provide source of medicine to humans.

Question 8.
Draw a flow chart depicting the classification of fruits.
Answer:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 4 Reproductive Morphology 9

Question 9.
Explain in detail about any two special inflorescence.
Answer:
The inflorescences do not show any of the development pattern types are classified under special type of inflorescence.
1. Cyathium: Cyathium inflorescence consists of small unisexual flowers enclosed by a common involucre which mimics a single flower. Male flowers are organized in a scorpioid manner. Female flower is solitary and centrally located on a long pedicel. Male flower is represented only by stamens and female flower is represented only by pistil. Cyathium may be actinomorphic (Example: Euphorbia) or zygomorphic (Example: Pedilanthus) Nectar is present in involucre.

2. Hypanthodium: Receptacle is a hollow, globose structure consisting unisexual flowers present on the inner wall of the receptacle. Receptacle is closed except a small opening called ostiole which is covered by a series of bracts. Male flowers are present nearer to the ostiole, female and neutral flowers are found in a mixed manner from middle below. Example: Ficus sp. (Banyan and Pipal).

V. Higher Order Thinking Skills (HOTs)

Question 1.
Brinjal fruit has persistent calyx. Have you ever noticed the same in any other fruits? Name them.
Answer:
Tomato, Lady’s finger, guava and chilli also have persistent calyx.

Question 2.
Whether parthenocarpic fruits develop endosperm? Why?
Answer:
No, parthenocarpic fruits are developed without fertilization, but endosperm will form only after fertilization. So parthenocarpic fruits do not have endosperm.

Question 3.
Ovary develops into fruit after fertilization. While eating an Apple which part do you eat? Explain.
Answer:
Apple belongs to false fruit. In false fruits, apart from the ovary, non – carpellary parts also develop into fruit. In apple, the thalamus develops into fleshy edible part.

Samacheer Kalvi 11th Bio Botany Solutions 4 Reproductive Morphology

Question 4.
Cremocarp and Carcerulus both are schizocarpic fruits yet they differ. How?
Answer:
Cremocarp:

  1. Fruit develops from syncarpous inferior ovary.
  2. Ripened fruit split into two, one – seeded segments called mericarps.

Carcerulus:

  1. Fruit develops from syncarpous superior ovary.
  2. Ripened fruit split into four, one – seeded segments called nutlets.

Question 5.
Mango and coconut are ‘drupe’ type of fruits. In Mango, the edible part is fleshy mesocarp. What does the milk of tender coconut represent?
Answer:
Endosperm is the liquid (milk) potable part of tender coconut, which is rich in nutrients and is formed as a result of triple fusion.

Question 6.
Pick out correct ratio of the male flower to female flower in cyathium inflorescence and explain it?
(a) one : one
(b) one : many
(c) many : many and
(d) many : one.
Answer:
(a) One : one – Cyathium has single female flower represented by pistil and male flower represented by stamen.

Samacheer Kalvi 11th Bio Botany Solutions 4 Reproductive Morphology

Question 7.
Pollen differs from pollinium. How?
Answer:
Pollen are microspores which produces male gametes, whereas pollinium refers to the single mass of fused pollen grains.

Question 8.
Sunflower is not a flower – Justify your answer.
Answer:
Sunflower is actually an inflorescence not a single flower. The inflorescence of sunflower is capitulum composed of disc florets and ray florets.

Samacheer Kalvi 11th Bio Botany Solutions 4 Reproductive Morphology

Question 9.
Flower is a modified shoot for reproduction – Give possible evidence.
Answer:
Modified shoot for reproduction:

  • Floral leaves (sepals and petals) are modified leaves.
  • Floral and vegetative buds both emerge either in terminal or axillary position.
  • Foliage leaves and floral leaves have identical arrangement on stem.

Question 10.
Is tomato a fruit or vegetable? Explain.
Answer:
Yes, tomato is a fruit. Because it develops from the ripened ovary and bears seeds, whereas vegetable refers to all other plant parts like root, stem and leaves.

Question 11.
What is caruncle? Where it is seen? How it helps the plant?
Answer:
Caruncle is the fleshy outgrowth at the base of seed. Usually caruncle helps in the seed dispersal, particularly by ants (Myrmecophily).

Question 12.
Both the prefixes (Uni – and Mono -) have the same meaning i.e. one in number. Does it mean that unisexual and monoecious are the same?
Answer:
That unisexual and monoecious:

  1. Unisexual refers to the sex of flower (i.e. whether it has anther or carpel).
  2. Monoecious refers to the plant bearing both the sexes in their flowers.

Samacheer Kalvi 11th Bio Botany Solutions Chapter 14 Respiration

Students can Download Bio Botany Chapter 14 Respiration Questions and Answers, Notes Pdf, Samacheer Kalvi 11th Bio Botany Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Bio Botany Solutions Chapter 14 Respiration

Samacheer Kalvi 11th Bio Botany Respiration Text Book Back Questions and Answers

Question 1.
The number of ATP molecules formed by complete oxidation of one molecule of pyruvic acid is:
(a) 12
(b) 13
(c) 14
(d) 15
Answer:
(a) 12

Question 2.
During oxidation of two molecules of cytosolic NADH + H+, number of ATP molecules produced in plants are:
(a) 3
(b) 4
(c) 6
(d) 8
Answer:
(c) 6

Samacheer Kalvi 11th Bio Botany Solutions 14 Respiration

Question 3.
The compound which links glycolysis and Krebs cycle is:
(a) succinic acid
(b) pyruvic acid
(c) acetyl CoA
(d) citric acid
Answer:
(c) acetyl CoA

Question 4.
Assertion (A): Oxidative phosphorylation takes place during the electron transport chain in mitochondria.
Reason (R): Succinyl CoA is phosphorylated into succinic acid by substrate phosphorylation.
(a) A and R is correct. R is correct explanation of A
(b) A and R is correct but R is not the correct explanation of A
(c) A is correct but R is wrong
(d) A and R is wrong.
Answer:
(a) A and R is correct. R is correct explanation of A

Question 5.
Which of the following reaction is not . involved in Krebs cycle.
(a) Shifting of phosphate from 3C to 2C
(b) Splitting of Fructose 1,6 bisphosphate of into two molecules 3C compounds.
(c) Dephosphorylation from the substrates
(d) All of these
Answer:
(d) All of these

Question 6.
What are enzymes involved in phosphorylation and dephosphorylation reactions in EMP pathway?
Answer:
(i) Enzymes involved phosphorylation in EMP pathway:

  • Hexokinase
  • Phospho – fructokinase
  • Glyceraldehyde – 3 – phosphate dehydrogenase

(ii) Enzymes involved in dephosphorylation in EMP pathway:

  • Phospho glycerate kinase,
  • Pyruvate kinase

Question 7.
Respiratory quotient is zero in succulent plants. Why?
Answer:
In some succulent plants like Opuntia, Bryophyllum carbohydrates are partially oxidised to organic acid, particularly malic acid without corresponding release of CO2 but O2 is consumed hence the RQ value will be zero.

Samacheer Kalvi 11th Bio Botany Solutions 14 Respiration

Question 8.
Explain the reactions taking place in mitochondrial inner membrane.
Answer:
In plants, an additional NADH dehydrogenase (External) complex is present on the outer surface of inner membrane of mitochondria which can oxidise cytosolic NADH + H+. Ubiquinone (UQ) or Coenzyme Quinone (Co Q) is a small, lipid soluble electron, proton carrier located within the inner membrane of mitochondria.

Question 9.
What is the name of alternate way of glucose breakdown? Explain the process involved in it?
Answer:
During respiration breakdown of glucose in cytosol occurs both by glycolysis (about 2 / 3) as well as by oxidative pentose phosphate pathway (about 1 / 3). Pentose phosphate pathway was described by Warburg, Dickens and Lipmann (1938). Hence, it is also called Warburg – Dickens – Lipmann pathway. It takes place in cytoplasm of mature plant cells. It is an alternate way for breakdown of glucose.

It is also known as Hexose monophosphate shunt (HMP Shunt) or Direct Oxidative Pathway. It consists of two phases, oxidative phase and non – oxidative phase. The oxidative events convert six molecules of six carbon Glucose – 6 – phosphate to 6 molecules of five carbon sugar Ribulose – 5 phosphate with loss of 6CO2 molecules and generation of 12 NADPH + H+ (not NADH). The remaining reactions known as non – oxidative pathway, convert Ribulose – 5 – phosphate molecules to various intermediates such as Ribose – 5 – phosphate(5C), Xylulose – 5 – phosphate(5C), Glyceraldehyde – 3 – phosphate(3C), Sedoheptulose – 7 – Phosphate (7C), and Erythrose – 4 – phosphate (4C). Finally, five molecules of glucose – 6 – phosphate is regenerated. The overall reaction is:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 14 Respiration 1
The net result of complete oxidation of one glucose – 6 – phosphate yield 6CO2 and 12 NADPH + H+. The oxidative pentose phosphate pathway is controlled by glucose – 6 – phosphate dehydrogenase enzyme which is inhibited by high ratio of NADPH to NADP+.

Question 10.
How will you calculate net products of one sucrose molecule upon complete oxidation during aerobic respiration as per recent view?
Answer:
When the cost of transport of ATPs from matrix into the cytosol is considered, the number will be 2.5 ATPs for each NADH + H+ and 1.5 ATPs for each FADH2 oxidised during electron transport system. Therefore, in plant cells net yield of 30 ATP molecules for complete aerobic oxidation of one molecule of glucose. But in those animal cells (showing malate shuttle mechanism) net yield will be 32 ATP molecules. Since sucrose molecule gives, two molecules of glucose and net ATP in plant cell will be 30 × 2 = 60. In animal cell it will be 32 × 2 = 64.

Samacheer Kalvi 11th Bio Botany Respiration Additional Questions & Answers

I. Choose the correct answer (1 Mark)
Question 1.
The term respiration was coined by:
(a) Lamark
(b) Kerb
(c) Pepys
(d) Blackman
Answer:
(c) Pepys

Samacheer Kalvi 11th Bio Botany Solutions 14 Respiration

Question 2.
In floating respiration the substrates are:
(a) carbohydrate or protein
(b) carbohydrate or fat
(c) protein or fat
(d) none of the above
Answer:
(b) carbohydrate or fat

Question 3.
The discovery of ATP was made by:
(a) Lipman
(b) Hans Adolt
(c) Warburg
(d) Karl Lohman
Answer:
(d) Karl Lohman

Question 4.
The end product of glycolysis is:
(a) pyruvate
(b) ethanol
(c) malate
(d) succinate
Answer:
(a) pyruvate

Samacheer Kalvi 11th Bio Botany Solutions 14 Respiration

Question 5.
On hydrolysis, one molecule of ATP releases energy of:
(a) 8.2 K cal
(b) 32.3 kJ
(c) 7.3 K cal
(d) 7.8 K cal
Answer:
(c) 7.3 K cal

Question 6.
Which of the following is known as terminal oxidation:
(a) glycolysis
(b) electron transport chain
(c) Kreb’s cycle
(d) pyruvate oxidation
Answer:
(b) electron transport chain

Question 7.
Identify the link reaction:
(a) conversion of glucose into pyruvic acid
(b) conversion of glucose into ethanol
(c) conversion of acetyl CoA into CO2 and water
(d) conversion of pyruvic acid into acetyl coenzyme – A
Answer:
(d) conversion of pyruvic acid into acetyl coenzyme – A

Samacheer Kalvi 11th Bio Botany Solutions 14 Respiration

Question 8.
Who was awarded Nobel prize in 1953 for the discovery of TCA cycle?
(a) Lipmann
(b) Hans Adolf Kreb
(c) Petermitchell
(d) Dickens
Answer:
(b) Hans Adolf Kreb

Question 9.
Kreb’s cycle is a:
(a) catabolic pathway
(b) anabolic pathway
(c) amphibolic pathway
(d) hydrolytic pathway
Answer:
(c) amphibolic pathway

Question 10.
Electron transport system during aerobic respiration takes place in:
(a) cytoplasm
(b) mitochondria
(c) chloroplast
(d) golgi apparatus
Answer:
(b) mitochondria

Question 11.
The oxidation of one molecule of NADH + H+ gives rise to:
(a) 2 ATP
(b) 3 ATP
(c) 4 ATP
(d) 2.5 ATP
Answer:
(b) 3 ATP

Samacheer Kalvi 11th Bio Botany Solutions 14 Respiration

Question 12.
In aerobic prokaryotes each molecule of glucose produces:
(a) 36 ATP
(b) 32 ATP
(c) 34 ATP
(d) 38 ATP
Answer:
(d) 38 ATP

Question 13.
Cyanide acts as electron transport chain inhibitor by preventing:
(a) synthesis of ATP from ADP
(b) flow of electrons from NADH + H+
(c) flow of electrons from cytochrome a3 to O2
(d) oxidative phosphorylation
Answer:
(c) flow of electrons from cytochrome a3 to O2

Question 14.
Respiratory quotient for oleic acid is:
(a) 0.69
(b) 0.71
(c) 0.80
(d) 0.36
Answer:
(b) 0.71

Question 15.
End products of fermentation in yeast is:
(a) pyruvic acid and CO2
(b) lactic acid qnd CO2
(c) ethyl alcohol and CO2
(d) mixed acid and CO2
Answer:
(c) ethyl alcohol and CO2

Question 16.
The end products of mixed acid fermentation in enterobacteriaceae are:
(a) lactic acid, ethanol, formic acid, CO2 and H2
(b) lactic acid, formic acid and CO2
(c) lactic acid, ethanol, CO2 and O2
(d) ethanol, formic acid, CO2 and H2
Answer:
(a) lactic acid, ethanol, formic acid, CO2 and H2

Samacheer Kalvi 11th Bio Botany Solutions 14 Respiration

Question 17.
The external factors that affect the respiration are:
(a) temperature, insufficient O2 and amount of protoplasm
(b) temperature, insufficient O2 and high concentration of CO2
(c) temperature, high concentration of CO2 and respiratory substrate
(d) temperature, high concentration of CO2 and amount of protoplasm
Answer:
(b) temperature, insufficient O2 and high concentration of CO2

Question 18.
Pentose phosphate pathway was described by:
(a) Pepys and Black man
(b) Kreb and Embden
(c) Warburg, Dickens and Lipmann
(d) Warburg and Pamas
Answer:
(c) Warburg, Dickens and Lipmann

Question 19.
The oxidative pentose phosphate pathway is controlled by the enzyme:
(a) glucose, 1, 6 diphosphate dehydrogenase
(b) glucose 6 phosphate dehydrogenase
(c) fructose – 6 – phosphate dehydrogenase
(d) none of the above
Answer:
(b) glucose 6 phosphate dehydrogenase

Question 20.
In pentose phosphate pathway the glucose – 6 – phosphate dehydrogenase enzyme is inhibited by high ratio of:
(a) FADH to FAD
(b) glucose to glucose – 6 – phosphate
(c) NADPH to NADP
(d) GTPH to GTP
Answer:
(c) NADPH to NADP

Samacheer Kalvi 11th Bio Botany Solutions 14 Respiration

Question 21.
In plant tissue erythrose is used for the synthesis of:
(a) Erythromycin
(b) Xanthophill
(c) Erythrocin
(d) Arithocyanin
Answer:
(d) Arithocyanin

Question 22.
As per the recent view, when a glucose molecule is completely aerobically oxidised, the net yield of ATP in plant cell is:
(a) 38
(b) 36
(c) 30
(d) 32
Answer:
(c) 30

Question 23.
Identify the electron transport inhibitor:
(a) phosphophenol
(b) dinitrophenol
(c) xylene
(d) indol acetic acid
Answer:
(b) dinitrophenol

Question 24.
The phenomenon of climacteric is present in:
(a) banana
(b) coconut
(c) cauli flower
(d) brinjal
Answer:
(a) banana

Samacheer Kalvi 11th Bio Botany Solutions 14 Respiration

Question 25.
Cyanide resistant respiration is known to generate heat in thermogenic tissues as high as:
(a) 35° C
(b) 38° C
(c) 40° C
(d) 51° C
Answer:
(d) 51° C

Question 26.
Match the following:

Substrate

RQ

A. Palmitic acid (i) 1.6
B. Oleic acid (ii) 4.0
C. Tartaric acid (iii) 0.36
D. Oxalic acid (iv) 0.71

(a) A – (ii), B – (iii); C – (i); D – (iv)
(b) A – (iii), B – (iv); C – (i); D – (ii)
(c) A – (ii); B – (iv); C – (i); D – (iii)
(d) A – (iii); B – (i); C – (iv); D – (ii)
Answer:
(b) A – (iii), B – (iv); C – (i); D – (ii)

Question 27.
Indicate the correct statement:
(a) In Bryophyllum, carbohydrates are partially oxidised to organic acid
(b) In opuntia, the Respiratory Quotient value is 0.5
(c) Alcoholic fermentation takes place in enterobacteriaceae
(d) Muscles of vertebrate does not have lactate dehydrogenase enzyme
Answer:
(a) In Bryophyllum, carbohydrates are partially oxidised to organic acid

Samacheer Kalvi 11th Bio Botany Solutions 14 Respiration

Question 28.
The order of aerobic respiration in plant cell is:
(a) glycolysis, Kreb’s cycle, pyruvate oxidation and electron transport chain
(b) glycolysis, pyruvate oxidate, Kreb’s cycle, electron transport chain
(c) pyruvate oxidation, glycolysis, Kreb’s cycle, electron transport chain
(d) none of the above order
Answer:
(b) glycolysis, pyruvate oxidate, Kreb’s cycle, electron transport chain

Question 29.
The complete reactions of glycolysis take place in:
(a) mitochondria
(b) cristae
(c) cytoplasm
(d) outer membrane of mitochondria
Answer:
(c) cytoplasm

Question 30.
The Co – enzyme quinone is a proton carrier located within:
(a) outer membrane of mitochondria
(b) cytoplasm
(c) inner membrane of mitochondria
(d) matrix of mitochondria
Answer:
(c) inner membrane of mitochondria

Samacheer Kalvi 11th Bio Botany Solutions 14 Respiration

Question 31.
How many molecules of CO2 are produced during link reaction:
(a) 1
(b) 6
(c) 4
(d) 2
Answer:
(d) 2

Question 32.
In the case of ground nut, during seed germination they use:
(a) carbohydrate as respiratory substrate
(b) fat alone as respiratory substrate
(c) fat and protein as respiratory substrate
(d) protein alone as respiratory substrate
Answer:
(c) fat and protein as respiratory substrate

Question 33.
Lactic acid fermentation takes place in:
(a) yeast
(b) bacillus
(c) enterobacteriaceae
(d) none of the above
Answer:
(b) bacillus

Samacheer Kalvi 11th Bio Botany Solutions 14 Respiration

Question 34.
The net result of complete oxidation of one glucose-6-phosphate in pentose phosphate pathway yield:
(a) 6 CO2 and 12 NADPH + H+
(b) 6 CO2 and 10 NADPH + H+
(c) 8 CO2 and 16 NADPH + H+
(d) 8 CO2 and 14 NADPH + H
Answer:
(a) 6 CO2 and 12 NADPH + H+

Question 35.
Ribose – 5 – phosphate and its derivatives are used in the synthesis of:
(a) lignin
(b) coenzyme A
(c) anthocyanin
(d) xanthophyll
Answer:
(b) coenzyme A

II. Answer the following (2 Marks)

Question 1.
Define respiration?
Answer:
Respiration is a biological process in which oxidation of various food substances like carbohydrates, proteins and fats take place and as a result of this, energy is produced where O2 is taken in and CO2 is liberated.

Question 2.
What is meant by protoplasmic respiration?
Answer:
Respiration utilizing protein as a respiratory substrate, it is called protoplasmic respiration. Protoplasmic respiration is rare and it depletes structural and functional proteins of protoplasm and liberates toxic ammonia.

Samacheer Kalvi 11th Bio Botany Solutions 14 Respiration

Question 3.
What do you understand by compensation of point?
Answer:
The point at which CO2 released in respiration is exactly compensated by CO2 fixed in photosynthesis that means no net gaseous exchange takes place, it is called compensation point.

Question 4.
Explain briefly about aerobic respiration.
Answer:
Respiration occurring in the presence of oxygen is called aerobic respiration. During aerobic respiration, food materials like carbohydrates, fats and proteins are completely oxidised into CO2, H2O and energy is released.

Question 5.
What is anaerobic respiration?
Answer:
In the absence of molecular oxygen glucose is incompletely degraded into either ethyl alcohol or lactic acid. It includes two steps:

  1. Glycolysis
  2. Fermentation

Question 6.
What do you know about transition reaction?
Answer:
In aerobic respiration the pyruvate with coenzyme A is oxidatively decarboxylated into acetyl CoA by pyruvate dehydrogenase complex. This reaction is irreversible and produces two molecules of NADH + H+ and 2CO2. It is also called transition reaction or Link reaction.

Samacheer Kalvi 11th Bio Botany Solutions 14 Respiration

Question 7.
Who is Sir Hans Adolf Krebs?
Answer:
Sir Hans Adolf Krebs was born in Germany on 25th August 1900. He was awarded Nobel Prize for his discovery of Citric acid cycle in Physiology in 1953.

Question 8.
Explain briefly about amphibolic pathway.
Answer:
Krebs cycle is primarily a catabolic pathway, but it provides precursors for various biosynthetic pathways thereby an anabolic pathway too. Hence, it is called amphibolic pathway.

Question 9.
Mention the role of NADH dehydrogenase enzyme in electron transport system.
Answer:
NADH dehydrogenase contains a flavoprotein (FMN) and associated with non – heme iron Sulphur protein (Fe – S). This complex is responsible for passing electrons and protons from mitochondrial NADH (Internal) to Ubiquinone (UQ).

Samacheer Kalvi 11th Bio Botany Solutions 14 Respiration

Question 10.
What is oxidative phosphorylation?
Answer:
The transfer of electrons from reduced coenzyme NADH to oxygen via complexes I to IV is coupled to the synthesis of ATP from ADP and inorganic phosphate (Pi) which is called Oxidative phosphorylation.

Question 11.
Mention any two electron transport chain inhibitors.
Answer:
Two electron transport chain inhibitors:

  1. 2, 4 DNP (Dinitrophenol) – It prevents synthesis of ATP from ADP, as it directs electrons from Co Q to O2.
  2. Cyanide – It prevents flow of electrons from Cytochrome a3 to O2.

Question 12.
Define respiratory quotient.
Answer:
The ratio of volume of carbon dioxide given out and volume of oxygen taken in during respiration is called Respiratory Quotient.

Question 13.
What are the significances of Respiratory Quotient?
Answer:
The significances of Respiratory Quotient:

  1. RQ value indicates which type of respiration occurs in living cells, either aerobic or anaerobic.
  2. It also helps to know which type of respiratory substrate is involved.

Samacheer Kalvi 11th Bio Botany Solutions 14 Respiration

Question 14.
Explain the term alcoholic fermentation.
Answer:
The cells of roots in water logged soil respire by alcoholic fermentation because of lack of oxygen by converting pyruvic acid into ethyl alcohol and CO2. Many species of yeast (Saccharomyces) also respire anaerobically. This process takes place in two steps:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 14 Respiration 2

Question 15.
Mention any two industrial uses of alcoholic fermentation.
Answer:
Two industrial uses of alcoholic fermentation:

  1. In bakeries, it is used for preparing bread, cakes, biscuits.
  2. In beverage industries for preparing wine and alcoholic drinks.

Question 16.
What do you understand by the term mixed acid fermentation?
Answer:
This type of fermentation is a characteristic feature of Enterobacteriaceae and results in the formation of lactic acid, ethanol, formic acid and gases like CO2 and H2.

Samacheer Kalvi 11th Bio Botany Solutions 14 Respiration

Question 17.
Mention any two internal factors, that affect the rate of respiration in plants.
Answer:
Two internal factors, that affect the rate of respiration in plants:

  1. The amount of protoplasm and its state of activity influence the rate of respiration.
  2. Concentration of respiratory substrate is proportional to the rate of respiration.

Question 18.
What is the control mechanism of pentose phosphate pathway?
Answer:
The oxidative pentose phosphate pathway is controlled by glucose-6-phosphate dehydrogenase enzyme which is inhibited by high ratio of NADPH to NADP+.

Question 19.
Write down any two significance of pentose phosphate pathway.
Answer:
Two significance of pentose phosphate pathway:

  1. HMP shunt is associated with the generation of two important products,
  2. Coenzyme NADPH generated is used for reductive biosynthesis and counter damaging the effects of oxygen free radicals.

III. Answer the following (3 Marks)

Question 1.
In biosphere how do plants and animals are complementary systems, which are integrated to sustain life?
Answer:
In plants, oxygen enters through the stomata and it is transported to cells, where oxygen is utilized for energy production. Plants require carbon dioxide to survive, to produce carbohydrates and to release oxygen through photosynthesis, these oxygen molecules are inhaled by human through the nose, which reaches the lungs where oxygen is transported through the blood and it reaches cells. Cellular respiration takes place inside or the cell for obtaining energy.

Samacheer Kalvi 11th Bio Botany Solutions 14 Respiration

Question 2.
What will happen, when you sleep under a tree during night time?
Answer:
If you are sleeping under a tree during night time you will feel difficulty in breathing. During night, plants take up oxygen and release carbon dioxide and as a result carbon dioxide will be abundant around the tree

Question 3.
What are the factors associated with compensation point in respiration?
Answer:
The two common factors associated with compensation point are CO2 and light. Based on this there are two types of compensation point. They are CO2 compensation point and light compensation point. C3 plants have compensation points ranging from 40 – 60 ppm (parts per million) CO2 while those of C4 plants ranges from 1 – 5 ppm CO2.

Question 4.
Why do you call ATP as universal energy currency of cell?
Answer:
ATP is a nucleotide consisting of a base- adenine, a pentose sugar – ribose and three phosphate groups. Out of three phosphate groups the last two are attached by high energy rich bonds. On hydrolysis, it releases energy (7.3 K cal or 30.6 KJ / ATP) and it is found in all living cells and hence it is called universal energy currency of the cell.

Question 5.
What is a redox reaction?
Answer:
NAD+ + 2e + 2H+ → NADH + H+
FAD + 2e + 2H+ → FADH2
When NAD+ (Nicotinamide Adenine Dinucleotide – oxidised form) and FAD (Flavin Adenine Dinucleotide) pick up electrons and one or two hydrogen ions (protons), they get reduced to NADH + H+ and FADH2 respectively. When they drop electrons and hydrogen off they go back to their original form. The reaction in which NAD+ and FAD gain (reduction) or f lose (oxidation) electrons are called redox reaction (Oxidation reduction reaction). These reactions are important in cellular respiration.

Samacheer Kalvi 11th Bio Botany Solutions 14 Respiration

Question 6.
Write down any three differences between aerobic and anaerobic respiration.
Answer:
Aerobic respiration:

  • It occurs in all living cells of higher organisms.
  • It requires oxygen for breaking the respiratory substrate.
  • The end products are CO2 and H2O.

Anaerobic Respiration:

  • It occurs yeast and some bacteria.
  • Oxygen is not required for breaking the respiratory substrate.
  • The end products are alcohol, and CO2 (or) lactic acid.

Question 7.
Mention the significance of Kreb’s cycle.
Answer:
The significance of Kreb’s cycle:

  1. TCA cycle is to provide energy in the form of ATP for metabolism in plants.
  2. It provides carbon skeleton or raw material for various anabolic processes.
  3. Many intermediates of TCA cycle are further metabolised to produce amino acids, proteins and nucleic acids.
  4. Succinyl CoA is raw material for formation of chlorophylls, cytochrome, phytochrome and other pyrrole substances.
  5. α – ketoglutarate and oxaloacetate undergo reductive amination and produce amino acids.
  6. It acts as metabolic sink which plays a central role in intermediary metabolism.

Question 8.
Derive the respiratory quotient for carbohydrate as substrate in oxidative metabolism.
Answer:
The respiratory substrate is a carbohydrate, it will be completely oxidised in aerobic respiration and the value of the RQ will be equal to unity.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 14 Respiration 3

Question 9.
Write down the characteristic of Anaerobic respiration.
Answer:
The characteristic of Anaerobic respiration:

  1. Anaerobic respiration is less efficient than the aerobic respiration.
  2. Limited number of ATP molecules is generated per glucose molecule.
  3. It is characterized by the production of CO2 and it is used for Carbon fixation in photosynthesis.

Question 10.
Distinguish between glycolysis and fermentation.
Answer:
Glycolysis:

  1. Glucose is converted into pyruvic acid.
  2. It takes place in the presence or absence of oxygen.
  3. Net gain is 2ATR
  4. 2 NADH + H+ molecules are produced.

Fermentation:

  1. Starts from pyruvic acid and is converted into alcohol or lactic acid.
  2. It takes place in the absence of oxygen.
  3. No net gain of ATP molecules.
  4. 2 NADH + H+ molecules are utilised.

Samacheer Kalvi 11th Bio Botany Solutions 14 Respiration

Question 11.
Write down any three external factors, that affect respiration in plants.
Answer:
Three external factors, that affect respiration in plants:

  1. Optimum temperature for respiration is 30°C. At low temperatures and very high temperatures rate of respiration decreases.
  2. When sufficient amount of O2 is available the rate of aerobic respiration will be optimum and anaerobic respiration is completely stopped. This is called Extinction point.
  3. High concentration of CO2 reduces the rate of respiration

Question 12.
How alcoholic beverages like beer and wine is made?
Answer:
The conversion of pyruvate to ethanol takes place in malted barley and grapes through fermentation. Yeasts carryout this process under anaerobic conditions and this Conversion increases ethanol concentration. If the concentration increases, it’s toxic effect kills yeast cells .and the left out is called beer and wine respectively.

IV. Answer the following (5 Marks)

Question 1.
Give the schematic representation of glycolysis or EMP pathway.
Answer:
The schematic representation of glycolysis or EMP pathway:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 14 Respiration 4

Question 2.
Write down the biochemical events in Kreb’s cycle.
Answer:
The biochemical events in Kreb’s cycle:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 14 Respiration 5

Question 3.
Mention the schematic diagram of the various steps involved in pentose phosphate pathway.
Answer:
The schematic diagram of the various steps involved in pentose phosphate pathway:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 14 Respiration 6

Question 4.
Describe the events in electron transport chain in plant cell.
Answer:
During glycolysis, link reaction and Krebs cycle the respiratory substrates are oxidised at several steps and as a result many reduced coenzymes NADH + H+ and FADH2 are produced. These reduced coenzymes are transported to inner membrane of mitochondria and are converted back to their oxidised forms produce electrons and protons. In mitochondria, the inner membrane is folded in the form of finger projections towards the matrix called cristae.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 14 Respiration 14
In cristae many oxysomes (F1 particles) are present which have election transport carriers are present. According to Peter Mitchell’s Chemiosmotic theory this electron transport is coupled to ATP synthesis. Electron and hydrogen (proton) transport takes place across four multiprotein complexes (I – IV). They are

(i) Complex – I (NADH dehydrogenase: It contains a flavoprotein (FMN) and associated with non – heme iron Sulphur protein (Fe – S). This complex is responsible for passing electrons and protons from mitochondrial NADH (Internal) to Ubiquinone (UQ).
NADH + H+ + UQ ⇌ NAD+ + UQH2
In plants, an additional NADH dehydrogenase (External) complex is present on the outer surface of inner membrane of mitochondria which can oxidise cytosolic NADH + H+ Ubiquinone (UQ) or Coenzyme Quinone (Co Q) is a small, lipid soluble electron, proton carrier located within the inner membrane of mitochondria.

(ii) Complex – II (Succinic dehydrogenase): It contains FAD flavoprotein is associated with non – heme iron Sulphur (Fe – S) protein. This complex receives electrons and protons from succinate in Krebs cycle and is converted into fumarate and passes to ubiquinone.
Succinate + UQ → Fumarate + UQH2

(iii) Complex – III (Cytochrome bc1 complex): This complex oxidises reduced ubiquinone (ubiquinol) and transfers the electrons through Cytochrome bc1 Complex (Iron Sulphur center bc1 complex) to cytochrome c. Cytochrome c is a small protein attached to the outer surface of inner membrane and act as a. mobile carrier to transfer electrons between complex III to complex IV.
UQH2 + 2Cyt coxidised  ⇌  UQ + 2Cyt creduced  + 2H+

(iv) Complex IV (Cytochrome c oxidase): This complex contains two copper centers (A and B) and cytochromes a and as. Complex IV is the terminal oxidase and brings about the reduction of 1/2 O2 to H2O. Two protons are needed to form a molecule of H2O (terminal oxidation).
2Cyt coxidised + 2H+ + 1/2 O⇌  2Cyt creduced + H2O

The transfer of electrons from reduced coenzyme NADH to oxygen via complexes I to IV is coupled to the synthesis of ATP from ADP and inorganic phosphate (Pi) which is called Oxidative phosphorylation. The F0F1 – ATP synthase (also called complex V) consists of F0 and F1. F1 converts ADP and Pi to ATP and is attached to the matrix side of the inner membrane. F0 is present in inner membrane and acts as a channel through which protons come into matrix.

Oxidation of one molecule of NADH + H+ gives rise to 3 molecules of ATP and oxidation of one molecule FADH2 produces 2 molecules of ATP within a mitochondrion. But cytoplasmic NADH + H+ yields only two ATPs through external NADH dehydrogenase. Therefore, two reduced coenzyme (NADH + H+) molecules from glycolysis being extra mitochondrial will yield 2 × 2 = 4 ATP molecules instead of 6 ATPs. The Mechanism of mitochondrial ATP synthesis is based on Chemiosmotic hypothesis.

According to this theory electron carriers present in the inner mitochondrial membrane allow for the transfer of protons (H+). For the production of single ATP, 3 protons (H+) are needed. The terminal oxidation of external NADH bypasses the first phosphorylation site and hence only two ATP molecules are produced per external NADH oxidised through However, in those animal tissues in which malate shuttle mechanism is present, the oxidation of external NADH will yield almost 3 ATP molecules.

Complete oxidation of a glucose molecule in aerobic respiration results in the net gain of 36 ATP molecules in plants. Since huge amount of energy is generated in mitochondria in the form of ATP molecules they are called ‘power house of the cell’. In the case of aerobic prokaryotes due to lack of mitochondria each molecule of glucose produces 38 ATP molecules.

Question 5.
Define respiratory quotient. Explain the derivation of respiratory quotient for various substrates oxidised :
Answer:
The ratio of volume of carbon dioxide given out and volume of oxygen taken in during respiration is called Respiratory Quotient or Respiratory ratio. RQ value depends, upon respiratory substrates and their oxidation.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 14 Respiration 13
(i) The respiratory substrate is a carbohydrate, it will be completely oxidised in aerobic respiration and the value of the RQ will be equal to unity.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 14 Respiration 7
(ii) If the respiratory substrate is . a carbohydrate it will be incompletely oxidised when it goes through anaerobic respiration and the RQ value will be infinity.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 14 Respiration 8
(iii) In some succulent plants like Opuntia, Bryophyllum carbohydrates are partially oxidised to organic acid, particularly malic acid without corresponding release of CO2 but O2 is consumed hence the RQ value will be zero.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 14 Respiration 9
(iv) When respiratory substrate is protein or fat, then RQ will be less than unity.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 14 Respiration 10
(v) When respiratory substrate is an organic acid the value of RQ will be more than unity.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 14 Respiration 11

Question 6.
Describe an experiment to demonstrate the production of CO2 in aerobic respiration.
Answer:
Take small quantity of any seed (groundnut or bean seeds) and allow them to germinate by imbibing them. While they are germinating place them in a conical flask. A small glass tube containing 4 ml of freshly prepared Potassium hydroxide (KOH) solution is hung into the conical flask with the help of a thread and tightly close the one holed cork. Take a bent glass tube, the shorter end of which is inserted into the conical flask through the hole in the cork, while the longer end is dipped in a beaker containing water.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 14 Respiration 12
Observe the position of initial water level in bent glass tube. This experimental setup is kept for two hours and the seeds were allowed to germinate. After two hours, the level of water rises in the glass tube. It is because, the CO2 evolved during aerobic respiration by germinating seeds will be absorbed by KOH solution and the level of water will rise in the glass tube.
CO2 + 2KOH → K2CO3 + H2O
In the case of groundnut or bean seeds, the rise of water is relatively lesser because these seeds use fat and proteins as respiratory substrate and release a very small amount of CO2. But in the case of wheat grains, the rise in water level is greater because they use carbohydrate as respiratory substrate. When carbohydrates are used as substrate, equal amounts of CO2 and O2 are evolved and consumed.

Textbook Page No. 145

Question 1.
How many ATP molecules are produced from one sucrose molecule?
Answer:
One sucrose molecules gives rise to two glucose molecules. The net production of ATP during complete oxidation of one glucose molecule in plant cell is 36 ATP. Therefore one sucrose molecule yields 36 x 2 = 72 ATP molecules.
As per recent view in plants cells, one molecules of glucose, after complete aerobic oxidation yields only 30 ATP molecules and hence one sucrose molecule yield only 30 x 2 = 60 ATP molecules.

Textbook Page No. 156

Question 1.
Why Microorganisms respire anaerobically?
Answer:
Some of the microorganism live in environments devoid of oxygen and they have to adopt themselves in anoxic condition. Hence they respire anaerobically and they are called anaerobic microbes.

Samacheer Kalvi 11th Bio Botany Solutions 14 Respiration

Question 2.
Does anaerobic respiration take place in higher plants?
Answer:
Anaerobic respiration some time occur in the root of some water – logged plants.

Samacheer Kalvi 11th Bio Botany Solutions Chapter 15 Plant Growth and Development

Students can Download Bio Botany Chapter 15 Plant Growth and Development Questions and Answers, Notes Pdf, Samacheer Kalvi 11th Bio Botany Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Bio Botany Solutions Chapter 15 Plant Growth and Development

Samacheer Kalvi 11th Bio Botany Plant Growth and Development Text Book Back Questions and Answers

Question 1.
Select the wrong statement from the following:
(a) Formative phase of the cells retain the capability of cell division.
(b) In elongation phase development of central vacuole takes place.
(c) In maturation phase thickening and differentiation takes place.
(d) In maturation phase, the cells grow further.
Answer:
(d) In maturation phase, the cells grow further.

Question 2.
If the diameter of the pulley is 6 inches, length of pointer is 10 inches and distance travelled by pointer is 5 inches. Calculate the actual growth in length of plant:
(a) 3 inches
(b) 6 inches
(c) 12 inches
(d) 30 inches
Answer:
(a) 3 inches

Question 3.
In uni sexual plants, sex can be changed by the application of:
(a) ethanol
(b) cytokinins
(c) ABA
(d) auxin
Answer:
(c) ABA

Question 4.
Select the correctly matched one:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 15 Plant Growth and Development
(a) A – (iii), B – (iv), C – (v), D – (vi), E – (i), F – (ii)
(b) A – (v), C – (ii), D – (iv), E – (vi), F – (iii)
(c) A – (iii), B – (v), C – (vi),D – (i), E – (ii), F – (iv)
(d) A – (ii), B – (iii), C – (v), D – (vi), E – (iv), F – (i)
Answer:
(b) A – (v), C – (ii), D – (iv), E – (vi), F – (iii)

Question 5.
Seed dormancy allows the plants to:
(a) overcome un favorable climatic conditions
(b) develop healthy seeds
(c) reduce viability
(d) prevent deterioration of seeds
Answer:
(a) overcome unfavorable climatic conditions

Samacheer Kalvi 11th Bio Botany Solutions Chapter 15 Plant Growth and Development

Question 6.
What are the parameters used to measure growth of plants?
Answer:

  • Increase in length or girth (roots and stems)
  • Increase in fresh or dry weight
  • Increase in area or volume (fruits and leaves)
  • Increase in number of cells produced.

Question 7.
What is plasticity?
Answer:
Plants follow different pathways in response to environment or phases of life to form different kinds of structures. This ability is called plasticity,
eg : Heterophylly in cotton and coriander. In such plants, the leaves of the juvenile plant are different in shape from those in mature plants.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 15 Plant Growth and Development
On the other hand, the difference in shapes of leaves produced in air and those produced in water in buttercup also represent he heterophyllous development due to the environment. This phenomenon of heterophylly is an example of plasticity.

Question 8.
Write the physiological effects of Cytokinins.
Answer:

  1. Cytokinin promotes cell division in the presence of auxin (IAA).
  2. Induces cell enlargement associated with IAA and gibberellins
  3. Cytokinin can break the dormancy of certain light-sensitive seeds like tobacco and induces seed germination.
  4. Cytokinin promotes the growth of lateral bud in the presence of apical bud.
  5. Application of cytokinin delays the process of aging by nutrient mobilization. It is known as Richmond Lang effect.
  6. Cytokinin:
    • increases rate protein synthesis
    • induces the formation of inter-fascicular cambium
    • overcomes apical dominance
    • induces formation of new leaves, chloroplast and lateral shoots.
  7. Plants accumulate solutes very actively with the help of cytokinins.

Question 9.
Describe the mechanism of photoperiodic induction of flowering.
Answer:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 15 Plant Growth and Development
Photoperiodic stimulus is perceived by the leaves. Floral hormone is synthesised in leaves and translocated to the apical tip to promote flowering. This can be explained by a simple experiment on Cocklebur (Xanthium pensylvanicum), a short day plant. Usually Xanthium will flower under short day conditions. If the plant is defoliated and kept under short day conditions it will not flower.

Flowering will occur even when all the leaves are removed except one leaf. If a cocklebur plant is defoliated and kept under long day conditions, it will not flower. If one of its leaves is exposed to short day condition and rest are in long day condition, flowering will occur.

Samacheer Kalvi 11th Bio Botany Solutions Chapter 15 Plant Growth and Development

Question 10.
Give a brief account on Pr grammed Cell Death (PCD).
Answer:
Senescence is controlled by plants own genetic program and death of the plant or plant part consequent to senescence is called Programmed Cell Death. In short senescence of an individual cell is called PCD. The proteolytic enzymes involving PCD in plants are phytases and in animals are caspases. The nutrients and other substrates from senescing cells and tissues are remobilized and reallocated to other parts of the plant that survives.

The protoplasts of developing xylem vessels and tracheids die and disappear at maturity to make them functionally efficient to conduct water for transport. In aquatic plants, aerenchyma is normally formed in different parts of the plant such as roots and stems which encloses large air spaces that are created through PCD. In the development of unisexual flowers, male and female flowers are present in earlier stages, but only one of these two completes its development while other aborts through PCD.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 15 Plant Growth and Development

Samacheer Kalvi 11th Bio Botany Plant Growth and Development Additional Questions & Answers

I. Choose T he correct answer (1 Mark)
Question 1.
Open form of the growth occurs in:
(a) leaves and flowers
(b) stem and root
(c) leaves and stem
(d) stem and flowers
Answer:
(b) stem and root

Question 2.
Bamboo is classified under:
(a) monocarpi c annual plants
(b) polycarpic perennials
(c) monocarpic perennials
(d) polycarpic annual plants
Answer:
(c) monocarpic perennials

Question 3.
Primary growth of the plant is due to the activity of:
(a) phloem parenchyma
(b) phloem meristem
(c) vascular cambium
(d) apical meristem
Answer:
(d) apical meristem

Question 4.
One single maize root apical meristem can give rise to more than:
(a) 17,500 hew cells per hour
(b) 18,500 new cells per hour
(c) 19,000 new cells per hour
(d) 500 new cells per hour
Answer:
(a) 17,500 hew cells per hour

Question 5.
Thickening and differentiation of cells take place during:
(a) elongation phase
(b) formative phase
(c) maturation phase
(d) flowering phase
Answer:
(c) maturation phase

Samacheer Kalvi 11th Bio Botany Solutions Chapter 15 Plant Growth and Development

Question 6.
When die total growth of a plant is plotted against time, the shape of the curve obtained is:
(a) hyperbolic curve
(b) ‘S’ shaped sigmoid curve
(c) linear curve
(d) none of the above
Answer:
(b) ‘S’ shaped sigmoid curve

Question 7.
The total growth of the plant consists of four phases in the following order.
(a) Log phase, lag phase, decelerating phase and maturation phase
(b) Log phase, lag phase, maturation phase and decelerating phase
(c) Lag phase, log phase, maturation phase and decelerating phase
(d) Lag phase, log phase, decelerating phase and maturation phase
Answer:
(d) Lag phase, log phase, decelerating phase and maturation phase

Question 8.
Internal factors, that influences the growth of the plant is:
(a) nutrition
(b) light
(c) C / N ratio
(d) oxygen
Answer:
(c) C / N ratio

Question 9.
Absence of light may lead to yellowish in color in plants and this is called:
(a) venation
(b) etiolation
(c) estivation
(d) vernation
Answer:
(b) etiolation

Question 10.
Differentiated cells, after multiplication again lose the ability to divide and mature to perform specific functions. This is called:
(a) plasticity
(b) differentiation
(c) dedifferentiation
(d) redifferentiation
Answer:
(d) redifferentiation

Question 11.
Indicate a plant growth regulator from the following:
(a) cytocin
(b) cytokinins
(c) acetic acid
(d) methylene
Answer:
(b) cytokinins

Samacheer Kalvi 11th Bio Botany Solutions Chapter 15 Plant Growth and Development

Question 12.
Some of the polyamines are known to behave like:
(a) growth inhibitors
(b) plant hormones
(c) flowering inhibitors
(d) fruit ripening agent
Answer:
(b) plant hormones

Question 13.
The activity of synergistic effect involves the activity of:
(a) auxin and gibberellins
(b) auxin and ethylene
(c) ABA and gibberellins
(d) none of the above
Answer:
(a) auxin and gibberellins

Question 14.
Phytohormones are usually produced to in tips of:
(a) root alone
(b) stem alone
(c) leaves alone
(d) root, stem and leaves
Answer:
(d) root, stem and leaves

Question 15.
The term auxin was first coined by:
(a) Charles Darwin
(b) Kogl
(c) F.W. Went
(d) Smith
Answer:
(c) F.W. Went

Question 16.
Indole Acetic Acid (IAA) is a:
(a) growth inhibitor
(b) hetero auxin
(c) root inhibitor
(d) synthetic auxin
Answer:
(b) hetero auxin

Question 17.
Indicate a synthetic auxin.
(a) Indole Acetic Acid
(b) Phenyl Acetic Acid
(c) Indole Butyric Acid
(d) Napthalene Acetic Acid
Answer:
(d) Napthalene Acetic Acid

Question 18.
Auxin has a similar chemical structure of:
(a) Indole acetic acid
(b) Napthalene acetic acid
(c) Phenyl acetic acid
(d) 2,4 – Dichloro phenoxy
Answer:
(a) Indole acetic acid

Question 19.
Auxin stimulates:
(a) transpiration
(b) respiration
(c) flowering
(d) none of the above
Answer:
(b) respiration

Samacheer Kalvi 11th Bio Botany Solutions Chapter 15 Plant Growth and Development

Question 20.
The term gibberllin was named by:
(a) Brain
(b) Yabuta
(c) Sumiki
(d) kurosawa
Answer:
(b) Yabuta

Question 21.
Who established the structure of gibberellic acid?
(a) Brain etal
(b) Kurosawa
(c) Cross et al
(d) Yabuta and Sumiki
Answer:
(c) Cross et al

Question 22.
Formation of seedless fruits without fertilization is induced by:
(a) auxin
(b) cytokinin
(c) ethylene
(d) gibberellin
Answer:
(d) gibberellin

Question 23.
Cytokinins inducing cell division was first demonstrated by:
(a) Haberlandt
(b) Charles Darwin
(c) Clarke
(d) Hubert
Answer:
(a) Haberlandt

Question 24.
Zeatin is first isolated from unripe grains of:
(a) paddy
(b) wheat
(c) maize
(d) com
Answer:
(c) maize

Question 25.
Indicate correct statements.
(i) Genes are intracellular factors for growth.
(ii) Temperature has no role in the growth of plant.
(iii) Oxygen has a vital role in the growth of plants.
(iv) CIN ratio of soil does not affect the growth of plant.
(a) (i) and (iv)
(b) (ii) and (iv)
(c) (i) and (iii)
(d) (ii) and (iii)
Answer:
(c) (i) and (iii)

Question 26.
Aspartic acid is classified under:
(a) freeauxin
(b) precursor of auxin
(c) chemical structure of auxin
(d) bound auxin
Answer:
(d) bound auxin

Question 27.
The stress phytohormones (Abscisic acid) was first isolated by:
(a) Linn et al
(b) Addicott et al
(c) Edward et al
(d) Stone and Black
Answer:
(b) Addicott et al

Question 28.
The chemical structure of abscisic acid resembles the structure of:
(a) indole Acetic Acid
(b) malanic acid
(c) carotenoid
(d) xanthophyll
Answer:
(c) carotenoid

Question 29.
Pick out the correct statement from the following:
(i) Abscisic acid is found abundantly inside the chloroplast of green cells.
(ii) ABA is a powerful growth promotor.
(iii) ABA is formed from pentose phosphate pathway.
(iv) ABA has anti-auxih and anti-gibberellin property.
(a) (i) and (iv)
(b) (i) and (ii)
(c) (ii) and (iii)
(d) (ii) and (iv)
Answer:
(a) (i) and (iv)

Samacheer Kalvi 11th Bio Botany Solutions Chapter 15 Plant Growth and Development

Question 30.
Abscisic acid induces male flower formation on female plants of:
(a) potato
(b) Cannabis sativa
(c) Vinca rosea
(d) Delomix regia
Answer:
(b) Cannabis sativa

Question 31.
Pea and barley are classified under:
(a) short day plants
(b) short long day plants
(c) long day plants
(d) long short day plants
Answer:
(c) long day plants

Question 32.
The term ‘photoperiodism’ was coined by:
(a) Miller and Amald
(b) Gamer and Allard
(c) Michael and Edward
(d) Darwin and Lamark
Answer:
(b) Gamer and Allard

Question 33.
Usually Xanthiumpensylvanicum will flower under:
(a) long day condition
(b) short long day condition
(c) photoneutral condition
(d) short day condition
Answer:
(d) short day condition

Question 34.
Phytochrome is a:
(a) reddish xanthophyll pigment
(b) bluish biliprotein pigment
(c) rodopsin pigment
(d) none of the above
Answer:
(b) bluish biliprotein pigment

Question 35.
Who found out the phytochrome in plants?
(a) Butler et al
(b) Michell et al
(c) Boumick et al
(d) Gamers and Allard
Answer:
(a) Butler et al

Question 36.
The term “vernalization” was first used by:
(a) Gamer
(b) Michell
(c) Lysenko
(d) Kawasacki
Answer:
(c) Lysenko

Question 37.
Pick out the wrong statement from the following:
(a) Vernalization increases the cold resistance of plants
(b) It increase the resistance of plants to fungal disease
(c) Vemalizatiqn increase the vegetative period of the plant
(d) It accelerates the plant breeding
Answer:
(c) Vemalizatiqn increase the vegetative period of the plant

Question 38.
In Oxalis, the seed viability ranges from:
(a) 10 to 15 years
(b) a few days
(c) more than 100 years
(d) upto 100 years
Answer:
(b) a few days

Question 39.
In apple and plum, the method of breaking seed dormancy involves the process of:
(a) impaction
(b) Scarification
(c) exposing to red light
(d) Stratification
Answer:
(d) Stratification

Question 40.
The proteolytic enzymes involved in – programmed cell death in plants are:
(a) phytochrome
(b) caspases
(c) phytaspases
(d) protolysis
Answer:
(c) phytaspases

II. Answer the following (2 Marks)

Question 1.
Define closed form of growth in plants.
Answer:
Leaves, flowers and fruits are limited in growth or of determinate or closed form growth.

Question 2.
What is meant by grand period of growth in plants?
Answer:
The total period from initial to the final stage of growth is called the grand period of growth. The total growth is plotted against time and ‘S’ shaped sigmoid curve (Grand period curve) is obtained.

Question 3.
Name the phases of growth in ‘S’ shaped growth curve.
Answer:

  • Lag phase
  • Log phase
  • Decelerating phase
  • Maturation phase

Question 4.
Define arithmetic growth rate in plant organ.
Answer:
If the length of a plant organ is plotted against time, it shows a linear curve and this growth is called arithmetic growth.

Samacheer Kalvi 11th Bio Botany Solutions Chapter 15 Plant Growth and Development

Question 5.
Distinguish between absolute growth rate and relative growth, rate.
Answer:
Absolute growth rate:
Increase in total growth of two organs measured and compared per unit time is called absolute growth rate.

Relative growth rate:
The growth of the given system per unit time expressed per unit initial parameter is called relative growth rate.

Question 6.
Define the term etiolation
Answer:
Light has its own contribution in the growth of the plant. Light is important for growth and photosynthesis. Light stimulates healthy growth. Absence of light may lead to yellowish in colour. This is called etiolation.

Question 7.
What is meant by redifferentiation of plant cells?
Answer:
Differentiated cells, after multiplication again lose the ability to divide and mature to perform specific functions. This is called redifferentiation, eg: Secondary xylem and Secondary phloem.

Question 9.
Mention any two synthetic auxins.
Answer:

  • 2,4 – Dichloro Phenoxy Acetic Acid (2,4 – D)
  • 2, 4, 5 – Trichloro Phenoxy Acetic Acid (2,4,5 – T)

Question 10.
Explain the synergistic effect of phytochromes.
Answer:
The effect of one or more substance in such a way that both promote each others activity, eg: Activity of auxin and gibberellins or cytokinins.

Question 11.
Name the natural auxins present in plants.
Answer:

  • Indole Acetic Acid (IAA)
  • Indole Propionic Acid (IPA)
  • Indole Butyric Acid (IBA)
  • Phenyl Acetic Acid (PAA)

Question 12.
Mention any two physiological effect of auxins in plant.
Answer:

  • They promote cell elongation in stem and coleoptile.
  • At higher concentrations auxins inhibit the elongation of roots but induce more lateral roots. Promotes growth of root only at extremely low concentrations.

Question 13.
Match the following.

(i) Indole acetic acid (a) bolting
(ii) Napthalene acetic acid (b) anti-auxin
(iii) Gibberellins (c) synthetic auxin
(iv) Abscisic acid (d) Natural auxin

Answer:
(i) – (d) Natural auxin
(ii) – (c) synthetic auxin
(iii) – (a) bolting
(iv) – (b) anti-auxin

Question 14.
Where do you find cytokinin hormone in plants?
Answer:
The distribution of cytokinin in plants is not as wide as those of auxin and gibberellins but found mostly in roots. Cytokinins appear to be translocated through xylem.

Question 15.
What is Richmond Lang effect?
Answer:
Application of cytokinin delays the process of aging by nutrient mobilization. It is known as Richmond Lang effect.

Samacheer Kalvi 11th Bio Botany Solutions Chapter 15 Plant Growth and Development

Question 16.
What is meant by non-climacteric fruits?
Answer:
All fruits cannot be ripened by exposure to ethylene. Such fruits are called nonclimacteric fruits and are insensitive to ethylene, eg: Grapes, Watermelon, Orange.

Question 17.
Why do you call Abscisic acid (ABA) as stress hormone?
Answer:
It inhibits the shoot growth and promotes growth of root system. This character protect the plants from water stress. Hence, ABA is called as stress hormone.

Question 18.
What is meant by short day plants?
Answer:
The plants that require a short critical day length for flowering are called short day plants or long night plants, eg: Tobacco, Cocklebur, Soybean, Rice and Chrysanthemum.

Question 19.
Write down the importance of photoperiodism in plants.
Answer:

  • The knowledge of photoperiodism plays an important role in hybridisation experiments.
  • Photoperiodism is an excellent example of physiological pre-conditioning that is using an external factor to induce physiological changes in the plant.

Question 20.
Define the term vernalization.
Answer:
Besides photoperiod certain plants require a low temperature exposure in their earlier stages for flowering. Many species of biennials and perennials are induced to flower by low temperature exposure (0°C to 5°C). This process is called Vernalization.

Question 21.
What is meant by Epigeal germination?
Answer:
During epigeal germination cotyledons are pushed out of the soil. This happens due to the elongation of the hypocotyl. eg: Castor and Bean.

Samacheer Kalvi 11th Bio Botany Solutions Chapter 15 Plant Growth and Development

Question 22.
Explain the term seed dormancy.
Answer:
The condition of a seed when it fails to germinate even in suitable environmental condition is called seed dormancy.

Question 23.
Define the term phytogerontology.
Answer:
The branch of botany which deals with ageing, abscission and senescence is called Phytogerontology.

Question 24.
Explain the term programmed cell death.
Answer:
Senescence is controlled by plants own genetic programme and death of the plant or plant part consequent to senescence is called Programmed Cell Death. In short senescence of an individual cell is called PCD.

Question 25.
Define the term “Abscission”.
Answer:
Abscission is a physiological process of shedding of organs like leaves, flowers, fruits and seeds from the parent plant body.

III. Answer the following (3 Marks)

Question 1.
Explain the different phases of growth in plants.
Answer:
There are three phases of growth.
1. Formative phase:
Growth in this phase occurs in. meristematic cells of shoot and root tips. These cells are small in size, have dense protoplasm, large nucleus and small vacuoles. Cells divide continuously by mitotic cell division. Some cells retain capability of cell division while other cells enter the next phase of growth.

2. Elongation Phase:
Newly formed daughter cells are pushed out of the meristematic zone and increases the volume. It requires auxin and food supply, deposition of new cell wall materials (intussusception), addition of protoplasm and development of central vacuole take place.

3. Maturation Phase:
During this stage cells attain mature form and size. Thickening and differentiation takes place. After differentiation, the cells do not grow further.

Question 2.
Draw the ‘S’ shaped growth curve and mark the different phases of growth
Answer:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 15 Plant Growth and Development

Question 3.
Mention the internal factors, that affect the growth of plant.
Answer:

  • Genes are intracellular factors for growth.
  • Phytohormones are intracellular factors for growth, eg: auxin, gibberellin, cytokinin.
  • C/N ratio.

Question 4.
What are the characteristic features of phytohormones?
Answer:

  • Usually produced in tips of roots, stems and leaves.
  • Transfer of hormones “from one place to another takes part through conductive systems.
  • They are required in trace quantities.
  • AH hormones are organic in nature.
  • There are no specialized cells or organs for their secretion.
  • They are capable of influencing physiological activities leading to promotion, inhibition and modification of growth.

Question 5.
List out the agricultural applications of auxins.
Answer:

  • It is used to eradicate weeds, eg: 2,4 – D and 2,4,5 – T.
  • Synthetic auxins are used in the formation of seedless fruits (Parthenocarpic fruit).
  • It is used to break the dormancy in seeds.
  • Induce flowering in Pineapple by NAA & 2,4 – D.
  • Increase the number of female flowers and fruits in cucurbits.

Question 6.
Mention any three physiological effects of gibberellins.
Answer:

  • It produces extraordinary elongation of stem caused by cell division and cell elongation.
  • Rosette plants (genetic dwarfism) plants exhibit excessive inter modal growth when they are treated with gibberellins. This sudden elongation of stem followed by flowering is called bolting.
  • Gibberellin breaks dormancy in potato tubers.

Question 7.
What are the uses of ethylene in agriculture?
Answer:

  • Ethylene normally reduces flowering in plants except in Pine apple and Mango.
  • It increases the number of female flowers and decreases the number of male flowers.
  • Ethylene spray in cucumber crop produces female flowers and increases the yield.

Question 8.
What is meant by climacteric fruits?
Answer:
In most of the plants, there is sharp rise in respiration rate near the end of the development of fruit, called climacteric  rise. Such fruits are called climacteric fruits. The ripening on demand can be induced in these fruits by exposing them to normal air containing about 1 ppm of ethylene. A liquid called ethephon is being used in fruit ripening as it continuously releases ethylene, eg: Tomato, Apples, Banana, Mango.

Samacheer Kalvi 11th Bio Botany Solutions Chapter 15 Plant Growth and Development

Question 9.
Give the classification of plants based on photoperiodism.
Answer:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 15 Plant Growth and Development

Question 10.
Explain the term photoperiodic induction.
Answer:
An appropriate photoperiod in 24 hours cycle constitutes one inductive cycle. Plants , may require one or more inductive cycles for flowering. The phenomenon of conversion of leaf primordia into flower primordia under the influence of suitable inductive cycles is called photoperiodic induction.
eg: Xanthium (SDP) – 1 inductive cycle and Plantago (LDP) – 25 inductive cycles.

Question 11.
What are the practical applications of vernalization in plants?
Answer:

  • Vernalization shortens’ the vegetative period and induces the plant to flower earlier.
  •  It increases the cold resistance of the plants.
  • It increases the resistance of plants to fungal disease.
  • Plant breeding can be accelerated.

Question 12.
Write down the internal factors, that affect seed germination.
Answer:
1. Maturity of embryo:
The seeds of some plants, when shed will contain immature embryo. Such seeds germinate only after maturation of embryo.

2. Viability:
Usually seeds remain viable or living only for a particular period. Viability of seeds range from a few days (eg: Oxalis) to more than hundred years. Maximum viability (1000 years) has been recorded in lotus seeds. Seeds germinate only within the period of viability.

3. Dormancy:
Seeds of many plants are dormant at the time of shedding.

Samacheer Kalvi 11th Bio Botany Solutions Chapter 15 Plant Growth and Development

Question 13.
Mention the factors causing dormancy of seeds.
Answer:

  • Hard, tough seed coat causes barrier effect as impermeability of water, gas and restriction of the expansion of embryo prevents seed germination.
  • Many species of seeds produce imperfectly developed embryos called rudimentary embryos which promotes dormancy.
  • Lack of specific light requirement leads to seed dormancy.
  • A range of temperatures either higher or lower cause dormancy.
  • The presence of inhibitors like phenolic compounds which inhibits seed germination cause dormancy.

Question 14.
What are the significances of abscission?
Answer:

  • Abscission separates dead parts of the plant, like old leaves and ripe fruits.
  • It helps in dispersal of fruits and continuing the life cycle of the plant.
  • Abscission of leaves in deciduous plants helps in water conservation during summer.
  • In lower plants, shedding of vegetative parts like gemmae or plantlets help in vegetative reproduction.

IV. Answer the following (5 Marks)

Question 1.
Describe the geometric growth rate in plants with suitable diagram.
Answer:
Geometric growth rate:
This growth occurs in many higher plants and plant organs and is measured in size or weight. In plant growth, geometric cell division results if all cells of an organism or tissue are active mitotically. eg: Round three in the given figure, produces 8 cells as 23 – 8 and after round 20 there are 220 = 1,048,576 cells. The large plant or animal parts are produced this way. In fact, it is common in animals but rare in plants except when they are young and small. Exponential growth curve can be expressed as,
Samacheer Kalvi 11th Bio Botany Solutions Chapter 15 Plant Growth and Development

W1 = W0ert
W1 = Final size (weight, height and number)
W0 = Initial size at the beginning of the period
r = Growth rate
t = Tittle of growth
e = Base of the natural logarithms

Here V is the relative growth rate and also a measure of the ability of the plant to produce new plant material, referred to as efficiency index. Hence, the final size of W1 depends on the initial size W0.

Question 2.
Describe the experiment to measure the increase in length of the stem tip using an arc auxanometer.
Answer:
The increase in the length of the stem tip can easily be measured by an arc auxanometer which consists of a small pulley to the axis of which is attached a long pointer sliding over a graduated arc. A thread one end of which is tied to the stem tip and another end to a weight passes over the pulley tightly.

As soon as the stem tip increases in length, the pulley moves and the pointer slide over the graduated arc (Figure). The reading is taken. The actual increase in the length of the stem is then calculated by knowing the length of the pointer and the radius of the pulley. If the radius of the pulley is 4 inches and the length of pointer 20 inches the actual growth is measured as follows:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 15 Plant Growth and Development
Samacheer Kalvi 11th Bio Botany Solutions Chapter 15 Plant Growth and Development

Question 3.
Write an essay on the phytochrome, Gibberelf ns in plants.
Answer:
1. Discovery:
The effect of gibberellins had been known in Japan since early 1800 where certain rice plants were found to suffer from ‘Bakanae’ or foolish seedling disease. This disease was found . by Kurosawa (1926) to be caused by a fungus Gibberella fujikuroi. The active substance was separated from fungus and named as gibberellin by Yabuta (1935). These are more than 100 gibberellins reported from both fungi and higher plants. They are noted as GA1, GA2, GA3 and so on. GA3 is the first discovered gibberellin. In 1938, Yabuta and Sumiki isolated gibberellin in crystalline form. In 1955, Brain et al., gave the name gibberellic acid. In 1961, Cross et al., established its structure.

2. Occurrence:
The major site of gibberellin production in plants is parts like embryo, roots and young leaves near the tip. Immature seeds are rich in gibberellins.

3. Precursors:
The gibberellins are chemically related to terpenoids (natural rubber, carotenoids and steroids) formed by 5 – C precursor, an Isoprenoid unit called Iso Pentenyl Pyrophosphate (IPP) through a number of intermediates. The primary precursor is acetate.

4. Chemical structure:
All gibberellins have gibbane ring structure.

5. Transport in plants:
The transport of gibberellins in plants is non-polar. Gibberellins are translocated through phloem and also occur in xylem due to lateral movement between vascular bundles.

6. Bioassay (Dwarf Pea. assay):
Seeds of dwarf pea are allowed to germinate till the formation of the coleoptile. GA solution is applied to some seedlings. Others are kept under control. Epicotyle length is measured and as such, GA stimulating epicotyle growth can be seen.

7. Physiological Effects:
It produces extraordinary elongation of stem caused by cell division and cell elongation.

  • Rosette plants (genetic dwarfism) plants exhibit excessive intermodal growth when they are treated with gibberellins. This sudden elongation of stem followed by flowering is called bolting.
    Samacheer Kalvi 11th Bio Botany Solutions Chapter 15 Plant Growth and Development
  • Gibberellin breaks dormancy in potato tubers.
  • Many biennials usually flower during second year of their growth. For flowering to take place, these plants should be exposed to cold season. Such plants could be made to flower without exposure to cold season in the first year itself, when they are treated with gibberellins.

8. Agricultural role:

  • Formation of seedless fruits without fertilization is induced by gibberellins eg: Seedless tomato, apple and cucumber.
  • It promotes the formation of male flowers in cucurbitaceae. It helps in crop improvement.
    Uniform bolting and increased uniform seed production.
  • Improves number and size of fruits in grapes. It increase yield.
  • Promotes elongation of inter-node in sugarcane without decreasing sugar content.
  • Promotion of flowering in long day plants even under short day conditions. .
  • It stimulates the seed germination

Question 4.
What are their physiological effects of Abscisic acid in – plants and its role in agriculture?
Answer:
Physiological effects:

  • It helps in reducing transpiration rate by closing stomata. It inhibits K+ uptake by guard cells and promotes the leakage of malic acid. It results in closure of stomata;
  • It spoils chlorophylls, proteins and nucleic acids of leaves making them yellow.
  •  Inhibition of cell division and cell elongation.
  •  ABA is a powerful growth inhibitor. It causes 50% inhibition of growth in Oat coleoptile.
  • It induces bud and seed dormancy.
  • If promotes, the abscission of leaves, flowers and fruits by forming abscission layers.
  • ABA plays an important role in plants dtiring water stress -and during drought conditions. It results in loss of turgor and closure of stomata,
  • It has anti-auxin and anti-gibberellin property.
  • Abscisic acid promotes senescence in leaves’ by causing loss of chlorophyll pigment decreasing the rate of photosynthesis and changing the rate of proteins and nucleic acid synthesis

Agricultural Role:

  • In Cannabis sativa, induces male flower formation on female plants.
  • Induction of flowers in short day plants.
  • It promotes sprouting in storage organs like Potato.
  • ABA plays an important role in plants during water stress drought conditions.
  • It inhibits the shoot growth and promotes growth of root system. This character protect the plants from water stress. Hence, ABA is called as stress hormone.

Question 5.
Write an essay on the role of ethylene on plant physiology and agriculture.
Answer:
Almost all plant tissues produce ethylene gas in minute quantities.
1. Discovery:
In 1924, Denny found that ethylene stimulates the ripening of lemons. In 1934, R. Gane found that ripe bananas contain abundant ethylene. In 1935, Cocken et al., identified ethylene as a natural plant hormone.

2. Occurrence:
Maximum synthesis occurs during climacteric ripening of fruits and tissues undergoing senescence. It is formed in almost all plant parts like roots, leaves, flowers, fruits and seeds.

3. Transport in plants:
Ethylene can easily diffuse inside the plant through intercellular spaces.

4. Precursor:
It is a derivative of amino acid methionine, linolenic acid and fumaric acid.

5. Bioassay (Gas Chromatography):
Ethylene can be measured by gas chromatography. This technique helps in the detection of exact amount of ethylene from different plant tissues like lemon and orange.

6. Physiological Effects:

  • Ethylene stimulates respiration and ripening in fruits.
  • It stimulates radial growth in sterft and roof and inhibits linear growth.
  • It breaks the dormancy of buds, seeds and storage organs.
  • It stimulates formation of abscission zone in leaves, flowers and fruits. This makes the leaves to shed prematurely.
  • Inhibition of stem elongation (shortening the intemode).
  • In low concentration, ethylene helps in root initiation.
  • Growth of lateral roots and root hairs. This increases the absorption surface of the plant roots.
  • The growth of fruits is stimulated by ethylene in some plants. It is more marked in climacteric fruits.
  • Ethylene causes epinasty.

7. Agricultural role:

  • Ethylene normally reduces flowering in plants except in Pine apple and Mango.
  • It increases the number of female flowers and decreases the number of male flowers.
  • Ethylene spray in cucumber crop produces female flowers and increases the yield.

Question 6.
Explain the two hypothesis of explaining the mechanism of vernalization.
Answer:
Two main theories to explain the mechanism of vernalization are:

1. Hypothesis of phasic development:
According to Lysenko, development of an annual seed plant consists of two phases. First phase is thermostage, which is vegetative phase requiring low temperature and suitable moisture. Next phase is photo stage which requires high temperature for synthesis of florigen (flowering hormone).

Samacheer Kalvi 11th Bio Botany Solutions Chapter 15 Plant Growth and Development

2. Hypothesis of hormonal involvement:
According to Purvis (1961), formation of a substance A from its precursor is converted into B after chilling. The substance B is unstable. At suitable temperature B is converted into stable compound D called Vemalin. Vernalin is converted to F (Florigen). Florigen induces flower formation. At high temperature B is converted to C and devemalization occurs.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 15 Plant Growth and Development

Question 7.
Write an essay on the types of senescence, its physiology and the factors affecting senescence.
Answer:
1. Types of Senescence:
Leopold (1961) has recognised four types of senescence.

(a) Overall senescence:
This kind of senescence occurs in annual plants when entire plant gets affected and dies, eg: Wheat and Soybean. It also occurs in few perennials also, eg: Agave and Bamboo.

(b) Top senescence:
It occurs in aerial parts of plants. It is common in perennials, underground and root system remains viable, eg: Banana and Gladiolus.

(c) Deciduous senescence:
It is common in deciduous plants and occurs only in leaves of plants, bulk of the stem and root system remains alive, eg: Elm and Maple.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 15 Plant Growth and Development

(d) Progressive senescence:
This kind of senescence is gradual. First it occurs in old leaves followed by new leaves f then stem and finally root system. It is common in annuals.

2. Physiology of Senescence:

  • Cells undergo changes in structure.
  • Vacuole of the cell acts as lysosome and secretes hydrolytic enzymes.
  • The starch content is decreased in the cells.
  • Photosynthesis is reduced due to loss of chlorophyll accompanied by synthesis and accumulation of anthocyanin pigments, therefore the leaf becomes red.
  • There is a marked decrease in protein content in the senescing organ.
  • RNA content of the leaf particularly rRNA level is decreased in the cells due to increased activity of the enzyme RNAase.
  • DNA molecules in senescencing leaves degenerate by the increased activity of enzyme DNAase.

3. Factors affecting Senescence:

  •  ABA and ethylene accelerate senescence while auxin and cytokinin retard senescence.
  • Nitrogen deficiency increases senescence whereas nitrogen supply retards, senescence. High temperature senescence but low retards senescence.
  • Senescence is rapid in dark than in light.
  • Water stress leads to accumulation of ABA leading to senescence.

Question 8.
Describe the methods of breaking dormancy of seeds in plants.
Answer:
The dormancy of seeds can be broken by different methods. These are:
1. Scarification:
Mechanical and chemical treatments like cutting or chipping of hard tough seed coat and use of organic solvents to remove waxy or fatty compounds are called as Scarification.

2. imp-action:
in some seeds water and oxygen are unable to penetrate micropyle due to blockage by cork cells. These seeds are shaken vigorously to remove the plug which is called Imp-action.

3. Stratification:
Seeds of rosaceous plants (Apple, Plum, Peach and Cherry) will not germinate until they have been exposed to well aerated, moist condition under low temperature (0°C to 10°C) for weeks to months. Such treatment is called Stratification.

4. Alternating temperatures: Germination of some seeds is strongly promoted by alternating daily temperatures. An alternation of low and high temperature improves the germination of seeds.

5. Light:
The dormancy of photoblastic seeds can be broken by exposing them to red light.

Samacheer Kalvi 11th Bio Botany Solutions Chapter 15 Plant Growth and Development

Solution To Activity

Textbook Page No : 164

Question 1.
Demonstration of phases of growth.
Answer:
To demonstrate and study the phases of growth, germinate a few seeds of bean on a circular filter paper soaked with water in a petridish. After two days of growth, select a few seedlings with straight radical of 2 to 3 cm length. Dry the surface of radical with a blotting paper and mark the radical from tip to base with at least 2 mm gap using water proof ink. Replace the seedlings in filter paper and observe further growth.

Observation:
The marked area in the radical will grow and increase in length and hence the marked area of 2mm is found to be grow beyond 2mm size due to the growth in the radical.

Textbook Page No: 169

Question 2.
Measurement of growth by direct method.
Answer:
Step 1: Take ordinary scale.
Step 2: Measure ground stem up to the growing point of the plant.
Step 3: Use Indian ink and mark at regular intervals to measure the length of root, stem, and girth of the trunk.

Observation:
At regular intervals measure the increase in length and girth of the trunk and it can be observed that the length of the root, stem and girth of the trunk increased with the increase in the period of growth.

 

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