CBSE Class 11

CBSE Class 11

NCERT Exemplar Solutions for Class 11 Biology Chapter 13 Photosynthesis

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NCERT Exemplar Solutions for Class 11 Biology Chapter 13 Photosynthesis

These Solutions are part of NCERT Exemplar Solutions for Class 11 Biology. Here we have given NCERT Exemplar Solutions for Class 11 Biology Chapter 13 Photosynthesis.

VERY SHORT ANSWER QUESTIONS

Question 1.
Examine the figure.
NCERT Exemplar Solutions for Class 11 Biology Chapter 13 Photosynthesis 1
(b) Can these be passed on to the progeny? How?
(c) Name the metabolic process taking place in the places marked (A) and (B).
Solution:
(a) Figure shows the chloroplast, which is green
in colour and performs photosynthesis in plants. The structure is present in plant cell.
(b) Yes, chloroplast has the power of self replication because of presence of extra nuclear DNA. Hence, known as semi- autonomous organelle.
(c) The metabolic processes that occurs in the marked places are as follows.
A-It is the stroma of chloroplast, where dark reaction of photosynthesis takes place.
B-It is the structure of extra nuclear DNA that is responsible for replication of chloroplast, when it is required in the photo synthesising cells.

Question 2.
2H2O—>4H+ + O2 + 4e–
Based on the above equation, answer the following questions
(a) Where does this reaction take place in plants?
(b) What is the significance of this reaction?
Solution:
(a) This reaction takes place in reaction centre
PS II, that is located on the inner surface of thylakoid membrane. It is known as water splitting centre, where electrons are extracted from water and the reaction is catalysed by Mn+ and Cl– ions.
(b) Spliting of water is an important event in photosynthesis are
(i) It liberates molecular oxygen as byproduct of photosynthesis and is the significant source of oxygen in air, or is essential for all living beings on earth.
(ii) Hydrogen ions produced takes part in reducing NADP to NADPH. It is a strong reducing agent.
(iii) The electrons released are transferred from PS II to PS I through a series of electron carriers thus, creating a gradient for the ATP synthesis.

Question 3.
Cyanobacteria and some other photosynthetic bacteria don’t have chloroplasts. How do they conduct photosynthesis?
Solution:

  1. In Cyanobacteria complex lamellar system (thylakoids) are present instead of chloroplast. These thylakoids are functionally analogous to the plastids of eukaryotic cells.
  2. Pigment like chlorophyll-a, C-phycocyanin, C-phycoerythrin embedded in these lamellar system and they trap solar energy and perform photosynthesis. They perform oxygenic photosynthesis.
  3. Photosynthetic bacteria possess related pigments called bacterichlorophyll which are of different types (a,b,c,d,e,f and g). Groups that contain chlorophyll, perform photosynthesis, but do not evolve oxygen.
  4. Bacteriochlorophyll, perform photosynthesis but do not evolve oxygen.
  5. Bacteriochlorophylls are photoreceptors similar to chlorophylls except for the reduction of an additional pyrrole ring and other minor differences that shift their absorption maxima to near infrared, to wavelength as long as 1000 nm.
  6. Thus, they utilize light wavelengths not used by green plants or cyanobacteria.
  7. Bacteriopheophytin is a variant of bacteriochlorophyll that has two protons are present instead of magnesium ion at its centre.

Question 4.
(a) NADP reductase enzyme is located on ……..
(b) Breakdown of proton gradient leads to release of ……….. .
(b) ATP molecules
Solution:
(a) NADP reductase enzyme is located on the outer side of thylakoid membrane.
(b) ATP molecules

Question 5.
Can girdling experiments be done in monocots? If yes, how? If no, why note?
Solution:
The girdling experiment cannot be done in monocots. The monocots vascular bundles are scattered all over the width of stem, so we cannot get the specific band of the phloem tissue which we get in dicot.

Question 6.
3CO2, + 9ATP + 6NADPH + water –>Glyceraldehyde 3-phosphate + 9ADP + 6NADP+ + 8Pi.
Analyse the above reaction and answer the following questions
(a) How many molecules of ATP and NADPH are required to fix one molecule of CO2?
(b) Where in the chloroplast does this process occur?
Solution:
(a) 2 molecules of ATP for phosphorylation and
two molecules ofNADPH for reduction are required to fix one molecule of CO2 (b) The calvin cycle occurs in the stroma of the chloroplast.

Question 7.
Does moonlight support photosynthesis? Find out.
Solution:
Moonlight does not carry enough energy to excite chlorophyll molecules, i.e; reaction centre PSI and PSII, so light dependent reactions are not initiated. Thus, photosynthesis cannot occur in moonlight.

Question 8.
ATPase enzyme consists of two parts. What are those parts? How are they arranged in the thylakoid membrane? Conformational change occur in which part of the enzyme?
Solution:
ATP synthase enzyme consists of two parts:
(a) F1– head piece is a peripheral membrane protein complex and contain the site for synthesis of ATP from ADP + pi (inorganic phosphate).
(b) F0-integral membrane protein complex that form the channel through which proton cross the inner membrane.
The arrangement of F1 and F0 in thylakoid membrane is as follows.
F0– is a portion present within the thylakoid membrane.
F1 is a portion of ATP synthase enzyme present in the stroma of chloroplast.
The conformational change occurs in F1 portion of ATP synthase thus, it facilitates the ATP synthesis.

SHORT ANSWER QUESTIONS

Question 1.
Succulents are known to keep their stomata closed during the day to check transpiration. How do they meet their photosynthetic C02 requirements?
Solution:
Succulent plants grow in dry and xeric conditions so, to prevent water loss through transpiration the stomata remains closed during day time. So that the gaseous exchange does not take place.

Thus plants have developed the mechanism to fix C02 during night in the form of malic acid, which is a 4 carbon compound and are released during the day, inside the photosynthetic cells.

Question 2.
Chlorophyll-‘a’ is the primary pigment for the light reaction. What are accessory pigments? What is their role in photosynthesis?
Solution:
Accessory pigments are also photosynthetic pigments, like chlorophyll-/), xanthophyll and carotenoids which are not directly involved in emission of excited electron, but they help in harvesting solar radiation and pass it on to chlorophyll-o.

This pigment itself absorbs maximum radiation at blue and red region. So the chief pigment of photosynthesis is chlorophyll and others (i.e, chlorophyll-/; xanthophyll and carotenoid) are accessory pigments.

Question 3.
Do reactions of photosynthesis called, as ‘Dark Reaction’ need light? Explain.
Solution:
Dark reactions is a type of independent reactions. Through various biochemical reactions CO2 is reduced to produce C6H12O6 (glucose) which does not need light. But they depend on the products formed during light reactions, i.e., NADPH2 and ATP.

Question 4.
How are photosynthesis and respiration related to each other?
Solution:
Photosynthesis and respiration are related to each other as in both mechanisms, the plants gain energy.
In photosynthesis, plants gain energy from solar radiations whereas, in respiration, they break down glucose molecule to get energy in the form of ATP molecules.

The product of photosynthesis i.e., glucose (food) is utilised in respiration to yield energy in the form of ATP. While doing so, it release many other simple molecules (CO2 + H2O) that are utilised in photosynthesis to produce more sugar.

Question 5.
If a green plant is kept in dark with proper ventilation, can this plant carry out photosynthesis? Can anything be given as supplement to maintain its growth or survival?
Solution:
The plant in given conditions cannot carry out photosynthesis. Light is necessary for any green plant to make its own food. The plant should be watered properly for its survival.

Question 6.
Photosynthetic organisms occur at different depths in the ocean. Do they receive qualitatively and quantitatively the same light? How do they adapt to carry out photosynthesis under these conditions.
Solution:
Mostly algae are present at various depth in ocean. These show great variations in its photosynthetic pigment. These can absorb different wave lengths of light and could perform photosynthesis.
Green algae-chlorophyll-o, (absorbs red) and b(absorbs blue violet).
Brown algae-chlorophyll-o, c and fucoxanthin (absorbs yellow).
Rhodophyceae-chlorophyll-o, d and phyocoerythrin.

Question 7.
What conditions enable RuBisCO to function as an oxygenase? Explain the ensuing process.
(a) Can we artificially induce the property of nitrogen-fixation in a plant, leguminous or non leguminous?
(b) What kind of relationship is observed between mycorrhiza and pine trees?
(c) Is it necessary for a microbe to be in close association with a plant to provide mineral nutrition? Explain with the help of one example.
Solution:
This is an enzyme that has dual nature. When CO2 concentration is good enough in atmosphere. It acts as carboxylase. But if concentration of O2 increase, its nature changes and it binds with O2 and acts as oxygenase enzyme that forces CO2 to enter in C2 cycle that leads to photorespiration and loss of CO2.

Question 8.
Why does the rate of photosynthesis decrease at higher temperatures?
Solution:
Photosynthesis is an enzyme specific process. All enzymes works at an optimum temperature (/. e., 25-35°C). As temperature increases, enzyme gets denatured thus leading to fall in the rate of photosynthesis.

Question 9.
Explain how during light reaction of photosynthesis, ATP Synthesis is a chemiosmotic phenomenon.
Solution:

  1. In light reaction plants solar radiation is trapped by photosynthetic pigments, which converts light energy into chemical energy.
  2. Photophosphorylation is the main event of light reaction i.e., formation of ATP from ADP + Pi by using energy of excited electron movement through electron transport chain, that is present in thylakoid membrane.
  3. The movement of ions across a selectively permeable membrane, down the electrochemical/ proton gradient is known as chemiosmosis.
  4. Chemiosmosis hypothesis of ATP formation was first proposed by Mitchell (1961), according to ATP generated by enzyme ATP synthase via a membrane, proton pump and proton gradient.
  5. ATP synthase allows ions 02 protons to pass through membrane and proton pump.
  6. Which creates a high concentration of protons (H+) in the lumen and hence diffuses across the membrane to activate ATPase, releasing ATP molecules. One molecule of ATP is released for every two (H+) ions passing through ATPase.
    NCERT Exemplar Solutions for Class 11 Biology Chapter 13 Photosynthesis 2

Question 10.
In What kind of plants do you come across ‘Kranz anatomy’? To which conditions are those plants better adapted? How are these plants better adapted than the plants, which lack this anatomy?
Solution:

    1. Kranz anatomy how dimorphism in the chloroplast structure. It is found in C4 plants. The cells of leaves consists two types of chloroplast in them.
    2. Granal Chloroplast is found in the mesophyll cells of leaves.
    3. Chloroplast have well developed grana in them. These chloroplast fixes C02 effectively even if it is present in lower concentrations.
    4. PEP carboxylase fixes CO2 to form oxaloacetic acid (4 carbon compound).
    5. Agranal Chloroplast is found in bundle sheath cells of the leaves. C3 cycle occurs in these cells in the presence of RuBisCo enzyme.
    6. The C4 plants are well adapted to high O2 concentrations and high temperature.
      C4 plants can absorb CO2 even when CO2 concentration in much low thus C4 plants can perform high rate of photosynthesis even the stomata are closed or there is the shortage of water thus, they can conserve water.
    7. Since, PEP-carboxylase is insensitive to O2 thus excess O2 has inhibitory effect in C4 pathway and no photosynthesis occurs in C4 plant.
    8. Thus, C4 plants are better adapted to tropical and desert (hot acid habitats) areas than the plants, that lack kranz anatomy.

Question 11.
Tomatoes, carrots and chilies are red in colour due to the presence of one pigment. Name the pigment. Is it a photosynthetic pigment?
Solution:
The pigments are chromoplasts, these are fat soluble carotinoid pigments like carotenes and xanthophylls. These are called accessory pigments, they absorb light and transfer energy to Chlorophyll a.

Question 12.
Observe the diagram and answer the following.
NCERT Exemplar Solutions for Class 11 Biology Chapter 13 Photosynthesis 3
(a) Which group of plants exhibit these two types of cells?
(b) What is the first product of C4 cycle?
(c) Which enzyme is there in bundle sheath cells and mesophyll cells?
Solution:
(a) Monocot plants that belongs to Graminae/
Poaceae family, e.g., sugarcane, maize etc., possess these two types of cells, i.e., bundle sheath and mesophyll cell (in kranz anatomy).
(b) First product of C4 cycle is 4-carbon compound oxaloacetic acid.
(c) Mesophyll cells consists PEP carboxylase enzyme to fix atmospheric C02 to form 4-carbon compound oxalo acetic acid, whereas bundle sheath cells consists RuBP carboxylase that fixes C02 to form 3-carbon compound 3 PGA (3 phosphoglyceric acid).

LONG ANSWER QUESTIONS

Question 1.
In the figure given below, the back line (upper) indicates action spectrum for photosynthesis and the lighter line (lower) indicates the absorption spectrum of chlorophyll-a, answer the following
NCERT Exemplar Solutions for Class 11 Biology Chapter 13 Photosynthesis 4
(a) What does the action spectrum indicate? How can we plot an action spectrum? Explain with an example.
(b) How can we derive an absorption spectrum for any substances?
(c) If chlorophyll-a is responsible for light reaction of photosynthesis, why do the action spectrum and absorption spectrum not overlap?
Solution:
(a) Action spectrum depicts the relative rates of
photosynthesis at different wavelenghths of light. Action spectrum of photosynthesis can be plotted by measurement of oxygen evolution at different wavelength. Englemann (1882) by using a green algae plotted action spectrum.
(b) Absorption spectrum of a substance can be derived by calculating amount of energy of different wavelength of light absorbed.
(c) Chlorophyll a is responsible for light reaction of photosynthesis, but the action spectrum and absorption spectrum do not overlap because, though chlorophyll a is the main pigment responsible for the absorption of light, other thylakoids pigment like chlorophyll b, xanthophylls, carotenoids,

which are accessory pigments also absorb and transfer the energy to chlorophyll a. Indeed they not only enable a wider range of wavelength of incoming light to be utilized for photosynthesis but also protect chlorophyll from photo-oxidation.

Question 2.
What are the important events and end products of the light reaction?
Solution:
Following are important events of light reaction:
(i) Excitation of chlorophyll molecule to release a pair of electrons and use their energy in the formation of ATP from ADP + Pi. This process is known as photophosphorylation.
(ii) Splitting of water molecule
(a) 2H2O —» 4H+ + 4e– + 0+↑
(b) NADP + 2H+ ->NADPH2
End products of light reaction are NADPH and ATP. Reducing power is produced in the light reaction i.e., ATP and NADPH2 molecules that are used up in dark reaction and O2 is evolved as a by product by the splitting of water.

Question 3.
Why does not photorespiration take place in C4 plants?
Solution:

  • Photorespiration is associated with C3 cycle, where plant lose CO2 fixation due to the increase in concentrate ion of O2 change in the nature of activity of RuBP carboxylase-oxygenase.
  • While C4 plants have evolved a mechanism to avoid loss of CO2.
  • There is not a direct involvement of RuBP carboxylase-oxygenase as C3 cycle operates in bundle sheath cells, where both temperature and oxygen level low.
  • CO2 fixation occurs by another enzyme PEP carboxylase in mesophyll cells andoxaloacetate is formed, that is later converted into malic acid and transported to bundle sheath cells.
  • There, it liberates CO2, which is used in Calvin cycle, operating in bundle sheath cells of C4 plants.
  • We hope the NCERT Exemplar Solutions for Class 11 Biology at Work Chapter 13 Photosynthesis, help you.
  • If you have any query regarding NCERT Exemplar Solutions for Class 11 Biology at Work Chapter 13 Photosynthesis, drop a comment below and we will get back to you at the earliest.

 

We hope the NCERT Exemplar Solutions for Class 11 Biology at Work Chapter 13 Photosynthesis, help you. If you have any query regarding NCERT Exemplar Solutions for Class 11 Biology at Work Chapter 13 Photosynthesis, drop a comment below and we will get back to you at the earliest.

NCERT Exemplar Solutions for Class 11 Biology Chapter 12 Mineral Nutrition

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NCERT Exemplar Solutions for Class 11 Biology Chapter 12 Mineral Nutrition

These Solutions are part of NCERT Exemplar Solutions for Class 11 Biology. Here we have given NCERT Exemplar Solutions for Class 11 Biology Chapter 12 Mineral Nutrition.

VERY SHORT ANSWER QUESTIONS

Question 1.
Name a plant, which accumulate silicon.
Solution:
Oryza sativa and Triticum aestivum are the plants that accumulates silicon. These plants absorbs silicon actively and accumulate them in their biomass.

Question 2.
Mycorrhiza is a mutualistic association. How do the organisms involved in this association gain from each other?
Solution:
Mycorrhiza is a mutualistic (symbiotic) association between fungus and roots of plants. The roots provide shelter and food to the fungus and the fungus helps plants in absorption of minerals, water uptake and protection against fungus.

Question 3.
Nitrogen fixation is shown by prokaryotes and not eukaryotes. Comment.
Solution:
Prokaryotes like Rhizobium and Anabaena are capable of nitrogen fixation as they contain enzyme nitrogenase but eukaryotes lack this enzyme.

Question 4.
A farmer adds Azotobacter culture to soil before sowing maize. Which mineral element is being replenished?
Solution:
Azotobacter provides nitrogen fixing bacteria which converts free nitrogen into nitrate and nitrites. It increases soil fertility.

Question 5.
What type of conditions are created by leghaemoglobin in the root nodule of a legume?
Solution:
Leghaemoglobin present in the root nodules of leguminous plants is responsible for creating anaerobic conditions and hence acts as an oxygen scavenger, protecting enzyme nitrogenase to come in contact with oxygen and help in the proper functioning of enzyme, i. e., conversion of atmospheric nitrogen to ammonia (NHj).

Question 6.
Yellowish edges appear in leaves deficient in.
Solution:
Yellowish edges or chlorosis appears in the leaves due to the deficiency of nitrogen. Its deficiency also causes delaying of flowering, interference in protein synthesis and dormancy of lateral buds.

Question 7.
Name the macronutrient which is a component of all organic compounds but it not obtained from soil.
Solution:
Carbon is an essential macronutrient, which is a component of all organic compounds but is not obtained by soil. Plant take it from atmosphere in the form of C02. Its concentration in atmosphere is about 0.03%. Plants use C02 for photosynthesis (as a source of carbon) to synthesises glucose.

Question 8.
Name one non-symbiotic nitrogen fixing prokaryote.
Solution:
Azotobacter is a non-symbotic nitrogen fixing prokaryote. It flourishs in the rice fields.

Question 9.
Complete the equation for reductive amination …….. .
NCERT Exemplar Solutions for Class 11 Biology Chapter 12 Mineral Nutrition 1
Solution:
NCERT Exemplar Solutions for Class 11 Biology Chapter 12 Mineral Nutrition 2

Question 10
Excess of Mn in soil leads to deficiency of Ca, Mg and Fe. Justify.
Solution:
When higher amounts of Mn2+ is absorbed by plants. The toxicity expressed in the form of brown sports surrounded by chlorotic vein.
It is due to the following reasons
(i) Reduction in uptake of Fe3+ and Mn2+.
(ii) Inhibition of binding of Mn2+ to specific enzymes.
(iii) Inhibition of Ca2+ translocation in shoot apex.
Thus, excess of Mn2+ causes deficiency of iron, magnesium and calcium.

SHORT ANSWER QUESTIONS

Question 1.
How is sulphur important for plants? Name the amino acids in which it is present.
Solution:
Sulphur is a macronutrient that is important for normal plant growth and development. It is also an integral part of some amino acids, proteins and helps in deciding the secondary structure of proteins as it forms disulphide bonds.

It is absorbed by the plants as SO42- ion. It is present in vitamins (biotin, thiamine), proteins, coenzyme-A, amino acid (cystein and methionine) etc. It is also an essential component of plants like (onion, garlic) and mustard.

Its deficiency causes chlorosis in young leaves, extensive root growth, formation of hard and woody stem. It also causes the reduction in juice content of citrus fruit and tea yellow disease of tea.

Question 2.
How are organisms like Pseudomonas and Thiobacillus of great significance in nitrogen cycle?
Solution:
In biological nitrogen fixation, the atmospheric N2 gets reduced to NH3 by the help of enzyme nitrogenase reductase present in some prokaryotes. NH3 is then oxidised in to N02 and NO3 by some other bacteria (Nitrosomonas and Nitrobacter) following are the various steps involved in nitrogen fixation.
NCERT Exemplar Solutions for Class 11 Biology Chapter 12 Mineral Nutrition 3
Pseudomonas and Thiobacillus are involved in the process of denitrification. They convert nitrate (NO3–) and nitrite (NO2–) into free nitrogen (N2), that is released into the atmosphere.

Question 3.
Carefully observe the following figure
NCERT Exemplar Solutions for Class 11 Biology Chapter 12 Mineral Nutrition 4
(a) Name the technique shown in the figure and the scientist who demonstrated this technique for the first time.
(b) Name atleast three plants for which this technique can be employed for their commercial production.
(c) What is the significance of aerating tube and feeding funnel in this setup?
Solution:
(a) Hydroponics,’Julius Von Sachs (1860)
(b) (i) Solanum lycopersicum (tomato)
(ii) Hibiscus asculentus (ladiesfinger)
(iii) Solanum melongena (brinjal)
(c) Aerating tube provides oxygen for the normal growth and development of the roots growing in the liquid solution. Feeding funnel is used to add water and nutrients in the hydroponic system when required.

Question 4.
Name of most crucial enzyme found in root nodules for N2-fixation? Does it require a special pink coloured pigment for its functioning? Elaborate.
Solution:
The most crucial enzyme found in the root nodules for N2-fixation is nitrogenase. It is a Mo – Fe protein that catalyses the conversion of atmospheric nitrogen to ammonia. Pink colour 8.
pigment present is root nodules of leguminous plants is called leghaemoglobin creates * anaerobic conditions for the functioning of nitrogenase enzyme.

Question 5.
Carnivorous plants exhibit nutritional adaption. Citing an example explain this fact.
Solution:
Carnivorous plants fulfill their nutritional requirements by feeding on small animals, like insects or protozoans, e.g. Nepenthes, Venus fly trap, Utricularia etc. Carnivorous plants grow in soil deficient in nitrogen.

In pitcher plant leaves are modified into pitcher which stores the juice to lure an insect. When the insect come to suck this juice, chemicals present in nectar dissolve the skin of the prey and the plant obtains nutrients (mainly nitrogen) from its skin.

Question 6.
A farmer adds/supplied Na, Ca, Mg and Fe regularly to his field and yet he observes that the plants show deficiency of Ca, Mg and Fe. Give a valid reason and suggest a way to help the farmer improve the growth of plants.
Solution:
Plant can tolerate a specific amount of micronutrients. A lesser amount of micro- nutrient can cause deficiency symptoms and higher amount can cause toxicity.

The concentration of mineral ion which reduces the dry weight of the tissues by 10% is called toxic concentration. This concentration is different for different micronutrients as well as for different plants e.g., Mn2+ is toxic beyond 600 mgg -1; (for soyabean) and (for sunflower) and beyond 5300 μgg-1.

It has also been observed that the toxicity of one micronutrient causes the deficiency of other nutrients.
To overcome such problems, farmers should use these nutrients in prescribed concentration so that the excess uptake of one element do not reduce the uptake of the element.

Question 7.
We find that Rhizobium forms nodules on the roots of leguminous plants. Also Frankia another microbe forms nitrogen fixing nodules on the roots of non-leguminous plant Alnus.
(a) Can we artificially induce the property of nitrogen-fixation in a plant, leguminous or non leguminous?
(b) What kind of relationship is observed between mycorrhiza and pine trees?
(c) Is it necessary for a microbe to be in close association with a plant to provide mineral nutrition? Explain with the help of one example.
Solution:
(a) Artificial induction in leguminous and non- leguminous plants have been tried by scientists. It’s success rate is very low because expression of gene is highly specific phenomenon. When it desired gene is introduced that may not work because conditions for its expressions are very specific.
(b) Symbiotic mutualistic relationship (mutualism) is observed between the pine roots and mycorrhiza as both are benefitted mutually.
(c) Yes it is necessary for a microbe to be in close association with plant to provide mineral nutrition, to develop a physical relationship for example Rhizobium gets into the root and involve root tissues, then only helps in nitrogen-fixation.

Question 8.
With the help of examples describe the classification of essential elements based on the function they perform.
Solution:
Based on the diverse functions of essential elements, these are categorised into following categories given below:
(i) Constituent of biomolecules: These are the essential component of biomolecules. Hence, known as structural elements of cells, e.g., carbon, hydrogen, oxygen and nitrogen.
(ii) Energy related Chemical compound: Some elements also function in providing energy to the cell e.g. phosphorus is a component of ATP which function as energy currency -of all the living system in which magne¬sium is a component of chlorophyll, which is involved in the conversion of light en¬ergy to chemical energy.
(iii) Enzyme showing catalytic effects: Many of the essential elements are required in the form of cofactors by enzymes. They function as the activator or inhibitor of enzymes, e.g., Mg2+ acts as an activator of several enzymes in both photosynthesis e.g., Ribulose bisphosphate(RuBP), Car boxylase , Phosphoenol pymvate carboxy¬lase and respiration (e.g., hexokinase and phosphofructokinase). While Zn2+ acts as an activator of alcohol dehydrogenase while Mo of nitrogenase during the course of nitrogen fixation.

Question 9.
Trace the events starting from the coming in contact of Rhizobium to a leguminous root till nodule formation. Add a note on importance of leghaemoglobin.
Solution:
Formation of Root Nodule The coordinated activities of the Rhizobiam bacteria regume depends on the chemical interaction between these symbiotic partners.
In the following diagram the principle stages in the nodule formation are summarised.
NCERT Exemplar Solutions for Class 11 Biology Chapter 12 Mineral Nutrition 5
Leghaemoglobin is an oxygen scavenger, that protects nitrogen enzyme from 02 and also creates anaerobic conditions for the reduction of N2 to NH3 by Rhizobium bacteria.

Question 10.
Give the biochemical events occurring in the root nodule of a pulse plant. What is the end product? What is its fate?
Solution:
Formation of root nodule in pulse plant is the result of infection of roots by Rhizobium. The following figure shows the process of nodule formation.
NCERT Exemplar Solutions for Class 11 Biology Chapter 12 Mineral Nutrition 6
(b) Successful infection of the root hair causes it to curl
NCERT Exemplar Solutions for Class 11 Biology Chapter 12 Mineral Nutrition 7
(c) Infected thread carries the bacteria to enter the cortex. Bacteria cause cortical and pericycle cells to divide, lead to nodule formation
NCERT Exemplar Solutions for Class 11 Biology Chapter 12 Mineral Nutrition 8
(d) Mature nodule with vascular tissues continuous with those of the roots.
The chemical reaction is as follows
N2 + 8e–+ 8H++ 16 ATP ->2NH3 + H2 + 16ADP + P1i
The reaction takes place in the presence of enzyme nitrogenase that acts in anaerobic conditions, which is created by leghaemoglobin.
Fate of Ammonia
There are two ways by which ammonia is further used.
NCERT Exemplar Solutions for Class 11 Biology Chapter 12 Mineral Nutrition 9
This reaction and transfer of NH2 group take place for amino acid to other amino acid catalysed by enzyme transaminase.

Question 11.
Hydroponics have been shown to be a successful technique for growing of plants. Yet most of the crops are still grown on land. Why?
Solution:
Hydroponics is a soil less culture and successful technique for plants, still many crops are grown on land because
(i) The major concern is its cost. The setting and handling of hydroponics requires much more investment than that of the soil based production.
(ii) Sanitization is extremely important, because especially with indoor hydroponic environments. Water borne disease can spread quickly through some methods of hydroponic production.
(iii) It is relatively a new technique and not used by the traditional farmers due to lack of knowledge.
(iv) Plants are less adaptable to the surrounding atmosphere. However weather and narrow oxygenation may minimise the production and quality of plant yield.

We hope the NCERT Exemplar Solutions for Class 11 Biology at Work Chapter 12 Mineral Nutrition, help you. If you have any query regarding NCERT Exemplar Solutions for Class 11 Biology at Work Chapter 12 Mineral Nutrition, drop a comment below and we will get back to you at the earliest.

NCERT Exemplar Solutions for Class 11 Biology Chapter 11 Transport in Plants

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NCERT Exemplar Solutions for Class 11 Biology Chapter 11 Transport in Plants

These Solutions are part of NCERT Exemplar Solutions for Class 11 Biology. Here we have given NCERT Exemplar Solutions for Class 11 Biology Chapter 11 Transport in Plants.

VERY SHORT ANSWER QUESTIONS

Question 1.
Smaller, lipid soluble molecules diffuse faster through cell membrane, but the movement of hydrophilic substances are facilitated by certain transporters which are chemically…….. .
Solution:
The movement of hydrophilic substances are facilitated by transporters which are chemically proteins. These proteins form porins, which are huge pores in the outer membranes of the plastids, mitochondria and some bacteria. These porins allow passage of small molecules through the membrane.

Question 2.
In a passive transport across a membrane. When two protein molecules move in opposite direction and independent of each other, it is called as………… .
NCERT Exemplar Solutions for Class 11 Biology Chapter 11 Transport in Plants 1.1

Solution:
Antiport which facilitates transport of molecules in both the directions across the membrane and their movement is independent of each other.

Question 3.
Osmosis is a special kind of diffusion, in which water diffuses across the cell membrane. The rate and direction of osmosis depends upon both………. .
Solution:
The rate and direction of osmosis is dependent upon the pressure and concentration gradient.

Question 4.
A flowering plant is planted in a earthen pot and irrigated. Urea is added to make the plant grow faster, but after sometime the plant dies. This may be due to ………… .
Solution:
The solution outside the plant is an hypertonic solution, and the plant cells are hypotonic in nature, so there is a gradual movement of water from plant cell to outside urea solution leading to plasmolysis of root cells and plant dies gradually due to exosmosis.

Question 5.
Absorption of water from soil by dry seeds increases the, thus helping seedlings to come out of soil.

Solution:
Imbibition of water by seed materials as starch and protein, pushes the seedlings out of the soil causing the seed to swell and increase of imbibition pressure inside the seed, contributes for germination of seeds.

Question 6.
Water moves up against gravity and even for a tree of 20 m height, the tip receives water within two hours. The most important physiological phenomenon which is responsible for the upward movement of water is………. .
Solution:
Transpiration pull is the physiological phenomenon which is responsible for the upward movement of water in tall trees the water molecules transpire from stomata, which pulls water molecules upward to the leaf from the continuous chain of water molecules carried by xylem

Question 7.
The plant cell cytoplasm is surrounded by both cell wall and cell membrane. The specificity of transport of substances are mostly across the cell membrane, because……… .
Solution:
Transpiration pull is the physiological phenomenon which is responsible for the upward movement of water in tall trees the water molecules transpire from stomata, which pulls water molecules upward to the leaf from the continuous chain of water molecules carried by xylem

Question 8.
The C4 plants are twice as efficient as C3 plants in terms of fixing C02 but lose only…. as much water C3 plants for the same amount of C02 fixed.
Solution:
C4 plants are twice as efficient as C3 plants in terms of fixing carbon in the form of glucose, but lose only half as much water as a C3 plant for the same amount of C2 fixed.

Question 9.
Movement of substances in xylem is unidirectional while in phloem it is bidirectional. Explain
Cell cycle is under genetic control and is a sequential event. Every cell prepares itself before it starts dividing. This preparation takes place in interphase stage of the cell cycle.
Solution:
Xylem is involved in the one way transport of water and minerals from soil to root ’ —> stem —> leaves. Several forces like imbibition, root pressure and finally transpiration pull. Act in this mechanism, It is a undirectional process as there is continuous loss of water at me body surface of plants.

The main function of Phloem is to transport food from source to sink where source is the part of plant responsible for food synthesis and sink are the organs requiring food for their growth and development.

These source and sink parts of a plants may vary in different phases of growth, thus the food needs to travel in both upwards and downward direction. So, phloem shows bidirectional movement of substances.

Question 10
Define water potential and solute potential.
Solution:
Water potential is a measure of free energy associated with water per unit volume (JM -3).
The water potential of pure (φw) at atmosp-heric Pressure is zero.
Additional of solutes reduce water potential (to a negative value). This reduces the of water concentration. Solutions thus have a lower water potential than pure water, the magnitude of this lowering due to dissolution of solute is called
solute potential of φs.

Question 11
An onion peel was taken and
(a) placed in salt solution for five minutes.
(b) after that it was placed in distilled water. When seen under the microscope what would be observed in (a) and (b) ?
Solution:
(a) When placed in salt solution an onion peel shrinks as water from cytoplasm of cell moves out of the cell to wards hypertonic solution.
(b) When again placed back in distilld water, cell regains it’s shape and absorbs water and become turgid.

Question 12
How does most of the water moves within the root?
Solution:
Water mostly flows in the roots via the apoplast pathway as the cortical cells are loosly packed and hence offer no resistance to water movement, through mass flow. This mass flow of water occurs due to adhesive and cohesive properties of water.
Like, symplast pathway is also involved in the movement of water molecules within the root (like, via endodermis to xylem).

Question 13
Transpiration is a necessary evil in plants. Explain.
Solution:
Loss of water in the form of water vapours from the surface of leaves of plant is called transpiration.
Transpiration a necessary evil because the plant continuously lose water in the vapour form from its body surfaces, Which creates a transpiration pull to absorb more and more water from soil through roots.
If water is not available to plants in soil, even then loss through transpiration does not ceasle, so plants sometimes sbfrws wilting.

Question 14
Describe briefly the three physical properties of water which helps in ascent of water in xylem.
Solution:
The following are physical properties of water that helps in ascent up to xylem.
Cohesive properties — Provider mutual attraction between molecules
Adhesive properties — Causes attraction of water molecules to polar surfaces (of tracheids)
Surface tension — Water molecules get attracted to each other more in liquid phase than in gas phase.

Question 15
Identify a type of molecular movement which is highly selective and requires special membrane proteins, but does not require energy.
Solution:
Facilitated diffusion’s is a highly selective passive process. Facilitated diffusion cause net transport of molecules from a low to high concentration. In facilitated diffusion special proteins help in movement of substances across the membrane without expenditure of ATP energy.

Question 16
Correct the statements.
(a) Cells shrink in hypotonic solutions and swell in hypertonic solutions.
(b) Imbibition is special type of diffusion when water is absorbed ‘*y living cells.
(c) Most of the water flow in the roots occurs via the symplast.
Solution:
(a) The cell swellSHORT ANSWER QUESTIONSter is adsorbed by living cells.
(c) Most of the water flow in roots occurs via the apoplast way.

SHORT ANSWER QUESTIONS

Question 1.
Minerals absorbed by the roots travel up the xylem. How do they reach the parts where they are needed most? Do all the parts of the plant get the same amount of the minerals?
Solution:

  1. The sabsorbed mineral are transported through the transpiration steam up the stem, to all parts of plant.
  2. The growing region of the plant, such as the apical and lateral meristems, young leaves, developing flowers, fruits, seeds and the storage organs are the chief sinks for the mineral elements.
  3. Uploading of the mineral ions occurs via fine vein endings through diffusion and active uptake by the cells.
  4. Xylem are involved in transport of inorganic nutrients where phloem transport only organic materials in plants.
  5. Mineral ions are frequently remobilised from older parts of plant like leaves to the younger regions.
  6. Most readily mobilised elements are phosphorus, sulphur, nitrogen, potassium, and some elements like calcium that forms the structural component are not remobilised.

Question 2.
Water is indispensable for life. What properties of water make it useful for all biological process on the earth?
Solution:
Following are the properties of water that make it useful for all biological processes.
(i) Water is the major solvent through which mineral nutrients enter a Plant from the soil solution.
(ii) It is an ideal solvent with neutral pH.
(iii) Water is the major constituent of protoplasm, it constitutes approximately 90% of the protoplasm.
(iv) Water acts as a medium for translocation of nutritive substances. Mineral nutrients are absorbed by the roots. Carbohydrates that are formed during photosynthesis are transported by water from cell to cell, tissue to tissue and organ to organ.
(v) Water is involved in photosynthesis in plants, as it incorporates hydrogen atom into carbohydrate and releases oxygen atoms as O2.
(vi) Water acts as an agent for temperature control. The specific heat of water helps plant in maintaining a relatively stable internal temperature.
(vii) Water is necessary for pollination in some plants in bryophytes and pteridophytes, water are essentially requires for the fertilisation process.

Question 3.
How is it that the intracellular levels of K+ are higher than extracellular levels in animal cells?
Solution:
The excitability of sensory cells, neurons and muscles is dependent on ion channels, signal transducers that provide a regulated path for the movement of inorganic ions such as Na+, K+, Ca2+, and Cl– across the plasma membrane in response to various stimuli.
Ion channels are ‘gated’ mplying that they may be open or closed. The Na+, K+, ATPase create a charge imbalance across the plasma membrane by carrying 3Na+ out of the cell for every 2K+ ion carried inside making the inside relatively negative outside.
The membrane is said to be polarised. That is the reason the intracellular levels ofK+ are higher than extracellular levels in animals cells.
NCERT Exemplar Solutions for Class 11 Biology Chapter 11 Transport in Plants 1.2

Question 4.
In a girdled plant, when water is supplied to the leaves above the girdle, leaves may remain green for sometime then wilt and ultimately die. What does it indicate?
Solution:
When water is supplied in a girdle plant to the leaves above the girdle, leaves may remain green for sometime because leaves can synthesise their own carbohydrate food through photosynthesis, they however, gradually wilt due to non-availability of water.
The system of xylem vessels from root to the leaf vein can supply the needed water during girding there is a possible loss of xylem vessels and the water supply is cut off, resulting in death of the plant.

Question 5.
Various types of transport mechanisms are needed to fulfil the mineral requirements of a plant. Why are they not fulfilled by diffusion alone?
Solution:
Ions, minerals and organic compound are transported in plants in various ways which include.
(i) Food substances ways which include v synthesised in leaves are translocated
downward towards root and stem.
(ii) Food is translocated upwards to the developing leaves, buds and fruits.
(iii) Radial transport of food occurs across the stem from the cells of pith, from cortex etc, towards epidermis.
(iv) Ions and minerals are transported upwards through xylem.
Diffusion is a slow process and allows movement of molecules only for short distances, so it cannot carry out the movements of organic and inorganic substances mentioned above. Therefore, a need arises for special long distance transport systems that permits and moves substances at a much faster rate, i.e., mas of bulk flow system through conducting tissues (translocation).

Question 6.
Will the ascent of sap be possible without the cohesion and adhesion of the water molecules? Explain.
Solution:
Ascent of sap is not possible without the cohesive and adhesive properties of water they play an important role in transport of water due to the following reasons
(i) Cohesion forces hold the water molecule together in the conducting channels, so vaccum is not created.
(ii) Adhesive forces acting between the water molecule and cellulose of cell wall make a thin film of water along the channels so that this film is pulled up by transpiration pull drawing more and more water upwards in the conducting channels from the root.

Question 7.
When a freshly collected Spirogyra filament is kept in a 10% potassium nitrate solution, it is observed that the protoplasm shrinks in size
(a) What is this phenomenon called?
(b) What will happen if the filament is replaced in distilled water?
Solution:
(a) The phenomenon, occurring is Spirogyra filament when placed in 10% potassium nitrate solution (hypertonic solution) is Plasmolysis. It occurs as water from the cell is drawn put to extracellular fluid causing the protoplast to shrink away from cell wall.
(b) The Spirogyra upon reabsorption of water, causes the protoplast to regain its original shape. This phenomenon is known as  deplasmolysis.

Question 8.
What are ‘aquaporins’? How does presence of aquaporins affect osmosis?
Solution:
Aquaporins are integral membrane proteins which form pores or channels in the membrane.
The water flows is more rapid through these pores to inside of the cell, as compared to the process of diffusion.
These are plumbing systems of the cells. They selectively conduct water in and out of the cells, while preventing the passage of ions and other solutes.

Question 9.
ABA (Abscicis Acid) is called a stress hormone.
A.How does this hormone overcome stress conditions?
B. From where does this hormone get released in leave?
Solution:
A. Stress hormone ABA (Abscisic Acid) induces closing of stomata, whenever there is scarcity of water available to the plant. This prevents the loss of water through transpiration by leaves. It also increases the tolerance of plants to various kinds of stresses.
B. (ABA) is released or transported from the stem apices to leaves.

Question 10.
How is facilitated diffusion different from diffusion?
Solution:
Difference between cytokinesis in plant cell and animal cell is as follows.
NCERT Exemplar Solutions for Class 11 Biology Chapter 11 Transport in Plants 1.3

Question 11.
Observe the diagram and answer the following.
NCERT Exemplar Solutions for Class 11 Biology Chapter 11 Transport in Plants 1.4

(a) Are these types of guard cells found in monocots or dicots?
(b) Which of these shows a higher water content (i) or (ii)?
(c) Which element plays an important role in the opening and closing of stomata?
Solution:
(a) The guard cells that are bean-shaped are
found in dicot plants.
(b) The guards cells in figure (i) are turgid as, they pull the inner wall of the cell outside thus, they have more water in figure (ii) cells are flaccid, this condition results when cells lose water and close stomatal pore.
(c) The K+ ions move from neighbouring cells to guards cells, lowering their water potential and as a result the water moves inside making them turgid and thus opening stomata.

Question 12.
Define uniport, symport and antiport. Do they require energy?
Solution:

  1. For movement of substances the biological membranes have many mechanism.
  2. Some are active and some are passive. Specific membrane proteins are also involved for special types of transport mechanisms. These mechanisms include:
  3. Uniport is a membrane transport system by an integral membrane protein that is involved in facilitated diffusion.
  4. These channels open in response to a stimulus for free flow of specific molecules in a specific direction. These channels transport molecule with a solute gradient without energy expenditure.
  5. Symport involves the movement of two or more different molecules or ions, across the membrane in the same direction, with no expenditure of energy.
  6. Antiport is called exchanger. This integral membrane protein is involved in secondary active transport of two or more different molecules or ions across the membrane in opposite directions, without affecting the transport of other molecules.
    NCERT Exemplar Solutions for Class 11 Biology Chapter 11 Transport in Plants 1.5

LONG ANSWER QUESTIONS

Question 1.
Minerals are present in the soil in sufficient amounts. Do plants need to adjust the type of solutes that reach the xylem? Which molecules help to adjust this? How do plants regulate the type and quantity of solutes that reach xylem?
Solution:

  • Plants do need to adjust the type and quantity of solutes that reach the xylem.
  • The transport of proteins in endodermal celr help in maintaining and adjusting solute movement.
  • The minerals are present in soil as charged particles with a very low concentration compared to that of roots, and thus cannot be completely transported passively across the cell membranes of roots hairs.
  • Minerals are thus transported both by active and passive processes, to the xylem.
  • Upon reaching xylem, they are further transported, upwards to sinks through transpiration stream.
  • At the sink regions mineral ions are unloaded through diffusion and active uptake by receptor cells. The mineral ions moving frequently through xylem include.

(i) Sulphur and Phosphorus in small amounts are carried in organic forms.
(ii) Njtrogen travels in plants as inorganic ions N02 and N03 but much of the nitrogen moves in the form of amino acids and related organic compounds.
(iii) Mineral ions are frequently remobilised particularly from older senescing parts. Older dying leaves export much of their mineral content to younger leaves. Similarly, before leaf fall in deciduous plants, minerals are removed to other parts.
The most readily mobilised elements are phosphorus, sulphur, nitrogen and potassium. Structural components elements like calcium are not remobilised.

Question 2.
Plants show temporary and permanent wilting. Differentiate between the two. Do any of them indicate the water status of the soil?
Solution:
The loss of turgidity of leaves and other soft aerial parts of a plant causing dropping, folding and rolling of non-woody plants is wilting. It occurs when rate of loss of water is higher than the rate of absorption.
NCERT Exemplar Solutions for Class 11 Biology Chapter 11 Transport in Plants 1.6

We hope the NCERT Exemplar Solutions for Class 11 Biology at Work Chapter 11 Transport in Plants, help you. If you have any query regarding NCERT Exemplar Solutions for Class 11 Biology at Work Chapter 11 Transport in Plants, drop a comment below and we will get back to you at the earliest.

NCERT Exemplar Solutions for Class 11 Biology Chapter 10 Cell Cycle and Cell Division

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NCERT Exemplar Solutions for Class 11 Biology Chapter 10 Cell Cycle and Cell Division

These Solutions are part of NCERT Exemplar Solutions for Class 11 Biology. Here we have given NCERT Exemplar Solutions for Class 11 Biology Chapter 10 Cell Cycle and Cell Division.

VERY SHORT ANSWER QUESTIONS

Question 1.
Between a prokaryote and a eukaryote, which cell has a shorter cell division time?
Solution:
Prokaryotic cell has simple cell structure and cellular organisation. It’s nucleus does not contain nuclear membrane. Prokaryotic cell thus has shorter cell cycle than the eukaryotic cell.

Question 2.
Name a stain commonly used to colour chromosomes.
Solution:
The chromosomes are the thickest and the shortest at metaphase. Acetocarmine and Giemsa stain can be used to stain the chromosomes. They are stained for karyo-typing for further study of chromosomes.

Question 3.
Which tissue of animals and plants exhibits meiosis?
Solution:
Meiosis is also called as reduction division, it is a special kind of cell division which occurs in germ cells or sex cells of male and female reproductive organs of plants and animals. They produce male (($) and female (C^) gametes that take part in sexual reproduction.

Question 4.
Which part of the human body should one use to demonstrate stages in mitosis?
Solution:
All the cells in the human body except germinal cells in the male and female reproductive organs are somatic cells. The somatic cells divide by mitotic cell division for growth and regeneration and can be used to demonstrate mitosis.

Question 5.
The diagram shows a bivalent at prophase-I of meiosis. Which of the four chromatids can cross over?
NCERT Exemplar Solutions for Class 11 Biology Chapter 10 Cell Cycle and Cell Division 1.1
Solution:
In prophase-I of meiosis, the homologous chromosomes lie parallel to each other in leptotene stage. Each chromosome has four chromatids and are bivalent. The non-sister chromatids of homologous chromosomes cross over in pachytene stage of prophase-I.
NCERT Exemplar Solutions for Class 11 Biology Chapter 10 Cell Cycle and Cell Division 1.2

Question 6.
If a tissue has at a given time 1024 cells, how many cycles of mitosis had the original parental single cell undergone?
Solution:
To give 1024 cells the parental cell undergoes 10 divisions of mitotic cycle.

Question 7.
An anther has 1200 pollen grains. How many pollen mother cells must have been there to produce them?
Solution:
The pollen mother cell (2n) undergoes meiotic cell divisions, each such cell produces four daughter cells with haploid (n) number of chromosomes. Three hundred pollen mother cells would have to be there to produce 1200 pollen grains, because one pollen mother cell will produce four pollen grains.

Question 8.
At what stage of cell cycle does DNA synthesis take place?
Solution:
The stage of cell cycle where DNA synthesis or replication takes place is Synthetic phase or S- phase of interphase.

Question 9.
It is said that the one cycle of cell division in human cells (eukaryotic cells) takes 24 hours. Which phase of the cycle, do you think occupies the maximum part of cell cycle?
Cell cycle is under genetic control and is a sequential event. Every cell prepares itself before it starts dividing. This preparation takes place in interphase stage of the cell cycle.
Solution:
If a cell takes 24 hours to divide, it spends 18-20 hours time in interphase stage to prepare itself to undergo cell division.

Question 10
It is observed that heart cells do not exhibit cell division. Such cells do not divide further and exit… phase to enter in inactive stage called…. of cell cycle. Fill in the blanks
Solution:
It is observed that heart cells do not exhibit cell division. Such cells do not divide further and exit G, phase to enter an inactive stage called quiescent stage (GQ) of cell cycle.
Muscle cells when reach a level of maturity, no longer divide and just perform their function all through it life.

SHORT ANSWER QUESTIONS

Question 1.
State the role of centrioles other than spindle formation.
Solution:
The animal cell are present in few membrane less cell organelles. Centrosome is one of them. Two cylindrical structures called centrioles are the part of centrosome.
In the centrosome the two centrioles lie perpendicular to each other. Each has organisation lie a cart wheel. These form the basal body of cilia and flagella of plant/animal cells besides forming spindle fibre in animal cell division. It also helps in the formation of microtubules and sperm tail.

Question 2.
Label the diagram and also determine the stage at which this structure is visible.
NCERT Exemplar Solutions for Class 11 Biology Chapter 10 Cell Cycle and Cell Division 1.3
Solution:
The transition stage between prophase and metaphase stage of mitotic cell division is shown in the diagram.
NCERT Exemplar Solutions for Class 11 Biology Chapter 10 Cell Cycle and Cell Division 1.4

Question 3.
A cell has 32 chromosomes. It undergoes mitotic division. What will be the chromosome number (n) during metaphase? What would be the DNA content (C) during anaphase?
Solution:
Mitosis helps in the growth of organism and its development. It also plays a vital role in a sexually reproducing organisms. The mitotic cell division occurs in somatic cells of an organism.

The chromosome number in the daughter cells remains same as that of the parent (dividing) cell, so even at metaphase or anaphase, the chromosome number does not change.

The DNA content gets doubled at the synthetic phase of interphase and gets divided at anaphase but the chromosome number remains same

Question 4.
While examining the mitotic stage in a tissue, one finds some cells with 16 chromosomes and some with 32 chromosomes. What possible reasons could you assign to this difference in chromosome number. Do you think cells with 16 chromosomes could have arisen from cells with 32 chromosomes orvice-versa?
Solution:
A condition as such, may arise in case of a mosaic, which denotes presence of two or more populations of cells in one individual with varying genotypes.
It can result from variou: mechanisms including non-disjunction, anaphase lagging and end replication. It may also result from a mutation during development, which is propagated to only a subset of the adult cells. In this case, cells with 16 chromosomes could have arisen from cells with 32 chromosomes.

Question 5.
The following events occur during the various phases of the cell cycle. Name the phase against each of the events.
(a) Disintegration of nuclear membrane ………
(b) Appearance of nucleolus ………
(c) Division of centromere ……..
(d) Replication of DNA ……….
Solution:
(a) Prophase
(b) Telophase
(c) Anaphase
(d) S-phase

Question 6.
Mitosis results in producing two cells which are similar to each other. What would be the consequence if each of the following irregularities occur during mitosis?
(a) Nuclear membrane fails to disintegrate
(b) Duplication of DNA does not occur
(c) Centromeres do not divide
(d) Cytokinesis does not occur
Solution:
(a) The spindle fibres would not be able to reach chromosomes if nuclear membrane fails to disintegrate and they would not move towards opposite poles of the cell.
In certain protozoans, such as Amoeba, the ‘ spindle is formed within the nucleus and this is called intra nuclear mitosis or premitosis.
(b) The cell might not be able to surpass S-phase of cell-cycles. If DNA duplication does not occur as no chromosome formation will take place, and cell will not be able to enter M-(mitotic phase) in case it enters mitosis, the cycle will cease.
(c) If the centromeres do not divide as it may result in trisomy, one of the daughter cell will receive a complete pair of chromosomes and other cell would not get any of them.
(d) Multinucleate condition called coenocyte, syncytium is produced. If cytokinesis does not occur as in Rhizopus and Vaucheria, etc

Question 7.
Both unicellular and multicellular organisms undergo mitosis. What are the difference, if any, observed in the process between the two?
Solution:
The type of cell divisions in unicellular organisms is known as amitosis in which somatic cell is directly divided into the parts. Occurs curs. In multicellular organisms.
In both unicellular and multicellular organisms. The difference between mitosis include:
NCERT Exemplar Solutions for Class 11 Biology Chapter 10 Cell Cycle and Cell Division 1.5

Question 8.
Comment on the statement-meiosis enables the conservation of specific chromosome number of each species even through the process per se results in reduction of chromosome number.
Solution:
Meiosis is the mechansim of conservation of specific chromosome number of each species across generations in organisms reproducing sexually. The process results in reduction of chromosome number by half, which is gradually conserved by union of male gamete 9n) and female gamete (n) in next generation. Meiosis also increases the genetic variability in the population of organisms from one generation to the next.

Question 9.
Name a cell that is found arrested in diplotene stage for months and years. Comment.
Solution:

  1. In mammalian occytes, meiotic arrest at diplotene stage usually occurs.
  2. In females, meiosis starts in the embryo and proceeds as for as diplotene, when the chromosomes become diffused and the cells are referred to as being in the dictyate stage. This arest is under hormonal control.
  3. In many amphibian oocyles, birds and insects with a long period of immaturity, the oocyte may be arrested in the dictyate stage for many years and spend a prolonged period in diplotene.
  4. This stage is characterised by formation of lampbrush chromosomes where intense RNA synthesis occurs and most of the genes in the DNA loops are actively transcribed and expressed.

Question 10.
How does cytokinesis in plant cells differ from that in animal cells?
Solution:
Difference between cytokinesis in plant cell and animal cell is as follows.
NCERT Exemplar Solutions for Class 11 Biology Chapter 10 Cell Cycle and Cell Division 1.6

LONG ANSWER QUESTIONS

Question 1.
Comment on the statement- Telophase is reverse of prophase.
Solution:
The following contrasting differences reveals that telophase is reverse of prophase, in cell division.
NCERT Exemplar Solutions for Class 11 Biology Chapter 10 Cell Cycle and Cell Division 1.7

Question 2.
An organisms has two pair of chromosomes (i.e., chromosome number = 4), Diagrammatically represent the chromosomal arrangement during different phases of meiosis-II.
Solution:
Meiosis is reduction division in which chromosome number reduces tc half in daughter cells. The number reduces as half set of chromosomes move to 2 daughter cells in meiosis-I. Thus two cells with half set of chromosomes again re-enter meiosis-II which is similar to mitotic cell division.
NCERT Exemplar Solutions for Class 11 Biology Chapter 10 Cell Cycle and Cell Division 1.8

We hope the NCERT Exemplar Solutions for Class 11 Biology at Work Chapter 10 Cell Cycle and Cell Division, help you. If you have any query regarding NCERT Exemplar Solutions for Class 11 Biology at Work Chapter 10 Cell Cycle and Cell Division, drop a comment below and we will get back to you at the earliest.

NCERT Exemplar Solutions for Class 11 Biology Chapter 9 Biomolecules

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NCERT Exemplar Solutions for Class 11 Biology Chapter 9 Biomolecules

These Solutions are part of NCERT Exemplar Solutions for Class 11 Biology. Here we have given NCERT Exemplar Solutions for Class 11 Biology Chapter 9 Biomolecules.

VERY SHORT ANSWER QUESTIONS

Question 1.
Medicines are either man made (i.e. synthetic) or obtained from living organisms like plants, bacteria, animals, etc., and hence, the latter are called natural products. Sometimes, natural products are chemically altered by man to reduce toxicity or side effects. Write against each of the following whether they were initially obtained as a natural product or as 3 synthetic chemical.
(a) Penicillin
(b) Sulphonamide
(c) Vitamin-C
(d) Growth hormone
Solution:
(a) Penicillin is a group oT antibiotics derived from fungi Penicillium obtained naturally.
(b) Sulphonamide an antimirobial agent is a synthetic chemical.
(c) Vitamin-C or L-ascorbic acid or ascorbate is a natural product and an essential nutrient for humans. It is present in citrus fruits.
(d) Growth hormone also known as somatotropin or somatropin is a peptide hormone occurring naturally in the body it stimulates growth.

Question 2.
Write the name of any one amino acid, sugar, nucleotide and fatty acid.
Solution:
(a) Amino acid — Leucine
(b) Sugar — Lactose
(c) Nucleotide — Adenosine triphosphate
(d) Fatty acid — Palmitic acid

Question 3.
Reaction given below is catalysed by oxidoreductase between two substrates A and
A’, complete the reaction.
A reduced + A’ oxidised —>
Solution:
Oxidoreductase is an enzyme that catalyses oxidation reduction reactions. This enzyme is associated in catalysing the transfer of electron from one molecule (the reduction), also called as electron donor, to another molecule (the oxidant), also called as electron acceptor.
The complete reaction is
NCERT Exemplar Solutions for Class 11 Biology Chapter 9 Biomolecules 1.1

Question 4.
How are prosthetic groups different from co¬factors?
Solution:
Organic compounds that are tightly bound to the apoenzyme, (an enzyme without cofactor) by covalent or non-covalent bonds are prosthetic groups e.g., peroxidase and
catalase catalyse the breakdown of hydrogen peroxide to water and oxygen where haeme is the prosthetic group and it is a part of the active site of the enzyme.
Co-factor is small, heat stable and non-protein part of conjugate enzyme. It may be inorganic or organic in nature. Co-factors when loosely bound to an enzyme is called coenzyme and when tightly bound to apoenzyme is called prosthetic group.

Question 5.
Glycine and alanine are different with respect to one substituent on the a-carbon. What are 4. the other common substituent groups?
Solution:
The common substituted groups in both the amino acids are NH2 COOH and H.
NCERT Exemplar Solutions for Class 11 Biology Chapter 9 Biomolecules 1.2

SHORT ANSWER QUESTIONS

Question 1.
Enzymes are proteins. Proteins are long chains of amino acids linked to each, other by peptide bonds. Amino acids have many functional groups in their structure.
NCERT Exemplar Solutions for Class 11 Biology Chapter 9 Biomolecules 1.3
These functional groups are many of them at least, ionisable. As they are peak acids and bases in chemical nature, this ionisation is influenced by pH of the solution. For many enzymes, activity is influenced by surrounding pH. This is depicted in the curve below, explain briefly.
Solution:
Enzymes, generally function in a narrow range of pH. Most of the enzymes show their highest activity at a particular pH called optimum pH- activity declines below and above this value. Extremely high or low pH values generally results in complete loss of activity for most enzymes. The given graph represents the maximum enzyme activity at the optimum pH.

Question 2.
Explain the association of carbohydrate to the plasma membrane and its significance.
Solution:
Secondary metabolites are chemicals produced by plants which do not play any [role] in growth, photosynthesis reproduction or other primary functions of the plant. Rubber (cis 1,4- polyisopyrene) is a secondary metabolite.
(i) Rubber is extracted from Hevea brasiliensis (rubber tree)
(ii) It is a byproduct of the lactiferous tissue of the vessels that are in the form of latex.
(iii) It contains over 400 isoprene units and thus is the largest of the terpenoids.
(iv) It is elastic, water proof and a good conductor of electricity.

Question 3.
Nucleic acids exhibit secondary structure, justify with example.
Solution:

  1. Nucleic acids are large biological molecules, essential for all known forms of life.
  2. The secondary structure of a nucleic acid molecule refers to the base pairing interactions within a single molecule or set of interacting molecules.
  3. DNA and RNA represent two main nucleic acids, their secondary structures however differ the secondary structure of DNA comprises of two complementary strands of polydeoxyribonucleotide, spirally coiled on a common axis forming a helical structure.
  4. This double helical structure of DNA is stabilized by phosphodiester bonds (between 5’ of sugar of one nucleotide and 3 sugar of another nucleotide), hydrogen bonds (between bases, and ionic interactions.

Question 4.
Comment on the statement ‘living state is a non-equilibrium steady state to be able to perform work’
Solution:

  1. Living organism are not in equilibrium because work cannot be performed by a system at equilibrium.
  2. The living organisms exist in a steady state characterised by concentration of each of the biomoleculSs.
  3. These biomolecules are in a metabolic flux. Any chemical or physical process moves simultaneously to equilibrium.
  4. Living organisms work continuously and they cannot afford to reach equilibrium.
  5. The living state thus is an a non-equilibrium steady-state to be able to perform work. This achieved by energy input provided by metabolism.

LONG ANSWER QUESTIONS

Question 1.
What are different classes of enzymes? Explain any two with the type of reactions they catalyse.
Solution:
Enzymes are divided into six classes each with
4-13 sub-classes and named accordingly by a number comparising of four digits.
(i) Oxidoreductases/dehydrogenases : These enzymes take part in oxidation, reduction or transfer of electrons,
(ii) Transferase : These enzymes transfer a functional group (other than hydrogen).
from one molecule to another. The transfer , chemical group does not occur in free state.
(iii) Hydrolases : These enzymes catalyse the hydrolysis of bonds like ester, ether,
peptide, glycosidic C-C, C-halide, P-N etc.
NCERT Exemplar Solutions for Class 11 Biology Chapter 9 Biomolecules 1.4
(iv) Lyases cause cleavage, removal of groups without hydrolysis and addition of groups to double bonds or removal of groups producing double bonds.
NCERT Exemplar Solutions for Class 11 Biology Chapter 9 Biomolecules 1.5
(v) Isomerases rearrangement of molecular structure to effect isomeric changes. They are of three types isomerases, epimerases and mutases.
NCERT Exemplar Solutions for Class 11 Biology Chapter 9 Biomolecules 1.6
(vi) Ligases catalyse bonding of two chemicals with the help of energy obtained from ATP resulting formation of bonds such as C—O, C—S, C—N and P—O e.g., pyruvate carboxylase
Pyruvric acid + C02 + ATP + H20
NCERT Exemplar Solutions for Class 11 Biology Chapter 9 Biomolecules 1.7

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