Category: Class 11 Biology

NCERT Exemplar Class 11 Biology Chapter 10 Cell Cycle and Cell Division are part of NCERT Exemplar Class 11 Biology. Here we have given NCERT Exemplar Class 11 Biology Chapter 10 Cell Cycle and Cell Division. https://www.cbselabs.com/ncert-exemplar-problems-class-11-chapter-10-cell-cycle-and-cell-division/

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

Multiple Choice Questions

Q1. Meiosis in diploid organisms results in
(a) Production of gametes
(b) Reduction in the number of chromosomes
(c) Introduction of variation
(d) All of the above
Ans: (d)‘ Meiosis in diploid organisms results in production of gametes, reduction in the number of chromosomes and introduction of variation.

Q2. At which stage of meiosis does the genetic constitution of gametes is finally
decided? –
(a) Metaphase-I (b) Anaphase-II (c) Metaphase-II (d) Anaphase-I
Ans: (d) At anaphase-I, stage of meiosis the genetic constitution of gametes is finally decided.

Q3. Meiosis occurs in organisms during
(a) Sexual reproduction
(b) Vegetative reproduction
(c) Both sexual and vegetative reproduction
(d) None of these
Ans: (a) Meiosis occurs in organisms during sexual reproduction. The production of offspring by sexual reproduction includes the fusion of two gametes, each with a complete haploid set of chromosomes. Gametes are produced through meiosis.

Q4. During anaphase-I of meiosis
(a) Homologous chromosomes separate
(b) Non-homologous chromosomes separate
(c) Sister chromatids chromosomes separate
(d) Non Sister chromatids chromosomes separate
Ans: (a) The homologous chromosomes separate, while sister chromatids remain associated at their centromeres. Separation of homologous chromosomes at anaphase is called disjunction.

Q5. Mitosis is characterised by
(a) Reduction division
(b) Equal division
(c) Both reduction and equal division
(d) Pairing of homologous chromosomes
Ans: (b) Mitosis is the most dramatic period of the cell cycle, involving a major reorganisation of virtually all components of the cell. Since the number of chromosomes in the parent and progeny cells is the same, it is also called as equational division.

Q6. A bivalent of meiosis-I consists of
(a) Two chromatids and one centromere
(b) Two chromatids and two centromeres
(c) Four chromatids and two centromeres
(d) Four chromatids and four centromeres.
Ans: (c) A bivalent of meiosis-I consists of four chromatids and two centromeres.

Q7. Cells which are not dividing are likely to be at
(a) G, (b) G2 (C) G0 (d) S phase
Ans: (c) These cells that do not divide further exit G{ phase to enter an inactive stage is called quiescent stage (G0) of the cell cycle. G0 stage of cell denotes exit “of cell from cell cycle. During G0 stage of cell cycle, cell decides to undergo differentiation. Cells in G0 stage remain metabolically active but no longer proliferate unless called on to do so depending on the requirement of the organism.

Q8. Which of the events listed below is not observed during mitosis?
(a) Chromatin condensation
(b) Movement of centrioles to opposite poles
(c) Appearance of chromosomes with two chromatids joined together at the centromere
(d) Crossing over
Ans: (d) Crossing over occurs in pachytene (it is a phase of meiosis-I). Crossing over is the exchange of genetic material (genes) between two homologous chromosomes. Crossing over is also an enzyme-mediated process and the enzyme involved is called recombinase. Crossing over leads to recombination of genetic material on the two chromosomes. Exchange of paternal and maternal chromosome material during pachytene is called crossing over.

Q9. Identify the wrong statement about meiosis.
(a) Pairing of homologous chromosomes
(b) Four haploid cells are formed
(c) At the end of meiosis number of chromosomes are reduced to half
(d) Two cycles of DNA replication occur.
Ans: (d) Meiosis involves two sequential cycles of nuclear and cell division called meiosis-I and meiosis-II but only a single cycle of DNA replication.

Q10. Select the correct statement about G1 phase.
(a) Cell is metabolically inactive
(b) DNA in the cell does not replicate
(c) It is not a phase of synthesis of macromolecules
(d) Cell stops growing.
Ans: (b) During Gj phase the ceil is metabolically active and continuously grows but does not replicate its DNA but proteins and RNA are synthesized.

Very Short Answer Type Questions
Q1. Between a prokaryote and an eukaryote, which cell has a shorter cell division time?
Ans: Prokaryotic cells has shorter cell division time than eukaryotic cells. A typical eukaryotic cell cycle is illustrated by human cells in culture. These cells divide once in approximately every 24 hours. In bacteria (E.coli) cell cycle is of 20 minutes.

Q2. Which of the phases of cell cycle is of longest duration?
Ans: Interphase

Q3. Name a stain commonly used to colour chromosomes.
Ans: Basic fuchsin, acetocarmine etc.

Q4. Which tissue of animals and plants exhibits meiosis?
Ans: Gohads (testes and ovary) in animals and sporangium in plants.

Q5. Given that the average duplication time of E.coli is 20 minutes, how much time will two E.coli cells take to become 32 cells?

For formation of 32 cells, two E.coli cells takes 4 cycles. So total time will be 4 x 20 = 80 minutes

Q6. Which part of the human body should one use to demonstrate stages in mitosis?
Ans: Nail base or any somatic cell (diploid cell).

Q7. What attributes does a chromatid require to be classified as a chromosome?
Ans: Centromere

Q8. The diagram shows a bivalent at prophase-I of meiosis. Which of the four chromatids can cross over?

Ans: Sister chromatids of homologous chromosome.

Q9. If a tissue has at a given time 1024 cells, how many cycles of mitosis had the original parental single cell undergone?

Q10. An anther has 1200 pollen grains. How many pollen mother cells (pmc) must have been there to produce them?
Ans: 4 pollen grains are produced by 1 pmc
1200 pollen grains are produced by = 1200/4
= 300 pmc

Q11. At what stage of cell cycle does DNA synthesis take place?
Ans: S-phase (interphase)

Q12. 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?
Ans: It is significant to note that in the 24 hour average duration of cell cycle of a human cell, cell division proper lasts for only about an hour. The interphase lasts more than 95% of the duration of cell cycle.

Q13. It is observed that heart cells do not exhibit cell division. Such cells do not
divide further and exit phase to enter an inactive stage called of cell cycle. Fill in the blanks.
Ans: It is observed that heart cells do not exhibit cell division. Such cells do not divide further and exit Gt phase to enter an inactive stage called G0 of cell cycle.

Q14. In which phase of meiosis are the following formed? Choose the answers from hint points given below.
a. Synaptonemal complex
b. Recombination nodules
c. Appearance/activation of enzyme recombinase
d. Termination of chiasmata
e. Interkinesis
f. Formation of dyad of cells
[Hints: (1) Zygotene, (2) Pachytene, (3) Pachytene, (4) Diakinesis, (5) After Telophase-I /before Meosis-II, (6) Telophase-I /after Meiosis-I]
Ans: a. Synaptonemal complex: zygotene
b. Recombination nodules: pachytene
c. Appearance/activation of enzyme recombinase: pachytene
d. Termination of chiasmata: diakinesis
e. Interkinesis: after Telophase-I /before Meosis-II
f. Formation of dyad of cells: Telophase-I /after Meiosis-I.

Short Answer Type Questions
Q1. State the role of centrioles other than spindle formation.
Ans: The centrioles form the basal body of cilia or flagella.

Q2. Mitochondria and plastids have their own DNA (genetic material). What is known about their fate during nuclear division like mitosis?
Ans: At the time of cytoplasmic division, organelles like mitochondria and plastids get distributed between the two daughter cells.

Q3. Label the diagram and also determine the stage at which this structure is visible.

Ans: This is a transition to metaphase

Q4. A cell has 32 chromosomes. It undergoes mitotic division. What will be the chromosome number (TV) during metaphase? What would be the DNA content (Q during anaphase?
Ans: Chromosome number (N) during metaphase = 32 (N)
DNA content (C) during anaphase = 2C

Q5. 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 or vice versa?
Ans: Cells with 16 chromosomes are produced by meiosis while that with 32 chromosomes are produced by mitosis.
• Cells with 16 chromosomes could have arisen from cells with 32 chromosomes through meiosis.
• Cells with 32 chromosomes could have arisen from cells with 16 chromosomes through fertilisation or syngamy.

Q6. 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 ________
Ans: a. Disintegration of nuclear membrane: Late prophase
b. Appearance of nucleolus: Telophase
c. Division of centromere: Anaphase
d. Replication of DNA: S-phase

Q7. 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 divid
d. Cytokinesis does not occu;
Ans: a. Nuclear membrane fails to disintegrate: In this condition, mitosis
takes place within nucleus. This is called endoduplication.
b. Duplication of DNA does not occur: There will be no mitosis
c. Centromeres do not divide: Polyploidy appears
d. Cytokinesis does not occur: In some organisms karyokinesis is not followed by cytokinesis as a result of which multinucleate condition arises leading to the formation of syncytium (e.g., liquid endosperm in coconut)

Q8. Both unicellular and multicellular organisms undergo mitosis. What are the differences, if any, observed in the process between the two?
Ans:
• The growth of multicellular organisms is due to mitosis.
• The reproduction of unicellular organisms is due to mitosis.

Q9. Name the pathological condition when uncontrolled cell division occurs.
Ans: Cancer

Q10. Two key events take place, during S phase in animal cells, DNA replication and duplication of centriole. In which parts of the cell do events occur?
Ans: In animal cells, during the S phase, DNA replication begins in the nucleus, and the centriole duplicates in the cytoplasm.

Q11. Comment on the statement—Meiosis enables the conservation of specific chromosome number of each species even though the process per se, results in a reduction of chromosome number.
Ans: Meiosis is the mechanism by which conservation of specific chromosome number of each species is achieved across generations in sexually reproducing organisms, even though the process, per se, paradoxically, results in reduction of chromosome number by half. But fertilisation restores the chromosome number.

Q12. Name a cell that is found arrested in diplotene stage for months and years. Comment in 2-3 lines how it completes cell cycle?
Ans: In oocytes of some vertebrates, diplotene can last for months or years.
• Lampbrush chromosomes or diplotene chromosome are found in diplotene
stage of most animal oocytes of frog or amphibians.
• Lampbrush chromosomes are observed in meiotic prophase. These chromosomes become normal after growth and thus completing the cell cycle.

Q13. How does cytokinesis in plant cells differ from that in animal cells?
Ans: In an animal cell, cytokinesis is achieved by the appearance of a furrow in the plasma membrane. The furrow gradually deepens and ultimately joins in the centre dividing the cell cytoplasm into two.
Plant cells however, are enclosed by a relatively inexte’nsible cell wall, therefore they undergo cytokinesis by a different mechanism. In plant cells, wall formation starts in the centre of the cell and grows outward to meet the existing lateral walls. The formation of the new cell wall begins with the formation of a simple precursor, called the cell-plate that represents the middle lamella between the walls of two adjacent cells.

Long Answer Type Questions
Q1. Comment on the statement— Telophase is reverse of prophase.
Ans: Prophase is marked by the initiation of condensation of chromosomal material. The chromosomal material becomes untangled during the process of chromatin condensation. At the beginning of the final stage of mitosis, i.e. telophase, the chromosomes that have reached their respective poles decondense and lose their individuality.
Cells at the end of prophase, when viewed under the microscope, do not show golgi complexes, endoplasmic reticulum, nucleolus and the nuclear envelope. In the telophase stage nuclear envelope assembles around the chromosome clusters. Nucleolus, golgi complex and ER reform.

Q2. What are the various stages of meiotic prophase-I? Enumerate the chromosomal events during each stage?
Ans: Meiosis-I:
Prophase-I: Prophase of the first meiotic division is typically longer and more complex when compared to the prophase of mitosis. It has been further subdivided into the following five phases based on chromosomal behaviour, i.e. Leptotene, Zygotene, Pachytene, Diplotene and Diakinesis. During leptotene stage, the chromosomes become gradually visible under the light microscope.
• The compaction of chromosomes continues throughout leptotene. This is followed by the second stage of prophase-I called zygotene. During this stage chromosomes start pairing together and this process of association is called synapsis. Such paired chromosomes are called homologous chromosomes. Electron micrographs of this stage indicate that chromosome synapsis is accompanied by the formation of complex structure called synaptonemal complex.
• The complex formed by a pair of synapsed homologous chromosomes is called a bivalent or a tetrad. However, these are more clearly visible at the next stage. The first two stages of prophase-I are relatively short-lived compared to the next stage that is pachytene. During this stage bivalent
. chromosomes now clearly appears as tetrads. This stage is characterised by the appearance of recombination nodules, the sites at which crossing over occurs between non-sister chromatids of the homologous chromosomes. Crossing over is the exchange of genetic material between two homologous chromosomes.
• Crossing over is also an enzyme-mediated process and the enzyme involved is called recombinase. Crossing over leads to the recombination of genetic material on the two chromosomes. Recombination between homologous chromosomes is completed by the end of pachytene, leaving the chromosomes linked at the sites of crossing over.
• The beginning of diploteneis recognised by the dissolution of the synaptonemal complex and the tendency of the recombined homologous chromosomes of the bivalents to separate from each other except at the sites of crossovers. These X-shaped structures, are called chiasmata. In oocytes of some vertebrates, diplotene can last for months or years.
• The final stage of meiotic prophase-I is diakinesis. This is marked by terminalisation of chiasmata. During this phase the chromosomes are fully condensed and the meiotic spindle is assembled to prepare the homologous chromosomes for separation. By the end of diakinesis, the nucleolus disappears and the nuclear envelope also breaks down.

Q3. Differentiate between the events of mitosis and meiosis

	Mitosis		Meiosis
1.	Take place in the somatic cells of the body.	1.	Take place in the germ cells
2.	Occurs in both sexually as well as asexually reproducing organisms.	2.	Occurs only in sexually reproducing organisms.
3.	Mitosis involves only one cycle of nuclear and cell division.	3.	Meiosis          involves         two sequential cycles of nuclear and cell division called meiosis-I and meiosis-II.
4.	The DNA replicates once for one cell division.	4.	The DNA replicates once for two cell divisions.
5.	The prophase is shorter.	5.	Prophase is typically longer
6.	Prophase is comparatively simple.	6.	Prophase of the first meiotic division is more complex when compared to prophase of mitosis.
7.	The cell divides only once and the chromosomes also divide only once. .	7.	There are two cell divisions but the chromosomes divide only once.
8.	Mitosis does not involves pairing                     of        homologous chromosomes                       and recombination between them.	8.	Meiosis involves pairing of homologous chromosomes and recombination between them.
9.	Two cells are formed at the end of mitosis.	9.	Four haploid cells are formed at the end of meiosis.

Q4. Write brief note on the following:
a. Synaptonemal complex
b. Metaphase plate
Ans: a. Synaptonemal complex: During zygotene stage chromosomes start pairing together and this process of association is called synapsis. Such paired chromosomes are called homologous chromosomes. Electron micrographs of this stage indicate that chromosome synapsis is accompanied by the formation of complex structure called synaptonemal complex. The complex formed by a pair of synapsed homologous chromosomes is called a bivalent or a tetrad. However, these are more clearly visible at the next stage.
b. Metaphase plate: At this stage, metaphase chromosome is made up of two sister chromatids, which are held together by the centromere. Small disc-shaped structures at the surface of the centromeres are called kinetochores. These structures serve as the sites of attachment of spindle fibres (formed by the spindle fibres) to the chromosomes that are moved into position at the centre of the cell. Hence, the metaphase is characterised by all the chromosomes coming to lie at the equator with one chromatid of each chromosome connected by its kinetochore to spindle fibres from one pole and its sister chromatid connected by its kinetochore to spindle fibres from the opposite pole. The plane of alignment of the chromosomes at metaphase is referred to as the metaphase plate.

Q5. Write briefly the significance of mitosis and meiosis in multicellular organism.
Ans: Significance of Mitosis:
Mitosis or the equational division is usually restricted to the diploid cells only. However, in some lower plants mitosis usually results in the production of diploid daughter cells with identical genetic complement. The growth of multicellular organisms is due to mitosis. A very significant contribution of mitosis is cell repair. The cells of the upper layer of the epidermis, cells of the lining of the gut, and blood cells are being constantly replaced. Mitotic divisions in the meristematic tissues —the apical and the lateral cambium, result in a continuous growth of plants throughout their life.
Significance of Meiosis:
Meiosis is the mechanism by which conservation of specific chromosome number of each species is achieved across generations in sexually reproducing organisms, even though the process, per se, paradoxically, results in reduction of chromosome number by half. It also increases the genetic variability in the population of organisms from one generation to the next. Variations are very important for the process of evolution.

Q6. An organism has two pair of chromosomes (i.e., chromosome number = 4). Diagrammatically represent the chromosomal arrangement during different phases of meiosis-II.

NCERT Exemplar ProblemsMathsPhysicsChemistryBiology

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

NCERT Exemplar Class 11 Biology Chapter 11 Transport in Plants are part of NCERT Exemplar Class 11 Biology. Here we have given NCERT Exemplar Class 11 Biology Chapter 11 Transport in Plants. https://www.cbselabs.com/ncert-exemplar-problems-class-11-chapter-11-transport-in-plants/

NCERT Exemplar Class 11 Biology Chapter 11 Transport in Plants

Multiple Choice Questions

Q1. Which of the following statements does not apply to reverse osmosis?
(a) It is used for water purification.
(b) In this technique, pressure greater than osmotic pressure is applied to the system.
(c) It is a passive process.
(d) It is an active process.
Ans: (c) If pressure greater than the osmotic pressure is applied to the higher concentration, the direction of water flow through the membrane can be reverse. This is called reverse osmosis. Reverse osmosis occurs when water is moved across the membrane against the concentration gradient, from lower concentration to higher concentration. Reverse osmosis is an active process.

Q2. Which one of the following will not directly affect transpiration?
(a) Temperature
(b) Light
(c) Wind speed
(d) Chlorophyll content of leaves
Ans: (d) The chlorophyll content of leaves will not directly affect transpiration, while temperature, light and wind speed directly affect the transpiration.

Q3. The lower surface of leaf will have more number of stomata in a
(a) Dorsiventral leaf
(b) Isobilateral leaf
(c) Both (a) and (b)
(d) None of the above
Ans: (a) Usually, the lower surface of a dorsiventral (dicotyledonous) leaf has a greater number of stomata. On the upper surface, stomata may be even absent sometimes.

Q4. The form of sugar transported through phloem is
(a) Glucose (b) Fructose (c) Sucrose (d) Ribose
Ans: (c) Food, primarily sucrose, is transported by the vascular tissue phloem from a source to a sink.

Q5. The process of guttation takes place
(a) when the root pressure is high and the rate of transpiration is low
(b) when the root pressure is low and the rate of transpiration is high
(c) when the root pressure equals the rate of transpiration
(d) when the root pressure as well as rate of transpiration are high.
Ans: (a) The effect of root pressure is observable at night as well as early morning when evaporation is low. Excess water gets collected in the form of droplets around special openings of veins near the tip of grass blades and leaves of many herbaceous parts of plants such as Tropaeolum, Balsam and grasses. Such water loss in its liquid phase is known as guttation.

Q6. Which of the following is an example of imbibition?
(a) Uptake of water by root hair (b) Exchange of gases in stomata (c) Swelling of seed when put in soil (d) Opening of stomata
Ans: (c) Imbibition is a special type of diffusion. A classic example of imbibition is absorption of water by seeds and dry wood.

Q7. When a plant undergoes senescence, the nutrients may be
(a) accumulated
(b) bound to cell wall
(c) translocated
(d) None of the above
Ans: (c) Mineral ions are frequently remobilized (translocation), particularly from older senescing parts. Before the leaf fall in deciduous plants, minerals are translocated to other parts.

Q8. Water potential of pure water at standard temperature is equal to
(a) 10
(b) 20
(c) Zero
(d) None of these
Ans: (c) The water potential of pure water at standard temperature is equal to zero.

Q9. Choose the correct option. Mycorrhiza is a symbiotic association of fungus with root system which helps in A. Absorption of water B. Mineral nutrition
C. Translocation D. Gaseous exchange
(a) Only A
(b) Only B –
(c) Both A and B
(d) Both B and C
Ans: (c) Mycorrhiza is a symbiotic association of fungus with root system which helps in absorption of water and mineral nutrition.

Q10. Based on the figure given below, which of the following statements is not correct?

(a) Movement of solvent molecules will take place from chamber AtoB
(b) Movement of solute will take place from A to B
(c) Presence of a semipermeable is a prerequisite for this process to occur
(d) The direction and rate of osmosis depends on both the pressure gradient and concentration gradient.
Ans: (b) The movement of solute will take place from B to A

Q11. Match the following and choose the correct option.

A.	Leaves	(i)	Anti-transpirant
B.	Seed	(ii)	Transpiration
C.	Roots	(iii)	Negative osmotic potential
D.	Aspirin	(iv)	Imbibition ‘
E.	Plasmolyzed cell	(v)	Absorption

Options:
(a) A—(ii), B—(iv), C—(v), D—(i), E—(iii)
(b) A—(iii), B—(ii), C—(iv), D—(i), E—(v)
(c) A—(i), B—(ii), C—(iii), D—(iv), E—(v)
(d) A—(v), B—(iv), C—(iii), D—(ii), E—(i)
Ans: (a)

A.	Leaves	(ii)	Transpiration
B.	Seed	(iv)	Imbibition
e.	Roots	(v)	Absorption
D.	Aspirin	(i)	Anti-transpirant
E.	Plasmolyzed cell	(ii)	Negative osmotic potential

 

Q12. Mark the mismatched pair.
(a) Amyloplast—Store protein granule
(b) Elaioplast—Store oils or fats
(c) Chloroplasts—Contain chlorophyll pigments
(d) Chromoplasts—Contain coloured pigments other than chlorophyll
Ans: (a) Aleuroplasts—Store proteins
Amyloplast—Store carbohydrate (starch)
Very Short Answer Type Questions .
Q1. Smaller, lipid soluble molecules diffuse faster through cell membrane, but the movement of hydrophilic substances are facilitated by certain transporters which are chemically ________.
Ans: Protein

Q2. In a passive transport across a membrane, when two protein molecules move in opposite direction and independent of each other, it is called as ________
Ans: Antiport

Q3. Osmosis is a special kind of’diffusion, in which water diffuses across the cell membrane. The rate and direction of osmosis depends upon both ________
Ans: Pressure and concentration gradient

Q4. A flowering plant is planted in an earthen pot and irrigated. Urea is added to make the plant grow faster, but after some time the plant dies. This may be due to ________.
Ans: Exosmosis

Q5. Absorption of water from soil by dry seeds increases the ________ thus helping seedlings to come out of soil.
Ans: Pressure

Q6. 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 _________
Ans: Transpiration pull

Q7. 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 _________ .
Ans: The cell wall is freely permeable to water and substances in solutions but membrane is selectively permeable.

Q8. The C4 plants are twice as efficient as C3 plants in terms of fixing C02 but lose only _________ as much water as C3 plants for the same amount of C02 fixed.
Ans: Half

Q9. Movement of substances in xylem is unidirectional while in phloem it is bidirectional. Explain.
Ans: The direction of movement in the phloem can be upwards or downwards, i.e. bi-directional. This contrasts with that of the xylem where the movement is always unidirectional, i.e. upwards. Hence, unlike one-way flow of water in transpiration, food in phloem sap can be transported in any required direction, as it is a source of sugar and works as a sink to use, store or remove the sugar

Q10. Identify the process occurring in I, II and III.

Ans: I—Uniport facilitated diffusion
II— Antiport facilitated diffusion
III— Symport facilitated diffusion

Q11. Given below is a table. Fill in the gaps.

	Property	Simple diffusion	Facilitated transport	Active transport
i.	Highly selective		Yes
ii.	Uphill transport			Yes
iii.	Requires ATP

Ans:

	Property	Simple diffusion	Facilitated transport	Active transport
i.	Highly selective	No	Yes	Yes
ii.	Uphill transport	No	No ,	Yes
iii.	Requires ATP	No	No	Yes

Q12. Define water potential and solute potential.
Ans: Water potential is considered as the potential energy of water. It is also taken as a measure of the difference between the potential energy in a given sample of wafer and pure water.
If a solute is dissolved in pure water, the solution is considered having fewer free water and hence the concentration of water decreases. Thus, all solutions have lower water potential than pure water. The magnitude of lowering in water potential due to dissolution of solute is called solute potential.

NCERT Exemplar ProblemsMathsPhysicsChemistryBiology

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

NCERT Exemplar Class 11 Biology Chapter 12 Mineral Nutrition are part of NCERT Exemplar Class 11 Biology. Here we have given NCERT Exemplar Class 11 Biology Chapter 12 Mineral Nutrition. https://www.cbselabs.com/ncert-exemplar-problems-class-11-chapter-12-mineral-nutrition/

NCERT Exemplar Class 11 Biology Chapter 12 Mineral Nutrition

Multiple Choice Questions

Q1. Which one of the following roles is not a characteristic of an essential element?
(a) Being a component of biomolecules
(b) Changing the chemistry of soil
(c) Being a structural component of energy-related chemical compounds
(d) – Activation or inhibition of enzymes
Ans: (b)
(i) Essential elements as components of biomolecules and hence structural elements of cells.
(ii) Essential elements that are components of energy-related chemical compounds in plants.
(iii) Essential elements that activate or inhibit enzymes.
(iv) Some essential elements can alter the osmotic potential of a cell.

Q2. Which one of the following statements can best explain the term critical concentration of an essential element?
(a) Essential element concentration below which plant growth is retarded
(b) Essential element concentration below which plant growth becomes enhanced
(c) Essential element concentration below which plant remains in the vegetative phase
(d) None of the above
Ans: (a) The concentration of the essential element below which plant growth is retarded is termed as critical concentration.

Q3. Deficiency symptoms of an element tend to appear first in young leaves. It indicates that the element is relatively immobile. Which one of the following elemental deficiency would show such symptoms?
(a) Sulphur (b) Magnesium (c) Nitrogen (d) Potassium
Ans: (a) The deficiency symptoms tend to appear first in the young tissues whenever the elements are relatively immobile and are not transported out of the mature organs, e.g., S and Ca.

Q4. Which one of the following symptoms is not due to manganese toxicity in plants?
(a) Calcium translocation in shoot apex is inhibited
(b) Deficiency in both Iron and Nitrogen is induced
(c) Appearance of brown spot surrounded by chlorotic veins
(d) None of the above
Ans: (b) Excess of manganese may, in fact, induce deficiencies of iron, magnesium and calcium.

Q5. Reaction carried out by N2 fixing microbes include

Which of the following statements about those equations is not true?
(a) Step (i) is carried out by Nitrosomonas or Nitrococcus
(b) Step (ii) is carried out by Nitrobacter
(c) Both steps (i) and (ii) can be called nitrification
(d) Bacteria carrying out these steps are usually photoautotrophs
Ans:(d) Bacteria carrying out these steps are usually chemoautotrophs.

Q6. With regard to the Biological Nitrogen Fixation by Rhizobium in association with soyabean, which one of the following statement/statements does not hold true?
(a) Nitrogenase may require oxygen to its functioning
(b) Nitrogenase is Mo-Fe protein
(c) Leghaemoglobin is a pink coloured pigment
(d) Nitrogenase helps to convert N2 gas into two molecules of ammonia
Ans:(a) Nitrogenase is highly sensitive to molecular oxygen (02), thus requires
anaerobic conditions. Nodules have adaptations that ensure that the enzyme is protected from 02. To protect nitrogehase, nodule contains an 02-scavenger celled leghaemoglobin.

Q7. Match the element with its associated functions/roles and choose the correct option among the given below.

A.	Boron	(i) V	Splitting of H20 to liberate 02 during photosynthesis
B.	Manganese	(ii)	Needed for synthesis of auxins
C.	Molybdenum	(iii)	Component of nitrogenase
D.	Zinc	(iv)	Pollen germination
E.	Iron	(v)	Component of ferredoxin

Options:
(a) A—(i), B—(ii), C—(iii). D—(iv), E—(v)
(b) A—(iv), B—(i), C—(iii), D—(ii), E—(v)
(c) . A—(iii), B—(ii), C—(iv), D—(v), E—(i)
(d) A—(ii), B—(iii), C—(v), D—(i), E—(iv)
Ans: (b)

A.	Boron	(iv)	Pollen germination
B.	Manganese	(0	Splitting of H20 to liberate 02 during photosynthesis
C.	Molybdenum	(iii)	Component of nitrogenase
D.	Zinc	(ii)	Needed for synthesis of auxins
E.	Iron	(v)	Component of ferredoxin

Q8. Plants can be grown in (Tick the incorrect option).
(a) Soil with essential nutrients
(b) Water with essential nutrients
(c) Either water or soil with essential nutrients
(d) Water or soil without essential nutrients
Ans: (d) Plants can be grown in soil with essential nutrients, water with essential nutrients and either water or soil with essential nutrients.

Very Short Answer Type Questions
Q1. Name a plant, which accumulates silicon.
Ans: Rice, sugarcane, etc.

Q2. Mycorrohiza is a mutualistic association. How do the organisms involved in this association gain from each other?
Ans: Mycorrhiza is a symbiotic association between a fungus and the roots of a vascular plant. Through mycorrhization, the plant obtains phosphate and other minerals, such as zinc and copper, from the soil. The fungus obtains nutrients, such as sugars, from the plant root.

Q3. Nitrogen fixation is shown by prokaryotes and not eukaryotes. Comment?
Ans: Very few living organisms can utilise the nitrogen in the form N2, available
abundantly in the air. Only certain prokaryotic species are capable of fixing nitrogen. The enzyme, nitrogenase which is capable of nitrogen reduction is present exclusively in prokaryotes. Such microbes are called N2-fixers.

Q4. Carnivorous plants like Nepenthes and Venus fly trap have nutritional adaptations. Which nutrient do they especially obtain and from where?
Ans: Carnivorous plants grow in nitrogen deficient soil but they utilise their nitrogen by killing the insect by some special structure.

Q5. Think of a plant which lacks chlorophyll. From where will it obtain nutrition? Give an example of such a type of plant.
Ans: Cuscuta, a parasitic plant that is commonly found growing on hedge plants, has lost its chlorophyll and leaves in the course of evolution. It derives its nutrition from the host plant which it parasitises.

Q6. Name an insectivorous angiosperm.
Ans: Nepenthes, Utricularia, Drosera, Dionea, etc.

Q7. A farmer adds Azotobacter culture to soil before sowing maize. Which mineral element is being replenished?
Ans: Nitrogen

Q8. What type of conditions are created by leghaemoglobin in the root nodule of a legume?
Ans: Anaerobic condition

Q9. What is common to Nepenthes, Utricularia and Drosera with regard to mode of nutrition?
Ans: All are carnivorous plant (angiosperms).

Q10. Plants with zinc deficiency show reduced biosynthesis of .
Ans: Auxin

Q11. Yellowish edges appear in leaves deficient in .
Ans: K (potassium)

Q12. Name the macronutrient which is a component of all organic compounds but is not obtained from soil.
Ans: Carbon

Q13. Name one non-symbiotic nitrogen fixing prokaryote.
Ans: (i) Free-living (non-symbiotic) and non photosynthetic aerobic N2-fixing microbes: Azotobacter and Beijernickia.
(ii) Free-living and anaerobic N2-fixing microbes: Rhodospirillum, Bacillus polymyxa and Clostridium.

Q14. Rice fields produce an important green house gas. Name it.
Ans: CH4 (methane) .

Q15. Complete the equation for reductive amination

Q16. Excess of Mn in soil leads to deficiency of Ca, Mg and Fe. Justify.
Ans: Manganese competes with iron and magnesium for uptake and with magnesium for binding with enzymes. Manganese also inhibits calcium translocation in shoot apex. Therefore, excess of manganese may, in fact, induce deficiencies of iron, magnesium and calcium. Thus, what appears as symptoms of manganese toxicity may actually be the deficiency symptoms of iron, magnesium and calcium.

Short Answer Type Questions
Q1. How is sulphur important for plants? Name the amino acids in which it is present.
Ans: Sulphur, besides being present in some amino acids essential for protein synthesis, is also a constituent of several coenzymes, vitamins and ferredoxin which are involved in some biochemical pathway.

Q2. How are organisms like Pseudomonas and Thiobacillus of great significance in nitrogen cycle?
Ans: Pseudomonas and Thiobacillus carry out denitrification process wherein the nitrate present in the soil is reduced to nitrogen thus contributing to the atmospheric nitrogen.

Q3. Carefully observe the following figure:

a. Name the technique shown in the figure and the scientist who demonstrated this technique for the first time.
b. Name at least 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?
Ans: a. Hydroponics, Julius von Sachs.
b. Tomato, seedless cucumber, lettuce.
c. Aerating tube ensures adequate aeration of the root for optimum growth of the plant. The funnel is used to release water and nutrients into the container with nutrient solution. This solution needs to be replaced every day or two for maximum growth.

Q4. Name the most crucial enzyme found in root nodules for N2 fixation. Does it require a special pink coloured pigment for its functioning? Elaborate.
Ans: Nitrogenase. Yes, it does require the presence of a pink-coloured pigment in the nodule called leghaemoglobin for its functioning. This pigment helps in scavenging oxygen as nitrogenase functions under anaerobic condition.

Q5. How are the terms ‘critical concentration’ and ‘deficient’ different from each other in terms of concentration of an essential element in plants? Can you find the values of ‘critical concentration’ and ‘deficient’ for minerals – Fe and Zn?
Ans: The concentration of the essential element below which plant growth is retarded is termed as critical concentration. The element is said to be deficient when present below the critical concentration. Yes. one can find the values of ‘critical concentration’ and ‘deficient’ for minerals – Fe and Zn through the hydroponics technique.

Q6. Carnivorous plants exhibit nutritional adaptation. Citing an example explain this fact.
Ans: Carnivorous plants have green leaves so they are autotrophic but they grow in nitrogen deficient soil. For nitrogen requirement they capture and digest the insects so they are partially heterotrophic nature.

Q7. A farrper adds/supplies 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.
Ans: This is due to the manganese toxicity. Many a times, excess of an element may inhibit the uptake of another element. For example, the prominent symptom of manganese toxicity is the appearance of brown spots surrounded by chlorotic veins. Manganese competes with iron and magnesium for uptake and with magnesium for binding with enzymes. Manganese also inhibits calcium translocation in shoot apex. Therefore, excess of manganese may, in fact, induce deficiencies of iron, magnesium and calcium.
• The farmer should not supplies Mn to his field.

Long Answer Type Questions
Q1. It is observed that deficiency of a particular element showed its symptoms initially in older leaves and then in younger leaves.
a. Does it indicate that the element is actively mobilised or relatively immobile?
b. Name two elements which are highly mobile and two which are relatively immobile.
c. How is the aspect of mobility of elements important to horticulture and agriculture?
Ans: a. It is actively mobilised.
b. Highly mobile—nitrogen, magnesium Relatively immobile—calcium, boron
c. Symptoms of deficiency of mobile elements are more pronounced in older leaves and symptoms of deficiency of relatively immobile element appear first in younger leaves. This information can be utilised by horticulturist and agriculturist to get a broad idea of the deficiency elements in plants.

Q2. 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 mycorrihiza 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.
Ans:  a. Yes, one can artificially induce the property of nitrogen fixation in a plant—leguminous or non-leguminous through genetic engineering which involves introduction of specific genes to the host plant that synthesises nitrogenase enzymes.
b. Symbiotic relationship
c. Yes, it is necessary for a microbe to be in close association with a plant to provide mineral nutrition as seen in leguminous plants. Species of rod-shaped Rhizobium has such relationship with the roots of several legumes such as alfalfa, sweet clover, sweet pea, lentils, garden pea, broad bean, clover beans, etc. The most common association on roots is as nodules. The nodule contains all the necessary biochemical components, such as the enzyme nitrogenase and leghaemoglobin. The enzyme nitrogenase is an Mo-Fe protein and catalyses the conversion of atmospheric nitrogen to ammonia.

Q3. What are essential elements for plants? Give the criteria of essentiality. How are minerals classified depending upon the amount in which they are needed by the plants?
Ans: Essential elements: carbon, hydrogen, oxygen, nitrogen, phosphorous, sulphur, potassium, calcium, magnesium, iron, manganese, copper, molybdenum, zinc, boron, chlorine and nickel.
Criteria for Essentiality:
The criteria for essentiality of an element are given below:
(a) The element must be absolutely necessary for supporting normal growth and reproduction. In the absence of the element, the plants do not complete their life cycle or set the seeds.
(b) The requirement of the element must be specific and not replaceable by another element. In other words, deficiency of any one element cannot be met by supplying some other element.
(c) The element must be directly involved in the metabolism of the plant. Based upon the above criteria only a few elements have been found to be absolutely essential for plant growth and metabolism. These elements are further divided into two broad categories based on their quantitative requirements.
(i) Macronutrients and
(ii) Micronutrients
Macronutrients are generally present in plant tissues in large amounts (in excess of 10 mmole kg~’ of dry matter). The macronutrients include carbon, hydrogen, oxygen, nitrogen, phosphorous, sulphur, potassium, calcium and magnesium. Of these, carbon, hydrogen and oxygen are mainly obtained from C02 and H20, while the others are absorbed from the soil as mineral nutrition.
Micronutrients or trace elements are needed in very small amounts (less than 10 mmole kg-1 of dry matter). These include iron, manganese, copper, molybdenum, zinc, boron, chlorine and nickel.

Q4. With the help of examples describe the classification of essential elements based on the function they perform.
Ans: Essential elements can also be grouped into four broad categories on the basis of their diverse functions. These categories are:
(i) Essential elements as components of biomolecules and hence structural elements of cells (e.g., carbon, hydrogen, oxygen and nitrogen).
(ii) Essential elements that are components of energy-related chemical compounds in plants (e.g., magnesium in chlorophyll and phosphorous in ATP).
(iii) Essential elements that activate or inhibit enzymes, for example Mg2+ is an activator for both ribulose bisphosphate carboxylase/oxygenase and phosphoenol pyruvate carboxylase, both of which are critical enzymes in photosynthetic carbon fixation; Zn +is an activator of alcohol dehydrogenase and Mo of nitrogenase during nitrogen metabolism.
(iv) Some essential elements can alter the osmotic potential of a cell. Potassium plays an important role in the opening and closing of stomata. Minerals also play role as solutes in determining the water potential of a cell.

Q5. We know that plants require nutrients. If we supply these in excess, will it be beneficial to the plants? If yes, how/ If no, why?
Ans: No, excess supply of nutrients is not beneficial for the plants. It is toxic to the plants. Any mineral ion concentration in tissues that reduces the dry weight of tissues by about 10% is considered toxic. Such critical concentrations vary widely among different micronutrients. The toxicity symptoms are difficult to identify. Toxicity levels for any element also vary for different plants. Many a times, excess of an element may inhibit the uptake of another element. For example, the prominent symptom of manganese toxicity is the appearance of brown spots surrounded by chlorotic veins. It is important to know that manganese competes with iron and magnesium for uptake and with magnesium for binding with enzymes. Manganese also inhibits calcium translocation in shoot apex. Therefore, excess of manganese may, in fact, induce deficiencies of iron, magnesium and calcium.

Q6. Trace the events starting from the coming in contact of Rhizobiwn to a leguminous root till nodule formation. Add a note on importance of leghaemoglobin.
Ans: Nodule Formation: Nodule formation involves a sequence of multiple interactions between Rhizobium and roots of the host plant. Principal stages in the nodule formation are summarised as follows: Rhizobia multiply and colonise the surroundings of roots and get attached to epidermal and root-hair cells. The root-hairs curl and the bacteria invade the root-hairs. An infection thread is produced carrying the bacteria into the cortex of the root, where they initiate the nodule formation in the cortex of the root. Then the bacteria are released from the thread into the cells which leads to the differentiation of specialised nitrogen fixing cells. The nodule thus formed, establishes a direct vascular connection with the host for exchange of nutrients.
Importance of leghemoglobin: The enzyme nitrogenase is highly sensitive to the molecular oxygen; it requires anaerobic conditions. The nodules have adaptations that ensure that the enzyme is protected from oxygen. To protect these enzymes, the nodule contains an oxygen scavenger called leghaemoglobin. It is interesting to note that these microbes live as aerobes under free-living conditions (where nitrogenase is not operational), but during nitrogen-fixing events, they become anaerobic (thus protecting the nitrogenase enzyme).

Q7. Give the biochemical events occurring in the root nodule of a pulse plant. What is the end product? What is its fate?
Ans: The nodule contains all the necessary biochemical components, such as the enzyme nitrogenase and leghaemoglobin. The enzyme nitrogenase is an Mo-Fe protein and catalyses the conversion of atmospheric nitrogen to ammonia, the first stable product of nitrogen fixation.

• Glutamic acid is the main amino acid from which transfer of NH2 (amino group) takes place and other amino acids are formed through transamination. Enzyme transaminase catalyses all such reactions.
For example, the two most important amides (asparagine and glutamine) found in plants are a structural part of proteins.
• Asparagine formed from aspartic acid and glutamine is formed from glutamic acid by addition of amino group to each. The hydroxyl part of the acid is replaced by another NH2 radicle.

Q8. Hydroponics have been shown to be a successful technique for growing of plants. Yet most of the crops are still grown on land. Why?
Ans: The technique of growing plants in anutrient solution is known as hydroponics. Since, then a number of improvised methods have been employed to try and determine the mineral nutrients essential for plants. The essence of all these methods involves the culture of plants in a soil-free, defined mineral solution. These methods require purified water and mineral nutrient salts. Hydroponics has been successfully employed as a technique for the commercial production of vegetables such as tomato, seedless cucumber and lettuce.
Yet most of the crops are still grown on land because it must be emphasised that the nutrient solutions must be adequately aerated to obtain the optimum growth. On land no such conditions are required.

NCERT Exemplar ProblemsMathsPhysicsChemistryBiology

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

NCERT Exemplar Class 11 Biology Chapter 13 Photosynthesis in Higher Plants are part of NCERT Exemplar Class 11 Biology. Here we have given NCERT Exemplar Class 11 Biology Chapter 13 Photosynthesis in Higher Plants. https://www.cbselabs.com/ncert-exemplar-problems-class-11-chapter-13-photosynthesis-higher-plants/

NCERT Exemplar Class 11 Biology Chapter 13 Photosynthesis in Higher Plants

Multiple Choice Questions

Q1. Which metal ion is a constituent of chlorophyll?
(a) Iron (b) Copper (c) Magnesium (d) Zinc
Ans: (c) Magnesium ion is a constituent of chlorophyll.

Q2. Which pigment acts directly to convert light energy to chemical energy?
(a) Chlorophyll a (b) Chlorophyll b
(c) Xanthophyll (d) Carotenoid
Ans: (a) Chlorophyll a pigment acts directly to convert light energy to chemical energy.

Q3. Which range of wavelength (in nm) is called photosynthetically active radiation (PAR)?
(a) 100-390 (b) 390-430 (c) 400-700 (d) 760-10000
Ans: (c) 400-700 range of wavelength (in nm) is called photosynthetically active radiation (PAR).

Q4. Which light range is least effective in photosynthesis?
(a) Blue (b) Green (c) Red (d) Violet
Ans: (b) Green light range is least effective in photosynthesis.

Q5. Chemosynthetic bacteria obtain energy from
(a) Sun (b) Infrared rays
(c) Organic substances (d) Inorganic chemicals
Ans: (d) Chemosynthetic bacteria obtain energy from inorganic chemicals.

Q6. Energy required for ATP synthesis in PSII comes from
(a) Proton gradient (b) Electron gradient
(c) Reduction of glucose (d) Oxidation of glucose
Ans: (a) Energy required for ATP synthesis in PSII comes from proton gradient.

Q7. During light reaction in photosynthesis, the following are formed
(a) ATP and sugar                                 
(b) Hydrogen, 0₂ and sugar
(c) ATP, hydrogen donor and 0₂         
(d) ATP, hydrogen and 0₂ donor
Ans: (c) During light reaction in photosynthesis the following are formed ATP, hydrogen donor/(NADPH) and 0₂.

Q8. Dark reaction in photosynthesis is called so because
(a) It can occur in dark also
(b) It does not directly depend on light energy
(c) It cannot occur during day light
(d) It occurs more rapidly at night
Ans: (b) Dark reaction in photosynthesis is called so because it does not directly depend on light energy.

Q9. PEP is primary C0₂ acceptor in
(a) C₄ plants                                            
(b) C₃ plants
(c) C₂-plants                
 (d) Both C₃ and C₄ plants
Ans: (a) PEP is primary C0₂ acceptor in C₄ plants.

Q10. Splitting of water is associated with
(a) Photosystem I
(b) Lumen of thylakoid
(c) Both Photosystem I and II
(d) Inner surface of thylakoid membrane
Ans: (d) Splitting of water is associated with inner surface of thylakoid membrane.

Q11. The correct sequence of flow of electrons in the light reaction is
(a) PSII, plastoquinone, cytochromes, PSI, ferredoxin
(b) PSI, plastoquinone, cytochromes, PSII, ferredoxin
(c) PSI, ferredoxin, PSII
(d) PSI, plastoquinone, cytochromes, PSII, ferredoxin
Ans: (a) The correct sequence of flow of electrons in the light reaction is PSII, plastoquinone, cytochromes, PSI and ferredoxin.

Q12. The enzyme that is not found in a C₃ plant is
(a)    RuBP Carboxylase                      
(b)     PEP Carboxylase
(c)     NADP reductase                           
(d)     ATP synthase
Ans: (b) The enzyme that is not found in a C₃ plant is PEP Carboxylase.

Q13. The reaction that is responsible for the primary fixation C0₂ is catalysed by
(a) RuBP carboxylase
(b) PEP carboxylase
(c) RuBP carboxylase and PEP carboxylase
(d) PGA synthase
Ans: (c) The reaction that is responsible for the primary fixation of C0₂ is catalysed by RuBP carboxylase and PEP carboxylase.

Q14. When C0₂ is added to PEP, the first stable product synthesised is
(a)    Pyruvate                                        
(b)     Glyceraldehyde-3-phosphate
(c)     Phosphoglycerate                        
 (d)     Oxaloacetate
Ans: (d) When C0₂ is added to PEP, the first stable product synthesised is oxaloacetate.

Very Short Answer Type Questions

Q1. Examine the figure.

a. Is this structure present in animal cell or plant cell?
b. Can these be passed on to the progeny? How?
c. Name the metabolic processes taking place in the places marked (1) and (2).
Ans: a. Plant cell.
b. Yes, through female gametes.
c. In part (1)— Photophosphorylation. In part (2)—Calvin cycle.

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?
Ans: a. Lumen of the thylakoids.
b. 0₂ is evolved during this reaction; moreover, electrons are made available to PS-II continuously.

Q3. Cyanobacteria and some other photosynthetic bacteria do not have chloroplasts. How do they conduct photosynthesis?
Ans: Cyanobacteria and other photosynthetic bacteria have thylakoids suspended freely in the cytoplasm (i.e., they are not enclosed in membrane), and they have bacteriochlorophyll.

Q4. a. NADP reductase enzyme is located on
b. Breakdown of proton gradient leads to release of
Ans: a. Grana-lamellae.
b. Energy.

Q5. Can girdling experiments be done in monocots? If yes, how? If no, why not?
Ans: No, because vascular bundles are scattered in monocot.

Analyse the above reaction and’answer the following questions:
a. How many molecules of ATP and NADPH are required to fix one molecule of C02?
b. Where in the chloroplast does this process occur?
Ans: a. Three molecules of ATP and two molecules of NADPH are required to fix one molecule of C02
b. Stroma of chloroplast

Q7. Does moonlight support photosynthesis? Find out.
Ans: As the intensity of moonlight is much less than the sunlight, so it does not support photosynthesis.

Q8. Some of these terms/chemicals are associated with the C₄ Explain.
a. Hatch and Slack pathway
b. Calvin cycle
c. PEP carboxylase
d. Bundle sheath cells
Ans: Though C₄ plants have C₄ oxaloacetic acid as is the first C0₂ fixation product they use the C₃ pathway or Calvin cycle as the main biosynthetic pathway. C₄ pathway is also called Hatch and Slack Pathway

• 1° C0₂ acceptor in C₄ plants is a 3-C molecule PEP (phosphoenol pyruvate) and is present in the mesophyll cells. The enzyme responsible for the fixation is PEPcase (PEP carboxylase) is found only in mesophyll cells. Bundle sheath cells lack PEPcase enzyme.
• C₄ acid (OAA) is formed by carboxylation is mesophyll cells; therefore, initial carboxylation reaction occurs in mesophyll cells (also in C₃ pathway). OAA forms other 4-C compounds like malic acid or aspartic acid in the mesophyll cells itself, which are transported to the bundle sheath cells.
• C0₂ released in the bundle sheath cells enters the C₃ or the Calvin pathway, a pathway common to all plants. The bundle sheath cells are rich in RuBisCO enzyme (necessary for the C₃ or the Calvin cycle), but lack PEPcase.
• Calvin pathway in C₄ plants takes place only in bundle sheath cells (because RuBisCO is present) but does not take place in the mesophyll cells because lack of RuBisCO enzyme in mesophyll cells of C₄ plants like maize, sorghum, sugarcane, Jowar, Euphorbia, Atriplex,

Q9. Where is NADP reductase enzyme located in the chloroplast? What is the role of this enzyme in proton gradient development?
Ans: The NADP reductase enzyme is located on the stroma side of the membrane. Along with electron that come from the acceptor of electrons of PSI protons are necessary for the reduction of NADP⁺ to NADPH + H⁺. These protons are also removed from the stroma.

Q10. ATPase enzyme consists of two parts. What are those parts? How are they arranged in the thylakoid membrane? Conformational change occurs in which part of the enzyme?
Ans: ATPase enzyme consists of two parts:
i.One portion called F₀ is imbedded in the membrane and forms a transmembrane channel that carries out facilitated diffusion of protons across the membrane.
ii.The other portion is called ‘F_l and protrudes on the outer surface-of the thylakoid membrane on the side that faces stroma.

The breakdown of the gradient provides enough energy to cause a conformational change in the F, particle of the ATPase, which makes the enzyme synthesise several molecules of energy-packed ATP.

Q11. Which products formed during the light reaction of photosynthesis are used to drive the dark reaction?
Ans: ATP and NADPH

Q12.What is the basis for designating C₃ and C₄ pathways of photosynthesis?
Ans:The number of carbon atoms in first stable product of carbondioxide fixation is the basis for designating C₃ and C₄ pathways of photosynthesis.

Short Answer Type Questions
Q1.Succulents are known to keep their stomata closed during the day to check transpiration. How do they meet their photosynthetic C0₂ requirements?
Ans: Succulent (water storing) plants such as cacti, euphorbias fix C0₂ into organic compound using PEP carboxylase at night, when the stomata are open.

The. organic compound (malic acid) accumulates throughout the night and is decarboxylated during the day to produce C0₂.

Q2.Chlorophyll a is the primary pigment for light reaction. What are accessory pigments? What is their role in photosynthesis?
Ans: Accessory pigments are those pigments, which assist in photosynthesis by capturing energy from light of different wavelengths, e.g., chlorophyll b, Xanthophylls and carotenoids.

Role in Photosynthesis:

1. They absorb wavelength of light not absorbed by chlorophyll a and transfer the energy to chlorophyll.
2. They also protect chlorophyll a from photo-oxidation.

Q3. Do reactions of photosynthesis called, as ‘Dark Reaction’ need light? Explain.
Ans :ATP and NADPH are used to drive the processes leading to the synthesis of food, more accurately, sugars. This is the biosynthetic phase or dark reaction of photosynthesis. This process does not directly depend on the presence of light but is dependent on the products of the light reaction, i.e., ATP and NADPH, besides C0₂ and H₂0.

Q4. How are photosynthesis and respiration related to each other?
Ans: Photosynthesis and respiration are related to each other as

1. Both processes take place in double membrane bound organelles.
2. In both processes ATP synthesis takes place.
3. In both processes electron transport system requires.

Q5.   If a green plant is kept in dark with proper ventilation, can this plant carry out photosynthesis? Can anything be given as a supplement to maintain its growth or survival?
Ans: No, this plant cannot photosynthesise in the absence of light. Only sunlight can be given as supplement to maintain its growth or survival.

Q6.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?\
Ans: Photosynthetic organisms occur at different depths in the ocean. They do not receive qualitatively and quantitatively the same light. The spectral quality of solar radiation is also important for life. The UV component of the spectrum is harmful to many organisms while not all the colour components of the visible spectrum are available for marine plants living at different depths of the ocean. Plants at great depth contains some accessory pigments that can easily capture the light.

Q7. In tropical rain forests, the canopy is thick and shorter plants growing below it, receive filtered light. How are they able to carry out photosynthesis?
Ans: In tropical rain forests, the canopy is thick and shorter plants growing below it called sciophytes (shade loving plants). They can photosynthesise in very low light conditions. They have larger photosynthetic units and hence they are able to carry out photosynthesis in filtered light.

Q8. What conditions enable RuBisCO to function as an oxygenase? Explain the ensuing process.
Ans: In the first step of the Calvin pathway RuBP combines with C0₂ to form 2 molecules of 3PGA, that is catalysed by RuBisCO.

Q9.Why does the rate of photosynthesis decrease at higher temperatures?
Ans: The rate of photosynthesis decreases at higher temperatures because at high temperatures the enzymes become denatured (destroy).

Q10. Explain how during light reaction of photosynthesis, ATP synthesis is a chemiosmotic phenomenon.
Ans: In the light reaction within the chloroplast, protons in the stroma decrease in number, while in the lumen there is accumulation of protons. This creates a proton gradient across the thylakoid membrane as well as a measurable decrease in pH in the lumen. This gradient is important because it is the breakdown of this gradient that leads to release of energy. The gradient is broken down due to the movement of protons across the membrane to the stroma through the transmembrane channel of the F₀ of the ATPase. The ATPase enzyme consists of two parts: one part called the F₀ is embedded in the membrane and forms a transmembrane channel that carries out facilitated diffusion of protons across the membrane. The other portion is called F1 and protrudes on the outer surface of the thylakoid membrane on the side that faces the stroma. The breakdown of the gradient provides enough energy to cause a conformational change in the F_i particle of the ATPase, which makes the enzyme synthesise several molecules of energy-packed ATP.

Q11. Find out how Melvin Calvin worked out the complete biosynthetic pathway for synthesis of sugar.
Ans: Just after World War II, among the several efforts to put radioisotopes to beneficial use, the work of Melvin Calvin is exemplary. The use of radioactive C¹⁴ by him in algal photosynthesis studies led to the discovery that the first C0₂ fixation product was a 3-carbon organic acid. He also contributed to working out the complete biosynthetic pathway; hence, it was called Calvin cycle after him. The first product identified was 3-phosphoglyceric acid or in short PGA.

Q12. Six turns of Calvin cycle are required to generate one mole of glucose. Explain.
Ans: The fixation of 6 molecules of C0₂ and 6 turns of the cycle are required for the removal of one molecule of glucose from the pathway. Hence for every C0₂ molecule entering the Calvin cycle, 3 molecules of ATP and 2 of NADPH are required. To make one molecule of glucose 6 turns of the cycle are required.

In	Out
Six C02	One glucose
18 ATP	18ADP
12 NADPH	12 NADP

 

Q13. Complete the flow chart for cyclic photophosphorylation of the photosystem-I

Q14. 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?
Ans: On studying vertical sections of leaves, one of a C₃ plant and the other of a C₄ plant. The particularly large cells around the vascular bundles of the C₄ pathway plants are called bundle sheath cells, and the leaves which have such anatomy are said to have ‘Kranz’ anatomy. ‘Kranz’ means ‘wreath’ and is a reflection of the arrangement of cells. The bundle sheath cells may form several layers around the vascular bundles; they are characterised by having a large number of chloroplasts, thick walls impervious to gaseous exchange and no intercellular spaces. Leaves of C₄ plants – maize or sorghum lack photorespiration. In addition these plants show tolerance to higher temperatures. Plants that are adapted to dry tropical regions have the C₄ pathway.

Q15. A process is occurring throughout the day, in ‘X’ organism. Cells are participating in this process. During this process ATP, C0₂ and water are evolved. It is not a light-dependent process.
a.Name the process.
b. Is it a catabolic or an anabolic process?
c .What could be the raw material of this process?
Ans: ‘ a. Respiration
b.Catabolic process (actually amphibolic pathway)
c.Glucose

Q16. Tomatoes, carrots and chillies are red in colour due to the presence of one pigment. Name the pigment. Is it a photosynthetic pigment?
Ans: Tomatoes, carrots and chillies are red in colour due to the presence of carotene pigment. It is an accessory photosynthetic pigment.

Q17. Why do we believe chloroplast and mitochondria to be semi-autonomous organelle?
Ans: Mitochondria and Chloroplast are semi-autonomous organelles or endosymbionts of cells because they
i. Possess their own nucleic acid (DNA molecule).
ii. Can form some of the required protein but for most of the proteins these are dependent on nuclear DNA and cytoplasmic ribosome.
iii. Do not arise de novo.
iv. Have membrane similar to those of bacteria.

Q18. Observe the diagram and answer the following.

a. Which group of plants exhibits 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?
Ans: a. C4 plants
b. OAA (Oxaloacetic acid)
c. Phosphoenol pyruvate (PEP) is present in the mesophyll cells. Enzyme Ribulose bisphosphate carboxylase-oxygenase (RuBisCO) is present in bundle sheath cells.

Q19. A cyclic process is occurring in C3 plant, which is light dependent, and needs O2 This process does not produce energy rather it consumes energy.
a. Can you name the given process?
b. Is it essential for survival?
c. What are the end products of this process?
d. Where does it occur?
Ans: a. Photorespiration
b. No
c. C02 and NH3
d. Photorespiration involves a complex network of enzyme reactions that exchange metabolites between chloroplasts, leaf peroxisomes and mitochondria.

Q20. Suppose Euphorbia and maize are grown in the tropical area.
a. Which one of them do you think will be able to survive under such conditions?
b. Which one of them is more efficient in terms of photosynthetic activity?
c. What difference do you think are there in their leaf anatomy?
Ans: a. Euphorbia is a CAM plant while maize is a C4 plant. Both of them will be able to survive in the tropical areas.
b. Maize (as it is a C4 plant)
c. Leaves of maize plant show Kranz anatomy which is absent in Euphorbia leaves.

Long Answer Type Questions

Q1. Is it correct to say that photosynthesis occurs only in leaves of a plant? Besides leaves, what are the other parts that may be capable of carrying out photosynthesis? Justify.
Ans: Photosynthesis does take place in the green leaves of plants but it does so also in other green parts of the plants. The mesophyll cells in the leaves, have a large number of chloroplasts. Usually the chloroplasts align themselves along the walls of the mesophyll cells, such that they get the optimum quantity of the incident light.
• Photosynthetic or Assimilatory roots: They are green roots which are capable of PHS, e.g., Trapa bispinosa (water chestnut = Singhara), Tmospora (Gillow or Gurcha), Podostemum.
• Some plants of arid regions modify their stems into flattened (Opuntia), or fleshy cylindrical (Euphorbia) structures. These modified stems of indefinite growth are called phylloclades. They contain chlorophyll and carry out photosynthesis.
• One intemode long phylloclade or stem which is leaf like is called cladode. Cladode is capable of photosynthesis. Cladode is found in certain xerophytes, e.g., Ruscus and Asparagus.

Q2.The entire process of photosynthesis consists of a number of reactions. Where in the cell do each of these take place?
a. Synthesis of ATP and NADPH _________
b. Photolysis of water _________
c. Fixation of C02 _________
d. Synthesis of sugar molecule _________
e. Synthesis of starch _________
Ans: a. Synthesis of ATP and NADPH: Membrane system (Grana)
b. Photolysis of water: Inner side of the membrane of thylakoid
c. Fixation of C02: Stroma of chloroplast
d. Synthesis of sugar molecule: Stroma of chloroplast
e. Synthesis of starch: Stroma of chloroplast

Q3. Which property of the pigment is responsible for its ability to initiate the process of photosynthesis? Why is the rate of photosynthesis higher in the red and blue regions of the spectrum of light?
Ans: Pigments are substances that have an ability to absorb light, at specific wavelengths. This property of the pigment is responsible for its ability to initiate the process of photosynthesis.
The wavelengths at which there is maximum absorption by chlorophyll a, i. e. in the blue and the red regions, also shows higher rate of photosynthesis.

Q4. What can we conclude from the statement that the action and absorption spectrum of photosynthesis overlap? At which wavelength do they show peaks?
Ans: The wavelengths at which therfe is maximum absorption by chlorophyll a, i.e. in the blue and the red regions, also shows higher rate of photosynthesis. Hence, we can conclude that chlorophyll a is the chief pigment associated with photosynthesis.
The action spectrum of photosynthesis superimposed on absorption spectrum of chlorophyll a.

Q5. Under what conditions are C4 plants superior to C3?
Ans: C4 plants are special:
i. They have a special type of leaf anatomy. ‘
ii: They tolerate higher temperatures. _
iii. They show a response to high light intensities.
iv. They lack a process called photorespiration.
v. They have greater productivity of biomass.

Q6. In the figure given below, the light line indicates action spectrum for photosynthesis and the black line indicates the absorption spectrum of chlorophyll a, answer the following:

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 substance?
c. If chlorophyll a is responsible for light reaction of photosynthesis, why do the action spectrum and absorption spectrum not overlap?
Ans: a. Action spectrum of photosynthesis superimposed on absorption spectrum of chlorophyll a. The wavelengths at which there is maximum . absorption by chlorophyll a, i.e. in the blue and the red regions, also shows higher rate of photosynthesis. Hence, one can conclude that chlorophyll a is the chief pigment associated with photosynthesis.
b. Absorption spectrum for any substance can be derived by plotting the different wavelengths of light.
c. Though chlorophyll a is the major pigment responsible for trapping light, other thylakoid pigments like chlorophyll b, xanthophylls and carotenoids, which are called accessory pigments, also absorb light and transfer the energy to chlorophyll a. Indeed, they not only enable a wider range of wavelength of incoming light to be utilised for photosynthesis but also protect chlorophyll a from photo-oxidation. Hence, the action spectrum and absorption spectrum not overlap.

Q7. What are the important events and end products of the light reaction?
Ans. Light reactions or the ‘Photochemical’ phase include light absorption, water splitting, oxygen release, and the formation of high-energy chemical intermediates, ATP and NADPH. Several complexes are involved in the process. The pigments are organised into two discrete photochemical light harvesting complexes (LHC) within the Photosystem I (PS I) and Photosystem II (PS II). These are named in the sequence of their discovery, and not in the sequence in which they function during the light reaction. The LHC are made up of hundreds of pigment molecules bound to proteins. Each photosystem has all the pigments (except one molecule of chlorophyll a) forming a light harvesting system also called antennae. These pigments help to make photosynthesis more efficient by absorbing different wavelengths of light. The single chlorophyll a molecule forms the reaction centre. The reaction centre is different in both the photosystems. In PS I the reaction centre chlorophyll a has an absorption peak at 700 nm, hence is called P700,while in PS II it has absorption maxima at 680 nm, and is called P680.

Q8. In the diagram shown below label A, B, C. What type of phosphorylation is possible in this?


Q9. Why is the RuBisCo enzyme more appropriately called RUBP Carboxylase- Oxygenase and what important role does it play in photosynthesis?
Ans: For ease of understanding, the Calvin cycle can be described under three stages: carboxylation, reduction and regeneration. Carboxylation is the fixation of C02 into a stable organic intermediate. Carboxylation is the most crucial step of the Calvin cycle where C02 is utilised for the carboxylation of RuBP. This reaction is catalysed by the enzyme RuBP carboxylase which results in the formation of two molecules of 3-PGA. Since this enzyme also has an oxygenation activity it would be more correct to call it RuBP carboxylase-oxygenase or RuBisCO.

Q10. What special anatomical features are displayed by leaves of C4 plants? How do they provide advantage over the structure of C3 plants?
Ans: Study vertical sections of leaves, one of a C3 plant and the other of a C4 plant. The particularly large cells around the vascular bundles of the C4 pathway plants are called bundle sheath cells, and the leaves which have such anatomy are said to have ‘Kranz’ anatomy. ‘Kranz’ means ‘wreath’ and is a reflection of the arrangement of cells. The bundle sheath cells may form several layers around the vascular bundles; they are characterised by having a large number of chloroplasts, thick walls impervious to gaseous exchange and no intercellular spaces. C4 plants lack photorespiration. In addition these plants show tolerance to higher temperatures. Plants that are adapted to dry tropical regions have the C4 pathway.
C4 plants are special:
i. They have a special type of leaf anatomy.
ii. They tolerate higher temperatures.
iii. They show a response to high light intensities.
iv. They lack a process called photorespiration.
v. They have greater productivity of biomass.

Q11. Name the two important enzynies of C3 and C4 pathway, respectively? What important role do they play in fixing C02?
Ans: The important enzyme of C3 pathway is RuBisCO and that of C4 pathway is PEPcase.
Carboxylation in the C3 pathway is the fixation of C02 into a stable organic intermediate. Carboxylation is the most crucial step of the Calvin cycle, where C02 is utilised for the carboxylation of RuBP. This reaction is catalysed by the enzyme RuBP carboxylase which results in the formation of two molecules of 3-PGA.
The primary C02 acceptor in the C4 pathway is a 3-carbon molecule phosphoenol pyruvate (PEP) and is present in the mesophyll cells. The enzyme responsible for this fixation is PEP carboxylase or PEPcase.

Q12. Why is RuBisCo enzyme the most abundant enzyme in the world?
Ans: RuBisCo enzyme is the most abundant enzyme in the world because this enzyme is responsible for photosynthesis and present in all green parts of the plants including leaves.

Q13. Why does not photorespiration take place in C4 plants?
Ans: In C4 plants photorespiration does not occur. This is because they have a mechanism that increases the concentration of C02 at the enzyme site. This takes place when the C4 acid from the mesophyll is broken down in the bundle sheath cells to release C02 – this results in increasing the intracellular concentration of C02. In turn, this ensures that the RuBisCO functions as a carboxylase minimising the oxygenase activity.

NCERT Exemplar ProblemsMathsPhysicsChemistryBiology

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

NCERT Exemplar Class 11 Biology Chapter 9 Biomolecules are part of NCERT Exemplar Class 11 Biology. Here we have given NCERT Exemplar Class 11 Biology Chapter 9 Biomolecules. https://www.cbselabs.com/ncert-exemplar-problems-class-11-biology-chapter-9-biomolecules/

NCERT Exemplar Class 11 Biology Chapter 9 Biomolecules

Multiple Choice Questions

Q1. It is said that elemental composition of living organisms and that of inanimate objects (like earth’s crust) are similar in the sense that all the major elements are present in both. Then what would be the difference between these two . groups? Choose a correct answer from the following.
(a) Living organisms have more gold in them than inanimate objects
(b) Living organisms have more water in their body than inanimate objects
(c) Living organisms have more carbon, oxygen and hydrogen per unit mass than inanimate objects
(d) Living organisms have more calcium in them than inanimate objects.
Ans: (c)

Element	% Weight of
	Earth’s Crust	Human Body
Hydrogen (H)	0.14	0.5
Carbon (C)	0.03	18.5
Oxygen(0)	46.6	65.0
Nitrogen (N)	very little	3.3
Sulphur (S)	0.03	0.3
Sodium (Na)	2.8	0.2
Calcium (Ca)	3.6	1.5
Magnesium (Mg)	2.1	0.1
Silicon (Si)	27.7	negligible

Q2. Many elements are found in living organisms either free or in the form of compounds. One of the following is not found in living organisms.
(a) Silicon (b) Magnesium (c) Iron (d) Sodium
Ans: (a) See Answer 2.
Q3. Aminoacids, as the name suggests, have both an amino group and a carboxyl group in their structure. In addition, all naturally occurring aminoacids (those which are found in proteins) are called L-aminoacids. From this, can you guess from which compound can the simplest aminoacid be made?
(a) Formic acid (b) Methane (c) Phenol acid (d) Glycine
Ans: (d) Glycine is an amino acid (which have both an amino group and a carboxyl group in their structure).

Q4. Many organic substances are negatively charged, e.g., acetic acid, while others are positively charged e.g., ammonium ion. An aminoacid under certain conditions would have both positive and negative charges simultaneously in the same molecule. Such a form of aminoacid is called
(a) Positively charged form (b) Negatively charged form
(c) Neutral form (d) Zwitterionic form
Ans: (d) In aqueous solution, the carboxyl group can lose a proton and amino group can accept a proton, giving rise to a dipolar ion called Zwitter ion. Zwitter ion is neutral but contains both positive and negative charges.

Q5. Sugars are technically called carbohydrates, referring to the fact that their formulae are only multiple of C(H20). Hexoses therefore have six carbons, twelve hydrogens and six oxygen atoms. Glucose is a hexose. Choose from among the following another hexose.
(a) Fructose (b) Erythrose ~(c) Ribulose (d) Ribose
Ans: (a) Sugars are technically called carbohydrates, referring to the fact that their formulae are only multiple of C(H20). Hexoses therefore have six carbons, twelve hydrogens and six oxygen atoms. E.g., glucose and fructose.

Q6. When you take cells or tissue pieces and grind them with an acid in a mortar and pestle, all the small biomolecules dissolve in the acid. Proteins polysaccharides and nucleic acids are insoluble in mineral acid and get precipitated. The acid soluble compounds include amino acids, nucleosides, small sugars etc. When one adds a phosphate group to a nucleoside one gets another acid soluble biomolecule called
(a) Nitrogen base
(b) Adenine
(c) Sugar phosphate
(d) Nucleotide
Ans: (d) Neucliotide = base + sugar + phosphate

Q7. When we homogenise any tissue in an acid, the acid soluble pool represents
(a) Cytoplasm (b) Cell membrane
(c) Nucleus (d) Mitochondria
Ans: (a) When we homogenise any tissue in an acid, the acid soluble pool represents cytoplasm.

Q8. The most abundant chemical in living organisms could be
(a) Protein (b) Water (c) Sugar (d) Nucleic acid
Ans: (b) Most abundant component of cell is water.

Component	% of the total Cellular Mass
Water	70-90
Proteins	10-15
Nucleic acids	5-7
Carbohydrates	3
Lipids	2
Ions	1

 

Q9. A homopolymer has only one type of building block called monomer repeated V number of times. A heteropolymer has more than one type of monomer. Proteins are heteropolymers usually made of aminoacids. While a nucleic acid like DNA or RNA is made up of only 4 types of nucleotide monomers, proteins are made of
(a) 20 types of monomers (b) 40 types of monomers
(c) 30 types of monomers (d) only one type of monomer
Ans: (a) A homopolymer has only one type of building block called monomer repeated V number of times. A heteropolymer has more than one type of monomer. Proteins are heteropolymers usually made of amino acids. While a nucleic acid like DNA or RNA is made of of only 4 types of nucleotide monomers, proteins are made of 20 types of monomers.

Q10. Proteins perform many physiological functions. For example, some proteins function as enzymes. One of the following represents an additional function that some proteins discharge
(a) Antibiotics
(b) Pigment conferring colour to skin
(c) Pigment making colours of flowers
(d) Hormones
Ans: (d) Proteins perform many physiological functions. For example, some proteins function as enzymes. Hormones represents an additional function that some proteins discharge (like insulin).

Q11. Glycogen is a homopolymer made of
(a) Glucose units (b) Galactose units
(c) Ribose units (d) Amino acids
Ans: (a) Glycogen is a homopolymer made of glucose units.

Q12. The number of ‘ends’ in a glycogen molecule would be
(a) Equal to the number of branches plus one
(b) Equal to the number of branch points
(c) One ‘
(d) Two, one on the left side and another on the right side
Ans: (d) In a polysaccharide chain (say glycogen), the right end is called the reducing end and the left end is called the non-reducing end.

Q13. The primary structure of a protein molecule has
(a) Two ends (b) One end (c) Three ends (d) No ends
Ans: (a) The primary structure of a protein molecule has two ends.
A protein is imagined as a line, the left end represented by the first amino acid and the right end is represented by the last amino acid. The first amino acid is also called as N-terminal amino acid. The last amino acid is called the C-terminal amino acid.

Q14. Enzymes are biocatalysts. They catalyse biochemical reactions. In general they reduce activation energy of reactions. Many physico-chemical processes are enzyme mediated.’ Some examples of enzyme mediated reactions are given below. Tick the wrong entry.
(a) Dissolving C02 in water
(b) Unwinding the two strands of DNA .
(c) Hydrolysis of sucrose
(d) Formation of peptide bond
Ans: (a) Dissolving C02 in water is a physical process.

Very Short Answer Type Questions
Q1. 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 a synthetic chemical.
a. Penicillin
b. Sulfonamide
c. Vitamin C
d. Growth Hormone
Ans: a. Penicillin: Natural product
b. Sulfonamide: Synthetic chemical
c. Vitamin C: Natural product
d. Growth Hormone: Natural product

Q2. Select an appropriate chemical bond among ester bond, glycosidic bond, peptide bond and hydrogen bond and write against each of the following.
a. Polysaccharide
b. Protein
c. Fat
d. Water
Ans: a. Polysaccharide: Glycosidic bond
b. Protein: Peptide bond
c. Fat: Ester bond
d. Water: Hydrogen bond

Q3. Write the name of any one aminoacid, sugar, nucleotide and fatty acid.
Ans: Glycine (amino acid), Ribose (sugar), Cytidylic acid (nucleotide) and
Arachidonic acid (fatty acid).

Q4. Reaction given below is catalysed by oxidoreductase between two substrates A and A’, complete the reaction. A reduced + A‘ oxidised -»
Ans: A reduced + A’ oxidised —»A oxidised + A’ reduced

Q5. How are prosthetic groups different from co-factors?
Ans: Prosthetic groups are organic compounds and are distinguished from other cofactors in that they are tightly bound to the apoenzyme. For example, in peroxidase and catalase, which catalyze the breakdown of hydrogen peroxide to water and oxygen, haem is the prosthetic group and it is a part of the active site of the enzyme.
Cofactor may be organic or inorganic (metal ions).

Q6. Glycine and Alanine are different with respect to one substituent on the a-carbon. What are the other common substituent groups?
Ans: The R-group in these proteinaceous amino acids could be a hydrogen (the amino acid is called glycine), a methyl group (alanine), hydroxy methyl (serine), etc.

 

Q7. Starch, Cellulose, Glycogen, Chitin are polysaccharides found among the following. Choose the one appropriate and write against each.
Cotton fibre __________
Exoskeleton of Cockroach __________
Liver __________
Peeled potato __________
Ans: Cotton fibre : Cellulose
Exoskeleton of Cockroach : Chitin
Liver: Glycogen
Peeled potato: Starch

Q4. Nucleic acids exhibit secondary structure, justify with example.
Ans: Nucleic acids exhibit a wide variety of secondary structures. For example, one of the secondary structures exhibited by DNA is the famous Watson—Crick model. This model says that DNA exists as a double helix. The two strands of polynucleotides are antiparallel, i.e. run in the opposite direction. The backbone is formed by the sugar-phosphate-sugar chain. The nitrogen bases are projected more or less perpendicular to this backbone but face inside. A and G of one strand compulsorily base pairs with T and C, respectively, on the other strand. There are two hydrogen bonds between A and T and three hydrogen bonds between G and C. Each strand appears like a helical staircase. Each step of ascent is represented by a pair of bases. At each step of ascent, the strand turns 36°. One full turn of the helical strand would involve ten steps or ten base pairs. Attempt drawing a line diagram. The pitch would be 34 A. The rise per base pair would be 3.4 A. This form of DNA with the above mentioned salient features is called B-DNA.

Q5. Comment on the statement “living state is a non-equilibrium steady state to be able to perform work”.
Ans: The most important fact of biological systems is that all living organisms exist in a steady-state characterised by concentrations of each of these biomolecules. These biomolecules are in a metabolic flux. Any chemical or physical process moves spontaneously to equilibrium. The steady state is a non-equilibrium state. One should remember from physics that systems at equilibrium cannot perform work. As living organisms work continuously, they cannot afford to reach equilibrium. Hence the living state is a non-equilibrium steady-state to be able to perform work; living process is a constant effort to prevent falling into equilibrium. This is achieved by energy input. Metabolism provides a mechanism for the production of energy. Hence the living state and metabolism are synonymous. Without metabolism there cannot be a living state.

Long Answer Type Questions
Q1. Formation of enzyme-substrate complex (ES) is the first step in catalysed reactions. Describe the other steps till the formation of product.
Ans: The catalytic cycle of an enzyme action can be described in the following steps:
(1) First, the substrate binds to the active site of the enzyme, fitting into the active site.
(2) The binding of the substrate induces the enzyme to alter its shape, fitting more tightly around the substrate.
(3) The active site of the enzyme, now in close proximity of the substrate breaks the chemical bonds of the substrate and the new enzyme-product complex is formed.
(4) The enzyme releases the products of the reaction and the free enzyme is ready to bind to another molecule of the substrate and run through the catalytic cycle once again.

Q2. What are different classes of enzymes? Explain any two with the type of reaction they catalyse.
Ans: Enzymes are divided into 6 classes each with 4—13 subclasses and named accordingly by a four-digit number.

Q3. Nucleic acids exhibit secondary structure. Describe through Watson-Crick Model.
Ans: Nucleic acids exhibit a wide .variety of secondary structures. For example, one of the secondary structures exhibited by DNA is the famous Watson- Crick model. This model says that DNA exists as a double helix. The two strands of polynucleotides are antiparallel, i.e. run in the opposite direction. The backbone is formed by the sugar—phosphate—sugar chain. The nitrogen bases are projected more or less perpendicular to this backbone but face inside. A and G of one strand compulsorily base pairs with T and C, respectively, on the other strand. There are two hydrogen bonds between A

and T and three hydrogen bonds between G and C. Each strand appears like a helical staircase. Each step of ascent is represented by a pair of bases. At each step of ascent, the strand turns 36°. One full turn of the helical strand would involve ten steps or ten base pairs. Attempt drawing a line diagram. The pitch would be 34 A. The rise per base pair would be 3.4 A. This form of DNA with the above mentioned salient features is called B-DNA.

Q4. What is the difference between a nucleotide and nucleoside? Give two examples of each with their structure.
Ans: Living organisms have a number of carbon compounds in which heterocyclic rings can be found. Some of these are nitrogen bases—adenine, guanine, cytosine, uracil and thymine. When found attached to a sugar, they are called nucleosides. If a phosphate group is also found esterified to the sugar they are called nucleotides. Adenosine, guanosine, thymidine, uridine and cytidine are nucleosides. Adenylic acid, thymidylic acid, guanylic acid, uridylic acid and cytidylic acid are nucleotides.

Q5. Describe various forms of lipid with a few examples.
Ans: Lipids are generally water insoluble. They could be simple fatty acids. A fatty acid has a carboxyl group attached to an R-group. The R-group could be a methyl (-CH3), or ethyl (—C2H5) or higher number of-CH2 groups (1 carbon to 19 carbons). For example, palmitic acid has 16 carbons including carboxyl carbon. Arachidonic acid has 20 carbon atoms including the carboxyl carbon. Fatty acids could be saturated (without double bond) or unsaturated (with one or more C=C double bonds). Another simple lipid is glycerol which is trihydroxy propane.
• Many lipids have both glycerol and fatty acids. Here the fatty acids are found esterified with glycerol. They can be then monoglycerides,
diglycerides and triglycerides. These are also called fats and oils based on melting point. Oils have lower melting point (e.g., gingely oil) and hence remain as oil in winters.
• Some lipids have phosphorous and a phosphorylated organic compound in them. These are phospholipids. They are found in cell membrane. Lecithin is one example. Some tissues especially the neural tissues have lipids with more complex structures.

NCERT Exemplar ProblemsMathsPhysicsChemistryBiology

We hope the NCERT Exemplar Class 11 Biology Chapter 9 Biomolecules help you. If you have any query regarding NCERT Solutions for NCERT Exemplar Class 11 Biology Chapter 9 Biomolecules, drop a comment below and we will get back to you at the earliest.