NCERT Solutions for Class 9th Hindi Chapter 4 माटी वाली 

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NCERT Exemplar Problems Class 12 Biology Chapter 1 Reproduction in Organisms

Multiple Choice Questions
Single Correct Answer Type
1. A few statements describing certain features of reproduction are given below
i. Gametic fusion takes place
ii. Transfer of genetic material takes place
iii. Reduction division takes place
iv. Progeny have some resemblance with parents
Select the options that are true for both asexual and sexual reproduction from the options given below:
(a) i and ii (b) ii and iii
(c) ii and iv (d) i and iii
Answer. (c) Transfer of genetic material and progeny have some resemblance with parents are the phenomenon common in’both asexual and sexual reproduction while gametic fusion and reduction division takes place in sexual reproduction only.

2. The term ‘ clone ’ cannot be applied to offspring formed by sexual reproduction because
(a) Offspring do not possess exact copies of parental DNA
(b) DNA of only one parent is copied and passed on to the offspring
(c) Offspring are formed at different times
(d) DNA of parent and offspring are completely different
Answer. (a)
• In asexual reproduction, a single individual (parent) is capable of producing offspring which are not only identical to one another but are also exact copies of their parent. The term clone is used to describe such morphologically and genetically similar individuals.
• In sexual reproduction because of the fusion of male and female gametes (either by same individual or by different individual of the opposite sex), sexual reproduction results in offspring that are not identical to the parents or amongst themselves.

3. Amoeba and Yeast reproduce asexually by fission and budding respectively, because they are
(a) Microscopic organisms
(b) Heterotrophic organisms
(c) Unicellular organisms
(d) Uninucleate organisms
Answer. (c) Many single-celled organisms reproduce by binary fission (e.g., Amoeba, Paramecium), where a cell divides into two halves and each rapidly grows into an adult.
In yeast, the division is unequal and small buds are produced that remain attached initially to the parent cell which eventually gets separated and mature into new yeast organism (cells). Budding is also found in Hydra.

4. A few statements with regard to sexual reproduction are given below
i. Sexual reproduction does not always require two individuals
ii. Sexual reproduction generally involves gametic fusion
iii. Meiosis never occurs during sexual reproduction
iv. External fertilisation is a rule during sexual reproduction
Choose the correct statements from the options below:
(a) i and iv , (b) i and ii
(c) ii and iii (d) i and iv
Answer. (b)
• Sexual reproduction requires male and female gametes (either by same individual or by different individual of the opposite sex).
• Sexual reproduction generally involves gametic fusion.
• Meiosis occurs during sexual reproduction in dipoloid organisms.
• External fertilisation is not a rule during sexual reproduction, internal fertilization also takes place

5. A multicellular, filamentous alga exhibits a type of sexual life cycle in which the meiotic division occurs after the formation of zygote. The adult filament of this alga has
(a) Haploid vegetative cells and diploid gametangia
(b) Diploid vegetative cells and diploid gametangia
(c) Diploid vegetative cells and haploid gametangia
(d) Haploid vegetative cells and haploid gametangia
Answer. (d) Adult filament of a multicellular, filamentous alga have haplontic life cycle in which the meiotic division occurs after the formation of zygote. So, the filament of this alga have haploid vegetative cells and haploid gametangia.

6. The male gametes of rice plant have 12 chromosomes in their nucleus. The chromosome number in the female gamete, zygote and the cells of the seedling will be, respectively,
(a) 12,24,12 . (b) 24,12,12
(c) 12,24,24 (d) 24,12,24
Answer. (c) Gametophytic structure (n) of rice plant contain 12 chromosomes and sporophytic structure (2n) of rice contain 24 chromosomes.
Female gamete (n) =12,
Zygote (2n) = 24,
The cells of the seedling (2n) = 24.

7. Given below are a few statements related to external fertilization. Choose the correct statements.
i. The male and female gametes are formed and released simultaneously.
ii. Only a few gametes are released into the medium.
iii. Water is the medium in a majority of organisms exhibiting external fertilization.
iv. Offspring formed as a result of external fertilization have better chance of survival than those formed inside an organism.
(a) iii and iv (b) i and iii
(c) ii and iv (d) i and iv .
Answer. (b) In most aquatic organisms, such as a majority of algae and fishes as well as amphibians, syngamy occurs in the external medium (water), i.e., outside the body of the organism. This type of gametic fusion is called external fertilisation. Organisms exhibiting external fertilisation show great synchrony between the sexes and release a number of gametes into the surrounding medium (water) in order to enhance the chances of syngamy. This happens in the bony fishes and frogs where a large number of offspring are produced. A major disadvantage is that the offspring are extremely vulnerable to predators threatening their survival up to adulthood.

8. The statements given below describe certain features that are observed in the pistil of flowers.
i. Pistil may have many carpels
ii. Each carpel may have more than one ovule
iii. Each carpel has only one ovule
iv. Pistil have only one carpel
Choose the statements that are true from the options below:
(a) i and ii (b) i and iii
(c) ii and iv (d) iii and iv
Answer. (a)
• Pistil may have many carpels (multicapillary pistil like Papaver)
• Each carpel may have more than one ovule (like Watermelon,.Papaya etc.)

9. Which of the following situations correctly describe the similarity between an angiosperm egg and a human egg?
i. Eggs of both are formed only once in a lifetime
ii. Both the angiosperm egg and human egg are stationary
iii. Both the angiosperm egg and human egg are motile transported
iv. Syngamy in both results in the formation of zygote
Choose the correct answer from the options given below:
(a) ii and iv (b) iv only
(c) iii and iv (d) i and iv
Answer. (b) Syngamy in both results in the formation of zygote is similarity between an angiosperm egg and a human egg.

10. Appearance of vegetative propagules from the nodes of plants such as sugarcane and ginger is mainly because
(a) Nodes are shorter than intemodes
(b) Nodes have meristematic cells
(c) Nodes are located near the soil
(d) Nodes have non-photosynthetic cells
Answer. (b) Appearance of vegetative propagules from the nodes of plants such as sugarcane and ginger is mainly because nodes have meristematic cells. Examples of vegetative propagules: (i) Leaf buds of bryophyllum, (ii) Eyes of potato, (iii) Bulbifof Agave, (iv) Offset of water hyacinth, (v) Rhizome of ginger.

11. Which of the following statements, support the view that elaborate sexual reproductive process appeared much later in the organic evolution?
i. Lower groups of organisms have simpler body design
ii. Asexual reproduction is common in lower groups
iii. Asexual reproduction is common in higher groups of organisms
iv. The high incidence of sexual reproduction in angiosperms and vertebrates
Choose the correct answer from the options given below:
(a) i, ii and iii (b) i, iii and iv
(c) i, ii and iv (d) ii, iii and iv
Answer. (c) Elaborate sexual reproductive process appeared much later in the organic evolution because of
• Lower groups of organisms have simpler body design.
• Asexual reproduction is common in lower groups of organisms.
• High incidence of sexual reproduction in angiosperms and vertebrates.

12. Offspring formed by sexual reproduction exhibit more variation than those formed by asexual reproduction because
(a) Sexual reproduction is a lengthy process
(b) Gametes of parents have qualitatively different genetic composition
(c) Genetic material comes from parents of two different species
(d) Greater amount of DNA is involved in sexual reproduction
Ans. (b)
• Offspring formed by sexual reproduction exhibit more variation than those formed by asexual reproduction because gametes of parents have qualitatively different genetic composition.
• In asexual reproduction due to involvement of only one parent, so there is no chance of variation.

13. Choose the correct statement from amongst the following:
(a) Dioecious (hermaphrodite) organisms are seen only in animals.
(b) Dioecious organisms are seen only in plants.
(c) Dioecious organisms are seen in both plants and animals.
(d) Dioecious organisms are seen only in vertebrates.
Answer. (c) Dioecious organisms are seen in both plants (like papaya) and animals (like cockroach).

14. There is no natural death in single celled organisms like Amoeba and bacteria because
(a) They cannot reproduce sexually
(b) They reproduce by binary fission
(c) Parental body is distributed among the offspring
(d) They are microscopic
Answer. (c) There is no natural death in single celled organisms like Amoeba and bacteria because the parental body is distributed among the offspring.

15. There are various types of reproduction. The type of reproduction adopted by an organism depends on
(a) The habitat and morphology of the organism
(b) Morphology of the organism
(c) Morphology and physiology of the organism
(d) The organism’s habitat, physiology and genetic make up
Answer. (d) The organism’s habitat, its internal physiology and several other factors (genetic make up) are collectively responsible for how it reproduces. When offspring is produced by a single parent with or without the involvement of gamete formation, the reproduction is asexual.

16. Identify the incorrect statement.
(a) In asexual reproduction, the offspring produced are morphologically and genetically identical to the parent.
(b) Zoospores are sexual reproductive structures.
(c) In asexual reproduction, a single parent produces offspring with or without the formation of gametes.
(d) Conidia are asexual structures in Penicillium.
Answer. (b) Zoospores are asexual reproductive structures.

17.Which of the following is a post-fertilisation event in flowering plants?
(a) Transfer of pollen grains
(b) Embryo development
(c) Formation of flower
(d) Formation of pollen grains
Answer. (b)

18. The number of chromosomes in the shoot tip cells of a maize plant is 20. The number of chromosomes in the micro spore mother cells of the same plant shall be
(a) 20 (b) 10 (c) 40 (d) 15
Answer. (a) Shoot tip cells of a maize plant is a sporophytic structure (2n) and microspore mother cells of maize plant is also a sporophytic structure (2n). So, microspore mother cells (MMC) contain 20 chromosomes.

Very Short Answer Type Questions
1. Mention two inherent characteristics of Amoeba and yeast that enable them to reproduce asexually.
Answer. a. They are unicellular organisms.
b. They have a very simple body structure.

2. Why do we refer to’offspring formed by asexual method of reproduction as clones?
Answer. Offspring formed by asexual reproduction are called clones because they are morphologically and genetically similar to the parent.

3. Although potato tuber is an underground part, it is considered as a stem. Give two reasons.
Answer. a. The tuber has nodes and intemodes (as stem),
b. Leafy shoots appear from the nodes.

4. Between an annual and a perennial plant, which one has a shorter juvenile phase? Give one reason.
Answer. An annual has a shorter juvenile phase. Since its entire life cycle has to be completed in one growing season, its juvenile phase is shorter.

5. Rearrange the following events of sexual reproduction in the sequence in which they occur in a flowering plant: embryogenesis, fertilisation, gametogenesis, pollination.
Answer. Gametogenesis, Pollination, Fertilisation, Embryogenesis

6. The probability of fruit set in a self-pollinated bisexual flower of a plant is far greater than a dioecious plant. Explain.
Answer. There is assured fruit set in self pollinated bisexual flower even in the absence of pollinators. In dioecious plants, there is male and female flowers present on different plants, so external pollinating agent is required for pollination.

7. Is the presence of large number of chromosomes in an organism a hindrance to sexual reproduction? Justify your answer by giving suitable reasons.
Answer. Presence of large number of chromosomes in an organism is not a hindrance to sexual reproduction. Butterfly has 380 chromosomes but it can reproduce sexually.

8. Is there a relationship between the size of an organism and its life span? Give two examples in support of your answer.
Answer. Life spans of organisms are not necessarily correlated with their sizes. The sizes of crows and parrots are not very different yet their life spans show a wide difference. Live span of crow is 15 year and of parrot is 140 years. A mango tree has a much shorter life span as compared to a peepal tree.

9. In the figure given below, the plant bears two different types of flowers marked ‘A’ and ‘B\ Identify the types of flowers and state the type of pollination that will occur in them.

Answer. ‘A’ is chasmogamous flower while ‘B’ is cleistogamous flower. A bisexual flower which normally open is called chasmogamous flower. Cleistogamous flowers do not open at all.
Cleistogamous flowers are invariably autogamous as there is no chance of cross-pollen landing on the stigma.
In a normal flower which opens and exposes the anthers and stigma complete autogamy is rather rare. Chasmogamous flower may show autogamy, geitonogamy or xenogamy.

10. Give reasons as to why cell division cannot be a type of reproduction in multicellular organisms.
Answer. Cell division cannot be a type of reproduction in multicellular organisms because cell division only increases the number of cells in an organism which leads to the growth of body.

11. In the figure given below, mark the ovule and pericarp.

Answer.

12. Why do gametes produced in large numbers in organisms exhibit external fertilisation?
Answer. Organisms exhibiting external fertilisation release a number of gametes into the surrounding medium (water) in order to enhance the chances of syngamy because there are few’ chances of fusion between male and female gametes.

13. Which of the followings are monoecious and dioecious organisms?
a. Earthworm ——————–
b. Chara ——————–
c. Marchantia ——————-
d. Cockroach ——————–
Answer. a. Earthworm—Monoecious
b. Chara—Monoecious
c. Marchantia—Dioecious
d. Cockroach—Dioecious

14. Match the organisms given in Column ‘A’ with the vegetative propagules given in column ‘B’.

Answer. Bryophyllum—leaf buds Agave—bulbils Potato—eyes
Water hyacinth—offset

15. What do the following parts of a flower develop into after fertilisation?
a. Ovary
b. Ovules
Answer. a. Ovary—Fruit
b. Ovules—Seeds

Short Answer Type Questions
1. In haploid organisms that undergo sexual reproduction, name the stage in the life cycle when meiosis occurs. Give reasons for your answer.
Answer. Meiosis takes place during its post-zygotic stage. Since the organism is haploid, meiosis cannot occur during gametogenesis.

2. The number of taxa exhibiting asexual reproduction is drastically reduced in higher plants (angiosperms) and higher animals (vertebrates) as compared with lower groups of plants and animals. Analyse the possible reasons for this situation.
Answer. Both angiosperms and vertebrates have a more complex structural organisation. They have evolved very efficient mechanism of sexual reproduction. Since asexual reproduction does not create new genetic pools in the offspring and consequently hampers their adaptability to external conditions, these groups have resorted to reproduction by the sexual method.

3. Honeybees produce their young ones only by sexual reproduction. Inspite of this, in a colony of bees we find both haploid and diploid individuals. Name the haploid and diploid individuals in the colony and analyse the reasons behind their formation.
Solution.
• The colony of honey bees has three types of members: (i) Diploid queen are fertile females, (ii) Worker bees are sterile females and (iii) Drones are haploid males.
• An offspring formed from the union of a sperm and an egg develops as a female (queen or worker), and an unfertilized egg develops as a male (drone) by means of parthenogenesis. This means that the males have half the number of chromosomes than that of a female.

4. With which type of reproduction do we associate the reduction division? Analyse the reasons for it.
Answer. Reduction division (meiosis) is associated with sexual reproduction. The reasons for this are:
a. Since sexual reproduction involves the fusion of two types of gametes (male and female), they must have haploid number of chromosomes.
b. The cell (meiocyte) which gives rise to gametes often has diploid number of chromosomes and it is only by reducing the number by half that we can get haploid gametes.
c. Reduction division also ensures maintenance of constancy of chromosome number from generation to generation.

5. Is it possible to consider vegetative propagation observed in certain plants like Bryophyllum, water hyacinth, ginger etc., as a type of asexual reproduction? Give two/three reasons.
Answer. Vegetative propagation is considered as a type of asexual reproduction because
(i) This is uniparental.
(ii) Clone formation takes place.
(iii) There is no fertilisation.

6. ‘Fertilisation is not an obligatory event for fruit production in certains plants’. Explain the statement.
Answer. Yes, it is observed in parthenocarpic fruits. The ‘seedless fruits’ that are available in the market such as pomegranate, grapes etc. are in fact good examples. Flowers of these plants are sprayed by a growth hormone that induces fruit development even though fertilisation has not occurred. The ovules of such fruits, however, fail to develop into seeds.

7. In a developing embryo, analyse the consequences if cell divisions are not followed by cell differentiation.
Answer. During embryogenesis, zygote undergoes cell-division (mitosis) and cell differentiation. While cell divisions increase the number of cells in the developing embryo; Cell differentiation helps groups of cells to undergo certain modifications to form specialised tissues and organs to form an organism.
If cell divisions are not followed by cell differentiation then there will be no formation of tissues or organs, so a new organisms cannot be formed.

8. List the changes observed in an angiosperm flower subsequent to pollination and fertilisation.
Answer. Post-fertilisation modifications

9. Suggest a possible explanation why the seeds in a pea pod are arranged in a row, whereas those in tomato are scattered in the juicy pulp.
Answer. In a fruit, seed arrangement depends on type of placentation. Pea and tomato shows different placentation. Pea shows marginal placentation while tomato shows axile placentation.

10. Draw the sketches of a zoospore and a conidium. Mention two dissimilarities between them and alt least one feature common to both structures.
Answer.

11. Justify the statement ‘Vegetative reproduction is also a type of asexual reproduction’.
Answer. Vegetative propagation is also a type of asexual reproduction because
(i) This is uniparental.
(ii) Clone formation takes place.
(iii) There is no fertilisation.
(iv) There is no gamete formation.

Long Answer Type Question
1. Enumerate the differences between asexual and sexual reproduction. Describe the types of asexual reproduction exhibited by unicellular organisms.
Answer.

The types of asexual reproduction exhibited by unicellular organisms:
• Many single-celled organisms reproduce by binary fission (e.g., Amoeba, Paramecium), where a cell divides into two halves and each rapidly grows into an adult.
• In yeast, the division is unequal and small buds are produced that remain attached initially to the parent cell which eventually gets separated and mature into new yeast organism (cells).

2. Do all the gametes formed from a parent organism have the same genetic composition (identical DNA copies of the parental genome)? Analyse the situation with the background of gametogenesis and provide or give suitable explanation.
Answer. The gametes of a parent do not have the same genetic composition because they do not have identical copies of DNA. In the pachytene and diplotene stages of meiosis-I, the phenomenon of crossing over and chiasma formation take place between homologous chromosomes. This shifts segments of DNA from one chromatid to another (homologous chromosomes) in a random manner resulting in several new combinations of DNA sequences. As a result, when meiotic division is completed, gametes possess DNA with varying degree of variations.

3. Although sexual reproduction is a long drawn, energy-intensive complex form of reproduction, many groups of organisms in Kingdom Animalia and Plantae prefer this mode of reproduction. Give at least three reasons for this.
Answer. a. Sexual reproduction brings about variation in the offspring.
b. Since gamete formation is preceded by meiosis, genetic recombination occurring during crossing over (meiosis-I), leads to a great deal of variation in the DNA of gametes.
c. The organism has better chances survival in a changing environment.

4. Differentiate between (a) oestrus arid menstrual cycles; (b) ovipary and vivipary. Cite an example for each type.
Answer. Differences between oestrus and menstrual cycles

5. Rose plants produce large, attractive bisexual flowers but they seldom produce Suits. On the other hand a tomato plant produces plenty of fruits though they have small flowers. Analyse the reasons Tor failure of fruit formation in rose.
Answer. Failure of fruit formation in rose may be due to several reasons. Some of the likely reasons are
a. Rose plants may not produce viable pollen.
b. Rose plants may not have functional egg.
c. Rose plants may have abortive ovules.
d. Being hybrids, the meiotic process may be abnormal resulting in non-viable gametes. ‘
e. There may be self-incompatibility.
f. There may be internal barriers for pollen tube growth and/or fertilisation.

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NCERT Solutions for Class 9th Hindi Chapter 1 निबंध -लेखन 

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NCERT Solutions for Class 9th Hindi Chapter 2 पत्र -लेखन 

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NCERT Solutions for Class 9th Hindi Chapter 3 प्रतिवेदन 

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NCERT Exemplar Problems Class 12 Biology Chapter 2 Sexual Reproduction in Flowering Plants

Multiple Choice Questions
Single Correct Answer Type
Question.1. Among the terms listed below, those that are of not technically correct names for a floral whorl are
i. Androecium ii. Carpel
iii. Corolla iv. Sepal
(a) i and iv (b) iii and iv
(c) ii andiv (d) i and ii
Answer. (c)
• There are 4 floral whorls viz., calyx, corolla, androecium and gynoecium. Calyx and corolla are accessory organs or non-essential whorl, while androecium and gynoecium are reproductive organs.
• The calyx is the outermost whorl of the flower and the members are called sepals.
• Gynoecium is the female reproductive part of the flower and is made up of one or more carpels.

Question.2. Embryo sac is to ovule as is to an anther.
(a) Stamen .(b) Filament
(c) Pollen grain (d) Androecium
Answer. (c)
• Embryo sac (female gametophyte) • Ovule (megasporangium)
• Pollen grain (male gametophyte) • Anther (microsporangium)

Question.3. In a typical complete, bisexual and hypogynous flower, the arrangement of floral whorls on the thalamus from the outermost to the innermost is
(a) Calyx, corolla, androecium and gynoecium
(b) Calyx, corolla, gynoecium and androecium
(c) Gynoecium, androecium, corolla and calyx
(d) Androecium, gynoecium, corolla and calyx
Answer. (a) Arrangement of floral whorls on the thalamus from the outermost to the innermost is calyx, corolla, androecium and gynoecium.

Question.4. A dicotyledonous plant bears flowers but never produces fruits and seeds. The most probable cause for the above situation is
(a) Plant is dioecious and bears only pistillate flowers
(b) Plant is dioecious and bears both pistillate and staminate flowers
(c) Plant is monoecious
(d) Plant is dioecious and bears only staminate flowers.
Answer. (d) A dicotyledonous plant bears flowers but never produces fruits and seeds because plant is dioecious and bears only staminate flowers.

Question.5. The outermost and innermost wall layers of microsporangium in an anther are respectively
(a) Endothecium and tapetum (b) Epidermis and endodermis
(c) Epidermis and middle layer (d) Epidermis and tapetum
Answer. (d) Wall layers of microsporangium in an anther are:

Question.6. During microsporogenesis, meiosis occurs in
(a) Endothecium (b) Microspore mother cells
(c) Microspore tetrads (d) Pollen grains
Answer. (b) During microsporogenesis, meiosis occurs in microspore mother cells.

Question.7. From among the sets of terms given below, identify those that are associated with the gynoecium.
(a) Stigma, ovule, embryo sac, placenta
(b) Thalamus, pistil, style, ovule
(c) Ovule, ovary, embryo sac, tapetum
(d) Ovule, stamen, ovary, qpibryo sac
Answer. (a) Stigma, ovule, embryo sac and placenta are associated with the gynoecium.

Question.8. Starting from the innermost part, the correct sequence of parts in an ovule are
(a) Egg, nucellus, embryo sac, integument .
(b) Egg, embryo sac, nucellus, integument.
(c) Embryo sac, nucellus, integument, egg
(d) Egg, integument, embryo sac, nucellus
Answer. (b) The correct sequence of parts in an ovule are

Question.9. From the statements given below, choose the option that are true for a typical female gametophyte of a flowering plant.
i. It is 8-nucleate and 7-celled at maturity
ii. It is free-nuclear during the development
iii. It is situated inside the integument but outside the nucellus
iv. It has an egg apparatus situated at the chalazal end
(a) i and iv (b) ii and iii
(c) i and ii (d) ii and iv
Answer. (c) A typical female gametophyte of a flowering plant is 8-nucleate and 7-celled at maturity and free-nuclear during the development.

Question.10. Autogamy can occur in a chasmogamous flower if
(a) Pollen matures before maturity of ovule
(b) Ovules mature before maturity of pollen
(c) Both pollen and ovules mature simultaneously
(d) Both anther and stigma are of equal lengths
Answer. (c) In a normal flower which opens and exposes the anthers and stigma complete autogamy is rather rare. Autogamy in such flowers requires synchrony in pollen release and stigma receptivity and also, the anthers and the stigma should lie close to each other so that self-pollination can occur.

Question.11. Choose the correct statement from the following:
(a) Cleistogamous flowers always exhibit autogamy
(b) Chasmogamous’flowers always exhibit geitonogamy
(c) Cleistogamous flowers exhibit both autogamy and geitonogamy
(d) Chasmogamous flowers never exhibit autogamy
Answer. (a) Cleistogamous flowers are invariably autogamous as there is no chance of cross-pollen landing on the stigma. Cleistogamous flowers produce assured seed-set even in absence of pollinators.
E.g. of cleistogamous flowers are Viola (common pansy), Oxalis, Commelina, Arachis hypogea and Oryza sativa.

Question.12. A particular species of plant produces light, non-sticky pollen in large numbers and its stigmas are long and feathery. These modifications facilitate pollination by
(a) Insects (b) Water (c) Wind (d) Animals
Answer. (c) Pollination by wind is called anemophily. Anemophilous flowers are small, in conspicuous non-scented without bright colours, nectar and fragrance. Wind pollination also requires that the pollen grains are light and non-sticky which is in large numbers and its stigmas are long and feathery.

Question.13. From among the situations given below, choose the one that prevents both autogamy and geitonogamy.
(a) Monoecious plant bearing unisexual flowers
(b) Dioecious plant bearing only male or female flowers
(c) Monoecious plant with bisexual flowers
(d) Dioecious plant with bisexual flowers
Answer. (b)
• Autogamy (same flower); geitonogamy (different flowers of same plants; xenogamy (different plant’s flower)
• Dioecious plant bearing only male or female flowers prevents both autogamy and geitonogamy.

Question.14. In a fertilised embryo sac, the haploid, diploid and triploid structures are
(a) Synergid, zygote and primary endosperm nucleus
(b) Synergid, antipodal and polar nuclei
(c) Antipodal, synergid and primary endosperm nucleus
(d) Synergid, polar nuclei and zygote
Answer. (a) In a fertilised embryo sac, the haploid, diploid and triploid structures are synergid, zygote and primary endosperm nucleus, respectively.

Question.15. In an embryo sac, the cells that degenerate after fertilisation are
(a) Synergids and primary endosperm cell
(b) Synergids and antipodals
(c) Antipodals and primary endosperm cell
(d) Egg and antipodals
Answer. (b) In an embryo sac, synergids and antipodals degenerate after fertilisation.

Question.16. While planning for an artificial hybridization programme involving dioecious plants, which of the following steps would not be relevant?
(a) Bagging of female flower
(b) Dusting of pollen on stigma
(c) Emasculation
(d) Collection of pollen
Answer. (c)
• Artificial hybridisation is one of the major approaches of crop improvement programme. In such crossing experiments it is important to make sure that only the desired pollen grains are used for pollination and the stigma is protected from contamination (from unwanted pollen). This is achieved by emasculation and bagging techniques.
• Emasculation is relevant in monoecious plants.

Question.17. In the embryos of a typical dicot and a grass, true homologous structures are
(a) Coleorhiza and coleoptile
(b) Coleoptile and scutellum
(c) Cotyledons and scutellum
(d) Hypocotyl and radicle
Answer. (c) Cotyledons of the typical dicot embryo are simple structures generally thick and swollen due to storage of food reserves (as in legumes) and embryo of monocots consists of one large and shield shaped cotyledon known as scutellum situated towards one side (lateral) of the embryonal axis. E.g.: Grass family, Sorghum.

Question.18. The phenomenon observed in some plants wherein parts of the sexual apparatus is used for forming embryos without fertilisation is called
(a) Parthenocarpy (b) Apomixis
(c) Vegetative propagation (d) Sexual reproduction
Answer. (b)
• Apomixis is the special mechanism, to produce seed without fertilisation. It is observed in few flowering plants such as some species of asteraceae and grasses.
• Apomixis is a form of asexual reproduction that mimics sexual reproduction. If a fruit is formed without fertilisation of ovary, it is called a parthenocarpic fruit, e.g., banana and grape.

Question.19. In a flower, if the megaspore mother cell forms megaspores without undergoing meiosis and if one of the megaspores develops into an embryo sac, its nuclei would be
(a) Haploid
(b) Diploid
(c) A few haploid and a few diploid
(d) With varying ploidy
Answer. (b) In a flower, if the megaspore mother cell forms megaspores without undergoing meiosis and if one of the megaspores develops into an embryo sac, its nuclei would be diploid.

Question.20. The phenomenon wherein, the ovary develops into a fruit without fertilisation is called
(a) Parthenocarpy
(b) Apomixis
(c) Asexual reproduction
(d) Sexual reproduction’
Answer. (a) If a fruit is formed without fertilisation of ovaty, it is called a parthenocarpic fruit, e.g., banana and grape.

Very Short Answer Type Questions
Question.1. Name the component cells of the ‘egg apparatus’ in an embryo sac.
Answer. Egg apparatus have three cells—one egg cell and two synergids.

Question.2. Name the part of gynoecium that determines the compatible nature of pollen grain.
Answer. Compatible nature of pollen grain is determined by the stigma of carpel/ pistil.

Question.3. Name the common function that cotyledons and nucellus perform.
Answer. Both cotyledons and nucellus provide nourishment.

Question.4.Complete the following flow chart:

Answer.

Question.5.Indicate the stages where meiosis and mitosis occur (1, 2 or 3) in the flow ?

Answer.

Question.6. In the diagram given below, show the path of a pollen tube from the pollen on the stigma into the embryo sac. Name the components of egg apparatus.

Answer. Components of egg apparatus: one egg cell and two synergids.

Question.7. Name the parts of pistil which develop into fruit and seeds.
Answer. Ovaiy of pistil develops into fruit while ovules develop into seeds.

Question.8.In case of polyembryony, if an embryo develops from the synergid and another from the nucellus which is haploid and which is diploid?
Answer.Synergid embryo is haploid and nucellar embryo is diploid.

Question.9.Can an unfertilised, apomictic embryo sac give rise to a diploid embryo? If yes, then how?
Answer.Yes, if the embryo develops from the cells of nucellus or integument it will be diploid.

Question.10.Which are the three cells found in a pollen grain when it is shed at the three- celled stage?
Answer.One vegetative cell and two male gametes.

Question.11.What is self-incompatibility?
Answer.The device to prevent inbreeding is self-incompatibility or self-sterlity. This is a genetic mechanism and prevents self-pollen (from the same flower or other flowers of the same plant) from fertilising the ovules by inhibiting pollen germination or pollen tube growth in the pistil.

Question.12.Name the type of pollination in self-incompatible plants.
Answer.Xenogamy

Question.13.Draw the diagram of a mature embryo sac and show its 8-nucleate, 7-celled nature. Show the following parts: antipodals, synergids, egg, central cell, polar nuclei.
Answer.

Question.14. Which is the triploid tissue in a fertilised ovule? How is the triploid condition achieved?
Answer. The triploid tissue in the ovule is the endosperm. Its triploid condition is . attained due to the fusion of two polar nuclei and one nucleus of male gamete (also referred to as triple fusion).

Question.15. Are pollination and fertilization necessary in apotnixis? Give reasons.
Answer. No, they are not necessary. Apomixis is actually an alternative to sexual
reproduction although the female sexual apparatus is used in the process. In apomicts, embryos can develop directly from the nucellus or synergid or egg. Therefore, there is no need for either pollination or fertilisation.

Question.16.Identify the type of carpel with the help of diagrams given below:

Answer.

Question.17. How is pollination carried out in water plants?
Answer. Pollination by water is called hydrophily. Some examples of water pollinated plants are Vallisneria and Hydrilla (both are angiospermic hydrophytes) which grow in fresh water and several marine sea-grasses such as Zostera. Not all aquatic plants use water for pollination. In a majority of aquatic plants such as water hyacinth and water lily, the flowers emerge above the level of water and pollinated by insects or wind as in most of the land plants.

Question.18. What is the function of the two male gametes produced by each pollen grain in angiosperms?
Answer. After entering one of the synergids, the pollen tube releases the two male gametes into the cytoplasm of the synergid. One of the male gametes moves towards the egg cell or oosphere and fuses with its nucleus, thus completing the syngamy. This results in the formation of a diploid cell, the zygote. The other male gamete moves towards the two polar nuclei located in the central cell and fuses with them to produce a triploid primary endosperm nucleus (PEN).

Short Answer Type Questions
Question.1. List three strategies that a bisexual chasmogamous flower can evolve to prevent self-pollination (autogamy).
Answer. Flowering plants have developed many devices to discourage self-pollination and to encourage cross-pollination.
• First device: In some species, pollen release and stigma receptivity are not synchronised. Either the pollen is released before the stigma becomes receptive or stigma becomes receptive much before the release of pollen. This condition is called dichogamy in which stigma and anther matures at different time.
• Second device: In some species, the anther and stigma are placed at different positions so that the-pollen cannot come in contact with the stigma of the same flower. This condition is called heterostyly.
• Herkogamy: Non-transfer of pollen from anther to stigma of the same flower due to a mechanical barrier is present between anther and stigma. E.g.: Calotropis (Asclepiadaceae), Aristolochia, Gloriosa superba.
The third device to prevent inbreeding is self-incompatibility or self-sterility. This is a genetic mechanism and prevents self-pollen (from the same flower or other flowers of the same plant) from fertilising the ovules by inhibiting pollen germination or pollen tube growth in the pistil.

Question.2. Given below are the events that are observed in an artificial hybridization programme. Arrange them in the correct sequential order in which they are followed in the hybridisation programme.
(a) Re-bagging, (b) Selection of parents, (c) Bagging, (d) Dusting the pollen on stigma, (e) Emasculation, (f) Collection of pollen from male parent.
Answer. (b) Selection of parents –>(e) Emasculation —> (c) Bagging –> (f) Collection of pollen from male parent –>(d) Dusting the pollen on stigma –>(e) Rebagging.

Question.3. Vivipary automatically limits the number of offspring in a litter. How?
Answer. In viviparous animals (majority of mammals including human beings), the
zygote develops into a young one inside the body of the female organism. After attaining a certain stage of growth, the young ones are delivered out of the body of the female organism. Vivipary automatically limits the number of offspring in a litter because female have limited space for the development of embryo.

Question.4. Does self-incompatibility impose any restrictions on autogamy? Give reasons and suggest the method of pollination in such plants.
Answer. Self-incompatibility imposes restriction to autogamy. The device to prevent inbreeding is self-incompatibility or self-sterility. This is a genetic mechanism and prevents self-pollen (from the same flower or other flowers of the same plant) from fertilising the ovules by inhibiting pollen germination or pollen tube growth in the pistil. Self-incompatiblity is overcome by mixed pollination.

Question.5. In the given diagram, write the names of parts shown with lines.

Answer.

Question.6. What is polyembryony and how can it be commercially exploited?
Answer. As in many Citrus and Mango varieties some of the nucellar cells surrounding the embryo sac start dividing, protrude into the embryo sac and develop into the embryos. In such species each ovule contains many embryos. Occurrence of more than one embryo in a seed is referred as polyembryony.
If hybrids are made into apomicts, there is no segregation of characters in the hybrid progeny. Then the farmers can keep on using the hybrid seeds to raise new crop year after year and he does not have to buy hybrid seeds every year.

Question.7. Are parthenocarpy and apomixis different phenomena? Discuss their benefits.
Answer. Yes, they are different. Parthenocarpy leads to development of seedless fruits.
Apomixis leads to embryo development.

Question.8. Why does the zygote begin to divide only after the division of Primary Endosperm Cell (PEC)?
Answer. The zygote needs nourishment during its development. As the mature, fertilised embryo sac offers very little nourishment to the zygote, the PEC divides and generates the endosperm tissue which nourishes the zygote. Hence, the zygote always divides after division of PEC.

Question.9. The generative cell of a two-celled pollen divides in the pollen tube but not in a three-celled pollen. Give reasons.
Answer. In a 3-celled pollen, as the generative cell has already been divided and formed 2 male gametes, it will not divide again in the pollen tube. But in a 2-celled pollen, as the generative cell has not divided, it divides in the pollen tube.

Question.10. In the figure given below, label the following parts: male gametes, egg cell, polar nuclei, synergid and pollen tube

Answer.

Long Answer Type Questions
Question.1. Starting with the zygote, draw the diagrams of the different stages of embryo development in a dicot.
Answer.

Question.2. What are the possible types of pollinations in chasmogamous flowers? Give reasons.
Answer. A bisexual flower which normally open is called chasmogamous flower.
Kinds of Pollination
1. Autogamy: In this type, pollination is achieved within the same flower. Transfer of pollen graift from the anther to the stigma of the same flower. In a normal flower which opens and exposes the anthers and stigma complete autogamy is rather rare.
Majority of flowering plants produce hermaphrodite flowers and pollen grains are likely to come in contact with the stigma of the same flower. Continued self-pollination result in inbreeding depression. Flowering plants have developed many devices to discourage self-pollination and to encourage cross-pollination.
2. Geitonogamy: Transfer of pollen grains from the anther to the stigma of another flower of the same plant.
3. Xenogamy: Transfer of pollen grains from anther to the stigma of a different plant.

Question.3. With a neat, labelled diagram, describe the parts of a mature angiosperm embryo sac. Mention the role of synergids.
Answer.

Filiform apparatus present at the micropylar part of the synergids guides the entry of pollen tube.

Question.4. Draw the diagram of a microsporangium and label its wall layers. Write briefly on the role of the endothecium.
Answer.

Endothecium performs the function of protection and help in dehiscence of anther to release the pollen.

Question.5. Embryo sacs of some apomictic species appear normal but contain diploid cells. Suggest a suitable explanation for the condition.
Answer. It is true that many apomicts possess normal looking embryo sacs. The only possibility of the embryo sac possessing diploid cells is due to failure of meiotic division at the megaspore mother cell stage. Since, the megaspore mother cell has a diploid nucleus, if it undergoes mitosis instead of meiosis, all the resulting nuclei and cells will be diploid in nature.

The post NCERT Exemplar Problems Class 12 Biology Sexual Reproduction in Flowering Plants appeared first on Learn CBSE.

NCERT Exemplar Problems Class 12 Biology Chapter 3 Human Reproduction

Multiple Choice Questions
Single Correct Answer Type
Question.1. Choose the incorrect statement from the following:
(a) In birds and mammals internal fertilisation takes place.
(b) Colostrum contains antibodies and nutrients.
(c) Polyspermy in mammals is prevented by the chemical changes in the egg surface.
(d) In the human female implantation occurs almost seven days after fertilisation.
Answer. (c) During fertilisation, a sperm comes in contact with the zona pellucida layer of the ovum and induces changes in the membrane that block the entry of additional sperms. Thus, it ensures that only one sperm can fertilise an ovum.

Question.2. Identify the correct statement from the following:
(a) High levels of estrogen triggers the ovulatory’ surge
(b) Oogonial cells start to proliferate and give rise to functional ova in regular cycles from puberty onwards.
(c) Sperms released from seminiferous tubules are highly motile.
(d) Progesterone level is high during the post ovulatory phase of menstrual cycle.
Answer. (d)
• Rapid secretion of LH leading to its maximum level during the mid-cycle called LH surge.
• Oogenesis is initiated during the embryonic development stage when a couple of million gamete mother cells (oogonia) are formed within each fetal ovary (about third month of foetal ovary), no more oogonia are formed and added after birth.
• In the head of epididymis sperms undergo physiological maturation, acquiring increased motility and fertilizing capacity.

Question.3. Spot the odd one out from the following structures with reference to the male reproductive system
(a) Rete testis (b) Epididymis
(c) Vasa efferentia (d) Isthmus
Answer. (d)
• Male reproductive system contain rete testis, vasa efferentia, Epididymis and vas deferens etc.
• Female reproductive system contain isthmus, ampulla, infundibulum and uterus etc.

Question.4. Seminal plasma, the fluid part of semen, is contributed by
i. Seminal vesicle
ii. Prostate
iii. Urethra
iv. Bulbourethral gland
(a) i and ii (b) i, ii and iv
(c) ii, iii and iv (d) i and iv
Answer. (b) Secretion of seminal vesicle (paired), prostate gland (unpaired) and bulbourethral glands or Cowper’s glands (paired) constitute the seminal plasma which is rich in fructose, calcium and certain enzymes.

Question.5. Spermiation is the process of the release of sperms from
(a) Seminiferous tubules (b) Vas deferens
(c) Epididymis (d) Prostate gland
Answer. (a)

Question.6. Mature Graafian follicle is generally present in the ovary of a healthy human female around
(a) 5-8 day of menstrual cycle
(b) 11-17 day of menstrual cycle
(c) 18-23 day of menstrual cycle .
(d) 24-28 day of menstrual cycle
Answer. (b) Mature Graafian follicle is generally present in the ovary of a healthy human female around 11-17 day of menstrual cycle.

Question.7. Acrosomal reaction of the sperm occurs due to
(a) Its contact with zona pellucida of the ova
(b) Reactions within the uterine environment of the female
(c) Reactions within the epididymal environment of the male
(d) Androgens produced in the uterus
Answer. (a) Acrosome reaction in sperm is triggered by release of fertilizin. The secretions of the acrosome help the sperm enter into the cytoplasm of the ovum through the zona pellucida and the plasma membrane.

Question.8. Which one of the following is not a male accessory gland?
(a) Seminal vesicle (b) Ampulla
(c) Prostate (d) Bulbourethral gland
Answer. (b)
• Male accessory gland are seminal vesicle, prostate gland and bulbourethral gland.
• Ampulla is a part of .female reproductive system.

Question.9. The immature male germ cell undergo division to produce sperms by the process of spermatogenesis. Choose the correct one with reference to above.
(a) Spermatogonia have 46 .chromosomes and always undergo meiotic cell division.
(b) Primary spermatocytes divide by mitotic cell division.
(c) Secondary spermatocytes have 23 chromosomes and undergo second meiotic division.
(d) Spermatozoa are transformed into spermatids.
Answer. (c)
• Secondary spermatocytes have 23 chromosomes and undergo second meiotic division.
• Spermatogonia have 46 chromosomes and always undergo mitotic cell division.
• Primary spermatocytes divide by meiotic cell division.
• Spermatids are transformed into spermatozoa.

Question.10. Match between the following representing parts of the sperm and their functions and choose the correct option.

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

Question.11. Which among the following has 23 chromosomes?
(a) Spermatogonia (b) Zygote
(c) Secondary oocyte (d) Oogonia
Answer. (c) Secondary oocyte (n = 23)

Question.12. Match the following and choose the correct options:

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

Question.13. Which of the following hormones is not secreted by human placenta?
(a) hCG (b) Estrogens (c) Progesterone (d) LH
Answer. (d)
• Placenta acts as an endocrine tissue and produces several hormones like human chorionic gonadotropin (hCG), human placental lactogen (hPL), estrogens, progestogens, etc.
• LH is secreted by pituitary gland.

Question.14. The vas deferens receives duct from the seminal vesicle and opens into urethra as
(a) Epididymis (b) Ejaculatory duct
(c) Efferent ductule (d) Ureter
Answer. (b) The vas deferens receives duct from the seminal vesicle and opens into urethra as ejaculatory duct.

Question.15. Urethral meatus refers to the
(a) Urinogenital duct
(b) Opening of vas deferens into urethra
(c) External opening of the urinogenital duct
(d) Muscles surrounding the urinogenial duct
Answer. (c) The urethra originates from the urinary bladder and extends through the penis to its external opening called urethral meatus (external opening of the urinogenital duct).

Question.16. Morula is a developmental stage
(a) Between the zygote and blastocyst
(b) Between the blastocyst and gastrula
(c) After the implantation
(d) Between implantation and parturition
Answer. (a) The embryo with 8 to 16 blastomeres is called a morula. Morula is the solid mass of cells and is mulberry like. Morula is a developmental stage between the zygote and blastocyst.

Question.17. The membranous cover of the ovum at ovulation is
(a) Corona radiata (b) Zonaradiata
(c) Zona pellucida (d) Chorion
Answer. (a) The membranous cover of the ovum at ovulation is corona radiata.

Question.18. Identify the odd one from the following:
(a) Labia minora (b) Fimbriae
(c) Infundibulum (d) Isthmus ,
Answer. (a) Fimbriae, infundibulum‘isthmus and ampulla are the part of fallopian duct while labia minora is female external genitalia.

Very Short Answer Type Questions
Question.1. Given below are the events in human reproduction. Write them in correct
sequential order: Insemination, gametogenesis, fertilisation, parturition, gestation, implantation
Answer. Gametogenesis —> Insemination -> Fertilisation —>Implantation —> Gestation —> Parturition

Question.2. The path of sperm transport is given below. Provide the missing.steps in blank boxes.

Answer.

Question.3. What is the role of cervix in the human female reproductive system?
Answer. Cervix helps in regulating the passage of sperms into the uterus and forms the birth canal to facilitate parturition.

Question.4. Why are menstrual cycles absent during pregnancy?
Answer. The high levels of progesterone and estrogens during pregnancy suppress the gonadotropins which is required for the development of new follicles. Therefore, a new cycle cannot be initiated.

Question.5. Female reproductive organs and associated functions are given below in column A and B. Fill in the blank boxes.

Answer.

Question.6. From where the parturition signals arise—mother or foetus? Mention the main hormone involved in parturition.
Answer. The signals for parturition originate from the fully developed foetus and the placenta which induce mild uterine contractions called foetal ejection reflex. This triggers release of oxytocin from the maternal pituitary.
The main hormone involved in parturition is oxytocin.

Question.7. What is the significance of epididymis in male fertility?
Answer. In the epididymis sperms undergo physiological maturation, acquiring increased motility and fertilizing capacity.

Question.8. Give the names and functions of the hormones involved in the process of spermatogenesis. Write the names of the endocrine glands from where they are released.
Answer. Spermatogenesis starts at the age of puberty due to significant increase in the secretion of gonadotropin releasing hormone (GnRH). This is a hypothalamic hormone.
• The increased levels of GnRH then acts at the anterior pituitary gland and stimulates secretion of two gonadotropins leutinising hormone (LH) and follicle stimulating hormone (FSH).
• LH acts at the Leydig cells and stimulates synthesis and secretion of androgens. Androgens, in turn, stimulate the process of spermatogenesis.
• FSH acts on the Sertoli cells and stimulates secretion of some factors which help in the process of spermiogenesis.

Question.9. The mother germ cells are transformed into a mature follicle through series of steps. Provide the missing steps in the blank boxes.

Answer.

Question.10. During reproduction, the chromosome number (2n) reduces to half (n) in the gametes and again the original number (2n) is restored in the offspring. What are the processes through which these events take place?
Answer. Halving of chromosomal number takes place during gametogenesis and regaining the 2n number occur as a result of fertilisation.

Question.11. What is the difference between a primary oocyte and a secondary oocyte?
Answer.

Question.12. What is the significance of ampullary-isthmic junction in the female reproductive tract?
Answer. In mammals, fertilization takes place in the ampullary-isthmic junction.

Question.13. How does zona pellucida of ovum help in preventing polyspermy?
Answer. Polyspermy means an egg has been fertilized by more than one sperms. The zona pellucida is modified by proteases. The proteases destroy the protein link between the cell membrane and the vitelline membrane, remove any receptors that other sperm have bound to, and help to form the fertilization membrane that prevents polyspermy.

Question.14. Mention the importance of LH surge during menstrual cycle.
Answer. LH surge is essential for the events leading to ovulation.

Question.15. Which type of cell division forms spermatids from the secondary spermatocytes? 
Answer. Second meiotic division

Short Answer Type Questions
Question.1. A human female experiences two major changes, menarche and menopause during her life. Mention the significance of both the events.
Answer. Menarche represents the beginning of menstrual cycle which is an indication of attainment of sexual maturity. Menopause, on the other hand, refers to the cessation of menstruation which in turn means stoppage of gamete production, i.e., it marks the end of reproductive/ fertile life of the female.

Question.2. (a) How many spermatozoa are formed from one secondary spermatocyte? (b) Where does the first cleavage division of zygote take place?
Answer. (a) Two spermatozoa are formed from one secondary spermatocyte.
(b) First cleavage division of zygote take place in fallopian tube or oviduct.

Question.3. Corpus luteum in pregnancy has a long life. However, if fertilisation does not take place, it remains active only for 10-12 days. Explain.
Answer. This is because of a neural signal given by the maternal endometrium to its hypothalamus in presence of a zygote to sustain the gonadotropin (LH) secretion, so as to maintain the corpus luteum as long as the embryo remains there. In the absence of a zygote, therefore, the corpus luteum cannot be maintained longer.

Question.4. What is foetal ejection reflex? Explain how it leads to parturition.
Answer. Vigorous contraction of the uterus at the end of the pregnancy causes expulsion/delivery of the foetus. This process of delivery of the foetus (childbirth) is called parturition. .
• Parturition is induced by a complex neuroendocrine mechanism. The signals for parturition originate from the fully developed foetus and the placenta which induce mild uterine contractions called foetal ejection reflex. This triggers release of oxytocin from the maternal pituitary. Oxytocin acts on the uterine muscle and causes stronger uterine contractions, which in turn stimulates further secretion of oxytocin.
• The stimulatory reflex between the uterine contraction and oxytocin secretion continues resulting in stronger and stronger contractions. This leads to expulsion of the baby out of the uterus through the birth canal- parturition. Soon after the infant is delivered, the placenta is also expelled out of the uterus.

Question.5. Except endocrine function, what are the other functions of placenta?
Answer. Placenta facilitates the supply of oxygen and nutrients to the embryo. It also removes C02 and excretory wastes produced by embryo.

Question.6. Why doctors recommend breast feeding during initial period of infant growth?
Answer. The mammary glands of the female undergo differentiation during pregnancy and starts producing milk towards the end of pregnancy by the process called lactation.
The milk produced during the initial few days of lactation is called colostrum which contains several antibodies absolutely essential to develop resistance for the new bom babies. Breast-feeding during the initial period of infant growth is recommended by doctors for bringing up a healthy baby.

Question.7. What are the events that take place in the ovary and uterus during follicular phase of the menstrual cycle?
Answer. 1. The primary follicle .grows and becomes fully mature graafian follicles.
2. Secretion of estrogen hormone.
3. Endometrium of uterus regenerates through proliferation.

Question.8. Given below is a flow chart showing ovarian changes during menstrual cycle. Fill in the spaces giving the name of the hormones responsible for the events shown.

Answer.

Question.9. Give a schematic labelled diagram to represent oogenesis (without descriptions).
Answer.

Question.10. What are the changes in the oogonia during the transition of a primary follicle to Graafian follicle?
Answer. Oogenesis is initiated during the embryonic development stage when a couple of million gamete mother cells (oogonia) are formed within each fetal ovary (about third month of foetal ovary), no more oogonia are formed and added after birth.
• These cells start division and enter into prophase-I of the meiotic division and get temporarily arrested at that stage, called primary oocytes. Each primary oocyte then gets surrounded by a layer of granulosa cells and then called the primary follicle. 1° follicle contains 1° oocyte. A large number of these follicles degenerate during the phase from birth to puberty.
• The primary follicles get surrounded by more layers of granulosa cells and a new theca and called secondary follicles. The theca layer is organised into an inner theca interna and an outer theca externa. Theca interna secretes estrogen hormone. The secondary follicle soon transforms into a tertiary follicle which is characterised by a fluid filled cavity called antrum.
• It is important to draw your attention that it is at this stage that the primary oocyte within the tertiary follicle grows in size and completes its first meiotic division. It is an unequal division resulting in the formation of a large haploid secondary oocyte and a tiny first polar body. The tertiary follicle further changes into the mature follicle or Graafian follicle.

Long Answer Type Questions
Question.1. What role does pituitary gonadotropins play during follicular and ovulatory phases of menstrual cycle? Explain the shifts in steroidal secretions.
Answer. Menstrual cycle is regulated by hypothalamus through the pituitary gland. At the end of menstrual phase, the pituitary FSH gradually increases resulting in follicular development within the ovaries. As the follicles mature, Estrogen secretion increases resulting in a surge in (FSHe and LH).The surge of LH responsible for ovulation. LH also induces luteinisation. This leading to the formation of corpus luteum. Corpus luteum secretes progesterone and some estrogen which help in maintaining the uterine endometrium for implantation.

Question.2. Meiotic division during oogenesis is different from that in spermatogenesis. Explain how and why?
Answer. Unequal cytoplasmic division of the oocyte is to ensure the retention of bulk of cytoplasm in one cell, instead of sharing with two. It has to provide nourishment for the developing embryo during early stages, so it is essential to retain as much cytoplasmic materials it could in a single daughter cell.

Question.3. The zygote passes through several developmental stages till implantation, Describe each stage briefly with suitable diagrams.
Answer. The mitotic division starts as the zygote moves through the isthmus of the oviduct called cleavage towards the uterus and forms 2, 4, 8, 16 daughter cells called blastomeres. The embryo with 8 to 16 blastomeres is called a morula. The morula continues to divide and transforms into blastocysts, it moves further into the uterus.

• The blastomeres in the blastocyst are arranged into an outer layer called trophoblast and an inner group of cells attached to trophoblast called the inner cell mass. The trophoblast layer then gets attached to the endometrium and the inner cell mass gets differentiated as the embryo.
• After attachment, the uterine cells divide rapidly and covers the blastocyst. As a result the blastocyst becomes embedded in the endometrium of the uterus. This is called implantation and it leads to pregnancy.

Question.4. Draw a neat diagram of the female reproductive system and label the parts associated with the following: (a) production of gametes, (b) site of fertilization (c) site of implantation and (d) birth canal.
Answer.

Question.5. With a suitable diagram, describe the organisation of mammary gland.
Answer. A functional mammary gland is characteristic of all female mammals. The mammary glands are paired structures (breasts) that contain glandular tissue and variable amount of fat. The glandular tissue of each breast is divided into 15—20 mammary lobes containing clusters of cells called alveoli. The cells of alveoli secrete milk, which is stored in the cavities (lumens) of alveoli.

The alveoli open into mammary tubules. The tubules of each lobe join to form a mammary duct. Several mammary ducts join to form a wider mammary ampulla which is connected to lactiferous ducts through which milk is sucked out.

The post NCERT Exemplar Problems Class 12 Biology Human Reproduction appeared first on Learn CBSE.

NCERT Exemplar Problems Class 12 Biology Chapter 4 Reproductive Health

Multiple Choice Questions
Single Correct Answer Type
1. The method of directly injecting a sperm into ovum is assisted by reproductive technology called
(a) GIFT (b) ZIFT (c) ICSI (d) ET
Answer. (c) Intra cytoplasmic sperm injection (ICSI) is a method in which embryo is formed in the laboratory in which a sperm is directly injected into the ovum.

2. Increased IMR and decreased MMR in a population will
(a) Cause rapid increase in growth rate
(b) Result in decline in grcfwth rate
(c) Not cause significant change in growth rate
(d) Result in an explosive population/exp
Answer. (b) Increased IMR and decreased MMR in a population will result in decline in growth rate.

3. Intensely lactating mothers do not generally conceive due to the
(a) Suppression of gonadotropins
(b) Hypersecretion of gonadotropins
(c) Suppression of gametic transport
(d) Suppression of fertilisation
Answer. will not lead to conception.

4. Sterilisation techniques are generally fool proof methods of contraception with least side effects. Yet, this is the last option for the couples because
i. It is almost irreversible.
ii. Of the misconception that it will- reduce sexual urge/drive
iii. It is a surgical procedure
iv. Of lack of sufficient facilities in many parts of the country
Choose the correct option:
(a) i and iii (b) ii and iii
(c) ii and iv (d) i, ii, iii and iv
Answer. (d) Sterilisation techniques are generally fool proof methods of contraception with least side effects. Yet, this is the last option for the couples because of (i),
(ii) , (iii) and (iv).

5. A national level approach to build up a reproductively healthy society was taken up in our country in (a) 1950s (b) 1960s (c) 1980s (d) 1990s
Answer. (a) A national level approach to build up a reproductively healthy society was taken up in our country in 1950s.

6. Emergency contraceptives are effective if used within
(a) 72 hrs of coitus (b) 72 hrs of ovulation
(c) 72 hrs of menstruation (d) 72 hrs of implantation
Answer. (a) Combination or IUDs within 72 hours of coitus have been found to be very effective as emergency contraceptives as they could be used to avoid possible pregnancy due to rape or casual unprotected intercourse.

7. Choose the right one among the statements given below:
(a) IUDs are generally inserted by the user herself
(b) IUDs increase phagocytosis reaction in the uterus
(c) IUDs suppress gametogenesis
(d) IUDs once inserted need not be replaced
Answer. (b)

• Hormone- releasing IUDs: Progestasert, LNG-20
• IUDs increases phagocytes of sperm within the uterus

8. Following statements are given regarding MTP. Choose the correct options given below:
i. MTPs are generally advised during first trimester.
ii. MTPs are used as a contraceptive method.
iii. MTPs are always surgical.
iv. MTPs require the assistance of qualified medical personnel.
(a) i and iii (b) ii and iii (c) i and iv (d) i and ii
Answer. (c)

• MTPs are considered relatively safe during the first trimester (up to 12 weeks of pregnancy).
• 2nd trimester abortions are much more risky.
•  MTPs require the assistance of qualified medical personnel.

9. From the sexually transmitted diseases mentioned below, identify the one which does not specifically affect the sex organs.
(a) Syphilis (b) AIDS (c) Gonorrhea (d) Genital warts
Answer. (b) Diseases or infections whieh are transmitted through sexual intercourse called sexually transmitted diseases (STD) or VD (Venereal diseases) or RTI (Reproductive tract infections).
Examples of STDs:

1. HIV (AIDS)
2. Hepatitis-B
3. Genital herpes
4. Chlamydiasis ,
5. Gonorrhoea
6. Genital warts
7.  Syphilis
8. Trichomoniasis

10. Condoms are one of the most popular contraceptives because of the following reasons:
(a) These are effective barriers for insemination
(b) They do not interfere with coital act
(c) These help in reducing the risk of STDs
(d) All of the above
Answer. (d) Condoms are barriers made of thin rubber/latex sheath. Condoms are used to cover the penis in male or vagina and cervix in the female, just before coitus so that the ejaculated semen would not enter into the female reproductive tract. This can prevent conception. Use of condoms has increased in recent years due to its additional benefit of protecting the user from contracting STDs and AIDS. Both male and female condoms are disposable and gives privacy to the user. ‘Nirodh’ is a popular brand of condom for the male. Condom is the most widely used contraceptive by males in India as it is cheap and easily available. It is sirrtple and effective and has no side effects.

11. Choose the correct statement regarding the ZIFT procedure:
(a) Ova collected from a female donor are transferred to the fallopian tube to facilitate zygote formation.
(b) Zygote is collected from a female donor and transferred to the fallopian tube
(c) Zygote is collected from a female donor and transferred to the uterus
(d) Ova collected from a female donor and transferred to the uterus
Answer. (b)

•  ZIFT (Zygote Intra Fallopian Transfer): The zygote or early embryos (up to 8 blastomeres) could then be transferred into the fallopian tube.
• Transfer of an ovum collected from a donor into the fallopian tube (GIFT: Gamete Intra Fallopian Transfer) of another female who cannot produce one but can provide suitable environment for fertilization and further development.

12. The correct surgical procedure as a contraceptive method is
(a) Ovariectomy (b) Hysterectomy
(c) Vasectomy (d) Castration
Answer. (c) Vasectomy and tubectomy are surgical procedure as a contraceptive method.

13. Diaphragms are contraceptive devices used by the females. Choose the correct option from the statements given below:
i. They are introduced into the uterus
ii. They are placed to cover the cervical region
iii. They act as physical barriers for sperm entry
iv. They act as spermicidal agents
(a) i and ii (b) i and iii (c) ii and iii (d) iii and iv
Answer. (c) Diaphragms, Cervical caps and Vaults: These are also barrier made of rubber that are inserted into the female reproductive tract to cover the cervix during coitus. They prevent conception by blocking the entry of sperms through the cervix. They are reusable. Spermicidal creams, jellies and foams are usually used along with these barrier to increase their contraceptive efficiency.

Very Short Answer Type Questions
1. Reproductive health refers only to healthy reproductive functions. Comment.
Answer. Reproductive health refers to a total well-being in all aspects of reproduction,
i.e., physical, behavioural, psychological, social and physiological.

2. Comment on the Reproductive and Child Health Care programme of the government to improve the reproductive health of the people.
Answer. Creating public awareness regarding reproduction related aspects and providing facilities to build up a healthy society with added emphasis on the health of mother and child are the basic aims of the RCH programmes.

3. The present population growth rate “in India is alarming. Suggest ways to check it.
Answer. The most important step to overcome this problem is to motivate smaller families by using various contraceptive methods.

• Government of India advertises in the media as well as in posters, showing a happy couple with two children with a slogan “Hum Do Hamare Do” (We two, our two).
• Many couples, mostly the young, urban working ones have even adopted a one child norm.
• Statutory raising of marriageable age of the female to 18 years and that of males to 21 years and incentives given to couples with small families are two of the other measure taken to tackle this problem.

4. STDs can be considered as self-invitedrdiseases. Comment.
Answer. STDs can be considered as self-invited diseases caused due to the conscious behaviour of person. STDs occur in those individuals that have unprotected sex with multiple or unknown partners.

5. Suggest the reproduction-related aspects in which counselling should be provided at the school level.
Answer.

• Introduction of sex education in schools should also be encouraged to provide right information to the young so as to discourage children from believing in myths and having misconceptions about sex-related aspects.
• Proper information about reproductive organs, adolescence and related changes, safe and hygienic sexual practices, sexually transmitted diseases (STD), AIDS, etc., would help people, especially those in the adolescent age group to lead a reproductively healthy life.

6. Mention the primary-aim of the “Assisted Reproductive Technology” (ART) programme.
Answer. The primary aim of the “Assisted Reproductive Technology” (ART) programme is the couples could be assisted to have children through certain special techniques.

7. What is the significance of progesterone-estrogen combination as a contraceptive measure?
Answer. Oral administration of small doses of progesterone-estrogen combination is a contraceptive measure which inhibit ovulation and implantation as well as alter the quality of cervical mucus to prevent/retard the entry of sperms.

8. Strict conditions are to be followed in medical termination of pregnancy (MTP) procedures. Mention two reasons.
Answer. Government of India legalised MTP in 1971 with some strict conditions to avoid its misuse.
Such restrictions are all the more important to check indiscriminate and illegal female foeticides which reported to be high in India.

9. Males in whom testes fail to descend to the scrotum are generally infertile. Why?
Answer. If the testes fail to descend to the scrotum, gametogenesis could be inhibited, the process of spermatogenesis require a marginally lesser ambient temperature than that in the abdominal cavity.

10. Mention two advantages of lactational amenorrhea as a contraceptive method.
Answer. (i) There is no ovulation and therefore the menstrual cycle do not take place.
(ii) As no medicines or devices are used in these methods, side effects are almost nil. .

Short Answer Type Questions
1. Suggest some important steps that you would recommend to be taken to improve the reproductive health standards in India.
Answer.

• Creating awareness among people about various reproduction related aspects and providing facilities and support for building up a reproductively healthy society are the major tasks under these programmes. With the help of audio-visual and the print-media, governmental and non-governmental agencies have taken various steps to create awareness among the people about reproduction-related aspects.
• Parents, other close relatives, teachers and friends, also have a major role in the dissemination of the above information. Educating people, especially fertile couples and those in marriageable age group, about available birth control options, care of pregnant mothers, post-natal care of the mother and child, importance of breast feeding, equal opportunities for the male and the female child, etc., would address the importance of bringing up socially conscious healthy families of desired size.

2. The procedure of GIFT involves the transfer of female gamete to the fallopian tube. Can gametes be transferred to the uterus to achieve the same result? Explain.
Answer. The uterine environment is not congenial for the survival of the gamete. If, directly transferred to the uterus they will undergo degeneration or could be phagocytosed and hence viable zygote would not be formed.

3. Copper ions-releasing IUDs are more efficient than non-medicated methods. Why?
Answer. Unlike non-medicated IUEs, the copper releasing IUDs releases Cu ions that suppress sperm motility and fertilising capacity of sperms.

4. What are the probable factors that contributed to population explosion in India?
Answer. Probable reasons for population explosion are:

1.  A rapid decline in death rate.
2. Decline in MMR (Maternal mortality rate).
3. Decline in IMR (Infant mortality rate).
4. An increase in number of people in reproducible age.

5. Briefly explain IVF and ET. What are the conditions in which these methods are advised? 
Answer. IVF and ET refer to In Vitro Fertilisation and Embryo Transfer. Gametes from the male and female are collected hygienically and induced to fuse in the laboratory set up under simulated conditions. The zygote formed is collected and is introduced into the uterine region of a host dr surrogate mother at an appropriate time (secretory phase). Early embryos (up to 8 cell) are generally transferred to the fallopian tube whereas embryos with more than 8 cells are transferred to the uterus.

6. What are the advantages of natural methods of contraception over artificial methods?
Ans. Advantages of natural methods of contraception over artificial methods:

1. As no medicines or devices are used in these methods, side effects are almost nil.
2.  There is no surgical intervention, so natural method is reversible in nature.

7. What are the conditions in which medical termination of pregnancy is advised?
Answer.

• MTP is used to get rid of unwanted pregnancies either due to casual unprotected intercourse or failure of the contraceptive used during coitus or rapes.
• MTPs are also essential in certain cases where continuation of the pregnancy could be harmful or even fatal either to the mother or to the foetus of both.

8. Comment on the essential features required for an ideal contraceptive.
Answer. An ideal contraceptive should be:

1. User-friendly
2. Easily available
3. Effective
4. Reversible
5. No or least side-effects.
6. Should not interfere with the sexual drive, desire or sexual act of the user

9. All reproductive tract infections RTJs are STDs, but all STDs are not RTIs. Justify with example.
Answer. The common STDs are gonorrhea, syphilis, genital herpes, chlamydiasis, hepatitis-B, AIDs etc. Hepatitis-B, and AIDs are not infections of the reproductive organs though their mode of transmission could be through sexual contact also. All other diseases are transmitted through sexual contact and are also infections of the reproductive tract.

Long Answer Type Questions
1. What are the Assisted Reproductive Techniques practised to help infertile couples? Describe any three techniques.
Answer. Inability to conceive or produce children even after 2 years of unprotected sexual cohabitation is called infertility. In India, often the female is blamed ‘ for the couple being childless, but more often than not the problem lies in male partner.
Specialised health care units (infertility, clinics) could help in diagnosis and corrective treatment of some of these disorders and enable these couples to have children. However, where such corrections are not possible, the couples could be assisted to have children through certain special techniques commonly known as ART (Assisted Reproductive Technologies).
1. Test Tube Baby Programme
In vitro fertilisation (IVF) followed by embryo transfer (ET) is a method to treat infertility and commonly known as the ‘Test tube baby’ programme. IVF-Fertilisation outside the body in almost similar conditions as that in the body. In this method ova from the wife/donor (female) and sperms from the husband/donor (male) are collected and are induced to form zygote under simulated conditions in the laboratory.

• ZIFT (Zygote Intra Fallopian Transfer): The zygote or early embryos (up to 8 blastomeres) could then be transferred into the fallopian tube.
• IUT (Intra Uterine Transfer): Embryos with more than 8 blastomeres could be transferred into the uterus, to complete its further development.
• Embryos formed by in vivo fertilization (fusion of gametes within the female) also could be used for such transfer to assist those females who cannot conceive.

2. GIFT
Transfer of an ovum collected from a donor into the fallopian tube (GIFT; Gamete Intra Fallopian Transfer) of another female who cannot produce one but can provide suitable environment for fertilization and further development.
3. AI Technique

• Infertility cases either due to inability of the male partner to inseminate the female or due to very low sperm counts in the ejaculates, could be corrected by artificial insemination (AI) technique.
• In this technique, the semen collected either from the husband or a healthy donor is artificially introduced, either into the vagina or into the uterus (IUI-Intra Uterine Insemination) of the female.

2. Discuss the mode of action and advantages/disadvantages of hormonal contraceptives.
Answer.

• Oral administration of small doses of either progestagens or progestogen- estrogen combinations is another contraceptive method used by the females.
• Most important component of oral contraceptive is progesterone. They are used in the form of tablets and hence are popularly called the pills. Pills have to be taken daily for a period of 21 days starting preferably within the first 5 days of menstrual cycle. After a gap of 7 days (during which menstruation occurs) it has to be repeated in the same pattern.
•  Oral contraceptive pills inhibit ovulation and implantation as well as alter (change) the quality of cervical mucus to prevent or retard entry of
  sperms. Progesterone present in OCP- is meant for checking ovulation. Pills are very effective with lesser side effects and are well accepted by the females. Combination or IUDs within 72 hours of coitus have been found to be very effective as emergency contraceptives as they could be used to avoid possible pregnancy due to rape or casual unprotected intercourse.

3. STDs are a threat to reproductive health. Describe any two such diseases and suggest preventive measures.
Answer. Diseases or infections which are transmitted through sexual intercourse called sexually transmitted diseases (STD) or VD (Venereal diseases) or RTI
(Reproductive tract infections).
Examples of STDs:
(i) HIV (AIDS)
(ii) Hepatitis-B
(iii) Genital herpes
(iv) Chlamydiasis
(v) Gonorrhoea
(vi) Genital warts
(vii) Syphilis
(viii) Trichomoniasis
AIDS
• AIDS is caused by HIV virus (Human Immuno deficiency virus or Human T-cell leukemia virus). There is always a time-lag between the infection and appearance of AIDS symptoms (incubation period). Incubation period may vary from a few month to many years (usually 5-10 years).
• ARC (AIDS Related Complex) is a mild or initial form of AIDS which develop after a few mon|h of infection. In AIDS patient, a reduction of 10% weight indicates ARC. After infection, HIV enters in macrophages where RNA of virus replicates to form viral DNA with the help of enzyme reverse transcriptase. Viral DNA incorporated into host cell DNA and directs the infected cell to produce new virus particles. The macrophages continue to produce virus therefore macrophages are called HIV Factory.
• Simultaneously HIV enters into helper T-lymphocytes (TH) or T4 replicates and produce progeny viruses. The progeny viruses released in the blood attack other helper T-lymphocytes. This is repeated leading to a progressive decrease in the number of helper T-lymphocytes in the body of infected person. During this period person suffers from: (i) Bouts of fever, (ii) Diarrhoea, (iii) Weight loss
• Due to the decrease in TH cells, person start suffering from infections that could have been otherwise overcome such as those due to bacteria especially Mycobacterium, virus, fungi and even parasites like Toxoplasma. The patient becomes so immuno-deficient that he/she is unable to protect himself/herself against these infections. Diagnostic test for AIDS is ELISA (Enzyme Linked Immuno Sorbent Assay).Werstem blot is used as confirmatory or supplemental test for AIDS. Routine test of AIDS is PCR.
Hepatitis-B
Hepatitis-B viras is horizontally transmitted by blood transfusions, contaminated needles, body fluids like semen, saliva, sweat, tear and breast milk. Hepatitis-B is also transmitted vertically from infected mother to foetus through placenta. For prevention and control hepatitis-B vaccine is now available. Hepatitis-B vaccine was developed by Blumberg, for which he was awarded Nobel Prize in 1976.

4. Do you justify the statutory ban on amniocentesis in our country? Give reasons.
Answer.

• Intentional or voluntary termination of pregnancy before full term is called MTP or induced abortion. Nearly 45 to 50 millions MTPs are performed in a year all over the world which accounts to l/5th (20%) of the total number of conceived pregnancies in a year. MTP has a significant role in decreasing the population though it is not meant for this purpose.
•  Government of India legalized MTP in 1971 with some strict conditions
  to avoid its misuse. Such restrictions are more important to check indiscriminate and illegal female foeticides which reported to be high in India. ‘ .
• MTP is used to get rid of unwanted pregnancies either due to casual unprotected intercourse or failure of the contraceptive used during coitus or rapes. MTPs are also essential in certain cases where continuation of the pregnancy could be harmful or even fatal either to the mother or to the foetus or both. MTPs are considered relatively safe during the first trimester (up to 12 weeks of pregnancy). 2nd trimester abortions are much more risky. Another dangerous trend is the misuse of amniocentesis to determine the sex of unborn child.
• Amniocentesis is a foetal sex determination test based on the chromosomal pattern in the amniotic fluid surrounding the developing embryo. Amniocentesis is employed for determining hereditary abnormality in embryo. Statutory ban on amniocentesis for sex-determination to legally Check: (i) Increasing female foeticides, (ii) Massive child immunisation.

5. Enumerate and describe any five reasons for introducing sex education to school-going children.
Answer. Proper information about reproductive organs-physiology and its functioning; discourage myths and misconceptions about sex-related aspects; knowledge about safe and hygienic sexual practices; adolescence and related changes, prevention of STDs, AlDs etc.

The post NCERT Exemplar Problems Class 12 Biology Reproductive Health appeared first on Learn CBSE.

NCERT Exemplar Problems Class 12 Biology Chapter 5 Principles of Inheritance and Variation

Multiple Choice Questions
Single Correct Answer Type
1. All genes located on the same chromosome
(a) Form different groups depending upon their relative distance
(b) Form one linkage group
(c) Will not form any linkage groups
(d) Form interactive groups that affect the phenotype.
Answer. (b) Linkage groups: A linkage group is a group of linked gene (on the same chromosome) and corresponds to the genome of organism, like human has 23 linkage groups, pea and Ngurospora has 7 linkage groups, Drosophila has 4 linkage groups.

2. Conditions of a karyotype 2n ± 1 and 2n ± 2 are called
(a) Aneuploidy (b) Polyploidy
(c) Allopolyploidy (d) Monosomy
Answer. (a) Failure of segregation of chromatids during cell division cycle results in the gain or loss of a chromosome(s), called aneuploidy. Aneuploidy arises due to non-disjunction of homologous chromosome.
Aneuploidy is of four types:
1. Monosomy = 2n – 1
2. Nullisomy = 2n – 2
3. Trisomy = 2n + 1
4. Tetrasomy = 2n + 2

3. Distance between the genes and percentage of recombination shows
(a) a direct relationship (b) an inverse relationship
(c) a parallel relationship (d) no relationship
Answer. (a) Distance between the genes and percentage of recombination shows
direct relationship.

4. If a genetic disease is transferred from a phenotypically normal but carrier female to only some of the male progeny, the disease is
(a) Autosomal dominant (b) Autosomal recessive
(c) Sex-linked dominant (d) Sex-linked recessive
Answer. (d) Sex-linked recessive disease: A genetic disease is transferred from a phenotypically normal but carrier female to only some of the male progeny.
E.g.: 1. Haemophilia/Bleeder’s disease
2. Colour-blindness
3. Congenital night blindness
4. DMD (Duchenne’s Muscular Dystrophy)
5. G-6-P dehydrogenase deficiency

5. In sickle cell anaemia glutamic acid is replaced by valine. Which one of the following are triplets codes for valine? (a) GGG (b) AAG (c) GAA (d) GUG
Answer. (d) The substitution of amino acid in the globin protein results due to the single base substitution at the sixth codon of the betaglobin gene from GAG (Glutamic acid) to GUG (Valine).

6. Person having genotype IA IB would show the blood group as AB. This is because of
(a) Pleiotropy (c) Segregation
Answer. (b)

7. ZZ/ZW type of sex determination is seen in
(a) Platypus (b) Snails
(c) Cockroach (d) Peacock
Answer. (d)

8. A cross between two tall plants resulted in offspring having few dwarf plants. What would be the genotypes of both the parents?
(a) TT and Tt (b) Tt and Tt ’
(c) TT and TT (d) Tt and tt
Answer. (b)

9. In a dihybrid cross, if you get 9 : 3 : 3 : 1 ratio, it denotes that
(a) The alleles of two genes are interacting with each other
(b) It is a multigenic inheritance
(c) It is a case of multiple allelism
(d) The alleles of two’genes are segregating independently
Answer. (d) Based upon such observations on dihybrid crosses (crosses between plants differing in two traits) Mendel proposed a second set of generalisations that we call Mendel’s Law of Independent Assortment. The law states that “When two pairs of traits are combined in a hybrid, segregation of one pair of characters is independent of the other pair of characters”.

10. Which of the following will not result in variations among siblings?
(a) Independent assortment of genes
(b) Crossing over
(c) Linkage
(d) Mutation
Answer. (c)
• Linkage is not result in variations among siblings.
• A sibling is one of two or more individuals having one or both parents in common. The emotional bond between siblings is often complicated and is influenced by factors such as parental treatment, birth order, personality, and personal experiences outside the family. Identical twins share 100% of their DNA. Full siblings are first-degree relatives and, on average, share 50% of their genes out of those that vary among humans. Half-siblings are second-degree relatives and have, on average, a 25% overlap in their human genetic variation.

11. Mendel’s Law of independent assortment holds good for genes situated on the
(a) Non-homologous chromosomes (b) Homologous chromosomes
(c) Extra nuclear genetic element (d) Same chromosome
Answer. (b) Mendel’s Law of independent assortment holds good for genes situated on the homologous chromosomes.

12. Occasionally, a single gene, may express more than one effect. The phenomenon is called
(a) Multiple allelism (b) Mosaicism
(c) Pleiotropy (d) Polygeny
Answer. (c) Occasionally, a single gene may express more than one effect. The phenomenon is called pleiotropy.

13. In a certain taxon of insects, some have 17 chromosomes and the others have 18
chromosomes. The 17 and 18 chromosome-bearing organisms are
(a) Males and females, respectively (b) Females and males, respectively (c) All males (d) All females
Answer. (a) The 17 and 18 chromosome-bearing organisms are males and females respectively because it comes under XO-type of sex determination.
Males have only 1 chromosome besides the autosomes whereas females have a pair of X chromosomes besides the autosomes.

14. The inheritance pattern of a gene over generations among humans is studied by the pedigree analysis. Character studied in the pedigree analysis is I equivalent to
(a) Quantitative trait . (b) Mendelian trait
(c) Polygenic trait (d) Maternal trait
Answer. (b) Character studied in the pedigree analysis is equivalent to Mendelian trait.

15. It is said that Mendel proposed that the factor controlling any character is discrete and independent. His proposition was based on the
(a) Results of F₃ generation of a cross
(b) Observations that the offspring of a cross made between the plants having two contrasting characters shows only one character without any blending.
(c) Self pollination of F₁  off springs
(d) Cross pollination of F₁  generations with recessive parental
Answer. (b) Mendel proposed that the factor Controlling any character is discrete and independent. His proposition was based on the observations that the offspring of a cross made between the plants having two contrasting characters shows only one character without any blending.

16. Two genes ‘A’ and ‘B’ are linked. In a dihybrid cross involving these two genes, the F1 heterozygote is crossed with homozygous recessive parental type (aa bb). What would be the ratio of offspring in the next generation?
(a) 1 : 1 : 1: 1 (b) 9 : 3 : 3 : 1 (c) 3:1 (d) 1:1
Answer. (d) Two genes ‘A’ and ‘B’ are linked. In a dihybrid cross involving these two genes, the Fj heterozygote is crossed with homozygous recessive parental type (aa bb). The ratio of offspring in the next generation will be 1 :1.

17. In the F₂ generation of a Mendelian dihybrid cross, the number of phenotypes and genotypes are
(a) Phenotypes—4; genotypes—16 (b) Phenotypes—9; genotypes—4 (c) Phenotypes—4; genotypes—8 (d) Phenotypes—4; genotypes—9
Answer. (d) In the F₂ generation of a Mendelian dihybrid cross the number of phenotypes and genotypes are 4, 9 respectively.

18. Mother and father of a person with ‘O’ blood group have ‘A’ and ‘B’ blood group respectively. What would be the genotype of both mother and father?
(a) Mother is homozygous for ‘A’ blood group and father is heterozygous for ‘B’
(b) Mother is heterozygous for ‘A’ blood group and father is homozygous for ‘B’
(c) Both mother and father are heterozygous for ‘A’ and ‘B’ blood group, respectively
(d) Both mother and father are homozygous for ‘A’ and ‘B’ blood group, respectively.
Answer. (c) See Answer 2 (Short Answer Type Questions)
Genotype of mother — I^A i
Genotype of father — I^B i
Both mother and father are heterozygous for ‘A’ and ‘B’ blood group, respectively.

Very Short Answer Type Questions
1. What is the cross between the progeny of F₁  and the homozygous recessive parent called? How is it useful?
Answer. When a progeny of F₁  is crossed with the homozygous recessive parent, it is called test cross. Such a cross is useful to determine the genotype of an unknown, i.e., whether it is heterozygous, or homozygous dominant for the trait.

2. Do you think Mendel’s laws of inheritance would have been different if the
characters that he choose were located on the same chromosome?
Answer. If Mendel choose characters that were located on the same chromosome then Mendel would not find the law of independent assortment.

3. Enlist the steps of controlled cross pollination. Would emasculation be needed in a cucurbit plant? Give reasons for your answer.
Answer. Steps of controlled cross pollination: Selection of parents —> Emasculation —> Bagging —>Collection of pollen from male parent —>Dusting the pollen on stigma —>Re-bagging
Emasculation could not be needed in a cucurbit plant because it has unisexual flowers.

4. A person has to perform crosses for the purpose of studying inheritance of a few traits/characters. What should be the criteria for selecting the organisms?
Answer. Criteria for selecting the organism are:
(i) Shorter life span
(ii) To produce large number of progeny-
(iii) Clear differentiation of sex/traits
(iv) Many type of hereditary variations

5. The pedigree chart given below shows a particular trait which is absent in parents but present in the next generation irrespective of sexes. Draw your conclusion on the basis of the pedigree.

Answer. The trait is auto some linked and recessive in nature. Both the parents are carrier (i.e., heterozygous) hence among offspring few show the trait irrespective of sex. The other off springs are either normal or carrier.

6. In order to obtain the. F₁  generation Mendel pollinated a pure-breeding tall plant with a pure breeding dwarf plant. But for getting the F₂  generation, he simply self-pollinated the tall F₁  plants. Why?
Answer. Genotype of 50% of the offspring would resemble one parent and the rest the other parent. All the F₁ , off springs of the cross are heterozygous so allowing self-pollination is sufficient to raise F₂  offspring. Also Mendel intended to understand the inheritance of the selected trait over generations.

7. “Genes contain the information that is required to express a particular trait.” Explain.
Answer. The genes present in an organism show a particular trait by way of forming certain product. This is facilitated by the process of transcription and translation (according to central dogma of genetics).

8. How are alleles of particular gene differ from each other? Explain its significance.
Answer. Alleles of a particular gene differ from each other on the basis of certain changes (i.e., mutations) in the genetic material (segment of DNA or RNA). Different alleles of a gene increases the variability or variation among the organisms.

9. In a monohybrid cross of plants with red and white flowered.plants, Mendel got only red flowered plants. On self-pollinating these F₁  plants got both red and white flowered plants in 3:1 ratio. Explain the basis of using RR and rr symbols to represent the geno type of plants of parental generation.
Answer. On crossing red and white flower only red colour flower appeared in the F₁  generation. But the white colour flower again appear in the F₂  generation
which is raised out of the F₁  individuals only. Mendel reasoned that there is a factor of each and every character. Accordingly, there has to be one factor (R) for red flower and other one factor (r) for white flower. In case, an organism possess only one copy of the gene then the possibility of reappearance of white flower in the F₂  generation of the given cross is not there. Also the ratio (3 : 1 of red and white) indicates that each organism must possess two copies of a particular gene.

10. For the expression of traits genes provide only the potentiality and the environment provides the opportunity. Comment on the veracity of the statement.
Answer.

11. A, B, D are three independently assorting genes with their recessive alleles a, b, d, respectively. A cross was made between individuals of Aa bb DD genotype with aa bb dd. Find out the type of genotypes of the off spring produced.
Answer. The given cross is Aa bb dd X aa bb dd. Accordingly the type of offspring produced would .

12. In our society a woman is o|ten blamed for not bearing male child. Do you think it is right? Justify.
Answer. 50 per cent of sperms carry the X chromosome while the other 50 percent carry the Y. After fusion of the male and female gametes the zygote would carry either XX or YY depending on whether the sperm carrying X or Y fertilised the ovum. Sex of the baby is determined by the father and not by the mother.
It is unfortunate that in our society women are blamed for producing female children and have been ostracised and ill-treated because of this false notion.

13. Discuss the genetic basis of wrinkled phenotype of pea seed.
Answer. Wrinkled phenotype of pea seed is due to small grain size produced by double recessive allele (bb).

14. Even if a character shows multiple allelism, an individual will only have two alleles for that character. Why?
Answer. A diploid organism has only two alleles of a character, although it has more than two alleles. Multiple alleles can be found only when population studies are made.

15. How does a mutagen induce mutation? Explain with example.
Answer. A mutagen is a physical or chemical agent that changes the genetic material, usually DNA. Different mutagens act on the DNA differently. Powerful mutagens may result in chromosomal instability, causing chromosomal breakages and rearrangement of the chromosomes such as translocation, deletion, and inversion.
Ionizing radiations such as X-rays, gamma rays and alpha particles may cause DNA breakage and other damages. Ultraviolet radiations with wavelength above 260 nm are absorbed strongly by bases, producing pyrimidine dimers. Radioactive decay, such as 14C in DNA which decays into nitrogen.

Short Answer Type Questions
1. In a Mendelian monohybrid cross, the F₂  generation shows identical genotypic and phenotypic ratios. What does it tell us about the nature of alleles involved? Justify your answer.
Answer. In a monohybrid cross, starting with parents which homozygous dominant and homozygous recessive, F₁ , would be heterozygous for the trait and would express the dominant allele. But in case of incomplete dominance, a monohybrid cross shows the result as follows.

Phenotypic ratio —> Red : Pink : White :: 1 : 2 : 1
Genotypic ratio —> RR : Rr: rr : .1:2:1
Here the genotypic and phenotypic ratios are the same. So, we can conclude that when genotypic and phenotypic ratios are the same, the alleles show incomplete dominance.

2. Can a child have blood group O if his parents have blood group ‘A’ and ‘B’? Explain.
Answer. If parents are heterozygous then child can have O blood group but if parents are homozygous then child cannot have O blood group.

3. What is Down’s syndrome? Give its symptoms and cause. Why is it that the chances of having a child with Down’s syndrome increases if the age of the mother exceeds forty years?
Answer. Down’s syndrome is a human genetic disorder caused due to trisomy of chromosome no. 21. Such individuals are aneuploid and have 47 chromosomes (2n + 1). The symptoms include mental retardation, growth abnormalities, constantly open mouth, dwarfness etc. The reason for the disorder is the non-disjunction (failure to separate) of homologous chromosome of pair 21 during meiotic division in the ovum. The chances of having a child with Down’s syndrome increase with the age of the mother (+40) because ova are present in females. Since their birth and therefore older cells are more prone to chromosomal non-disjunction because of various physico-chemical exposures during the mother’s life-time.

4. How was it concluded that genes are located on chromosomes?
Answer. Morgan confirmed Mendelian laws of inheritance and the hypothesis that genes are located on chromosomes. Morgan had discovered that eye colour in Drosophila expressed a sex-linked trait. All first-generation offspring of a mutant white-eyed male and a normal red-eyed female would have red eyes because every chromosome pair would contain at least one copy of the X chromosome with the dominant trait.

5. A plant with red flowers was crossed with another plant with yellow flowers. If Fj showed all flowers orange in colour, explain the inheritance.
Answer. This is due to the incomplete dominance as the F, do not resemble either of the two parents and is in between the two.

6. What are the characteristic features of a true-breeding line?
Answer. A true-breeding line for a trait is one that, has undergone continuous self¬pollination or brother-sister mating, showing a stability in the inheritance of the trait for several generations.

7. In peas, tallness is dominant over dwarfness, and red colour of flowers is dominant over the white colour. When a tall plant bearing red flowers was pollinated with a dwarf plant bearing white flowers, the different phenotypic groups were obtained in the progeny in numbers mentioned against them: Tall, Red =138
Tall, White = 132
Dwarf, Red =136
Dwarf, White =128
Mention the genotypes of the two parents and of the four offspring types.
Answer. The result shows that the four types of offspring are in a ratio of 1:1:1:1. Such a result is observed in a test-cross progeny of a dihybrid cross.
The cross can be represented as:
Tall & Red (Tt Rr) x Dwarf & White (ttrr)

8. Why is the frequency of red-green colour blindness is many times higher in males than that in the females?
Answer. For becoming colourblind, the female must have the allele for it in both her X-chromosomes; but males develop colourblindness when their sole X-chromosome has the allele for it.

9. If a father and son are both defective in red-green colour vision, is it likely that the son inherited the trait from his father? Comment.
Answer. Gene for colour blindness is X-chromosome linked, and sons receive their sole from their mother, not from their father. Male-to-male inheritances is not possible for X-linked traits in humans. In the given case the mother of the child must be a carrier (heterozygous) for colour blindness gene.

10. Discuss why Drosophila has been used extensively for genetical studies.
Answer. Morgan worked with the tiny fruit files, Drosophila melanogaster, which were found very suitable for such studies, as:

1. They could be grown on simple synthetic medium (ripe banana) in the laboratory.
2. They complete their life cycle in about two weeks.
3.  A single mating could produce a large number of progeny flies.
4. There was a clear differentiation of the sexes—the male and female flies are easily distinguishable.
5.  It has many types of hereditary, variations that can be seen with low power microscopes.

11. How do genes and chromosomes share: similarity from the point of view of genetical studies?
Answer. Similarities between Chromosomes and Genes:

1. Both occur in pairs.
2. Both segregate at the time of gamete formation such that only one of each pair is transmitted to a gamete.
3. Independent pairs segregate independently of each other in both.

12. What is recombination? Discuss the applications of recombination from the point of view of genetic engineering.
Answer. The formation of new combinations of genes, either by crossing over or independent assortment is called recombination. Alfred Sturtevant used the frequency of recombination between gene pairs on the same chromosome as a measure of the distance between genes and ‘mapped’ their position on the chromosome. Today genetic maps are extensively used as a starting point in the sequencing of whole genomes as was done in the case of the Human Genome Sequencing Project.

13. What is artificial selection? Do you think it affects the process of natural selection? How?
Answer. A process in the breeding of organisms by which the scientist chooses to only those forms having certain desirable characteristics is called artificial selection.

• Natural selection is the differential survival and reproduction of individuals due to differences in phenotype. In natural selection, the difference in reproductive success based on a particular trait is driven by natural processes like predators, weather conditions, and environmental constraints.
•  Artificial selection is imposed by humans on other organisms. It affects the process of natural selection through change in environmental conditions.

14. With the help of an example, differentiate between incomplete dominance and co-dominance.
Answer.

15. It is said, that the harmful alleles get eliminated from population over a period of time, yet sickle cell anaemia is persisting in human population. Why?
Answer. The harmful alleles get eliminated from population over a period of time, yet sickle cell anaemia is persisting in human population because SCA is a harmful condition which is also a potential saviour from malaria.
Those with the benign sickle trait possess a resistance to malarial infection. The pathogen that causes the disease spends part of its cycle in the red blood cells and triggers an abnormal drop in oxygen levels in the cell. In carriers, this drop is sufficient to trigger the full sickle-cell reaction, which leads to infected cells being rapidly removed from circulation and strongly limiting the infection’s progress. These individuals have a great resistance to infection and have a greater chance of surviving outbreaks. This resistance to infection is the main reason the SCA allele and SCA disease still exist. It is found in greatest frequency in populations where malaria was and often still is a serious problem.

Long Answer Type Questions
1. In a plant tallness is dominant over dwarfness and red flower is dominant over white. Starting with the parents work out a dihybrid cross. What is standard dihybrid ratio? Do you think the values would deviate if the two genes in question are interacting with each other?
Answer.

Yes, the ratio will deviate if the two genes are interacting with each other. In this condition the genes do not independently assort with each other. So, ratio will be changed from 9:3:3: 1.

2. (a) In humans, males are heterogametic and females are homogametic. Explain.
Are there any examples where males are homogametic and females heterogametic?
(b) Also describe as to who determines the sex of an unborn child?
Mention whether temperature has a role in sex determination.
Answer. (a) The term homogametic and heterogametic refer to the organism depending upon whether all the gametes contain one type of sex chromosome (Homo = same) or two different types of sex chromosomes (Hetero = different). Humans show XX/XY type of sex determination, i.e. Females contain two copies of X chromosome and males contain one X and one Y chromosome. Therefore, ova produced by females contain the same sex chromosome, i.e. X. On the other hand the sperms contain two different types of chromosomes,
i. e. 50% sperms have X and 50% have Y chromosome open from half the autosomes (Meiosis). Therefore, the sperms are different with respect to the composition of sex chromosome. In case of humans, females are considered to be homogametic while males are heterogametic. Yes, there are examples where males are homogametic and females are heterogametic. In some birds the mode of sex determination is denoted by ZZ (males) and ZW (females), (b) As a rule the heterogametic organism determines the sex of the unborn child. In case of humans, since males are heterogametic it is the father and not the mother who decides the sex of the child. In some animals like crocodiles, lower temperature favour hatching of female offsprings and higher temperatures lead to hatching of male off springs.

3. A normal visioned woman, whose father is colour blind, marries a normal visioned man. What would be probability of her sons and daughters to be colour blind? Explain with the help of a pedigree chart.
Answer.

All daughters are normal visioned; 50% of sons are likely to be colour blind.

4, Discuss in detail the contributions of Morgan and Sturtervant in the area of genetics.
Answer. Experimental verification of the chromosomal theory of inheritance by Thomas Hunt Morgan (Father of experimental genetics) and his colleagues, led to discovering the basis .for the variation that sexual reproduction produced. Morgan worked with the tiny fruit files, Drosophila melanogaster, which were found very suitable for such studies.

• Morgan carried out several dihybrid crosses in Drosophila to study genes that were sex-linked. The crosses were similar to the dihybrid crosses carried out by Mendel in peas. For example, Morgan hybridised yellow-bodied, white-eyed females to brown-bodied, red-eyed males and intercrossed their F, progeny.
• He observed that the two genes did not segregate independently of each other and the F2 ratio deviated very significantly from the. 9 : 3 : 3 : 1 ratio (expected when the two genes are independent). Morgan and his group knew that the genes were located on the X chromosome and saw quickly that when the two genes in a dihybrid cross were situated on the same chromosome, the proportion of parental gene combinations were much higher than the non-parental type.
• Morgan attributed this due to the physical association or linkage of the two genes and coined the term linkage to describe this physical association of genes on a chromosome and the term recombination to describe the generation of non-parental gene combinations.
• His student Alfred Sturtevant used the frequency of recombination between gene pairs on the same chromosome as a measure of the distance between genes and ‘mapped’ their position on the chromosome. Today genetic maps are extensively used as a starting point in the sequencing of whole genomes as was done in the case of the Human Genome Sequencing Project.

5. Define aneuploidy. How is it different from polyploidy? Describe the individuals having the following chromosomal abnormalities.
a. Trisomy of 21st chromosome
b. XXY
c. XO
Answer. Failure of segregation of chromatids during cell division cycle results in the gain or loss of a chromosome(s), called aneuploidy. – Difference between aneuploidy and polyploidy
1. Failure of segregation of chromatids during cell division cycle results in the gain or loss of a chromosome(s), called aneuploidy. Failure of cytokinesis after telophase stage of cell division results in an increase in a whole set of chromosomes in an organism and,this phenomenon is known as polyploidy.
2. Polyploidy occurs due to altering set of chromosome number such as 2n, 3n, 5n, whereas aneuploidy occurs due to altering particular chromosome or part of a chromosome such as 2n + 1 (trisomic) and 2n – 1 (monosomic).
3. Aneuploidy can be seen in human as genetic disorders; for example, Tuner syndrome, Klinefelter syndrome and Down syndrome, whereas polyploidy is common in plants.
(i) Down’s Syndrome (Mongolism)
• The cause of this genetic disorder-is the presence of an additional copy of the chromosome number 21 (trisomy of 21) due to non-disjunction of chromosomes during sperm or ova formation.
• The affected individual is short statured with small round head, furrowed tongue and partially open mouth. Palm is broad with characteristic palm crease.
• Physical, psychomotor and mental development is retarded.
(ii) Klinefelter’s Syndrome
• This genetic disorder is also caused due to the presence of an additional copy of X-chromosome resulting into a karyotype of 47, XXY.
• Such an individual has overall masculine development, however, the feminine development (development of breast, i.e., Gynaecomastia) is also expressed. Such individuals are sterile male.
(iii) Turner’s Syndrome
Such a disorder is caused due to the absence of one of the X chromosomes, i.e., 45 with XO. Such females are sterile as ovaries are rudimentary besides other features including lack of other secondary sexual characters.

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NCERT Exemplar Problems Class 12 Biology Chapter 6 Molecular Basis of Inheritance

Multiple Choice Questions
Single Correct Answer Type
1. In a DNA strand the nucleotides are linked together by
(a) Glycosidic bonds (b) Phosphodiester bonds
(c) Peptide bonds (d) Hydrogen bonds
Answer. (b) In a DNA strand the nucleotides are linked together by Phosphodiester bonds.

2. A nucleoside differs from a nucleotide. It lacks the
(a) Base (b) Sugar
(c) Phosphate group t (d) Hydroxyl group
Answer. (b) Nucleoside = Base + sugar
Nucleotide = Base + sugar + phosphate group

3. Both deoxyribose and ribose belong to a class of sugars called
(a) Trioses (b) Hexoses
(c) Pentoses (d) Polysaccharides
Answer. (c) Both deoxyribose and ribose belong to a pentoses class of sugars.

4. The fact that a purine base always paired through hydrogen bonds with a pyrimidine base leads to, in the DNA double helix
(a) The antiparallel nature (b) The semiconservative nature
(c) Uniform width throughout DNA (d) Uniform length in all DNA
Answer. (c) A purine base always paired through hydrogen bonds with a pyrimidine base leads to, in the DNA double helix, uniform width throughout DNA.

5. The net electric charge on DNA and histones is
(a) Both positive
(b) Both negative
(c) Negative and positive, respectively
(d) Zero
Answer. (c) The net electric charge on DNA and.histones is negative and positive, respectively.

6. The promoter site and the terminator site for transcription are located at
(a) 3′ (downstream) end and 5′ (upstream) end, respectively of the transcription unit
(b) 5′ (upstream) end and 3′ (downstream) end, respectively of the transcription unit
(c) The 5′ (upstream) end
(d) The 3′ (downstream) end
Answer. (b) The promoter site and the terminator site for transcription are located at 5′ (upstream) end and 3′ (downstream) end, respectively of the transcription unit.

7. Which of the following statements is the most appropriate for sickle cell anaemia?
(a) It cannot be treated with iron supplements
(b) It is a molecular disease
(c) It confers resistance to acquiring malaria
(d) All of the above
Answer. (d) Sick/e cell anaemia: It is a molecular disease, autosomal recessive disorder and an example of pleiotropy. It cannot be treated with iron supplements and it confers resistance to acquiring malaria.

8. One of the following true with respect to AUG is
(a) It codes for methionine only 
(b) It is also an initiation codon
(c) It codes for methionine in both prokdryotes and eukaryotes
(d) All of the above
Answer. (d) AUG—It codes for methionine only in both prokaryotes and eukaryotes, it is also an initiation codon.

9. The first genetic material could be
(a) Protein (b) Carbohydrates (c) DNA (d) RNA
Answer. (d) The first genetic material could be RNA but ideal genetic material is DNA.

10. With regard to mature mRNA in eukaryotes,
(a) Exons and introns do not appear in the mature RNA
(b) Exons appear but introns do not appear in the mature RNA
(c) Introns appear but exons do not appear in the mature RNA
(d) Both exons and introns appear in the mature RNA
Answer. (b) In eukaryotes exons appear but introns do not appear in the mature RNA.

11. The human chromosome with the highest and least number of genes in them are respectively
(a) Chromosome 21 and Y (b) Chromosome 1 and X
(c) Chromosome 1 and Y (d) Chromosome X and Y
66 NCERT Exemplar Biology: Class XU
Answer. (c)

• Chromosome one have 2968 genes (highest)
• Chromosome Y have 231 genes (lowest)

12. Who amongst the following scientists had no contribution in the development of the double helix model for the structure of DNA?
(a) Rosalind Franklin (b) Maurice Wilkins
(c) Erwin Chargaff (d) Meselson and Stahl
Answer. (d) Meselson and Stahl give the experimental proof of semiconservative DNA replication. He had no contribution to the development of the double helix model for the. structure of DNA.

13. DNA is a polymer of nucleotides which are linked to each other by 3′-5′ phosphodiester bond. To prevent polymerisation of nucleotides, which of the following modifications would you choose?
(a) Replace purine with pyrimidines
(b) Remove/Replace 3′ OH group in deoxy ribose
(c) Remove/Replace 2′ OH group with some other group in deoxy ribose
(d) Both‘b’and‘c’.
Answer. (b) DNA is a polymer of nucleotides which are linked to each other by 3′-5′ phosphodiester bond. Remove/replace 3’OH group in deoxyribose, to prevent polymerisation of nucleotides. .

14. Discontinuous synthesis of DNA occurs in one strand, because
(a) DNA molecule being synthesised is very long
(b) DNA dependent DNA polymerase catalyses polymerisation only in one direction (5′ —> 3′)
(c) It is a more efficient process
(d) DNA ligase has to have a role
Answer. (b) Discontinuous synthesis of DNA occurs in one strand, because DNA dependent DNA polymerase catalyses polymerisation only in one direction (5′ —> 3′).

15. Which of the following steps in transcription is catalysed by RNA polymerase?
(a) Initiation (b) Elongation
(c) Termination (d) All of the above
Answer. (b) Elongation step in transcription is catalysed by RNA polymerase.

16. Control of gene expression takes place at the level of
(a) DNA-replication (b) Transcription
(c) Translation (d) None of the above
Answer. (b) Control of gene expression takes place at the level of transcription.

17. Regulatory proteins are the accessory proteins that interact with RNA polymerase and affect its role in transcription. Which of the following statements is correct about regulatory protein?
(a) They only increase expression
(b) They only decrease expression
(c) They interact with RNA polymerase but do not affect the expression
(d) They can act both as activators and as repressors
Answer. (d) Regulatory proteins are the accessory proteins that interact with RNA polymerase and affect its role in transcription. They can act both as activators and as repressors.

18. Which was the last human chromosome to be completely sequenced?
(a) Chromosome 1 (b) Chromosome 11
(c) Chromosome 21 (d) Chromosome x
Answer. (a) The Human genome project was completed in 2003. The sequence of chromosome 1 was completed only in May 2006.

19. Which of the following are the functions of RNA?
(a) It is a carrier of genetic information from DNAto ribosomes synthesising polypeptides.
(b) It carries amino acids to ribosomes.
(c) It is a constituent component of ribosomes.
(d) All of the above.
Answer. (d) Functions of mRNA:

• Carrier of genetic information from DNA to ribosomes synthesising polypeptides.
• Carries amino acids to ribosomes.
• Constituent component of ribosomes. .

20. While analysing the DNA of an organism a total number of 5386 nucleotides were found out of which the proportion of different bases were:
Adenine = 29%, Guanine = 17%, Cytosine = 32%, Thymine = 17% Considering the Chargaff’s rule it can be concluded that
(a) It is a double stranded circular DNA
(b) It is single stranded DNA
(c) It is a double stranded linear DNA
(d) No conclusion can be drawn. ‘
Answer. (b) In the DNA of an organism a total number of5386 nucleotides were found out of which the proportion of different bases were: Adenine = 29%, Guanine = 17%, Cytosine = 32%, Thymine = 17%. Considering the Chargaff’s rule it can be concluded that it is a single stranded DNA.

21. In some viruses, DNA is synthesised by using RNA as template. Such a DNA is called
(a) A-DNA (b) B-DNA (c) c DNA (d) r DNA
Answer. (c) In some viruses, DNA is synthesised by using RNA as template. Such a DNA is called c-DNA (Complementary DNA).

22. If Meselson and Stahl’s experiment is continued for four generations in bacteria, the ratio of ₁₅N/₁₅N: ₁₅N/₁₄N: ₁₄N/₁₄N containing DNA in the fourth generation would be
(a) 1 : 1 : 0 (b) 1 : 4 : 0 (c) 0:1:3 (d) 0:1:7
Answer. (d)

•  If Meselson and Stahl’s experiment is continued for four generations in bacteria, the ratio of 15_N/15_N : 15_N/14_N : 14_N/14_N containing DNA in the fourth generation would be 0 : 1 : 7.
• After third generation (60 min.) bacteria contains 25% hybrid (15_N14_N) in 1 :3 ratio.
• After 4th generation (80 min.) bacteria contains 12.5% hybrid and 87.5% light DNA in 1 : 7 ratio.

23. If the sequence of nitrogen bases of the coding strand of DNA in a transcription unit is 5′- ATGAATG-3′, The sequence of bases in its RNA transcript would be
(a) 5′ – AU G AAU G – 3′ (b) 5′- U AC U U AC – 3′
(c) 5′-CAUUCAU-3′ (d) 5;-GUAAGUA-3′
Answer. (a)
Coding strand: 5′-AT G A AT G – 3′
Template strand: 3′ – TACTTAC-5″
RNA transcript: 5′-AUGAAUG-3′

24. The RNA polymerase holoenzyme transcribes
(a) The promoter, structural gene and the terminator region
(b) The promoter and the terminator region
(c) The structural gene and the terminator regions
(d) The structural gene only
Answer. (d) The RNA polymerase holoenzyme transcribes the structural gene only.

25. If the base sequence of a codon in mRNA is 5′-AUG-3′, the sequence of tRNA pairing with it must be
(a) 5′ -UAC – 3′ (b) 5′-CAU-3′
(c) 5′-AUG-3′ (d) 5′ – GUA – 3′
Answer. (b) Base sequence of a codon in mRNA is 5′-AUG-3′, the sequence of tRNA pairing with it must be cqpiplementary, i.e., 3′ – UAC – 5′

26. The amino acid attaches to the tRNA at its
(a) 5′-end (b) 3′-end
(c) Anti codon site – (d) DHU loop
Answer. (b) The amino acid attaches to the tRNA at its 3′-end.

27. To initiate translation, the mRNA first binds to
(a) The smaller ribosomal sub-unit (b) The larger ribosomal sub-unit (c) The whole ribosome (d) No such specificity exists
Answer. (a) To initiate translation, the mRNA first binds to the smaller ribosomal sub-unit.

28. In E.coli, the lac operon gets switched on when
(a) Lactose is present and it binds to the repressor
(b) Repressor binds to operator
(c) RNA polymerase binds to the operator
(d) Lactose is present and it binds to RNA polymerase
Answer. (a) In E.coli, the lac operon gets switched on when lactose is present and it binds to the repressor.

Very Short Answer Type Questions
1. What is the function of histones in DNA packaging?
Answer. Histone are the basis protein that are rich in arginine and lysine. Histones form octamer on which DNA is wrapped and form nucleosome that means it helps in packaging of DNA in eukaryotes.

2. Distinguish between heterochromatin and euchromatin. Which of the two is transcriptionally active?
Answer.

3. The enzyme DNA polymerase in E.coli is a DNA dependent polymerase and also has the ability to proofread the DNA strand being synthesised. Explain. Discuss the dual polymerase.
Answer. DNA polymerase helps in the replication of DNA molecules. DNA polymerase also helps in the proofreading, if any error/mutation occurred during replication.

4. What is the cause of discontinuous synthesis of DNA on one of the parental strands of DNA? What happens to these short stretches of synthesised DNA?
Answer. The DNA-dependent DNA polymerases catalyse polymerisation only in one direction, that is 5′ —> 3′. This creates some additional complications at the replicating fork. Consequently, on leading strand (the template with polarity 3′ —> 5′), the replication is continuous, while on the lagging strain (the template with polarity 5′ —> 3′), it is discontinuous. The discontinuously synthesised fragments are later joined by the enzyme DNA ligase.

5. Given below is the sequence of coding strand of DNA in a transcription unit. 3’AATGCAGCTATTAGG’ Write the sequence of
(a) its complementary strand
(b) the mRNA
Answer. (a) Complementary strand:
5′ – TTACGTCGATAA T C C – 3′
(b) The mRNA
5′ – AAU GCAGCUAUUAGG-3′

6. What is DNA polymorphism? Why is it important to study it?
Answer. Polymorphism (variation at genetic level) arises due to mutations. Allelic sequence variation has traditionally been described as a DNA polymorphism if more than one variant (allele) at a locus occurs in human population with a frequency greater than 0.01. In simple terms, if an inheritable mutation is observed in a population at high frequency, it is referred to as DNA polymorphism.
Polymorphism become very useful identification tool in forensic applications. Further, as the polymorphisms are inheritable from parents to children, DNA fingerprinting is the basis of paternity testing, in case of disputes.

7. Based on your understanding of genetic code, explain the formation of any abnormal hemoglobin molecule. What are the known consequences of such a change?
Answer. The defect is caused by the substitution (trans version) of Glutamic acid (Glu) by Valine (Val) at the sixth position of the betaglobin chain of the haemoglobin molecule. The substitution of amino acid in the glob in protein results due to the single base substitution at the sixth codon of the betaglobin gene from GAG to GUG. The mutant haemoglobin molecule undergoes polymerisation under low oxygen tension causing the change in the shape of the RBC from biconcave disc to elongated sickle like structure.
Sickle-shaped red blood cells that obstruct capillaries and restrict blood flow to an organ resulting in ischaemia, pain, necrosis, and often organ damage.

8. Sometimes cattle or even human beings give birth to their young ones that are having extremely different sets of organs like limbs/position of eye(s) etc. Comment.
Answer. There is a disturbance in co-ordinated regulation of expression of sets of genes associated with organ development.

9. In a nucleus, the number of ribonucleoside triphosphates is 10 times the number of deoxy ribonucleoside triphosphates, but only deoxy ribonucleotides are added during the DNA replication. Suggest a mechanism.
Answer. DNA polymerase is highly specific to recognise only deoxy ribonucleoside triphosphates. Therefore it cannot hold RNA nucleotides.

10. Name a few enzymes involved in DNA replication other than DNA polymerase and ligase. Name the key functions for each of them.
Answer. (i) Helicase—opens the helix
(ii) Topoisomerases—removes the super coiling of DNA
(iii) Primase—synthesises RNA primer
(iv) Telomerase—to synthesises the DNA of telomeric end of chromosomes

11. Name any three viruses which have RNA as the genetic material.
Answer. (i) TMV (Tobacco Mosaic virus)
(ii) HIV (Human Immuno Deficiency virus)
(iii) QB bacteriophage

Short Answer Type Questions
1. Define transformation in Griffith’s experiment. Discuss how it helps in the identification of DNA as the genetic material.
Answer. Transforming Principle

• In 1928, Frederick Griffith, in a series of experiments with Streptococcus pneumoniae (bacterium responsible for pneumonia), witnessed a miraculous transformation, in the bacteria. During the course of his . experiment, a living organism (bacteria) had changed in physical form.
• When Streptococcus pneumoniae (pneumococcus) bacteria are grown on a culture plate, some produce smooth shiny colonies (S) while others produce rough colonies (R).
• This is because the S strain bacteria have a mucous (polysaccharide) coat, while R strain does not. Mice infected with the S strain (virulent) die from pneumonia infection but mice infected with the R strain do not develop pneumonia. .
  S strain -»Inject into mice -» Mice die R strain —»Inject into mice —> Mice live
• Griffith was able to kill bacteria by heating them. He observed that heat- killed S strain bacteria injected into mice did not kill them.
  S strain (heat killed) —> Inject into mice —> Mice live
• When he injected a mixture of heat-killed S and live R bacteria, the mice died. Moreover, he recovered living S bacteria from the dead mice.
  S strain (heat killed ) + —> Inject into mice —> Mice die R strain (live)
• He concluded that the R strain bacteria had somehow been transformed by the heat-killed S strain bacteria. Some ‘transforming principle’, transferred from the heat-killed S strain, had enabled the R strain to synthesise a smooth polysaccharide coat and become virulent. This must be due to the transfer of the genetic material. However, the biochemical nature of genetic material was not defined from his experiments.

2. Who revealed biochemical nature of the transforming principle? How was it done?
Answer. Prior to the work of Oswald Avery, Colin MacLeod and Maclyn McCarty (1933-44), the genetic material was thought to be a protein. They worked th determine the biochemical nature of ‘transforming principle’ in Griffith’s experiment. They purified biochemicals (proteins, DNA, RNA, etc.) from the heat-killed S cells to see which ones could transform live R cells into S cells. They discovered that DNA alone from S bacteria caused R bacteria to become transformed.
They also discovered that protein-digesting enzymes (proteases) and RNA-digesting enzymes (RNases) did not affect transformation, so the transforming substance was not a protein or RNA. Digestion with DNase did inhibit transformation, suggesting that the DNA caused the transformation. They concluded that DNA is the hereditary material, but not all biologists were convinced.

3. Discuss the significance of heavy isotope of nitrogen in the Meselson and Stahl’s experiment.
Answer. Matthew Meselson and Franklin Stahl performed the following experiment in 1958. They grew E. coli in a medium containing 15NH4C1 (15N is the heavy isotope of nitrogen) as the only nitrogen source for many generations. The result was that l5N was incorporated into newly synthesised DNA (as well as other nitrogen containing compounds).
This heavy DNA molecule could be .distinguished from the normal DNA by centrifugation in a cesium chloride (CsCl) density gradient 15N is not a radioactive isotope, and it can be separated from 14N only based on densities).

4. Define a cistron. Giving examples differentiate between monocistronic and polycistronic transcription unit.
Answer. Portion of DNA having information for an entire polypeptide or trait is called cistron. However by defining a cistron as a segment of DNA coding for a polypeptide, the structural gene in a transcription unit jcould be said as monocistronic (mostly in eukaryotes) or polycistronic (mostly in bacteria or prokaryotes). In eukaryotes, the monocistronic structural genes have interrupted coding sequences—the genes in eukaryotes are split.
The coding sequences or expressed sequences are defined as exons. Exons are said to be those sequence that appear in mature or processed RNA. The exons are interrupted by introns. Introns or intervening sequences do not appear in mature or processed RNA.

5. Give any six features of the human genome.
Answer. Some of the salient Observations drawn from human genome project are as follows:

1. The human genome contains 3164.7 million nucleotide bases.
2. The average gene consists of3000 bases, but sizes vary greatly, with the largest known human gene being dystrophin at 2.4 million bases.
3. The total number of genes is estimated at 30,000-much lower than previous estimates of 80,000 to 1,40,000 genes. Almost all (99.9 per cent) nucleotide bases are exactly the same in all people.
4. The functions are unknown for over 50 per cent of the discovered genes.
5. Less than 2 per cent of the genome codes for proteins.
6. Repeated sequences make up a very large portion of the human genome.

6. During DNA replication, why is it that the entire molecule does not open in one go? Explain replication fork. What are the two functions that the monomers (dNTPs) play?
Answer. For long DNA molecules, since the two strands of DNA cannot be separated in its entire length (due to very high energy requirement), the replication occur within a small opening of the DNA helix, referred to as replication fork.

7. Retroviruses do not follow central Dogma. Comment.
Answer. Genetic material of retrovirus is RNA. At the time of synthesis of protein, RNA is ‘reverse transcribed’ to its complementary DNA first, which is opposite to the central dogma. Hence, retrovirus are not known to follow central dogma.

8. In an experiment, DNA is treated with a compound which tends to place itself amongst the stacks of nitrogenous base pairs. As a result of this, the distance between two consecutive base increases from 0.34 nm to 0.44 nm. Calculate the length of DNA double helix (which has 2 x 10⁹ bp) in the presence of saturating amount of this compound.
Answer. 2 x 10⁹ x 0.44 x 10^-9/bp = 0.88 m.

9. What would happen if histones were to be mutated and made rich in acidic amino acids such as aspartic acid and glutamic acid in place of basic amino acids such as lysine and arginine?
Answer. If histone proteins were rich in acidic amino acids instead of basic amino acids then they may not have any role in DNA packaging in eukaryotes as DNA is also negatively charged molecule. The packaging of DNA around the nucleosome would not happen. Consequently, the chromatin fibre would not be formed.

10. Recall the experiments done by Frederick Griffith, Avery, MacLeod and McCarty, where DNA was speculated to be the genetic material. If RNA, instead of DNA was the genetic material, would the heat killed strain of Pneumococcus have transformed the R-strain into virulent strain? Explain.
Answer. RNA is more labile and prone to degradation (owing to the presence of 2′ OH group in its ribose). Hence heat-killed S-strain may not have retained its ability to transform the R-strain into virulent form if RNA was its genetic material.

11. You are repeating the Hershey-Chase experiment and are provided with two isotopes: ³²P and ¹⁵N (in place of ³⁵S in the original experiment). How do you expect your results to be different?
Answer. Use of ¹⁵N will be inappropriate because method of detection of ³⁵P and ¹⁵N is different (³²P being a radioactive isotope while ¹⁵N is not radioactive but is the heavier isotope of Nitrogen). Even if ¹⁵N was radioactive then its presence would have been detected both inside the cell (¹⁵N incorporated as nitrogenous base in DNA) as well as in the supernatant because ¹⁵N would also get incorporated in amino group of amino acids in proteins). Hence the use of ¹⁵N would not give any conclusive results.

12. There is only one possible Sequence of amino acids when deduced from a given nucleotides. But multiple nucleotides sequence can be deduced from a single amino acid sequence. Explain this phenomena.
Answer. Some amino acids are coded by more than one codon (known as degeneracy of codons), hence on deducing a nucleotide sequence from an amino acid sequence, multiple nucleotide sequence will be obtained.
For example, lie has three codous: AUU, AUC AUA hence a depeptide. Met-Ile can have the following nucleotide sequence:
(i) AUG-AUU
(ii) AUG-AUC
(iii) AUG-AUA – And if, we deduce amino acid sequence the above nucleotide sequences, all the three will code for Met-IIe

13. A single base mutation in a gene may not ‘always’ result in loss or gain of function. Do you think the statement is correct? Defend your answer.
Answer. The statement is correct. Because of degeneracy of codons, mutations at third base of codon, usually does not result into any change in phenotype. This is called silent mutations.

14. A low level of expression of lac operon occurs at all the time. Can you explain the logic behind this phenomena?
Answer. In the complete absence of expression of lac operon, permease will not be synthesised which is essential for transport of lactose from medium into the cells. And if lactose cannot be transported into the cell, then it cannot act as inducers, hence cannot relieve the lac operon from its repressed state.

15. How has the sequencing of human genome opened new windows for treatment of various genetic disorders. Discuss amongst your classmates.
Answer. Knowledge about the effects of DNA variations among individuals can lead to revolutionary new ways to diagnose, treat and someday prevent the thousands of disorders that affect human beings.

• Besides providing clues to understanding human biology, learning about non-human organisms DNA sequences can lead to an understanding of their natural capabilities that can be applied toward solving challenges in health care, agriculture, energy production, environmental remediation. Deriving meaningful knowledge from the DNA sequences will define research through the coming decades leading to our understanding of biological systems. .
• This enormous task will require the expertise and creativity of tens of thousands of scientists from varied disciplines in both the public and private sectors worldwide. One of the greatest impacts of having the HG sequence may well be enabling a radically new approach to biological research. In the past, researchers studied one or a few genes at a time. With whole-genome sequences and new high-throughput technologies, we can approach questions systematically and on a much broader scale.
• They can study all the genes in a genome, for example, all the transcripts in a particular tissue or organ or tumor, or how tens of thousands of genes and proteins work together in interconnected networks to orchestrate the chemistry of life.

16. The total number of genes in humansds far less (< 25,000) than the previous estimate (upto 1,40,000 gene). Comment.
Answer. Repeated sequences make up very large portion of the human genome. Repetitive sequences are stretches of DNA sequences that are repeated many times, sometimes hundred to thousand times. They are thought to have no direct coding functions, but they shed light on chromosome structure, dynamics and evolution. Less than 2 per cent of the genome codes for proteins.

17. Now, sequencing of total genomes getting is less expensive day by the day. Soon it may be affordable for a common man to get his genome sequenced. What in your opinion could be the advantage and disadvantage of this development?
Answer. Advantages:
1.Knowledge about the effects of DNA variations among individuals can lead to revolutionary new ways to diagnose, treat and some day prevent the thousands of disorders that affect human beings.
2. Besides providing clues to understanding human biology, learning about non-human organisms DNA sequences can lead to an understanding of their natural capabilities that can be applied toward solving challenges in health care, agriculture, energy production, environmental remediation.
3. Deriving meaningful knowledge from the DNA sequences will define research through the coming decades leading to our understanding of biological systems.
4. They can study all the genes in a genome, for example, all the transcripts in a particular tissue or organ or tumor, or how tens of thousands of genes and proteins work together in interconnected networks to orchestrate the chemistry of life.
Disadvantages:
1. It creates the problem of patenting of genes.
2. Persons becomes more careless.

18. Would it be appropriate to use DNA probes such as VNTR in DNA finger printing of a bacteriophage?
Answer. Bacteriophage does not have repetitive sequences such as VNTRs in its genome as its genome is very small and have all the coding sequence. DNA finger printing is not done for phages.

19. During in vitro synthesis of DNA, a researcher used 2′, 3′ – dideoxy cytidine triphosphate as raw nucleotide in place of 2′-deoxy cytidine. What would be the consequence?
Answer. Further polymerisation would not occur, as the 3′-OH on sugar is not there to add a new nucleotide for forming ester bond.

20. What background information did Watson and Crick have made available for developing a model of DNA What was their contribution?
Answer. Watson and Crick had the following information which helped them to develop a model of DNA.
(i) Chargaffs’ law suggesting A = T and C = G.
(ii) Wilkins and Rosalind Franklin’s work on DNA crystal’s X-ray diffraction studies about DNA’s physical structure.
(iii) Watson and crick proposed
a. How complementary bases may pair
b. Semi conservative replication and ‘
c. Mutation through tautomerism

21. What are the functions of (i) methylated guanosine cap, (ii) poly-A “tail” in a mature on RNA?
Answer. Methylated Guanine cap helps in binding of mRNA to smaller ribosomal sub-unit during initiation of translation. Poly-A tail provides longevity to mRNA’s life. Tail length and longevity of mRNA are positively correlated.

22. Do you think that the alternate splicing of exons may enable a structural gene to code for several isoproteins from one and the same gene? If yes, how? If not, why so?
Answer. Functional mRN A of structural genes need not always include all of its exons. This alternate splicing of exons is sex-specific, tissue-specific and even developmental stage-specific. By such alternate splicing of exons, a single gene may encode for several isoproteins and/or proteins of similar class. In absence of such a kind of splicing, there should have been new genes for every protein/isoprotein. Such an extravagancy has been avoided in natural phenomena by way of alternative splicing.

23. Comment on the utility of variability in number of tandem repeats during DNA finger printing.
Answer. Tandemness in repeats provides many copies of the sequence for fingerprinting, and variability in nitrogen base sequence in them. Being individual-specific, this proves to be useful in the process of DNA fingerprinting.

Long Answer Type Questions
1. Give an account of Hershey and Chase experiment. What did it conclusively prove? If both DNA and proteins contained phosphorus and sulphur do you think the result would have been the same?
Answer. The unequivocal proof that DNA is the genetic material came from the experiments of Alfred Hershey and Martha Chase (1952). They worked with viruses that infect bacteria called bacteriophages. The bacteriophage attaches to the bacteria and its genetic material then enters the bacterial cell. The bacterial cell treats the viral genetic material as if it was its own and subsequently manufacture^ more virus particles.

• Hershey and Chase worked to discover whether it was protein or DNA from the viruses that entered the bacteria. They grew some viruses on a medium that contained radioactive phosphorus and some others on medium that contained radioactive sulfur. Viruses grown in the presence of radioactive phosphorus contained radioactive DNA but not radioactive protein because DNA contains phosphorus but protein does not
• Similarly, viruses grown on radioactive sulfur contained radioactive protein but not radioactive DNA because DNA does not contain sulfur. Radioactive phages were allowed to attach to E. coli bacteria. Then, as the infection proceeded, the viral coats were removed from the bacteria by agitating them in a blender.
• Radioactive phages were allowed to attach to E. Coli bacteria. Then, as the infection proceeded, the viral coats were removed from the bacteria by agitating them in a blender. The virus particles were separated from the bacteria by spinning them in a centrifuge. Bacteria which was infected with viruses that had radioactive DNA were radioactive, indicating that DNA was the material that passed from the virus to the bacteria.
• Bacteria that were infected with viruses that had radioactive proteins were not radioactive. This indicates that proteins did not enter the bacteria from the viruses. DNA is therefore the genetic material that is passed from virus to bacteria.
  

2. During the course of evolution why DNA was chosen over RNA as genetic material? Give reasons by first discussing the desired criteria in a molecule that can act as genetic material and in the light of biochemical differences between DNA and RNA.
Answer. A molecule that can act as a genetic material must fulfill the following criteria:
(i) It should be able to generate its replica (Replication):
• Because of rule of base pairing and complementarity, both the nucleic acids (DNA and RNA) have the ability to direct their duplications. The other molecules in the living system, such as proteins fail to fulfill first criteria itself.
(ii) It should chemically and structurally be stable.
• The genetic material should be stable enough not to change with different stages of life cycle, age or with change in physiology of the organism. Stability as one of the properties of genetic material was very evident in Griffith’s ‘transforming principle’ itself that heat, which killed the. bacteria,- at least did not destroy some of the properties of genetic material. This now can easily be explained in light of the DNA that the two strands being complementary if separated by heating come together, when appropriate conditions are provided.
• Further, 2′-OR group present at every nucleotide in RNA is a reactive group and makes RNA labile and easily degradable. RNA is also now known to be catalytic, hence reactive. Therefore, DNA chemically is less reactive and structurally more stable when compared to RNA. Therefore, among the two nucleic acids, the DNA is a better genetic material. In fact, the presence of thymine at the place of uracil also confers additional stability to DNA.
(iii) It should provide the scope for slow changes (mutation) that are required
for evolution.
• Both DNA and RNA are able to mutate. In fact, RNA being unstable, mutate at a faster rate. Consequently, viruses having RNA genome and having shorter life span mutate and evolve faster.
(iv) It should be able to express itself in the form of ‘Mendelian Characters’.
• RNA can directly code for the synthesis of proteins, hence can easily express the characters. DNA, however, is dependent on RNA for synthesis of proteins. The protein synthesising machinery has evolved around RNA. The above discussion indicate that both RNA and DNA can function as genetic material, but DNA being more stable is preferred for storage of genetic information. For the transmission of genetic information, RNA is better.
• RNA being a catalyst was reactive and hence unstable. Therefore, DNA has evolved from RNA with chemical modifications that make it more stable.

3. Give an account of post transcriptional modifications of an eukaryotic mRNA. “
Answer. The primary transcripts (hn-RNA) contain both the exons and the introns and are non-functional. Hence, it is subjected to a process called splicing where the introns are removed and exons are joined in a defined order. Intron is the portion of gene which is transcribed but not translated. In prokaryotes hnRNA is absent so splicing in not required. hnRNA undergoes additional processing called as capping and tailing.
In capping an unusual nucleotide (methyl guanosine triphosphate) is added to the 5′-end of hnRNA. In tailing, adenylate residues (200-300) are added at 3′-end in a template independent manner. It is the fully processed hnRNA, . now called mRNA, that is transported out of the nucleus for translation.

4. Discuss the process of translation in detail.
Answer. Translation refers to the process of polymerisation of amino acids to form a polypeptide. Ribosome is the site of protein synthesis. The amino acids are joined by a bond which is known as a peptide bond. Formation of a peptide bond requires energy. Therefore, in the first phase itself amino acids are activated in the presence of ATP and linked to their cognate tRNA—a process commonly called as activajjon or charging of tRNA or aminoacylation of tRNA to be more specific.

•  If two such charged tRNAs are brought close enough, the formation of peptide bond between them would be favoured energetically. The presence of a catalyst would enhance the rate of peptide bond formation. The cellular factory responsible for synthesising proteins is the ribosome. The ribosome consists of structural RNAs and about 80 different proteins. In its inactive state, it exists as two subunits; a large subunit and a small subunit.
• When the small subunit encounters an mRNA, the process of translation of the mRNA to protein begins. There are two sites in the large subunit, for subsequent amino acids to bind to and thus, be close enough to each other for the formation of a peptide bond. The ribosome also acts as a catalyst (23S rRNA in bacteria is the enzyme- ribozyme) for the formation of peptide bond or polyribonucleotides.
• A translational unit in mRNA is the sequence of RNA that is flanked by the start codon (AUG) and the stop codon and codes for a polypeptide. An mRNA also has some additional sequences that are not translated and are referred as untranslated regions (UTR). The UTRs are present at both 5′- end (before start codon) and at 3′-end (after stop codon). They are required for efficient translation process.
• For initiation, the ribosome binds to the mRNA at the start codon (AUG) that is recognised only by the initiator tRNA. The ribosome proceeds to the elongation phase of protein synthesis. During this stage, complexes composed of an amino acid linked to tRNA, sequentially bind to the appropriate codon in mRNA by forming complementary base pairs with the tRNA anticodon. The ribosome moves from codon to codon along the mRNA.
• Amino acids are added one by one, translated into polypeptide sequences dictated by DNA and represented by mRNA. At the end, a release factor binds to the stop codon, terminating translation and releasing the complete polypeptide from the ribosome.
  

5. Define an operon. Giving an example, explain an Inducible operon.
Answer. Lac Operon

• The elucidation of the lac operon was also a result of a close association between a geneticist, Francois Jacob and a biochemist, Jacque Monod in 1961. They were the first to elucidate a transcriptionally regulated system. In lac operon (here lac refers to lactose), a polycistronic structural gene is regulated by a common promoter and regulatory genes.
• Such arrangement is very common in bacteria and is referred to as operon. To name few such examples, lac operon, trp operon, ara operon, his operon, val operon, etc. Lac operon is a type of inducible operon. In inducible operon, presence of a chemical switch on the operon. The lac operon consists of one regulatory gene (the i gene – here the term i does not refer to inducer, rather it Is derived from the word inhibitor) and three structural genes (z, y and a).
• The z gene codes for beta-galactosidase (p-gal), which is primarily responsible for the hydrolysis of the disaccharide, lactose into its monomeric units, galactose and glucose. The y-gene codes for permease, which increases permeability of the cell to p-galactosides. The a-gene encodes a transacetylase.
• Hence, all the three gene products in lac operon are required for metabolism of lactose. In most other operons as well, the genes present in the operon are needed together to function in the same or related metabolic pathway. Operator gene is switched off in the presence of a repressor. RNA polymerase binds with the promoter gene. Lactose is the substrate for the enzyme beta-galactosidase and it regulates switching on and off of the operon. Hence, it is termed as inducer.
• In the absence of a preferred carbon source such as glucose,; if lactose is provided in the grcfwth medium of the bacteria, the lactose is transported into the cells through the action of permease (Remember, a very low level of expression of lac operon has to be present in the cell all the time, otherwise lactose cannot enter the cells). The lactose then induces the operon in the following manner.
  
• The repressor of the operon is synthesised (all-the-time constitutively) from the i gene.
  Constitutive genes or housekeeping genes: These genes are constantly expressing themselves because their product is required by cell all the time. E.g.: Genes for ATPalse and glycolysis. The repressor protein binds to the operator region of the operon and prevents RNA polymerase from transcribing the operon.
• In the presence of an inducer, such as lactose or allolactose, the repressor is inactivated by interaction with the inducer. This allows RNA polymerase access to the promoter and transcription proceeds. Essentially, regulation of lac operon can also be visualised as regulation of enzyme synthesis by its substrate.

6. ‘There is a paternity dispute for a child’. Which technique can solve the problem? Discuss the principle involved.
Answer. DNA finger printing is used to solve the paternity dispute. DNA fingerprinting involves identifying differences in some specific regions in DNA sequence called as repetitive DNA, because in these sequences, a small stretch of DNA is repeated many times. These repetitive DNA are separated from bulk genomic DNA as different peaks during density gradient centrifugation.
• The bulk DNA forms a major peak and the other small peaks are referred to as satellite DNA. Depending on base composition (A : T rich or G : C rich), length of segment, and number of repetitive units, the satellite DNA
• is classified into many categories, such as micro-satellites, mini-satellites etc. These sequences normally do not code for any proteins, but they form a large portion of human genome.
• These sequence show high degree of polymorphism and form the basis of DNA fingerprinting. Since DNA from every tissue (such as blood, hair- follicle, skin, bone, saliva, sperm etc.), from an individual show the same degree of polymorphism, they become very useful identification tool in forensic applications. Further, as the polymorphisms are inheritable from parents to children, DNA fingerprinting is the basis of paternity testing, in case of disputes. *
• The technique of DNA fingerprinting was initially developed by Alec
Jeffreys. Lalji Singh is called father of Indian DNA fingerprinting or DNA profiling or DNA typing. He used a satellite DNA as probe that shows very high degree of polymorphism. It was called as Variable Number of Tandem Repeats (VNTR).
• The technique, as used earlier, involved Southern blot hybridisation using radiolabelled VNTR as a probe. It included
(i) Isolation of DNA,
(ii) Digestion of DNA by restriction endonucleases,
(iii) Separation of DNA fragments by electrophoresis,
(iv) Transferring (blotting) of separated DNA fragments to synthetic membranes, such as nitrocellulose or nylon,
(v) Hybridisation using labelled VNTR probe, and
(vi) Detectionof hybridised DNA fragments by autoradiography.

7. Give an account of the methods used in sequencing the human genome.
Answer. Methodologies:
• The methods involved two major approaches. One approach focused on identifying all the genes that are expressed as RNA (referred to as Expressed Sequence Tags (ESTs). The other took the blind approach of simply sequencing the whole set of genome that contained all the coding and non-coding sequence, and later assigning different regions in the sequence with functions (a term referred to as Sequence Annotation).
• For sequencing, the total DNA from a cell is isolated and converted into random fragments of relatively smaller sizes (recall DNA is a very long polymer, and there are technical limitations in sequencing very long pieces of DNA) and cloned in suitable host using specialised vectors. The cloning resulted into amplification of each piece of DNA fragment so that it subsequently could be sequenced with ease.
• The commonly used hosts were bacteria and yeast, and the vectors were called as BAC (bacterial artificial chromosomes), and YAC (yeast artificial chromosomes). The fragments were sequenced using automated DNA sequencers that worked on the principle of a method developed by Frederick Sanger. Sanger is also credited for developing method for determination of amino acid’sequences in proteins. These sequences were then arranged based on some overlapping regions present in them.
• This required generation of overlapping fragments for sequencing. Alignment of these sequences was humanly not possible. Therefore, specialised computer based programs were developed. These sequences were subsequently annotated and were assigned to each chromosome. Another challenging task was assigning the genetic and physical maps on the genome. This was’-generated using information on polymorphism of restriction endonuclease recognition sites, and some repetitive DNA sequences known as microsatellites.

8. List the various markers that are used in DNA finger printing.
Answer. Different DNA marker systems such as Restriction Fragment Length Polymorphisms (RFLPs), Random Amplified Polymorphic DNAs (RAPDs), Amplified Fragment Length Polymorphisms (AFLPs), Simple Sequence Repeats (SSRs) which also called as microsatellites, Single Nucleotide Polymorphisms (SNPs) and others have been developed.

9. Replication was allowed to take place in the presence of radioactive deoxynucleotides precursors in E. coli that was a mutant for DNA ligase. Newly synthesised radioactive DNA was purified and strands were separated by denaturation. These were centrifuged using density gradient centrifugation. Which of the following would be a correct result?

Answer. (d)

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NCERT Exemplar Problems Class 12 Biology Chaper 7 Evolution

Multiple Choice Questions
Single Correct Answer Type
1. Which of the following is used as an atmospheric pollution indicator?
(a) Lepidoptera (b) Lichens
(c) Lycopersicon (d) Lycopodium
Answer. (b) Lichens are sensitive to SO₂ pollution. Lichens are very good pollution indicators, they do not grow in polluted areas.

2. The theory of spontaneous generation stated that:
(a) Life arose from living forms only
(b) Life can arise from both living and non-living
(c) Life can arise from non-living things only.
(d) Life arises spontaneously, neither from living nor from the non-living.
Answer. (c)

• For a long time it was also believed that life came out of decaying and rotting matter like straw, mud, etc.
• According to theory of abiogenesis, life originates from non-living.

3. Animal husbandry and plant breeding programmes are the examples of
(a) Reverse evolution (b) Artificial selection
(c) Mutation (d) Natural selection
Answer. (b) Animal husbandry and plant breeding programmes are the examples of artificial selection.

4. Paleontological evidences for evolution refer to the
(a) Development of embryo (b) Homologous organs
(c) Fossils (d) Analogous organs
Answer. (c) Palaeritological evidences for evolution refer to the fossils. Fossils provide direct and solid evidence in favour of organic evolution through ages. Fossils are studied for knowing about extinct organisms. Birbal Sahni Institute of Palaeobotany is situated at Lucknow. Birbal Sahni is called ‘Father of Indian Palaeobotany’. Age of fossils is determined by Uranium-Lead method (U²³⁸) or Potassium-Argon method or Radioactive Carbon dating (C¹⁴). ESR (electron-spin-resonance method) is the most accurate method of dating of fossils.

5. The bones of forelimbs of whale, bat, cheetah and man are similar in structure, because
(a) One organism has given rise to another
(b) They share a common ancestor
(c) They perform the same function
(d) The have biochemical similarities
Answer. (b) Organs having same origin but different functions are called homologous organs. For example, whales, bats, Cheetah and human (all mammals) share similarities in the pattern of bones of forelimbs. Homology indicates common ancestry.

6. Analogous organs arise due to
(a) Divergent evolution (b) Artificial selection
(c) Genetic drift (d) Convergent evolution
Answer. (d) Organs having different origin but similar function are called analogous. Wings of butterfly and of birds look alike. They are not anatomically similar structures though they perform similar functions. Hence, analogous structures are a result of convergent evolution (different structures evolving for the same function and hence having similarity).

7. (p + q)²= p² + 2pq + q² = 1 represents an equation used in
(a) Population genetics (b) Mendelian genetics
(c) Biometrics (d) Molecular genetics
Answer. (a) (p + q)²= p²+ 2pq + q² = 1 represents an equation used in population genetics.
Above equation is known as Hardy-Weinberg principle in which
p²—Dominant homozygous
2pq — Heterozygous
q²— Recessive homozygous

8. Appearance of antibiotic-resistant bacteria is an example of
(a) Adaptive radiation
(b) Transduction
(c) Pre-existing variation in the population
(d) Divergent evolution
Answer. (c) Appearance of antibiotic-resistant bacteria is an example of pre-existing variation in the population.

9. Evolution of life shows that life forms had a trend of moving from
(a) Land to water (b) Dryland to wet land
(c) Freshwater to sea water (d) Water to land
Answer. (d) Evolution of life shows that life forms had a trend of moving from water to land.

10. Viviparity is considered to be more evolved because
(a) The young ones are left on their own
(b) The young ones are protected by a thick shell
(c) The young ones are protected inside the mother’s body and are looked after they are bom leading to more chances of survival
(d) The embryo takes a long time to develop
Answer. (c) Viviparity is considered to be more evolved because the young ones are protected inside the mother’s body and are looked after they are bom leading to more chances of survival.

11. Fossils are generally found in
(a) Sedimentary rocks (b) Igneous rocks
(c) Metamorphic rocks (d) Any type of rock
Answer. (a) Fossils are generally found in sedimentary rocks.

12. For the MN-blood group system, the frequencies of M and N alleles are 0.7 and 0.3, respectively. The expected frequency of MN-blood group bearing organisms is likely to be (a) 42% (b) 49% (c) 9% (d) 58%
Answer.

13. The most accepted line of descent in human evolution is
(a) Australopithecus —> Ramapithecus —> Homo sapiens —> Homo habilis
(b) Homo erectus —>Homo habilis —> Homo sapiens
(c) Ramapithecus —> Homo habilis —>Homo erectus —>Homo sapiens
(d) Australopithecus —> Ramapithecus —> Homo erectus —>Homo habilis —> Homo sapiens
Answer. (c) Ramapithecus —> Homo habilis —> Homo erectus —>Homo sapiens

14. Which of the following is an example for link species?
(a) Lobe fish (b) Dodo bird
(c) Seaweed (d) Chimpanzee
Answer. (d) Chimpanzee is an example for link species.

15. Which type of selection is industrial melanism observed in moth, Bistonbitularia?
(a) Stabilising (b) Directional
(c) Disruptive (d) Artificial
Answer. (b) Directional type of selection is industrial melanism observed in moth, Bistonbitularia. (See figure on next page)

16. Match the scientists listed under column ‘A’ with ideas listed column ‘B’

(a) i—M; ii—P; iii—N; iv—O (b) i—P; ii—M; iii—N; iv—O
(c) i—N; ii—P; iii—O; iv—M (d) i—p; ii—O; iii—N; iv—M
Answer. (b)

17. In 1953 S. L. Miller created primitive earth conditions in the laboratory and gave experimental evidence for origin of first form of life from pre-existing non-living organic molecules. The primitive earth conditions created include
(a) Low temperature, volcanic storms, atmosphere rich in oxygen
(b) Low temperature, volcanic storms, reducing atmosphere
(c) High temperature, volcanic storms, non-reducing atmosphere
(d) High temperature, volcanic storms, reducing atmosphere containing CH₄ , NH₃ etc.
Answer. (d) In 1953 S. L. Miller created primitive earth conditions in the laboratory and gave experimental evidence for origin of first form of life from pre-existing non-living organic molecules. The primitive earth conditions created include high temperature, volcanic storms, reducing atmosphere containing CH₄ , NH₃ etc.

18. Variations during mutations of meiotic recombinations are
(a) Random and direction less (b) Random and directional
(c) Random and small (d) Random, small and directional.
Answer. (a) Variations during mutations of meiotic recombinations are random and direction less.

Very Short Answer Type Questions
1. What were the characteristics of life forms that had been fossilised?
Answer. Fossils are remains of hard parts (like bones, teeth, etc.) of life-forms found in rocks.

2. Did aquatic life forms get fossilised? If, yes where do we come across such fossils?
Answer. Yes, aquatic life forms get fossilised in the sediments of the water bodies. Later, sediments form the part of sedimentary rocks in which fossils, are deposited.

3. What are we referring to? When we say ‘simple organisms’ or ‘complex organisms’.
Answer. When we say simple or complex organisms we are talking in terms of evolutionary history of an Organism. A ‘simple organism’ is considered to be primitive and has simple thallus organisation; The level of complexity of metabolism is also low. On the other hand a ‘complex organism’ refers to a more evolved form forming higher levels of structural and functional complexities. They are believed to have arisen from simple organisms.

4. How do we compute the age of a living tree?
Answer. Age of the living tree is calculated by counting the number of annual rings or by the radioactive carbon dating,

5. Give an example for convergent evolution and identify the features towards which they are converging.
Answer. Presence of wings in birds and butterfly is an example of convergent evolution. They are adapted for flying (volant mode).

6. How do we compute the age of a fossil?
Answer. To compute the age of a fossil, we use radiocarbon dating.

7. What is the most important pre-condition for adaptive radiation?
Answer. Pre-existing variations are the most important pre-condition for adaptive radiation.

8. How do we compute the age of.a rock?
Answer. Age of rock is computed by radiocarbon dating, potassium-argon dating, uranium-lead dating and rubidium-strontium dating method.

9. When we talk of functional macro-molecules (e.g., proteins as enzymes, hormones, receptors, antibodies etc.), towards what are they evolving?
Answer. Similarities in proteins and genes performing a given function among diverse organisms give clues to common ancestry. These biochemical similarities point to the same shared ancestry as structural similarities among diverse organisms. Trypsin (ancient enzyme) is present from protozoa to mammals.

10. In a certain population, the frequency of three genotypes is as follows:
Genotypes: BB, Bbbb
Frequency: 22%, 62%, 16%
What is the likely frequency of B and b alleles?
Answer. Frequency of B allele = all of BB + 1/2 of Bb = 22 + 31 = 53%
Frequency of b allele = all of bb + 1/2 of Bb = 16 + 31 = 47%

11. Among the five factors that are known to affect Hardy-Weinberg equilibrium, ‘ three factors are gene flow, genetic drift and genetic recombination. What are the other two factors?
Answer. Natural selection and mutation.

12. What is founder effect?
Answer. Sometimes the change in allele frequency is so different in the new sample of population that they become a different species. Small group of population called founders left their habitat and goes into new habitat. In new habitat this population (founders) shows different genotype frequency from that of the original population and leads to variation. The original drifted population becomes founders and the*effect is called founder effect.

13. Who among the Dryopithecus and Ramapithecus was more man-like?
Answer. Ramopithecus was more man-like wille Dryopithecus was more a ape-like

14. By what Latin name the first hominid was known?
Answer. Homo habilis

15. Among Ramapithecus, Australopithecus and Homo habilis, who probably did not eat meat?
Answer. Homo habilis

Short Answer Type Questions
1. Louis Pasteur’s experiments, if you recall, proved that life can arise from only pre-existing life. Can we correct this as life evolves from pre-existent life or otherwise we will never answer the question as to how the first forms of life arose? Comment.
Answer. We can correct the statement of Louis Pasteur’s because Oparin of Russia and Haldane of England proposed that the first form of life could have come
from pre-existing non-living organic molecules (e.g., RNA, protein, etc.) and that formation of life was preceded by chemical evolution, i.e., formation of diverse organic molecules from inorganic constituents.

2. The scientists believe that evolution is gradual. But extinction, part of evolutionary story, are ‘sudden’ and ‘abrupt’ and also group-specific. Comment whether a natural disaster can be the cause for extinction of species.
Answer. Natural disaster like earth quake can be the cause for extinction of species. During the long period since the origin and diversification of life on earth there were five episodes of mass extinction of species.

3. Why is nascent oxygen supported to be toxic to aerobic life forms?
Answer. Nascent oxygen is highly reactive. It can react readily with different kinds of molecules, including DNA, proteins present in the cells of aerobic life forms. This may lead to mutations and undesirable metabolic changes.

4. While creation and presence of variation is direction less, natural selection is directional as it is in the context of adaptation. Comment.
Answer. Creation and variation occur in a sexually reproducing population as a result of crossing over during meiosis and random fusion of gametes. It is however the organisms that are selected over a period of time which are determined by the environmental conditions. In other words, the environment provides the direction with respect to adaptations so that the organisms are more and more fit in terms of survival.

5. The evolutionary story of moths in England during industrialisation reveals, that ‘evolution is apparently reversible’. Clarify this statement.
Answer. In a collection of moths made in 1850s, i.e., before industrialisation set in, it was observed that there were more white-winged peppered moths (Biston betularia) on trees than dark-winged or melanised moths (Biston carbonaria). However, in the collection carried out from the same area, but after industrialisation, i.e., in 1920, there were more dark-winged moths in the same area, i.e., the proportion was reversed.

• Before industrialisation set in, thick growth of almost white-coloured lichen covered the trees—in that background the white winged moth survived but the dark-coloured moth were picked out by predators. Lichens can be used as industrial pollution indicators. They will not grow in areas that are polluted. During post-industrialisation period, the tree trunks became dark due to industrial smoke and soots.
• Under this condition the white-winged moth did not survive due to predators, dark-winged or melanised moth survived. Hence, moths that were able to camouflage themselves, i.e., hide in the background, survived. This understanding is supported by the fact that in areas where industrialisation did not occur, e.g., in rural areas, the count of melanic moths was low. This showed that in a mixed population, those that can better-adapt, survive and increase in population size.

6. Comment on the statement that “evolution and natural selection are end result or consequence of some other processes but themselves are not processes”.
Answer. The world we see, inanimate .and animate, is only the success stories of evolution. When we describe the story of this world we describe evolution as a process. On the other hand when we describe the story of life on earth, we treat evolution as a consequence of a process called natural selection. We are still not very clear whether to regard evolution and natural selection as processes or end result of unknown processes.

7. State and explain any three factors affecting allele frequency in populations.
Answer. (i) Gene migration or gene flow: When migration of a section of population to another place and population occurs, gene frequencies change in the original as well as in the new population. New genes/alleles are added to the new population and these are lost from the old population. There would be a gene flow if this gene migrationhappens multiple times.
(ii) Genetic drift: If the same change occurs by chance, it is called genetic drift. Sometimes the change in allele frequency is so different in the new sample of population that they become a different species. The original drifted population becomes founders and the effect is called founder effect.
(iii) Mutation: Microbial experiments show that pre-existing advantageous mutations when selected will result in observation of new phenotypes. Over few generations, this would result in speciation. Natural selection is a process in which heritable variations enabling better survival are enabled to reproduce and leave greater number of progeny.

8. Gene flow occurs through generations. Gene flow can occur across language barriers in humans. If we have a technique of measuring specific allele frequencies in different population of the world, can we not predict human migratory patterns in pre-history and history? Do you agree or disagree? Provide explanation to your answer.
Answer. Yes, I agree. Gene flow occurs through generations. By studying specific allele frequencies, we can predict the human migratory patterns in prehistory and history. Studies have used specific genes/chromosomes/mitochondrial DNA to trace the evolutionary history and migratory patterns of humans. (The project is known as the Human Genographics Project).

9. How do you express the meaning of words like race, breed, cultivars or variety?
Answer.

•  Race is a group of people who share similar and distinct physical characteristics.
• A group of animals related by descent and similar in most characters like general appearance, features, size, configuration, etc., are said to belong to a breed.
• A cultivar is a plant or grouping of plants selected for desirable characteristics that can be maintained by propagation.
• A taxonomic category that ranks below species, its members differing from others of the same species in minor but heritable characteristics is called variety.

10. When we say “survival of the fittest”, does it mean that
a. those which are fit only survive, or
b. those that survive are called fit? Comment.
Answer. Those individuals which survive and reproduce in their respective environment are called fit.

11. Enumerate three most characteristic criteria for designating a Mendelian population.
Answer. Population must be sufficiently large with potentialities for free flow of genetic material among individuals (through sexual reproduction). Migration should either be nil or negligible.

12. “Migration may enhance or blur the effects of selection”. Comment.
Answer. Migration may cause enrichment of the gene pool of such alleles that are being selected for, or blur the effects of selection through replenishment of alleles that were selected against by nature.

Long Answer Type Questions
1. Name the law that states that the sum of allelic frequencies in a population remains constant. What are the five factors that influence these values?
Answer. Hardy-Weinberg principle states that the sum of allelic frequencies in a population remains constant. Five factors are known to affect Hardy-Weinberg equilibrium. These are gene migration or gene flow, genetic drift, mutation, genetic recombination and natural selection.

2. Explain divergent evolution in detail. What is the driving force behind it?
Answer. Whales, bats, Cheetah and human (all mammals) share similarities in the
pattern of bones of forelimbs. Though these forelimbs perform different functions in these animals, they have similar anatomical structure. All of them have humerus, radius, ulna, carpals, metacarpals and phalanges in their forelimbs.
Hence, in these animals, the same structure developed along different directions due to adaptations to different needs. This is divergent evolution and these structures are homologous. Homology indicates common ancestry. Driving force behind the divergent evolution is adaptation in different environments.

3. You have studied the story of Pepper moths in England. Had the industries been removed, what impact could it have on the moth population? Discuss.
Answer. In the population of Peppermoth, two variants were already existing in the population, the black and the grey. In the absence of industrialisation the grey moths were prevalent because they blended very well with the lichen and moss covered trees camouflage and the predators cannot spot them. The black ones were easily spotted and killed by predators and therefore were fewer in numbers. With industrialisation the stems got covered with black
soot. This provided better camouflage tb the black variant and their number increased. If the industries had been removed the population of black moths would have declined because as stated earlier they would have been spotted better by predators and therefore be eaten more frequently.

4. What are the key concepts in the evolution theory of Darwin?
Answer. Branching descent and natural selection are the two key concepts of Darwinian Theory of Evolution.

• The novelty and brilliant insight of Darwin was this: He asserted that variations, which are heritable and which make resource utilisation better for few (adapted to habitat better) will enable only those to reproduce and leave more progeny. Hence for a period of time, over many generations, survivors will leave more progeny and there would be a change in population characteristic and hence new forms appear to arise.
• The fitness, according to Darwin, refers ultimately and only to reproductive fitness. Hence, those who are better fit in an environment, leave more progeny than others. These, therefore, will survive more and hence are selected by nature. He called it natural selection and implied it as a mechanism of evolution.

5. Two organisms occupying a particular geographical area (say desert) show similar adaptive strategies. Taking examples, describe the phenomenon.
Answer. One can say that it is the similar habitat that has resulted in selection of similar adaptive features in different groups of organisms but toward the same function. Spins of Opuntia and cactus are modifications of leaves to prevent the loss of water in desert.

6. We are told that evolution is a continuing phenomenon for all living things. Are humans also evolving? Justify your answer.
Answer. Yes, human is also evolving. Fossils give the evidences that evolution is a continuous phenomenon. Our ancestors like Ramapithecus, Australopithecus, Homo habilis, Homo erectus, Neanderthal man and Cro-magnon man continuously evolving and modem man (Homo sapiens sapiens) arises from certain modifications.

7. Had Darwin been aware of Mendel’s work, would he been able to explain the origin of variations. Discuss.
Answer. Yes, Darwin has been aware of Mendel’s work. Even though Mendel had talked of inheritable ‘factors’ influencing phenotype, Darwin either ignored these observations or kept silence.
Darwin would have been able to explain the origin of variations. He asserted that variations, which are heritable and which make resource utilisation better for few (adapted to habitat better) will enable only those to reproduce and leave more progeny. Hence for a period of time, over many generations, survivors will leave more progeny and there would be a change in population characteristic and hence new forms appear to arise.

The post NCERT Exemplar Problems Class 12 Biology Evolution appeared first on Learn CBSE.

NCERT Exemplar Problems Class 12 Biology Chapter 8 Human Health and Diseases

Multiple Choice Questions
Single Correct Answer Type
1. The term ‘Health’ is defined in many ways. The most accurate definition of the health would be
(a) Health is the state of body and mind in a balanced condition
(b) Health is the reflection of a smiling face
(c) Health is a state of complete physical, mental and social well-being
(d) Health is the symbol of economic prosperity.
Answer. (c) Health is not just the absence of disease. It is a state of complete physical, mental, social and psychological well-being.

2. The organisms which cause diseases in plants and animals are called
(a) Pathogens (b) Vectors (c) Insects (d) Worms
Answer. (a) A wide range of organisms belonging to bacteria, viruses, fungi protozoans, helminthes, etc., could cause diseases in man. Such disease causing organisms are called pathogens.

3. The chemical test that is used for diagnosis of typhoid is
(a) ELISA-Test (b) ESR-Test
(c) PCR-Test (d) Widal-Test
Answer. (d) The chemical test that is used for diagnosis of typhoid is Widal-Test.

4. Diseases are broadly grouped into infectious and non-infectious diseases. In the list given below, identify the infectious diseases.
i. Cancer ii. Influenza
iii. Allergy iv. Small pox
(a) i and ii (b) ii and iii –
(c) iii and iv (d) ii and iv
Answer. (d)

5. The sporozoites that cause infection, when a female Anopheles mosquito bites a person being are formed in
(a) Liver of person (b) RBCs of mosquito .
(c) Salivary glands of mosquito (d) Intestine of mosquito
Answer. (d) The sporozoites that cause infection, when a female Anopheles mosquito bites a person being are formed in intestine of mosquito.

6. The disease chikunguniya is transmitted by
(a) House flies (b) Aedes mosquitoes
(c) Cockroach (d) Female Anopheles
Answer. (b) Dengue and chikunguniya are transmitted by Aedes mosquitoes.

7. Many diseases can’be diagnosed by observing the symptoms in the patient. Which group of symptoms are indicative of pneumonia?
(a) Difficulty in respiration, fever, chills, cough, headache
(b) Constipation, abdominal pain, cramps, blood clots
(c) Nasal congestion and discharge, cough, sorethroat, headache
(d) High fever, weakness, stomach pain, loss of appetite and constipation.
Answer. (a)

• Difficulty in respiration, fever, chills, cough, headache: Pneumonia
• Constipation, abdominal pain, cramps, blood clots: Amoebiasis
• Nasal congestion and discharge, cough, sorethroat, headache: Common cold
• High fever, weakness, stomach pain, loss of appetite and constipation: Typhoid

8. The genes causing cancer are
(a) Structural genes (b) Expressor genes
(c) Oncogenes ’ (d) Regulatory genes
Answer. (c) Oncogenes are the cancer causing genes.

9. In malignant tumors, the cells proliferate, grow rapidly and move to other . parts of the body to form new tumors. This stage of disease is called
(a) Metagenesis (b) Metastasis
(c) Teratogenesis (d) Mitosis
Answer. (b) In malignant tumors, the cells proliferate, grow rapidly and move to other parts of the body to form new tumors. This stage of disease is called metastasis.

10. When an apparently healthy person is diagnosed as unhealthy by a psychiatrist, the reason could be that
(a) The patient was not efficient at his work
(b) The patient was not economically prosperous
(c) The patient shows behavioural and social maladjustment
(d) He does not take interest in sports
Answer. (c) When an apparently healthy person- is diagnosed as unhealthy by a psychiatrist, the reason could be that the patient shows behavioural and social maladjustment.

11. Which of the following are the reason(s) for Rheumatoid arthritis? Choose the correct option.
i. The ability to differentiate pathogens or foreign molecules from self cells increases
ii. Body attacks self cells
iii. More afitibodies are produced in the body
iv. The ability to differentiate pathogens or foreign molecules from self cells is lost
(a) i and ii (b) ii and iv (c) iii and iv (d) i and iii
Answer. (b) Rheumatoid arthritis is an autoimmune diseases in which

• Body attacks self cells
• The ability to differentiate pathogens or foreign molecules from self cells is lost

12. AIDS is caused by HIV. Among the following, which one is not a mode of transmission of HIV?
(a) Transfusion of contaminated blood
(b) Sharing the infected needles
(c) Shaking hands with infected persons
(d) Sexual contact with infected persons
Answer. (c) Mode of transmission of HIV:

• Transfusion of contaminated blood
• Sharing the infected needles
• Sexual contact with infected persons

13. ‘Smack’ is a drug obtained fipm the
(a) Latex of Papaver somniferum (b) Leaves of Cannabis sativa
(c) Flowers of Datura (d) Fruits of Erythroxylum coca
Answer. (a)

• ‘ Smack’ is a drug obtained from the latex of Papaver somniferum.
• Smack are brown sugar, the common name of Heroin.
• Heroin is obtained by acetylation of morphine.

14. The substance produced by a cell in viral infection that can protect other cells from further infection is N
(a) Serotonin (b) Colostrum
(c) Interferon (d) Histamine
Answer. (c) The substance produced by a cell in viral infection that can protect other cells from further infection is interferon.

15. Transplantation-of tissues/organs to save certain patients often fails due to rejection of such tissues/organs by the patient. Which type of immune response is responsible for such rejections?
(a) Auto-immune response (b) Humoral immune response
(c) Physiological immune response (d) Cell-mediated,immune response
Answer. (d) Transplantation of tissues/organs to-save certain patients often fails due to rejection of such tissues/organs by the patient. Cell-mediated immune response is responsible for such rejections.

16. Antibodies present in colostrum which protect the new bom from certain diseases is of
(a) Ig G type (b) Ig A type (c) Ig D type (d) Ig E type
Answer. (b) Antibodies present in colostrum which protect the new bom from certain diseases is of Ig A type.

17. Tobacco consumption is known to stimulate secretion of adrenaline and nor¬adrenaline. The component causing this could be
(a) Nicotine (b) Tannic acid (c) Curaimin (d) Catechin
Answer. (a) Tobacco consumption is known to stimulate secretion of adrenaline and nor-adrenaline. The component causing this could be nicotine. Morphine, cocaine, codeine and nicotine are all alkaloids.

18. Anti-venom against’snake poison contains
(a) Antigens (b) Antigen-antibody complexes
(c) Antibodies (d) Enzymes
Answer. (c) Anti-venom against snake poison contains antibodies.

19. Which of the following is not a lymphoid tissue?
(a) Spleen (b) Tonsils (c) Pancreas (d) Thymus
Answer. (c) Spleen, Tonsils and Thymus are lymphoid tissue while pancreas is mixed gland.

20. Which of the following glands is large sized at birth but reduces in size with ageing?
(a) Pineal (b) Pituitary (c) Thymus (d) Thyroid
Answer. (c) Thymus glands is large sized at birth but reduces in size with ageing.

21. Haemozoin is a
(a) Precursor of hemoglobin
(b) Toxin released from Streptococcus infected cells
(c) Toxin released from Plasmodium infected cells
(d) Toxin released from Haemophilus infected cells ’
Answer. (c) Haemozoin is a toxin released from Plasmodium infected cells.

22. One Of the following is not the causal organism for ringworm.
(a) Microsporum (b) Trichophyton
(c) Epidermophyton (d) Macrosporum
Answer. (d) Microsporum, Trichophyton and Epidermophytonare the causal organism for ringworm while Macrosporum is a brown alga.

23. A person with sickle cell anemia is
(a) More prone to malaria (b) More prone to typhoid
(c) Less prone to malaria (d) Less prone to typhoid.
Answer. (c) A person with sickle cell anemia is less prone to malaria.

Very Short Answer Type Questions
1. Certain pathogens are tissue/organ specific. Justify the statement with suitable examples.
Answer. Salmonella typhi causes typhoid and infects small intestine while intestinal endoparasite causes amoebic dysentery and infects large intestine.

2.The immune system of a person is suppressed. In the ELISA test, he was found positive to a pathogen.
a. Name the disease the patient is suffering from.
b. What is the causative organism?
c. Which cells of body are affected by the pathogen?
Answer. a. Acquired Immuno Deficiency Syndrome (AIDS)
b. Human Immuno deficiency virus (HIV)
c. Helper T-lymphocytes (T_H or T₄)

3. Where are B-cells and T-cells formed? How do they differ from each other?
Answer. Both B-cells and T-cells are formed in bone marrow. B-cells matures in bone marrow while T-cells matures in thymus. B-cells provides humoral immunity and T-cells provides cell mediated immunity (CMI).

4. Given below are the pairs of pathogens and the diseases caused by them. Which out of these is not a matching pair and why?
(a) Virus – common cold (b) Salmonella – typhoid
(c) Microsporum – filariasis (d) Plasmodium – malaria
Answer. Pair is mismatched. Microsporum causes ringworm disease.

5. What would happen to immune system, if thymus gland is removed from the body of a person?
Answer. Thymus is the primary lymphoid organ. In thymus gland, immature lymphocytes differentiate into antigen-sensitive lymphocytes. If thymus gland is removed from the body of a person, his immune system becomes weak. As a result the person’s body becomes prone to infectious diseases,

6. Many microbial pathogens enter the gut of humans along with food. What are the preventive barriers to protect the body from such pathogens? What type of immunity do you observe in this case?
Answer. (i) The mucus coating of the epithelium lining of the gut helps in trapping microbes entering the body.
(ii) Saliva in the mouth and hydrochloric acid in gastric juice secreted by stomach prevent microbial growth. This type of immunity is innate immunity.

7. Why is mother’s milk considered the most appropriate food for a new born infant?
Answer. Mother’s milk is considered as the most appropriate food for a new born infant because the yellowish fluid colostrum secreted by mother during the initial days of lactation has abundant antibodies (IgA) to protect the infant.

8. What are interferons? How do interferons check infection of new cells?
Answer. Interferons are natural proteins produced by the cells of immune system in
response to foreign agents such as viruses, tumor cells and parasites and protect non-infected cells from further infection. Interferons inhibit the viral replication within host cells, activate natural killer cells and macrophages, increases antigen presentation to lymphocytes, and induce the resistance of host cells to viral infection. When the antigen is presented to matching T-cells’ and B-cells, these cells multiply and remove the foreign substance.

9. In the figure, structure of an antibody molecule is shown. Name the parts A, B and C. Show A, B and C in the diagram.

Answer. A:—Constant region of heavy chain,
B—Constant region of light chain,
C—Variable region of light and heavy chain

10. If a regular dose of drug or alcohol is not provided to an addicted person, he shows some withdrawal symptoms. List any four such withdrawal symptoms.
Answer. The withdrawal symptoms are:
a. Anxiety b. Shakiness c. Nausea d. Sweating

11. Why is it that during changing weather, one is advised to avoid closed, crowded and air-conditioned places like cinema halls etc.?
Answer. During changing weather, one is advised to avoid closed, crowded and air- conditioned placed like cinema halls, etc., because during this period the infectious agents are more humerous and prevalent to which we are more vulnerable.

12. The harmful allele of sickle cell anemia has not been eliminated from human population. Such afflicted people derive some other benefits. Discuss.
Answer. The harmful alleles get eliminated from population over a period of time, yet sickle cell anaemia is persisting in human population because SCA is a harmful condition which is also a potential saviour from malaria.
Those with the benign sickle trait possess a resistance to malarial infection. The pathogen that causes the disease spends part of its cycle in the red blood cells and triggers an abnormal drop in oxygen levels in the cell. In carriers, this drop is sufficient to trigger the full sickle-cell reaction, which leads to infected cells being rapidly removed from circulation and strongly limiting the infection’s progress. These individuals have a great resistance to infection and have a greater chance of surviving outbreaks. This resistance to infection is the main reason the SCA allele and SCA disease still exist. It is found in greatest frequency in populations where malaria was and is still often a serious problem.

13. Lymph nodes are secondary lymphoid organs. Explain the role of lymph nodes in our immune response.
Answer. Lymph nodes are small solid structures located at different points along the lymphatic system. Lymph nodes trap the microorganisms or other antigens, which happen to get into the lymph and tissue fluid. Antigens trapped in the lymph nodes are responsible for the activation of lymphocytes present there and cause the immune response.

14. Why is an antibody molecule represented as H₂ L₂ ?
Answer. Each antibody molecule is made of the two heavy chains (H₂ ) and two light chains (L₂ ), hence represented as H₂ L₂ .

15. What does the term ‘memory’ of the Immune system mean?
Answer. When body encounters a pathogen for the first time produce a response called 1° response. 1° response is of low intensity. When body encounters the same pathogen subsequently then body elicits 2° response. 2° response is highly intensified. This is due to the fact that our body have memory of the first encounter.

16. If a patient is advised Anti Retroviral Therapy, which infection is he suffering from? Name the causative organism.
Answer. The patient is suffering from AIDS. The causative organism for AIDS is HIV (Human Immuno deficiency Virus).

Short Answer Type Questions
1. Differentiate between active immunity and passive immunity.
Answer.

2. Differentiate between benign tumor and malignant tumor.
Answer.

• Benign tumors normally remain confined to their original location and do not spread to other parts of the body and cause little damage.
• Malignant tumors are a mass of proliferating cells called neoplastic or tumor cells. Neoplastic cells grow very rapidly, invading and damaging the surrounding normal tissues. As these cells actively divide and grow they also starve the normal cells by competing for vital nutrients.
• Cells sloughed from such tumors reach distant sites through blood, and wherever they get lodged in the body, they start a new tumor there. This property called metastasis is the most feared property of malignant tumors.

3. Do you consider passive smoking is more dangerous than active smoking? Why?
Answer. Yes, passive smoking is also dangerous as the active smoking because the person is exposed to the same harmful effects of smoking like emphysema, bronchitis, lung cancer, urinary bladder cancer or even peptic ulcer.

4. “Prevention is better than cure”. Comment.
Answer. Prevention is better than cure is true as in same cases the disease is non- curable like AIDS and Hepatitis-B, and in some cases’ the treatment causes financial problems in the family.

5. Explain any three preventive measures to control microbial infections.
Answer. (i) Maintenance of personal and public hygiene is very important for
prevention and control of many infectious diseases. Measures for personal hygiene include keeping the body clean; consumption of clean drinking water, food, vegetables, fruits, etc. Public hygiene includes proper disposal of waste and excreta; periodic cleaning and disinfection of water reservoirs, pools, cesspools and tanks and observing standard practices of hygiene in public catering.
(ii) In cases of air-borne diseases such as pneumonia and common cold, in addition to the above measures, close contact with the infected persons or their belongings should be avoided.
(iii) For diseases such as malaria and filariasis that are transmitted through insect vectors, the most important measure is to control or eliminate the vectors and their breeding places. This can be achieved by avoiding stagnation of water in and around residential areas, regular cleaning of household coolers, use of mosquito nets, introducing fishes like Gambusia in ponds that feed on mosquito larvae, spraying of insecticides in ditches, drainage areas and swamps, etc. In addition, doors and windows should be provided with wire mesh to prevent the entry of mosquitoes.

6. In the given flow diagram, the*replication of retrovirus in a host is shown. Observe and answer the following questions.
(a) Fill in (1) and (2).
(b) Why is the virus called retrovirus?
(c) Can the infected cell survive while viruses are being replicated and released?

Answer. (a) 1. Viral DNA is produced by reverse transcriptase.
2. New viral RNA is produced by infected cell.
(b) HIV is called retrovirus because it forms DNA from RNA by reverse transcription.
(c) Yes, infected cell can survive while viruses are being replicated and released.

7. “Maintenance of personal and public hygiene is necessary for prevention and control of many infectious diseases”. Justify’ the statement giving suitable examples.
Answer. Measures for personal hygiene include keeping the body clean; consumption of clean drinking water, food, vegetables, fruits, etc. Public hygiene includes proper disposal of waste and excreta; periodic cleaning and disinfection of water reservoirs, pools, cesspools and tanks, and observing standard practices of hygiene in public catering. These measures are particularly essential where the infectious agents are transmitted through food and water such as typhoid, amoebiasis and ascariasis.

8. The following table shows certain diseases, their causative organisms and symptoms. Fill the gaps.

Answer.

9. The outline structure of a drug is given below.

(a) Which group of drugs does this represent?
(b) What are the modes of consumption of these drugs?
(c) Name the organ of the body which is affected by consumption of these drugs.
Answer. (a) Cannabinoids
(b) Generally taken by inhalation and oral ingestion
(c) Affect the cardiovascular system of the body

10. Give the full form of CT and MRI. How are they different from each other? Where are they used?
Answer. CT (computed tomography) and MRI (magnetic resonance imaging) are very useful to detect cancers of the internal organs. Computed tomography uses X-rays to generate a three-dimensional image of the internals of an object. MRI uses strong magnetic fields and non-ionising radiations to accurately detect pathological and physiological changes in the living tissue.

11. Many secondary metabolites of plants have medicinal properties. It is their misuse that creates problems. Justify the statement with an example.
Answer. Several plants, fruits and seeds having hallucinogenic properties and have been used for hundreds of years in folk-medicine, religious ceremonies and rituals all over the globe. When these are taken for a purpose other than medicinal use or in amounts/frequency that impairs one’s physical, physiological or psychological functions, it constitutes drug abuse.

12. Why cannabinoids are banned in sports and games?
Answer. As these days cannabinoids are being abused by some sports persons to increase their performance, that is why cannabinoids are banned in sports and games.

13. What is secondary metabolism?
Answer. Secondary metabolism is a term for pathways and small molecule products of metabolism that are not absolutely required for the survival of the organism. Examples of the products include antibiotics and pigments.

14. Drugs and alcohol give short-term ‘high’ and long-term ‘damages’. Discuss.
Answer. Curiosity, need for adventure and excitement, and experimentation, constitute
common causes, which motivate youngsters towards drug and alcohol use.
A child’s natural curiosity motivates him/her to experiment. This is complicated further by effects that might be perceived as benefits, of alcohol or drug use. Thus, the first use of drugs or alcohol may be out of curiosity or experimentation, but later the child starts using these to escape facing problems. Of late, stress, from pressures to excel in academics or examinations, has played a significant role in persuading the youngsters to try alcohol and drugs. The perception among youth that it is ‘cool’ or progressive to smoke, use drugs or alcohol, is also in a way a major cause for youth to start these habits. Television, movies, newspapers, internet also help to promote this perception. Other factors that have been seen to be associated with drug and alcohol abuse among adolescents are unstable or unsupportive family structures and peer pressure.

15. Diseases like dysentery, cholera, typhoid etc., are more common in overcrowded human settlements. Why?
Answer. Diseases like dysentery, cholera, typhoid etc., are more common in overcrowded human settlements because these are infectious diseases that can transmitted from one person to another. In overcrowded settlements there is more chances of transmission of disease from one person to other.

16. From which plant cannabinoids are obtained? Name any two cannabinoids. Which part of the body is effected by consuming these substances?
Answer. Cannabinoids are obtained from the inflorescence of the plant Cannabis sativa. Marijuana, hashish, charas, ganja are some of the cannabinoids. These chemicals interact with cannabinoid receptors of the body, mainly present in the brain. Cardiovascular system is affected adversely.

17. In the metropolitan cities of India, many children are suffering from allergy/ asthma. What are the main causes of this problem? Give some symptoms of allergic reactions.
Answer. Allergy is the exaggerated response of the immune system of certain antigens present in the environment. In metropolitan cities life style is responsible for lowering of immunity and sensitivity to allergens. More polluted environment increases the chances of allergy in children. Some symptoms of allergic reactions are sneezing, watery eyes, running nose and difficulty in breathing.

18. What is the basic principle of vaccination? How do vaccines prevent microbial infections? Name the organism from which hepatitis B vaccine is produced.
Answer. The principle of vaccination is based on the property of ’memory’ of the immune system. In vaccination, a preparation of antigenic proteins of pathogens or inactivated/live but weakened pathogens is introduced into the body. The antigens generate the primary immune response by producing antibodies. The vaccines also generate the memory B-cells and T-cells. When the vaccinated person is attacked by the same pathogens, the existing memory B-cells or T-cells recognise the antigen quickly and overwhelm the invaders with massive production of lymphocytes and antibodies. Hepatitis B vaccine is produced from yeast.

19. What is cancer? How is a cancer cell different from the normal cell? How do normal cells attain cancerous nature?
Answer. An abnormal and uncontrolled division of cells is termed as Cancer. The cancerous cells are different from the normal cells in the following ways.

In our body, the growth and differentiation of cells is highly controlled and regulated. The normal cells show a property called contact inhibition.
The surrounding cells inhibits uncontrolled growth and division of cells. The normal cells lose this property and become cancerous cells giving rise to masses of cells called tumors. Transformation of normal cells into cancerous cells is induced by some physical, chemical and biological agents (carcinogens).

20. A person shows strong unusual hypersensitive reactions when exposed to certain substances present in the air. Identify the condition. Name the cells responsible for such reactions. What precaution should be taken to avoid such reactions?
Answer. Allergy. Mast Cells are responsible for such reactions. To avoid such reactions following precautions must be taken:
(i) The use of drugs like antihistamine, adrenalin and steroids quickly reduce the symptoms of allergy.
(ii) Avoid contact with substances to which a person is hypersensitive.

21. For an organ transplant, it is an advantage to have an identical twin. Why?
Answer. Very often, when some human organs like heart, eye, liver, kidney fail to
function satisfactorily, transplantation is the only remedy to enable the patient to live a normal life. Then a search begins—to find a suitable donor.
Grafts from just any source—an animal, another primate, or any human beings cannot be made since the grafts would be rejected sooner or later. Tissue matching, blood group matching are essential before undertaking any graft/transplant and even after this the patient has to take immuno-suppresants all his/her life. The body is able to differentiate ‘self ’ and ‘nonself’ and the cell-mediated immune response is responsible for the graft rejection. In an identical twin there is no chance of rejection of transplanted organ, so it is advantageous.

22. What are lifestyle diseases? How. are they caused? Name any two such diseases.
Answer. Lifestyle diseases are defined as diseases linked with the way people live their life. This is commonly caused by alcohol, drug and smoking abuse as well as lack of physical activity and unhealthy eating. Diseases that impact on our lifestyle are heart disease, stroke and obesity.

23. If there are two pathogenic viruses, one with DNA and other with RNA, which would mutate faster? And why?
Answer. Both DNA and RNA are able to mutate. In fact, RNA being unstable, mutate at a faster rate. Consequently, viruses having RNA genome and having shorter life span mutate and evolve faster.

Long Answer Type Questions
1. Represent schematically the life cycle of a malarial parasite.
Answer.

2. Compare the life style of people living in the urban areas with those of rural areas and briefly describe how the life style affects their health.
Answer. Urban areas
The social environment: Urban environments are more likely to see higher rates of crime and violencfc. ‘
The physical environment: In densely populated urban areas, there is often a lack of facilities and outdoor areas for exercise. In addition, air quality is often lower in urban environments which can contribute to chronic diseases such as asthma. In the developing world, urban dwellers often live in large slums which lack basic sanitation and utilities such as water and electricity. Access to health and social service: Persons of lower socioeconomic status are more likely to live in urban areas and are more likely to lack health insurance. The high prevalence of individuals without health insurance or citizenship creates a greater burden on available systems.
Rural areas
The social environment: Rural dwellers have significantly poorer health status than urban elders. Also, rural residents smoke more, exercise less, have less nutritional diets.
The physical environment: Rural women especially less educated women, are more sedentary than urban women. While poor air quality and crime rates are likely to be less of an issue in rural areas, insufficiencies in the built environment make it difficult for rural residents to exercise and maintain healthy habits.
Access to health and social seryice: Evidence indicates that rural residents have limited access to health care. Some rural areas have a higher proportion of uninsured and individually insured residents than urban areas.

3. Why do some adolescents start taking drugs? How can this be avoided?
Answer. The reasons why adolescents and youngsters take to consumption of drugs are:
(i) Curiosity of child motivates him/her to experiment.
(ii) Need for adventure and excitement.
(iii) Peer group pressure
(iv) Desire to do more physical and mental work.
(v) To overcome frustration and depression, due to failure in examinations or in other activities.
(vi) Unstable or unsupportive family structures.
The following measures can be taken to avoid taking drugs:
(i) Avoid undue pressure on child to perform beyond his/her capability in studies, sports ox any other activities.
(ii) Education and counselling are very important to face problem of stress and failure in life.
(iii) Seeking help from parents, elders and peers. This would help the young to share their feelings and concern.
(iv) Looking for danger signs and taking appropriate measures to treat them.
(v) Seeking professional and medical help for de-addiction and rehabilitation.

4. In your locality, if a person is addicted to alcohol, what kind of behavioural changes do you observe in that person? Suggest measures to overcome the problem.
Answer. The immediate adverse effects of drugs and alcohol abuse are manifested in the form of reckless behaviour, vandalism and violence. Excessive doses of drugs may lead to coma and death due to respiratory failure, heart failure or cerebral hemorrhage. A Combination of drugs or their intake along with alcohol generally results in overdosing and even deaths. The most common warning signs of drug and alcohol abuse among youth include drop in academic performance, unexplained absence from school/college, lack of interest in personal hygiene, withdrawal, isolation, depression,, fatigue, aggressive and rebellious behaviour, deteriorating relationships with family and friends, loss of interest in hobbies, change in sleeping and eating habits, fluctuations in w’eight, appetite, etc. There may even be some far-reaching implications of drug/alcohol abuse. If an abuser is unable to get money to buy drugs/alcohol he/she may turn to stealing. The adverse effects are just not restricted to the person who is using drugs or alcohol. At times, a drug/ alcohol addict becomes the cause of mental and financial distress to his/her entire family and friends.
The age-old adage of‘prevention is better than cure’ holds true here also. It is also true that habits such as smoking, taking drug or alcohol are more likely to be taken up at a young age, more during adolescence. Hence, it is best to identify the situations that may push an adolescent towards use of drugs or alcohol, and to take remedial measures well in time. In this regard, the parents and the teachers have a special responsibility. Parenting that combines with high levels of nurturance and consistent discipline, has been associated with lowered risk of substance (alcohol/drugs/tobacco) abuse. Some of the measures mentioned here would be particularly useful for prevention and control of alcohol and drugs abuse among adolescents.

5. What are the methods of cancer detection? Describe the common approaches for treatment of cancer.
Answer. Cancer detection and diagnosis: Early detection of cancers, is essential as it allows the disease to be treated successfully in many cases. Cancer detection is based on biopsy and histopathological studies of the tissue and blood and bone marrow tests for increased cell counts in the case of leukemias. In biopsy, a piece of the suspected tissue cut into thin sections is stained and examined under microscope (histopathological studies) by a pathologist. Techniques like radiography (use of X-rays), CT (computed tomography) and MRI (magnetic resonance imaging) are very useful to detect cancers of the internal organs. Computed tomography uses X-rays to generate a three-dimensional image of the internals of an object. MRI uses strong magnetic fields and non-ionising radiations to accurately detect pathological and physiological changes in the living tissue.

• Antibodies against cancer-specific antigens are also used for detection of certain cancers. Techniques of molecular biology can be applied to detect genes in individuals with inherited susceptibility to certain cancers. Identification of such genes, which predispose an individual to certain cancers, may be very helpful in prevention of cancers. Such individuals may be advised to avoid exposure to particular carcinogens to which they are susceptible (e.g., tobacco smoke in case of lung cancer).
• Treatment of cancer: The common approaches for treatment of cancer are surgery, radiation therapy and immunotherapy. In radiotherapy, tumor cells are irradiated lethally, taking proper care of the normal tissues surrounding the-tumor mass. Several chemotherapeutic drugs are used to kill cancerous cells. Some of these are specific for particular tumors. Majority of drugs have side effects like hair loss; anemia, etc. Most cancers are treated by combination of surgery, radiotherapy and chemotherapy. Tumor cells have been shown to avoid detection and destruction by immune system. Therefore, the patients are given substances called biological response modifiers such as a-interferon which activates their immune system and helps in destroying the tumor.

6. Drugs like LSD, barbiturates, amphetamines, etc., are used as medicines to help patients with mental illness. However, excessive doses and abusive usage are harmful. Enumerate the major adverse effects of such drugs in humans.
Answer. Drugs like barbiturates, amphetamines, benzodiazepines, and other similar drugs, that are normally used as medicines to help patients cope with mental illnesses like depression and insomnia, are often abused. Morphine is a very effective sedative and painkiller, and is very useful in patients who have undergone surgery. Several plants, fruits and seeds having hallucinogenic properties have been used for hundreds of years in folk-medicine, religious
ceremonies and rituals all over the globe. When these are taken for a purpose other than medicinal use or in amounts/frequency that impairs one’s physical, physiological or psychological functions, it constitutes drug abuse. ,

7. What is Pulse Polio-Programme of Government of India? What is OPV? Why is it that India is yet to eradicate Polio?
Answer. Pulse Polio is an immunisation campaign established by the government of India to eliminate poliomyelitis (polio) in India by vaccinating all children under the age of five years against the polio virus. The project fights poliomyelitis through a large-scale pulse vaccination programme and monitoring for polio cases.

•  In 1995, following the Global Polio Eradication Initiative of the World Health Organization (1988), India launched Pulse Polio immunisation program with Universal Immunization Program which aimed at 100% coverage.
• The last reported cases of wild polio in India were in West Bengal and Gujarat on 13 January 2011. On 27 March 2014, the World Health Organization (WHO) declared India a polio free country, since no cases of wild polio had been reported in for three years.
•  Polio vaccines are the vaccines used to prevent poliomyelitis (polio).One type uses inactivated poliovirus and is given by injection (IPV), while the other type uses weakened poliovirus and is given by mouth (OPV). The World Health Organization recommends all children be vaccinated against polio. The two vaccines have eliminated polio from most of the world. The oral polio vaccine was developed by Albert Sabin and came into commercial use in 1961. _

8. What are recombinant DNA vaccines? Give two examples of such vaccines. Discuss their advantages.
Answer. A recombinant vaccine is a vaccine produced through recombinant DNA technology. This involves inserting the DNA encoding an antigen that stimulates an immune response into bacterial or mammalian cells.
Recombinant DNA technology has allowed the production of antigenic polypeptides of pathogen in bacteria or yeast. Vaccines produced using jthis approach allow large scale production and hence greater availability for immunisation, e.g., hepatitis B vaccine (Recombivax HB) produced from yeast. As of June 2015 one human DNA vaccine had been approved for human use, the single-dose Japanese encephalitis vaccine called IMOJEV, released in 2010 in Australia.
Advantages of recombinant DNA vaccines:
1. No risk for infection
2. Ease of development and production
3. Stability for storage and shipping
4. Cost-effectiveness
5. Expression and purification of recombinant proteins
6. Long-term persistence of immunogen
7. In vivo expression ensures protein more closely resembles normal eukaryotic structure, with accompanying post-translational modifications.

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NCERT Exemplar Problems Class 12 Biology Chapter 9 Strategies for Enhancement in Food Production

Multiple Choice Questions
Single Correct Answer Type
1. The chances of contacting bird flu from a properly cooked (above 100°C) chicken and egg are
(a) Very high (b) High
(c) Moderate (d) None
Answer. (d) Properly cooked (above 100°C) chicken and egg have no chances of contacting bird flu.

2. A group of animals which are related by descent and share many similarities are referred to as
(a) Breed (b) Race
(c) Variety (d) Species
Answer. (a) A group of animals which are related by descent and share many similarities are referred to as breed.

3. Inbreeding is carried out in animal husbandry because it
(a) Increases vigour (b) Improves the breed
(c) Increases heterozygosity (d) Increases homozygosity
Answer. (d) Inbreeding is carried out in animal husbandry because it increases homozygosity.

4. Sonalika and Kalyan Sona are varieties of
(a) Wheat (b) Rice
(c) Millet (d) Tobacco
Answer. (a) Sonalika and Kalyan Sona are varieties of wheat.

5. Which one of the following is not a fungal disease?
(a) Rust of wheat (b) SmutofBajra
(c) Black rot of crucifers (d) Red rot of sugarcane
Answer. (c) Black rot of crucifers is a bacterial disease.

6. In virus-infected plants the meristematic tissues in both apical and axillary buds are free of virus because
(a) The dividing cells are virus resistant
(b) Meristems have antiviral compounds
(c) The cell division of meristems are faster than the rate of viral multiplication
(d) Viruses cannot multiply within meristem cell(s).
Answer. (c) In virus-infected plants the meristematic tissues in both apical and axillary buds are free of virus because the cell division of meristems are faster than the rate of viral multiplication.

7. Several South Indian states raise 2-3 crops of rice annually. The agronomic feature that makes this possible is because of
(a) Shorter rice plant (b) Better irrigation facilities
(c) Early yielding rice variety (d) Disease resistant rice variety
Answer. (c) Several South Indian states raise 2-3 crops of rice annually. The agronomic feature that makes this possible is because of early yielding rice variety.

8. Which one of the following combination would a sugarcane farmer look for in the sugarcane crop?
(a) Thick stem, long intemodes, high sugar content and disease resistant
(b) Thick stem, high sugar content and profuse flowering
(c) Thick stem, short internodes, high sugar content, disease resistant
(d) Thick stem, low sugar content, disease resistant.
Answer. (a) Thick stem, long internodes, high sugar content and disease resistant, combination would a sugarcane farmer look for in the sugarcane crop.

9. Fungicides and antibiotics are chemicals that
(a) Enhance yield and disease resistance
(b) Kill pathogenic fungi and bacteria, respectively
(c) Kill all pathogenic microbes
(d) Kill pathogenic bacteria and fungi respectively.
Answer. (b) Fungicides and antibiotics are chemicals that kill pathogenic fungi and bacteria, respectively.

10. Use of certain chemicals and radiation to change the base sequences of genes of crop plants is termed
(a) Recombinant DNA technology (b) Transgenic mechanism (c) Mutation breeding (d) Gene therapy
Answer. (c) Use of certain chemicals and radiation to change the base sequences of genes of crop plants is termed as mutation breeding. Mutations is induced artificially through use of chemicals or radiations like gamma radiations. In mung bean, resistance to yellow mosaic virus and powdery mildew were induced by mutations.

11. The scientific process by which crop plants are enriched with certain desirable nutrients is called
(a) Crop protection (b) Breeding
(c) Bio-fortification (d) Bio-remediation
Answer. (c) Bio-fortification: It is the most practical means to improve public health. In bio-fortification breeding of crops with higher levels of vitamins and minerals or higher protein and healthier fats. In 2000, maize hybrids were developed that had twice the amount of amino acid lysine and tryptophan (MALT). Atlas 66 (Wheat variety), having high protein content has been used as a donor for improving cultivated wheat. Iron fortified rice variety contains over 5 times much Fe.

12. The term‘totipotency’refers to the capacity of a
(a) Cell to generate whole plant (b) Bud to generate whole plant
(c) Seed to germinate (d) Cell to enlarge in size
Answer. (a) The capacity of explant or any cell to generate whole plant is called cellular totipotency. Cellular totipotency is shown by all plant cells. Cellulartotipotency is as demonstrated by F.C. Steward.

13. Given below are a few statements regarding somatic hybridization. Choose the correct statements.
(i) Protoplasts of different cells of the same plant are fused.
(ii) Protoplasts from cells of different species can be fused.
(iii) Treatment of cells with cellulase and pectinase is mandatory.
(iv) The hybrid protoplast contains characters of only one parental protoplast,
(a) (i) and (iii) (b) (i) and (ii)
(c) (i) and (iv) (d) (ii) and (iii)
Answer. (d) Isolated protoplast (surrounded by plasma membranes) from two different varieties can be fused to get hybrid protoplasts which grow to form new plant. The enzyme required to obtain wall free or naked protoplasts are cellulase, hemiceilulase and pectinase, which dissolves cell wall. Somatic cell hybrid (protoplast hybrid) are produced with the help of polyethylene glycol (PEG) and sodium nitrate which promote fusion. In protoplast fusion, chemofusion and electrofusion are employed.

14. An explant is
(a) Dead plant
(b) Part of the plant
(c) Part of the plant used in tissue culture
(d) Part of the plant that expresses a specific gene.
Answer. (c) An explant is a part of the plant used in tissue culture.

15. The biggest constraint of plant breeding is
(a) Availability of desirable gene in the crop and its wild relatives
(b) Infrastructure
(c) Trained manpower
(d) Transfer of genes from unrelated sources
Answer. (a) The biggest constraint of plant breeding is availability of desirable gene in the crop and its wild relatives.

16. Lysine and tryptophan are
(a) Proteins (b) Non-essential amino acids
(c) Essential amino acids (d) Aromatic amino acids
Answer. (c) Lysine and tryptophan are essential amino acids.

17. Micro-propagation is
(a) Propagation of microbes in vitro
(b) Propagation of plants in vitro
(c) Propagation of cells in vitro
(d) Growing plants on smaller scale
Answer. (b) This method of producing thousands of plants through tissue culture is called micro-propagation. Banana, Apple and Tomato are produced on commercial scale using this method.

18. Protoplast is
(a) Another name for protoplasm
(b) An animal cell 
(c) A plant cell without a cell wall
(d) A plant cell
Answer. (c) Protoplast is a plant cell.without a cell wall.

19. To isolate protoplast, one needs
(a) Pectinase (b) Cellulase
(c) Both pectinase and cellulase (d) Chitinase
Answer. (c) The enzyme required to obtain wall free or naked protoplasts are cellulase, hemiceliulase and pectinase, which dissolves cell wall.

20. Which one of the following is a marine fish?
(a) Rohu (b) Hilsa
(c) Catla (d) Common Carp
Answer. (b) Freshwater edible fished: Rohu, Catla and Common carp. Marine edible fishes: Sardines, Hilsa, Mackerel and Pomfrets.

21. Which one of the following products of apiculture is used in cosmetics and polishes?
(a) Honey (b) Oil
(c) Wax (d) Royal jelly
Answer. (c) Wax is a product of apiculture used in cosmetics and polishes.

22. More than 70 per cent of livestock population is in
(a) Denmark . (b) India
(c) China (d) India and China
Answer. (d) More than 70% of the world livestock population is in India and China, but contribution to the world farm produce is only 25% (means productivity per unit is very low).

23. The agriculture sector of India employs about
(a) 50 per cent of the population (b) 70 per cent of the population
(c) 30 per cent of the population (d) 60 per cent of the population
Answer. (d) The agriculture sector of India employs about 60 per cent of the population.

24. 33 percent of India’s (Gross Domestic Product) comes from
(a) Industry (b) Agriculture
(c) Export (d) Small-scale cottage industries.
Answer. (b) 33 percent of India’s (Gross Domestic Product) corner from agriculture.

25. A collection of all the alleles of all the genes of a crop plant is called
(a) Germplasm collection (b) Protoplasm collection
(c) Herbarium (d) Somaclonal collection
Answer. (a) A collection of all the alleles of all the genes of a crop plant is called germplasm collection.

Very Short Answer Type Questions 
1. Millions of chicken were killed in West Bengal, Assam, Orissa and Maharashtra recently, What was the reason?
Answer. Millions of chicken were killed (culled) in West Bengal, Orissa and Maharashtra because they were found to be infected with H5N1 virus, the causal organism of Bird Flu.

2. Can gamma rays used for crop improvement programmes prove to be harmful for health? Discuss.
Answer. Gamma rays used for crop improvement programmes cannot be harmful for health if they are not exposed directly to the human beings. It only induces mutation in the crop varieties.

3. In animal husbandry, if two closely related animals are mated for a few generations, it results in loss of fertility and vigour. Why is this so?
Answer. The phenomenon being referred to is called ‘Inbreeding Depression’ and results in loss of fertility and vigour. This happens because the recessive alleles tend to get together and express harmful effects in the progeny.

4. In the area of plant breeding, it is important not only to preserve the seeds of the variety being cultivated, but also to preserve all its wild relatives. Explain with a suitable example.
Answer. In the area of plant breeding, it is important not only to preserve the seeds of the variety being cultivated, but also to preserve all its wild relatives because several wild relatives of different cultivated species of plants seem to have certain resistant characters but their yields are very low. Hence, there is a need to introduce the resistant genes into the high-yielding cultivated varieties.

5. Name a man-made cereal. Trace how it was developed and where it is used?
Answer. Triticale. It was developed by crossing Triticum aestivum (wheat) and Secale cerale (rye).

6. Fill in the blanks:

Answer. I. Cellulase; II. Somatic hybridization; III. Pomato; IV. Somatic hybrid

7. A few statements are given below followed by a set of terms in a box. Pick the correct term and write it against the appropriate statement.
a. Mating of closely related individuals within the same breed
b. Mating of animals of same breed but having no common ancestors on either side for 4-6 generations
c. Mating of animals of two different species
d. Breeding of animals belonging to different breeds
(i) Cross breeding, (ii) Inter-specific hybridization,
(iii) Out breeding, (iv) Out crossing, (v) Inbreeding
Answer. a. Mating of closely related individuals within the same breed—Inbreeding
b. Mating of animals of same breed but having no common ancestors on either side for 4-6 generations—Out crossing
c. Mating of animals of two different species—Inter-specific hybridization
d. Breeding of animals belonging to different breeds—Cross breeding

8. What is meant by ‘hidden hunger’?
Answer. Consumption of food deficient in nutrients particularly, micronutrients, proteins and vitamins is called hidden hunger.

9. Why are plants obtained by protoplast culture called somatic hybrids?
Answer. Plants obtained by protoplast culture are called somatic hybrids because these are obtained by fusion of somatic cells of two different plants.

10. What is protoplast fusion?
Answer. The ability of protoplasts obtained from two different cells to fuse and form a hybrid protoplast is called protoplast fusion.

11. Why is it easier to culture meristems Compared to permanent tissues?
Answer. Meristems have the capacity of division while permanent tissue have lost the capacity of division.

12. Why are proteins synthesised from SpiruJina called single cell proteins?
Answer. Proteins synthesised from Spirulina are called single cell proteins (SCP) because it is a single celled organism that produce large quantities of food rich in protein, minerals, fats, carbohydrate and vitamins.

13. A person who is allergic to pulses was advised to take a capsule of Spirulina daily. Give the reasons for the advise.
Answer. Humans derive protein from pulses. If a person is allergic to pulses was advised to take a capsule of Spirulina daily because Spirulina is rich in protein.

14. What is aquaculture? Give example of an animal that can be multiplied by aquaculture.
Answer. The rearing of aquatic animals or the cultivation of aquatic plants for food is called aquaculture. Fishes, shell-fish and crustaceans (prawns, crabs, etc.) are the animals that can be multiplied by aquaculture.

15. What are the duties of a veterinary doctor in management of a poultry farm?
Answer. Duties of a veterinary doctor:
1. Veterinary doctor treat disease or injury in animals, which includes diagnosis and treatment.
2. Veterinary doctor is concerned with prophylactic treatment, in order to prevent problems occurring in the future.
3. Veterinary doctor uses vaccination against common animal diseases, such as rabies.
4. Veterinary doctor has important roles in public health and the prevention of zoonoses.

16. Would it be wrong to call plants obtained through micropropagation as ‘clones’? Comment.
Answer. No, it is not wrong because each of the plant will be genetically identical to the original plant from which they were grown, i.e., they are clones/somaclones.

17. How is a somatic hybrid different from a hybrid?
Answer. Isolated’protoplasts from two different varieties of plants can be fused to get hybrid protoplasts, which can be further grown to form a new hybrid plant that is called somatic hybrid. –

18. What is emasculation? Why and when is it done?
Answer. Removal of anther from a bisexual flower is called emasculation. It is used to avoid self-pollination and is used in artificial hybridisation programme. It is done before the dehiscence of anther.

19. Discuss the two main limitations of plant hybridization programme.
Answer. (i) Plant hybridization programme is often constrained by the availability of limited number of disease resistance genes that are present in various crop varieties or wild relatives.
(ii) Several wild relatives of different cultivated species of plants have been shown to have certain resistant character but have very low yield.

20. Interspecific crosses are rare in nature and intergeneric crosses almost unknown. Why?
Answer. The crosses between two different species are called interspecific crosses. As two species are reproductively isolated, the interspecific hybrid is sterile (like mule). Intergenic hybrids are almost non-viable hence intergenic hybrids are almost unknown.

21. Differentiate between pisciculture and aquaculture.
Answer. Pisciculture or fishery is an industry devoted to the catching, processing or selling of fish, shellfish or other aquatic animals.
The rearing of aquatic animals or the cultivation of aquatic plants for food is called aquaculture. .

22. Give two important contributions of Dr. M. S. Swaminathan.
Answer. (i) M.S. Swaminathan developed short-duration high-yielding varieties of rice including scented Basmati.
(ii) He is also known for the development of the concept of crop cafeteria, crop scheduling and genetically improving the yield and quality.

23. The term ‘desirable trait’ can mean different things for different plants.. Justify the statement with suitable examples.
Answer. In millets, resistance to water stress is desirable trait while in mung bean, resistance to yellow mosaic*virus and powdery mildew are desirable traits.

Short Answer Type Questions
1. You are planning to set up a Dairy Farm. Describe the various aspects you would consider before you start the venture.
Answer. Dairying is the management of animals for milk and its products for human consumption.
1. In dairy farm management, we deal with the processes and systems that increase yield and improve quality of milk. Milk yield is primarily dependent on the quality of breeds in the farm.
2. Selection of good breeds having high yielding potential (under the climatic conditions of the area), combined with resistance to diseases is very important.
3. For the yield potential to be realised the cattle have to be well looked after they have to be housed well, should have adequate water and be maintained disease free.
4. The feeding of cattle should be carried out in a scientific manner – with special emphasis on the quality and quantity of fodder.
5. Besides, stringent cleanliness and hygiene (both of the cattle and the handlers) are of paramount importance while milking, storage and transport of the milk and its products.

2. It is said, that diseases are spreading faster due to globalisation and increased movement of people. Justify the statement taking the example of H5N1 virus.
Answer. Diseases are spreading faster due to globalisation and increased movement of people because it is responsible for rapid spreading of infectious diseases.

3. Explain the concept of the Blue Revolution.
Answer. Rapid increase in the fishery industry in recent years is called blue revolution. For example, through aquaculture and pisciculture we have been able to increase the production of aquatic plants and animals, both fresh-water and marine. This has led to the development and flourishing of the fishery industry, and it has brought a lot of income to the farmers in particular and the country in general. We now talk about the ‘Blue Revolution’ which is being implemented along the same lines as ‘Green Revolution’.

4. A farmer was facing the problem of low yield from his farm. He was advised to keep a beehive in the vicinity. Why? How would the beehive help in enhancing yield? .
Answer. Bees are the pollinators of many of our crop species such as sunflower, Brassica, apple and pear. Keeping beehives in crop fields during flowering period increases pollination efficiency and improves the yield beneficial both from the point of view of crop yield and honey yield.

5. Life style diseases are increasing alarmingly in India. We are also dealing with large scale malnutrition in the population. Is there any method by which we can address both of these problems together?
Answer. Lifestyle disorders can be prevented by improving the lifestyle of a person. Taking too much of a stress can lead to heart stroke and paralysis which can be avoided by living in a healthy environment.
Malnutrition is very common in poor people who do not have access to nutritious food and India has a large scale population who are malnutritious. Proper distribution of food and less wasting can help to address both of these problems together.

6. How can we improve the success rate of fertilisation during artificial insemination in animal husbandry programmes?
Answer. The technology is called MOET or Multiple Ovulation Embryo Transfer. During the procedure, a cow is given hormonal treatment so that more than one ovule (6-8 eggs) is produced per cycle. After mating or artificial insemination the embryos at 8-32 celled state are transferred to different surrogate mother cows. The method has been successfully used for cattle, sheep, buffalo etc.

7. What is meant by germplasm collection? What are its benefits?
Answer. The collection of diverse alleles of all the genes of a crop plant is called germplasm collection. It is of great benefits in plant breeding programmes it offers to the breeders, the entire of genes and alleles and the characteristics which they express. The breeder selects the most favourable characters of a particular gene and manipulates its transfer to a desirable parent.

8. Name the improved characteristics of wheat that helped India to achieve green revolution.
Answer. i. Semi-dwarf nature ii. Quick yielding feature iii. High yielding feature iv. Disease resistant feature

9. Suggest some of the features of plants that will prevent insect and pest infestation.
Answer. i. Increasing hair growth on aerial parts of plants.
ii. Rendering the flowers nectarless.
iii. Enabling plants to secrete insect killing chemicals (toxins).

10. It is easier to culture plant cells in vitro as compared to animal cells. Why?
Answer. Plant have meristematic cells that have the capacity of division, so can grow easily in in vitro culture. Animals have mainly differentiated cells that have lost the capacity of division.

11. The culture medium (nutrient medium) can be referred to as a ‘highly enriched laboratory soil’. Justify the statement.
Answer. The culture medium (nutrient medium) can be referred to as a ‘highly enriched laboratory soil’ because in culture medium we must provide a carhon source such as sucrose and also inorganic salts, vitamins, amino acids and growth regulators like auxins, cytpkinins etc. .

12. Is there any relationship between dedifferentiation and the higher degree of success achieved in plant tissue culture experiments?
Answer. Yes, the cells that has lost the capability of division can regain the capacity of division through dedifferentiation. So, in plant tissue culture experiments more success is achieved.

13. “Give me a living cell of any plant and I will give you a thousand plants of the same type”. Is this only a slogan or is it scientifically possible? Write your comments and justify them. 
Answer. It is scientifically possible. By application of tissue culture it is possible to achieve propagation of a large number of plants in very short durations. This method of producing thousands of plants through tissue culture is called micro propagation. Each of these plants will be genetically identical to the original plant from which they were grown, i.e., they are somaclones.

14. What is the difference between n breed and a species? Give an example for each category.
Answer. A group of animals related by descent and similar in most characters like general appearance, features, size, configuration, etc., are said to belong to a breed. Hisardale is a new breed of sheep.
A group of individual that is reproductively isolated from other such group is called species. For examples, Panthera leo (lion) and Mangifera indica are different species.

15. Plants raised through tissue cultures are clones of the ‘parent’ plant. Discuss the utility of these plants.
Answer. Plant raised through tissue cultures are clones of the parent plant that means they are genetically identical to the original plant. They are utilised for maintaining a desirable trait of parent.

16. Discuss the importance of testing of new plant varieties in a geographically vast country like India.
Answer. The newly selected lines are evaluated for their yield and other agronomic traits of quality, disease resistance, etc. This evaluation is done by growing these in the research fields and recording their performance under ideal fertilizer application, irrigation, and other crop management practices. The evaluation in research fields is followed by testing the materials.in fanners’ fields, for at least three growing seasons at several locations in the country, representing all the agroclimatic zones where the crop is usually grown. The material is evaluated in comparison to the best available local crop cultivar-a check or reference cultivar.

17. Define the term ‘stress’ for plants. Discuss briefly the two types of stress encountered by plants.
Answer. Any unfavourable condition for the plant growth and development is called stress. For example, drought, salinity, cold, pest and diseases are different type of stresses.

18. Discuss natural selection and artificial selection. What are the implications of the latter on the process of evolution?
Answer. Artificial selection is the intentional breeding of plants or animals. Selective breeding is a technique used when breeding domesticated animals, such as dogs, pigeons or cattle.
Natural selection is the differential survival and reproduction of individuals due to differences in phenotype. It is a key mechanism of evolution, the change in heritable traits of a population over time.

19. Discuss briefly how pure lines are created in animal husbandry.
Answer. Inbreeding is necessary if we want to evolve a pure line in any animal. Inbreeding refers to the mating of more closely related individuals within the same breed for 4-6 generations.

20. What are the physical barriers of a cell in-the protoplast fusion experiment? How are the barriers overcome?
Answer. Cell wall is the most important physical barrier in such experiments. This can be overcomed by treatment with enzymes like cellulase and pectinase which have the ability to digest the cell wall and liberate the naked protoplast surrounded only by the dell membrane.

21. Give few examples of biofortified crops. What benefits do they offer to the society?
Answer. Maize, wheat, rice, bathua, spinach, pulses have biofortified varieties. Maize hybrids have twice the amount of amino acids, fortified wheat variety has high protein content, and fortified rice has high quantity of iron. Consumption of such biofortified foods will enrich the nutritive value of our common foods and will vastly improve public health. Instead of consuming different food items for obtaining different nutrients, if 2 or 3 nutrients can be incorporated into a single crop it offers enormous benefits to human beings and may even help overcome several nutrient deficiency disorders latent in our country.

Long Answer Type Questions
1. You are a Botanist working in the area of plant breeding. Describe the various steps that you will undertake to release a new variety.
Answer. The main steps in breeding a new genetic variety of a crop are:
(i) Collection of variability: Genetic variability is the root of any breeding programme. In many crops pre-existing genetic variability is available from wild relatives of the crop. Collection and preservation of all the different wild varieties, species and relatives of the cultivated species (followed by their evaluation for their characteristics) is a pre-requisite for effective exploitation of natural genes available in the populations. The entire collection (of plants/seeds) having all the diverse alleles for all genes in a given crop is called germplasm collection.
(ii) Evaluation and selection of parents: The germplasm is evaluated so as to identify plants with desirable combination of characters. The selected plants are multiplied and used in the process of hybridisation. Pure lines are created wherever desirable and possible. ‘
(iii) Cross hybridisation among the selected parents: The desired characters have very often to be combined from two different plants (parents), for example high protein quality of one parent may need to be combined with disease resistance from another parent. This is possible by cross hybridising the two parents to produce hybrids that genetically combine the desired characters in one plant. This is a very time-consuming and tedious process since the pollen grains from the desirable plant chosen as male parent have to be collected and placed on the stigma of the flowers selected as female parent (In chapter 2 details on how to make crosses have been described). Also, it is not necessary that the hybrids do combine the desirable characters; usually only one in few hundred to a thousand crosses shows the desirable combination.
(iv) Selection and testing of superior, recombinants: This step consists of selecting, among the progeny of the hybrids, those plants that have the desired character combination. The selection process is crucial to the success of the breeding objective and requires careful scientific evaluation of the progeny. This step yields plants that are superior to both of the parents (very often more than one superior progeny plant may become available). These are self-pollinated for several generations till they reach a state of uniformity (homozygosity), so that the characters will not segregate in the progeny.
(v) Testing, release and commercialisation of new cultivars: The newly selected lines are evaluated for their yield and other agronomic traits of quality, disease resistance, etc. This evaluation is done by growing these in the research fields and recording their performance under ideal fertilizer application, irrigation, and other crop management practices. The evaluation in research fields is followed by testing the materials in farmers’ fields:, for at least three growing seasons at several locations in the country, representing all the agro climatic zones where the crop is usually grown. The material is evaluated in comparison to the best available local crop cultivar a check or reference cultivar.

2. (a) The shift from grain to meat diets creates more demands for cereals. Why? (b) A250 kg cow produces 200 g of protein per day but 250 g of Methylophillus methylotrophus can produce 25 tonnes of protein. Name this emerging area of research. Explain its benefits.
Answer. (a) It takes 3-10 kg of grain to produce 1 kg of meat using animal farming. That is why cereals demand increases.
(b) Production of single cell proteins (SCP) by microbes. Microbes are being grown on an Industrial scale. Spirulina can be easily grown on starch, molasses etc., and can make food which is rich in proteins, minerals, fats, carbohydrates and vitamins. This could be a good alternative for dealing with the problem of malnutrition.

3. What are the advantages of tissue culture methods over conventional method of plant breeding in crop improvement programmes?
Answer. The advantages of tissue culture methods are:
By application of these methods it is possible to achieve propagation of a large number ofplants in very short durations. This method of producing thousands of plants through tissue culture is called micro-propagation. Each of these plants will be genetically identical to the original plant from which they were grown, i.e., they are somaclones. Many important food plants like tonjato, banana, apple, etc., have been produced on commercial scale using this method. Try to visit a tissue culture laboratory with your teacher to better understand and . appreciate the process. Another important application of the method is the recovery of healthy plants from diseased plants. Although the plant is infected with a virus, the meristem (apical and axillary) is free of virus. Hence, one can remove the meristem and grow it in vitro to obtain virus-free plants. Scientists have succeeded in culturing meristems of banana, sugarcane, potato, etc.

4. ‘Modem methods of breeding animals and plants can alleviate the global food shortage’. Comment on the statement and give suitable examples.-
Answer. The main steps in breeding a new genetic variety of a crop are:
(i) Collection of variability: Genetic variability is the root of any breeding programme. In many crops pre-existing genetic variability is available from wild relatives of the crop. Collection and preservation of all the different wild varieties, species and relatives of the cultivated species (followed by their evaluation for their characteristics) is a pre-requisite for effective exploitation of natural genes available in the populations. The entire collection (of plants/seeds) having all the diverse alleles for all genes in a given crop is called germplasm collection.
(ii) Evaluation and selection of parents: The germplasm is evaluated so as
to identify plants with desirable combination of characters. The selected plants are multiplied and used in the process of hybridisation. Pure lines are created wherever desirable and possible. ‘
(iii) Cross hybridisation among the selected parents: The desired characters have very often to be combined from two different plants (parents), for example high protein quality of one parent may need to be combined with disease resistance from another parent. This is possible by cross hybridising the two parents to produce hybrids that genetically combine the desired characters in one plant. This is a very time-consuming and tedious process since the pollen grains from the desirable plant chosen as male parent have to be collected and placed on the stigma of the flowers selected as female parent (In chapter 2 details on how to make crosses have been described). Also, it is not necessary that the hybrids do combine the desirable characters; usually only one in few hundred to a thousand crosses shows the desirable combination.
(iv) Selection and testing of superior, recombinants: This step consists of selecting, among the progeny of the hybrids, those plants that have the desired character combination. The selection process is crucial to the success of the breeding objective and requires careful scientific evaluation of the progeny. This step yields plants that are superior to both of the parents (very often more than one superior progeny plant may become available). These are self-pollinated for several generations till they reach a state of uniformity (homozygosity), so that the characters will not segregate in the progeny.
(v) Testing, release and commercialisation of new cultivars: The newly selected lines are evaluated for their yield and other agronomic traits of quality, disease resistance, etc. This evaluation is done by growing these in the research fields and recording their performance under ideal fertilizer application, irrigation, and other crop management practices. The evaluation in research fields is followed by testing the materials in farmers’ fields:, for at least three growing seasons at several locations in the country, representing all the agro climatic zones where the crop is usually grown. The material is evaluated in comparison to the best available local crop cultivar a check or reference cultivar.
(a) It takes 3-10 kg of grain to produce 1 kg of meat using animal farming. That is why cereals demand increases.
(b) Production of single cell proteins (SCP) by microbes. Microbes are being grown on an Industrial scale. Spirulina can be easily grown on starch, molasses etc., and can make food which is rich in proteins, minerals, fats, carbohydrates and vitamins. This could be a good alternative for dealing with the problem of malnutrition.
The advantages of tissue culture methods are:
By application of these methods it is possible to achieve propagation of a large number ofplants in very short durations. This method of producing thousands of plants through tissue culture is called micro-propagation. Each of these plants will be genetically identical to the original plant from which they were grown, i.e., they are somaclones. Many important food plants like tonjato, banana, apple, etc., have been produced on commercial scale using this method. Try to visit a tissue culture laboratory with your teacher to better understand and . appreciate the process. Another important application of the method is the recovery of healthy plants from diseased plants. Although the plant is infected with a virus, the meristem (apical and axillary) is free of virus. Hence, one can remove the meristem and grow it in vitro to obtain virus-free plants. Scientists have succeeded in culturing meristems of banana, sugarcane, potato, etc.

5. Does apiculture offer multiple advantages to farmers? List its advantages if it is located near a place of commercial flower Cultivation.
Answer. Bee-keeping or apiculture is the maintenance of hives of honeybees for the production of honey, ft has been an age-old cottage industry. Honey is a food of high nutritive value and also find its use in the indigenous systems of medicine. Honeybee also produces beeswax, which finds many uses in industry, such as in the preparation of cosmetics and polishes of various kinds. The increased demand of honey has led to large-scale beekeeping practices; it has become an established income generating industry, whether practiced on a small or on a large scale.
Bees are the pollinators of many of our crop species such as sunflower, Brass tea, apple and pear. Keeping beehives in crop fields during flowering period increases pollination efficiency and improves the yield-beneficial both from the point of view of crop yield and honey yield.

6. (a) Mutations are beneficial for plant breeding. Taking an example, justify the statement.
(b) Discuss briefly the technology that made us self-sufficient in food production.
Answer. (a) Mutation is the process by which genetic variations are created through changes, in the base sequence within genes resulting in the creation of a new character or trait not found in the parental type. It is possible to induce mutations artificially through use of chemicals or radiations (like gamma radiations), and selecting and using the plants that have the desirable character as a source in breeding. This process is called mutation breeding. In mung bean, resistance to yellow mosaic virus and powdery mildew were induced by mutations.
(b) Plant breeding is the purposeful manipulation of plant species in order to create desired plant types that are better suited for cultivation, give better yields and are disease resistant. With advancements in genetics, molecular biology and tissue culture, plant breeding is now increasingly being carried out by using molecular genetic tools.

7. Discuss how the property of plant cell totipotency has been utilised for plant propagation and improvement
Answer. The capacity to generate a whole plant from any cell/explant is called totipotency. By application of these methods it is possible to achieve propagation of a large number of plants in very short durations. This method of producing thousands of plants through tissue culture is called micro propagation. Each of these plants will be genetically identical to the original plant from which they were grown, i.e., they are somaclones. Many important food plants like tomato, banana, apple, etc., have been produced on a commercial scale using this method. Try to visit a tissue culture laboratory with your teacher to better understand and appreciate the process.
Another important application ef the method is the recovery of healthy plants from diseased plants. Although the plant is infected with a virus, the meristem (apical and axillary) is free of virus. Hence, one can remove the meristem and grow it in vitro to obtain virus-free plants. Scientists have succeeded in culturing meristems of banana, sugarcane, potato, etc.

8. What are three options to increase food production? Discuss each giving the salient features, merits and demerits.
Answer. With ever increasing population of the world, enhancement of food production is a major necessity. Biological principles as applied to animal husbandry (See Ans 1 of short answer type questions) and plant breeding (See Ans 1) have a major role in our efforts to increase food production. Several new techniques like embryo transfer technology and tissue culture tecbniques(See Ans 7) are going to play a pivotal role in further enhancing food production.

The post NCERT Exemplar Problems Class 12 Biology Strategies for Enhancement in Food Production appeared first on Learn CBSE.

NCERT Exemplar Problems Class 12 Biology Chapter 10 Microbes in Human Welfare

Multiple Choice Questions
Single Correct Answer Type
1. The vitamin whose content increases following the conversion of milk into curd by lactic acid bacteria is
(a) Vitamin C (b) Vitamin D
(c) Vitamin B₁₂ (d) Vitamin E
Answer. (c) The vitamin B₁₂ content increases following the conversion of milk into curd by lactic acid bacteria (LAB).

2. Wastewater treatment generates a large quantity of sludge, which can be treated by
(a) Anaerobic digesters (b) Floe
(c) Chemicals (d) Oxidation pond
Answer. (a) Wastewater treatment generates a large quantity of sludge, which can be treated by anaerobic sludge digesters.

3. Methanogenic bacteria are not found in
(a) Rumen of cattle (b) Gobar gas plant
(c) Bottom of water-logged paddy fields (d) Activated sludge
Answer. (d) Methanogenic bacteria are found in rumen of cattle, gobar gas plant and bottom of,water-logged paddy fields.

4. Match the following list of bacteria and their commercially important products.

Choose the correct match:
(a) i—B, ii—C, iii—D, iv—A (b) i—B, ii—D, iii—C, iv—A
(c) i—D, ii—C, iii—B, iv—A (d) i—D, ii—A, iii—C, iv—B
Answer. (c)

5. Match the following list of bio active substances and their roles:

Choose the correct match:
(a) i—B, ii—C, iii—A, iv—D (b) i—D, ii—B, iii—A, iv—C
(c) i—D, ii—A, iii—B, iv—C (d) i—C, ii—D, iii—B, iv—A
Answer. (d)

6. The primary treatment of waste water involves the removal of
(a) Dissolved impurities (b) Stable particles
(c) Toxic substances (d) Harmful bacteria
Answer. (b) 10 treatment basically involves physical removal of particles from sewage through: (i) Filtration and (ii) Sedimentation
These are removed in stages:
(i) Initially floating debris is removed by sequential filtration.
(ii) Then the grit (soil and pebbles) are removed by sedimentation

7. BOD of waste water is estimated by measuring the amount of
(a) Total organic matter (b) Biodegradable organic matter
(c) Oxygen evolution (d) Oxygen consumption
Answer. (d) BOD of waste water is estimated by measuring the amount of oxygen consumption.

8. Which one of the following alcoholic drinks is produced without distillation?
(a) Wine (b) Whisky
(c) Rum (d) Brandy
Answer. (a) Wine and beer are produced, withont distillation. Whisky, brandy and rum are produced by distillation.

9. The technology of biogas production from cow dung was developed in India largely due to the efforts of
(a) Gas Authority of India
(b) Oil and Natural Gas Commission
(c) Indian Agricultural Research Institute and Khadi and Village Industries Commission
(d) Indian Oil Corporation
Answer. (c) The technology of biogas production from cow dung was developed in India largely due to the efforts of Indian Agricultural Research Institute and Khadi and Village Industries Commission.

10. The free-living fungus Trichoderma can be used for
(a) Killing insects
(b) Biological control_of plant diseases
(c) Controlling butterfly caterpillars
(d) Producing antibiotics
Answer. (b) The free-living fungus Trichoderma can be used for biological control of plant diseases.

11. What would happen if oxygen availability to activated sludge floes is reduced?
(a) It will slow down the rate of degradation of organic matter.
(b) The center of floes will become anoxic, which would cause death of bacteria and eventually breakage of floes.
(c) Floes would increase in size as anaerobic bacteria would grow around floes.
(d) Protozoa would grow in large numbers.
Answer. (b) If oxygen availability to activated sludge floes is reduced the center of floes will become anoxic, which would cause death of bacteria and eventually breakage of floes.

12. Mycorrhiza does not help the host plant in
(a) Enhancing its phosphorus uptake capacity
(b) Increasing its tolerance to drought
(c) Enhancing its resistance to root pathogens
(d) Increasing its resistance to insects
Answer. (d) Mycorrhiza help the host plant in enhancing its phosphorus uptake capacity, increasing its tolerance to drought and enhancing its resistance to root pathogens.

13. Which one of the following is not a nitrogen-fixing organism?
(a) Anabaena (b) Nostoc
(c) Azdtobacter (d) Pseudomonas
Answer. (d) Anabaena, Nostoc and Azotobacter are the cyanobacteria. All are nitrogen-fixing organism.

14. Big holes in Swiss cheese are made by a
(a) machine
(b) bacterium that produces methane gas
(c) bacterium producing a large amount of carbon dioxide
(d) fungus that releases a lot of gases during its metabolic activities.
Answer. (c) Large holes in Swiss cheese are due to production of large amount of C02
by a bacterium named Proponibacterium sharmanii.

15. The residue left after methane production from cattle dung is
(a) Burnt (b) Buried in landfills
(c) Used as manure (d) Used in civil construction
Answer. (c)

• Thus gobar (excreta/dung) of cattle is rich in these bacteria. Dung can be used for generation of biogas, commonly called gobar gas.
•  The residue left after methane production from cattle dung is used as manure.

16. Methanogens do not produce
(a) Oxygen (b) Methane
(c) Hydrogen sulfide (d) Carbon dioxide
Answer. (a) Methanogens are obligate anaerobes, which produce methane, hydrogen sulfide and carbon dioxide but do not produce oxygen.

17. Activated sludge should have the ability to settle quickly so that it can
(a) Be rapidly pumped back from sedimentation tank to aeration tank
(b) Absorb pathogenic bacteria present in waste water while sinking to the bottom of the settling tank
(c) Be discarded and anaerobically digested
(d) Absorb colloidal organic matter
Answer. (a) Activated sludge should have the ability to settle quickly so that it can be rapidly pumped back from sedimentation tank to aeration tank.

18. Match the items in Column ‘A’ and Column ‘B’ and choose the correct answer.

The correct answer is
(a) i—B, ii—D, iii—C, iv—A (b) i—C, ii—D, iii—B, iv—A
(c) i—D, ii—A, iii—B, iv—C (d) i—C, ii—B, iii—A, iv—D
Answer. (b)

Very Short Answer Type Questions
1.Why does ‘Swiss cheese’ have big holes?
Answer. The large holes in ‘Swiss cheese’ are due to the production of a large amount of C02 by a bacterium named Propionibacterium sharmanii.

2.What are fermentors?
Answer. Even in industry, microbes are used to synthesise a number of products valuable to human beings. Beverages and antibiotics are some of the examples. Production on an industrial scale, requires growing microbes in very large vessels called fermentors.

3.Name a microbe used for statin production. How do statins lower blood cholesterol level?
Answer. Monascus purpureus statins lower blood cholesterol level by competitively inhibiting the enzyme responsible for the synthesis of cholesterol.

4. Why do we prefer to call secondary waste water treatment as biological treatment? .
Answer. The secondary treatment is also called biological treatment because in secondary treatment living organisms like bacteria and fungi are used.

5. What for Nucleopolyhydro viruses are being used now-a-days?
Answer. Necleopolyhydroviruses are used for the biological control of insect pesto.

6. How has the discovery of antibiotics helped mankind in the field of medicine?
Answer. If antibiotics were not discovered bacterial and fungal diseases would not have been controllable.

7. Why is distillation required for producing certain alcoholic drinks?
Answer. Distillation increases the alcohol content in alcoholic drinks.

8. Write the most important characteristic that Aspergillus niger, Clostridium butylicum and Lactobacillus share.
Answer. These all are acid producers. Examples of acid producers are Aspergillus niger (a fungus) of citric acid, Clostridium butylicum (a bacterium) of butyric acid and Lactobacillus (a bacterium) of lactic acid.

9. What would happen if our intestine harbours microbial flora exactly similar to that found in the rumen of cattle?
Answer. If our intestine harbours microbial flora exactly similar to that found in the rumen of cattle then we would be able to digest the cellulose present in our food.

10. Give any two microbes that are useful in biotechnology.
Answer. E.coli and Saccharomyces cerevisae

11. What is the source organism for EcoRI, restriction endonuclease?
Answer. Escherichia coli RY 13.

12. Name any genetically modified crop.
Answer. Bt cotton.

13. Why are blue green algae not popular as biofertilisers?
Answer. Blue green algae are not popular as biofertilisers because they causes algal bloom in polluted water bodies.

14. Which species of Penicillium produces Roquefort cheese?
Answer. Roquefort cheese produced by Penicillium roqueforti.

15. Name the states involved in Ganga action plan.
Answer. Uttarakhand, Uttar Pradesh, Bihar, West Bengal and Jharkhand.

16. Name any two industrially important enzymes.
Answer. Lipase, Amylase.

17. Name an immune immunosuppressive agent.
Answer. Cyclosporin A

18. Give an example of a rod shaped virus.
Answer. Tobacco mosaic virus (TMV).

19. What is the group of bacteria found in both the rumen of cattle and sludge of sewage treatment?
Answer. Methanogens

20. Name a microbe used for the production of Swiss cheese.
Answer. Propionibacterium shaynanii.

Short Answer Type Questions
1. Why are floes important in biological treatment of waste water?
Answer. Secondary treatment or Biological treatment: The primary effluent is passed into large aeration tanks where it is constantly agitated mechanically and air is pumped into it. This allows vigorous growth of useful aerobic microbes into floes (masses of bacteria associated with fungal filaments to form mesh like structures). While growing, these microbes consume the major part of the organic matter in the effluent. This significantly reduces the BOD (biochemical oxygen demand) of the effluent.

2. How has the bacterium Bacillus thuringiensis helped us in controlling caterpillars of insect pests?
Answer. Bacillus thuringiensis products are endotoxin which when ingested and released in the gut of the larvae of insect pest disrupts the insect gut lining thereby killing them.

3. How do mycorrhizal fungi help the plants harbouring them?
Answer. The mycorrhizal fungi absorb phosphorus from the soil and transfer them to the host cells. They also’impart resistance to host plants against root pathogens. They also help plant tolerate salinity and draught.

4. Why are cyanobacteria considered useful in paddy fields?
Answer. Cyanobacteria are autotrophic microbes widely distributed in aquatic and terrestrial environments many of which can fix atmospheric nitrogen, e.g. Anabaena, Nostoc, Oscillatoria, etc. In paddy fields, cyanobacteria serve as an important biofertiliser. Blue green algae also add organic matter to the soil and increase its fertility.

5. How was penicillin discovered?
Answer. Penicillin was an accidental discovery. Sir Alexander Fleming observed that in unwashed culture plates of Staphylococcus, a mould Penicillium was growing. This mould inhibited the-growth of Staphylococcus. Later the antibiotic Penicillin was isolated from this fungus.

6. Name the scientists who were credited for showing the role of Penicillin as an antibiotic.
Answer. Its full potential as an effective antibiotic was established much later by Ernest Chain and Howard Florey. This antibiotic was extensively used to treat American soldiers wounded in World War II. Fleming, Chain and Florey were awarded the Nobel Prize in 1945, for this discovery.

7. How do bioactive molecules of fungal origin help in restoring good health of humans?
Answer. A bioactive molecule, cyclosporin A, that is used as an immunosuppressive agent in organ-transplant patients, is produced by the fungus Trichoderma polysporum. Statins produced by the yeast Monascus purpureus have been commercialised as blood-cholesterol lowering agents. It acts by competitively inhibiting the enzyme responsible for synthesis of cholesterol.

8. What roles do enzymes play in detergents that we use for washing clothes? Are these enzymes produced from some unique microorganisms?
Answer. Microbes are also used for production of enzymes. Lipases are used in detergent formulations and are helpful in removing oily stains from the laundry. Lipases are produced by Candida lipolytica (fungus).

9. What is the chemical nature of biogas? Name an organism which is involved in biogas production.
Answer. The chemical nature of Biogas is methane, CO₂ and H₂. Methanobacterium, a type of methanogen is employed for biogas productioh.

10. How do microbes reduce the environmental degradation caused by chemicals?
Answer. Chakravarthy Bug is a super bug of Pseudomonas with multiple plasmid.
They are helpful in removing oil spills.

11. What is a broad spectrum antibiotic? Name one such antibiotic.
Answer. A broad spectrum antibiotic is one which can inhibit the growth of both G +ve and G -ve bacteria.

12. What are viruses parasitising bacteria called? Draw a well labelled diagram of the same.
Answer. Viruses parasitising bacteria are called bacteriophages.

13. Which bacterium has been used as a clot buster? What is its mode of action?
Answer. Streptokinase produced by the bacterium Streptococcus and modified by
genetic engineering is used as a ‘clot buster’ for removing clots from the blood vessels of patients who have undergone myocardial infarction leading to heart attack.

14. What are biofertilisers? Give two examples.
Answer. Biofertilisers are organisms that enrich the nutrient quality of the soil. The main sources of biofertilisers are bacteria, fungi and cyanobacteria.

Long Answer Questions
1. Why is aerobic degradation more important than anaerobic degradation for the treatment of large volumes of waste waters rich in organic matter? Discuss.
Answer. Secondary treatment or Biological treatment: The primary effluent is passed into large aeration tanks where it is constantly agitated mechanically and air is pumped into it. This allows vigorous growth of useful aerobic microbes into floes (masses of bacteria associated with fungal filaments to form mesh like structures). While growing, these microbes consume the major part of the organic matter in the effluent. This significantly reduces the BOD (biochemical oxygen demand) of the effluent. BOD refers to the amount of the oxygen that would be consumed if all the organic matter in one litre of water were oxidised by bacteria. The sewage water is treated till the BOD is reduced. The BOD test measures the rate of uptake of oxygen by micro¬organisms in a sample of water and thus, indirectly, BOD is a measure of the organic matter present in the water. The greater the BOD of waste water, more is its polluting potential.
Once the BOD of sewage or waste water is reduced significantly, the effluent is then passed into a settling tank where the bacterial ‘floes’ are allowed to sediment. This sediment is called activated sludge. A small part of the activated sludge is pumped back into the aeration tank to serve as the inoculum. The remaining major part of the sludge is pumped into large tanks called anaerobic sludge digesters. Here, other kinds of bacteria, which grow anaerobically, digest the bacteria and the fungi in the sludge. During this digestion, bacteria produce a mixture of gases such as methane, hydrogen sulphide and carbon dioxide. These gases form biogas and can be used as source of energy as it is inflammable.

2. (a) Discuss about the major programs that the Ministry of Environment and Forests, Government of India, has initiated for saving major Indian rivers from pollution.
(b) Ganga has recently been declared the national river. Discuss the implication with respect to pollution of this river.
Answer. (a) The untreated sewage is often discharged directly into rivers leading to their pollution and increase in water-borne diseases. The Ministry of Environment and Forests has initiated Ganga Action Plan and Yamuna Action Plan to save these major rivers of our country from pollution. Under these plans, it is proposed to build a large number of sewage treatment plants so that only treated sewage may be discharged in the rivers.
(b) The Ganga is the largest river in India with an extraordinary religious ‘ importance for Hindus. Situated along its banks are some of the world’s oldest inhabited cities like Varanasi and Patna. It provides water to about 40% of India’s population across 11 states, serving an estimated population of 500 million people or more, which is larger than any other river in the world.
A number of initiatives have been undertaken to clean the river but failed to deliver desired results. After getting elected, India’s Prime minister Narendra Modi affirmed to work for cleaning the river and controlling pollution. Subsequently, Namami Ganga project was announced by the Government in July 2014 budget. An estimated Rs 2,958 crores have been spent till July 2016 in various efforts to clean up the river.

3. Draw a diagrammatic sketch of biogas plant and label its various components given below: Gas Holder, Sludge Chamber, Digester, Dung+water chamber
Answer.

4. Describe the main ideas behind the biological control of pests and diseases.
Answer. Biological control means life against life. It is a natural and ecofriendly
concept. It employs the natural organisms to control the population of pathogens and pests in an ecosystem. Classical examples are Trichoderma which is antagonist against many soil borne plant pathogens. Similarly, Penicillium inhibits the growth of Staphylococcus and therefore has been successfully used in the production of Penicillin antibiotic to control many human bacterial pathogens.

5. (a) What would happen if a large volume of untreated sewage is discharged into a river?
(b) In what way anaerobic sludge digestion is important in sewage treatments?
Answer. (a) Due to increasing urbanisation, sewage is being produced in much larger quantities than ever before. However the number of sewage treatment plants has not increased enough to treat such large quantities. So the untreated sewage is often discharged directly into rivers leading to their pollution and increase in water-borne diseases.
(b) Once the BOD of sewage or waste water is reduced significantly, the effluent is then passed into a settling tank where the bacterial ‘floes’ are allowed to sediment. This sediment is called activated sludge. A small part of the activated sludge is pumped back into the aeration tank to • serve as the inoculum. The remaining major part of the sludge is pumped
into large tanks called anaerobic sludge digesters. Here, other kinds of bacteria, which grow anaerobically, digest the bacteria and the fungi in the sludge. During this digestion, bacteria produce a mixture of gases such as methane, hydrogen sulphide and carbon dioxide. These gases form biogas and can be used as a source of energy as it is inflammable.

6. Which type of food would have lactic acid bacteria? Discuss their useful application.
Answer. Curd. Micro organisms such as lactic acid bacteria (LAB) grow in milk and convert it to curd. During growth, the LAB produce acids that coagulate and partially digest the milk proteins. A small amount of curd added to the fresh milk as inoculum or starter contain millions of LAB, which at suitable temperatures multiply, thus converting milk into curd, which also improves its nutritional quality by increasing vitamin B₁₂. In our stomach too, the LAB play very beneficial role in checking disease causing microbes.

The post NCERT Exemplar Problems Class 12 Biology Microbes in Human Welfare appeared first on Learn CBSE.

NCER Exemplar Problems Class 12 Biology Chapter 11 Biotechnology: Principles and Processes

Multiple Choice Questions
Single Correct Answer Type
1. Rising of dough is due to
(a) Multiplication of yeast (b) Production of CO₂
(c) Emulsification (d) Hydrolysis of wheat flour starch into sugars.
Answer. (b) Dough, which is used for making dosa and idli is fermented by bacteria. The puffed up appearance of dough is due to the production of CO₂ gas. Dough which is used for making bread, is fermented by fungi Saccharomyces cerevisiae (Baker’s yeast)

2. An enzyme catalysing the removal of nucleotides from the ends of DNA is
(a) Endonuclease (b) Exonuclease
(c) DNAligase (d) Hind-II
Answer. (b) Exonucleases remove nucleotides from the ends of the DNA whereas endonucleases make cuts at specific positions within the DNA.

3. The transfer of genetic material from one bacterium to another through the mediation of a vector like virus is termed as
(a) Transduction (b) Conjugation
(c) Transformation (d) Translation
Answer. (a) The transfer of genetic material from one bacterium to another through the mediation of a vector like virus is termed as transduction.

4. Which of the given statement is correct in the context of observing DNA separated by agarose gel electrophoresis?
(a) DNA can be seen in visible light
(b) DNA can be seen without staining in visible light
(c) Ethidium bromide stained DNA can be seen in visible light
(d) Ethidium bromide stained DNA can be seen under exposure to UV light
Answer. (d) The separated DNA fragments can he visualised only after staining the DNA with a compound known as ethidium bromide followed by exposure to ultraviolet radiation (we cannot see pure DNA fragments in the visible light and without staining).

5. ‘Restriction’ in Restriction enzyme refers to
(a) Cleaving of phosphodiester bond in DNA by the enzyme
(b) Cutting of DNA at specific position only
(c) Prevention of the multiplication of bacteriophage in bacteria
(d) All of the above
Answer. (c)‘Restriction’inRestrictionenzymereferstopreventionofthemultiplication of bacteriophage in bacteria.

6. Which of the following is not required in the preparation of recombinant DNA molecule?
(a) Restriction endonuclease (b) DNA ligase
(c) DNA fragments (d) E. coli
Answer. (d) Genetic engineering or r-DNA technology can be accomplished only if we have the key tools:
(1) Restriction enzymes (2) Polymerase enzymes
(3) Ligases (4) Vectors
(5) Host organism

7. In agarose gel electrophoresis, DNA molecules are separated on the basis of their
(a) Charge only (b) Size only
(c) Charge to size ratio (d) Both charge and size
Answer. (d) The differential mobility of DNA depends upon charge and size of DNA.

8. The most important feature in a plasmid to be used as a vector is
(a) Origin of replication (on)
(b) Presence of a selectable marker
(c) Presence of sites for restriction endpnuclease
(d) Its size
Answer. (a) The most important feature in a plasmid to be used as a vector is origin of replication (ori).

9. While isolating DNA from bacteria, which of the following enzymes is not used?
(a) Lysozyme (b) Ribonuclease
(c) Deoxyribonuclease (d) Protease
Answer. (c) While isolating DNA from bacteria lysozyme, ribonuclease and protease, enzymes is used.

10. Which of the following has popularised the PCR (polymerase chain reactions)?
(a) Easy availability of DNA template
(b) Availability of synthetic primers
(c) Availability of cheap deoxyribonucleotides
(d) Availability of ‘Thermostable’ DNA polymerase
Answer. (d) If the process of replication ‘of DNA is repeated many times, the segment of DNA can be amplified to approximately billion times i.e., 1 billion copies are made. Such repeated amplification is achieved by the use of a thermostable DNA polymerase (isolated from a thermophilic bacterium Thermits aquaticus), which is active during the high temperature induced denaturation of double stranded DNA.

11. An antibiotic resistance gene in a vector usually helps in the selection of
(a) Competent cells (b) Transformed cells
(c) Recombinant cells (d) None of the above
Answer. (b) An antibiotic resistance gene in a vector usually helps in the selection of transformed cells.

12. Significance of ‘heat shock’ method in bacterial transformation is to facilitate
(a) Binding of DNA to the cell wall
(b) Uptake of DNA through membrane transport proteins
(c) Uptake of DNA through transient pores in the bacterial cell wall
(d) Expression of antibiotic resistance gene
Answer. (c) Significance of ‘heat shock’ method in bacterial transformation is to facilitate uptake of DNA through transient pores in the bacterial cell wall.

13. The role of DNA ligase in the construction of a recombinant DNA molecule is
(a) Formation of phosphodiester bond between two DNA fragments
(b) Formation of hydrogen bonds between sticky ends of DNA fragments
(c) Ligation of all purine and pyrimidine bases .
(d) None of the above
Answer. (a) The role of DNA ligase in the construction of a recombinant DNA molecule is formation of phosphodiester bond between two DNA fragments.

14. Which of the following is not a source of restriction endonuclease?
(a) Haemophilus influenzae (b) Escherichia coli
(c) Entamoeba coli . (d) Bacilliusquifacieus
Answer. (c) Haemophilus influenzae, Escherichia coli and Bacillius quifacieus are source of restriction endonuclease.

15. Which of the following steps are catalysed by Taq polymerase in a PCR reaction?
(a) Denaturation of template DNA
(b) Annealing of primers to template DNA
(c) Extension of primer end on the template DNA
(d) All of the above
Answer. (c) Taq polymerase is used between annealing and extension, help in extension of primer end on the template DNA.

16. A bacterial cell was transformed with a recombinant DNA that was generated using a human gene. However, the transformed cells did not produce the desired protein. Reasons could be
(a) Human gene may have intron which bacteria cannot process
(b) Amino acid codons for humans and bacteria are different
(c) Human protein is formed but degraded by bacteria
(d) All of the above .
Answer. (a) A bacterial cell was transformed with a recombinant DNA that was generated using a human gene However, the transformed cells did not produce the desired protein because human gene may have intron which bacteria cannot process.

17. Which of the following should be chosen for best yield if one were to produce a recombinant protein in large amounts?
(a) Laboratory flask of largest capacity
(b) A stirred-tank bioreactor without inlets and outlets
(c) A continuous culture system
(d) Any of the above
Answer. (c)

• After having cloned the gene of interest and having optimised the conditions to induce the expression of the target protein one has to consider producing it on a large scale. Small volume cultures cannot yield appreciable quantities of products.
•  To produce in large quantities the development of bioreactors, where large volume (100-1000 litres) of culture (continuous) can be processed, was required.

18. Who among the following was awarded the Nobel Prize for the development of PCR technique?
(a) Herbert Boyer (b) Hargovind Khurana
(c) KaryMullis (d) Arthur Komberg
Answer. (c) Karv Mullis was awarded the Nobel Prize for the development of PCR technique.

19. Which of the following statements does not hold true for restriction enzyme?
(a) It recognises a palindromic nucleotide sequence
(b) It is an endonuclease .
(c) It is isolated from viruses
(d) It produces the same kind of sticky ends in different DNA molecules
Answer. (c) Restriction enzyme:

• recognises a palindromic nucleotide sequence
• is an endonuclease isolated from bacteria
• produces the same kind of sticky ends in different DNA molecules

Very Short Answer Type Questions
1. How is copy number of the plasmid vector related to yield of recombinant protein?
Answer. Higher the copy number of vector plasmid, higher the copy number of gene and consequently, protein coded by the gene is produced in high amount.

2. Would you choose an exonuclease while producing a recombinant DNA molecule?
Answer. No, as exonuclease acts on the free ends of linear DNA molecule. Therefore, instead of producing DNA fragments with sticky ends, it will shorten or completely degrade the DNA fragment containing the gene of interest, and the circular plasmid (vector) will not get cut as it lacks free ends.

3. What does ‘H’ in ‘d’ and ‘III’ refer to in the enzyme Hind III?
Answer. ‘H’ from name of genus Haemophilus, ‘d’ from name of strain Rd. TIT indicates the order in which the enzyme were isolated from Rd strain of bacteria.

4. Restriction enzymes should not have more than one site of action in the cloning site of a vector. Comment.
Answer. If the restriction enzymes have more than one recognition site in a vector, than the vector itself will get fragmented on treatment with the restriction enzyme.

5. What does ‘competent’ refer to in competent cells used in transformation experiments?
Answer. Competent means bacterial cells, on treatment with CaCl₂, are made capable of taking up foreign DNA.

6. What is the significance of adding proteases at the time of isolation of genetic material (DNA)?
Answer. Role of proteases is to degrade the proteins present inside a cell (from which DNA is being isolated). If the proteins are not removed from DNA preparation then they could interfere with any downstream treatment of DNA (such action of restriction endonuclease, DNA ligase etc).

7. While doing a PCR, ‘denaturation’ step is missed. What will be its effect on the process?
Answer. If denaturation of double-stranded DNA does not take place, then primers will not be able to anneal to the template, no extension will take place, hence no amplification will occur.

8. Name a recombinant vaccine that is currently being used in vaccination program.
Answer. Hepatitis B recombinant vaccine-engerix is used for vaccination of hepatitis virus.

9. Do biomolecules (DNA, protein) exhibit biological activity in anhydrous conditions?
Answer. No, biomolecules like DNA and protein cannot exhibit biological activity in anhydrous conditions. Hence, life is unsustainable without water.

10. What modification is done on the Ti plasmid of Agrobacterium tumifaciens to convert it into a cloning vector?
Answer. The tumor inducing (Ti) plasmid of Agrobacterium tumifaciens has now been modified (disarmed) into a cloning vector which is no more pathogenic to the plants but is still able to use the mechanisms to deliver genes of our interest into a variety of plants.

Short Answer Type Questions
1. What is meant by gene cloning?
Answer. Gene cloning refers to a process in which a gene of interest is ligated to a vector. The recombinant DNA thus produced is introduced in a host cell by transformation. Each cell gets one DNA molecule and when the transformed cell grows to a bacterial colony, each cell in the colony has a copy of the gene. This is precisely gene cloning.

2. Both a wine maker and a molecular biologist who had developed a recombinant vaccine claim to be biotechnologists. Who in your opinion is correct?
Answer. Both. As biotechnology is a very wide area which deals with techniques of using a ‘natural’ organism (or its parts) as well as genetically modified organism to produce products and processes useful for mankind. A wine maker employs a strain-of yeast to produce wine by fermentation (a natural phenomenon), while the molecular biologist has cloned gene for the antigen (that is used as vaccine) in an organism which allows the production of the antigen in large amount. .

3. A recombinant DNA molecule was created by ligating a gene to a plasmid vector. By mistake, an exonuclease was added to the tube containing the recombinant DNA. How does this affect the next step in the experiment, i.e. bacterial transformation?
Answer. The experiment will not likely to be affected as recombinant DNA molecule is circular closed, with no free ends. Hence, it will not be a substrate for exonuclease enzyme which removes nucleotides from the free ends of DNA.

4. Restriction enzymes that are used in the construction of recombinant DNA
are endonucleases which cut* the DNA at ‘specific-recognition sequence’. What would be the disadvantage if they do not cut the DNA at specific- recognition sequence? .
Answer. If the restriction enzymes would cut DNA at random sites instead of at specific sites, then the DNA fragments obtained will not have ‘sticky ends’. In the absence of sticky ends, construction of recombinant DNA molecule would not be possible.

5. A plasmid DNA and a linear DNA (both are of the same size) have one site for a restriction endonuclease. When cut and separated on agarose gel electrophoresis, plasmid shows one DNA band while linear DNA shows two fragments. Explain.
Answer. It is because plasmid is a circular DNA molecule. When cut with enzyme, it becomes linear but does not get fragmented. Whereas, a linear DNA molecule gets cut into two fragments. Hence, a single DNA band is observed for plasmid while two DNA bands are observed for linear DNA in agarose gel.

6. How does one visualise DNA on an agarose gel?
Answer. A compound called Ethidium Bromide stains DNA, which on irradiating with Ultraviolet, fluoresce gives orange light. Hence, DNA fragments appear as orange band in the presence of Ethidium Bromide and UV.

7. A plasmid without a selectable marker was chosen as vector for cloning a gene. How does this affect the experiment?
Answer. In a gene, cloning experiment, first a recombinant DNA molecule is constructed, where the gene of interest is ligated to the vector (the step would not be affected) and introduced inside the host cell (transformation). Since, not all the cells get transformed with the recombinant/plasmid DNA, in the absence of selectable marker, it will be difficult to distinguish between transformants and non-transformant, because role of selectable marker is in the selection of transformants.

8. A mixture of fragmented DNA was electrophoresed in an agarose gel. After staining the gel with ethidium bromide, no DNA bands were observed. What could be the reason?
Answer. The reasons are as follows:
(i) DNA sample that was loaded on the gel may have got contaminated with nuclease (exo-or endo-or both) and completely degraded.
(ii) Electrodes were put in opposite orientation in the gel assembly that is anode towards the wells (where DNA sample is loaded). Since DNA molecules are-negatively charged, they move towards anode and hence move out of the gel instead of moving into the matrix of gel.
(iii) Ethidium bromide was not added at all or was not added in sufficient concentration and so DNA was not visible.

9. Describe the role of CaCl₂ in the preparation of competent cells.
Answer. CaCl₂ is known to increase the efficiency of DNA uptake to produce transformed bacterial cells. The divalent Ca²⁺ ions supposedly create transient pores on the bacterial cell wall by which the entry of foreign DNA is facilitated into the bacterial cells.

10. What would happen when one grows a recombinant bacterium in a bioreactor but forget to add antibiotic to the medium in which the recombinant is growing?
Answer. In the absence of antibiotic, there will be no pressure on recombinants to retain the plasmid (containing the gene of your interest). Since, maintaining a high copy number of plasmids is a metabolic burden to the microbial cells, will thus tend to lose the plasmid.

11. Identify and explain steps ‘A’, ‘B’ and ‘C’ in the PCR diagram given below.

Answer. A—Denaturation
B—Annealing
C—Extension of primers

12. Name the regions marked A, B and C.

Answer. A—Bam HI
B—Pst I .
C—Ampicillin resistance gene (amp^R)

Long Answer Type Questions
1. For selection of recombinants, insertional inactivation of antibiotic marker has been superseded by insertional inactivation of a marker gene coding for a chromogenic substrate. Give reasons.
Answer. Selection of recombinants due to inactivation of antibiotics is a laborious process as it requires:
(i) a vector with two antibiotic resistance marker
(ii) preparation of two kinds of media plate, with one antibiotic each.
Transformed cells are first plated on that antibiotic plate which has not been insertionally inactivated (ampicillin) and incubated overnight for growth of transformants. For selection of recombinants, these transformants are Replica plated on second antibiotic (tetracycline) plate (which got inactivated due to insertion of gene). Non-recombinants grow on both the plates (one carrying ampicillin and the other carrying tetracycline) while recombinants will grow only on ampicillin plate.
This entire exercise is laborious and takes more time (two overnight incubation) as well. However, if we choose the second option (insertional inactivation of a marker that produces colour in the presence of a chromogenic compound), we can distinguish between the recombinants and non-recombinants on a single medium plate (containing one antibiotic and the chromogenic compound) after overnight growth.
Hence I would choose a marker which produces a coloured compound but gets inactivated due to insertion of foreign DNA.

2. Describe the role of Agrobacterium tumafaciens in transforming a plant cell.
Answer. Agrobacterium tumafaciens harbours a mega plasmid called Ti-plasmid.
This has a T-DNA region flanked by left border and right border sequence. The T-DNA gets transferred and integrates with the host plant DNA. This property of Ti-plasmid has been exploited for cloning of gene of interest and stably integrating them in the plant genese. Therefore, by using Ti-plasmid or its derivatives, recombinant plant cells with desired genes of interest stably integrated in the plant genome has been successfully produced.

3. Illustrate the design of a bioreactor. Highlight the difference between a flask in your laboratory and a bioreactor which allows cells to grow in a continuous culture system.
Answer. Small volume cultures cannot yield appreciable quantities of products. To produce in large quantities, the development of bioreactors, where large volumes (100-1000 litres) of culture can be processed, was required. Thus, bioreactors can be thought of as vessels in which raw materials are biologically converted into specific products, individual enzymes, etc., using microbial plant, animal or human cells. A bioreactor provides the optimal conditions for achieving the desired product by providing optimum growth conditions (temperature, pH, substrate”, salts, vitamins, oxygen).

The post NCERT Exemplar Problems Class 12 Biology Biotechnology: Principles and Processes appeared first on Learn CBSE.

NCERT Exemplar Problems Class 12 Biology Chapter 12 Biotechnology and its Applications

Multiple Choice Questions
Single Correct Answer Type
1. Bt cotton is not
(a) A GM plant
(b) Insect resistant
(c) A bacterial gene expressing system
(d) Resistant to all pesticides
Answer. (d) Bt cotton is a GM plant, insect resistant and a bacterial gene expressing system.

2. C-peptide of human insulin is
(a) A part of mature insulin molecule
(b) Responsible for formation of disulphide bridges
(c) Removed during maturation of pro-insulin to insulin
(d) Responsible for its biological activity.
Answer. (c) C-peptide of human insulin is removed during maturation of pro-insulin to insulin. •

3. GEAC stands for
(a) Genome Engineering Action Committee
(b) Ground Environment Action Committee
(c) Genetic Engineering Approval Committee .
(d) Genetic and Environment Approval Committee
Answer. (c) GEAC stands for Genetic Engineering Approval Committee.

4. α -1 antitrypsin is
(a) An antacid
(b) An enzyme
(c) Used to treat arthritis
(d) Used to treat emphysema. 
Answer. (d) α -1 antitrypsin is used to treat emphysema.

5. A probe which is a molecule used to locate specific sequences in a mixture of, DNA or RNA molecules could be
(a) A single stranded RNA
(b) A single stranded DNA
(c) Either RNA or DNA
(d) Can be ssDNA but not ssRNA
Answer. (c) A single stranded DNA or RNA, tagged with a radioactive molecule is called Probe. .

6. Choose the correct option regarding Retrovirus:
(a) An RNA virus that can synthesise DNA during infection
(b) A DNA virus that can synthesise RNA during infection
(c) A ssDNA virus
(d) AdsRNAvirus
Answer. (a) Retrovirus is an RNA virus that can synthesise DNA during infection.

7. The site of production of ADA in the body is
(a) Erythrocytes ‘
(b) Lymphocytes
(c) Blood plasma
(d) Osteocytes
Answer. (b) The site of production of ADA in the body is lymphocytes.

8. A protoxin is
(a) A primitive toxin .
(b) A denatured toxin
(c) Toxin produced by protozoa
(d) Inactive toxin
Answer. (d) Bt toxin is a protein. This toxin does not kill the Bacillus because the Bt toxin protein exists as inactive protoxins but once an insect ingest the inactive toxin, it is converted into an active form of toxin due to the alkaline pH of the gut which solubilise the crystals.

9. Pathophysiology is the
(a) Study of physiology of pathogen
(b) Study of normal physiology of host
(c) Study of altered physiology of host
(d) None of the above
Answer. (c) Pathophysiology is the study of altered physiology of host.

10. The trigger for activation of toxin of Bacillus thuringiensis is
(a) Acidic pH of stomach
(b) High temperature
(c) Alkaline pH of gut
(d) Mechanical action in the insect gut
Answer. (c) The trigger for activation of-toxin of Bacillus thuringiensis is alkaline pH of gut.

11. Golden rice is
(a) A variety of rice grown along the yellow river in China
(b) Long stored rice having yellow colour tint
(c) A transgenic rice having gene for b-carotene
(d) Wild variety of rice with yellow coloured grains.
Answer. (c) Golden rice is a transgenic rice having gene for β-carotene.

12. In RNAi, genes are silenced using
(a) ss DNA (b) ds DNA
(c) dsRNA (d) ssRNA
Answer. (c) RNAi takes place in all eukaryotic organisms as a method of cellular defense. This method involves silencing of a specific mRNA due to the complementary dsRNA molecule that binds to and prevents translation of the mRNA (silencing).

13. The first clinical gene therapy was done for the treatment of
(a) AIDS . (b) Cancer
(c) Cystic fibrosis 
(d) SCID (Severe Combined Immuno Deficiency resulting from deficiency of ADA)
Answer. (d) A first clinical gene therapy was given in 1990 to a 4 year old girl with ADA (Adenosine deaminase) deficiency. ADA deficiency causes SCID (Severe Combined Immuno Deficiency) in which B-lymphocytes and T-lymphocytes are not formed. .

14. ADA is an enzyme which i§ deficient in a genetic disorder SCID. What is the full form of ADA?
(a) Adenosine deoxy aminase (b) Adenosine deaminase
(c) Aspartate deaminase  (d) Arginine deaminase
Answer. (b) ADA stands for Adenosine deaminase.

15. Silencing of a gene could be achieved through the use of
(a) (RNAi) only .(b) Antisense RNA only
(c) By both (d) None of the above
Answer. (c) Silencing of a gene could be achieved through the use of (RNAi) only and Antisense RNA only.

Very Short Answer Type Questions
1. In view of the current food crisis, it is said, that we need another green revolution. Highlight the major limitations of the earlier green revolution.
Answer. Limitations of green revolution:

1. The Green Revolution succeeded in tripling the food supply but yet it was not enough to feed the growing human population. Increased yields have partly been due to the use of improved crop varieties, but mainly due to the use of better management practices and use of agrochemicals (fertilisers and pesticides). However, for farmers in the developing world, agrochemicals are often too expensive, and further increases in yield with existing varieties are not possible using conventional breeding.
2. Use of agrochemicals causes eutrophication in aquatic ecosystems and biomagnification in terrestrial ecosystems.
3. Water logging and soil salinity are some of the problems that have come in the wake of green revolution.

2. Expand GMO. How is it different from a hybrid?
Answer. GMO stands for Genetically Modified Organism. It differs from a hybrid because in a hybrid,cross is done between total genomes of two species or strains, where as in a GMO, foreign gene(s) is introduced in the organism and is usually maintained as extra-chromosomal entity or is integrated into the genome of the organism. .

3. Differentiate between diagnostics and therapeutics. Give one example and for each category.
Answer. A diagnostic technique helps us to identify a disease. Example: ELISA is a test for HIV.
A therapeutic agent on the other hand helps in the treatment of a disease. Example: Antibiotics for bacterial infections.

4. Give the full form of ELISA. Which disease can be detected using it? Discuss the principle underlying the test.
Answer. Enzyme Linked Immuno-sorbent Assay (ELISA) is used for the detection of AIDS. ELISA is based on the principle of antigen-antibody interaction. Infection by pathogen can be detected by the presence of antigens (proteins, glycoproteins, etc.) or by detecting the antibodies synthesised against the pathogen.

5. Can a disease be detected before its symptoms appear? Explain the principle involved.
Answer. Presence of a pathogen (bacteria, viruses, etc.) is normally suspected only when the pathogen has produced a disease symptom. By this time the concentration of pathogen is already very high in the body. However, very low concentration of a bacteria or virus (at a time when the symptoms of the disease are not yet visible) can be detected by amplification of their nucleic acid by PCR.

6. Write a short note on biopiracy highlighting the exploitation of developing countries by the developed countries.
Answer. Biopiracy is the term used to refer to the use of bio-resources by multinational companies and other organisations without proper authorisation from the countries and people concerned without compensatory payment. Most of the industrialised nations are rich financially but poor in biodiversity arid traditional knowledge. In contrast the developing and the underdeveloped world is rich in biodiversity and traditional knowledge related to bio-resources. Traditional knowledge related to bio-resources can be exploited to develop modem applications and can also be used to save time, effort and expenditure during their commercialisation.

7. Many proteins are secreted in their inactive form. This is also hue of many toxic proteins produced by micro organisms. Explain how the mechanism is useful for the organism producing the toxin.
Answer. Many proteins including certain toxins are secreted in their inactive form. They get activated, only when exposed to a specific trigger (pH. temperature etc.). It is advantageous to the bacteria producing it because the bacteria does not get killed due to the action of protein.

8. While creating genetically modified organisms, genetic barriers are not respected. How can this be dangerous in the long run?
Answer. The manipulation of living organisms by the human race cannot go on any further, without regulation. Some ethical standards are required to evaluate the morality of all human activities that might help or harm living organisms. Going beyond the morality of such issues, the biological significance of such things is also important. Genetic modification of organisms can have unpredictable results when such organisms are introduced into the ecosystem.

9. Why has the Indian Parliament cleared the second amendment of the country’s patents bill?
Answer. The Indian Parliament has recently cleared the second amendment of the Indian Patents Bill, that take such issues into consideration, including patent terms emergency provisions and research and development initiative.

10. Give any two reasons why the patent on Basmati should not have gone to an American Company.
Answer. (i) Rice is an important food grain, the presence of which goes back ‘ thousands of years in Asia’s agricultural history. There is an estimated 200,000 varieties of rice in India alone. The diversity of rice in India is one of the richest in the world. Basmati rice is distinct for its unique aroma and flavour and 27 documented varieties of Basmati are grown in India.
(ii) There is a reference to Basmati in ancient texts, folklore and poetry.

11. How was Insulin obtained before the advent of rDNA technology? What were the problems encountered?
Answer. Insulin used for diabetes was earlier extracted from pancreas of slaughtered cattle and pigs. Insulin from an animal source, though caused some patients to develop allergy or other types of reactions to the foreign protein.

12. With respect to understanding diseases,’discuss the importance of transgenic animal models.
Answer. Many transgenic animals are designed to increase our understanding of how genes contribute to the development of disease. These are specially made to serve as models for human diseases so that investigation of new treatments for diseases is made possible. Today transgenic models exist for many human diseases such as cancer, cystic fibrosis, rheumatoid arthritis and Alzheimer’s.

13. Name the first transgenic cow. Which gene was introduced in this cow?
Answer. Rosie was the name of the first transgenic cow. Gene for human alpha lactalbumin was introduced in its gene, which made the milk nutritionally richer.

14. PCR is a useful tool for early diagnosis of an infectious disease. Elaborate.
Answer. PCR is a very sensitive technique which enables the specific amplification of desired DNA from a limited amount of DNA template. Hence, it can detect the presence of an infectious organism in the infected patient at an early stage of infection (even before the infectious organism has multiplied to large number).

15. What is GEAC and what are its objectives?
Answer. GEAC (Genetic Energy Approval Committee) is an Indian government organisation. Its objective are to:
(a) examine the validity of GM (Genetic modification of organism) research.
(b) inspect the safety of introducing GM for public services.

16. For which variety of Indian rice, the patent was filed by a USA Company?
Answer. Indian Basmati was crossed with semi-dwarf variety and was claimed as a new variety for which the patent was filed by a USA company.

17. Discuss the advantages of GMO.
Answer. Plants, bacteria, fungi and animals whose genes have been altered by manipulation are called Genetically Modified Organisms (GMO). GM plants have been useful in many ways. Genetic modification has:

1.  Made crops more tolerant to abiotic stresses (cold, drought, salt, heat).
2.  Reduced reliance on chemical pesticides (pest-resistant crops).
3. Helped to reduce post-harvest losses.
4. Increased efficiency of mineral usage by plants (this prevents early exhaustion of fertility of soil).
5. Enhanced nutritional value of food, e.g., Vitamin ‘A’ enriched rice.

Short Answer Type Questions
1. Gene expression can be controlled with the help of RNA. Explain the method with an example.
Answer. Gene expression can be controlled by using RNA molecule. The technology is called RNA interference or RNAi. It is used to block the expression of certain genes and also referred to as gene silencing. During this process a complementary RNA to the mRN A being produced by the gene is introduced into the cell. This RNA binds to the mRNA making it double stranded and therefore stops translation. Resistance to nematode Meloidegyne incognita in tomato has been achieved by this method.

2. Ignoring our traditional knowledge can we prove costly in the area of biological patenting. Justify.
Answer. Most of the industrialised nations are rich financially but poor in biodiversity and traditional knowledge. In contrast the developing and the underdeveloped world is rich in biodiversity and traditional knowledge related to bio-resources. Traditional knowledge related to bio-resources can be exploited to develop modem applications and can also be used to save time, effort and expenditure during their commercialisation. There has been” growing realisation of the injustice, inadequate compensation and benefit sharing between developed and developing countries. Therefore, some nations are developing laws to prevent such unauthorised exploitation of their bio-resources and traditional knowledge.

3. Highlight any four areas where genetic modification of plants has been useful.
Answer. Plants, bacteria, fungi and animals whose genes have been altered by manipulation_are called Genetically Modified Organisms (GMO). GM plants have been useful in many ways. Genetic modification has
(i) Made crops more tolerant to abiotic stresses (cold, drought, salt, heat).
(ii) Reduced reliance on chemical pesticides (pest-resistant crops).
(iii) Helped to reduce post-harvest losses.
(iv) Increased efficiency of mineral usage by plants (this prevents early exhaustion of fertility of soil).

4. What is a recombinant DNA vaccine? Give two examples.
Answer. A recombinant vaccine is a vaccine produced through recombinant DNA technology. This involves inserting the DNA encoding an antigen that stimulates an immune response into bacterial or mammalian cells. Recombinant DNA technology has allowed the production of antigenic polypeptides of pathogen in bacteria or yeast. Vaccines produced using this .approach allow large scale production and hence greater availability for immunisation, e.g., hepatitis B vaccine (Recombivax HB) produced from yeast. As of June 2015 one human DNA vaccine had been approved for human use, the single-dose Japanese encephalitis vaccine called IMOJEV, released in 2010 in Australia.

5. Why is it that the line of treatment for a genetic disease is different from infectious diseases?
Answer. If a person is born with a hereditary disease, can a corrective therapy be taken for such a disease? Gene therapy is an attempt to do this. Gene therapy is a collection of methods that allows correction of a gene defect that has been diagnosed in a child/embryo.

6. Discuss briefly how a probe is used in molecular diagnostics.
Answer. A single stranded DNA or RNA, tagged with a radioactive molecule (probe) is allowed to hybridise to its complementary DNA in a clone of cells followed by detection using autoradiography. The clone having the mutated gene will hence not appear on the photographic film, because the probe will not have complementarity with the mutated gene.

7. Who was the first patient who was given gene therapy? Why was the given treatment recurrent in nature?
Answer. The first clinical gene therapy was given in 1990 to a 4-year old girl with adenosine deaminase (ADA) deficiency. This enzyme is crucial for the immune system to function. The disorder is caused due to the deletion of the gene for adenosine deaminase. In some children ADA deficiency can be cured by bone marrow transplantation; in others it can be treated by – enzyme replacement therapy, in which functional ADA is given to the patient by injection. But the problem- with both of these approaches that they are not completely curative. As a first step towards gene therapy, lymphocytes from the blood of the patient are grown in a culture outside the body. A functional ADA cDNA (using a retroviral vector) is then introduced into these lymphocytes, which are subsequently returned to the patient. However, as these cells are not immortal, the patient requires “periodic infusion of such genetically engineered lymphocytes. However, if the gene isolate from marrow cells producing ADA is introduced into cells at early embryonic stages, it could be a permanent cure.

8. Taking examples under each category, discuss upstream and downstream processing.
Answer. Upstream processing: Biotechnological processes can be separated into upstream processes and dpwnstream processes. The upstream process is defined as the entire process from DNA isolation and culture expansion of the cells until final product.
Downstream processing: After completion of the biosynthetic stage, the product has to be subjected through a series of processes before it Is ready for marketing as a finished product. The processes include separation and purification, which are collectively referred to as downstream processing. The product has to be formulated with suitable preservatives. Such formulation has to undergo through clinical trials as in case of drugs. Strict quality control testing for each product is also required. The downstream processing and quality control testing vary from product to product.

9. Define Antigen and Antibody. Name any two diagnostic kits based upon them.
Answer. An antigen is a foreign substance that elicits the formation of an antibody. Antibody is a protein that is synthesised in response to an antigen. Antigen and antibody show high degree of specificity in binding each other. Two diagnostic kits based on antigen-antibody interaction are:
(a) ELISA for HIV.
(b) Pregnancy test kits.

10. ELISA technique is based on the principles of antigen-antibody interaction. Can this techique be used in the molecular diagnosis of a genetic disorder, such as phenyketonuria?
Answer. Yes. One can use antibody against the enzyme (that is responsible for the metabolism of phenylalanine) to develop ELISA based diagnostic technique. The patient where the enzyme protein is absent would give negative result in ELISA when compared to normal individual.

11. How is a mature, functional insulin hormone different from its prohormone form?
Answer. Mature functional insulin is obtained by processing of pro-hormone which contains extra peptide called C-peptide. This C-peptide is removed during maturation of pro-insulin to insulin.

12. Gene therapy is an attempt to correct a genetic defect by providing a normal gene into the individual. By this’the normal function can be restored. An alternate method would be to provide the gene product (protein/enzyme) known as enzyme replacement therapy, which would also restore the function. Which in your opinion is a better option? Give reason for your answer.
Answer. Gene therapy is an attempt to correct a genetic defect by providing a normal gene into the individual. By this the normal function can be restored. Alternate method would be to provide the gene product (protein/ enzyme) know as enzyme replacement therapy, which would also restore the function. Which in your opinion is a better option? Give reason for your answer.

13. Transgenic animals are the animals in which a foreign gene is expressed. Such animals can be used to study the fundamental biological process, phenomenon as well as for producing products useful for mankind. Give one example for each type.
Answer. Study of basic biological process—how genes are regulated, how they affect the normal functions of the body and its development. Transgenic cow, Rosie is an example for the second category.

14. When a foreign DNA is introduced into an organism, how is it maintained in the host and how is it transferred to the progeny of the organism?
Answer. Foreign gene is usually ligated to a plasmid vector and introduced in the host. As plasmid replicates, and makes multiple copies of itself, so does the foreign gene gets replicated and its copes are made. When the host organism divides, its progeny also receives the plasmid DNA containing the foreign gene.

15. Bt cotton is resistant to pest, such as lepidopteron, dipterans and coleopterans. Is Bt cotton also resistant to other pests as well?
Answer. Bt cotton is made resistant to only certain specific taxa of pests. It is quite likely that in future,some other pests may infest this Bt cotton plants. It has similar immunisation against small-pox which does not provide immunity against other pathogens like those that cause cholera, typhoid etc.

Long Answer Type Questions
1. A patient is suffering from ADA deficiency. Can he be cured. How?
Answer. The first clinical gene therapy was given in 1990 to a 4-year old girl with adenosine deaminase (ADA) deficiency. This enzyme is crucial for the immune system to function. The disorder is caused due to the deletion of the gene for adenosine deaminase. In some children ADA deficiency can be cured by bone marrow transplantation; in others it can be treated by enzyme replacement therapy, in which functional ADA is given to the patient by injection. But the problem with both of these approaches that they are not completely curative. As a first step towards gene therapy, lymphocytes from the blood of the patient are grown in a culture outside the body. A functional ADA cDNA (using a retroviral vector) is then introduced into these lymphocytes, which are subsequently returned to the patient. However, as these cells are not immortal, the patient requires periodic infusion of such genetically engineered lymphocytes. However, if the gene isolate from marrow cells producing ADA is introduced into cells at early embryonic stages, it could be a permanent cure.

2. Define transgenic animals. Explain in detail any four areas where they can be utilised.
Answer. Transgenic animals are the products of genetic engineering and express specific gene(s) from totally unrelated source. Following are the four main areas where they can be utilised: .
(1) To study normal physiology and development these animals can be used to study as to which factor/gene products are needed at what time of development. By expression of certain genes, they help scientists to understand the normal gene expression at various stages of growth and development. 4
(2) Study of Diseases: Transgenic animals can be created to serve as models for various human diseases. They also help us understand the involvement of various genes in diseases like cancer, Parkinson’s disease etc.
(3) Vaccine safety: Transgenic animals can be used to test vaccines like polio vaccine. Transgenic mice have shown promising results in this area and would replace the vaccine testing on monkeys in the years to come.
(4) Chemical safety testing: Transgenic animals are created which are more sensitive to certain chemicals/drugs. These are used to study the toxicity or side effects of that chemical/drug. The advantage is that we get results faster.

3. You have identified a useful gene in bacteria. Make a flow chart of the steps that you would follow to transfer this gene to a plant.
Answer. After identifying a useful gene in bacteria, following steps should be undertaken:

4. Highlight five areas where biotechnology has influenced our lives.
Answer. The applications of biotechnology include:

1. therapeutics and diagnostics
2. genetically modified crops for agriculture
3. processed food
4. bioremediation
5. waste treatment and energy production.

5. What are the various advantages of using genetically modified plants to increase the overall yield of the crop?
Answer. GM plants have been useful in many ways. Genetic modification has:

1. Made crops more tolerant to abiotic stresses (cold, drought, salt, heat).
2.  Reduced reliance on chemical pesticides (pest-resistant crops).
3. Helped to reduce post-harvest losses.
4. Increased efficiency of mineral usage by plants (this prevents early exhaustion of fertility of soil).
5. Enhanced nutritional value of food>e.g., Vitamin ‘A’ enriched rice.

In addition to these uses, GM has been used to create tailor-made plants to supply alternative resources to industries, in the form of starches, fuels and pharmaceuticals.

6. Explain with the help of one example how genetically modified plants can:
(a) Reduce usage of chemical pesticides .
(b) Enhance nutritional value of food crops
Answer. (a) Reduce usage of chemical pesticides: Bt toxin is produced by a bacterium called Bacillus thuringiensis (Btfor short). Bt toxin gene has been cloned from the bacteria and been expressed in plants to provide resistance to insects without the need for insecticides; in effect created a bio-pesticide. Examples are Bt cotton, Bt com, rice, tomato, potato and soyabean etc.
(b) Enhance nutritional value of food crops: Golden rice is the transgenic variety of basmati rice which gives high yield and rich in vitamin A,
so it is used in the deficiency of vitamin-A causing night blindness and skin disorder.

7. List the disadvantages of insulin obtained from the pancreas of slaughtered cows and pigs:
Answer.

1. Insulin being a hormone is produced in very little amounts in the body.
   Hence, a large number of animals need to be sacrificed for obtaining small quantities of insulin. This makes the cost of insulin very high. [Demand being many fold higher than supply].
2. Slaughtering of animals is also not ethical.
3. There is potential of immune response in humans against the administered insulin which is derived from animals.
4.  There is possibility of slaughtered animals being infested with some infectious micro organism which may contaminate insulin.

8. List the advantages of recombinant insulin.
Answer. Insulin used for diabetes was earlier extracted from pancreas of slaughtered cattle and pigs. Insulin from an animal source, though caused some patients to develop allergy or other types of reactions to the foreign protein. Insulin consists of two short polypeptide chains: chain A and chain B, that are linked together by disulphide bridges. In mammals, including humans, insulin is synthesised as a pro-hormone (like a pro-enzyme, the pro-hormone also needs to be processed before it becomes a fully mature and functional hormone) which contains an extra stretch called the C peptide. This C peptide is not present in the mature insulin and is removed during maturation into insulin. The main challenge for production of insulin using rDNA techniques was getting insulin assembled into a mature form. In 1983, Eli Lilly an American company prepared two DNA sequences corresponding to A and B, chains of human insulin and introduced them in plasmids of E. coli to produce insulin chains. Chains A and B were produced separately, extracted and combined by creating disulfide bonds to form human insulin.

9. What is meant by the term bio-pesticide? Name and explain the mode of action of a popular bio-pesticide.
Answer. Biopesticide is a pesticide which is
(a) not chemical in nature
(b) more specific in action against the pest
(c) safer for environment than chemical pesticides
A popularly known bio-pesticide is Bt toxin, which is produced by a bacterium called Bacillus thuringiensis. Bt toxin gene has been cloned from this bacterium and expressed in plants. Bt toxin protein when ingested by the insect, gets converted to its active form due to the alkaline pH of the gut. The activated toxin binds to the surface of midgut epithelial cells and create pores that cause cell swelling and lysis and eventually kills the insect.

10. Name the five key tools for accomplishing the tasks of recombinant DNA technology. Also mention the functions of each tool.
Answer.

1. Restriction endonucleases: for cutting the desired DNA at desired places
2. Gel electrophoresis: for separating the desired DNA fragments
3. Ligase enzyme: for creating recombinant DNA molecule.
4. DNA delivery system: like electroporation, microinjection, gene gun method.
5. Competent host (usually bacteria/yeast): to take up recombinant DNA.

The post NCERT Exemplar Problems Class 12 Biology Biotechnology and its Applications appeared first on Learn CBSE.

NCERT Exemplar Problems Class 12 Biology Chapter 13 Organisms and Populations

Multiple Choice Questions
Single Correct Answer Type
1. Autecology is the
(a) Relation of a population to its environment
(b) Relation of an individual to its environment
(c) Relation of a community to its environment
(d) Relation of a biome to its environment
Answer. (b) Autecology is the relation of an individual to its environment.

2. Ecotone is
(a) A polluted area
(b) The bottom of a lake
(c) A zone of transition between two communities
(d) A zone of developing community
Answer. (c) Ecotone is a zone of transition between two communities.

3. Biosphere is
(a) A component in the ecosystem
(b) Composed of the plants present in the soil
(c) Life in the outer space
(d) Composed of all living organisms present on earth which interact with the physical environment.
Answer. (d) Biosphere is composed of all living organisms present on earth which interact with the physical environment.

4. Ecological niche is
(a) The surface area of the ocean
(b) An ecologically adapted zone
(c) The physical position and functional role of a species within the community
(d) Formed of all plants and animals living at the bottom of a lake.
Answer. (c) Ecological niche is the physical position and functional role of a species within the community.

5. According to Allen’s Rule, the mammals from colder climates have
(a) Shorter ears and longer limbs (b) Longer ears and shorter limbs
(c) Longer ears and longer limbs (d) Shorter ears and shorter limbs
Answer. (d) According to Allen’s Rule, the mammals from colder climates have shorter ears and shorter limbs.

6. Salt concentration (Salinity) of the sea measured in parts per thousand is
(a) 10-15 (b) 30-70 (c) 0-5 (d) 30-35
Answer. (d) The salt concentrations is measured as salinity in parts per thousand, is less than 5 in inland waters, 30-35 in the sea and greater than 100 percent in some hypersaline lagoons.

7. Formation of tropical forests needs mean annual temperature and mean annual precipitation as
(a) 18-25°C and 150-400 cm (b) 5-15°C and 50-100 cm
(c) 30-50°C and 100-150 cm (d) 5-15°C and 100-200 cm
Answer. (a)

8. Which of the following forest plants controls the light conditions at the ground?
(a) Lianas and climbers  (b) Shrubs
(c) Tall trees (d) Herbs
Answer. (c) Tall trees control the light conditions at the ground.

9. What will happen to a well growing herbaceous plant in the forest if it is transplanted outside the forest in a park?
(a) It will grow normally.
(b) It will grow well because it is planted in the same locality.
(c) It may not survive because of change in its micro climate.
(d) It grows very well because the plant gets more sunlight.
Answer. (c) A well growing herbaceous plant in the forest if it is transplanted outside the forest in a park. It may not survive because of change in its micro climate.

10. If a population of 50 paramoecium present in a pool increases to 150 after an hour, what would be the growth rate of population?
(a) 50 per hour (b) 200 per hour
(c) 5 per hour (d) 100 per hour
Answer. (d)
Initial population of paramecium Pi = 50
After one hour population of paramecium Pf = 150
Growth rate after one hour = Pf – Pi
= 150-50= 100

11. What would be the per cent growth or birth rate per individual per hour for the same population mentioned in the previous question (Question 10)?
(a) 100 (b) 200 (c) 50 (d) 150
Answer. (b)

12. A population has more young individuals compared to the older individuals. What would be the status of the population after some years?
(a) It will decline
(b) It will stabilise
(c) It will increase
(d) It will first decline and then stabilise
Answer. (c) A population has more young individuals compared to the older individuals. It will increase the status of the population after some years.

13. What parameters are used for tiger census in our country’s national parks and sanctuaries?
(a) Pug marks only (b) Pug marks and faecal pellets .
(c) Faecal pellets only (d) Actual head counts
Answer. (b) Sometimes population size is indirectly estimated without actually counting them or seeing them. E.g. ; The tiger census in our National Parks and Tiger Reserves is often based on pug marks and faceal pellets.

14. Which of the following would necessarily decrease the density of a population in a given habitat?
(a) Natality > mortality (b) Immigration > emigration
(c) Mortality and emigration (d) Natality and immigration
Answer. (c) Mortality and emigration would necessarily decrease the density of a population in a given habitat.

15. A protozoan reproduces by binary fission. What will be the number of protozoans in its population after six generations?
(a) 128 (b) 24 (c) 64 (d) 32
Answer. (c)
Population after nth generations = 2n
Population after 6th generations = 26 =64

16. In 2005, for each of the 14 million people present in a country, 0.028 were born and 0.008 died during the year. Using exponential equation, the number of people present in 2015 is predicted as
(a) 25 millions (b) 17 millions
(c) 20 millions (d) 18 millions
Answer. (b)

17. Amensalism is an association between two species where
(a) One species is harmed and other is benefitted
(b) One species is harmed and other is unaffected
(c) One species is benefitted and other is unaffected
(d) Both the species are harmed.
Answer. (b)

18. Lichens are the associations of
(a) Bacteria and fungus (b) Algae and bacterium
(c) Fungus and algae (d) Fungus and virus
Answer. (c) Lichens are the associations of fungus and algae.

19. Which of the following is a partial root parasite? .
(a) Sandal wood (b) Mistletoe
(c) Orobanche (d) Ganoderma
Answer. (a) Sandal Wood is a partial root parasite.

20. Which one of the following organisms reproduces sexually only once in its life time?
(a) Banana plant (b) Mango
(c) Tomato (d) Eucalyptus
Answer. (a) Banana plant organisms reproduce sexually only once in its life time.

Very Short Answer Type Questions
1. Species that can tolerate narrow range of temperature are called
Answer. Stenothermic

2. What are Eurythermic species?
Answer. Species that tolerate wide range of temperature are called Eurythermic . species.

3. Species that can tolerate wide range of salinity are called .
Answer. Euryhaline

4. Define stenohaline species.
Answer. Species that tolerate narrow range of salinity are called stenohaline species.

5. What is the interaction between two species called?
Answer. Interspecific interaction

6. What is commensalism?
Answer. Commensalism is the interaction in which one species benefits and the other is neither harmed nor benefited.

7. Name the association in which one species produces poisonous substance or a change in environmental conditions that is harmful to another species.
Answer. Amensalism

8. What is Mycorrhiza?
Answer. Mycorrhiza is a symbiotic association between a fungus and the roots of higher plants.

9. Emergent land plants that can tolerate the salinities of the sea are called 
Answer. Mangroves

10. Why do high altitude areas have brighter sunlight and lower temperatures as compared to the plains?
Answer. High altitude areas have brighter sunlight because at high altitude there is a very low concentration of dust particles and atmospheric gases which absorbs the sunlight. There is a low atmospheric pressure at high altitudes. Lower atmospheric pressure results in lower temperatures at high altitudes.

11. What is homeostasis?
Answer. To maintain the constancy of internal environment despite varying external environmental conditions is called homeostasis.

12. Define aestivation.
Answer. Aestivation is a state of dormancy characterized by inactivity and a lowered metabolic rate in response to high temperatures and arid conditions.

13. What is diapause and its significance?
Answer. Under unfavourable conditions many zooplankton species in lakes and ponds are known to enter diapause, a stage of suspended development.

14. What would be the growth rate pattern, when the resources are unlimited?
Answer. Exponential.

15. What are the organisms that feed on plant sap and other plant parts called?
Answer. Phytophagous

16. What is high altitude sickness? Write its symptoms.
Answer. If one had ever been to any high altitude place (>3,500m like Rohtang Pass near Manali and Mansarovar .in Tibetan Autonomous Region), thepathological effect caused by acute exposure to low partial pressure of oxygen at high altitude is called altitude sickness. Its symptoms include nausea, fatigue and heart palpitations.

17. Give a suitable example for commensalism.
Answer. Cattle egret and grazing cattle.

18. Define ectoparasite and endoparasite, and give suitable examples.
Answer.

• Parasites that feed on the external surface of the host organism are called ectoparasites. The most familiar examples of this group are the lice on humans and ticks on dogs.
• Endoparasitesare those that live inside the host body at different sites (liver, kidney, lungs, red blood cells, etc.).'”The human liver fluke (a trematode parasite) is an endoparasite.

19. What is brood parasitism? Explain with the help of an example.
Answer. Brood parasitism in birds is a fascinating example of parasitism in which the parasitic bird lays its eggs in the nest of its host and let the host incubate them. During the course of evolution, the eggs of the parasitic bird have evolved to resemble the host’s egg in size and colour to reduce the chances of the host bird detecting the foreign eggs and ejecting them from the nest.

Short Answer Type Questions
1. Why are coral reefs not found in the regions from. West Bengal to Andhra Pradesh but are found in Tamil Nadu and on the east coast of India?
Answer. High salinity, optimal temperature and less siltation are essential to colonise corals. If siltation and fresh water inflow are very high, the corals don’t colonise. In contrast when the siltation and fresh water inflow by the rivers are very less, the coral do colonise.

2. If a fresh water fish is placed in an aquarium containing sea water, will the fish be able to survive? Explain giving reasons.
Answer. If a fresh water fish is placed in an aquarium containing sea water, it will not be able to survive because of the osmotic problems, they would face. Sea water is hypertonic as compared to fish, so it lost water through exosmosis and die due to dehydration.

3. Why do all the fresh water organisms have contractile vacuoles whereas , majority of marine organisms lack them?
Answer. In majority of fresh water organisms, contractile vacuoles is present which help in osmoregulation (remove excess water from body). In marine organism there is no need of removal of water from body (due to hyptonic condition), hence contractile vacuoles are absent.

4. Define heliophytes and sciophytes. Name a plant from your locality that is either heliophyte or sciophyte.
Answer.

• Heliophytes also called sun-loving plants, are those that require for their optimum growth full exposure to the sun. E.g., Mango
• Sciophytes also called shade-loving plants, are those plants that require reduced light intensity. E.g., Lycopodium

5. Why do submerged plants receive weaker illumination than exposed floating plants in a lake?
Answer. Submerged plants receive weaker illumination than exposed floating plants
in a lake because on passing of light through water much more amount of light is lost.

6. In a sea shore, the benthic animals live in sandy, muddy and rocky substrata
and accordingly developed the following adaptations.
a. Burrowing 
b. Building cubes
c. Holdfasts/peduncle
Find the suitable substratum against each adaptation.
Answer. a. Sandy, b. Muddy, c. Rocky

7. Categorise the following plants into hydrophytes, halophytes, mesophytes and xerophytes. Give reasons for your answers.
a. Salvinia b. Opuntia
c. Rhizophora d. Mangifera
Answer. a. Hydrophyte, b. Xerophyte, c. Halophyte, d. Mesophyte

8. In a pond, we see plants which are free-floating; rooted-submerged; footed emergent; rooted with floating leaves. Write the type of plants against each of them.

Answer. a. Submerged, b. Rooted emergent, c. Rooted with floating leaves, d. Free- floating, e. Rooted submerged

9. The density of a population in a habitat per unit area is measured in different units. Write the unit of measurement against the following:
a. Bacteria …………..
b. Banyan …………..
c. Deer …………..
d. Fish …………..
Answer. a. Nos. / Vol; b. Coverage / area; c. Biomass / area; d. Nos. / area; e. Wt. / area

10.

a. Label the three tiers 1, 2, 3 given in the above age pyramid.
b. What type of population growth is represented by the above age pyramid?
Answer. (a)
1. Pre-reproductive pogulation
2. Reproductive population
3. Post-reproductive population
(b) Expanding or growing population

11. In an association of two animal species, one is a termite which feeds on wood and the other is a protozoan Trichonympha present in the gut of the termite. What type of association they establish?
Answer. They shows mutualism.

12. Lianas are vascular plants rooted in the ground and maintain erectness of their stem by making use of other trees for support. They do not maintain direct relation with those trees. Discuss the type of association the lianas have with the trees.
Answer. This association is called commensalism.

13. Give the scientific names of any two micro organisms inhabiting the human intestine.
Answer. 1. Escherichia coli
2. Enterococcusfaecalis

14. What is a tree line?
Answer. When we go up the altitude, beyond a particular height no trees are found and the vegetation comprises only of shrubs and herbs. The altitude beyond which no tree is seen is known as tree line.

15. Define ‘zero population growth rate’. Draw an age pyramid for the same.
Answer. Yes. An inverted bell shaped age pyramid is obtained. The young of pre- reproductive age group individuals are less in number and both pre- reproductive and reproductive stages are in the same level.

16. List any four characters that are employed in human population census.
Answer. 1. Birthrates
2. Death rates
3. Sex ratio
4. Age distribution

17. Give one example for each of the following types.
(a) Migratory animal (b) Camouflaged animal
(c) Predator animal (d) Biological control agent
(e) Phytophagous animal (f) Chemical defense agent
Answer. (a) Migratory animal—Siberian crane, Salmon
(b) Camouflaged animal—Frog, insects
(c) Predator animal—Tiger, sparrow
(d) Biological control agent—Moth (against prickly pear cactus)
(e) Phytophagous animal—Insects like Locusta
(f) Chemical defense agent—Cardiac glycosides produced by Calotropis

18. Fill in the blanks:

Answer.

19. Observe the set of 4 figures A, B, C and D, and answer the following questions: •
(i) Which one of the figures shows mutualism?
(ii) What kind of association is shown in D?
(iii) Name the organisms and the association in C.
(iv) What role is the insect performing in B?

Answer. (i) Figure ‘A’ shows’ mutualism (plant-animal relationship).
(ii) Figure ‘D’ shows predation (leopard killing deer and eating it)
(iii) Figure ‘C’ shows commensalism (cattle egret and grazing cattle)
(iv) In figure ‘B’ insect is phytophagous that feed on sap of the flower.

Long Answer Type Questions
1. Comment on the following figures 1, 2 and 3:
A, B, C, D, G, P, Q, R, S are species

Answer. Fig. 1: It is a single population.and all individuals are of the same species, i.e. A—Individual interact among themselves and their environment.
Fig. 2: It is a community and it contains three populations of species A, B and C. They interact with each other and their environment.
Fig. 3: It is a biome. It contains three communities of which one is in climax and other two are in different stages of development. All three communities are in the same environment and they interact with each other and their environment.

2. An individual and a population has certain characteristics. Name these attributes with definitions.
Answer. A population has certain attributes that an individual organism does not. An individual may have births and deaths, but a population has birth rates and death rates. In a population these rates refer to per capita births and deaths, respectively. The rates, hence, expressed as change in numbers (increase or decrease) with respect to members of the population.
• Another attribute characteristic of a population is sex ratio. An individual is either a male or a female but a population has a sex ratio (e.g., 60 per cent of the population are females and 40 per cent males).
• A population at any given time is composed of individuals of different ages. If the age distribution (per cent individuals of a given age or age group) is plotted for the population, the resulting structure is called an age pyramid. For human population, the age pyramids generally show age distribution of males and females in a combined diagram. The shape of the pyramids reflects the growth status of the population (a) whether it is growing, (b) stable or (c) declining.

3. The following diagrams are the age pyramids of different populations. Comment on the status of these populations.

Answer. Fig. A: It is a pyramid shaped age pyramid. In this figure, the base, i.e., pre-reproductive stage is very large when compared with the reproductive
and past reproductive stages of the population. This type of age structure indicates that the population would increase rapidly.
Fig. B: It is an inverted bell shaped pyramid. In this figure, the pre- reproductive and reproductive stages are same. This type of age structure indicates that the population is stable.
Fig. C: It is ‘Urn’ shaped pyramid. In this figure, the pre-reproductive and reproductive stages are less than the post reproductive stage of this population. In this population, more older people are present. This type of age structure indicates that the population definitely is declining,

4. Comment on the growth curve given below.

Answer. A population growing in a habitat with limited resources show initially a lag phase, followed by phases of acceleration and deceleration and finally an asymptote, when the population density reaches the carrying capacity. A’ plot of N in relation to time (t) results in a sigmoid curve. This type of population growth is called Verhulst-Pearl Logistic Growth and is described by the following equation:

5. A population of Paramoecium caudatum was grown in a culture medium. After 5 days the culture medium became pvercrowded with Paramoecium and had depleted nutrients. What will happen to the population and what type of growth curve will the population attain? Draw the growth curve.
Answer. It shows logistic growth. (See Ans no. 4)

6. Discuss the various types of positive interactions between species.
Answer. Both the species benefit in mutualism. The interaction where one species is benefitted and the other is neither benefitted nor harmed is called commensalism.

7. In an aquarium two herbivorous species of fish are living together and feeding
on phytoplanktons. As per the Gause’s Principle, one of the species is to be eliminated in due course of time, but both are surviving well in the aquarium. Give possible reasons.
Answer. Each species has a specific position or functional role within the community, called niche. According to the Gausse’s principle, no two species can live in the same niche. In this case, two herbivorous species are living in the same niche and feeding on phytoplanktons. It may be because of the availability of sufficient phytoplanktons and or less number of individuals of the fish species. Of the two species might have occurred and though neither of the species have been eliminated, niche overlapping may effect the growth and development of individuals of the species.

8. While living in and on the host species, the animal parasite has evolved certain adaptations. Describe these adaptations with examples.
Answer. In accordance with their life styles, parasites evolved special adaptations such as the loss of unnecessary sense organs, presence of adhesive organs or suckers to cling on to the host, loss of digestive system and high reproductive capacity. The life cycles of parasites are often complex, involving one or two intermediate hosts or vectors to facilitate parasitisation of its primary host.

9. Do you agree that regional and local variations exist within each.biome? Substantiate your answer with suitable example.
Answer. Yes, regional and local variations exist within each biome. Regional and local variations within each biome lead to the formation of a wide variety of habitats. On planet Earth, life exists not just in a few favourable habitats but even in extreme and harsh habitats-^scorching Rajasthan desert, perpetually rain-soaked Meghalaya forests, deep ocean trenches, torrential streams, permafrost polar regions, high mountain tops, boiling thermal springs, and stinking compost pits, to name a few. Even our intestine is a unique habitat for hundreds of species of microbes..

10. Which element is responsible for causing soil salinity? At what concentration does the soil become saline?
Answer. Soil salinity is the salt content in the soil. Salts are a natural component in soils and water. The ions responsible for salination are: Na⁺, K⁺, Ca²⁺, Mg²⁺ and Cl^–.

11. Does light factor affect the distribution of organisms? Write a brief note giving suitable examples of either plants or animals.
Answer. Since plants produce food through photosynthesis, a process which is only possible when sunlight is available as a source of energy, we can quickly understand the importance of light for living organisms, particularly autotrophs. Many species of small plants (herbs and shrubs) growing in forests are adapted to photosynthesise optimally under very low light conditions because they are constantly overshadowed by tall, canopied trees. Many plants are also dependent on sunlight to meet their photoperiodic requirement for flowering. For many animals too, light is important in that they use the diurnal and seasonal variations in light intensity and duration (photoperiod) as cues for timing their foraging, reproductive and migratory activities.

12. Give one example for each of the following:
(i) Eurythermal plant species ………………
(ii) A hot water spring organism ………………
(iii) An organism seen in deep ocean trenches ………………
(iv) An organism seen in compost pit ………………
(v) A parasitic angiosperm ………………
(vi) A stenothermal plant species ………………
(vii) Soil organism ………………
(viii) A benthic animal ………………
(ix) Antifreeze compound seen in antarcticfish ………………
(x) An organism which can conform ………………
Answer. (i) Eurythermal plant species—Red algae
(ii) A hot water spring organism—Thermus aquaticus
(iii) An organism seen in deep ocean trenches—Sea cucumbers
(iv) An organism seen in compost pit—Earthworm
(v) A parasitic angiosperm—Cuscuta reflexa
(vi) A stenothermal plant species—Conifers
(vii) Soil organism—Earthworm
(viii) A benthic animal—Crabs, Sponges
(ix) Antifreeze compound seen in Antarctic fish—Antifreeze glycoproteins orAFGPs
(x) An organism which can conform—Frog

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NCERT Exemplar Problems Class 12 Biology Chapter 14 Ecosystem

Multiple Choice Questions
Single Correct Answer Type
1. Decomposers like fungi and bacteria are
i. Autotrophs ii. Heterotrophs
iii. Saprotrophs iv. Chemo-autptrophs
Choose the correct answer:
(a) i and iii (b) i and iv (c) ii and iii (d) i and ii
Answer. (c) Decomposers like fungi and bacteria are heterotrophs and saprotrophs.

2. The process of mineralisation by micro organisms helps in the release of
(a) Inorganic nutrients from humus
(b) Both organic and inorganic nutrients from detritus
(c) Organic nutrients from humus
(d) Inorganic nutrients from detritus and formation of humus.
Answer. (a)

• The process of mineralisation by microorganisms helps in the release of inorganic nutrients from humus.
• The humus is further degraded by some microbes and release of inorganic nutrients occur by the process called mineralisation.

3. Productivity is the rate of production of biomass expressed in terms of
i. (kcalm^-3 )yr^-1 ii. g^-2 y^-1 r
iii. g^-1yr^-1 iy. (kcal m^-2) yr^-1
(a) ii (b) iii (c) iiandiv (d) i and iii
Answer. (c) Productivity is the rate of production of biomass expressed in terms of g^-1yr^-1 and (heal m^-2) yr^-1.

4. An inverted pyramid of biomass can be found in which ecosystem?
(a) Forest (b) Marine (c) Grassland (d) Tundra
Answer. (b) In terrestrial ecosystem pyramid of biomass is generally upright. The pyramid of biomass in sea or lake is generally inverted because the biomass of fishes for exceeds that of phytoplankton.

5. Which of the following is not a producer?
(a) Spirogyra (b) Agaricus (c) Volvox (d) Nostoc
Answer. (b) Spirogyra, Volvox and Nostoc are producers. Agaricus is a fungus.

6. Which of the following ecosystems is most productive in terms of net primary . production?
(a) Deserts (b) Tropical rain forests
(c) Oceans (d) Estuaries
Answer. (b) Tropical rain forests ecosystems is most productive in terms of net primary production.

7. Pyramid of numbers is
(a) Always upright (b) Always inverted
(c) Ether upright or inverted (d) Neither upright nor inverted
Answer. (c) Pyramid of number in grassland or forest or pond ecosystem is upright. In parasitic food chain, the pyramid of number is inverted. In an ecosystem dominated by trees, the pyramid of number is spindle shaped/inverted.

8. Approximately how much of the solar energy that falls on the leaves of a plant is converted to chemical energy by photosynthesis?
(a) Less than 1% (b) 2-10% (c) 30% (d) 50%
Answer. (b) Approximately how much of the solar energy that falls on the leaves of a plant is converted to chemical energy by photosynthesis 2-10%.

9. Among the following, where do you think the process of decomposition would be the fastest?
(a) Tropical rain forest (b) Antarctic
(c) Dry arid region (d) Alpine region
Answer. (a)

• Warm and moist environment favour decomposition. Decomposition is largely an oxygen requiring process. Low temperature and anaerobiosis inhibit decomposition resulting in build up of organic materials.
• The process of decomposition would be the fastest is tropical rain forest.

10. How much of the net primary productivity of a terrestrial ecosystem is eaten and digested by herbivores?
(a) 1% (b) 10% (c) 40% (d) 90%
Answer. (b) 10% of the net primary productivity of a terrestrial ecosystem is eaten and digested by herbivores.

11. During the process of ecological .succession the changes that take place in communities are
(a) Orderly and sequential
(b) Random
(c) Very quick
(d) Not influenced by the physical environment
Answer. (a) The gradual and fairly predictable change in the species composition of a given area is called ecological succession. An important characteristic of all the communities is that their composition and structure constantly change in response to the changing environmental conditions. These changes are orderly and sequential parallel with the changes in the physical environment.

12. Climax community is in a state of
(a) Non-equilibrium (b) Equilibrium
(c) Disorder (d) Constant change
Answer. (b) Climax community is in a state of equilibrium.

13. Among the following bio-geo-chemical cycles which one does not have losses due to respiration?
(a) Phosphorus (b) Nitrogen
(c) Sulphur (d) All of the above
Answer. (d) Phosphorus, nitrogen and sulphur bio-geo-chemical cycles does not have losses due to respiration.

14. The sequence of communities of primary succession in water is
(a) Phytoplankton, sedges, free-floating hydrophytes, rooted hydrophytes, grasses and trees
(b) Phytoplankton, free-floating hydrophytes, rooted hydrophytes, sedges, grasses and trees
(c) Free-floating hydrophytes, sedges, phytoplankton, rooted hydrophytes, grasses and trees
(d) Phytoplankton, rooted submerged hydrophytes, floating hydrophytes, reed swamp, sedges, meadow and trees
Answer. (b) The sequence of communities of primary succession in water is phytoplankton, free-floating hydrophytes, rooted hydrophytes, sedges, grasses and trees

15. The reservoir for the gaseous type of bio-geo chemical cycle exists in
(a) Stratosphere (b) Atmosphere (c) Ionosphere (d) Lithosphere
Answer. (b) The reservoir for the gaseous- type of bio-geo-chemical cycle exists in atmosphere.

16. Which of the following type of ecosystem is expected in an area where evaporation exceeds precipitation, and mean annual rainfall is below 100 mm
(a) Grassland (b) Shrubby forest
(c) Desert (d) Mangrove
Answer. (c) Desert ecosystem is expected in an area where evaporation exceeds precipitation, and mean annual rainfall is below 100 mm.

17. The zone at the edge of a. lake or ocean which is alternatively exposed to air and immersed in water is called
(a) Pelagic zone (b) Benthic zone
(c) Lentic one (d) Littoral zone
Answer. (d) The zone at the edge of a lake or ocean which is alternatively exposed to air and immersed in water is called littoral zone.

18. If the carbon atoms fixed by producers already have passed through three species, the trophic level of the last species would be
(a) Scavenger (b) Tertiary producer
(c) Tertiary consumer (d) Secondary consumer
Answer. (c)

If the carbon atoms fixed by producers already have passed through three species, the trophic level of the last species would be tertiary consumer.

19. Edaphic factor refers to
(a) Water (b) Soil
(c) Relative humidity (d) Altitude
Answer. (b) Edaphic factor refers to soil.

20. Which of the following is <an ecosystem service provided by a natural ecosystem?
(a) Cycling of nutrients
(b) Prevention of soil erosion
(c) Pollutant absorption and reduction of the threat of global warming
(d) All of the above
Answer. (d) Cycling of nutrients, prevention of soil erosion and pollutant absorption and reduction of the threat of global warming is an ecosystem service provided by a natural ecosystem.

Very Short Answer Type Questions
1. Name an organism found as secondary carnivore in an aquatic ecosystem.
Answer. Catfish/water snake etc.

2. What does the base tier of the ecological pyramid represent?
Answer. Producers

3. Under what conditions would a particular stage in the process of succession revert back to an earlier stage?
Answer. Natural or human induced disturbances like fire, deforestation etc.

4. Arrange the following as observed in vertical stratification of a forest: Grass, Shrubby plants, Teak, Amaranths
Answer. Grass, Amaranths, Shrubby plants, Teak

5. Name an omnivore which occurs in both grazing food chain and the decomposer food chain.
Answer. Sparrow/crow

6. Justify the pitcher plant as a producer.
Answer. It is chlorophyllous and is thus capable of photosynthesis.

7. Name any two organisms which can occupy more than one trophic level in an ecosystem.
Answer. Man and sparrow ete.

8. In the North East region of India, during the process of Jhum cultivation, forests are cleared by burning and left for regrowth after a year of cultivation. How would you explain the regrowth of forest in ecological term?
Answer. The regrowth of forest is called secondary succession.

9. Climax stage is achieved quickly in secondary succession as compared to primary succession. Why?
Answer. The rate of succession is much faster in secondary succession as the substratum (soil) is already present as compared to primary succession where the process starts from a bare area (rock).

10. Among bryophytes, lichens and fern which one is a pioneer species in a xeric succession?
Answer. Crustose lichens.

11. What is the ultimate source of energy for the ecosystems?
Answer. Sunlight (solar radiation)

12. Is the common edible mushroom an autotroph or a heterotroph?
Answer. Heterotroph (as it does not have chlorophyll)

13. Why are oceans least productive?
Answer. Oceans are least productive because they do not receive sufficient sunlight which is necessary for 1° productivity.

14. Why is the rate of assimilation of energy at the herbivore level called secondary productivity?
Answer. It is because the biomass available to the consumer for consumption is a resultant of the primary productivity from plants.

15. Why are nutrient cycles in nature called biogeochemical cycles?
Answer. Nutrient cycles in nature called biogeochemical cycles because the cycling of nutrients occurs between living organisms (bio) and non-living things-geo- (rocks, air and water, etc.)

16. Give any two examples of xerarch succession.
Answer. (a) Bare rocks (xerosere)
(b) Sandy areas (psammosere)

17. Define self sustainability.
Answer. The degree at which the system can sustain itself without external support is called self sustainability.

18. Given below is a figure of an ecosystem. Answer the following questions.

(i) What type of ecosystem is shown in the figure?
(ii) Name any plant that is characteristic of such ecosystem.
Answer. (i) Desert ecosystem
(ii) Opuntia, Cacti and Acacia

19. What is common to earthworm, mushroom, soil mites and dung beetle in an ecosystem.
Answer. They are all detritivores i.e., decomposing organisms which feed on dead remains of plants and animals.

Short Answer Type Questions
1. Organisms at a higher trophic level have less energy available. Comment.
Answer. The transfer of energy follows 10 per cent law – only 10 per cent of the
energy is transferred to each trophic level from the lower trophic level. So, higher trophic level have less energy as- compared to lower trophic level.

2. The number of trophic levels in an ecosystem are limited. Comment.
Answer. In a food chain, only 10% of the total amount of energy is passed on to the next trophic level from the previous trophic level. So, there is a decrease in the amount of energy available at the successive trophic levels. As we move higher up in the food chain the amount of energy diminishes to a level at which it cannot sustain any trophic level, thereby limiting the number of trophic levels.

3. Is an aquarium a complete ecosystem?
Answer. Aquarium has all the biotic and abiotic components of an ecosystem, so, aquarium is a complete ecosystem.

4. What could be the reason for the faster rate of decomposition in the tropics?
Answer. The rate of decomposition is regulated by climatic factors like temperature and soil moisture as they have an effect on the activities of soil microbes. The
tropics with its hot and humid.climaticcondition provides an environment which is ideal for the microbes to speed up the process of decomposition.

5. Human activities interfere with carbon cycle. List any two such activities.
Answer. Human activities have significantly influenced the carbon cycle. Rapid
deforestation and massive burning of fossil fuel for energy and transport have significantly increased the rate of release of carbon dioxide into the atmosphere.

6. Flow of energy through various trophic levels in an ecosystem is unidirectional and non-cyclic. Explain.
Answer. The energy from the sun reaches the food chain through the primary producers (plants). This energy is passed on through successive trophic levels in the food chain. The energy transfer in the food chain follows the 10 percent law where in only 10% of the energy is transferred from one trophic level to the next successively. So, the movement of energy is only in one direction from lower to higher trophic level.

7. Apart from plants and animals, microbes form a permanent biotic component in an ecosystem. While plants have been referred to as autotrophs and animals as heterotrophs, what are microbes referred to as? How do the microbes fulfil their energy requirements?
Answer. Microbes are referred to as heterotrophs and saprotrophs. They fulfil their energy requirement by feeding on dead remains of plants and animals through the process of decomsposition.

8. Poaching of tiger is a burning issue in today’s world. What implication would this activity have on the functioning of the ecosystem of which the tigers are an integral part?
Answer. Poaching (hunting) of tiger disturbs the ecological balance in ecosystem. Tigers are important predator of forest. Predator keeps prey population under control. Due to hunting of tiger its population decreases, hence the population of deer (prey) increases and interspecific competition occurs between them. Due to increase in prey population (that feeds on vegetation/grass), vegetation destroyed rapidly which causes ecological disturbance.

9. In relation to energy transfer in ecosystem, explain the statement “10 kg of deer’s meat is equivalent to 1 kg of lion’s flesh”.
Answer. Transfer of energy in an ecosystem follow 10% law, i.e., only 10% energy is transferred from one trophic level to higher trophic level. So, if lion feeds on deer than 10 kg of deer’s meat will form 1 kg of lion’s flesh.

10. Primary productivity varies from ecosystem to ecosystem. Explain.
Answer. Primary productivity varies from ecosystem to ecosystem because it depends on the plant species inhabiting the area and their photosynthetic activity. It also depend on various environmental factors.

11. Sometimes due to biotic/abiotic factor the climax remain in a particular serai stage (pre climax) without reaching climax. Do you agree with this statement? If yes, give a suitable example.
Answer. It is true that any change in the. abiotie/biotic factor will arrest a particular serai stage leading to a pre-climax condition before the climax stage is achieved. This can happen in cases of forest fires, landslides, changes in soil characteristics, increase in herbivore population leading to overgrazing.

12. What is an incomplete ecosystem? Explain with the help of suitable example.
Answer. An ecosystem is a functional unit with biotic and abiotic factors interacting
with one another resulting in a physical structure. Absence of any component will make an ecosystem incomplete as it will hinder the functioning of the ecosystem. Examples of such an ecosystem can be a fish tank or deep aphotic zone of the oceans where producers are absent.

13. What are the shortcomings of ecological pyramids in the study of ecosystem?
Answer. The ecological pyramid assumes a simple food chain and does not
accommodate food webs. Thereby,’ it does not take irlto account the fact that species may belong to two or more trophic levels at a time. Also saprophytes despite their vital rolg in ecosystem are given no place in the ecological pyramids.

14. How do you distinguish between humification and mineralisation?
Answer. Humification leads to accumulation of a dark coloured amorphous substance called humus that is highly resistant to microbial action and undergoes decomposition at an extremely slow rate. Being colloidal in nature it serves as a reservoir of nutrients . The humus is further degraded by some microbes and release of inorganic nutrients occur by the process known as mineralisation.

15. Fill in the trophic levels (1, 2, 3 and 4) in the boxes provided in the figure.

Answer.

16. The rate of decomposition of detritus is affected by the abiotic factors like availability of oxygen, pH of the soil substratum, temperature etc. Discuss.
Answer. The decomposition of detritus is due to activities of micro-organisms. The rate of growth of microbes is affected by temperature. The pH of substratum affects the composition of microbes (acidophiles / basophiles) which degrade the dead organic matter. If oxygen is present, aerobic degradation occurs. In the absence of oxygen anaerobiosis sets in and there will be incomplete degradation. Also, the degradation is due to activity of exo enzymes secreted by the microbes and the activity of enzyme is affected by factors such as temperature etc.

Long Answer Type Questions
1. A farmer harvests his crop and expresses his harvest in three different ways.
(a) I have harvested 10 quintals of wheat.
(b) I have harvested 10 quintals of wheat today in one acre of land.
(c) I have harvested 10 quintals of wheat in one acre of land, 6 months after sowing.
Do the above statements mean one and the same thing. If your answer is yes, give reasons. And if your answer is ‘no’ explain the meaning of each expression.
Answer. No, the meaning of above statements is different
(a) In this statement farmer gives information of only the quantity of the wheat but area and time period is not given.
(b) In this statement farmer gives information of primary production that is quantity and area but not the time period.
(c) In this statement farmer gives information of primary productivity that is quantity, area and time period all are given.

2. Justify the following statement in terms of ecosystem dynamics. “Nature tends to increase the gross primary productivity, while man tends to increase the net primary productivity”.
Answer. Gross primary productivity of an ecosystem is the rate of production of organic matter during photosynthesis. A considerable amount of GPP is utilised by plants in respiration. Gross primary productivity minus respiration losses (R), is the net primary productivity (NPP). Net primary productivity is the available biomass for the consumption to heterotrophs (herbivores and decomposers).
• Nature tends to increase the gross primary productivity so that it can utilise much of the solar energy and convert it into food.
• Man tends to increase the net primary productivity so that sufficient food is provided to the growing human population.

3. Which of the following ecosystems will be more productive in terms of primary productivity? Justify your answer. A young forest, a natural old forest, a shallow polluted lake, alpine meadow.
Answer. Among these a young forest will be more productive because primary productivity depends on the plant species inhabiting a particular area. A young forest grows more rapidly than old forest hence it has more primary productivity.
A shallow polluted lake and alpine meadow has lower number of producers hence shows lower primary productivity.

4. What are. the three types of ecological pyramids? What information is conveyed by each pyramid with regard to structure, function and energy in the ecosystem?
Answer. The three ecological pyramids that are usually studied are (a) pyramid of number (b) pyramid of biortiass and (c) pyramid of energy.
• Any calculations of energy content, biomass, or numbers has to include all organisms at that trophic level. No generalisations we make will be true if we take only a few individuals at any trophic level into account. Also a given organism may occupy more than one trophic level simultaneously. One must remember that the trophic level represents a functional level, not a species as such. A given species may occupy more than one trophic level in the same ecosystem at the same time; for example, a sparrow is a primary consumer when it eats seeds, fruits, peas, and a secondary consumer when it eats insects and worms.
• In most ecosystems, all the pyramids, of number, of energy and biomass are upright, i.e., producers are more in number and biomass than the herbivores, and herbivores are more in number and biomass than the carnivores. Also energy at a lower trophic level is always more than at a higher level.

5. Write a short note on pyramid of numbers and pyramid of biomass.
Answer. Pyramid of Number
• Pyramid of number deals with number of individuals in a trophic level.

• In Grassland ecosystem only 3 top carnivores are supported in an ecosystem based on production of nearly 6 million plants.
• Pyramid of number in grassland or forest or pond ecosystem is upright.

• In parasitic food chain, the pyramid of number is inverted.
• In an ecosystem dominated by trees the pyramid of number shaped/inverted.

• Among the grassland, pond and forest ecosystem, the number of primary producers per unit area would be maximum in pond ecosystem.
Pyramid of Biomass
• Total amount of living material at various trophic level of a food chain is depicted by pyramid of biomass.

• In most of the ecosystem, the pyramids of number and biomass are upright.
• In terrestrial ecosystem pyramid of biomass is generally upright.
• The pyramid of biomass in sea or lake is generally inverted because the biomass of fishes for exceeds that of phytoplankton.

6. Given below is a list of autotrophs and heterotrophs. With your knowledge about food chain, establish various linkages between the organisms on the principle of ‘eating and being eaten’. What is this inter-linkage established known as?
Algae, Hydrilla, grasshopper, rat, squirrel, crow, maize plant, deer, rabbit, lizard, wolf, snake, peacock, phytoplankton, crustaceans, whale, tiger, lion, sparrow, duck, crane, cockroach, spider, toad, fish, leopard, elephant, goat, Nymphaea, Spirogyra.
Answer. Primary Producer’s (First trophic level): Algae, Hydrilla, maize plant, Phytoplankton, Nymphaea, Spirogyra.
Primary Consumer (Second trophic level): Crustaceans, grasshopper, deer, mouse, squirrel, rabbit, elephant, goat.
Secondly Consumer (Third trophic level): Spider, cockroach, lizard, wolf, snake, toad, fish, crane.
Top carnivore (Fourth trophic level): Lion, Tiger.

7. “The energy flow in the ecosystem follows the second law of thermodynamics.” Explain.
Answer. Ecosystems need a constant supply of energy to synthesise the molecules they require, to counteract the universal tendency toward increasing disorderliness. Hence, ecosystems are not exempt from the Second Law of thermodynamics.

8. What will happen to an ecosystem if:
(a) All producers are removed;
(b) All organisms of herbivore level are eliminated; and
(c) All top carnivore population is removed
Answer. (a) Reduction in primary productivity. No biomass available for consumption by higher trophic levels / heterotrophs
(b) Increase in primary productivity and biomass of producers.
Carnivores population will subsequently dwindle due to food shortage.
(c) Increase in number of herbivores
• Overgrazing by herbivores
• Desertification

9. Give two examples of artificial or mail made ecosystems. List the salient features by which they differ from natural ecosystems.
Answer. Crop fields and an aquarium considered as man-made ecosystems. In artificial ecosystems the biotic and abiotic components of ecosytems are maintained artificially by regular feeding and clearing of aquarium and regular irrigation and sowing of seeds in crop field ecosystem. Natural ecosystem self sustains biotic and abiotic components of an ecosystem.

10. The biodiversity increases when one moves from the pioneer to the climax stage. What could be the explanation?
Answer. In the successive serai stages there is a change in the diversity of species of organisms, increase in the number of species and organisms as well as an • increase in the total biomass. Description of ecological succession usually focuses on changes in vegetation. However, these vegetational changes in turn affect food and shelter for various types of animals. Thus, as succession proceeds, the numbers and types of animals and decomposers also change.

11. What is a biogeochemical cycle? What is the role of the reservoir in a biogeochemical cycle? Give an example of a sedimentary cycle with reservoir located in earth’s crust.
Answer. The movement of nutrient elements through the various components of an ecosystem is called nutrient cycling. Another name of nutrient cycling is biogeochemical cycles (bio: living organism, geo: rocks, air, water). Nutrient cycles are of two types: (a) gaseous and (b) sedimentary.
The function of the reservoir is to meet with the deficit which occurs due to imbalance in the rate of influx and efflux. For the sedimentary cycle (e.g., sulphur and phosphorus cycle), the reservoir is located in Earth’s crust.

12. What will be the P/R ratio of a climax community and a pioneer community? What explanation could you offer for the changes seen in P/R ratio of a pioneer community and the climax community?
Answer. P/R (Production/Respiration) ratio shows the relationship between gross production and respiration.
In climax community the P/R ratio will be 1. In pioneer community the P/R ratio will be more than one.

The post NCERT Exemplar Problems Class 12 Biology Ecosystem appeared first on Learn CBSE.

NCERT Exemplar Problems Class 12 Biology Chapter 15 Biodiversity and Conservation

Multiple Choice Questions
Single Correct Answer Type
1. Which of the following countries has the highest biodiversity?
(a) Brazil (b) South Africa (c) Russia (d) India
Answer. (a) Brazil has the highest biodiversity because it is located near equator.

2. Which of the following is not a cause for loss of biodiversity?
(a) Destruction of habitat
(b) Invasion by alien species
(c) Keeping animals in zoological parks
(d) Over-exploitation of natural resources
Answer. (c) Destruction of habitat, invasion by alien species, and over-exploitation, of natural resources are cause for loss of biodiversity.

3. Which of the following is not an invasive alien species in the Indian context?
(a) Lantana (b) Cynodon (c) Parthenium (d) Eichhomia
Answer. (b) Carrot grass (Parthenium), Lantana and water hyacinth (Eichhomia crassipe) are invasive weeds that cause environment damage.

4. Where among the following will you find pitcher plant?
(a) Rain forest of North-East India
(b) Sunderbans
(c) Thar Desert
(d) Western Ghats
Answer. (a) Pitcher plants are found at rain forest’ of North-East India.

5. Which one of the following is not a characteristic feature of biodiversity hot spots?
(a) Large number of species
(b) Abundance of endemic species
(c) Mostly located in the polar regions
(d) Mostly located in the tropics
Answer. (c) Characteristic feature of biodiversity hot spots are large number of species, abundance of endemic species and mostly located in the tropics.

6. Match the animals given in column A with their location in column B.

Choose the correct match from the following:
(a) i—A, ii—C, iii—B, iv—D (b) i—D, ii—C, iii—A, iv—B
(c) i—C, ii—A, iii—B, iv—D (d) i—C, ii—A, iii—D, iv—B
Answer. (d)

7. What is common to the following plants: Nepenthes, Psilotum, Rauwoljia and Aconitum?
(a) All are ornamental plants
(b) All are phylogenic link species
(c) All are prone to over exploitation
(d) All are exclusively present in the Eastern Himalayas.
Answer. (c) Nepenthes, Psilotum, Rauwolfta md Aconitum, all are prone to over-exploitation.

8. The one-homed rhinoceros is specific to which of the following sanctuary
(a) Bhitar Kanika (b) Bandipur (c) Kaziranga (d) Corbett park
Answer. (c)

9. Amongst the animal groups given below, which one has the highest percentage of endangered species?
(a) Insects (b) Mammals (c) Amphibians (d) Reptiles
Answer. (c) Presently, 12% all the birds species, 23% all mammals species, 31% all gymnosperms species and 32% all amphibian species in world face the threat of extinctions.

10. Which one of the following is an endangered plant species of India?
(a) Rauwolfia serpentina  (b) Santalum album (Sandal wood)
(c) Cycas beddonei (d) All of the above
Answer. (d) Endangered (ER): It is facing a high risk of extinction in the wild in the near future if conservation measures are not promptly taken.
E.g.: Red panda (AHums fulgens)

• Giant panda
• Largest lemur Idri idri of Madagascar
• Asiatic Wild Ass (Asinus hemionus khur now restricted to Rann of Kutch)-
•  Lion Tailed Macaque
• Bald Eagle
•  Asiatic lion .
• Drosera indica
• Nepenthes
•  Hombill
• Indian Aconite
•  Bentinckia nicobarica
• Snow leopard
• Rauwolfia serpentina
• Santalum album (Sandal wood)
•  Cycas beddonei

11. What is common to Lantana, Eichhomia and African catfish?
(a) All are endangered species of India.
(b) All are key stone species.
(c) All are mammals found in India.
(d) All the species are neither threatened nor indigenous species of India.
Answer. (d) Lantana, Eichhomia and African catfish, all the species are neither threatened nor indigenous species of India.

12. The extinction of passenger pigeon was due to
(a) Increased number of predatory birds
(b) Over exploitation by humans
(c) Non-availability of the food
(d) Bird flu virus infection
Answer. (b) Humans have always depended on nature for food and shelter but when ‘need’ turns to ‘greed’ it leads to over-exploitation of natural resources. In the last 500 years many species extinctions (Steller’s sea cow, passenger’s pigeon) were due to over-exploitation by human.

13. Which of the following statements is correct?
(a) Parthenium is an endemic species of our country
(b) African catfish is not a threat to indigenous catfishes.
(c) Steller’s sea cow is an extinct animal.
(d) Lantana is popularly known as carrot grass.
Answer. (c) Steller’s sea cow is an extinct animal.

14. Among the ecosystem mentioned below, where can one find maximum biodiversity?
(a) Mangroves (b) Desert
(c) Coral reefs (d) Alpine meadows
Answer. (c) Coral reefs has maximum biodiversity among above options.

15. Which of the following forests is known as the ‘lungs of the planet Earth’?
(a) Tiaga forest (b) Tundra forest
(c) Amazon rain forest (d) Rain forests of North East India
Answer. (c) Amazon rain forest is known as the ‘lungs of the planet Earth’.

16. The active chemical drug reserpine is obtained from
(a) Datura (b) Rauwolfia (c) Atropa (d) Papaver
Answer. (b) The genetic variation shown by the medicinal plant Rauwolfia vomitoria growing in different Himalayan ranges might be in terms of potency and concentration of the active chemical (Reserpine) obtained from roots of plants.

17. Which of the following group exhibit more species diversity?
(a) Gymnosperms (b) Algae
(c) Bryophytes (d) Fungi
Answer. (d)

18. Which of the below mentioned regions exhibit less seasonal variations?
(a) Tropics (b) Temperates
(c) Alpines (d) Both (a) and (b)
Answer. (a) Tropical regions exhibit less seasonal variations.

19. The historic convention on Biological Diversity held in Rio de Janeiro in 1992 is known as:
(a) CITES Convention (b) The Earth Summit
(c) G-16 Summit (d) MAB Programme
Answer. (b) The Earth Summit: This historical convention on biological diversity held in Rio-de-Janeiro, Brazil in 1992. Attending nations take appropriate measure for conservation of biodiversity and sustainable utilization of its benefits.

20. What is common to the techniques (i) in vitro fertilisation, (ii) Cryo preservation and (iii) tissue culture?
(a) All are in situ conservation methods.
(b) All are ex situ conservation methods.
(c) All require ultra modem equipment and large space.
(d) All are methods of conservation of extinct organisms.
Answer. (b)

Very Short Answer Type Questions
1. What characteristics make a community stable?
Answer. Characteristics that make a community stable are:
(i) A stable community should not show too much variation in productivity from year to year
(ii) It must be either resistant or resilient to occasional disturbances (natural or man-made)
(iii) It must also be resistant to invasions by alien species.

2. What could have triggered mass extinctions of species in the past?
Answer. (i) Volcanic eruption
(ii) Earthquake
(iii) Extremes of temperatures
(iv) Glaciations
(v) Continental drift

3. What accounts for the greater ecological diversity of India?
Answer. India with its deserts, rain forests, mangroves, coral reefs, wetlands, estuaries, and alpine meadows has a greater ecosystem diversity.

4. According to David Tilman, greater the diversity, greater is the primary productivity. Can you think of a very low diversity man-made ecosystem that has high productivity?
Answer. Agricultural fields like wheat field / paddy field which are also examples of monoculture practices.

5. What does ‘Red’ indicate in the IUCN Red list (2004)?
Answer. Red indicates the species that are at the verge of extinction or threatened species.