CBSE previous Year Solved  Papers  Class 12 Biology Delhi 2013

Time allowed : 3 hours                                                                                           Maximum Marks: 70

General Instructions :

  1.  There are a total of 26 questions and five sections in the question paper, All questions are compulsory.
  2. Section A contains question number 1 to 5, Very Short Answer type questions of one mark each.
  3.  Section B contains question number 6 to 10, Short Answer type I questions of two marks each.
  4.  Section C contains question number 11 to 22, Short Answer type II questions of three marks each.
  5.  Section D contains question number 23, Value Based Question of four marks.
  6. Section E contains question number 24 to 26, Long Answer type questions of five marks each.
  7. There is no overall choice in the question paper, however, an internal choice is provided in one question of two marks, one question of three marks and all three questions of five marks. An examined is to attempt any one of the questions out of two given in the question paper with the same question number.

SET-I

SECTION-A

Question.1. An anther with malfunctioning tapetum often fails to produceviable male gametophytes. Give any one reason.
Answer : The tapetum is responsible for nourishing the pollen grains. Tapetum malfunctioning results in insufficient nvitrition of the pollen grains, and will not be able to produce viable male gametophytes.

Question.2. Why sharing of injection needles between two individuals is not recommended ?
Answer: Sharing of injection needle between two individuals could lead to the transfer of blood borne diseases viz. HIV from infected person to a healthy person.

Question.3. Name the enzyme and state its property that is responsible for continuous and discontinuous replication of the two strands of a DNA molecule.
Answer : DNA dependent DNA polymerase is the enzyme. The property t hat is enzyme catalys polymerization only in one direction, i..e., 5′ —> 3′. As a result on template strand with 3’ —> 5′ the replication is continuous while on the template strand with polarity 5′ —> 3′ it is discontinuous.

Question.4. Identify the examples of convergent evolution from the following:
(i) Flippers of penguins and dolphins
(ii) Eyes of octopus and mammals
(iii) Vertebrate brains
Answer : The examples of convergent evolution are :
(i) Flippers of penguins and dolphins.
(ii) Eyes of octopus and mammals.

Question.5. Write the importance of MOET.
Answer : The technique is used to increase the successful rate of production of hybrids in short duration of time. e.g. In MOET cow is given hormones with FSH like activity. These hormones induce follicular maturation and super-ovulation which produces 6-8 eggs per cycle instead of one egg.

Question.6. Why is the enzyme cellulase needed for isolating genetic material from plant ct’Us and not from the animal cells ?
Answer : Plant cells have cell wall of cellulose but animal cells have no cell wall and do not have cellulose. So the enzyme cellulase is required for isolating genetic material from plant cells only.

Question.7. Name the type of biodiversity represented by the following:
(a) 50,000 different strains of rice in India.
(b) Estuaries and alpine meadows in India.
Answer : (a) Genetic diversity (b) Ecological diversity

Question.8. Write the equation that helps in deriving the net primary productivity of an ecosystem.
Answer : Net primary productivity (NPP) in an ecosystem can be derived using the following equation,
NPP = GPP (Gross primary productivity) – R (Respiratory losses)

SECTION-B

Question.9. Geitonogamous flowering plants are genetically autogamous but functionally cross-pollinated. Justify.
Answer : Geitonogamy is a pollination where the pollen grains are transferred from the anther of a flower to the stigma of another flower on the same plant.
Geitonogamous flowering plants are usually autogamous as the gametes come from the same parent plant, but since the pollen grains are being passed to a different flower which requires a pollinating agent, it is an example of cross pollination.

Question.10. When and where do chorionic villi appear in humans ? State their function.
Answer : Chorionic villi appear as finger like projections that arise from the trophoblast layer of blastocyst when it is undergoing implantation.
Functions of Chorionic villi:

  1.  It inter-digitates with projections from uterine tissue to form a structure called the placenta, which is the connecting link between the mother and the foetus.
  2.  It facilitates the supply of oxygen and nutrients to the embryo.

Question.11. In a cross between two tall pea plants some of the off springs produced were dwarf. Show with the help of Punett Square how this is possible. 
Answer : In a cross between 2 phenotypically tall plants, some of the progeny may become to be phenotypically dwarf. It defines both parents are heterozygous (Tt). The cross is demonstrated below by a Punnett Square.
Genotype of Parents = Tt.
Genotypic ratios of progeny after F] cross= TT, Tt, Tt, tt. Phenotypic ratios= 1 TT+ 2 Tt: 1 tt or 3 Tall: 1 dwarf .
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Question.12. A student on a school trip started sneezing and wheezing soon after reaching the hill station for no explained reasons. But, on return to the plains, the symptoms disappeared. What is such a response called ? How does the body produce it?
Answer : The responses such as sneezing and wheezing for no explained reasons is called Allergy. The substance against which the immune system show such exaggerated response are termed as allergens. Allergy„is produced in a body due to release of chemicals like histamine and serotonin from mast cells. Body produces IgE type of antibody against allergens as a response. Most common example of-such allergens are pollen grains, dust etc.

Question.13. Name two commonly used bioreactors. State the importance of using a bioreactor.
Answer : The two most widely used bioreactors are simple stirred – tank bioreactor and sparged stirred- tank bioreactor. The importance of using bioreactors :

  1. It gives huge amount for cultures. So, products are pro-duced in large quantity.
  2.  It gives optimal conditions like temperature and pH for growth of desired product.

Question.14. Write the function of adenosine deaminase enzyme. State the cause of ADA deficiency in humans. Mention a possible cure for a ADA deficiency patient.
Answer : The enzyme adenosine deaminase (ADA) is very important for the proper functioning of our immune system. The cause of ADA deficiency in humans is deletion of the gene which codes. for ADA. ADA deficiency can be cured permanendy by gene therapy. Then functional ADA gene is inserted in the cells at early embryonic age for permanent cure.

Question.15. Expand the following and mention one application of each :
(i) P.CR (ii) ELISA
Answer : (i) PCR-Polymerase chain reaction is a technique in molecular biology to amplify a gene or a piece of DNA to obtain its several copies. Each cycle of PCR has three steps :
(a) Denaturation of DNA Strand
(b) Primer annealing
(c) Extension of primers
Use : It is extensively used in the process of gene manipulation,
(ii) ELISA (Enzyme Linked Immunosorbent Assay) is a method in molecular biology which utilizes antigens and . antibodies to find infectious diseases. An infection is found by the presence of antigens like proteins or by the synthesis of antibodies against the infection.
This method is broadly used for finding AIDS.
OR
(a) Mention the difference in the mode of action of exonuclease and endonuclease.
(b) How does restriction endonuclease function ?
Answer : (a) Exonuclease removes the nucleotides from the ends of the DNA chain while endonuclease cuts the DNA at the specific positions within the DNA strand.
(b) Each restriction endonuclease finds its specific pallindromic • nucleotide sequences in the DNA and cut the DNA at these specific sites. It does so by binding to the DNA at these sites and cutting both the strands at specific points in their sugar- phosphate backbones, e.g. E CORI cuts the DNA at following palindromic sequences.
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Question.16. Name any two source of e-waste and write two different ways for their disposal.
Answer : Two sources of e-waste are :

  1. Irreparable computers.
  2.  Electronic items like mobile phones, television sets etc. Two different ways of disposal of e-waste :
    (i) Dumping the e-wastes into landfills.
    (ii) Incinerating e-waste i.e. burning the e-wastes com pletely into ashes.
    These ways of disposal pose threat to the environment by, releasing toxic substances into it. So, recycling, of e-waste in an environment-friendly manner is the only solution for its disposal.

Question.17. Why the pyramid of energy is always upright ? Explain.
Answer : The pyramid of energy shows the total quantity of energy used by each trophic level in a given food chain. An energy pyramid is always upright becau.se the total quantity of energy available for utilization at the upper trophic level less than the energy available at the lower levels. This occurs because according to the 10% law of energy transfer, only 10% of the total energy is transmitted from one trophic level to another and when energy transmits frotn a one trophic level to next trophic level, some amount is always lost as heat at each step.
Eventually it is lost to atmosphere and never goes back to Sun.
Question.18. Explain why very small animals are rarely found in polar region ? .
Answer: Loss of body heat is direcdy proportional to the surface area. So, small animals have larger surface area compare to their volume. They lose body heat rapidly in colder areas. Loss of body heat in colder areas can create challenge to their survival. That’s why very small animals are rarely found in polar areas.

SECTION-C

Question.19. Draw a diagram of the microscopic structure of human sperm. Label the following parts in it and write their Functions : (a) Acrosome (b) Nucleus (c) Middle piece
Answer : Structure of human sperm :
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Question.20. With the help of any two suitable examples explain the effect of anthropogenic action or organic evolution
Answer: The anthropogenic activities results in increased rate of organic evolution. For example :

  1.  The extreme use of herbicides and pesticides causes selection of resistant variety of pests and insects in a short time period. The change in the environment faciliate resistant pests and insects results in their evolution.
  2.  The excessive use of antibiotics causes selection of drug resistant microbes. The micro-organisms which are sensitive , to specific antibiotics died but few variants of micro-organisms which develop resistance against the specific antibiotics are survived. This assist in the evolution of deadly lethal micro-organisms.

Question.21. (a) Why is human ABO blood group gene considered a good example of multiple alleles ?
(b) Work out a cross up to F generation only, between a mother with blood group A (Homozygous) and the father with blood group B (Homozygous). Explain the pattern of inheritance exhibited.
Answer : (a) A gene is represented by two alleles. But, for the blood group in humans there are three alleles namely IA, IB and i governing a same character. Thus, it is an example of multiple allele. IA & IB alleles are dominant over 1°. These alleles decide the blood group of person.
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Question.22. Describe the structure of a RNA polynucleotide chain having four different types of nucleotides.
Answer : A RNA nucleotide has three main components a nitrogenous base, a ribose sugar and a phosphate group.

  1.  The ribose sugar and the phosphates form the backbone of a polynucleotide, chain with nitrogenous bases linked to sugar moiety and projecting from the backbone.
  2.  Two types of nitrogenous bases are present i.e. Purines . (Adenine (A) and Guanine (G)) and Pyrimidines
    (Cytosine (C) and Uracil (U)).
  3. A nitrogenous base is linked to the ribose sugar through N-glycosidic linkages to form a nucleoside (like adenosine, guanosine or cytidine and uridine).
  4.  A phosphate group is linked to 5’-OH of a nucleoside through phosphoester linkage to form a corresponding nucleotide.
  5.  Every nucleotide residue has an additional -OH group present at 2’ -position in the ribose.
  6.  Many nucleotides are linked through 3 -5′ phospho- diester linkages to each other to form the polynucleotide chain.
  7.  The end of the chain which has a free phosphate moiety at 5′- end of ribose sugar is referred to as 5 – end and the other end of the chain having a free 3 – OH group at the ribose sugar is referred to as 3′- end of the polynucleotide chain.
    cbse-previous-year-solved-papers-class-12-biology-delhi-2013-5

Question.23. Differentiate between inbreeding and out-breeding in catde. State one advantage and one disadvantage for each one of them.
Answer : Difference between inbreeding and out-breeding
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Inbreeding:
Advantage : Pure breed of progeny.
Disadvantage : It reduces in the fertility and productivity of an organism because of constant inbreeding. This is also called as Inbreeding depression.
Outbreeding:
Advantage : Progeny has proffered features of both the parents.
Disadvantage : The hybrid animal produced is not every time fertile.

Question.24. (a) Why are the fruit juices bought from market clearer as compared to those made at home ?
(b) Name the bioactive molecules produced by Trichoderma polysporum and Monascus purpureas.
Answer: (a) The fruit juices available in the market (the bottled ones) are made clear by treating them with the enzymes – pectinases and proteases. So, they are clearer as compare to those made at home.
(b) Trichoderma polysporum is used to produce immunosuppressive agent cyclosporin A.
Monascus purpureus is used to produce blood-cholesterol lowering agent called statin.

Question.25. (a) Why are transgenic animals so called ?
(b) Explain the role of transgenic animals in (i) Vaccine safety and (ii). Biological products with the help of an example each.
Answer : Transgenic animals are called so as these animals 1 acquire foreign gene that is intentionally inserted into the genome. The foreign gene is inserted in the genome of the organisms by recombinant DNA technology.
Role of transgenic animal in vaccine safety : Now a days transgenic mice are being used in testing the safety of vaccines before they are available to humans use.
Example : Transgenic mice are being used to test the safety of the polio vaccine. If the transgenic mice found doing well and trustworthy, they could substitute the use of monkeys to test the safety of batches of the vaccine.
Role of transgenic animal in production of biological products : Transgenic cow, Rosie is used for the production of human protein-enriched milk, which contained human a-lactalbumin and it was having increased nutrition more suitable for human babies.

Question.26. How have human activities caused desertification ? Explain.
Answer : Following human activities contribute to desertification :

  1. Deforestation : To construct road, buildings, new city etc. the trees are being cut down which results in desertification.
  2.  Improper farming practices : If same crop is growing again and again over a longer peroid of time makes the soil deprived resulting in the loss of fertility of soil.
  3.  Excessive ploughing of field.
  4. Soil erosion : Soil erosion by deforestation due to indu- serialization and construction of houses.
  5.  Mining actions and leaching of minerals further oblite-rate soil quality results’ in infertile soil.

OR
How does algal bloom destroy the quality of a fresh water body ? Explain.
Answer : An algal bloom is the result of excessive growth of planktons forms in a highly nutrient rich water body. When the planktonic species grow repetitively on the surface of water body, at results in a layer that finally covers the whole surface of the water body. They prevent sunlight to reach to the bottom and make unavailable to submerged aquatic species which is having role in delievering essential nutrients to other aquatic life forms.
Some algal species discharge molecules that are toxic and ‘ lethal for other aquatic organisms. Because of high respiratory rate the biological oxygen demand. (BOD) of the water body increases which causes death of a number of aquatic organisms. Their remains after their death again contribute to the deterioration of water quality.

Question.27. Explain mutualism with the help of any two examples. How is it different from commensalism ?
Answer : Mutualism is a kind of population interaction in which both the participating species derive a benefit from each other’s presence. Examples of mutualism are given below:

  1.  Associations between fungi and plants, known as mycorrhizae : The plant gets benefits by soil nutrients that the fungus absorbs and transmits to the plant by its roots. The fungus then derives the benefit of getting energy yielding carbohydrates from plant.
  2. Pollination : The flowers provide sweet, mucilaginous nectar to birds or insects on the contrary for getting benefit from the bird or insect in passing their pollen grains onto other flowers. The plant – pollinator’ pair often goes co-evolution to safeguard against the use of the nectar by other non-important organisms.
    Mutualism is different from commensalism in that the latter gives benefit to just one of the participating species, the benefitted species being called a commensal.

SECTION-D

Question.28. (a) Draw a diagrammatic sectional view of a mature anatropous ovule and label the following parts in it:
(1) that develops into seed coat.
(2) that develops into an embryo after fertilization.
(3) that develops into an endosperm in an albuminous seed.
(4) through which the pollen tube gain entry into the embryo sac.
(5) that attaches the ovule to the placenta.
(b) Describe the characteristic features of wind pollinated flowers.
Answer: (a)
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  1. The part that develops into seed coat- Integument
  2.  The part that develops into an embryo after fertilization Embryo sac/ovule
  3. The part that develops into an endosperm in an albuminous seed – Nucellus
  4.  The part through which the pollen tube gains entry into the embryo sac – Micropyle .
  5.  The pollens are dry and unwettable.

OR
(a) Draw a diagrammatic sectional view of the female reproductive system of human and label the parts :
(1) Where the secondary oocytes develop ?
(2) Which helps in collection of ovum after ovulation ?
(3) Where fertilization occurs
(4) Where implantation of embryo occurs ?
(5) Explain the role of pituitary and the ovarian hormones in menstrual cycle in human females.
Answer:(a) The female Reproductive System Uterine fundus

  1.  The Ovary
  2.  Fimbriae
  3.  Fallopian tubes
  4. Uterus
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(B) The role of pituitary and the ovarian hormones in menstrual cycle in human females :
The Menstrual cycle lasts for about 4 -5 days. If fertilization does not occur, the corpus luteum degenerate, and the inner lining of uterus and fallopian tubes which contain soft tissue and blood vesseles sheds off, resulting in menstrual flow. In this phase, the primary follicles grow into the GrafAan follicles results in the regeneration of the endometrium.
Ovarian and pituitary hormones are responsible for following changes:

  1.  The release of gonadotropins (LH and FSH) increases . which facilitate follicular growth and the growing follicles produce oestrogen.
  2. In the middle of the menstruation cycle (14th day) the LH and FSH are at their peak, assist rupture of the Graffian follicles to discharge ovum. This phase is known as ovulatory phase.
  3.  The remains of the Graffian follicles get transformed into the corpus luteum, which secretes progesterone for the maintenance of the endometrium.

Question.29. Describe the asexual and sexual phase of life cycle of Plasmodium that causes malaria in humans.
Answer: Life Cycle of Plasmodium :

  1. Plasmodium requires two hosts to complete its life cycle.
  2. When female Anopheles mosquito bites a healthy human being, it inject Plasmodium, which lives in its Anopheles mosquito as sporozoite the infectious form.
  3. The parasites Plasmodium multiply (asexual reproduction) after transportation to the liver cells and finally rupture the liver cells. Sporozoites parasite are then released in to blood stream.
  4.  Bursting of RBCs is accompanied by release of a toxic substance called haemozoin (associated with fever and chills).
  5.  In the RBCs, only sporozoites change into gametocytes (sexual stage).
  6.  When the diseased person is bitten by a female Anop¬heles mosquito, gametocytes are introduced into the mosquito.
  7. Gametocytes fertilise and develop inside the intestine of mosquito to form sporozoites.
  8. Sporozoites are stored in the salivary glands of mosquito and are released into the healthy person who is bitten by this mosquito.
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OR
(a) What is plant breeding ? List the two steps the classical . plant breeding involves.
(b) How has the mutation breeding helped in improving crop varieties ? Give one example where this technique has helped.
(c) How has the breeding programme helped in improving the public nutritional health ? State two example in support of your answer.
Answer : (a) Plant breeding is a method that involves in the crossing of two plants to create the progeny with certain traits in their genes and transfer on to the future generation to create specific plant types, which are more suitable for cultivation, give better yields, and are disease resistant.
Two steps involved in Classical plant breeding are :

  1.  Crossing of superior pure lines and
  2. Selection of plants with desired characteristics.

(b)Mutation breeding : In this, genetic variations are made, which then creates traits, not found in the parental type. It has benefited in making disease resistant plants by giving resistance against bacterial, fungal and viral diseases.
(c) Breeding helped in improvising the public nutritional health by growing crops that are good in nutrients. This is called bio-fortification of crops. Benefits of bio-fortification are to help :

  1. Protein content and quality
  2.  Oil content and quality
  3. Vitamin content
  4.  Micronutrient and mineral content Two examples are :
    (i) Maize hybrids grows 2000 and have 2 times the quantity of lysine and tryptophan compared to other maize hybrids.
    (ii) Atlas 66 (a wheat variety having higher protein content).

Question.30. A child suffering from Thalassemia is born to a normal couple. But the mother is being blamed by the family for delivering a sick baby.
(a) What is Thalassemia ?
(b) How would you counsel the family not to blame the mother for delivering a child suffering from this disease ? Explain.
(c) List the values your counselling can propagate in the families.
Answer : (a) Thalassemia is a autosomal recessive blood disorder. Thalassemia is illustrated by severe anaemia due to production of imperfect haemoglobin chains. Thalassemia is caused by mutations in the genes coding either the alpha, beta or delta chains constituting haemoglobin lead to the synthesis of incorrectly folded haemoglobin that is unable to transport , .oxygen proficiently.
(b) Thalassemia is an autosomal recessive disease, which means the mutation is carried on one of the autosomes, so the carrier can be any one of the two parents. It has an equal chance of coming from the mother or the father, so to just blame the mother for the child’s defect is unfair.
(c) The values of councelling can propagate in the families are :

  1.  Provide healthy diet to the child
  2. Accepting their child with all his/her positives and negatives
  3. None of the parents is responsible for giving birth to a sick baby.
  4.  The defect in the gene is caused by a random change in the genes of the child.
  5.  Encouraging the child to follow his/her treatment regularly * and lead a happy and normal life.

SET- II

SECTION-A

Question.5. Identify the examples of homologous structures from the following:
(i) Vertebrate hearts
(ii) Thoms in Bougainvillea and tendrils of Cucurbita.
(iii) Food storage organs in sweet potato and potato.
Answer: (ii) Thorns in Bougainvillea and tendrils of Cucurbita

SECTION-B

Question.9. Describe the gene therapy procedure for an ADA-deficient patient.
Answer : Gene therapy of ADA deficiency :

  1.  The term gene therapy describes as genetic modification of cells. The material transferred into patient’s cell through genes, gene segment or oligonucleotides.
  2.  Lymphocytes isolated from patients blood are cultured in vitro. Functional ADA are then introduced into the cultured lymphocytes.
  3.  These lymphocytes are returned back to the patient’s body.
  4. The gene must be delivered inside the target cells and
    work properly without causing adverse effects to cure the disease. .
  5.  The introduction of isolated gene from bone marrow cell producing ADA into cells at early embryonic stages used can be, a permanent cure for this disorder.

Question.14. (a) How does cleistogamy ensure autogamy ?
(b) State one advantage and one disadvantage of cleistogamy to the plant.
Answer : (a) Cleistogamous flowers do not open at all and pollen from other plants cahnot fall on the stigma of these flowers. In this situation cross pollination cannot be possible and only autogamy occurs. Therefore, clesistogamy ensures autogamy.
(b) Cleistogamy has this advantage that the plant can propagate itself under unfavorable conditions and disadvantage is that, no chances of cross pollination, self pollination occurs therefore chances of variation and evolution of genetically superior progeny is reduced.

Question.17. A young boy when brought a pet dog home started to complain of watery eyes and running nose. The symptoms disappeared when the boy was kept away from the pet.
(a) Name of type of antibody and the chemicals responsible for such a response in the boy.
(b) Mention the name of any one drug that could be given to the boy for immediate relief from such a response.
Answer : (a) The IgE type of antibody and chemicals like histamine and serotonin released from mast cells are responsible for this type of response.
(b) (i) Antihistamine
(ii) Adrenalin and
(iii) Steroids can be given to the boy for immediate relief from such a response.

Question.18. (a) Explain how to find whether an E coli bacterium has transformed or not when a recombinant DNA bearing ampicillin resistant gene is transferred into it.
(b) What does the ampicillin resistant gene act as in the above case ?
Answer : (a) When E. coli with a recombinant DNA will be grown on culture media having ampicillin the transformed cells will survive as they carry recombinant DNA with ampicillin resistant gene; while the non transformed cells will die as they are ampicillin sensitive.
(b) Antibiotic resistant genes such as ampR (ampicillin resistant), acts as selectable markers.

SECTION-C

Question.22. (a) Explain how to overcome inbreeding depression in cattle.
(b) List three advantages of inbreeding in catde.
(c) Name an improved breed of catde.
Answer : (a) Inbreeding depression in cattle can be made overcome by mating selected animals with unrelated superior animals of the same breed. It is called out-breeding.
(b) Advantages of inbreeding in catde :

  1.  Maintaining the accumulation of superior quality of breed of organisms.
  2. Evolve a pure breed of progeny.
  3. Elimination of deleterious alleles as they are not passed to future generations.

(c) Jersey cow is an improved breed of cattle.

Question.27. (a) Draw a diagram of the structure of a human ovum surrounded by corona radiate. Label the following parts : (i) Ovum (ii) Plasma Membrane (iii) Zona Pellucida (b) State the function of Zona Pellucida.
Answer : (a) Diagram of a mature ovum
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(b) Zona pellucida layer of the ovum-prevent entry of more than one sperm and inhibit polyspermy in humans. It also facilitate changes in the membrane of the ovum to block the entry of additional sperms, thereby makes sure the entry of only one sperm inside the ovum.

SECTION-D

Question.30. (a) Draw a labeled schematic diagram of the transverse
section of a mature anther of an angiosperm plant.
(b) Describe the characteristic feature of an insect pollinated flower.
Answer: (a) Diagram of transverse section of a mature anther of an angiospermic plant
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(b) Characteristic features of insect pollinated flower:

  1.  Contains nectar to attract the,animal pollinators
  2. Large, colourful and fragrant flowers
  3.  Ovule may be one or more
  4.  Sticky pollen grains

OR
(a) Describe the events of spermatogenesis with the help of a schematic representation.
(b) Write two differences between spermatogenesis and oogenesis.
Answer : (a) Events of spermatogenesis with the help of schematic diagram

  1. Spermatogenesis starts at the age of puberty.
  2.  Spermatogonia are present on the inside wall of semniferous tubules multiply by mitotic divisions and increase in number.
  3. Each spermatogonium is diploid and contain 46 chromosomes.
  4. Some of spermatogonia called primary spermatocytes
    undergo meiosis I to produce 2 equal haploid cells called – secondary spermatocytes with 23 chromosomes each.
  5.  Secondary spermatocyte undergo second meiotic division to produce 4 equal haploid spermatids.
  6.  Spermatid later transform into spermatozoa.
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SET- III

SECTION-A

Question.6. “Sweet potato tubers and potato tubers are the result of convergent evolution.” Justify the statement.
Answer : Convergent evolution is the process where some diverse organisms individually develop the traits which are similar in function, i.e, sweet potato tubers and potato tubers – are not anatomically similar structures even though they perform similar function, e.g. storage of food and vegetative reproduction but both are different in origin i.e., sweet potato is an adventitious root and potato tuber is an underground stem.

SECTION-B

Question.13. Explain the steps that ensure cross pollination in an autogamous flower.
Answer: Cross pollination in an autogamous flower represents by following steps :

  1.  Pollen grains and stigma receptivity should not be synchronized.
  2. Anther and stigma are placed at different positions so that pollen cannot come in contact with stigma of the same flower.
  3. Self incompatibility also prevents inbreeding.

Question.16. A student on a picnic to a park on a windy day started sneezing and having difficulty in breathing on reaching the park. The teacher enquired whether the student was allergic to something.
(a) What is an allergy ?
(b) Write the two unique characteristics of the system involved in the response observed in the student ?
Answer: (a) Allergy is a hypersensitive reaction of foreign substances by the immune system. The body system of defense against these substances particularly pathogens (the agents of infection) caused allergy. The inflammation of the tissues inside the nose causes allergy after allergens are inhaled.
(b) The two unique characteristics of the system involved in the response observed in the students are :

  1.  The immune system recognizes foreign antigens and responds to them.
  2.  It has a memory so it remembers antigens.

Question.17. Why and how bacteria can be made ‘competent’?
Answer : The bacterial cells are made competent by treating them with a specific concentration of divalent cations like calcium or magnesium e.g., CaCl2 or MgCl2 This makes the cell wall permeable and bacterial cell takes up the plasmid DNA.

Question.18. (a) Name the deficiency for which first clinical gene therapy was given.
(b) Mention the cause of and one cure for this deficiency.
Answer : (a) ADA (Adenosine deaminase) deficiency.
(b) Cause : The gene coding for enzyme ADA gets deleted leading to deficiency of ADA and problems in immune system. Cure : It can be treated by isolating the gene for ADA from bone marrow cells at embryonic stage.

SECTION-C

Question.23 Draw the following diagrams related to human reproduction and label them.
(a) The zygote after the first cleavage division
(b) Morula stage
(c) Blastocyst stage (sectional view)
Answer : (a) The zygote after the first cleavage division.
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Question.29. Draw a diagram of a mature embryo sac of an angiosperm and label the following parts in it.
(i) Filiform apparatus (ii) Synergids
(iii) Central cell (iv) Egg cell
(v) Polar nuclei (vi) Antopodals
(b) Write the fate of egg cell and polar nuclei after fertilization.
Answer: (a)
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(b) Fate of Egg cell and polar nuclei: Egg cell fuses with male gametes to form zygote. Zygote finally give rise to Embryo while the polar nuclei fuses with the other male gamete to produce a triploid Primary Endosperm Nucleus (PEN). PEN develops into endosperm. Since two kinds of fusion—syngamy and triple fusion—take place, the process is known as double
fertilisation, and is a characteristic of flowering plants.