CBSE previous Year Solved  Papers  Class 12 Biology Outside Delhi 2012

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. Why is banana considered a good example of parthenocarpy ? 
Answer: Banana is considered a good example of parthenocarpy because formation of fruit in banana occurs without fertilization (parthenocarpy), i.e., there is no formation of seeds.

Question.2. State two different roles of spleen in the human body.
Answer : The roles of spleen in the human body is that the spleen is the secondary lymphoid organ that produce lymphocytes and the red pulp of spleen removes the old or damaged red blood cells from the body.

Question.3. A garden pea plant produced axial white flowers. Another of the same species produced terminal violet flowers. Identify the dominant traits.
Answer : The dominant traits are : Axial, violet flower.

Question.4. Why is it desirable to use unleaded petrol in vehicles fitted with catalytic converters ?
Answer: It is desirable to use unleaded petrol in vehicles fitted with catalytic converters because lead in petrol inactivates the catalysts which convert harmful pollutants (CO, unburnt hydrocarbons, nitric oxide) to lesser harmful pollutants (C02,H2o,N2).

Question.5. Where is acrosome present in humans ? Write its function.
Answer: The anterior portion of the sperm head in human beings is covered by a cap-like structure called acrosome. Function of Acrosome :
(i) Acrosome is filled with hydrolytic enzyme-Hyaluronidase that aids in the entry of sperm into the ovum.

Question.6. Write the name of the following :
(a) The most common species of bees suitable for apiculture
(b) An improved breed of chicken
Answer : (a) Apis indica is the most common species of bees for apiculture.
(b) Leghorn is an improved breed of chicken.

Question.7. Comment on the similarity between the wing of a cockroach
and the wing of a bird. What do you infer from the above, with reference to evolution ?
Answer : The wing of a cockroach and the wing of a bird are not similar anatomically i.e., not similar in structure but similar in function. Thus we infer that these organs are analogous which has resulted in convergent evolution.

Question.8. Mention the role of cyanobacteria as a biofertiliser.
Answer: The role of cyanobacteria as a biofertiliser : Cyanobacteria (Anabaena and- Nostoc) are free-living in the root nodules of leguminous plants and they fix atmospheric nitrogen. They act as a bio-fertilizers especially in paddy fields.

SECTION-B

Question.9. (a) Draw a neat labelled diagram of a nudeosome.
(b) Mention what enables histones to acquire a positive charge.
Answer: (a)
cbse-previous-year-solved-papers-class-12-biology-outside-delhi-2012-1
(b) Depend on the abundance of basic amino acid residues of lysines and arginines with charged side chain.

Question.10. State one advantage and one disadvantage of cleistogamy.
Answer: Advantage of cleistogamy: Self-pollination is assured, seed production is also assured in the absence of pollinators., Disadvantage of cleistogamy: In cleistogamous flowers due to self-pollination least variations observed.

Question.11. (a) Where do the signals for parturition originate from in humans ?
(b) Why is it important to feed the newborn babies on colostrum ?
Answer : (a) The signals for parturition in humans originate from the fully developed foetus and the placenta which include mild uterine contraction.
(b) The colostrum or first milk is important to feed the new born babies because it contains antibodies (IgA), to provide passive immunity to the baby.

Question.12. (a) A recombinant vector with a gene of interest inserted
within the gene of a-galactosidase enzyme, is introduced into a bacterium. Explain the method that would help in selection of recombinant colonies from non-recombinant ones.
(b) Why is this method of selection referred to as “insertional inactivation” ?
Answer: (a) Bacteria is grown in a medium with chromOgenic substrate, blue coloured colonies with no recombinations and colonies with no blue colour show presence of recombinants.
Chromogenic substrate used to identify recombinants and non-recombinants
(b) Gene for the enzyme is inactivated by insertion which is referred to as insertional inactivation.

Question.13. Explain brood parasitism with the help of an example.
Answer: Brood parasitism involves the use of host individuals of the same or different species to raise the young of the brood parasite. This relieves the parasitic parent from the investment of rassing young or building nests, producing offspring etc.
eg : The cowbird family is the bird family in North America is an brood parasite (because are not capable of building a nest). While most brood parasites have eggs that mimic the hosts eggs the cowbird is again different.

Question.14. Give reasons for the following :
(a) The human testes are located outside the abdominal cavity.
(b) Some organisms like honey-bees are called parthenogenetic animals.
Answer : (a) To maintain the temperature (2-2.5°C) lower than the normal internal body temperature,, which is essential for spermatogenesis.
(b) The phenomenon of development of female gamete directly into an individual without fertilization is called parthenogenesis. Example : The drones/males develop from unfertilised eggs of honey bees.

Question.15. Name the plant source of ganja. How does it affect the body of the abuser ?
Answer : Plant source : Cannabis Sativalhemp plant.
Affect: It damages cardio-vascular system of the body
OR
Name the two special types of lymphocytes in humans. How do they differ in their roles in immune response ?
Answer: B lymphocytes, T lymphocytes.
B-cells produce pathogen specific antibodies called humoral immune response.
T-cells help the B-cells to produce antibodies and are responsible for direct cell mediated immunity.

Question.16. (a) Mention the cause and the body system affected by ADA deficiency in humans.
(b) Name the vector used for transferring ADA-DNA into the recipient cells in humans. Name the recipient cells.
Answer : (a) The body system affected by ADA deficiency in humans is immune system. ADA deficiency is caused due to lack the gene coding for adenosine deaminase.
(b) A retroviral vector is used to transfer ADA-DNA into recipient cells. The recipient cells are lymphocytes.

Question.17. How did Ahmed Khan, plastic sacks manufacturer from Bangalore, solve the ever – increasing problem of accumulating plastic waste ?
Answer : Ahmed Khan, a plastic sacks manufacturer solve the ever increasing problem and accumulation of waste. Polyblend fine powder of recycled modified plastic can be used to lay roads that will increased road life. When blended with bitumen, it enhances the bitumen’s water repellent properties and increase the life of road.

Question.18. Name the bacterium that causes typhoid. Mention two diagnostic symptoms. How is this disease transmitted to others ?
Answer : Bacterium: Salmonella typhi.
Diagnostic Symptom: Constipation, stomach pain, headache, weakness, loss of appetite, high fever.
The disease is transmitted through contaminated food and water.

SECTION-C

Question.19. (a) Explain the phenomena of multiple allelism and codominance taking ABO blood group as an example.
(b) What is the phenotype of the following:
(i) \({ I }^{ A }\)i (ii) ii
Answer: Multiple allelism : (a) In humans, the ABO blood groups are controlled by a gene called gene ‘I’. It has three alleles, \({ I }^{ A }\), \({ I }^{ B }\) and i, hence, referred to as multiple allelism. Co-dominance : If \({ I }^{ A }\) and \({ I }^{ B }\) both are present in an individual, and they both are expressed because of the phenomenon of co-dominance.
(b) (i) Phenotype of \({ I }^{ A }\) / : A blood group.
(ii) Phenotype of i i: O blood group,

Question.20. How does industrial melanism support Darwins theory of Natural Selection ? Explain.
Answer: In England, before industrial revolution the environment was unpolluted. The white-winged moths were more and lichens on the barks of trees were pale. The white-winged moths could easily camouflage, while the darkwinged were spotted out by the birds for food. Hence, they could not survive. After industrial revolution the lichens became dark (due to soot deposit). This favoured the dark-winged moths while the white-winged were picked by birds. The population of the former which was naturally selected increased.

Question.21. (a) What is the programme called that is involved in
improving success rate of production of desired hybrid and herd size of catde ?
(b) Explain the method used for carrying this programme for cows.
Answer: (a) Multiple ovulation embryo transfer Technology/ MOET is used for improving success rate of production of desired’hybrid and herd size of cattle.
(b) Methods used for carrying this programme for cows :

  1.  The cow is administered with FSH to induce follicular maturation and super – ovulation to produce 6 to 8 eggs.
  2.  The animal is either mated with an elite bull or artificially inseminated.
  3.  The fertilised eggs 8-32 cells stage are recovered non- surgically and transferred to surrogate mother where they develop into an improved variety.

Question.22. Explain the function of each of the following :
(a) Coleorhiza (b) Umbilical cord (c) Germ pores
Answer: The function of each of the following:
(a) Goleorhiza : Protects the radical of (monocot) embryo.
(b) Umbilical cord : Transports nutrients and respiratory gases and metabolic wastes to and from mother and foetus!
(c) Germ pores : Allow germination of pollen grain and formation of pollen tubes.

Question.23. How is the amplification of a gene sample of interest carried out using Polymerase Chain Reaction (PCR) ?
Answer: Polymerase chain reaction (PCR) is a method in which the desired gene is synthesised in vitro in following steps:

  1.  Denaturation : The double-stranded DNA is denatured by applying high temperature of 95°C for 15 seconds. Each separated single stranded strand now acts as template for DNA synthesis.
  2.  Annealing: Two sets of primers are added which anneal to the 3′ end of each separated strand. Primers act as initiators of replication.
  3.  Extension : DNA polymefase extends the primers by adding nucleotides complementary to the template provided in the reaction. A thermostable DNA polymerase (Taq polymerase) is used in the reaction which can tolerate the high temperature of the reaction. All these steps are repeated many times to obtain several copies of desired DNA.
    cbse-previous-year-solved-papers-class-12-biology-outside-delhi-2012-2

Question.24. Trace the life-cycle of malarial parasite in the human body when bitten by an infected female Anopheles.
Answer : Plasmodium requires two hosts to complete its life cycle.When female Anopheles mosquito bites a healthy person. Sporozoite of Plasmodium gets into human blood through the bite of female Anopheles mosquito. The parasite multiply in liver cells and finally burst in liver cells and released in blood, then they get into red blood cells, where they further multiply asexually and burst in RBCs also and released toxic substance haemozoin (associated with fever and chills). After a while they change into gametocytes, which are picked up by the mosquitoes and the entire cycle occurs again.
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Question.25. List the salient features of double helix structure of DNA.
Answer: The salient features of Double Helix structure of DNA:

  1.  It is made of two polynucleotido chains where backbone is sugar phosphats constituted and bases project inside.
  2.  There is complementary base pairing between the two strands of DNA. The amount of adenine is equal to thymine and the amount of guanine is equal to cytosine.
  3.  The two strands are coiled in right-handed fashion and are anti-parallel in orientation. One chain has a 5’—>3′ polarity while the other has 3’—>5′ polarity.
  4.  The diameter of the strand is always constant due to pairing of purine and pyrimidine, i. e., adenine is complementary
    to thymine while guanine is complementary to cytosine.
  5.  The distance between the base pairs in a helix is 0.34 nm and a complete turn contains approximately ten base pairs. The pitch of the helix is 3.4 nm and the two strands are right-handed coiled.

OR
How are the structural genes activated in the lac operon in E. coli ?
Answer: The structural genes activated in the lac operonin E. coil in the following manner :
Lactose consists of the genes lac z, y and a. Lactose acts as the inducer that binds with repressor protein and frees the operator gene. RNA polymerase freely moves over the structural genes, transcribing lac mRNA, which in turn – produces the enzymes responsible for the digestion of lactose.

Question.26. Alien species are highly invasive and are a threat to indigenous species. Substantiate this statement with any three examples.
Answer : The three examples of the above statement are :

  1.  Nile perch introduced into Lake Victoria in East Africa led to the extinction of Cichlid fish.
  2. Invasive plants like Parthenium /Lantana /Eichhomia caused environmental damage and posed a threat to indigenous species.
  3.  Introduction of African catfish (Clarias gariepinus) to aquaculture is a threat to indigenous Indian catfishes.

Question.27. (a) Tobacco plants are damaged severely when infested with
Meloidegyne incognitia. Name and explain the strategy that is adopted to stop this infestation.
(b) Name the vector used for introducing the nematode specific gene in tobacco plant.
Answer: (a) The infestation jyas prevented by strategy which was based on RNA interference or RNAi or gene silencing. During this process nematode specific gene is introduced into host plant (using Agrobacterium) which produce dsRNA. This specific mRNA of the nematode silenced and parasite dies.
(b) Agrobacterium tumifaciens vector are used for introducing the nematode specific gene in tobacco plant.

SECTION – D

Question.28. (a) Taking one example each of habitat loss and fragmentation, explain how are the two responsible for biodiversity loss.
(b) Explain two different ways of biodiversity conservation.
Answer : (a) Habitat loss and fragmentation are responsible for biodiversity loss are :

  1. Habitat loss : The Amazon rainforest (called the “lungs of the planet”) is being cut and cleared for cultivation of soya beans and for conversion into grasslands for raising beef catde.
  2. Fragmentation : When large-sized habitats are broken or fragmented due to human setdements, building of roads, digging of canals, etc., the population of animals requiring large territories and some animals with migratory habitats declines.

(b) The two different ways of biodiversity conservation are :

  1.  Ex situ
  2.  In situ conservation.

Ex situ conservation : In this conservation threatened organism are taken out from the natural habitat and placed in special setting with care and protected, eg., zoological park, botanical garden, wild safari.
In situ conservation: In this conservation threatened organisms are conserved in their natural habitat, e.g, national , parks, biosphere reserves.
OR
(a) What depletes ozone in the stratosphere? How does this affect human life?
(b) Explain biomagnification of DDT in an aquatic food chain. How does it affect the bird population?
Answer: (a) Chlorofluorocarbons (CFCs) released from the refrigerators air conditioners deplete ozene in the stratosphere. Ozone acts as a shield and protects the earth from, the harmful UV rays of the sun.
Effect on Human life : Chlorofluorocarbons depletes ozone layer causing UV rays to reach to earth which damages DNA causing mutation, skin cancer, inflammation of cornea, cataract, aging of skin, snow blindness.
(b) If DDT leaches from the agricultural field, it gets into the water body (the concentration is 0.0003 ppm) and enters the food chain:
zooplanktons (0.04 ppm) —> small fish (0.05 ppm) —» large fish (2 ppm) —» any fish eating bird (5 ppm). Concentration of DDT increases along the food chain, reaching a high level in the top carnivore bird.
Effect on Bird population : DDT concentration disturbs Ca++ metabolism, egg shells become thin, premature breaking resulting in decline in bird population.

Question.29. The following is the illustration of the sequence of ovarian events “a” to “i” in a human female:
cbse-previous-year-solved-papers-class-12-biology-outside-delhi-2012-4
(a) Identify the figure that illustrates corpus luteum and name the pituitary hormone that influences its formation.
(b) Specify the endocrine function of corpus luteum. How
does it influence the uterus ? Why is it essential ?
(c) What is the difference between “d” and “e” ?
(d) Draw a neat labelled sketch of Graafian follicle.
Answer: (a) Corpus luteum is illustrated by ‘g’ and the hormone influencing its formation is luteininsing hormone (LH).
(b) Corpus luteum produces the hormone progesterone, causes proliferation of the endometrium which gets highly vascularised. It is essential for the implantation of the fertilized ovum and maintains the same during pregnancy.
(c) “d” is the developing tertiary follicle, “e” is the Graafian follicle.
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OR
(a) Why is fertilisation in an angiosperm referred to as double fertilisation ? Mention the ploidy of the cells involved.
(b) Draw a neat labelled sketch ofL.S. of an endospermous monocot seed.
Answer: (a) Fertilisation of haploid egg cell by one haploid male gamete to form diploid zygote is called syngamy. Fertilisation of two (diploid) polar nuclei by the other haploid male gamete to form triploid primary endosperm nucleus is called triple fusion.
(b) L.S. of an endospermic monocot seed.
cbse-previous-year-solved-papers-class-12-biology-outside-delhi-2012-6

Question.30. Describe Frederick Griffiths experiment on Streptococcus pneumoniae. Discuss the conclusion he arrived at.
Answer : Frederick Griffiths experiment Transforming Principle conducted by him in 1928.

  1. Frederick Griffith (1928) conducted experiments with Streptococcuspneumonia (bacterium causing pneumonia).
  2.  He observed two strains of this bacterium—one forming smooth shiny colonies (S-type) with capsule, while other forming rough colonies (R-type) without capsule.
  3.  When live S-type cells were injected into mice, they died due to pneumonia.
  4.  When live R-type cells were injected into mice, they survived and he arrived at this conclusion
  5. When heat-killed S-type cells were injected into mice, they survived and there were no symptoms of pnuemonia.
  6. When, heat-killed S-type cells were mixed with live R-type cells and injected into mice, they died due to unexpected symptoms of pneumonia.
  7. He concluded that heat-killed S-type bacteria caused a transformation of the R-type bacteria into S-type bacteria but he was not able to understand the cause of this bacterial transformation.

OR
(a) Explain a monohybrid cross taking seed coat colour as a trait in Pisum sativum. Work out the cross up to F2 generation.
(b) State the laws of inheritance that can be derived from such a cross.
(c) How is the phenotypic ratio of F2 generation different in a dihybrid cross ?
Answer : (a) A monohybrid cross taking seed coat colour as a trait in Pisum sativum, for e.g if pea plant with yellow seed coat is crossed with pea plant having green seed coat then in the Fj generation all the plants are yellow seeds.
cbse-previous-year-solved-papers-class-12-biology-outside-delhi-2012-7
F2 Phenotypic ratio = 3:1
F2 Genotypic ratio =1:2:1
The law of inheritance can be derived from such a cross :
(b) (i) Law of Dominance: Factor occurs in pairs. In a contrasting pair of factors one member of the pair dominates (dominant) the other (recessive).
(ii) Law of Segregation : Factors or allele of pair segregate from each other such that gamete receives only one of the two factors. Paired condition is restored at the time of zygote’ formation.
(c) Phenotypic ratio of F2 generation in monohybrid cross is 3 :1 whereas in a dihybrid cross the phenotypic ratio is 9 : 3 : 3 : 1.

SET-II

SECTION – A

Question.1. How do the pollen grains of Vallisneria protect themselves ?
Answer : The pollen grains of Vallisneria have mucilaginous covering to prevent them from getting wet.

Question.2. Name the respective pattern of inheritance where Ft phenotype.
(a) does not resemble either of the two parents and is in between the two.
(b) resembles only one of the two parents.
Answer : (a) The respective pattern of inheritance where phenotype not resemble either of the two parents and is in
between the two known as Incomplete dominance.
(b) resembles only one of the two parents Law of Dominance.

Question.5. How is the entry of only one sperm and not many ensured into an ovum during fertilisation in humans ?
Answer : During fertilisation in humans: At the sperm head there is an enzyme to dissolve the follicles of ovum and facilitate entry of the sperm nucleus for fertilization and help the sperm enter into the cytoplasm of the ovum.

Question.7. State the significance of Coelacanth in evolution.
Answer : It is an ancestor of amphibians. The latest analysis shows that the genes of modern coelacanths’can-themselves be considered living fossils.

SECTION – B

Question.12. Name the source organism that possesses Taq polymerase.
What is so special about the function of this enzyme ?
Answer: Thermus aquaticus.
The enzyme can tolerate high temperature and is thermostable. It does not get denatured during PCR at high temperature.

Question.13. State one advantage and one disadvantage of cleistogamy.
Answer : Advantage : Cleistogamy flowers produce assured seed set even in the absence of pollinators to increase genetic variations. Disadvantage: The disadvantage of Cleistogamy is the offspring produced have limited genetic diversity.

Question.15. Name the source of cyclosporin-A. How does this bioactive molecule function in our body ?
Answer: The source of cyclosporin-A Trichoderma polysporum.
Bioactive molecule function in our body: It is used as an immuno-suppressant agent in organ transplant patient.
OR
(a) Name the group of viruses responsible for causing AIDS in humans. Why are these viruses so named ?
(b) List any two ways of transmission of HIY infection in humans, other than sexual contact.
Answer : (a) The group of viruses responsible for causing AIDS in humans iis Retrovirus. These are named so because they (have RNA g;enome) have reverse transcriptase enzyme which carries on die processes RNA —> DNA —» RNA.
(b) (i) Transfusion , of infected blood.
(ii) Sharing infect ed syringes and needles. .
(iii) Children bor n to HIV mother through placents.

Question.17. Name any two or ganisms that are responsible for ringworms in humans. Mention two diagnostic symptoms. Name the specific parts of the human body where these organisms thrive and explain why.
Answer : Micro sporum and Trichophyton are two organisms that are responsible for ringworms in humans.
Symptoms : Dry and scaly lesion on skin, nails, scalp, intense itching These thrive in body groin, between toes, thrive better in heat, moisture, perspiration.

SECTION – C

Question.19. Differentiate between perisperm and endosperm giving one example of each.
Answer:
cbse-previous-year-solved-papers-class-12-biology-outside-delhi-2012-8

Question.25. (a) List any three ways of measuring population density of a habitat.
(b) Mention the essential information that can. be obtained by studying the population density of an organism.
Answer : There are different ways of measuring population density of a habitat are as follows :

  1.  Quadrat method: This method involves the use of square of particular dimensions to measure, no. of organisms.
  2. Direct observation : This method is used for counting  of organism.
  3. Indirect method : This method is used for the number fish caught per trap gives the measure of their total t density in a given water bod y.

(b) The population density of an organism provides us the status of habitat, whether competition for survival exists or not, whether population is increas ing or declining, natality, mortality, emigration, immigration.

SECTION – D

Question.28. (a) Explain the significance of ecological pyramids with the help of an example.
(b) Why are the pyramids referred to as1 upright’ or ‘inverted’ ?
Answer : (a) Ecological pyramid expresses the relationship between the organisms at differen t trophic levels with reference to their number, energy and biomass.
cbse-previous-year-solved-papers-class-12-biology-outside-delhi-2012-9
Upright Pyramid : In this pyramid producers are more in number and in biomass than the herbivores, and herbivores are more in number and biomass than the carnivores. The Pyramid of energy is always upright as only 10% energy is transferred from one trophic level to the next.
Inverted pyramid: It shows less number/biomass of producers when compared to primary consumers. For e.g. large number of insects feeding on a big tree give inverted pyramid of number.
OR
(a) Explain giving reasons why the tourists visiting Rohtang Pass or Mansarovar are advised to resume normal ‘active life’ only after a few days of reaching there.
(b) It is impossible to find small animals in the polar regions. Give reasons.
Answer: (a) The tourists visiting Rohtang Pass or Mansarovar are advised to resume normal active life’ only after a few days of reaching there because initially the person suffers from altitude sickness, nausea, fatigue and heart palpitation because of low oxygen availability and low atmospheric pressure. Gradually the body increases RBC production, decreasing binding capacity of Hb and increases the breathing rate to get acclimatised. •
(b) Small animals are rarely found in polar regions because small birds have larger surface area relative to their volume, so they lose heat much faster spend more energy to generate body heat. They have to expand much energy to generate body heat through metabolism.

SET -III

SECTION-A

Question.1.Identify the figure given below and the part Labelled “A”.
cbse-previous-year-solved-papers-class-12-biology-outside-delhi-2012-10
Answer:The figure is of blastula/blastocyst.
A—Trophob’last.

Question.2. How do interferons protect us ?
Answer : Interferons protect uninfected cells from further viral infections, by creating cytokine barriers.

Question.3. Name the interaction between a whale and the bar nacles growing on its back?
Answer : Commensal ism is the interaction between a wdiale and the barnacles grow ing on its back.

Question.7.In a dihybrid cross, when would the proportion of parental . gene combinations be much higher than non-parental types, as experimentally shown by Morgan and his group ?
Answer: When the genes are present on a same chromosome or incomplete linkage is present then the proportion of parental gene combinations be much higher than non-parental types,* according to Morgan and his group.

SECTION—B

Question.12. Name the cells that act as HIV factory in humans when infected by HIV. Explain the events that occur in the infected cell.
Answer : Macrophages/Helper T-cells act as HIV factory in humans.
The events occur in the infected cell:

  1.  HIV attached to CD-4 cell with GP-120 to CD-4 protein.
  2.  Reverse transcription by reverse transcriptase, enzyme.
  3.  Its DNA attached with host cell DNA.
  4.  Multiplication of HIV.
  5.  Lysis of infected cell.

Question.15. Name and explain the two types of immune responses in humans.
Answer: The two types of immunity are active immunity and passive immunity.
Active immunity: Immunity developed in the host body due to production of antibodies in response to antigens which of low intensity and produce memory cells.
Passive immunity : When ready-made antibodies are direcdy given to protect the body against foreign agents which is of very high intensity.

Question.16. How does the study of different parts of a flower help in identifying in wind as its pollinating agent ?
Answer: Pollination in which wind distributed the pollens is called anemophily.
Wind pollinated flowers have light and non sticky pollen grains, well exposed stamens to disperse pollen easily, large and often feathery stigma for easy trapping of pollen, single ovule in each ovary, numerous flowers packed into an inflorescence.

SECTION-C

Question.22. Explain how do the following act as contraceptives :
(a) CuT (b) “Saheli”
Answer: (a) CuT release Cu ions when inserted into the uterus which suppress sperm motility, lowers the fertilising capacity of sperms. These devices inserted by doctors as experts in the uterus through vagina
(b) Oral contraceptive for the female! contain a non-steroidal preperation. It is once a week pill with very few side effects and high contraceptive value inhibit ovulation, implantation, as well as alter the quality of cervical mucus to prevent or retard the entry of sperms.

Question.26. Name and explain the evolutionary concept represented in the illustration given below:
cbse-previous-year-solved-papers-class-12-biology-outside-delhi-2012-11
Answer : The illustration represents adaptive radiation or divergent evolution.
It is the example of adaptive radiation in placental animals of Australia.

  1.  A variety of placental mammals have evolved which appear similar to a corresponding marsupial.
  2.  When more than one adaptive radiation appeal; to have occurred in an isolated geographical area, and two or more
    groups of unrelated animals come to resemble each other for similar mode of life or habitat, it is called convergent evolution.

OR
(a) Why is it that the father never passes on the gene for haemophilia to his sons? Explain.
(b) State the functions of the following in a prokaryote :
(i) tRNA (ii)rRNA
Answer: (a) The father never passes on the gene for haemophilia to his sons because haemophilia is a sex-linked recessive disease and the defective.Gene is present on X chromosome only and not on Y chromosome.
(b) Function of t-RNA in prokaryotes –
(i) tRNA reads the genetic codes, carries amino acids to the site of protein synthesis and act as an adap tor molecule.
(ii) rRNA plays structural and catalytic role during translation .

SECTION – D

Question.30. (a) A garden pea plant bearing terminal, violet flowers, when crossed with another pea plant bearing aixial, violet flowers, produced axial, violet flower and axtial, white flowers in the ratio of 3 : 1. Work out the cross showing the genotypes of the parent pea plants and their progeny, (b) Name and state the law that can be derived from this cross and not from a monohybrid cross.
Answer: (a)
cbse-previous-year-solved-papers-class-12-biology-outside-delhi-2012-12
(b) This cross is based on Law of Independent Assortment. This 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 character.
OR
(a) Describe the process of synthesis of fully functional mRNA in a eukaryotic cell.
(b) How is this process of mRNA synthesis different from that in prokaryotes?
Answer : (a) The process of synthesis of fully functional mRNA in a eukaryoic cell.

  1. The primery transcrips are non-functional, containing both the coding region, exon, and non-coding reigion, irron, in RNA and are called heterogenous RNA or hnRNA.
  2.  The hnRNA undergoes two additional process called cappint and tailing.
  3.  In cappint, an unusual nucleotide, methyl guanosine triphosphate, is added to the 5′-end of hnRNA.
  4. In tailing, adenylate rersidues (about 200-300) are added at 3–end in a template independent manner.
  5.  Now the hnRNA undergoes a process where the introns are removed and exons are joined to form mRNA by the process called splicing.
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(b)In prokaryotes, there is a single DNA-dependent RNA polymerase that catalyses transcription of all types of RNA in bacteria. In bac teria, mRNA does not require any processing as it does not have any introns.