CBSE previous Year Solved  Papers  Class 12 Biology Delhi 2014

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. Write the name of the organism that is referred to as the ‘Terror of Bengal’.
Answer : Water hyacinth (Eichhomia Cmssipes)

Question.2. What are ‘true breeding lines’ that are used to study inheritance pattern of traits in plants ?
Answer : Breeding line having undergone with a number of repeated self pollination and shows stable trait inheritance and expression for several generations called a true breeding line.

Question.3. Name any two types of cells which acts as a cellular barriers to provide innate immunity in humans.
Answer : (i) Monocytes (natural killer)
(ii) PMNL-Neutrophils (polymorpho- nuclear leukocytes)

Question.4. Mention the type of host cells suitable for the gene guns to introduce an alien DNA.
Answer : The type of host cells suitable for the gene guns to introduce an alien DNA is plant cells.

Question.5. How is‘stratification represented in a forest ecosystem?
Answer : Stratification is the way in which different species occupying different levels or arranged in a habitat. It is a way of minimizing competition for limited but vital resources, for survival.
Vertical stratification in a forest ecosystem is represented by following strata of plantss starting from the lowest layrer :
(i) The herb layer               (ii) The shrub layer
(iii) The small tree layer   (iv) The canopy layer

Question.6. Give an example of an organism that enters ‘diapause’ and why ?
Answer : Bombyx mori (silk moth) is an insect that enters diapauses to avoid adverse environmental conditions such as drought, extreme temperature, reduced food availability, which, in turn, delays the overall development. The physiological and metabolic activities are also diminished at this particular time.

Question.7. Identify ‘a’ and ‘b’ in the figure given below representing proportionate number of major vertebrate taxa.
cbse-previous-year-solved-papers-class-12-biology-delhi-2014-1
Answer: ‘a’ represents mammals and ‘b’ represents amphibians.

Question.8. State the cause of Accelerated Eutrophication.
Answer: Accelerated Eutrophication is the aging of a water body due to nutrient enrichment of its water. Release of nutrient-rich sewage and industrial effluents containing nutrients such as nitrogen and phosphorus causes accelerated eutrophication.

SECTION-B

Question.9. Why do algae and fungi shift to sexual mode of reproduction just before’the onset of adverse conditions ?
Answer : The fungi and algae switch to the sexual mode of reproduction during adverse conditions because sexual reproduction brings variation into the individuals. In algae and fungi, the zygote develops a thick wall that is resistant to dessication and damage. This ensures the continuity of species.

Question.10. A cross was carried out between two pea plants showing the contrasting traits of height of the plants. The result of the cross showed 50% parental characters, (i) Work out the cross with the help of a Punnett square.
(ii) Name the type of the cross
Answer: carried out.
cbse-previous-year-solved-papers-class-12-biology-delhi-2014-2

Question.11. How does the gene ‘I’ control ABO blood groups in humans ? Write the effect the gene has on the structure of red blood cells.
Answer: In humans, the ABO blood groups are controlled by a gene called gene ‘I’. Sugar polymers protrude from the surface of red blood cells and I controls the kind of sugar. It has three alleles, IA, IB and i. A person possesses any two of the three.
Table: The Genetic Basis of Blood Groups in Human Population alleles. IA and P dominate over i . But with each other, Tand P are co-dominant.
cbse-previous-year-solved-papers-class-12-biology-delhi-2014-3
The red blood cells have sugar polymers protrude from the plasma membrane surface and it is regulated by the gene ‘I’ of ABO blood group. The alleles IA and P produce A and B types of sugar, while allele i does not produce any sugar.
OR
Write the types of sex-determination mechanisms the following crosses show. Give an example of each type.
(i) Female XX with Male XO .
(ii) Female ZW with Male ZZ
Answer : The types of sex – determination mechanisms in the following crosses are :
(1) Female XX with Male XO : In this case females has a pair of X chromosomes and males have only one X chromosomes (the O indicates absence of chromosome), so it is the case of male 0 hetrogamety eg: grasshopper (2) Female ZW with Male ZZ : This cross shows ZW type of sex determination. In this case female has one Z and one W chromosome and these chromosomes produces different kinds of gametes, so it is the case of female hetrogamety. eg: birds

Question.12. (i) Name the scientist who suggested that the genetic code should be made of a combination of three nucleotides.
(ii) Explain the basis on which he arrived at this conclusion.
Answer : (i) George Gamow suggested that the genetic code should be made up of a combination of three nucleotides.
(ii) He arrived at this conclusion by giving the explanation if a single nucleotide code for one amino acid, then only four amino acid could be provided. Alternatively, if two nucleotides specified one amino acid, then these could be a maximum number of 16 possible arrangements of three nucleotides code for one amino acid, then there could be 64 possible combinations. Later he suggested that every amino acid is coded by atleast one nucleotide triplet or codon.

Question.13. State the disadvantage of inbreeding among cattle. How it can be overcome ?
Answer : Continuous inbreeding among catde causes inbreeding depression, decreases the fertility and, even, the productivity of an animal. It can be overcome by applying outbreeding, in which mating is done between different breeds or individuals of the same breed with unrelated superior animals.

Question.14. Explain with the help of a suitable example the naming of a restriction endonuclease.
Answer : The nomenclature of a restriction endonuclease follows a rule as like :

  1.  Ist letter of the name represents the genus of the organism from which the enzyme is derived.
  2.  IInd and IIIrd letters represent the species of the organism,
    from wher it is isolated.
  3. IVth letter represents the name of the strain.
  4.  Last is the Roman number which represents order of isolation.
    For example, In EcoRI – Derived from E.coli, strain R. It : is the first to be isolated from the bacteria.

Question.15. State how has Agrobacterium tumifaciens been made a useful cloning vector to transfer DNA to plant cells.
Answer : Agrobacterium is a bacterium that transfers a piece of DNA to plant tissues by transferring its plasmid T-DNA to the plant genome. The plasmid T-DNA of Agrobacterium is cut with restriction endonuclease and the desired gene that has to be to transferred to a particular plant is inserted with the help of ligase enzyme. Then, this Agrobacterium plasmid is allowed to infect that
” particular plant, so that it can transfer the desired gene in to the plant genome through its T-DNA. ”

Question.16. Construct an age pyramid which reflects a stable growth status of human population.
Answer : The age pyramid that reflects a stable growth status of human population can be represented as follows :
cbse-previous-year-solved-papers-class-12-biology-delhi-2014-4

Question.17. Apart from being a part of the food chain, predators play other important roles. Mention any two such roles supported by examples.
Answer : Predators play an important role in :

  1.  Maintaining the prey population under control, this regulates intra-species competition.
    For example, if tigers are removed from a forest, spotted deer will multiply rapidly. This would result in rapid destruction of herbs and grasses in the forest and ultimately the whole forest.
  2.  Predators also help in maintaining species diversity in a community, by reducing the intensity of competition among prey species.
    For Example : the starfish pisaster is an important predator on the rocky intertidal communities of the American Pacific Coast. In a field experiment, when all the starfish were removed from the area, more than 10 species of invertebrates became extinct within a year, because of inter-specific competition.

Question.18. How are ‘sticky ends’ formed on a DNA strand ? Why are they so called ?
Answer : Sticky ends in DNA strands are produced with the help of restriction enzymes. These enzymes cut the strand of DNA a little away from the centre of the palindrome sites but between the two same bases on the opposite strands. This leaves a single stranded portions at the ends. There are overhanging stretches called ‘sticky ends’ on each. strand,
cbse-previous-year-solved-papers-class-12-biology-delhi-2014-5
These are called sticky ends because they form hydrogen bonds with their complementary cut counterparts. This stickiness of the ends facilitates the action of the enzyme DNA ligase.

SECTION-C

Question.19. Explain any three advantages at seeds offer to angiosperms.
Answer: Seeds offer the following advantages to angiosperms :

  1. They provide nourishment and parental care to the developing embryo.
  2.  They protect the embryo from harsh environmental conditions.
  3. The dispersal of seeds to far-off places prevents competition among the members of the same species, thus preventing their extinction.

Question.20. Name and explain the role of the inner and middle walls of the human uterus.
Answer : The inner glandular wall of the uterus is known as endometrium.
Role : During the menstrual cycle, the endometrium wall grows into a thick, blood vessel-rich, glandular layer. It is the where embryo gets implanted. If fertilisation does not occur, the endometrium is shed during the hemorrhagic phase of the menstrual cycle.
The middle wall of the uterus is known as myometrium.
Role: It consists of smooth muscles. It brings strong contraction during delivery of the baby.

Question.21. A color-blind child is bom to a normal couple. Work out a cross to show how it is possible. Mention the sex of this child.
Answer : Color blindness is a sex-linked disease. The gene for this disorder is present on the X chromosome. The color blind child is son with genotype XCY. and sex of the child is male and carrier female.
cbse-previous-year-solved-papers-class-12-biology-delhi-2014-6
OR
Mendel published his work on inheritance of characters in 1865, but it remained unrecognised till 1900. Give three reasons for the delay in accepting his work.
Answer: The three reasons that delay in accepting Mendels work:

  1.  Lack of communication and publicity. He published it in a journal that had limited circulation.
  2.  His concept of factors (genes) as a discrete units that did not blend with each other was not accepted in the light of variations occurring continuously in nature.
  3. Mathematical approach : Mendel’s approach to explain biological phenomenon with the help of mathematics was also not accepted.

Question.22. Women are often blamed for producing female children. Consequently, they are ill-treated and ostracized. How will you address this issue scientifically if you were to conduct an awareness programme to highlight the values involved ?
Answer: Women are not responsible for determination of the gender of a child. It is absolutely wrong to ill-treat a woman for giving birth to a girl child.
In human there are 22 pairs of autosomal chromosomes and one pair of sex chromosome. In human sperm (haploid) has 22 autosomes and one of the two types of sex chromosomes, i.e. either X or Y. While human females ova (haploid) have 22 autosomes and contaih only X chromosomes. The gender of a child is determined by the type of the sex chromosome (X or Y) carried by sperm that fuses with the ovum at the time of fertilization. If the fertilising sperm has an X chromosome, then the baby would be a female and if a sperm with Y chromosomefuses with the ovum, it will develop into a male child. Thus scientifically it is correct that to say that males are responsible for determination of the gender of a child. Both males and females are equally important in every respect for the balance of nature and continuity of our species and it should because of equal joy to parents. .

Question.23. (a) Name the tropical sugar cane variety grown in South India. How has it helped in improving the sugar cane quality grown in North India ?
(b) Identify ‘a’, ‘b’ and c’ in the following table :
cbse-previous-year-solved-papers-class-12-biology-delhi-2014-7
Answer : (a) Saccharum ojficinarum variety grown in South India, which has a thicker stem and high sugar content but, it didnot grow well in Northern India. Saccharum barberi is a natively grown in Northern India. These two varieties vtere crossed to get the desirable qualities of both (high yield, higher sugar content, thicker stem and the ability to grow in Northern India).
(b) (i) Aphids (ii) Jassids and fruit borer (iii) Okra (Bhindi)

Question.24. Why are beehives kept in a crop field during flowering period ? Name any two crop fields where this is practised.
Answer: Beehives are kept in a crop field during flowering period to increase the pollination efficiency of the crop, which increases the crop yield. Also, bees collect huge amounts of nectar from the flowers of the crop in a close reach without much foraging without difficulty. This increases honey yield. Crop fields where this is practiced: Apple, sunflower and watermelon fields

Question.25. How did the process ofRNA interference help to control the nematode from infecting roots of tobacco plants ? Explain.
Answer: RNA Interference (RNAi) is a gene-silencing process that blocks the expression of genes in the parasite when it enters the host s body.

  1. Meloidegyne incognitia infects roots of tobacco plants and cause a severe loss by causing reduction in yield.
  2.  RNAi is a method to prevent infestation of roots of tobacco plants by a nematode Meloidegyne incognitia.
  3.  It is a defense in all eukaryotic organisms.
  4.  In RNAi, a complementary RNA binds to mRNA to form a double strand RNA(dsRNA) that cannot translate and blocked the expression mRNA.
  5. In this process, initially nematode-specific genes (DNA) are introduced in the host plant.
  6.  This introduced DNA forms both sense and antisense RNA.
  7.  These two strands, being complementary to each # other so that they form dsRNA.
  8. This dsRNA results in RNA interference and finally silenced the specific mRNA of nematode.
  9.  Thus and the parasite cannot survive in the transgenic host expressing specific RNAi. The transgenic plant therefore got itself protected from the parasite.

Question.26. Study the graph given below and answer the questions that follow:
cbse-previous-year-solved-papers-class-12-biology-delhi-2014-8
(i) Write the status of food and space in the curves (a) and (b).
(ii) In the absence of predators, which one of the two curves would appropriately depict the prey population?
(iii) Time has been shown on X-axis and there is a parallel dotted line above it. Give the significance of this dotted line.
Answer : (i) This curve show sample food and space for the population depicted by curve :
a : When the food and space are unlimited available
b :When the resources are limiting, the curve becomes sigmoid.
(ii) In the absence of predators, curve b would appropriately depict the prey population.
(iii) The dotted line represents the carrying capacity. The carrying capacity represents the size of population that the environment can hold by providing necessary resources.

Question.27. (i) What is primary productivity ? .Why does it vary in different types of ecosystems ?
(ii) State the relation between gross and net primary productivity.
Answer : (i) Primary productivity is the amount of biomass produced per unit area in a certain time period by plants during photosynthesis. It is expressed in terms of g/m2 or Kcal/m2.
Primary productivity depends upon the type of plant species associated with an ecosystem, photosynthetic capacity of these plants and nutrient availability. This is the reason why it varies in different types of ecosystems.
(ii) The relation between the gross and net primary productivity can be shown as :
cbse-previous-year-solved-papers-class-12-biology-delhi-2014-11
It is the rate of production of organic matter during photosynthesis, where,
Pn = Net primary productivity (NPP)
Pg = Gross primary productivity (GPP)
R = Respiration losses.

SECTION-D

Question.28. (a) Coconut palm is monoecious, while date palm is dioecious. Why are they so called ?
(b) Draw a labelled diagram of sectional view of a mature embryo sac of an angiosperm.
Answer : (a) Cocunut palm is monoecious, while date palm is dioecious because in coconut palm both as male and female flowers are borne on the same plant, while date palm bears exclusively either male flowers or female flowers.
(b) Diagram showing sectional view of a mature embryo sac of an angiosperm
cbse-previous-year-solved-papers-class-12-biology-delhi-2014-9
OR
(a) How is ‘oogenesis’ markedly different from ‘spermatogenesis’ with respect to the growth till puberty in the humans?
(b) Draw a sectional view of human ovary and label the different follicular stages, ovum and Corpus luteum.
cbse-previous-year-solved-papers-class-12-biology-delhi-2014-10
cbse-previous-year-solved-papers-class-12-biology-delhi-2014-12

Question.29. (a) Explain the process of DNA replication with the help of a schematic diagram.
(b) In which phase of the cell cycle does replication occur in
Eukaryotes ? Whast would happen if cell-division is not
followed after DNA replication.
Answer : (a) Mechanism of DNA Replication. The following
steps are involved in replication of DNA’:

  1.  Origin of Replication : Replication begins at a particular region of DNA which has a particular nucleotide sequence called autonomic replicating dioxyribonucleo tides:
  2. Oxynation of dioxyribonuleotides:
    cbse-previous-year-solved-papers-class-12-biology-delhi-2014-13
  3.  Initation of Replication: Origin of replication is recognised by complex. The unwinding of DNA molecules starts at specific points called Initiation point. These are identified by specific initiator proteins.
  4.  Unwinding of Helix :
    (a) The enzyme Helicase unwinds the DNA helix & unzips the double strands of DNA. This process is ATP dependent. It takes place by breaking of H-bonds.
    (b) Unwinding of DNA molecule into two strands results in the formation of DNA replication bubble which later extend as a Y-shaped structure called Replication fork.
    (c) The separated strands become stablised in this condition with the help of single strand binding proteins (SSBPs).
    (d) Due to unwinding, a supercoiling & tenstion is created, which is released by enzymes Topoisomerases I & II. Topoisomerase II of prokaryotes is also called Gyrase, which functions both as Helicase & Topoisomerases .
  5. Formation of Primer Strand:
    (a)Once the primer strand formed,DNA replication occurs in 5”->3′ direction i.e during synthesis of new strand dioxyribonuleotides (dATP,dGTP,dTTP,dCTP) are added in the free 3′ OHead.
    cbse-previous-year-solved-papers-class-12-biology-delhi-2014-14
    cbse-previous-year-solved-papers-class-12-biology-delhi-2014-15
    The -release of 2 phosphate molecule & energy aid in formation of H-bonds.
    cbse-previous-year-solved-papers-class-12-biology-delhi-2014-16
    (b) DNA replication occurs in S phase of cell cycle in eukaryotes. If cell division is not followed aftre DNA. replication then cell enters in G0 phase and becomes permanent and specialized.
    (c) As the DNA replication proceeds on the two parental strands, synthesis of daughter parent 3’—> 5’ strand. It is called Leading Daughter strand.
    (d) Synthesis of the other daughter strand along the other parental strand, takes place in the form of short pieces because of the opposite arrangement of nucleotides. A new RNA primer is formed everytime, where new DNA strand is built in small segments. These RNA primers are removed by polymerase I & a enzymes in prokaryotes .
    (e) Since replication is continuous over one strand & discontinuous over the other, it is called semi-continuous replication.
    (f) Discontinuous pieces of the lagging strand are joined Overall Direction of Replication.

(b) DNA replication occurs in S phase of cell cycle in eukaryotes.If cell division is not followed after DNA replication then cell enters in G0 phase and becomes permanent and specialized.
OR
(a)Explain Darwinian theory of evolution with the help of one suitable example. State the two key concept of the theory.
(b) Mention any three characteristics of Neanderthal man that lived in near east and central Asia.
Answer: (a) Darwinian Theory of Evolution took place by natural selection. The number of life forms depends upon their life span and their ability to multiply. Another aspect of Darwinian Theory is natural selection, the survival of the fittest where nature selects the individuals, which are most fit to adapt to their environment. Example: Selection of the antibiotic resistant in bacteria. When a bacterial population grow on an agar plate containing antibiotic penicillin, the colonies that are sensitive to penicillin die, whereas one or few bacterial colonies that are resistant to penicillin survive. This is because these bacteria have undergone mutation results in evolution of a gene that made them resistant to penicillin drug. Hence, the advantage of an individual over the other helps in the struggle for existence.The two key concepts of the theory are :

  1. Branching descent : According to this concept, various species have come into existence from a common ancestor.
  2. Natural selection : According to this concept, nature selects the individuals, which are most fit to adapt to their environment.

(b) Characteristics of Neanderthal man :

  1.  They possess a brain size of 1400 cc.
  2.  They were short but very strong with upwards curved thigh bones. .
  3.  They used the hides to protect their body and to bury the dead.

Question.30. (a) Name the technology that has helped scientists to propagate on a large scale the desired crops in a short duration. List the steps carried out to propagate the crops by the said technique.
(b) How are somatic hybrids obtained?
Answer : (a) Plant tissue culture is the technique of in vitro maintenance and growth of plant cells, tissues & organs on a suitable culture medium. The technique of tissue culture was first suggested by Gottleib Haberlandt in 1902.
cbse-previous-year-solved-papers-class-12-biology-delhi-2014-17
The following steps are carried out to propagate crops by tissue culture :

  1.  Preparation of suitable medium : Suitable medium, containing a carbon source, such as sucrose, and inorganic
    salts, vitamins, amino acids and growth regulators like auxin, cytokinin etc.
  2. Selection of Explant : Any part of the plant, especially apical and axillary meristem can be used as explant.
  3.  Incubation : Growing the explant in the test tube, under sterile conditions.
  4.  Regeneration : Since explant show the property of totipotency, new plantlet can be regererated.
  5. Hardening : Regenerated plants are grown in pots, to expose them to environmental conditions.
  6.  Plantlet transfer: After hardening, plantlets are transferred to field.

(b) Isolated protoplasts from two different varieties of plants, each having a desirable character, can be further gronw to form a new plant. These hybrids are called somatic hybrid while the process is called somatic hybridisation.
OR
(a) Cancer is one of the most dreaded diseases of humans.
Explain ‘Contact inhibition and ‘Metastasis’ with respect to the disease. .
(b) Name the group of genes which have been identified in normal cells that could lead to cancer and How they do so?
(c) Name any two techniques which are useful to detect cancers of internal organs.
(d) Why are cancer patients often given a-interferon as part of the treatment ?
Answer : (a) Cancer is one of the most dreaded disease of humans. Normal cells show a property called contact1 inhibition, by virtue of which contact with other cells and inhibits their uncontrolled growth but cancer cells lose this property. As a result, cancer cells divide continously to give rise to a mass of cells (tumours).
Metastasis is a property of malignant tumours. Some cancer cells from tumors gets sloughed from the tumor and the reach distant sites through the blood and wherever they reach initiate the formation of new tumours by dividing actively. This property is called metastasis.
(b) Cellular oncogenes (c-onc) and proto-oncogenes are the group of genes that have been identified in normal cells. These genes when activated under certain conditions, would lead to oncogenic transformation of the cells.
(c) Techniques such as radiography, CT (Computed Tomography) and MRI (Magnetic Resonance Imaging) are useful to detect cancers of internal organs.
(d) The biological response modifiers such as a-interferoris are given to cancer patients as part of their treatment because it activates a patient s immune system and helps in destroying the tumour.

SET -II

SECTION-A

Question.4. Name the two intermediate hosts which the human liver fluke depends on to complete its life cycle so as to facilitate parasitization of its primary host.
Answer: Terrestrial snail and Fish, are two intermediate hosts on which the human liver fluke depends on to complete its life cycle so as to facilitate parasitization of its primary host.

Question.7. Mention how does DNA polymorphism arise in a population.
Answer : DNA polymorphism is a genetic variant and intro-duced in a population by mutation and genetic drift.

SECTION-B

Question.9. Name the organic materials the exine and intine of an angiosperm pollen grains are made up of. Explain the role of exine.
Answer: Exine is made of sporopollenin. Intine is made of cellulose and pectin. Sporopollenin (exine) is most resistant organic material which can with stand high temperature and acids and alkalies.

Question.13. How can healthy potato plants be obtained from a desired potato variety which is viral infected ? Explain.
Answer: Tissue culture can get us disease free potato plants from viral infected plants. The apical and axillary meristems of virus infected plant are free of virus. So meristems can be removed and their culture can give us virus free plants.

Question.15. What is Biopiracy ? State the initiative taken by the Indian Parliament towards it.
Answer : Biopiracy is defined as the use of bio-resources by mitltinational companies or other organisations without
proper authorisation from concerned country or people. Indian Parliament has declared second amendment of Indian Patents bill Act (1970) this bill considers patenting, its terns emergency provisions and research and development initiatives.

Question.18. Write the role of‘Ori’ and ‘restriction site in a cloning vector pBR322.
Answer: Role of Ori sequence and ristriction site in pBR322: Ori is a genetic sequence that acts as the initiation site for replication of DNA, when any fragment linked .to this sequence ‘ can be initiated to replicate within host cells.
Recognition site is the specific DNA sequences which contains different palindromic sequence, as recognized by respective restriction enzymes (such as EcoRI, Hind III, Pvul, BamHI etc.). Recognition sites are sequences where the restriction enzymes cut the DNA.

SECTION-C

Question.20. A cross between a normal couple resulted in a son who was haemophilic and a normal daughter. In course of time, when the daughter was married to a normal man, to their surprise, the grandson was also haemophilic.
(a) Represent this cross in the form of a pedigree chart. Give the genotypes of the daughter and her husband.
(b) Write the conclusion you draw of the inheritance pattern of this disease.
Answer: (a)
Genotype of daughter- XX
Genotype of husband- XY
cbse-previous-year-solved-papers-class-12-biology-delhi-2014-18
(b) Conclusion: Haemophilia is a sex-linked recessive disease which shows it is transmitted from the carrier female to the sons. From the above pedigree chart, it can be observed that the disease is being transmitted from the carrier female to her daughter (carrier) and son (affected). The carrier daughter transmits this disease to the grandson and the possibility of a female becoming a haemophilic is extremely rare. This inheritance is called crisscross inheritance.

Question.22. Draw a labelled diagram of the sectional view of a Human seminiferous tubule, (six parts to be labelled)
Answer:
cbse-previous-year-solved-papers-class-12-biology-delhi-2014-19

SECTION-D

Question.30. Explain the ovarian and uterine events that occur during a menstrual cycle in a human female, under the influence of Pituitary and Ovarian hormones respectively.
Answer : Menstrual Cycle : Menstrual cycle is the reproductive cycle in all primates and begins at puberty (menarche). It involves cyclic changes in females reproductive tract culminating in menstruation that is flow of cast off uterine and fallopian tube lining along with blood and tissue fluid through the vagina. In human females, menstruation occurs once in 28 to 29 days. The cycle of events starting from one menstruation till the next one is called the menstrual cycle.
It consist of three phases:

  1. Proliferating phases : It last for about 14 days. Lining of the uterus and fallopian tubes proliferates and its vascularization increases. A graafian follicle grows, matures and secretes oestrogen. It ruptures to release its egg (secondary oocyte) after about 14 days. The LH and FSH are at their peak in the middle before release of oocyte. This phase is also called the ovulatory phase.
  2.  Secretory phase : It lasts for-about 10 days. The empty graafian follicle forms in it corpus luteum which secretes progesterone. The lining of uterus and fallopian tubes undergoes further hypertrophy. Endometrial glands of the uterus secrete a nutritive fluid for the foetus.
  3.  Menstrual (Bleeding) phases : It lasts for about 4 days. If fertilization does not occur, the corpus luteum regresses, and the linning of uterus and fallopian tubes breakdown, resulting in menstrual flow. This occurs after 25 days and continues 3 to 5 days. The basal part of endometrial lining remain intact during menstruation and produces new uterine lining.
    cbse-previous-year-solved-papers-class-12-biology-delhi-2014-20

OR
(a) Why does endosperm development precede embryo development in angiosperm seeds ? State the role of endosperm in mature albuminous seeds.
(b) Describe with the help of three labelled diagrams the different embryonic stages that include mature embryo of dicot plants.
Answer : (a) Endosperm development precedes embryo development in angiosperm seeds because primary endosperm cell divides repeatedly and forms a triploid endosperm tissue. The cells of this tissue are filled with reserve food materials which provide nutrition to the developing embryo.
Albuminous seeds stores starch and fat to retain a part of endosperm as it is not completely used up during embryo development (e.g. wheat, maize, barley, castor, sunflower).
(b) Development of embryo :
The Zygote formed after fertilization of egg cell starts dividing and gives rise to proembryo. This proembryo further divides, forming a globular, heart-shaped and mature embryo.
cbse-previous-year-solved-papers-class-12-biology-delhi-2014-21
Following are the steps that occur during the development of embryo:

  1.  In dicots zygote elongates and divides into upper and lower cell.
  2.  The lower one lying toward micropyle further divides in one direction into a row of cell called suspensor.
  3.  The upper cell towards the antipodal end is called embryo cell.
  4.  The first cell of the suspensor often enlarges and acts as , haustorium or absorbing organ while its terminal cell called hypophysis cell divides giving rise to the apex of the radicle.
  5. The upper cell or embryo cell enlarges and divides repeatedly to form eight cells that are arranged in two tiers – epibasal (terminal) and hypobasal (near the suspensor). A typical dicot embryo consists of an embryonal axis and two cotyledons.
  6.  The portion of the embryonal axis above the level of cotyledons is called epicotyl. It contains the plumule (shoot tip). The portion below the axis is called hypocotyl. It contains the radicle (root tip). The root tip is covered by the root cap.
  7.  Subsequent divisions give rise to globular heart-shaped embryo.

SET -III

SECTION-A

Question.3. How is repetitive/satellite DNA separated from bulk genomic DNA for various genetic experiments ?
Answer : Satellite DNA is separated from bulk genomic DNA by density-gradient centrifugation technique. They are separated as different peaks. The bulk DNA forms a Major peak and the other small peaks are referred to as satellite DNA.

Question.5. Name the Green House gases that contribute to total global warming.
Answer: Carbon dioxide, nitrous oxide, ozone and methane are the Greenhouse gases that are responsible for global warming.

SECTION-B

Question.12. What is gene therapy ? Name the first clinical case where it was used ?
Answer : Gene therapy is a experimental technique in which genes can be used for treating and preventing disease. A four year old girl became the first gene therapy patient on September 14, 1990 at the NIH clinical center. She was suffering from adenosine deaminase (ADA) deficiency.

Question.15. Why does Bt toxin not kill the bacterium that produces it, but kill the insect that ingests it ?
Answer : Bt toxin protein is produced by a soil bacterium called Bacillus thuringiensis in inactive prototoxin and crystalline form. The prototoxin form does not kill the bacteria. It becomes active and toxic when it is consumed by insects such as lepidopterans (armyworm), coleopterans (beedes) and dipterans (flies/mosquitoes) due to presence of alkaline pH in the gut. The activated toxin (delta endotoxins) binds to the epithelial cells in the midgut of an insect and creates pores that cause lyses and swelling, eventually killing the insect.

Question.17. Identify the following pairs as homologous of analogous organs1:
(i) Sweet potato and potato
(ii) Eye of octopus and eye of mammals ,
(iii) Thoms of Bougainvillaea and tendrils of Cucurbits
(iv) Fore limbs of bat and whale
Answer : (i) Analogous organs, (ii) Analogous organs,
(iii) Homologous organs, (iv) Homologous organs

Question.18. List the post – fertilisation events in angiosperms.
Answer : The various post-fertilisation events occurring in angiosperms are:

  1. Sepals, petals and stamens of the flower dry up alid fall off.
  2. The Zygote develops into an embryo.
  3.  Ovules develop into the seed.
  4.  The Ovary develops into the fruit.

SECTION-C

Question.21. What are Methanogens ? Name the animals they are present in and the role they play there.
Answer: Methanogens are anaerobic bacteria grow anaerobically on cellulose material that produce large amounts of methane.
Ex : Methanobacterium.
Methanogens are commonly found in the rumen of catde and help in cellulose digestion. Hence, excreta of cattle (gobar) is rich in methanogens.

Question.27. There are many animals that have become extinct in the wild but continue to be maintained in Zoological parks.
(i) What type of biodiversity conservation is observed in this case ?
(ii) Explain any other two ways which help in this type of conservation.
Answer : (i) It is an example of ex-situ conservation.
(ii) Cryopreservation and tissue culture are two ways that helps in ex-situ conservation.
Cryopreservation : The preservation of gamates of threatened species in viable and fertile conditions at sub-zero temperatures which help in preserving these cells for longer periods.
Tissue culture : Plants are propagated from a small mass of tissue, called callus.

SECTION-D

Question.29. (a) Draw a labelled diagrammatic view of human male reproductive system.
(b) Differentiate between :
(i) Vas deferens and vasa efferentia
(ii) Spermatogenesis and spermiogenesis
Answer: (a) Diagrammatic view of a human male reproductive
OR
(a) Explain the phenomenon of double fertilization.
(b) Draw a labelled diagram of a typical anatropous ovule.
Answer : (a) Phenomenon of double fertilisation :
Pollen grains gets transferred from the anther to the stigma, and then the pollen tube enters one of the synergids and released two male gametes. One gamete moves towards the egg cell and fuses to form the zygote to complete the syngamy. The other gamete fuses with the two polar nuclei and forms triploid Primary Endosperm Nucleus (PEN). This fusion of one male gamete with two polar nuclei is termed as triple fusion. Since, , two kinds of fusion—syngamy and triple fusion—take place during fertilization in a flower, the process is known as double fertilization. It is a characteristic of flowering plants.
cbse-previous-year-solved-papers-class-12-biology-delhi-2014-22
(b) Labelled diagram of a typical anatropous ovule
cbse-previous-year-solved-papers-class-12-biology-delhi-2014-23

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