CBSE previous Year Solved Papers Class 12 Biology Delhi 2015

CBSE previous Year Solved  Papers  Class 12 Biology Delhi 2015

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. A geneticist interested in studying variations and patterns of inheritance in living beings prefers to choose organisms
for experiments with shorter life cycle. Provide a reason. 5.
Answer : A geneticist interested in studying variations & pattern of inheritance in Living beings. Prefers organisms having short life cycle as then he will be able to study to detect variations that occur & inherit from one generation to another.

Question.2. Name the transcriptionally active region of chromatin in a nucleus.
Answer : The transcriptionally active region of chromatin in
a nucleus is Euchromatin.

Question.3. State a reason for the increased population of dark coloured moths coinciding with the loss of lichens (on tree barks) during industrialization period in England.
Answer: ‘Deposition of soot and smoke causes the tree trunks to become darker (after industrialization); hence the number of dark moths increased as they were not easily visible to their predators while the white-winged ones were easily picked up by the predators. Thus, light coloured ones fail to survive and dark ones were selected by nature (natural selection) .

Question.4. Indiscriminate diagnostic practices using X-rays etc., should be avoided. Give one reason.
Answer: X-rays are ionizing radiations that may cause adversed effects in children in the form of mutations. The mutations can alter the genetic make-up of an organisation so the usage of X-rays causing these mutations should be avoided.

Question.5.What is Biopiracy ?
Answer: Biopiracy is defined as the illegal removal of biological material of a country by organizations or multinational companies without proper authorization from the concerned countries.

SECTION – B

Question.6.After a brief medical examination a healthy couple came to know that both of them are unable to produce functional gametes and should look for an ART’ (Assisted Reproductive Technique). Name the ART’ and the procedure involved that you can suggest to them to help them bear a child.
Answer: The assisted reproductive technique (ART) for such couples is ZIFT, which stands for Zygote Intra Fallopian transfer.
In this technique, the sperm and ovum are collected from the donor male and the donor female respectively. The sperm and the ovum are fused in laboratory conditions and developed till the 8-blastomere stage. This early embryo or zygote is then transferred to the fallopian tube of the mother for further development.

Question.7.Differentiate between male and female heterogamety.
Answer:

SECTION – C

Question.12. Describe the process of Parturition in humans.
Answer : Parturition is the process of expelling the fully developed foetus from mother’s uterus at the end of the gestation period. Parturition involves forceful muscular contraction of uterine wall called labour. Expel thebaby from the uterus.
Signal of parturition is controlled by a complex neuroendocrine mechanism. Signals originate from fully formed foetus & secreting certain hormones which diffuse into mother’s blood & cause secretion of oxytocin. Oxytocin stimulate uterine contractions. These are called foetal ejection reflexes. These reflexes become stronger & more frequent to push the foetus out of the uterus. These contractions called labour are induced by placental & foetal hormones Labour period is divided into three stages.

Question.13. A teacher wants his/her students to find the genotype of pea plants bearing purple coloured flowers in their school garden. Name and explain the cross that will make it possible.
Answer : Purple colour of flowers is a dominant phenotype in pea plants. The genotype of such plants can be determined by a test cross.
Test Cross
When an F, hybrid is crossed with its homolygous recessive parent, it is known as test cross. Test cross is used to determine that the genotype of a plant with the dominant phenotype is homozygous or heterozygous, e.g. purple flower coming from PP or Pw.
In this cross, plant with purple-coloured flowers is crossed with plant with white-coloured flowers, which will always have homozygous recessive genotype.
If the offspring produced is all purple-flowered plants, then the genotype of the parent purple-coloured plant is PP (homozygous). But if the offspring produced are purple & white flowered plants in equal proportions, then genotype of parent purple flowered plant is Pw (heterozygous).
Thus, the genotype of pea plant bearing purple-coloured flowers in the school garden can be Pw or PP.

Question.14. (a) A DNA segment has a total of 1000 nucleotides, out of which 240 of them are adenine containing nucleotides. How many pyrimidine bases this DNA segment possesses?
(b) Draw a diagrammatic sketch of a portion of DNA segment to support your answer.
Answer : (a) Cytosine and thymine are pyrimidines. According to Chargaff’s rule, ratio of purines to pyrimidines is equal.
Thus, the number of adenine (A) will be equal to the number of thymine (T). .
Number of adenine (A) containing nucleotides = 240
Thus, A = T = 240
Therefore, A + T = 240+240 = 480
The number of cytosine (C) will be equal to the number of guanine (G).
Thus, G + C = Total number of nucleotides — Nucleotides containing A and T nitrogenous bases = 1000 — 480
= 520
Therefore, G = 260, C = 260
Number of guanine will be equal to number of cytosine, which will be 260.
Therefore, the number of pyrimidines that the segment possess = C + T
= 260 + 240
= 500
(b) A diagrammatic sketch of a portion of DNA segment is given below:

Question.15. Explain adaptive radiation with the help of a suitable example.
Answer : Adaptive radiation also referred as divergent evolution is the formation of a number of divergent species from a common ancestor, with new species moving out of the native habit at & adapting to new ecological niches.
The best example of adaptive radiation is the finches of Galapagos islands. Finches are sparrow-like small black- coloured birds. There are around 14 species of these finches 13 of which are on the galapagos island & 1 species is found on the nearby cocos island. The species on the cocus island resemble the mainland finches in plumge body structure & short tails. But these show differences amongst themselves & those on the mainland finches in shape & size of beaks, food habits, body size & feather colour.

Question.16. A team of students are preparing to participate in the interschool sports meet. During a practice session you find some vials with labels of certain cannabinoids.
(a) Will you report to the authorities ? Why ?
(b) Name of a plant from which such chemicals are obtained.
(c) Write the effect of these chemicals on human body.
Answer: (a) Yes, I would report the matter to the authorities because cannabinoids are classified under drugs and drug abuse1 is an illegal practice.
(b) Cannabinoids can be obtained from a plant called Cannabis sativa.
(c) The cannabinoids bind to the brain which has more cannabinoid receptors and they affect the cardiovascular system.

Question.17. Enlist the steps involved in inbreeding of cattle. Suggest two disadvantages of this practice.
Answer: In breeding involves mating of closely related individuals within the same breed for 4-6 generations.
The process of inbreeding of cattle is done when, Superior males and superior females of the same breed are identified and are made to mate in pairs. Then, evaluation of obtained progeny is done to identify the superior males and females for further mating. Following are the two disadvantages of the process of inbreeding:
(i) Inbreeding depression used by continuous inbreeding among cattle.
(ii) It decreases the fertility and also the productivity of an animal.

Question.18. Choose the three microbes, from the following which are suited for organic farming which is in great demand these days for various reasons. Mention one application of each one chosen.
Mvcorrhiza ; Monascus ; Anabaena; Rhizobium: Methanobacterium: Trichoderma.
Answer: Following three microbes can be chosen for organic farming.
Mycorrhiza : Absorb phosphorous from soil.
. Anabaena : It is involved in the process of atmospheric nitrogen fixation.
Rhizobium : It also plays a role in nitrogen fixation in legumhour plants.
Monascus : It produces a group of drugs called statins that are used to lower the cholesterol in body.
Methanobacterium : It is used in the biological generation of methane by anaerobic processes.
Trichoderma : It produces cyclosporin A which is used as an immunosuppressive agent.

Question.19. Recombination DNA – technology is of great importance in
the field of medicine. With the help of a flow chart, show how this technology has been used in preparing genetically engineered human insulins.
Answer : Preparation of Human Insulin Using Recombinant DNA Technology

Question.20. Draw a labelled sketch of sparged-stirred-tank bioreactor. Write its application.
Answer:

A bioreactor is an apparatus in which a biological process is carried out. Following are the applications of
bioreactors :

1.  They are also used to produce beverages like alcohol.
2.  They are used to produce large quantities of proteins.

Question.21. Following the collision of two trains a large number of passengers are killed. A majority of them are beyond recognition. Authorities want to hand over the dead bodies to their relatives. Name a modern scientific method and * write the procedure that would help in the identification of kinship. 
Answer: DNA fingerprinting is the modern scientific method used for the identification of kinship.
DNA Fingerprinting process can be carried out by follpwing point:

1.  Extract and purify DNA from cells (A drop of blood or semen or a piece of hair root or any tissues can be used to isolate DNA.)
2. DNA is digested or restricted with enzymes (restriction endonucleases) without causing cut in the mini satellite region.
3. DNA fragments are separated by electrophoresis.
4.  Denature DNA.
5.  Separated DNA fragments are transferred to synthetic membranes like nitrocellulose or nylon.
6. Add labelled VNTR (Variable Number of Tandem Repeats or Mini Satellite) probe for hybridization to take place.
7.  Wash off unbound probe.
8. Ffybridised DNA fragments are detected by autora-diography.
9.  Matching banding pattern of DNA of the passengers killed and that of relatives.

Question.22. Many plants and animals species are on the verge of their extinction because of loss of forest land by indiscriminates use by the humans. As biology student what method would you suggest along with its advantages that can protect such threatened species from getting extinct ?
OR
“Determination of Biological oxygen Demand (BOD) can help in suggesting the quality of a water body”. Explain.
Answer: Ex-situ conservation is the preservation of components of biological diversity outside their natural habitats. This involves conservation of genetic resources, as well as wild and cultivated or species and draws on a diverse body of techniques and facilities.
Some of these include :

1. Gene banks, e.g. seed banks, sperm and ova banks, field banks.
2. In vitro plant tissue and microbial culture collections.
3. Captive breeding of animals and artificial propagation of plants, with possible reintroduction into the wild, and
4.  Collecting living organisms for zoos, aquaria, and botanical gardens for research and public awareness.

OR
Biochemical oxygen demand or B.O.D. is a chemical procedure for determining the amount of dissolved oxygen needed by aerobic biological organisms in a body of water to break down organic material present in a given water sample at certain temperature over a specific time period. It is widely used as an indication of the organic quality of water. The greater the BOD, the more rapidly oxygen is depleted in the stream. This . means less oxygen is available to higher forms of aquatic life. Lower BOD of water body indicates less number of micro organisms in water, quality of water, less polluting potential and aquatic life fluorishes.

SECTION-D.

Question.23 . Since October 02, 2014 “Swachh Bharat Abhiyan” has been launched in our country.
(a) Write your views on this initiative giving justification.
(b) As a biologist name two problems that you may face while implementing the programme in your locality.
(c) Suggest two remedial methods to overcome these problems.
Answer:
(a) “Swachh Bharat Abhiyan” is an initiative started on 2nd October, 2014 by India’s Prime Minister, Narendra Modi. It is India’s biggest cleanliness drive since independence. On a personal note, I am completely in support for this movement. It is our main duty to clean our nation as wastes are the biggest evil that hinders the development and progress of a country. An unclean surrounding will lead to a lot of problems, starting from the health of citizens to the shortage of land. On a large scale, it is responsible for the air and water pollution, which creates problem both on economic aspects and development of the country.
(b) As a biologist, two problem faced while implementing the programme in my locality are as follows :

1. Problem of proper sanitation.
2. Separation of biodegradable and non-biodegradable wastes.

(c) Remedies to overcome the problems :

1.  To overcome sanitation problems, we should make people aware regarding the benefits of proper sanitation and encourage the local people to make proper toilets.
2.  There should be separate bins for both biodegradable and non-biodegradable wastes so that they can be disposed and recycled accordingly.

SECTION-E

Question.24. A flower of tomato plant following the process of sexual reproduction produces 240 viable seeds.
Answer the following questions giving reasons :
(a) What is the minimum number of pollen grains that must have been involved in the pollination of its pistil ?
(b) What would have been the minimum number of ovules present in the ovary ?
(c) How many megaspore mother cells were involved ?
(d) What is the minimum number of microspore mother cells involved in the above case ?
(e) How many male gametes were involved in this case ?
OR
During the reproductive cycle of a human female, when,
where and how does a placenta develop ? What is the function of placenta during pregnancy and embryo development ?
Answer : The number of viable seeds produced by the tomato plant through sexual reproduction = 240
(a) The minimum number of pollen grains that must have been involved in the pollination of its pistil are 240 because each pollen grain contains two male gametes. Out of theses two gametes, one fuses with polar nuclei and forms endosperm. While, the other male gamete fuses with the egg cell to form the zygote that eventually give rise to seeds. Therefore, in order to obtain 240 seeds, number of pollen grains needed would be 240.
(b) The number of ovules minimally involved in this process would be 240, as the number of viable seeds are 240. After fertilization, the ovary turns into fruit and the ovules turn into seeds. Therefore, the number of ovules are corresponding to the number of seeds formed.
(c) During the process of gametogenesis, 240 megaspore mother cells are involved as only one megaspore of the tetrad becomes functional and develops further and the rest three megaspores get degenerated.
(d) In the above case, 60 microspore mother cells must have undergone reduction division prior to dehiscence of anther, as each microspore mother cell would give rise to 4 microspores. Since 1 microspore mother cell would produce 4 microspores, therefore, to obtain 240 microspores, there must be 60 microspore mother cells.
(e) The number of male gametes involved in seed formation would be 240 as each male gamete will fuse with one egg nuclei to form zygote, which will further give rise to the seed.
OR
Placenta is a foetal maternal connective that develops during pregnancy and forms a temporar y association between foetal and mother tissues. It supports the foetus during its development. The foetus is connected to the placenta by a long, flexible string called umbilical cord. It is made up of allonto is after fertilisation, the zygote divides and leads to the formation of blastocyst. The blastocyst comes into contact with the endometrium. The outer layer of blastocyst (trophoblast) secrete lytic enzymes which cause wearing away of endometrial lining. These give rise to finger-like projections called chorionic villi. This forms the foetal part of placenta. These villi extend into the material part of placenta called- Decidua. The villi are immersed into the blood sinuses found in decidua region which is divided into chree distinct regions.
(A) Decidua Basalis
(B) Decidua capsularies
(C) Decidua Parietalis
Decidua Basalis is the part of decidua underlying the chorionic „ villi and overlying the myometrium. Decidua capsularis is the part lying between the embryo and lumen of uterus and decidua parietalis line the uterus at places other than the site of attachment of embryo.

Functions of placenta during pregnancy and embryonic development are as follows :

1.  Placenta serves as a medium to provide nutrition required by the foetus. The nutrients like amino acids, minerals,sugars, lipids, vitamins etc. are transferred from mothers blood to foetal blood.
2.  Placenta aids in exchange of respiratory gases-oxygen & carbon dioxide (oxygen from mother & carbon dioxide from foetus).
3. Placenta acts as an important endocrine gland that produces hormones-human chorionic gonadotropin (hCG) chorionic thyrotropin, chorionic corticotropin, human placental lactogen (hPL), progesterone, estrogen & relaxin. Relaxin, hCG & hPL are secreted only during . pregnancy.

Question.25. Explain the genetic basis of blood grouping in human population.
OR
How did Hershey and Chase established that DNA is transferred from virus to bacteria ?
Answer : Blood groups in human beings : Blood groups in human beings is a character that is controlled by three different alleles, namely, IA, IB and Ic or i. The letter I is related to their is agglutination. The phenomenon of more than two alleles controlling a single character is referred as multiple allelism and the alleles are called multiple alleles. The three different alleles combine differently to form four blood groups. A, B, AB&O.

1.  A person will have blood group as A if he has the presence of allele IA in homozygous condition (IAIA) or alleles IA & 1° (i) in heterozygous condition (IAi),
2.  A person will have blood group as B, if he has the presence of allele IB in homozygous condition (IBIB) or alleles P & i in heterozygous condition (Pi).
3.  Blood group AB is characterised by thje presence of
   both IA & P alleles. This indicated that IA & IB are co-dominant when these occur together.
4. Blood group O is present in an individual when allele ‘i’ occurs in homozygous condition (ii or I°I°). It indicates that ‘i’ allele can express only in the absence of either IA or IB allele. It is thus recessive in nature.
   

OR
Alfred Hershey & Martha Chase in 1952 conducted
Transduction experiments (Bacteriophage infecting bacteria
to prove that DNA is the genetic material.

1.  They first took T2 bacteriophages and made them infect two separate bacterial colonies. One bacterial colony was having radioactive phosphorus 32P & the other was impregnated with radioactive sulphur 35S.
2. When T2 bacteriophages infected these two colonies, radioactive sulphur (35S) got incorporated in the capsid proteins of bacteriophage. While radioactive phosphorus (32P) became a component of phage DNA.
3.  The two types of bacteriophages were then introduces to infect two different bacterial colonies of E.coli.
4.  The two culture were then centrifuged independence. The phage capsids got separated from the bacterial. The bacterial cells were present at the bottom of centrifuge tube as pellets.
5. Hershey & Chase observed that in experiment using 35S the supernatant continuing capsids showed presence of radioactivity. The bacterial cells in this case, were without radioactive whereas in the experiment using 32P. the supernatant showed no signs of radioactivity. It was present in the bacteria & the phages, which multiplied in them.
   Thus, it was concluded that the DNA which is able to enter the bacteria is the genetic material. It aided in phage multiplication. The protein part had no such function.
   

Question.26. “Analysis of age-pyramids for human population can provide important inputs for long-term planning strategies”. Explain.
OR
Describe the advantages for keeping the ecosystems healthy.
Answer : Analysis of age pyramids for human population can provide important inputs for long-term planning strategies. The different age groups present in a population determine its reproductive status. Distribution of age groups highly influences the growth of the population. Each population displays following three ecological ages or age groups :

1.  Pre-reproductive
2.  Reproductive
3.  Post-reproductive

Population having large number of young members grow rapidly, while, the population bearing more number of post-reproductive members tends to be declining. There are basically three types of age pyramids found to be present in human population. These are as follows :

Therefore, through the analysis of age pyramids of a particular population, the distribution of resources can be done more efficiently. A better fanning strategy can be adopted considering the demand of the resource;/thus, long-term management of resources can be done in Such a way that the population can derive maximum bendfhfvvith minimum effects on nature, leading the population to flourish efficiently.
The advantages of keeping the ecosystems healthy are as follows :

1.  The products of ecosystem processes are named as ecosystem services, as they are of great help to the organisms living within an ecosystem.
2.  Healthy ecosystem is the base for a wide range of economic, environmental and aesthetic goods and services.
3.  It also mitigates droughts and floods and cycle nutrients.
4.  Healthy forest ecosystem purify air and water.
5. It also provides aesthetic, cultural and spiritual values.
6. Maintenance of biodiversity is also an important aspect of healthy ecosystem.
7. Healthy ecosystem generates fertile soil and provides wildlife habitat.

SET-II

SECTION-B

Question.7. Name any two common Indian millet crops. State one characteristic of millets that has been improved as a result of hybrid breeding so as to produce high yielding millet crops.
Answer : The common Indian millet crops : Maize and Jowar Hybrid breeding has resulted in the production of high yielding millet varieties that are resistant to water stress.

Question.9. Explain mechanism of sex-determination in birds.
Answer : In birds, the method of sex determination is s ZZ-ZW type. In this system, both the sexes have two sex chromosomes. The females are heteromorphic with two different sex chromosomes (Z & W) whereas the males are homomorphic with two identifical sex chromosomes (Z). The female produces two types of eggs, one with Z chromosome. Thus, in birds the egg determines the sex of the offspririg. Fusion of a sperm with egg having z chromosome, will give rise to male offspring while fusion of egg with w’ chromosome, will give rise to female offspring. It is called female heterogamety.

Question.10. After a brief medical examination a healthy couple come to know that both of them are unable to produce functional gametes and should look for an ART’ (Assisted Reproductive Technique). Name the ART’ and the procedure involved that you can suggest them to help them bear a child.
Answer: The Assisted Reproductive Technique (ART) for such couples is ZIFT, which stands for zygote intrafallopian transfer.
In this technique, The sperms and the eggs are collected from the donor male and donor female respectively. The sperm and the ovum are fused in laboratory conditions to make embryos. These zygotes or early embryos are then transferred to the fallopian tube of the mothers for further development.

SECTION – C

Question.13. (a) A DNA segment has a total of 1,500 nucleotides, out of which 410 are Guanine containing nucleotides. How many pyrimidine bases this segment possesses ? (b) Draw a diagrammatic sketch of a portion of DNA segment to support your answer.
Answer : Cytosine and thymine are pyrimidines.
According to ChargafFs rule, the ratio of purines to pyrimidines is equal.
Thus, the number of cytosine (C) will be equal to the number of guanine (G).
Number of guanine-containing nucleotides = 410
Thus,
G = C = 410 Now,
G + C = 410 + 410 = 820
The number of adenine (A) will be equal to the number of thymine (T).
Thus,
A + T= Total number of nucleotides — Nucleotides containing G and C nitrogenous bases = 1500-820 = 680
Therefore,
A = 340 and T = 340
The number of adenine will be equal to number of thymine, which is 340.
Therefore,
Number of pyrimidines that the segment possess = C + T = 410 + 340 = 750

Question.14. Name the stage of human embryo at which it gets implanted. Explain the process of implantation.
Answer : Blastocyst is that stage of the human embryo in which it gets implanted in the uterus.
The process of implantation can be explained as follows :

1. Fertilization results in the formation of a diploid zygote.
2.  Mitotic divisions occur in the zygote as it moves through the uterus and 2, 4, 8, 16 daughter cells are formed by this divisions which are called as blastomers.
3.  Embryo with 8 to 16 blastomers is known as morula.
4.  Morula undergoes further cleavage and develops into blastocyst, blastomeres of blastocyst get arranged into an inner cell mass where as outer layer called trophoblast.
5.  The trophoblast does not take part in the formation of the embryo & gives rise to extra-embryonic membranes-Chorion & Amnion for nourishment of embryo & protection. The inner cell mass is referred as ‘precursor of embryo’.
6.  The blastocyst comos in contact with the endometrium
   in the region of embryonal knob. The surface cells of trophoblast secrete lytic enzymes which cause corrosion of endometrial lining. These give rise to finger-like projections called chorionic villi, which assists in fixation & absorption of nutrients.
   

Question.15. A non biology person is quite shocked to know that apple is a false fruit, mango is a true fruit and banana is a seedless fruit. As a biology student how would you satisfy this person ?
Answer : Fruits that are derived from matured ovaries of flowers are called true fruits and fruits that develop from other floral parts of plants are called false fruits. For example apple is a false fruit as in apple, the thalamus produces the fleshy edible part. Mango, on the other hand is a true fruit as it develops from the ovary of the flower, Fruits that develop without fertilization are called parthenocarpic fruits. Since no fertilization takes place, such fruits are seedless. Banana is an example of a seedless fruit.

SECTION -E

Question.25. A flower of brinjal plant following the process of sexual reproduction produces 360 viable seeds. Answer the following questions giving reasons.
(a) How many ovules are minimally involved ?
(b) How many megaspore mother cells are involved ?
(c) What is the minimum number of pollen grains that must land on stigma for pollination ?
(d) How many male gametes are involved in the above case ?
(e) How many microspore mother cells must have undergone reduction division prior to dehiscence of anther in the above case ?
Or
Describe the changes that occur in ovaries and uterus in human female during the reproductive cycle.
Answer: The number of viable seeds produced by the brinjal plant through sexual reproduction = 360
(a) The number of ovules minimally involved in this process would be 360, as the number of viable seeds are 360. After fertilization, the ovary turns into fruit and the ovules turn into seeds. Therefore, the number of ovules are corresponding to the number of seeds formed.
(b) During the process of garnetogenesis, 360 megaspore mother cells are involved as only one megaspore of the tetrad becomes functional and develops further and the rest three megaspores get degenerated.
(c) The minimum number of pollen grains that must land on stigma for pollination are 360 becauseeach pollen grain contains two male gametes. Out of these two gametes,
one fuses with polar nuclei and forms endosperm, while, the other male gamete fuses with the egg cell to form the zygote that eventually give rise to seeds.
(d) 720, each pollen grain carries two male gametes which participate in double fertilization.
(e) In the above case, 90 microspore mother cells must have undergone reduction division prior to dehiscence of anther, as each microspore mother cell would give rise to 4 microspores. Since 1 microspore mother cell would produce 4 microspores, therefore, to obtain 360 microspores, there must be 90 microspore mother cells.
OR
Menstrual cycle is the reproductive cycle in all primates , and begins at puberty (menarche). In human females, menstruation occurs once in 28 to 29 days. The cycle of 1 events starting from one menstruation till the next one is
called the menstrual cycle. These changes are brought about by ovarian and pituitary hormones.

Changes occurring in the Ovary :

1. During menstrual phase the levels of estrogen and progesterone falls considerably.
2.  This induces adenohypophysis to secrete FSH and LH.
3.  Increased levels of FSH stimulates the graafian follicle to mature and secrete estrogens. The rising level of estrogen, causes the endometrium to become thicker and more richly supplied with blood vessels. The level of estrogens in blood increases gradually for a few days and is at the peak on the 12th day of the cycle.
4. The estrogen surge reduces FSH secretion and this in turn introduces LH surge within 12 hours i.e. on the 13th day of the cycle.
5.  LH causes ovulation and formation of corpus futeum.
6.  During post ovulatory phase corpus luteum secretes progesterone.
7.  Corpus luteum secretes progesterone, which facilitates the preparation of the endometrium of the uterus for receiving the blastocyst and its implantation and it also inhibits the contraction of the uterus and any further development of a new follicle.

Change occurring in the uterus :

1.  If fertilization does not occur, the rising progesterone level inhibits the release of Gonadotropin releasing hormone (GnRH), which in turn inhibits the production of FSH, LH and progesterone.
2.  Once the progesterone level drops, the corpus luteum begins to degenerate resulting in its transformation into a white body called the corpus albicans.
3.  These hormonal changes, further, cause the breakdown of the endometrium, inhibition of uterine contraction ceases and the menstrual bleeding begins.
4. The low level of estrogens and progesterone stimulates secretion of FSH and LH from anterior pituitary initiating the next ovarian cycle.
5.  If fertilization occurs, corpus luteum persists and secretes progesterone and estrogens during pregnancy. Fertilised egg starts developing and simultaneously travels down and gets implanted in the uterus. Ovarian cycle comes to a temporary halt.

SET-III

SECTION-B

Question.6. Differentiate between ‘ZZ’ and ‘XY’ type of sex-determination mechanisms
Answer : The differences between ZZ and XY type of sex- determination mechanisms are listed below :

Question.7. An infertile couple is advised to adopt test-tube baby programme. Describe two principle procedures adopted for such technologies.
Answer : In-Vitro fertilization (IVF): In this process, the fertilization takes place outside the body (test tube baby). The ‘ following techniques are included in IVF :

1.  Zygote Intrafallopian Transfer (ZIFT) – In ZIFT, the sperm front a male donor and the ovum from a female donor are fused in the laboratory. The zygote so formed is transferred into the fallopian tube at the 8 blastomeres stage.
2.  Gamete Intrafallopian Transfer (GIFT) – In GIFT, females who cannot produce ovum, but can provide suitable conditions for the fertilization of ovum, are provided with ovum from a donor, which is transferred to the follopian tube for fertilisation.

Question.9. Enumerate four objectives for improving the nutritional quality of different crops for the health benefits of the human population by the process of “Biofortification”.
Answer : Biofortification involves the breeding of crops to increase their nutritional value. The objectives for biofortification are as follows :

1.  To improve protein content and quality
2.  To improve oil content and quality
3.  To improve vitamin content
4.  To improve micronutrient and mineral content

SECTION-C

Question.12. Describe the development of endosperm after double fertilization in an angiosperm. Why does endosperm development preceeds that of zygote ?
Answer : Endosperm is the nutritive tissue formed as a result of triple fusion in the angiosperms Endosperm is generally triploid meant for nourishing the embryo. The formation of endosperm starts with degeneration of the unclear tissue. Based on the mode of development there are three types of . endosperms, (i) Nuclear (ii) Cellular (iii) Helobial.

1. Nuclear type : Primary endosperm nucleus divides ‘ repeatedly to form a large number of free nuclei. No cell plate formation takes place at this stage. A central vacuole appears later.
2. Cellular type : In this case, there is cytokinesis after each nuclear division of endosperm nucleus. The endosperm, thus, has a cellular form, from the very beginning because first and subsequent divisions are all accompanied by wall formation e.g. Petunia, Datura, Adoxa, etc.
3.  Helobial type : It is an intermediate type between the nuclear and cellular types. The first division is accompanied by cytokinesis but the subsequent ones are free nuclear. The chamber towards micropylar end_ of embryo sac is usually much larger than the chamber towards chalazal and.
   The endosperm develops before embryo because the cells of endosperm provide nutrition to the developing embryos.