CBSE Class 12 Biology Question Paper (Delhi 2019) with Solutions

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CBSE Class 12 Biology Question Paper (Delhi 2019) with Solutions

Time Allowed: 3 Hours
Maximum Marks : 70

General Instructions:

1. There are total 27 questions and four sections in the question paper. All questions are compulsory.
2. Section A, 1 to 5 one marks each.
3. Section B, 6 to 10, two marks each.
4. Section C, 11 to 22, three marks each.
5. Section D, 23, four marks.
6. Section E, 24 to 27, five marks each.
7. There is no overall choice in the question paper, however, an internal choice is provided in two questions of one mark, two questions of two marks, four questions of three marks and all the three questions of five marks. In these questions, an examinee is to attempt any one of the two given alternatives.
8. Wherever necessary, the diagram drawn should be neat and properly labelled.

Section – A

Question 1.
What do ‘X’ and ‘Y’ represent in the transcription unit of the DNA molecule shown?

Answer:
The ‘X’ and ‘ Y’ in the given figure represents template strand and terminator region. (1 Mark)
Template strand: The strand that has polarity 3 ’ → 5 ’ acts as a template strand.
Terminator region: The terminator region is located towards 3’-end (downstream) of the coding strand and it terminates the process of transcription.

Note
Template strand (3′ → 5′) is the strand used by DNA dependent KNA polymerase to attach complementary bases during the process of replication or transcription.

Question 2.
Humans are categorised as “regulators”. Explain how they maintain a constant normal body temperature.
Answer:
Glomus forms mycorrhiza. It is symbiotic association between the fungi and root nodules of leguminous plants. They absorb phosphorus from the soil and transport it to the plants.

Question 3.
The diploid number of chromosomes in an angiospermic plant is 16. What will be the number of chromosomes in its endosperm and antipodal cells?
OR
State the reason why pollen grains lose their viability when the tapetum in the anther is malfunctioning.
Answer:
The diploid number of chromosomes in an angiosperm plant
(2n) = 16.
The haploid number (n) will be = 8
Endosperm of an angiosperm is triploid (3n), so the number of chromosome present in endosperm 3n = 8 × 3 = 24.
Whereas antipodal cells are haploid (n) in nature. So, the number of chromosome in antipodal cells will be =8
OR
Tapetum provides nourishment to the developing pollen grain. V/hen the tapetum in anther is malfunctioning the pollen grain will not get enough nourishment and will lose its viability.

Question 4.
Biotechnologic techniques can help to diagnose the pathogen much before the symptoms of the disease appear in the patient. Suggest any two such techniques.

OR

Mention the form in which inactive protein toxin is produced by Bacillus thuringiensis. How does it get activated in the pest body to kill it?
Answer:
Two techniques used to diagnose pathogen are

1. PCR
2. Gel electrophoresis

PCR (Polymerase Chain Reaction) is molecular biology technique that is used for the early diagnosis of symptoms of diseases. It focuses on the actual segment of DNA of its own interest and from a segment of DNA, the genotypes can be determined. In this way, genetic disorders can be diagnosed and detected. This technique is used for the detection of HIV and cancer.

Note
DNA polymerase used in PCR is Taq DNA polymerase which is isolated from a bacterium. Thermus aquaticus. This bacterium is found in hot springs and hydrothermal vents. The Taq Polymerase remains active at high temperature during denaturation.

Gel electrophoresis is a molecular biology technique and is also used for the early diagnosis of symptoms of diseases. In this technique, DNA sequences are separated according to their size.

OR

Inactive protein toxin of Bacillus thuringiensis is present in the form of crystals. This inactive form gets activated by the alkaline pH of the alimentary canal of the insect. The activated toxin binds to the surface of midgut epithelial cells and creates pores which can cause cell swelling and lysis of midgut wall and finally causes death of the insect.

Question 5.
Name the disorder in humans with the following karyotype:
(a) 22 pairs of autosomes + XO
(b) 22 pairs of autosomes + 21st chromosome + XY
Answer:
(i) 22 pairs of autosome + XO occurs in case of turner syndrome. The total number of chromosomes is 45 with XO. Females with turner syndrome are sterile as their ovaries are rudimentary. Such females also lack other secondary sexual characters.

(ii) Down syndrome is a disorder that results in 22 pairs of autosomes along with an extra chromosome 21 (22 pairs of autpsomes + 21st chromosome + XY). Because this condition results due to an extia copy of the 21st chromosome, it is also called trisomy of chromosome no. 21. Individuals with Down syndrome have a flattened face, a short neck and a tongue that protrude outwards.

Section – B

Question 6.
Humans are categorised as “regulators”. Explain how they maintain a constant normal body temperature.
Answer:
Regulators: Regulators are those organisms that are able to maintain homeostasis according to external environment. They have ability to maintain their constant body temperature and osmotic concentration. For example: Birds and mammals.

Regulators during winters when the outside the temperature is less than body temperature, the body start shivering. Shivering is a way to maintain the internal body as this practice produces heat. (2 Marks)

Note
The tendency to maintain a stable, relatively constant internal environment is called homeostasis.

Question 7.
You are given a tall pea plant and asked to find its genotype. How would you find its genotype? Explain.
Answer:
A test cross is used to determine the genotype of pea plant.
In this a tall plant is crossed with dwarf plant. In the F₁ generation, all the progeny obtained are tall. So the genotype of the tall plant is TT. In the F₂ generation, 50% tall plants and 50% dwarf plants are obtained and the genotype of the tall plant is T₁.
Diagrammatic Representation of Test cross:

Question 8.
Scientists are trying to solve the issues of malnutrition and hunger by using microbes. By taking one suitable example, explain how they have been able to help.
Answer:
Scientist are trying to solve the issues related to malnutrition and hunger by using microbes by using a technique called Single Cell Protein (SCP). In this, microbes such as Spirulina are grown on an industrial scale as good source of protein. Spirulina can be grown on materials like waste water from potato processing plants that contains starch, straw, molasses, animal manure and even sewage in order to produce a food that contains large quantities of protein, minerals, fats, carbohydrate and vitamins. This technique also reduces environmental pollution.

Question 9.
MOET is a programme for herd improvement. Write the steps in correct sequence that are carried in the programme.
OR
Why is tobacco smoking associated with rise in blood pressure and emphysema? Explain.
Answer:
MOET is Multiple Ovulation Embryo Transfer Technology is one such programme for herd improvement. The following steps are involved in this method:

• A cow is administered with hormone such as FSH (follicle stimulating hormone) to induce follicular maturation and super ovulation instead of one egg that is produced normally per cycle.
• In this process, they produce 6-8 eggs.
• Then the animal is either mated with an elite bull or artificially insemination.
• The fertilised eggs are then at 8-32 celled stage.
• The 8-32 egg cell stage are recovered non-surgically and transferred to surrogate mother.
• This genetic mother is then available for another round of super ovulation. (2 Marks)

Note
This technology is used for cattle, sheep, rabbits, buffaloes, mares and so on. In this technique, high milk yielding breeds of females and high quality meat yielding bulls have been bred successfully in order to increase the herd size in shorter time.

OR

Tobacco contains nicotine that tends to stimulate the release of adrenaline and non-adrenaline. It increases the blood pressure and heart rate. Smoking increases carbon monoxide content in the blood and reduces the concentration of haem-bound oxygen in the blood. It leads to cause oxygen deficiency in the body. Emphysema is caused because of the over-inflammation of the alveolar sacs that impairs gaseous exchange in the lungs.

Question 10.
(a) How will you measure population density of fish in a lake? (1 Mark)
(b) In a pond there are 100 frogs. 20 more were bom in a year. Calculate the birth rate of this population.
OR
Draw a “stable” human age pyramid. Comment on the population growth rate that is depicted by it.
Answer:
(a) The population density of a fishes in a lake is determined by the fish caught per trap. Population density refers to the measure to determine the number of individuals of a species in a particular unit area. The relative densities are used to determine the population size in a specific area.

(b) To calculate the birth rate of population
\(=\frac{\text { Number of births }}{\text { Total population }}\) × 100
The birth rate of a frog population = \(\frac{20}{120}\) × 100
= 16.66 per year (1 Mark)

OR

In a stable or Bell-shaped age pyramid the number of pre- reproductive and reproductive individuals is almost equal. If the post-reproductive individuals are comparatively fewer than the population size remains stable as it is neither growing nor diminishing.

Note
Age pyramid is defined as a way for representing the age-sex structure of population.

Diagrammatic Representation of Stable Age pyramid:

Section – C

Question 11.
What is cryopreservation? Mention how it is used in conservation of biodiversity.
Answer:
Cryopreservation: It is a method used for the preservation of cells and tissue structures in an extremely cold temperature such as at -196°C in a liquid nitrogen. It inhibits metabolism in cells.

Cryopreservation of biodiversity in a following ways: The gametes of endangered plants and animals are kept viable by preserving them at a very low temperature (-196°C) in a liquid nitrogen.

Question 12.
How can childless couples be helped by the followed assisted reproductive technologies:
(a) GIFT
(b) Cytoplasmic Sperm Injection
Answer:
(a) GIFT (Gamete Intra Fallopian Transfer): It is an In vitro fertilisation technique that involves the transfer of ovum collected from a donor into the fallopian tube of another female who is unable to produce eggs. This technique also provide suitable environment for fertilisation and further development.

(b) Cytopiasmic sperm injection: It is another specialised In vitro technique in which an embyo is formed in a laboratory in which a sperm is directly injected into the ovum.

Note
Assisted Reproductive Technology (ARTs) is used for the treatment of infertility. This technology involves the mating of egg and sperm outside of the body (In vitro) under sterilized condition.

Question 13.
Draw a diagram of LS of Maize grain and label its any six parts.
Answer:
Diagrammatic Representation of L.S of Maize grain:

Question 14.
Study the figure of vector pBR322 given below:

Identify A, B, C and D and explain their roles in cloning a vector.
Answer:
In the given figure, A is ampicillin resistance gene, B is origin of replication, C is repressor of primer, D is tetracycline resistance gene.

(a) Ampicillin resistance gene (ampR): It acts as a selectable marker that provides resistance against ampicillin antibiotic.
(b) ori: It is origin of replication. A sequence that initiates the process of replication. It controls the copy number of linked DNA.
(c) rop (repressor of primer): This site is responsible for the restricting the plasmid copy number.
(d) tet^R: It is tetracyclin resistance gene that acts as a selectable marker which provides resistance against tetracyclin antibiotic.

Question 15.
Compare the mechanism of evolution as put forth by Charles Darwin and de Vries.
Answer:
The Darwin’s theory of evolution is based on Natural selection. According to Darwin’s, variation results in evolution. Variations are inheritable and were small as well as directional. Natural selection favours survival of the fittest one as nature selects individuals that are fit and able to adapt as well as survive in changing environment.

The theory of Hugo deVaries is based on mutation. He believed that large differences occurs suddenly in a population. He also said that mutation results in evolution. According to Hugo deVaries mutation are random and directionless. Mutation leads to cause speciation and hence is called as saltation or single step large mutation.

Question 16.
(a) A patient had suffered myocardial infarction and clots were found in his blood vessels. Name a ‘clot buster’ that can be used to dissolve the clots and the microorganism from which it is obtained.
(b) A woman had just undergone a kidney transplant. A bioactive molecular drug is administered to oppose kidney rejection by the body. What is the bioactive molecule? Name the microbe from which this is extracted.
(c) What do doctors prescribe to lower the blood cholesterol level in patients with high blood cholesterol? Name the source organism from which this drug can be obtained.
Answer:
(a) Streptokinase is used as clot buster for patients suffering from myocardial infraction. It is produced from bacterium Streptococcus.
(b) Cyclosporin A is a bioactive molecule that is used as an immunosuppressive agent in organ-transplant patients. It is produced by a fungus Trichoderma polysporum.
(c) Statins is a blood-cholesterol lowering agents produced by yeast Monascuspurpureus. It competitively inhibits the enzyme that is responsible for the synthesis of cholesterol.

Question 17.
Give reasons for the following:
(a) Antibody mediated immunity is called humoral immunity.
(b) How is a child protected from a disease for which he/ she is vaccinated?
(c) Name the type of cells the AIDS virus enters after getting into the human body.

OR

(a) Identify the nos. (i) to (iv) in the following table:

(b) Which one of the above mentioned diseases are transmitted through mechanical carriers?
Answer:
(a) Antibodies are also called immunoglobulins that are produced by B-lymphocytes in the blood and produces immune response. This type of immune response is called humoral immune response. (1 Mark)
(b) Vaccination is based on the memory of the immune system. Vaccination involves the preparation of inactivated or weakened pathogens or a preparation of antigenic proteins of pathogen in the host body. The antibodies produced in the body against these antigens neutralise the pathogenic agents during actual infections. (1 Mark)
(c) The AIDS virus after entering into the body get enters into the macrophages. Then the AIDS virus enters into the helper-T cells.

OR

(a)

1. Fever, chills, cough, cold and headache. In severe cases, the lips and fingernails may turn grey to bluish in color.
2. Salmonella typhi
3. Common cold
4. Internal bleeding, muscular pain, fever, anaemia and blockage of intestinal passage.

(b) None of the above mentioned diseases are transmitted through mechanical carriers.

Question 18.
Draw a diagram of the sectional view of a human seminiferous tubule and label any six of its parts.
Answer:
Diagrammatic Representation of seminiferous tubules:

Note
The seminiferous tubules is lined on its inside by two types of cells called male germ cells that undergo meiotic divisions results in sperm formation. Whereas sertoli cells provide nourishment to the germ cells.

Question 19.
A woman with ‘O blood group’ marries a man with ‘AB blood group’. Work out the cross to show all the possible phenotypes and genotypes of the progeny with respect to blood groups. Explain the pattern of inheritance observed in this cross.
Answer:
The inheritance of ABO blood grouping in humans is an example of Co-dominance and multiple alleles. ABO blood grouping in human beings are controlled by I gene. The plasma membrane of the red blood cells has sugar polymers that are found on the surface of RBCs and is controlled by this gene.
Diagrammatic representation of a Cross:

The above cross represents that in FI generation blood A and B are dominant over blood O group. So, the children’s have 50 % chances of getting A and B blood group.

Note
Co-dominance is a type of inheritance in which the alleles of a gene pair in a heterozygote are independently expressed themselves. Multiple alleles refer to the more than three alternative forms of allele of a gene.

Question 20.
(a) What is the breeding of crops for enhancing their nutritional value called? Why is the need felt for enhancing the nutritional value of the crops?
(b) Rice, wheat and maize are the most commonly used food grains the world over. How have these grains improved in their nutritional value in comparison to their conventional varieties?
OR
(a) Write the scientific names of the source plants from where opioids and canabinoids are extracted.
(b) Write their receptor sites in the human body. How do these drugs affect the human beings?
Answer:
(a) Biofortification is a technique used by farmers for increasing the nutrient content in the breeding crops. This breeding technique involves breeding crops with higher levels of vitamins and minerals or higher protein as well as healthier fats in order to improve public health.
Biofortification involves following objectives for improving nutritional quality of crops such as:

• Increasing protein content and quality
• Oil content and quality
• Vitamin content and
• Micronutrient and mineral content.

(b) Biofortification is a boon for crops such as maize, wheat and rice. As the maize hybrids contains twice the amount of the amino acids lysine and tryptophan as compared to existing maize hybrid.

Wheat variety such as Atlas 66, contains a high protein content and it has been used as a donor for improving cultivated wheat. Iron-fortified rice variety is also developed that contains five times iron than commonly used varieties of rice.

Several vegetables are also rich in Vitamins and minerals for example carrot enriched in vitamin A, spinach, pumpkin, vitamin C enriched bitter gourd, bathua, mustard, tomato. Iron and calcium enriched spinach and bathua, protein enriched beans such as broad, lablab, French and garden peas.

Note
All the biofortified vegetables are developed by Indian Agricultural Research Institute.

OR

(a) Opioids are the drugs that are obtained from the latex of poppy plant called Papaver somniferum. Whereas cannabinoids are the group of chemicals that are naturally obtained from the inflorescences of the plant Cannabis sativa. (1 Mark)
(b) Opioids are the drugs that get bids with the specific opioid receptors present in the central nervous system and gastrointestinal tract of human beings. Opioids such as heroin acts a depressant and slows down body functions.

Cannabinoids are a group of chemicals that interact with cannabinoid receptors present in the brain. Cannabinoids such as marijuana, hashish, charas and ganja effects on cardiovascular system of the body.

Question 21.
Write by taking a suitable example, the convention followed for naming the restriction enzymes.
Answer:
The convention for naming these enzymes is the first letter of the name that comes from the genes and the second two letters come from the species of the prokaryotic cell from which they were isolated for e.g., EcoRl comes from Escherichia coli RY13.

In ECoRI, in this the letter ‘R’ is derived from the name of strain. Roman numbers following the names indicate the order in which the enzymes were isolated from that strain of bacteria.

Question 22.
Hershey and Chase carried out their experiment under three steps:
(a) Infection,
(b) Blending, and
(c) Centrifugation. Explain each one of these steps that helped them to prove that DNA is the hereditary material.
OR
(a) Why does DNA replication occur within a replication fork and not in its entire length simultaneously?
(b) ” DNA replication is continuous and discontinuous on the two strands within the replication fork.” Give reasons.
Answer:
To proof that DNA is the genetic material an experiment was performed by Alfred Hershey and Martha Chase in 1952. They worked with viruses that infect bacteria called bacteriophage.
Following steps are involved in Hershey and Chase experiment:
(a) Infection

• They grow viruses on a different medium containing radioactive phosphorus and radioactive sulphur.
• Viruses grown in the presence of radioactive phosphorus contained radioactive DNA but not protein because DNA contains phosphorus but is absent in protein.
• Whereas vims grown in a medium containing radioactive sulphur contained radioactive protein but not radioactive DNA because sulphur is absent in DNA.
• Radioactive phages were allowed to attach to E.coli bacteria.

(b) Blending:

• As the infection proceeded, the viral coats were removed from the bacteria by agitating them in a blender.

(c) Centrifugation:

• Then the vims particles were separated from the bacteria by spinning them in a centrifuge.
• So the bacteria that infected with viruses that contain radioactive DNA were radioactive.
• This indicates that DNA was the material that can be passed from the virus to bacteria.
• Bacteria that were infected with viruses that contain radioactive proteins were not radioactive.
• This indicates that proteins did not enter the bacteria from the viruses.
• Hence, it is proved that DNA is the genetic material that is passed from virus to bacteria.

Diagrammatic representation of Hershey and Chase experiment:

OR

(a) DNA is a large and double stranded helical molecule. The process of DNA replication occurs only in small region of DNA not in its entire length in one time. As large amount of energy is required for opening a whole double helical DNA.

Note
DNA is more stable than RNA due to absence of 2¹ – OH group and presence of thymine in place of uracil in DNA.

(b) In DNA replication, DNA-dependent DNA polymerase is involved in the process of DNA replication. This enzyme catalyses the polymerisation of deoxynucleotides in 5′ → 3′ direction, this is called lagging strand of the DNA and replication is discontinuous whereas when the replication takes place in 3′ → 5′, this strand is called leading strand and replication is continuous.

Note
The discontinuously synthesised strand of the fragments are joined together by DNA ligase enzyme and such fragments are called OKAZAKI fragments.

Section – D

Question 23.
Restriction endonucleases have played a very significant role in rDNA technology. Explain the roles of ZscoRI and DNA ligase in formation of recombinant DNA. (3 Marks)
Answer:
In Recombinant DNA technology, EcoRI cuts the DNA between bases G and A only when the sequences GAATTC is present in the DNA. It acts as a molecular scissors that serves as tool for DNA at specif.c palindromic sites at specific points. DNA ligase enzyme is also called molecular gum as it is involved in joining the two segments of DNA by creating phosphodiester bond between the two fragments of DNA.

Note
A recombinant DNA molecule is formed from the segments of two or more different DNA molecules.

Question 24.
Describe the phosphorus cycle in an ecosystem. (3 Marks)
Answer:
Phosphorus is a major constituent of all biological membranes such as nucleic acids and cellular energy transfer systems. Animals also require a large amount of phosphorus to make shells, bones and teeth. Naturally phosphorus is found in rocks in the form of phosphates. Plants absorb phosphorus through roots from soil when the rocks are weathered, minute amount of phosphates that is dissolved in soil solution.

So, the herbivores and other animals obtain phosphorus from plants. Whereas the waste products and the dead organisms are decomposed by phosphate-solubilising bacteria releasing phosphorus.
Diagrammatic representation of Phosphorus cycle:

Question 25.
(a) Describe the experiment conducted by F. Griffith in 1928 with Streptococcus pneumoniae and write the conclusions he arrived at.
(b) State the contribution of Avery. MacLeod and McCarty in providing biochemical nature to the results as obtained by Griffith.
Answer:
(a) The transforming principle was proposed by Frederick Griffith in 1928. He performed his experiment with Streptococcus pneumonia a bacteria which is responsible for pneumonia.
The following steps are involved in his experiment:

• When Streptococcus pneumonia bacteria are grown on a culture plate, some bacteria produces smooth shiny colonies (s) whereas other produce rough colonies (R).
• It is because the S strain bacteria have a mucous polysaccharide coating whereas the R strain bacteria lack this coating.
• When the mice infected with the S strain or virulent strain, the mice die because of pneumonia infection.
• When the mice infected with the R strain do not develop pneumonia.
• Then, Griffith was able to kill bacteria by heating. He observed that heat-killed S strain bacteria injected into mice did not kill them.
• When he injected a mixture of heat-killed S and live R bacteria, the mice died. He recovered living S bacteria from the dead mice.

After this experiment, he concluded that the R strain bacterium has been transformed by the heat-killed S strain bacteria. As some ‘transforming principle’ was transferred from the heat-killed S strain and enabled the R strain to synthesise a smooth polysaccharide coating and become virulent. This is because of the transfer of the genetic material.

(b) The biochemical characterisation of Transforming principle was determined by Oswald Avery, Colin Macleod and Maclyn McCarty. Prior it was thought that the genetic material was protein.
The following steps are involved in his experiment:

• They purified biochemicals such as proteins, DNA and RNA from the heat-killed S cells to determine which one could transform live R cells into S cells.
• They discovered that DNA alone from S bacteria caused R bacteria to become transformed.
• They also discovered that protein-digesting enzymes such as proteases and RNA-digesting enzymes such as RNases did not affect the transformation.
• Hence, it was proved that the transforming substance was not protein and RNA.
• Digestion with DNase did inhibit transformation and the DNA caused the transformation.
• So, they concluded that DNA is the hereditary material.

Question 26.
(a) State what is an ecological succession.
(b) Write one similarity and one difference between hydarch and xerarch successions.
(c) Explain the mechanism of co-evolution as seen in orchid Ophrys and bee.
OR
(a) List any two ways the biodiversity loss affects any region.
(b) Explain any two causes the biodiversity loss affects any region.
Answer:
(a) Ecological succession is defined as the gradual and fairly predictable change in the species composition of a given area. During this process, some species colonies in an area and their populations become more numerous, while populations of other species decline and even disappear.

(b) Difference between Hydrarch succession and Xerarch succession:

Similarities between hydrarch succession and Xerarch succession:

1. Both the hydrarch and xerach succession conditions leads to medium water conditions (mesic)- neither too dry (xeric) nor too wet (hydric).
2. Both end in the forest and both contribute to mesic conditions.

(c) Mediterranean orchid Ophyrs employs ‘ sexual deceit ’ to get pollination done by species of bees. In this, one petal of its flower bears an uncanny resemblance to the female of the bee in size, colour and markings. The male bee is attracted to what is perceives as a female, then ‘pseudocopulates’ with the flower, and during that process is dusted with pollen from the flower.

When this same bee ‘pseudocopulates’ with another flower, it transfers pollen to it and thus pollinates the flower. This example represents the phenomena of co-evolution. In this case, if the female bee’s colour patterns change even slightly for any reason during evolution, the process of pollination success will be reduced unless the orchid flower co-evolves in order to maintain the resemblance of its petal to the female bee.

OR

(a) Loss of biodiversity in a region may lead to cause following effects such as:

• Decline in plant production,
• Lowered resistance to environmental perturbations such as drought and,
• Increased variability in certain ecosystem processes such as plant productivity, water use and pest and disease cycles.

(b) The causes of biodiversity losses are as follows:

(i) Habitat loss and fragmentation: This is the most important cause of extinction of plants and animals species. Habitat loss comes from tropical rain forests. The Amazon rain forest as it is also called the ‘lungs of the planet’ harbour millions of species is being cut and cleared for cultivating soya beans or for conversion to grasslands for raising beef cattle.

Population also leads to cause degradation of many habitats and threatens the survival of many species. Large habitats are broken into small fragments due to various human activities, mammals and birds require large territories and certain animals with migratory habits are badly affected results in population declines.

(ii) Over-exploitation: Humans are always depended on nature for food and shelter but their need turns into ‘greed’ and it leads to over-exploitation of natural resources. Many species become extinct in the last decades such as Steller’s sea cow, passenger pigeon because of overexploitation by humans.

Question 27.
(a) Draw the embryo sac of a flowering plant and label the following:
(i) Central cell
(ii) Chalazal end
(iii) Synergids
(b) Name the cell and explain the process it undergoes to develop into an embryo sac. (3 Marks)
(c) Explain the development of endosperm in coconut. (1 Mark)
OR
Write the duration and the events that occur in the ovary and the uterus during follicular and luteal phases of the menstural cycle in humans. How do pituitary and ovarian hormones influence these two phases?
Answer:
(a) Diagrammatic representation of embryo sac:

(b) Megspore mother cell undergoes the process of meiosis to develop embryosac. The process of formation of megaspores from the megaspore mother cell is called megasporogenesis. The megaspore mother cell undergoes the process of meiosis and forms a four haploid tetrad megaspores. The chalazal megaspore remains functional whereas the other 3 will degenerate. So, the functional megaspore is the first cell of the female gametophyte. The cell enlarges and undergoes three free nuclear mitotic divisions.

So the first meiotic division produces two nucleate embryo sac and two nuclei shift to the two ends and again gets divide and forms four nucleate. In this way, eight nucleate structures is formed. One nucleus from each side moves to the middles and they are called polar nuclei. Then the remaining three nuclei form cells at the two ends, 3-celled egg apparatus at the micropylar end and three antipodal cells at the chalazal end.

(c) The Primary Endosperm Nucleus (PEN) is triploid (3n) in nature that undergoes nuclear divisions and give rise to free nuclear endosperm. This free nuclear endosperm is a coconut water whereas its white kernel is the cellular endosperm that is formed when it undergoes cytokinesis.

OR

The menstrual phase is followed by follicular phase and during this phase the primary follicles in the ovary grow to become a fully mature Graafian follicle. Then simultaneously the endometrium of uterus regenerates through proliferation. Such changes in the ovary and uterus are induced by changes in the levels of pituitary and ovarian hormones. The secretion of gonadotropins such as luteinizing hormone and follicle stimulating hormone levels gradually increases during follicular phase.

This stimulates follicular development as well as secretion of estrogens hormone by the growing follicles. Both LH and FSH attain a peak level in the middle of cycle that is about 14th day. The rapid secretion of LH surge induces rupture of Graafian follicle and then induces the release of ovum results in ovulation.

The process of ovulation is followed by luteal phase. During this phase, the remaining parts of the Graafian follicle transforms as the corpus luteum. It secretes a large amount of progesterone hormone that is essential for the maintenance of the endometrium.

As, endometrium is necessary for the process of implantation of fertilised ovum and other events during pregnancy.

If the process of fertilisation does not take place, the corpus luteum degenerates results in the disintegration of the endometrium leading to menstruation. This marks a beginning of new cycle.

A decrease in level of ovarian hormone results in the constriction of arteries that lead to cause shredding of uterine lining and menstruation. The deficit in gonadotropins removes the negative feedback on the hypothalamus and menstrual cycle again begins with the GnRH release.

Note
In human females, menstruation is repeated at an average interval of about 28/29 days and the cycle of events starring from one menstruation till the next one is called the menstrual cycle.