BIOLOGY 2010 SET 3Question 1 ( 1.0 marks)Mention one positive and one negative application of amniocentesis.

Solution:Positive impact of amniocentesis:

It is used for detecting foetal abnormalities or genetic diseases during pregnancy.

Negative impact of amniocentesis:

It is used for determining the sex of foetus which, in turn, can lead to female foeticide.

Question 2 ( 1.0 marks)How do animals like fish and snails avoid summer related unfavourable conditions?

Solution:Fishes and snails undergo aestivation or summer sleep to avoid summer-related unfavourable conditions.

Question 3 ( 1.0 marks)In a pond there were 200 frogs. 40 more were born in a year. Calculate the birth rate of the population.

Solution:Birth rate = 40/200 = 0.2 off springs per frog per year

Question 4 ( 1.0 marks)Mention two functions of the codon AUG.

Solution:Two functions of codon AUG are:

It is the start-codon. It marks the beginning of translation. It codes for the amino acid methionine.

Question 5 ( 1.0 marks)Name a molecular diagnostic technique to detect the presence of a pathogen in its early stage of infection.

Solution:PCR (Polymerase Chain Reaction) is used for detecting the presence of pathogens in the early stage of infection since it amplifies a minute quantity of DNA.

Question 6 ( 1.0 marks)Name the scientist who disproved spontaneous generation theory.

Solution:Louis Pasteur disproved the spontaneous generation theory.

Question 7 ( 1.0 marks)What is it that prevents a child to suffer from a disease he/she is vaccinated against? Give one reason.

Solution:Vaccination consists of weakened germs responsible for causing a particular disease. These weakened germs stimulate the immune system to synthesise antibodies against the particular disease. When the active germs for the same disease attack a person in future, the synthesised antibodies fight against these germs.

Question 8 ( 1.0 marks)Why is the enzyme cellulose used for isolating genetic material from plant cells but not for animal cells?

Solution:

Enzyme cellulase is used for digesting cellulose, which is present in the cell wall of plant cells. Since the cell wall is absent in animal cells, no cellulose is found in animals. Hence, cellulase is essentially effective for digesting and isolating genetic material from plant cells not animal cells.

SECTION B

Question 9 ( 2.0 marks)Where does triple fusion take place in a flowering plant? Why is it so called? Mention its significance.

Solution:Triple fusion takes place inside the embryo sac.

The pollen tube releases two male gametes into the embryo sac. Out of the two male gametes, one gamete fuses with the nucleus of the egg cell and forms the diploid zygote (syngamy). The other male gamete fuses with two polar nuclei and results in the formation of triploid primary endosperm nucleus. Since this process involves the fusion of three haploid nuclei, it is known as triple fusion.

Triple fusion is significant as it results in the formation of endosperm. Endosperm is a nutritive tissue that surrounds the embryo and provides nutrition in the form of starch, though it can also contain oils and protein.

Question 10 ( 2.0 marks)Why certain regions have declared as biodiversity “hot spots” by environmentalists of the world? Name any two “hot spot” regions of India.

Solution:Certain regions have been declared as biodiversity hotspots by environmentalists of the world because these regions have

high species richness and also a high degree of endemism. Endemism is the indigenousness of a species to a particular region. Certain species are exclusive to certain regions and enrich the biodiversity of such regions. Such regions with exclusive and exotic species are declared as biodiversity hotspots. Western Ghats and Eastern Himalaya are two high biodiversity regions in India.

Question 11 ( 2.0 marks)A moss plant produces a large number of antherozoids but relatively only a few egg cells. Why?

Solution:Antherozoids produced by moss plant essentially require moisture or water for their transport and fertilization. For fertilizing the egg, antherozoids need to be carried to the egg cell. Out of all the antherozoids that are produced, very few reach the egg cell. The rest either get degenerated or are diverted from the path due to water current. This is why the number of antherozoids produced by a moss plant is much higher than that of the egg cells.

Question 12 ( 2.0 marks)Why is the introduction of genetically engineered lymphocytes into a ADA deficiency patient not a permanent cure? Suggest a possible permanent cure.

Solution:In case of ADA deficient patients, lymphocytes from their blood are grown in a culture.

A functional ADA cDNA (using a retroviral vector) is then introduced into the lymphocytes. These lymphocytes are subsequently returned to the patients. However, as these cells are not immortal, the patient requires periodic infusion. Thus,

infusion of genetically engineered lymphocytes is not a permanent cure.

A permanent cure for ADA deficiency is to introduce genes isolated from marrow cells that produce ADA into cells at early embryonic stages.

Question 13 ( 2.0 marks)Study the given food chain and answer the questions that follow:

(i) Give reasons why there is a continuous increase in the DDT content in different trophic levels of the chain.

(ii) Name the phenomenon responsible for the increase in DDT content.

Solution:(i) The continuous increase in the DDT content in different trophic levels of the food chain is because of continuous increase in the number of organisms eaten for food. As shown in the question, the DDT concentration in the pond was 0.003 ppm. Since this water is taken up by zooplanktons, the concentration of DDT in the bodies of the zooplanktons was found to be 0.04 ppm. Small fishes that feed on such zooplanktons accumulated more DDT in their bodies. Similarly, large fishes that feed on smaller fishes had even more concentration of DDT. Thus, fish-eating-birds (the top carnivore), which feed on large fishes, had the highest concentration of DDT, i.e., 5 ppm.

(ii) The phenomenon responsible for the increase in DDT content is known as biomagnification.

Question 14 ( 2.0 marks)Honey collection improves when beehives are kept in crop-fields during flowering season. Explain.

OR

How does addition of a small amount of curd to fresh milk help formation of curd? Mention a nutritional quality that gets added to the curd.

Solution:Bees act as pollinators of many crop species such as sunflower, Brassica, apple and pear. Keeping beehives in crop fields during flowering period increases the frequency of pollination by bees. Since bees visit the flowers in the crop

fields often, they collect more nectar and use this nectar to produce additional honey. Thus, keeping beehives in crop-fields during flowering season is beneficial both for crop yield and for honey yield.

OR

Even a small amount of curd contains millions of Lactic Acid Bacteria (LAB) that multiply at suitable temperatures when the curd is added to milk. During growth, LAB produce acids that coagulate and partially digest the milk proteins, thus converting the milk to curd.

This improves the nutritional quality of curd by increasing vitamin B12 levels in it.

Question 15 ( 2.0 marks)Name the type of food chains responsible for the flow of larger fraction of energy in an aquatic and a terrestrial ecosystem respectively. Mention one difference between the two food chains.

Solution:Grazing food chain is the major conduit for energy flow in aquatic ecosystem. On the other hand, a much larger fraction of energy flows through the detritus food chain in a terrestrial ecosystem.

The grazing food chain begins with producer and continues to primary and secondary consumers. However, the detritus food chain begins with dead organic matter and is made up of decomposers - mainly fungi and bacteria.

Question 16 ( 2.0 marks)Name the host and the site where the following occur in the life-cycle of a malarial parasite:

(a) Formation of gametocytes

(b) Fusion of gametocytes

Solution:(a) Human beings act as hosts for the formation of gametocytes. This formation takes place in human red blood cells.

(b) Female Anopheles mosquitoes act as hosts for fusion of gametocytes. The fusion takes place in the mosquitoes’ intestines.

Question 17 ( 2.0 marks)Why are F2 phenotypic and genotypic ratios same in a cross between red-flowered snapdragon and white-flowered snapdragon plants. Explain with the help of a cross.

Solution:A cross between red-flowered Snapdragon and white-flowered Snapdragon plants is shown below:

In a normal Mendelian cross, Rr would have corresponded to the red coloured flower only, because in the Mendelian inheritance pattern, only dominant allele is expressed in the phenotype. However, Snapdragon shows incomplete dominance in which Rr, instead of expressing dominant phenotype, in fact expresses a mixture of dominant (red) and recessive (white) traits that leads to the pink colour for Rr. Hence, the phenotypic ratio modifies to 1:2:1. Thus, both phenotypic and genotypic ratios are same for a Snapdragon flower.

Question 18 ( 2.0 marks)How does the floral pattern of Mediterranean orchid Ophrys guarantee cross pollination?

Solution:The Mediterranean orchid Ophrys employs ‘sexual deceit’ to get pollinated by a species of bees. One petal of its flower bears an uncanny resemblance to the female bee in size, colour and markings. The male bee is attracted to what it perceives as a female, ‘psuedocopulates’ with the flower and during that process, the male bee is dusted with pollen from the flower. When the bee ‘psuedocopulates’ with another flower, it transfers pollen to that flower. The flower thus, gets pollinated.

This is how the floral pattern of the Mediterranean orchid Ophrys guarantees cross pollination.

SECTION C

Question 19 ( 3.0 marks)During his studies on genes in Drosophila that were sex-linked T.H. Morgan found F2 − population phenotypic ratios

deviated from expected 9:3:3:1. Explain the conclusion he arrived at.

Solution:The phenotypic ratios deviated from expected Mendelian ratio in case of Drosophila because the two genes representing a single character did not segregate independently.

Morgan explained that the deviation from Mendelian ratio was because of linkage. Linkage is defined as the coexistence of two or more genes in the same chromosome that inherited together. If the genes are situated on the same chromosome and lie close to each other, then they are inherited together and are said to be linked genes. He also observed that some genes were more tightly linked than the others. He concluded that the occurrence of recombinants was further restricted in the case of the more tightly linked genes.

Question 20 ( 3.0 marks)Describe the termination process of transcription in bacteria.

Solution:Termination of transcription starts when the termination region is recognised. It is a particular sequence of nucleotide where the translation stops.

In bacteria, as soon as the polymerase enzyme encounters the terminator region, the nascent RNA so formed falls off. The polymerase enzyme associates transiently with termination factor (ρ) to terminate the transcription. Actually, the association with termination factor alters the specificity of RNA polymerase. In bacteria, the mRNA does not require any post transcriptional processing. Also, transcription and translation take place simultaneously in the same region of the

cell (i.e., translation begins before the mRNA is fully synthesised) in them.

Question 21 ( 3.0 marks)How does RNA interference help in developing resistance in tobacco plant against nematode infection?

Solution:RNA interference or RNAi is a method adopted to prevent infestation of roots of tobacco plants by a nematode Meloidogyne incognita. In this method, a complementary RNA binds to mRNA to form a ds RNA, which cannot translate. Hence, its expression is blocked. It is also known as gene silencing.

This complementary mRNA may come from either an infection by RNA viruses or through transposons (mobile genetic elements). These genes can be incorporated into the tobacco plant through vectors. The introduced DNA forms both sense and anti-sense RNA.

Two strands being complementary to each other bend and form ds RNA, leading to RNAi. Thus, the mRNA of nematode is silenced and the parasite cannot survive in the transgenic tobacco plant.

Question 22 ( 3.0 marks)Draw a longitudinal section of a post − pollinated Pistil showing entry of pollen tube into a mature embryo-sac. Label filiform apparatus, chalazal end, Hilum, antipodals, male gametes and secondary nucleus.

OR

Draw a labelled sectional view of seminiferous tubule of a human male.

Solution:

OR

Question 23 ( 3.0 marks)Explain the efforts which must be put in to improve health, hygiene and milk yield of cattle in a dairy farm.

Solution:Following efforts must be put in to improve the health, hygiene and milk yield of cattle in a dairy farm:

Development of high yielding varieties - The high yield of milk from the animals depends on the quality of breed selected. Quality encompasses yielding potential and disease resistance. Modern scientific procedures such as hybridization and genetic engineering can be used to develop better varieties.

Care of cattle - Care of cattle includes providing proper accommodation, adequate water, feeding in a scientific manner with special emphasis laid on the quality and quantity of the fodder provided.

Care of cattle health - Hygiene and regular visits by a veterinary doctor are must to improve the health of cattle. This help to identify and rectify the medical problems related to the cattle at early stages.

Question 24 ( 3.0 marks)Explain convergent and divergent evolution with the help of one example of each.

Solution:Convergent evolution

When more than one adaptive radiation occurs in an isolated geographical area, the phenomenon is called convergent evolution. The best example for convergent evolution is the case of Australian marsupials. A number of marsupial animals evolved from a single stock but they were restricted to the Australian continent.

Divergent evolution

The process of evolution starting from a single point and radiating in different directions is called adaptive radiation or divergent evolution. The best example for divergent evolution

is Darwin’s finches. During his exploration of the Galapagos Islands, he observed many varieties of finches in the same island. They varied from normal seed-eating varieties to those that ate insects. They all diverged from one original species of finch.

Question 25 ( 3.0 marks)Eco RI is used cut a segment of foreign DNA and that of a vector DNA to form a recombinant DNA. Show with the help of schematic diagrams.

(i)The set of palindronic nucleotide sequence of base pairs the Eco RI will recognise in both the DNA segments. Mark the site at which Eco RI will act and cut both the segments.

(ii)Sticky ends formed on both the segments where the two DNA segments will join later to form a recombinant DNA.

Solution:

The encircled small arrows, as shown in the figure, indicate the sites at which Eco R1 will act and cut the both segments.

Question 26 ( 3.0 marks)An antibody molecule is represented as H2L2. Explain.

Solution:Antibodies are specialized proteins with four peptide chains. Out of the four peptide chains, two are light chains and two are heavy chains. Hence, it is denoted as H2L2. The structure of antibody showing two light and heavy chains each can be represented as follows:

IgA, IgE, IgM and IgG are some examples of antibodies produced in the body.

Question 27 ( 3.0 marks)Identify a, b, c, d, e and f in the table given below:

Organism Bioactive molecule Use

1. Monascus perpureus (Yeast)

2. Antibiotic

3. Cyclosporin A

Solution: Organism Bioactive

moleculeUse

Monascus purpureus Statin Lowering of cholesterol

Penicillium notatum Penicillin Antibiotic

Trichoderma polysporum

Cyclosporin A Immuno-suppressive agent

SECTION D

Question 28 ( 5.0 marks)When and where are primary oocytes formed in a human female? Trace the development of these oocytes till ovulation (in menstrual cycle). How do gonadotropins influence this developmental process?

OR

(a)Explain the events taking place at the time of fertilization of an ovum in a human female.

(b)Trace the development of zygote upto its implantation in the uterus.

(c)Name and draw a labelled sectional view of the embryonic stage that gets implanted.

Solution:The primary oocytes are formed in the foetal ovary during the development of the foetus.

The ovum is formed by the process of oogenesis. During embryonic growth, millions of gamete mother cells (oogonia) are formed in the foetal ovary. These cells undergo meiosis, but get temporarily arrested at the prophase and are called primary oocytes.

Before reaching puberty, a large number of primary oocytes degenerate and the remaining ones get surrounded by layers of granulosa cells and a new theca. These are called secondary follicles. The secondary follicles are then converted into tertiary follicles that have characteristic fluid-filled cavity called antrum. At this stage, the primary oocyte present within the tertiary follicle completes meiosis, which results in the formation of haploid secondary oocyte and a tiny polar body. This tertiary follicle further changes into the Graafian follicle. The secondary oocyte in the Graffian follicle is surrounded by the zona pellucida. Then, the Graafian follicle ruptures to release the ovum by ovulation.

Role of gonadotropins:

The gonadotropins (Follicle Stimulating Hormone and the Luteinising Hormone) are released by the anterior portion of the pituitary gland. The maturation of the primary follicles into the Graffian follicles occurs during the follicular phase. The secretion of gonadotropins increases during this phase and cause follicular growth and the growing follicles produce oestrogen. The LH and FSH are at their peak in the middle of the cycle (14th day), and cause the rupture of the Graffian follicles to release ovum. This phase is called the ovulatory phase.

OR

(a) The process of fusion of the sperm and the ovum is known as fertilisation. During fertilisation, the sperm induces changes in the zona pellucida and blocks the entry of other sperms. This ensures that only one sperm fertilises an ovum. The enzymatic secretions of the acrosome help the sperm enter the cytoplasm of the ovum. This causes the completion of meiotic division of the secondary oocyte, resulting in the formation of a haploid ovum (ootid) and a secondary polar body. Then, the haploid sperm nucleus fuses with the haploid nucleus of the ovum to form a diploid zygote.

(b) The zygote formed after fertilization undergoes mitosis. Mitosis takes place in the isthmus of the oviduct or the fallopian tube. This process is known as cleavage. The cleavage results in the formation of 2, 4, 8, 16 daughter cells that are called blastomeres. The 8−16 blastomeres is called a morula, which continues to divide to form the blastocyst. The morula moves further into the uterus. The cells in the blastocyst are arranged into an outer trophoblastand an inner cell mass. The trophoblast gets attached to the uterine endometrium, and the process is called implantation. This leads to pregnancy.

(c) The stage of the embryo that gets implanted in the uterine wall is known as blastocyst. It can be represented as follows:

Blastocyst

Question 29 ( 5.0 marks)Draw and explain a logistic curve for a population of density (N) at time (t) whose intrinsic rate of natural increase is (r) and carrying capacity is (k).

OR

Describe the process of decomposition of detritus under the following heads: Fragmentation; leaching; catabolism; humification and mineralization.

Solution:The population tends to grow in a logistic manner, when the resources are limited leading to competition between individuals, and survival of the fittest occurs.

A logistic curve includes five phases, namely, lag phase, positive acceleration phase, exponential phase, negative acceleration phase and stationary phase. There is an initial lag phase followed by acceleration or deceleration phases and finally asymptote, when it reaches its carrying capacity (K).

When the relationship between N and t is plotted, a sigmoid curve called the Verhulst − Pearl logistic growth is obtained. It is determined by the given equation

Where:

N − Population density at time t

r − Intrinsic rate of natural increase

K − Carrying capacity

OR

Decomposition of detritus is the process of breakdown of complex organic matter into inorganic substances such as carbon dioxide, water, nutrients, etc. This process is carried out by decomposers such as earthworm and other microorganisms. Decomposers break down complex organic matter into inorganic matter. Raw materials for the process of

decomposition include detritus such as remains of dead plant, animal wastes such as, dead bodies and faecal matter, etc. Decomposition of detritus includes five steps explained as follows:

Fragmentation − It refers to the breaking down of detritus (dead plant and animal remains, faecal matter) into smaller particles by detritivores (decomposers).

Leaching − It is the process by which these inorganic matters enter the soil. The inorganic matter must be water soluble for getting leached.

Catabolism − The process by which detritus is degraded into simpler inorganic substances by bacterial and fungal enzymes is called catabolism.

Humification − Humification is the process of accumulation of humus in the soil. Humus is resistant to microbial action and decomposes at an extremely slow rate. It acts as a reservoir of nutrients. Humus enhances the quality of soil.

Mineralization − It refers to the process by which humus further degrades to release minerals into the soil. It is an oxygen-consuming process and is controlled by the chemical composition of detritus and climatic conditions.

Question 30 ( 5.0 marks)Write the symptoms of haemophilia and sickle-cell anaemia in humans.

Explain how the inheritance pattern of the two diseases differ from each other.

OR

(a) Write Hardy-Weinberg principle.

(b) Explain the three different ways the natural selection can affect the frequency of a heritable trait in a population shown in the graph given below.

Solution:Haemophilia is a sex-linked recessive disease that shows its transmission from unaffected carrier female to some of male progeny. The symptom of a haemophiliac patient includes non-stop bleeding even on a simple cut.

Sickle-cell anaemia is an autosome-linked recessive trait exhibiting change in shape of the red blood cells from biconcave disk to sickle shape under low oxygen tension.

It is characterised by low haemoglobin count and other symptoms of anaemia such as fatigue and irritability, etc.

The pattern of inheritance of haemophilia involves the passing on of haemophilia from the heterozygous female (carrier) to her sons. However in sickle-cell anaemia, the disease is controlled by a single pair of allele, HbA and HbS. Only homozygous individuals for haemophilia (HbS HbS) show the diseased phenotype. However, heterozygous (HbA HbS)

individuals appear apparently unaffected; they are the carriers of the disease.

OR

(a) Hardy-Weinberg principle states that the frequency of occurrence of alleles of a gene in a population remains constant through generations unless disturbances such as mutations, non-random mating, natural selection, etc. are introduced.

(b) Natural selection can affect the frequency of a heritable trait in a population in the following ways:

i. It can lead to stabilization (in which more individual acquire mean character value i.e. medium-sized individuals)

ii. It may result in directional change (more individuals acquire value other than the mean character value)

iii. It may result in disruption (more individuals acquire peripheral character value at both ends of the distribution curve)

The effect of natural selection on the frequency of a heritable trait in a population can be represented schematically as follows: