Class XII - Biology - Sample Paper - 3

Kindly share you feedback about the website – Click here

Biology (044)

Class XII Session 2023-24

Time: 3 Hours Max. Marks: 70

General Instructions:

1. All questions are compulsory.

2. The question paper has five sections and 33 questions. All questions are compulsory.

3. Section—A has 16 questions of 1 mark each;

Section—B has 5 questions of 2 marks each;

Section—C has 7 questions of 3 marks each:

Section—D has 2 case-based questions of 4 marks each; and

Section—E has 3 questions of 5 marks each.

4. There is no overall choice. However, internal choices have been provided in some questions. A student has to attempt only one of the alternatives in such questions.

5. Wherever necessary, neat and properly labeled diagrams should be drawn.

SECTION – A

1. Level of which hormones get elevated by the intake of nicotine?

(a) FSH, LH

(b) Thyroxine, progesterone

(d) Adrenaline, nor-adrenaline

Ans. (d) Adrenaline, nor-adrenaline

2. From bacteria to human nearly universal code for phenylalanine is

(a) UUU

(b) UUA

(d) CUU

Ans. (a) UUU

3. In a population of 500 zebras, average natality is 25, average mortality is 20, immigration is 30 and emigration is 35. What will be the population at the end of 10 years?

(a) 550

(b) 600

(d) 500

Ans. (d) 500

N_{t+ 1} = N_t+ [(B + I) – (D + E)]

500 + [(25 + 30) – (20 + 35)]

500 + [35 – 35] = 500

4. An urn shaped population age pyramid represents

(a) growing population

(b) static population

(d) extinct population

Ans. (c) declining population

In the urn shaped age pyramid the proportion of reproductive age group is higher than the individuals in pre-reproductive age group. It is declining or diminishing population with negative growth.

5. Select the option that correctly identifies A, B and C in the given table.

(a) A-Top carnivore, B-Detritus, C-Secondary consumer

(b) A-Top carnivore, B-Detritus, C-Primary consumer

(d) A-Scavanger, B-Grazing, C-Producer

Ans. (a) A-Top carnivore, B-Detritus, C-Secondary consumer

6. The main reason why antibiotics could not always treat the bacteria-mediated diseases is

(a) insensitivity of the individual following prolonged exposure to antibiotics

(b) inactivation of antibiotics by bacterial enzymes

(d) the development of mutant bacterial strains resistant to antibiotics.

Ans. (d) the development of mutant bacterial strains resistant to antibiotics.

In response to antibiotics, bacteria develop mutant strains that become resistant to the antibiotics. Thus, these antibiotics become incapable against bacterial medicated diseases.

7. Match column I with column II and select the correct option from the given codes.

(a) A-(ii), B-(iv), C-(iii), D-(i)

(b) A-(iv), B-(iii), C-(ii), D-(i)

(d) A-(iii), B-(iv), C-(i), D-(ii)

Ans. (d) A-(iii), B-(iv), C-(i), D-(ii)

8. Having become an expert on gel electrophoresis, you are asked to examine a gel. Where would you find the smallest segments of DNA?

(a) Near the positive electrode, farthest away from the wells

(b) Near the negative electrode, close to the wells

(d) Near the middle, they tend to slow down after the first few minutes

Ans. (a) Near the positive electrode, farthest away from the wells

Since DNA is itself negatively charged, it would move towards the positive electrode. In gel electrophoresis, DNA fragments are separated on the basis of charge and masses. Thus, smaller the DNA fragment farther it moves from the wall

9. Which among the following birth control measures is considered to be highly effective?

(a) The rhythm method

(b) The use of physical barriers

(d) Sterilisation techniques

Ans. (d) Sterilisation techniques

Sterilisation or surgical intervention blocks gamete transfer and there prevent conception. These techniques are highly effective.

10. Hugo de Vries gave his mutation theory on organic evolution while working on

(a) Pisum sativum

(b) Drosophila melanogaster

(d) Althea rosea

Ans. (c) Oenothera lamarckiana

Hugo de Vries stated that mutations are heritable and persist in successive generations. He conducted his experiments on Oenothera lamarckiana (evening primrose)

11. The most important human activity, leading to the extinction of wildlife, is

(a) pollution of air and water

(b) hunting for valuable wildlife products

(d) alteration and destruction of the natural habitats.

Ans. (d) alteration and destruction of the natural habitats.

12. Match column I with column II and select the correct option from the codes given below.

(a) A-(iii), B-(ii), C-(iv), D-(i)

(b) A-(i), B-(iii), C-(ii), D-(iv)

(d) A-(i), B-(ii), C-(iii), D-(iv)

Ans. (b) A-(i), B-(iii), C-(ii), D-(iv)

13. Assertion : Infundibulum is a funnel-shaped part closer to ovary.

Reason : The edges of infundibulum helps in collection of the ovum after ovulation.

(a) Both A and R are true and R is the correct explanation of A.

(b) Both A and R are true and R is not the correct explanation of A.

(d) A is false but R is true.

Ans. (b) Both A and R are true and R is not the correct explanation of A.

In human females each fallopian tube extends from the periphery of each ovary to the uterus, the part closer to the ovary to the uterus, the part closer to the ovary is the funnel shaped infundibulum. The edges of the infundibulum possess finger-like projections called fimbriae, which help in collection of the ovum after ovulation.

14. Assertion : Agrobacterium tumefaciens is called natural genetic engineer.

Reason : Agrobacterium tumefaciens infects all broad-leaved agricultural crops but does not infect cereal crops.

(a) Both A and R are true and R is the correct explanation of A.

(b) Both A and R are true and R is not the correct explanation of A.

(d) A is false but R is true.

Ans. (b) Both A and R are true and R is not the correct explanation of A.

Ti plasmid (tumor inducing) from Agrobacterium tumefaciens is effectively used as a verctor for gene transfer to plant cells. In nature, it induces tumors in broad leaf crops, e.g., tomato, soybean, cotton, etc.

15. Age sex structure of a population can be depicted in the form of a pyramid by plotting the percentage of population of each sex in each age class. Study this pyramid and comment upon the appropriateness of the Assertion and the Reason.

Assertion : It is a bell-shaped age pyramid.

Reason : In a stable population, proportion of individuals in reproductive age group is higher than the individuals in pre-reproductive age group.

(a) Both A and R are true and R is the correct explanation of A.

(b) Both A and R are true and R is not the correct explanation of A.

(d) A is false but R is true.

Ans. (c) A is true but R is false.

In a bell shaped age pyramid, the number of pre-reproductive and reproductive individuals is almost equal. Post-reproductive individuals are comparatively fewer. It represents a stable population.

16. Assertion : In eukaryotes, replication and transcription occur in the nucleus but translation occurs in the cytoplasm.

Reason : mRNA is transferred from the nucleus to the cytoplasm where ribosomes and amino acids are available for protein synthesis.

(a) Both A and R are true and R is the correct explanation of A.

(b) Both A and R are true and R is not the correct explanation of A.

(d) A is false but R is true.

Ans. (a) Both A and R are true and R is the correct explanation of A.

SECTION – B

17. A student on a school trip started sneezing and wheezing soon after reaching the hill station for no explained reasons. But, on return to the plains, the symptoms disappeared. What is such a response called? How does the body produce it?

Ans. Hill station and plains do have different weather conditions and environment. Sneezing and wheezing on hill station is due to exposure to different allergens, this response is called allergy. It is a hypersensitive response of a person to foreign substance coming in contact with or entering the body. Allergy involves IgE antibodies and release of chemicals like histamine and serotonin from mast cells.

18. Study the given diagram.

A is an embryonic stage that gets transformed into B, which in turn gets implanted in the endometrium in human females.

(a) Identify A, B and its parts C and D.

Ans. Morula is shown in figure A and blastocyst is shown in the figure B. ‘C’ is inner cell mass and ‘D’ is trophoblast.

(b) State the fate of C and D in the course of embryonic development in humans.

Ans. Inner cell mass (C) forms the fetus. The cells of trophoblast (D) helps to provide nutrition to the embryo. The cells of trophoblast later form the extra embryonic membranes.

19. In a typical monohybrid cross the F2 population ratio is written as 3 : 1 for phenotype but expressed as 1 : 2 : 1 for genotype. Discuss with the help of an example.

Ans. Monohybrid cross is the cross that considers only one trait, e.g., cross of a yellow seeded pea plant with green seeded pea plant. If pea plant with yellow seed coat is crossed with pea plant having green seed coat, then in the F₁ generation all the plants will produce yellow seeds. On selfing F₂ generation will be as shown below:

20. Draw the vector DNA and a foreign DNA showing the sites where EcoRI has acted to form the sticky ends.

Ans.

21. The given figure shows the different types of age pyramids for human population

(a) What does the parts ‘X, ‘Y’ and ‘Z’ represent?

Ans. X, Y and Z represent different age groups (male and female) found in a population. Here X is post-reproductive, Y is reproductive group and Z is pre-reproductive group.

(b) Which type of population is represented by pyramids A, B and C? Explain.

Ans. Pyramid A is triangular age pyramid which represents expanding population. Here, number of pre-reproductive individuals is very large, number of reproductive individuals is moderate and post- reproductive individuals are fewer.

B is bell-shaped age pyramid that represents stable population because number of pre-reproductive and reproductive individuals is almost equal but post-reproductive individuals are fewer.

C is urn-shaped age pyramid representing declining population with higher proportion of reproductive age group than pre and post- reproductive age group.

Why the pyramid of energy is always upright? Explain.

Ans. The pyramid of energy is always upright in shape as there is always a gradual decrease in the energy content at successive trophic levels from producers to various consumers. This is because some energy is used at each trophic level for various metabolic activities and some energy is lost as heat, so only 10% of the energy is available to the next trophic level (Lindeman’s 10% law).

SECTION – C

22. The sacred groves of Aravalli Hills and Ooty botanical garden both aim at biodiversity conservation. How do they differ in their approaches? Explain.

Ans. Sacred groves are undisturbed forest patches, surrounded by highly degraded landscapes where not even a single branch of tree is allowed to be cut. As a result, many endemic species which are rare or have become extinct, seen to flourish here. While botanical gardens are areas where many species of plants, are conserved outside their natural habitats. They help to restore endangered species; whose chances of survival are very small. Thus, sacred groves of Aravalli Hills are sites of in-situ conservation, where endangered species are protected in their natural habitat whereas Ooty Botanical gardens are sites of ex-situ conservation, where the endangered species are protected outside their natural habitats.

23. Write the mode of pollination in Vallisneria and water lily. Explain the mechanism of pollination in Vallisneria.

Ans. The mode of pollination in Vallisneria is water whereas mode of pollination in water lily is wind or insects. Vallisneria is a submerged dioecious aquatic plant in which pollination takes place by the agency of water. The mechanism of pollination in Vallisneria as follows: (i) The male plants produce a large number of male flowers. The male flowers abscise and rise to the surface where they float.

(ii) The male flowers have two fertile stamens. Two of their tepals form a boat-shaped structure while the third one functions as a sail.

(iii) The female plants bear long stalked solitary pistillate flowers. The mature female flowers are brought to the surface of water by the elongation of their stalks. They have large sticky trifid stigmas.

(iv) While floating, the male flowers are drawn in the depression surrounding each female flower. One anther of a male or staminate flower comes in contact with the stigma of the female flower. The anther bursts and pollination is performed.

(v) Pollen grains are covered by mucilage which helps them in sticking to stigma as well as protects them from wetting by water.

(vi) After pollination, the female flower is pulled inside water by the coiling of its stalk.

24. Answer the following based on the dinucleotide shown below.

(a) Name the type of sugar to which guanine base is attached.

Ans. Guanine base is attached to a pentose sugar.

(b) Name the linkage connecting the two nucleotides.

Ans. Two nucleotides are linked through 3′-5′ phosphodiester linkage to form a dinucleotide.

Ans. ‘B’ is the 3′ end of the given dinucleotide. It is because 3′ end of a nucleotide molecule bears a free 3′ OH group of sugar residue and the 5′ end bears phosphate radical (5′ – P).

25. Briefly explain the implantation in an adult human female.

Ans. Implantation is the attachment of blastocyst to the uterine wall. It occurs after 7 days of fertilisation. As zygote moves towards the uterus, it undergoes series of mitotic divisions known as cleavage and forms 2,4,8,16 daughter cells called blastomeres. The embryo with 8 blastomeres is called morula. The morula transforms into blastocyst. In a blastocyst, the blastomeres are arranged into an outer layer called trophoblast and an inner group of cells called the inner cell mass. The trophoblast then gets attached to the endometrium and the inner cell mass gets differentiated as the embryo. After attachment the uterine cells divide rapidly and cover the blastocyst. As a result, the blastocyst becomes embedded in the endometrium of the uterus. This whole phenomenon is called implantation and it leads to pregnancy.

26. Refer to the given figure showing the variety of beaks of finches that Darwin found in Galapagos Island. Refer this figure and answer the following questions.

(a) What does the above figure depict?

Ans. Darwin finches show variations in the shape of their beaks. During the process of evolution, they have evolved from seed eating finches to vegetarian insectivorous finches, etc.

(b) How did Darwin explain the existence of different varieties of finches on Galapagos Islands?

Ans. Process of evolution of different species in a given geographical area starting from a point and literally radiating to other areas of geography (habitats) is called adaptive radiation. Darwin’s finches represent one of the best example of this phenomenon, as many varieties of finches were observed to be present in the same island. All the varieties, evolved on the island itself from the original seed-eating finches that lead to various modifications in the finches according to their food habits. This evolution in finches enabled the birds to become insectivorous, vegetarian, wood pecking, ground feeding, etc.

27. (a) Name the organism in which the vector shown is inserted to get the copies of the desired gene.

Ans. The vector shown is inserted into Escherichia coli to get the copies of the desired gene.

(b) Mention the area labelled in the vector responsible for controlling the copy number of the inserted gene.

Ans. Origin of replication (ori) is a sequence from where replication starts and is also responsible for controlling the copy number of the linked DNA.

Ans. The sequence amp and tet are considered useful selectable markers. The given vector shows ampR selectable marker. Vector requires a selectable marker, because it helps in identifying and eliminating non-transformants and selectively permitting the growth of the transformants.

28. Identify A, B and C in the schematic diagram of an antibody given above and answer the questions.

Identify A, B and C in the schematic diagram of an antibody given above and answer the questions.

(a) Write the chemical nature of an antibody.

(b) Name the cells that produce antibodies in humans.

Ans. In the given structure of an antibody molecule, ‘A’ is the antigen binding site, ‘B’ is constant region of light chain and ‘C’ is constant region of heavy chain.

(a) Antibodies are proteinaceous in nature.

(b) B-cells produce antibodies.

(c) Humoral immune response is an antibody mediated immune response.

(a) Why do the symptoms of malaria not appear immediately after the entry of sporozoites into the human body when bitten by female Anopheles? Explain.

Ans. The malarial parasite, Plasmodium enters the human body as sporozoites (infectious form) through the bite of infected female Anopheles mosquito (vector). From the human blood sporozoites enter liver cells and multiply here and then attack the red blood cells (RBCs) resulting in their rupture. The rupture of RBCs is associated with release of a toxic substance, haemozoin, which is responsible for the chill and high fever recurring every three to four days. The released parasites from the ruptured RBCs infect new RBCs and develop into gametocytes (male and female). When a female Anopheles mosquito sucks the blood of an infected human host, it receives RBCs containing gametocytes.

(b) Give the scientific name of the malarial parasite that causes malignant malaria in humans.

Ans. Plasmodium falciparum is the causative agent of the malignant malaria.

SECTION – D

DIRECTION : Question No. 29 and 30 are case based questions. Each question has subparts with internal choice in one subpart.

29. Biogas generation is done on a large scale in rural India. The given diagram shows a typical biogas plant.

(a) Identify A, B and C.

Ans. In the given diagram, ‘A’ represents sludge, ‘B’ represents gas holder and ‘C’ represents dung and water.

(b) Name the bacteria involved in the production of biogas.

Ans. Methanogens are used in generation of biogas plant, e.g., Methanococcus, Methanobacterium.

Ans. Methanogens are present in the rumen (a part of stomach of cattle) such as cow, buffalo. In rumen, these bacteria help in the breakdown of cellulose and play an important role in the nutrition of cattle. The excreta of cattle is rich in these bacteria (methanogens) and can be used for generation of biogas.

(d) How biogas is generated from activated sludge?

Ans. A small part of the activated sludge is pumped back into the aeration tank to serve as the inoculum. The remaining major part of the activated sludge is pumped into large tanks called anaerobic sludge digesters. The anaerobic bacteria digest the bacteria and fungi in a sludge and produce mixture of gases like methane, hydrogen sulphide and CO, which constitute biogas.

30. In 1958, Matthew Meselson and Franklin Stahl provided a strong experimental evidence to prove the template mechanism of DNA replication given by Watson and Crick. They used heavy (15N) and light (14N) isotopes of nitrogen to differentiate between parental and newly synthesised DNA strands. The experiment and results are diagrammatically shown below.

(a) What does the given experiment show?

Ans. The shown DNA replication is said to be semi-conservative as after the completion of replication, each DNA molecule would have one parental and one newly synthesized strand.

(b) Which bacteria are used in the above experiment?

Ans. E.coli bacteria cells are used in the given experiments.

Ans. According to the shown demonstration, after two generation, we will obtain 75% light (¹⁴N¹⁴N) DNA strands and 25% hybrid (¹⁵N¹⁵N) DNA strands. So, the ratio of ¹⁵N : ¹⁴N DNA strands will be 1 : 3

(d) If E. coli is allowed to grow in N15 medium for 20 minutes and in N14 medium for next 40 minutes, then what will be the number of hybrid and light double stranded DNA molecules respectively?

Ans. If E.coli culture is allowed to grow in 15N medium for 20 minutes and in 14N medium for next 40 minutes, then there would be 6 light dsDNA molecules and 2 hybrid dsDNA molecules.

SECTION – E

31. If a desired gene is identified in an organism for some experiments, explain the process of the following :

(a) Cutting of desired gene at specific locations.

Ans. Ans. Desirable DNA sequences are cut by the use of enzyme restriction endonuclease. The restriction enzymes cut the strand of DNA a little away from the centre of the palindromic sites, between the same two bases on the opposite strands, it leaves single stranded portions at the ends. This forms overhanging stretches called sticky ends on each strand. They are called sticky as they form hydrogen bonds with their complementary cut counterparts. The stickiness of the ends facilitates the action of the enzyme DNA ligase.

(b) Synthesis of multiple copies of the desired gene.

Ans. (b) Multiple copies of desirable gene is synthesised by using polymerase chain reaction (PCR)

The three steps involved in each cycle of PCR are :

(i) Denaturation

(ii) Annealing

(iii) Extension

PCR is based on the principle that a DNA molecule, when subjected to high temperature, splits into two strands due to denaturation. These single stranded DNA molecules are then converted to original double stranded molecules, in the presence of enzyme DNA polymerase. A double stranded molecule of DNA is duplicated in this way and multiple copies of original DNA sequence can be generated by repeating the process several times. Such repeated amplification is achieved by the use of thermostable DNA polymerase (isolated from Thermus aquaticus), which remain active during the high temperature induced denaturation of double stranded DNA.

GM crops especially Bt crops are known to have higher resistance to pest attacks. To substantiate this an experimental study was conducted in 4 different farmlands growing Bt and non Bt-Cotton crops. The farm lands had the same dimensions, fertility and were under similar climatic conditions. The histogram below shows the usage of pesticides on Bt crops and non-Bt crops in these farm lands.

(a) Which of the above 4 farm lands has successfully applied the concepts of biotechnology to show better management practices and use of agrochemicals? If you had to cultivate, which crop would you prefer (Bt or Non- Bt) and why?

Ans. Farm land II shows better management practices and use of agro chemicals. If I had to cultivate, I would personally prefer Bt crop, because the use of pesticides is highly reduced for Bt crops.

(b) Cotton bollworms were introduced in another experimental study on the above farm lands wherein no pesticide was used. Explain what effect would a Bt and Non-Bt crop have on the pest.

Ans. In Bt cotton a cry gene has been introduced from bacterium Bacillus thuringiensis (Bt) which causes synthesis of a toxic protein. This protein becomes active in the alkaline gut of bollworm feeding on cotton, punching holes in the lining causing death of the insect.

However; a non Bt crop will have no effect on the cotton bollworm and the yield of cotton will decrease as non Bt will succumb to pest attack.

32. Describe the post-zygotic events leading to implantation and placenta formation in humans. Mention any two functions of placenta.

Ans. After the formation of zygote, a rapid series of mitotic divisions (called cleavage) of the zygote occurs as it moves through the isthmus of the oviduct towards uterus. The new daughter cells formed after division are called blastomeres. The embryo with solid mass of 8-16 blastomeres is called a morula. The morula continues to divide further and transforms into blastocyst and reaches into the uterus. The blastomeres in the blastocyst get arranged into an outer layer called trophoblast and an inner group of cells attached to trophoblast called the inner cell mass. The trophoblast layer then gets attached to the endometrium and the inner cell mass gets differentiated as the embryo. After attachment, the uterine cells divide rapidly and covers the blastocyst. As a result, the blastocyst becomes embedded in the endometrium of the uterus. This is called implantation and it leads to pregnancy.

The outer surface of the chorion (fetal membrane) develops a number of finger like projections, called chorionic villi, which grow into the tissue of the uterus. These villi penetrate the tissues of the uterine wall in which they are embedded, to make up the organ known as placenta.

Two functions of placenta are as follows:

(i) The placenta facilitate the supply of oxygen

and nutrients to the embryo and also removal

of carbon dioxide and excretory/waste materials produced by the embryo.

(ii) Placenta also acts as an endocrine tissue and produces several hormones like human chorionic gonadotropin (hCG), human placental lactogen (hPL), oestrogens, progesterone, etc. for the maintenance of pregnancy.

(a) Draw a diagram of an enlarged view of T.S. of one microsporangium of an angiosperm and label the following parts:

(i) Tapetum

(ii) Middle layer

(iii) Endothecium

(iv) Microspore mother cells

Ans. Sectional view of microsporangium of an angiosperm is as follow:

(b) Mention the function of tapetum.

Ans. Role of tapetum in an anther is as follows:

(i) Nourishment of the developing microspore mother cells and pollen grains.

(ii) It produces lipid rich Ubisch granules containing sporopollenin for exine formation, pollenkitt in case of entomophilous plants, special proteins for the pollen grains to recognise compatibility and hormone IAA.

(iii) It secretes enzyme callase responsible for the degradation of callose wall around pollen tetrad.

(i) Pollen grains are well preserved as fossils.

Ans. Sporopollenin is present in exine layer of pollen grains. Sporopollenin is highly resistant fatty substance not degraded by any enzyme and not affected by high temperature, strong acid or strong alkali therefore pollen grains can be well preserved as microfossils.

(ii) Pollen tablets are in use by people these days.

Ans. Pollen tablets are used as food supplement by people to improve health.

33. Illustrated below is a DNA segment, which constitute a gene

(i) Will the whole gene be transcribed into RNA primarily? State ‘Yes’ or ‘No’.

Ans. Yes

(ii) Name the shaded and unshaded parts of the gene.

Ans. Shaded regions are called introns while the unshaded regions are called exons.

(iii) Explain how these genes are expressed.

Ans. The primary RNA contains both introns and exons, by splicing mechanism introns are removed and exons are joined to frame functional mRNA after capping and tailing.

(iv) How is this gene different from prokaryotic gene in its expression?

Ans. In prokaryotic genes, introns are not found.

Study the schematic representation of the genes involved in the lac operon given below and answer the questions that follow:

(a) Identify and name the regulatory gene in this operon. Explain its role in ‘switching off’ the operon.

Ans. The given schematic representation is of lac operon. In lac operon, the regulatory gene is called i-gene because it produces an inhibitor or repressor. The repressor binds to operator gene and stops the operator from working.

In the absence of an inducer (i.e., lactose), the repressor binds to the operator gene making it non-functional. RNA polymerase enzyme cannot move over it to reach the structural genes. Thus, structural genes are inactivated and transcription cannot take place.

(b) Why is the lac operon’s regulation referred to as negative regulation?

Ans. As regulatory gene exerts a negative control over the working of structural genes, therefore regulation of lac operon is called negative regulation.

(c) Name the inducer molecule and the products of the genes ‘z’ and ‘y’ of the operon. Write the functions of these gene products.

Ans. ‘Inducer’ for the given operon is ‘lactose’. Its role is to bind with repressor, change the latter into non-DNA binding state so as to free the operator gene and switch on the lac operon.

The products of the genes ‘2’ and ‘y’ of the lac operon are B-galactosidase and permease respectively. B-galactosidase brings about hydrolysis of lactose to form glucose and galactose. Galactoside permease is required for entry of lactose into the bacterium.

Class XII – Biology – Sample Paper – 3