Class XII - Biology - Sample Paper - 2

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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. The structure in chromatin seen as ‘beads-on string’ when viewed under electron microscope are called

(a) nucleotides

(b) nucleosides

(d) nucleosomes

Ans. (d) nucleosomes

Nucleosomes constitute the repeating unit of a structure in nucleus called chromatin, thread like stained (coloured) bodies seen in nucleus. The nucleosomes in chromatin are seen as ‘beads-on-string’ structure when viewed under electron microscope.

2. For which of the following cases, population density can be easily determined by not utilising biological-entities directly?

(a) Fish density

(b) Density of bacteria in bacterial culture

(d) Tiger census

Ans. (d) Tiger census

The tiger census in our National parks and tiger reserves is often based on pug marks (animal’s foot print) and faecal pellets.

3. Identify the palindromic sequence in the following.

(a) GAATTC/CTTUUG

(b) GGATCC/CCTAGG

(d) CGATA/GCTAA

Ans. (b) GGATCC/CCTAGG

The palindromes in DNA are base pair sequences that are the same when read forward (left to right) or backward (right to left) from a central axis of symmetry.

4. Identify the incorrect pair from the following with respect to angiosperms.

(a) Primary endosperm nucleus-3n

(b) Antipodals-2n

(d) Vegetative cell of male gametophyte-n

Ans. (b) Antipodals-2n

All the cells of the embryo sac are haploid except central cell which is first binucleate and then becomes diploid due to fusion of polar nuclei. The three cells on the chalazal end of the embryo sac are called antipodal cells. They are the vegetative cells of the embryo sac which may degenerate later.

5. Biochemical oxygen demand (BOD) in a river water

(a) has no relationship with concentration of oxygen in the water

(b) gives a measure of Salmonella in the water

(d) remains unchanged when algal bloom occurs

Ans. (c) increases when sewage added to river water

Discharge of domestic sewage into a river will result in the side of BOD because decomposer organisms consume a lot of oxygen.

6. In a given population of 2000 individuals, 80 births and 125 deaths were reported over a given period of time. Which of the following graphs will correspond to it?

Ans. (a) Number of deaths is more than the number of births, showing a declining population.

7. A plant native to South America, which produces cocaine is

(a) Erythroxylum coca

(b) Atropa belladonna

(d) Papaver somniferum

Ans. (a) Erythroxylum coca

8. Match column I with column II.

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

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

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

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

9. Replacement of the lighter-coloured variety of peppered moth (Biston betularia) to its darker variety (Biston carbonaria) in England is the example of

(a) natural selection

(b) regeneration

(d) temporal isolation

Ans. (a) natural selection

10. The inoculum is added to the fresh milk in order to convert milk into curd, the term ‘inoculum’ here refers to

(a) a starter rich in vitamin B12

(b) a starter rich in proteins

(d) an aerobic digester

Ans. (c) a starter containing millions of LAB

11. Given figure represents a pyramid of biomass in an aquatic ecosystem.

Identify A and B and select the correct answer.

(i) A is the crop which supports and B is the crop which is supported.

(ii) A is the crop which is supported and B is the crop which supports.

(iii) A is phytoplanktons and B is zooplanktons.

(iv) A is zooplanktons and B is phytoplanktons.

(a) (i) and (iv)

(b) (ii) and (iii)

(d) (ii) and (iv)

Ans. (c) (i) and (iii)

The given pyramid represents pyramid of biomass in an aquatic ecosystem. In this pyramid, biomass of phytoplanktons, i.e., producers (represented by A) is smaller than that of zooplanktons, i.e., primary consumers (represented by B). Phytoplanktons support zooplanktons as the latter feed upon them.

12. During insertional inactivation, the presence of a chromogenic substrate gives blue coloured colonies if the plasmid in the bacteria does not have an insert. The blue colour is produced by the enzyme

(a) α -glucosidase

(b) restriction endonuclease

(d) Taq polymerase

Ans. (c) β –galactosidase

Alternative selectable markers have been developed which differentiate recombinants from the non- recombinants on the basis of their ability to produce colour in the presence of a chromogenic substrate. In this, a recombinant DNA is inserted within the coding sequence of an enzyme, B-galactosidase. This results into inactivation of the enzyme, which is referred to as insertional inactivation. The presence of a chromogenic substrate gives blue coloured colonics if the plasmid in the bacteria does not have an insert. Presence of insert results into insertional inactivation of the -galactosidase and the colonies do not produce any colour, these are identified as recombinant colonies.

13. Assertion : The plant biomass which serves as the food of herbivores and decomposers is said to result from the net primary productivity.

Reason : Gross primary productivity is the rate of total production of organic material (biomass) during photosynthesis.

(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.

The rate of biomass production is called productivity which is of two kinds-primary and secondary. Primary productivity refers to the rate at which sunlight is captured by producers for the synthesis of energy rich organic compounds through photosynthesis. Productivity is a rate function, and is expressed in terms of dry matter produced or energy captured per unit area of land, per unit time. Primary productivity has two aspects, gross and net. The rate of total capture of energy, or the rate of total production of organic material (biomass) is known as gross primary productivity. However, while the energy capture process is operating in the green tissues, these as well as other tissues, are consuming energy in respiration. The balance energy or biomass remaining after meeting the cost of respiration of producers, is called net primary productivity, as shown below:

Net productivity = Gross productivity – Respiration rate

At the trophic level of consumers, the rate at which food energy is assimilated, is called secondary productivity.

14. Assertion : In a monohybrid cross, F1 generations indicate dominant characters.

Reason : Dominance occurs only in heterozygous state.

(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.

Monohybrid cross is a cross between two organisms of a species which is made to study the inheritance of a single pair of alleles or factors of a character. Dominant character is one of a pair of alleles which can express itself whether present in homozygous or heterozygous state. In F, generation, (the generation of hybrids produced from a cross between the genetically different homozygous individuals called parents) the progenies are heterozygous dominant.

15. Assertion : Many endemic species are seen to flourish in sacred forests.

Reason : Sacred forests are undisturbed forest patches and biodiversity rich areas.

(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.

Sacred forests are forest patches around places of worship which are held in esteem by tribal communities. These are the most undisturbed forest patches. As a result many endemic species can be seen to flourish here.

16. Assertion : The primary productivity of different ecosystems can be easily compared.

Reason : The magnitude of primary productivity depends on the photosynthetic capacity of producers and the prevailing environmental conditions.

(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.

Primary productivity refers to the rate at which sunlight is captured by producers for the synthesis of energy – rich organic compounds through photosynthesis. Productivity is a rate function, and is expressed in terms of dry matter produced or energy captured per unit area of land, per unit time. It is generally expressed in terms of gm^-2 year^-1 Hence, the productivity of different ecosystems can be easily compared.

The magnitude of primary productivity depends on the photosynthetic capacity of producers and the prevailing environmental conditions, particularly solar radiation, temperature and soil moisture.

SECTION – B

17. What could be the possible treatments for a patient exhibiting ADA deficiency?

Ans. The possible treatments that can be given to a patient exhibiting adenosine deaminase (ADA) deficiency are:

(i) bone marrow transplantation

(ii) enzyme replacement therapy.

18. Where is Sporopollenin present in plants? State its significance with reference to its chemical nature.

Ans. Sporopollenin is present in exine layer of pollen grains. Sporopollenin is highly resistant fatty substance which is 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.

19. Refer to the given below figure.

(a) Redraw the structure as a replicating fork and label the parts.

Ans.

(b) Write the source of energy for this replication.

Ans. the sources of energy for the replication of DNA are phosphorylated nucleotides or deoxyribonucleoside triphosphates i.e., dATP, dCTP, dGTP and dTTP.

20. Name the genus of baculovirus that acts as a biological control agent in spite of being a pathogen. Justify by giving three reasons that make it an excellent candidate for the job.

Ans. Nucleopolyhedrovirus, a genus of baculovirus that act as a biological control agent in spite of being a pathogen.

(i) They are species specific.

(ii) They are narrow spectrum bioinsecticides.

(iii) There is no side effect on plants, mammals, birds/ fish and non-targent insects. Beneficial insects are conserved.

21.

The above graph show species-area relationship. Write the equation of the curve ‘a’ and explain it.

Ans. The equation of curve ‘a’ is S = CA where,

S = Species richness

C = Y – intercept

A = Area

Z = Slope of the line (regression coefficient)

The graph of species – area relationship shows that within a region, species richness increases with increasing explorable area, but only upto a certain limit. The relation between species richness and area for a wide variety of taxa turns out to be rectangular hyperbola.

How does over-exploitation of beneficial species affect biodiversity? Explain with the help of one example.

Ans. Excessive exploitation of a species, whether a plant or animal, reduces size of its population so that it becomes vulnerable to extinction. For example, presently many marine fish populations around the world are declining due to over harvesting results in endangering the continued existence of some commercially important species.

SECTION – C

22. Although a prokaryotic cell has no defined nucleus, yet DNA is not scattered throughout the cell. Explain.

Ans. In prokaryotes, DNA lies in the cytoplasm which is supercoiled (coiled and recoiled) with the help of RNAs and non-histone basic proteins like polyamines. DNA being negatively charged is held in place with the help of these proteins that have positive charges in a region termed as nucleoid. The DNA in nucleoid is organised in large loops held by proteins.

23. A cross was carried out between two pea plants showing the contrasting traits of height of the plants. The result of the cross showed 50% parental characters.

(a) Work out the cross with the help of a Punnett square.

(b) Name the type of the cross carried out.

Ans. Two contrasting characters of height are tall and dwarf. In the given cross, if 50% of the progeny shows parental characters, then it must be a cross between a heterozygous tall and a homozygous recessive dwarf parent.

(b) This type of cross is known as test cross.

24. Prior to a sports event, blood and urine samples of sports persons are collected for drug tests.

(a) Why is there a need to conduct such tests?

Ans. It is necessary to conduct such tests on sportspersons because they take various drugs like cocaine/coca alkaloids and cannabinoids to increase their muscle tone and performance in sports.

(b) Name the drugs the authorities usually look for.

Ans. Cocaine/coca alkaloids and cannabinoids.

Ans. Cannabinoids are obtained from Cannabis sativa and cocaine is obtained from Erythroxylum coca.

25. Why is predation required in a community of different organisms?

Ans. Predators plays an important role in a community :

(i) They act as conduits for energy transfer across trophic levels.

(ii) Predators keep prey population under control. They are used for biological control of weeds and pests.

(iii) Predators help in maintaining species diversity.

(iv) They help in growth of vegetation by controlling

(a) Explain “birth rate” in a population by taking a suitable example.

Ans. Birth rate refers to per capita births, i.e., average number of individuals produced per unit time. For example, if in a pond there were 20 lotus plants last year and through reproduction 8 new plants are added, then taking the current population to 28, we calculate the birth rate as 8/20 = 0.4 offspring per lotus per year.

(b) Write the other two characteristics which only a population shows but an individual cannot.

Ans. Other attributes of population which individuals cannot show include –

(i) Death rate – An individual dies but a population has death rate. It refers to per capita deaths, i.e., average number of individuals that die per unit time.

(ii) Sex ratio – An individual has sex but a population has sex ratio, i.e., number of females and males per 1000 individuals.

26. Study the transverse section of human ovary given below and answer the questions that follow.

(a) Name the hormone that helps in growth of A — >B — > C

Ans. In the given figure A is primary follicle, B is tertiary follicle showing antrum and C is Graafian follicle.

Anterior lobe of pituitary gland secretes LH and FSH. FSH stimulates the growth of ovarian follicles i.e., from A –> B — > C

(b) Name the hormone secreted by A and B.

Ans. Hormone secreted by A and B is estrogen.

Ans. D in the given figure is corpus luteum. It secretes progesterone which helps in the maintenance of endometrium.

27. ‘Plasmid is a boon to biotechnology’. Justify this statement quoting the production of human insulin as an example.

Ans. Plasmids are extra-chromosomal, self-replicating, usually circular, double stranded, DNA molecules found naturally in many bacteria.

Plasmid is a boon to biotechnology. It has certain characteristics which make it a good vector in production of human insulin. These are discussed as follows:

(i) It has specific restriction sites where the enzyme

restriction endonucleases make a cut and segment

of DNA which codes for human insulin is inserted. (ii) It has number of orgin of replication (ori) where replication starts.

(iii) Recombinant plasmid is introduced into E.coli host cell where it replicates and produces large amount of insulin.

To determine the genotype of a plant i.e., whether the individual exhibiting dominant character is homozygous or heterozygous, a test cross is carried out by a geneticist. The individual having dominant phenotype is crossed with its homozygous recessive parent and all offspring’s are analysed. For example, a tall plant may be heterozygous or homozygous. If tall plant is heterozygous then on crossing it with homozygous recessive parent, tall and dwarf plants will be produced, in equal proportion while if tall plant is homozygous then the upcoming progenies will contain all tall plants.

28. When does a geneticist need to carry a test cross? How is it carried?

To determine the genotype of a plant i.c., whether the individual exhibiting dominant character is homozygous or heterozygous, a test cross is carried out by a geneticist. The individual having dominant phenotype is crossed with its homozygous recessive parent and all offsprings are analysed. For example, a tall plant may be heterozygous or homozygous. If tall plant is heterozygous then on crossing it with homozygous recessive parent, tall and dwarf plants will be produced, in equal proportion while if tall plant is homozygous then the upcoming progenics will contain all tall plants.

SECTION – D

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

29. Study the given figure and answer the following questions.

(a) Identify A, B, C and D from the given figure.

Ans. The given cross shows Morgans’ experiment on Drosophila. ‘A’ is red eyed carrier female, ‘B’ is red eyed male, ‘C’ is Y-chromosome and ‘D’ is X-chromosome carrying eye colour gene.

(b) What kind of inheritance is shown in the given the figure?

Ans. Here, the eye colour gene is linked to sex chromosome and is present of X-chromosome showing crisscross inheritance.

Ans. In criss-cross inheritance, the trait is transferred from parent to grand child of same sex through offspring of the opposite sex. A father transmits his traits to his grandson through daughter while a mother transmits the traits to her granddaughter through her son.

(c) What would happen in the given cross if the parents phenotype be reversed i.e., white eyed female and red eyed male respectively?

Ans. In the given cross, if the parents phenotype will be reversed, i.e., cross between white eyed female and red eyed male then it will result into the red eyed female that are carrier and normal white eyed male.

30. Refer to the given below flow chart that shows the sewage treatment

(a) With reference to the above flow chart explain the role of step A in the given process.

Ans. Step A (mechanical agitation) allows the vigourous growth of useful aerobic microbes into flocs

(b) Identify A, B, C and D in the given process.

Ans. A – Mechanical agitation

B – Reduced BOD

C – Activated sludge

D – Anaerobic sludge digesters

Ans. At step D, other kinds of bacteria, which grow anaerobically, digest the bacteria and the fungi in the sludge. During this digestion, bacteria produce a mixture of gases such as methane, hydrogen sulphide and carbon dioxide. These gases form biogas and can be used as source of energy as it is inflammable.

Ans. The low BOD level indicates that water is less polluted. Once the BOD of sewage or wastewater is reduced significantly, the effluent is then passed into a settling tank where the bacterial ‘flocs’ are allowed to sediment. This sediment is called activated sludge.

SECTION – E

31. An experiment ‘X’ provided evidence in support of ‘Y’. In this experiment, four gases were circulated ‘A’, ‘B’, ‘C, and ‘D’ in an air tight apparatus and electrical discharge from electrodes was passed at 800°C. The mixture of gases was passed through a condenser. After a week, the chemical composition of the liquid inside the apparatus was analysed. The results provided evidence through which ‘Y’ was more or less accepted.

(i) Identify gases A, B, C, D.

Ans. Gases A, B, C and D could be methane (CH₄), ammonia (NH₃), hydrogen (H₂) and water vapour (H₂O)

(ii) Which theory of origin of life is supported by the above experiment?

Ans. Miller’s experiment that (X) supported Oparin- Haldane theory which states that the life originated on early earth through physico-chemical processes of atoms combining to form molecules. These molecules in turn reacting to produce inorganic and organic compounds. Organic compounds interacting to produce all types of macromolecules which organised to form the first living system or cells.

(iii) Draw a diagrammatic representation of experiment X.

Ans.

(iv) What does A, B, C and D together produced in the experiment X?

Ans. A, B, C and D together produces amino acids within a variety of organic compound in Miller’s and Urey experiment (X)

Explain three different ways in which natural selection can affect the frequency of a heritable trait in a population.

Ans. Natural selection can produce three different types of results and hence is divided into following three types:

(i) Stabilising selection This type of selection favours average sized individuals while eliminates small sized individuals. It reduces variation and hence does not promote evolutionary change. However, it maintains the mean value from generation to generation. If we draw a graphical curve of population, it is bell-shaped. For example, there is an optimum wing length for a hawk of a particular size with a certain mode of life in a given environment. Stabilising selection, operating through differences in breeding potential, will climinate those hawks with wing spans larger or smaller than this optimum length.

(ii) Directional selection: In this selection, the population changes towards one particular

direction. It is a progressive selection. It favours small or large, sized individuals and more individuals of that type will be present in next generation. The mean size of the population changes. For example – evolution of DDT resistant mosquitoes, industrial melanism in peppered moth, etc.

(iii) Disruptive selection This type of selection simultaneously favours individuals at both extremes of the distribution curve. As a result, two peaks in distribution of a trait are produced. It is rare in occurrence but important for evolutionary changes.

32. Give reasons why:

(a) DNA cannot pass into a host cell through the cell membrane.

Ans. DNA is a hydrophilic molecule, so it cannot pass into a host cell through cell membrane. The cell membrane consists of lipid bilayers that are generally impermeable to hydrophilic molecules.

(b) Proteases are added during isolation of DNA for genetic engineering.

Ans. DNA is interwined with proteins like histones and RNA. To obtain purified DNA, proteases are added during isolation of DNA which convert proteins into amino acids. The purified DNA finally precipitates out after the addition of chilled ethanol.

Ans. In order to link the alien DNA, the vector needs to have very few, preferably single recognition sites for the commonly used restriction enzymes. Presence of more than one recognition sites within the vector will generate several fragments, which will complicate the gene cloning process.

(d) Maintenance of sterile conditions in biotechnological processes.

Ans. Sterile conditions enable growth of only the desired microbe/eukaryotic cell in large quantities for the biotechnological products like antibodies, enzymes, etc

(e) Genes encoding resistance to antibiotics considered as useful selectable markers for E.coli cloning vector.

Ans. Genes encoding resistance to antibiotics are considered useful selectable markers for E.coli cloning vector because they help in selecting transformant cell from non-transformant ones.

Causative agents of HIV-AIDS and COVID-19 belong to the same group of viruses. To diagnose and amplify the genetic material for further study of COVID-19 virus, ‘RT-PCR’ test is carried out.

(a) What does ‘RT-PCR’ stand for?

Ans. RT-PCR stands for Reverse transcriptase – ‘Polymerase chain reaction’.

(b) Explain the various steps of PCR technique.

Ans. The various steps of polymerase chain reaction are:

(i) Denaturation: The target DNA is heated to a high temperature of 92-94°C resulting in the separation of its two strands. Each single strand of the DNA then acts as a template for DNA synthesis.

(ii) Primer annealing: The two oligosaccharide primers hybridise to form each of the single stranded template DNA, since the sequence of the primer is complementary to the 3′ end of the template DNA.

(iii) Extension of primer The Taq DNA polymerase synthesises the DNA region between the primers, using DNTPs (deoxynucleoside triphosphate) and Mg²⁺

33. Study the graph given below related with menstrual cycle in females:

(a) Identify ovarian hormones X and Y mentioned in the graph and specify their source.

Ans. X – Estrogen secreted by growing follicles

Y – Progesterone secreted by corpus luteum

(b) Correlate and describe the uterine events that take place according to the ovarian hormone levels X and Y mentioned in the graph on –

(i) 6 – 15 days

(ii) 16 – 25 days

(iii) 26 – 28 days (when ovum is not fertilised)

Ans. Uterine events that take place according to the ovarian hormone levels X and Y are as follows:-

(i) 6 – 15 days: Endometrium of the uterus regenerates by proliferation under the influence of estrogen.

(ii) 16-25 days: Under the influence of progesterone the endometrium of the uterus is maintained for implantation of fertilized ovum and other events of pregnancy.

(iii) 26 – 28 days (when ovum is not fertilized): In the absence of fertilization, corpus luteum degenerates which causes disintegration of endometrium leading to menstruation, making a new cycle

Refer the given below figure and answer the questions that follows:

(i) What phenomenon is represented in the above given figure ?

Ans. The given figure represents L.S of pistil showing path of pollen tube growth.

(ii) What is the path of entry of pollen tube ?

Ans. The path of entry of pollen tube is as follow:

(iii) Label the parts marked as A to E.

Ans. A – Pollen tube, B – Antipodal, C – Polar nuclei, D – egg cell, E – Synergid.

(iv) What will happen after entering of pollen into one of the synergids?

Ans. After entering of pollen tube into one of the synergids, the pollen tube releases the two male gametes into the cytoplasm of the synergid. One of the male gametes moves towards the egg cell and fuses with its nucleus thus completing the syngamy (generative fertilisation). This results in the formation of a diploid cell, the zygote. The other male gamete moves towards the two polar nuclei located in the central cell and fuses with them to produce a triploid primary endosperm nucleus (PEN). As this involves the fusion of three haploid nuclei it is termed triple fusion (vegetative fertilisation).

Class XII – Biology – Sample Paper – 2