Class XII - Biology - Sample Paper - 1

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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. If most individuals in a population are young, why is the population likely to grow rapidly in the future?

(a) Many individuals will begin to reproduce soon

(b) Death rates will be low

(d) All of these

Ans. (a) Many individuals will begin to reproduce soon

Different age groups have different reproductive capabilities. Pre-reproductive individuals are the young individuals which will enter the reproductive age after some time. They are the potential source of increase in population. Reproductive individuals are the ones which are actually adding new members to the population.

2. Primary endosperm nucleus (PEN) is formed by the fusion of

(a) 2 polar nuclei + 1 synergid cell nucleus

(b) 1 polar nucleus + 1 antipodal cell nucleus + 1 synergid cell nucleus

(d) 2 antipodal cell nuclei + 1 male gamete nucleus.

Ans. (c) 2 polar nuclei + 1 male gamete nucleus

During the event of double fertilisation in angiosperms, the second male gamete fuses with the two haploid polar nuclei or diploid secondary nucleus of the central cell to form a triploid primary endosperm nucleus (PEN). This second fertilisation is called vegetative fertilisation.

3. Which enzyme helps in removing oil stains from clothes?

(a) Streptokinase

(b) Trypsin

(d) Amylase

Ans. (c) Lipase

A large number of microorganisms are capable of using natural oils and fats as carbon source for their growth. The enzyme responsible for this work is known as lipases. Some microorganisms like Candida cylindracae, Candida rugosa, Aspergillus niger etc. have been reported for extracellular lipase production. This microbial lipases are used to remove oil stains from clothes.

4. Which of the following is a cause of transmission of HIV?

(a) Multiple sexual partners

(b) Transfusion of contaminated blood

(d) All of these

Ans. (d) All of these

5. Hardy-Weinberg equilibrium is known to be essentially affected by factors like, gene flow, genetic drift, mutation, genetic recombination and

(a) evolution

(b) limiting factors

(d) natural selection

Ans. (d) natural selection

Hardy-Weinbeg principle describes a theoretical situation in which a population is undergoing no evolutionary change. It states that allele frequencies in a population are stable and constant from generation to generation. There are five factors that affect Hardy-Weinberg Principle. These are – mutation, gene flow, genetic drift, genetic recombination and natural selection pressure.

6. Plasmid used to construct the first recombinant DNA was isolated from which bacterium species?

(a) Escherichia coli

(b) Salmonella typhimurium

(d) Thermus aquaticus

Ans. (b) Salmonella typhimurium

The first recombinant DNA was constructed by Stanley Cohen and Herbert Boyer in 1972. They cut the piece of DNA from a plasmid carrying antibiotic resistance gene in the bacterium Salmonella typhimurium and linked it to the plasmid of Escherichia coli.

7. Microbes are used in

I. primary treatment of sewage

II. secondary treatment of sewage

III. anaerobic sludge digesters

IV. production of biogas.

Choose the correct option.

(a) I, II and III

(b) I, III and IV

(d) I, II, III and IV

Ans. (c) II, III and IV

Primary treatment is the physical removal of large and small particles from sewage.

Secondary treatment of the liquid effluent from the primary setting-tank is purely a biological treatment involving microbial activity.

In the anaerobic sludge digesters, heterotrophic microbes anaerobically digest organic matter of sludge. They produce mixture of gases such as methane, hydrogen sulphide and CO2, which form the biogas.

8. If a double stranded DNA has 20% of cytosine, what will be the percentage of adenine in it?

(a) 20%

(b) 40%

(d) 60%

Ans. (c) 30%

According to Chargaff’s rule, the amount of adenine is always equal to that of thymine and the amount of guanine is always equal to that of cytosine, i.c., A = T and G = C. Also, the purines and pyrimidines are always in equal amounts, i.c., A + GT+ C. Now, given dsDNA has 20% cytosine and hence guanine will be also 20%.

So, AT must be 60%. Therefore, percentage of adenine would be 60/2 = 30%.

9. The density of a population in a given habitat during a given period, fluctuates due to changes in certain basic processes. On this basis, fill up boxes A and B in the given flow chart with correct option.

(a) A-Natality, B-Mortality

(b) A-Immigration, B-Emigration

(d) Both (a) and (b)

Ans. (d) Both (a) and (b)

Natality and immigration add to the population density. Mortality and emigration decrease the population density.

10. The given Punnett’s square represents the pattern of inheritance in a dihybrid cross where yellow (Y) and round (R) seed condition is dominant over white (y) and wrinkled (r) seed condition.

A plant of type ‘H’ will produce seeds with the genotype identical to seeds produced by the plants of

(a) type M

(b) type J

(d) type N

Ans. (d) type N

Plant H is formed by fusion of gametes yR and YR and hence has the genotype YYRR. Plant N is formed by fusion of gametes YR and yR and hence will have the same genotype as plant. H i.e., YYRR.

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

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

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

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

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

12. The given table shows differences between spermatogenesis and spermiogenesis. Select the incorrect option.

Ans. (b) Spermatogenesis involves conversion of a diploid structure (spermatogonia) into haploid structures (spermatozoa). Spermiogenesis changes a haploid structure (spermatid) into another haploid structure (spermatozoon).

13. Assertion : Elimination of a competitively inferior species in a closely related or otherwise similar group is known as competitive exclusion principle.

Reason : If two species compete for the same resource, they could avoid competition by choosing different times for feeding or different foraging patterns.

(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.

Ans. Gause’s ‘competitive exclusion principle’ states that two closely related species competing for the same resources cannot co-exist indefinitely and the competitively inferior one is eliminated eventually. This may be true if resources are limiting, but not otherwise. Mechanism of ‘Resource partitioning’ states that if two species compete for the same resource, they could avoid competition by choosing, for instance, different times for feeding or different foraging patterns.

14. Assertion : Mouse is the most preferred mammal for studies on gene transfers.

Reason : Mouse possesses features like short oestrous cycle and gestation period, relatively short generation time, production of several offspring per pregnancy, etc.

(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.

Mouse is the most preferred mammal for studies on gene transfers due to its many favourable features like (i) short oestrous cycle and gestation period, (ii) relatively short generation time, (iii) production of several offspring per pregnancy, (iv) convenient in vitro fertilisation, successful culture of embryos in vitro.

15. Given below is the diagram of a normal 28 day menstrual cycle in a human female. It depicts phase I, II and III. Study this diagram and comment upon the appropriateness of the Assertion and the Reason.

Assertion : The hormone secreted in large amounts in phase III is also responsible for maintaining pregnancy in human females.

Reason : Corpus luteum secretes progesterone in phase I however it degenerates completely in phase III.

(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.

The corpus luteum secretes large amount of progesterone during luteal or secretory phase (phase III)

16. Assertion : Emigration is outward movement of some individuals from local population.

Reason : Emigration is caused by occurrence of deficiencies and calamities.

(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.

SECTION – B

17. Write the role of on and ‘restriction’ site in a cloning vector pBR322.

Ans. Origin of replication (ori) site in cloning vector pBR322 is a sequence from where replication starts. Any piece of DNA when linked to this sequence can be made to replicate within host cell. Restriction site within the markers tetR and amp genes permit an easy selection for cells transformed or the recombinant pBR322.

18. (a) It is generally observed that the children who had suffered from chicken-pox in their childhood may not contract the same disease in their adulthood. Explain giving reasons the basis of such an immunity in an individual. Name this kind of immunity.

Ans. Children who had suffered from chicken-pox may not contract the same disease in their adulthood because of development of memory cells. These type of cells develop during first encounter with the pathogen. Memory cells are highly specific and may remain in body for decades. This type of immunity is known as acquired natural active immunity.

(b) What are interferons? Mention their role.

Ans. Interferons are the proteins produced by virus infected cells. They protect non-infected cells from further viral infection.

19. 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) gets differentiated as the embryo. The cells of trophoblast (D) gets attached to the endometrium and helps to provide nutrition to the embryo. The cells of trophoblast later form the extra embryonic membranes.

20. Draw a pyramid of numbers considering a big banyan tree supporting a population of insects, small birds and their predators.

Ans. The pyramid of numbers considering a big banyan tree supporting a population of insects, small birds and their predators will be spindle-shaped shown below:

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.

Ans. Grazing food chain and detritus food chain are the major conduits for flow of energy in aquatic and terrestrial ecosystems, respectively.

Differences between grazing food chains and detritus food chain are as follow:

21. Two children one with blood group ‘AB’ and other with blood group ‘0’ are born to parents where the father has blood group W and the mother has blood group ‘B’. Work out a cross to show how is it possible?

Ans. Father has blood group A and mother has blood group B whereas their children have blood groups AB and O. This indicates that the parents are heterozygotes.

Therefore, Genotype of father = I^A I^O

And genotype of mother = I^B I^O

Hence, all the four types of blood groups viz. AB, A, B and O are possible in their offspring.

SECTION – C

22. Carefully examine structures A and B of pentose sugar given below. Which one of the two is more reactive? Give reasons.

Ans. In the given structures A represents ribose sugar and B represents deoxyribose sugar.

Ribose sugar (A) is more reactive than deoxyribose sugar (B) as 2′-OH group present in the ribose sugar that makes it more labile and easily degradable. DNA has evolved from RNA with chemical modification that makes it more stable. DNA being double stranded and having complementary strand further resists changes.

23. Explain the role of pituitary and ovarian hormones in the menstrual cycle of human females.

Ans. Menstrual cycle is regulated by certain hormones, some of which are secreted by the pituitary gland. The pituitary gland is stimulated by releasing factors produced in hypothalamus. Anterior pituitary gland secretes two hormones FSH and LH. FSH stimulates maturation of follicle and stimulate it to secrete estrogens. Rapid secretion of LH (LH surge) induces rupturing of Graafian follicle, thereby leading to release of ovum (ovulation).

Ovary secretes two hormones estrogen and progesterone. Estrogen stimulates follicular development and proliferation of the endometrium of the uterine wall. Progesterone produced by corpus luteum helps to maintain endometrium which is required for implantation of the fertilised ovum and other events of pregnancy.

24. Differentiate between autogamy, geitonogamy and xenogamy.

Ans. Difference between autogamy, geitonogamy and xenogamy is as follow:

25. Convergent evolution and divergent evolution are the two concepts explaining organic evolution. Explain each one with the help of an example.

Ans. Differences between divergent evolution and convergent evolution are as follow:

26. (i) Name the plant source of the drug commonly called ‘smack. How does it affect the body of the abuser?

Ans. Smack is an opioid narcotic, obtained from poppy plant, Papaver somniferum. It is a powerful analgesic and relieves the pain by acting on central nervous system.

(ii) What is humoral immunity?

Ans. Humoral immunity refers to immunity provided by the antibodies that are circulating in the body fluids (humors).

(i) What is colostrum? Why are breast-fed babies likely to be healthy?

Ans. Colostrum (mother’s first milk) is rich in IgA antibodies. It is secreted by mother during the initial days of lactation. It provides passive immunity to new born and protects it from various diseases and therefore, this milk is considered very essential for the infants.

(ii) Malaria, typhoid, pneumonia and amoebiasis are some of the human infectious diseases. Which one of these are transmitted through mechanical carriers?

Ans. Mechanical carrier is one that simply carries pathogens to a susceptible individual and is not essential to the development of the pathogen. The pathogens are simply carried on the mouthparts, legs, body surface of the carrier from an infected to a susceptible host. Amoebiasis and typhoid are carried through mechanical carriers like housefly.

27. (a) Write the palindromic nucleotide for the following DNA segment : 5l-GAATTC- 3l

Ans. Palindromic nucleotide sequence for the given DNA segment is: 3′-CTTAAG-5′

(b) Name the restriction endonuclease that recognises this sequence.

Ans. EcoRI is a restriction endonuclease enzyme it recognises base sequences 5′-GAATTC-3′ / 3′-CTTAAG in DNA duplex and cuts each of the two strands between G and A. On the other hand, exonuclease remove nucleotide from the terminal ends of DNA in one strand of duplex. Hence, EcoRI cut each of the two strand of DNA duplex at specific point whereas exonuclease remove nucleotide from the terminal ends (either 5′ or 3′) of DNA in one strand of duplex.

Ans. When 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.

28. A particular species of wild cat is endangered. In order to save them from extinction, which is a desirable in situ or ex situ? Justify your answer and explain the difference between the two approaches.

Ans. Endangered species of wild cat can be saved from extinction by ex situ conservation. In ex situ approach, endangered species are taken out from their natural habitat and kept under human supervision, and are protected from all adverse factors. They are provided with all essentials and are also protected from predators. More offspring are produced by captive breeding and are released in their natural habitat to acclimatise, which helps in maintaining their population.

Differences between in situ and ex situ conservation are:

SECTION – D

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

29. Look at the diagram and answer the following questions.

(a) Write the genotypes of A, B, C, D.

Ans. A – Tt

B – TT

C – Tt

D – tt

(b) Write the phenotypes of A, B, C, D.

Ans. A – Tall

B – Tall

C- Tall

D – Dwarf

Ans. Phenotypic ratio is 3:1; Tall:Dwarf

Genotypic ratio of progeny is

TT Tt tt

1 : 2 : 1

Homozygous : Heterozygous : Dwarf

Tall tall

(c) Write the conclusions Mendel arrived at on dominance of traits on the basis of monohybrid crosses that he carried out in pea plants.

Ans. Whenever Mendel carried out a cross between plants for a contrasting trait he found that only one trait out of the two appears in the F1 generation. He concluded that the trait which is expressed in F1 is dominant while the one which remains hidden is recessive. He also said that characters are controlled by discrete unit called factors which occur in pair.

(a) Write the genotypes of A, B, C, D. (b) Write the phenotypes of A, B, C, D. (c) Write phenotypic and genotypic ratio of progeny. O (c) Write the conclusions Mendel arrived at on dominance of traits on the basis of monohybrid crosses that he carried out in pea plants.

(a) Mention the major component of biogas.

Ans. Biogas is a mixture of gases such as methane, hydrogen sulphide and carbon dioxide. It is used as source of energy as it is inflammable.

(b) Identify ‘B’ in the flow chart?

Ans. In the second stage of biogas production, the simple soluble compounds or monomers are acted upon by fermentation causing microbes and converted into organic acids especially acetic acid.

Ans. A is methanogenic bacteria that converts organic acids as well as carbon dioxide into methane. The gas thus formed is stored in tanks for supply.

Ans. Biogas is used for cooking and lightning.

(ii) It provides both energy and manure.

(iii) It is a storable form of energy which can be used more efficiently and economically.

SECTION – E

31. (a) Describe the events of oogenesis with the help of schematic representation.

Ans. The schematic representation of events of oogenesis is as follow:

(b) Write two differences between oogenesis and spermatogenesis.

Ans. Difference between spermatogenesis and oogenesis are as follow:

(a) When a seed of an orange is squeezed, many embryos, instead of one are observed. Explain how it is possible.

Ans. When a seed of an orange is squeezed, many embryos instead of one are observed, this is due to polyembryony. Polyembryony is the occurrence of more than one embryo in a seed. This takes place in many Citrus fruits such as orange and mango varieties when some of the nucellar cells surrounding the embryo sac start dividing and protrude into the embryo sac. These cells develop into the embryos and as a result in such species each ovule contains many embryos.

(b) Are these embryos genetically similar or different? Comment.

Ans. The embryos formed as a result of polyembryony may or may not be genetically similar. In polyembryony, there may be more than one egg cell in an embryo sac or more than one embryo sac in an ovule. All the egg cells may get fertilised. The embryos may be developed directly from a diploid cell other than egg, e.g., nucellus, integuments or they may also develop from synergids and antipodal cells. If the polyembryony occurs due to the fertilisation of more than one egg, then the formed embryo will not be of same genetic constitution. But, if all the embryos develop directly from diploid cell, then they will have same genetic makeup.

32. (a) Describe the process of synthesis of fully functional mRNA in a eukaryotic cell.

Ans. Transcription is the process of copying genetic information from one strands of DNA into RNA. A transcription unit of a DNA has three regions a promoter, a structural gene and a terminator. Transcription requires a DNA dependent RNA polymerase. Prokaryotes have only one DNA dependent RNA polymerase which synthesises all types of RNA. RNA polymerase binds to promoter and initiates transcription (Initiation). It uses nucleoside triphosphates as substrate and polymerises the mRNA strand in a template depended fashion following the rule of complementarity. It also facilitates opening of the helix and continues elongation. Only a short stretch of RNA remains bound to the enzyme. Once the polymerases reaches the terminator region, the nascent RNA falls off, so also the RNA polymerase. This results n termination of transcription.

(b) How is the process of mRNA synthesis different from that in prokaryotes?

Ans.

(a) The given flow chart highlighting the steps in DNA fingerprinting technique. Identify a, b, c, d, e and f

Ans. a — > Restriction endonuclease: DNA is extracted from the cells in high speed centrifuge. DNA is cut into fragments by restriction endonuclease.

b — > Agarose gel DNA are separated by gel electrophoresis using agarose gel. The DNA fragments separate according to their size through sieving effect provided by the agarose gel.

c — > Nitro cellulose membrane: The fragments of separated DNA are (like nylon or nitrocellulose) transferred to synthetic membranes.

d – > VNTR : DNA splits into single strand and with the help of VNTR probe the single strands are hybridised.

e — > Hybridisation: Hybridisation is done with the help of VNTR probe which gets attached to single stranded VNTR having complementary nucleotide sequences.

f — > Autoradiography The hybridised DNA fragments are detected by autoradiography. This shows many dark bands of different sizes.

(b) List any two applications of DNA fingerprinting technique.

Ans. (b) Applications of DNA fingerprinting are as follows:

(i) Paternity or maternity disputes can be solved by DNA fingerprinting as it can identify the real genetic mother, father and the offspring.

(ii) This technique is being used to identify genes connected with hereditary diseases.

(iii) It is useful in detection of crime and legal pursuits.

(iv) It can identify racial groups, their origin, historical migrations and invasions.

33. (a) Explain how to find whether an E. coli bacterium has transformed or not when a recombinant DNA bearing ampicillin resistant gene is transferred into it.

Ans. If the recombinant DNA bearing ampicillin resistant is transferred into E.coli bacteria, it will confer resistance to ampicillin and therefore will grow on ampicillin containing medium. However, the non-transformants will not grow on same medium. The recombinants or transformed E.coli can be selected by use of selectable markers. the transformants must be plated on a plate bearing ampicillin containing medium. hence, transformants can be selected from non- transformants.

(b) What does the ampicillin resistant gene act as in the above case?

Ans. Ampicillin resistant gene acts as selectable marker and helps in selecting the transformant in the above case.

There are two different farm lands, one where Bt-cotton crop was cultivated and the other where non Bt-cotton crop (indigenous) was cultivated. Farmers responsible for this experimental cultivation were free to use the farming practices of their choice. During the cultivation period, the data was collected with respect to the amount of pesticide used, water required for irrigation and at harvesting time, the crop productivity. Based on the data collected, a bar graph was plotted which is shown below.

Answer the following questions:

(i) Write your interpretation, with reason, on the basis of the three parameters plotted in the graph.

Ans. Ans. On the basis of three parameters, the following interpretations can be drawn by looking at the graph:

(a) Average crop productivity was much higher in case of Bt crop (120 tonnes/Ha) as compared to the Non-Bt crop (80 tonnes/ Ha). This is due to the fact that Bt crop is resistant to insects, nematodes, lepidopterans etc. and hence leads to higher productivity.

(b) The input of pesticide in case of Bt crop is less as compared to Non-Bt crops. This is due to the fact that Bt crops are resistant to insects. The Bt toxin genes are incorporated

(c) The Bt crops needs more water for their growth hence the amount of irrigation in litres/Ha is more in Bt crops as compared to non Bt crop.

(ii) Which one of the crops would you like to cultivate in your farm and why?

Ans. Looking at the higher productivity and lesser input of pesticides in Bt crops, I would like to cultivate Bt crop in my farm as plant productivity is higher and input of pesticides is lesser in Bt crops.

(iii) Which one out of these two crops would a farmer from West Bengal like to cultivate and why?

Ans. From the given graph we can observe that Bt crop requires more water for their growth. Therefore, farmers from West Bengal prefer to grow Bt crops because there is no scarcity of water there for irrigation.

Class XII – Biology – Sample Paper – 1