Biology: CBSE Sample Question Paper (2020-21) - 2 | Sample Papers for Class 12 Medical and Non-Medical PDF Download

Class -XII
Biology
Time - 3 hrs
M.M. - 70 Marks

General Instructions :

(i) All questions are compulsory.
(ii) The question paper has four sections: Section A, Section B, Section C and Section D. There are 33 questions in the question paper.
(iii) Section–A has 14 questions of 1 mark each and 02 case-based questions. Section–B has 9 questions of 2 marks each. Section–C has 5 questions of 3 marks each and Section–D has 3 questions of 5 marks each.
(iv) 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.
(v) Wherever necessary, neat and properly labelled diagrams should be drawn.

Section 'A'

Q.1. British geneticist R.C. Punnett developed a graphical representation of a genetic cross called "Punnett Square". Mention the possible result this reproduction predicts of the genetic cross carried.
Ans. (Probability of) all genotypes / genotypic ratio.

Q.2. Name an IUD that you would recommend to promote the cervix hostility to the sperms.
Ans. Cu-T (Copper releasing IUD), suppress sperm motility and fertilizing capacity of sperms.

Q.3. Indiscriminate diagnostic practices using X-rays etc., should be avoided. Give one reason.
Ans. Indiscriminate diagnostic practices using X-rays, gamma rays etc. are ionizing radiations which usually produce breaks in the chromosomes and chromatids and abnormal mitosis in the irradiated cells. They cause abnormal functioning of the cells, mutations resulting in the development of various types of cancers specially blood cancer or leukemia, etc.

Q.4. Name the glands that contribute to human seminal plasma.
Ans. Prostate, Seminal vesicle and Bulbourethral gland.

Q.5. Name the nutrient that gets enhanced while curdling of milk by Lactobacillus.
Ans. Vitamin B₁₂.

Q.6. Name the condition in vertebrates where the body attacks self-cells.
Ans. Auto immune disorder or auto-immune disease.

Q.7. A patient is down with Amoebiasis. List the symptoms that confirm this infection. Name the causative pathogen.
Ans. Constipation, abdominal pain, stools with mucous, blood clot.
Entamoeba histolytica.

Q.8. Differentiate between pro-insulin and mature insulin.
Ans. Pro- insulin contains an extra stretch called the C peptide which is not present in the mature insulin.

Q.9. Name the type of immune response which is responsible for the rejection of tissues/organs in the patient's body post transplantation ?
Ans. The body is able to differentiate self and non-self and the cell-mediated immune response is responsible for the graft rejection.

Q.10. Name one amino acid, which is coded by only one codon.
Ans. Methionine / Tryptophan

Q.11. Assertion : Banana is a parthenocarpic fruit.
Reason : Parthenocarpic fruit develops without fertilization.
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.
Ans. (a) The banana is a parthenocarpic fruit as it develops without fertilization. The parthenocarpic fruits never contain seeds.

Q.12. Assertion : E. coli having pBR322 with DNA insert at Bam HI site cannot grow in medium containing tetracycline.
Reason : Recognition site for Bam HI is present in TetR region of pBR322.
(a) Both assertion and reason are true, and the reason is the correct explanation of the assertion.
(b) Both assertion and reason are true, but the reason is not the correct explanation of the assertion.
(c) Assertion is true but reason is false.
(d) Both assertion and reason are false.
Ans. (a) Both assertion and reason are true, and the reason is the correct explanation of the assertion.

Q.13. Assertion (A) : Sacred groves are highly protected.
Reason (R) : They are of religious importance to the communities.
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.
Ans. (a) Sacred groves are highly protected by certain communities because they are of religious importance to the communities. They have a significant role in in-situ conservation.

Q.14. Assertion : Haemophilia is an autosomal disorder.
Reason : A haemophilic father can never pass the gene for haemophilia to his son.
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.
Ans. (d) Haemophilia is a sex linked recessive disorder in which X-chromosome has the haemophilic gene. A haemophilic father can never pass the gene for haemophilia to his son.

Q.15. Reason: Over exploitation is a major cause of biodiversity loss.
Study the table given below and answer any of the four questions.

(a) Identify (i), (ii), (iii) and (iv).
(b) A polypeptide consists of 14 different amino acids.  How many base pairs must be there in the processed mRNA that codes for this polypeptide?
(c) How many different types of tRNA are needed for the synthesis of this polypeptide?
(d) Which one of the following pairs of codons is correctly matched with their function or the signal for the particular amino acid ?
(i) GUU, GCU – Alanine
(ii) UAG, UGA – stop
(iii) AUG, ACG – Start/methionine
(iv) UUA, UCA – Leucine
(e) One of the salient features of the genetic code is that it is nearly universal from bacteria to humans. Mention two exceptions to this rule.
Ans. (a)

(b) A polypeptide containing 14 different amino acids = 14 × 3 = 42 base pairs.
(c) 14 different types of RNA are needed for the synthesis of polypeptide.
(d) (ii) Three codons UAG, UAA and UGA are the stop or termination codons.
(e) (i) Mitochondrial codons.
(ii) Some protozoans.

Q.16. Read the following passage and answer any four questions from 16(i) to 16(v) given below :
Sickle cell anaemia is a genetic disorder where the body produces an abnormal haemoglobin called haemoglobin S. Red blood cells are normally flexible and round, but when the haemoglobin is defective, blood cells take on a “sickle” or crescent shape. Sickle cell anaemia is caused by mutations in a gene called HBB.
It is an inherited blood disorder that occurs if both the maternal and paternal copies of the HBB gene are defective. In other words, if an individual receives just one copy of the defective HBB gene, either from mother or father, then the individual has no sickle cell anaemia but has what is called “sickle cell trait”. People with sickle cell trait usually do not have any symptoms or problems but they can pass the mutated gene onto their children. There are three inheritance scenarios that can lead to a child having sickle cell anaemia :
— Both parents have sickle cell trait
— One parent has sickle cell anaemia and the other has sickle cell trait
— Both parents have sickle cell anaemia
(i) Sickle cell anaemia is a/an ________________ disease.
(a) X linked
(b) autosomal dominant
(c) autosomal recessive
(d) Y linked
Ans. (i) (c) autosomal recessive

(ii) If both parents have sickle cell trait, then there is _______________of the child having sickle cell anaemia.
(a) 25 % risk
(b) 50 % risk
(c) 75% risk
(d) No risk
Ans. (ii) (a) 25 % risk

(iii) If both parents have sickle cell trait, then there is _______________of the child having sickle cell trait.
(a) 25 % risk
(b) 50 % risk
(c) 75% risk
(d) No risk
Ans. (iii) (b) 50 % risk

(iv) If one parent has sickle cell anaemia and the other has sickle cell trait, there is __________that their children will have sickle cell anaemia and ___________will have sickle cell trait.
(a) 25 % risk, 75% risk
(b) 50 % risk, 50% risk
(c) 75% risk, 25% risk
(d) No risk
Ans. (iv) (b) 50 % risk, 50% risk

(v)

The following statements are drawn as conclusions from the above data (Kenya).
I. Patients with SCD (Sickle Cell Disease) are less likely to be infected with malaria.
II. Patients with SCD (Sickle Cell Disease) are more likely to be infected with malaria.
III. Over the years the percentage of people infected with malaria has been decreasing.
IV. Year 2000 saw the largest percentage difference between malaria patients with and without SCD.
Choose from below the correct alternative.
(a) only I is true
(b) I and IV are true
(c) III and II are true
(d) I and III are true
Ans. (v) (d) I and III are true
Section ‘B’ 

Q.17. Name any two copper releasing IUD’s. State how they act as a contraceptive.
Ans. CuT, Cu7, Multiload 375
Cu ions released suppresses sperm motility and the fertilizing capacity of sperms.

Q.18. Describe the structure of a nucleosome.
Ans. A unit of eight molecules of positively charged histones, negatively charged DNA , wrapped around the histone octamer, contains 200 bp of DNA helix.
// In lieu of the above explanation the following diagram along with the following statement can be considered.

DNA is negatively charged, histone is positively charged, 200 bp of DNA helix.

Q.19. Explain co-dominance with the help of one example.
OR
A segment of DNA molecule comprises of 546 nucleotides. How many cytosine nucleotides would be present in it if the number of adenine nucleotides is 96?
Ans. When the dominant alleles of the same gene which are contributed by both parents are expressed, it is called co-dominance // F₁ generation resembles both the parents.
In human blood group

In human red blood cells, alleles I^A and I^B of gene I are both dominant, when I^A &  I^B are present together in an individual both are expressed as  I^AI^B, (AB blood group).

Q.20. Explain the method to increase the competency of the bacterial cell membrane to take up recombinant DNA.
Ans. The recombinant DNA can be forced into the bacterial cell treated with divalent cations and incubating it with recombinant DNA on ice. This is to be followed by placing it briefly at 42°C (heat shock), and then putting it back on ice. This process would enable the bacteria to take up the recombinant DNA.
OR
What are bioreactors ? How are large volumes of cultures maintained and processed in them ?
Ans. Bioreactors are vessels in which raw materials are biologically converted into specific products such as enzymes using microbial, plant, animal or human cells. A bioreactor provides the optimal conditions for achieving the desired product by providing optimum growth conditions like temperature, pH, substrate, salts, vitamins and oxygen.

Q.21. What kind of biodiversity of more than a thousand varieties of mangoes in India represent ? How is it possible?
Ans. Genetic diversity is represented by a large varieties of mangoes found in India.
Reason for genetic diversity : Use of horticulture techniques likes grafting, breeding etc. by humans.

Q.22. Give two reasons as to why a weed such as Calotropis flourishes in abandoned fields.
Ans. Dry hairy seeds helps in dissemination/having xerophytic adaptations (thick hair on leaves and stems) not grazed by animals as it produces poisonous substances/cardiac glycosides.

Q.23. Write the binomials of two fungi and mention the products/bioactive molecules they help to produce.
Ans. (a) Trichoderma polysporum, cyclosporin A.
(b) Aspergillus niger, citric acid
(c) Monascus purpureus, statin.
(d) Saccharomyces cerevisiae, ethanol / alcohol.
(e) Penicillium notatum, Penicillin.

Q.24. Define interference competition. Give one example that supports competitive exclusion occurring in nature.
Ans. Interference competition is the feeding efficiency of one species which might be reduced due to the interfering and inhibitory presence of the other species, even if resources (food and space) are abundant. Examples that support competitive exclusion occurring in nature are:

• The Abingdon tortoise became extinct within a decade after goats were introduced on the island, apparently due to the greater browsing efficiency of the goats.
• The larger and competitively superior barnacle Balanus dominates the intertidal area and excludes the smaller barnacle Chathamalus from that zone. (Any one example)

Q.25. What are the two approaches for conserving biodiversity ? Name them. Give one example of each.
Ans. Two approaches for conserving biodiversity are : In-situ conservation and ex-situ conservation. E.g. of in-situ conservation : wildlife sanctuaries/ national parks / biosphere reserves/ sacred groves E.g. of ex-situ Conservation : in botanical gardens/ zoological gardens/ seed/pollen/gene banks.
OR
There are many animals that have become extinct in the wild but continue to be maintained in Zoological parks.  What type of biodiversity conservation is observed in this case ?
Ans. It is an example of ex-situ conservation (off–site conservation). In this approach, threatened plants and animals are taken out of their natural habitat and placed into suitable settings and given special care.

Q.26. (a) Differentiate between spermatogenesis and spermiogenesis.
(b) Mention the function of mitochondria in sperm.
Ans. (a) Difference between spermatogenesis and  spermiogenesis : Spermatogenesis is a process of formation of haploid spermatozoa from germinal cells while spermiogenesis is a process of differentiation of spermatozoa into a spermatid. Here, a spermatid forms a single spermatozoon.
(b) Mitochondria provide energy for the movement of sperm.

Q.27. How does the activity of each one of the following help in organic farming ?
(a) Mycorrhiza
(b) Cyanobacteria
(c) Rhizobium
Ans. (a) Mycorrhiza : The fungal symbiont in these association absorb phosphorous from soil and pass it to plant. Plants in return provide food to fungus. Plants also show resistance to root borne pathogens,  tolerance to salinity / drought , an overall increase in plant growth and development.
(b) Cyanobacteria : Serve as an important biofertiliser by fixing atmospheric nitrogen, also add organic matter to the soil, and increase its fertility.
(c) Rhizobium : Fix atmospheric nitrogen into organic forms, which is used by plant as nutrient/ increase soil fertility / symbiotic association in root nodules of leguminous plants.
Detailed Answer :
(a) Mycorrhiza is a symbiotic association of fungi (e.g. the genus of Glomus) with the roots of higher plants.
The fungal symbiont in these association absorbs phosphorus from soil and passes it to the plant. This association help the plant to absorb phosphorus in natural way thus helping in organic farming.
(b) Cyanobacteria are autotrophic microbes, which can fix atmospheric nitrogen, e.g. Anabaena, Nostoc, Oscillatoria, etc. They are capable of converting atmospheric nitrogen into soluble form like nitrites or nitrates.
They enrich the nitrogen content of the soil, acting as a natural fertiliser.
(c) Rhizobium bacteria live in the root nodules of the leguminous plants. They help in fixing atmospheric nitrogen into organic forms, which is used by the plant as nutrient. Thus, it enriches the soil and makes it fertile thereby helping farmers.

Q.28. Define flocs and state their importance in biological treatment of waste water.
Ans.

• Flocs are masses of semi - decayed organic matter along with decomposer microbes which are surrounded by slime. They separate the organic matter from waste water.
• Flocs settle down in secondary tanks and take part in the formation of sludge.
• They can be used as inoculum in biological treatment of waste water as well as source of biogas and manure.

Section ‘C’ 

Q.29.“Specific Bt Toxin gene is incorporated into cotton plant so as to control infestation of Bollworm”. Mention the organism from which the gene was isolated and explain its mode of action.
Ans. Specific Bt toxin genes isolated from Bacillus thuringiensis is incorporated into cotton is coded by the genes cryIAc and cryIIAb that control the cotton bollworms.

• Bacillus forms protein crystals that contain a toxic insecticidal protein.
• Once an insect ingest the inactive toxin, it is converted into an active form.
• The toxin in the form of crystals gets solubilised due to alkaline pH in the gut.
• The activated toxin binds to the surface of gut epithelial cells and perforate the walls causing the death of insect larva.

Q.30. Highlight the differences and a similarity between the following population interactions: competition, predation and commensalism.
Ans. 

Section 'D'

Q.31. (a) Explain one application of each one of the following :
(A) Amniocentesis
(B) Lactational amenorrhoea
(C) ZIFT
(b) Prepare a poster for the school programme depicting the objectives of : “Reproductive and Child Health Care Programme”.
Ans. (a) (A) To detect chromosomal disorders / sex determination (legally banned) / detect genetic disorder / Karyotyping.
(B) To prevent pregnancy / means of natural contraception.
(C) To assist an infertile couple to have children by transferring the zygote / early embryo / embryo at eight blastomere stage into fallopian tube.

(b) A poster made on RCH - Any relevant slogan or sketch made should be awarded marks.
Example: Hum Do Hamare Do, Do Boond Zindagi Ke, Beti Bachao Beti Padhao, Stop STD, Gender selection and detection is punishable, (Any other relevant theme)
Detailed Answer :
(a) (A) Amniocentesis is a medical procedure used in prenatal diagnosis of genetic disorders like thalassemia, Down's syndrome etc. In this process, amniotic fluid from the uterus of pregnant lady is taken out by a needle to test the developmental abnormalities of the baby inside the mother's womb.
(B) Lactational amenorrhoea (absence of menstruation) is a temporary contraceptive method based on the fact that ovulation and therefore the cycle do not occur during the period of intense lactation following parturition.
(C) Zygote Intra Fallopian Transfer (ZIFT) is an assisted reproductive technology used for infertility treatment.
In this, the sperm from a donor male and ova from a donor female are fused in the laboratory. The formed zygote is then transferred into the fallopian tube at 8 blastomeres stage.
(b) Poster for RCH programme :

OR
(i) Name the hormones secreted and write their functions :
(a) By corpus luteum and placenta (any two).
(b) During Follicular phase and parturition.
(ii) Name the stages in a human female where :
(a) Corpus luteum and placenta co-exist.
(b) Corpus luteum temporarily ceases to exist.
Ans. (i) (a) Corpus luteum : progesterone, essential for maintenance of the endometrium.
Placenta : hCG / human chorionic gonadotropin, produce during pregnancy / stimulates and maintains the corpus luteum / to secrete progesterone / growth of mammary glands.
HPL / human placental lactogen, produced during pregnancy.
Estrogen, maintenance of pregnancy /supporting foetal growth / metabolic changes in mother (Any two).
(b) Follicular phase : LH / FSH, stimulates follicular development / secretion of estrogen by growing follicles.
Parturition : ox ytocin causes stronger uterine contractions / relaxin, secreted during (later stage) the pregnancy / softens symphysis pubis.
(ii) (a) Pregnancy / gestation.
(b) Menstruation / proliferative phase / ovulatory phase / follicular phase.

Q.32. Evaluate the suitability of DNA and RNA as genetic material and justify the suitability of the one that is preferred as an ideal genetic material.
Ans. Evaluation of DNA and RNA on the basis of the properties of the genetic material :
(i) It should be able to generate its replica (Replication) : As per the rule of base pairing and complementarity, both the nucleic acids (DNA and RNA) have the ability to direct their duplications.
(ii) The genetic material should be chemically and structurally stable enough not to change with different stages of life cycle, age or with change in physiology of the organism.
Presence of 2'-OH group and uracil make RNA more reactive and structurally less stable than DNA. Therefore, DNA is a better genetic material than RNA.
(iii) It should provide the scope for slow changes (mutation) that are required for evolution : Both DNA and RNA are able to mutate. In fact, RNA being unstable, mutates at a faster rate. Consequently, viruses having RNA genome and having shorter life span mutate and evolve faster.
(iv) It should be able to express itself in the form of 'Mendelian Characters’ : RNA can directly code for the synthesis of proteins, hence can easily express the characters. DNA, however, is dependent on RNA for synthesis of proteins. The protein synthesising machinery has evolved around RNA.
(v) The above discussion indicates that both RNA and DNA can function as genetic material, but DNA being more stable is preferred for storage of genetic information.
OR
Explain the mechanism of DNA replication as suggested by Watson and Crick.
Ans. Mechanism of Replication of DNA suggested by Watson and Crick

• The two strands of DNA would separate and act as a template for the synthesis of new complementary strands. After the completion of replication, each DNA molecule would have one parental and one newly synthesised strand. This scheme was termed as semi conservative replication of DNA.
• In living cells, such as E. coli, the process of replication requires a set of catalysts (enzymes). The main enzyme is referred to as DNA-dependent DNA polymerase, since it uses a DNA template to catalyse the polymerization of deoxynucleotides.
• Furthermore, energetically, replication is a very expensive process. Deoxyribonucleoside triphosphates serve dual purposes. In addition to acting as substrates, they provide energy for polymerisation reaction.
• For long DNA molecules, since the two strands of DNA cannot be separated in its entire length (due to very high energy requirement), the replication occurs within a small opening of the DNA helix, referred to as replication fork. l The DNA-dependent DNA polymerases catalyse polymerization only in one direction, that is 5' → 3'.
• Consequently, on one strand (the template with polarity 3' → 5'), the replication is continuous, while on the other (the template with polarity 5' → 3'), it is discontinuous. The discontinuously synthesised fragments are later joined by the enzyme DNA ligase.

• The DNA polymerase on their own cannot initiate the process of replication.
• There is a definite region in E. coli DNA where the replication originates, such regions are termed as origin of replication.
• In eukaryotes, the replication of DNA takes place at S-phase of the cell-cycle.
• The replication of DNA and cell - division cycle should be highly coordinated.
• A failure in cell division after DNA replication results into polyploidy.

Q.33. (a) What is an age-pyramid ?
(b) Name three representative kinds of age-pyramids for human population and list the characteristics for each one of them.
Ans. (a) If the age distribution (per cent individuals of a given age or age group) is plotted for the population, the resulting structure is called the age pyramid.
(b) (i) Expanding: Pre-reproductive population is greater than reproductive or post reproductive population / growing with maximum no. of individuals in pre-reproductive phase and least no. in post-reproductive phase
(ii) Stable: Pre-reproductive & reproductive population are almost similar / ideal for population / maintains balanced continuity / no. of individuals in reproductive and pre- reproductive phase is almost same and less no. of individuals in post-reproductive phase
(iii) Declining: Pre-reproductive population is less than reproductive population /less no. of individuals in pre reproductive phase than reproduction phase
In lieu of the above explanation the following diagram can be considered

OR

(a) Which of the above represents the increase or decrease of population?
(b) If N is the population density at time t, then what would be its density at time (t+1)? Give the formula.
(c) In a barn there were 30 rats. 5 more rats enter the barn and 6 out of the total rats were eaten by the cats. If 8 rats were born during the time period under consideration and 7 rats left the barn, find out the resultant population at time (t+1).
(d) If a new habitat is just being colonized, out of the four factors affecting the population growth which factor contributes the most?
Ans. (a) a and d represents increase of population and b and c represent decrease of population.
(b) N_t+1 = N_t + [( B+I) - ( D + E)]
(c) Here N_t=30; I=5; E=7 ;D=6 ; B=8
Putting the value in N_t+1 = N_t + [( B+I) - ( D + E) ]
N_t+1 = 30+ [(8+5) - (6+7)]
=30 + [13 -13]
= 30 + 0 = 30 rats
(d) Immigration contributes the most.