Biology: CBSE Sample Question Paper (2020-21) - 3 Notes | Study Sample Papers for Class 12 Medical and Non-Medical - Class 12

Class 12: Biology: CBSE Sample Question Paper (2020-21) - 3 Notes | Study Sample Papers for Class 12 Medical and Non-Medical - Class 12

The document Biology: CBSE Sample Question Paper (2020-21) - 3 Notes | Study Sample Papers for Class 12 Medical and Non-Medical - Class 12 is a part of the Class 12 Course Sample Papers for Class 12 Medical and Non-Medical.
All you need of Class 12 at this link: Class 12

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. Name the type of fruit apple is categorised under and why ? Mention two other examples which belong to the same category as apple.
Ans. False fruit, thalamus contributes to fruit formation : Strawberry, Cashew (any other correct examples)
OR
In case of polyembryony, an embryo A develops from the synergids and the embryo B develops from the nucellus. State the ploidy of embryo A and B.

Ans. A- Haploid ; B- Diploid

Q.2. Write the physiological reason, why a woman generally cannot conceive a child after 50 years of age ?
Ans. In human beings, menstrual cycle ceases around at 50 years of age and there will be no production of egg. It is the phase in a woman's life when menopause occurs, ovulation and menstruation stop. Hence, a woman cannot conceive a child after 50 years of age.

Q.3. How does pollination takes place in water hyacinth and water lily ?
Ans. In water hyacinth and water lily, the flowers emerge above the level of water and are pollinated by insects or wind as in most of the land plants.

Q.4. Suggest a method to ensure an anamnestic response in humans.
Ans. Vaccination or Immunization (Active / passive) or weakened or inactive microbes or pathogens or proteins or antibodies introduced into the host body.

Q.5. Who developed a graphical representation of a genetic cross called "Punnett Square".
Ans. R.C. Punnett.

Q.6. Suggest a molecular diagnostic procedure that detects HIV in a suspected AIDS patient.
Ans. PCR/ELISA
Detailed Answer :
PCR- Polymerase Chain Reaction.
ELISA test-(Enzyme Linked Immuno Sorbent Assay).

Q.7. Predict the effect if, the codon UAU coding for an amino acid at the 25th position of a polypeptide of 50 amino acids, is mutated to UAA.
Ans. A polypeptide of 24 amino acids will be formed as UAA is a stop codon which will prevent further translation.

Q.8. What makes RNA more reactive in comparison to DNA?
Ans. 2'–OH present in RNA (in every nucleotide) makes it more reactive.

Q.9. Why it is not possible for an alien DNA to become part of a chromosome anywhere along its length and replicate normally ?
Ans. Alien DNA must be linked to ori / origin of replication / site to start replication.

Q.10. What does nature’s carrying capacity for a species indicate ?
Ans. In nature a given habitat has enough or limited resources to support a maximum possible number and nature’s carrying capacity indicates that no further growth in population is possible.

Q.11. Assertion : Primary transcripts in eukaryotes are non-functional.
Reason : Methyl guanosine triphosphate is attached to 5’ – end of hnRNA.
(a) Both assertion and reason are true, and reason is the correct explanation of assertion. (b) Both assertion and reason are true, but reason is not the correct explanation of assertion.
(c) Assertion is true but reason is false.
(d) Both assertion and reason are false.
Ans. (b) Both Assertion and Reason are true, but Reason is not the correct explanation of Assertion.

OR

Assertion : An organism with lethal mutation may not even develop beyond the zygote stage.

Reason : All types of gene mutations are lethal.
(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. (b) Both Assertion and Reason are true but Reason, is not the correct explanation of Assertion.

Q.12. Assertion (A) : EcoRI is a restriction endonuclease enzyme.
Reason (R) : Exonuclease removes nucleotides from the ends of DNA.

(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. (b) EcoRI is restriction endonuclease enzyme.  Exonuclease removes nucleotides from the ends of DNA.  EcoRI makes cuts at specific position within the DNA.

Q.13. Assertion (A): ‘Saheli’ is considered as an improved form of contraceptive for human females.
Reason : It is a non-steroidal preparation and is once a week pill.
(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) ‘Saheli’ is considered as an improved form of contraceptive for human females because :
(i) It is a non-steroidal preparation.
(ii) It has lesser or no side effects.
(iii) It has a high contraceptive value.
(iv) It is once a week pill.

Q.14. Assertion : Many species like Stellar ’s sea cow, Passenger pigeon etc became extinct due to over exploitation.
Reason : Over exploitation is a major cause of biodiversity loss.

(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. (b) Many species like Stellar ’s sea cow, Passenger pigeon etc-became extinct due to over exploitation. The dependence of humans on nature for food and shelter led to over-exploitation of natural resources. It is one of the major causes of biodiversity loss.

Q.15. Read the following Passage and answer any four questions from 15(i) to 15(v) given below :
Ecological Indicators :

The presence of dragonflies can reveal changes in the water ecosystems more quickly than studying other animals or plants. In fact, from the nymph to the adult stage, the dragonfly has a significant, positive ecological impact. Dragonfly eggs are laid and hatched in or near water, so their lives show impact on both water and land ecosystems. Once hatched, dragonfly nymphs can breathe in the air or underwater which enables them to eat mosquito larvae, other aquatic insects and worms, and even small aquatic vertebrates like tadpoles and small fish. Adult dragonflies capture and eat adult mosquitoes.
Community wide mosquito control programs that spray insecticides to kill adult mosquitoes also kill dragonflies.
(i) The approach to biological control includes:
(a) Import and release of an insect pest to a new area to provide hosts for natural enemies.
(b) Import and release of natural enemies from the native home of an alien insect pest that has invaded a new area.
(c) Preservation of natural enemies (predators & parasites) that are already established in an area.
(d) Use of insecticides to reduce alien insect pests to establish new equilibrium position.
Ans. (i) (a) Preservation of natural enemies (predators & parasites) vectors.

(ii) Two diseases less likely to occur in a region with plenty of dragonflies are_____
(a)
Yellow fever and amoebic dysentery
(b) Malaria and Yellow fever
(c) Anthrax and typhoid
(d) Cholera and typhoid
Ans. (ii) (b) Malaria and yellow fever

(iii) Dragonflies indicate positive ecological impact as.
(a) 
The presence of dragonflies indicates polluted water.
(b) Dragonfly nymphs selectively eat mosquito larvae.
(c) They help to decrease the probability of diseases spread by vectors.
(d) Dragonfly do not cause any harm to beneficial species.
Ans. (iii) (c) They help to decrease the probability of diseases spread by mosquitoes, horseflies and deer flies.

(iv) The most effective stages in the life cycle of dragonfly that eradicate mosquitoes area.
(a)
Larvae and adult
(b) Caterpillar and adult
(c) Nymph and adult
(d) Pupa and adult
Ans. (iv) (c) Nymph and Adult

(v) Assertion : Releasing dragonflies in areas where there is an outbreak of malarial diseases can be an environment friendly method of control.
Reason : Dragon flies are dominant species and will not allow mosquitoes to reproduce
(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. (v) (c) Assertion is true statement, but reason is false.

Q.16. With respect to Meselson and Stahl’s experiment, answer any of the four questions.
(a) Meselson and Stahl demonstrated _____________ mode of replication of DNA molecule.
Ans. (a) Meselson and Stahl demonstrated semi conservative mode of replication of a DNA molecule.

(b) Identify the method used to distinguish between heavy and light isotopes of nitrogen.
Ans. (b) Centrifugation in a CsCl density gradient.

(c) With the help of diagrams, compare the results for the DNA isolated after 20 minutes of experiment with the DNA which was isolated after 40 minutes.
Ans. (c) Since E. coli divides every 20 minutes, the DNA extracted after 20 minutes of experiment had a hybrid density (14N15N).
The DNA extracted after 40 minutes had equal amounts of hybrid and light densities (i.e. 50% hybrid 14N15N, 50% light 14N14N).

(d) If Meselson and Stahl’s experiment is continued for four generations in bacteria, the ratio of 15N/15N: 15N/14N: 14N/14N containing DNA in the fourth generation would be
(i) 1 : 1 : 0
(ii) 1 : 4 : 0
(iii) 0:1:3
(iv) 0:1:7
Ans. (d) If Meselson and Stahl’s experiment is continued for four generations in bacteria, the ratio of 15N/15N : 15N/14N : 14N/14N containing DNA in the fourth generation would be 0 : 1 : 7. After third generation (60 min.) bacteria contains 25% hybrid (15N14N) in 1 : 3 ratio. After 4th generation (80 min.) bacteria contains 12.5% hybrid and 87.5% light DNA in 1 : 7 ratio.

(e) Semi-conservative replication of DNA was first demonstrated in :
(i) 
Drosophila melanogaster
(ii) 
Escherichia coli
(iii) 
Streptococcus pneumoniae
(iv) 
Salmonella typhimurium
 Ans. 
(e) (ii)  Meselson and Stahl experimentally proved semi-conservative replication of DNA in E.coli.
Section ‘B’
Q.17. A mature embryo-sac in a flowering plant may possess 7-cells, but 8-nuclei. Explain with the help of a diagram only.

Ans.
Biology: CBSE Sample Question Paper (2020-21) - 3 Notes | Study Sample Papers for Class 12 Medical and Non-Medical - Class 12
Q.18. Why does the lac operon shut down some time after the addition of lactose in the medium where E.coli was growing? Why low level expression of lac operon is always required?
Ans. After addition of lactose, complete breakdown of lactose to glucose and galactose takes place. Therefore, there is no more lactose to bind to the repressor protein and the lac operon shuts down.1 A very low level of expression of lac operon has to be present in the cell all the time, otherwise lactose cannot enter the cells.

Q.19. Explain four advantages of mycorrhizal association to plants.
Ans. The fungal symbiont in mycorrhizal associations with plants:
(i) absorbs phosphorus from soil and passes it to the plant.
(ii) provides resistance to root-borne pathogens,
(iii) enhances tolerance to salinity and drought,
(iv) induces an overall increase in plant growth and development.

Q.20. How are sticky ends formed on a DNA strand? Why are they so called?
Ans. Sticky ends on a DNA strand are formed by the action of enzymes restriction endonucleases. These enzymes cut the strand of DNA a little away from the centre of the palindromic sequence between the same two bases on both the strands. This results in single stranded stretches on both the complementary strands at their end.
These overhanging stretches are called sticky ends as they form hydrogen bonds with the complementary base pair sequences.

Q.21. A mature embryo-sac in a flowering plant may possess 7-cells, but 8-nuclei. Explain with the help of a diagram only.
Ans. “Altitude sickness” is because of low atmospheric pressure at high altitude, the body does not get sufficient oxygen.
The body compensates low oxygen availability by increasing RBCs production, decreasing the binding capacity of haemoglobin, by increasing breathing rate.

Q.22. Substantiate with the help of one example that in an ecosystem mutualists (i) tend to co-evolve and (ii) are also one of the major causes of biodiversity loss.
Ans. Shark : Tolerates wide range of temperature so its species is wide spread / survives in all waters.
Polar bear : Restricted occurrence in narrow range of temperature so it is constraint to live in very cold icy environment.

Q.23. Explain how advanced ex-situ conservation techniques assist in preserving threatened species of plants and animals.
Ans.

  • Advanced techniques are being used now for ex-situ conservation. Gametes of threatened species can be preserved in viable and fertile condition for long periods using cryopreservation techniques. Eggs can, thus, be fertilized in vitro.
  • In plants, the explants can be propagated using tissue culture methods and can be kept for long periods in seed banks.

Q.24. What is bioreactor? How has the development of bioreactor helped in biotechnology ?
Ans. Bioreactors are vessels in which raw materials are biologically converted into specific products, enzymes etc. using microbial plants, animal or human cells. Importance : Larger biomass / large volume of culture can be processed leading to higher yields of desired specific products (protein / enzymes), under controlled condition.

Q.25. Discuss the role the enzyme DNA ligase plays during DNA replication.
Ans. (Discontinuous) DNA fragments are joined/sealed by them//sticky ends of vector and foreign DNA, joined by them. (The following diagram can be considered in lieu of explanation).
Biology: CBSE Sample Question Paper (2020-21) - 3 Notes | Study Sample Papers for Class 12 Medical and Non-Medical - Class 12
Q.26. (a) Given below is a single stranded DNA molecule. Frame and label its sense and antisense RNA molecule. 5’ ATGGGGCTC 3’ sense
(b) How the RNA molecules made from above DNA strand help in silencing of the specific RNA molecules?

Ans. (a) 5’ ATGGGGCTC 3’ sense
3’ TACCCCGAG 5’ antisense
RNA 5’AUGGGGCUC 3’ sense
3’UACCCCGAG 5’ antisense
(b) The two strands of RNA (i.e. sense and antisense) being complementary will bind with each other and form double stranded RNA. As a result, its translation and protein expression would be inhibited.
OR
Explain the mechanism of DNA replication with the help of a replication fork. What role does the enzyme DNA ligase play in a DNA replication fork ?

Ans. (a) The process of DNA replication begins at a point called the origin of replication (ori), to form a replication fork.
(b) The separated strands act as templates for the synthesis of new strands.
(c) DNA replicates in the 5' ® 3' direction.
(d) dNTPs (Deoxyribonucleotide triphosphate) act as substrate and also provide energy for polymerization of nucleotides.
(e) DNA polymerase is an enzyme that assembles a new DNA strand that is complementary to the template strand.
(f) DNA polymerase continues to move along the template strand and add new nucleotides to the growing or complementary strand until the entire genome is replicated.
(g) The DNA polymerase forms one new strand (leading strand) in a continuous stretch in the 5' → 3' direction (Continuous synthesis).
(h) The other new strand is formed in small stretches (Okazaki fragments) in 5' ® 3' direction (discontinuous synthesis).
(i) The Okazaki fragment are then joined together to form a new strand by an enzyme, DNA ligase. This new strand is called lagging strand.
Biology: CBSE Sample Question Paper (2020-21) - 3 Notes | Study Sample Papers for Class 12 Medical and Non-Medical - Class 12The function of DNA ligase is to join two nucleotides. During the DNA replication process, it joins Okazaki fragments together to form the complete DNA strand.

Q.27. How would you find out the genotype of a pea plant with violet flowers? Explain with the help of Punnett square showing crosses.
Ans.

Biology: CBSE Sample Question Paper (2020-21) - 3 Notes | Study Sample Papers for Class 12 Medical and Non-Medical - Class 12Q.28. (a) Draw the figure of vector pBR322 and label the following:

  • Origin of replication
  • Ampicillin resistance site
  • Tetracycline resistance site
  • Bam H1 restriction site

(b) Identify the significance of Origin of replication.
Ans. (a)
Biology: CBSE Sample Question Paper (2020-21) - 3 Notes | Study Sample Papers for Class 12 Medical and Non-Medical - Class 12
(b) Origin of replication is responsible for controlling the copy number of the DNA sequence inserted.

Section 'C'

Q.29. Name and describe any three causes of biodiversity losses.
Ans. Habitat loss and fragmentation, Habitat loss from tropical rainforest / The Amazon rain forest is being cut and cleared / for raising cattle / for conversion to grasslands / for cultivating soyabeans / large habitats are broken up into small fragments due to human activities / mammals and birds requiring large territories are badly affected leading to decline in population.
Over exploitation, when ‘need’ turns ‘greed’ lead to over exploitation of natural resources / Steller’s sea cow / passenger pigeon were over exploited / marine fish populations around the world are over exploited / endangering existence of commercially important species
Alien species invasions, when introduced unintentionally or deliberately for any purpose some of them turn invasive and decline indigenous species / carrot grass / Parthenium / African cat fish / Clarias gariepinus poses threat to indigenous cat fishes of our river
Co-extinctions, when a species becomes extinct the plant or animal species associated with it (an obligate way) become extinct / when a host species becomes extinct (its unique assemblage of) parasites meets the same fate / extinction of any member in plant pollinator mutualism leads to extinction of other

Q.30. Explain three basic steps to be followed during genetic modification of an organism.
Ans. (i) Identification of DNA with desirable genes, so that the genetically modified organism has largely desirable genes.
(ii) Introduction of the DNA with desirable genes, into the host using vector.
(iii) Maintenance of introduced DNA in the host, and transfer of the DNA to its progeny through cloning.

Section 'D'

Q.31. 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.
Biology: CBSE Sample Question Paper (2020-21) - 3 Notes | Study Sample Papers for Class 12 Medical and Non-Medical - Class 12

(b) Corelate 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 fertilized)
Ans. (a) X- Estrogen secreted by growing follicles;
Y – Progesterone secreted by corpus luteum
(b) Uterine events that take place according to the ovarian hormone levels X and Y on—
(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 fertilised ovum and other events of pregnancy.
(iii) 26 – 28 days (when ovum is not fertilized) : in the absence of fertilisation, corpus luteum degenerates which causes disintegration of endometrium leading to menstruation, marking a new cycle.
OR
The following figure shows a foetus within the uterus. On the basis of the given figure, answer the questions that follow :

Biology: CBSE Sample Question Paper (2020-21) - 3 Notes | Study Sample Papers for Class 12 Medical and Non-Medical - Class 12

(a) In the above figure, choose and name the correct part (A, B, C or D) that act as a temporary endocrine gland and substantiate your answer. Why is it also called the functional junction ?
(b) Mention the role of B in the development of the embryo.
(c) Name the fluid surrounding the developing embryo. How is it misused for sex-determination ?

Ans. (a) Part labelled A -Placenta. It acts as an endocrine tissue as it produces several hormones like Human Chorionic Gonadotropin (hCG), Human Placental Lactogen (hPL), estrogens, progestogens, etc. It is also called the functional junction because it facilitates the supply of oxygen and nutrients to the embryo and removes carbon dioxide and excretory/waste materials produced by the embryo.
(b) The placenta is connected to the embryo through an umbilical cord which helps in the transport of substances to and from the embryo.
(c) Amniotic fluid; a foetal sex determination test is based on the chromosomal pattern of the cells in the amniotic fluid surrounding the developing embryo.

Q.32. Certain phenotypes in human population are spread over a gradient and reflect the contribution of more than two genes. Mention the term used for the type of inheritance? Describe it with the help of an example in human population.
Ans.
Polygenic inheritance
If we assume skin colour is controlled by three genes A, B, C.
Dominant forms (A, B, C) are responsible for dark skin colour and recessive form (a, b, c) for light skin colour.
The genotype with all dominant alleles (AABBCC) will be darkest skin colour and with recessive alleles (aabbcc) will be light skin colour.
OR
Summarize the process by which the sequence of DNA bases in Human Genome Project was determined using the method developed by Frederick Sanger. Name a free living non-pathogenic nematode whose DNA has been completely sequenced.

Ans. The sequences were arranged based on some overlapping regions present in them (Alignment of these sequences was not humanly possible).
Therefore, specialized computer based programme was developed.
These sequences were subsequently annotated and were assigned to each chromosome
Caenorhabditis elegans

Q.33. Give the scientific name of the parasite that causes malignant malaria in humans. At what stage does the parasite enter the human body? Trace its life cycle in human body.
Ans. 
Plasmodium falciparum causes malaria.
It enters into human body in sporozoite form.

Life cycle of Malaria parasite in human body is as follows: The malarial parasite requires two hosts – human and Anopheles, to complete their life cycle.
(a) Life cycle of plasmodium starts with inoculation of sporozoites (infective stage) through the bite of infected female Anopheles mosquitoes.
(b) The parasite is initially multiplied within the liver cells and then attack the red blood cells (RBCs) resulting in their rupture.
(c) There is release of a toxic substance called hemozoin from the ruptured RBCs which is responsible for the chill and high fever.
(d) From the infected human, the parasite enters into the body of Anopheles mosquito during biting and sucking blood.
(e) Further development takes place in the body of Anopheles mosquitoes.
(f) The female mosquito takes up gametocytes with the blood meal.
(g) Formation of gametes and fertilization takes place in the intestine of mosquito.
(h) The zygote develops further and forms thousands of sporozoites which migrates into the salivary gland of mosquito. When the mosquito bite another human, sporozoites are injected.
OR

(a) If a patient is advised anti-retroviral drug, name the possible infection he/she is likely to be suffering from. Name the causative organism.
(b) How do vaccines prevent subsequent microbial infection by the same pathogen?
(c) How does a cancerous cell differ from a normal cell?
(d) Many microbial pathogens enter the gut of humans along with food. Name the physiological barrier that protects the body from such pathogens.

Ans.
(a) AIDS caused by the Human Immuno deficiency Virus.
(b) Vaccines prevent microbial infections by initiating production of antibodies against these antigens to neutralise the pathogenic agents during later actual infection.
The vaccines also generate memory – B and T-cells that recognize the pathogen quickly on subsequent exposure.
(c) Normal cells show a property called contact inhibition by virtue of which contact with other cells which inhibits their uncontrolled growth.
Cancer cells appear to have lost this property.
These cells grow very rapidly, invading and damaging the surrounding normal tissues.
Cells sloughed from such tumors reach distant sites through blood, and wherever they get lodged in the body, they start a new tumor there. This property is called metastasis.
(d) Physiological barriers : Acid in the stomach and saliva in the mouth.

The document Biology: CBSE Sample Question Paper (2020-21) - 3 Notes | Study Sample Papers for Class 12 Medical and Non-Medical - Class 12 is a part of the Class 12 Course Sample Papers for Class 12 Medical and Non-Medical.
All you need of Class 12 at this link: Class 12

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