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CUET 2022 Biology Question Paper with Answers & Solutions

50 questions with answer key & explanations

Q1.
Select the statements that are CORRECT regarding patterns of biodiversity. (A) Species diversity increases as we move away from the equator towards the poles (B) The tropical Amazon rain forest in South America has the greatest biodiversity on earth. (C) There is more solar energy available in the temperate region than tropics. (D) Tropical environments are less seasonal relatively more constant and predictable (E) A. Humboldt observed that within a region, species richness decreases with increasing explored area, up to a limit. Choose the correct answer from the options given below :
A. (A) and (E) only
B. (B) and (C) only
C. (B) and (D) only
D. (C), (D) and (E) only
Show answer & explanation

Correct answer: C

Species diversity decreases (not increases) from equator to poles, so (A) is wrong. Tropics receive more solar energy, so (C) is wrong. Humboldt showed species richness increases with area, so (E) is wrong. The Amazon rainforest has the greatest biodiversity (B correct) and tropical environments are less seasonal, more constant and predictable (D correct). Hence (B) and (D) only.

Q2.
Flowers are a fascinating organ of Angiosperms. Flowers are considered the object of aesthetic, ornamental, social and cultural value. Biologically flower is a __________.
A. Modified root
B. Modified shoot
C. Modified leaf
D. Modified tuber
Show answer & explanation

Correct answer: B

A flower is biologically a modified shoot, where the shoot apical meristem changes to a floral meristem bearing modified leaves (sepals, petals, stamens, carpels).

Q3.
Which of the following hormone is not produced by placenta ?
A. hCG
B. hPL
C. estrogen
D. androgens
Show answer & explanation

Correct answer: D

The placenta secretes hCG, hPL, estrogens, progestogens and relaxin. Androgens are not produced by the placenta.

Q4.
Individuals with karyotype of 44+XXY having overall masculine development with few feminine development like Gynaecomastia has __________ chromosomal disorder.
A. Klinefelter's Syndrome
B. Turner's Syndrome
C. Down's Syndrome
D. Acquired Immuno Deficiency Syndrome
Show answer & explanation

Correct answer: A

Karyotype 44+XXY (47, XXY) with masculine development plus gynaecomastia (feminine breast development) is characteristic of Klinefelter's Syndrome.

Q5.
Which one of the following enzyme brings about hydrolysis of lactose to glucose and galactose ?
A. Transacetylase
B. Amylase
C. Permease
D. β-galactosidase
Show answer & explanation

Correct answer: D

β-galactosidase (lactase) hydrolyses lactose into glucose and galactose. It is coded by the lacZ gene of the lac operon.

Q6.
Match List - I with List - II. List - I: (A) Streptokinase (B) Cyclosporin A (C) Statins (D) Swiss Cheese List - II: (I) Blood-Cholestrol lowering agents (II) Clot Buster (III) Propionibacterium sharmanii (IV) Immuno suppressive agent Choose the correct answer from the options given below :
A. (A) - (II), (B) - (IV), (C) - (III), (D) - (I)
B. (A) - (II), (B) - (IV), (C) - (I), (D) - (III)
C. (A) - (IV), (B) - (I), (C) - (II), (D) - (III)
D. (A) - (IV), (B) - (II), (C) - (III), (D) - (I)
Show answer & explanation

Correct answer: B

Streptokinase is a clot buster (II); Cyclosporin A is an immunosuppressive agent (IV); Statins are blood-cholesterol lowering agents (I); Swiss cheese is associated with Propionibacterium sharmanii (III). Hence A-II, B-IV, C-I, D-III.

Q7.
Which of the following option determines percolation and water holding capacity of soils ?
A. Climate
B. Grain size, soil composition and aggregation
C. Weathering process
D. Soil development
Show answer & explanation

Correct answer: B

Soil texture, grain size, composition and aggregation determine the percolation and water holding capacity of soil (NCERT, Organisms and Populations).

Q8.
Which of the following is not a barrier method of birth control.
A. Voults
B. Diaphragms
C. Sterilization
D. Cervical caps
Show answer & explanation

Correct answer: C

Vaults, diaphragms and cervical caps are barrier methods. Sterilization (tubectomy/vasectomy) is a surgical method, not a barrier method.

Q9.
'Golden rice' variety of rice shows :
A. enhanced nutritional value of food
B. less post harvest loss
C. tolerence to abiotic stress
D. pest resistance
Show answer & explanation

Correct answer: A

Golden rice is a transgenic rice enriched with provitamin A (beta-carotene), enhancing the nutritional value of food.

Q10.
Arrange the stages of bio magnification of DDT. (A) Fish eating birds (PPT 25 PPm) (B) Small fish (DDT 0.5 PPm) (C) Zooplankton (DDT 0.04 PPm) (D) Water (DDT 0.003 PPb) (E) Large fish (DDT 2 PPm) Choose the correct answer from the options given below :
A. (B) → (C) → (E) → (A) → (D)
B. (D) → (C) → (B) → (E) → (A)
C. (D) → (B) → (C) → (E) → (A)
D. (E) → (D) → (B) → (C) → (A)
Show answer & explanation

Correct answer: B

Biomagnification increases up the trophic levels: Water (0.003 ppb) → Zooplankton (0.04 ppm) → Small fish (0.5 ppm) → Large fish (2 ppm) → Fish-eating birds (25 ppm). This is D → C → B → E → A.

Q11.
Which of the following is not is not the cause of biodiversity loss ?
A. Co-Extinction
B. Over-exploitation
C. Endemism
D. Alien species invasions
Show answer & explanation

Correct answer: C

The 'evil quartet' causes of biodiversity loss are habitat loss, over-exploitation, alien species invasions and co-extinctions. Endemism (species restricted to a region) is not a cause of loss.

Q12.
Which of the following statements are correct ? (A) Certain mass of living material at each trophic level is called as standing crop. (B) The crop that can withstand adverse conditions is called standing crop (C) The amount of nutrients in soil is called Biomass (D) Only Biotic components make an Ecosystem (E) Most of Phytoplanktons are member of algae Choose the correct answer from the options given below :
A. (A) and (E) only
B. (A), (C), (D) only
C. (A), (C) only
D. (B), (D) only
Show answer & explanation

Correct answer: A

Standing crop is the mass of living material at each trophic level (A correct). Phytoplankton are mostly algae (E correct). (B) is wrong (that is not standing crop), (C) is wrong (nutrients in soil = standing state, not biomass), (D) is wrong (ecosystem has both biotic and abiotic components). Hence (A) and (E) only.

Q13.
The shape of pyramids gives the reflection of growth status of a population. Identify the growth status depicted by the given pyramid.
[Figure in original paper — see source PDF]
A. Expanding
B. Declining
C. Exploding
D. Stable
Show answer & explanation

Correct answer: B

An age pyramid with a narrow pre-reproductive base and a broad post-reproductive top indicates fewer young individuals and a shrinking population, i.e. a declining (diminishing) population.

Q14.
Which of the following is NOT an application of PCR ?
A. Paternity testing
B. Detection of mutations of genes in suspected cancer patient
C. Powerful technique to identify genetic disorder
D. To cure ADA (adenosine diaminase) deficiency
Show answer & explanation

Correct answer: D

PCR is used in paternity testing, detecting gene mutations and identifying genetic disorders. Curing ADA deficiency is done by gene therapy, not PCR.

Q15.
Complementary ds RNA which prevents translation is formed in __________.
A. PCR
B. RNA interference
C. Gene therapy
D. ELISA
Show answer & explanation

Correct answer: B

RNA interference (RNAi) involves complementary dsRNA that silences a specific mRNA, preventing its translation.

Q16.
Which enzymes are used for clarification of bottled fruit juices ? (A) Amylases (B) Pectinases (C) Proteases (D) Lipases Choose the correct answer from the options given below :
A. (A) only
B. (A) and (B) only
C. (B) and (C) only
D. (C) and (D) only
Show answer & explanation

Correct answer: C

Bottled fruit juices are clarified using pectinases and proteases, which make the juice clearer. Hence (B) and (C) only.

Q17.
New breed of sheep is developed by __________.
A. Cross breeding
B. Inbreeding
C. Outcrossing
D. Interspecific hybridisation
Show answer & explanation

Correct answer: A

Cross breeding (mating superior males of one breed with superior females of another) is used to develop new breeds; the Hisardale sheep was developed by cross breeding Bikaneri ewes and Marino rams.

Q18.
Which of the following is incorrect about oral contraceptive pill 'Saheli' ?
A. It is taken by females
B. It has very few side effects
C. It is a steroidal preparation.
D. It is 'Once a week' pill
Show answer & explanation

Correct answer: C

Saheli is a non-steroidal 'once a week' oral pill for females with very few side effects. Hence stating it is a steroidal preparation is incorrect.

Q19.
Vertical distribution of different species occupying different level is called :
A. Fragmentation
B. Stratification
C. Humification
D. Primary production
Show answer & explanation

Correct answer: B

The vertical distribution of different species occupying different levels (e.g. trees, shrubs, herbs) is called stratification.

Q20.
Match List - I with List - II. List - I: (A) Initiation factor (B) Introns (C) Termination factor (D) Adenylate residue List - II: (I) Tailing (II) Rho(p) (III) Sigma(σ) (IV) Splicing Choose the correct answer from the options given below :
A. (A) - (IV), (B) - (III), (C) - (II), (D) - (I)
B. (A) - (III), (B) - (II), (C) - (IV), (D) - (I)
C. (A) - (IV), (B) - (III), (C) - (I), (D) - (II)
D. (A) - (III), (B) - (IV), (C) - (II), (D) - (I)
Show answer & explanation

Correct answer: D

Initiation factor in prokaryotes is the sigma (σ) factor (III); introns are removed by splicing (IV); termination is by Rho (ρ) factor (II); adenylate residues are added during tailing/polyadenylation (I). Hence A-III, B-IV, C-II, D-I.

Q21.
The reason for deviation from Mendel's dihybrid cross in T.H. Morgan's experiment is __________.
A. Pleiotropy
B. Linkage
C. Overlapping
D. Polygenic Inheritance
Show answer & explanation

Correct answer: B

Morgan's experiments in Drosophila showed that genes located on the same chromosome are linked and do not assort independently, causing deviation from Mendel's dihybrid ratio. This is due to linkage.

Q22.
Given below are two statements : Statement I : Phenylketonuria is an example of Pleiotropy Statement II : Affected individuals lack an enzyme which converts pheylalanine into tyrosine In the light of the above statements, choose the most appropriate answer from the options given below :
A. Both Statement I and Statement II are correct
B. Both Statement I and Statement II are incorrect
C. Statement I is correct but Statement II is incorrect
D. Statement I is incorrect but Statement II is correct
Show answer & explanation

Correct answer: A

Phenylketonuria is an inborn error of metabolism; affected individuals lack the enzyme phenylalanine hydroxylase that converts phenylalanine to tyrosine, leading to multiple effects (mental retardation, etc.) - hence an example of pleiotropy. Both statements are correct.

Q23.
Sequentially arrange the developmental stages of human spermatozoon : (A) Secondary Spermatocyte (B) Spermatid (C) Spermatozoa (D) Spermatogonia (E) Primary spermatocytes Choose the correct answer from the options given below :
A. (E), (A), (D), (B), (C)
B. (E), (A), (D), (C), (B)
C. (E), (A), (C), (D), (B)
D. (D), (E), (A), (B), (C)
Show answer & explanation

Correct answer: D

Spermatogenesis sequence: Spermatogonia (D) → Primary spermatocyte (E) → Secondary spermatocyte (A) → Spermatid (B) → Spermatozoa (C). This is D, E, A, B, C.

Q24.
Which part of the sperm help in its entry into the cytoplasm of the ovum ?
A. Plasma membrane
B. Nucleus
C. Neck
D. Acrosome
Show answer & explanation

Correct answer: D

The acrosome of the sperm contains enzymes (hyaluronidase) that help the sperm penetrate the zona pellucida and enter the cytoplasm of the ovum.

Q25.
Types of cells in 2 celled pollen grains are __________.
A. Central cell and Synergids
B. Antipodals and Megaspore
C. Micropylar and Filliform
D. Vegetative and Generative
Show answer & explanation

Correct answer: D

A 2-celled pollen grain consists of a large vegetative (tube) cell and a smaller generative cell.

Q26.
Among the animal on the planet, the species rich group making more than 70% is __________.
A. Fishes
B. Mammals
C. Insects
D. Reptiles
Show answer & explanation

Correct answer: C

Insects constitute more than 70% of all animal species, making them the most species-rich group (NCERT, Biodiversity and Conservation).

Q27.
Replication of DNA is characterised by : (A) The direction of replication is 5' → 3' (B) Only template with 5' → 3' polarity is replicated. (C) Replication is initiated at ori. (D) DNA polymerase catalyses the process. (E) The daughter molecule formed has one parental strand Choose the correct answer from the options given below :
A. (A), (B), (D), (E)
B. (A), (B), (C), (D)
C. (A), (C), (D), (E)
D. (B), (C), (D), (E)
Show answer & explanation

Correct answer: C

DNA synthesis proceeds 5'→3' (A correct); replication begins at the origin 'ori' (C correct); DNA polymerase catalyses it (D correct); replication is semiconservative so each daughter molecule has one parental strand (E correct). Statement (B) is wrong because both template strands are replicated. Hence (A), (C), (D), (E).

Q28.
Match List - I with List - II. List - I: (A) Detritus food chain (B) Standing state (C) Standing crop (D) Net Primary Productivity List - II: (I) Available biomass for consumption (II) Dead organic matter (III) Amount of nutrients in soil (IV) Mass of living material Choose the correct answer from the options given below :
A. (A) - (I), (B) - (II), (C) - (III), (D) - (IV)
B. (A) - (IV), (B) - (II), (C) - (III), (D) - (I)
C. (A) - (II), (B) - (III), (C) - (I), (D) - (IV)
D. (A) - (II), (B) - (III), (C) - (IV), (D) - (I)
Show answer & explanation

Correct answer: D

Detritus food chain begins with dead organic matter (II); standing state is the amount of nutrients in soil (III); standing crop is the mass of living material (IV); net primary productivity is the available biomass for consumption by heterotrophs (I). Hence A-II, B-III, C-IV, D-I.

Q29.
In the technology called MOET, which one of the following hormones is used ?
A. LH
B. ACTH
C. FSH
D. TSH
Show answer & explanation

Correct answer: C

In MOET (Multiple Ovulation Embryo Transfer), FSH-like hormones are administered to induce superovulation (follicular maturation and multiple ovulation).

Q30.
Vegetative propagation in Eicchornia and Pistia occurs by __________.
A. Sucker
B. Offset
C. Runner
D. Stolon
Show answer & explanation

Correct answer: B

Water hyacinth (Eichhornia) and Pistia propagate vegetatively by offsets (short, thick lateral branches arising in aquatic plants).

Q31.
The term "Clone" is used to describe the offspring that are :
A. Morphologically identical only
B. Morphologically and genetically identical
C. Morphologically identical but genetically different
D. Genetically identical only
Show answer & explanation

Correct answer: B

A clone refers to offspring that are morphologically and genetically identical to the parent (e.g. produced by asexual reproduction).

Q32.
Why was Drosophila melanogaster used for studies in Genetics ? (A) They could be cultured easily in the monastery (B) They showed many contrasting traits (C) The generation time was one year (D) There was clear differentiation of sexes (E) Very few progeny were produced in a single mating Choose the correct answer from the options given below :
A. (B) and (D) only
B. (A) and (D) only
C. (B) and (C) only
D. (D) and (E) only
Show answer & explanation

Correct answer: A

Drosophila were used because they could be grown easily in the lab (not monastery), have a short life cycle of about two weeks (not one year), produce large progeny, show many hereditary contrasting traits, and have clear sexual dimorphism. The valid reasons here are (B) showing contrasting traits and (D) clear differentiation of sexes. Hence (B) and (D) only.

Q33.
Given below are two statements : Statement I : Ladybird and Dragonflies are useful to get rid of aphids and mosquitoes. Trichoderma fungus are effective bio control agents of several plant pathogens. Statement II : The biological control of plant diseases and pest can control increasing use of insecticides and pesticides, thus saving our environment from being getting polluted. In the light of the above statements, choose the most appropriate answer from the options given below :
A. Both Statement I and Statement II are correct
B. Both Statement I and Statement II are incorrect
C. Statement I is correct but Statement II is incorrect
D. Statement I is incorrect but Statement II is correct
Show answer & explanation

Correct answer: A

Ladybirds and dragonflies control aphids and mosquitoes, and Trichoderma is a biocontrol agent against plant pathogens (Statement I correct). Biological control reduces dependence on chemical pesticides, reducing pollution (Statement II correct). Both are correct.

Q34.
To prove theory of mutation, Hugo de Vries used __________ plant.
A. Snap dragon or Antirrhinum
B. Evening primrose
C. Dog flower
D. Pisum Sativum
Show answer & explanation

Correct answer: B

Hugo de Vries proposed the theory of mutation based on his work on the evening primrose (Oenothera lamarckiana).

Q35.
Match List - I with List - II. List - I: (A) ZIFT (B) IUI (C) GIFT (D) ART List - II: (I) Semen is artificially introduced into female (II) Couples are assisted to have children by corrective treatment (III) Zygote can be transferred into Fallopian tube (IV) Ovum can be transferred to Fallopian tube of another female Choose the correct answer from the options given below :
A. (A) - (IV), (B) - (I), (C) - (III), (D) - (II)
B. (A) - (III), (B) - (I), (C) - (IV), (D) - (II)
C. (A) - (I), (B) - (II), (C) - (III), (D) - (IV)
D. (A) - (II), (B) - (III), (C) - (IV), (D) - (I)
Show answer & explanation

Correct answer: B

ZIFT transfers the zygote/early embryo into the fallopian tube (III); IUI is intra-uterine insemination where semen is artificially introduced (I); GIFT transfers an ovum from a donor into the fallopian tube of another female (IV); ART (assisted reproductive technologies) assist couples to have children (II). Hence A-III, B-I, C-IV, D-II.

Q36.
Human activities like over cultivation, unrestricted grazing, deforestation and poor irrigation practices result into __________.
A. Water Logging
B. Soil erosion and desertification
C. Biomagnification
D. Eutrophication
Show answer & explanation

Correct answer: B

Overcultivation, unrestricted grazing, deforestation and poor irrigation lead to soil erosion and desertification (NCERT, Environmental Issues).

Q37.
Match List - I with List - II. List - I: (A) Natality (B) Mortality (C) Immigration (D) Emigration List - II: (I) Number of individuals coming into the habitat from elsewhere (II) Number of births in the population (III) Number of individuals leaving the habitat (IV) Number of deaths in the population Choose the correct answer from the options given below :
A. (A) - (II), (B) - (IV), (C) - (I), (D) - (III)
B. (A) - (IV), (B) - (II), (C) - (I), (D) - (III)
C. (A) - (I), (B) - (IV), (C) - (III), (D) - (II)
D. (A) - (III), (B) - (II), (C) - (I), (D) - (IV)
Show answer & explanation

Correct answer: A

Natality = number of births (II); Mortality = number of deaths (IV); Immigration = individuals coming into the habitat (I); Emigration = individuals leaving the habitat (III). Hence A-II, B-IV, C-I, D-III.

Q38.
A students was repeating the experiments of Alfred Hershey and Maltha Chase (1952). Results obtained by him are shown in the given figure. Select the biomolecule that was radio actively labelled by the students for his experiment.
[Figure in original paper — see source PDF]
A. DNA of bacteriophage with radio active Phosphorous
B. Plasmid DNA of E. coli with radio active Sulphur
C. Proteins of bacteriophage with radio active Sulphur
D. Chromosomal mol DNA of E. coli with radio active Sulphur
Show answer & explanation

Correct answer: A

In the Hershey-Chase experiment, radioactivity found in the pellet (bacterial cells) indicates that the labelled molecule entered the bacteria. Since DNA enters the bacterium, the radioactive label is in the pellet only when DNA is labelled with radioactive phosphorus (32P). Hence DNA of bacteriophage labelled with radioactive phosphorus.

Q39.
Match List - I with List - II. List - I: (A) Prostate gland (B) Leydig cells (C) Ejaculatory duct (D) Penis List - II: (I) Store and transport sperms (II) Male external genitilia (III) Male accessory gland (IV) Testicular hormones Choose the correct answer from the options given below :
A. (A) - (III), (B) - (IV), (C) - (I), (D) - (II)
B. (A) - (IV), (B) - (III), (C) - (I), (D) - (II)
C. (A) - (I), (B) - (III), (C) - (IV), (D) - (II)
D. (A) - (IV), (B) - (II), (C) - (III), (D) - (I)
Show answer & explanation

Correct answer: A

Prostate gland is a male accessory gland (III); Leydig cells secrete testicular hormones/androgens (IV); ejaculatory duct stores and transports sperms (I); penis is the male external genitalia (II). Hence A-III, B-IV, C-I, D-II.

Q40.
Which one of the following is not an example of adaptive radiation ?
A. Australian Marsupials
B. Australian Placental Mammals
C. Moths in England
D. Darwins Finches
Show answer & explanation

Correct answer: C

Australian marsupials, Australian placental mammals and Darwin's finches are classic examples of adaptive radiation. The peppered moth (industrial melanism) in England is an example of natural selection, not adaptive radiation.

The primary lymphoid organs are bone marrow and thymus, where immature lymphocytes differentiate into antigen sensitive lymphocytes. They migrate to secondary lymphoid organ like spleen, lymph nodes, tonsils, payer's patches of small intestine and appendix. All blood cells including lymphocytes are produced in bone marrow which is main lymphoid organ. Spleen is large reservoir of erythrocytes. It is large bean shaped organ mainly containing lymphocyte and phagocyte. It acts as filter of blood by trapping blood borne micro organism. Lymph nodes are small solid structures located at different points along lymphatic system. Antigens trapped in lymph node are responsible for activation of lymphocytes present there and cause the immune response. There is lymphoid tissue located within. Lining of major tracts like respiratory, digestive ad urinogenital tracts called mucous associated lymphoid tissue (MALT).
Q41.
The primary lymphoid organ is __________.
A. Lymph nodes
B. Bone marrow
C. Tonsils
D. Spleen
Show answer & explanation

Correct answer: B

Per the passage, 'The primary lymphoid organs are bone marrow and thymus.' Among the options, bone marrow is the primary lymphoid organ; lymph nodes, tonsils and spleen are secondary lymphoid organs.

Q42.
Which of the following is not feature of spleen ?
A. It is large reservoir of erythrocyte.
B. It acts as filter of blood.
C. It shows reduction in size from birth to puberty
D. It is a bean shaped organ containing Lymphocyte and Phagocyte
Show answer & explanation

Correct answer: C

The passage states the spleen is a large reservoir of erythrocytes, a bean-shaped organ containing lymphocytes and phagocytes, and acts as a blood filter. It does not mention reduction in size from birth to puberty (that is the thymus). Hence option C is not a feature of the spleen.

Q43.
Which of the following is NOT secondary lymphoid organ ?
A. Tonsils
B. Thymus
C. Appendix
D. Payer's Patches
Show answer & explanation

Correct answer: B

Per the passage, secondary lymphoid organs include spleen, lymph nodes, tonsils, Peyer's patches and appendix. The thymus is a primary lymphoid organ, hence NOT a secondary lymphoid organ.

Q44.
Mucous associated lymphoid tissue (MALT) is NOT found in :
A. Urinogenital tract
B. Digestive tract
C. Tonsils
D. Respiratory tract
Show answer & explanation

Correct answer: C

Per the passage, MALT lines the major tracts - respiratory, digestive and urinogenital. Tonsils are a discrete secondary lymphoid organ, not part of MALT tract lining. Hence MALT is NOT found in tonsils.

The primary lymphoid organs are bone marrow and thymus. Where immature lymphocytes differentiate into antigen sensitive lymphocytes. They migrate to secondary lymphoid organ like spleen, lymph nodes, tonsils, payer's patches of small intestine and appendix. All blood cells including lymphocytes are produced in bone marrow which is main lymphoid organ. Spleen is large reservoir of erythrocytes. It is large bean shaped organ mainly containing lymphocyte and phagocyte. It acts as filter of blood by trapping blood borne micro organism. Lymph nodes are small solid structures located at different points along lymphatic system. Antigens trapped in lymph node are responsible for activation of lymphocytes present there and cause the immune response. There is lymphoid tissue located within. Lining of major tracts like respiratory, digestive ad urinogenital tracts called mucous associated lymphoid tissue (MALT).
Q45.
Which of the following is incorrect statement. (A) All blood cells including lymphocytes are produced in bone marrow (B) Lymph nodes are small solid structures located at different points along lymphatic system (C) Payer's patches of small intestine is primary lymphoid organ. (D) Antigen trapped in lymph node are responsible for activation of lymphocytes present in lymph node and causes immune response Choose the correct answer from the options given below :
A. (C) only
B. (B) only
C. (A) only
D. (B) and (D) only
Show answer & explanation

Correct answer: A

Statements A, B and D match the passage and are correct. Statement C is wrong because Peyer's patches of the small intestine are a secondary lymphoid organ, not a primary lymphoid organ. Hence the incorrect statement is (C) only.

The cutting of DNA by restriction endonucleases results in the fragments of DNA. These fragments can be separated by a technique known as gel electrophoresis. Since DNA fragments are negatively charged molecules, they can be separated by forcing them to move towards the anode under an electric field through a medium/matrix. The commonly used matrix is agarose which is a natural polymer extracted from sea weeds. The DNA fragments separate according to their size through sieving effect provided by the agarose gel. Hence, the smaller fragments move farther in the agarose gel. The separated DNA fragments can be visualised only after staining the DNA with ethidium bromide followed by exposure to UV radiation. Bright orange coloured bands of DNA can be observed. The separated bands of DNA are cut out from the agarose gel and extracted from the gel piece. This step is known as elution. The DNA fragments purified in this way are used in constructing recombinant DNA by joining them with cloning vectors.
Q46.
Agarose is extracted from __________.
A. Bacteria
B. Marine animals
C. Sea weeds
D. Fungi
Show answer & explanation

Correct answer: C

Per the passage, agarose is a natural polymer extracted from sea weeds.

Q47.
What is elution ?
A. It is movement of negatively charged DNA fragments through agarose gel.
B. Extraction of DNA from the host
C. Extraction of DNA and treatment with restriction endonuclease
D. Cutting of separated DNA fragments from agarose gel and extraction of DNA fragment
Show answer & explanation

Correct answer: D

Per the passage, elution is the step where the separated bands of DNA are cut out from the agarose gel and extracted from the gel piece. Hence option D.

Q48.
The DNA fragments formed by treatment with endonuclease are separated by __________.
A. PCR
B. Gel Electrophoresis
C. Cloning
D. Restriction digestion
Show answer & explanation

Correct answer: B

Per the passage, DNA fragments produced by restriction endonucleases are separated by the technique known as gel electrophoresis.

Q49.
The separated DNA fragments can be visualised after staining with __________, followed by exposure to __________.
A. β-galactosidase, UV radiation
B. β-galactosidase, Gamma radiation
C. Ethidium bromide, UV radiation
D. Ethidium bromide, Gamma radiation
Show answer & explanation

Correct answer: C

Per the passage, DNA is visualised after staining with ethidium bromide followed by exposure to UV radiation.

Q50.
The fragments of DNA separate on agarose gel, based on the __________.
A. Size of fragments
B. Charge of each fragment
C. Colour of fragment
D. Type of DNA
Show answer & explanation

Correct answer: A

Per the passage, DNA fragments separate according to their size through the sieving effect of the agarose gel, with smaller fragments moving farther.

Original question paper source: National Testing Agency (NTA), CUET (UG) 2022. Reproduced for educational use. Answers & explanations by UniDrill.