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CUET 2024 Biology Question Paper with Answers & Solutions

50 questions with answer key & explanations

Q1.
Analogous structures are a result of :
A. Convergent evolution
B. Divergent evolution
C. Parallel evolution
D. Retrogressive evolution
Show answer & explanation

Correct answer: A

Analogous structures have different structural origins but similar functions, arising due to similar environmental pressures. This is the result of convergent evolution (e.g., wings of butterfly and birds).

Q2.
Which of the following does not affect the Hardy-Weinberg equilibrium ?
A. Natural selection
B. Genetic drift
C. Gene pool
D. Gene migration
Show answer & explanation

Correct answer: C

Factors disturbing Hardy-Weinberg equilibrium are: gene migration (gene flow), genetic drift, mutation, genetic recombination and natural selection. Gene pool is simply the total of all alleles in a population and is not a disturbing factor.

Q3.
Which of the following primates was more like an ape ?
A. Homo erectus
B. Dryopithecus
C. Australopithecines
D. Ramapithecus
Show answer & explanation

Correct answer: B

Dryopithecus was more ape-like, while Ramapithecus was more man-like (more like humans). Dryopithecus was hairy and walked like gorillas and chimpanzees.

Q4.
Nucleosome is associated with _______ molecules of histones.
A. Four
B. Nine
C. Two
D. Eight
Show answer & explanation

Correct answer: D

A nucleosome contains a histone octamer (8 histone molecules: two each of H2A, H2B, H3 and H4) around which DNA is wrapped.

Q5.
Select the observations drawn from the human genome project which are correct. (A) The human genome contains 3164.7 million bp. (B) The average gene consists of 3000 bases. (C) Total number of genes is estimated at 30,000. (D) The functions are unknown for over 50% of discovered genes. (E) Less than 2% of the genome codes for proteins.
A. (A), (B), (C) and (D) only
B. (A), (C), (D) and (E) only
C. (A), (C) and (E) only
D. (A), (B), (C), (D) and (E)
Show answer & explanation

Correct answer: D

All five statements are correct salient features of the Human Genome Project: total genome ~3164.7 million bp; average gene ~3000 bases; ~30,000 genes; functions unknown for over 50% of genes; less than 2% of the genome codes for proteins.

Q6.
Match List-I with List-II : List-I (Placental mammals): (A) Anteater, (B) Bobcat, (C) Lemur, (D) Flying squirrel List-II (Counterpart Marsupials): (I) Spotted cuscus, (II) Numbat, (III) Flying Phalanger, (IV) Tasmanian tiger cat
A. (A) - (II), (B) - (IV), (C) - (I), (D) - (III)
B. (A) - (II), (B) - (I), (C) - (IV), (D) - (III)
C. (A) - (IV), (B) - (I), (C) - (II), (D) - (III)
D. (A) - (IV), (B) - (I), (C) - (III), (D) - (II)
Show answer & explanation

Correct answer: A

Placental–marsupial convergent counterparts: Anteater–Numbat (II), Bobcat–Tasmanian tiger cat (IV), Lemur–Spotted cuscus (I), Flying squirrel–Flying Phalanger (III). So A-II, B-IV, C-I, D-III.

Q7.
Identify the incorrect statement/s : (A) Intestinal perforation and death may occur in severe cases of typhoid infection. (B) Common cold is caused by Rhinoviruses. (C) Lips and fingernails may turn grey to bluish colour in severe cases of pneumonia. (D) Pneumonia is caused by Salmonella. (E) Typhoid fever could be confirmed by Widal test.
A. (A), (C) and (D) only
B. (B) and (E) only
C. (D) only
D. (A) and (D) only
Show answer & explanation

Correct answer: C

Pneumonia is caused by Streptococcus pneumoniae and Haemophilus influenzae, not Salmonella (Salmonella causes typhoid). So statement (D) is the only incorrect one. All others are correct.

Q8.
Match List-I with List-II : List-I (Types of barriers): (A) Cytokine barriers, (B) Physical barriers, (C) Cellular barriers, (D) Physiological barriers List-II (Examples): (I) Mucus coating, (II) Tears from eyes, (III) Phagocytosis, (IV) Interferons
A. (A) - (IV), (B) - (III), (C) - (I), (D) - (II)
B. (A) - (II), (B) - (IV), (C) - (III), (D) - (I)
C. (A) - (II), (B) - (I), (C) - (IV), (D) - (III)
D. (A) - (IV), (B) - (I), (C) - (III), (D) - (II)
Show answer & explanation

Correct answer: D

Cytokine barriers–Interferons (IV); Physical barriers–Mucus coating (I); Cellular barriers–Phagocytosis (III) by PMNL/monocytes; Physiological barriers–Tears from eyes/acid in stomach, saliva (II). So A-IV, B-I, C-III, D-II.

Q9.
Smack is chemically :
A. Diacetyl morphine
B. Cocaine
C. Benzodiazepine
D. Amphetamine
Show answer & explanation

Correct answer: A

Smack (heroin) is chemically diacetylmorphine, obtained by acetylation of morphine, extracted from the latex of Papaver somniferum.

Q10.
Antibodies are secreted by :
A. T-Cells
B. B-Cells
C. α-Cells
D. β-Cells
Show answer & explanation

Correct answer: B

Antibodies are produced by plasma cells which are derived from B-lymphocytes (B-cells). This constitutes humoral (antibody-mediated) immunity.

Q11.
In sewage treatment, flocs are :
A. the solids that settle during sedimentation.
B. the supernatant that is formed above the primary sludge.
C. the masses of bacteria associated with fungal filaments.
D. the bacteria which grow anaerobically and are also called anaerobic sludge digesters.
Show answer & explanation

Correct answer: C

In secondary (biological) treatment of sewage, flocs are masses of aerobic bacteria associated with fungal filaments forming mesh-like structures.

Q12.
Match List-I with List-II : List-I (Products): (A) Statin, (B) Clot buster, (C) Swiss cheese, (D) Cyclosporin-A List-II (Organisms): (I) Streptococcus, (II) Trichoderma, (III) Monascus, (IV) Propionibacterium
A. (A) - (II), (B) - (I), (C) - (IV), (D) - (III)
B. (A) - (III), (B) - (I), (C) - (IV), (D) - (II)
C. (A) - (III), (B) - (IV), (C) - (II), (D) - (I)
D. (A) - (II), (B) - (III), (C) - (I), (D) - (IV)
Show answer & explanation

Correct answer: B

Statin–Monascus purpureus (III); Clot buster (streptokinase)–Streptococcus (I); Swiss cheese (large holes)–Propionibacterium sharmanii (IV); Cyclosporin-A–Trichoderma polysporum (II). So A-III, B-I, C-IV, D-II.

Q13.
The beetle used as a biocontrol agent for aphids and mosquitoes is :
A. Trichoderma
B. Dragonflies
C. Ladybird
D. Silver fish
Show answer & explanation

Correct answer: C

Ladybird (a beetle) is used as a biocontrol agent against aphids. (Dragonflies control mosquitoes but are not beetles.) Among the options the beetle is the Ladybird.

Q14.
Downstream processing method involves :
A. Identification
B. Amplification
C. Fermentation
D. Purification
Show answer & explanation

Correct answer: D

Downstream processing comprises separation and purification of the product, followed by formulation, quality control and clinical trials. It occurs after fermentation/biosynthetic phase.

Q15.
Which of the following is not the correctly matched pair of organism and its respective cell wall degrading enzyme ?
A. Fungi – Chitinase
B. Algae – Methylase
C. Plant cells – Cellulase
D. Bacteria – Lysozyme
Show answer & explanation

Correct answer: B

Algal cell walls are degraded by methylase is incorrect; algae cell walls are degraded by methylcellulase/methylase per NCERT, but the correct enzyme set is Fungi–chitinase, Plant–cellulase, Bacteria–lysozyme, Algae–methylase. NCERT lists Algae–methylase, but the question asks the NOT correctly matched. The genuinely incorrect pair is Algae – Methylase since algal walls (cellulose) are degraded by cellulase, not methylase.

Q16.
Arrange the following steps involved in transformation of bacteria in a sequence from initiation to end. (A) Incubation of rDNA with bacterial cell on ice (B) Treatment with divalent cations (C) Heat shock treatment (D) Selection on antibiotic containing agar plate (E) Placed them again on ice
A. (A), (B), (D), (C), (E)
B. (B), (A), (C), (E), (D)
C. (B), (C), (D), (A), (E)
D. (A), (C), (B), (D), (E)
Show answer & explanation

Correct answer: B

Transformation sequence: treat cells with divalent cations (B) to make them competent, incubate rDNA with cells on ice (A), give heat shock at 42°C (C), place again on ice (E), then select transformants on antibiotic-containing agar plate (D). So B, A, C, E, D.

Q17.
Which of the following statements are incorrect ? (A) Fragments of DNA can be separated by ELISA. (B) Transformation is a procedure through which a piece of DNA is introduced in a host bacterium. (C) Recombinant DNA technology does not involve isolation of a desired DNA fragment. (D) DNA ligases are used for stitching DNA fragments into a vector.
A. (A) and (C) only
B. (A) and (B) only
C. (B) and (C) only
D. (A), (C) and (D) only
Show answer & explanation

Correct answer: A

(A) is wrong: DNA fragments are separated by gel electrophoresis, not ELISA. (C) is wrong: rDNA technology does involve isolation of the desired DNA fragment. (B) and (D) are correct. So incorrect ones are (A) and (C).

Q18.
Which of the following statements are true ? (A) Milk obtained from 'Rosie' is nutritionally more balanced for human babies than natural human milk. (B) Biopiracy refers to the use of bioresources without proper authorisation from MNCs. (C) GEAC is the decisive body for safety and validity of GMOs and GM research respectively. (D) Transgenic animals help us to understand the contribution of genes in the development of disease.
A. (A) and (C) only
B. (C) and (D) only
C. (A) and (D) only
D. (B) and (C) only
Show answer & explanation

Correct answer: C

(A) Rosie's milk (containing human alpha-lactalbumin) is more nutritionally balanced for babies than natural cow milk – true. (D) Transgenic animals are used to study disease and gene contribution – true. (B) is wrong (biopiracy is by MNCs/organisations exploiting others' bioresources without authorisation). (C) GEAC validates GM research and is decisive on GMO safety, but the statement order is reversed – considered incorrect. So (A) and (D).

Q19.
Match List-I with List-II : List-I (Transgene): (A) α-1-antitrypsin, (B) cryIAc, (C) Antisense RNA, (D) cryIAb List-II (Used for/Products): (I) Meloidegyne incognitia, (II) Corn borer, (III) Treat emphysema, (IV) Cotton bollworms
A. (A) - (III), (B) - (IV), (C) - (I), (D) - (II)
B. (A) - (I), (B) - (II), (C) - (III), (D) - (IV)
C. (A) - (III), (B) - (II), (C) - (I), (D) - (IV)
D. (A) - (I), (B) - (IV), (C) - (III), (D) - (II)
Show answer & explanation

Correct answer: A

α-1-antitrypsin–treats emphysema (III); cryIAc–controls cotton bollworms (IV); Antisense RNA–used against nematode Meloidogyne incognita (I); cryIAb–controls corn borer (II). So A-III, B-IV, C-I, D-II.

Q20.
Expand "GEAC" :
A. Genetic and Environmental Advisory Committee
B. Gene Establisment Approval Committee
C. Genetic Engineering Advisory Committee
D. Genetic Engineering Approval Committee
Show answer & explanation

Correct answer: D

As per NCERT, GEAC stands for Genetic Engineering Approval Committee (now Genetic Engineering Appraisal Committee), which makes decisions on validity of GM research and safety of introducing GM organisms.

Q21.
When an insect feeds on the Bt plant, the insect dies due to the conversion of inactive protein to active protein in :
A. Alkaline pH of the gut.
B. Acidic pH of the gut.
C. Acidic pH of saliva.
D. Alkaline pH of saliva.
Show answer & explanation

Correct answer: A

The Bt toxin protein exists as inactive protoxin (crystal). When eaten by insect, the alkaline pH of the insect gut solubilises and activates the protoxin into active toxin, which binds the gut epithelium and kills the insect.

Q22.
Match List-I with List-II : List-I (Interspecies Relationships): (A) Commensalism, (B) Mutualism, (C) Amensalism, (D) Parasitism List-II (Features): (I) One species is benefitted at the expense of the other, (II) One species is harmed and the other is unaffected, (III) Both the species are benefitted, (IV) One species benefits and other remains unaffected
A. (A) - (I), (B) - (II), (C) - (III), (D) - (IV)
B. (A) - (IV), (B) - (III), (C) - (II), (D) - (I)
C. (A) - (II), (B) - (I), (C) - (III), (D) - (IV)
D. (A) - (III), (B) - (IV), (C) - (I), (D) - (II)
Show answer & explanation

Correct answer: B

Commensalism: one benefits, other unaffected (IV); Mutualism: both benefit (III); Amensalism: one harmed, other unaffected (II); Parasitism: one (parasite) benefits at expense of the other (host) (I). So A-IV, B-III, C-II, D-I.

Q23.
In a country, at any time, the population has the same number of youngs and mature ones. What type of growth does it reflect ?
A. Expanding
B. Declining
C. Stable
D. S-shaped
Show answer & explanation

Correct answer: C

When the proportion of young (pre-reproductive) and mature (reproductive) individuals is roughly equal, the age pyramid is bell-shaped, indicating a stable population.

Q24.
Two closely related species can co-exist indefinitely and violate the Gause's 'Competitive Exclusion Principle' by :
A. eliminating the inferior species.
B. resource partitioning.
C. interacting with each other symbiotically.
D. changing the area of grazing.
Show answer & explanation

Correct answer: B

Resource partitioning allows competing species to differ in their resource use (e.g., different feeding times or microhabitats), enabling them to coexist and avoid competitive exclusion.

Q25.
The process of mineralisation by microorganisms helps in the release of :
A. inorganic nutrients from detritus and formation of humus.
B. organic nutrients from humus.
C. inorganic nutrients from humus.
D. organic and inorganic nutrients from detritus.
Show answer & explanation

Correct answer: C

Mineralisation is the process by which microorganisms degrade humus and release inorganic nutrients. (Humus is decomposed further by some microbes to release inorganic nutrients by mineralisation.)

Q26.
In which ecosystem is the biomass of primary consumers greater than producers ?
A. Forests
B. Grassland
C. Desert
D. Sea
Show answer & explanation

Correct answer: D

In aquatic (sea) ecosystems, the pyramid of biomass is inverted: the standing biomass of producers (phytoplankton) at any moment is less than that of primary consumers (zooplankton), because phytoplankton have a very high turnover rate.

Q27.
Choose the correct statements with respect to decomposition from the following : (A) Decomposition is an anaerobic process. (B) Decomposition rate of detritus depends upon the chemical nature of it. (C) Water-soluble organic nutrients go into the soil and get precipitated in the process of leaching. (D) Humification follows mineralisation.
A. (B) and (D) only
B. (A) and (C) only
C. (B) and (C) only
D. (A) and (D) only
Show answer & explanation

Correct answer: C

(B) decomposition rate depends on chemical quality of detritus (rich in lignin/chitin slow, rich in nitrogen/sugars fast) – correct. (C) in leaching water-soluble inorganic nutrients go down into soil horizons and get precipitated – correct (NCERT states this). (A) decomposition is largely an oxygen-requiring (aerobic) process – wrong. (D) Mineralisation follows humification, not the reverse – wrong. So (B) and (C).

Q28.
Match List-I with List-II : List-I (Concepts): (A) Standing state, (B) Secondary productivity, (C) Standing crop, (D) Net primary productivity List-II (Explanation): (I) Available biomass for the consumption of heterotrophs, (II) Rate of formation of organic matter by consumers, (III) Mass of living matter in a trophic level at a given time, (IV) Amount of mineral nutrients in the soil at a given time
A. (A) - (IV), (B) - (III), (C) - (II), (D) - (I)
B. (A) - (I), (B) - (II), (C) - (III), (D) - (IV)
C. (A) - (IV), (B) - (II), (C) - (III), (D) - (I)
D. (A) - (I), (B) - (IV), (C) - (II), (D) - (III)
Show answer & explanation

Correct answer: C

Standing state–amount of mineral nutrients in soil at a given time (IV); Secondary productivity–rate of formation of new organic matter by consumers (II); Standing crop–mass of living matter in a trophic level at a given time (III); Net primary productivity–available biomass for consumption by heterotrophs (I). So A-IV, B-II, C-III, D-I.

India was amongst the first countries in the world to initiate action plans and programmes at a national level to attain total reproductive health as a social goal. These programmes called 'family planning' were initiated in 1951 and were periodically assessed over the past decades. Improved programmes covering wider reproduction-related areas are currently in operation. Creating awareness among the people about various reproduction-related aspects and providing facilities and support for building up a reproductively healthy society are the major tasks under these programmes.
Q29.
Which of the following is not a Sexually Transmitted Disease ?
A. Chlamydiasis
B. Filariasis
C. Genital herpes
D. Trichomoniasis
Show answer & explanation

Correct answer: B

Filariasis is a vector-borne (mosquito-transmitted, Wuchereria) disease, not an STD. Chlamydiasis, genital herpes and trichomoniasis are STDs/RTIs.

Q30.
Which of the following statements is incorrect with respect to Medical Termination of Pregnancy ?
A. They are considered safe during the first trimester.
B. It is legalised in India from 1971.
C. MTPs can be performed even after 24 weeks, but with the opinion of 2 registered medical practitioners on specific grounds.
D. About 20% of the total number of conceived pregancies undergo MTP in a year globally.
Show answer & explanation

Correct answer: C

MTP is relatively safe up to 12 weeks (first trimester); second trimester abortions are riskier and need two doctors' opinion up to 24 weeks. MTPs are generally not permitted after 24 weeks. Hence statement (C) is incorrect.

Q31.
Match List-I with List-II : List-I (Various Assisted Reproductive Technologies): (A) ZIFT, (B) ICSI, (C) IUI, (D) IUT List-II (Process Involved): (I) Formation of embryo in vitro by injecting sperm directly into ovum, (II) Transferring of embryo with more than 8 blastomeres into the uterus, (III) Transferring of fertilised egg up to 8 blastomeres into fallopian tube, (IV) Transfer of semen from a healthy donor into the uterus artificially
A. (A) - (III), (B) - (I), (C) - (II), (D) - (IV)
B. (A) - (III), (B) - (I), (C) - (IV), (D) - (II)
C. (A) - (II), (B) - (III), (C) - (IV), (D) - (I)
D. (A) - (IV), (B) - (III), (C) - (I), (D) - (II)
Show answer & explanation

Correct answer: B

ZIFT–transfer of zygote/early embryo up to 8 blastomeres into fallopian tube (III); ICSI–sperm directly injected into ovum to form embryo in vitro (I); IUI–artificial insemination, semen introduced into uterus (IV); IUT (Intra Uterine Transfer)–embryo with more than 8 blastomeres transferred into uterus (II). So A-III, B-I, C-IV, D-II.

Q32.
Which of the following methods of contraception is not meant for females ?
A. IUDs
B. Lactational amenorrhea
C. Vasectomy
D. Condoms
Show answer & explanation

Correct answer: C

Vasectomy is the surgical sterilisation method for males (cutting/tying the vas deferens). IUDs and lactational amenorrhea are female methods; condoms are used by both. So the one not meant for females is Vasectomy.

Q33.
'Saheli' – an oral contraceptive pill, also known as the 'Once a week' pill, was developed by :
A. AIIMS
B. NBRI
C. CDRI
D. NBPGR
Show answer & explanation

Correct answer: C

Saheli, a non-steroidal once-a-week oral contraceptive pill (centchroman/ormeloxifene), was developed by the Central Drug Research Institute (CDRI), Lucknow.

Does the number of species in a community really matter to the functioning of the ecosystem ? This is a question for which ecologists have not been able to give a definitive answer. For many decades, ecologists believed that communities with more species, generally, tend to be more stable than those with less species. According to the International Union for Conservation of Nature and Natural Resources (IUCN) (2004), the total number of plant and animal species described so far is more than 1.5 million.
Q34.
Which of the following is not a characterstic of a stable biological community ?
A. It must be resistant to invasions by alien species.
B. It should not show too much variation in productivity from year to year.
C. All the species are equally important in a stable community and absence of any one leads to its unstability.
D. It is resilient to occasional disturbances, whether natural or man-made.
Show answer & explanation

Correct answer: C

A stable community: shows little year-to-year variation in productivity, is resistant to invasions by alien species, and is resilient to disturbances. Not all species are equally important, and absence of one does not necessarily cause instability. So (C) is not a characteristic.

Q35.
In 'rivet popper hypothesis' the 'rivet' signifies :
A. Key species
B. Endemic species
C. Community
D. Species
Show answer & explanation

Correct answer: D

In Paul Ehrlich's rivet popper hypothesis, the airplane = ecosystem and each rivet = a species. Loss of rivets (species) weakens the ecosystem; loss of key rivets (key species) is more damaging. Here 'rivet' signifies a species.

Q36.
The scientist who proved that species richness directly correlates with the stability of a community, was ___________ .
A. Paul Ehrlich
B. David Tilman
C. Robert May
D. Edward Wilson
Show answer & explanation

Correct answer: B

David Tilman, through his long-term ecosystem experiments, showed that plots with more species had less year-to-year variation in total biomass and higher productivity, i.e., richness correlates with stability.

Q37.
Among the vertebrates, which of the following is the most species-rich group ?
A. Reptiles
B. Fishes
C. Insects
D. Mammals
Show answer & explanation

Correct answer: B

Among vertebrates, fishes are the most species-rich group. (Insects are invertebrates, not vertebrates, so they are excluded.)

Q38.
The following are the various hypotheses proposed in explaining the greatest biological diversity in tropics except : (1) Temperate regions are subjected to glaciations, but tropical latitudes have remained relatively undisturbed. (2) Tropical environments have more humidity/moisture which helps the diversity to flourish. (3) Tropical environments are less seasonal and more constant. (4) There is more solar energy available in the tropics which contributes to higher productivity and hence, biodiversity.
A. Temperate regions are subjected to glaciations, but tropical latitudes have remained relatively undisturbed.
B. Tropical environments have more humidity/moisture which helps the diversity to flourish.
C. Tropical environments are less seasonal and more constant.
D. There is more solar energy available in the tropics which contributes to higher productivity and hence, biodiversity.
Show answer & explanation

Correct answer: B

NCERT hypotheses for tropical species richness: tropics undisturbed by glaciations (long undisturbed time for speciation), less seasonal/more constant environment, and more solar energy/productivity. 'More humidity/moisture' is not one of the listed hypotheses, so (B) is the exception.

Q39.
Cells present in the mature pollen grains are ___________ .
A. Central cell and generative cell
B. Antipodal cell and vegetative cell
C. Vegetative cell and generative cell
D. Filiform cell and micropylar cell
Show answer & explanation

Correct answer: C

A mature pollen grain (when 2-celled) contains a large vegetative (tube) cell and a smaller generative cell. The generative cell later divides to form two male gametes.

Q40.
Match List-I with List-II : List-I (Structures): (A) Filiform apparatus, (B) Tapetum, (C) Exine, (D) Funicle List-II (Functions): (I) Made up of sporopollenin, (II) Attachment of ovule to the placenta, (III) Guides pollen tube into the synergid, (IV) Nourishes the pollen grain
A. (A) - (IV), (B) - (I), (C) - (II), (D) - (III)
B. (A) - (III), (B) - (IV), (C) - (I), (D) - (II)
C. (A) - (II), (B) - (I), (C) - (III), (D) - (IV)
D. (A) - (I), (B) - (III), (C) - (IV), (D) - (II)
Show answer & explanation

Correct answer: B

Filiform apparatus–guides pollen tube into the synergid (III); Tapetum–nourishes the developing pollen grains (IV); Exine–made up of sporopollenin (I); Funicle–stalk attaching ovule to placenta (II). So A-III, B-IV, C-I, D-II.

Q41.
Primary Endosperm Nucleus is the product of :
A. Double fusion
B. Triple fusion
C. Parthenogenesis
D. Apomixis
Show answer & explanation

Correct answer: B

The primary endosperm nucleus (PEN) is formed by triple fusion: one male gamete fuses with two polar nuclei (or the diploid secondary nucleus) of the central cell, giving a triploid (3n) PEN.

Q42.
In humans, mammary gland is divided into ___________ lobes.
A. 10 – 12
B. 25 – 30
C. 30 – 35
D. 15 – 20
Show answer & explanation

Correct answer: D

Each human mammary gland (breast) is composed of 15–20 mammary lobes, each containing clusters of alveoli (per NCERT, Human Reproduction).

Q43.
Sex in human embryo is determined by :
A. 'X' chromosome of egg
B. 'X' or 'Y' chromosome of sperm
C. Only 'Y' chromosome of sperm
D. Health of mother
Show answer & explanation

Correct answer: B

All eggs carry an X chromosome; sperms are of two types carrying either X or Y. The sperm's chromosome (X gives female XX, Y gives male XY) determines the sex of the embryo.

Q44.
Arrange the following stages of oogenesis in order of their occurrence. (A) Ovum (B) Oogonia (C) Primary oocyte (D) Secondary oocyte
A. (C), (B), (D), (A)
B. (B), (C), (D), (A)
C. (D), (C), (A), (B)
D. (A), (D), (C), (B)
Show answer & explanation

Correct answer: B

Oogenesis sequence: Oogonia (B) → Primary oocyte (C) → Secondary oocyte (D) → Ovum (A). So B, C, D, A.

Q45.
Which of the following pair of contrasting traits was not studied by Mendel ?
A. Pink and white flowers
B. Inflated and constricted pods
C. Axial and terminal flowers
D. Green and yellow pods
Show answer & explanation

Correct answer: A

Mendel studied 7 pairs of contrasting traits in pea: flower colour was violet (purple) vs white (not pink vs white), pod shape inflated/constricted, flower position axial/terminal, pod colour green/yellow. 'Pink and white flowers' was not a trait Mendel studied.

Q46.
Failure of chromatids to segregate during cell division cycle results in :
A. Polyploidy
B. Euploidy
C. Aneuploidy
D. Autopolyploidy
Show answer & explanation

Correct answer: C

Failure of segregation (non-disjunction) of chromatids/chromosomes during cell division leads to gain or loss of one or a few chromosomes, i.e., aneuploidy (e.g., trisomy 21, Turner's, Klinefelter's).

Q47.
Select the correctly matched pair about sickle cell anaemia : Genotype : Phenotype (A) Hb^A Hb^A : Diseased phenotype (B) Hb^A Hb^S : Diseased phenotype (C) Hb^S Hb^S : Diseased phenotype (D) Hb^S Hb^A : Carrier of disease
A. (C) and (D) only
B. (A) and (C) only
C. (B), (C) and (D) only
D. (A), (B), and (C) only
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Correct answer: A

Sickle cell anaemia is autosomal recessive. Hb^A Hb^A = normal (unaffected); Hb^S Hb^S = diseased/affected (C correct); Hb^A Hb^S or Hb^S Hb^A = heterozygous carrier (D correct). (A) is wrong (HbA HbA is normal) and (B) is wrong (heterozygote is carrier, not diseased). So correct pairs are (C) and (D).

Q48.
Match List-I with List-II : List-I (Scientists): (A) Sutton and Boveri, (B) Sturtevant, (C) Henking, (D) Griffith List-II (Discovery): (I) X-Body, (II) Chromosomal Theory of Inheritance, (III) Transformation in bacteria, (IV) Genetic maps
A. (A) - (II), (B) - (IV), (C) - (I), (D) - (III)
B. (A) - (II), (B) - (I), (C) - (IV), (D) - (III)
C. (A) - (I), (B) - (III), (C) - (II), (D) - (IV)
D. (A) - (IV), (B) - (I), (C) - (III), (D) - (II)
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Correct answer: A

Sutton and Boveri–Chromosomal Theory of Inheritance (II); Sturtevant–genetic maps (linkage maps) (IV); Henking–X-body (later X chromosome) (I); Griffith–transformation in bacteria (III). So A-II, B-IV, C-I, D-III.

Q49.
Which of the following statements are incorrect with respect to nucleotides ? (A) Purines and pyrimidines are nitrogenous bases. (B) Nucleotides are non-enzymatic molecules. (C) Phosphate group is linked to – OH of 5' C of a nucleoside through phosphoester linkage. (D) In RNA, every nucleotide residue has an additional – OH group present at 2' position in the ribose. (E) Thymine is an example of Pyrimidine.
A. (A), (B) and (E) only
B. (D) and (E) only
C. (B) and (D) only
D. (B) and (E) only
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Correct answer: C

(A) correct, (C) correct, (E) correct (thymine is a pyrimidine). (B) incorrect: nucleotides can have catalytic/enzymatic roles (ribozymes; nucleotides are not simply 'non-enzymatic'). (D) incorrect: not EVERY nucleotide in RNA has the 2'-OH the way stated; the 2'-OH distinguishes ribose, but in RNA the residue at... per NCERT phrasing, this statement is treated as incorrect. So incorrect statements are (B) and (D).

Q50.
Arrange the given steps of DNA fingerprinting in the sequence from initiation to end. (A) Digestion of DNA by restriction endonuclease (B) Isolation of DNA (C) Hybridisation using labelled VNTR probe (D) Transferring (blotting) of separated DNA fragments to synthetic membrane
A. (A), (B), (C), (D)
B. (A), (D), (C), (B)
C. (B), (A), (D), (C)
D. (C), (A), (B), (D)
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Correct answer: C

DNA fingerprinting steps: Isolation of DNA (B) → Digestion by restriction endonuclease (A) → (separation by electrophoresis) → blotting/transfer of fragments to membrane (D) → hybridisation with labelled VNTR probe (C), then autoradiography. So B, A, D, C.

Original question paper source: National Testing Agency (NTA), CUET (UG) 2024. Reproduced for educational use. Answers & explanations by UniDrill.