CUET Biology Question Paper 2024 SET D Download Solutions and Answer Key
| Updated On - Dec 7, 2024
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CUET 2024 Biology Question Paper (SET D) with Solution PDF
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CUET 2024 BIOLOGY Answer Key (Set D)
| Question | Answer | Detailed Solution |
| Q1. In a country, at any time, the population has the same number of young and mature ones. What type of growth does it reflect? • (1) Expanding • (2) Declining • (3) Stable • (4) S-shaped |
Correct answer: 3. Stable | Explanation: A stable population shows equal proportions of young and mature individuals, which reflects balanced birth and death rates over time. |
| Q2. Two closely related species can co-exist indefinitely and violate Gause’s ‘Competitive Exclusion Principle’ by: • (1) Eliminating the inferior species • (2) Resource partitioning • (3) Interacting with each other symbiotically • (4) Changing the area of grazing |
Correct answer: 2. Resource partitioning | Explanation: Resource partitioning allows closely related species to divide resources or use them in different ways, reducing competition and enabling coexistence. |
| Q3. The process of mineralisation by microorganisms helps in the release of: • (1) Inorganic nutrients from detritus and formation of humus • (2) Organic nutrients from humus • (3) Inorganic nutrients from humus • (4) Organic and inorganic nutrients from detritus |
Correct answer: 1. Inorganic nutrients from detritus and formation of humus | Explanation: Mineralisation involves the breakdown of detritus by microorganisms, releasing inorganic nutrients essential for plant growth and forming humus. |
| Q4. In which ecosystem is the biomass of primary consumers greater than producers? • (1) Forests • (2) Grassland • (3) Desert • (4) Sea |
Correct answer: 4. Sea | Explanation: In aquatic ecosystems like seas, primary consumers such as zooplankton often have a greater biomass than phytoplankton due to the rapid turnover of phytoplankton. |
| Q5. Match List-I with List-II: List-I (Relationships) | List-II (Features) (A) Commensalism | (I) One species is benefitted at the expense of the other (B) Mutualism | (II) One species is harmed, the other is unaffected (C) Amensalism | (III) Both species benefit (D) Parasitism | (IV) One species benefits, the other remains unaffected Options: (1) (A) - (I), (B) - (II), (C) - (III), (D) - (IV) (2) (A) - (IV), (B) - (III), (C) - (II), (D) - (I) (3) (A) - (II), (B) - (I), (C) - (III), (D) - (IV) (4) (A) - (III), (B) - (IV), (C) - (I), (D) - (II) |
Correct answer: 2. (A) - (IV), (B) - (III), (C) - (II), (D) - (I) | Explanation: - Commensalism: One species benefits, the other remains unaffected. - Mutualism: Both species benefit. - Amensalism: One species is harmed, the other unaffected. - Parasitism: One species benefits at the expense of the other. |
| Q6. Choose the correct statements with respect to decomposition from the following: (A) Decomposition is an anaerobic process. (B) Decomposition rate of detritus depends upon its chemical nature. (C) Water-soluble organic nutrients go into the soil and get precipitated in the process of leaching. (D) Humification follows mineralisation. Options: • (1) (B) and (D) only • (2) (A) and (C) only • (3) (B) and (C) only • (4) (A) and (D) only |
Correct answer: 1. (B) and (D) only | Explanation: - (A) Incorrect: Decomposition is primarily an aerobic process. - (B) Correct: Decomposition depends on the chemical nature of detritus (e.g., lignin content). - (C) Incorrect: Water-soluble nutrients are leached into the soil but do not precipitate. - (D) Correct: Humification occurs after mineralisation. |
| Q7. Match List-I with List-II: List-I (Concepts) | List-II (Explanation) (A) Standing state | (I) Available biomass for heterotrophs (B) Secondary productivity | (II) Rate of organic matter formation by consumers (C) Standing crop | (III) Mass of living matter in a trophic level (D) Net primary productivity | (IV) Amount of mineral nutrients in soil Options: (1) (A) - (IV), (B) - (III), (C) - (II), (D) - (I) (2) (A) - (I), (B) - (II), (C) - (III), (D) - (IV) (3) (A) - (IV), (B) - (II), (C) - (III), (D) - (I) (4) (A) - (I), (B) - (IV), (C) - (II), (D) - (III) |
Correct answer: 3. (A) - (IV), (B) - (II), (C) - (III), (D) - (I) | Explanation: - (A) Standing state: Refers to the amount of nutrients in the soil. - (B) Secondary productivity: Organic matter production rate by consumers. - (C) Standing crop: Mass of living matter in a trophic level. - (D) Net primary productivity: Biomass available for heterotroph consumption. |
| Q8: Which of the following is not a Sexually Transmitted Disease (STD)? Options: 1. Chlamydiasis 2. Filariasis 3. Genital herpes 4. Trichomoniasis |
2. Filariasis | Solution: Filariasis is a mosquito-borne disease caused by parasitic worms, while the others listed are sexually transmitted diseases. |
| Q9: Which of the following statements is incorrect with respect to Medical Termination of Pregnancy (MTP)? Options: 1. They are considered safe during the first trimester. 2. It is legalized in India from 1971. 3. MTPs can be performed even after 24 weeks, but with the opinion of 2 registered medical practitioners on specific grounds. 4. About 20% of the total number of conceived pregnancies undergo MTP in a year globally. |
4. About 20% of the total number of conceived pregnancies undergo MTP in a year globally. | Solution: MTPs are generally safe during the first trimester and regulated in India under the Medical Termination of Pregnancy Act (1971). However, the global figure of 20% is incorrect. |
| Q10: Match List-I with List-II: List-I (ART Techniques) List-II (Processes) (A) ZIFT (I) Formation of embryo in vitro by injecting sperm into ovum (B) ICSI (II) Transferring embryos with ≤8 blastomeres into uterus (C) IUI (III) Transfer of fertilized egg (up to 8 blastomeres) into fallopian tube (D) IUT (IV) Artificial transfer of semen into uterus |
2. (A) - (III), (B) - (I), (C) - (IV), (D) - (II) | Solution: - ZIFT: Transfer of fertilized eggs (up to 8 blastomeres) into the fallopian tube. - ICSI: Formation of embryos by injecting sperm into the ovum in vitro. - IUI: Artificial insemination of semen into the uterus. - IUT: Transfer of embryos with more than 8 blastomeres into the uterus. |
| Q11: Which of the following methods of contraception is not meant for females? Options: 1. IUDs 2. Lactational amenorrhea 3. Vasectomy 4. Condoms |
3. Vasectomy | Solution: Vasectomy is a permanent contraceptive method performed on males. It involves cutting or sealing the vas deferens to prevent the release of sperm. |
| Q12: Saheli—an oral contraceptive pill, also known as the "Once a week" pill, was developed by: Options: 1. AIIMS 2. NBRI 3. CDRI 4. NBPGR |
3. CDRI | Solution: Saheli, developed by the Central Drug Research Institute (CDRI), Lucknow, is a non-steroidal contraceptive pill taken once a week. It is recognized as a safe and effective method of contraception. |
| Q13: Which of the following is not a characteristic of a stable biological community? Options: 1. It must be resistant to invasions by alien species. 2. It should not show too much variation in productivity from year to year. 3. All the species are equally important in a stable community, and absence of any one leads to its instability. 4. It is resilient to occasional disturbances, whether natural or man-made. |
3. All the species are equally important in a stable community, and absence of any one leads to its instability. | Solution: In a stable biological community, not all species are equally important. Keystone species have a disproportionate impact on community stability, while other species play minor roles. |
| Q14: In the ’rivet popper hypothesis,’ the ’rivet’ signifies: Options: 1. Key species 2. Endemic species 3. Community 4. Species |
4. Species | Solution: The rivet popper hypothesis compares species in an ecosystem to rivets in an airplane. Each species contributes to ecosystem stability, and losing too many species may lead to ecosystem collapse. |
| Q15: The scientist who proved that species richness directly correlates with the stability of a community was: Options: 1. Paul Ehrlich 2. David Tilman 3. Robert May 4. Edward Wilson |
2. David Tilman | Solution: David Tilman’s experiments demonstrated that ecosystems with greater species richness are more stable and resilient to environmental disturbances. His findings highlight the role of biodiversity in maintaining ecosystem functionality. |
| Q16: Among vertebrates, which of the following is the most species-rich group? Options: 1. Reptiles 2. Fishes 3. Insects 4. Mammals |
2. Fishes | Solution: Fishes are the most species-rich group among vertebrates, with over 33,000 species identified. They inhabit diverse aquatic environments worldwide, contributing significantly to global biodiversity. |
| Q17: The following are the various hypotheses proposed in explaining the greatest biological diversity in the tropics, except: Options: 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. |
2. Tropical environments have more humidity/moisture which helps the diversity to flourish. | Solution: While humidity and moisture may support tropical diversity, other factors such as stability, constant climate, and solar energy are more significant contributors to high biodiversity. |
| Q18: Cells present in the mature pollen grains are . Options: 1. Central cell and generative cell 2. Antipodal cell and vegetative cell 3. Vegetative cell and generative cell 4. Filiform cell and micropylar cell |
3. Vegetative cell and generative cell | Solution: Mature pollen grains consist of two cells: a large vegetative cell and a smaller generative cell, which divides to form sperm cells during fertilization. |
| Q19: Match List-I with List-II: List-I (Structures) List-II (Functions) (A) Filiform apparatus (I) Made up of sporopollenin (B) Tapetum (II) Attachment of ovule to the placenta (C) Exine (III) Guides pollen tube into the synergid (D) Funicle (IV) Nourishes the pollen grain Options: 1. (A) - (IV), (B) - (I), (C) - (II), (D) - (III) 2. (A) - (III), (B) - (IV), (C) - (I), (D) - (II) 3. (A) - (II), (B) - (I), (C) - (III), (D) - (IV) 4. (A) - (I), (B) - (III), (C) - (IV), (D) - (II) |
2. (A) - (III), (B) - (IV), (C) - (I), (D) - (II) | Solution: - Filiform apparatus: Guides pollen tube to the synergid. - Tapetum: Provides nourishment to the developing pollen grains. - Exine: Made of sporopollenin, providing durability. - Funicle: Connects the ovule to the placenta. |
| Q20: Primary Endosperm Nucleus is the product of: Options: 1. Double fusion 2. Triple fusion 3. Parthenogenesis 4. Apomixis |
2. Triple fusion | Solution: The primary endosperm nucleus is formed by the fusion of one sperm nucleus with the two polar nuclei in the embryo sac, a process called triple fusion. |
| Q21: In humans, mammary gland is divided into lobes. Options: 1. 10 – 12 2. 25 – 30 3. 30 – 35 4. 15 – 20 |
4. 15 – 20 | Solution: The human mammary gland is divided into 15–20 lobes, each containing milk-producing alveoli. |
| Q22: Sex in human embryo is determined by: Options: 1. ‘X’ chromosome of egg 2. ‘X’ or ‘Y’ chromosome of sperm 3. Only ‘Y’ chromosome of sperm 4. Health of mother |
2. ‘X’ or ‘Y’ chromosome of sperm | Solution: Sex in human embryos is determined by the type of sperm that fertilizes the egg. Sperm carrying an X chromosome results in a female, while sperm carrying a Y chromosome results in a male. |
| Q23: Arrange the following stages of oogenesis in order of their occurrence: (A) Ovum (B) Oogonia (C) Primary oocyte (D) Secondary oocyte Options: 1. (C), (B), (D), (A) 2. (B), (C), (D), (A) 3. (D), (C), (A), (B) 4. (A), (D), (C), (B) |
2. (B), (C), (D), (A) | Solution: The correct sequence of oogenesis is: 1. Oogonia (immature germ cells). 2. Primary oocyte (formed during fetal development). 3. Secondary oocyte (formed during meiosis I). 4. Ovum (formed after meiosis II and fertilization). |
| Q24: Which of the following pair of contrasting traits was not studied by Mendel? Options: 1. Pink and white flowers 2. Inflated and constricted pods 3. Axial and terminal flowers 4. Green and yellow pods |
1. Pink and white flowers | Solution: Mendel studied traits like seed shape, seed color, flower position, and pod shape, but he did not study flower color as pink and white (he studied purple and white). |
| Q25: Failure of chromatids to segregate during cell division cycle results in: Options: 1. Polyploidy 2. Euploidy 3. Aneuploidy 4. Autopolyploidy |
3. Aneuploidy | Solution: Aneuploidy occurs when there is an addition or loss of chromosomes due to improper segregation of chromatids during cell division. For example, Down syndrome results from an extra chromosome 21. |
| Q26: Select the correctly matched pair about sickle cell anaemia: Genotype : Phenotype (A) HbAHbA : Diseased phenotype (B) HbAHbS : Diseased phenotype (C) HbSHbS : Diseased phenotype (D) HbSHbA : Carrier of disease Options: 1. (C) and (D) only 2. (A) and (C) only 3. (B), (C) and (D) only 4. (A), (B), and (C) only |
1. (C) and (D) only | Solution: - HbAHbA: Normal phenotype (not diseased). - HbAHbS: Carrier of sickle cell anaemia. - HbSHbS: Diseased phenotype (sickle cell anaemia). - HbSHbA: Carrier of disease. |
| Q27: Match List-I with List-II: List-I (Scientists) List-II (Discovery) (A) Sutton and Boveri (B) Sturtevant (C) Henking (D) Griffith Options: 1. (A) - (II), (B) - (IV), (C) - (I), (D) - (III) 2. (A) - (III), (B) - (I), (C) - (IV), (D) - (II) 3. (A) - (I), (B) - (III), (C) - (II), (D) - (IV) 4. (A) - (IV), (B) - (I), (C) - (III), (D) - (II) |
1. (A) - (II), (B) - (IV), (C) - (I), (D) - (III) | Solution: - Sutton and Boveri: Proposed the Chromosomal Theory of Inheritance. - Sturtevant: Developed genetic maps. - Henking: Discovered the X-Body, now known as the X chromosome. - Griffith: Demonstrated bacterial transformation in his experiments. |
| Q28: 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. Options: 1. (A), (B), and (E) only 2. (D) and (E) only 3. (B) and (D) only 4. (B) and (E) only |
2. (D) and (E) only | Solution: - (A) Correct: Purines (adenine, guanine) and pyrimidines (cytosine, thymine, uracil) are nitrogenous bases. - (B) Correct: Nucleotides are not enzymes. - (C) Correct: Phosphate attaches to 5’ carbon of nucleoside. - (D) Incorrect: In RNA, the ribose sugar has a 2’ –OH group, but not every nucleotide residue has this group. - (E) Incorrect: Thymine is a pyrimidine found in DNA, but not in RNA. |
| Q29: 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 Options: 1. (A), (B), (C), (D) 2. (A), (D), (B), (C) 3. (B), (A), (D), (C) 4. (C), (D), (A), (B) |
3. (B), (A), (D), (C) | Solution: The correct sequence for DNA fingerprinting is: 1. Isolation of DNA (B): DNA is extracted from the sample. 2. Digestion by restriction endonucleases (A): DNA is cut into fragments. 3. Blotting to synthetic membrane (D): DNA fragments are transferred to a nylon or nitrocellulose membrane. 4. Hybridisation with VNTR probe (C): Probes hybridize with complementary DNA sequences. |
| Q30: Nucleosome is associated with molecules of histones. Options: 1. Four 2. Nine 3. Two 4. Eight |
4. Eight | Solution: A nucleosome consists of eight histone proteins, forming an octamer (two each of H2A, H2B, H3, and H4), around which DNA is wrapped. |
| Q31: 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. Options: 1. (A), (B), (C) and (D) only 2. (A), (C), (D) and (E) only 3. (A), (C) and (E) only 4. (A), (B), (C), (D) and (E) |
4. (A), (B), (C), (D) and (E) | Solution: All listed statements are true observations from the Human Genome Project, which revealed the detailed composition and structure of the human genome. |
| Q32: Analogous structures are a result of: Options: 1. Convergent evolution 2. Divergent evolution 3. Parallel evolution 4. Retrogressive evolution |
1. Convergent evolution | Solution: Analogous structures arise from convergent evolution, where different organisms independently evolve similar traits to adapt to similar environments. |
| Q33: Which of the following does not affect the Hardy-Weinberg equilibrium? Options: 1. Natural selection 2. Genetic drift 3. Gene pool 4. Gene migration |
3. Gene pool | Solution: The Hardy-Weinberg equilibrium is disturbed by factors like natural selection, genetic drift, gene flow, and mutation. The gene pool itself does not directly affect the equilibrium unless it undergoes change. |
| Q34: Which of the following primates was more like an ape? Options: 1. Homo erectus 2. Dryopithecus 3. Australopithecines 4. Ramapithecus |
2. Dryopithecus | Solution: Dryopithecus is considered an early ape-like primate. It was quadrupedal and arboreal, more similar to modern apes than humans. |
| Q35: Match List-I with List-II: List-I (Placental mammals) List-II (Counterpart Marsupials) (A) Anteater (I) Spotted cuscus (B) Bobcat (II) Numbat (C) Lemur (III) Flying Phalanger (D) Flying squirrel (IV) Tasmanian tiger cat Options: 1. (A) - (II), (B) - (IV), (C) - (I), (D) - (III) 2. (A) - (III), (B) - (I), (C) - (IV), (D) - (II) 3. (A) - (IV), (B) - (I), (C) - (II), (D) - (III) 4. (A) - (IV), (B) - (I), (C) - (III), (D) - (II) |
1. (A) - (II), (B) - (IV), (C) - (I), (D) - (III) | Solution: - Anteater matches with the marsupial Numbat. - Bobcat corresponds to Tasmanian tiger cat. - Lemur pairs with Spotted cuscus. - Flying squirrel is matched with Flying Phalanger. |
| Q36: 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 color in severe cases of pneumonia. (D) Pneumonia is caused by Salmonella. (E) Typhoid fever could be confirmed by Widal test. Options: 1. (A), (C), and (D) only 2. (B) and (E) only 3. (D) only 4. (A) and (D) only |
3. (D) only | Solution: - (A) Correct: Severe typhoid can cause intestinal perforation. - (B) Correct: Common cold is caused by Rhinoviruses. - (C) Correct: Bluish lips and fingernails indicate hypoxia in severe pneumonia. - (D) Incorrect: Pneumonia is caused by Streptococcus pneumoniae, not Salmonella. - (E) Correct: Widal test diagnoses typhoid fever. |
| Q37: Match List-I with List-II: List-I (Types of barriers) List-II (Examples) (A) Cytokine barriers (I) Mucus coating (B) Physical barriers (II) Tears from eyes (C) Cellular barriers (III) Phagocytosis (D) Physiological barriers (IV) Interferons Options: 1. (A) - (IV), (B) - (III), (C) - (I), (D) - (II) 2. (A) - (IV), (B) - (II), (C) - (III), (D) - (I) 3. (D) - (I), (B) - (C), (C) - (IV), (A) - (III) 4. (A) - (IV), (B) - (I), (C) - (II), (D) - (III) |
2. (A) - (IV), (B) - (II), (C) - (III), (D) - (I) | Solution: - (A) Cytokine barriers: Interferons act as antiviral proteins. - (B) Physical barriers: Tears protect from infections. - (C) Cellular barriers: Phagocytosis by macrophages. - (D) Physiological barriers: Mucus traps pathogens. |
| Q38: Smack is chemically: Options: 1. Diacetyl morphine 2. Cocaine 3. Benzodiazepine 4. Amphetamine |
1. Diacetyl morphine | Solution: Smack, commonly known as heroin, is chemically diacetyl morphine, a semisynthetic opioid. |
| Q39: Antibodies are secreted by: Options: 1. T-Cells 2. B-Cells 3. α-Cells 4. β-Cells |
2. B-Cells | Solution: B-cells are specialized white blood cells that produce and secrete antibodies as part of the adaptive immune response. |
| Q40: In sewage treatment, flocs are: Options: 1. The solids that settle during sedimentation. 2. The supernatant formed above primary sludge. 3. The masses of bacteria associated with fungal filaments. 4. The bacteria which grow anaerobically and are also called anaerobic sludge digesters. |
3. The masses of bacteria associated with fungal filaments. | Solution: Flocs are aggregates of bacteria and fungi that aid in decomposing organic matter during the secondary treatment of sewage. |
| Q41: Match List-I with List-II: List-I (Products) List-II (Organisms) (A) Statin (I) Streptococcus (B) Clot buster (II) Trichoderma (C) Swiss cheese (III) Monascus (D) Cyclosporin-A (IV) Propionibacterium Options: 1. (A) - (II), (B) - (I), (C) - (IV), (D) - (III) 2. (A) - (III), (B) - (I), (C) - (IV), (D) - (II) 3. (A) - (III), (B) - (IV), (C) - (II), (D) - (I) 4. (A) - (II), (B) - (III), (C) - (I), (D) - (IV) |
2. (A) - (III), (B) - (I), (C) - (IV), (D) - (II) | Solution: Product Organism/Explanation: - Statin: Monascus - Lowers cholesterol. - Clot buster: Streptococcus - Produces streptokinase. - Swiss cheese: Propionibacterium - Used in fermentation. - Cyclosporin-A: Trichoderma - Acts as an immunosuppressant. |
| Q42: The beetle used as a biocontrol agent for aphids and mosquitoes is: Options: 1. Trichoderma 2. Dragonflies 3. Ladybird 4. Silverfish |
3. Ladybird | Solution: Ladybird beetles are natural predators of aphids and other pests, making them effective biocontrol agents. |
| Q43: Downstream processing method involves: Options: 1. Identification 2. Amplification 3. Fermentation 4. Purification |
4. Purification | Solution: Downstream processing refers to the recovery and purification of biosynthetic products such as proteins, which ensures their usability. |
| Q44: Which of the following is not the correctly matched pair of organism and its respective cell wall degrading enzyme? Options: 1. Fungi – Chitinase 2. Algae – Methylase 3. Plant cells – Cellulase 4. Bacteria – Lysozyme |
2. Algae – Methylase | Solution: Methylase modifies nucleic acids rather than degrading algal cell walls. Other pairs (Chitinase, Cellulase, Lysozyme) are correct matches. |
| Q45: Arrange the following steps involved in the transformation of bacteria in sequence from initiation to end: (A) Incubation of rDNA with bacterial cells on ice (B) Treatment with divalent cations (C) Heat shock treatment (D) Selection on antibiotic-containing agar plate (E) Placed them again on ice Options: 1. (A), (B), (D), (C), (E) 2. (B), (A), (C), (E), (D) 3. (B), (C), (D), (A), (E) 4. (A), (C), (B), (D), (E) |
2. (B), (A), (C), (E), (D) | Solution: Steps in bacterial transformation: 1. Treatment with divalent cations (e.g., Ca2+) increases cell permeability. 2. Incubation with rDNA on ice aids DNA uptake. 3. Heat shock creates conditions for DNA entry. 4. Ice stabilization follows. 5. Antibiotic selection identifies transformed cells. |
| Q46: Which of the following statements are incorrect? (A) Fragments of DNA can be separated by ELISA. (B) Transformation introduces DNA into 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. Options: 1. (A) and (C) only 2. (A) and (B) only 3. (B) and (C) only 4. (A), (C), and (D) only |
1. (A) and (C) only | Solution: - (A) Incorrect: ELISA is used to detect antigens or antibodies, not DNA fragments. - (C) Incorrect: Isolation of the desired DNA fragment is an essential step in recombinant DNA technology. - (B) and (D) are correct: Transformation introduces DNA, and ligases join DNA fragments. |
| Q47: Which of the following statements are true? (A) Milk from ‘Rosie’ is nutritionally more balanced for human babies than natural human milk. (B) Biopiracy refers to the use of bioresources without proper authorization. (C) GEAC is the decisive body for GMO safety and research. (D) Transgenic animals help us study the contribution of genes in the development of diseases. Options: 1. (A) and (C) only 2. (C) and (D) only 3. (B) and (C) only 4. (A), (B), and (C) only |
2. (C) and (D) only | Solution: - (A) False: While ‘Rosie’ produces milk enriched with human proteins, it is not more balanced than human milk. - (B) False: Biopiracy involves unauthorized use of bioresources but is not specific to MNCs. - (C) True: GEAC regulates GMO safety. - (D) True: Transgenic animals help in genetic research for disease understanding. |
| Q48: Match List-I with List-II: List-I (Transgene) List-II (Used for/Products) (A) α-1-antitrypsin (I) Meloidogyne incognita (B) cryIAc (II) Corn borer (C) Antisense RNA (III) Treat emphysema (D) cryIAb (IV) Cotton bollworms Options: 1. (A) - (III), (B) - (IV), (C) - (I), (D) - (II) 2. (A) - (I), (B) - (III), (C) - (III), (D) - (IV) 3. (A) - (III), (B) - (II), (C) - (I), (D) - (IV) 4. (A) - (I), (B) - (IV), (C) - (III), (D) - (II) |
3. (A) - (III), (B) - (II), (C) - (I), (D) - (IV) | Solution: Transgene Explanation: - α-1-antitrypsin: Treats emphysema caused by genetic disorders. - cryIAc: Effective against the corn borer pest. - Antisense RNA: Used to silence genes in nematode pests (Meloidogyne incognita). - cryIAb: Targets cotton bollworms. |
| Q49: Expand “GEAC”: 1. Genetic and Environmental Advisory Committee 2. Gene Establishment Approval Committee 3. Genetic Engineering Advisory Committee 4. Genetic Engineering Approval Committee |
4. Genetic Engineering Approval Committee | Solution: The Genetic Engineering Approval Committee (GEAC) is responsible for the safety and regulation of genetically modified organisms (GMOs) and related research in India. |
| Q50: When an insect feeds on the Bt plant, the insect dies due to the conversion of inactive protein to active protein in: 1. Alkaline pH of the gut. 2. Acidic pH of the gut. 3. Acidic pH of saliva. 4. Alkaline pH of saliva. |
1. Alkaline pH of the gut | Solution: Bt toxin, produced by Bacillus thuringiensis, is an inactive protoxin. In an insect’s gut, the alkaline pH activates the toxin, which binds to gut cells, creating pores that kill the insect. |
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