Q1. Four daughter cells formed after meiosis are
Correct Answer: (b)
Due to crossing over and independent assortment of homologous chromosomes, the four daughter cells are genetically unique.
Due to crossing over and independent assortment of homologous chromosomes, the four daughter cells are genetically unique.
Q2. Regulation by small RNAs and alternative splicing are similar in that both
Correct Answer: (d)
Both alternative splicing and small RNA regulation (like miRNA) are post-transcriptional events. Splicing involves the spliceosome (an RNA/protein complex), and small RNAs function as part of a larger protein complex (such as RISC) to inhibit gene expression.
Both alternative splicing and small RNA regulation (like miRNA) are post-transcriptional events. Splicing involves the spliceosome (an RNA/protein complex), and small RNAs function as part of a larger protein complex (such as RISC) to inhibit gene expression.
Q3. Match the following and choose the correct option.
| Column - I | Column - II |
|---|---|
| A. Phosphorus | I. Atmosphere |
| B. Carbon | II. Producers |
| C. Goat | III. Rock |
| D. Grasses | IV. T2 |
Correct Answer: (b)
A-III: Phosphorus reservoir is Rock; B-I: Carbon reservoir is Atmosphere; C-IV: Goat is at the second trophic level (T2); D-II: Grasses are Producers.
A-III: Phosphorus reservoir is Rock; B-I: Carbon reservoir is Atmosphere; C-IV: Goat is at the second trophic level (T2); D-II: Grasses are Producers.
Q4. In the given columns, column-I contain complexes and column-II contain their alternative names. Select the correct match from the option given below.
| Column - I | Column - II |
|---|---|
| A. Complex I | I. Cytochrome bc1 complex |
| B. Complex II | II. NADH dehydrogenase |
| C. Complex III | III. ATP synthetase |
| D. Complex IV | IV. FADH2 dehydrogenase |
| E. Complex V | V. Cytochrome c oxidase |
Correct Answer: (a)
Complex I: NADH Dehydrogenase. II: FADH2 Dehydrogenase (Succinate Dehydrogenase). III: bc1 complex. IV: Cyt c oxidase. V: ATP synthase.
Complex I: NADH Dehydrogenase. II: FADH2 Dehydrogenase (Succinate Dehydrogenase). III: bc1 complex. IV: Cyt c oxidase. V: ATP synthase.
Q5. Which ecosystem has the maximum biomass?
Correct Answer: (d)
Forests house massive woody trees, resulting in the highest standing crop biomass compared to grasslands or aquatic systems.
Forests house massive woody trees, resulting in the highest standing crop biomass compared to grasslands or aquatic systems.
Q6. Match the contraceptive methods with their categories:
| Column-I (Method) | Column-II (Category) |
|---|---|
| A. Periodic abstinence | (I) Barrier |
| B. Condoms | (II) Natural |
| C. Vasectomy | (III) IUD |
| D. CuT | (IV) Surgical |
Correct Answer: (a)
Contraceptive classifications as per the text.
Contraceptive classifications as per the text.
Q7. The figure given below is the characteristic structure of the phylum in which animals are aquatic, free swimming or sessile, mostly marine, radially symmetrical. Identify the phylum and correct function of the structure.
Correct Answer: (c)
The structure is a cnidocyte/cnidoblast, used by Cnidarians for defense and prey capture.
The structure is a cnidocyte/cnidoblast, used by Cnidarians for defense and prey capture.
Q8. The F1 generation of the monohybrid cross purple (PP) × white (pp) flower pea plants should
Correct Answer: (c)
The cross between a homozygous dominant individual (PP) and a homozygous recessive individual (pp) produces 100% heterozygous offspring (Pp). Since purple is completely dominant over white in Mendel's peas, all F1 plants will express the purple phenotype.
The cross between a homozygous dominant individual (PP) and a homozygous recessive individual (pp) produces 100% heterozygous offspring (Pp). Since purple is completely dominant over white in Mendel's peas, all F1 plants will express the purple phenotype.
Q9. The large holes in 'Swiss cheese' are due to the production of a large amount of CO2 by:
Correct Answer: (a)
Page 3 attributes the large holes in Swiss cheese to Propionibacterium sharmanii.
Page 3 attributes the large holes in Swiss cheese to Propionibacterium sharmanii.
Q10. Fermentation takes place
Correct Answer: (a)
Fermentation is an anaerobic process used by various prokaryotes and unicellular eukaryotes (like yeast) to regenerate NAD+ in the absence of oxygen.
Fermentation is an anaerobic process used by various prokaryotes and unicellular eukaryotes (like yeast) to regenerate NAD+ in the absence of oxygen.
Q11. Match the secondary metabolite with its category (Table 9.3):
| Column I (Metabolite) | Column II (Category) |
|---|---|
| A. Morphine | (I) Toxin |
| B. Monoterpenes | (II) Alkaloid |
| C. Abrin | (III) Terpenoides |
| D. Curcumin | (IV) Drug |
Correct Answer: (a)
Data from the summary table in Unit VIII.
Data from the summary table in Unit VIII.
Q12. In angiosperms, a mature male gametophyte is derived from a pollen mother cell by
Correct Answer: (b)
A pollen mother cell (2n) undergoes meiosis to produce 4 microspores (n). Each microspore then undergoes mitosis to form a generative cell and vegetative cell (pollen grain).
A pollen mother cell (2n) undergoes meiosis to produce 4 microspores (n). Each microspore then undergoes mitosis to form a generative cell and vegetative cell (pollen grain).
Q13. Heartwood differs from sapwood in:
Correct Answer: (c)
Heartwood is the central, non-conducting wood characterized by depositions that make it dark and durable, and its elements are dead.
Heartwood is the central, non-conducting wood characterized by depositions that make it dark and durable, and its elements are dead.
Q14. Pinus seed cannot germinate and establish without fungal association. This is because:
Correct Answer: (b)
Pinus has an obligate symbiotic relationship with mycorrhizal fungi for nutrient absorption.
Pinus has an obligate symbiotic relationship with mycorrhizal fungi for nutrient absorption.
Q15. Difference between virus and viroid is:
Correct Answer: (a)
Viroids are infectious agents composed only of RNA, lacking the protein coat (capsid) of a virus.
Viroids are infectious agents composed only of RNA, lacking the protein coat (capsid) of a virus.
Q16. With regard to the mineral absorption, consider the following statements:
(1) The entry of ions to the symplast is an active process while exit from it is passive.
(2) The movement of ions into the apoplast occurs through the trans-membrane proteins.
(1) The entry of ions to the symplast is an active process while exit from it is passive.
(2) The movement of ions into the apoplast occurs through the trans-membrane proteins.
Correct Answer: (a)
Ion entry into the symplast is active. However, movement into the apoplast is primarily a passive process (diffusion/mass flow) through cell walls, not requiring trans-membrane proteins.
Ion entry into the symplast is active. However, movement into the apoplast is primarily a passive process (diffusion/mass flow) through cell walls, not requiring trans-membrane proteins.
Q17. Fick’s Law of Diffusion states that the rate of diffusion is directly proportional to
Correct Answer: (b)
According to Fick's Law (R = DAΔp/d), the rate of diffusion (R) is directly proportional to the surface area (A) available for exchange and the difference in partial pressure (Δp) of the gas across the membrane.
According to Fick's Law (R = DAΔp/d), the rate of diffusion (R) is directly proportional to the surface area (A) available for exchange and the difference in partial pressure (Δp) of the gas across the membrane.
Q18. Studies that demonstrate that species living in an ecological community change independently of one another in space and time
Correct Answer: (b)
Independent changes of species across space and time imply that each species responds to different environmental factors. This supports the view that realized niches are regulated by different aspects of the environment rather than by a single controlling interaction.
Independent changes of species across space and time imply that each species responds to different environmental factors. This supports the view that realized niches are regulated by different aspects of the environment rather than by a single controlling interaction.
Q19. The amount of trace elements per kg dry matter is:
Correct Answer: (c)
Micronutrients or trace elements are needed in very small amounts (less than 10 mmole kg–1 of dry matter).
Micronutrients or trace elements are needed in very small amounts (less than 10 mmole kg–1 of dry matter).
Q20. Conversion of glucose to glucose-6-phosphate, the first irreversible reaction of glycolysis, is catalysed by:
Correct Answer: (b)
Hexokinase catalyzes the irreversible first step of glycolysis.
Hexokinase catalyzes the irreversible first step of glycolysis.
Q21. Match the followings and choose the correct option from below.
| A. Meristem | (i) photosynthesis, storage |
| B. Parenchyma | (ii) mechanical support |
| C. Collenchyma | (iii) actively dividing cells |
| D. Sclerenchyma | (iv) stomata |
| E. Epidermal tissue | (v) sclereids |
Correct Answer: (b)
Meristem divides; Parenchyma is metabolic; Collenchyma provides support; Sclerenchyma contains sclereids; Epidermis contains stomata.
Meristem divides; Parenchyma is metabolic; Collenchyma provides support; Sclerenchyma contains sclereids; Epidermis contains stomata.
Q22. According to the idea of coevolution between predator and prey, when a prey species evolves a novel defense against a predator,
Correct Answer: (d)
Coevolution involves reciprocal evolutionary change. When prey evolve a new defense, natural selection favors predators that can overcome that defense, leading to evolutionary counter-adaptations.
Coevolution involves reciprocal evolutionary change. When prey evolve a new defense, natural selection favors predators that can overcome that defense, leading to evolutionary counter-adaptations.
Q23. Humans excrete their excess nitrogenous wastes as
Correct Answer: (d)
Humans are ureotelic organisms, meaning they convert toxic ammonia into urea in the liver. Urea is less toxic and can be safely transported in the blood and excreted by the kidneys.
Humans are ureotelic organisms, meaning they convert toxic ammonia into urea in the liver. Urea is less toxic and can be safely transported in the blood and excreted by the kidneys.
Q24. Most carbon dioxide is
Correct Answer: (c)
Once carbon dioxide enters red blood cells, the enzyme carbonic anhydrase rapidly catalyzes the reaction between CO2 and H2O to form carbonic acid (H2CO3), which is a key intermediate step in the formation of bicarbonate for transport.
Once carbon dioxide enters red blood cells, the enzyme carbonic anhydrase rapidly catalyzes the reaction between CO2 and H2O to form carbonic acid (H2CO3), which is a key intermediate step in the formation of bicarbonate for transport.
Q25. Which of the following best describes the electrical state of a neuron at rest?
Correct Answer: (a)
The resting membrane potential of a neuron is approximately –70 mV, meaning the intracellular environment is more negatively charged relative to the extracellular fluid. This state is maintained by the Na+/K+ pump and the differential permeability of the membrane to ions.
The resting membrane potential of a neuron is approximately –70 mV, meaning the intracellular environment is more negatively charged relative to the extracellular fluid. This state is maintained by the Na+/K+ pump and the differential permeability of the membrane to ions.
Q26. How would a loss-of-function mutation in the α-amylase gene affect seed germination?
Correct Answer: (b)
α-amylase is responsible for breaking down the starch in the endosperm into sugars. If this enzyme is absent, the developing embryo has no fuel source and will starve before it can establish itself.
α-amylase is responsible for breaking down the starch in the endosperm into sugars. If this enzyme is absent, the developing embryo has no fuel source and will starve before it can establish itself.
Q27. Assertion: Fertilisation restores the diploid phase.
Reason: Haploid gametes fuse to form a diploid zygote.
Correct Answer: (a)
Fusion of n + n = 2n.
Fusion of n + n = 2n.
Q28. The Meselson and Stahl experiment used a density label to be able to
Correct Answer: (c)
By growing bacteria in heavy nitrogen (15N) and then switching them to light nitrogen (14N), Meselson and Stahl could distinguish original parental DNA from newly synthesized DNA based on their different densities when centrifuged in a cesium chloride gradient.
By growing bacteria in heavy nitrogen (15N) and then switching them to light nitrogen (14N), Meselson and Stahl could distinguish original parental DNA from newly synthesized DNA based on their different densities when centrifuged in a cesium chloride gradient.
Q29. Countercurrent flow systems do not occur in lungs because they
Correct Answer: (a)
Countercurrent exchange requires a continuous, unidirectional flow of two different fluids in opposite directions (e.g., water and blood in gills). Lungs utilize a tidal (in-and-out) flow of air, which does not provide the steady unidirectional stream needed for a countercurrent system.
Countercurrent exchange requires a continuous, unidirectional flow of two different fluids in opposite directions (e.g., water and blood in gills). Lungs utilize a tidal (in-and-out) flow of air, which does not provide the steady unidirectional stream needed for a countercurrent system.
Q30. A molecule of CO₂ that is generated in the cardiac muscle of the left ventricle would not pass through which of the following structures before leaving the body?
Correct Answer: (b)
CO₂ produced in the left ventricle enters the coronary circulation, returns to the right atrium, then passes through the right ventricle and lungs before being exhaled. It does not re-enter the left atrium before leaving the body.
CO₂ produced in the left ventricle enters the coronary circulation, returns to the right atrium, then passes through the right ventricle and lungs before being exhaled. It does not re-enter the left atrium before leaving the body.
Q31. For bacteria to continue growing rapidly when they are shifted from an environment containing O2 to an anaerobic environment, they must:
Correct Answer: (c)
Anaerobic respiration is much less efficient (2 ATP vs 36/38 ATP). To maintain the same energy levels, the rate of glucose consumption (glycolysis) must increase significantly (Pasteur effect).
Anaerobic respiration is much less efficient (2 ATP vs 36/38 ATP). To maintain the same energy levels, the rate of glucose consumption (glycolysis) must increase significantly (Pasteur effect).
Q32. The correct order of the process of decomposition is
Correct Answer: (d)
The steps are fragmentation of detritus, leaching, catabolism, humification, and mineralization.
The steps are fragmentation of detritus, leaching, catabolism, humification, and mineralization.
Q33. Select the incorrect statement.
Correct Answer: (c)
Lysine is a basic amino acid, not neutral, as it contains an additional amino group in its side chain.
Lysine is a basic amino acid, not neutral, as it contains an additional amino group in its side chain.
Q34. Flocs in secondary sewage treatment consist of:
Correct Answer: (b)
Page 6 defines flocs as masses of bacteria associated with fungal filaments.
Page 6 defines flocs as masses of bacteria associated with fungal filaments.
Q35. Which of the following statement(s) is/are correct about biogeochemical cycle?
(i) Gaseous exchanges of phosphorus between organism and environment are negligible
(ii) All biogeochemical cycles include both organisms and non-living components.
(iii) Most elements remain longest in the living portion of their cycle.
(iv) The chemical elements used by organisms in large quantities cycle back and forth between organisms and environment.
(i) Gaseous exchanges of phosphorus between organism and environment are negligible
(ii) All biogeochemical cycles include both organisms and non-living components.
(iii) Most elements remain longest in the living portion of their cycle.
(iv) The chemical elements used by organisms in large quantities cycle back and forth between organisms and environment.
Correct Answer: (b)
Statements (i), (ii) and (iv) are correct. (iii) is false because reservoirs (soil/air/rock) usually hold nutrients for much longer periods than the biotic phase.
Statements (i), (ii) and (iv) are correct. (iii) is false because reservoirs (soil/air/rock) usually hold nutrients for much longer periods than the biotic phase.
Q36. As you sit quietly reading this sentence, the part of the nervous system that is most active is the
Correct Answer: (c)
The parasympathetic nervous system is the 'rest and digest' division of the autonomic nervous system. It predominates during periods of relaxation and low stress, maintaining homeostatic functions and conserving energy.
The parasympathetic nervous system is the 'rest and digest' division of the autonomic nervous system. It predominates during periods of relaxation and low stress, maintaining homeostatic functions and conserving energy.
Q37. Which of the following statement(s) is/are incorrect ?
(a) H2S, not H2O, is involved in photosynthesis of purple sulphur bacteria.
(b) Light and dark reactions are stopped in the absence of light.
(c) Calvin cycle occurs in the grana of chloroplast.
(a) H2S, not H2O, is involved in photosynthesis of purple sulphur bacteria.
(b) Light and dark reactions are stopped in the absence of light.
(c) Calvin cycle occurs in the grana of chloroplast.
Correct Answer: (d)
The dark reaction occurs in the stroma, not grana. Both (b) and (c) are incorrect in this biological context.
The dark reaction occurs in the stroma, not grana. Both (b) and (c) are incorrect in this biological context.
Q38. Diadelphous stamens are the characteristic features of
Correct Answer: (b)
In Fabaceae (pea family), stamens are diadelphous, meaning they are united in two bundles (usually 9+1).
In Fabaceae (pea family), stamens are diadelphous, meaning they are united in two bundles (usually 9+1).
Q39. Receptors that trigger innate immune responses
Correct Answer: (c)
Innate immune receptors, such as Toll-like receptors, recognize conserved pathogen-associated molecular patterns (PAMPs) that are common to many pathogens, allowing a rapid, non-specific immune response.
Innate immune receptors, such as Toll-like receptors, recognize conserved pathogen-associated molecular patterns (PAMPs) that are common to many pathogens, allowing a rapid, non-specific immune response.
Q40. Which of the following effects is brought about by gibberellins but not by auxins?
Correct Answer: (d)
Gibberellins are uniquely effective in breaking dormancy of seeds and buds, whereas auxins generally promote apical dominance (maintaining bud inhibition).
Gibberellins are uniquely effective in breaking dormancy of seeds and buds, whereas auxins generally promote apical dominance (maintaining bud inhibition).
Q41. Ustilago causes plant diseases (called smuts) because
Correct Answer: (c)
The name 'smut' refers to the thick-walled, black, sooty teliospores produced by fungi like Ustilago.
The name 'smut' refers to the thick-walled, black, sooty teliospores produced by fungi like Ustilago.
Q42. Suppose that you stick your finger with a sharp pin. The area affected is very small and only one pain receptor fires. However, it fires repeatedly at a rapid rate (it hurts!). This is an example of
Correct Answer: (a)
Temporal summation occurs when a single presynaptic neuron (or receptor) fires multiple action potentials in rapid succession, resulting in the additive effect of postsynaptic potentials at the same location over time.
Temporal summation occurs when a single presynaptic neuron (or receptor) fires multiple action potentials in rapid succession, resulting in the additive effect of postsynaptic potentials at the same location over time.
Q43. If you wanted to study the use of ATP during a single contraction cycle within a muscle cell, which of the following processes would you use?
Correct Answer: (b)
A twitch is a single, brief contraction and relaxation cycle of a muscle fiber in response to a single action potential. This makes it the ideal unit for studying the molecular events and energy consumption of one contraction cycle.
A twitch is a single, brief contraction and relaxation cycle of a muscle fiber in response to a single action potential. This makes it the ideal unit for studying the molecular events and energy consumption of one contraction cycle.
Q44. Experiments done by Charles Darwin and his son on plant phototropism showed that:
Correct Answer: (b)
The Darwins observed that the coleoptile of canary grass responds to unilateral light only if the tip is present, concluding the tip senses the light.
The Darwins observed that the coleoptile of canary grass responds to unilateral light only if the tip is present, concluding the tip senses the light.
Q45. Sex determination in birds follows the ZW system, where males have two copies of the Z chromosome (ZZ) and females are heterozygous (ZW). A male bird (who is heterozygous for a trait on the Z chromosome) is mated with a female bird (who expresses the recessive phenotype for that trait). Which of the following best describes their most likely offspring?
Correct Answer: (a)
(A) The following Punnett square shows the cross between a heterozygous male (ZAZa) and a female with the recessive trait (ZaW):
So 1/4 of the offspring would be males with the dominant phenotype, 1/4 would be males with the recessive phenotype, 1/4 would be females with the dominant phenotype, and 1/4 would be females with the recessive phenotype.
(A) The following Punnett square shows the cross between a heterozygous male (ZAZa) and a female with the recessive trait (ZaW):
| ZA | Za | Results | |
| Za | ZAZa | ZaZa | 1/4 males with dominant phenotype, 1/4 males with recessive phenotype |
| W | ZAW | ZaW | 1/4 females with dominant phenotype, 1/4 females with recessive phenotype |
So 1/4 of the offspring would be males with the dominant phenotype, 1/4 would be males with the recessive phenotype, 1/4 would be females with the dominant phenotype, and 1/4 would be females with the recessive phenotype.
Q46. Identify the incorrect statement.
Correct Answer: (b)
Sapwood is the peripheral (outermost) part of the secondary xylem; heartwood is the innermost part.
Sapwood is the peripheral (outermost) part of the secondary xylem; heartwood is the innermost part.
Q47. A pea plant that is heterozygous for round peas and heterozygous for smooth pods is crossed with a pea plant that is heterozygous for round peas and has constricted pods. What fraction of the offspring are expected to have round peas and constricted pods? (Remember that round peas are dominant to wrinkled peas, and smooth pods are dominant to constricted pods.)
Correct Answer: (b)
The genotypes of the plants described in the problem are RrSs and Rrss. First, calculate the probability for each trait separately. The probability of two plants that are heterozygous for round peas producing an offspring with round peas is 3/4. The probability of a plant with heterozygous smooth pods crossed with a plant with constricted pods producing an offspring with constricted pods is 1/2. Multiplying the two probabilities together gives the probability of a plant with both round peas and constricted pods: 3/4 × 1/2 = 3/8.
The genotypes of the plants described in the problem are RrSs and Rrss. First, calculate the probability for each trait separately. The probability of two plants that are heterozygous for round peas producing an offspring with round peas is 3/4. The probability of a plant with heterozygous smooth pods crossed with a plant with constricted pods producing an offspring with constricted pods is 1/2. Multiplying the two probabilities together gives the probability of a plant with both round peas and constricted pods: 3/4 × 1/2 = 3/8.
Q48. An organ is classified as part of the endocrine system if it
Correct Answer: (d)
The defining functional characteristic of an endocrine organ or gland is that it secretes its chemical products (hormones) directly into the circulatory system (bloodstream) rather than through a duct system.
The defining functional characteristic of an endocrine organ or gland is that it secretes its chemical products (hormones) directly into the circulatory system (bloodstream) rather than through a duct system.
Q49. Which of the following is responsible for the formation of an embryonic shoot called axillary bud?
Correct Answer: (b)
The shoot apical meristem is the source of cells that form axillary buds in the axils of leaves.
The shoot apical meristem is the source of cells that form axillary buds in the axils of leaves.
Q50. Mild vitamin D deficiency can lead to osteoporosis, or reduced bone mineral density. This is thought to be due to an association with increased levels of
Correct Answer: (b)
Vitamin D is necessary for calcium absorption from the gut. When Vitamin D is low, blood calcium drops, triggering the parathyroid glands to release more Parathyroid Hormone (PTH). PTH increases blood calcium by breaking down bone tissue, leading to a loss of bone density over time.
Vitamin D is necessary for calcium absorption from the gut. When Vitamin D is low, blood calcium drops, triggering the parathyroid glands to release more Parathyroid Hormone (PTH). PTH increases blood calcium by breaking down bone tissue, leading to a loss of bone density over time.
Q51. Identify correct facts about 'Saheli':
(i) Oral pill.
(ii) High contraceptive value.
(iii) Few side effects.
(i) Oral pill.
(ii) High contraceptive value.
(iii) Few side effects.
Correct Answer: (a)
Features of Saheli (Page 45).
Features of Saheli (Page 45).
Q52. Which of the following class is being correctly described by given statements (i - iv)?
(i) All living members of this class are ectoparasites on some fishes.
(ii) They have a sucking and circular mouth without jaws.
(iii) Circulation is of closed type.
(iv) They are marine but migrate for spawning to fresh water. After spawning, within a few days they die.
(i) All living members of this class are ectoparasites on some fishes.
(ii) They have a sucking and circular mouth without jaws.
(iii) Circulation is of closed type.
(iv) They are marine but migrate for spawning to fresh water. After spawning, within a few days they die.
Correct Answer: (a)
Description matches class Cyclostomata.
Description matches class Cyclostomata.
Q53. Refer the functions of the growth hormones given below: I. Cell division, II. Cell enlargement, III. Pattern formation, IV. Tropic growth, V Flowering, VI. Fruiting, VII. Seed germination, VIII.Response to wound, IX. Response to stresses of biotic and abiotic origin. Identify the functions of growth promoters and growth inhibitors from the above.
Correct Answer: (c)
Growth promoters (Auxin, GA, Cytokinin) handle division, flowering, etc. Growth inhibitors (ABA, Ethylene) handle stress responses and dormancy.
Growth promoters (Auxin, GA, Cytokinin) handle division, flowering, etc. Growth inhibitors (ABA, Ethylene) handle stress responses and dormancy.
Q54. The given figure shows the diagrammatic representation of the Hatch & Slack pathway few labelling are marked as P, Q and R.
Which of the following option shows the correct labeling of P, Q, and R?
Which of the following option shows the correct labeling of P, Q, and R?
Correct Answer: (b)
In C4 plants: Atmospheric CO2 is fixed (P) into a C4 acid (Q). After transport and decarboxylation, the residue returns to complete regeneration (R).
In C4 plants: Atmospheric CO2 is fixed (P) into a C4 acid (Q). After transport and decarboxylation, the residue returns to complete regeneration (R).
Q55. In the tropical rainforest, the majority of trees have showy animal-pollinated flowers. In temperate forests the majority of trees are wind pollinated. Which of the following explains these contrasting patterns?
(i) Wind is rare in tropical forests
(ii) Because of high species diversity in the tropics, individuals of tree species are often widely separated making wind an inefficient means of pollen dispersal.
(iii) More opportunities for coevolved mutualisms exist in tropical forests because of the high diversity of animal species.
(iv) Trees in tropical forests are mostly evergreen and year round leaf canopies impede pollen dispersal by wind.
(v) Flowering in tropical forests occurs over a short period of time when wind is absent.
(i) Wind is rare in tropical forests
(ii) Because of high species diversity in the tropics, individuals of tree species are often widely separated making wind an inefficient means of pollen dispersal.
(iii) More opportunities for coevolved mutualisms exist in tropical forests because of the high diversity of animal species.
(iv) Trees in tropical forests are mostly evergreen and year round leaf canopies impede pollen dispersal by wind.
(v) Flowering in tropical forests occurs over a short period of time when wind is absent.
Correct Answer: (c)
High diversity means individuals of the same species are far apart (ii), leading to co-evolution with animal pollinators (iii), while dense evergreen canopies block wind (iv).
High diversity means individuals of the same species are far apart (ii), leading to co-evolution with animal pollinators (iii), while dense evergreen canopies block wind (iv).
Q56. If the diploid number of a flowering plant is 36. What would be the chromosome number in its endosperm?
Correct Answer: (c)
Endosperm in angiosperms is triploid (3n). If 2n = 36, then n = 18. Thus, 3n = 3 × 18 = 54.
Endosperm in angiosperms is triploid (3n). If 2n = 36, then n = 18. Thus, 3n = 3 × 18 = 54.
Q57. Which of the following statements on phytohormones & their action are correct?
(i) Cytokinins specially help in delaying senescence.
(ii) Auxins are involved in regulating apical dominance.
(iii) Ethylene is specially useful in enhancing seed germination.
(iv) Gibberellins are responsible for immature falling of leaves.
(i) Cytokinins specially help in delaying senescence.
(ii) Auxins are involved in regulating apical dominance.
(iii) Ethylene is specially useful in enhancing seed germination.
(iv) Gibberellins are responsible for immature falling of leaves.
Correct Answer: (c)
Statements (i) and (ii) are correct. Ethylene breaks seed dormancy (initiation) but isn't the primary 'germination enhancer'. Gibberellins prevent premature fall.
Statements (i) and (ii) are correct. Ethylene breaks seed dormancy (initiation) but isn't the primary 'germination enhancer'. Gibberellins prevent premature fall.
Q58. Select the correctly matched pair.
I. S phase – DNA replication
II. Zygotene – Synapsis
III. Diplotene – Crossing over
IV. Meiosis – Both haploid and diploid cells
I. S phase – DNA replication
II. Zygotene – Synapsis
III. Diplotene – Crossing over
IV. Meiosis – Both haploid and diploid cells
Correct Answer: (a)
S-phase is for DNA replication and Zygotene is for Synapsis. Crossing over occurs in Pachytene.
S-phase is for DNA replication and Zygotene is for Synapsis. Crossing over occurs in Pachytene.
Q59. In the Irish Setter, the overall red coat color is controlled by the melanocortin receptor. When this receptor is bound by MSH, dark eumelanin is produced. When bound by an antagonist, red pheomelanin is made. The red color is likely due to a mutation in the
Correct Answer: (b)
If the red coat color is caused by the antagonist successfully out-competing MSH or MSH failing to bind, a mutation that prevents the receptor from binding MSH would leave the receptor constantly available to the antagonist, leading to the permanent production of red pigment.
If the red coat color is caused by the antagonist successfully out-competing MSH or MSH failing to bind, a mutation that prevents the receptor from binding MSH would leave the receptor constantly available to the antagonist, leading to the permanent production of red pigment.
Q60. Down syndrome is the result of trisomy for chromosome 21. Why is this trisomy viable whereas trisomy for most other chromosomes is not?
Correct Answer: (c)
Viability of aneuploidy is generally linked to the amount of genetic imbalance. Chromosome 21 is the smallest human autosome with the lowest gene count. Extra copies of larger chromosomes create a 'gene dosage' imbalance so severe that the embryo typically cannot develop.
Viability of aneuploidy is generally linked to the amount of genetic imbalance. Chromosome 21 is the smallest human autosome with the lowest gene count. Extra copies of larger chromosomes create a 'gene dosage' imbalance so severe that the embryo typically cannot develop.
Q61. Morphogenesis is the development of
Correct Answer: (a)
Morphogenesis refers to the biological process that causes an organism to develop its shape. In plants, it specifically describes the development of the growth form through orchestrated cell division and expansion.
Morphogenesis refers to the biological process that causes an organism to develop its shape. In plants, it specifically describes the development of the growth form through orchestrated cell division and expansion.
Q62. Sweet potato is a modified
Correct Answer: (b)
Sweet potato (Ipomoea batatas) is an adventitious root modified for food storage.
Sweet potato (Ipomoea batatas) is an adventitious root modified for food storage.
Q63. Which of the following statements are correct?
(i) Magnesium competes with iron and manganese for uptake and with iron for binding with enzymes.
(ii) Magnesium inhibit calcium translocation in shoot apex.
(iii) Excess of manganese may induce deficiencies of iron, magnesium and calcium.
(iv) Symptoms of manganese toxicity may actually be the deficiency symptoms of iron, magnesium and calcium.
(i) Magnesium competes with iron and manganese for uptake and with iron for binding with enzymes.
(ii) Magnesium inhibit calcium translocation in shoot apex.
(iii) Excess of manganese may induce deficiencies of iron, magnesium and calcium.
(iv) Symptoms of manganese toxicity may actually be the deficiency symptoms of iron, magnesium and calcium.
Correct Answer: (c)
Manganese competes with iron and magnesium for uptake and with magnesium for enzyme binding. It also inhibits calcium translocation. Thus, Mn toxicity mimics Fe, Mg, and Ca deficiency.
Manganese competes with iron and magnesium for uptake and with magnesium for enzyme binding. It also inhibits calcium translocation. Thus, Mn toxicity mimics Fe, Mg, and Ca deficiency.
Q64. Chlorophyll is suited for the capture of light energy because:
Correct Answer: (d)
Chlorophyll absorbs specific light energy, which excites its electrons, and its hydrophobic tail keeps it anchored in the thylakoid membrane.
Chlorophyll absorbs specific light energy, which excites its electrons, and its hydrophobic tail keeps it anchored in the thylakoid membrane.
Q65. In terms of studying gene function, what is the main benefit that genome editing has over RNAi?
Correct Answer: (a)
RNA interference (RNAi) is a 'knockdown' technology that degrades mRNA or blocks translation, resulting in reduced protein levels but often leaving some function intact. Genome editing (like CRISPR/Cas9) creates permanent changes in the DNA sequence, allowing for a complete 'knockout' or total elimination of the gene's function.
RNA interference (RNAi) is a 'knockdown' technology that degrades mRNA or blocks translation, resulting in reduced protein levels but often leaving some function intact. Genome editing (like CRISPR/Cas9) creates permanent changes in the DNA sequence, allowing for a complete 'knockout' or total elimination of the gene's function.
Q66. Genetic drift and natural selection can both lead to rapid rates of evolution. However,
Correct Answer: (d)
Gene flow (migration) involves the movement of alleles between populations. It acts as a homogenizing force that can counteract the effects of local natural selection (by introducing alleles from a different environment) or genetic drift (by replenishing alleles lost by chance), thereby slowing the rate of evolutionary divergence between populations.
Gene flow (migration) involves the movement of alleles between populations. It acts as a homogenizing force that can counteract the effects of local natural selection (by introducing alleles from a different environment) or genetic drift (by replenishing alleles lost by chance), thereby slowing the rate of evolutionary divergence between populations.
Q67. How do the muscles move your hand through space?
Correct Answer: (d)
Muscles generate movement by contracting and by being attached to bones across joints. When a muscle contracts, it pulls one bone toward another, resulting in movement of the hand through space. Muscles do not actively lengthen; they relax.
Muscles generate movement by contracting and by being attached to bones across joints. When a muscle contracts, it pulls one bone toward another, resulting in movement of the hand through space. Muscles do not actively lengthen; they relax.
Q68. Natural selection can
Correct Answer: (d)
Natural selection plays a central role in speciation. It can drive populations to adapt to different environments (divergence), favor traits that prevent mating between groups (reinforcement of isolation), and select against hybrids if they are less fit than the parent species.
Natural selection plays a central role in speciation. It can drive populations to adapt to different environments (divergence), favor traits that prevent mating between groups (reinforcement of isolation), and select against hybrids if they are less fit than the parent species.
Q69. An osmoregulator would maintain its internal fluids at a concentration that is ____ relative to its surroundings.
Correct Answer: (d)
Osmoregulators actively control their internal osmotic concentration and may maintain body fluids that are hypertonic, hypotonic, or isotonic relative to the external environment, depending on ecological conditions.
Osmoregulators actively control their internal osmotic concentration and may maintain body fluids that are hypertonic, hypotonic, or isotonic relative to the external environment, depending on ecological conditions.
Q70. Which of the following statements is wrong for viroids?
Correct Answer: (d)
Viroids have RNA of low molecular weight.
Viroids have RNA of low molecular weight.
Q71. Fertilisation in humans usually takes place in the:
Correct Answer: (c)
The ovum released by the ovary is also transported to the ampullary region where fertilisation takes place.
The ovum released by the ovary is also transported to the ampullary region where fertilisation takes place.
Q72. Place the following events in the correct order:
1. Sarcoplasmic reticulum releases Ca2+.
2. Myosin binds to actin.
3. Action potential arrives from neuron.
4. Ca2+ binds to troponin.
1. Sarcoplasmic reticulum releases Ca2+.
2. Myosin binds to actin.
3. Action potential arrives from neuron.
4. Ca2+ binds to troponin.
Correct Answer: (d)
The correct sequence starts with the stimulus (Action potential arrives, 3), which triggers the release of calcium from storage (SR releases Ca2+, 1). The calcium then binds to the regulatory protein (Ca2+ binds to troponin, 4), which reveals the binding sites, allowing the cross-bridge to form (Myosin binds to actin, 2).
The correct sequence starts with the stimulus (Action potential arrives, 3), which triggers the release of calcium from storage (SR releases Ca2+, 1). The calcium then binds to the regulatory protein (Ca2+ binds to troponin, 4), which reveals the binding sites, allowing the cross-bridge to form (Myosin binds to actin, 2).
Q73. Which one of the following is related to Ex-situ conservation of threatened animals and plants?
Correct Answer: (d)
Safari parks are off-site human-managed areas for animal protection.
Safari parks are off-site human-managed areas for animal protection.
Q74. A scientist at a pharmaceutical company wants to create bacteria that will synthesize human growth hormone.
What should the scientist use to add the DNA code for human growth hormone to a plasmid?
What should the scientist use to add the DNA code for human growth hormone to a plasmid?
Correct Answer: (b)
DNA ligase attaches DNA fragments together, joining them with phosphodiester bonds. Choice (A) is incorrect because bacterial transformation is used to insert naked, foreign DNA into a cell. Gel electrophoresis separates DNA fragments by size, so choice (C) is incorrect. Polymerase chain reaction is used to amplify the number of copies of a specific sequence of DNA, so choice (D) is incorrect.
DNA ligase attaches DNA fragments together, joining them with phosphodiester bonds. Choice (A) is incorrect because bacterial transformation is used to insert naked, foreign DNA into a cell. Gel electrophoresis separates DNA fragments by size, so choice (C) is incorrect. Polymerase chain reaction is used to amplify the number of copies of a specific sequence of DNA, so choice (D) is incorrect.
Q75. Wildlife management introduces 135 ducks to a reservoir in an effort to restore the duck population in the area. The growth rate of the population (rmax) is 0.074 ducks/year. Which of the following is the best estimate of the size of this duck population after one year?
Correct Answer: (c)
(C) dN/dt = rmaxN = (0.074)(135) = 10 ducks in one year. Since the starting population size is 135 ducks, after one year, there will be 135 + 10 = 145 ducks.
(C) dN/dt = rmaxN = (0.074)(135) = 10 ducks in one year. Since the starting population size is 135 ducks, after one year, there will be 135 + 10 = 145 ducks.
Q76. The adjoining graph shows change in concentration of substrate on enzyme activity. Identify A, B and C.
Correct Answer: (d)
In an enzyme saturation graph: A is the Maximum Velocity (Vmax), B is the substrate concentration at half Vmax (Km), and C is the point of Half Maximum Velocity (Vmax/2).
In an enzyme saturation graph: A is the Maximum Velocity (Vmax), B is the substrate concentration at half Vmax (Km), and C is the point of Half Maximum Velocity (Vmax/2).
Q77. Which of the following is not a function of cytoskeleton in a cell?
Correct Answer: (a)
While it provides the track, 'intracellular transport' itself is usually assigned to membrane systems like the ER/Golgi or motor proteins, but NCERT emphasizes support, motility, and shape.
While it provides the track, 'intracellular transport' itself is usually assigned to membrane systems like the ER/Golgi or motor proteins, but NCERT emphasizes support, motility, and shape.
Q78. Reproductive isolation is
Correct Answer: (c)
The biological species concept defines a species as a group of actually or potentially interbreeding natural populations that are reproductively isolated from other such groups. Therefore, reproductive isolation is the primary criterion for defining species boundaries under this concept.
The biological species concept defines a species as a group of actually or potentially interbreeding natural populations that are reproductively isolated from other such groups. Therefore, reproductive isolation is the primary criterion for defining species boundaries under this concept.
Q79. An ECG measures
Correct Answer: (a)
An electrocardiogram (ECG) records the electrical activity of the heart by detecting changes in electrical potential produced during depolarization and repolarization of cardiac muscle fibers throughout the cardiac cycle. It does not measure ion concentration, force of contraction, or blood volume directly.
An electrocardiogram (ECG) records the electrical activity of the heart by detecting changes in electrical potential produced during depolarization and repolarization of cardiac muscle fibers throughout the cardiac cycle. It does not measure ion concentration, force of contraction, or blood volume directly.
Q80. In sewage treatment, primary treatment involves:
Correct Answer: (b)
Page 6 defines primary treatment as the physical removal of large and small particles through filtration and sedimentation.
Page 6 defines primary treatment as the physical removal of large and small particles through filtration and sedimentation.
Q81. The lac repressor, the trp repressor, and CAP are all
Correct Answer: (c)
These regulatory proteins are allosteric, meaning they change their conformation and DNA-binding affinity when they bind to a specific small effector molecule (like lactose, tryptophan, or cAMP). This allows the proteins to toggle between 'on' and 'off' states in response to environmental conditions.
These regulatory proteins are allosteric, meaning they change their conformation and DNA-binding affinity when they bind to a specific small effector molecule (like lactose, tryptophan, or cAMP). This allows the proteins to toggle between 'on' and 'off' states in response to environmental conditions.
Q82. Rh positive (R) is a trait that is dominant over Rh negative (r). A woman, who is Rh positive, has a son, who is Rh negative.
Which of the following is a true statement about the father of the woman’s son?
Which of the following is a true statement about the father of the woman’s son?
Correct Answer: (d)
Since the son is Rh negative, his genotype must be rr. He must have inherited one r allele from each parent. So the father (whose Rh phenotype is unknown) could be either Rr or rr. Choice (A) is incorrect because an RR father would not be able to pass on an r allele to his offspring. Choices (B) and (C) are incorrect because while both are possible genotypes for the father, it is not possible to say with certainty which genotype the father has without knowing the father’s phenotype.
Since the son is Rh negative, his genotype must be rr. He must have inherited one r allele from each parent. So the father (whose Rh phenotype is unknown) could be either Rr or rr. Choice (A) is incorrect because an RR father would not be able to pass on an r allele to his offspring. Choices (B) and (C) are incorrect because while both are possible genotypes for the father, it is not possible to say with certainty which genotype the father has without knowing the father’s phenotype.
Q83. Cells that target and kill body cells infected by viruses are
Correct Answer: (b)
Natural killer (NK) cells are part of the innate immune system and specialize in detecting and destroying virus-infected and cancerous cells without prior sensitization. They recognize altered or missing MHC-I molecules on infected cells and induce apoptosis.
Natural killer (NK) cells are part of the innate immune system and specialize in detecting and destroying virus-infected and cancerous cells without prior sensitization. They recognize altered or missing MHC-I molecules on infected cells and induce apoptosis.
Q84. Which of the following statements (i - v) are correct ?
(i) The pelvic fins of female sharks bear claspers.
(ii) In Obelia, polyps produce medusae sexually and medusae form the polyps asexually.
(iii) Flame cells in platyhelminthes help in osmoregulation and excretion.
(iv) In non-chordates, central nervous system is ventral, solid and double.
(v) Pinnae are present in mammals.
(i) The pelvic fins of female sharks bear claspers.
(ii) In Obelia, polyps produce medusae sexually and medusae form the polyps asexually.
(iii) Flame cells in platyhelminthes help in osmoregulation and excretion.
(iv) In non-chordates, central nervous system is ventral, solid and double.
(v) Pinnae are present in mammals.
Correct Answer: (c)
Statement (i) is false (males have claspers). (ii) is false (polyp to medusa is asexual). (iii), (iv), (v) are correct.
Statement (i) is false (males have claspers). (ii) is false (polyp to medusa is asexual). (iii), (iv), (v) are correct.
Q85. A stamen contains a
Correct Answer: (c)
The stamen is the male reproductive organ of the flower and is comprised of an anther (pollen-bearing part) and a stalk known as the filament. Style and stigma are parts of the carpel.
The stamen is the male reproductive organ of the flower and is comprised of an anther (pollen-bearing part) and a stalk known as the filament. Style and stigma are parts of the carpel.
Q86. An important essential element which is required by plants in the greatest amount is:
Correct Answer: (a)
Nitrogen is the essential nutrient element required by plants in the largest amount.
Nitrogen is the essential nutrient element required by plants in the largest amount.
Q87. A forensic scientist cuts DNA from a crime scene and DNA from a suspect with the same enzyme. Which tool should the scientist use to separate the DNA fragments according to their size?
Correct Answer: (c)
Gel electrophoresis separates fragments of DNA according to their size. Bacterial transformation involves the uptake and expression of foreign DNA by bacteria, so choice (A) is incorrect. CRISPR-Cas9 is used for gene editing, so choice (B) is incorrect. PCR amplifies specific sequences of DNA, so choice (D) is also incorrect.
Gel electrophoresis separates fragments of DNA according to their size. Bacterial transformation involves the uptake and expression of foreign DNA by bacteria, so choice (A) is incorrect. CRISPR-Cas9 is used for gene editing, so choice (B) is incorrect. PCR amplifies specific sequences of DNA, so choice (D) is also incorrect.
Q88. Which of the following best describes hormones?
Correct Answer: (b)
Hormones are chemical signaling molecules produced by endocrine glands. Unlike local paracrine signals, hormones are secreted into the bloodstream to reach distant target organs and generally remain active in the body for a longer duration to regulate physiological processes.
Hormones are chemical signaling molecules produced by endocrine glands. Unlike local paracrine signals, hormones are secreted into the bloodstream to reach distant target organs and generally remain active in the body for a longer duration to regulate physiological processes.
Q89. The middle thick muscular layer of the uterus is called:
Correct Answer: (c)
The middle thick layer of smooth muscle is the myometrium.
The middle thick layer of smooth muscle is the myometrium.
Q90. If you wanted to cure allergies by bioengineering an antibody that would bind and disable the antibody responsible for allergic reactions, which would you target?
Correct Answer: (c)
IgE antibodies are responsible for allergic reactions by binding to mast cells and triggering histamine release. Neutralizing IgE would therefore reduce or prevent allergic responses.
IgE antibodies are responsible for allergic reactions by binding to mast cells and triggering histamine release. Neutralizing IgE would therefore reduce or prevent allergic responses.