Q1. A scientist is studying the allele frequencies for a particular gene in a population of wild prairie dogs over a five-year period. The allele frequencies are shown in the table.
Which of the following is the most likely cause of the observed change in allele frequencies?
| Year | Frequency of Allele A1 | Frequency of Allele A2 |
|---|---|---|
| 1 | 0.00 | 1.00 |
| 2 | 0.00 | 1.00 |
| 3 | 0.10 | 0.90 |
| 4 | 0.13 | 0.87 |
| 5 | 0.15 | 0.85 |
Which of the following is the most likely cause of the observed change in allele frequencies?
Correct Answer: (d)
(D) Initially, the population of wild prairie dogs did not possess the A1 allele, so it is likely that it arrived in an individual prairie dog that migrated into the population sometime between years 2 and 3. Therefore, gene flow is the correct answer. Choice (A) is incorrect because these are wild prairie dogs, and there is no evidence of selective breeding of prairie dogs by humans. No information is given in the question about whether or not alleles A1 or A2 confer a survival advantage, so choice (B) is incorrect. No allele was lost in the population, so choice (C) is incorrect.
(D) Initially, the population of wild prairie dogs did not possess the A1 allele, so it is likely that it arrived in an individual prairie dog that migrated into the population sometime between years 2 and 3. Therefore, gene flow is the correct answer. Choice (A) is incorrect because these are wild prairie dogs, and there is no evidence of selective breeding of prairie dogs by humans. No information is given in the question about whether or not alleles A1 or A2 confer a survival advantage, so choice (B) is incorrect. No allele was lost in the population, so choice (C) is incorrect.
Q2. Chloroplasts are disrupted and the stroma separated from the lamellae. The isolated stroma will fix CO2 if it is supplied with:
Correct Answer: (d)
Stroma contains the enzymes for the dark reaction (Calvin cycle). If provided with the products of light reaction (ATP and NADPH), it can perform carbon fixation in vitro.
Stroma contains the enzymes for the dark reaction (Calvin cycle). If provided with the products of light reaction (ATP and NADPH), it can perform carbon fixation in vitro.
Q3. 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.
Q4. An example of a trait that involves multiple gene inheritance in plants is seed size. If seed size involves three genes (A, B, and C), in which each dominant allele contributes to increased seed size, which of the following genotypes would result in the smallest seed?
Correct Answer: (d)
(D) AaBbCc has the least number of dominant alleles (three) out of all the answer choices given. So the additive effect of the alleles in that genotype would be the least, and this genotype would result in the smallest seed. Since each dominant allele contributes to increased seed size, choice (A) is incorrect because it has the highest number of dominant alleles out of all the answer choices given. So it would produce the largest seeds. Choices (B) and (C) are incorrect because they both have four dominant alleles, so they both would produce larger seeds than choice (D).
(D) AaBbCc has the least number of dominant alleles (three) out of all the answer choices given. So the additive effect of the alleles in that genotype would be the least, and this genotype would result in the smallest seed. Since each dominant allele contributes to increased seed size, choice (A) is incorrect because it has the highest number of dominant alleles out of all the answer choices given. So it would produce the largest seeds. Choices (B) and (C) are incorrect because they both have four dominant alleles, so they both would produce larger seeds than choice (D).
Q5. Generally, which type of data is considered most reliable when constructing phylogenies?
Correct Answer: (d)
(D) The DNA sequence of an organism is not affected by its environment or geological events, so it provides reliable evidence for ancestry of organisms. Fossils do not necessarily show when a species originated, so choice (A) is incorrect. Habitats and morphological characteristics can change during an organism’s lifetime, so choices (B) and (C) are incorrect.
(D) The DNA sequence of an organism is not affected by its environment or geological events, so it provides reliable evidence for ancestry of organisms. Fossils do not necessarily show when a species originated, so choice (A) is incorrect. Habitats and morphological characteristics can change during an organism’s lifetime, so choices (B) and (C) are incorrect.
Q6. Which of the following statement is wrong regarding chitin?
Correct Answer: (a)
Chitin is a structural polysaccharide, not a storage one. It provides rigidity to the fungal cell walls and the exoskeletons of arthropods.
Chitin is a structural polysaccharide, not a storage one. It provides rigidity to the fungal cell walls and the exoskeletons of arthropods.
Q7. Cytological information like chromosome number, structure, behaviour are related with
Correct Answer: (b)
Cytotaxonomy is based on cytological information such as chromosome number, structure, and behavior during meiosis.
Cytotaxonomy is based on cytological information such as chromosome number, structure, and behavior during meiosis.
Q8. Contraction of the smooth muscle layers of the arterioles
Correct Answer: (a)
Contraction of arteriole smooth muscle causes vasoconstriction, narrowing the lumen and increasing resistance to blood flow. It generally reduces, rather than increases, blood flow to an organ and decreases heat loss through the skin.
Contraction of arteriole smooth muscle causes vasoconstriction, narrowing the lumen and increasing resistance to blood flow. It generally reduces, rather than increases, blood flow to an organ and decreases heat loss through the skin.
Q9. ADP is phosphorylated and NADP is reduced, this happens during.
Correct Answer: (b)
The production of ATP (phosphorylation) and NADPH (reduction) are the primary results of the light-dependent reactions.
The production of ATP (phosphorylation) and NADPH (reduction) are the primary results of the light-dependent reactions.
Q10. Under which of the following conditions would pollen from an S2S5 plant successfully pollinate an S1S5 flower?
Correct Answer: (b)
In gametophytic SI, the compatibility of the pollen is determined by its own haploid genotype. Since the pollen of an S2S5 plant is either S2 or S5, the S2 pollen will be compatible with an S1S5 carpel, while the S5 pollen will not. Thus, 50% (half) is successful.
In gametophytic SI, the compatibility of the pollen is determined by its own haploid genotype. Since the pollen of an S2S5 plant is either S2 or S5, the S2 pollen will be compatible with an S1S5 carpel, while the S5 pollen will not. Thus, 50% (half) is successful.
Q11. Which of the following statements is not correct for viruses?
Correct Answer: (c)
Viruses are extremely small and can pass through bacterial-proof filters; this was how Ivanowsky discovered them.
Viruses are extremely small and can pass through bacterial-proof filters; this was how Ivanowsky discovered them.
Q12. Unlike bee-pollinated flowers, bird-pollinated flowers
Correct Answer: (d)
Birds have highly developed color vision, particularly in the red portion of the spectrum, and a poor sense of smell. Consequently, bird-pollinated flowers are often red and lack the strong scents found in bee-pollinated flowers.
Birds have highly developed color vision, particularly in the red portion of the spectrum, and a poor sense of smell. Consequently, bird-pollinated flowers are often red and lack the strong scents found in bee-pollinated flowers.
Q13. Identify the given diagrams and mark the correct option:
Correct Answer: (b)
Based on the visual evidence from the diagrams in the textbook context, simple leaves (B, D) have a single blade while compound leaves (A, C) have blades divided into leaflets.
Based on the visual evidence from the diagrams in the textbook context, simple leaves (B, D) have a single blade while compound leaves (A, C) have blades divided into leaflets.
Q14. A shark’s blood is isotonic to the surrounding seawater because of the reabsorption of ____ in its blood.
Correct Answer: (c)
Sharks retain high concentrations of urea in their blood. This raises the osmolarity of their body fluids, making them isotonic with seawater and preventing excessive water loss by osmosis.
Sharks retain high concentrations of urea in their blood. This raises the osmolarity of their body fluids, making them isotonic with seawater and preventing excessive water loss by osmosis.
Q15. Which one of the following process help the water-soluble inorganic nutrients go down into the soil horizon and get precipitated as unavailable salts?
Correct Answer: (b)
By the process of leaching, water-soluble inorganic nutrients go down into the soil horizon and get precipitated as unavailable salts.
By the process of leaching, water-soluble inorganic nutrients go down into the soil horizon and get precipitated as unavailable salts.
Q16. 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.
Q17. During the process of respiration, which of the followings are released as products?
Correct Answer: (d)
The complete combustion of glucose in aerobic respiration yields carbon dioxide, water, and energy.
The complete combustion of glucose in aerobic respiration yields carbon dioxide, water, and energy.
Q18. Caffeine inhibits the secretion of ADH. Prior to an exam, you have a large coffee. During the exam, you can expect
Correct Answer: (b)
ADH normally increases water reabsorption in the collecting duct. By inhibiting ADH, caffeine reduces water permeability of the collecting duct, resulting in less water reabsorption and increased urine production.
ADH normally increases water reabsorption in the collecting duct. By inhibiting ADH, caffeine reduces water permeability of the collecting duct, resulting in less water reabsorption and increased urine production.
Q19. Select the correct match of the feature present in column I with its respective terms given in column II.
| Column-I (features) | Column-II (term) |
|---|---|
| A. Presence of tap roots and coralloid roots | I. Bryophyte |
| B. The synergids and antipodal cells degenerates after fertilization | II. Pteridophytes |
| C. The food is stored as floridean starch which is very similar to amylopectin and glycogen in structure | III. Red algae |
| D. Presence of sporophyte which is not free living but attached to the photosynthetic gametophytes and derives nourishment from it | IV. Angiosperms |
| E. Members of this group are used for medicinal purposes, as soil binders and frequently grown as ornamentals | V. Gymnosperms |
Correct Answer: (d)
A-V (Gymnosperms); B-IV (Angiosperms); C-III (Red algae); D-I (Bryophyte); E-II (Pteridophytes).
A-V (Gymnosperms); B-IV (Angiosperms); C-III (Red algae); D-I (Bryophyte); E-II (Pteridophytes).
Q20. An important function of the excretory system is to eliminate excess nitrogen produced by metabolic processes. Which of the following organisms is most efficient at packaging nitrogen for excretion?
Correct Answer: (c)
Iguanas excrete nitrogen primarily as uric acid, which is relatively non-toxic and requires very little water for elimination. This makes uric acid the most efficient form of nitrogen packaging among the options given.
Iguanas excrete nitrogen primarily as uric acid, which is relatively non-toxic and requires very little water for elimination. This makes uric acid the most efficient form of nitrogen packaging among the options given.
Q21. Regarding the placenta:
(i) It supplies oxygen and nutrients to the embryo.
(ii) It removes carbon dioxide and waste.
(iii) It secretes hCG, hPL, and estrogens.
(i) It supplies oxygen and nutrients to the embryo.
(ii) It removes carbon dioxide and waste.
(iii) It secretes hCG, hPL, and estrogens.
Correct Answer: (a)
Placenta is vital for both metabolic exchange and endocrine support.
Placenta is vital for both metabolic exchange and endocrine support.
Q22. Which of the following characteristic(s) is/are used by Whittaker for the classification of organisms ?
Correct Answer: (d)
R.H. Whittaker's five-kingdom classification was based on cell structure, thallus organization, mode of nutrition, reproduction, and phylogenetic relationships.
R.H. Whittaker's five-kingdom classification was based on cell structure, thallus organization, mode of nutrition, reproduction, and phylogenetic relationships.
Q23. Which one of the following is incorrectly matched?
Correct Answer: (d)
Ethylene ($) is a simple gaseous hydrocarbon. Adenine derivatives are cytokinins.
Ethylene ($) is a simple gaseous hydrocarbon. Adenine derivatives are cytokinins.
Q24. If you have type AB blood, which of the following results would be expected?
Correct Answer: (c)
Individuals with AB blood type express both A and B antigens on their red blood cells. Therefore, their blood will agglutinate when exposed to either anti-A or anti-B antibodies.
Individuals with AB blood type express both A and B antigens on their red blood cells. Therefore, their blood will agglutinate when exposed to either anti-A or anti-B antibodies.
Q25. Alternate type of phyllotaxy is found in
Correct Answer: (d)
In alternate type of phyllotaxy, a single leaf arises at each node in alternate manner, as in china rose, mustard and sunflower plants.
In alternate type of phyllotaxy, a single leaf arises at each node in alternate manner, as in china rose, mustard and sunflower plants.
Q26. Which of the following ability enables the phenomenon of heterophylly in cotton, coriander and larkspur?
Correct Answer: (d)
Plasticity is the ability of plants to follow different pathways in response to environment or life stages, resulting in different leaf forms (heterophylly).
Plasticity is the ability of plants to follow different pathways in response to environment or life stages, resulting in different leaf forms (heterophylly).
Q27. Cooperation of the two photosystems of the chloroplast is required for:
Correct Answer: (b)
Non-cyclic electron transport (Z-scheme) requires both PSII (to split water and start electron flow) and PSI (to ultimately reduce NADP+ to NADPH).
Non-cyclic electron transport (Z-scheme) requires both PSII (to split water and start electron flow) and PSI (to ultimately reduce NADP+ to NADPH).
Q28. Which of the following is true about lipophilic hormones?
Correct Answer: (b)
Lipophilic hormones do not dissolve well in the aqueous environment of the blood plasma. Therefore, they must travel through the circulatory system bound to specialized transport proteins to remain in solution until they reach their target cells.
Lipophilic hormones do not dissolve well in the aqueous environment of the blood plasma. Therefore, they must travel through the circulatory system bound to specialized transport proteins to remain in solution until they reach their target cells.
Q29. Huntington’s disease is caused by a short tandem CAG repeat in the HTT gene. Individuals with fewer than 35 CAG repeats in the HTT gene do not develop Huntington’s disease. Individuals with 40 or more CAG repeats will develop Huntington’s disease.
Which tool would be most useful for estimating the size of the HTT gene isolated?
Which tool would be most useful for estimating the size of the HTT gene isolated?
Correct Answer: (b)
Gel electrophoresis separates DNA fragments by size and would be most useful for estimating the size of the HTT gene isolated. DNA ligase attaches DNA fragments together, so choice (A) is incorrect. Polymerase chain reaction amplifies specific sequences of DNA, so choice (C) is incorrect. Restriction enzymes cut DNA at specific base pair sequences, so choice (D) is also incorrect.
Gel electrophoresis separates DNA fragments by size and would be most useful for estimating the size of the HTT gene isolated. DNA ligase attaches DNA fragments together, so choice (A) is incorrect. Polymerase chain reaction amplifies specific sequences of DNA, so choice (C) is incorrect. Restriction enzymes cut DNA at specific base pair sequences, so choice (D) is also incorrect.
Q30. 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).
Q31. Root hairs develop from
Correct Answer: (a)
In the region of maturation, some of the epidermal cells form very fine and delicate, thread-like structures called root hairs.
In the region of maturation, some of the epidermal cells form very fine and delicate, thread-like structures called root hairs.
Q32. If an ecosystem is composed of only three trophic levels, then how much energy will be conserved at the IIIrd trophic level?
Correct Answer: (d)
According to the 10% law: $100\% (T1) → 10\% (T2) → 1\% (T3)$.
According to the 10% law: $100\% (T1) → 10\% (T2) → 1\% (T3)$.
Q33. An isolated population in the United States are descendants of approximately 200 immigrants who arrived in the 1700s. Some individuals in this group of immigrants carried the allele for Ellis-van Creveld syndrome. Today, this allele occurs at a much higher frequency in the descendants of this group than in the general population in the United States. This difference in the frequency of the Ellis-van Creveld allele is an example of
Correct Answer: (b)
(B) This scenario is an example of the founder effect. Choice (A) is incorrect because the original population did not undergo a rapid decrease in population size. There is no evidence in the question that Ellis-van Creveld syndrome leads to decreased survival, so choice (C) is incorrect. Choice (D) is incorrect because it would be extremely unlikely that random mutations over several generations would lead to the appearance of the allele that causes Ellis-van Creveld syndrome.
(B) This scenario is an example of the founder effect. Choice (A) is incorrect because the original population did not undergo a rapid decrease in population size. There is no evidence in the question that Ellis-van Creveld syndrome leads to decreased survival, so choice (C) is incorrect. Choice (D) is incorrect because it would be extremely unlikely that random mutations over several generations would lead to the appearance of the allele that causes Ellis-van Creveld syndrome.
Q34. Vernalization induces flowering following exposure to
Correct Answer: (c)
Vernalization is the acquisition of a plant's ability to flower in the spring by exposure to the prolonged cold of winter. This ensure that reproductive development occurs at an appropriate time.
Vernalization is the acquisition of a plant's ability to flower in the spring by exposure to the prolonged cold of winter. This ensure that reproductive development occurs at an appropriate time.
Q35. Identify the correct and incorrect statements from the following. (i) 17,500 new cells are produced per hour by a single maize root apical meristem. (ii) With the help of length, growth of pollen tube is measured. (iii) The growth of the leaf is measured in term of volume. (iv) Cells in a watermelon may increase in size by upto 3,500,000 times.
Correct Answer: (b)
Statement (iii) is wrong because leaf growth is typically measured in terms of surface area, not volume. The others are standard quantitative examples from NCERT.
Statement (iii) is wrong because leaf growth is typically measured in terms of surface area, not volume. The others are standard quantitative examples from NCERT.
Q36. Blood clots are made of
Correct Answer: (a)
Blood clots consist primarily of fibrin, an insoluble protein formed from fibrinogen during the clotting cascade. Prothrombin is a clotting factor but is not a structural component of the clot itself.
Blood clots consist primarily of fibrin, an insoluble protein formed from fibrinogen during the clotting cascade. Prothrombin is a clotting factor but is not a structural component of the clot itself.
Q37. If a PCR is started using 10 pieces of template DNA, how many pieces of DNA would there be after 10 cycles?
Correct Answer: (c)
PCR amplification follows an exponential growth pattern defined by the formula N = N0 × 2n, where N0 is the starting number and n is the number of cycles. In this case, 10 × 210 = 10 × 1024 = 10,240, which is approximately 10,000.
PCR amplification follows an exponential growth pattern defined by the formula N = N0 × 2n, where N0 is the starting number and n is the number of cycles. In this case, 10 × 210 = 10 × 1024 = 10,240, which is approximately 10,000.
Q38. Source–sink metapopulations are distinct from other types of metapopulations because
Correct Answer: (b)
In source–sink metapopulations, some populations (sinks) have negative growth rates and would go extinct without immigration from source populations. This presence of populations with negative intrinsic growth distinguishes source–sink systems from other metapopulations.
In source–sink metapopulations, some populations (sinks) have negative growth rates and would go extinct without immigration from source populations. This presence of populations with negative intrinsic growth distinguishes source–sink systems from other metapopulations.
Q39. Refer the given diagrammatic representation of trophic levels in an ecosystem some spaces are marked A, B, C and D. Identify A, B, C and D.
Correct Answer: (b)
A is the third trophic level (Secondary consumer). B is the first trophic level (Primary producer). C is the top level (Tertiary consumer: Man/Lion). D is a T1 example (Plants).
A is the third trophic level (Secondary consumer). B is the first trophic level (Primary producer). C is the top level (Tertiary consumer: Man/Lion). D is a T1 example (Plants).
Q40. Which of the following features is common to prokaryotes and many eukaryotes?
Correct Answer: (b)
Cell walls are found in prokaryotes and in many eukaryotes like plants, fungi, and algae.
Cell walls are found in prokaryotes and in many eukaryotes like plants, fungi, and algae.
Q41. What potential problems must be considered in creating a transgenic bacterium with the human insulin gene isolated from genomic DNA to produce insulin?
Correct Answer: (b)
Human insulin is synthesized as a precursor (proinsulin) that requires eukaryotic-specific post-translational modifications, including the formation of disulfide bonds and the enzymatic cleavage of the C-peptide in the Golgi apparatus. Bacteria lack these organelles and specialized processing enzymes, so they cannot produce active insulin from a simple genomic gene sequence without additional engineering.
Human insulin is synthesized as a precursor (proinsulin) that requires eukaryotic-specific post-translational modifications, including the formation of disulfide bonds and the enzymatic cleavage of the C-peptide in the Golgi apparatus. Bacteria lack these organelles and specialized processing enzymes, so they cannot produce active insulin from a simple genomic gene sequence without additional engineering.
Q42. 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.
Q43. ATP is:
Correct Answer: (d)
ATP (Adenosine Triphosphate) acts as the chemical energy currency of the cell, is structurally a nucleotide, and is synthesized during both photosynthesis (photophosphorylation) and respiration (oxidative phosphorylation).
ATP (Adenosine Triphosphate) acts as the chemical energy currency of the cell, is structurally a nucleotide, and is synthesized during both photosynthesis (photophosphorylation) and respiration (oxidative phosphorylation).
Q44. If you hold your breath for a long time, body CO2 levels are likely to ________, and the pH of body fluids is likely to ________.
Correct Answer: (c)
When breathing is suspended, carbon dioxide (CO2) accumulates in the blood. CO2 reacts with water to form carbonic acid, which dissociates into hydrogen ions (H+) and bicarbonate. An increase in H+ concentration directly causes the pH to decrease (become more acidic).
When breathing is suspended, carbon dioxide (CO2) accumulates in the blood. CO2 reacts with water to form carbonic acid, which dissociates into hydrogen ions (H+) and bicarbonate. An increase in H+ concentration directly causes the pH to decrease (become more acidic).
Q45. The graph given below shows a geometrical growth rate. Which of the following statement regarding the above graph is incorrect?
Correct Answer: (c)
The growth rate slows down (reaching stationary phase) when nutrient supply becomes limited, it doesn't increase rapidly.
The growth rate slows down (reaching stationary phase) when nutrient supply becomes limited, it doesn't increase rapidly.
Q46. Read the following statement and answer the question: Infected thread carries the bacteria to the inner ‘X’ cells. The bacteria get modified into rod-shaped bacteroids and cause inner ‘X’ and ‘Y’ cells to divide. Division and growth of ‘X’ and ‘Y’ cells lead to nodule formation. Identify ‘X’ and ‘Y’:
Correct Answer: (b)
Rhizobial bacteria released from the infection thread cause division in both the cortical and pericycle cells of the root, leading to nodule formation.
Rhizobial bacteria released from the infection thread cause division in both the cortical and pericycle cells of the root, leading to nodule formation.
Q47. 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.
Q48. Exoskeletons and endoskeletons differ in that
Correct Answer: (d)
The fundamental structural difference is placement. An exoskeleton (like that of an arthropod) provides a hard outer covering for soft internal tissues, while an endoskeleton (like that of a human) is a framework located inside the body's soft tissues.
The fundamental structural difference is placement. An exoskeleton (like that of an arthropod) provides a hard outer covering for soft internal tissues, while an endoskeleton (like that of a human) is a framework located inside the body's soft tissues.
Q49. Steroid hormones
Correct Answer: (d)
Steroid hormones are lipophilic (lipid-soluble). This property allows them to diffuse directly across the plasma membrane of target cells. Once inside, they typically bind to intracellular receptors that move into the nucleus to act as transcription factors, directly influencing gene expression.
Steroid hormones are lipophilic (lipid-soluble). This property allows them to diffuse directly across the plasma membrane of target cells. Once inside, they typically bind to intracellular receptors that move into the nucleus to act as transcription factors, directly influencing gene expression.
Q50. Keel is the characteristic feature of flower of
Correct Answer: (a)
Indigofera belongs to Fabaceae, which has papilionaceous corolla with a 'keel' (two fused anterior petals).
Indigofera belongs to Fabaceae, which has papilionaceous corolla with a 'keel' (two fused anterior petals).
Q51. What would be the minimum number of nucleotides required to code for a protein made of 12 amino acids?
Correct Answer: (c)
(C) Each amino acid is coded by a three base pair codon, so a protein made of 12 amino acids would require a minimum of 12 × 3 = 36 nucleotides. Choice (A) is incorrect because 6 nucleotides would only contain two codons. Since 3 nucleotides are required to make a codon, 12 nucleotides would only be sufficient to code for four amino acids, so choice (B) is incorrect. While 48 nucleotides might code for 12 amino acids if there were introns to be removed, 48 is not the minimum number of nucleotides required, so choice (D) is incorrect.
(C) Each amino acid is coded by a three base pair codon, so a protein made of 12 amino acids would require a minimum of 12 × 3 = 36 nucleotides. Choice (A) is incorrect because 6 nucleotides would only contain two codons. Since 3 nucleotides are required to make a codon, 12 nucleotides would only be sufficient to code for four amino acids, so choice (B) is incorrect. While 48 nucleotides might code for 12 amino acids if there were introns to be removed, 48 is not the minimum number of nucleotides required, so choice (D) is incorrect.
Q52. Match the following Columns.
| Column - I (Organism) | Column - II (Connecting link) |
|---|---|
| A. Echidna | I. between Annelida and Mollusca |
| B. Peripatus | II. between Reptiles and Mammals |
| C. Neopilina | III. between Annelida and Arthropoda |
| D. Protopterus | IV. between Pisces and Amphibian |
Correct Answer: (b)
A-II; B-III; C-I; D-IV.
A-II; B-III; C-I; D-IV.
Q53. Cats with two X chromosomes are female, and cats with one X and one Y chromosome are male. The gene for fur color in cats is on the X chromosome: XB is the black allele and XO is the orange allele. Cats who are heterozygous express the calico phenotype, with patches of black fur and patches of orange fur. A female calico cat is mated with a black male cat. Which of the following best describes the predicted ratios of their offspring?
Correct Answer: (b)
(B) As shown in the figure, 1/4 of the offspring would be black female cats, 1/4 would be calico female cats, 1/4 would be black male cats, and 1/4 would be orange male cats.
(B) As shown in the figure, 1/4 of the offspring would be black female cats, 1/4 would be calico female cats, 1/4 would be black male cats, and 1/4 would be orange male cats.
| XB | XO | Results | |
| XB | XBXB | XBXO | 1/4 black female cats, 1/4 calico female cats |
| Y | XBY | XOY | 1/4 black male cats, 1/4 orange male cats |
Q54. Botox, a derivative of the botulinum toxin that causes food poisoning, inhibits the release of acetylcholine at the neuromuscular junction. How could this strange-sounding treatment produce desired cosmetic effects?
Correct Answer: (c)
Acetylcholine is the primary neurotransmitter that signals skeletal muscles to contract. By blocking its release, Botox induces a temporary paralysis of the facial muscles responsible for creating frown lines and wrinkles, resulting in smoother skin.
Acetylcholine is the primary neurotransmitter that signals skeletal muscles to contract. By blocking its release, Botox induces a temporary paralysis of the facial muscles responsible for creating frown lines and wrinkles, resulting in smoother skin.
Q55. Why do retroviruses have a high mutation rate?
Correct Answer: (d)
(D) Retroviruses use reverse transcriptase to make a DNA copy of their RNA genome. Reverse transcriptase has a very high error rate, so many mutations occur.
(D) Retroviruses use reverse transcriptase to make a DNA copy of their RNA genome. Reverse transcriptase has a very high error rate, so many mutations occur.
Q56. The given figures (A and B) show the modification of roots. Which of the following statements regarding the figures is correct ?
Correct Answer: (a)
Figure A shows storage roots. Carrot and turnip are modified tap roots, while sweet potato is a modified adventitious root, all of which store food.
Figure A shows storage roots. Carrot and turnip are modified tap roots, while sweet potato is a modified adventitious root, all of which store food.
Q57. Second messengers are activated in response to
Correct Answer: (c)
Hydrophilic (water-soluble) hormones cannot cross the phospholipid bilayer of the cell membrane. To exert their effects, they bind to cell-surface receptors, which then trigger the production of intracellular second messengers (like cAMP) to relay the signal.
Hydrophilic (water-soluble) hormones cannot cross the phospholipid bilayer of the cell membrane. To exert their effects, they bind to cell-surface receptors, which then trigger the production of intracellular second messengers (like cAMP) to relay the signal.
Q58. Which of the following statement(s) is/are correct regarding phylum mollusca?
Correct Answer: (d)
All statements correctly describe Phylum Mollusca.
All statements correctly describe Phylum Mollusca.
Q59. Which of following ratio is correct for the production of one molecule of glucose through 6 rounds of Calvin cycle?
| CO2 | ATP | NADPH2 | |
|---|---|---|---|
| (a) | 1 | 2 | 2 |
| (b) | 6 | 18 | 12 |
| (c) | 6 | 12 | 18 |
| (d) | 5 | 6 | 9 |
Correct Answer: (b)
To produce 1 molecule of glucose (6C), 6 cycles are needed, consuming $6 imes 3 = 18$ ATP and $6 imes 2 = 12$ NADPH.
To produce 1 molecule of glucose (6C), 6 cycles are needed, consuming $6 imes 3 = 18$ ATP and $6 imes 2 = 12$ NADPH.
Q60. 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.
Q61. The elimination of a predator by humans
Correct Answer: (c)
If a prey population was regulated by predation, removing the predator can allow prey numbers to increase, though other limiting factors may eventually constrain growth.
If a prey population was regulated by predation, removing the predator can allow prey numbers to increase, though other limiting factors may eventually constrain growth.
Q62. A specimen of fungus is brought by a student for identification. Upon close examination, he discovered that its hyphae are completely septate and it has gills on the underside of the pileus. To which fungal group does it most likely belong?
Correct Answer: (a)
Gills under a pileus (cap) and septate hyphae are characteristic of mushrooms, which belong to Basidiomycetes.
Gills under a pileus (cap) and septate hyphae are characteristic of mushrooms, which belong to Basidiomycetes.
Q63. Microtubules, motor proteins, and actin filaments are all part of the
Correct Answer: (c)
These are all fundamental components of the eukaryotic cytoskeleton.
These are all fundamental components of the eukaryotic cytoskeleton.
Q64. 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.
Q65. Bone develops by one of two mechanisms depending on the underlying scaffold. Which pairing correctly describes these mechanisms?
Correct Answer: (d)
Intramembranous ossification occurs when bone forms directly from embryonic mesenchymal tissue (common in flat bones of the skull). Endochondral ossification occurs when bone replaces a pre-existing hyaline cartilage model (common in long bones like the femur).
Intramembranous ossification occurs when bone forms directly from embryonic mesenchymal tissue (common in flat bones of the skull). Endochondral ossification occurs when bone replaces a pre-existing hyaline cartilage model (common in long bones like the femur).
Q66. 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.
Q67. A virus can be considered a living organism because it
Correct Answer: (c)
Reproduction is a defining characteristic of life, and although viruses require a host to do so, they exhibit genetic replication.
Reproduction is a defining characteristic of life, and although viruses require a host to do so, they exhibit genetic replication.
Q68. In eukaryotes, binding of RNA polymerase to a promoter requires the action of
Correct Answer: (b)
Eukaryotic RNA polymerase cannot bind to a promoter on its own. It requires a set of general (or basal) transcription factors to assemble a transcription initiation complex at the TATA box, which then recruits and positions the polymerase correctly.
Eukaryotic RNA polymerase cannot bind to a promoter on its own. It requires a set of general (or basal) transcription factors to assemble a transcription initiation complex at the TATA box, which then recruits and positions the polymerase correctly.
Q69. The statement, 'Tiger is the apex of food chain', indicates
Correct Answer: (d)
As a top carnivore, the tiger sits at the highest trophic level and its existence depends on the health and abundance of the lower trophic levels (herbivores and producers).
As a top carnivore, the tiger sits at the highest trophic level and its existence depends on the health and abundance of the lower trophic levels (herbivores and producers).
Q70. During the formation of leaves and elongation of stem, some cells ‘left behind’ from the shoot apical meristem, constitute the
Correct Answer: (b)
Axillary buds are formed from cells that were part of the shoot apical meristem but were left behind during primary growth of the stem.
Axillary buds are formed from cells that were part of the shoot apical meristem but were left behind during primary growth of the stem.
Q71. Conversion of phosphenol pyruvic acid to pyruvic acid and ADP to ATP are examples of:
Correct Answer: (d)
When a phosphate group is transferred directly from a substrate (PEP) to ADP to form ATP, it is called substrate-level phosphorylation.
When a phosphate group is transferred directly from a substrate (PEP) to ADP to form ATP, it is called substrate-level phosphorylation.
Q72. Which of the following does not affect the phenotype of an organism?
Correct Answer: (d)
(D) The phenotype of an organism is determined by the genes that are expressed, the level of expression of those genes, and the timing of the expression of those genes. The phenotype is not affected by the number of genes in the organism, so choice (D) is the only choice that does not affect the phenotype of an organism.
(D) The phenotype of an organism is determined by the genes that are expressed, the level of expression of those genes, and the timing of the expression of those genes. The phenotype is not affected by the number of genes in the organism, so choice (D) is the only choice that does not affect the phenotype of an organism.
Q73. Match the cells with their ploidy:
| Column-I | Column-II |
|---|---|
| A. Spermatogonia | (I) Haploid (n) |
| B. Primary spermatocytes | (II) Diploid (2n) |
| C. Spermatids | (III) Diploid (2n) |
Correct Answer: (a)
Note: Both Spermatogonia and Primary Spermatocytes are 2n; Spermatids are n.
Note: Both Spermatogonia and Primary Spermatocytes are 2n; Spermatids are n.
Q74. A scientist at a pharmaceutical company wants to create bacteria that will synthesize human growth hormone.
The scientist has successfully created a plasmid that contains the DNA code for the human growth hormone gene. Which technique should the scientist use to insert the plasmid into a cell?
The scientist has successfully created a plasmid that contains the DNA code for the human growth hormone gene. Which technique should the scientist use to insert the plasmid into a cell?
Correct Answer: (a)
Bacterial transformation inserts naked, foreign DNA into a bacterial cell. CRISPR-Cas9 is used for gene editing, not inserting plasmids into a cell, so choice (B) is incorrect. Choice (C) is incorrect because gel electrophoresis is used to separate DNA fragments according to their size. Polymerase chain reaction is used to amplify the number of copies of a specific DNA sequence, so choice (D) is incorrect.
Bacterial transformation inserts naked, foreign DNA into a bacterial cell. CRISPR-Cas9 is used for gene editing, not inserting plasmids into a cell, so choice (B) is incorrect. Choice (C) is incorrect because gel electrophoresis is used to separate DNA fragments according to their size. Polymerase chain reaction is used to amplify the number of copies of a specific DNA sequence, so choice (D) is incorrect.
Q75. Marine mammals are able to hold their breath for extended periods underwater because
Correct Answer: (c)
Marine mammals have high concentrations of myoglobin in their muscles. Myoglobin has a higher affinity for oxygen than hemoglobin, acting as an internal oxygen 'scuba tank' that allows them to remain submerged for long durations without access to atmospheric air.
Marine mammals have high concentrations of myoglobin in their muscles. Myoglobin has a higher affinity for oxygen than hemoglobin, acting as an internal oxygen 'scuba tank' that allows them to remain submerged for long durations without access to atmospheric air.
Q76. Match the embryonic layers with their types:
| Column-I (Layer) | Column-II (Type) |
|---|---|
| A. Ectoderm | (I) Inner layer |
| B. Mesoderm | (II) Outer layer |
| C. Endoderm | (III) Middle layer |
Correct Answer: (a)
Standard embryonic layer positions.
Standard embryonic layer positions.
Q77. Swimming underwater using forelimbs for propulsion is similar to flying through the air because
Correct Answer: (d)
Both swimming and flying rely on generating lift by pushing against a fluid medium (water or air). By exerting a downward force on the surrounding fluid, an upward lift force is generated, allowing the organism to remain suspended or move efficiently against gravity.
Both swimming and flying rely on generating lift by pushing against a fluid medium (water or air). By exerting a downward force on the surrounding fluid, an upward lift force is generated, allowing the organism to remain suspended or move efficiently against gravity.
Q78. Deficiency symptoms of nitrogen and potassium are visible first in
Correct Answer: (a)
Elements like N and K are highly mobile and are translocated from old (senescent) leaves to young leaves during deficiency.
Elements like N and K are highly mobile and are translocated from old (senescent) leaves to young leaves during deficiency.
Q79. A bivalent meiosis-I consists of
Correct Answer: (c)
A bivalent is a pair of homologous chromosomes. Since each chromosome has 2 chromatids and 1 centromere, the pair has 4 chromatids and 2 centromeres.
A bivalent is a pair of homologous chromosomes. Since each chromosome has 2 chromatids and 1 centromere, the pair has 4 chromatids and 2 centromeres.
Q80. An egg cell has 5 pico gram of DNA in its nucleus. How much amount of DNA will be at the end of G2-phase of mitosis?
Correct Answer: (d)
An egg cell is haploid (n), so DNA = 5pg (1C). A somatic cell in G1 (2n) has 10pg (2C). After S-phase ($), DNA content doubles to 20pg (4C).
An egg cell is haploid (n), so DNA = 5pg (1C). A somatic cell in G1 (2n) has 10pg (2C). After S-phase ($), DNA content doubles to 20pg (4C).
Q81. Relative fitness
Correct Answer: (a)
Fitness is a measure of the relative ability of an individual (or a specific phenotype/genotype) to survive and reproduce in a given environment. It is usually expressed relative to the most successful phenotype in the population, which is assigned a fitness value of 1.0.
Fitness is a measure of the relative ability of an individual (or a specific phenotype/genotype) to survive and reproduce in a given environment. It is usually expressed relative to the most successful phenotype in the population, which is assigned a fitness value of 1.0.
Q82. 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.
Q83. Match the following biofertilisers with their type:
| Column I | Column II |
|---|---|
| A. Rhizobium | (I) Symbiotic fungus |
| B. Azospirillum | (II) Symbiotic bacterium |
| C. Glomus | (III) Free-living bacterium |
Correct Answer: (a)
Biofertiliser categories from section 8.6.
Biofertiliser categories from section 8.6.
Q84. A researcher isolated the plasma membrane from animal cells and them put them in a solution that contains chemical to stabilised the membranes. When he added a small amount of salt solution in it, he discovered that although the membranes seemed intact, the amount of protein in the stabilising solution had increased. These new proteins in the stabilising solution were probably
Correct Answer: (d)
Peripheral proteins are loosely bound to the membrane surface by ionic interactions; adding a salt solution disrupts these charges, releasing the proteins without destroying the lipid bilayer.
Peripheral proteins are loosely bound to the membrane surface by ionic interactions; adding a salt solution disrupts these charges, releasing the proteins without destroying the lipid bilayer.
Q85. Maximal size in terms of wall thickening and protoplasmic modification are achieved by:
Correct Answer: (b)
Cells in the zone of maturation undergo their final structural changes, reaching their maximum size and complexity to perform specific functions.
Cells in the zone of maturation undergo their final structural changes, reaching their maximum size and complexity to perform specific functions.
Q86. The ‘Earth Summit’ held at Rio de Janeiro in 1992 resulted into
Correct Answer: (c)
The Earth Summit (1992) called upon all nations to take appropriate measures for conservation of biodiversity and sustainable utilization of its benefits (Convention on Biological Diversity).
The Earth Summit (1992) called upon all nations to take appropriate measures for conservation of biodiversity and sustainable utilization of its benefits (Convention on Biological Diversity).
Q87. How many ATP will be produced during the production of 1 molecule of acetyl CoA from 1 molecule of pyruvic acid?
Correct Answer: (a)
The oxidative decarboxylation of 1 pyruvate yields 1 NADH. In the ETS, 1 NADH produces 3 ATP molecules.
The oxidative decarboxylation of 1 pyruvate yields 1 NADH. In the ETS, 1 NADH produces 3 ATP molecules.
Q88. Which of the following statements is incorrect ?
Correct Answer: (d)
Porifera includes fresh water species like Spongilla.
Porifera includes fresh water species like Spongilla.
Q89. The one-horned rhinoceros is specific to which of the following sanctuary ?
Correct Answer: (c)
Kaziranga National Park in Assam is famous for its successful protection of the Indian one-horned rhinoceros.
Kaziranga National Park in Assam is famous for its successful protection of the Indian one-horned rhinoceros.
Q90. Assertion: Roquefort cheese is ripened by fungi.
Reason: The fungi give the cheese a particular flavour.
Correct Answer: (a)
The purpose of ripening with fungi is to develop the specific flavour.
The purpose of ripening with fungi is to develop the specific flavour.