Q1. The given figure shows the diagrammatic representation of a section of chloroplast. Few parts are marked as A, B, C, D & E.
Combination of which parts is responsible for trapping the light energy & also the synthesis of ATP and NADPH?
Combination of which parts is responsible for trapping the light energy & also the synthesis of ATP and NADPH?
Correct Answer: (a)
A (Grana), B (Stroma lamellae), and C (Thylakoids) represent the membrane system of the chloroplast where the light reaction occurs.
A (Grana), B (Stroma lamellae), and C (Thylakoids) represent the membrane system of the chloroplast where the light reaction occurs.
Q2. A patient with a pituitary tumor exhibits fatigue, weight loss, and low blood sugar. This is probably due to a lack of production of
Correct Answer: (b)
Adrenocorticotropic hormone (ACTH) triggers the adrenal cortex to release glucocorticoids like cortisol. Cortisol is vital for maintaining blood glucose levels. A lack of ACTH leads to low cortisol, causing hypoglycemia (low blood sugar) and chronic fatigue.
Adrenocorticotropic hormone (ACTH) triggers the adrenal cortex to release glucocorticoids like cortisol. Cortisol is vital for maintaining blood glucose levels. A lack of ACTH leads to low cortisol, causing hypoglycemia (low blood sugar) and chronic fatigue.
Q3. Consider the following statements:
(1) Ammonium ions toxic to plants and hence cannot accumulate in them.
(2) In soyabean, the fixed nitrogen is exported as ureides.
(1) Ammonium ions toxic to plants and hence cannot accumulate in them.
(2) In soyabean, the fixed nitrogen is exported as ureides.
Correct Answer: (c)
NH4+ is indeed toxic at high levels and is quickly assimilated. Ureides are the primary transport form of fixed nitrogen in certain legumes like soybean.
NH4+ is indeed toxic at high levels and is quickly assimilated. Ureides are the primary transport form of fixed nitrogen in certain legumes like soybean.
Q4. Consider the following statements:
(1) In Blackman’s Law, if a chemical process is affected by more than one factor, then its rate will be determined by the factor which is nearest to its minimal value.
(2) For photosynthesis, light is rarely a limiting factor in nature.
(1) In Blackman’s Law, if a chemical process is affected by more than one factor, then its rate will be determined by the factor which is nearest to its minimal value.
(2) For photosynthesis, light is rarely a limiting factor in nature.
Correct Answer: (c)
Both statements are correct. Blackman's law relies on the limiting (minimal) factor, and CO2 is usually the limiting factor in nature, not light.
Both statements are correct. Blackman's law relies on the limiting (minimal) factor, and CO2 is usually the limiting factor in nature, not light.
Q5. Match the following column I with column II.
| Column - I | Column - II |
|---|---|
| A. Carcinogen | I. Tumour |
| B. Anaphase-I | II. Disjunction |
| C. Mitosis | III. Synapse |
| D. Zygotene | IV. Plectonemic coiling |
Correct Answer: (a)
Carcinogens cause tumors. Anaphase I involves disjunction. Mitosis shows plectonemic coiling (tight twisting). Zygotene involves synapse.
Carcinogens cause tumors. Anaphase I involves disjunction. Mitosis shows plectonemic coiling (tight twisting). Zygotene involves synapse.
Q6. 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.
Q7. Which of the following organisms are known as chief producers in the oceans?
Correct Answer: (b)
Diatoms are the chief producers in the ocean due to their high photosynthetic activity.
Diatoms are the chief producers in the ocean due to their high photosynthetic activity.
Q8. Which of the following elements are constituents of protein?
Correct Answer: (a)
Nitrogen is a major constituent of amino acids and proteins. Phosphorus is present in certain proteins (phosphoproteins), though sulfur is also a common structural constituent of proteins (cysteine/methionine).
Nitrogen is a major constituent of amino acids and proteins. Phosphorus is present in certain proteins (phosphoproteins), though sulfur is also a common structural constituent of proteins (cysteine/methionine).
Q9. In a chloroplast, the highest number of protons are found in
Correct Answer: (b)
During light reactions, protons are actively pumped into the thylakoid lumen, making it the area with the highest concentration and lowest pH.
During light reactions, protons are actively pumped into the thylakoid lumen, making it the area with the highest concentration and lowest pH.
Q10. Choose the correct option for the following events of meiosis in correct sequence
(A) Crossing over
(B) Synapsis
(C) Terminalisation of chiasmata
(D) Disappearance of nucleolus
(A) Crossing over
(B) Synapsis
(C) Terminalisation of chiasmata
(D) Disappearance of nucleolus
Correct Answer: (a)
The correct sequence within Prophase I is pairing (Synapsis) → exchange (Crossing over) → end-movement (Terminalization) → nuclear breakdown (Diakinesis/Disappearance).
The correct sequence within Prophase I is pairing (Synapsis) → exchange (Crossing over) → end-movement (Terminalization) → nuclear breakdown (Diakinesis/Disappearance).
Q11. The essential distinction between long-day and short-day plants is that flowering in long day plants is promoted only when the day length exceeds a certain duration, called the ______________.
Correct Answer: (a)
Critical day length is the specific threshold of light/dark duration that determines the flowering response in photoperiodic plants.
Critical day length is the specific threshold of light/dark duration that determines the flowering response in photoperiodic plants.
Q12. Match the follicle stage with features:
| Column-I | Column-II |
|---|---|
| A. Primary follicle | (I) Antrum present |
| B. Secondary follicle | (II) Primary oocyte + granulosa |
| C. Tertiary follicle | (III) More granulosa + new theca |
Correct Answer: (a)
Follicular development descriptions in Section 2.3.
Follicular development descriptions in Section 2.3.
Q13. 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.
Q14. Which of the following statement(s) is correct for both blue-green algae and bacteria ?
Correct Answer: (c)
Both blue-green algae (Cyanobacteria) and bacteria are prokaryotes, meaning they lack a true nucleus and membrane-bound organelles.
Both blue-green algae (Cyanobacteria) and bacteria are prokaryotes, meaning they lack a true nucleus and membrane-bound organelles.
Q15. You take X-rays of two individuals. Ray has been a weight lifter and body builder for 30 years; Ben has led a mostly sedentary life. What differences would you expect in their X-rays?
Correct Answer: (c)
Bone is a dynamic tissue that responds to mechanical stress. Weight-bearing and resistance exercises stimulate osteoblast activity, leading to increased bone density and thickness. Therefore, Ray’s bones would appear thicker and stronger due to long-term physical stress, unlike Ben’s sedentary lifestyle.
Bone is a dynamic tissue that responds to mechanical stress. Weight-bearing and resistance exercises stimulate osteoblast activity, leading to increased bone density and thickness. Therefore, Ray’s bones would appear thicker and stronger due to long-term physical stress, unlike Ben’s sedentary lifestyle.
Q16. Japanese four o’clocks that are red and tall are crossed to white short ones, producing an F1 that is pink and tall. If these genes assort independently, and the F1 is self-crossed, what would you predict for the ratio of F2 phenotypes?
Correct Answer: (d)
Color shows incomplete dominance (1 red : 2 pink : 1 white) and height shows simple dominance (3 tall : 1 short). Using the product rule: (1 Red + 2 Pink + 1 White) × (3 Tall + 1 Short) = 3 Red Tall, 1 Red Short, 6 Pink Tall, 2 Pink Short, 3 White Tall, 1 White Short. Sorting these leads to choice D.
Color shows incomplete dominance (1 red : 2 pink : 1 white) and height shows simple dominance (3 tall : 1 short). Using the product rule: (1 Red + 2 Pink + 1 White) × (3 Tall + 1 Short) = 3 Red Tall, 1 Red Short, 6 Pink Tall, 2 Pink Short, 3 White Tall, 1 White Short. Sorting these leads to choice D.
Q17. Plastids storing fat are called
Correct Answer: (a)
Elaioplasts store oils and fats; aleuroplasts store proteins, and amyloplasts store carbohydrates.
Elaioplasts store oils and fats; aleuroplasts store proteins, and amyloplasts store carbohydrates.
Q18. Suppose that your research mentor has decided to do a project on the filtering capabilities of Malpighian tubules. Which of the following creatures will you be spending your summer studying?
Correct Answer: (a)
Malpighian tubules are the primary excretory structures found in insects. Ants, being insects, possess Malpighian tubules, whereas birds, mammals, and earthworms use different excretory mechanisms.
Malpighian tubules are the primary excretory structures found in insects. Ants, being insects, possess Malpighian tubules, whereas birds, mammals, and earthworms use different excretory mechanisms.
Q19. 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).
Q20. 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.
Q21. Which pairing of structure and function is incorrect?
Correct Answer: (d)
All listed pairings are correct: erythrocytes transport oxygen via hemoglobin, platelets function in clotting, and plasma transports wastes, nutrients, and hormones. Therefore, none of the pairings is incorrect.
All listed pairings are correct: erythrocytes transport oxygen via hemoglobin, platelets function in clotting, and plasma transports wastes, nutrients, and hormones. Therefore, none of the pairings is incorrect.
Q22. 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.
Q23. The following is a list of the components of a chemical synapse. A mutation in the structure of which of these would affect only the reception of the message, not its release or the response?
Correct Answer: (a)
Membrane proteins on the postsynaptic cell include the receptors that specifically bind to neurotransmitters. A mutation here would specifically block the 'reception' of the chemical signal from the cleft, without preventing the presynaptic cell from releasing the neurotransmitter or the postsynaptic cell from responding internally once a signal is received.
Membrane proteins on the postsynaptic cell include the receptors that specifically bind to neurotransmitters. A mutation here would specifically block the 'reception' of the chemical signal from the cleft, without preventing the presynaptic cell from releasing the neurotransmitter or the postsynaptic cell from responding internally once a signal is received.
Q24. Which one of the following is well developed and present in hydrophytes?
Correct Answer: (a)
Aerenchyma provides buoyancy and allows internal air circulation, which is vital for aquatic plants (hydrophytes).
Aerenchyma provides buoyancy and allows internal air circulation, which is vital for aquatic plants (hydrophytes).
Q25. Which of the following statements is accurate for DNA replication in your cells, but not PCR?
Correct Answer: (c)
During in vivo DNA replication, the lagging strand is synthesized in short Okazaki fragments that must be joined together by the enzyme DNA ligase. In PCR, the target sequence is typically small enough that the polymerase completes the entire strand from a single primer, and there are no fragments to seal. Furthermore, PCR uses heat-stable Taq polymerase, whereas human cells use heat-labile polymerases.
During in vivo DNA replication, the lagging strand is synthesized in short Okazaki fragments that must be joined together by the enzyme DNA ligase. In PCR, the target sequence is typically small enough that the polymerase completes the entire strand from a single primer, and there are no fragments to seal. Furthermore, PCR uses heat-stable Taq polymerase, whereas human cells use heat-labile polymerases.
Q26. Consider the following statements:
(1) Cytochrome c is a large protein complex attached to the inner surface of the outer membrane
(2) It transfers electrons between complex III and IV.
(1) Cytochrome c is a large protein complex attached to the inner surface of the outer membrane
(2) It transfers electrons between complex III and IV.
Correct Answer: (b)
Statement (1) is wrong: Cytochrome c is a small protein attached to the outer surface of the inner membrane. Statement (2) correctly identifies its shuttle role.
Statement (1) is wrong: Cytochrome c is a small protein attached to the outer surface of the inner membrane. Statement (2) correctly identifies its shuttle role.
Q27. In comparing T-cell receptors and immunoglobulins
Correct Answer: (c)
T-cell receptors and immunoglobulins share similar structural features and both achieve diversity through gene rearrangement processes such as V(D)J recombination.
T-cell receptors and immunoglobulins share similar structural features and both achieve diversity through gene rearrangement processes such as V(D)J recombination.
Q28. 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.
Q29. Assertion: The morula continues to divide and transforms into a blastocyst.
Reason: The blastocyst moves further into the uterus.
Correct Answer: (a)
Both are sequential events in embryonic passage.
Both are sequential events in embryonic passage.
Q30. The human TAS2R38 gene encodes a cell membrane protein that influences the ability to taste bitter compounds. Individuals who possess at least one TAS2R38 allele have the “taster” phenotype and can taste certain types of bitter compounds. It is estimated that about 70% of humans have the taster phenotype. Which of the following best explains the frequency of the taster phenotype?
Correct Answer: (a)
(A) Since TAS2R38 is present in the population at a higher frequency, it probably did provide a survival advantage and allowed individuals who possessed this allele to survive and reproduce at a greater rate than individuals who did not possess the allele. Choice (B) is incorrect because TAS2R38 does not cause individuals to consume bitter foods; it just influences their ability to taste bitter foods. Bitter foods do not necessarily have a higher nutrient content than other foods, so choice (C) is incorrect. There is no evidence in the question that the TAS2R38 allele lowered an individual’s chance of survival, so choice (D) is incorrect.
(A) Since TAS2R38 is present in the population at a higher frequency, it probably did provide a survival advantage and allowed individuals who possessed this allele to survive and reproduce at a greater rate than individuals who did not possess the allele. Choice (B) is incorrect because TAS2R38 does not cause individuals to consume bitter foods; it just influences their ability to taste bitter foods. Bitter foods do not necessarily have a higher nutrient content than other foods, so choice (C) is incorrect. There is no evidence in the question that the TAS2R38 allele lowered an individual’s chance of survival, so choice (D) is incorrect.
Q31. ATP synthesis, in photosynthesis involves the:
Correct Answer: (a)
ATP synthesis is coupled to the establishment of a proton (H+) gradient across the thylakoid membrane.
ATP synthesis is coupled to the establishment of a proton (H+) gradient across the thylakoid membrane.
Q32. Which of the following best supports the existence of a common ancestor of all three domains of living organisms (Archaea, Bacteria, and Eukarya)?
Correct Answer: (a)
(A) All living organisms perform glycolysis in their cytoplasm, indicating a common ancestor of all life-forms. Choice (B) is incorrect because not all living organisms have membrane-bound organelles. Prokaryotes do not have linear chromosomes or introns, so choices (C) and (D) are incorrect.
(A) All living organisms perform glycolysis in their cytoplasm, indicating a common ancestor of all life-forms. Choice (B) is incorrect because not all living organisms have membrane-bound organelles. Prokaryotes do not have linear chromosomes or introns, so choices (C) and (D) are incorrect.
Q33. A functional reflex requires
Correct Answer: (a)
The simplest functional reflex arc is monosynaptic, consisting of a single sensory neuron and a single motor neuron. While many reflexes involve interneurons, the brain (cerebral cortex or thalamus) is not required for the basic execution of a reflex action.
The simplest functional reflex arc is monosynaptic, consisting of a single sensory neuron and a single motor neuron. While many reflexes involve interneurons, the brain (cerebral cortex or thalamus) is not required for the basic execution of a reflex action.
Q34. Match the Following for Chromosomes:
| Cell | Chromosome Count |
|---|---|
| A. Human Somatic Cell | (I) 46 |
| B. Human Gamete | (II) 23 |
Correct Answer: (a)
Diploids have 46; haploids have 23.
Diploids have 46; haploids have 23.
Q35. The enzyme ________ unwinds the double helix of DNA, and the enzyme ________ relieves the supercoiling created by this unwinding.
Correct Answer: (a)
(A) Helicase unwinds the double helix of DNA, and topoisomerase relieves the stress caused by the supercoiling (created when helicase unwinds the DNA). Choices (B) and (C) are incorrect because DNA polymerase adds new nucleotides; it does not have an unwinding function. Choices (C) and (D) are incorrect because ligase joins together short segments of DNA created on the lagging strand by discontinuous replication.
(A) Helicase unwinds the double helix of DNA, and topoisomerase relieves the stress caused by the supercoiling (created when helicase unwinds the DNA). Choices (B) and (C) are incorrect because DNA polymerase adds new nucleotides; it does not have an unwinding function. Choices (C) and (D) are incorrect because ligase joins together short segments of DNA created on the lagging strand by discontinuous replication.
Q36. The given characters are seen in which of the following group?
(i) Unicellular, colonial, filamentous, marine or terrestrial forms.
(ii) The colonies are surrounded by a gelatinous sheath.
(iii) Some can fix atmospheric nitrogen in specialized cells called heterocysts.
(iv) They often form blooms in polluted water bodies.
(i) Unicellular, colonial, filamentous, marine or terrestrial forms.
(ii) The colonies are surrounded by a gelatinous sheath.
(iii) Some can fix atmospheric nitrogen in specialized cells called heterocysts.
(iv) They often form blooms in polluted water bodies.
Correct Answer: (b)
These features (gelatinous sheath, heterocysts for N2 fixation, and water blooms) are characteristic of Cyanobacteria (blue-green algae) like Nostoc and Anabaena.
These features (gelatinous sheath, heterocysts for N2 fixation, and water blooms) are characteristic of Cyanobacteria (blue-green algae) like Nostoc and Anabaena.
Q37. Which of the following is correct regarding 'Microbes in Household Products'?
(i) Lactobacillus is used for making curd.
(ii) Baker's yeast is used for bread making.
(iii) Toddy is a fermented sap from palms.
(i) Lactobacillus is used for making curd.
(ii) Baker's yeast is used for bread making.
(iii) Toddy is a fermented sap from palms.
Correct Answer: (a)
All three points are correctly described in section 8.1.
All three points are correctly described in section 8.1.
Q38. Tigers (Panthera tigris) and leopards (Panthera pardus) can mate and produce a zygote. The zygote will undergo cell division a few times, but this fails to result in the production of a viable embryo. This is an example of which type of reproductive isolation?
Correct Answer: (b)
(B) Since tigers and leopards can mate and produce a zygote, but doing so does not result in any viable offspring, this is an example of reduced hybrid viability. It is not gametic isolation since their gametes can form a zygote, so choice (A) is incorrect. There is no infertile adult hybrid in this scenario, so choice (C) is also incorrect. No viable hybrid was produced, so there is no chance of hybrid breakdown, making choice (D) incorrect.
(B) Since tigers and leopards can mate and produce a zygote, but doing so does not result in any viable offspring, this is an example of reduced hybrid viability. It is not gametic isolation since their gametes can form a zygote, so choice (A) is incorrect. There is no infertile adult hybrid in this scenario, so choice (C) is also incorrect. No viable hybrid was produced, so there is no chance of hybrid breakdown, making choice (D) incorrect.
Q39. 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.
Q40. When a plant is not reproducing, most of the cytokinins are produced in its
Correct Answer: (c)
The roots are the major site of cytokinin synthesis, from where they are transported upward to the rest of the plant.
The roots are the major site of cytokinin synthesis, from where they are transported upward to the rest of the plant.
Q41. In relation to Gross primary productivity and Net primary productivity of an ecosystem, which one of the following statements is correct? [NEET 2020, S]
Correct Answer: (a)
Since = GPP - Respiration$, and respiration is a loss, GPP is always greater than NPP.
Since = GPP - Respiration$, and respiration is a loss, GPP is always greater than NPP.
Q42. Endosperm is produced by the union of
Correct Answer: (a)
Angiosperms undergo double fertilization. While one sperm fuses with the egg, the second sperm fuses with the two polar nuclei in the central cell of the embryo sac to form the triploid endosperm.
Angiosperms undergo double fertilization. While one sperm fuses with the egg, the second sperm fuses with the two polar nuclei in the central cell of the embryo sac to form the triploid endosperm.
Q43. Specialised epidermal cells surrounding the guard cells are called
Correct Answer: (b)
Subsidiary cells are modified epidermal cells that assist in the opening and closing of the stomatal pore.
Subsidiary cells are modified epidermal cells that assist in the opening and closing of the stomatal pore.
Q44. In Iowa, a company called Team Corn works to ensure that fields of seed corn outcross... you would like to develop genetically engineered corn plants that
Correct Answer: (b)
By introducing S genes (self-incompatibility genes), the plants would naturally reject their own pollen, making the manual removal of staminate flowers (detasseling) unnecessary and making the process more efficient.
By introducing S genes (self-incompatibility genes), the plants would naturally reject their own pollen, making the manual removal of staminate flowers (detasseling) unnecessary and making the process more efficient.
Q45. Which of the following describes a population that could experience exponential growth?
Correct Answer: (c)
(C) Populations grow exponentially if there are no limiting resources or predators that would limit the population size. If the population was limited by density-dependent factors, its growth would be logistic, so choice (A) is incorrect. Choice (B) is incorrect because if the size of the population was beyond the carrying capacity of the environment, its numbers would be declining, not growing. K-selected populations experience logistic growth, so choice (D) is also incorrect.
(C) Populations grow exponentially if there are no limiting resources or predators that would limit the population size. If the population was limited by density-dependent factors, its growth would be logistic, so choice (A) is incorrect. Choice (B) is incorrect because if the size of the population was beyond the carrying capacity of the environment, its numbers would be declining, not growing. K-selected populations experience logistic growth, so choice (D) is also incorrect.
Q46. Refer to the figure, which shows a pedigree of an inherited trait.
If individual III-3 has a child with a woman who does not have the allele, what is the most likely probability that their child will have the trait?
If individual III-3 has a child with a woman who does not have the allele, what is the most likely probability that their child will have the trait?
Correct Answer: (a)
(A) This is most likely a mitochondrial trait (see the explanation for Question 9). Individual III-3 does not have the trait, so he does not carry the allele. Even if he did carry the allele, males cannot pass on mitochondrial traits to the next generation. If he has a child with a woman who also does not have the allele, there is 0% chance of them having a child with the trait since neither of them have the allele for the trait. Choices (B), (C), and (D) are incorrect because if the woman does not have the allele for a mitochondrial trait, her offspring will not have the allele for the trait.
(A) This is most likely a mitochondrial trait (see the explanation for Question 9). Individual III-3 does not have the trait, so he does not carry the allele. Even if he did carry the allele, males cannot pass on mitochondrial traits to the next generation. If he has a child with a woman who also does not have the allele, there is 0% chance of them having a child with the trait since neither of them have the allele for the trait. Choices (B), (C), and (D) are incorrect because if the woman does not have the allele for a mitochondrial trait, her offspring will not have the allele for the trait.
Q47. 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.
Q48. According to Robert May, the global species diversity is about
Correct Answer: (c)
Robert May provided a more conservative and scientifically based estimate of about 7 million species globally.
Robert May provided a more conservative and scientifically based estimate of about 7 million species globally.
Q49. Phenotypes like height in humans, which show a continuous distribution, are usually the result of
Correct Answer: (d)
Continuous variation, where traits do not fall into discrete categories but show a range (like a bell curve), is typically caused by polygenic inheritance. This occurs when multiple independent genes have an additive effect on a single phenotypic character.
Continuous variation, where traits do not fall into discrete categories but show a range (like a bell curve), is typically caused by polygenic inheritance. This occurs when multiple independent genes have an additive effect on a single phenotypic character.
Q50. You study a gene known to be important in the early stages of heart development. Loss of the gene is also suspected to play a role in triggering lung cancer. What kind of transgenic mouse would you use to study your gene in vivo?
Correct Answer: (c)
A traditional knockout mouse lacks the gene from conception, which would likely result in embryonic lethality if the gene is essential for heart development. A conditional knockout allows the researcher to delete the gene at a specific time or in a specific tissue (such as the lungs of an adult mouse), allowing the study of its role in cancer without stopping development.
A traditional knockout mouse lacks the gene from conception, which would likely result in embryonic lethality if the gene is essential for heart development. A conditional knockout allows the researcher to delete the gene at a specific time or in a specific tissue (such as the lungs of an adult mouse), allowing the study of its role in cancer without stopping development.
Q51. Protonema is
Correct Answer: (a)
Protonema is the first stage of the haploid gametophytic generation in mosses.
Protonema is the first stage of the haploid gametophytic generation in mosses.
Q52. G protein–coupled receptors are involved in the nervous system by
Correct Answer: (d)
G protein–coupled receptors (metabotropic receptors) act as postsynaptic receptors that, upon binding neurotransmitters, initiate complex intracellular signaling cascades. These are distinct from ionotropic receptors, which are ligand-gated ion channels themselves.
G protein–coupled receptors (metabotropic receptors) act as postsynaptic receptors that, upon binding neurotransmitters, initiate complex intracellular signaling cascades. These are distinct from ionotropic receptors, which are ligand-gated ion channels themselves.
Q53. Which one of the following have vessels as their characteristic feature?
Correct Answer: (a)
The presence of vessels in the xylem is a diagnostic and characteristic feature of the flowering plants (Angiosperms).
The presence of vessels in the xylem is a diagnostic and characteristic feature of the flowering plants (Angiosperms).
Q54. 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.
Q55. Which of the following phylum is being described by the given statements?
(i) They are bilaterally symmetrical, triploblastic, segmented and coelomate animals.
(ii) The body consists of head, thorax, abdomen and have jointed appendages.
(iii) Circulatory system is of open type.
(iv) Excretion takes place through malphigian tubules.
(i) They are bilaterally symmetrical, triploblastic, segmented and coelomate animals.
(ii) The body consists of head, thorax, abdomen and have jointed appendages.
(iii) Circulatory system is of open type.
(iv) Excretion takes place through malphigian tubules.
Correct Answer: (a)
Description matches Phylum Arthropoda.
Description matches Phylum Arthropoda.
Q56. Assertion: Secondary sewage treatment is also called biological treatment.
Reason: It involves the use of heterotrophic aerobic microbes to consume organic matter.
Correct Answer: (a)
The reliance on living organisms for treatment is why it is termed 'biological'.
The reliance on living organisms for treatment is why it is termed 'biological'.
Q57. N-15, also known as heavy nitrogen, is an isotope of nitrogen that is heavier than the isotope that is typically found in nature, N-14. Meselson and Stahl allowed parent DNA (containing N-15) to replicate in the presence of N-14.
After two rounds of DNA replication, which of the following results would support the statement “DNA replication is semiconservative”?
After two rounds of DNA replication, which of the following results would support the statement “DNA replication is semiconservative”?
Correct Answer: (d)
(D) As shown in Figure 15.3, after two rounds of DNA replication, 50% of the DNA molecules would contain strands with only N-14. The other 50% of the DNA molecules would contain one strand with only N-15 and one strand with only N-14. Choice (A) describes the result of only one round of DNA replication if DNA replication was semiconservative, not the result of two rounds of DNA replication. So (A) is incorrect. Conservative DNA replication would produce the result described in choice (B), so (B) is incorrect. Choice (C) is incorrect because it describes the result if no replication was happening at all.
(D) As shown in Figure 15.3, after two rounds of DNA replication, 50% of the DNA molecules would contain strands with only N-14. The other 50% of the DNA molecules would contain one strand with only N-15 and one strand with only N-14. Choice (A) describes the result of only one round of DNA replication if DNA replication was semiconservative, not the result of two rounds of DNA replication. So (A) is incorrect. Conservative DNA replication would produce the result described in choice (B), so (B) is incorrect. Choice (C) is incorrect because it describes the result if no replication was happening at all.
Q58. 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.
Q59. Which of the following statements with respect to gymnosperms and angiosperms is/are correct?
(i) The process of double fertilization is present in gymnosperms.
(ii) Angiosperms range in size from microscopic Wolffia to tall trees of Sequoia.
(iii) In gymnosperms, the seeds are not covered.
(iv) In gymnosperms, the male and female gametophytes have an independent free living existence.
(i) The process of double fertilization is present in gymnosperms.
(ii) Angiosperms range in size from microscopic Wolffia to tall trees of Sequoia.
(iii) In gymnosperms, the seeds are not covered.
(iv) In gymnosperms, the male and female gametophytes have an independent free living existence.
Correct Answer: (b)
Only statement (iii) is correct. Double fertilization is in angiosperms; Sequoia is a gymnosperm; and gymnosperm gametophytes are dependent on the sporophyte.
Only statement (iii) is correct. Double fertilization is in angiosperms; Sequoia is a gymnosperm; and gymnosperm gametophytes are dependent on the sporophyte.
Q60. In pea plants, smooth (S) pods are dominant and constricted (s) pods are recessive. Which of the following is a correct statement?
Correct Answer: (c)
A genotype of ss would result in the constricted pod phenotype. Choices (A) and (B) are incorrect because both SS and Ss would result in the smooth phenotype. Choice (D) is incorrect because smooth pods would not always have the SS genotype; smooth pods could also have the Ss genotype.
A genotype of ss would result in the constricted pod phenotype. Choices (A) and (B) are incorrect because both SS and Ss would result in the smooth phenotype. Choice (D) is incorrect because smooth pods would not always have the SS genotype; smooth pods could also have the Ss genotype.
Q61. Where does glycosylation of protein occur?
Correct Answer: (a)
Glycosylation (addition of sugars to proteins) begins in the RER and is completed in the Golgi apparatus.
Glycosylation (addition of sugars to proteins) begins in the RER and is completed in the Golgi apparatus.
Q62. A sterile stamen is known as
Correct Answer: (a)
A staminode is a stamen that is sterile and does not produce fertile pollen grains.
A staminode is a stamen that is sterile and does not produce fertile pollen grains.
Q63. Structures on invading cells recognized by the adaptive immune system are known as
Correct Answer: (a)
Antigens are specific molecular structures, often proteins or polysaccharides, present on pathogens that are recognized by B-cell receptors, T-cell receptors, and antibodies, thereby activating the adaptive immune response.
Antigens are specific molecular structures, often proteins or polysaccharides, present on pathogens that are recognized by B-cell receptors, T-cell receptors, and antibodies, thereby activating the adaptive immune response.
Q64. The arabinose operon is an inducible operon that codes for the genes required to digest the sugar arabinose. Arabinose functions as an inducer molecule for the operon. If arabinose is present in the bacteria’s environment, which of the following is most likely?
Correct Answer: (a)
(A) In inducible operons, the inducer molecule binds to the repressor protein, which prevents the repressor protein from binding to the operator, increasing the level of expression of the operon. Choice (B) is incorrect because inducible operons are usually catabolic, not anabolic, and would not be involved in the production of arabinose. The presence of the inducer does not affect the production of the repressor protein, so choice (C) is incorrect. The presence of the inducer results in decreased binding of the repressor protein to the operator, so choice (D) is also incorrect.
(A) In inducible operons, the inducer molecule binds to the repressor protein, which prevents the repressor protein from binding to the operator, increasing the level of expression of the operon. Choice (B) is incorrect because inducible operons are usually catabolic, not anabolic, and would not be involved in the production of arabinose. The presence of the inducer does not affect the production of the repressor protein, so choice (C) is incorrect. The presence of the inducer results in decreased binding of the repressor protein to the operator, so choice (D) is also incorrect.
Q65. 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 of the following tools would be most useful in amplifying the number of copies of the HTT gene so that more DNA would be available for analysis?
Which of the following tools would be most useful in amplifying the number of copies of the HTT gene so that more DNA would be available for analysis?
Correct Answer: (c)
Polymerase chain reaction makes multiple copies of a specific DNA sequence and would be the best choice for amplifying the number of copies of the HTT gene. Choice (A) is incorrect because CRISPR-Cas9 is used for gene editing. Gel electrophoresis separates fragments of DNA by size, so choice (B) is incorrect. Restriction enzymes cut DNA at specific sequences, so choice (D) is incorrect.
Polymerase chain reaction makes multiple copies of a specific DNA sequence and would be the best choice for amplifying the number of copies of the HTT gene. Choice (A) is incorrect because CRISPR-Cas9 is used for gene editing. Gel electrophoresis separates fragments of DNA by size, so choice (B) is incorrect. Restriction enzymes cut DNA at specific sequences, so choice (D) is incorrect.
Q66. 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.
Q67. Which of the following is not the characteristic features of Fabaceae?
Correct Answer: (b)
Fabaceae flowers are zygomorphic, have vexillary aestivation, and are polypetalous (except the keel which is slightly fused).
Fabaceae flowers are zygomorphic, have vexillary aestivation, and are polypetalous (except the keel which is slightly fused).
Q68. A young dicot seedling (e.g. soyabean) is laid horizontally on a surface and is subjected to gravity stimulus. Which out of the following is the possible reason for its movement?
Correct Answer: (a)
Gravity causes auxin to accumulate on the lower side. In shoots, high auxin stimulates growth (bending up); in roots, high auxin inhibits growth (bending down).
Gravity causes auxin to accumulate on the lower side. In shoots, high auxin stimulates growth (bending up); in roots, high auxin inhibits growth (bending down).
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. Identify the basic amino acid from the following.
Correct Answer: (b)
Lysine, along with Arginine and Histidine, are classified as basic amino acids due to their side chain properties.
Lysine, along with Arginine and Histidine, are classified as basic amino acids due to their side chain properties.
Q71. Which of the following is a function of the kidneys?
Correct Answer: (d)
Kidneys perform multiple vital roles including removal of metabolic wastes, regulation of water balance, and maintenance of electrolyte (salt) levels. Therefore, all listed functions are correct.
Kidneys perform multiple vital roles including removal of metabolic wastes, regulation of water balance, and maintenance of electrolyte (salt) levels. Therefore, all listed functions are correct.
Q72. Regarding 'ICSI':
(i) Intra cytoplasmic sperm injection.
(ii) Sperm is injected into the fallopian tube.
(iii) Sperm is injected directly into the ovum.
(i) Intra cytoplasmic sperm injection.
(ii) Sperm is injected into the fallopian tube.
(iii) Sperm is injected directly into the ovum.
Correct Answer: (a)
Mechanism of ICSI (Page 48).
Mechanism of ICSI (Page 48).
Q73. 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.
Q74. 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.
Q75. Match the following for cheese specificity:
| Cheese | Feature/Microbe |
|---|---|
| A. Swiss Cheese | (I) Specific fungi |
| B. Roquefort Cheese | (II) Bacterium / large holes |
Correct Answer: (a)
Details from section 8.1.
Details from section 8.1.
Q76. Match the medical procedures with their use:
| Procedure | Primary Use |
|---|---|
| A. Amniocentesis | (I) Voluntary abortion |
| B. MTP | (II) Genetic disorder test |
| C. AI | (III) Low sperm count treatment |
Correct Answer: (a)
Clinical procedures described in the text.
Clinical procedures described in the text.
Q77. The area where wild populations, traditional life-styles and genetic resources are protected is called
Correct Answer: (c)
Biosphere reserves are multipurpose protected areas designed to preserve genetic diversity in representative ecosystems and support sustainable development.
Biosphere reserves are multipurpose protected areas designed to preserve genetic diversity in representative ecosystems and support sustainable development.
Q78. Which one of the following is not an abiotic component?
Correct Answer: (b)
Abiotic components are non-living parts like temperature, water, and soil. Decomposers are biotic (living) components.
Abiotic components are non-living parts like temperature, water, and soil. Decomposers are biotic (living) components.
Q79. Biological nitrogen fixation is the:
Correct Answer: (a)
Biological nitrogen fixation is specifically the reduction of atmospheric N2 to NH3 by certain living prokaryotes.
Biological nitrogen fixation is specifically the reduction of atmospheric N2 to NH3 by certain living prokaryotes.
Q80. Two main structural features of an ecosystem are
Correct Answer: (a)
Identification and enumeration of plant and animal species of an ecosystem gives its species composition. Vertical distribution of different species occupying different levels is called stratification.
Identification and enumeration of plant and animal species of an ecosystem gives its species composition. Vertical distribution of different species occupying different levels is called stratification.
Q81. If you wanted to design an artificial cell that could safely carry drugs inside the body, which molecule would you need to mimic to deter the immune system?
Correct Answer: (a)
MHC-I molecules signal to the immune system that a cell is ‘self.’ Mimicking MHC-I would help an artificial cell avoid immune detection and destruction, allowing safe drug delivery.
MHC-I molecules signal to the immune system that a cell is ‘self.’ Mimicking MHC-I would help an artificial cell avoid immune detection and destruction, allowing safe drug delivery.
Q82. 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.
Q83. Character displacement
Correct Answer: (a)
Character displacement occurs when two similar species compete for the same resources in the same area (sympatry). Natural selection favors individuals in each species that use different resources, leading to the divergence of physical traits, which can further enhance reproductive isolation.
Character displacement occurs when two similar species compete for the same resources in the same area (sympatry). Natural selection favors individuals in each species that use different resources, leading to the divergence of physical traits, which can further enhance reproductive isolation.
Q84. Which cells are analyzed in amniocentesis to detect genetic disorders?
Correct Answer: (b)
In amniocentesis some of the amniotic fluid... is taken to analyse the fetal cells.
In amniocentesis some of the amniotic fluid... is taken to analyse the fetal cells.
Q85. Which of the following is not currently a major cause of the global reduction in biodiversity?
Correct Answer: (b)
While global warming is a threat, the four 'major' causes (Evil Quartet) are habitat loss, overexploitation, alien species, and co-extinctions. Habitat destruction is the single most important cause.
While global warming is a threat, the four 'major' causes (Evil Quartet) are habitat loss, overexploitation, alien species, and co-extinctions. Habitat destruction is the single most important cause.
Q86. White matter is ________, and gray matter is ________.
Correct Answer: (d)
White matter consists primarily of myelinated axons, giving it a white appearance. Gray matter consists of neuronal cell bodies, dendrites, and unmyelinated axons. Both are essential components of the central nervous system (CNS).
White matter consists primarily of myelinated axons, giving it a white appearance. Gray matter consists of neuronal cell bodies, dendrites, and unmyelinated axons. Both are essential components of the central nervous system (CNS).
Q87. Crossing over takes place between which chromatids and in which stage of the cell cycle?
Correct Answer: (b)
Crossing over specifically occurs between the non-sister chromatids of a homologous pair during the Pachytene stage.
Crossing over specifically occurs between the non-sister chromatids of a homologous pair during the Pachytene stage.
Q88. Match the male accessory ducts:
| Column-I | Column-II |
|---|---|
| A. Rete testis | (I) Opens into epididymis |
| B. Vasa efferentia | (II) Network of tubules |
| C. Epididymis | (III) Leads to vas deferens |
Correct Answer: (a)
Ductal pathway sequence.
Ductal pathway sequence.
Q89. Read the statements given below with regard to the functions performed by Golgi apparatus ?
(i) Transport and chemically modify the materials contained within it.
(ii) Performs the function of packaging materials.
(iii) Important site of formation of glycoproteins and glycolipids.
(i) Transport and chemically modify the materials contained within it.
(ii) Performs the function of packaging materials.
(iii) Important site of formation of glycoproteins and glycolipids.
Correct Answer: (d)
The Golgi apparatus is involved in processing (modification), packaging, and glycosylation.
The Golgi apparatus is involved in processing (modification), packaging, and glycosylation.
Q90. Which one of the following is correct explanation for the given floral formula ?
% ⚥ K(5) C1+2+(2) A(9)+1 G1
% ⚥ K(5) C1+2+(2) A(9)+1 G1
Correct Answer: (c)
The formula represents Fabaceae: Zygomorphic (%), bisexual (♀), fused sepals (K(5)), papilionaceous petals (1+2+(2)), diadelphous stamens ((9)+1), and superior monocarpellary ovary (G1).
The formula represents Fabaceae: Zygomorphic (%), bisexual (♀), fused sepals (K(5)), papilionaceous petals (1+2+(2)), diadelphous stamens ((9)+1), and superior monocarpellary ovary (G1).