Q1. In which of the following gametophyte is not independent free living?
Correct Answer: (c)
In gymnosperms like Pinus, the gametophyte is highly reduced and remains dependent on the sporophyte.
In gymnosperms like Pinus, the gametophyte is highly reduced and remains dependent on the sporophyte.
Q2. Most carbon dioxide is
Correct Answer: (c)
Once carbon dioxide enters red blood cells, the enzyme carbonic anhydrase rapidly catalyzes the reaction between CO2 and H2O to form carbonic acid (H2CO3), which is a key intermediate step in the formation of bicarbonate for transport.
Once carbon dioxide enters red blood cells, the enzyme carbonic anhydrase rapidly catalyzes the reaction between CO2 and H2O to form carbonic acid (H2CO3), which is a key intermediate step in the formation of bicarbonate for transport.
Q3. The difference between the amphibian and mammal hearts is that
Correct Answer: (c)
Amphibians have a three-chambered heart with a single ventricle, but internal ridges and channels partially separate oxygenated and deoxygenated blood, reducing mixing. Mammals, in contrast, have complete separation with a four-chambered heart.
Amphibians have a three-chambered heart with a single ventricle, but internal ridges and channels partially separate oxygenated and deoxygenated blood, reducing mixing. Mammals, in contrast, have complete separation with a four-chambered heart.
Q4. The oxygenation activity of RuBisCo enzyme in photorespiration leads to the formation of
Correct Answer: (a)
RuBisCO oxygenase activity splits RuBP into 1 molecule of 3-PGA (3-C) and 1 molecule of 2-phosphoglycolate (2-C).
RuBisCO oxygenase activity splits RuBP into 1 molecule of 3-PGA (3-C) and 1 molecule of 2-phosphoglycolate (2-C).
Q5. When mutations that affected DNA replication were isolated, two kinds were found. One class put an immediate halt to replication, whereas the other put a much slower stop to the process. The first class affects functions at the replication fork. The second class affects functions necessary for
Correct Answer: (c)
Mutation in elongation factors (like polymerase) stops replication immediately because the fork can no longer move. Mutations in initiation factors allow the cell to finish the round of replication it has already started but prevent it from beginning any subsequent rounds, leading to a slower cessation of the process in a population.
Mutation in elongation factors (like polymerase) stops replication immediately because the fork can no longer move. Mutations in initiation factors allow the cell to finish the round of replication it has already started but prevent it from beginning any subsequent rounds, leading to a slower cessation of the process in a population.
Q6. In conifers fibres are likely to be absent in
Correct Answer: (b)
Gymnosperm (conifer) wood is typically 'non-porous' and 'soft' because it lacks vessels and xylem fibres.
Gymnosperm (conifer) wood is typically 'non-porous' and 'soft' because it lacks vessels and xylem fibres.
Q7. 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.
Q8. Km is related to
Correct Answer: (b)
Km is a measure of the affinity of an enzyme for its substrate, describing the stability and formation of the Enzyme-Substrate (ES) complex.
Km is a measure of the affinity of an enzyme for its substrate, describing the stability and formation of the Enzyme-Substrate (ES) complex.
Q9. Select the correct statement related to mitosis.
Correct Answer: (b)
In mitosis, DNA content doubles in S-phase and then is equally split into two daughter cells, maintaining the parent cell's DNA level (2C).
In mitosis, DNA content doubles in S-phase and then is equally split into two daughter cells, maintaining the parent cell's DNA level (2C).
Q10. 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.
Q11. If 20 J of energy is trapped at producer level, then how much energy will be available to peacock as food in the following chain?
plant → mice → snake → peacock
plant → mice → snake → peacock
Correct Answer: (a)
According to the 10% law: 20J (Producer) → 2J (Mice) → 0.2J (Snake) → 0.02J (Peacock).
According to the 10% law: 20J (Producer) → 2J (Mice) → 0.2J (Snake) → 0.02J (Peacock).
Q12. MMR stands for:
Correct Answer: (a)
The text lists maternal mortality rate (MMR) as a reason for population trends.
The text lists maternal mortality rate (MMR) as a reason for population trends.
Q13. Which type of plant tissue is being described by the given statements?
(i) It consists of long, narrow cells with thick and lignified cell walls having a few or numerous pits.
(ii) They are dead and without protoplasts.
(iii) On the basis of variation in form, structure, origin and development, it may be either fibres or sclereids.
(iv) It provides mechanical support to organs.
(i) It consists of long, narrow cells with thick and lignified cell walls having a few or numerous pits.
(ii) They are dead and without protoplasts.
(iii) On the basis of variation in form, structure, origin and development, it may be either fibres or sclereids.
(iv) It provides mechanical support to organs.
Correct Answer: (b)
Lignified walls, absence of protoplasm at maturity, and providing mechanical support are definitive traits of sclerenchyma.
Lignified walls, absence of protoplasm at maturity, and providing mechanical support are definitive traits of sclerenchyma.
Q14. Read the following statements and answer the question.
(i) Cambium is very active and produces a large number of xylary elements having vessels with wider cavities.
(ii) It is also called early wood.
(iii) It is lighter in colour and has lower density.
Which type of wood is described by the above statements?
(i) Cambium is very active and produces a large number of xylary elements having vessels with wider cavities.
(ii) It is also called early wood.
(iii) It is lighter in colour and has lower density.
Which type of wood is described by the above statements?
Correct Answer: (c)
These characteristics define spring wood (early wood), which is produced when environmental conditions favor vigorous growth.
These characteristics define spring wood (early wood), which is produced when environmental conditions favor vigorous growth.
Q15. Match the following for Male Glands:
| Gland | Count |
|---|---|
| A. Seminal vesicle | (I) Paired |
| B. Prostate | (II) Paired |
| C. Bulbourethral | (III) Single (Unpaired) |
Correct Answer: (a)
Prostate is single; others are paired.
Prostate is single; others are paired.
Q16. Phospholipids are important cell membrane constituents, because they
Correct Answer: (d)
Phospholipids are amphipathic molecules, meaning they have both a hydrophilic polar head and hydrophobic non-polar tails, allowing them to form stable bilayers in aqueous environments.
Phospholipids are amphipathic molecules, meaning they have both a hydrophilic polar head and hydrophobic non-polar tails, allowing them to form stable bilayers in aqueous environments.
Q17. Strobili or cones are found in
Correct Answer: (c)
In some pteridophytes like Equisetum and Selaginella, sporophylls form distinct compact structures called strobili or cones.
In some pteridophytes like Equisetum and Selaginella, sporophylls form distinct compact structures called strobili or cones.
Q18. The following diagram shows some of the missing structures in a plant cell marked as A, B, C, D and E. Choose the option with their correct names.
Correct Answer: (a)
A: Connections between cells (Plasmodesmata); B: ER with dots (Rough ER); C: Stacked cisternae (Golgi); D: Powerhouse (Mitochondrion); E: Small dots (Ribosomes).
A: Connections between cells (Plasmodesmata); B: ER with dots (Rough ER); C: Stacked cisternae (Golgi); D: Powerhouse (Mitochondrion); E: Small dots (Ribosomes).
Q19. Select the correct statements for oogenesis:
(i) Initiated during embryonic development.
(ii) Tertiary follicle has a fluid-filled cavity (antrum).
(iii) Graafian follicle releases a secondary oocyte.
(i) Initiated during embryonic development.
(ii) Tertiary follicle has a fluid-filled cavity (antrum).
(iii) Graafian follicle releases a secondary oocyte.
Correct Answer: (a)
All points are characteristically true for female gamete development.
All points are characteristically true for female gamete development.
Q20. Which one of the following acts as the “alarm signal” to activate the body’s adaptive immune system by stimulating helper T cells?
Correct Answer: (b)
Interleukin-1 is a cytokine released by macrophages and antigen-presenting cells. It functions as an alarm signal by activating helper T cells, thereby linking innate immunity to adaptive immunity.
Interleukin-1 is a cytokine released by macrophages and antigen-presenting cells. It functions as an alarm signal by activating helper T cells, thereby linking innate immunity to adaptive immunity.
Q21. Which one of the following statement is incorrect?
Correct Answer:
This is reversed: guard cells are bean-shaped in most dicots and dumbbell-shaped in many monocots like grasses.
This is reversed: guard cells are bean-shaped in most dicots and dumbbell-shaped in many monocots like grasses.
Q22. 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.
Q23. Triclosan is an antibacterial chemical that is used in some household products to reduce bacteria levels. Which of the following is the most likely result of increased use of products that contain triclosan over time?
Correct Answer: (b)
(B) Increased use of products that contain triclosan will create an environment in which triclosan-resistant bacteria have a survival advantage, so over time the relative numbers of triclosan-resistant bacteria will increase. Choice (A) is incorrect because there is no evidence that triclosan kills all bacteria. Individuals do not evolve, populations evolve, so choice (C) is incorrect. Choice (D) is incorrect because there is no evidence in the question that the use of triclosan increases the frequency of mutations in bacteria.
(B) Increased use of products that contain triclosan will create an environment in which triclosan-resistant bacteria have a survival advantage, so over time the relative numbers of triclosan-resistant bacteria will increase. Choice (A) is incorrect because there is no evidence that triclosan kills all bacteria. Individuals do not evolve, populations evolve, so choice (C) is incorrect. Choice (D) is incorrect because there is no evidence in the question that the use of triclosan increases the frequency of mutations in bacteria.
Q24. Imagine that you are doing an experiment on the movement of ions across neural membranes. Which of the following plays a role in determining the equilibrium concentration of ions across these membranes?
Correct Answer: (d)
The electrochemical gradient is the combined force of the chemical concentration gradient and the electrical gradient (charge difference) across a membrane. Together, these factors determine the direction and equilibrium of ion movement.
The electrochemical gradient is the combined force of the chemical concentration gradient and the electrical gradient (charge difference) across a membrane. Together, these factors determine the direction and equilibrium of ion movement.
Q25. Assertion: Only one sperm can fertilise an ovum.
Reason: During fertilisation, a sperm induces changes in the zona pellucida that block the entry of additional sperms.
Correct Answer: (a)
The block to polyspermy ensures a diploid zygote.
The block to polyspermy ensures a diploid zygote.
Q26. Which one is wrong statement?
Correct Answer: (a)
Mucor (Phycomycetes) produces non-motile aplanospores or motile zoospores, but Mucor specifically produces non-motile spores.
Mucor (Phycomycetes) produces non-motile aplanospores or motile zoospores, but Mucor specifically produces non-motile spores.
Q27. In which class of Echinodermata, Pluteus larva is formed
Correct Answer: (b)
Pluteus larva is found in class Echinoidea.
Pluteus larva is found in class Echinoidea.
Q28. An organ is classified as part of the endocrine system if it
Correct Answer: (d)
The defining functional characteristic of an endocrine organ or gland is that it secretes its chemical products (hormones) directly into the circulatory system (bloodstream) rather than through a duct system.
The defining functional characteristic of an endocrine organ or gland is that it secretes its chemical products (hormones) directly into the circulatory system (bloodstream) rather than through a duct system.
Q29. Which two of the following statements regarding food chains are correct ?
(i) Removal of 80% tigers resulted in increased growth of vegetation.
(ii) Removal of most carnivores resulted in increased population of deer.
(iii) Length of food chain is limited to 3 – 4 trophic levels due to energy loss.
(iv) Length of food chain may vary from 2 – 3 trophic levels.
(i) Removal of 80% tigers resulted in increased growth of vegetation.
(ii) Removal of most carnivores resulted in increased population of deer.
(iii) Length of food chain is limited to 3 – 4 trophic levels due to energy loss.
(iv) Length of food chain may vary from 2 – 3 trophic levels.
Correct Answer: (c)
(ii) is true due to release from predation. (iii) is true because of energy loss at each step (10% law), there's rarely enough energy to support more than 4-5 levels.
(ii) is true due to release from predation. (iii) is true because of energy loss at each step (10% law), there's rarely enough energy to support more than 4-5 levels.
Q30. 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.
Q31. Select the correct statement with respect to mitosis.
Correct Answer: (d)
Alignment at the equatorial plate is the hallmark of metaphase. Chromatids separate in anaphase, and organelles disappear by the end of prophase.
Alignment at the equatorial plate is the hallmark of metaphase. Chromatids separate in anaphase, and organelles disappear by the end of prophase.
Q32. You are designing a pesticide to disrupt the endocrine systems of arthropods without harming mammals. Which of the following substances should be the target of your investigations?
Correct Answer: (c)
Juvenile hormone is a crucial regulator of development, molting, and metamorphosis specifically in insects and other arthropods. Because mammals do not utilize this hormone, targeting it allows for a pesticide that is highly specific to pests with minimal risk to humans.
Juvenile hormone is a crucial regulator of development, molting, and metamorphosis specifically in insects and other arthropods. Because mammals do not utilize this hormone, targeting it allows for a pesticide that is highly specific to pests with minimal risk to humans.
Q33. Assertion: Male germ cells undergo meiotic divisions to form sperms.
Reason: Sertoli cells provide nutrition to the germ cells.
Correct Answer: (b)
Both are true facts about seminiferous tubules, but the nutrition provided by Sertoli cells is not the reason why germ cells undergo meiosis.
Both are true facts about seminiferous tubules, but the nutrition provided by Sertoli cells is not the reason why germ cells undergo meiosis.
Q34. Abscisic acid is a:
Correct Answer: (b)
ABA is called the stress hormone because it increases plant tolerance to various stresses, particularly by regulating water loss through stomata.
ABA is called the stress hormone because it increases plant tolerance to various stresses, particularly by regulating water loss through stomata.
Q35. Which of the following represents the correct combination without any exception?
| Characteristics | Class |
|---|---|
| (a) Mouth ventral, gills without operculum; skin with placoid scales; persistent notochord | Chondrichthyes |
| (b) Sucking and circular mouth; jaws absent, integument without scales; paired appendages | Cyclostomata |
| (c) Body covered with feathers; skin moist and glandular; fore-limbs form wings; lungs with air sacs | Aves |
| (d) Mammary gland; hair on body; pinnae; two pairs of Limbs | Mammalia |
Correct Answer: (a)
Combination (a) is correct without exceptions. Cyclostomes can lack paired appendages; birds have air sacs but not 'moist skin'; some mammals lack pinnae.
Combination (a) is correct without exceptions. Cyclostomes can lack paired appendages; birds have air sacs but not 'moist skin'; some mammals lack pinnae.
Q36. Which contraceptive method also protects the user from contracting STIs and AIDS?
Correct Answer: (c)
Use of condoms has increased... due to its additional benefit of protecting the user from contracting STIs and AIDS.
Use of condoms has increased... due to its additional benefit of protecting the user from contracting STIs and AIDS.
Q37. 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.
Q38. What is a proteome?
Correct Answer: (c)
While the genome is the complete set of genetic material, the proteome is the entire set of proteins expressed by a cell, tissue, or organism at a given time under specific conditions. It is much more complex and dynamic than the genome.
While the genome is the complete set of genetic material, the proteome is the entire set of proteins expressed by a cell, tissue, or organism at a given time under specific conditions. It is much more complex and dynamic than the genome.
Q39. Bananas can be prevented from over ripening by:
Correct Answer: (c)
Antioxidants like ascorbic acid can sometimes be used to slow down the oxidative processes associated with ripening and browning. (Wait, check Key: 71(c)).
Antioxidants like ascorbic acid can sometimes be used to slow down the oxidative processes associated with ripening and browning. (Wait, check Key: 71(c)).
Q40. 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.
Q41. During exercise more oxygen is delivered to the muscles because
Correct Answer: (c)
Exercise triggers the Bohr effect: increased CO2 production lowers blood pH, which decreases hemoglobin's affinity for oxygen, causing it to 'unload' O2 more readily. Additionally, increased muscle temperature further reduces hemoglobin affinity, facilitating even more oxygen delivery to the working tissues.
Exercise triggers the Bohr effect: increased CO2 production lowers blood pH, which decreases hemoglobin's affinity for oxygen, causing it to 'unload' O2 more readily. Additionally, increased muscle temperature further reduces hemoglobin affinity, facilitating even more oxygen delivery to the working tissues.
Q42. Worms use a hydrostatic skeleton to generate movement. How do they do this?
Correct Answer: (b)
A hydrostatic skeleton consists of a fluid-filled cavity (coelom) surrounded by muscles. When these muscles contract, they put pressure on the incompressible fluid, changing the animal's shape and allowing for movement like peristalsis.
A hydrostatic skeleton consists of a fluid-filled cavity (coelom) surrounded by muscles. When these muscles contract, they put pressure on the incompressible fluid, changing the animal's shape and allowing for movement like peristalsis.
Q43. Which of the following statement(s) about algae is/are correct?
(i) Algae are chlorophyll bearing simple, thalloid, heterotrophic and aquatic (both fresh water and marine) organisms.
(ii) Algae reproduce by vegetative means only.
(iii) Fusion of two gametes dissimilar in size is termed as oogamous.
(iv) A few of the massive forms of algae such as kelps, form massive plant bodies.
(i) Algae are chlorophyll bearing simple, thalloid, heterotrophic and aquatic (both fresh water and marine) organisms.
(ii) Algae reproduce by vegetative means only.
(iii) Fusion of two gametes dissimilar in size is termed as oogamous.
(iv) A few of the massive forms of algae such as kelps, form massive plant bodies.
Correct Answer: (c)
Only statement (iv) is correct. Algae are autotrophic (not heterotrophic), reproduce sexually/asexually as well, and oogamy is a specific type of anisogamy involving a large non-motile female and small motile male.
Only statement (iv) is correct. Algae are autotrophic (not heterotrophic), reproduce sexually/asexually as well, and oogamy is a specific type of anisogamy involving a large non-motile female and small motile male.
Q44. 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.
Q45. 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.
Q46. Motor neurons stimulate muscle contraction via the release of
Correct Answer: (c)
Motor neurons release the neurotransmitter acetylcholine (ACh) into the synaptic cleft at the neuromuscular junction. ACh binds to receptors on the muscle fiber's sarcolemma, triggering an action potential that leads to contraction.
Motor neurons release the neurotransmitter acetylcholine (ACh) into the synaptic cleft at the neuromuscular junction. ACh binds to receptors on the muscle fiber's sarcolemma, triggering an action potential that leads to contraction.
Q47. Mountain climbers may have difficulty at high elevations because
Correct Answer: (a)
At high altitudes, the total atmospheric pressure is lower. Since oxygen still makes up about 21% of the air, its partial pressure (PO2) also decreases. This reduction in PO2 decreases the pressure gradient necessary to drive oxygen diffusion from the lungs into the blood.
At high altitudes, the total atmospheric pressure is lower. Since oxygen still makes up about 21% of the air, its partial pressure (PO2) also decreases. This reduction in PO2 decreases the pressure gradient necessary to drive oxygen diffusion from the lungs into the blood.
Q48. Suppose that a new disease is discovered that suppresses the immune system. Which of the following would indicate that the disease specifically affects the B cells rather than the helper or cytotoxic T cells?
Correct Answer: (c)
B cells differentiate into plasma cells that secrete antibodies. A reduction in plasma cells directly indicates impaired B-cell function rather than defects in helper or cytotoxic T cells.
B cells differentiate into plasma cells that secrete antibodies. A reduction in plasma cells directly indicates impaired B-cell function rather than defects in helper or cytotoxic T cells.
Q49. Diuretics are drugs that can be used to treat high blood pressure by increasing urinary output. Possible mechanisms of action in the kidney include
Correct Answer: (b)
Many diuretics work by inhibiting NaCl reabsorption in the loop of Henle or proximal tubule. Reduced salt reabsorption decreases water reabsorption, leading to increased urine output and lowered blood pressure.
Many diuretics work by inhibiting NaCl reabsorption in the loop of Henle or proximal tubule. Reduced salt reabsorption decreases water reabsorption, leading to increased urine output and lowered blood pressure.
Q50. Morphogenesis is the development of
Correct Answer: (a)
Morphogenesis refers to the biological process that causes an organism to develop its shape. In plants, it specifically describes the development of the growth form through orchestrated cell division and expansion.
Morphogenesis refers to the biological process that causes an organism to develop its shape. In plants, it specifically describes the development of the growth form through orchestrated cell division and expansion.
Q51. Yeast is not included in protozoans but are placed fungi because
Correct Answer: (b)
Yeasts are classified as fungi because they share fungal characteristics such as chitinous cell walls and the ability to form chains of cells called pseudomycelium during budding.
Yeasts are classified as fungi because they share fungal characteristics such as chitinous cell walls and the ability to form chains of cells called pseudomycelium during budding.
Q52. Match the contraceptive brand with its feature:
| Column-I (Brand) | Column-II (Feature) |
|---|---|
| A. Saheli | (I) Popular condom for male |
| B. Nirodh | (II) Once-a-week oral pill |
| C. Multiload 375 | (III) Copper releasing IUD |
Correct Answer: (a)
Standard brand descriptions in the chapter.
Standard brand descriptions in the chapter.
Q53. Phosphorus is absorbed by the plants in the form of
(i) H2PO4–
(ii) HPO42–
(iii) HPO42–
(iv) H2PO4–
(i) H2PO4–
(ii) HPO42–
(iii) HPO42–
(iv) H2PO4–
Correct Answer: (c)
Phosphorus is absorbed from the soil by plants primarily as phosphate ions: H2PO4– or HPO42–.
Phosphorus is absorbed from the soil by plants primarily as phosphate ions: H2PO4– or HPO42–.
Q54. 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.
Q55. Which type of function is performed by the fleshy leaves of onion and garlic?
Correct Answer: (a)
In onion and garlic, the fleshy leaves store food material.
In onion and garlic, the fleshy leaves store food material.
Q56. When the margins of sepals or petals overlap one another without any particular direction, the condition is termed as
Correct Answer: (b)
In imbricate aestivation, the margins overlap but not in any specific uniform direction, unlike twisted aestivation.
In imbricate aestivation, the margins overlap but not in any specific uniform direction, unlike twisted aestivation.
Q57. A dikaryon is formed when
Correct Answer: (b)
A dikaryon (n+n) stage results when plasmogamy occurs but karyogamy is delayed.
A dikaryon (n+n) stage results when plasmogamy occurs but karyogamy is delayed.
Q58. Leghaemoglobin helps in:
Correct Answer: (c)
Nitrogenase is highly sensitive to oxygen. Leghaemoglobin acts as an oxygen scavenger to maintain anaerobic conditions in the nodule.
Nitrogenase is highly sensitive to oxygen. Leghaemoglobin acts as an oxygen scavenger to maintain anaerobic conditions in the nodule.
Q59. What are those structures that appear as beads-on-string in the chromosomes when viewed under electron microscope?
Correct Answer: (c)
The basic unit of chromatin packaging is the nucleosome, which appears as 'beads-on-a-string' structure.
The basic unit of chromatin packaging is the nucleosome, which appears as 'beads-on-a-string' structure.
Q60. Which of the following statements (i - v) are correct ?
(i) The pelvic fins of female sharks bear claspers.
(ii) In Obelia, polyps produce medusae sexually and medusae form the polyps asexually.
(iii) Flame cells in platyhelminthes help in osmoregulation and excretion.
(iv) In non-chordates, central nervous system is ventral, solid and double.
(v) Pinnae are present in mammals.
(i) The pelvic fins of female sharks bear claspers.
(ii) In Obelia, polyps produce medusae sexually and medusae form the polyps asexually.
(iii) Flame cells in platyhelminthes help in osmoregulation and excretion.
(iv) In non-chordates, central nervous system is ventral, solid and double.
(v) Pinnae are present in mammals.
Correct Answer: (c)
Statement (i) is false (males have claspers). (ii) is false (polyp to medusa is asexual). (iii), (iv), (v) are correct.
Statement (i) is false (males have claspers). (ii) is false (polyp to medusa is asexual). (iii), (iv), (v) are correct.
Q61. Which are correct for 'Azospirillum' and 'Azotobacter'?
(i) They are symbiotic nitrogen fixers.
(ii) They are free-living in the soil.
(iii) They enrich soil nitrogen content.
(i) They are symbiotic nitrogen fixers.
(ii) They are free-living in the soil.
(iii) They enrich soil nitrogen content.
Correct Answer: (a)
They are free-living, not symbiotic (Page 10).
They are free-living, not symbiotic (Page 10).
Q62. T.S. of monocot leaf is given below, certain parts have been marked by alphabets (A – G). Which one is the option showing there correct labelling?
Correct Answer: (a)
A is upper (adaxial) epidermis; xylem is towards the upper surface; mesophyll is undifferentiated; abaxial epidermis is lower with stomata.
A is upper (adaxial) epidermis; xylem is towards the upper surface; mesophyll is undifferentiated; abaxial epidermis is lower with stomata.
Q63. Which of the following pair is mismatched?
Correct Answer: (d)
The dark reaction (Calvin cycle) takes place in the stroma, not the grana.
The dark reaction (Calvin cycle) takes place in the stroma, not the grana.
Q64. Hyperventilation occurs
Correct Answer: (d)
Hyperventilation is breathing that is faster or deeper than necessary, which results in excessive elimination of carbon dioxide. Consequently, a hallmark state of hyperventilation is a low partial pressure of CO2 (hypocapnia) in the blood.
Hyperventilation is breathing that is faster or deeper than necessary, which results in excessive elimination of carbon dioxide. Consequently, a hallmark state of hyperventilation is a low partial pressure of CO2 (hypocapnia) in the blood.
Q65. Which of the following class of fungi is being described by the given statements?
(i) They are found in aquatic habitats and on decaying wood in moist and damp places.
(ii) Mycelium is aseptate and coenocytic.
(iii) Asexual reproduction takes place by zoospores (motile) or by aplanospores (non-motile).
(iv) Some common examples are Mucor, Rhizopus and Albugo.
(i) They are found in aquatic habitats and on decaying wood in moist and damp places.
(ii) Mycelium is aseptate and coenocytic.
(iii) Asexual reproduction takes place by zoospores (motile) or by aplanospores (non-motile).
(iv) Some common examples are Mucor, Rhizopus and Albugo.
Correct Answer: (b)
Phycomycetes are characterized by coenocytic (non-septate) mycelium and aquatic/moist habitats. Examples include Rhizopus and Mucor.
Phycomycetes are characterized by coenocytic (non-septate) mycelium and aquatic/moist habitats. Examples include Rhizopus and Mucor.
Q66. Select correct statements for 'MTP Act 2017':
(i) Enacted to reduce illegal abortions.
(ii) Ground for termination includes risk to the life of the woman.
(iii) Legal only if the foetus is male.
(i) Enacted to reduce illegal abortions.
(ii) Ground for termination includes risk to the life of the woman.
(iii) Legal only if the foetus is male.
Correct Answer: (a)
MTP for sex selection (foeticide) is illegal.
MTP for sex selection (foeticide) is illegal.
Q67. 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.
Q68. Parasitism differs from predation because
Correct Answer: (d)
All listed statements are incorrect: parasitism can drive strong defensive adaptations, involve coevolutionary arms races, and significantly affect host population dynamics.
All listed statements are incorrect: parasitism can drive strong defensive adaptations, involve coevolutionary arms races, and significantly affect host population dynamics.
Q69. Match Column-I (Discovery) and Column-II (Scientists) and select the correct option.
| Column-I (Discovery) | Column-II (Name of scientists) |
|---|---|
| A. Foolish seedling disease of rice | I. Cousins |
| B. Crystallized the Kinetic | II. F.W. Went |
| C. Release of ethylene gas | III. Skoog and Miller |
| D. Bioassay of Auxin | IV. E. Kurosawa |
Correct Answer: (c)
Kurosawa discovered GA in foolish rice. Skoog/Miller found kinetin. Cousins found ethylene in oranges. Went did the Avena test for auxin.
Kurosawa discovered GA in foolish rice. Skoog/Miller found kinetin. Cousins found ethylene in oranges. Went did the Avena test for auxin.
Q70. 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.
Q71. Site of photosynthesis in C4 plant is:
Correct Answer: (c)
In C4 plants, the reactions are split between mesophyll cells (initial fixation) and bundle sheath cells (Calvin cycle).
In C4 plants, the reactions are split between mesophyll cells (initial fixation) and bundle sheath cells (Calvin cycle).
Q72. Assertion: LAB play a very beneficial role in our stomach.
Reason: They help in checking disease-causing microbes in the stomach.
Correct Answer: (a)
This is a direct fact from the NCERT text (Page 3).
This is a direct fact from the NCERT text (Page 3).
Q73. FISH analysis of a breast tumor biopsy for HER2 gene reveals two spots of fluorescence in cells. What conclusion do these data support about the tumor?
Correct Answer: (d)
Fluorescence In Situ Hybridization (FISH) is used to visualize specific DNA sequences. Normal diploid cells have two copies of every gene. Since the analysis shows only two spots of fluorescence, the HER2 gene has not been duplicated or amplified. Because targeted therapies like Herceptin are only effective in tumors where the gene is amplified and overexpressing protein, these treatments would not be suitable for this patient.
Fluorescence In Situ Hybridization (FISH) is used to visualize specific DNA sequences. Normal diploid cells have two copies of every gene. Since the analysis shows only two spots of fluorescence, the HER2 gene has not been duplicated or amplified. Because targeted therapies like Herceptin are only effective in tumors where the gene is amplified and overexpressing protein, these treatments would not be suitable for this patient.
Q74. The F1 plants from the previous question are allowed to self-fertilize. The phenotypic ratio for the F2 should be
Correct Answer: (c)
In the F2 generation of a monohybrid cross (Pp × Pp), the genotypes produced are 1 PP, 2 Pp, and 1 pp. Because the dominant allele (P) masks the recessive allele (p), the resulting phenotypic ratio is 3 dominant (purple) to 1 recessive (white).
In the F2 generation of a monohybrid cross (Pp × Pp), the genotypes produced are 1 PP, 2 Pp, and 1 pp. Because the dominant allele (P) masks the recessive allele (p), the resulting phenotypic ratio is 3 dominant (purple) to 1 recessive (white).
Q75. Nucleoside is :
Correct Answer: (b)
In the diagrams: A is a nitrogenous base only, B and C are base + sugar (nucleosides), and D is base + sugar + phosphate (nucleotide).
In the diagrams: A is a nitrogenous base only, B and C are base + sugar (nucleosides), and D is base + sugar + phosphate (nucleotide).
Q76. 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.
Q77. 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.
Q78. The presence of one species (A) in a community may benefit another species (B) if
Correct Answer: (d)
All listed scenarios describe direct or indirect positive effects of one species on another, demonstrating how complex interactions can benefit species within communities.
All listed scenarios describe direct or indirect positive effects of one species on another, demonstrating how complex interactions can benefit species within communities.
Q79. Which of the following ecosystems is most productive in terms of net primary production?
Correct Answer: (b)
Tropical rain forests have ideal temperature and moisture conditions, resulting in the highest annual net primary productivity.
Tropical rain forests have ideal temperature and moisture conditions, resulting in the highest annual net primary productivity.
Q80. A forensic scientist recovers a very small amount of evidence at a crime scene. The scientist would like to amplify the number of copies of DNA in the evidence sample. Which of the following techniques should the scientist use?
Correct Answer: (d)
Polymerase chain reaction is used to make multiple copies of (amplify) specific DNA sequences. Choice (A) is incorrect because bacterial transformation is the process in which bacteria absorb and express foreign DNA. CRISPR-Cas9 is used in gene editing, so choice (B) is incorrect. Gel electrophoresis is used to separate fragments of DNA by size, so choice (C) is incorrect.
Polymerase chain reaction is used to make multiple copies of (amplify) specific DNA sequences. Choice (A) is incorrect because bacterial transformation is the process in which bacteria absorb and express foreign DNA. CRISPR-Cas9 is used in gene editing, so choice (B) is incorrect. Gel electrophoresis is used to separate fragments of DNA by size, so choice (C) is incorrect.
Q81. Match the terms (given in column I) with their explanation (given in column II) and choose the correct combination from the options given below.
| Column I (Terms) | Column II (Explanation) |
|---|---|
| A. Terminalization | I. Pairing of homologous chromosomes. |
| B. Synapsis | II. Point of attachment between homologous chromosomes. |
| C. Chiasmata | III. Nuclear protein complex that helps in adherence of sister chromatids and then homologous chromosomes. |
| D. Synaptonemal complex | IV. Shifting of chiasmata outwards towards the ends of a bivalent. |
Correct Answer: (a)
A matches IV; B matches I; C matches II; D matches III based on biological definitions.
A matches IV; B matches I; C matches II; D matches III based on biological definitions.
Q82. 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.
Q83. Which of the following types of mammalian cell does not carry out oxidative phosphorylation?
Correct Answer: (b)
Mature mammalian erythrocytes (RBCs) lack mitochondria and therefore cannot perform the TCA cycle or oxidative phosphorylation; they rely solely on glycolysis.
Mature mammalian erythrocytes (RBCs) lack mitochondria and therefore cannot perform the TCA cycle or oxidative phosphorylation; they rely solely on glycolysis.
Q84. Why is a heat-stable DNA polymerase required for the polymerase chain reaction (PCR)?
Correct Answer: (b)
The denaturation stage that occurs during each cycle of PCR requires high temperatures that would denature most enzymes, so a heat-stable DNA polymerase is required for PCR. Heat-stable DNA polymerases do not add DNA nucleotides at a faster rate than other DNA polymerases, so choice (A) is incorrect. PCR primers anneal to the target DNA sequence, not to the DNA polymerase, so choice (C) is incorrect. Choice (D) is incorrect because heat-stable DNA polymerases are not more accurate than other DNA polymerases.
The denaturation stage that occurs during each cycle of PCR requires high temperatures that would denature most enzymes, so a heat-stable DNA polymerase is required for PCR. Heat-stable DNA polymerases do not add DNA nucleotides at a faster rate than other DNA polymerases, so choice (A) is incorrect. PCR primers anneal to the target DNA sequence, not to the DNA polymerase, so choice (C) is incorrect. Choice (D) is incorrect because heat-stable DNA polymerases are not more accurate than other DNA polymerases.
Q85. Which is not a stem modification ?
Correct Answer: (c)
The pitcher of Nepenthes (pitcher plant) is a modification of the leaf, not the stem.
The pitcher of Nepenthes (pitcher plant) is a modification of the leaf, not the stem.
Q86. The antibiotic tetracycline inhibits bacterial growth by blocking the binding of tRNAs to the ribosome. Which of the following processes is most directly affected by tetracycline?
Correct Answer: (c)
(C) If tRNAs could not bind to the ribosome, the initiation of translation could not occur. Choices (A) and (B) are incorrect because transcription and exon splicing do not involve the ribosome, so an antibiotic that interferes with ribosomes would not affect those processes. The export of proteins from the cell involves the Golgi bodies and vesicles, so choice (D) is incorrect.
(C) If tRNAs could not bind to the ribosome, the initiation of translation could not occur. Choices (A) and (B) are incorrect because transcription and exon splicing do not involve the ribosome, so an antibiotic that interferes with ribosomes would not affect those processes. The export of proteins from the cell involves the Golgi bodies and vesicles, so choice (D) is incorrect.
Q87. Match the following stem modifications given in column I with their examples given in column II and select the correct combination from the options given below.
| Column-I (Stem Modifications) | Column-II (Found in) |
|---|---|
| A. Underground stem | I. Euphorbia |
| B. Stem tendril | II. Opuntia |
| C. Stem thorns | III. Potato |
| D. Flattened stem | IV. Citrus |
| E. Fleshy cylindrical stem | V. Cucumber |
Correct Answer: (d)
Potato is an underground stem. Cucumber has stem tendrils. Citrus has thorns. Opuntia has flattened stems. Euphorbia has fleshy cylindrical stems.
Potato is an underground stem. Cucumber has stem tendrils. Citrus has thorns. Opuntia has flattened stems. Euphorbia has fleshy cylindrical stems.
Q88. Artificial selection experiments in the laboratory such as in figure 21.5 are an example of
Correct Answer: (d)
According to the provided answer key, these experiments represent disruptive selection. In the context of the Drosophila experiments mentioned (selecting for both very high and very low bristle counts simultaneously), the population is being pushed toward two extreme phenotypes, which is the definition of disruptive selection.
According to the provided answer key, these experiments represent disruptive selection. In the context of the Drosophila experiments mentioned (selecting for both very high and very low bristle counts simultaneously), the population is being pushed toward two extreme phenotypes, which is the definition of disruptive selection.
Q89. Suppose that you stick your finger with a sharp pin. The area affected is very small and only one pain receptor fires. However, it fires repeatedly at a rapid rate (it hurts!). This is an example of
Correct Answer: (a)
Temporal summation occurs when a single presynaptic neuron (or receptor) fires multiple action potentials in rapid succession, resulting in the additive effect of postsynaptic potentials at the same location over time.
Temporal summation occurs when a single presynaptic neuron (or receptor) fires multiple action potentials in rapid succession, resulting in the additive effect of postsynaptic potentials at the same location over time.
Q90. 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.
A transgenic mouse has the HTT gene. A scientist wants to investigate whether inserting a stop codon into the HTT gene might prevent the mouse from developing the disease. Which of the following tools would be most useful in inserting a stop codon into the HTT gene?
A transgenic mouse has the HTT gene. A scientist wants to investigate whether inserting a stop codon into the HTT gene might prevent the mouse from developing the disease. Which of the following tools would be most useful in inserting a stop codon into the HTT gene?
Correct Answer: (a)
CRISPR-Cas9 is used for gene editing, and with the correct donor DNA, it could be used to insert a stop codon into a gene. Gel electrophoresis separates DNA fragments by size, so choice (B) is incorrect. Choice (C) is incorrect because PCR amplifies the number of copies of a specific DNA sequence. Restriction enzymes cut DNA at specific sequences, so choice (D) is incorrect.
CRISPR-Cas9 is used for gene editing, and with the correct donor DNA, it could be used to insert a stop codon into a gene. Gel electrophoresis separates DNA fragments by size, so choice (B) is incorrect. Choice (C) is incorrect because PCR amplifies the number of copies of a specific DNA sequence. Restriction enzymes cut DNA at specific sequences, so choice (D) is incorrect.