Meraki 1 (Full Syllabus Biology Practice Paper for NEET 2025)

1. Which of the following accurately describes the steps involved in the systematic approach to classifying a newly discovered organism, and how it differs from taxonomy?

1.	Systematics includes observing physical traits, placing organisms in existing classifications, and naming the organism, whereas taxonomy focuses on identifying genetic markers only.
2.	Systematics incorporates gathering and analysing evolutionary data, constructing phylogenetic trees, and determining evolutionary relationships, while taxonomy is limited to naming and organizing species based on observable characteristics.
3.	In systematics, researchers categorize the organism by habitat and ecological function, whereas taxonomy considers only the reproductive behaviour of the species.
4.	Systematics disregards hierarchical classification and uses random assignment based on ancestry, whereas taxonomy places organisms strictly by morphological similarity.

2. What is a distinguishing characteristic of sexual spore formation among the fungal groups Phycomycetes, Ascomycetes, and Basidiomycetes?

1.	Phycomycetes produce motile spores, while Ascomycetes and Basidiomycetes do not form sexual spores.
2.	Both Ascomycetes and Basidiomycetes produce conidiospores, while Phycomycetes produce zygospores externally.
3.	In Ascomycetes and Phycomycetes, spores are formed on specialized stalks, while Basidiomycetes form spores in spherical structures.
4.	Phycomycetes form zygospores, Ascomycetes form ascospores within sac-like structures, and Basidiomycetes form basidiospores externally on club-shaped structures.

3. Consider the following statements about infectious agents:

1.	Prions are unique infectious agents that lack nucleic acids and propagate by inducing misfolding in host proteins.
2.	Viroids consist solely of a short strand of circular RNA without a protein coat and exclusively infect animals.
3.	Viruses contain both DNA and RNA within a protein coat and rely on host cells for replication.

How many of the above statements are correct?
1. Only one
2. Only two
3. All three
4. None

4. Consider the information given in the following statements as true:

Statement I:	Nitrosomonas bacteria are nitrifying bacteria that oxidize ammonia (NH₃) to nitrite (NO₂⁻) to obtain energy.
Statement II:	Thiobacillus bacteria oxidize sulfur compounds like hydrogen sulfide (H₂S), sulfur (S), or thiosulfate (S₂O₃²⁻) to generate energy, converting these compounds into sulfate (SO₄²⁻).

Based on the above statements, it can be concluded that:

Conclusion I:	Nitrosomonas and Thiobacillus must be classified as chemoheterotrophs.
Conclusion II:	Nitrosomonas and Thiobacillus play important roles in nutrient cycling in an ecosystem.

1. Only Conclusion I
2. Only Conclusion II
3. Both Conclusion I and Conclusion II
4. Neither Conclusion I nor Conclusion II

5. Rhodophyceae, or red algae, are adapted to deep ocean habitats. Which of the following best describes the physiological and structural adaptations in Rhodophyceae that allow them to inhabit low-light, deep-sea environments?

1.	The presence of multiple chlorophyll a variants, enhancing light capture across different depths.
2.	Reduced dependence on photosynthesis, using chemosynthetic pathways for survival in low-light conditions.
3.	Enhanced structural rigidity through calcified cell walls, allowing red algae to withstand high-pressure environments.
4.	High levels of phycoerythrin, which absorbs blue light wavelengths that penetrate deeper waters.

6. In gymnosperms, the process of fertilization and subsequent seed formation shows several adaptations compared to bryophytes and most pteridophytes. Which of the following best describes how gymnosperms exhibit an advanced reproductive adaptation?

1.	They produce seeds without the need for an ovary, allowing faster fertilization and seed development.
2.	Gametophytes develop into free-living structures that support embryo formation independently of the sporophyte.
3.	Male and female gametophytes are retained within sporangia on sporophytes, with the pollen tube facilitating direct fertilization within the ovule.
4.	Pollen grains develop fully outside the microsporangium and fertilize ovules directly through airborne dispersal.

7. Consider the following table enumerating differences between bryophytes and pteridophytes. Based on your knowledge, identify how many of these statements are incorrect:

Character	Bryophytes	Pteridophytes
Dominant Generation	Gametophyte is dominant	Sporophyte is dominant
Vascular Tissue	Absent	Present
Habitat	Restricted mainly to moist environments	Highly adapted to xeric terrestrial environments
Reproductive Structures	Archegonia and antheridia are present	Only archegonia present
Seed Formation	Seeds are formed pre-fertilization	Seeds are formed post-fertilization
Fertilization Requirement	Water is necessary for fertilization	Fertilization does not require water

Options:
1. 2
2. 3
3. 4
4. 5

8. Match each item in Column I with one item in Column II and select the best match from the codes given:

	Column I		Column II
A.	Stem tendril	P.	Gloriosa
B.	Leaf tendril	Q.	Grapevine
C.	Racemose inflorescence	R.	Indigofera
D.	Cymose inflorescence	S.	Solanum

Options	A	B	C	D
1.	Q	P	R	S
2.	Q	P	S	R
3.	P	Q	R	S
4.	S	R	Q	P

9. The given floral diagram describes members of which angiosperm family?

1. Compositae
2. Malvaceae
3. Solanaceae
4. Brassicaceae

10. In the context of seed anatomy, which of the following statements correctly differentiates monocot and dicot seeds?

1.	Dicot seeds, such as castor, are exclusively endospermic, while monocots like maize lack a protective seed coat.
2.	Monocot seeds have a single cotyledon called the scutellum that absorbs nutrients from the endosperm, whereas dicots may or may not retain the endosperm in mature seeds.
3.	The plumule is encased in the coleorhiza in monocots, providing an initial protective sheath during germination.
4.	Only dicot seeds possess a micropyle, facilitating water entry and gas exchange, while monocots rely on direct absorption through the scutellum.

11. Consider the following statements regarding the structure and growth regions of a root:

A.	The root cap protects the apex of the root as it grows through the soil.
B.	The region of meristematic activity is located just above the region of elongation.
C.	Cells in the region of elongation are primarily responsible for increasing the length of the root.
D.	Root hairs are formed from some epidermal cells in the region of maturation, aiding in the absorption of water and minerals.

How many of the above statements are correct?
1. Two statements are correct
2. Three statements are correct
3. All four statements are correct
4. Only one statement is correct

12. In the given figure A and B respectively represent transverse section of a:

A	B

1. Monocot root and Monocot stem
2. Monocot root and Dicot stem
3. Dicot root and Monocot stem
4. Dicot root and Dicot stem

13. Consider the following statements regarding vascular bundles and their structure in dicotyledonous and monocotyledonous plants:

A.	In dicotyledonous stems, the presence of cambium between xylem and phloem allows for the formation of secondary tissues, making these vascular bundles open.
B.	Monocotyledonous vascular bundles lack cambium, which prevents the formation of secondary xylem and phloem, thus making them closed.
C.	A radial arrangement of xylem and phloem in vascular bundles, where they are aligned on alternate radii, is commonly found in stems.
D.	In conjoint vascular bundles, xylem and phloem are arranged along the same radius, with phloem typically positioned on the outer side of xylem.

How many of the above statements are correct?
1. Two statements are correct
2. Three statements are correct
3. All four statements are correct
4. Only one statement is correct

14. Which of the following accurately describes the significance of the bacterial glycocalyx?

1.	The glycocalyx is a rigid structure that provides protection against osmotic pressure and is found exclusively in Gram-negative bacteria.
2.	It serves as a storage compartment for nutrients and metabolic waste, aiding in bacterial metabolism under nutrient-limited conditions.
3.	The glycocalyx forms a capsule or slime layer that aids in attachment to surfaces, protection against desiccation, and resistance to phagocytosis.
4.	It houses the electron transport chain components, crucial for ATP production in aerobic bacterial cells.

15. Chloroplasts have their own DNA and ribosomes. Which of the following statements best explains the evolutionary advantage of these structures being retained within chloroplasts?

1.	The DNA and ribosomes in chloroplasts allow them to replicate independently of the cell, supporting rapid energy production during photosynthesis.
2.	These components enable chloroplasts to produce specific proteins locally, especially those essential for photosynthesis, reducing dependence on nuclear DNA and allowing efficient response to light changes.
3.	The DNA in chloroplasts encodes all the genes needed for cellular respiration, allowing chloroplasts to maintain energy production independently of mitochondria.
4.	The presence of DNA and ribosomes prevents chloroplasts from being degraded by the cell during periods of low light availability.

16. Which of the following statements best describes the structural complexity and functional significance of the plant cell wall, particularly in relation to growth and environmental interactions?

1.	The plant cell wall is composed primarily of cellulose microfibrils embedded in a matrix of lignin, providing flexibility for cell expansion while strengthening cells against environmental stress.
2.	Hemicellulose and pectin form the bulk of the primary cell wall, binding tightly to cellulose fibres, which allows for rigidity but restricts the flexibility needed during cell growth.
3.	The primary cell wall contains cellulose, hemicellulose, and pectin, forming a dynamic and flexible structure that allows for cell expansion, while the middle lamella rich in pectin facilitates adhesion between adjacent cells.
4.	The secondary cell wall, found in all plant cells, is rich in suberin, which protects cells from pathogen invasion but limits water transport, a key feature in xylem tissue.

17. Why do lipids, whose molecular weights do not exceed 800 Da, come under the acid insoluble fraction, i.e., macromolecular fraction, even though they are not strictly macromolecules?

1.	Lipids are inherently large molecules that always belong to the macromolecular fraction.
2.	Lipids form vesicles from membrane fragments during tissue grinding, which are not water soluble and get separated along with the acid insoluble fraction.
3.	Lipids increase in molecular weight during the grinding process, making them part of the macromolecular fraction.
4.	Lipids dissolve in water and therefore remain in the acid soluble fraction.

18. Which of the following statements accurately describes a key difference in the process of cytokinesis in plant cells compared to animal cells?

1.	In plant cells, a contractile ring of actin filaments pinches the cell membrane to form two daughter cells, while in animal cells, vesicles fuse at the centre to build a new cell wall.
2.	In plant cells, the cell wall prevents the complete separation of daughter cells, resulting in cells that remain connected, whereas animal cells undergo complete separation.
3.	Cytokinesis in plant cells involves the formation of a cell plate that ultimately dissolves, while in animal cells, cytokinesis forms a stable new membrane boundary without requiring vesicle fusion.
4.	Plant cells form a cell plate during cytokinesis, where vesicles from the Golgi apparatus coalesce at the cell’s centre to create a new cell wall, whereas animal cells use a contractile ring that forms a cleavage furrow.

19. In cyclic photophosphorylation, electrons cycle back to Photosystem I instead of moving to NADP⁺. The outcome and importance of is that:

1.	It generates only ATP without producing NADPH, which is crucial when the Calvin cycle requires more ATP than NADPH for carbon fixation.
2.	It produces both ATP and NADPH, balancing the energy requirements of the Calvin cycle in high-light conditions.
3.	Cyclic photophosphorylation leads to the reduction of water molecules, providing an additional source of protons for the lumen.
4.	It produces a proton gradient without generating ATP, solely to prevent over-reduction of Photosystem II under low-light conditions.

20. C4 photosynthesis is an adaptation that allows certain plants to minimize photorespiration and improve efficiency in hot, dry environments. Which of the following best explains the biochemical and anatomical adaptations involved in C4 photosynthesis?

1.	C4 plants fix CO₂ directly in bundle sheath cells, creating a high concentration of CO₂ around Rubisco, which reduces the oxygenation reaction in the Calvin cycle.
2.	PEP carboxylase in C4 plants fixes CO₂ in the bundle sheath cells into a three-carbon compound, which then enters the Calvin cycle directly, avoiding the mesophyll cells.
3.	In C4 plants, both the Calvin cycle and initial CO₂ fixation occur in the mesophyll cells, allowing rapid regeneration of CO₂ through the breakdown of stored organic acids.
4.	In C4 plants, CO₂ is initially fixed into a four-carbon compound in mesophyll cells by PEP carboxylase, which has a higher affinity for CO₂ than Rubisco, then transported to bundle sheath cells for release and subsequent fixation by Rubisco in the Calvin cycle.

21. In Engelmann's experiment, which used a filamentous green alga and aerobic bacteria to study the action spectrum of photosynthesis, what was the primary conclusion about the wavelengths of light and their effectiveness for photosynthesis?

1.	Engelmann observed that aerobic bacteria clustered in regions exposed to green light, demonstrating that green light is least effective for photosynthesis.
2.	The experiment showed that aerobic bacteria concentrated around areas with red and blue light, indicating these wavelengths are most effective in driving photosynthesis.
3.	Engelmann concluded that all wavelengths of light are equally effective in photosynthesis, as bacteria were evenly distributed along the spectrum.
4.	The experiment demonstrated that only red light drives photosynthesis effectively, as bacteria showed no movement towards other wavelengths.

22. During aerobic cellular respiration, ATP production varies depending on the entry point of intermediates into the pathway. Suppose 2 molecules of NADH are produced in the cytoplasm during glycolysis, while 8 molecules of NADH, 2 molecules of FADH₂, and 2 molecules of GTP are generated in the mitochondria during the citric acid cycle. Assuming the mitochondrial NADH produces 2.5 ATP, FADH₂ produces 1.5 ATP, and cytoplasmic NADH produces 1.5 ATP due to the shuttle system, what is the total ATP yield from these intermediates?
1. 38 ATP
2. 30 ATP
3. 28 ATP
4. 24 ATP

23. Consider the given two statements:

Assertion (A):	Fermentation is a crucial process for cells in anaerobic conditions, allowing them to produce ATP without using oxygen.
Reason (R):	During fermentation, cells regenerate NAD⁺ by reducing pyruvate, which is essential to sustain glycolysis and continue producing ATP in the absence of oxygen.

Select the correct option:

1.	Both (A) and (R) are True and (R) is the correct explanation of (A).
2.	Both (A) and (R) are True but (R) is not the correct explanation of (A).
3.	(A) is True but (R) is False.
4.	(A) is False but (R) is True.

24. Match each item in Column I with one item in Column II and select the best match from the codes given:

	Column I		Column II
A.	Complex I	P.	Cytochrome c oxidase
B.	Complex II	Q.	NADH dehydrogenase
C.	Complex III	R.	Succinate dehydrogenase
D.	Complex IV	S.	Cytochrome bc₁ complex

Options	A	B	C	D
1.	Q	R	S	P
2.	R	Q	P	S
3.	S	P	Q	R
4.	P	S	R	Q

25. Match each term in Column I with the correct example in Column II.

	Column I		Column II
A.	Differentiation	P.	Formation of vascular cambium from parenchyma cells in dicot stems
B.	Dedifferentiation	Q.	Formation of parenchyma from meristem
C.	Redifferentiation	R.	Formation of secondary xylem and phloem from cambium cells

Codes:

Options	A	B	C
1.	P	R	Q
2.	R	Q	P
3.	Q	P	R
4.	Q	R	P

26. Consider the given two statements:

Assertion (A):	Phenotypic plasticity is generally more important in plants than in animals.
Reason (R):	Plants cell are surrounded by a rigid cell wall that is lacking in an animal cell making change in cell shape easier in animal cell.

Select the correct option:

1.	Both (A) and (R) are True and (R) is the correct explanation of (A).
2.	Both (A) and (R) are True but (R) is not the correct explanation of (A).
3.	(A) is True but (R) is False.
4.	(A) is False but (R) is True.

27. Consider the following statements about the applications of plant growth regulators in horticulture:

A.	Gibberellins are used on grapevines to increase berry size and elongate clusters, leading to higher-quality fruit presentation and increased yield.
B.	Ethylene is applied to fruits like tomatoes and bananas to accelerate ripening, which helps in coordinating the timing of harvest and enhancing market readiness.
C.	Cytokinins are sprayed on mature fruit trees to induce early flowering, helping to synchronize crop production and reduce the growing period.
D.	Auxins are commonly applied to promote lateral root growth in saplings and seedlings, enhancing structural stability but delaying flowering onset.

Select the correct option:

1.	Only Statement A and Statement B are correct.
2.	Only Statement A and Statement D are correct.
3.	Only Statement B and Statement C are correct.
4.	Only Statement C and Statement D are correct.

28. In some flowering plants, fruits are produced without fertilization in a process known as parthenocarpy. Which of the following is a potential evolutionary advantage of parthenocarpy?

1.	Parthenocarpy ensures genetic diversity in offspring by allowing the fruit to develop without fertilization.
2.	It enables fruit production under conditions where pollinators are scarce or unreliable, increasing reproductive success.
3.	Parthenocarpy produces seeds that are more resistant to environmental stress than those formed through fertilization.
4.	It reduces energy expenditure by eliminating the need for flowering structures, conserving resources.

29. Which of the following statements accurately describes the process of double fertilization in angiosperms and its significance?

1.	Double fertilization results in two identical zygotes, providing redundancy and ensuring seed development even if one fails.
2.	In double fertilization, both sperm cells fuse with the egg cell, enhancing genetic diversity in the resulting seed.
3.	The two fertilization events form twin embryos that develop independently within the same seed coat.
4.	One sperm fertilizes the egg cell to form a diploid zygote, while the other fuses with two polar nuclei to form a triploid endosperm, which nourishes the developing embryo.

30. In which scenario would a high degree of self-incompatibility provide a significant reproductive advantage to a flowering plant population?

1.	In a dense population with many plants of the same species, where cross-pollination can easily occur to increase genetic diversity.
2.	In a remote location with a limited number of the same species, ensuring that at least some seeds are produced through self-pollination.
3.	In a controlled agricultural setting where high genetic uniformity is desired to maintain crop consistency.
4.	In an environment with no pollinators, to ensure that self-pollination compensates for the lack of cross-pollination.

31. In an experiment, you observe that a certain flowering plant produces seeds in the absence of fertilization. Which of the following mechanisms is most likely responsible for this observation, and what is its primary benefit to the plant?

1.	Apomixis, which allows the plant to produce genetically diverse seeds without pollination, increasing adaptability to changing environments.
2.	Parthenocarpy, which promotes seed production by allowing the ovules to develop autonomously, ensuring reproduction in pollinator-scarce areas.
3.	Apomixis, which produces seeds genetically identical to the parent plant, ensuring successful reproduction in stable environments without genetic recombination.
4.	Polyploidy, which doubles the chromosome number in seeds, enhancing the plant’s genetic diversity without the need for fertilization.

32. In a flowering plant, a mutation causes one of the synergids to lose its ability to produce attractant chemicals. What is the likely impact of this mutation on fertilization?

1.	The egg cell would still be fertilized, as synergids are not essential for guiding the pollen tube.
2.	Fertilization would likely be disrupted, as synergids release attractants that guide the pollen tube to the egg.
3.	The pollen tube would be unable to penetrate the embryo sac, but fertilization could proceed once the tube reaches the ovule.
4.	Multiple pollen tubes would be attracted to the same ovule, leading to polyploidy in the zygote.

33. In a dihybrid cross (RrYy × RrYy), what is the probability that an offspring will have a dominant phenotype for one trait and a recessive phenotype for the other?
1. 4/16
2. 9/16
3. 6/16
4. 8/16

34. In a genetic experiment, two different types of flower colour inheritance are observed in plants:

I.	In Plant A, crossing a red-flowered plant (RR) with a white-flowered plant (WW) produces all pink-flowered offspring (RW).
II.	In Plant B, crossing a black-coated cow (BB) with a white-coated cow (WW) produces offspring with both black and white patches (BW).

Which of the following best describes the difference in inheritance patterns between Plant A and Plant B, and what conclusion can be drawn?

1.	Both Plant A and Plant B show co-dominance, as each set of parents produces an intermediate phenotype in the offspring.
2.	Plant A shows codominance, as both colours contribute equally to create a new phenotype, while Plant B shows incomplete dominance with mixed patches.
3.	Both Plant A and Plant B show incomplete dominance, as each set of parents produces an intermediate phenotype in the offspring.
4.	Plant A shows incomplete dominance, as the offspring show a blended phenotype, while Plant B shows codominance, where both parental traits are fully expressed in the offspring.

35. If a mutation caused DNA to pair A with G and C with T, what effect would this have on the structure of the DNA double helix?

1.	The DNA helix would widen at certain points, disrupting the uniform structure necessary for replication and stability.
2.	The DNA structure would remain unchanged, as all nucleotide pairs have the same spacing.
3.	The new pairings would enhance genetic diversity without affecting structural stability.
4.	The helix would shrink in size, as A-G and C-T pairings would bring the strands closer together.

36. Consider the following statements about types of RNA molecules and their functions:

I.	Messenger RNA (mRNA) carries the genetic code from DNA to the ribosomes, where it is translated into a protein.
II.	Transfer RNA (tRNA) helps in assembling amino acids into a polypeptide chain by carrying them to the ribosome and matching them to the codons on mRNA.
III.	Ribosomal RNA (rRNA) is a structural and catalytic component of ribosomes.

Select the correct option:
1. Only Statement I and Statement II are correct.
2. Only Statement I and Statement III are correct.
3. Only Statement II and Statement III are correct.
4. All statements are correct.

37. In a bacterial cell, a mutation causes the lacI gene (which encodes the lac repressor protein) to be nonfunctional, resulting in no repressor protein production. If this mutant cell is grown in a medium without lactose, what would be the expected behaviour of the lac operon genes (lacZ, lacY, and lacA), and why?

1.	The lac operon genes would be expressed at high levels even in the absence of lactose, because without the repressor, nothing prevents RNA polymerase from binding to the promoter.
2.	The lac operon genes would remain off, as the lack of lactose prevents RNA polymerase from binding to the promoter.
3.	Expression of the lac operon genes would fluctuate, as the cell needs the repressor to regulate the operon even in the absence of lactose.
4.	The lac operon genes would be expressed at low levels, as RNA polymerase only partially binds to the promoter without lactose.

38. In the process of sewage treatment, which phase involves the formation of activated sludge and how does this contribute to the purification of wastewater?

1.	The primary treatment phase, where large solids settle at the bottom and form activated sludge.
2.	The secondary treatment phase, where aerobic bacteria grow in the aeration tank and form activated sludge, which removes organic matter from wastewater.
3.	The tertiary treatment phase, where chemicals are added to form activated sludge, helping to neutralize pathogenic microbes.
4.	The anaerobic digestion phase, where methanogens form activated sludge, which releases methane as a byproduct.

39. Match each microbe in Column I with the product it produces in Column II.

	Column I		Column II
A.	Monascus purpureus	P.	Citric acid
B.	Trichoderma polysporum	Q.	Immunosuppressive cyclosporin A
C.	Streptococcus pyogenes	R.	Streptokinase
D.	Aspergillus niger	S.	Statins

Options	A	B	C	D
1.	Q	P	R	S
2.	P	R	S	Q
3.	S	Q	R	P
4.	R	Q	P	S

40. Consider the given two statements:

Assertion (A):	RNA interference (RNAi) is used to produce GM tobacco plants resistant to Meloidogyne nematodes by introducing single-stranded DNA that silences essential nematode genes.
Reason (R):	The single-stranded DNA introduced into the tobacco plant directly disrupts the nematode’s DNA, preventing it from infecting the plant roots.

Select the correct option:

1.	Both (A) and (R) are True and (R) is the correct explanation of (A).
2.	Both (A) and (R) are True but (R) is not the correct explanation of (A).
3.	(A) is False but (R) is True.
4.	Both (A) and (R) are False.

41. Bt cotton is genetically engineered to produce toxins that provide resistance to specific pests. Which of the following statements best describes the mechanism and implications of Bt cotton’s pest resistance and addresses one of its primary challenges?

1.	Bt cotton produces the Cry protein, which binds to the gut cells of all insects, forming pores that lead to cell lysis and the insect’s death, offering broad-spectrum pest resistance.
2.	The Cry proteins in Bt cotton target specific pests, such as the cotton bollworm, by disrupting their digestive systems; however, continuous exposure has led to resistance development in some pest populations.
3.	Bt cotton releases the Cry toxin, which protects the plant from all soil pathogens by inhibiting bacterial and fungal growth, making it suitable for diverse environmental conditions.
4.	Bt cotton expresses the cry gene only in the presence of pests, making the toxin harmless to non-target organisms and beneficial insects, preventing any resistance development in pests.

42. Consider the given two statements:

Assertion (A):	In a mutualistic symbiotic relationship, plants provide nectar as a food source to insect pollinators, and in return, the insects help plants with seed dispersal.
Reason (R):	The relationship between plants and insect pollinators is crucial for maintaining genetic diversity in plant populations through cross-pollination.

Select the correct option:

1.	Both (A) and (R) are True and (R) is the correct explanation of (A).
2.	Both (A) and (R) are True but (R) is not the correct explanation of (A).
3.	(A) is True but (R) is False.
4.	(A) is False but (R) is True.

43. What is the primary reason for inverted biomass pyramid in aquatic ecosystems?

1.	Primary producers in aquatic ecosystems, like phytoplankton, have a high turnover rate, producing biomass rapidly but maintaining a relatively small standing crop compared to higher trophic levels.
2.	Aquatic ecosystems have lower energy transfer efficiency between trophic levels, resulting in larger biomass at higher trophic levels.
3.	Herbivores in aquatic ecosystems consume primary producers less efficiently, leading to a larger biomass of herbivores than primary producers.
4.	Nutrient cycling in aquatic ecosystems is slower, allowing herbivores to accumulate more biomass over time.

44. Which of the following conditions is most likely to slow down decomposition rates in a forest floor environment?

1.	High moisture levels that increase microbial activity, leading to faster breakdown of organic matter.
2.	Lower temperatures, which decrease enzyme activity, slowing down the metabolic processes of decomposers.
3.	High nitrogen content in organic matter, which accelerates decomposer activity and the rate of organic matter breakdown.
4.	Presence of lignin and cellulose, which provide easily digestible compounds that decomposers can quickly break down.

45. Which of the following strategies aligns most closely with in-situ conservation and its benefits for preserving biodiversity?

1.	Establishing a seed bank to preserve genetic material of endangered plant species from different regions.
2.	Developing wildlife sanctuaries and national parks to protect species within their natural ecosystems and maintain ecological interactions.
3.	Conducting captive breeding programs in controlled environments for critically endangered species.
4.	Setting up gene banks and tissue culture facilities for endangered flora to enable reintroduction into the wild.

46. The importance of biodiversity hotspots for global conservation efforts is best explained by:

1.	Biodiversity hotspots contain a high level of endemic species and the species face a high threat of extinction, making them critical for conservation.
2.	Biodiversity hotspots are regions with high human population density and minimal natural resources, identified as low priority for conservation efforts.
3.	Biodiversity hotspots represent areas with the highest number of plant species globally, regardless of habitat loss or species endemism.
4.	Biodiversity hotspots are areas identified by the UN for reforestation projects, as they have completely lost their natural vegetation.

47. Cyanobacteria, such as Anabaena and Nostoc, are commonly used in paddy fields as biofertilizers. Which of the following statements best describes the role of cyanobacteria in enhancing soil fertility and crop yield in paddy fields?

1.	Cyanobacteria perform photosynthesis to release oxygen into the soil, which increases soil aeration and enhances root growth in rice plants.
2.	They fix atmospheric nitrogen through specialized cells called heterocysts, enriching the soil with bioavailable nitrogen and reducing the need for synthetic fertilizers.
3.	Cyanobacteria release phosphorus into the soil, which is crucial for rice plant growth, especially in waterlogged conditions.
4.	They decompose organic matter in the soil, producing humus that enhances the soil structure and nutrient availability.

48. How do mycorrhizal associations benefit plants in nutrient-poor soils, and what type of symbiotic relationship do they exhibit?

1.	Mycorrhizae provide nitrogen directly to plants, and in return, the plant supplies carbon, forming a parasitic relationship.
2.	Mycorrhizae decompose organic matter to release nitrogen, which the plant then absorbs, forming a commensal relationship.
3.	Mycorrhizae fix carbon in the soil, improving plant growth without requiring anything in return from the plant.
4.	Mycorrhizae increase the plant's absorption of water and essential nutrients like phosphorus, forming a mutualistic relationship.

49. Seed dormancy is an important evolutionary adaptation in many flowering plants. Which of the following best describes the role of seed dormancy in plant survival and evolution?

1.	Seed dormancy allows seeds to germinate only under favourable conditions, improving survival rates and reducing competition among seedlings.
2.	Dormant seeds immediately start growing, ensuring rapid coverage of the area and dominance over other plant species.
3.	Seed dormancy eliminates the need for flowering, conserving energy and increasing reproductive efficiency.
4.	Dormant seeds are highly resistant to predation, ensuring that all seeds germinate regardless of environmental conditions.

50. Which of the following features is most commonly found in wind-pollinated plants, and how does it enhance the efficiency of wind pollination?

1.	Brightly coloured flowers that attract wind currents and help direct pollen to other flowers more effectively.
2.	Sticky and large pollen grains that easily adhere to wind currents, ensuring pollen reaches other flowers at a higher rate.
3.	Long and feathery stigmas that increase the surface area for catching pollen grains carried by the wind, enhancing the chances of successful pollination.
4.	Nectar-rich flowers that attract pollinators, indirectly assisting in the wind dispersal of pollen.

51. In the context of body cavity (coelom), which of the following pairs is correctly matched?
1. Annelids – Acoelomates
2. Aschelminthes – Pseudocoelomates
3. Platyhelminthes – Coelomates
4. Echinoderms – Pseudocoelomates

52. What is metamerism in animals?

1.	Metamerism is a body arrangement seen only in aquatic animals, where the body is divided into repeated segments with identical organs.
2.	Metamerism refers to the formation of a notochord along the body axis, providing structural support in segmented animals.
3.	Metamerism is characteristic of animals with an incomplete digestive system, where segments develop based on nutrient availability.
4.	In metamerism, the body is segmented both externally and internally, with a serial repetition of some organs, as seen in animals like earthworms.

53. Match each animal group in Column I with its characteristic feature in Column II.

Column I	Column II
A. Sponges	P. Water vascular system
B. Echinoderms	Q. Radula
C. Mollusks	R. Canal system
D. Flatworms	S. Flame cells

Options	A	B	C	D
1.	R	P	Q	S
2.	S	Q	R	P
3.	P	S	R	Q
4.	Q	R	S	P

54. The organism shown in the image has a circular, toothed mouth adapted for attaching to host fish and feeding on their body fluids. Which of the following characteristics is NOT associated with this organism?

1.	It belongs to the class Agnatha and lacks true jaws.
2.	It has a cartilaginous skeleton and no paired fins.
3.	It exhibits parasitic feeding habits by attaching to host fish.
4.	It has an operculum covering its gills, similar to most bony fishes.

55. Match each type of epithelial tissue in Column I with its location in Column II.

	Column I		Column II
A.	Columnar brush-bordered epithelium	P.	Proximal convoluted tubule
B.	Cuboidal ciliated epithelium	Q.	Small intestine
C.	Cuboidal brush-bordered epithelium	R.	Fallopian tubes
D.	Columnar ciliated epithelium	S.	Respiratory bronchioles

Options	A	B	C	D
1.	Q	R	P	S
2.	P	S	Q	R
3.	R	P	S	Q
4.	Q	S	P	R

56. Consider the following statements regarding respiration in frogs:

1.	Frogs utilize cutaneous respiration in water, where oxygen is absorbed through the skin by diffusion.
2.	On land, frogs rely exclusively on their lungs for respiration, with air entering through the nostrils and reaching the lungs directly.
3.	During periods of aestivation and hibernation, frogs exchange gases primarily through the skin.

Select the correct option:
1. Only Statement 1 and Statement 2 are correct.
2. Only Statement 1 and Statement 3 are correct.
3. Only Statement 2 and Statement 3 are correct.
4. All statements are correct.

57. The accurate description of the structural organization and function of the eukaryotic flagellum will be:

1.	The basal body of the eukaryotic flagellum is identical to the axoneme and serves only as an anchor, without involvement in flagellar movement.
2.	The dynein arms in the eukaryotic flagellum are positioned to generate sliding between adjacent microtubule pairs, resulting in a rotating movement rather than an undulating movement as seen in prokaryotic flagella.
3.	The eukaryotic flagellum contains a 9 + 2 arrangement of microtubules, where nine pairs of microtubules surround a central pair, allowing for coordinated bending movements powered by ATP.
4.	Eukaryotic flagella are structurally identical to prokaryotic flagella and can only perform small, rhythmic motions without large-scale undulations.

58. What functional characteristic of the plasma membrane arises due to the unique properties of its phospholipid bilayer?

1.	The plasma membrane allows free passage of all molecules due to the hydrophilic phosphate heads facing the extracellular environment.
2.	The fluid mosaic model suggests that membrane proteins can move freely within the bilayer, providing flexibility, though lipid mobility is restricted to maintain stability.
3.	Due to its semi-permeable nature, the plasma membrane allows hydrophobic molecules to pass through easily, while hydrophilic molecules require specific transport proteins.
4.	Cholesterol molecules in the plasma membrane reduce fluidity at high temperatures, making the membrane more rigid and impermeable to all types of solutes.

59. Consider the given two statements:

Assertion (A):	In competitive enzyme inhibition, the inhibitor binds to the enzyme-substrate complex, reducing the enzyme's affinity for its substrate and thereby lowering the reaction rate.
Reason (R):	Competitive inhibitors have a structure similar to the substrate, allowing them to compete for the enzyme's active site, and their effect can be overcome by increasing substrate concentration.

Select the correct option:

1.	Both (A) and (R) are True and (R) is the correct explanation of (A).
2.	Both (A) and (R) are True but (R) is not the correct explanation of (A).
3.	(A) is True but (R) is False.
4.	(A) is False but (R) is True.

60. Which of the following statements accurately describes the primary structure of a protein and its importance in protein function?

1.	The primary structure is the unique sequence of amino acids in a protein, which determines its final three-dimensional shape and functional properties.
2.	The primary structure refers to the alpha helices and beta sheets formed by hydrogen bonding between the backbone atoms of the amino acids.
3.	The primary structure is stabilized by hydrophobic interactions between non-polar amino acid side chains, which are critical for maintaining protein solubility.
4.	The primary structure is disrupted in denaturation, but the protein can refold correctly once normal conditions are restored, as only weak bonds are broken.

61. The cell shown in the given figure is at mitotic:

1. Prometaphase
2. Metaphase
3. Anaphase
4. Telophase

62. Consider the following statements regarding the significance of mitosis:

Statement 1:	Mitosis is crucial for growth and development in multicellular organisms, as it produces genetically identical cells that contribute to tissue formation and repair.
Statement 2:	Mitosis introduces genetic variability within the cell population of an organism, which is essential for adapting to environmental changes.

Which of the following is correct?

1.	Both statements are True and statement 2 is the correct explanation of statement 1.
2.	Both statements are True but statement 2 is not the correct explanation of statement 1.
3.	Statement 1 is true but statement 2 is False.
4.	Both statements are false.

63. During the process of inspiration in humans, which of the following events occurs, and how does it facilitate air entry into the lungs?

1.	The diaphragm relaxes and moves upward, increasing the volume of the thoracic cavity and creating negative pressure that draws air in.
2.	The external intercostal muscles contract, expanding the rib cage outward, which decreases the intrapulmonary pressure, allowing air to flow into the lungs.
3.	The alveolar pressure becomes higher than atmospheric pressure as the thoracic cavity expands, forcing air into the lungs.
4.	The internal intercostal muscles contract, raising the volume of the thoracic cavity and facilitating air entry.

64. Which of the following factors would most likely increase the release of oxygen from haemoglobin to tissues, and why?

1.	Decrease in temperature and increase in blood pH, which shift the oxygen-haemoglobin dissociation curve to the left, favouring oxygen release at the tissues.
2.	Increase in partial pressure of carbon dioxide (pCO₂) and decrease in blood pH, which reduce haemoglobin’s affinity for oxygen, facilitating oxygen release.
3.	High oxygen levels at the tissue site and increase in blood pH, which enhance haemoglobin’s oxygen affinity, allowing more oxygen to be released to the tissues.
4.	Decrease in carbon dioxide levels and increase in temperature, which shift the dissociation curve to the right, reducing oxygen release.

65. A patient with blood type B- is in need of an urgent blood transfusion. Which of the following donor blood types can safely be given to this patient, and what precautions must be considered in this transfusion?

1.	AB- and B+, as both have B antigens, which the patient’s immune system will recognize as compatible.
2.	A- and O+, because the lack of Rh factor compatibility is not essential in emergency transfusions.
3.	AB+ and B-, because both have B antigens and the presence of Rh antigen in AB+ will not affect the patient negatively.
4.	O- and B-, as O- lacks both A and B antigens and Rh factor, and B- matches the patient’s blood type.

66. During the cardiac cycle, which of the following accurately describes the sequence of events that occur during ventricular systole?

1.	The atrioventricular (AV) valves open, allowing blood to enter the ventricles, while the semilunar valves remain closed to maintain blood pressure in the ventricles.
2.	The semilunar valves close to prevent backflow from the arteries, while the ventricles relax, allowing blood to flow in from the atria.
3.	The atria contract, increasing pressure on the AV valves, forcing them closed and initiating ventricular filling.
4.	The ventricles contract, closing the AV valves to prevent backflow into the atria, and opening the semilunar valves, allowing blood to be ejected into the aorta and pulmonary artery.

67. Which of the following accurately describes the role of checkpoints in the regulation of the cell cycle, and what would be the potential consequence of a malfunction in the G1 checkpoint?

1.	The G1 checkpoint ensures that DNA replication is completed before the cell enters mitosis; a malfunction here would lead to the cell dividing without adequate DNA, potentially causing cancerous growth.
2.	The G2 checkpoint ensures the cell has grown enough to sustain two daughter cells; failure here could lead to incomplete DNA synthesis, resulting in mutations passed on to daughter cells.
3.	The G1 checkpoint verifies that the cell has adequate resources and no DNA damage before proceeding to the S phase; a malfunction in this checkpoint could result in cells with damaged DNA entering replication, increasing the risk of uncontrolled cell division.
4.	The metaphase checkpoint prevents cells with chromosome misalignment from progressing to anaphase; malfunction here would allow cells with an incorrect chromosome number to continue dividing, leading to immediate cell death.

68. Which of the following best explains the difference between heart failure and a heart attack?

1.	Heart failure is the sudden blockage of blood flow to the heart muscle, often due to a clot, whereas a heart attack is the inability of the heart to pump blood efficiently, usually due to weakened cardiac muscle.
2.	Heart attack results from the narrowing of arteries, reducing blood flow, while heart failure refers to a complete cessation of blood flow in the coronary arteries.
3.	Heart failure is a chronic condition where the heart cannot pump blood effectively to meet the body’s needs, while a heart attack is an acute event caused by blocked coronary arteries, resulting in damage to heart muscle.
4.	Heart failure occurs due to the rupture of a blood vessel in the heart, whereas a heart attack occurs due to valve malfunction, restricting blood flow.

69. Consider the following statements regarding the absorption capabilities of the two limbs of the Loop of Henle:

Statement 1:	The descending limb of the Loop of Henle is permeable to water but relatively impermeable to solutes, which allows water to be reabsorbed and the filtrate to become increasingly concentrated as it moves down the limb.
Statement 2:	The ascending limb of the Loop of Henle is impermeable to water but actively transports sodium and chloride ions out, causing the filtrate to become diluted as it ascends.

Select the correct option:
1. Both Statement 1 and Statement 2 are correct.
2. Only Statement 1 is correct.
3. Only Statement 2 is correct.
4. Both Statement 1 and Statement 2 are incorrect.

70. Which of the following statements best describes the antagonistic roles of Angiotensin II and Atrial Natriuretic Factor (ANF) in the regulation of blood pressure and fluid balance?

1.	Angiotensin II constricts blood vessels and promotes sodium and water retention, increasing blood pressure, while ANF dilates blood vessels and enhances sodium and water excretion, reducing blood pressure.
2.	Angiotensin II acts by dilating blood vessels and decreasing blood pressure, while ANF causes vasoconstriction and water retention, leading to increased blood pressure.
3.	Angiotensin II increases blood flow to the kidneys, promoting sodium excretion, whereas ANF acts by retaining sodium and increasing blood volume.
4.	Angiotensin II and ANF both function to decrease blood volume but differ in their effects on blood vessel diameter, with ANF promoting vasodilation and Angiotensin II promoting vasoconstriction.

71. Arrange the following events in the correct chronological order for skeletal muscle contraction:

1.	Release of calcium ions from the sarcoplasmic reticulum.
2.	Binding of ATP to the myosin head, causing it to detach from actin.
3.	Hydrolysis of ATP, re-cocking the myosin head into a high-energy state.
4.	Depolarization of the sarcolemma triggered by an action potential. 5. Cross-bridge formation between actin and myosin. 6. Power stroke movement of the myosin head, pulling the actin filament.

Select the correct option:
1. 1 → 4 → 6 → 2 → 3 → 5
2. 4 → 1 → 5 → 2 → 6 → 3
3. 4 → 5 → 1 → 6 → 3 → 2
4. 4 → 1 → 5 → 6 → 2 → 3

72. What type of synovial joint is the carpometacarpal joint shown in the shown in the given figure:

1. Hinge
2. Saddle
3. Gliding
4. Condyloid

73. During the depolarization phase of an action potential in a neuron, which of the following events occurs, and how does it facilitate the propagation of the action potential along the axon?

1.	Sodium (Na⁺) channels open, allowing Na⁺ ions to rush out of the axon, making the inside of the membrane more negative and propagating the action potential.
2.	Potassium (K⁺) channels close, trapping K⁺ ions inside the axon and causing the membrane to reach the threshold for depolarization.
3.	Sodium (Na⁺) channels open, allowing Na⁺ ions to rush into the axon, reversing the membrane polarity and creating a positive charge inside the axon.
4.	Chloride (Cl⁻) channels open, allowing Cl⁻ ions to enter the axon and depolarize the membrane by increasing the positive charge inside the neuron.

74. Which of the following is NOT a function of the specified part of the brain?

1.	Cerebellum – Coordinates voluntary movements and maintains balance and posture.
2.	Medulla oblongata – Regulates heart rate, breathing, and other autonomic functions.
3.	Hypothalamus – Responsible for controlling language and speech processing.
4.	Frontal lobe – Involved in decision-making, problem-solving, and personality expression.

75. Which of the following best describes the mechanism of action of steroid hormones in target cells?

1.	Steroid hormones bind to receptors on the cell membrane, activating second messenger systems that stimulate gene transcription directly in the nucleus.
2.	Steroid hormones bind to specific intracellular receptors in the cytoplasm or nucleus, forming a hormone-receptor complex that directly interacts with DNA to regulate gene expression.
3.	Steroid hormones activate adenylate cyclase on the cell membrane, resulting in increased cAMP levels and subsequent protein phosphorylation.
4.	Steroid hormones interact with G-protein coupled receptors on the cell surface, initiating a signal transduction pathway that leads to ion channel modulation

76. Identify the correct statement regarding the antagonistic roles of insulin and glucagon in regulating blood glucose levels:

1.	Insulin promotes gluconeogenesis in the liver, increasing blood glucose levels, while glucagon promotes glycogenesis, reducing blood glucose levels.
2.	Insulin is released in response to low blood glucose levels, while glucagon is released in response to high blood glucose levels to maintain balance.
3.	Both insulin and glucagon function by promoting glycogenesis, but insulin works at a faster rate to regulate blood glucose.
4.	Insulin facilitates the uptake of glucose into cells and stimulates glycogenesis, lowering blood glucose levels, whereas glucagon stimulates glycogenolysis and gluconeogenesis in the liver, increasing blood glucose levels.

77. Match each hormonal disorder in Column I with the abnormality in hormone secretion in Column II.

	Column I		Column II
A.	Acromegaly	P.	Hyposecretion of ADH
B.	Diabetes insipidus	Q.	Hypersecretion of GH
C.	Graves' disease	R.	Hyposecretion of cortisol
D.	Addison's disease	S.	Hypersecretion of thyroid hormones

Options	A	B	C	D
1.	Q	P	S	R
2.	P	Q	R	S
3.	S	P	Q	R
4.	R	S	Q	P

78. If the seminal vesicles failed to secrete their fluid, what would be the most likely consequence on male fertility?

1.	Decreased motility of sperm due to lack of fructose
2.	Impaired capacitation of sperm in the female reproductive tract
3.	Failure of spermatogenesis due to lack of hormonal support
4.	Disruption in sperm transport from testes to urethra

79. Consider the following statements regarding hormonal regulation in the menstrual cycle:

Statement 1:	Follicle-stimulating hormone (FSH) promotes the growth of ovarian follicles, and as these follicles mature, they produce increasing levels of oestrogen.
Statement 2:	A sharp increase in luteinizing hormone (LH) triggers ovulation, after which the ruptured follicle transforms into the corpus luteum, which secretes both progesterone and oestrogen.
Statement 3:	High levels of progesterone during the luteal phase provide positive feedback on the hypothalamus, maintaining high levels of GnRH and preventing the next cycle from starting.

Select the correct option:
1. Only Statement 1 and Statement 2 are correct.
2. Only Statement 2 and Statement 3 are correct.
3. Only Statement 3 is correct.
4. All statements are correct.

80. Which of the following best describes a key difference between oogenesis and spermatogenesis in humans?

1.	Oogenesis involves one meiotic division, while spermatogenesis requires two complete meiotic divisions.
2.	Oogenesis results in a single viable gamete and polar bodies due to asymmetric cytoplasmic division, while spermatogenesis produces four viable gametes.
3.	Oogenesis occurs continuously throughout life, whereas spermatogenesis halts at puberty.
4.	In oogenesis, meiosis results in four haploid cells of equal size, while in spermatogenesis, only two of the resulting cells are functional.

81. Consider the given two statements:

Assertion (A):	The population growth rate in India has been significantly reduced through the Reproductive and Child Health Care (RCH) Programmes.
Reason (R):	The RCH Programmes have led to a rapid decline in death rate, maternal mortality rate, and infant mortality rate.

Select the correct option:

1.	Both (A) and (R) are True and (R) is the correct explanation of (A).
2.	Both (A) and (R) are True but (R) is not the correct explanation of (A).
3.	(A) is True but (R) is False.
4.	(A) is False but (R) is True.

82. If the distance between two genes on the same chromosome is 70 map units, what would be the approximate recombination frequency between them?
1. 35%
2. 50%
3. 70%
4. 100%

83. Which of the following correctly describe autosomal recessive inheritance?

1.	Affected individuals are more common in the offspring of consanguineous marriages.
2.	Heterozygous carriers are usually asymptomatic.
3.	Two affected parents cannot have an unaffected child.
4.	The trait appears more frequently in females than males.

Select the correct answer from the options below:
1. 1, 2, 3
2. 1, 2
3. 2, 3, 4
4. 1, 3

84. Which of the following best differentiates thalassemia from sickle cell anaemia?

1.	Thalassemia involves abnormal haemoglobin shape, while sickle cell anaemia results from reduced haemoglobin production.
2.	Sickle cell anaemia produces abnormally shaped red blood cells, while thalassemia reduces the quantity of functional haemoglobin.
3.	Thalassemia affects oxygen-carrying capacity due to misshapen red blood cells, while sickle cell anaemia impacts red blood cell production.
4.	Both conditions result solely from abnormal haemoglobin production, without any shape alterations in red blood cells.

85. Why can male honeybees have grandsons but not sons?

1.	Male honeybees are haploid and do not produce male offspring through their sperm.
2.	Male honeybees are sterile and cannot produce any offspring.
3.	Male honeybees produce only female offspring because females develop from fertilized egg.
4.	Male honeybees are diploid but only pass genetic material to grandsons indirectly.

86. How would the results of Avery, MacLeod, and McCarty’s experiment have changed if DNase (an enzyme that digests DNA) had been used during their experiment on the transforming principle?

1.	The transformation from R strain to S strain would have still occurred because protein was responsible for transformation.
2.	The transformation would have been unaffected because RNA would have taken over the role of DNA.
3.	The transformation from R strain to S strain would not have occurred because DNA would have been degraded by DNase, proving that DNA is the genetic material.
4.	The transformation from R strain to S strain would still have occurred, indicating that proteins are the genetic material.

87. Which of the following properties make DNA a suitable molecule for the storage of genetic information?

1.	DNA is more stable than RNA due to the absence of the 2'-OH group.
2.	DNA can mutate more easily than RNA, allowing rapid evolution.
3.	DNA has the ability to generate identical replicas through replication.
4.	DNA can directly code for protein synthesis without the need for RNA.

Select the correct answer from the options below:
1. 1, 2
2. 1, 3
3. 2, 4
4. 3, 4

88. Which of the following best describes the role of Variable Number Tandem Repeats (VNTRs) in RFLP analysis for genetic fingerprinting?

1.	VNTRs are regions of DNA that vary between individuals and are cut by restriction enzymes to produce fragment patterns unique to each individual, allowing for identification.
2.	VNTRs are single-copy DNA sequences that provide a uniform pattern in RFLP analysis across all individuals, making them ideal for determining ancestry.
3.	VNTRs require PCR amplification in RFLP analysis to produce a sufficient quantity of fragments for gel electrophoresis, ensuring accurate results.
4.	VNTRs are short sequences of base pairs repeated a fixed number of times, resulting in identical RFLP patterns for family members.

89. Which of the following accurately describes the environmental conditions on early Earth that contributed to the origin of life?

1.	Oxygen was abundant, which provided the necessary energy for the formation of complex organic molecules.
2.	The atmosphere was highly reductive, composed mainly of methane, ammonia, water vapor, and hydrogen, creating conditions favourable for the synthesis of organic compounds.
3.	High concentrations of nitrogen and carbon dioxide in a neutral atmosphere provided a stable environment that protected organic molecules from degradation.
4.	The Earth’s surface was cold and dry, allowing for the slow, stable formation of organic molecules over time.

90. Consider the given two statements:

Assertion (A):	The concept of "survival of the fittest" can sometimes be misleading in describing natural selection.
Reason (R):	"Fitness" in an evolutionary context refers solely to physical strength and dominance, traits that ensure survival in all environments.

Select the correct option:

1.	Both (A) and (R) are True and (R) is the correct explanation of (A).
2.	Both (A) and (R) are True but (R) is not the correct explanation of (A).
3.	(A) is True but (R) is False.
4.	(A) is False but (R) is True.

91. Consider the following statements regarding homologous, analogous, and vestigial organs in the context of evolution:

I.	Homologous organs indicate common ancestry, as they arise from similar structures but have diversified functions.
II.	Analogous organs result from convergent evolution, where unrelated species evolve similar structures due to similar environmental pressures.
III.	Vestigial organs are functionally essential in present-day organisms, showcasing evolutionary stability rather than change.

Select the correct option:

1.	Only Statement I and Statement II are correct.
2.	Only Statement III is correct.
3.	Only Statement II and Statement III are correct.
4.	All statements are correct.

92. Which of the following statements accurately describes the evolutionary significance of coelacanths?

1.	Coelacanths are an ancient species of fish that developed lungs, showing the direct evolutionary transition to terrestrial life.
2.	Coelacanths are closely related to modern reptile, showing both gills and primitive legs, which provide insight into reptilian evolution.
3.	Coelacanths lost the ability to use gills and rely solely on lungs, illustrating the adaptation to a fully terrestrial environment.
4.	Coelacanths represent a lineage that branched off early from other fish, with features like lobed fins that may resemble the ancestors of amphibians.

93. Match each disease in Column I with its primary mode of transmission in Column II.

	Column I		Column II
A.	AIDS	P.	Droplet transmission
B.	Dengue	Q.	Vector-borne transmission
C.	Pneumonia	R.	Faecal-oral transmission
D.	Typhoid	S.	Direct contact with infected body fluids

Options	A	B	C	D
1.	S	Q	P	R
2.	Q	S	R	P
3.	S	P	Q	R
4.	P	R	S	Q

94. Which of the following best describes a key difference between humoral and cell-mediated immunity?

1.	Humoral immunity relies on cytotoxic T cells for pathogen destruction, while cell-mediated immunity uses plasma cells to release antibodies.
2.	Both humoral and cell-mediated immunity involve antibody production by B cells but differ in their mechanisms of pathogen targeting.
3.	Cell-mediated immunity is the body’s first line of defence against infection, whereas humoral immunity only responds to viruses.
4.	Humoral immunity targets extracellular pathogens using antibodies, whereas cell-mediated immunity targets intracellular pathogens through cytotoxic T cells.

95. Arrange the following events in the correct sequence for the life cycle of the malarial parasite (Plasmodium) in human hosts and the mosquito vector:

1.	Sporozoites are injected into the human bloodstream by an infected mosquito.
2.	Merozoites are released into the bloodstream and invade red blood cells, where they multiply.
3.	Gametocytes form in human blood and are taken up by a mosquito during a blood meal.
4.	Sporozoites develop in the mosquito’s salivary glands, ready to infect the next host. 5. Schizonts form in the liver and burst, releasing merozoites into the bloodstream.

Select the correct option:
1. 1 → 5 → 2 → 3 → 4
2. 1 → 2 → 5 → 3 → 4
3. 5 → 2 → 3 → 1 → 4
4. 2 → 1 → 4 → 5 → 3

96. A researcher is using a plasmid vector containing two selectable markers, ampR (ampicillin resistance gene) and lacZ (which encodes the enzyme β-galactosidase), to clone a foreign gene into E. coli. The foreign gene is inserted into the plasmid at the BamHI restriction site, which is located within the lacZ gene. After transformation, the researcher plates the E. coli cells on an agar medium containing both ampicillin and X-gal (a chromogenic substrate that turns blue when cleaved by β-galactosidase).
Which of the following best describes the expected results and the correct method for selecting recombinant transformants?

1.	Blue colonies: Recombinant E. coli cells will form blue colonies because the foreign gene will enhance the expression of β-galactosidase, causing the colonies to turn blue in the presence of X-gal.
2.	White colonies: Non-recombinant E. coli cells will form white colonies because they lack the foreign gene, and the cells will remain sensitive to ampicillin, but will still inactivate lacZ.
3.	No colonies: Non-recombinant E. coli cells will not grow because the lacZ gene will be activated, and ampicillin resistance will be lost after insertional inactivation of the plasmid.
4.	White colonies: Recombinant E. coli cells will form white colonies because the insertion of the foreign DNA disrupts the lacZ gene, inactivating β-galactosidase, and the cells will still be resistant to ampicillin.

97. Which of the following best describes a critical step in the production of recombinant human insulin?

1.	The insulin gene is directly inserted into an E. coli plasmid, which then synthesizes functional insulin as a single protein chain.
2.	The insulin gene is inserted into yeast cells, where post-translational modifications occur naturally to yield functional insulin.
3.	Human insulin can be synthesized in E. coli without the need for any post-translational modification steps.
4.	E. coli is engineered to produce two separate insulin polypeptide chains, which are then chemically combined to form active insulin.

98. Given that the E. coli genome is approximately 4.6 million base pairs in size and that EcoRI recognizes a 6-base-pair palindromic sequence (GAATTC), how many EcoRI restriction sites would you expect to find on the E. coli genome, assuming random distribution of nucleotides and ignoring variances in sequence distribution?
1. Approximately 100
2. Approximately 750
3. Approximately 1122
4. Approximately 4096

99. Consider the following statements regarding the roles of predators in an ecosystem:

1.	Predators help regulate prey populations, which prevents overgrazing and maintains balance within the food web.
2.	Predators contribute to natural selection by removing weaker, sick, or genetically less fit individuals, which can lead to stronger, more resilient prey populations over time.
3.	Predators typically drive prey populations toward extinction, reducing biodiversity in an ecosystem and destabilizing food chains.
4.	Predators enhance nutrient cycling by breaking down biomass, allowing nutrients to return to the ecosystem and supporting plant growth.

Select the correct option:
1. Only statements 1, 2, and 4 are correct.
2. Only statements 1 and 3 are correct.
3. Only statements 2 and 4 are correct.
4. All statements are correct.

100. Which of the following best explains why habitat fragmentation poses a threat to biodiversity comparable to that of habitat loss?

1.	Fragmentation reduces the total area of habitats, forcing species to adapt rapidly, which may lead to increased genetic diversity in isolated populations.
2.	Habitat fragmentation isolates populations, limiting gene flow and making species more vulnerable to local extinction due to factors such as inbreeding, disease, and environmental fluctuations.
3.	Fragmentation results in isolated patches where resources are concentrated, allowing predator-prey dynamics to balance more effectively in limited areas.
4.	Habitat fragmentation increases edge effects, creating more favourable microhabitats for native species to thrive near human-modified environments.

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