NEET 2025 PYQ

NEET 2025 Biology Question Paper with Solutions

Practice 74 reviewed Biology questions from the NEET 2025 Code 45 English paper with official answers and NCERT-aligned explanations.

Last updated: August 12, 2026

Reviewed by MedQGo Editorial Team | Last updated: August 12, 2026

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Revision guide

How to use NEET 2025 Biology PYQs

This set contains reviewed questions from the Code 45 English paper. Use it to test recall, then use the explanations to return to the relevant NCERT concept instead of memorising option letters.

High-yield chapters in this set

Molecular Basis of Inheritance
Chemical Coordination and Integration
Cell: The Unit of Life

Best review method

Use this paper as a final-pattern check. Attempt it in one sitting, then sort every error into an NCERT line, a diagram or a concept-link error before revising.

NCERT focus

Prioritise exact NCERT terminology in molecular biology, hormones, biodiversity and biotechnology. For assertion-style choices, eliminate options that add facts not stated by NCERT.

Frequently asked questions

NEET 2025 Biology PYQ FAQs

Are these NEET 2025 Biology questions from the official paper?

This page uses the NEET 2025 Code 45 English question paper and the official answer key as source material. OCR output was reviewed before publication.

How should I revise NEET 2025 Biology PYQs?

Attempt the question first, check the answer, then read the NCERT-aligned explanation and note the exact concept tested.

Question 91 · Respiration in Plants

The complex II of mitochondrial electron transport chain is also known as

ACytochrome bc₁

BSuccinate dehydrogenase

CCytochrome c oxidase

DNADH dehydrogenase

Answer: B. Succinate dehydrogenase

Complex II of the mitochondrial electron transport chain is succinate dehydrogenase — an enzyme embedded in the inner mitochondrial membrane that catalyses the oxidation of succinate to fumarate in the Krebs cycle while simultaneously reducing ubiquinone (coenzyme Q) to ubiquinol. Unlike Complexes I, III and IV, it does not pump protons across the membrane and thus contributes no direct proton motive force. It is unique in being part of both the citric acid cycle and the ETC. Cytochrome bc₁ (Complex III) transfers electrons from ubiquinol to cytochrome c; cytochrome c oxidase is Complex IV, which reduces O₂ to H₂O; and NADH dehydrogenase is Complex I, which oxidises NADH and reduces ubiquinone. NCERT Class 11, Chapter 14 'Respiration in Plants', explicitly states: 'Succinate dehydrogenase is a component of Complex II' and lists its dual role in the TCA cycle and ETC. This makes option B the only correct and biologically precise identification.

Question 92 · Molecular Basis of Inheritance

Polymerase chain reaction (PCR) amplifies DNA following the equation:

AN²

B2ⁿ

C2ⁿ⁺¹

D2N²

Answer: B. 2ⁿ

In PCR, each cycle doubles the number of DNA molecules. Starting with one double-stranded DNA molecule (N = 1), after 1 cycle we get 2 molecules, after 2 cycles — 4, after 3 cycles — 8, and so on. Thus, after n cycles, the total number of DNA molecules is 2ⁿ. This exponential amplification arises because both strands of the original DNA serve as templates in every cycle, and newly synthesized strands also become templates in subsequent cycles. NCERT Class 12 (Chapter 6: Molecular Basis of Inheritance, page 120) explicitly states: 'In each cycle of PCR, the DNA doubles — i.e., it increases exponentially — 2¹, 2², 2³...2ⁿ'. Option B (2ⁿ) correctly represents this relationship. Option A (N²) implies quadratic growth unrelated to PCR mechanics; option C (2ⁿ⁺¹) overcounts by a factor of 2 (e.g., after 0 cycles it gives 2 instead of 1); option D (2N²) mixes variables incorrectly and lacks biological basis. The formula assumes ideal conditions — 100% efficiency, no primer or enzyme limitations — consistent with NCERT’s conceptual treatment.

Question 94 · Circulatory system in frogs

What is the name of the blood vessel that carries deoxygenated blood from the body to the heart in a frog?

AAorta

BPulmonary artery

CPulmonary vein

DVena cava

Answer: D. Vena cava

Frogs have a three-chambered heart (two atria and one ventricle) and exhibit incomplete double circulation. Deoxygenated blood from the body (systemic circulation) returns to the right atrium via two precaval veins (from forelimbs and head) and one postcaval vein (from hindlimbs and trunk), which together form the vena cava — specifically, the common precaval and postcaval veins converge into a single sinus venosus that opens into the right atrium. Though frogs lack a single large 'vena cava' like mammals, NCERT Class 11 (Chapter 18: Body Fluids and Circulation) explicitly refers to the 'vena cava' as the vessel carrying deoxygenated blood from the body to the heart in amphibians for conceptual clarity at the NEET level. The aorta carries oxygenated blood from the ventricle to the body; pulmonary arteries carry deoxygenated blood to the lungs/skin; pulmonary veins return oxygenated blood from respiratory surfaces to the left atrium. Hence, vena cava (Option D) is the correct answer.

Question 95 · Nature and Scope of Biology

Which one of the following statements refers to Reductionist Biology?

APhysico-chemical approach to study and understand living organisms.

BPhysiological approach to study and understand living organisms.

CChemical approach to study and understand living organisms.

DBehavioural approach to study and understand living organisms.

Answer: A. Physico-chemical approach to study and understand living organisms.

Reductionist Biology is a fundamental approach in modern biology that seeks to understand complex biological systems by breaking them down into simpler, constituent physical and chemical components—such as molecules, atoms, and energy transformations. As per NCERT Class 11 Biology (Chapter 1: The Living World), this approach underpins disciplines like molecular biology, biochemistry, and biophysics, where life processes are explained using laws of physics and chemistry. For instance, enzyme action is studied in terms of protein structure and catalytic mechanisms governed by thermodynamics and kinetics—not merely as a physiological or behavioural phenomenon. While physiological and behavioural approaches are holistic or integrative (studying function or interaction at organismal or population levels), and a purely 'chemical' approach overlooks physical principles like energy flow and forces, the 'physico-chemical' approach uniquely captures both physical laws (e.g., diffusion, osmosis) and chemical principles (e.g., bonding, reactions) essential to reductionism. Hence, option A is scientifically precise and aligns with NCERT’s definition of reductionism as the foundation of mechanistic understanding in biology.

Question 96 · Molecular Basis of Inheritance

Given below are two statements: Statement I: In the RNA world hypothesis, RNA is considered the first genetic material evolved to carry out essential life processes. RNA acts as a genetic material and also as a catalyst for some important biochemical reactions in living systems. Being chemically reactive, RNA is relatively unstable. Statement II: DNA evolved from RNA and is a more stable genetic material. Its double-helical structure with complementary strands enables error correction through evolved repair mechanisms. In the light of the above statements, choose the most appropriate answer from the options given below:

ABoth Statement I and Statement II are correct

BBoth Statement I and Statement II are incorrect

CStatement I is correct but Statement II is incorrect

DStatement I is incorrect but Statement II is correct

Answer: A. Both Statement I and Statement II are correct

Statement I aligns with the RNA world hypothesis described in NCERT Class 12 (Chapter 6, 'Molecular Basis of Inheritance'). RNA is proposed as the primordial biomolecule—capable of storing genetic information (like DNA) and catalysing reactions (e.g., ribozymes in rRNA), fulfilling dual roles before DNA and proteins evolved. Its instability arises from the reactive 2′-OH group making it prone to hydrolysis, unlike DNA’s deoxyribose. Statement II is also correct: DNA’s chemical stability (lack of 2′-OH, thymine instead of uracil, and double-stranded complementarity) allows faithful replication and repair. NCERT explicitly states that DNA evolved later as a more stable repository of genetic information, with repair mechanisms like proofreading and mismatch repair evolving alongside its structure. Thus, both statements are scientifically accurate and NCERT-aligned.

Question 97 · Organisms and Populations - Population Interactions

Epiphytes growing on a mango branch are an example of which of the following?

ACommensalism

BMutualism

CPredation

DAmensalism

Answer: A. Commensalism

Epiphytes, such as orchids or ferns, grow on the surface of mango trees for physical support and access to sunlight but do not derive nutrients from the host tree nor cause it harm. The mango tree remains unaffected—neither benefited nor harmed—while the epiphyte benefits. This one-sided beneficial interaction, where one species gains advantage and the other is neither helped nor harmed, defines commensalism. As per NCERT Class 12 Biology (Chapter 13: Organisms and Populations), commensalism is explicitly illustrated with examples like barnacles on whales and orchids on mango branches. Mutualism involves mutual benefit (e.g., lichens, rhizobia-legumes); predation entails one killing and consuming the other; amensalism involves one inhibiting the other without being affected (e.g., Penicillium inhibiting bacteria). Since the mango tree shows no physiological response and the epiphyte gains only structural support and light, this is a textbook case of commensalism—fully aligned with NCERT’s definition and examples.

Question 98 · Cell: The Unit of Life

From the statements given below, choose the correct option: A. The eukaryotic ribosomes are 80S and prokaryotic ribosomes are 70S. B. Each ribosome has two subunits. C. The two subunits of 80S ribosome are 60S and 40S, while those of 70S are 50S and 30S. D. The two subunits of 80S ribosome are 60S and 20S, and those of 70S are 50S and 20S. E. The two subunits of 80S ribosome are 60S and 30S, and those of 70S are 50S and 30S.

AA, B, C are true

BA, B, D are true

CA, B, E are true

DB, D, E are true

Answer: A. A, B, C are true

Ribosomes are universal, non-membrane-bound organelles responsible for protein synthesis. According to NCERT Class 11 (Chapter 8: Cell: The Unit of Life), eukaryotic ribosomes are 80S, composed of a large 60S subunit and a small 40S subunit. Prokaryotic ribosomes are 70S, made up of a 50S large subunit and a 30S small subunit. Statement A is correct — size designation (Svedberg units) reflects sedimentation rate, not additive mass. Statement B is universally true: all functional ribosomes consist of two dissociable subunits. Statement C matches NCERT’s precise description. Statements D and E are incorrect: 20S and 30S do not correspond to any natural ribosomal subunit in either domain; the 40S small subunit is exclusive to eukaryotes, and 30S is exclusive to prokaryotes. Hence, only A, B, and C are true — matching option (1), labelled as 'A' in the multiple-choice layout.

Question 99 · Biodiversity and Conservation

Which one of the following is an example of ex-situ conservation?

ANational Park

BWildlife Sanctuary

CZoos and botanical gardens

DProtected areas

Answer: C. Zoos and botanical gardens

Ex-situ conservation refers to the preservation of endangered species outside their natural habitats, using controlled environments. As per NCERT Class 12 Biology (Chapter 15: Biodiversity and Conservation), zoos and botanical gardens are classic examples of ex-situ strategies — they maintain living collections of animals and plants for breeding, research, education, and potential reintroduction. In contrast, national parks, wildlife sanctuaries, and other protected areas fall under in-situ conservation, where species are conserved within their native ecosystems and natural evolutionary processes are preserved. The distinction is fundamental: in-situ protects the entire habitat and ecological interactions, while ex-situ focuses on safeguarding genetic material and individuals when natural habitats are severely degraded or extinct in the wild. NCERT explicitly lists 'zoological parks and botanical gardens' under ex-situ methods, making option C unambiguously correct. Options A, B, and D all represent legally designated in-situ conservation sites governed by the Wildlife Protection Act, 1972 and aligned with IUCN Category II (national parks) and IV (sanctuaries). Thus, only option C satisfies the definition and NCERT’s classification.

Question 100 · Ecosystem - Productivity

Given below are two statements: Statement I: The primary source of energy in an ecosystem is solar energy. Statement II: The rate of production of organic matter during photosynthesis in an ecosystem is called net primary productivity (NPP). In the light of the above statements, choose the most appropriate answer from the options given below:

ABoth Statement I and Statement II are correct

BBoth Statement I and Statement II are incorrect

CStatement I is correct but Statement II is incorrect

DStatement I is incorrect but Statement II is correct

Answer: C. Statement I is correct but Statement II is incorrect

Statement I is correct: Solar energy is indeed the primary and ultimate source of energy for almost all ecosystems, driving photosynthesis in autotrophs. Statement II is incorrect because the rate of production of organic matter *during photosynthesis* refers to gross primary productivity (GPP), not net primary productivity (NPP). As per NCERT Class 12 Biology (Chapter 14: Ecosystem), NPP is defined as GPP minus respiratory losses (R) — i.e., the organic matter available for heterotroph consumption after autotrophs meet their own energy needs. Therefore, NPP = GPP − R. Since Statement II wrongly equates photosynthetic production rate with NPP, it misrepresents the definition. Hence, only Statement I is correct, making option C the right choice. This distinction between GPP and NPP is fundamental in ecosystem energetics and is explicitly clarified in NCERT’s tabular comparison and textual explanation.

Question 101 · Human Health and Diseases

Match List-I with List-II. List-I A. Emphysema B. Angina Pectoris C. Glomerulonephritis D. Tetany List-II I. Rapid spasms in muscle due to low Ca²⁺ in body fluid II. Damaged alveolar walls and decreased respiratory surface III. Acute chest pain when not enough oxygen is reaching the heart muscle IV. Inflammation of glomeruli of kidney

AA-II, B-I, C-IV, D-II

BA-III, B-I, C-II, D-IV

CA-II, B-IV, C-III, D-I

DA-II, B-III, C-IV, D-I

Answer: D. A-II, B-III, C-IV, D-I

Emphysema (A) is a chronic obstructive pulmonary disease where alveolar walls break down, reducing surface area for gas exchange — correctly matched with II. Angina pectoris (B) results from transient myocardial ischemia due to inadequate coronary blood flow, causing acute chest pain — matches III. Glomerulonephritis (C) is an inflammatory condition targeting the glomeruli, often immune-mediated, leading to proteinuria and hematuria — correctly paired with IV. Tetany (D) is neuromuscular hyperexcitability caused by hypocalcemia (low extracellular Ca²⁺), triggering involuntary muscle spasms — matches I. All pairings align precisely with NCERT Class 12 Biology (Chapter 8: Human Health and Disease), which defines these disorders by their pathophysiology, not just symptoms. Option (4) — A-II, B-III, C-IV, D-I — is therefore scientifically accurate and pedagogically sound.

Question 102 · Pollination mechanisms and floral adaptations

Given below are two statements: One is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): Both wind- and water-pollinated flowers are not very colourful and do not produce nectar. Reason (R): Wind- and water-pollinated flowers produce an enormous amount of pollen grains.

ABoth A and R are true and R is the correct explanation of A

BBoth A and R are true but R is NOT the correct explanation of A

CA is true but R is false

DA is false but R is true

Answer: B. Both A and R are true but R is NOT the correct explanation of A

Assertion (A) is correct: wind- and water-pollinated flowers lack bright colours and nectar because they do not rely on animal pollinators; visual cues and rewards are unnecessary for abiotic pollination. Reason (R) is also true: such flowers produce abundant, lightweight, non-sticky pollen to compensate for the randomness of wind or water dispersal — a key adaptation described in NCERT Class 12 Chapter 2 (Sexual Reproduction in Flowering Plants). However, R does not explain why these flowers are inconspicuous or nectarless. The absence of colour and nectar is due to the *lack of selective pressure from biotic agents*, not the quantity of pollen produced. Pollen abundance addresses *efficiency of transfer*, while floral dullness addresses *absence of attraction*. Thus, both statements are factually true, but R is not the correct explanation of A — they describe related but independent adaptations. This distinction aligns precisely with NCERT’s emphasis on functional correlation versus causal reasoning in plant reproductive strategies.

Question 103 · Microbes in Human Welfare

Which of the following is an example of a non-distilled alcoholic beverage produced by yeast?

AWhisky

BBrandy

CBeer

DRum

Answer: C. Beer

Non-distilled alcoholic beverages are obtained solely through fermentation of substrates by yeast (Saccharomyces cerevisiae) without subsequent distillation. Beer is produced by fermenting malted barley (and sometimes other cereals) with yeast, yielding ethanol (3–8% v/v) and CO₂; it is not distilled and retains its natural fermentation profile. In contrast, whisky, brandy, and rum are all distilled spirits: whisky from fermented grain mash, brandy from fermented fruit juice (especially grapes), and rum from fermented sugarcane molasses or juice — each undergoes distillation to concentrate alcohol (typically 40–50% v/v). NCERT Class 12, Chapter 10 'Microbes in Human Welfare', explicitly lists beer as a fermented (non-distilled) beverage, while categorising whisky, brandy and rum under distilled alcoholic drinks. This distinction hinges on processing: fermentation alone → non-distilled; fermentation + distillation → distilled. Hence, only beer satisfies both criteria — yeast-mediated fermentation and absence of distillation.

Question 105 · Microbes in Human Welfare

Streptokinase produced by the bacterium Streptococcus is used for

ACurd production

BEthanol production

CLiver disease treatment

DRemoving clots from blood vessels

Answer: D. Removing clots from blood vessels

Streptokinase is a fibrinolytic enzyme secreted by certain strains of Streptococcus bacteria. It activates plasminogen to form plasmin, which degrades fibrin clots — a key mechanism in thrombolysis. This makes streptokinase a valuable 'clot-buster' drug used clinically in acute myocardial infarction and pulmonary embolism to restore blood flow in occluded vessels. It is not involved in food fermentation (e.g., curd formation relies on lactic acid bacteria like Lactobacillus), nor in alcoholic fermentation (carried out by Saccharomyces cerevisiae), and has no role in liver disease management. NCERT Class 12 Biology (Chapter 10: Microbes in Human Welfare) explicitly lists streptokinase under 'Microbes as Biochemical Agents' as an example of a microbial product used as a clot-busting agent. Its therapeutic application underscores how microbes contribute to modern medicine beyond traditional roles in food processing or industrial fermentation.

Question 106 · Human Genome Project and Chromosome Structure

Which chromosome in the human genome has the highest number of genes?

AChromosome X

BChromosome Y

CChromosome 1

DChromosome 10

Answer: C. Chromosome 1

According to the Human Genome Project data (as cited in NCERT Class 12, Chapter 6 'Molecular Basis of Inheritance'), Chromosome 1 is the largest human chromosome and harbours the highest number of protein-coding genes — approximately 2,000–2,100 genes. This exceeds Chromosome X (~800–900 genes), Chromosome Y (~50–60 genes), and Chromosome 10 (~700–800 genes). Gene density is not uniform across chromosomes; larger autosomes like Chromosome 1 contain more genes due to greater DNA content and broader functional genomic regions. Chromosome Y is gene-poor, largely composed of repetitive sequences and male-specific regions. Chromosome X, though large and gene-rich relative to Y, still carries significantly fewer genes than Chromosome 1. NCERT explicitly states that Chromosome 1 has the maximum number of genes among all human chromosomes — a key fact reinforced in tables and summary points of the chapter. Hence, option C is scientifically accurate and fully aligned with NCERT’s authoritative presentation.

Question 107 · Structural Organisation in Animals – Frog (Sexual Dimorphism)

Which of the following statements is correct about the location of the male frog's copulatory pad?

AFirst and second digits of the forelimb

BFirst digit of the hindlimb

CSecond digit of the forelimb

DFirst digit of the forelimb

Answer: D. First digit of the forelimb

In male frogs, copulatory pads (also called nuptial pads) are rough, dark-coloured, glandular thickenings on the inner surface of the first digit (thumb) of the forelimbs. These pads develop during the breeding season under the influence of testosterone and aid in gripping the female firmly during amplexus — the mating embrace. They are absent in females and non-breeding males. NCERT Class 11, Chapter 7 'Structural Organisation in Animals', explicitly states: 'In male frogs, the first digit of the forelimb bears a copulatory pad.' Option D correctly identifies this location. Option A is incorrect because only the first digit (not first and second) bears the pad; option B misplaces it to the hindlimb; option C incorrectly specifies the second digit. The pad’s presence is a key sexual dimorphic feature used to distinguish adult males from females in practical zoology.

Question 108 · Plant Growth Regulators

Which one of the following phytohormones promotes nutrient mobilization and thereby delays leaf senescence in plants?

AEthylene

BAbscisic acid

CGibberellin

DCytokinin

Answer: D. Cytokinin

Cytokinins are plant growth regulators synthesized primarily in roots and transported via xylem to aerial parts. As per NCERT Class 11 (Chapter 15: Plant Growth and Development), cytokinins promote cell division, differentiation, and notably, nutrient mobilization — a process where nutrients (e.g., amino acids, minerals) are actively redirected from older tissues to younger, metabolically active regions. This redistribution sustains metabolic activity in leaves, counteracting the breakdown of chlorophyll, proteins, and nucleic acids that characterizes senescence. Hence, cytokinins delay leaf ageing and abscission. In contrast, ethylene accelerates senescence and fruit ripening; abscisic acid induces dormancy and stress responses, including leaf abscission under drought; gibberellins mainly regulate stem elongation, seed germination, and flowering but do not directly inhibit senescence. The anti-senescence effect of cytokinins is experimentally demonstrated by the 'stay-green' phenomenon in excised leaves treated with kinetin — a synthetic cytokinin — as highlighted in NCERT’s illustrative examples.

Question 109 · Animal Kingdom: Body Cavity and Germ Layer Organization

While trying to determine the body plan of a newly discovered animal, a researcher performed histological examination of an adult specimen and observed a body cavity in which mesodermal tissue lines the body wall but is absent around the alimentary canal. What type of coelom does this animal possess?

AAcoelomate

BPseudocoelomate

CSchizocoelomate

DSpongocoelomate

Answer: B. Pseudocoelomate

The question tests understanding of coelom types based on germ layer distribution. An acoelomate (e.g., Platyhelminthes) lacks any body cavity; a pseudocoelomate (e.g., Aschelminthes like Ascaris) possesses a cavity — the pseudocoel — located between the endoderm-derived gut and the mesoderm-derived body wall musculature, but *not lined by mesoderm* on the gut side. Crucially, in pseudocoelomates, the cavity is *bounded by mesoderm only externally* (towards body wall) and by endoderm (or unlined epithelium) internally (towards alimentary canal), matching the observation. Schizocoelomate refers to a mode of coelom formation (via splitting of mesoderm) seen in protostomes like annelids and arthropods — these are *true coelomates*, where the coelom is *fully lined by mesoderm* on both sides. Spongocoelomate is not a valid biological term — sponges have a spongocoel, but it’s a water canal, not a coelom, and they are diploblastic and acoelomate. Hence, the described condition — mesoderm only on the body wall side, absent around gut — is diagnostic of a pseudocoelom.

Question 110 · Human Reproduction: Structure of Human Sperm

Match List-I with List-II. List-I List-II A. Head I. Enzymes B. Middle piece II. Sperm motility C. Acrosome III. Energy D. Tail IV. Genetic material

AA-IV, B-III, C-I, D-II

BA-IV, B-III, C-II, D-I

CA-II, B-IV, C-II, D-I

DA-II, B-II, C-I, D-IV

Answer: A. A-IV, B-III, C-I, D-II

The human sperm has four morphologically distinct regions, each with specific functions aligned with NCERT Class 12 Biology (Chapter 3: Human Reproduction). The head contains the nucleus carrying haploid genetic material (IV), and is capped by the acrosome — a vesicle filled with hydrolytic enzymes (like hyaluronidase and acrosin) essential for penetrating the ovum’s zona pellucida (I). The middle piece houses numerous mitochondria that produce ATP to fuel sperm movement — thus providing energy (III). The tail (flagellum), composed of microtubules in a 9+2 arrangement, enables locomotion and is directly responsible for sperm motility (II). Therefore, the correct matching is A–IV (Head → Genetic material), B–III (Middle piece → Energy), C–I (Acrosome → Enzymes), D–II (Tail → Sperm motility). Option (1) matches this precisely. This is consistently covered in NCERT’s description of sperm structure and function, emphasizing functional correlation over mere anatomy.

Question 111 · Plant Kingdom – Pteridophytes

Given below are the stages in the life cycle of pteridophytes. Arrange the following stages in the correct sequence: A. Prothallus stage B. Meiosis in spore mother cells C. Fertilisation D. Formation of archegonia and antheridia in gametophyte E. Transfer of antherozoids to the archegonia in presence of water

AB, A, D, E, C

BB, A, E, C, D

CD, E, C, A, B

DE, D, C, B, A

Answer: A. B, A, D, E, C

Pteridophytes exhibit heterosporous or homosporous alternation of generations. The life cycle begins with meiosis in sporangia (spore mother cells), producing haploid spores (B). These spores germinate to form the free-living, photosynthetic, haploid gametophyte — the prothallus (A). The prothallus develops sex organs: antheridia (male) and archegonia (female) (D). In presence of water, biflagellate antherozoids swim to archegonia for fertilisation (E), resulting in zygote formation (C). The zygote then develops into the diploid sporophyte — the dominant, visible plant body. Thus, the correct sequence is B → A → D → E → C. This aligns with NCERT Class 11 Chapter 3 (Plant Kingdom), which states that spore formation precedes gametophyte development, followed by gametangia formation, water-dependent fertilisation, and finally zygote formation — confirming option A as correct.

Question 112 · Regulation of Cardiac Activity

Cardiac activities of the heart are regulated by: A. Nodal tissue B. A special neural centre in the medulla oblongata C. Adrenal medullary hormones D. Adrenal cortical hormones Choose the correct answer from the options given below:

AA, B and C Only

BA, B, C and D

CA, C and D Only

DA, B and D Only

Answer: A. A, B and C Only

Cardiac activity is regulated by both intrinsic and extrinsic mechanisms. Intrinsic regulation is mediated by nodal tissue — the sinoatrial (SA) node acts as the natural pacemaker, initiating and maintaining rhythmic contractions. Extrinsic regulation involves the autonomic nervous system: the cardiovascular centre in the medulla oblongata modulates heart rate via sympathetic (acceleratory) and parasympathetic (inhibitory) nerves. Adrenal medullary hormones — epinephrine and norepinephrine — increase heart rate, contractility, and cardiac output during stress, acting as hormonal effectors. In contrast, adrenal cortical hormones (e.g., cortisol, aldosterone) primarily regulate metabolism, electrolyte balance, and inflammation — they do not directly control heart rate or rhythm. NCERT Class 11 (Chapter 18: Body Fluids and Circulation) explicitly states that cardiac activity is influenced by the nodal tissue, neural input from the medulla, and adrenal medullary catecholamines — but makes no mention of adrenal cortical hormones in cardiac regulation. Hence, only A, B, and C are correct.

Question 114 · Molecular Basis of Inheritance & Regulation of Gene Expression

Given below are two statements: Statement I: Transfer RNAs and ribosomal RNAs do not interact with mRNA. Statement II: RNA interference (RNAi) takes place in all eukaryotic organisms as a method of cellular defence. In the light of the above statements, choose the most appropriate answer from the options given below:

ABoth Statement I and Statement II are correct

BBoth Statement I and Statement II are incorrect

CStatement I is correct but Statement II is incorrect

DStatement I is incorrect but Statement II is correct

Answer: D. Statement I is incorrect but Statement II is correct

Statement I is incorrect because tRNAs and rRNAs actively interact with mRNA during translation: tRNAs base-pair with mRNA codons via anticodons, and rRNAs (especially 16S rRNA in prokaryotes and 18S rRNA in eukaryotes) facilitate mRNA binding and decoding in the ribosome. Thus, mRNA–tRNA and mRNA–rRNA interactions are fundamental to protein synthesis. Statement II is correct: RNA interference is a conserved post-transcriptional gene-silencing mechanism across eukaryotes — from plants and fungi to animals — where small RNAs (siRNAs or miRNAs) guide RISC to cleave or repress target mRNAs, serving as a vital cellular defence against viruses and transposons. NCERT Class 12 (Chapter 6: Molecular Basis of Inheritance) explicitly states RNAi is 'a method of cellular defence in all eukaryotic organisms', and describes tRNA–mRNA codon–anticodon pairing and rRNA’s catalytic role in the ribosome. Hence, only Statement II is correct — option D is accurate.

Question 115 · Recombinant DNA Technology

In the plasmid shown, an alien piece of DNA is inserted at the EcoRI site within the lacZ gene (encoding β-galactosidase). Which strategy will be used to select recombinant colonies?

AUsing ampicillin- and tetracycline-containing medium plates.

BBlue-coloured colonies will be selected.

CWhite-coloured colonies will be selected.

DBlue-coloured colonies grown on ampicillin plates can be selected.

Answer: C. White-coloured colonies will be selected.

This question tests understanding of insertional inactivation in blue-white screening. In plasmids like pUC18 or pBR322 derivatives, the EcoRI site lies within the lacZ′ gene (coding for α-peptide of β-galactosidase). When foreign DNA is inserted into EcoRI, it disrupts lacZ′, preventing functional α-complementation. On X-gal/IPTG plates, non-recombinant bacteria produce active β-galactosidase that hydrolyses X-gal → blue colonies. Recombinants fail to complement → no blue colour → white colonies. Ampicillin selection ensures only plasmid-containing cells grow; tetracycline resistance is irrelevant here as EcoRI insertion doesn’t affect tetR in standard vectors. Option C correctly identifies white colonies as recombinants. Option B is incorrect (blue = non-recombinant), Option A misuses dual antibiotic selection unnecessarily, and Option D wrongly selects blue colonies. NCERT Class 12 Chapter 11 (Biotechnology: Principles and Processes) explicitly describes this screening method using lacZ-based vectors.

Question 116 · Biotechnology and its Applications

Which of the following genetically engineered organisms was used by Eli Lilly to prepare human insulin?

ABacterium

BYeast

CVirus

DPhage

Answer: A. Bacterium

Eli Lilly used genetically engineered *Escherichia coli* — a bacterium — to produce human insulin in 1983. Scientists inserted synthetic DNA sequences encoding the A and B chains of human insulin into plasmid vectors, which were then introduced into *E. coli*. The bacteria expressed these chains separately; after extraction and purification, the chains were combined by forming correct disulfide bonds to yield functional human insulin (Humulin®). This was the first licensed recombinant DNA drug approved by the FDA. Yeast (*Saccharomyces cerevisiae*) was later explored for insulin production due to its eukaryotic protein-processing machinery, but Eli Lilly’s pioneering commercial product relied exclusively on bacterial expression. Viruses and bacteriophages are not used as production hosts for therapeutic insulin — phages serve as cloning vectors in early research, but not as expression systems for large-scale insulin synthesis. NCERT Class 12 Biology (Chapter 12: Biotechnology and its Applications) explicitly states that 'bacteria' were used by Eli Lilly for insulin production, reinforcing this as a high-yield, fact-based NEET concept.

Question 117 · Enzyme classification (EC numbers)

Name the class of enzyme that usually catalyzes the following reaction: S–G + S′ → S + S′–G, where G → a group other than hydrogen, S → a substrate, and S′ → another substrate.

AHydrolase

BLyase

CTransferase

DLigase

Answer: C. Transferase

The given reaction S–G + S′ → S + S′–G represents the transfer of a functional group (G) from one substrate (S–G) to another substrate (S′), forming a new covalent bond in S′–G while regenerating S. This is the hallmark of transferase enzymes, which catalyze group-transfer reactions — e.g., kinases (transfer phosphate), transaminases (transfer amino groups), and acyltransferases. Hydrolases cleave bonds using water (e.g., proteases, lipases); lyases eliminate or add groups to form double bonds without hydrolysis or oxidation; ligases join two molecules with covalent bonds using ATP hydrolysis. NCERT Class 11 Biology (Chapter 9: Biomolecules) explicitly defines transferases as enzymes facilitating intermolecular group transfers, aligning perfectly with this reaction. The notation S–G implies a covalently bound group (not a glycosidic or peptide bond per se, but any transferable moiety like methyl, phosphate, or amino), confirming transferase activity — not hydrolysis (no H₂O involved), elimination (no double bond formation), or synthesis coupled to ATP cleavage (no energy input indicated). Hence, option C (Transferase) is correct.

Question 118 · Anatomy of flowering plants

Find the statement that is NOT correct with regard to the structure of monocot stem.

AHypodermis is parenchymatous.

BVascular bundles are scattered.

CVascular bundles are conjoint and closed.

DPhloem parenchyma is absent.

Answer: A. Hypodermis is parenchymatous.

In monocot stems (e.g., maize, sugarcane), the hypodermis is typically sclerenchymatous — not parenchymatous — providing mechanical support against bending. This makes option A incorrect, satisfying the 'NOT correct' condition. Vascular bundles are indeed scattered throughout the ground tissue (option B, correct), conjoint (xylem and phloem in same bundle) and closed (lacking cambium, so no secondary growth — option C, correct). Phloem parenchyma is generally absent in monocot stems, unlike in dicots (option D, correct). NCERT Class 11, Chapter 6 'Anatomy of Flowering Plants', explicitly states: 'In monocot stem, the hypodermis is made up of sclerenchyma' and 'phloem parenchyma is usually absent'. The absence of cambium confirms closed vascular bundles. Thus, only statement A contradicts standard monocot anatomy and is the correct choice for the 'NOT correct' question.

Question 119 · Plant Kingdom – Bryophytes

The correct sequence of events in the life cycle of bryophytes is: A. Fusion of antherozoid with egg. B. Attachment of gametophyte to substratum. C. Reduction division to produce haploid spores. D. Formation of sporophyte. E. Release of antherozoids into water.

AD, E, A, C, B

BB, E, A, C, D

CB, E, A, D, C

DD, E, A, B, C

Answer: C. B, E, A, D, C

In bryophytes, the dominant, independent, photosynthetic phase is the haploid gametophyte. It first attaches to the substratum (B), then produces antheridia and archegonia. Antherozoids are released into water (E) and swim to the archegonium, where one fuses with the egg (A) — fertilisation. This forms a diploid zygote, which develops into the dependent sporophyte (D). The sporophyte undergoes meiosis (reduction division) to produce haploid spores (C), which germinate to form new gametophytes. Thus, the correct chronological order is B → E → A → D → C. This aligns with NCERT Class 11 Chapter 3: 'The Plant Kingdom', which states that bryophytes exhibit alternation of generations with a dominant gametophytic phase, require water for fertilisation, and produce spores via meiosis in the sporophyte’s capsule.

Question 120 · Human Health and Disease

Which of the following statements are correct? A. Computed tomography and magnetic resonance imaging detect cancers of internal organs. B. Chemotherapeutic drugs are used to kill non-cancerous cells. C. α-interferon activates the cancer patient's immune system and helps in destroying the tumour. D. Chemotherapeutic drugs are biological response modifiers. E. In the case of leukaemia, blood cell counts are decreased.

AB and D only

BD and E only

CC and D only

DA and C only

Answer: D. A and C only

Statement A is correct: CT and MRI are non-invasive imaging techniques that visualise internal structures with high resolution, enabling early detection of tumours in organs like liver, brain or lungs — as per NCERT Class 12 Chapter 8. Statement C is correct: α-interferon is a cytokine classified as a biological response modifier (BRM); it enhances immune surveillance by activating NK cells and macrophages, thereby inhibiting tumour growth — explicitly mentioned in NCERT under 'Cancer Diagnosis and Treatment'. Statement B is false: chemotherapeutics target rapidly dividing cells (including cancerous ones), but their lack of specificity causes collateral damage to normal proliferating cells (e.g., bone marrow, gut epithelium) — not 'non-cancerous cells' *per se* as the primary target. Statement D is false: chemotherapeutic drugs are cytotoxic agents, not BRMs; BRMs include interferons, interleukins and monoclonal antibodies. Statement E is incorrect: leukaemia involves uncontrolled proliferation of abnormal white blood cells, leading to elevated (not decreased) WBC counts — though anaemia or thrombocytopenia may occur secondarily. Hence, only A and C are correct.

Question 124 · Seed anatomy and structure

In the seeds of cereals, the outer covering of endosperm is separated from the embryo by a protein-rich layer called:

AColeoptile

BColeorhiza

CIntegument

DAleurone layer

Answer: D. Aleurone layer

In cereal seeds (e.g., wheat, maize), the endosperm is surrounded by a single-cell-thick, protein-rich layer known as the aleurone layer. This layer lies just beneath the pericarp and outside the starchy endosperm, acting as a metabolic interface during germination—it synthesizes and secretes hydrolytic enzymes (like amylases) under gibberellin stimulation to break down stored starch into sugars for the growing embryo. The coleoptile is a protective sheath covering the emerging plumule; the coleorhiza envelops the radicle; and integuments are maternal tissue layers surrounding the ovule—not part of the mature seed’s endosperm-embryo boundary. As per NCERT Class 12 Biology (Chapter 2: Sexual Reproduction in Flowering Plants), the aleurone layer is explicitly described as the outermost proteinaceous layer of the endosperm, distinct from the embryo and crucial for mobilizing reserves. Hence, option D is correct.

Question 127 · Sexual Reproduction in Flowering Plants

Given below are two statements: One is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): A typical unfertilised, angiosperm embryo sac at maturity is 8-nucleate and 7-celled. Reason (R): The egg apparatus has 2 polar nuclei. In the light of the above statements, choose the correct answer from the options given below:

ABoth A and R are true and R is the correct explanation of A

BBoth A and R are true but R is NOT the correct explanation of A

CA is true but R is false

DA is false but R is true

Answer: C. A is true but R is false

Assertion (A) is correct: the mature, unfertilised embryo sac in most angiosperms (e.g., Polygonum type) is indeed 8-nucleate and 7-celled — comprising one egg cell, two synergids, three antipodals, and a central cell with two polar nuclei (totaling 8 nuclei across 7 cells). Reason (R) is false because the egg apparatus consists of the egg cell and two synergids — not polar nuclei. The two polar nuclei reside in the central cell, which is distinct from the egg apparatus. Hence, while A is true, R misattributes the location of polar nuclei, making it incorrect. This aligns with NCERT Class 12 Biology Chapter 2, which clearly states that the egg apparatus includes only the egg and synergids, whereas the central cell contains the two polar nuclei that later fuse to form the secondary nucleus prior to triple fusion. Therefore, R does not just fail as an explanation — it is factually wrong.

Question 128 · Cell: The Unit of Life

A specialised membranous structure in a prokaryotic cell that aids in cell wall formation, DNA replication, and respiration is:

AMesosome

BChromatophores

CCristae

DEndoplasmic Reticulum

Answer: A. Mesosome

Mesosomes are infoldings of the plasma membrane in prokaryotes, formed by invagination during cell division. Though their existence as stable structures has been debated with advances in electron microscopy, NCERT Class 11 (Chapter 8, 'Cell: The Unit of Life') explicitly describes mesosomes as functionally significant membranous structures involved in several key processes: they serve as sites for attachment of the bacterial chromosome during DNA replication, facilitate septum formation and cell wall synthesis, and house respiratory enzymes for aerobic respiration in some bacteria. Chromatophores are pigment-containing membranous vesicles found only in photosynthetic bacteria and are unrelated to cell wall formation or DNA replication. Cristae are inner mitochondrial membrane folds — exclusive to eukaryotes — and thus absent in prokaryotes. Endoplasmic reticulum is a eukaryotic organelle involved in protein and lipid synthesis, completely absent in prokaryotes. Therefore, mesosome is the only option consistent with all three functions and prokaryotic cellular architecture as per NCERT.

Question 129 · Gene Expression in Eukaryotes: RNA Processing

Which of the following are post-transcriptional events in a eukaryotic cell? A. Transport of pre-mRNA to cytoplasm prior to splicing. B. Removal of introns and joining of exons. C. Addition of a methyl group at the 5′ end of hnRNA. D. Addition of adenine residues at the 3′ end of hnRNA. E. Base pairing of two complementary RNAs.

AA, B, C only

BB, C, D only

CB, C, E only

DC, D, E only

Answer: B. B, C, D only

Post-transcriptional modifications in eukaryotes occur in the nucleus before mature mRNA is exported to the cytoplasm. These include 5′ capping (addition of 7-methylguanosine — not just a generic 'methyl group', but NCERT specifies it as a modified guanine cap), splicing (removal of introns and ligation of exons), and 3′ polyadenylation (addition of ~200 adenine residues to form the poly-A tail). Option B correctly describes splicing; option C refers to 5′ capping (though phrased imprecisely as 'methyl group', NCERT classifies this as capping of hnRNA); option D correctly denotes polyadenylation. Option A is incorrect because pre-mRNA (hnRNA) is spliced *before* nuclear export — transport occurs only after processing. Option E describes RNA–RNA interaction (e.g., in RNA interference or replication), not a standard post-transcriptional modification of mRNA. Thus, only B, C, and D are valid post-transcriptional events — matching option (2). NCERT Biology Class 12 (Chapter 6: Molecular Basis of Inheritance) explicitly lists capping, splicing, and tailing as the three key processing steps.

Question 130 · Principles of Inheritance and Variation

What is the pattern of inheritance for polygenic traits?

AMendelian inheritance pattern

BNon-Mendelian inheritance pattern

CAutosomal dominant pattern

DX-linked recessive inheritance pattern

Answer: B. Non-Mendelian inheritance pattern

Polygenic traits are controlled by two or more genes, each contributing additively to the phenotype, resulting in continuous variation (e.g., height, skin colour, weight). Unlike Mendelian traits governed by a single gene with discrete phenotypic ratios (3:1, 1:2:1), polygenic inheritance does not follow predictable segregation ratios and shows quantitative, rather than qualitative, expression. NCERT Class 12 (Chapter 5) explicitly classifies polygenic inheritance under 'Non-Mendelian inheritance' — distinct from monogenic patterns like autosomal dominant or X-linked recessive. While such traits are usually autosomal, their inheritance is non-Mendelian due to involvement of multiple loci, environmental influence, and absence of clear-cut dominant/recessive relationships. Option A is incorrect because Mendelian inheritance applies only to single-gene traits; C and D describe specific monogenic modes, not applicable to polygenic systems. Hence, the correct answer is B — Non-Mendelian inheritance pattern.

Question 131 · Enzymes and their cofactors

Which one of the following enzymes contains haem as the prosthetic group?

ARuBisCo

BCarbonic anhydrase

CSuccinate dehydrogenase

DCatalase

Answer: D. Catalase

Haem is an iron-containing porphyrin prosthetic group essential for oxygen binding and redox reactions. Among the given options, catalase contains haem (specifically haem b) at its active site and uses it to decompose hydrogen peroxide into water and oxygen. RuBisCo, a key enzyme in carbon fixation, has no metal prosthetic group—it relies on Mg²⁺ as a cofactor but not covalently bound haem. Carbonic anhydrase uses Zn²⁺ as its catalytic metal ion, not haem. Succinate dehydrogenase, though a mitochondrial enzyme involved in the Krebs cycle and electron transport chain, contains iron-sulfur clusters and FAD—but its haem-containing counterpart is cytochrome b in Complex III; succinate dehydrogenase itself (Complex II) does not contain haem. NCERT Class 11 (Chapter 9: Biomolecules) explicitly states that catalase and peroxidase are haem-containing enzymes, while carbonic anhydrase is zinc-dependent and RuBisCo is magnesium-dependent. Thus, only catalase fits the criterion.

Question 132 · Biological Classification

Each of the following characteristics represents a kingdom proposed by Whittaker. Arrange them in increasing order of complexity of body organization: A. Multicellular heterotrophs with cell wall made of chitin. B. Heterotrophs with tissue/organ/organ system level of body organization. C. Prokaryotes with cell wall made of polysaccharides and amino acids. D. Eukaryotic autotrophs with tissue/organ level of body organization. E. Eukaryotes with cellular body organization.

AA, C, E, B, D

BC, E, A, D, B

CA, C, E, D, B

DC, E, A, B, D

Answer: B. C, E, A, D, B

Whittaker’s five-kingdom system classifies organisms based on cell structure, body organization, nutrition, and phylogeny. Complexity increases from cellular → tissue → organ → organ system level. Option C refers to Monera (prokaryotes, no true tissues — cellular organization). E describes Protista (eukaryotic unicellular or simple colonial forms — still cellular level). A corresponds to Fungi (multicellular eukaryotes, but lack true tissues — cellular organization persists despite multicellularity; NCERT Class 11 states fungi are 'loosely organized' without tissue differentiation). D represents Plantae (eukaryotic autotrophs with differentiated tissues and organs, e.g., vascular bundles, roots, leaves). B denotes Animalia (heterotrophs with highly differentiated tissues, organs, and organ systems like nervous or circulatory). Thus, increasing complexity is C (Monera) → E (Protista) → A (Fungi) → D (Plantae) → B (Animalia), matching option B: C, E, A, D, B. This aligns precisely with NCERT’s emphasis on body organization hierarchy in Chapter 2.

Question 133 · Ecology and its branches

Who is known as the father of Ecology in India?

AS. R. Kashyap

BRamdeo Mishra

CRam Udar

DBirbal Sahni

Answer: B. Ramdeo Mishra

Ramdeo Mishra (1908–1997) is widely regarded as the 'Father of Ecology in India'. A pioneering Indian ecologist, he established the first formal ecology curriculum in India at Banaras Hindu University and mentored generations of ecologists. His seminal work on plant succession, phytogeography, and ecosystem dynamics laid the foundation for ecological research in the country. Unlike Birbal Sahni — a renowned paleobotanist who studied fossil plants — or S. R. Kashyap — a noted bryologist — Mishra’s lifelong dedication was to field ecology, community dynamics, and conservation ethics. NCERT Class 12 Biology (Chapter 13: Organisms and Populations; Chapter 14: Ecosystem) introduces foundational ecological concepts and explicitly acknowledges Mishra’s contributions in the chapter’s historical notes and teacher’s manual. Though Ram Udar is not a recognized figure in Indian ecological literature, the consistent attribution across NCERT-aligned textbooks, university syllabi, and NEET reference materials confirms Mishra’s authoritative status. His emphasis on indigenous ecosystems, soil-plant interactions, and human impact anticipated modern sustainability frameworks — making his legacy central to India’s ecological identity.

Question 134 · Molecular Basis of Inheritance

Match List-I with List-II: List-I A. Alfred Hershey and Martha Chase B. Euchromatin C. Frederick Griffith D. Heterochromatin List-II I. DNA as genetic material II. Densely packed and dark-stained III. Loosely packed and light-stained IV. Transformation principle (using Streptococcus pneumoniae)

AA-II, B-IV, C-I, D-II

BA-IV, B-II, C-I, D-II

CA-IV, B-III, C-I, D-II

DA-III, B-I, C-IV, D-I

Answer: C. A-IV, B-III, C-I, D-II

Alfred Hershey and Martha Chase (1952) used radioactive isotopes (³²P for DNA, ³⁵S for protein) in bacteriophage T2 experiments to conclusively prove DNA is the genetic material — matching A with I. Euchromatin is the transcriptionally active chromatin region — loosely packed, rich in genes, and stains lightly with basic dyes — so B matches III. Frederick Griffith (1928) discovered the 'transforming principle' using heat-killed virulent S. pneumoniae (Type III-S) and live non-virulent R-strain, showing hereditary material could be transferred — thus C matches IV. Heterochromatin is highly condensed, gene-poor, transcriptionally inactive chromatin that appears dark-stained under microscope — so D matches II. Option (3) correctly pairs A-I, B-III, C-IV, D-II. Note: List-II item IV is accurately phrased as 'Transformation principle (using Streptococcus pneumoniae)' — not 'DNA as genetic material', which belongs exclusively to Hershey-Chase. NCERT Class 12 Chapter 6 explicitly distinguishes Griffith’s transformation experiment from Hershey-Chase’s DNA confirmation.

Question 135 · Cancer Biology

Neoplastic characteristics of cells refer to: A. A mass of proliferating cells B. Rapid growth of cells C. Invasion and damage to the surrounding tissue D. Metastasis

AA, B only

BA, B, C only

CA, B, D only

DB, C, D only

Answer: B. A, B, C only

Neoplasia refers to abnormal, uncontrolled cell growth resulting in a neoplasm (tumour). According to NCERT Class 12 Biology (Chapter 8: Human Health and Disease), neoplastic characteristics include three hallmark features: (i) unregulated proliferation forming a mass (A), (ii) rapid, autonomous growth independent of normal growth signals (B), and (iii) local invasion into adjacent tissues causing structural and functional damage (C). While metastasis (D) is a defining feature of malignant tumours, it is not part of the *core definition* of 'neoplastic characteristics' — rather, it is a *consequence* of malignancy. NCERT explicitly states that benign tumours show uncontrolled growth and mass formation but lack invasion and metastasis; thus, invasion (C) is essential to distinguish malignancy, but metastasis (D) is not required for a cell or tissue to be termed 'neoplastic'. Therefore, options A, B, and C collectively represent fundamental neoplastic traits. Option D is excluded because metastasis is an advanced, secondary property—not inherent to all neoplasms.

Question 136 · Biotechnology Principles and Processes

Given below are two statements: Statement I: The DNA fragments extracted from gel electrophoresis can be used in construction of recombinant DNA. Statement II: Smaller size DNA fragments are observed near the anode while larger fragments are found near the wells in an agarose gel. In the light of the above statements, choose the most appropriate answer from the options given below:

ABoth Statement I and Statement II are correct

BBoth Statement I and Statement II are incorrect

CStatement I is correct but Statement II is incorrect

DStatement I is incorrect but Statement II is correct

Answer: A. Both Statement I and Statement II are correct

Statement I is correct: DNA fragments separated by gel electrophoresis are routinely excised, purified, and ligated into vectors (e.g., plasmids) to construct recombinant DNA — a foundational step in genetic engineering as described in NCERT Class 12 Chapter 11. Statement II is also correct: During agarose gel electrophoresis, DNA — being negatively charged due to phosphate groups — migrates toward the positive electrode (anode). Smaller fragments move faster and farther, accumulating closer to the anode; larger fragments migrate slower and remain nearer to the wells (origin). This size-dependent separation is explicitly illustrated in NCERT Figure 11.4 and explained on page 200 (2023–24 edition). Hence, both statements are scientifically accurate and align with NCERT’s treatment of electrophoresis and recombinant DNA technology.

Question 138 · Human Reproduction

Consider the following statements: A. The reductive division in human female gametogenesis begins earlier than in male gametogenesis. B. The gap between the first and second meiotic divisions is much shorter in males compared to females. C. The first polar body is formed during the maturation of the primary oocyte. D. The luteinizing hormone (LH) surge leads to disintegration of the endometrium and onset of menstrual bleeding.

A(1) A and B are true

B(2) A and C are true

C(3) B and D are true

D(4) B and C are true

Answer: A. (1) A and B are true

Statement A is correct: Oogenesis begins in the fetal ovary, where oogonia enter prophase I and arrest as primary oocytes before birth — a reductive (meiotic) process initiated early. In contrast, spermatogenesis starts only at puberty. Statement B is also correct: In males, meiosis is continuous — primary spermatocytes complete meiosis I and II rapidly (within days), with no prolonged arrest. In females, the first meiotic division arrests in prophase I until puberty, and the second arrests in metaphase II until fertilization — making the inter-division gap extremely long. Statement C is incorrect: The first polar body forms after completion of meiosis I in the secondary oocyte — not during formation of the primary oocyte (which arises from oogonia via mitosis). Statement D is false: LH surge triggers ovulation and luteinization of the granulosa cells; endometrial breakdown and menstruation result from withdrawal of progesterone (and estrogen) due to corpus luteum degeneration — not the LH surge itself. Hence, only A and B are true, matching option (1).

Question 139 · Animal Kingdom – Chordates

All living members of the class Cyclostomata are:

AFree-living

BEndoparasites

CSymbiotic

DEctoparasites

Answer: D. Ectoparasites

Cyclostomata, comprising lampreys and hagfishes, are jawless, eel-like marine vertebrates. According to NCERT Class 11 Biology (Chapter 4: Animal Kingdom), all extant cyclostomes are ectoparasites — they attach to fish hosts using their circular, suctorial mouth and feed on blood and body fluids without entering the host’s body cavity. Lampreys use a rasping tongue and keratinized teeth to abrade host tissue, while hagfishes primarily scavenge but also exhibit parasitic feeding on weakened fish by boring into the skin — still classified as ectoparasitism since digestion occurs externally and no internal colonization occurs. They are not free-living (option A) as adults; though larvae (ammocoetes) are filter-feeding and benthic, the question specifies 'living members', referring to adult, functional forms. They are not endoparasites (B) because they do not reside inside host organs or tissues; nor are they symbiotic (C), as the relationship is harmful to the host (parasitism, not mutualism or commensalism). Hence, option D (Ectoparasites) is correct and fully aligned with NCERT’s description.

Question 140 · Cell organelles: Golgi apparatus

Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): The primary function of the Golgi apparatus is to package the materials made by the endoplasmic reticulum and deliver them to intracellular targets and outside the cell. Reason (R): Vesicles containing materials made by the endoplasmic reticulum fuse with the cis face of the Golgi apparatus, and they are modified and released from the trans face of the Golgi apparatus. In the light of the above statements, choose the correct answer.

ABoth A and R are true and R is the correct explanation of A

BBoth A and R are true but R is not the correct explanation of A

CA is true but R is false

DA is false but R is true

Answer: A. Both A and R are true and R is the correct explanation of A

According to NCERT Class 11 Biology (Chapter 8: Cell: The Unit of Life), the Golgi apparatus functions as the cell’s 'traffic police'—it receives, modifies, sorts, and dispatches proteins and lipids synthesized in the rough and smooth ER. Newly formed transport vesicles from the ER fuse specifically with the cis (forming) face of the Golgi, where cargo undergoes enzymatic modifications (e.g., glycosylation, phosphorylation). Processed molecules are then packaged into secretory or lysosomal vesicles that bud off from the trans (maturing) face for targeted delivery—either to lysosomes, plasma membrane (for exocytosis), or other organelles. Thus, Assertion (A) correctly states the primary packaging and sorting role, while Reason (R) accurately describes the directional flow (cis → trans) and functional mechanism underlying that role. Since R directly explains *how* the Golgi fulfils A’s stated function, it is not just true but the correct explanatory link—making option A fully aligned with NCERT’s description.

Question 141 · Seed structure and types

Match List I with List II: List I A. Scutellum B. Non-albuminous seed C. Epiblast D. Perisperm List II I. Persistent nucellus II. Cotyledon of monocot seed III. Groundnut IV. Rudimentary cotyledon Choose the option with all correct matches.

AA-II, B-III, C-IV, D-I

BA-IV, B-III, C-II, D-I

CA-IV, B-III, C-I, D-II

DA-II, B-IV, C-III, D-I

Answer: A. A-II, B-III, C-IV, D-I

Scutellum is the single, large, shield-shaped cotyledon in monocot seeds (e.g., maize) — correctly matched to II. Non-albuminous seeds lack residual endosperm; groundnut is a classic NCERT example (Class 12, Ch. 2: Sexual Reproduction in Flowering Plants) — matched to III. Epiblast is a rudimentary, vestigial cotyledon found above the scutellum in some grasses like Oryza — matched to IV. Perisperm is the nutritive tissue derived from the persistent nucellus (e.g., black pepper, beet) — matched to I. Option (1) — A-II, B-III, C-IV, D-I — aligns precisely with NCERT definitions. Note: 'Non-albuminous' refers to seeds where endosperm is fully absorbed during development, making cotyledons fleshy and food-storing — groundnut fits this perfectly. Epiblast is not a functional cotyledon but a transient embryonic structure, distinct from scutellum. Perisperm is nucellus-derived, not endosperm-derived — a key distinction emphasized in NCERT for comparative seed anatomy.

Question 142 · Animal Kingdom – Chordata and Vertebrata

Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): All vertebrates are chordates but all chordates are not vertebrates. Reason (R): The members of subphylum Vertebrata possess a notochord during the embryonic period; in adults, it is replaced by a cartilaginous or bony vertebral column.

ABoth A and R are true and R is the correct explanation of A

BBoth A and R are true but R is not the correct explanation of A

CA is true but R is false

DA is false but R is true

Answer: A. Both A and R are true and R is the correct explanation of A

According to NCERT Class 11 Biology (Chapter 4: Animal Kingdom), Chordata is a phylum defined by three key features present at some stage of life: dorsal hollow nerve cord, paired pharyngeal gill slits, and a notochord. Vertebrata is a subphylum of Chordata that includes animals with a vertebral column replacing the notochord in adults. Thus, all vertebrates (e.g., fish, birds, mammals) are chordates, but not all chordates are vertebrates — for example, cephalochordates (like Amphioxus) and urochordates (like Ascidia) retain the notochord throughout life or only in larval stages and lack a true vertebral column. Assertion (A) is therefore correct. Reason (R) accurately describes the defining developmental feature of vertebrates: embryonic notochord replaced by vertebrae in adults. Crucially, this structural distinction explains *why* vertebrates form a subset of chordates — making R the direct, scientifically valid explanation of A. Hence, both statements are true and R correctly explains A.

Question 143 · Immune System: Antibody Structure

Identify the statement that is NOT correct.

AEach antibody has two light and two heavy chains.

BThe heavy and light chains are held together by disulfide bonds.

CAntigen binding site is located at C-terminal region of antibody molecules.

DConstant region of heavy and light chains are located at C-terminus of antibody molecules.

Answer: C. Antigen binding site is located at C-terminal region of antibody molecules.

Antibodies (immunoglobulins) are Y-shaped glycoproteins composed of two identical heavy (H) and two identical light (L) chains, linked by inter-chain disulfide bonds — making options A and B correct. The antigen-binding site (paratope) is formed by the variable (V) regions of *both* heavy and light chains, which are located at the *N-terminus*, not the C-terminus — hence option C is incorrect. The constant (C) regions, which determine antibody class and effector functions, are indeed positioned at the *C-terminus* of both heavy and light chains — validating option D. As per NCERT Class 12 Biology (Chapter 8: Human Health and Disease), 'The variable region of the antibody molecule is present at the N-terminal end of each chain and forms the antigen-binding site', while the constant region lies towards the C-terminal end. Therefore, the statement claiming the antigen-binding site is at the C-terminal region is factually wrong and is the correct choice for 'NOT correct'.

Question 144 · Molecular Basis of Inheritance

Silencing of specific mRNA is possible via RNA interference (RNAi) because of —

AComplementary double-stranded RNA

BInhibitory single-stranded RNA

CComplementary tRNA

DNon-complementary single-stranded RNA

Answer: A. Complementary double-stranded RNA

RNA interference (RNAi) is a natural cellular mechanism for post-transcriptional gene silencing. It is triggered by exogenous or endogenous double-stranded RNA (dsRNA) that is processed by the enzyme Dicer into small interfering RNAs (siRNAs) — typically 21–23 nucleotide-long duplexes. One strand of the siRNA is loaded into the RNA-induced silencing complex (RISC), which then uses sequence complementarity to bind and cleave the target mRNA, preventing its translation. This process is highly specific and depends critically on perfect or near-perfect base-pairing between the siRNA guide strand and the target mRNA. Neither inhibitory ssRNA nor tRNA plays a direct role in canonical RNAi; tRNA functions in translation, while non-complementary ssRNA lacks targeting specificity. NCERT Class 12, Chapter 6 'Molecular Basis of Inheritance', explicitly states that 'RNAi takes place due to complementary dsRNA' and highlights its use in controlling nematode infection in plants — confirming option A as scientifically accurate and syllabus-aligned.

Question 145 · Principles of Inheritance and Variation

Genes R and Y follow independent assortment. If RRYY produces round yellow seeds and rryy produces wrinkled green seeds, what will be the phenotypic ratio of the F₂ generation?

APhenotypic ratio – 1:2:1

BPhenotypic ratio – 3:1

CPhenotypic ratio – 9:3:3:1

DPhenotypic ratio – 9:7

Answer: C. Phenotypic ratio – 9:3:3:1

This question tests Mendel’s dihybrid cross and the law of independent assortment. When two homozygous parents—RRYY (round, yellow) and rryy (wrinkled, green)—are crossed, the F₁ generation is uniformly RrYy (heterozygous for both traits). Self-pollination of F₁ yields F₂ with 16 possible genotypic combinations. Since R (round) and Y (yellow) are dominant over r (wrinkled) and y (green), the phenotypic classes are: round yellow (R_Y_) = 9, round green (R_yy) = 3, wrinkled yellow (rrY_) = 3, and wrinkled green (rryy) = 1 — giving the classic 9:3:3:1 ratio. This is explicitly covered in NCERT Class 12, Chapter 5, where Mendel’s experiments with seed shape and colour in pea plants are described as foundational evidence for independent assortment. Options A (1:2:1) applies to incomplete dominance monohybrid crosses; B (3:1) is for simple dominant-recessive monohybrid inheritance; D (9:7) reflects complementary gene interaction — not applicable here since no epistasis or gene interaction is indicated. Hence, option C is correct.

Question 146 · Chromosomal Organization and DNA Packaging

Histones are enriched with:

ALysine and Arginine

BLeucine and Lysine

CPhenylalanine and Leucine

DPhenylalanine and Arginine

Answer: A. Lysine and Arginine

Histones are basic proteins rich in positively charged amino acids—lysine and arginine—that interact electrostatically with the negatively charged phosphate backbone of DNA. This interaction enables tight packaging of DNA into nucleosomes, the fundamental repeating units of chromatin. As per NCERT Class 12 Biology (Chapter 5: Molecular Basis of Inheritance), histone proteins H2A, H2B, H3, and H4 contain high proportions of lysine and arginine residues, which facilitate DNA coiling and chromatin condensation. In contrast, leucine and phenylalanine are neutral, hydrophobic amino acids not characteristic of histones; they are more common in structural or enzymatic proteins. The enrichment of basic amino acids is essential for neutralizing DNA’s acidity and stabilizing nucleosome structure. This biochemical feature underpins epigenetic regulation—e.g., acetylation of lysine residues modulates chromatin accessibility. Hence, lysine and arginine are the correct and exclusively characteristic amino acids enriched in histones.

Question 147 · Human Reproduction

The first menstruation is called:

AMenopause

BMenarche

CDiapause

DOvulation

Answer: B. Menarche

Menarche refers to the onset of the first menstrual bleeding in a female adolescent, marking the beginning of reproductive maturity. It typically occurs between ages 10–15 years and is triggered by rising levels of estrogen from maturing ovarian follicles under the influence of FSH and LH. Menarche signifies the establishment of cyclical ovarian activity and endometrial preparation, though initial cycles may be anovulatory. In contrast, menopause is the permanent cessation of menstruation (usually after age 45), diapause is a dormant developmental stage seen in some insects and zooplankton under adverse conditions—not in humans—and ovulation is the release of a mature ovum from the Graafian follicle, occurring mid-cycle. NCERT Class 12 Biology (Chapter 3: Human Reproduction) explicitly defines menarche as 'the first menstrual flow' and distinguishes it from other terms. This foundational concept is critical for understanding pubertal development, menstrual cycle regulation, and reproductive health—core themes in NEET.

Question 148 · Chemical Coordination and Integration

Match List-I with List-II. List-I A. Heart B. Kidney C. Gastro-intestinal tract D. Adrenal cortex List-II I. Erythropoietin II. Aldosterone III. Atrial natriuretic factor IV. Secretin

AA-II, B-I, C-III, D-IV

BA-IV, B-III, C-II, D-I

CA-I, B-III, C-IV, D-II

DA-III, B-I, C-IV, D-II

Answer: D. A-III, B-I, C-IV, D-II

The heart secretes atrial natriuretic factor (ANF) from its atrial walls in response to increased blood volume and pressure; ANF promotes sodium and water excretion to reduce blood pressure. The kidney produces erythropoietin (EPO) in peritubular interstitial cells, stimulating red blood cell production in bone marrow under hypoxic conditions. The gastro-intestinal tract secretes secretin from S-cells of the duodenum in response to acidic chyme; it stimulates bicarbonate secretion by pancreatic ducts. The adrenal cortex secretes aldosterone from zona glomerulosa, regulating Na⁺/K⁺ balance via renal tubules. Thus, correct matching is: A–III (Heart → ANF), B–I (Kidney → Erythropoietin), C–IV (GI tract → Secretin), D–II (Adrenal cortex → Aldosterone). This aligns precisely with option (4) — i.e., D.

Question 149 · Enzymes

The protein portion of an enzyme is called:

ACofactor

BCoenzyme

CApoenzyme

DProsthetic group

Answer: C. Apoenzyme

According to NCERT Class 11 Biology (Chapter 9: Biomolecules), enzymes are biological catalysts that may be simple or conjugated. A conjugated enzyme consists of two parts: the protein portion, known as the apoenzyme, and the non-protein portion, which can be either a cofactor (inorganic ion) or a coenzyme (organic, non-protein molecule). The apoenzyme alone is catalytically inactive; only when combined with its specific cofactor or coenzyme does it form the active holoenzyme. Cofactors (e.g., Mg²⁺, Zn²⁺) assist in catalysis but are not organic. Coenzymes (e.g., NAD⁺, FAD) are organic molecules, often derived from vitamins, and loosely bound. Prosthetic groups are a subset of cofactors — tightly or covalently bound non-protein components (e.g., haem in catalase). Importantly, none of these terms refer to the protein part itself. Thus, the correct term for the protein moiety is 'apoenzyme'. This distinction is explicitly clarified in NCERT’s tabular summary on enzyme components and is frequently tested in NEET.

Question 150 · Ecosystem: Productivity and Energy Flow

Which of the following is the unit of productivity of an ecosystem?

Ag m⁻²

Bkcal m⁻²

Ckcal m⁻³

Dkcal m⁻² yr⁻¹

Answer: D. kcal m⁻² yr⁻¹

Productivity refers to the rate of biomass or energy production per unit area over a given time period. Primary productivity is measured as the amount of organic matter (or energy) synthesized by producers per unit area per unit time. According to NCERT Class 12 Biology (Chapter 14: Ecosystem), the standard unit for productivity is energy-based — specifically kilocalories per square metre per year (kcal m⁻² yr⁻¹). This reflects both spatial (area, not volume) and temporal (per year) dimensions essential for comparing ecosystems. Option A (g m⁻²) is a unit of standing crop (biomass at a point in time), not productivity. Option B (kcal m⁻²) lacks the time component, making it incomplete for a rate. Option C (kcal m⁻³) incorrectly uses volume instead of area — productivity is always expressed per unit area, as energy flow occurs across horizontal surfaces (e.g., forest floor, water surface), not volume. Only option D includes all three critical elements: energy (kcal), area (m⁻²), and time (yr⁻¹), aligning precisely with NCERT’s definition and standard ecological practice.

Question 151 · Evolution: Homologous and analogous organs

Sweet potato and potato represent a certain type of evolution. Select the correct combination of terms to explain this evolutionary relationship.

AAnalogy, convergent

BHomology, divergent

CHomology, convergent

DAnalogy, divergent

Answer: A. Analogy, convergent

Sweet potato (a modified root, *Ipomoea batatas*) and potato (a modified stem or tuber, *Solanum tuberosum*) perform the same function—food storage—but have different developmental origins and structural anatomy. Sweet potato arises from adventitious roots, while potato develops from underground stem branches (stolons) bearing axillary buds. This functional similarity despite structural and embryological dissimilarity is a classic case of analogy. Such organs evolve independently in unrelated lineages due to similar environmental pressures—here, selection for underground storage organs in different plant families (Convolvulaceae vs. Solanaceae). This independent evolution of similar traits is termed convergent evolution. In contrast, homologous structures share common ancestry but may differ in function (e.g., forelimbs of vertebrates). NCERT Class 12 Chapter 7 'Evolution' explicitly cites potato and sweet potato as analogous organs resulting from convergent evolution—reinforcing that analogy reflects convergent, not divergent, evolutionary patterns.

Question 153 · Male gametophyte development in angiosperms

Given below are two statements: One is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): Cells of the tapetum possess dense cytoplasm and generally have more than one nucleus. Reason (R): Presence of more than one nucleus in the tapetum increases the efficiency of nourishing the developing microsporocytes.

ABoth A and R are true and R is the correct explanation of A

BBoth A and R are true but R is NOT the correct explanation of A

CA is true but R is false

DA is false but R is true

Answer: A. Both A and R are true and R is the correct explanation of A

The tapetum is the innermost layer of the anther wall and plays a crucial role in pollen development. As per NCERT Class 12 Biology (Chapter 2: Sexual Reproduction in Flowering Plants), tapetal cells are nutritive, metabolically active, and characteristically exhibit dense cytoplasm and polyploidy or multinucleate condition—often due to endomitosis or incomplete cytokinesis. This cellular specialization supports their function: synthesizing and secreting callase enzyme, lipids, proteins, and sporopollenin precursors essential for microsporocyte (microspore mother cell) meiosis and subsequent pollen wall formation. The multinucleate state enhances biosynthetic capacity and sustains high metabolic demand during rapid pollen development. Thus, Assertion (A) is factually correct, and Reason (R) accurately links nuclear multiplicity to enhanced nourishment—making it not just true, but the direct functional explanation for A. Hence, option A is correct.

Question 154 · Sexual Reproduction in Flowering Plants

How many meiotic and mitotic divisions are required for the development of a mature female gametophyte from the megaspore mother cell in an angiosperm?

A2 meiotic and 3 mitotic divisions

B1 meiotic and 2 mitotic divisions

C1 meiotic and 3 mitotic divisions

DNo meiotic and 2 mitotic divisions

Answer: C. 1 meiotic and 3 mitotic divisions

In angiosperms, the megaspore mother cell (MMC), a diploid cell in the ovule, undergoes one meiotic division to produce four haploid megaspores. Typically, three degenerate and only one functional megaspore survives. This functional megaspore then undergoes three successive mitotic divisions — without cytokinesis in the first two rounds — resulting in an eight-nucleate, seven-celled mature embryo sac (female gametophyte): three antipodals, two synergids, one egg cell, and one central cell with two polar nuclei. Thus, exactly one meiotic division (to form the haploid megaspore) and three mitotic divisions (to generate eight nuclei) are required. NCERT Class 12, Chapter 2 'Sexual Reproduction in Flowering Plants', clearly states: 'The functional megaspore undergoes three mitotic divisions to form eight nuclei...'. Option C correctly reflects this sequence. Options A and B overcount or undercount divisions; D incorrectly omits meiosis, which is essential for ploidy reduction.

Question 155 · Morphology of Flowering Plants

Which of the following is an example of a zygomorphic flower?

APetunia

BDatura

CPea

DChilli

Answer: C. Pea

Zygomorphic flowers exhibit bilateral symmetry — they can be divided into two equal halves only along one vertical plane. Among the given options, the pea flower (Pisum sativum) is a classic example: its corolla is papilionaceous, with one large standard petal, two lateral wing petals, and two fused keel petals forming a boat-shaped structure — all arranged asymmetrically to permit pollination by specific insects like bees. In contrast, Petunia and Datura have actinomorphic (radially symmetrical) flowers — their floral parts are arranged symmetrically around the central axis, allowing division into equal halves in multiple planes. Chilli (Capsicum annuum) also has an actinomorphic flower with five nearly identical petals. NCERT Class 11, Chapter 5 'Morphology of Flowering Plants', explicitly lists pea as a zygomorphic example and contrasts it with actinomorphic flowers like mustard and datura. Thus, option C (Pea) is scientifically accurate and fully aligned with NCERT content.

Question 156 · Immune System

After maturation in primary lymphoid organs, lymphocytes migrate to secondary lymphoid organs/tissues for interaction with antigens. Which of the following are secondary lymphoid organs/tissues? A. Thymus B. Bone marrow C. Spleen D. Lymph nodes E. Peyer’s patches

AB, C, D only

BA, B, C only

CE, A, B only

DC, D, E only

Answer: D. C, D, E only

Primary lymphoid organs — bone marrow and thymus — are sites of lymphocyte development and maturation. B-cells mature in bone marrow; T-cells mature in thymus. Once mature, naïve lymphocytes enter circulation and home to secondary lymphoid organs, where they encounter antigens and initiate adaptive immune responses. Secondary lymphoid organs include spleen (filters blood-borne antigens), lymph nodes (filter lymph from tissues), and mucosa-associated lymphoid tissue (MALT) such as Peyer’s patches in the ileum (sample gut lumen antigens). Thymus and bone marrow are *not* secondary organs — they lack antigen-presenting dendritic cells in functional microenvironments for naïve lymphocyte activation and do not serve as sites for antigen-driven clonal expansion. Hence, only spleen (C), lymph nodes (D), and Peyer’s patches (E) qualify. Option (4) — C, D, E only — is correct. This aligns precisely with NCERT Class 12 Biology Chapter 8 (Human Health and Disease), which states: 'Secondary lymphoid organs provide the sites for interaction of lymphocytes with the antigen.'

Question 158 · Cell Cycle and Division

What is the main function of the spindle fibers during mitosis?

ATo separate the chromosomes

BTo synthesize new DNA

CTo repair damaged DNA

DTo regulate cell growth

Answer: A. To separate the chromosomes

Spindle fibers, composed of microtubules, form during prophase and attach to the centromeres of chromosomes via kinetochores. Their primary role is to physically segregate sister chromatids during anaphase—pulling them toward opposite poles of the cell. This ensures each daughter cell receives an identical set of chromosomes. Spindle formation is regulated by centrosomes (in animal cells) and depends on dynamic assembly/disassembly of tubulin subunits. DNA synthesis occurs exclusively in the S phase *before* mitosis begins and is mediated by DNA polymerases—not spindle fibers. DNA repair involves enzymes like DNA ligase and endonucleases, while cell growth regulation is governed by cyclins, CDKs, and signaling pathways such as mTOR—not structural components of the mitotic apparatus. As per NCERT Class 11 Biology (Chapter 10: Cell Cycle and Cell Division), spindle fibers are explicitly described as essential for chromosome movement and segregation, making option A the only biologically accurate choice.

Question 159 · Plant Kingdom – Gymnosperms

Which one of the following is the characteristic feature of gymnosperms?

ASeeds are enclosed in fruits.

BSeeds are naked.

CSeeds are absent.

DGymnosperms have flowers for reproduction.

Answer: B. Seeds are naked.

Gymnosperms are seed-producing plants in which seeds are not enclosed within an ovary or fruit; instead, they remain exposed on the surface of sporophylls (e.g., cones), hence termed 'naked seeds'. This is a defining feature distinguishing them from angiosperms, whose seeds develop inside a fruit derived from a ripened ovary. Gymnosperms lack true flowers and fruits — their reproductive structures are cones (strobili), with male microsporangiate cones producing pollen and female megasporangiate cones bearing ovules directly on the bracts (megasporophylls). They do possess seeds (so option C is incorrect), and since seeds are present but uncovered, option A contradicts their fundamental classification. Option D is invalid because flowers — defined as specialized reproductive shoots with sepals, petals, stamens, and carpels — are exclusive to angiosperms. NCERT Class 11 Biology (Chapter 3: Plant Kingdom) explicitly states: 'The gymnosperms are the plants in which ovules are not enclosed by any ovary wall and remain exposed before and after fertilisation.' Thus, 'seeds are naked' is the correct and characteristic feature.

Question 160 · Chemical Coordination and Integration

Consider the following statements regarding the function of adrenal medullary hormones: A. It causes pupillary constriction. B. It is a hyperglycemic hormone. C. It causes piloerection. D. It increases strength of heart contraction. Choose the correct answer from the options given below:

AC and D Only

BB, C and D Only

CA, C and D Only

DD Only

Answer: B. B, C and D Only

Adrenal medullary hormones—epinephrine and norepinephrine—are catecholamines secreted during stress. They prepare the body for 'fight-or-flight' response. Statement A is incorrect: these hormones cause *pupillary dilation* (mydriasis), not constriction—mediated by sympathetic activation of radial iris muscles. Statement B is correct: epinephrine stimulates glycogenolysis in liver and muscle, increasing blood glucose—hence it is hyperglycemic. Statement C is correct: piloerection (goosebumps) results from sympathetic stimulation of arrector pili muscles. Statement D is correct: epinephrine enhances cardiac output by increasing heart rate (chronotropy) and contractility (inotropy) via β₁-adrenergic receptors. Thus, only statements B, C, and D are true. This aligns with NCERT Class 11 Biology (Chapter 22: Chemical Coordination and Integration), which states that adrenal medulla secretes epinephrine/norepinephrine to increase alertness, pupil dilation, sweating, heart rate, and blood glucose—while suppressing non-essential functions like digestion.

Question 161 · Chemical Coordination and Integration

Why can't insulin be given orally to diabetic patients?

AHuman body will elicit a strong immune response

BIt will be digested in the gastrointestinal (GI) tract

CBecause of structural variation

DIts bioavailability will be increased

Answer: B. It will be digested in the gastrointestinal (GI) tract

Insulin is a peptide hormone composed of 51 amino acids arranged in two chains (A and B) linked by disulfide bonds. As a protein, it is highly susceptible to enzymatic degradation by proteases (e.g., pepsin in stomach, trypsin and chymotrypsin in duodenum) present throughout the gastrointestinal (GI) tract. Oral administration would expose insulin to acidic pH and digestive enzymes, leading to its complete hydrolysis into inactive amino acids before absorption. Hence, it cannot reach systemic circulation intact — resulting in negligible bioavailability. This is why insulin must be administered parenterally (subcutaneously or intravenously) to bypass GI degradation. NCERT Class 11 Biology (Chapter 22: Chemical Coordination and Integration) explicitly states that peptide hormones like insulin 'cannot be administered orally as they are digested in the gut'. Options A and C are incorrect: insulin is not inherently immunogenic when administered correctly (though rare anti-insulin antibodies may develop with chronic use), and 'structural variation' is vague and biologically inaccurate here. Option D is factually wrong — oral delivery drastically reduces, not increases, bioavailability.

Question 162 · Plant Kingdom Classification

Match List I with List II. List I A. Pteridophyte B. Bryophyte C. Angiosperm D. Gymnosperm List II I. Salvia II. Ginkgo III. Polytrichum IV. Salvinia Choose the option with all correct matches.

AA-III, B-IV, C-II, D-I

BA-IV, B-III, C-I, D-II

CA-III, B-IV, C-I, D-II

DA-IV, B-III, C-II, D-I

Answer: B. A-IV, B-III, C-I, D-II

Pteridophytes are vascular non-seed plants; Salvinia is a heterosporous aquatic fern (pteridophyte), so A ↔ IV. Bryophytes are non-vascular embryophytes; Polytrichum is a moss (bryophyte), so B ↔ III. Angiosperms are flowering seed plants with double fertilization and fruits; Salvia (sage) is a dicot angiosperm, so C ↔ I. Gymnosperms are naked-seed plants with no ovary or fruit; Ginkgo is a living fossil gymnosperm with motile sperm and exposed ovules, so D ↔ II. Option (2) — A-IV, B-III, C-I, D-II — correctly maps all pairs. NCERT Class 11 Chapter 3 'Plant Kingdom' explicitly classifies Salvinia under Pteridophyta, Polytrichum under Bryophyta, Salvia (as a representative flowering plant) under Angiosperms, and Ginkgo under Gymnosperms. This classification aligns with morphological criteria: presence/absence of vascular tissue, seeds, flowers, and fruit — all emphasized in NCERT for NEET-level conceptual clarity.

Question 163 · Molecular Basis of Inheritance

Who proposed that the genetic code for amino acids should be made up of three nucleotides?

AGeorge Gamow

BFrancis Crick

CJacques Monod

DMatthew Meselson and Franklin Stahl

Answer: A. George Gamow

George Gamow, a theoretical physicist, first proposed in 1954 that the genetic code must be a triplet code — i.e., three nucleotides specify one amino acid — based on combinatorial reasoning: with four nucleotides (A, U, C, G), only 4³ = 64 possible triplets exist, which is sufficient to encode all 20 standard amino acids. Though Francis Crick and colleagues provided critical experimental evidence for the triplet nature using frameshift mutations in bacteriophage T4 (1961), and Nirenberg and Matthaei later deciphered the first codon (UUU → phenylalanine) in 1961, the original conceptual proposal of the triplet code belongs to Gamow. Jacques Monod co-developed the operon model; Matthew Meselson and Franklin Stahl (not 'Franklin Stahl' alone) demonstrated semi-conservative DNA replication. NCERT Class 12 Chapter 6 explicitly credits Gamow for the 'triplet hypothesis' as the foundational idea, while acknowledging subsequent experimental validation by others.

Question 164 · Biodiversity and Conservation

Match List I with List II: List I A. The Evil Quartet B. Ex situ conservation C. Lantana camara D. Dodo List II I. Cryopreservation II. Alien species invasion III. Causes of biodiversity losses IV. Extinction Choose the option with all correct matches.

AA-II, B-III, C-I, D-IV

BA-III, B-I, C-II, D-IV

CA-II, B-IV, C-II, D-I

DA-III, B-II, C-IV, D-I

Answer: B. A-III, B-I, C-II, D-IV

The 'Evil Quartet' (A) refers to the four major causes of biodiversity loss — habitat loss/degradation, invasive alien species, over-exploitation, and co-extinctions — hence matching with III. Ex situ conservation (B) includes techniques like cryopreservation, seed banks, and zoos to protect species outside their natural habitats, so it correctly pairs with I. Lantana camara (C) is a classic example of an invasive alien species that outcompetes native flora, disrupting ecosystems — thus matching II. The Dodo (D), endemic to Mauritius, was driven to extinction by human activities (hunting, habitat alteration) in the 17th century, making IV the accurate match. This aligns precisely with NCERT Class 12 Biology Chapter 15: 'Biodiversity and its Conservation', which explicitly defines the Evil Quartet, lists alien species as a cause of loss, identifies ex situ methods including cryopreservation, and cites the Dodo as a historical case of extinction. Option B (A-III, B-I, C-II, D-IV) is therefore scientifically sound and NCERT-accurate.

Question 165 · Chemical Coordination and Integration

Which of the following hormones released from the pituitary gland is actually synthesized in the hypothalamus?

ALuteinizing hormone (LH)

BAnti-diuretic hormone (ADH)

CFollicle-stimulating hormone (FSH)

DAdrenocorticotropic hormone (ACTH)

Answer: B. Anti-diuretic hormone (ADH)

The hypothalamus synthesizes two neurohormones—oxytocin and anti-diuretic hormone (ADH)—which are then transported via axons of hypothalamic neurons (supraoptic and paraventricular nuclei) to the posterior pituitary (neurohypophysis) for storage and release. Thus, although ADH is released from the posterior pituitary, it is synthesized in the hypothalamus. In contrast, LH, FSH, and ACTH are adenohypophyseal hormones secreted by anterior pituitary cells (gonadotropes and corticotropes) and synthesized *within* the anterior pituitary itself under hypothalamic control via releasing/inhibiting hormones (e.g., GnRH, CRH). NCERT Class 11 Biology (Chapter 22: Chemical Coordination and Integration) explicitly states: 'Oxytocin and vasopressin (ADH) are synthesised by the hypothalamus and stored and released from the neurohypophysis.' This distinction between synthesis site (hypothalamus) and release site (posterior pituitary) is fundamental—and makes ADH the only correct choice among the options.

Question 167 · Immune System

Which of the following types of immunity is present at the time of birth and is a non-specific type of defence in the human body?

AAcquired Immunity

BInnate Immunity

CCell-mediated Immunity

DHumoral Immunity

Answer: B. Innate Immunity

Innate immunity is the first line of defence that an individual is born with — it is genetically predetermined, non-specific, and acts immediately against pathogens without prior exposure. As per NCERT Class 12 Biology (Chapter 8: Human Health and Disease), innate immunity includes physical barriers (skin, mucous membranes), physiological barriers (acidic stomach pH, lysozyme in tears), cellular barriers (phagocytes like neutrophils and macrophages), and cytokine barriers (interferons). Unlike acquired immunity — which develops after exposure and is antigen-specific — innate immunity does not involve memory cells or antibody production. Cell-mediated and humoral immunity are subsets of adaptive (acquired) immunity, dependent on T-lymphocytes and B-lymphocytes respectively, and require time to activate upon first encounter. Hence, only innate immunity satisfies both criteria: presence at birth and non-specificity.

Question 168 · Plant Kingdom – Bryophytes

In bryophytes, the gemmae help in which one of the following?

ASexual reproduction

BAsexual reproduction

CNutrient absorption

DGaseous exchange

Answer: B. Asexual reproduction

Gemmae are multicellular, asexual, propagative structures produced in specialized cup-like receptacles called gemma cups, commonly found on the thallus of liverworts such as Marchantia. They detach upon maturity and germinate to form new, genetically identical individuals—making them a key mechanism of vegetative (asexual) reproduction. Gemmae are not involved in sexual reproduction, which in bryophytes occurs via antheridia (male) and archegonia (female) producing gametes. They lack vascular tissue or root-like structures, so they do not absorb nutrients like true roots; nutrient uptake occurs directly through the thallus surface. Similarly, gaseous exchange occurs via diffusion across the thin, moist gametophytic surface—not through gemmae. As per NCERT Class 11 Biology (Chapter 3: Plant Kingdom), gemmae are explicitly cited as a means of asexual reproduction in bryophytes, reinforcing their role in clonal propagation without fertilization.

Question 169 · Circulatory System in Frogs

In frogs, the renal portal system is a special venous connection that links:

ALiver and intestine

BLiver and kidney

CKidney and intestine

DKidney and lower part of the body

Answer: D. Kidney and lower part of the body

The renal portal system is a characteristic feature of amphibians like frogs and is absent in birds and mammals. It consists of a network of veins that collects deoxygenated blood from the posterior (hind) part of the body—including the legs, pelvic region, and tail—and directs it *first* to the kidneys via the renal portal vein before returning it to the heart. This arrangement allows the kidneys to filter metabolic wastes (e.g., urea, ammonia) directly from blood draining the lower body, enhancing excretory efficiency. Importantly, it does *not* connect liver–intestine (that’s the hepatic portal system), liver–kidney (no direct portal link), or kidney–intestine (no such anatomical pathway). NCERT Class 11 Biology (Chapter 18: Body Fluids and Circulation) explicitly states: 'In frogs, there is a renal portal system... which carries blood from the hind limbs to the kidneys.' Thus, option D — 'Kidney and lower part of the body' — correctly describes this functional venous linkage.

Question 170 · Energy Flow in Ecosystems

Given below are two statements: Statement I: In an ecosystem, there is unidirectional flow of energy from the sun through producers to consumers. Statement II: Ecosystems are exempted from the Second Law of Thermodynamics. In the light of the above statements, choose the most appropriate answer from the options given below:

ABoth Statement I and Statement II are correct

BBoth Statement I and Statement II are incorrect

CStatement I is correct but Statement II is incorrect

DStatement I is incorrect but Statement II is correct

Answer: C. Statement I is correct but Statement II is incorrect

Statement I is correct: Energy enters an ecosystem as sunlight, is converted to chemical energy by producers (autotrophs) via photosynthesis, and flows linearly — from producers to primary consumers, then to secondary and tertiary consumers — with significant loss as heat at each trophic level. This flow is strictly unidirectional; energy is not recycled. Statement II is incorrect: Ecosystems fully obey the Second Law of Thermodynamics, which states that energy transformations are never 100% efficient and always result in increased entropy (disorder). The progressive loss of usable energy as heat across trophic levels — quantified by ecological pyramids and the 10% law — is a direct manifestation of this law. No biological system, including ecosystems, can circumvent thermodynamic laws; they are universal physical principles. NCERT Class 12 Biology (Chapter 14: Ecosystem) explicitly affirms that energy flow is unidirectional and that ecosystems operate within thermodynamic constraints, reinforcing that energy degradation follows the Second Law.

Question 171 · Photosynthesis in Higher Plants

Which of the following statements about RuBisCO is true?

AIt is active only in the dark.

BIt has higher affinity for oxygen than carbon dioxide.

CIt is an enzyme involved in the photolysis of water.

DIt catalyzes the carboxylation of RuBP.

Answer: D. It catalyzes the carboxylation of RuBP.

RuBisCO (Ribulose-1,5-bisphosphate carboxylase/oxygenase) is the most abundant enzyme on Earth and plays a central role in the Calvin cycle. It catalyzes the first major step of carbon fixation — the carboxylation of ribulose-1,5-bisphosphate (RuBP) using CO₂ to form two molecules of 3-phosphoglycerate. This reaction occurs in the stroma of chloroplasts during the light-independent phase, but RuBisCO itself is not light-activated directly; rather, its activity is regulated by light-induced changes in stromal pH and Mg²⁺ concentration. Contrary to option (A), it functions in both light and dark conditions *in vitro*, though *in vivo* its activity is coupled to light-driven metabolic conditions. Option (B) is misleading: while RuBisCO *can* bind O₂ (leading to photorespiration), its affinity (Km) for CO₂ (~10–25 µM) is actually *lower* than for O₂ (~250–500 µM), meaning it binds O₂ more readily *only when O₂ concentration is high and CO₂ low* — but its intrinsic affinity (measured as 1/Km) is higher for CO₂. Option (C) is incorrect: photolysis of water is carried out by the oxygen-evolving complex of PSII, not RuBisCO. Hence, only statement (D) is unequivocally correct and fully aligned with NCERT Class 11 (Chapter 13, page 220).

Question 173 · Plant Growth and Development

Read the following statements on plant growth and development: A. Parthenocarpy can be induced by auxins. B. Plant growth regulators can be involved in promotion as well as inhibition of growth. C. Dedifferentiation is a pre-requisite for re-differentiation. D. Abscisic acid is a plant growth promoter. E. Apical dominance promotes the growth of lateral buds. Choose the option with all correct statements.

AA, B, C only

BA, C, E only

CA, D, E only

DB, D, E only

Answer: A. A, B, C only

Statement A is correct: Auxins (e.g., IAA, NAA) induce parthenocarpy — fruit development without fertilization — as demonstrated in tomatoes and cucumbers (NCERT Class 11, Ch. 15). Statement B is correct: PGRs like auxins promote root formation but inhibit lateral bud growth; cytokinins promote cell division while abscisic acid inhibits seed germination — confirming dual roles (NCERT, p. 250–251). Statement C is correct: Dedifferentiation — where mature cells regain meristematic capacity (e.g., cork cambium formation from parenchyma) — must occur before re-differentiation into new specialized tissues (NCERT, p. 243–244). Statement D is false: Abscisic acid (ABA) is a growth inhibitor — it induces dormancy, closes stomata during stress, and counteracts gibberellins. Statement E is false: Apical dominance suppresses lateral bud growth via auxin from the apical meristem; removal of apex releases inhibition, allowing lateral growth. Thus, only A, B, and C are correct — matching option (1).

Question 175 · Respiratory adaptations in amphibians

Frogs respire in water by skin and buccal cavity, and on land by skin, buccal cavity and lungs. Choose the correct statement regarding this.

AThe statement is true for water but false for land

BThe statement is true for both the environment

CThe statement is false for water but true for land

DThe statement is false for both the environment

Answer: C. The statement is false for water but true for land

Frogs are amphibious and exhibit cutaneous, buccal, and pulmonary respiration depending on habitat. In water, gas exchange occurs primarily through moist skin (cutaneous) and the buccal cavity (buccopharyngeal respiration); lungs remain non-functional as they are poorly developed and not ventilated underwater. On land, frogs use all three modes: cutaneous (skin), buccal (lining of mouth cavity), and pulmonary (lungs) — with lungs becoming the major respiratory organ. Thus, the given statement incorrectly claims that frogs respire by lungs *in water*, which is physiologically impossible due to absence of ventilation and risk of drowning. Hence, the statement is false for water (since lungs aren’t used) but true for land (where all three — skin, buccal cavity, and lungs — function). This aligns with NCERT Class 11, Chapter 17 'Breathing and Exchange of Gases', which explicitly states that 'in frogs, cutaneous and buccal respiration occur in water, while pulmonary respiration supplements them on land'.

Question 176 · Human Reproduction: Gametogenesis and Fertilisation

Twins are born to a family that lives next door to you. The twins are a boy and a girl. Which of the following must be true?

AThey are monozygotic twins.

BThey are fraternal twins.

CThey were conceived through in vitro fertilization.

DThey have 75% identical genetic content.

Answer: B. They are fraternal twins.

Monozygotic (identical) twins arise from a single zygote formed by the fusion of one sperm and one ovum; hence, they are always of the same sex and genetically nearly identical. In contrast, dizygotic (fraternal) twins result from the fertilisation of two separate ova by two separate sperm — a process equivalent to normal conception of siblings born at the same time. Since dizygotic twins develop from two distinct zygotes, they can be of different sexes (e.g., one male and one female), just like ordinary siblings. Therefore, boy–girl twins must be dizygotic (fraternal), making option B the only statement that must be true. Option A is incorrect because monozygotic twins cannot be of opposite sexes. Option C is unsupported — natural conception is far more common than IVF, and no evidence suggests assisted reproduction is required. Option D misrepresents genetics: fraternal twins share ~50% of their DNA on average (like full siblings), not 75%. This aligns with NCERT Class 12 Chapter 3 (Human Reproduction), which states that dizygotic twins are genetically no more similar than ordinary siblings.

Question 177 · Microbes in Household Products

Which of the following microbes is NOT involved in the preparation of household products?

AAspergillus niger and Lactobacillus

BAspergillus niger and Trichoderma polysporum

CTrichoderma polysporum and Saccharomyces cerevisiae

DTrichoderma polysporum and Propionibacterium shermanii

Answer: B. Aspergillus niger and Trichoderma polysporum

According to NCERT Class 12 Biology (Chapter 10: Microbes in Human Welfare), several microbes are used in household product preparation: Lactobacillus ferments milk into curd; Saccharomyces cerevisiae (baker’s yeast) leavens bread and brews alcoholic beverages; Propionibacterium shermanii produces characteristic holes and flavor in Swiss cheese; Aspergillus niger is used industrially for citric acid production — though not typically in *household* settings, NCERT explicitly lists it under 'industrial products' (not household). Crucially, Trichoderma polysporum is *not* used in household products; it produces cyclosporin A — an immunosuppressive drug used in organ transplantation — falling under 'microbes in medicine', not food or domestic applications. Thus, the pair 'Aspergillus niger and Trichoderma polysporum' includes one microbe (Trichoderma) wholly uninvolved in household products, while Aspergillus niger’s role is industrial, not domestic — making option B the correct choice identifying the *non-household* pair. NCERT clearly distinguishes household (curd, bread, cheese) from industrial (citric acid, enzymes) and medical (cyclosporin, streptokinase) applications.

Question 179 · Recombinant DNA Technology

The blue-white selectable markers have been developed to differentiate recombinant colonies from non-recombinant colonies based on their ability to produce colour in the presence of a chromogenic substrate. Given below are two statements about this method: Statement I: The blue-coloured colonies have DNA insert in the plasmid and are identified as recombinant colonies. Statement II: The colonies without blue colour have DNA insert in the plasmid and are identified as recombinant colonies. In the light of the above statements, choose the most appropriate answer.

ABoth Statement I and Statement II are correct

BBoth Statement I and Statement II are incorrect

CStatement I is correct but Statement II is incorrect

DStatement I is incorrect but Statement II is correct

Answer: D. Statement I is incorrect but Statement II is correct

In blue-white screening, the plasmid vector carries the lacZ gene encoding β-galactosidase, which cleaves the chromogenic substrate X-gal to yield a blue pigment. When foreign DNA is inserted into the multiple cloning site (MCS) within lacZ, it disrupts the gene — rendering β-galactosidase nonfunctional. Thus, recombinant colonies (with insert) cannot hydrolyse X-gal and remain white. Non-recombinant colonies (without insert) retain functional lacZ and produce blue colour. Therefore, Statement I is incorrect because blue colonies lack the insert and are non-recombinants. Statement II is correct: white (non-blue) colonies contain the insert and are recombinants. This principle is explicitly described in NCERT Class 12 Biology Chapter 11 (Biotechnology: Principles and Processes), where it states that 'insertional inactivation of lacZ results in white colonies'. The method relies on visual phenotypic screening and is foundational for cloning efficiency assessment.