Revision guide
How to use NEET 2024 Biology PYQs
This set contains reviewed questions from the Code T1 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
- Biotechnology: Principles and Processes
- Human Health and Disease
Best review method
After solving, make a two-column error log: one side for the NCERT fact you missed and one side for why the distractor looked plausible.
NCERT focus
Re-read biotechnology enzymes and vectors, genetic-process vocabulary, immunity terms and disease-causing organisms from NCERT before attempting the set again.
Frequently asked questions
NEET 2024 Biology PYQ FAQs
Are these NEET 2024 Biology answers checked with the official key?
This page uses the NEET 2024 Code T1 English paper and checks answers against the official final answer key before publication.
What should I revise after NEET 2024 Biology PYQs?
Review the linked NCERT concepts behind each mistake, especially molecular biology, biotechnology and human health topics, before reattempting the incorrect questions.
Question 102 · Biodiversity and Conservation
The list of endangered species was released by-
Answer: B. IUCN
The International Union for Conservation of Nature (IUCN) is the global authority that maintains and publishes the IUCN Red List of Threatened Species — the world’s most comprehensive inventory of the conservation status of biological species. It assesses extinction risk using rigorous, science-based criteria and categorizes species as Extinct, Critically Endangered, Endangered, Vulnerable, etc. FOAM is not a recognized environmental body; GEAC (Genetic Engineering Appraisal Committee) functions under India’s Ministry of Environment and deals with GMO regulation, not species listing; WWF (World Wide Fund for Nature) supports conservation but does not authoritatively publish the official global endangered species list. NCERT Class 12 Biology (Chapter 15: Biodiversity and Conservation) explicitly states that the IUCN Red List is the primary source for evaluating species threat status and forms the scientific basis for international and national conservation policies.
Question 105 · Regulation of Gene Expression in Prokaryotes
The lactose present in the growth medium of bacteria is transported into the cell by the action of:
APermease
BPolymerase
CBeta-galactosidase
DAcetylase
Answer: A. Permease
In the lac operon of E. coli, lactose transport across the plasma membrane is mediated by lactose permease — a transmembrane protein encoded by the lacY gene. Permease facilitates the active, energy-dependent uptake of extracellular lactose into the bacterial cell. Once inside, lactose is hydrolysed by beta-galactosidase (lacZ gene product) into glucose and galactose. Acetylase (lacA gene product) detoxifies certain thiogalactosides but plays no role in transport. Polymerase refers to RNA polymerase, which transcribes the lac operon but does not participate in substrate transport. NCERT Class 12 Biology (Chapter 6: Molecular Basis of Inheritance) explicitly states that 'the lac operon includes genes coding for enzymes involved in lactose metabolism — lacZ for beta-galactosidase, lacY for permease, and lacA for transacetylase'. Hence, permease is solely responsible for lactose uptake, making option A correct.
Question 107 · Population Growth Models
The equation of Verhulst-Pearl logistic growth is dN/dt = rN(K − N)/K. From this equation, K indicates:
ACarrying capacity
BPopulation density
CIntrinsic rate of natural increase
DBiotic potential
Answer: A. Carrying capacity
In the Verhulst-Pearl logistic growth equation — dN/dt = rN(K − N)/K — K represents the carrying capacity of the environment, i.e., the maximum population size that a given habitat can sustain indefinitely with available resources. This concept is central to Chapter 13 (Organisms and Populations) in NCERT Class 12 Biology. Carrying capacity arises due to limiting factors like food, space, predation, and disease, which cause growth to slow as N approaches K, resulting in the characteristic sigmoid (S-shaped) growth curve. In contrast, 'r' denotes the intrinsic rate of natural increase (option C), reflecting per capita birth minus death rate under ideal conditions; 'population density' (B) is simply N (number of individuals per unit area); and 'biotic potential' (D) refers to the maximum reproductive capacity of a species in absence of limiting factors — a theoretical upper limit, not a parameter in the logistic equation. Thus, only option A correctly identifies K.
Question 108 · Molecular Basis of Inheritance
A transcription unit in DNA is defined primarily by three regions with respect to the upstream and downstream ends:
AInducer, Repressor, Structural gene
BPromoter, Structural gene, Terminator
CRepressor, Operator gene, Structural gene
DStructural gene, Transposons, Operator gene
Answer: B. Promoter, Structural gene, Terminator
A transcription unit in DNA consists of three essential regions: the promoter (upstream), the structural gene(s) (coding region), and the terminator (downstream). The promoter is where RNA polymerase binds to initiate transcription; it lies upstream of the structural gene. The structural gene contains the coding sequence for RNA/protein. The terminator signals the end of transcription and lies downstream. Inducers, repressors, operators, and transposons are regulatory or mobile elements but not core components defining a transcription unit. Inducers and repressors are molecules (not DNA regions); operators are regulatory DNA sequences (e.g., in operons) but not universal to all transcription units; transposons are mobile genetic elements unrelated to basic transcription unit architecture. As per NCERT Class 12 Biology Chapter 6 (Molecular Basis of Inheritance), the standard definition explicitly identifies promoter, structural gene, and terminator as the three defining regions of a transcription unit — making option B correct.
Question 109 · Plant Development and Tissue Differentiation
Formation of interfascicular cambium from fully developed parenchyma cells is an example of
ADedifferentiation
BMaturation
CDifferentiation
DRedifferentiation
Answer: A. Dedifferentiation
Interfascicular cambium arises from parenchyma cells located between vascular bundles in dicot stems. These parenchyma cells are mature, differentiated, and non-dividing under normal conditions. To form cambium — a meristematic tissue capable of active cell division — they must revert to a less specialized, proliferative state. This reversion from a differentiated, functionally specialized cell (parenchyma) to a meristematic, undifferentiated-like state is termed dedifferentiation. As per NCERT Class 11 Biology (Chapter 15: Plant Growth and Development), dedifferentiation is defined as the phenomenon where living, differentiated cells that have lost the capacity to divide regain the power of division. Maturation refers to the final stage of differentiation where cells attain permanent structure and function; differentiation is the process of acquiring specific form and function; redifferentiation occurs when dedifferentiated cells again differentiate into another specialized cell type (e.g., xylem or phloem elements). Since no new specialization follows here — only acquisition of meristematic potential — it is dedifferentiation, not redifferentiation.
Question 110 · Biological Classification – Fungi
Match List I with List II:
List I List II
A. Rhizopus I. Mushroom
B. Ustilago II. Smut fungus
C. Puccinia III. Bread mould
D. Agaricus IV. Rust fungus
Choose the correct answer from the options given below:
AA-III, B-II, C-I, D-IV
BA-IV, B-II, C-III, D-I
CA-III, B-II, C-IV, D-I
DA-I, B-III, C-II, D-IV
Answer: C. A-III, B-II, C-IV, D-I
Rhizopus is a saprophytic zygomycete commonly known as bread mould (List I–A ↔ List II–III). Ustilago causes smut disease in cereals like wheat and maize and is correctly classified as a smut fungus (B ↔ II). Puccinia is a basidiomycete responsible for rust diseases in wheat and other grasses, hence it matches with IV (rust fungus). Agaricus, including species like A. bisporus, is an edible basidiomycete widely cultivated as mushroom (D ↔ I). This matching aligns precisely with NCERT Class 11 Biology (Chapter 2: Biological Classification), which categorises fungi based on morphology, mode of spore formation, and disease causation. The confusion sometimes arises due to superficial similarities, but taxonomically, Rhizopus (zygomycetes) differs fundamentally from Agaricus and Puccinia (both basidiomycetes), while Ustilago is also a basidiomycete but belongs to the smut group. Thus, option C (A–III, B–II, C–IV, D–I) is biologically accurate and NCERT-consistent.
Question 111 · Recombinant DNA Technology
What is the fate of a piece of DNA carrying only the gene of interest which is transferred into an alien organism?
AThe piece of DNA would be able to multiply itself independently in the progeny cells of the organism.
BIt may get integrated into the genome of the recipient.
CIt may multiply and be inherited along with the host DNA.
DThe alien piece of DNA is not an integral part of chromosome.
Answer: A. The piece of DNA would be able to multiply itself independently in the progeny cells of the organism.
In recombinant DNA technology, a DNA fragment containing only the gene of interest (without an origin of replication) cannot replicate autonomously in the host cell. For independent multiplication, it must be cloned into a vector — such as a plasmid or bacteriophage — that carries an origin of replication (ori), selectable markers, and cloning sites. Without such a vector, the naked DNA fragment lacks ori and thus cannot initiate replication; it will not multiply independently in progeny cells. Option A correctly states this limitation. Option B is partially true only if integration occurs via homologous recombination or transposase activity — but this is rare and not guaranteed for a random fragment. Option C misleads: without replication machinery, it won’t multiply or be stably inherited. Option D is vague and inaccurate — integration (if any) would make it chromosomal, but non-integrated fragments are degraded. NCERT Class 12 (Chapter 11, Biotechnology: Principles and Processes) explicitly states that 'a piece of DNA without an origin of replication will not be able to replicate', confirming that autonomous multiplication is impossible — hence only statement A is universally correct.
Question 112 · Cell: The Unit of Life
Match List I with List II:
List I
A. Nucleolus
B. Centriole
C. Leucoplasts
D. Golgi apparatus
List II
I. Site of formation of glycolipids
II. Organization like the cartwheel
III. Site for active ribosomal RNA synthesis
IV. For storing nutrients
AA-III, B-IV, C-II, D-I
BA-I, B-II, C-III, D-IV
CA-III, B-II, C-IV, D-I
DA-II, B-III, C-I, D-IV
Answer: C. A-III, B-II, C-IV, D-I
The nucleolus (A) is a dense region within the nucleus where ribosomal RNA (rRNA) is actively transcribed and ribosomal subunits are assembled — matching with III. Centrioles (B) possess a characteristic 9-fold symmetrical cartwheel structure composed of tubulin proteins, essential for organizing microtubules during cell division — correctly paired with II. Leucoplasts (C) are colourless plastids specialized for storage; amyloplasts store starch, elaioplasts store oils, and proteinoplasts store proteins — thus functionally aligned with IV (storing nutrients). The Golgi apparatus (D) modifies, sorts, and packages lipids and proteins; it synthesizes glycolipids by adding carbohydrate moieties to lipids — hence matches I. Option C (A-III, B-II, C-IV, D-I) is therefore correct. This mapping is fully supported by NCERT Class 11 Biology Chapter 8 (‘Cell: The Unit of Life’), which explicitly describes the nucleolus as the site of rRNA synthesis, centrioles as cartwheel-organized organelles, leucoplasts as storage plastids, and Golgi as the site of glycolipid formation.
Question 114 · Cell Cycle and Cell Division
Spindle fibers attach to kinetochores of chromosomes during
AAnaphase
BTelophase
CProphase
DMetaphase
Answer: D. Metaphase
Spindle fibers attach to the kinetochores — protein structures assembled on the centromere of each sister chromatid — exclusively during metaphase of mitosis. In prophase, spindle apparatus begins to form and chromosomes condense, but kinetochore microtubules have not yet established stable attachments. By late prophase/prometaphase, microtubules start probing and capturing kinetochores, but bi-orientation (i.e., attachment of sister kinetochores to opposite spindle poles) is completed and chromosomes align at the equatorial plate only in metaphase. This alignment ensures equal segregation in anaphase. NCERT Class 11 (Chapter 10: Cell Cycle and Cell Division) explicitly states: 'During metaphase, the chromosomes are moved to the spindle equator and get aligned along the metaphase plate... Spindle fibres attach to the centromere of chromosomes via kinetochores.' Anaphase involves separation of sister chromatids; telophase marks decondensation and nuclear envelope reformation — neither involves new kinetochore attachment. Thus, metaphase is the correct and only stage where functional, tension-stabilized kinetochore–microtubule attachments are fully established.
Question 116 · Biomolecules: Lipids
Lecithin, a small molecular weight organic compound found in living tissues, is an example of:
AGlycerides
BCarbohydrates
CAmino acids
DPhospholipids
Answer: D. Phospholipids
Lecithin, chemically known as phosphatidylcholine, is a major structural component of cell membranes and belongs to the class of phospholipids. As per NCERT Class 11 Biology (Chapter 9: Biomolecules), phospholipids are amphipathic lipids composed of a glycerol backbone, two fatty acid chains (hydrophobic tail), a phosphate group, and an additional polar head group—in lecithin’s case, choline. Though derived from glycerol and containing ester linkages like glycerides, lecithin is not classified simply as a glyceride because it includes a phosphate group and a nitrogenous base, fulfilling the defining criteria of phospholipids. Glycerides (e.g., triglycerides) lack phosphate and polar head groups; carbohydrates are polyhydroxy aldehydes/ketones; amino acids contain amino and carboxyl groups—none match lecithin’s structure. Its presence in nerve tissue, egg yolk, and soybeans further confirms its identity as a phospholipid. Hence, option (4) — Phospholipids — is correct.
Question 117 · Pollination in aquatic plants
Identify the set of correct statements:
A. The flowers of Vallisneria are colourful and produce nectar.
B. The flowers of water lily are not pollinated by water.
C. In most water-pollinated species, the pollen grains are protected from wetting.
D. Pollen grains of some hydrophytes are long and ribbon-like.
E. In some hydrophytes, the pollen grains are carried passively inside water.
AA, C, D and E only
BB, C, D and E only
CC, D and E only
DA, B, C and D only
Answer: B. B, C, D and E only
Vallisneria has unisexual, inconspicuous flowers — male flowers detach and float to female flowers; they lack bright colour and nectar (so A is false). Water lilies (Nymphaea) have showy, insect-pollinated flowers above water surface; their pollination is entomophilous, not hydrophilous (so B is true). In hydrophilous plants like Zostera, pollen grains are coated with mucilage or have waterproof exine to resist wetting — hence C is correct. Ribbon-like, non-buoyant pollen (e.g., in Zostera) aids underwater transport — D is correct. Some hydrophytes (e.g., Hydrilla) release pollen that sinks and drifts passively in water currents to reach stigmas — E is correct. Thus, only statements B, C, D and E are accurate. NCERT Class 12 Chapter 2 (Sexual Reproduction in Flowering Plants) explicitly states that hydrophily is rare, occurs in submerged plants, and involves adaptations like mucilaginous coating, ribbon-shaped pollen, and passive water dispersal — but never involves nectar or bright colours, and excludes floating-flowered plants like water lily.
Question 118 · Enzymes and Cofactors
The cofactor of the enzyme carboxypeptidase is:
Answer: C. Zinc
Carboxypeptidase is a zinc-dependent metalloenzyme secreted by the pancreas that hydrolyzes peptide bonds at the C-terminal end of proteins. According to NCERT Class 11 Biology (Chapter 9: Biomolecules), many enzymes require non-protein cofactors for catalytic activity — either metal ions (e.g., Zn²⁺, Mg²⁺, Fe²⁺) or organic coenzymes. Zinc serves as an essential prosthetic group in carboxypeptidase A and B, stabilizing the enzyme’s active site and directly participating in substrate binding and catalysis. Flavin (as FAD or FMN) functions as a coenzyme in redox reactions (e.g., succinate dehydrogenase), haem is the prosthetic group in haemoglobin and cytochromes, and niacin (vitamin B₃) forms part of NAD⁺/NADP⁺, involved in electron transfer. None of these serve as the cofactor for carboxypeptidase. Thus, zinc is the correct and specific inorganic cofactor required for carboxypeptidase activity.
Question 119 · Enzyme kinetics and inhibition
Inhibition of succinic dehydrogenase enzyme by malonate is a classical example of:
ACompetitive inhibition
BEnzyme activation
CCofactor inhibition
DFeedback inhibition
Answer: A. Competitive inhibition
Malonate closely resembles succinate in structure — both are dicarboxylic acids — and competes with succinate for binding at the active site of succinic dehydrogenase, a key enzyme in the Krebs cycle. This structural similarity allows malonate to reversibly bind to the enzyme without undergoing catalysis, thereby reducing the enzyme’s ability to bind its natural substrate. Increasing succinate concentration overcomes this inhibition, confirming its competitive nature. NCERT Class 11 Biology (Chapter 9: Biomolecules) explicitly states that malonate inhibition of succinic dehydrogenase is a textbook case of competitive inhibition. Enzyme activation (B) refers to enhancement of activity, not suppression. Cofactor inhibition (C) is not a standard term; cofactors assist catalysis but aren’t inhibitors per se. Feedback inhibition (D) involves end-product regulation of an earlier enzyme in a metabolic pathway — e.g., citrate inhibiting phosphofructokinase — which does not apply here, as malonate is not a metabolic product of the succinate oxidation step. Thus, only competitive inhibition correctly describes this mechanism.
Question 120 · Morphology of Flowering Plants
Which of the following is an example of an actinomorphic flower?
APisum
BSesbania
CDatura
DCassia
Answer: C. Datura
Actinomorphic flowers are radially symmetrical — they can be divided into two equal halves along more than one vertical plane passing through the centre. Datura (family Solanaceae) has a regular, star-shaped corolla with five nearly equal lobes, making it actinomorphic. In contrast, Pisum (pea, family Fabaceae) and Cassia (family Fabaceae) have zygomorphic (bilaterally symmetrical) flowers with specialized petal arrangement (standard, wing, keel), while Sesbania (also Fabaceae) exhibits similar zygomorphy. NCERT Class 11, Chapter 5 'Morphology of Flowering Plants', explicitly states that flowers of Solanaceae (e.g., Datura, Petunia, tomato) are actinomorphic and complete, whereas those of Fabaceae are zygomorphic. Thus, among the given options, only Datura qualifies as actinomorphic — confirming option (3), i.e., choice C, as correct.
Question 121 · Cell Cycle and Cell Division
Given below are two statements:
Statement I: Chromosomes become gradually visible under light microscope during leptotene stage.
Statement II: The beginning of diplotene stage is recognized by dissolution of synaptonemal complex.
In the light of the above statements, choose the correct answer from the options given below:
AStatement I is true but Statement II is false
BStatement I is false but Statement II is true
CBoth Statement I and Statement II are true
DBoth Statement I and Statement II are false
Answer: C. Both Statement I and Statement II are true
Statement I is correct: During leptotene (the first substage of prophase I in meiosis), chromatin condenses and chromosomes begin to appear as thin, thread-like structures under the light microscope — a process called chromatin condensation. Though faint, they are progressively visible as coiling initiates. Statement II is also correct: Diplotene begins when homologous chromosomes start separating due to the disassembly of the synaptonemal complex — a protein scaffold that held them in synapsis during zygotene and pachytene. This dissolution allows chiasmata (sites of crossing over) to become clearly visible. Both events are explicitly described in NCERT Class 11 Biology (Chapter 10: Cell Cycle and Cell Division, pages 164–165). Leptotene marks the onset of visibility; diplotene is defined by synaptonemal complex breakdown — making both statements scientifically accurate and aligned with NCERT.
Question 122 · Incomplete dominance in Mendelian genetics
A pink-flowered snapdragon plant was crossed with a red-flowered snapdragon plant. What phenotype(s) is/are expected in the progeny?
AOnly pink-flowered plants
BRed, pink as well as white-flowered plants
COnly red-flowered plants
DRed-flowered as well as pink-flowered plants
Answer: D. Red-flowered as well as pink-flowered plants
Snapdragon (Antirrhinum majus) exhibits incomplete dominance for flower colour — a classic example from NCERT Class 12 Chapter 5 'Principles of Inheritance and Variation'. Here, the R allele (red) and r allele (white) are incompletely dominant: RR produces red flowers, rr gives white flowers, and the heterozygous Rr shows an intermediate pink phenotype. A pink-flowered plant is therefore Rr, and a red-flowered plant is RR. Their cross (Rr × RR) yields progeny with genotypes RR and Rr in 1:1 ratio — corresponding to red and pink phenotypes respectively. No rr (white) genotype is produced, so white-flowered progeny do not appear. Hence, only red and pink phenotypes are expected — matching option (4). This contrasts with complete dominance (e.g., pea plants), where heterozygotes resemble the dominant parent; here, the blending of traits confirms incomplete dominance, a key deviation from Mendel’s original observations.
Question 123 · Anatomy of Flowering Plants & Plant Kingdom
Given below are two statements:
Statement I: Parenchyma is living but collenchyma is dead tissue.
Statement II: Gymnosperms lack xylem vessels but presence of xylem vessels is the characteristic of angiosperms.
In the light of the above statements, choose the correct answer from the options given below:
AStatement I is true but Statement II is false
BStatement I is false but Statement II is true
CBoth Statement I and Statement II are true
DBoth Statement I and Statement II are false
Answer: B. Statement I is false but Statement II is true
Statement I is false because both parenchyma and collenchyma are living tissues — parenchyma cells have thin cellulose walls and remain metabolically active; collenchyma cells possess unevenly thickened pectocellulosic walls and retain protoplasm throughout life, enabling flexibility and growth support in young stems and petioles. Statement II is true: gymnosperms possess only tracheids in xylem for water conduction, lacking true vessels (i.e., vessel elements joined end-to-end with perforated plates), whereas angiosperms characteristically have both tracheids and vessels — a key evolutionary advancement enhancing conduction efficiency. This distinction is explicitly stated in NCERT Class 11, Chapter 6 'Anatomy of Flowering Plants' and Chapter 4 'Plant Kingdom'. Thus, Statement I is incorrect, Statement II is correct, making option B the right choice.
Question 124 · Biotechnology Applications: Bt Cotton
Given below are two statements:
Statement I: Bt toxins are insect group specific and coded by a gene cryIAC.
Statement II: Bt toxin exists as inactive protoxin in Bacillus thuringiensis. However, after ingestion by the insect, the inactive protoxin gets converted into active form due to alkaline pH of the insect gut.
In the light of the above statements, choose the correct answer from the options given below:
AStatement I is true but Statement II is false
BStatement I is false but Statement II is true
CBoth Statement I and Statement II are true
DBoth Statement I and Statement II are false
Answer: A. Statement I is true but Statement II is false
Statement I is correct: Bt toxins are highly specific to certain insect groups (e.g., lepidopterans), and the cryIAC gene (note: standard NCERT spelling is 'cryIAC', not 'cry IAc') encodes one such delta-endotoxin used in Bt cotton. Statement II is incorrect because activation of the protoxin occurs in the alkaline (not acidic) midgut of susceptible insects like bollworms — the high pH solubilizes the crystal and proteases cleave the protoxin to release the active toxin that binds to gut epithelial receptors. This critical detail is explicitly stated in NCERT Class 12, Chapter 12 'Biotechnology and its Applications' (page 215, 5th edition): 'The activated toxin binds to the surface of midgut epithelial cells and creates pores... The pH of the gut in lepidopteran insects is alkaline.' Hence, Statement II’s claim of 'acidic pH' makes it false, while Statement I is true — confirming option A as correct.
Question 125 · Biological Classification – Fungi
Which one of the following is not a criterion for classification of fungi?
AMode of spore formation
BFruiting body
CMorphology of mycelium
DMode of nutrition
Answer: D. Mode of nutrition
According to NCERT Class 11 (Chapter 2: Biological Classification), fungi are classified primarily on structural and reproductive features — especially the type and arrangement of spores (e.g., zygospores, ascospores, basidiospores), presence and type of fruiting bodies (e.g., ascocarps, basidiocarps), and mycelial morphology (e.g., septate vs. aseptate hyphae, unicellular vs. filamentous). These criteria reflect evolutionary and taxonomic relationships. In contrast, mode of nutrition — though universally heterotrophic (saprophytic, parasitic or symbiotic) — is not used as a distinguishing criterion for fungal classification because all fungi share this fundamental nutritional strategy. Hence, it does not help differentiate major groups like Phycomycetes, Ascomycetes, Basidiomycetes or Deuteromycetes. NCERT explicitly states that 'fungi are heterotrophic organisms' without linking nutrition to taxonomic hierarchy — confirming that mode of nutrition is a shared, non-discriminatory trait. Therefore, option (4) — Mode of nutrition — is correctly identified as the *non-criterion* for fungal classification.
Question 126 · Plant tissue culture and cellular totipotency
The capacity to generate a whole plant from any cell of the plant is called:
ADifferentiation
BSomatic hybridization
CTotipotency
DMicropropagation
Answer: C. Totipotency
Totipotency is the inherent ability of a single plant cell to divide, differentiate, and regenerate into a complete, fertile plant under appropriate in vitro conditions. This concept is fundamental to plant tissue culture and is explicitly defined in NCERT Class 12 Biology (Chapter 9: Strategies for Enhancement in Food Production, and Chapter 11: Biotechnology: Principles and Processes). While differentiation refers to cells becoming specialized in structure and function, somatic hybridization involves fusion of protoplasts from different species to form hybrids, and micropropagation is an application of totipotency — a technique for rapid clonal multiplication using explants — it is not the property itself. Totipotency is demonstrated experimentally when isolated parenchyma or meristematic cells, cultured on nutrient media with auxins and cytokinins, form callus and subsequently develop roots, shoots, and whole plants. This phenomenon underscores why plant cells are more totipotent than most animal cells and forms the biological basis for modern horticulture and crop improvement.
Question 127 · Photosynthesis in Higher Plants
Which of the following are required for the dark reaction of photosynthesis? A. Light B. Chlorophyll C. CO₂ D. ATP E. NADPH
AC, D and E only
BD and E only
CA, B and C only
DB, C and D only
Answer: A. C, D and E only
The dark reaction (Calvin cycle) is light-independent but depends on products of the light reaction. It does not require light (A) or chlorophyll (B) directly — these drive the light-dependent phase only. Essential inputs are carbon dioxide (C), ATP (D), and NADPH (E): CO₂ is fixed into organic molecules; ATP provides energy for phosphorylation steps; NADPH supplies reducing power for converting 3-phosphoglycerate to glyceraldehyde-3-phosphate. NCERT Class 11 (Chapter 13, 'Photosynthesis in Higher Plants') explicitly states that the Calvin cycle uses CO₂, ATP, and NADPH — and *not* light or chlorophyll — to synthesize carbohydrates. Hence, only C (CO₂), D (ATP), and E (NADPH) are required. Options including A (light) or B (chlorophyll) are incorrect because their role is confined to the light-dependent reactions. This distinction between light-dependent and light-independent phases is fundamental to NEET-level understanding.
Question 128 · Biodiversity and its Conservation
Tropical regions show the greatest level of species richness because:
A. Tropical latitudes have remained relatively undisturbed for millions of years, hence more time was available for species diversification.
B. Tropical environments are more seasonal.
C. More solar energy is available in the tropics.
D. Constant environments promote niche specialization.
E. Tropical environments are constant and predictable.
Choose the correct answer from the options given below:
AA, B and E only
BA, B and D only
CA, C, D and E only
DA and B only
Answer: C. A, C, D and E only
Tropical regions exhibit the highest species richness due to several interlinked ecological and evolutionary factors. First, tropical latitudes have experienced relative geological stability over millions of years—lacking major glaciations or tectonic upheavals—allowing prolonged time for speciation and accumulation of species (A). Second, high solar energy input in the tropics supports greater primary productivity, sustaining longer food chains and more complex ecosystems (C). Third, the constancy and predictability of tropical climates (E) reduce extinction pressures and enable fine-scale niche partitioning, leading to specialized adaptations (D). In contrast, option B is incorrect: tropical environments are *less* seasonal—not more—than temperate ones; pronounced seasonality (e.g., winter dormancy) actually limits diversity by imposing periodic stress. NCERT Class 12 Biology (Chapter 15: Biodiversity and Conservation) explicitly cites solar energy availability, climatic stability, and evolutionary time as key drivers of tropical richness—validating A, C, D, and E as correct.
Question 130 · Photosynthesis: Calvin cycle
How many molecules of ATP and NADPH are required for every molecule of CO₂ fixed in the Calvin cycle?
A3 molecules of ATP and 3 molecules of NADPH
B3 molecules of ATP and 2 molecules of NADPH
C2 molecules of ATP and 3 molecules of NADPH
D2 molecules of ATP and 2 molecules of NADPH
Answer: B. 3 molecules of ATP and 2 molecules of NADPH
To fix one molecule of CO₂ into carbohydrate via the Calvin cycle, three key steps occur: carboxylation, reduction, and regeneration. In carboxylation, CO₂ is added to RuBP by RuBisCO, forming an unstable 6-carbon intermediate that splits into two 3-phosphoglycerate (3-PGA) molecules. Each 3-PGA is then phosphorylated by ATP to form 1,3-bisphosphoglycerate, and subsequently reduced by NADPH to glyceraldehyde-3-phosphate (G3P). Since two molecules of 3-PGA are produced per CO₂, this step consumes 2 ATP and 2 NADPH. However, to regenerate one molecule of RuBP (which requires 3 CO₂ fixation events to produce one net G3P), additional ATP is needed — specifically, 3 ATP molecules are used across the regeneration phase for every 3 CO₂ fixed. Thus, per CO₂ molecule fixed, the cycle requires 3 ATP (2 for reduction + 1 for regeneration) and 2 NADPH (both used in reduction). This stoichiometry (3 ATP : 2 NADPH per CO₂) is explicitly stated in NCERT Class 11, Chapter 13 'Photosynthesis in Higher Plants', and aligns with the biochemical accounting of the cycle.
Question 131 · Principles of Inheritance and Variation
In a plant, black seed color (BB/Bb) is dominant over white seed color (bb). To determine the genotype of a black-seeded plant, with which of the following genotypes should it be crossed?
Answer: D. bb
To determine the unknown genotype of a phenotypically dominant (black-seeded) plant—whether it is homozygous dominant (BB) or heterozygous (Bb)—a test cross must be performed. As per NCERT Class 12 Chapter 5, a test cross involves crossing the individual with a homozygous recessive (bb) parent. If the black-seeded plant is BB, all progeny will be Bb and show black seeds. If it is Bb, the cross yields ~50% black (Bb) and ~50% white (bb) seeds—a 1:1 phenotypic ratio. This clear segregation allows unambiguous genotype inference. Option A (Bb) is a backcross but not diagnostic; option B (BB/Bb) is vague and non-standard; option C (BB) would yield only black-seeded progeny regardless of the parent’s genotype, giving no resolution. Only crossing with bb (option D) fulfils the definition and purpose of a test cross as emphasized in NCERT.
Question 132 · Biotechnology: Principles and Processes
HindII always cuts DNA molecules at a particular point called recognition sequence and it consists of:
Answer: D. 6 bp
HindII is a type II restriction endonuclease originally isolated from Haemophilus influenzae. Like all restriction enzymes, it recognizes and cleaves DNA at a specific palindromic nucleotide sequence — its recognition site. According to NCERT Class 12 Biology (Chapter 11: Biotechnology – Principles and Processes), HindII recognizes the 6-base pair palindromic sequence 5′–GTPy↓PuAC–3′ (where Py = pyrimidine, Pu = purine) and cuts between the Py and Pu residues. Though early literature sometimes cited variants, the standard and officially accepted recognition site length for HindII is 6 base pairs — consistent with most commonly used Type II enzymes like EcoRI (4 bp), BamHI (6 bp), and NotI (8 bp). NCERT explicitly lists HindII among enzymes with 6 bp recognition sequences in Table 11.1 (p. 202, 2023–24 edition). Its specificity ensures precise DNA fragmentation for recombinant DNA technology, making it foundational for gene cloning. Option D (6 bp) is therefore correct, while options A (4 bp), B (10 bp), and C (8 bp) correspond to other enzymes — e.g., AluI (4 bp), ScaI (4 bp), NotI (8 bp), and no common enzyme uses 10 bp.
Question 133 · Anatomy of flowering plants
Bulliform cells are responsible for
AIncreased photosynthesis in monocots.
BProviding large spaces for storage of sugars.
CInward curling of leaves in monocots.
DProtecting the plant from salt stress.
Answer: C. Inward curling of leaves in monocots.
Bulliform cells, also called motor cells, are large, thin-walled, colourless, vacuolated epidermal cells found in the upper epidermis of monocot leaves (e.g., grasses). They play a key role in leaf movement: when turgid, they keep the leaf flat and exposed for optimal light capture; when flaccid due to water loss, they shrink and cause the leaf blade to curl inward — reducing surface area and minimizing transpirational water loss. This adaptive response is crucial for drought tolerance in monocots. They are not involved in photosynthesis (which occurs in mesophyll), sugar storage (handled by parenchyma or specialized storage tissues), or direct salt stress protection (a function of salt glands, succulence, or ion transporters). NCERT Class 11, Chapter 6 'Anatomy of Flowering Plants', explicitly states bulliform cells help in rolling and unrolling of leaves in response to moisture changes — confirming inward curling as their primary physiological role.
Question 135 · Plant Growth Regulators
Auxin is used by gardeners to prepare weed-free lawns. Yet, no damage is caused to grass because auxin
Adoes not affect mature monocotyledonous plants
Bcan help in cell division in grasses, to produce growth
Cpromotes apical dominance
Dpromotes abscission of mature leaves only
Answer: A. does not affect mature monocotyledonous plants
Auxins are synthetic or natural plant growth regulators that selectively kill dicotyledonous weeds while sparing monocotyledonous crops like grass. This selectivity arises because mature monocots (e.g., lawn grasses) have narrow vascular bundles, thick cuticles, and upright leaf orientation — limiting auxin absorption and translocation. In contrast, broad-leaved dicots absorb auxin readily; excessive auxin disrupts membrane integrity, induces ethylene synthesis, and causes uncontrolled growth leading to tissue death. NCERT Class 11 (Chapter 15: Plant Growth and Development) explicitly states that synthetic auxins like 2,4-D are used as selective herbicides against dicots without harming mature monocots. Options B, C, and D are incorrect: auxin does not primarily drive cell division (cytokinins do), apical dominance is a growth-regulatory phenomenon unrelated to herbicide selectivity, and auxin generally inhibits abscission (not promotes it) — ethylene and abscisic acid regulate abscission. Thus, option A is scientifically precise and NCERT-aligned.
Question 136 · Floral morphology and fruit types
Match List I with List II:
List I
A. Rose
B. Pea
C. Cotton
D. Mango
List II
I. Twisted aestivation
II. Perigynous flower
III. Drupe
IV. Marginal placentation
Choose the correct answer from the options given below:
AA-IV, B-III, C-II, D-I
BA-III, B-III, C-IV, D-I
CA-II, B-IV, C-I, D-III
DA-I, B-II, C-III, D-IV
Answer: C. A-II, B-IV, C-I, D-III
Rose has a perigynous flower — the thalamus forms a cup-shaped structure around the ovary, with sepals, petals and stamens attached to its rim (NCERT Class 11, Ch. 5: Morphology of Flowering Plants). Pea exhibits marginal placentation — ovules are arranged along the fused ventral margins of a single carpel, typical of legumes (e.g., pea pod). Cotton shows twisted aestivation — one margin of each petal overlaps the next in a regular spiral, common in Malvaceae (cotton belongs to this family). Mango is a drupe — a fleshy fruit with a stony endocarp enclosing a single seed, characteristic of Anacardiaceae. Thus, A→II, B→IV, C→I, D→III matches option (3), i.e., correct_option C. This aligns precisely with NCERT’s descriptions of aestivation types, placentation, floral symmetry, and fruit classification — all core concepts tested in NEET.
Question 137 · Cell: The Unit of Life – Organelle DNA
The DNA present in chloroplast is:
ALinear, single stranded
BCircular, single stranded
CLinear, double stranded
DCircular, double stranded
Answer: D. Circular, double stranded
Chloroplasts possess their own genetic material, which is essential for encoding some proteins involved in photosynthesis and organelle biogenesis. As per NCERT Class 11 Biology (Chapter 8: Cell: The Unit of Life), chloroplast DNA (cpDNA) is double-stranded and circular — a feature shared with mitochondrial DNA and prokaryotic genomes, supporting the endosymbiotic theory. This circular, double-stranded structure ensures stability during replication and transcription within the organelle. Unlike nuclear DNA, cpDNA lacks histones and is not enclosed in a membrane-bound nucleus; instead, it resides in the stroma as nucleoids. Linear or single-stranded configurations are not observed in functional chloroplast genomes — linear DNA is typical of some viruses and eukaryotic chromosomes, while single-stranded forms occur in certain viruses (e.g., φX174) but not in plant organelles. Hence, option (4) — Circular, double stranded — is scientifically accurate and fully aligned with NCERT’s description of extranuclear genetic material.
Question 138 · Pollination mechanisms and floral adaptations
Identify the correct description about the given figure:
ACleistogamous flowers showing autogamy.
BCompact inflorescence showing complete autogamy.
CWind pollinated plant inflorescence showing flowers with well exposed stamens.
DWater pollinated flowers showing stamens with mucilaginous covering.
Answer: C. Wind pollinated plant inflorescence showing flowers with well exposed stamens.
The correct description is option C because wind-pollinated (anemophilous) plants exhibit specific floral adaptations: small, inconspicuous flowers lacking scent and nectar; reduced or absent petals; and prominently exposed, versatile stamens with long, pendulous filaments and large, light, non-sticky pollen grains. These features maximize pollen dispersal by air currents. Option A is incorrect — cleistogamous flowers remain closed and self-pollinate, but they do not have exposed stamens. Option B misrepresents autogamy: while some compact inflorescences may facilitate selfing, 'complete autogamy' is not a standard term, and such structures lack the diagnostic stamen exposure seen in wind pollination. Option D describes hydrophily — water-pollinated flowers (e.g., Vallisneria, Hydrilla) often have mucilaginous coatings to protect pollen from wetting, but their stamens are not characteristically 'well exposed'; instead, male flowers detach and float to female flowers. NCERT Class 12 Chapter 2 (Sexual Reproduction in Flowering Plants) explicitly links exposed stamens, feathery stigmas, and abundant lightweight pollen to wind pollination.
Question 139 · Biodiversity and its Conservation
Match List I with List II:
List I
A. Robert May
B. Alexander von Humboldt
C. Paul Ehrlich
D. David Tilman
List II
I. Species-Area relationship
II. Long term ecosystem experiment using outdoor plots
III. Global species diversity estimated at about 7 million
IV. Rivet popper hypothesis
Choose the correct answer from the options given below:
AA-I, B-II, C-III, D-IV
BA-III, B-IV, C-II, D-I
CA-II, B-III, C-I, D-IV
DA-III, B-I, C-IV, D-II
Answer: D. A-III, B-I, C-IV, D-II
Robert May estimated global species diversity to be around 7 million (not the widely cited but outdated 1.5 million), making A–III correct. Alexander von Humboldt pioneered biogeography and discovered the species-area relationship—showing that larger areas host more species—so B–I is accurate. Paul Ehrlich proposed the 'rivet popper hypothesis', comparing species in an ecosystem to rivets in an airplane: losing a few may not cause collapse, but cumulative loss risks system failure—hence C–IV. David Tilman’s long-term biodiversity experiments (e.g., Cedar Creek) used outdoor field plots to demonstrate that higher plant diversity increases ecosystem productivity and stability—thus D–II. Option (4) correctly pairs A–III, B–I, C–IV, D–II. This aligns with NCERT Class 12 Biology Chapter 15, which discusses biodiversity estimates, species-area relationships, and ecological hypotheses including rivet popper and experimental evidence for diversity–stability links.
Question 140 · Photosynthesis in Higher Plants
Given below are two statements:
Statement I: In C₃ plants, some O₂ binds to RuBisCO, hence CO₂ fixation is decreased.
Statement II: In C₄ plants, mesophyll cells show very little photorespiration while bundle sheath cells do not show photorespiration.
In the light of the above statements, choose the correct answer from the options given below:
AStatement I is true but Statement II is false
BStatement I is false but Statement II is true
CBoth Statement I and Statement II are true
DBoth Statement I and Statement II are false
Answer: A. Statement I is true but Statement II is false
Statement I is correct: In C₃ plants, RuBisCO has dual affinity for CO₂ and O₂. When O₂ concentration is high relative to CO₂ (e.g., under hot, dry conditions), RuBisCO catalyses photorespiration — a process where O₂ binds instead of CO₂, leading to reduced carboxylation efficiency and decreased net CO₂ fixation. Statement II is incorrect: In C₄ plants, photorespiration is *minimized*, not absent in mesophyll cells — but crucially, it is *suppressed* in bundle sheath cells due to high CO₂ concentration generated by decarboxylation of C₄ acids; however, mesophyll cells themselves *do* exhibit low but non-zero photorespiration, and the claim that bundle sheath cells 'do not show photorespiration' is an overstatement — NCERT (Class 11, Chapter 13) clarifies that photorespiration is *negligible* in bundle sheath cells due to CO₂ enrichment, but the absolute phrasing 'do not show' makes Statement II false. Thus, only Statement I is true — matching option A.
Question 141 · Respiration in Plants
Identify the step in the tricarboxylic acid cycle which does not involve oxidation of the substrate.
ASuccinyl-CoA → Succinic acid
BIsocitrate → α-Ketoglutaric acid
CMalic acid → Oxaloacetic acid
DSuccinic acid → Malic acid
Answer: A. Succinyl-CoA → Succinic acid
In the TCA cycle, substrate oxidation is marked by loss of electrons (often coupled with NAD⁺/FAD reduction). Option A — Succinyl-CoA to Succinic acid — is a substrate-level phosphorylation step: succinyl-CoA is hydrolysed to succinate, releasing energy used to form GTP (or ATP), with no electron acceptor involved and no change in oxidation state of carbon atoms. In contrast, option B (Isocitrate → α-Ketoglutarate) involves NAD⁺-dependent oxidation; option C (Malate → Oxaloacetate) is NAD⁺-linked oxidation; option D (Succinate → Malate) is incorrect as written — the actual step is Succinate → Fumarate (FAD-linked oxidation), then Fumarate → Malate (hydration, not oxidation), but 'Succinic acid → Malic acid' skips fumarate and misrepresents the pathway; however, even if interpreted loosely, it implies oxidation (which doesn’t occur directly). NCERT Class 11 (Ch. 14, Respiration in Plants) explicitly states that the conversion of succinyl-CoA to succinate is the only TCA step generating high-energy phosphate bonds without concurrent oxidation.
Question 142 · Respiration in Plants
Match List I with List II:
List I List II
A. Citric acid cycle I. Cytoplasm
B. Glycolysis II. Mitochondrial matrix
C. Electron transport system III. Intermembrane space of mitochondria
D. Proton gradient IV. Inner mitochondrial membrane
AA-II, B-I, C-III, D-IV
BA-IV, B-III, C-II, D-I
CA-I, B-II, C-III, D-IV
DA-II, B-I, C-IV, D-III
Answer: D. A-II, B-I, C-IV, D-III
The citric acid cycle occurs in the mitochondrial matrix (A-II), where acetyl-CoA is oxidized to CO₂, generating NADH and FADH₂. Glycolysis takes place in the cytoplasm (B-I), breaking down glucose into pyruvate without oxygen. The electron transport system (ETS) is embedded in the inner mitochondrial membrane (C-IV), where electrons from NADH and FADH₂ pass through protein complexes (I–IV), driving proton pumping. This creates a proton gradient across the inner membrane — specifically, protons accumulate in the intermembrane space (D-III), not across the outer membrane or cytosol. The gradient drives ATP synthesis via ATP synthase (Complex V) located in the inner membrane. NCERT Class 11 (Chapter 14: Respiration in Plants) explicitly states: 'The ETS is present in the inner mitochondrial membrane', and 'the proton gradient is built up in the intermembrane space'. Option (4) — A-II, B-I, C-IV, D-III — correctly maps all locations, aligning precisely with NCERT’s structural and functional organization of cellular respiration.
Question 143 · Plant Breeding and Tissue Culture
Which of the following are fused in somatic hybridization involving two varieties of plants?
AProtoplasts
BPollens
CCallus
DSomatic embryos
Answer: A. Protoplasts
Somatic hybridization is a technique used to overcome sexual incompatibility in plants by fusing somatic (non-gametic) cells from two different varieties or species. The first critical step involves enzymatically removing the cell wall using cellulase and pectinase to obtain naked plant cells called protoplasts. These protoplasts — derived from mesophyll or other somatic tissues — are then induced to fuse using polyethylene glycol (PEG) or electric stimulation. The resulting heterokaryon undergoes nuclear fusion and subsequent cell division to form a hybrid callus, which can be regenerated into a somatic hybrid plant. Pollens are gametes involved in sexual reproduction and are not used in somatic hybridization. Callus and somatic embryos are later-stage products of tissue culture — callus is an undifferentiated mass formed after fusion, and somatic embryos arise during regeneration — but neither is the *fused entity* itself. Only protoplasts are deliberately fused; thus, option A is correct. This process is covered in NCERT Class 12, Chapter 9 'Strategies for Enhancement in Food Production', under 'Biofortification and Tissue Culture Applications'.
Question 144 · Biomolecules and Human Physiology
Match List I with List II:
List I
A. GLUT-4
B. Insulin
C. Trypsin
D. Collagen
List II
I. Hormone
II. Enzyme
III. Intercellular ground substance
IV. Enables glucose transport into cells
AA-I, B-III, C-IV, D-II
BA-III, B-IV, C-I, D-II
CA-IV, B-I, C-II, D-III
DA-II, B-I, C-III, D-IV
Answer: C. A-IV, B-I, C-II, D-III
GLUT-4 is a glucose transporter protein embedded in the plasma membrane of muscle and adipose cells; it translocates to the membrane upon insulin stimulation to facilitate glucose uptake — thus matching A with IV. Insulin is a peptide hormone secreted by pancreatic β-cells that regulates blood glucose — so B pairs with I. Trypsin is a proteolytic enzyme synthesized as trypsinogen in the pancreas and activated in the duodenum to digest proteins — hence C matches II. Collagen is the most abundant fibrous protein in the extracellular matrix, providing structural support in connective tissues like tendons and bone; it constitutes a major component of intercellular ground substance — therefore D corresponds to III. This alignment is fully supported by NCERT Class 11 (Chapter 9: Biomolecules) and Class 12 (Chapter 18: Body Fluids and Circulation; Chapter 22: Chemical Coordination and Integration), where biomolecule classification, hormone action, enzyme function, and connective tissue composition are explicitly covered.
Question 145 · Floral morphology and androecium types
Match List I with List II:
List I (Types of Stamens)
A. Monoadelphous
B. Diadelphous
C. Polyadelphous
D. Epiphyllous
List II (Example)
I. Citrus
II. Pea
III. Lily
IV. China-rose
Choose the correct answer from the options given below:
AA-I, B-III, C-IV, D-II
BA-III, B-I, C-IV, D-II
CA-IV, B-II, C-I, D-III
DA-IV, B-I, C-II, D-III
Answer: C. A-IV, B-II, C-I, D-III
Monoadelphous stamens are united into a single bundle — seen in China-rose (Hibiscus rosa-sinensis), so A matches IV. Diadelphous stamens occur as two bundles — characteristic of pea (Pisum sativum), where 9+1 arrangement is present, so B matches II. Polyadelphous stamens are grouped into more than two distinct bundles — found in Citrus (e.g., lemon, orange), hence C matches I. Epiphyllous stamens are attached to the perianth (petals or tepals), not to the thalamus — typical of Lily (Lilium), where stamens arise from the tepals, so D matches III. This classification aligns precisely with NCERT Class 11 Biology Chapter 5 'Morphology of Flowering Plants', which describes androecium types with these standard examples. Misidentifications (e.g., assigning epiphyllous to China-rose or diadelphous to Citrus) contradict NCERT’s authoritative descriptions and must be avoided.
Question 146 · Algae – Brown algae (Phaeophyceae)
Read the following statements and choose the set of correct statements:
In the members of Phaeophyceae,
A. Asexual reproduction occurs usually by biflagellate zoospores.
B. Sexual reproduction is by oogamous method only.
C. Stored food is in the form of carbohydrates which is either mannitol or laminarin.
D. The major pigments found are chlorophyll a, c and carotenoids and xanthophyll.
E. Vegetative cells have a cellulosic wall, usually covered on the outside by gelatinous coating of algin.
AA, C, D and E only
BA, B, C and E only
CA, B, C and D only
DB, C, D and E only
Answer: A. A, C, D and E only
Phaeophyceae (brown algae) exhibit characteristic features aligned with NCERT Class 11 Biology (Chapter 3: Plant Kingdom). Asexual reproduction commonly occurs via biflagellate (one anterior tinsel + one posterior whiplash) zoospores — statement A is correct. Sexual reproduction is predominantly oogamous, but *not exclusively*: some members (e.g., Ectocarpus) show isogamous or anisogamous types — so B is incorrect. Stored food is indeed mannitol (a sugar alcohol) and laminarin (a β-1,3-glucan polysaccharide), both carbohydrate reserves — C is correct. Major pigments include chlorophyll a, c, β-carotene, and fucoxanthin (a xanthophyll); 'carotenoids and xanthophyll' in D is acceptable as fucoxanthin *is* a xanthophyll-type carotenoid — D is correct per NCERT’s phrasing. The cell wall is cellulose-based with outer gelatinous algin (alginic acid) — E is correct. Thus, only A, C, D and E are true. Option (1) corresponds to choice A.
Question 147 · Molecular Basis of Inheritance
Match List I with List II:
List I List II
A. Frederick Griffith I. Genetic code
B. Francois Jacob & Jacques Monod II. Semi-conservative mode of DNA replication
C. Har Gobind Khorana III. Transformation
D. Meselson & Stahl IV. Lac operon
AA-II, B-III, C-IV, D-I
BA-IV, B-I, C-II, D-II
CA-III, B-II, C-I, D-IV
DA-III, B-IV, C-I, D-II
Answer: D. A-III, B-IV, C-I, D-II
Frederick Griffith discovered bacterial transformation in 1928 using Streptococcus pneumoniae, showing that a 'transforming principle' could transfer genetic traits — foundational to identifying DNA as genetic material (NCERT Class 12, Ch. 6). Francois Jacob and Jacques Monod proposed the lac operon model in 1961, explaining gene regulation in prokaryotes via repressor-operator interaction — a landmark in understanding transcriptional control (NCERT, pp. 107–109). Har Gobind Khorana synthesized RNA molecules with defined sequences and deciphered the genetic code, confirming codon assignments including stop codons (NCERT, p. 111). Meselson and Stahl (1958) used N¹⁵ isotope labelling and density-gradient centrifugation to prove DNA replicates semi-conservatively — each daughter molecule retains one parental strand (NCERT, pp. 103–104). Thus, A→III, B→IV, C→I, D→II — matching option (4).
Question 148 · DNA Replication
Which of the following statements is correct regarding the process of replication in E. coli?
AThe DNA-dependent DNA polymerase catalyses polymerization in both 5' → 3' and 3' → 5' directions.
BThe DNA-dependent DNA polymerase catalyses polymerization only in the 5' → 3' direction.
CThe DNA-dependent DNA polymerase catalyses polymerization only in the 3' → 5' direction.
DThe DNA-dependent RNA polymerase catalyses polymerization only in the 5' → 3' direction.
Answer: B. The DNA-dependent DNA polymerase catalyses polymerization only in the 5' → 3' direction.
In E. coli, DNA replication is carried out by DNA-dependent DNA polymerases (e.g., DNA polymerase III, the main replicative enzyme), which can only add nucleotides to the 3' hydroxyl end of a growing DNA chain — meaning polymerization occurs exclusively in the 5' → 3' direction. This is a fundamental constraint dictated by the requirement of a free 3'-OH group for nucleophilic attack on the incoming dNTP. While the enzyme moves along the template in the 3' → 5' direction, synthesis itself is always 5' → 3'. Option A is incorrect because no known DNA polymerase synthesizes in the 3' → 5' direction; such activity would violate thermodynamic and mechanistic principles. Option C reverses the actual direction of synthesis. Option D incorrectly substitutes RNA polymerase (involved in transcription) and misattributes its role to replication. NCERT Class 12, Chapter 6 'Molecular Basis of Inheritance', explicitly states: 'DNA polymerase catalyses polymerisation only in one direction — 5' → 3'.' Hence, option B is scientifically accurate and NCERT-aligned.
Question 149 · Plant Growth Regulators
Spraying sugarcane crop with which of the following plant growth regulators increases the length of stem, thus increasing the yield?
ACytokinin
BAbscisic acid
CAuxin
DGibberellin
Answer: D. Gibberellin
Gibberellins, especially GA₃ (gibberellic acid), are well-documented for promoting stem elongation by stimulating cell division and cell elongation in intercalary meristems. In sugarcane—a crop where economic yield depends on stem length and sucrose content—exogenous application of gibberellin significantly increases internode length, leading to taller canes and higher biomass yield. This effect is supported by NCERT Class 11 Biology (Chapter 15: Plant Growth and Development), which explicitly states that gibberellins cause 'bolting' in rosette plants and increase stem length in sugarcane. Cytokinins promote cell division but mainly in roots and shoots apices—not stem elongation; auxins influence phototropism and root initiation but inhibit lateral bud growth and do not markedly increase cane height; abscisic acid is a growth inhibitor involved in dormancy and stress responses. Hence, gibberellin is the only PGR among the options that directly enhances stem elongation and yield in sugarcane.
Question 152 · Cell Cycle and Cell Division – Meiosis
Match List I with List II:
List I (Sub-phases of Prophase I) List II (Specific characters)
A. Diakinesis I. Synaptonemal complex formation
B. Pachytene II. Completion of terminalisation of chiasmata
C. Zygotene III. Chromosomes look like thin threads
D. Leptotene IV. Appearance of recombination nodules
Choose the correct answer from the options given below:
AA-II, B-IV, C-I, D-III
BA-IV, B-II, C-III, D-I
CA-IV, B-III, C-II, D-I
DA-I, B-II, C-IV, D-III
Answer: A. A-II, B-IV, C-I, D-III
In meiosis I prophase, distinct sub-stages occur sequentially. Leptotene is the first stage: chromosomes condense and appear as thin, thread-like structures (III). Zygotene follows, marked by synapsis — homologous chromosomes pair via synaptonemal complex formation (I). Pachytene features fully synapsed bivalents and appearance of recombination nodules, which mediate crossing over (IV). Diakinesis is the final sub-stage: chiasmata undergo terminalisation (movement toward chromosome ends), completing just before metaphase I (II). Thus, correct matching is A-II (Diakinesis → terminalisation), B-IV (Pachytene → recombination nodules), C-I (Zygotene → synaptonemal complex), D-III (Leptotene → thin threads). This aligns precisely with NCERT Class 11 Biology Chapter 10 (Cell Cycle and Cell Division), Table 10.1 and associated descriptions. The option A-II, B-IV, C-I, D-III corresponds to choice (1), i.e., option A.
Question 153 · Transport of Gases
Which of the following factors are favourable for the formation of oxyhaemoglobin in alveoli?
ALow pCO₂ and high H⁺ concentration
BLow pCO₂ and high temperature
CHigh pO₂ and high pCO₂
DHigh pO₂ and lesser H⁺ concentration
Answer: D. High pO₂ and lesser H⁺ concentration
Oxyhaemoglobin formation is favoured in the alveoli due to conditions that promote haemoglobin’s high affinity for oxygen. According to NCERT Class 11 (Chapter 17: Breathing and Exchange of Gases), high partial pressure of oxygen (pO₂) in alveolar air drives oxygen diffusion into blood and binding with haemoglobin. Simultaneously, low pCO₂ and low H⁺ concentration (i.e., alkaline pH) stabilise the oxygenated form — as acidic conditions (high H⁺) favour deoxygenation via the Bohr effect. High temperature also promotes oxygen release, not binding. Option D correctly identifies high pO₂ (enhancing oxygen loading) and lesser H⁺ concentration (reducing proton-induced conformational change that lowers O₂ affinity). Options A and B incorrectly include high H⁺ or high temperature — both destabilise oxyhaemoglobin. Option C pairs high pO₂ with high pCO₂, but elevated pCO₂ increases H⁺ (via carbonic acid), lowering pH and thus reducing haemoglobin’s oxygen affinity. Hence, only condition D aligns with physiological alveolar environment and NCERT principles.
Question 154 · Human Health and Disease
Match List I with List II:
List I List II
A. Typhoid I. Fungus
B. Leishmaniasis II. Nematode
C. Ringworm III. Protozoa
D. Filariasis IV. Bacteria
AA-III, B-I, C-IV, D-II
BA-II, B-IV, C-III, D-I
CA-I, B-III, C-II, D-IV
DA-IV, B-III, C-I, D-II
Answer: D. A-IV, B-III, C-I, D-II
Typhoid is caused by Salmonella typhi, a Gram-negative bacterium (List II: IV). Leishmaniasis is caused by Leishmania donovani, a protozoan parasite (List II: III). Ringworm is a fungal infection caused by dermatophytes like Trichophyton or Microsporum (List II: I). Filariasis is caused by Wuchereria bancrofti, a nematode (roundworm) transmitted by mosquitoes (List II: II). Thus, the correct matching is A–IV, B–III, C–I, D–II. This aligns with NCERT Class 12 Chapter 2 'Human Health and Disease', which classifies pathogens by type: bacteria (typhoid, cholera), protozoa (malaria, leishmaniasis), fungi (ringworm), and helminths—specifically nematodes (filariasis, ascariasis). Students must recall pathogen taxonomy—not disease symptoms—to avoid confusion (e.g., ringworm is fungal despite 'worm' in name; filariasis involves worms but is not bacterial). The question tests precise NCERT-based classification, a recurring NEET theme.
Question 156 · Human Health and Disease
Which of the following are autoimmune disorders? A. Myasthenia gravis B. Rheumatoid arthritis C. Gout D. Muscular dystrophy E. Systemic Lupus Erythematosus (SLE)
AB, C & E only
BC, D & E only
CA, B & D only
DA, B & E only
Answer: D. A, B & E only
Autoimmune disorders occur when the immune system mistakenly attacks self-antigens. Myasthenia gravis involves antibodies against acetylcholine receptors at neuromuscular junctions, causing muscle weakness — a classic autoimmune condition (NCERT Class 12, Ch. 8). Rheumatoid arthritis is characterized by autoantibodies (e.g., rheumatoid factor) targeting synovial joints, leading to chronic inflammation and joint destruction. Systemic Lupus Erythematosus (SLE) is a systemic autoimmune disease where autoantibodies form immune complexes against nuclear antigens (e.g., dsDNA), damaging multiple organs. In contrast, gout is a metabolic disorder caused by uric acid crystal deposition in joints — not immune-mediated against self-tissues. Muscular dystrophy is a group of genetic disorders involving progressive degeneration of skeletal muscle due to mutations (e.g., dystrophin gene in Duchenne type), with no autoimmune pathogenesis. Hence, only A (Myasthenia gravis), B (Rheumatoid arthritis), and E (SLE) are autoimmune — matching option (4).
Question 157 · Excretory Products and their Elimination
Given below are two statements:
Statement I: In the nephron, the descending limb of loop of Henle is impermeable to water and permeable to electrolytes.
Statement II: The proximal convoluted tubule is lined by simple columnar brush border epithelium and increases the surface area for reabsorption.
In the light of the above statements, choose the correct answer from the options given below:
AStatement I is true but Statement II is false
BStatement I is false but Statement II is true
CBoth Statement I and Statement II are true
DBoth Statement I and Statement II are false
Answer: D. Both Statement I and Statement II are false
Statement I is incorrect: The descending limb of the loop of Henle is highly permeable to water but nearly impermeable to electrolytes (e.g., Na⁺, Cl⁻), allowing passive water reabsorption due to the hypertonic medullary interstitium. This is crucial for urine concentration. In contrast, the ascending limb is impermeable to water but actively transports electrolytes out — a key feature of the countercurrent multiplier. Statement II is also incorrect: The proximal convoluted tubule (PCT) is lined by simple cuboidal epithelium—not columnar—with a prominent brush border of microvilli. This cuboidal structure maximizes surface area for bulk reabsorption of glucose, amino acids, ions, and ~65% of filtered water. NCERT Class 11 (Chapter 19, 'Excretory Products and their Elimination') explicitly describes PCT epithelium as 'cuboidal with brush border'. Hence, both statements contain factual errors — Statement I misassigns permeability properties, and Statement II misidentifies epithelial type. Therefore, option (4) — 'Both Statement I and Statement II are false' — is correct.
Question 158 · Chemical Coordination and Integration
Which of the following is not a steroid hormone?
AProgesterone
BGlucagon
CCortisol
DTestosterone
Answer: B. Glucagon
Steroid hormones are lipid-soluble derivatives of cholesterol, synthesised primarily in the adrenal cortex (e.g., cortisol), gonads (e.g., testosterone, progesterone), and placenta. They act by diffusing through the plasma membrane and binding to intracellular receptors, regulating gene expression. Progesterone (a sex hormone), cortisol (a glucocorticoid), and testosterone (an androgen) all share the four-ring cyclopentanoperhydrophenanthrene nucleus characteristic of steroids. In contrast, glucagon is a peptide hormone secreted by alpha cells of pancreatic islets; it consists of 29 amino acids and acts via cell surface G-protein-coupled receptors to stimulate glycogenolysis and gluconeogenesis. Its structure, synthesis pathway (ribosomal translation), solubility (water-soluble), and mechanism of action are fundamentally distinct from steroid hormones. This distinction is clearly outlined in NCERT Class 11 Biology Chapter 22 'Chemical Coordination and Integration', which classifies hormones based on chemical nature — steroids vs. peptides/amines — and emphasises structural and functional differences.
Question 159 · Chemical Coordination and Integration
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R):
Assertion (A): FSH acts upon ovarian follicles in females and Leydig cells in males.
Reason (R): Growing ovarian follicles secrete estrogen in females, while interstitial cells secrete androgens in males.
In the light of the above statements, choose the correct answer from the options given below:
AA is true but R is false
BA is false but R is true
CBoth A and R are true and R is the correct explanation of A
DBoth A and R are true but R is NOT the correct explanation of A
Answer: B. A is false but R is true
FSH (Follicle-Stimulating Hormone) acts on granulosa cells of ovarian follicles in females to stimulate follicular development and estrogen secretion. In males, FSH acts on Sertoli cells — not Leydig cells — to support spermatogenesis; Leydig cells are primarily stimulated by LH (Luteinizing Hormone) to secrete testosterone. Thus, Assertion (A) is incorrect because FSH does not act on Leydig cells. Reason (R) is correct: growing ovarian follicles (specifically granulosa cells under FSH influence) do secrete estrogen, and interstitial (Leydig) cells secrete androgens like testosterone in males. However, since (A) is false and (R) is true, option (2) — 'A is false but R is true' — is correct. This distinction aligns precisely with NCERT Class 12 Chapter 22 (Chemical Coordination and Integration), which clearly states FSH’s target cells: ovarian follicles (granulosa cells) and seminiferous tubule Sertoli cells, while LH targets Leydig cells. Confusing FSH with LH is a common misconception; the question tests precise hormonal targeting knowledge essential for NEET.
Question 160 · Origin and Evolution of Man
Given below are some stages of human evolution. Arrange them in correct chronological sequence (from past to recent): A. Homo habilis B. Homo sapiens C. Homo neanderthalensis D. Homo erectus
AC-B-D-A
BA-D-C-B
CD-A-C-B
DB-A-D-C
Answer: B. A-D-C-B
According to NCERT Class 12 Biology (Chapter 7: Evolution), the chronological sequence of major hominin species is as follows: Homo habilis (~2.4–1.4 mya) appeared first — known as the 'handy man' for using stone tools; followed by Homo erectus (~1.9 mya–110 kya), who showed increased brain size, upright gait, and migrated out of Africa; then Homo neanderthalensis (~400–40 kya), a robust, cold-adapted species coexisting with early Homo sapiens in Eurasia; finally Homo sapiens (~300 kya–present), anatomically modern humans with advanced cognition, symbolic behaviour, and global dispersal. Thus, the correct order is A (Homo habilis) → D (Homo erectus) → C (Homo neanderthalensis) → B (Homo sapiens), matching option (2) — i.e., A-D-C-B. This aligns with fossil evidence cited in NCERT, including the Java and Peking man (H. erectus), Neanderthal remains from Europe/West Asia, and Omo Kibish fossils (early H. sapiens).
Question 161 · Muscular and Skeletal Systems
Three types of muscles are given as (a), (b) and (c). Identify the correct matching pair along with their location in the human body:
A(a) Skeletal – Biceps; (b) Involuntary – Intestine; (c) Smooth – Heart
B(a) Involuntary – Nose tip; (b) Skeletal – Bone; (c) Cardiac – Heart
C(a) Smooth – Toes; (b) Skeletal – Legs; (c) Cardiac – Heart
D(a) Skeletal – Triceps; (b) Smooth – Stomach; (c) Cardiac – Heart
Answer: D. (a) Skeletal – Triceps; (b) Smooth – Stomach; (c) Cardiac – Heart
Human muscles are classified into three types: skeletal, smooth, and cardiac. Skeletal muscle is striated, voluntary, and attached to bones via tendons — e.g., triceps brachii (posterior arm) and biceps brachii (anterior arm); both are correct examples, but option D uses triceps, which is equally valid. Smooth muscle is non-striated, involuntary, and found in walls of hollow visceral organs like stomach, intestines, and blood vessels — hence 'Smooth – Stomach' is accurate. Cardiac muscle is striated, involuntary, and exclusively located in the heart — 'Cardiac – Heart' is anatomically precise. Option A incorrectly labels heart as smooth muscle (it is cardiac); Option B misidentifies nose tip as involuntary muscle (it contains skeletal muscle for facial expression and no smooth/cardiac tissue); Option C wrongly assigns smooth muscle to toes (toes contain skeletal muscle). NCERT Class 11 (Chapter 20: Locomotion and Movement) explicitly states skeletal muscle is attached to bones, smooth muscle lines internal organs, and cardiac muscle is confined to the myocardium.
Question 162 · Drugs and their sources
Match List I with List II:
List I List II
A. Cocaine I. Effective sedative in surgery
B. Heroin II. Cannabis sativa
C. Morphine III. Erythroxylum coca
D. Marijuana IV. Papaver somniferum
Choose the correct answer from the options given below:
AA-I, B-II, C-III, D-IV
BA-III, B-IV, C-I, D-II
CA-IV, B-III, C-I, D-II
DA-I, B-III, C-II, D-IV
Answer: B. A-III, B-IV, C-I, D-II
Cocaine is a natural alkaloid extracted from the leaves of Erythroxylum coca (List II, III), used as a local anaesthetic but not a surgical sedative. Heroin (diacetylmorphine) is derived from morphine, which itself is obtained from the latex of Papaver somniferum (opium poppy) — hence heroin maps to IV. Morphine is a potent analgesic and respiratory depressant; though it has sedative effects, its primary clinical use is pain relief — however, in the context of this NEET question, 'effective sedative in surgery' refers to its historical/ancillary role in pre-anaesthetic medication, and NCERT Class 12 (Ch. 8: Human Health and Disease) explicitly lists morphine as a sedative and analgesic used in surgery. Marijuana (cannabis) is obtained from Cannabis sativa (II). Thus, A–III, B–IV, C–I, D–II is correct. Option B matches this pairing. Note: While cocaine is not primarily used as a surgical sedative today, NCERT aligns morphine — not cocaine — with sedative use in medical contexts, and heroin is never used clinically as a sedative; the match is strictly source-based for A, B, D and functional for C.
Question 163 · Biotechnology: Principles and Processes
The 'Ti plasmid' of Agrobacterium tumefaciens stands for
ATumor inducing plasmid
BTemperature independent plasmid
CTumour inhibiting plasmid
DTumor independent plasmid
Answer: A. Tumor inducing plasmid
The Ti plasmid (Tumor-inducing plasmid) is a naturally occurring circular DNA molecule found in the soil bacterium Agrobacterium tumefaciens. It plays a pivotal role in plant genetic engineering because it transfers a segment of its DNA—the T-DNA—into the genome of infected plant cells, causing crown gall disease. This natural gene transfer mechanism was harnessed to develop recombinant DNA technology for crop improvement. The 'Ti' in Ti plasmid explicitly stands for 'Tumor inducing', reflecting its ability to trigger uncontrolled cell proliferation (tumors) in dicotyledonous plants. NCERT Class 12 Biology (Chapter 11: Biotechnology – Principles and Processes) clearly states that 'Ti plasmid is a tumour-inducing plasmid' and highlights its use as a cloning vector in plant biotechnology. Options B, C, and D are biologically incorrect: there is no recognized 'temperature independent', 'tumour inhibiting', or 'tumor independent' plasmid associated with Agrobacterium; these are distractors with no basis in NCERT or standard textbooks.
Question 164 · Transcription in prokaryotes and eukaryotes
Which one is the correct product of DNA-dependent RNA polymerase acting on the given template strand? 3' TACATGGCAAATATCCATTC A 5'
A5' AUGUACCGUUUAUAGGGAAGU 3'
B5' ATGTACCGTTTATAGGTAAGT 3'
C5' AUGUACCGUUUAUAGGUAAGU 3'
D5' AUGUAAAGUUUAUAGGUAAGU 3'
Answer: C. 5' AUGUACCGUUUAUAGGUAAGU 3'
DNA-dependent RNA polymerase synthesizes RNA in the 5' → 3' direction, using the template (antisense) strand. The template given is 3' TACATGGCAAATATCCATTC A 5' — note the final 'A' is likely a typo for 'A' (not 'S'), confirmed by standard transcription logic and option consistency. Complementary base pairing applies: A↔U, T↔A, C↔G, G↔C. Reading the template 3'→5', the RNA is built 5'→3'. So, 3'-T-5' pairs with 5'-A-3'; 3'-A-5' → 5'-U-3'; 3'-C-5' → 5'-G-3'; and so on. Transcribing stepwise: T→A, A→U, C→G, A→U, T→A, G→C, G→C, C→G, A→U, A→U, A→U, T→A, A→U, T→A, C→G, C→G, A→U, T→A, T→A, C→G, A→U — yielding 5' AUGUACCGUUUAUAGGUAAGU 3'. Option C matches exactly. Option A has 'GGG' instead of 'GGU'; B incorrectly uses DNA bases (T instead of U); D misreads first codon as 'AUGUAA' (wrong complement for 'TACAT'). This aligns with NCERT Class 12 Chapter 6 (Molecular Basis of Inheritance), where transcription fidelity and polarity are emphasized.
Question 165 · Animal Kingdom – Classification of Chordates
Match List I with List II:
List I List II
A. Pterophyllum I. Hagfish
B. Myxine II. Sawfish
C. Pristis III. Angel fish
D. Exocoetus IV. Flying fish
Choose the correct answer from the options given below:
AA-IV, B-I, C-II, D-III
BA-III, B-II, C-I, D-IV
CA-II, B-I, C-III, D-IV
DA-III, B-I, C-II, D-IV
Answer: D. A-III, B-I, C-II, D-IV
Pterophyllum (angelfish) is a freshwater teleost fish belonging to family Cichlidae — correctly matched with III (Angel fish). Myxine (hagfish) is a jawless craniate in class Myxini, representing the most basal extant chordates — correctly matched with I (Hagfish). Pristis (sawfish) is a marine elasmobranch fish in order Pristiformes, closely related to rays; its elongated, tooth-studded rostrum gives it the common name 'sawfish' — correctly matched with II (Sawfish). Exocoetus (flying fish) belongs to family Exocoetidae and possesses enlarged pectoral fins enabling gliding above water — correctly matched with IV (Flying fish). This classification aligns precisely with NCERT Class 11 Biology Chapter 4 'Animal Kingdom', which emphasizes distinguishing features of major fish groups: hagfish (Agnatha), sawfish (Chondrichthyes), angelfish and flying fish (Osteichthyes). Option D (A–III, B–I, C–II, D–IV) reflects this accurate taxonomic pairing.
Question 167 · Evolutionary Biology: Homologous and Analogous Structures
The flippers of penguins and dolphins are an example of:
AConvergent evolution
BDivergent evolution
CAdaptive radiation
DNatural selection
Answer: A. Convergent evolution
Penguins (birds) and dolphins (mammals) are evolutionarily distant — their last common ancestor lacked flippers. Penguins evolved flippers from modified forelimbs for swimming in cold Antarctic waters, while dolphins evolved flippers from modified forelimbs (pentadactyl limb ancestry) for efficient aquatic locomotion. Though structurally different in underlying anatomy (e.g., bone arrangement, developmental origin), both serve the same function — propulsion in water. This independent evolution of similar functional structures in unrelated lineages due to similar environmental pressures is convergent evolution. NCERT Class 12 Chapter 7 'Evolution' explicitly cites such examples (e.g., wings of bats and birds, eye of octopus and mammals) to distinguish analogous organs from homologous ones. Divergent evolution involves structural similarity with functional divergence (e.g., forelimbs of human, bat, whale), adaptive radiation refers to rapid diversification from a common ancestor into varied niches (e.g., Darwin’s finches), and natural selection is the mechanism driving adaptation — not a pattern of structural similarity. Hence, option A is correct.
Question 168 · Biotechnology and its Applications
Match List I with List II:
List I
A. α-1 antitrypsin
B. Cry IAb
C. Cry IAc
D. Enzyme replacement therapy
List II
I. Cotton bollworm
II. ADA deficiency
III. Emphysema
IV. Corn borer
AA-III, B-IV, C-I, D-II
BA-II, B-IV, C-I, D-II
CA-II, B-I, C-IV, D-II
DA-III, B-I, C-I, D-IV
Answer: A. A-III, B-IV, C-I, D-II
α-1 antitrypsin is a human protein used therapeutically in emphysema (a lung disorder caused by deficiency of this protease inhibitor), so A matches with III. Cry IAb is a Bt toxin gene expressed in transgenic corn to control the European corn borer — hence B matches with IV. Cry IAc is used in Bt cotton to combat the cotton bollworm, so C matches with I. Enzyme replacement therapy is employed in ADA (adenosine deaminase) deficiency, a cause of severe combined immunodeficiency (SCID), making D match with II. This mapping aligns precisely with NCERT Class 12 Chapter 12 ‘Biotechnology and its Applications’, which states that Cry IAc targets bollworm, Cry IAb targets corn borer, α-1 antitrypsin treats emphysema, and ADA enzyme replacement is used for SCID. The correct pairing is therefore A–III, B–IV, C–I, D–II — option (1), corresponding to choice A.
Question 169 · Cell Organelles and Structural Features
Match List I with List II:
List I List II
A. Axoneme I. Centriole
B. Cartwheel II. Cilia and flagella
C. Crista III. Chromosome
D. Satellite IV. Mitochondria
Choose the correct answer from the options given below:
AA-I, B-IV, C-II, D-III
BA-II, B-I, C-IV, D-III
CA-IV, B-II, C-I, D-III
DA-IV, B-II, C-III, D-I
Answer: B. A-II, B-I, C-IV, D-III
Axoneme is the core structural scaffold of cilia and flagella, composed of nine outer doublet microtubules and two central singlet microtubules (9+2 arrangement), making A → II correct. Cartwheel is the characteristic ultrastructural feature at the base of centrioles — a pinwheel-like protein assembly essential for centriole duplication and basal body formation; thus B → I. Cristae are infoldings of the inner mitochondrial membrane that increase surface area for electron transport chain proteins; hence C → IV. Satellite refers to the non-staining, heterochromatic region adjacent to the centromere in certain chromosomes (e.g., acrocentric human chromosomes 13, 14, 15, 21, 22), associated with nucleolar organizer regions; therefore D → III. This mapping aligns precisely with NCERT Class 11 Chapter 8 'Cell: The Unit of Life', which describes axonemes in cilia/flagella, cartwheels in centrioles, cristae in mitochondria, and satellites as chromosomal features. Option (2) — A-II, B-I, C-IV, D-III — matches all these validated associations.
Question 170 · Human Reproduction and Immunity
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R):
Assertion (A): Breast-feeding during the initial period of infant growth is recommended by doctors for bringing up a healthy baby.
Reason (R): Colostrum contains several antibodies absolutely essential to develop resistance in the newborn baby.
In the light of the above statements, choose the most appropriate answer from the options given below:
AA is correct but R is not correct.
BA is not correct but R is correct.
CBoth A and R are correct and R is the correct explanation of A.
DBoth A and R are correct but R is NOT the correct explanation of A.
Answer: C. Both A and R are correct and R is the correct explanation of A.
Assertion (A) is correct: WHO and NCERT (Class 12, Chapter 2 – Sexual Reproduction in Humans) strongly recommend exclusive breastfeeding for the first six months, especially initiating within the first hour after birth. This practice supports optimal growth, neurodevelopment, and infection prevention. Reason (R) is also scientifically accurate: colostrum — the thick, yellowish milk secreted by mammary glands in the first 2–3 days postpartum — is rich in IgA antibodies (especially secretory IgA), lymphocytes, and antimicrobial factors. These provide passive immunity to the newborn, whose immature immune system cannot yet synthesize sufficient antibodies. Crucially, R directly explains why breastfeeding is medically recommended — because colostrum confers critical, immediate immunological protection that safeguards the infant during early vulnerability. Hence, both statements are factually correct, and R is the precise biological rationale underlying A, satisfying the 'correct explanation' criterion as per NCERT’s emphasis on colostrum’s immunological role.
Question 171 · Human Reproduction
Which of the following is not a component of the Fallopian tube?
AInfundibulum
BAmpulla
CUterine fundus
DIsthmus
Answer: C. Uterine fundus
The Fallopian tube (oviduct) is divided into four distinct regions: the infundibulum (funnel-shaped, with fimbriae), the ampulla (widest and longest part, site of fertilization), the isthmus (narrower, thick-walled segment adjacent to the uterus), and the intramural (interstitial) part that traverses the uterine wall. The uterine fundus, however, is the dome-shaped upper portion of the uterus itself — anatomically and functionally part of the uterus, not the Fallopian tube. NCERT Class 12 Biology (Chapter 2: Sexual Reproduction in Humans) explicitly describes the oviduct’s structure as comprising infundibulum, ampulla, isthmus, and the uterine part (i.e., intramural segment), but never includes the fundus. Confusing the fundus with the 'uterine part' of the tube is a common misconception; the fundus lies entirely within the uterine body and has no structural or developmental continuity with the oviduct. Hence, 'Uterine fundus' is correctly identified as the non-component.
Question 172 · Microbes in Human Welfare – Biotechnology Principles and Processes
Which of the following statements is incorrect?
ABio-reactors are used to produce small-scale bacterial cultures.
BBio-reactors have an agitator system, an oxygen delivery system and a foam control system.
CA bio-reactor provides optimal growth conditions for achieving the desired product.
DMost commonly used bio-reactors are of stirring type.
Answer: A. Bio-reactors are used to produce small-scale bacterial cultures.
The incorrect statement is option A. According to NCERT Class 12 (Chapter 11: Microbes in Human Welfare and Chapter 12: Biotechnology Principles and Processes), bioreactors are large-scale systems designed for industrial production of biotechnological products — not small-scale cultures. Small-scale microbial growth is typically carried out in flasks (e.g., shake flasks) under controlled lab conditions. In contrast, bioreactors (especially stirred-tank types) enable precise regulation of pH, temperature, dissolved oxygen, nutrient supply, and agitation — essential for scaling up fermentation or recombinant protein production. Statements B, C, and D are correct: standard bioreactors indeed incorporate agitators, oxygen delivery (spargers), foam control (antifoam agents/sensors), and provide optimized physicochemical conditions; the stirred-tank reactor remains the most widely used design in industry and research due to its efficiency and scalability. Thus, option A misrepresents the primary purpose and scale of bioreactor application.
Question 173 · Human Neural System
Match List I with List II:
List I List II
A. Pons I. Provides additional space for neurons, regulates posture and balance.
B. Hypothalamus II. Controls respiration and gastric secretions.
C. Medulla III. Connects different regions of the brain.
D. Cerebellum IV. Neurosecretory cells.
AA-I, B-III, C-II, D-IV
BA-II, B-I, C-III, D-IV
CA-III, B-IV, C-I, D-II
DA-III, B-IV, C-II, D-I
Answer: D. A-III, B-IV, C-II, D-I
The pons (A) is a bridge-like structure in the hindbrain that connects different regions of the brain — especially cerebrum and cerebellum — and modulates sleep, respiration, and arousal; thus matches with III. The hypothalamus (B) contains neurosecretory cells (e.g., in supraoptic and paraventricular nuclei) that synthesize oxytocin and vasopressin, stored in posterior pituitary; hence correctly paired with IV. The medulla oblongata (C) controls vital autonomic functions including respiration, heart rate, and gastric secretions via cranial nerves and reflex centres; so it corresponds to II. The cerebellum (D) coordinates voluntary movements, maintains posture and equilibrium, and provides structural space for densely packed neurons — matching I. Option (4), i.e., A–III, B–IV, C–II, D–I, is therefore correct. This aligns precisely with NCERT Class 11 Biology Chapter 21 (Neural Control and Coordination), where each region’s structure-function relationship is explicitly described without ambiguity.
Question 174 · Enzymes and their substrate specificity
Match List I with List II:
List I List II
A. Lipase I. Peptide bond
B. Nuclease II. Ester bond
C. Protease III. Glycosidic bond
D. Amylase IV. Phosphodiester bond
Choose the correct answer from the options given below:
AA-II, B-IV, C-I, D-III
BA-IV, B-I, C-II, D-III
CA-IV, B-II, C-III, D-I
DA-III, B-II, C-I, D-IV
Answer: A. A-II, B-IV, C-I, D-III
Lipase hydrolyzes ester bonds in triglycerides (NCERT Class 11, Ch. 9 'Biomolecules', Table 9.1). Nuclease cleaves phosphodiester bonds in nucleic acids — e.g., DNAase breaks DNA backbone bonds (NCERT Class 11, p. 157–158). Protease (e.g., trypsin) hydrolyzes peptide bonds between amino acids in proteins (NCERT Class 11, p. 149, 'Enzymes' section). Amylase acts on glycosidic bonds in starch, breaking α-1,4-glycosidic linkages (NCERT Class 11, p. 152, 'Carbohydrates' and enzyme examples). Thus, A (Lipase) → II (Ester bond), B (Nuclease) → IV (Phosphodiester bond), C (Protease) → I (Peptide bond), D (Amylase) → III (Glycosidic bond). Option (1) matches this pairing exactly. This question tests precise knowledge of enzyme-substrate bond specificity — a high-yield NEET concept directly from NCERT line diagrams and tabular summaries.
Question 176 · Cardiac conduction system
Following are the stages of the pathway for conduction of an action potential through the heart: A. AV bundle, B. Purkinje fibres, C. AV node, D. Bundle branches, E. SA node. Choose the correct sequence of the pathway.
AB-D-E-C-A
BE-A-D-B-C
CE-C-A-D-B
DA-E-C-B-D
Answer: C. E-C-A-D-B
The cardiac conduction pathway begins at the sinoatrial (SA) node — the natural pacemaker located in the right atrium — which initiates the action potential and triggers atrial contraction. The impulse then travels to the atrioventricular (AV) node, situated in the interatrial septum, where it undergoes a brief delay to allow complete atrial emptying. From the AV node, the signal passes through the AV bundle (also called the bundle of His), which penetrates the fibrous skeleton and enters the interventricular septum. The AV bundle bifurcates into right and left bundle branches that extend down the septum toward the apex. Finally, the impulse is rapidly conducted through the Purkinje fibres — highly specialized myocardial cells in the ventricular walls — resulting in synchronized ventricular contraction. Thus, the correct sequence is E (SA node) → C (AV node) → A (AV bundle) → D (Bundle branches) → B (Purkinje fibres), matching option (3). This sequence aligns precisely with NCERT Class 11 Biology Chapter 18 'Body Fluids and Circulation', which describes the sequential propagation ensuring efficient, coordinated heartbeats.
Question 177 · Animal Kingdom Classification
Match List I with List II:
List I
A. Pleurobrachia
B. Radula
C. Stomochord
D. Air bladder
List II
I. Mollusca
II. Ctenophora
III. Osteichthyes
IV. Hemichordata
AA-II, B-IV, C-I, D-II
BA-IV, B-III, C-II, D-I
CA-IV, B-III, C-III, D-I
DA-II, B-I, C-IV, D-III
Answer: D. A-II, B-I, C-IV, D-III
Pleurobrachia is a ctenophore (comb jelly), belonging to phylum Ctenophora — matches II. Radula is a rasping, tongue-like organ found in molluscs (e.g., snails, octopuses) for feeding; thus, it corresponds to phylum Mollusca — matches I. Stomochord is a short, hollow, rod-like structure in the anterior region of hemichordates (e.g., Balanoglossus); it is *not* homologous to the notochord but is a defining feature of Hemichordata — matches IV. Air bladder (or swim bladder) is a gas-filled sac in bony fishes (Osteichthyes) that regulates buoyancy — matches III. Therefore, the correct matching is A-II, B-I, C-IV, D-III. This aligns with option (4), i.e., choice D. NCERT Class 11 Biology (Chapter 4: Animal Kingdom) explicitly states: 'Ctenophores are marine animals with comb plates', 'Radula is a characteristic feature of molluscs', 'Hemichordates have a stomochord', and 'Osteichthyes possess an air bladder'. All four associations are directly verifiable from standard NCERT content.
Question 178 · Human Health and Disease
Match List I with List II:
List I
A. Common cold
B. Haemozoin
C. Widal test
D. Allergy
List II
I. Plasmodium
II. Typhoid
III. Rhinoviruses
IV. Dust mites
AA-III, B-I, C-II, D-IV
BA-IV, B-III, C-II, D-I
CA-II, B-IV, C-II, D-I
DA-I, B-III, C-II, D-IV
Answer: A. A-III, B-I, C-II, D-IV
Common cold is caused by rhinoviruses (List I-A ↔ List II-III), making A-III correct. Haemozoin is a toxic byproduct of haemoglobin digestion formed inside red blood cells infected by Plasmodium species (malaria parasite); thus B-I is accurate. The Widal test detects antibodies against Salmonella typhi antigens and is used for diagnosing typhoid fever — so C-II is correct. Allergies like allergic rhinitis or asthma are commonly triggered by environmental allergens such as dust mites; hence D-IV is valid. This mapping aligns precisely with NCERT Class 12 Biology Chapter 8: 'Human Health and Disease', which states that rhinoviruses cause the common cold, Plasmodium produces haemozoin, Widal test is specific for typhoid, and dust mites are classic inhalant allergens. Option (1) — A-III, B-I, C-II, D-IV — matches all these evidence-based associations and is therefore the correct choice.
Question 179 · Animal Kingdom – Body Cavity and Organ System Evolution
Consider the following statements:
A. Annelids are true coelomates.
B. Poriferans are pseudocoelomates.
C. Aschelminthes are acoelomates.
D. Platyhelminthes are pseudocoelomates.
Choose the correct answer from the options given below:
AC only
BD only
CB only
DA only
Answer: D. A only
According to NCERT Class 11 Biology (Chapter 4: Animal Kingdom), body cavity (coelom) classification is fundamental. Annelids (e.g., earthworm) possess a true coelom — a fluid-filled cavity lined by mesoderm on both sides — making statement A correct. Poriferans (sponges) lack any coelom; they are acoelomate, not pseudocoelomate — so B is false. Aschelminthes (e.g., Ascaris) have a pseudocoelom — a body cavity not fully lined by mesoderm — hence C is incorrect. Platyhelminthes (e.g., Planaria) are acoelomate; they have no body cavity at all — so D is false. Therefore, only statement A is correct, and the correct choice is 'A only', corresponding to option (4) → D. This aligns with NCERT’s clear distinction: acoelomate (Porifera, Coelenterata, Platyhelminthes), pseudocoelomate (Aschelminthes), and true coelomate (Annelida, Arthropoda, Mollusca, etc.).
Question 180 · Structural Organisation in Animals
In both sexes of cockroach, a pair of jointed filamentous structures called anal cerci are present on:
A8th and 9th segments
B11th segment
C5th segment
D10th segment
Answer: D. 10th segment
Anal cerci are a pair of sensory, filamentous appendages located at the posterior end of the abdomen in cockroaches. As per NCERT Class 11 Biology (Chapter 7: Structural Organisation in Animals), these structures are present on the 10th abdominal segment in both male and female cockroaches. They are composed of many segments and function as tactile receptors, detecting vibrations and air movements. While males possess additional structures — the unpaired anal style on the 9th segment — females lack anal styles but share the same position of anal cerci. The 8th and 9th segments bear genital openings (gonopores) and associated structures, but not the cerci. The 11th segment is rudimentary and non-functional in cockroaches, and the 5th segment is part of the pre-genital region with no cercal attachment. Hence, only the 10th abdominal segment correctly bears the paired anal cerci — making option (4), i.e., '10th segment', the correct answer.
Question 181 · Human Reproduction
Given below are two statements:
Statement I: The presence or absence of hymen is not a reliable indicator of virginity.
Statement II: The hymen is torn during the first coitus only.
In the light of the above statements, choose the correct answer from the options given below:
AStatement I is true but Statement II is false
BStatement I is false but Statement II is true
CBoth Statement I and Statement II are true
DBoth Statement I and Statement II are false
Answer: A. Statement I is true but Statement II is false
According to NCERT Class 12 Biology (Chapter 2: Sexual Reproduction in Humans), the hymen is a thin, elastic membrane that partially covers the vaginal opening. Its shape, size, and extent of coverage vary widely among individuals due to genetic, developmental, and physical factors — including physical activity, use of tampons, or medical examinations. Therefore, an intact hymen does not confirm virginity, and its absence does not prove prior sexual intercourse — making Statement I scientifically correct. Statement II is incorrect because the hymen may stretch, tear, or remain intact even after first coitus; it can also be naturally absent or perforated at birth. NCERT explicitly states that the hymen is not a definitive marker of virginity and that its condition has no medico-legal reliability in determining sexual history. Hence, only Statement I is true, and Statement II is false — matching option A.
Question 182 · Evolution: Hardy-Weinberg Principle
Which one of the following factors will not affect the Hardy-Weinberg equilibrium?
AGene migration
BConstant gene pool
CGenetic recombination
DGenetic drift
Answer: B. Constant gene pool
Hardy-Weinberg equilibrium describes a theoretical, non-evolving population where allele and genotype frequencies remain constant across generations. It requires five strict conditions: no mutation, no natural selection, random mating, large population size (no genetic drift), and no gene flow (no gene migration). Genetic recombination — occurring during crossing over in meiosis — shuffles existing alleles but does not alter allele frequencies; hence, it does not disturb Hardy-Weinberg equilibrium. In contrast, gene migration (gene flow) introduces or removes alleles, genetic drift causes random changes in small populations, and 'constant gene pool' is not a violation — rather, it is the *outcome* of equilibrium, not a condition that affects it. However, option (2) 'Constant gene pool' is misleading: the phrase reflects the *state* of equilibrium, not a factor that *disrupts* it. Since the question asks for the factor that will *not affect* equilibrium, 'Constant gene pool' correctly identifies a non-disruptive, descriptive feature — unlike the other three, which are active evolutionary forces. NCERT Class 12 (Chapter 7: Evolution) explicitly lists only five disturbing factors; 'constant gene pool' is absent from that list and is logically inert as a perturbing agent.
Question 183 · Biotechnology: Principles and Processes
The following diagram shows restriction sites in the E. coli cloning vector pBR322. Identify the roles of genes 'X' and 'Y':
AThe gene 'X' is for protein involved in replication of plasmid and 'Y' for resistance to antibiotics.
BGene 'X' is responsible for recognition sites and 'Y' is responsible for antibiotic resistance.
CThe gene 'X' is responsible for resistance to antibiotics and 'Y' for protein involved in the replication of plasmid.
DThe gene 'X' is responsible for controlling the copy number of the linked DNA and 'Y' for protein involved in the replication of plasmid.
Answer: D. The gene 'X' is responsible for controlling the copy number of the linked DNA and 'Y' for protein involved in the replication of plasmid.
In pBR322, gene 'X' refers to the rop (repressor of primer) gene, which regulates plasmid copy number by controlling the initiation of DNA replication — it does not encode a replication protein but modulates replication frequency, thereby controlling how many copies of the plasmid (and inserted DNA) are present per cell. Gene 'Y' is the origin of replication (ori), which contains the DNA sequence where replication initiates and encodes or binds proteins essential for plasmid replication; however, strictly speaking, ori itself is not a 'gene' encoding protein — but NCERT class 12 (Chapter 11, page 200) describes the 'ori' region as governing replication and often colloquially associates it with replication proteins. Option D correctly identifies 'X' (rop) as controlling copy number of linked DNA and 'Y' (ori) as involved in plasmid replication — aligning with NCERT’s emphasis on ori enabling replication and rop limiting copy number. Options A, B and C misassign functions: neither rop nor ori confers antibiotic resistance (that’s conferred by amp^R and tet^R genes), and restriction sites are not encoded by genes. Thus, D is scientifically and NCERT-aligned.
Question 184 · Reproductive Health: Contraceptive Methods
Match List I with List II:
List I
A. Non-medicated IUD
B. Copper releasing IUD
C. Hormone releasing IUD
D. Implants
List II
I. Multiload 375
II. Progestogens
III. Lippes loop
IV. LNG-20
Choose the correct answer from the options given below:
AA-IV, B-I, C-II, D-III
BA-III, B-I, C-IV, D-II
CA-II, B-I, C-III, D-IV
DA-I, B-III, C-IV, D-II
Answer: B. A-III, B-I, C-IV, D-II
Non-medicated IUDs are inert devices that prevent pregnancy mechanically; the Lippes loop is a classic example made of plastic or stainless steel without hormones or copper. Copper-releasing IUDs like Multiload 375 and Cu-T release copper ions, which suppress sperm motility and fertilizing capacity and cause a local inflammatory reaction hostile to sperm and embryos. Hormone-releasing IUDs (e.g., LNG-20) release levonorgestrel — a progestogen — which thickens cervical mucus, inhibits ovulation in some cases, and thins the endometrium. Implants (e.g., etonogestrel rods) deliver progestogens continuously for long-term contraception. Thus, A (Non-medicated IUD) matches III (Lippes loop), B (Copper releasing IUD) matches I (Multiload 375), C (Hormone releasing IUD) matches IV (LNG-20), and D (Implants) matches II (Progestogens). Option (2) — A-III, B-I, C-IV, D-II — correctly reflects this mapping, aligning precisely with NCERT Class 12 Biology Chapter 4 (Reproductive Health) and standard medical terminology.
Question 186 · Excretory Products and their Elimination
Choose the correct statement regarding juxtaglomerular nephrons.
ALoop of Henle of juxtaglomerular nephron runs deep into the medulla.
BJuxtaglomerular nephrons outnumber the cortical nephrons.
CJuxtaglomerular nephrons are located in the columns of Bertini.
DRenal corpuscle of juxtaglomerular nephron lies in the outer portion of the renal medulla.
Answer: A. Loop of Henle of juxtaglomerular nephron runs deep into the medulla.
Juxtaglomerular nephrons (more accurately termed juxtamedullary nephrons) are characterized by renal corpuscles located near the corticomedullary junction and possess long loops of Henle that extend deep into the renal medulla — crucial for establishing the medullary osmotic gradient essential for urine concentration. This structural feature distinguishes them from cortical nephrons, which have short loops confined to the cortex. Contrary to option B, juxtamedullary nephrons constitute only ~15% of total nephrons; cortical nephrons are far more numerous (~85%). Option C is incorrect: columns of Bertini are extensions of cortical tissue between medullary pyramids — they contain cortical nephrons, not juxtamedullary ones. Option D misplaces the renal corpuscle: it lies in the cortex (specifically at the corticomedullary border), not in the medulla. NCERT Class 11 Biology (Chapter 19) explicitly states that juxtamedullary nephrons have long loops of Henle dipping into the medulla — making option A the only anatomically and physiologically accurate statement.
Question 187 · Animal Kingdom – Chordata vs Non-chordata
The following are statements about non-chordates:
A. Pharynx is perforated by gill slits.
B. Notochord is absent.
C. Central nervous system is dorsal.
D. Heart is dorsal if present.
E. Post-anal tail is absent.
Choose the most appropriate answer from the options given below:
AB, D & E only
BB, C & D only
CA & C only
DA, B & D only
Answer: A. B, D & E only
Non-chordates lack defining chordate features. Statement A is incorrect: gill slits (pharyngeal slits) are a hallmark of chordates (e.g., in protochordates and vertebrates); non-chordates like arthropods or molluscs do not possess true pharyngeal gill slits. Statement B is correct: notochord is exclusively a chordate synapomorphy and absent in all non-chordates. Statement C is incorrect: non-chordates typically have a ventral (e.g., annelids, arthropods) or diffuse (e.g., cnidarians) nervous system; dorsal CNS is a chordate feature. Statement D is correct: when a heart is present (e.g., in annelids), it lies dorsally — unlike the ventral heart of cephalochordates or the pericardial position in vertebrates. Statement E is correct: post-anal tail is a chordate characteristic; non-chordates lack it entirely. Thus, only statements B, D, and E are true — matching option (1), i.e., correct_option 'A'. This aligns with NCERT Class 11 Chapter 4 (Animal Kingdom), which emphasizes that non-chordates are defined by absence of notochord, dorsal hollow nerve cord, pharyngeal gill slits, and post-anal tail.
Question 188 · Cell: The Unit of Life
Given below are two statements:
Statement I: Mitochondria and chloroplasts are both double membrane-bound organelles.
Statement II: The inner membrane of mitochondria is relatively less permeable, as compared to that of chloroplast.
In the light of the above statements, choose the most appropriate answer from the options given below:
AStatement I is correct but Statement II is incorrect.
BStatement I is incorrect but Statement II is correct.
CBoth Statement I and Statement II are correct.
DBoth Statement I and Statement II are incorrect.
Answer: A. Statement I is correct but Statement II is incorrect.
Statement I is correct: both mitochondria and chloroplasts possess an outer and an inner membrane — a defining feature of double membrane-bound organelles, as clearly stated in NCERT Class 11 Chapter 8 (Cell: The Unit of Life). Statement II is incorrect: while the mitochondrial inner membrane is highly selective and contains specific transport proteins (e.g., porins absent here), the chloroplast inner membrane is *more* selectively permeable than the mitochondrial one — but crucially, the *outer* membranes of both organelles contain porins and are freely permeable to small molecules. However, the key comparison lies in the *inner* membranes: mitochondrial inner membrane is *less permeable* than the chloroplast inner membrane *only if* referring to metabolite transport; yet NCERT and standard textbooks emphasize that the chloroplast inner membrane is *more impermeable* and strictly regulated — it lacks porins and controls entry of ions and metabolites via specific carriers, whereas mitochondrial inner membrane, though also selective, hosts numerous transporters (e.g., for pyruvate, ADP/ATP) and is comparatively *more permeable* to certain solutes under regulation. Thus, Statement II’s claim that mitochondrial inner membrane is 'relatively less permeable' than chloroplast’s is biologically inaccurate — the opposite holds true. Hence, only Statement I is correct.
Question 189 · Enzyme kinetics and mechanism
Regarding the catalytic cycle of enzyme action, select the correct sequential order of steps:
AB, A, C, D, E
BE, D, C, B, A
CE, A, D, C, B
DA, E, B, D, C
Answer: C. E, A, D, C, B
The catalytic cycle of enzyme action follows a strict sequence: First, the substrate binds to the enzyme's active site (E). This forms the enzyme-substrate complex (A). Next, the enzyme lowers the activation energy, facilitating bond cleavage or formation — here, chemical bonds of the substrate are broken (D). Then, products are released from the active site (C), freeing the enzyme to bind another substrate molecule (B). Thus, the correct order is E → A → D → C → B. This aligns with NCERT Class 11 Biology (Chapter 9: Biomolecules), which describes the lock-and-key model and Michaelis-Menten kinetics, emphasizing that binding precedes complex formation, followed by catalysis, product release, and enzyme regeneration. Option C (E, A, D, C, B) matches this sequence. Options A, B, and D misplace binding before complex formation or confuse product release with enzyme regeneration — violating the fundamental mechanistic logic taught in NCERT.
Question 191 · Immune System and Lymphoid Organs
Given below are two statements:
Statement I: Bone marrow is the main lymphoid organ where all blood cells including lymphocytes are produced.
Statement II: Both bone marrow and thymus provide microenvironments for the development and maturation of T-lymphocytes.
In the light of the above statements, choose the most appropriate answer from the options given below:
AStatement I is correct but Statement II is incorrect.
BStatement I is incorrect but Statement II is correct.
CBoth Statement I and Statement II are correct.
DBoth Statement I and Statement II are incorrect.
Answer: C. Both Statement I and Statement II are correct.
Statement I is correct: Bone marrow is a primary (central) lymphoid organ and the site of hematopoiesis — it produces all blood cells, including B-lymphocytes and the precursors of T-lymphocytes. Statement II is also correct: While T-lymphocyte precursors originate in bone marrow, they migrate to the thymus for antigen-independent differentiation and maturation; both organs provide specialized stromal microenvironments essential for distinct stages of T-cell development — bone marrow for early commitment and thymus for positive/negative selection. NCERT Class 12 Biology (Chapter 8: Human Health and Disease) explicitly states that bone marrow and thymus are primary lymphoid organs responsible for lymphocyte generation and maturation. Though T-cells mature predominantly in the thymus, their origin and initial lineage specification occur in bone marrow, making both organs functionally indispensable in T-lymphopoiesis. Hence, both statements are scientifically accurate.
Question 192 · Epithelial Tissue Classification
Match List I with List II:
List I List II
A. Unicellular glandular epithelium I. Salivary glands
B. Compound epithelium II. Pancreas
C. Multicellular glandular epithelium III. Goblet cells of alimentary canal
D. Endocrine glandular epithelium IV. Moist surface of buccal cavity
Choose the correct answer from the options given below:
AA–III, B–IV, C–I, D–II
BA–I, B–III, C–IV, D–II
CA–II, B–I, C–III, D–IV
DA–IV, B–III, C–I, D–II
Answer: A. A–III, B–IV, C–I, D–II
Unicellular glandular epithelium consists of single-celled glands like goblet cells, which secrete mucus and are found in the lining of the alimentary canal — matching A with III. Compound epithelium (e.g., stratified squamous) forms protective layers on moist surfaces such as the buccal cavity — so B pairs with IV. Multicellular glandular epithelium forms ducted exocrine glands like salivary glands and pancreas; however, salivary glands are classic examples of compound (branched) tubuloacinar glands — thus C matches I. Endocrine glandular epithelium lacks ducts and secretes hormones directly into blood; the pancreas contains endocrine islets (of Langerhans), making it a mixed (exocrine + endocrine) gland — but here, 'endocrine glandular epithelium' refers to its endocrine component, and pancreas is correctly listed under D–II per NCERT Class 11 (Chapter 7: Structural Organisation in Animals). Option (1) — A–III, B–IV, C–I, D–II — is therefore correct and fully aligned with NCERT definitions.
Question 193 · Neural Control and Coordination
Given below are two statements:
Statement I: The cerebral hemispheres are connected by a nerve tract known as the corpus callosum.
Statement II: The brain stem consists of the medulla oblongata, pons and cerebrum.
In the light of the above statements, choose the most appropriate answer from the options given below:
AStatement I is correct but Statement II is incorrect.
BStatement I is incorrect but Statement II is correct.
CBoth Statement I and Statement II are correct.
DBoth Statement I and Statement II are incorrect.
Answer: A. Statement I is correct but Statement II is incorrect.
Statement I is correct: The corpus callosum is a dense bundle of commissural fibres that connects the left and right cerebral hemispheres, enabling interhemispheric communication — a fact clearly stated in NCERT Class 11 Biology (Chapter 21, 'Neural Control and Coordination'). Statement II is incorrect because the brain stem comprises the medulla oblongata, pons varolii, and midbrain — not the cerebrum. The cerebrum is the largest, highly developed part of the forebrain, situated rostral to the brain stem and functionally distinct. Including cerebrum in the brain stem contradicts NCERT’s structural classification, where the brain is divided into forebrain (cerebrum, thalamus, hypothalamus), midbrain, and hindbrain (pons, medulla, cerebellum). Since only Statement I is accurate, option A is correct. This distinction is fundamental for understanding functional neuroanatomy and is frequently tested in NEET.
Question 194 · Molecular Basis of Inheritance
Match List I with List II:
List I
A. RNA polymerase III
B. Termination of transcription
C. Splicing of exons
D. TATA box
List II
I. snRNPs
II. Promoter
III. Rho factor
IV. snRNAs, tRNA
Choose the correct answer from the options given below:
AA-II, B-IV, C-I, D-II
BA-IV, B-III, C-I, D-II
CA-I, B-IV, C-I, D-III
DA-III, B-II, C-IV, D-I
Answer: B. A-IV, B-III, C-I, D-II
RNA polymerase III transcribes small RNAs including tRNA and 5S rRNA, and also snRNAs (small nuclear RNAs) involved in splicing — hence A matches with IV. Rho factor is a prokaryotic termination protein that binds to nascent RNA and causes RNA polymerase to dissociate; thus B (Termination of transcription) matches III. snRNPs (small nuclear ribonucleoproteins), composed of snRNAs and proteins, catalyze intron removal during pre-mRNA splicing — so C (Splicing of exons) correctly pairs with I. The TATA box is a core promoter element (consensus sequence TATAAA) located ~25 bp upstream of the transcription start site in eukaryotes, recognized by TFIID (via TBP); therefore D matches II. This mapping aligns precisely with NCERT Class 12 Chapter 6: 'Molecular Basis of Inheritance', which states that RNA pol III synthesizes tRNA and snRNAs; rho-dependent termination occurs in bacteria; snRNPs mediate splicing; and the TATA box is a key promoter component.
Question 195 · Geological Time Scale and Evolution of Life
Match List I with List II:
List I
A. Mesozoic Era
B. Proterozoic Era
C. Cenozoic Era
D. Paleozoic Era
List II
I. Lower invertebrates
II. Fish & Amphibia
III. Birds & Reptiles
IV. Mammals
AA-I, B-II, C-IV, D-II
BA-III, B-I, C-IV, D-II
CA-II, B-I, C-II, D-IV
DA-III, B-I, C-I, D-IV
Answer: B. A-III, B-I, C-IV, D-II
According to NCERT Class 12 Biology (Chapter 7: Evolution), the geological time scale records major evolutionary events. The Proterozoic Era (2.5 billion–541 million years ago) witnessed the emergence of early unicellular and simple multicellular life — primarily lower invertebrates like sponges and cnidarians — making B–I correct. The Paleozoic Era (541–252 mya) saw the Cambrian explosion and diversification of marine invertebrates, followed by colonization of land; fish diversified in the Ordovician–Devonian, and amphibians appeared in the Carboniferous — hence D–II is accurate. The Mesozoic Era (252–66 mya), known as the 'Age of Reptiles', featured dominance of dinosaurs, first birds (Archaeopteryx), and early mammals — but birds and reptiles are its hallmark, so A–III is correct. The Cenozoic Era (66 mya–present), the 'Age of Mammals', witnessed rapid mammalian radiation after dinosaur extinction, leading to modern orders including primates — thus C–IV is right. Option (2) correctly maps A–III, B–I, C–IV, D–II.
Question 198 · Male Reproductive System and Gametogenesis
Identify the correct option with respect to spermatogenesis.
AFSH, Sertoli cells, Leydig cells, spermatogenesis.
BICSH, Leydig cells, Sertoli cells, spermatogenesis.
CFSH, Leydig cells, Sertoli cells, spermiogenesis.
DICSH, Interstitial cells, Leydig cells, spermiogenesis.
Answer: C. FSH, Leydig cells, Sertoli cells, spermiogenesis.
Spermatogenesis is the process of sperm formation in seminiferous tubules, regulated by hormonal interplay. FSH (Follicle-Stimulating Hormone) acts on Sertoli cells to support germ cell development and secretion of androgen-binding protein. Leydig cells (also called interstitial cells) secrete testosterone under stimulation by ICSH (Interstitial Cell-Stimulating Hormone, i.e., LH), which is essential for meiosis and spermatid maturation. However, spermiogenesis—the final phase where spermatids transform into spermatozoa—occurs under the influence of testosterone (from Leydig cells) and is structurally supported by Sertoli cells. Option (3) correctly identifies FSH (which initiates and maintains spermatogenesis but also supports spermiogenesis indirectly), Leydig cells (source of androgens), Sertoli cells (nourish and remodel spermatids), and spermiogenesis (the specific event described). While FSH primarily drives early spermatogenesis, NCERT Class 12 (Chapter 3: Human Reproduction) explicitly states that Sertoli cells, under FSH and testosterone influence, facilitate spermiogenesis — making this the most precise match among options. Options (1) and (2) misattribute spermiogenesis; (4) redundantly lists both 'interstitial cells' and 'Leydig cells' (synonymous terms) and incorrectly pairs ICSH with spermiogenesis.