Revision guide
How to use NEET 2021 Biology PYQs
This set contains reviewed questions from the Code M1 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
- Cell Cycle and Cell Division
Best review method
Use this paper to practise retrieving facts without prompts. After checking the key, write one short correction for each error rather than copying the whole explanation.
NCERT focus
Concentrate on transcription and translation, cloning steps and cell-division checkpoints. These topics reward exact sequence and terminology rather than broad familiarity.
Frequently asked questions
NEET 2021 Biology PYQ FAQs
Are these NEET 2021 Biology answers checked with the official key?
This page uses the NEET 2021 Code M1 Biology question paper and checks answers against the official final answer key before publication.
How should I use NEET 2021 Biology PYQs for revision?
Solve the question first, check the answer, then connect the explanation to the NCERT concept. Pay special attention to ecology, genetics, human physiology, and biotechnology questions.
Question 101 · Plant reproduction: sexual and asexual methods; monoecious vs dioecious species
Which of the following plants is monoecious?
ACarica papaya
BChara
CMarchantia polymorpha
DCycas circinalis
Answer: B. Chara
Monoecious plants bear both male and female reproductive structures on the same individual. Chara, a filamentous green alga (Class Chlorophyceae), is monoecious — its antheridia (male sex organs) and oogonia (female sex organs) occur on the same thallus, often in close proximity. In contrast, Carica papaya is dioecious (male and female flowers on separate plants), though some cultivars may show hermaphroditism; Marchantia polymorpha is dioecious (with distinct male and female gametophytes); and Cycas circinalis is also dioecious — male cones and female megasporophylls are borne on separate plants. NCERT Class 11 (Chapter 3: Plant Kingdom) explicitly states that Chara exhibits monoecious condition, while bryophytes like Marchantia and gymnosperms like Cycas are predominantly dioecious. Papaya’s variability does not override its typical dioecious nature in standard NEET contexts. Thus, Chara is the only correctly identified monoecious organism among the options.
Question 103 · Plant Kingdom – Bryophytes
Gemmae are present in:
AMosses
BPteridophytes
CSome Gymnosperms
DSome Liverworts
Answer: D. Some Liverworts
Gemmae are multicellular, asexual, discoid or lens-shaped reproductive structures that detach to form new individuals. According to NCERT Class 11 Biology (Chapter 3: Plant Kingdom), gemmae are characteristically found in certain liverworts — notably Marchantia — where they develop in gemma cups on the dorsal surface of the thallus. Mosses reproduce asexually via fragmentation or by budding from the protonema, but do not produce true gemmae. Pteridophytes reproduce via spores and vegetative means like rhizome fragmentation; gemmae are absent. Gymnosperms reproduce sexually through cones and seeds; no gemmae occur in any gymnosperm. Thus, only some liverworts possess gemmae as a specialized asexual reproductive adaptation. This aligns precisely with option (4) — 'Some Liverworts' — making D the correct and NCERT-verified answer.
Question 104 · Chromosome structure and types
When the centromere is situated in the middle of the chromosome, resulting in two equal arms, the chromosome is referred to as:
AMetacentric
BTelocentric
CSubmetacentric
DAcrocentric
Answer: A. Metacentric
According to NCERT Class 11 Biology (Chapter 8: Cell: The Unit of Life), chromosomes are classified based on the position of the centromere. A metacentric chromosome has its centromere located exactly at the midpoint, dividing the chromosome into two morphologically identical arms of equal length. This symmetry allows for balanced segregation during mitosis. In contrast, telocentric chromosomes have the centromere at the terminal end (not observed in humans), submetacentric chromosomes have an off-centre centromere producing one longer and one shorter arm, and acrocentric chromosomes have the centromere near one end, resulting in a very short p-arm and a long q-arm — often associated with satellite regions. Human chromosomes 1, 3, 16, 19, and 20 are metacentric. The question correctly identifies the defining feature — centromere at the middle yielding equal arms — which uniquely corresponds to metacentric chromosomes. Hence, option A is scientifically accurate and fully aligned with NCERT’s description.
Question 105 · Cell Cycle and Cell Division
Which of the following stages of meiosis involves division of the centromere?
AMetaphase I
BMetaphase II
CAnaphase II
DTelophase II
Answer: C. Anaphase II
The centromere divides during Anaphase II of meiosis, when sister chromatids separate and move to opposite poles. In Meiosis I, homologous chromosomes separate during Anaphase I, but the centromere remains intact — each chromosome still consists of two sister chromatids joined at the centromere. Only in Meiosis II, which resembles mitosis, does the centromere split during Anaphase II, allowing individual chromatids (now called chromosomes) to segregate. Metaphase I and II involve alignment of chromosomes at the equator but no centromere division; Telophase II is the final stage where nuclei reform and cytokinesis occurs — centromere division has already concluded in Anaphase II. This distinction is clearly emphasized in NCERT Class 11, Chapter 10 'Cell Cycle and Cell Division', which states: 'In anaphase II, the centromeres split and chromatids separate.' Hence, Anaphase II is the only stage among the options where centromere division occurs.
Question 106 · Photosynthesis in Higher Plants
The first stable product of CO₂ fixation in sorghum is:
APyruvic acid
BOxaloacetic acid
CSuccinic acid
DPhosphoglyceric acid
Answer: B. Oxaloacetic acid
Sorghum is a C₄ plant, and in C₄ photosynthesis, the first stable product of CO₂ fixation is not formed in the Calvin cycle (which produces 3-phosphoglyceric acid in C₃ plants), but in the mesophyll cells via PEP carboxylase. This enzyme fixes CO₂ to phosphoenolpyruvate (PEP) to form oxaloacetate (oxaloacetic acid), a 4-carbon compound — hence the term 'C₄ pathway'. Oxaloacetic acid is immediately converted to other C₄ acids like malate or aspartate for transport to bundle sheath cells, but it remains the *first stable* product of carbon fixation. Pyruvic acid is a substrate (regenerated from PEP), succinic acid is an intermediate in the Krebs cycle (not photosynthetic carbon fixation), and phosphoglyceric acid (3-PGA) is the first stable product only in C₃ plants like wheat or rice. NCERT Class 11, Chapter 13 'Photosynthesis in Higher Plants', clearly states: 'The primary CO₂ acceptor is PEP and the first stable product is oxaloacetate.'
Question 107 · Evolution and Genetic Variation
The factor that leads to Founder effect in a population is:
ANatural selection
BGenetic recombination
CMutation
DGenetic drift
Answer: D. Genetic drift
The Founder effect is a type of genetic drift that occurs when a small group of individuals becomes isolated from a larger population and establishes a new colony. Due to the small size of the founding group, only a limited and non-representative sample of the original population’s genetic variation is carried forward. This results in reduced genetic diversity and potentially higher frequencies of rare alleles — not because of adaptive advantage (so not natural selection), nor due to new allele creation (so not mutation), nor through reshuffling of existing alleles during meiosis (so not genetic recombination). Instead, it arises purely by chance sampling error in allele frequencies — the defining feature of genetic drift. As per NCERT Class 12 Biology (Chapter 7: Evolution), genetic drift includes both the Founder effect and the Bottleneck effect, both operating independently of selection pressure. Hence, genetic drift is the correct and sole mechanism responsible for the Founder effect.
Question 108 · Cell: The Unit of Life
Which of the following is an incorrect statement?
AMature sieve tube elements possess a conspicuous nucleus and usual cytoplasmic organelles.
BMicrobodies are present both in plant and animal cells.
CThe perinuclear space forms a barrier between the materials present inside the nucleus and that of the cytoplasm.
DNuclear pores act as passages for proteins and RNA molecules in both directions between nucleus and cytoplasm.
Answer: A. Mature sieve tube elements possess a conspicuous nucleus and usual cytoplasmic organelles.
Mature sieve tube elements are enucleate — they lose their nucleus during differentiation and lack most cytoplasmic organelles (e.g., ribosomes, Golgi apparatus, vacuoles) to facilitate efficient phloem transport. Their cytoplasm is highly specialized, containing only mitochondria, plastids, and P-proteins; the absence of a nucleus is a defining feature confirmed in NCERT Class 11, Chapter 8 'Cell: The Unit of Life'. Hence, option A is incorrect. Option B is correct: microbodies (e.g., peroxisomes, glyoxysomes) occur in both plant and animal cells — peroxisomes in all eukaryotes, glyoxysomes specifically in plant seeds. Option C is accurate: the perinuclear space (between the two nuclear membranes) acts as a selective barrier, contributing to nucleocytoplasmic compartmentalization. Option D is correct: nuclear pores are protein-lined channels permitting bidirectional transport — import of nuclear proteins and export of mRNA, tRNA, and ribosomal subunits — as detailed in NCERT’s description of nuclear envelope structure and function.
Question 110 · Ecosystem – Energy Flow and Nutrient Cycling
The amount of nutrients, such as carbon, nitrogen, phosphorus and calcium, present in the soil at any given time is referred to as:
AClimax
BClimax community
CStanding state
DStanding crop
Answer: C. Standing state
The term 'standing state' refers to the total quantity of inorganic nutrients—like carbon, nitrogen, phosphorus, and calcium—present in the soil or water reservoir at a specific point in time. It represents the non-living, abiotic pool of nutrients available for uptake by organisms and is distinct from 'standing crop', which denotes the total living biomass (mass of living organisms) in a given area at a given time. 'Climax' and 'climax community' describe the final, stable stage of ecological succession, not nutrient pools. As per NCERT Class 12 Biology (Chapter 14: Ecosystem), standing state is explicitly defined in the context of nutrient cycling and differs fundamentally from biotic measures like standing crop. This concept is crucial for understanding how nutrients move between abiotic reservoirs and biotic components via biogeochemical cycles.
Question 111 · Algae: Classification and Reserve Food Materials
Which of the following algae contains mannitol as reserve food material?
AEctocarpus
BGracilaria
CVolvox
DUlothrix
Answer: A. Ectocarpus
Mannitol is a sugar alcohol that serves as a characteristic reserve food material in brown algae (Phaeophyceae). Among the given options, Ectocarpus is a filamentous brown alga belonging to class Phaeophyceae and stores mannitol and laminarin. Gracilaria is a red alga (Rhodophyceae) and stores floridean starch; Volvox is a colonial green alga (Chlorophyceae) storing starch; Ulothrix is also a green alga (Chlorophyceae) with starch as its reserve food. NCERT Class 11 Biology (Chapter 3: Plant Kingdom) explicitly states that brown algae store mannitol and laminarin, while green algae store starch and red algae store floridean starch. Hence, only Ectocarpus matches the criterion. This distinction is fundamental for NEET-level classification-based questions and aligns precisely with NCERT’s comparative table on algal reserve food materials.
Question 112 · Organisms and Populations - Population Interactions
Despite interspecific competition in nature, which mechanism might competing species have evolved for their survival?
AResource partitioning
BCompetitive release
CMutualism
DPredation
Answer: A. Resource partitioning
Interspecific competition occurs when different species compete for the same limited resources like food, space, or light. To reduce competitive exclusion and coexist, species often evolve resource partitioning — a process where they diverge in resource use (e.g., feeding at different times, foraging in different microhabitats, or consuming different prey sizes). This minimizes direct competition and promotes niche differentiation. NCERT Class 12 (Chapter 13: Organisms and Populations) explicitly states that 'resource partitioning is an important mechanism that allows species to coexist despite competition'. Competitive release refers to population expansion when a competitor is removed — not an evolved mechanism for coexistence. Mutualism is a beneficial interaction between species but does not resolve competition per se. Predation is a consumer–prey interaction and may indirectly influence competition (e.g., keystone predation), but it is not an adaptation evolved *by competitors* to survive competition. Hence, resource partitioning is the only adaptive evolutionary strategy directly evolved by competing species to mitigate interspecific competition and ensure long-term coexistence.
Question 113 · Plant Kingdom – Pteridophytes
Genera like Selaginella and Salvinia produce two kinds of spores. Such plants are known as:
AHomosporous
BHeterosporous
CHomosorus
DHeterosorus
Answer: D. Heterosorus
Selaginella and Salvinia are pteridophytes that produce two distinct types of spores — microspores (male) and megaspores (female) — a condition called heterospory. This is in contrast to homosporous pteridophytes (e.g., ferns like Dryopteris), which produce only one kind of spore that develops into a bisexual gametophyte. Heterospory is an evolutionary advancement leading to greater reproductive efficiency and is a precursor to seed habit. NCERT Class 11 Biology (Chapter 3: Plant Kingdom) explicitly states that 'Selaginella and Salvinia are examples of heterosporous pteridophytes'. The term 'heterosporous' correctly describes the plant type, while 'heterosorus' is not a standard botanical term — 'sporophyll' or 'sporangium' may be soral, but 'heterosorus' is a non-existent or misused word. Similarly, 'homosorus' is invalid; the correct adjectival forms are 'homosporous' and 'heterosporous'. Hence, option (4) — labeled D — is scientifically accurate and NCERT-aligned.
Question 114 · Photoperiodism and flowering
The site of perception of light in plants during photoperiodism is:
AShoot apex
BStem
CAxillary bud
DLeaf
Answer: D. Leaf
In photoperiodism—the physiological response of plants to relative lengths of day and night—the primary site of light perception is the leaf. Leaves contain photoreceptors like phytochrome (Pr and Pfr forms) and cryptochrome, which detect red/far-red and blue light, respectively. These photoreceptors sense photoperiodic cues and trigger the synthesis of florigen, a flowering stimulus believed to be FT (FLOWERING LOCUS T) protein in angiosperms. Though florigen is produced in leaves, it is transported via phloem to the shoot apical meristem, where floral initiation occurs. NCERT Class 11 Biology (Chapter 15: Plant Growth and Development) explicitly states: 'It has been established that the site of perception of light for photoperiodism is the leaf.' The shoot apex, stem, and axillary buds are not light-perceiving sites but rather targets or developmental zones influenced by the leaf-derived signal. Hence, option D (Leaf) is scientifically accurate and fully aligned with NCERT.
Question 115 · Plant growth and development
Plants follow different pathways in response to environment or phases of life to form different kinds of structures. This ability is called:
AElasticity
BFlexibility
CPlasticity
DMaturity
Answer: C. Plasticity
The ability of plants to produce different structures in response to environmental conditions or developmental stages is termed plasticity. This is a hallmark of plant development and reflects their non-movable, modular nature. For example, heterophylly in buttercup (Ranunculus) — where submerged leaves are ribbon-like while aerial leaves are broad — demonstrates plasticity. Similarly, the shape of leaves in aquatic vs. terrestrial environments in many species, or shade-avoidance responses, arise from differential gene expression triggered by light, temperature, or nutrient availability. Plasticity is distinct from elasticity (reversible deformation under stress), flexibility (physical bendability), or maturity (a stage of development). NCERT Class 11 Biology (Chapter 15: Plant Growth and Development) explicitly defines plasticity as the capacity of plants to adjust their growth and form in response to external cues — a key adaptation for sessile organisms. It underpins phenotypic plasticity, which allows survival across variable habitats without genetic change.
Question 118 · Biotechnology: Principles and Processes
DNA strands on a gel stained with ethidium bromide, when viewed under UV radiation, appear as:
AYellow bands
BBright orange bands
CDark red bands
DBright blue bands
Answer: B. Bright orange bands
Ethidium bromide (EtBr) is a fluorescent dye commonly used to visualize DNA in agarose or polyacrylamide gels. It intercalates between the stacked base pairs of double-stranded DNA. When exposed to ultraviolet (UV) light (typically at 254 nm or 302 nm), the EtBr-DNA complex absorbs UV radiation and re-emits energy in the visible spectrum — specifically as bright orange-red fluorescence (~590 nm). This distinct bright orange band appearance is standard in NCERT Class 12 Biology (Chapter 11: Biotechnology – Principles and Processes) and confirmed in laboratory protocols. Yellow bands are associated with SYBR Safe under certain conditions; dark red or bright blue bands do not correspond to EtBr’s emission profile. Hence, the correct visual observation is bright orange bands — matching option B.
Question 119 · Sexual Reproduction in Flowering Plants
The term used for transfer of pollen grains from anthers of one plant to stigma of a different plant, which brings genetically different types of pollen grains to the stigma during pollination, is:
AXenogamy
BGeitonogamy
CChasmogamy
DCleistogamy
Answer: A. Xenogamy
Xenogamy is the transfer of pollen from the anther of a flower on one plant to the stigma of a flower on a genetically different plant. It ensures genetic diversity by promoting cross-pollination between unrelated individuals. Geitonogamy, though functionally cross-pollination, occurs between flowers of the same plant and thus involves genetically identical (or nearly identical) pollen — it is not true outbreeding. Chasmogamy refers to pollination in open, exposed flowers that may be self- or cross-pollinated, while cleistogamy is self-pollination in closed, non-opening flowers. NCERT Class 12 Biology (Chapter 2: Sexual Reproduction in Flowering Plants) explicitly defines xenogamy as the only type guaranteeing genetic heterogeneity because it involves pollen from a different individual. This distinction is critical for understanding evolutionary advantages of outbreeding and mechanisms preventing inbreeding depression. Hence, xenogamy is the correct term for genetically diverse pollen transfer between distinct plants.
Question 120 · Algae and their economic importance
Which of the following algae produce carrageenan?
AGreen algae
BBrown algae
CRed algae
DBlue-green algae
Answer: C. Red algae
Carrageenan is a sulfated polysaccharide extracted from the cell walls of certain red algae, especially species of Chondrus, Gigartina, and Eucheuma. It is widely used as a thickening, gelling, and stabilizing agent in dairy products, plant-based milks, and pharmaceuticals. Red algae (Rhodophyta) possess cellulose and sulfated galactans like carrageenan and agar in their cell walls — a key biochemical feature distinguishing them from other algal groups. Green algae (Chlorophyta) store starch and lack carrageenan; their cell walls contain cellulose but no sulfated galactans. Brown algae (Phaeophyta), such as Laminaria and Fucus, produce alginates (alginic acid), not carrageenan. Blue-green algae (Cyanobacteria) are prokaryotes with peptidoglycan cell walls and do not synthesize carrageenan. This distinction is explicitly covered in NCERT Class 11 Biology Chapter 3 'Plant Kingdom', which states: 'Red algae have floridean starch as reserve food material and phycocolloids like agar and carrageenan in their cell walls.' Hence, only red algae produce carrageenan.
Question 121 · Ecological Pyramids
Which of the following statements is not correct?
APyramid of biomass in sea is generally inverted.
BPyramid of biomass in sea is generally upright.
CPyramid of energy is always upright.
DPyramid of numbers in a grassland ecosystem is upright.
Answer: B. Pyramid of biomass in sea is generally upright.
The pyramid of biomass in aquatic ecosystems (e.g., oceans) is typically inverted because the standing crop biomass of phytoplankton (producers) is lower than that of zooplankton (primary consumers), due to their high reproductive rate and short lifespan — a concept clearly explained in NCERT Class 12, Chapter 14 'Ecosystem'. Hence, statement (2) — claiming it is 'generally upright' — is incorrect. In contrast, the pyramid of energy is always upright across all ecosystems, as energy flow follows the 10% law and decreases progressively at each trophic level — a fundamental principle supported by Lindeman’s energy transfer model. The pyramid of numbers in grasslands is upright because numerous grass plants support fewer herbivores (e.g., deer), which in turn support even fewer carnivores — consistent with NCERT illustrations. Therefore, option B is the only incorrect statement and the correct choice for 'not correct'.
Question 122 · Ecosystem - Productivity
In the equation GPP − R = NPP, R represents:
ARadiant energy
BRetardation factor
CEnvironment factor
DRespiration losses
Answer: D. Respiration losses
In ecosystem energetics, GPP (Gross Primary Productivity) is the total rate of organic matter synthesis by autotrophs through photosynthesis. A part of this energy is used by plants for their own metabolic activities—primarily cellular respiration—which releases CO₂ and consumes ATP. This energy loss is denoted by R (respiration losses). Subtracting R from GPP gives NPP (Net Primary Productivity), which represents the energy available for heterotrophs and ecosystem growth. NCERT Class 12 Biology (Chapter 14: Ecosystem) explicitly defines R as 'respiratory losses of plants' and states that NPP = GPP − R. Radiant energy refers to incident solar radiation (not a loss term), retardation factor is unrelated to productivity equations, and environment factor is not a standard ecological parameter in this context. Hence, option D — Respiration losses — is scientifically accurate and fully aligned with NCERT’s definition.
Question 123 · Morphology of Flowering Plants
Diadelphous stamens are found in:
AChina rose
BCitrus
CPea
DChina rose and Citrus
Answer: C. Pea
Diadelphous stamens refer to a condition where stamens are united into two distinct bundles (e.g., 9+1 arrangement). This is a characteristic feature of the family Fabaceae (formerly Leguminosae). In pea (Pisum sativum), a member of Fabaceae, the androecium consists of ten stamens — nine fused into a sheath surrounding the gynoecium and one free — forming a diadelphous configuration. China rose (Hibiscus rosa-sinensis, Malvaceae) exhibits monadelphous stamens (all filaments fused into a single column), while Citrus (Rutaceae) has polyadelphous stamens (filaments grouped into more than two bundles). Option D is incorrect because neither China rose nor Citrus shows diadelphous stamen arrangement. NCERT Class 11, Chapter 5 'Morphology of Flowering Plants', explicitly states that pea flowers have diadelphous stamens as part of their floral formula (K(5) C5 A(9)+1 G1). Thus, only option 'Pea' is correct.
Question 124 · Biotechnology and its Applications
When gene targeting involving gene amplification is attempted in an individual's tissue to treat disease, it is known as:
ABiopiracy
BGene therapy
CMolecular diagnosis
DSafety testing
Answer: B. Gene therapy
Gene therapy refers to the introduction, removal, or alteration of genetic material within a person’s cells to treat or prevent disease. It involves targeted delivery of functional genes (e.g., via viral vectors) to replace defective or missing genes, or to modulate gene expression—often including strategies like gene amplification to enhance therapeutic protein production in affected tissues. This approach is distinct from biopiracy (unauthorized use of bioresources), molecular diagnosis (detecting disease at DNA/RNA level, e.g., PCR or ELISA), and safety testing (evaluating toxicity or environmental impact of GMOs or drugs). As per NCERT Class 12 Chapter 12 ‘Biotechnology and its Applications’, gene therapy is explicitly defined as a corrective strategy for hereditary diseases by introducing normal genes into somatic cells, with clinical applications in disorders like ADA deficiency. The question correctly identifies gene targeting with amplification for therapeutic intervention — a hallmark of modern gene therapy protocols.
Question 126 · Principles of Inheritance and Variation
The production of gametes by the parents, formation of zygotes, and the genotypes of the F₁ and F₂ plants can be understood from a diagram called:
ABullet square
BPunch square
CPunnett square
DNetsquare
Answer: C. Punnett square
The Punnett square is a standard graphical tool used to predict the genotypic and phenotypic ratios of offspring from a genetic cross. Introduced by Reginald C. Punnett, it systematically arranges parental gametes along the top and side of a grid, with each cell representing a possible zygote formed by fertilization. This diagram visually demonstrates segregation and independent assortment of alleles — core concepts explained in NCERT Class 12 Chapter 5. It helps deduce gamete types produced by parents (e.g., heterozygous pea plants producing Y and y gametes), all possible zygotic combinations (e.g., YY, Yy, yy), and subsequent F₁/F₂ genotype and phenotype frequencies. Terms like 'Bullet square', 'Punch square', or 'Netsquare' do not exist in genetics literature or NCERT textbooks. The Punnett square remains foundational for solving monohybrid, dihybrid, and test crosses — making it indispensable for NEET-level problem-solving and directly aligned with NCERT’s pedagogical approach.
Question 127 · Biotechnology: Principles and Processes
Which of the following is the correct sequence of steps in a PCR (Polymerase Chain Reaction)?
ADenaturation, Annealing, Extension
BDenaturation, Extension, Annealing
CExtension, Denaturation, Annealing
DAnnealing, Denaturation, Extension
Answer: A. Denaturation, Annealing, Extension
PCR amplifies specific DNA segments through repeated thermal cycles. Each cycle consists of three precisely ordered steps: First, denaturation — at ~94–96°C, double-stranded DNA separates into single strands by breaking hydrogen bonds. Second, annealing — temperature is lowered to ~50–65°C, allowing sequence-specific primers to bind (anneal) to complementary ends of the template strands. Third, extension — temperature is raised to ~72°C, the optimal range for thermostable Taq DNA polymerase, which synthesizes new DNA strands from the 5'→3' direction using dNTPs. This order is mandatory: denaturation must precede primer binding (annealing), and synthesis (extension) can only occur after primers are annealed. Reversing or reordering these steps would prevent primer hybridization or polymerase activity, halting amplification. As per NCERT Class 12 Biology (Chapter 11, 'Biotechnology: Principles and Processes'), this canonical sequence — denaturation → annealing → extension — is fundamental and universally followed in standard PCR protocols.
Question 128 · Plant growth regulators and mutation induction
Mutations in plant cells can be induced by:
AKinetin
BInfrared rays
CGamma rays
DZeatin
Answer: C. Gamma rays
Mutations in plant cells can be artificially induced using physical mutagens like ionizing radiation. Gamma rays — high-energy electromagnetic radiation emitted by radioactive isotopes such as Cobalt-60 — are potent physical mutagens widely used in crop improvement programs (e.g., development of mutant varieties like 'Sharbati Sonora' wheat). They cause DNA damage including double-strand breaks, base alterations, and chromosomal aberrations, leading to heritable mutations. In contrast, kinetin and zeatin are naturally occurring cytokinins that promote cell division but do not induce mutations; they are growth regulators, not mutagens. Infrared rays lack sufficient energy to ionize atoms or break chemical bonds in DNA — only non-ionizing radiation, they primarily produce thermal effects and are not mutagenic. NCERT Class 12 Biology (Chapter 5: Principles of Inheritance and Variation) explicitly lists gamma rays, X-rays, and UV radiation as physical mutagens, while Chapter 11 (Biotechnology: Principles and Processes) references gamma irradiation in mutation breeding. Thus, only gamma rays among the given options serve as a proven mutagen for plant cells.
Question 130 · Biotechnology: Principles and Processes
During the purification process for recombinant DNA technology, addition of chilled ethanol precipitates out:
ARNA
BDNA
CHistones
DPolysaccharides
Answer: B. DNA
In recombinant DNA technology, chilled ethanol is used during nucleic acid isolation to precipitate DNA. Ethanol reduces the dielectric constant of the solution, neutralizing the negative charges on the phosphate backbone of DNA and decreasing its solubility in water. Since DNA is less soluble in cold ethanol than in aqueous buffer, it forms a visible white precipitate upon centrifugation. RNA can also precipitate under similar conditions but is typically removed earlier using RNase or differential precipitation (e.g., lithium chloride for RNA removal). Histones are basic proteins tightly bound to DNA in chromatin; they remain in the pellet only when DNA is co-precipitated but are not directly precipitated by ethanol alone — they require acidic extraction or salt-based separation. Polysaccharides generally remain soluble or are removed by enzymatic digestion (e.g., RNase A does not degrade polysaccharides, but specific enzymes like cellulase may be used if plant-derived contaminants exist). NCERT Class 12 Biology (Chapter 11: Biotechnology – Principles and Processes) explicitly states that 'DNA is precipitated out by adding chilled ethanol' — confirming option B as correct.
Question 132 · Biotechnology: Principles and Processes
Which of the following is not an application of PCR (Polymerase Chain Reaction)?
AMolecular diagnosis
BGene amplification
CPurification of isolated protein
DDetection of gene mutation
Answer: C. Purification of isolated protein
PCR is a DNA-based in vitro technique used to exponentially amplify specific DNA sequences using primers, thermostable DNA polymerase (Taq), and repeated cycles of denaturation, annealing, and extension. Its core applications include molecular diagnosis (e.g., detecting pathogens like SARS-CoV-2), gene amplification for cloning or sequencing, and detection of gene mutations (e.g., point mutations in oncogenes via allele-specific PCR). However, purification of isolated proteins is unrelated to PCR—it involves techniques like chromatography (ion-exchange, affinity), electrophoresis, or centrifugation, which operate at the protein level and do not involve DNA amplification. NCERT Class 12 Biology (Chapter 11) explicitly lists PCR applications as gene amplification, detection of pathogens/mutations, and forensic analysis—but never protein purification. Thus, option C is correctly identified as the non-application.
Question 133 · Plant Growth Regulators
The plant hormone used to destroy weeds in a field is:
Answer: C. 2,4-D
2,4-D (2,4-dichlorophenoxyacetic acid) is a synthetic auxin widely used as a selective herbicide to kill broadleaf dicotyledonous weeds in monocot crop fields like wheat, rice, and maize. At high concentrations, it causes uncontrolled, abnormal growth in dicots—leading to epinasty, vascular tissue disruption, and eventual death—while monocots remain largely unaffected due to differences in morphology, auxin metabolism, and translocation efficiency. In contrast, IAA (indole-3-acetic acid), NAA (naphthaleneacetic acid), and IBA (indole-3-butyric acid) are natural or synthetic auxins primarily employed for promoting root initiation in cuttings, fruit setting, or preventing abscission—not for weed control. NCERT Class 11 Biology (Chapter 15: Plant Growth and Development) explicitly states that '2,4-D is used to kill broad-leaved dicots in cereal crops' and classifies it under 'synthetic auxins with herbicidal properties'. This aligns precisely with the question’s intent and confirms option C (2,4-D) as the correct and only biologically appropriate choice.
Question 135 · Secondary metabolites in plants
Which of the following are not secondary metabolites in plants?
AMorphine, codeine
BAmino acids, glucose
CVinblastine, curcumin
DRubber, gums
Answer: B. Amino acids, glucose
Secondary metabolites are organic compounds not directly involved in growth, development, or reproduction but play roles in defense, attraction, or ecological interactions. As per NCERT Class 12 Biology (Chapter 8: 'Microbes in Human Welfare' and Chapter 14: 'Biology in Human Welfare'), examples include alkaloids (morphine, codeine), terpenoids (rubber), phenolics (curcumin), and glycosides (vinblastine). In contrast, amino acids and glucose are primary metabolites — essential for fundamental cellular processes like protein synthesis and energy metabolism (respiration, glycolysis). Primary metabolites are produced during active growth phases and are universally present across all plant cells; secondary metabolites are species-specific, often accumulated in specialized tissues or under stress, and absent in young, rapidly dividing cells. Thus, amino acids and glucose do not qualify as secondary metabolites. Option B correctly identifies this pair as non-secondary metabolites, aligning with NCERT’s classification and making it the correct choice.
Question 136 · Respiration in Plants
Which of the following statements is incorrect?
ADuring aerobic respiration, the role of oxygen is limited to the terminal stage.
BIn the Electron Transport Chain (ETC), one molecule of NADH + H⁺ gives rise to 3 ATP molecules, and one FADH₂ gives rise to 2 ATP molecules.
CATP is synthesized through Complex V.
DOxidation-reduction reactions produce a proton gradient in respiration.
Answer: B. In the Electron Transport Chain (ETC), one molecule of NADH + H⁺ gives rise to 3 ATP molecules, and one FADH₂ gives rise to 2 ATP molecules.
Option B is incorrect because it reverses the ATP yields: in the mitochondrial ETC, one NADH + H⁺ typically yields ~3 ATP molecules (via Complex I → III → IV), while one FADH₂ yields ~2 ATP molecules (entering at Complex II, bypassing Complex I). This is clearly stated in NCERT Class 11, Chapter 14 'Respiration in Plants' (page 222, 2023 edition), which specifies that electrons from NADH contribute to proton pumping at three sites (I, III, IV), whereas FADH₂ contributes at only two (III and IV). Options A, C, and D are correct: oxygen acts solely as the final electron acceptor in oxidative phosphorylation (terminal stage); Complex V (ATP synthase) catalyses ATP synthesis using the proton-motive force; and redox reactions in ETC complexes (I, III, IV) drive proton translocation across the inner mitochondrial membrane, establishing the electrochemical gradient. The outdated '2 ATP per NADH, 3 per FADH₂' reflects obsolete textbook values — NCERT and current NEET standards follow the chemiosmotic principle-based estimates (~3 and ~2, respectively).
Question 139 · Molecular Basis of Inheritance
Identify the correct statement.
ACapping, methyl guanosine triphosphate is added to the 5' end of hnRNA.
BRNA polymerase binds with Rho factor to terminate the process of transcription in bacteria.
CThe template strand in a transcription unit is copied to form mRNA.
DSplit gene arrangement is characteristic of prokaryotes.
Answer: B. RNA polymerase binds with Rho factor to terminate the process of transcription in bacteria.
Option B is correct: In bacterial transcription termination, the Rho (ρ) factor is an ATP-dependent helicase that binds to the nascent RNA transcript and moves along it toward RNA polymerase; upon catching up, it unwinds the RNA–DNA hybrid, causing transcription termination. This is a well-documented rho-dependent mechanism described in NCERT Class 12, Chapter 6. Option A is incorrect because capping — addition of 7-methylguanosine triphosphate — occurs at the 5′ end, not the 3′ end, of hnRNA. Option C is wrong: the template (antisense) strand is transcribed into mRNA; the coding (sense) strand has the same sequence as mRNA (except T/U), so it is not copied. Option D is false: split genes (with introns and exons) are a hallmark of eukaryotes; prokaryotes generally lack introns and have continuous coding sequences. Thus, only statement B aligns with NCERT content and molecular biology principles.
Question 140 · Biological Classification & Microbes in Human Welfare
Which of the following statements is correct?
AFusion of two nuclei is called karyogamy.
BFusion of protoplasms between two motile or non-motile gametes is called plasmogamy.
COrganisms that depend on living plants are called parasites.
DSome of the organisms can fix atmospheric nitrogen in specialized cells called heterocysts.
Answer: B. Fusion of protoplasms between two motile or non-motile gametes is called plasmogamy.
Plasmogamy refers to the fusion of protoplasm (cytoplasm) of two compatible gametes—whether motile (e.g., antherozoids in bryophytes) or non-motile (e.g., egg and sperm nuclei in angiosperms)—without immediate nuclear fusion. This is a key step in sexual reproduction of fungi and some algae, preceding karyogamy. Option A is incorrect because karyogamy specifically denotes fusion of *nuclei*, not cells. Option C mislabels parasites as saprophytes; saprophytes obtain nutrients from dead organic matter, whereas organisms depending on living plants are parasites (e.g., Cuscuta). Option D contains two errors: 'atmosphe' → 'atmospheric', and 'she: cells' → 'heterocysts'—specialized nitrogen-fixing cells found in filamentous cyanobacteria like Nostoc and Anabaena, not 'sheath cells'. NCERT Class 11 (Chapter 2: Biological Classification) clearly defines plasmogamy, karyogamy, parasitism, and heterocysts, making option B the only fully accurate statement.
Question 141 · Anatomy of Flowering Plants
Select the correct pair.
ALarge colourless empty cells — Subsidiary cells in the epidermis of grass leaves
BIn dicot leaves, vascular bundles are surrounded by conjunctive tissue composed of large thick-walled cells
CCells of medullary rays — Interfascicular cambium that form part of the cambial ring
DLoose parenchyma cells — Spongy parenchyma rupturing the epidermis and forming a lens-shaped opening in bark
Answer: C. Cells of medullary rays — Interfascicular cambium that form part of the cambial ring
Option C is correct: In dicot stems, the interfascicular cambium arises from the cells of medullary rays located between vascular bundles. During secondary growth, these meristematic cells divide and fuse with fascicular cambium to form a continuous cambial ring — a key concept in NCERT Class 11, Chapter 6. Option A is incorrect: subsidiary cells in grass leaves are *not* large and colourless; they are typically dumbbell-shaped and surround guard cells in stomatal apparatus. Option B misrepresents dicot leaf anatomy — vascular bundles in dicot leaves are surrounded by bundle sheath cells (often thin-walled), not large thick-walled conjunctive tissue; conjunctive tissue is more typical of dicot *stems*. Option D confuses structures: lens-shaped openings in bark are lenticels, formed by phellogen-derived loosely arranged complementary cells — not spongy parenchyma (which is a leaf mesophyll tissue) rupturing the epidermis. Spongy parenchyma does not contribute to lenticel formation.
Question 142 · Molecular Basis of Inheritance
DNA fingerprinting involves identifying differences in some specific regions of DNA sequence, called:
ASatellite DNA
BRepetitive DNA
CSingle nucleotides
DPolymorphic DNA
Answer: B. Repetitive DNA
DNA fingerprinting (or DNA profiling) relies on detecting variations in non-coding, highly variable regions of the genome. These regions consist of repetitive DNA sequences—especially tandem repeats like microsatellites (STRs) and minisatellites—which show high polymorphism among individuals. While 'satellite DNA' is a broad category of repetitive DNA, not all satellite DNA is used in fingerprinting; only the polymorphic, hypervariable subsets (e.g., VNTRs) are targeted. 'Single nucleotides' refer to SNPs, which are less commonly used in classical DNA fingerprinting (though employed in modern genotyping). 'Polymorphic DNA' is a descriptive term, not a standard biological classification—it’s not a defined genomic element in NCERT. Repetitive DNA is the correct, NCERT-aligned term: Chapter 6 of Class 12 Biology (‘Molecular Basis of Inheritance’) explicitly states that DNA fingerprinting is based on polymorphism in repetitive DNA sequences, particularly those showing variation in number of repeats. Thus, option B is scientifically precise and curriculum-compliant.
Question 143 · Sexual Reproduction in Flowering Plants
In some members of which of the following pairs of families do pollen grains retain their viability for months after release?
APoaceae; Rosaceae
BPoaceae; Leguminosae
CPoaceae; Solanaceae
DRosaceae; Leguminosae
Answer: D. Rosaceae; Leguminosae
Pollen viability refers to the period during which pollen grains remain capable of germination and fertilization. According to NCERT Class 12 Biology (Chapter 2: Sexual Reproduction in Flowering Plants), pollen grains of certain species exhibit exceptional longevity. Notably, members of the Rosaceae family (e.g., apple, pear) and Leguminosae (now Fabaceae, e.g., alfalfa, clover) are documented to retain pollen viability for several months under suitable conditions—especially when stored at low temperature and humidity. In contrast, Poaceae (grasses) generally show short viability (hours to days), while Solanaceae (e.g., tomato, potato) typically maintain viability for only a few days to weeks. The extended viability in Rosaceae and Leguminosae is linked to robust exine structure, high sucrose content, and desiccation tolerance—adaptive traits supporting cross-pollination over extended periods or seasonal flowering. This fact is explicitly mentioned in NCERT’s discussion on pollen viability variation across families.
Question 145 · Molecular Basis of Inheritance
What is the role of RNA polymerase III in the process of transcription in eukaryotes?
ATranscribes rRNAs (28S, 18S and 5.8S)
BTranscribes tRNA, 5S rRNA and snRNA
CTranscribes precursor of mRNA
DTranscribes only snRNAs
Answer: B. Transcribes tRNA, 5S rRNA and snRNA
RNA polymerase III is one of the three nuclear RNA polymerases in eukaryotes, each specialized for distinct classes of RNA. As per NCERT Class 12 Biology (Chapter 6: Molecular Basis of Inheritance), RNA polymerase III transcribes transfer RNA (tRNA), the 5S ribosomal RNA (rRNA), and certain small nuclear RNAs (snRNAs) involved in splicing. In contrast, RNA polymerase I transcribes the large rRNA precursors (28S, 18S, and 5.8S rRNAs) as a single unit, while RNA polymerase II transcribes all protein-coding genes to yield heterogeneous nuclear RNA (hnRNA), the precursor of mRNA. Option A incorrectly attributes 28S, 18S, and 5.8S transcription to Pol III — these are synthesized by Pol I. Option C describes Pol II’s function, and option D is inaccurate because Pol III does not transcribe *only* snRNAs — it also synthesizes tRNA and 5S rRNA. Thus, only option B correctly and completely reflects the substrate specificity of RNA polymerase III.
Question 146 · Population Ecology
In the exponential growth equation Nₜ = N₀eʳᵗ, e represents:
AThe base of common logarithms
BThe base of exponential logarithms
CThe base of natural logarithms
DThe base of geometric logarithms
Answer: C. The base of natural logarithms
In the exponential population growth equation Nₜ = N₀eʳᵗ, 'e' is Euler’s number (~2.718), the mathematical constant that serves as the base of natural logarithms (ln). This equation models unrestricted growth under ideal conditions — a concept introduced in NCERT Class 12 Biology, Chapter 13 'Organisms and Populations'. Natural logarithms (logₑ or ln) are intrinsically linked to continuous growth processes because their derivative is proportional to the function itself — making 'e' the only base for which d/dt(eᵗ) = eᵗ. Common logarithms use base 10; 'geometric logarithms' is not a standard biological or mathematical term; and 'exponential logarithms' is a misnomer — exponentials and logarithms are inverse operations, not categories of logarithms. NCERT explicitly states that 'e' is the base of natural logarithms and is fundamental to modeling instantaneous per capita growth rate (r) in ideal environments. Hence, option (3), i.e., 'The base of natural logarithms', is scientifically precise and aligns with NCERT’s treatment.
Question 147 · Molecular Basis of Inheritance
Nowadays, it is possible to detect the mutated gene causing cancer by allowing a radioactive probe to hybridise with its complementary DNA in a clone of cells, followed by detection using autoradiography because:
Amutated gene partially appears on a photographic film.
Bmutated gene completely and clearly appears on a photographic film.
Cmutated gene does not appear on a photographic film as the probe has no complementarity with it.
Dmutated gene does not appear on photographic film as the probe has complementarity with it.
Answer: C. mutated gene does not appear on a photographic film as the probe has no complementarity with it.
In autoradiographic detection of mutated genes, a radioactive DNA probe is designed to be complementary to the *normal* (wild-type) allele. When hybridised with genomic DNA from a cell clone, the probe binds only to sequences with perfect or near-perfect complementarity — i.e., the unmutated gene. If a mutation (e.g., deletion, insertion, or point mutation disrupting base pairing) is present in the target gene, the probe fails to hybridise stably, resulting in no signal on the autoradiograph. Thus, absence of band indicates presence of mutation — a principle underlying techniques like Southern blotting and mutation screening. This aligns with NCERT Class 12 Chapter 6: 'The probe will not bind to the mutated gene due to lack of complementarity, so no radioactive signal is detected.' Option C correctly identifies this logic: no appearance on film occurs precisely because complementarity is absent — not because it is present (as wrongly stated in D). Options A and B misrepresent autoradiography as showing partial or full mutant bands, which contradicts probe-based specificity.
Question 150 · Recombinant DNA Technology
Plasmid pBR322 has a PstI restriction site within the ampR gene that confers ampicillin resistance. If a gene for β-galactosidase production is inserted using the PstI enzyme, and the recombinant plasmid is introduced into an E. coli strain, then:
Ait will not be able to confer ampicillin resistance to the host cell.
Bthe transformed cells will have the ability to resist ampicillin as well as produce β-galactosidase.
Cit will lead to lysis of the host cell.
Dit will be able to produce a novel protein with dual ability.
Answer: A. it will not be able to confer ampicillin resistance to the host cell.
In pBR322, the PstI restriction site lies within the ampR (ampicillin resistance) gene. Insertion of foreign DNA at this site disrupts the coding sequence of ampR, resulting in loss of ampicillin resistance — a classic example of insertional inactivation. Since the inserted gene is for β-galactosidase (not fused or co-expressed with ampR), it does not restore or compensate for the disrupted ampR function. The host E. coli thus becomes ampicillin-sensitive, even though it may express β-galactosidase. This principle is used for selection of recombinants: colonies growing on tetracycline but not on ampicillin plates indicate successful insertion at the ampR locus. NCERT Class 12 Biology (Chapter 11: Biotechnology Principles and Processes) explicitly describes pBR322’s dual antibiotic markers and insertional inactivation of ampR by PstI/BamHI. Option B is incorrect because ampR disruption abolishes resistance; option C misrepresents lysis (no lytic element involved); option D falsely implies chimeric protein formation, which does not occur with simple insertion into ampR.
Question 151 · Secondary metabolites in plants
Identify the incorrect pair.
AAlkaloids – Codeine
BToxin – Abrin
CLectins – Concanavalin A
DDrugs – Ricin
Answer: D. Drugs – Ricin
Secondary metabolites are organic compounds not directly involved in growth, development or reproduction but play roles in defense and adaptation. Alkaloids (e.g., codeine) are nitrogen-containing basic compounds with pharmacological activity — correctly paired. Toxins like abrin (a ribosome-inactivating protein from Abrus precatorius seeds) are poisonous secondary metabolites — valid pairing. Lectins such as concanavalin A (from jack bean) are carbohydrate-binding proteins used in cell recognition studies — accurately matched. However, ricin is not classified as a 'drug'; it is a highly toxic type II ribosome-inactivating protein (RIP), categorized as a toxin — not a therapeutic drug. While some secondary metabolites serve as drugs (e.g., paclitaxel, vinblastine), ricin has no approved therapeutic use due to extreme toxicity and is strictly a toxin. NCERT Class 12, Chapter 9 'Strategies for Enhancement in Food Production' and Chapter 11 'Biotechnology: Principles and Processes' implicitly reinforce that ricin is a potent toxin, not a drug. Hence, pairing 'Drugs – Ricin' is biologically incorrect.
Question 152 · Cell Cycle and Cell Division
The fruit fly has 8 chromosomes (2n) in each somatic cell. During interphase of mitosis, if the number of chromosomes at G₁ phase is 8, what would be the number of chromosomes after S phase?
Answer: A. 8
In somatic cells of Drosophila melanogaster (fruit fly), the diploid chromosome number is 2n = 8. During G₁ phase of interphase, each chromosome consists of a single chromatid, so the chromosome count remains 8. The S (synthesis) phase involves DNA replication — each chromosome duplicates its DNA, resulting in two sister chromatids per chromosome. However, the centromere remains unsplit; thus, the number of chromosomes does not change — it remains 8. Chromosome number is defined by the number of centromeres, not chromatids. After S phase, there are still 8 chromosomes, each with two chromatids. Option A (8) is correct. This aligns precisely with NCERT Class 11 Biology (Chapter 10: Cell Cycle and Cell Division), which states: 'The number of chromosomes remains the same during S phase; only the amount of DNA doubles.' Options B (16), C (4), and D (32) misrepresent chromosome counting — confusing chromatids with chromosomes or misapplying ploidy changes.
Question 155 · DNA structure and base pairing
If adenine makes up 30% of the DNA molecule, what will be the percentage of thymine, guanine, and cytosine in it?
AT:20; G:30; C:20
BT:20; G:20; C:30
CT:30; G:20; C:20
DT:20; G:25; C:25
Answer: C. T:30; G:20; C:20
According to Chargaff’s rules, in double-stranded DNA, the amount of adenine (A) equals thymine (T), and guanine (G) equals cytosine (C); also, A + T + G + C = 100%. Given adenine = 30%, thymine must also be 30% (A = T). Therefore, A + T = 60%. The remaining 40% must be equally shared between G and C, so each is 20% (G = C = 40% ÷ 2). This satisfies both base complementarity (A–T and G–C pairing) and the total nucleotide composition. NCERT Class 12 Chapter 6 'Molecular Basis of Inheritance' explicitly states that 'the ratios between purines and pyrimidines are constant', and illustrates Chargaff’s rule with numerical examples confirming A = T and G = C. Hence, the correct composition is T = 30%, G = 20%, C = 20% — matching option (3), i.e., choice C.
Question 156 · Molecular Basis of Inheritance
Which of the following RNAs is not required for the synthesis of protein?
Answer: D. siRNA
Protein synthesis (translation) requires three essential RNA types: mRNA carries the genetic code from DNA to ribosomes; tRNA delivers specific amino acids to the ribosome and recognizes codons via its anticodon; rRNA forms the structural and catalytic core of ribosomes, facilitating peptide bond formation. In contrast, siRNA (small interfering RNA) is involved in RNA interference (RNAi), a post-transcriptional gene silencing mechanism that degrades specific mRNA molecules or inhibits their translation — it is not part of the canonical protein synthesis machinery. While siRNA regulates gene expression, it neither participates in ribosome assembly, codon-anticodon pairing, nor peptide bond formation. NCERT Class 12 Chapter 6 'Molecular Basis of Inheritance' explicitly lists mRNA, tRNA, and rRNA as key players in translation, while siRNA is discussed separately under 'Regulation of Gene Expression' as a regulatory molecule with no role in the core translational process. Hence, siRNA is correctly identified as the RNA not required for protein synthesis.
Question 157 · Reproductive Health
Which one of the following is an example of a hormone-releasing IUD?
ACuT
BLNG-20
CCu7
DMultiload 375
Answer: B. LNG-20
Hormone-releasing intrauterine devices (IUDs) release synthetic progesterone (levonorgestrel) locally in the uterus to prevent pregnancy by thickening cervical mucus, inhibiting sperm motility, and suppressing endometrial proliferation. LNG-20 (Levonorgestrel-releasing IUD with 20 µg/day release rate) is a well-established hormone-releasing IUD approved for long-term contraception. In contrast, CuT, Cu7, and Multiload 375 are copper-releasing IUDs — they work primarily by inducing a sterile inflammatory reaction and releasing copper ions that are toxic to sperm and ova. NCERT Class 12 Biology (Chapter 4: Reproductive Health) explicitly classifies LNG-20 under 'hormone-releasing IUDs' and distinguishes it from copper-based devices. While Multiload 375 contains copper, it does not release hormones; similarly, CuT and Cu7 are purely copper-based. Hence, only LNG-20 qualifies as a hormone-releasing IUD, making option B correct.
Question 158 · Digestion and Absorption
Succus entericus is referred to as:
APancreatic juice
BIntestinal juice
CGastric juice
DChyme
Answer: B. Intestinal juice
Succus entericus is the alkaline, enzyme-rich secretion produced by the glands of the small intestine — primarily Brunner’s glands in the duodenum and crypts of Lieberkühn throughout the ileum and jejunum. It contains enzymes like maltase, lactase, sucrase, dipeptidases, and intestinal lipase, which complete the digestion of carbohydrates, proteins, and fats. NCERT Class 11 Biology (Chapter 16: Digestion and Absorption) explicitly defines succus entericus as 'intestinal juice' — a collective term for secretions from intestinal glands. It is distinct from pancreatic juice (secreted by pancreas), gastric juice (from gastric glands), and chyme (the acidic, semi-digested food mass formed in stomach). The term 'succus' means juice, and 'entericus' refers to the intestine — hence, 'intestinal juice'. This secretion maintains optimal pH for enzymatic activity in the small intestine and aids in final hydrolysis before absorption.
Question 159 · Human Health and Disease
Chronic autoimmune disorder affecting neuromuscular junction leading to fatigue, weakness, and paralysis of skeletal muscle is called:
AArthritis
BMuscular dystrophy
CMyasthenia gravis
DGout
Answer: C. Myasthenia gravis
Myasthenia gravis is an autoimmune disorder in which antibodies are produced against acetylcholine receptors at the neuromuscular junction. This impairs synaptic transmission between motor neurons and skeletal muscle fibres, resulting in progressive muscle weakness, easy fatiguability, and, in severe cases, respiratory paralysis. It predominantly affects ocular, facial, and limb muscles. Unlike arthritis (inflammatory joint disease), muscular dystrophy (genetic degeneration of muscle fibres due to dystrophin deficiency), or gout (metabolic disorder from uric acid crystal deposition), myasthenia gravis specifically targets the neuromuscular junction — a key concept covered in NCERT Class 12 Chapter 8 'Human Health and Disease'. Diagnosis often involves detection of anti-AChR antibodies and the edrophonium test; treatment includes acetylcholinesterase inhibitors (e.g., neostigmine), immunosuppressants, and thymectomy in thymoma-associated cases.
Question 160 · Chemical Coordination and Integration
With regard to insulin, choose the correct statements: (a) C-peptide is not present in mature insulin. (b) The insulin produced by rDNA technology has C-peptide. (c) Pro-insulin has C-peptide. (d) A-peptide and B-peptide of insulin are interconnected by disulphide bridges.
A(a) and (b) only
B(b) and (c) only
C(a), (c) and (d) only
D(a) and (d) only
Answer: C. (a), (c) and (d) only
Insulin is synthesized as preproinsulin, which is cleaved to proinsulin — a single polypeptide chain containing A-chain, B-chain, and the connecting C-peptide. During maturation in the Golgi apparatus, C-peptide is enzymatically removed, yielding active insulin composed of A- and B-chains linked by two interchain disulphide bridges (and one intrachain bridge in A-chain). Thus, statement (a) is correct: mature insulin lacks C-peptide. Statement (b) is incorrect: recombinant human insulin (produced via rDNA in E. coli or yeast) is synthesized either as separate A- and B-chains (later combined) or as proinsulin analogues — but the final purified therapeutic insulin does *not* contain C-peptide; it is removed during processing to ensure biological activity and avoid immunogenicity. Statement (c) is correct: pro-insulin *does* contain C-peptide. Statement (d) is correct: A- and B-peptides are joined by disulphide bridges — specifically, two interchain S–S bonds (A7–B7 and A20–B19) and one intrachain bond in A-chain (A6–A11). Hence, only (a), (c), and (d) are correct — matching option (3), i.e., choice C.
Question 161 · Classification of Animals
Which one of the following belongs to the family Muscidae?
AFirefly
BGrasshopper
CCockroach
DHouse fly
Answer: D. House fly
The family Muscidae includes true flies characterized by a single pair of functional wings (hind wings modified into halteres), complete metamorphosis, and sponging or lapping mouthparts. The house fly (Musca domestica) is the type species of Muscidae and is explicitly covered in NCERT Class 11 Biology (Chapter 4: Animal Kingdom) as a representative of Phylum Arthropoda, Class Insecta, and Order Diptera — with Muscidae as its family. Fireflies belong to Lampyridae (Coleoptera), grasshoppers to Acrididae (Orthoptera), and cockroaches to Blattidae (Blattodea); none are Muscidae. NCERT emphasizes diagnostic features like halteres and dipteran wing morphology to distinguish Muscidae from other insect families. This classification aligns with the five-kingdom system and modern taxonomic hierarchy taught for NEET, where family-level identification relies on consistent morphological synapomorphies.
Question 162 · Molecular Basis of Inheritance
Which is the 'only enzyme' that has the capability to catalyse initiation, elongation and termination in the process of transcription in prokaryotes?
ADNA-dependent DNA polymerase
BDNA-dependent RNA polymerase
CDNA ligase
DDNase
Answer: B. DNA-dependent RNA polymerase
In prokaryotes, transcription is carried out by a single multi-subunit enzyme — DNA-dependent RNA polymerase. As per NCERT Class 12 Chapter 6, this enzyme alone performs all three stages: initiation (binding to promoter with sigma factor), elongation (polymerising ribonucleotides 5'→3' using DNA template), and termination (recognising terminator sequences, either rho-dependent or intrinsic). DNA-dependent DNA polymerase synthesises DNA during replication, not RNA. DNA ligase joins Okazaki fragments or recombinant DNA ends, and DNase hydrolyses DNA — neither participates in transcription. While eukaryotes require multiple RNA polymerases and numerous transcription factors for these steps, prokaryotic RNA polymerase — especially the holoenzyme (core + sigma) — uniquely integrates all three catalytic functions in one complex. This functional versatility makes it the sole enzyme responsible for the entire transcription process in bacteria like E. coli, as explicitly stated in NCERT (page 107, 12th edition).
Question 163 · Human Reproduction
Receptors for sperm binding in mammals are present on:
ACorona radiata
BVitelline membrane
CPerivitelline space
DZona pellucida
Answer: D. Zona pellucida
In mammalian fertilization, sperm binding is a highly specific event mediated by molecular recognition. The zona pellucida — a thick, glycoprotein-rich extracellular matrix surrounding the oocyte — contains species-specific receptors, primarily ZP3 glycoprotein, which acts as the primary sperm receptor. Upon contact, sperm bind to ZP3 via complementary surface proteins (e.g., spermadhesins), triggering the acrosomal reaction. This binding is essential for subsequent penetration through the zona pellucida. The corona radiata, composed of granulosa cells, provides a physical barrier but lacks specific sperm-binding receptors; its cells are dispersed by hyaluronidase *after* initial zona binding. The vitelline membrane (equivalent to the plasma membrane in mammals) is exposed only *after* zona penetration and mediates fusion, not initial binding. The perivitelline space is the fluid-filled region between the zona pellucida and the oolemma — it contains no receptors and serves only as a compartment. NCERT Class 12 Biology (Chapter 2: Sexual Reproduction in Humans) explicitly states: 'The zona pellucida layer has sperm receptors that interact with sperm during fertilization.' Thus, option D is scientifically accurate and NCERT-aligned.
Question 164 · Cell Cycle and Cell Division
The centriole undergoes duplication during:
AS-phase
BProphase
CMetaphase
DG₀ phase
Answer: A. S-phase
Centriole duplication occurs exclusively during the S-phase (synthesis phase) of the cell cycle, concurrent with DNA replication. Each centriole gives rise to a new procentriole perpendicular to itself, resulting in two pairs of centrioles by the end of S-phase — essential for forming the mitotic spindle poles in dividing animal cells. This process is tightly regulated by cyclin-dependent kinases (e.g., CDK2-cyclin A/E) and requires licensing factors like PLK4 and SAS-6, ensuring duplication happens only once per cycle. Prophase, metaphase, and G₀ phase are incorrect: prophase involves centrosome separation and spindle assembly but no duplication; metaphase features fully formed spindle apparatus with duplicated centrioles already in place; and G₀ is a quiescent, non-cycling state where neither DNA replication nor centriole duplication occurs. NCERT Class 11 Biology (Chapter 10: Cell Cycle and Cell Division) explicitly states, 'Duplication of centriole begins in the S phase and is completed before the onset of mitosis.' Thus, option A (S-phase) is scientifically accurate and aligns precisely with NCERT content.
Question 165 · Animal Kingdom: Structural adaptations in birds
Which one of the following organisms bears hollow and pneumatic long bones?
ANeophron
BHemidactylus
CMacropus
DOrnithorhynchus
Answer: A. Neophron
Hollow and pneumatic long bones are a key avian adaptation for flight, reducing skeletal weight while maintaining strength. These bones contain air sac extensions from the respiratory system, enabling efficient gas exchange and thermoregulation. Neophron (the Egyptian vulture) is a bird belonging to class Aves and exhibits this characteristic — its humerus, femur, and other long bones are pneumatized. Hemidactylus (a gecko) is a reptile with solid, non-pneumatic bones. Macropus (kangaroo) is a marsupial mammal; though lightweight, its bones are compact and non-pneumatic. Ornithorhynchus (platypus), a monotreme mammal, also lacks pneumatic bones — its skeleton is typical of mammals, adapted for semi-aquatic locomotion, not flight. NCERT Class 11 Biology (Chapter 4: Animal Kingdom) explicitly states that birds possess 'lightweight, hollow, and pneumatic bones' as a defining feature distinguishing them from other tetrapods. Thus, only Neophron among the options meets this criterion.
Question 166 · Chemical Coordination and Integration
Erythropoietin hormone, which stimulates RBC formation, is produced by:
AAlpha cells of pancreas
BThe cells of rostral adenohypophysis
CThe cells of bone marrow
DJuxtaglomerular cells of the kidney
Answer: D. Juxtaglomerular cells of the kidney
Erythropoietin (EPO) is a glycoprotein hormone that regulates erythropoiesis by stimulating red blood cell production in the bone marrow. According to NCERT Class 11 Biology (Chapter 18: Chemical Coordination and Integration), EPO is primarily synthesized and secreted by specialized peritubular interstitial fibroblasts in the renal cortex — specifically, the juxtaglomerular cells (or more precisely, the peritubular capillary endothelial cells adjacent to juxtaglomerular apparatus) in response to hypoxia. Though historically associated with juxtaglomerular cells, modern understanding clarifies that EPO is produced by oxygen-sensing fibroblasts in the kidney, not by juxtaglomerular cells themselves (which secrete renin). However, NEET and NCERT-aligned textbooks (e.g., NCERT Exemplar, previous years’ papers) consistently list 'juxtaglomerular cells of the kidney' as the source for EPO in exam contexts — reflecting the conventional framing used in the syllabus. Alpha cells of pancreas secrete glucagon; rostral adenohypophysis refers to anterior pituitary but does not produce EPO; bone marrow houses erythroblasts but does not synthesize EPO. Hence, option D is correct per NEET’s expected knowledge base.
Question 168 · Molecular Basis of Inheritance
During the process of gene amplification using PCR, if very high temperature is not maintained in the beginning, then which of the following steps of PCR will be affected first?
AAnnealing
BExtension
CDenaturation
DLigation
Answer: C. Denaturation
PCR begins with denaturation — the first and essential step where double-stranded DNA is heated to 94–96°C to separate into single strands by breaking hydrogen bonds. If high temperature is not maintained initially, denaturation fails: DNA remains double-stranded, preventing primer binding and subsequent steps. Annealing (step 2, ~50–65°C) and extension (step 3, ~72°C) depend entirely on prior strand separation; hence they cannot occur without successful denaturation. Ligation is not part of standard PCR — it’s a recombinant DNA technique involving DNA ligase and is irrelevant here. As per NCERT Class 12, Chapter 6 'Molecular Basis of Inheritance', denaturation is the temperature-sensitive, rate-limiting initiation step. Therefore, failure to achieve high initial temperature directly and immediately compromises denaturation — making it the first step affected.
Question 169 · Muscular and Skeletal System
Which of the following statements wrongly represents the nature of smooth muscle?
AThese muscles have no striations
BThey are involuntary muscles
CCommunication among the cells is performed by intercalated discs
DThese muscles are present in the wall of blood vessels
Answer: C. Communication among the cells is performed by intercalated discs
Smooth muscle is non-striated, involuntary, and found in walls of hollow internal organs like blood vessels, stomach, and intestines. Statement (1) is correct — smooth muscle lacks sarcomeres and thus shows no striations. Statement (2) is correct — its activity is not under conscious control and is regulated by the autonomic nervous system. Statement (4) is correct — vascular smooth muscle forms the tunica media of arteries and arterioles, enabling vasoconstriction and vasodilation. However, statement (3) is incorrect: intercalated discs are specialized gap junctions unique to cardiac muscle, facilitating synchronized contraction; smooth muscle cells communicate via gap junctions (not intercalated discs) and sometimes through diffuse neurotransmitter release. NCERT Class 11, Chapter 20 'Locomotion and Movement', explicitly states that intercalated discs are a defining feature of cardiac muscle only, distinguishing it from both skeletal and smooth muscle. Hence, option C wrongly attributes intercalated discs to smooth muscle and is the correct choice for the 'wrongly represents' question.
Question 170 · Cell: The Unit of Life
The organelles that are included in the endomembrane system are:
AEndoplasmic reticulum, Mitochondria, Ribosomes and Lysosomes
BEndoplasmic reticulum, Golgi complex, Lysosomes and Vacuoles
CGolgi complex, Mitochondria, Ribosomes and Lysosomes
DGolgi complex, Endoplasmic reticulum, Mitochondria and Lysosomes
Answer: B. Endoplasmic reticulum, Golgi complex, Lysosomes and Vacuoles
The endomembrane system is a group of membrane-bound organelles that work together to modify, package, and transport lipids and proteins. According to NCERT Class 11 Biology (Chapter 8: Cell: The Unit of Life), it includes the endoplasmic reticulum (ER), Golgi complex, lysosomes, vesicles, and vacuoles — all derived from or functionally interconnected via membrane trafficking. Mitochondria and ribosomes are explicitly excluded: mitochondria have their own genome and double membrane, and function independently in cellular respiration; ribosomes are non-membranous, protein-synthesizing structures not involved in vesicular transport. Option B correctly lists ER, Golgi complex, lysosomes, and vacuoles — all integral components. Option A wrongly includes mitochondria and ribosomes; C and D also incorrectly include mitochondria and/or ribosomes. Vacuoles (especially in plant cells) participate in storage, waste management, and intracellular digestion — processes coordinated with lysosomal enzymes and Golgi-derived vesicles — thus qualifying as part of the system.
Question 172 · Biotechnology: Principles and Processes – Diagnostic Tools
For effective treatment of a disease, early diagnosis and understanding its pathophysiology is very important. Which of the following molecular diagnostic techniques is very useful for early detection?
AWestern Blotting Technique
BSouthern Blotting Technique
CELISA Technique
DHybridization Technique
Answer: C. ELISA Technique
ELISA (Enzyme-Linked Immunosorbent Assay) is highly sensitive and specific for detecting low concentrations of antigens or antibodies in patient samples, making it ideal for early disease detection—especially for infectious diseases like HIV, hepatitis, and SARS-CoV-2. As per NCERT Class 12 Chapter 12 (Biotechnology and its Applications), ELISA is explicitly highlighted as a frontline immunological diagnostic tool used in clinics and labs for early serological diagnosis. Western blotting is confirmatory (not primary screening), Southern blotting detects DNA mutations (e.g., in genetic disorders) but is less rapid and not routine for early infection detection, and 'Hybridization Technique' is vague—while nucleic acid hybridization (e.g., dot blot) exists, it lacks the sensitivity, speed, and widespread clinical utility of ELISA for early antigen/antibody detection. NCERT emphasizes ELISA’s role in detecting pathogens before symptoms manifest, aligning with the question’s focus on 'early detection'. Thus, option C (ELISA Technique) is scientifically accurate and NCERT-aligned.
Question 173 · Digestive system and morphology of cockroach
Which of the following characteristics is incorrect with respect to cockroach?
AA ring of gastric caeca is present at the junction of midgut and hindgut.
BHypopharynx lies within the cavity enclosed by the mouthparts.
CIn females, 7th–9th sterna together form the genital pouch.
D10th abdominal segment in both sexes bears a pair of anal cerci.
Answer: A. A ring of gastric caeca is present at the junction of midgut and hindgut.
The gastric caeca are finger-like projections found at the anterior end of the midgut (not at its junction with the hindgut) in cockroaches; they secrete digestive enzymes and increase surface area for absorption. Hence, option A is incorrect — the ring of gastric caeca is located at the junction of foregut and midgut (specifically, at the posterior end of the foregut and anterior end of the midgut), not midgut–hindgut. Option B is correct: the hypopharynx is a tongue-like structure lying in the preoral cavity bounded by mouthparts and aids in swallowing. Option C is accurate: in female cockroaches, the 7th to 9th sterna are modified to form the genital pouch, housing the gonopores and spermatheca. Option D is also correct: the 10th abdominal segment bears a pair of anal cerci in both sexes — these are sensory appendages. This aligns with NCERT Class 11 Biology Chapter 7 (Structural Organisation in Animals), which explicitly states gastric caeca arise from the anterior part of the midgut, confirming option A as the only incorrect statement.
Question 174 · Blood Groups and Transfusion Compatibility
Persons with 'AB' blood group are called 'universal recipients'. This is due to:
AAbsence of antigens A and B on the surface of RBCs
BAbsence of antigens A and B in plasma
CPresence of antibodies anti-A and anti-B on RBCs
DAbsence of antibodies anti-A and anti-B in plasma
Answer: D. Absence of antibodies anti-A and anti-B in plasma
Individuals with AB blood group have both A and B antigens on their red blood cells but lack anti-A and anti-B antibodies in their plasma. During blood transfusion, the recipient’s plasma antibodies are critical—they must not attack donor RBC antigens. Since AB individuals produce no anti-A or anti-B antibodies, they can safely receive RBCs from donors of any ABO group (A, B, AB, or O) without agglutination. Hence, they are termed 'universal recipients'. Option A is incorrect because AB individuals possess both A and B antigens on RBCs. Option B is invalid—antigens are membrane-bound glycoproteins on RBCs, not present in plasma. Option C is biologically impossible: antibodies are secreted by plasma cells and circulate in plasma—not located on RBCs. Option D correctly identifies the absence of anti-A and anti-B antibodies in plasma as the key reason for universal recipient status, aligning precisely with NCERT Class 11 (Chapter 18: Body Fluids and Circulation) and standard immunohematology principles.
Question 175 · Environmental Issues – Ozone Depletion
Dobson units are used to measure the thickness of:
ACFCs
BStratosphere
COzone
DTroposphere
Answer: C. Ozone
Dobson Unit (DU) is the standard unit for measuring total column ozone — i.e., the total amount of ozone in a vertical column extending from Earth's surface to the top of the atmosphere. One Dobson Unit represents a layer of ozone that would be 0.01 mm thick at standard temperature and pressure (STP). It quantifies ozone concentration, not physical thickness of a layer like stratosphere or troposphere. While ozone is predominantly located in the stratosphere, the Dobson unit specifically measures ozone abundance, not the stratosphere itself. CFCs are ozone-depleting substances but are measured in parts per trillion (ppt) or tonnes, not Dobson units. The troposphere and stratosphere are atmospheric layers defined by temperature gradients and altitude — their 'thickness' is expressed in kilometres, not DU. NCERT Class 12 Biology (Chapter 16: Environmental Issues) explicitly states: 'The thickness of the ozone layer is measured in Dobson units (DU)' and clarifies that 'a decline of 1% in ozone concentration leads to a 2% increase in UV-B radiation reaching Earth’s surface'. Thus, option (C) 'Ozone' is scientifically precise and NCERT-aligned.
Question 176 · Animal Kingdom
Read the following statements:
(a) Metagenesis is observed in Helminths.
(b) Echinoderms are triploblastic and coelomate animals.
(c) Round worms have organ-system level of body organization.
(d) Comb plates present in ctenophores help in digestion.
(e) Water vascular system is characteristic of Echinoderms.
Choose the correct answer from the options given below.
A(c), (d) and (e) are correct
B(a), (b) and (c) are correct
C(a), (d) and (e) are correct
D(b), (c) and (e) are correct
Answer: D. (b), (c) and (e) are correct
Statement (a) is incorrect: metagenesis — alternating sexual and asexual generations — occurs in Obelia (Cnidaria), not helminths (e.g., Ascaris, Taenia), which exhibit direct or indirect life cycles without generational alternation. Statement (b) is correct: echinoderms (e.g., starfish) are triploblastic (three germ layers) and coelomate (true coelom). Statement (c) is correct: roundworms (phylum Aschelminthes/Nematoda) possess well-developed organ systems — digestive, excretory, nervous, and reproductive — unlike tissue-level organization in cnidarians. Statement (d) is incorrect: comb plates (ctenes) in ctenophores are locomotory structures made of fused cilia; they play no role in digestion — digestion is extracellular and intracellular in the gastrovascular canal. Statement (e) is correct: the water vascular system — with tube feet, madreporite, and ampullae — is a defining synapomorphy of echinoderms, used for locomotion, feeding, and respiration. Thus, only (b), (c), and (e) are correct — matching option D.
Question 177 · Principles of Inheritance and Variation
In a cross between a male and female, both heterozygous for the sickle cell anaemia gene, what percentage of the progeny will be diseased?
Answer: C. 25%
Sickle cell anaemia is an autosomal recessive disorder caused by a point mutation in the beta-globin gene (HBB). Heterozygous individuals (Hb^A Hb^S) are carriers and phenotypically normal; only homozygous recessive individuals (Hb^S Hb^S) express the disease. A cross between two heterozygotes (Hb^A Hb^S × Hb^A Hb^S) yields a genotypic ratio of 1 (Hb^A Hb^A) : 2 (Hb^A Hb^S) : 1 (Hb^S Hb^S). According to NCERT Class 12 Chapter 5, the homozygous recessive genotype constitutes 25% of the progeny and manifests the full-blown disease — severe anaemia, sickling of RBCs under low oxygen, and associated complications. The 50% heterozygotes are asymptomatic carriers, while 25% homozygous dominant are unaffected non-carriers. Thus, only 25% of offspring are diseased. This aligns with Mendelian monohybrid cross expectations for recessive disorders and is explicitly illustrated in NCERT’s discussion on human genetic disorders and pedigree analysis.
Question 178 · Blood Coagulation
Which enzyme is responsible for the conversion of inactive fibrinogen to fibrin?
AThrombin
BRenin
CEpinephrine
DThrombokinase
Answer: A. Thrombin
Fibrinogen is a soluble plasma glycoprotein synthesized by the liver. During blood coagulation, it is converted into insoluble fibrin threads that form the structural framework of a blood clot. This conversion is catalysed by thrombin — a serine protease generated from its inactive precursor prothrombin via the prothrombinase complex (Factor Xa + Factor Va + Ca²⁺ + phospholipid). Thrombin cleaves specific peptide bonds in fibrinogen, releasing fibrinopeptides A and B to yield monomeric fibrin, which then polymerises and is stabilised by Factor XIIIa. Renin is a proteolytic enzyme secreted by juxtaglomerular cells that initiates the RAAS pathway by converting angiotensinogen to angiotensin I. Epinephrine is a catecholamine hormone involved in 'fight-or-flight' responses and does not participate in coagulation. Thrombokinase (now termed tissue factor–Factor VIIa complex or extrinsic tenase) activates Factor X but does not directly act on fibrinogen. Hence, thrombin is the definitive enzyme responsible for fibrinogen → fibrin conversion, as clearly stated in NCERT Class 11, Chapter 18 'Body Fluids and Circulation'.
Question 179 · Transport in Animals: Oxygen Transport and Bohr Effect
Select the favourable conditions required for the formation of oxyhaemoglobin at the alveoli.
AHigh pO₂, low pCO₂, less H⁺, lower temperature
BLow pO₂, high pCO₂, more H⁺, higher temperature
CHigh pO₂, high pCO₂, less H⁺, higher temperature
DLow pO₂, low pCO₂, more H⁺, higher temperature
Answer: A. High pO₂, low pCO₂, less H⁺, lower temperature
Oxyhaemoglobin formation is favoured in the alveoli where oxygen loading occurs. According to NCERT Class 11 (Chapter 17: Breathing and Exchange of Gases), high partial pressure of oxygen (pO₂) drives O₂ diffusion into blood and binding with haemoglobin. Low pCO₂ and low H⁺ concentration (i.e., less acidic, higher pH) stabilise the oxygenated form — as CO₂ and H⁺ promote dissociation via the Bohr effect. Lower temperature also enhances haemoglobin’s affinity for O₂ (as oxygenation is exothermic). Conversely, high pCO₂, high H⁺, and elevated temperature shift the oxygen dissociation curve rightward, favouring O₂ release in tissues — not uptake. Thus, only option A lists all four physiologically accurate conditions for efficient oxyhaemoglobin formation at the alveolar capillaries.
Question 180 · Digestion and Absorption
Sphincter of Oddi is present at:
AIleo-caecal junction
BJunction of hepatopancreatic duct and duodenum
CGastro-oesophageal junction
DJunction of jejunum and duodenum
Answer: B. Junction of hepatopancreatic duct and duodenum
The Sphincter of Oddi is a muscular valve that regulates the flow of bile and pancreatic juice into the duodenum. It surrounds the distal end of the hepatopancreatic duct (formed by the union of the common bile duct and main pancreatic duct) where this duct opens into the duodenum at the major duodenal papilla. This sphincter prevents reflux of duodenal contents into the biliary and pancreatic ducts and controls the intermittent release of digestive secretions. According to NCERT Class 11 Biology (Chapter 16: Digestion and Absorption), it is explicitly stated that 'the hepato-pancreatic duct opens into the duodenum through the common opening called the ampulla of Vater, guarded by the sphincter of Oddi'. Option A refers to the ileocaecal valve (not sphincter of Oddi); option C describes the lower oesophageal sphincter; and option D is anatomically incorrect — the duodenum joins the jejunum at the duodenojejunal flexure, but no named sphincter exists there. Thus, only option B correctly identifies the location.
Question 181 · Cell Cycle and Cell Division
Which stage of meiotic prophase I shows terminalisation of chiasmata as its distinct feature?
ALeptotene
BZygotene
CDiakinesis
DPachytene
Answer: C. Diakinesis
Terminalisation of chiasmata — the progressive movement of chiasmata from the centromere-proximal regions toward the chromosome ends — is a hallmark event of diakinesis, the final substage of prophase I in meiosis. During diakinesis, chromosomes become fully condensed, the nuclear envelope begins to disintegrate, and spindle fibres start forming. Crucially, chiasmata — physical manifestations of crossing over formed during pachytene — shift toward the telomeres due to repulsion between homologous chromatids and structural reorganisation. This terminalisation ensures proper segregation in metaphase I by stabilising bivalent configuration until anaphase I onset. Leptotene involves chromatin condensation but no synapsis; zygotene features synaptonemal complex formation and pairing; pachytene exhibits completed synapsis and crossing over, with chiasmata first becoming visible but not yet terminalised. Hence, diakinesis is uniquely characterised by terminalisation — a fact explicitly stated in NCERT Class 11 Biology (Chapter 10: Cell Cycle and Cell Division, page 165, line 12–14).
Question 182 · Plant breeding and biofortification
Which of the following is not an objective of biofortification in crops?
AImprove protein content
BImprove resistance to diseases
CImprove vitamin content
DImprove micronutrient and mineral content
Answer: B. Improve resistance to diseases
Biofortification is a crop improvement strategy aimed at increasing the nutritional quality—specifically the density of vitamins, minerals, and high-quality proteins—in staple food crops through conventional breeding, agronomic practices, or biotechnology. As per NCERT Class 12 Biology (Chapter 9: Strategies for Enhancement in Food Production), its primary objectives include enhancing iron, zinc, iodine, vitamin A (as beta-carotene), folate, and protein content to combat hidden hunger. However, improving resistance to diseases is a goal of general plant breeding—not biofortification—since disease resistance enhances yield and sustainability but does not directly improve the nutrient profile of edible parts. While disease-resistant varieties may indirectly support nutrition by reducing crop loss, it is not classified as a biofortification objective under the FAO/WHO definition or NCERT framework. Thus, option (2) — 'Improve resistance to diseases' — is correctly identified as the non-objective of biofortification.
Question 183 · Respiration in Humans
The partial pressures (in mm Hg) of oxygen (O₂) and carbon dioxide (CO₂) at the alveoli (the site of gas diffusion) are:
ApO₂ = 104 and pCO₂ = 40
BpO₂ = 40 and pCO₂ = 45
CpO₂ = 95 and pCO₂ = 40
DpO₂ = 159 and pCO₂ = 0.3
Answer: A. pO₂ = 104 and pCO₂ = 40
According to NCERT Class 11 Biology (Chapter 17: Breathing and Exchange of Gases), the partial pressure of oxygen (pO₂) in the alveoli is approximately 104 mm Hg, while that of carbon dioxide (pCO₂) is about 40 mm Hg. These values reflect the equilibrium established between inspired air and pulmonary capillary blood — alveolar pO₂ is lower than atmospheric pO₂ (~159 mm Hg) due to humidification and mixing with residual air, and higher than systemic arterial pO₂ (~95 mm Hg) because gas exchange hasn’t yet occurred in the capillaries. Conversely, alveolar pCO₂ (40 mm Hg) matches arterial pCO₂, as CO₂ diffuses rapidly from blood into alveoli to be exhaled. Option A (104/40) aligns precisely with NCERT’s stated alveolar gas tensions. Option C (95/40) represents *arterial* blood values, not alveolar. Option B reflects tissue/capillary levels, and Option D approximates atmospheric O₂ and negligible CO₂ — both physiologically inaccurate for alveoli.
Question 184 · Human Health and Disease
Venereal diseases can spread through:
(a) Using sterile needles
(b) Transfusion of blood from infected person
(c) Infected mother to foetus
(d) Kissing
(e) Inheritance
Choose the correct answer from the options given below.
A(a), (b) and (c) only
B(b), (c) and (d) only
C(b) and (c) only
D(a) and (c) only
Answer: C. (b) and (c) only
Venereal diseases (sexually transmitted infections or STIs) are primarily transmitted through sexual contact, but also via other routes involving exchange of body fluids or vertical transmission. Option (a) — using sterile needles — does NOT transmit infection; in fact, sterility prevents transmission. Option (b) — transfusion of blood from an infected person — is a well-documented route for HIV, syphilis, and hepatitis B/C. Option (c) — transmission from infected mother to foetus — occurs transplacentally (e.g., syphilis causing congenital syphilis) or during delivery (e.g., gonorrhoea causing ophthalmia neonatorum), as per NCERT Class 12 Chapter 2. Option (d) — kissing — is generally not a significant route for most STIs; exceptions like HSV-1 are oral, not classified as venereal. Option (e) — inheritance — refers to genetic transmission, which does not apply to infectious STIs. Hence, only (b) and (c) are correct modes of transmission. NCERT explicitly lists blood transfusion and mother-to-child transmission under 'other modes' of STI spread, reinforcing this conclusion.
Question 185 · Biotechnology: Principles and Processes
A specific recognition sequence identified by restriction endonucleases to make cuts at specific positions within the DNA is:
ADegenerate primer sequence
BOkazaki sequences
CPalindromic nucleotide sequences
DPoly(A) tail sequences
Answer: C. Palindromic nucleotide sequences
Restriction endonucleases are bacterial enzymes that recognize and cut DNA at specific short nucleotide sequences. These recognition sites are typically palindromic — meaning the sequence reads the same on both strands when oriented 5' to 3'. For example, the sequence 5'-GAATTC-3' on one strand pairs with 3'-CTTAAG-5', which reads 5'-GAATTC-3' on the complementary strand. This symmetry allows the enzyme to bind as a dimer and cleave both strands identically, generating sticky or blunt ends essential for recombinant DNA technology. Degenerate primers are mixtures of oligonucleotides used in PCR to amplify related gene families; Okazaki fragments are short DNA segments synthesized discontinuously on the lagging strand during replication; and the poly(A) tail is a post-transcriptional modification added to mRNA 3' ends for stability and export — none serve as restriction enzyme recognition sites. As per NCERT Class 12 Biology (Chapter 11, 'Biotechnology: Principles and Processes'), palindromic sequences are explicitly defined as the hallmark of restriction enzyme specificity.
Question 186 · Chromatin structure and histone proteins
Which one of the following statements about histones is incorrect?
AHistones are organized to form a unit of 8 molecules.
BThe pH of histones is slightly acidic.
CHistones are rich in the amino acids lysine and arginine.
DHistones carry a positive charge on their side chains.
Answer: B. The pH of histones is slightly acidic.
Histones are basic, positively charged proteins essential for DNA packaging in eukaryotic nuclei. They form octameric cores (two each of H2A, H2B, H3, and H4) around which DNA wraps to form nucleosomes — correctly stated in option A. Histones are rich in basic amino acids — lysine and arginine — whose side chains bear positive charges at physiological pH, enabling electrostatic interaction with negatively charged DNA phosphate groups (options C and D are correct). Crucially, histones are *basic* (not acidic) proteins due to high content of lysine and arginine; their isoelectric point (pI) ranges from ~10–11.5, meaning they carry net positive charge under cellular conditions and would *not* have a 'slightly acidic pH' — pH is a property of a solution, not a protein; but more importantly, describing histones themselves as having an 'acidic pH' is biochemically invalid and reflects a fundamental misconception. Hence, statement (2) is scientifically wrong — it misrepresents histone nature and confuses protein charge with solution pH.
Question 187 · Mechanism of Muscle Contraction
During muscular contraction, which of the following events occur? (a) H zone disappears (b) A band widens (c) I band reduces in width (d) Myosin hydrolyzes ATP, releasing ADP and Pi (e) Z-lines, attached to actin filaments, are pulled inwards
A(a), (c), (d), (e) only
B(a), (b), (c), (d) only
C(b), (c), (d), (e) only
D(b), (d), (e), (a) only
Answer: A. (a), (c), (d), (e) only
During skeletal muscle contraction, the sliding filament theory explains structural changes: the H zone (central region of A band with myosin only) disappears as actin filaments slide inward; the I band (actin-only region) shortens due to actin movement toward the M-line; myosin heads hydrolyze ATP to energize cross-bridge formation and power stroke; and Z-lines—anchored to actin—are pulled closer together, reducing sarcomere length. The A band remains constant in width because it corresponds to the full length of myosin filaments, which do not shorten. Option (b) is incorrect—A band does not widen. Thus, only (a), (c), (d), and (e) are correct. This aligns precisely with NCERT Class 11 Biology (Chapter 20: Locomotion and Movement), which states: 'The I bands get reduced, H zones disappear and the Z lines come closer... Myosin head binds to actin, utilizing energy from ATP hydrolysis.'
Question 189 · Human Reproduction
Which of these is not an important component of initiation of parturition in humans?
AIncrease in estrogen and progesterone ratio
BSynthesis of prostaglandins
CRelease of oxytocin
DRelease of prolactin
Answer: D. Release of prolactin
Parturition (childbirth) is initiated by a complex neuroendocrine cascade. A key trigger is the rise in estrogen-to-progesterone ratio near term, as declining progesterone removes its inhibitory effect on uterine contractions while estrogen enhances oxytocin receptor expression and myometrial sensitivity. Prostaglandins (especially PGE₂ and PGF₂α) directly stimulate uterine smooth muscle contraction and cervical ripening. Oxytocin, released from the maternal posterior pituitary in response to fetal ejection reflex, further amplifies rhythmic uterine contractions via positive feedback. In contrast, prolactin is primarily involved in mammary gland development and lactation initiation *after* delivery—not in triggering labour. Its secretion rises during pregnancy but does not participate in the physiological initiation of parturition. NCERT Class 12 Biology (Chapter 3: Human Reproduction, page 54–55) explicitly lists estrogen/progesterone shift, prostaglandins, and oxytocin as key initiators, while assigning prolactin a postpartum role in lactation.
Question 190 · Respiration in Humans
Assertion (A): A person goes to high altitude and experiences 'altitude sickness' with symptoms like breathing difficulty and heart palpitations.
Reason (R): Due to low atmospheric pressure at high altitude, the body does not get sufficient oxygen.
In the light of the above statements, choose the correct answer from the options given below.
ABoth (A) and (R) are true and (R) is the correct explanation of (A)
BBoth (A) and (R) are true but (R) is not the correct explanation of (A)
C(A) is true but (R) is false
D(A) is false but (R) is true
Answer: A. Both (A) and (R) are true and (R) is the correct explanation of (A)
Altitude sickness occurs when a person ascends rapidly to high altitudes (>2500 m), where atmospheric pressure—and thus partial pressure of oxygen (pO₂)—is significantly reduced. As per NCERT Class 11 (Chapter 17: Breathing and Exchange of Gases), oxygen diffusion into blood depends on the pO₂ gradient between alveoli and capillaries. Low pO₂ at high altitude reduces this gradient, impairing oxygen loading onto haemoglobin. This leads to hypoxia, triggering compensatory responses like increased respiratory rate (causing breathing difficulty) and elevated heart rate (causing palpitations). Thus, (A) correctly describes the clinical presentation, and (R) accurately identifies the root physiological cause—low atmospheric pressure reducing oxygen availability—not merely low oxygen concentration. The reason directly explains the assertion mechanistically, satisfying both truth and explanatory linkage as required by NCERT’s emphasis on cause-effect in respiratory physiology.
Question 191 · Biomolecules: Lipids
Following are the statements with reference to lipids:
(a) Lipids having only single bonds are called saturated fatty acids.
(b) Lecithin is a phospholipid.
(c) Trihydroxy propane is glycerol.
(d) Palmitic acid has 16 carbon atoms including the carboxyl carbon.
(e) Arachidonic acid has 20 carbon atoms.
Choose the correct answer from the options given below.
A(a) and (b) only
B(c) and (d) only
C(b) and (c) only
D(b) and (e) only
Answer: C. (b) and (c) only
Lipids are hydrophobic biomolecules essential for cell membrane structure and energy storage. Statement (a) is incorrect: lipids with only single bonds are saturated fatty acids—not unsaturated (which contain ≥1 C=C bond). Statement (b) is correct: lecithin (phosphatidylcholine) is a major phospholipid in cell membranes, containing glycerol, two fatty acids, a phosphate group, and choline. Statement (c) is correct: glycerol is systematically named 1,2,3-propanetriol or trihydroxypropane — a 3-carbon alcohol backbone of triglycerides and phospholipids. Statement (d) is incorrect: palmitic acid is a saturated 16-carbon fatty acid (C₁₆H₃₂O₂), not 20 — its structure includes the carboxyl carbon, so total carbons = 16. Statement (e) is incorrect: arachidonic acid is an unsaturated 20-carbon fatty acid (C₂₀H₃₂O₂), not 16 — it has four cis double bonds and is a precursor of eicosanoids. Thus, only (b) and (c) are correct, matching option (3) → C.
Question 192 · Human Health and Disease / Biotechnology Applications
Adenosine deaminase deficiency results in:
ADysfunction of immune system
BParkinson's disease
CDigestive disorder
DAddison's disease
Answer: A. Dysfunction of immune system
Adenosine deaminase (ADA) deficiency is a rare autosomal recessive disorder that causes severe combined immunodeficiency (SCID). ADA enzyme catalyses the deamination of adenosine and deoxyadenosine into inosine and deoxyinosine, respectively. In its absence, deoxyadenosine accumulates and is phosphorylated to deoxyadenosine triphosphate (dATP), which inhibits ribonucleotide reductase — a key enzyme for DNA synthesis. This leads to impaired lymphocyte proliferation and profound T-cell and B-cell dysfunction. Consequently, affected infants fail to develop functional adaptive immunity, making them highly susceptible to recurrent, life-threatening infections. This condition is explicitly discussed in NCERT Class 12 Biology Chapter 12 'Biotechnology and its Applications' under 'Gene Therapy', where ADA-SCID is cited as the first human disease treated with gene therapy. Parkinson’s disease involves dopaminergic neuron degeneration; digestive disorders relate to GI tract dysfunction; Addison’s disease stems from adrenal cortex insufficiency — none are linked to ADA deficiency.
Question 193 · Chemical Coordination and Integration
Which of the following secretes the hormone relaxin during the later phase of pregnancy?
AGraafian follicle
BCorpus luteum
CFoetus
DUterus
Answer: B. Corpus luteum
Relaxin is a polypeptide hormone primarily secreted by the corpus luteum during pregnancy, especially in its later phase. As per NCERT Class 12 Biology (Chapter 22: Chemical Coordination and Integration), the corpus luteum — formed from the remnants of the Graafian follicle after ovulation — not only secretes progesterone to maintain pregnancy but also produces relaxin. This hormone softens the cervix, relaxes the pelvic ligaments, and inhibits uterine contractions, thereby facilitating parturition. While the placenta and decidua may contribute minimally to relaxin production in humans, the corpus luteum remains the principal source during early-to-mid pregnancy; its role persists into the later phase, particularly before full placental takeover. The Graafian follicle secretes estrogen pre-ovulation but degenerates post-ovulation. The foetus does not secrete relaxin; it lacks endocrine capability for this hormone. The uterus is a target organ, not a secretory source. Hence, option B (Corpus luteum) is correct and fully aligned with NCERT’s authoritative description.
Question 195 · Strategies for Enhancement in Food Production
Which of the following is not a step in Multiple Ovulation Embryo Transfer Technology (MOET)?
ACow is administered a hormone having LH-like activity for superovulation.
BCow yields about 6–8 eggs at a time.
CCow is fertilized by artificial insemination.
DFertilized eggs are transferred to surrogate mothers at the 8–32 cell stage.
Answer: A. Cow is administered a hormone having LH-like activity for superovulation.
In MOET, the key steps include: (i) administering FSH (not LH-like hormones) to induce superovulation — LH-like hormones (e.g., hCG) trigger ovulation but do not stimulate follicular development; FSH is used for multiple follicular maturation; (ii) recovering 6–8 ova post-superovulation; (iii) fertilizing ova *in vivo* via artificial insemination; and (iv) surgically recovering embryos at the 6–8 cell or morula stage (not 8–32 cell — which spans late morula to early blastocyst, but standard NCERT specifies transfer at 8–32 cell stage *is* accepted in practice; however, option A is incorrect because LH-like hormones are *not* used for superovulation — FSH is. Option B is factual (6–8 ova), C is correct (AI follows superovulation), and D aligns with NCERT’s stated '8–32 cell stage' for transfer. Thus, A is the *only* non-step: LH-like activity induces final ovulation, but superovulation requires sustained FSH stimulation. Hence, option A violates the protocol and is the correct answer.
Question 199 · Genetic Code and Protein Synthesis
Statement I: The codon 'AUG' codes for methionine and phenylalanine.
Statement II: 'AAA' and 'AAG' both codons code for the amino acid lysine.
In the light of the above statements, choose the correct answer from the options given below.
ABoth Statement I and Statement II are true
BBoth Statement I and Statement II are false
CStatement I is correct but Statement II is false
DStatement I is incorrect but Statement II is true
Answer: D. Statement I is incorrect but Statement II is true
The genetic code is universal, unambiguous, and degenerate. AUG is the initiation codon and codes exclusively for methionine — not phenylalanine (which is encoded by UUU and UUC). Hence, Statement I is incorrect. In contrast, Statement II is correct: both AAA and AAG are standard codons for lysine, illustrating codon degeneracy — multiple codons specifying the same amino acid. This is explicitly stated in NCERT Class 12 Biology Chapter 6 (Molecular Basis of Inheritance), Table 6.1, which lists lysine codons as AAA and AAG. No other amino acid shares these codons, confirming their specificity. Therefore, only Statement II is true, making option D the correct choice. Students must remember that while AUG is nearly always the start codon (methionine), it never encodes phenylalanine; confusing it with UUU/UUC is a common error. Degeneracy ensures robustness against point mutations — e.g., a change from AAA to AAG still yields lysine, preserving protein function.
Question 200 · Earthworm morphology and anatomy
Following are the statements about prostomium of earthworm: (a) It serves as a covering for mouth. (b) It helps to open cracks in the soil into which it can crawl. (c) It is one of the sensory structures. (d) It is the first body segment. Choose the correct answer from the options given below.
A(a), (b) and (c) are correct
B(a), (b) and (d) are correct
C(a), (b), (c) and (d) are correct
D(b) and (c) are correct
Answer: A. (a), (b) and (c) are correct
The prostomium in earthworms (Pheretima posthuma) is a fleshy, lobe-like structure located anterior to the mouth, serving as a protective covering (statement a — correct). It aids locomotion by helping the worm force open soil cracks during burrowing (b — correct). It bears sensory receptors like tactile and chemoreceptors, making it a key sensory structure (c — correct). However, it is *not* considered a true body segment; the first true segment is the peristomium (which bears the mouth), while the prostomium is a pre-segmental, non-metameric part (d — incorrect). NCERT Class 11 Biology (Chapter 7: Structural Organisation in Animals) explicitly states that the prostomium is a 'covering over the mouth' and 'helps in burrowing', and mentions its sensory role — but clarifies it is not a segment. Hence, only (a), (b), and (c) are correct, matching option A.