🧬 NEET Crackers — PYQ Biology Series

Top 50 NEET Biology
Questions & Answers

Curated from previous year NEET papers — the exact questions that decide your rank.

50 Questions
PYQ Based
NCERT Aligned
Detailed Explanations
12 Topics
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All 50 Questions
Ecology · Genetics · Cell Biology · Physiology · Biotechnology · Evolution · Plant Biology & more
Q 01
Ecology
Which of the following is the correct unit of productivity in an ecosystem?
✅ Correct Answer
A — KCal m⁻² yr⁻¹
💡 Explanation
Productivity is the rate of energy production per unit area per unit time, so it must include both area (m⁻²) and time (yr⁻¹). Options B and C lack the time component, and D uses volume instead of area — neither fits ecosystem energy flow measurement.
Q 02
Human Reproduction
The first occurrence of menstruation in females is known as:
✅ Correct Answer
C — Menarche
💡 Explanation
Menarche marks the onset of the very first menstrual cycle, typically between ages 11–13, signalling the start of reproductive maturity in females. Menopause is the permanent cessation of menstruation, and diapause is a state of suspended development seen in some animals — not humans.
Q 03
Genetics
Genes R and Y assort independently. RRYY produces round yellow seeds and rryy produces wrinkled green seeds. What is the phenotypic ratio in the F₂ generation?
✅ Correct Answer
D — 9:3:3:1
💡 Explanation
This is Mendel's classic dihybrid cross. When F₁ hybrids (RrYy) are self-crossed, Mendel's Law of Independent Assortment applies, generating 4 phenotypic classes — 9 round yellow : 3 round green : 3 wrinkled yellow : 1 wrinkled green. The 9:7 ratio appears only when there is epistasis.
Q 04
Cell Division
What is the main function of spindle fibres during mitosis?
✅ Correct Answer
B — Separate chromosomes
💡 Explanation
Spindle fibres (made of tubulin protein) attach to the kinetochores at centromeres of chromosomes and physically pull sister chromatids toward opposite poles during anaphase. They ensure each daughter cell receives an equal, accurate set of chromosomes.
Q 05
Plant Reproduction
How many meiotic and mitotic divisions are required to form a mature female gametophyte from a megaspore mother cell in angiosperms?
✅ Correct Answer
D — 1 meiosis and 3 mitosis
💡 Explanation
The megaspore mother cell undergoes 1 meiotic division to produce 4 haploid megaspores (3 degenerate). The functional megaspore then undergoes 3 successive mitotic divisions to produce the 7-celled, 8-nucleate embryo sac (the mature female gametophyte).
Q 06
Immunology
Which of the following statements about antibody structure is NOT correct?
✅ Correct Answer
D — Antigen binding site is at C-terminal (this is INCORRECT)
💡 Explanation
The antigen-binding site (paratope) is located at the variable (V) region at the N-terminal end of both heavy and light chains — not the C-terminal. The C-terminal carries the constant regions. An antibody has a Y-shaped structure with two identical antigen-binding sites.
Q 07
Plant Anatomy
Assertion: Tapetum cells have dense cytoplasm and are often multinucleate. Reason: Multiple nuclei increase efficiency in nourishing developing pollen.
✅ Correct Answer
D — Assertion is true, Reason is false
💡 Explanation
Tapetum cells are indeed multinucleate with dense cytoplasm — this is true per NCERT. However, the reason given (that multiple nuclei increase nourishing efficiency) is not the accepted NCERT explanation. Tapetum is the innermost layer of the anther wall and provides nutrition to developing microspores, but the multinucleate condition is not explained as the reason for this.
Q 08
Plant Kingdom
In bryophytes, gemmae are involved in:
✅ Correct Answer
C — Asexual reproduction
💡 Explanation
Gemmae are small, multicellular, green vegetative propagules found in special cups (gemma cups) on the thallus of liverworts like Marchantia. When they detach and land in suitable conditions, they germinate directly into new plants — a classic example of vegetative/asexual reproduction in bryophytes.
Q 09
Molecular Biology
Who proposed that the genetic code consists of triplet nucleotides?
✅ Correct Answer
B — George Gamow
💡 Explanation
George Gamow, a physicist-turned-biologist, mathematically proposed in 1954 that codons must consist of 3 nucleotides, since \(4^3 = 64\) combinations are enough to encode all 20 amino acids (whereas \(4^2 = 16\) is insufficient). Francis Crick later experimentally proved the triplet nature of the code.
Q 10
Evolution
Sweet potato (root) and potato (stem) represent which type of evolution?
✅ Correct Answer
B — Analogy, Convergent Evolution
💡 Explanation
Sweet potato (modified root) and potato (modified stem) perform the same function (food storage) but have different structural origins — this is analogy. When unrelated structures converge to serve the same function due to similar environmental pressures, it is called convergent evolution. Homology implies similar origin but different functions (e.g., forelimbs of vertebrates).
Q 11
Molecular Biology
Histone proteins are rich in which amino acids?
✅ Correct Answer
B — Lysine and Arginine
💡 Explanation
Histone proteins are positively charged due to their abundance of basic amino acids — lysine (Lys) and arginine (Arg). This positive charge enables strong electrostatic attraction with the negatively charged phosphate backbone of DNA, allowing 147 bp of DNA to wrap around each histone octamer to form a nucleosome.
Q 12
Ecology
Which equation correctly represents logistic population growth?
✅ Correct Answer
C — \(dN/dt = rN\,(K - N)/K\)
💡 Explanation
The Verhulst-Pearl logistic growth equation is \[ \frac{dN}{dt} = rN \cdot \frac{K - N}{K} \] Here, \(r\) = intrinsic rate of natural increase, \(N\) = current population size, \(K\) = carrying capacity. As \(N \to K\), the term \((K-N)/K \to 0\), slowing growth and producing the characteristic S-shaped (sigmoid) curve.
Q 13
Photosynthesis
Which statement about RuBisCO is correct?
✅ Correct Answer
A — Catalyzes carboxylation of RuBP
💡 Explanation
RuBisCO (Ribulose-1,5-bisphosphate carboxylase/oxygenase) is the key enzyme of the Calvin cycle. It catalyses the fixation of CO₂ onto RuBP (a 5-carbon compound), producing two molecules of 3-PGA (3-carbon). It is the most abundant protein on Earth. It also has oxygenase activity (leading to photorespiration in C3 plants), but its primary role is carboxylation.
Q 14
Biomolecules
The protein part of an enzyme is called:
✅ Correct Answer
D — Apoenzyme
💡 Explanation
An enzyme consists of a protein portion called the apoenzyme and a non-protein component called the cofactor (which can be a metal ion or organic molecule). Together they form the holoenzyme (fully active enzyme). Coenzymes are loosely bound organic cofactors (e.g., NAD⁺, FAD), and prosthetic groups are tightly/permanently bound cofactors.
Q 15
Molecular Biology
Which factor is responsible for termination of transcription in prokaryotes?
✅ Correct Answer
D — Rho (ρ) factor
💡 Explanation
In prokaryotes, transcription terminates by two mechanisms — Rho-dependent (requires the Rho protein, which unwinds RNA-DNA hybrid) and Rho-independent (intrinsic termination via hairpin loops). Sigma (σ) factor is needed for promoter recognition and initiation — not termination. Rho is an ATP-dependent RNA-DNA helicase.
Q 16
Human Physiology
Which hormone is released by the pituitary but synthesised in the hypothalamus?
✅ Correct Answer
C — ADH (Antidiuretic Hormone / Vasopressin)
💡 Explanation
ADH (vasopressin) and oxytocin are both synthesised by neurosecretory cells in the hypothalamus, then transported via axons to the posterior pituitary, where they are stored and released. ACTH, LH, FSH and other tropic hormones are both synthesised AND secreted by the anterior pituitary.
Q 17
Biotechnology
Why is insulin not given orally to diabetic patients?
✅ Correct Answer
C — It gets digested in the GI tract
💡 Explanation
Insulin is a polypeptide hormone made of two chains (A and B) linked by disulfide bonds. If taken orally, digestive enzymes (proteases like pepsin and trypsin) in the gastrointestinal tract would break it down into amino acids before it reaches the bloodstream — rendering it ineffective. Hence insulin must be administered subcutaneously (injection).
Q 18
Biotechnology
RNA interference (RNAi) works due to:
✅ Correct Answer
B — Complementary dsRNA (double-stranded RNA)
💡 Explanation
RNAi is a gene-silencing mechanism in which complementary double-stranded RNA (dsRNA) triggers sequence-specific degradation of the corresponding mRNA. The dsRNA is processed into small interfering RNAs (siRNAs) by the enzyme Dicer. This is used in agriculture (e.g., nematode-resistant tobacco) and forms the basis of many gene therapy approaches.
Q 19
Biomolecules
Which enzyme contains haem as a prosthetic group?
✅ Correct Answer
A — Catalase
💡 Explanation
Catalase contains an iron-containing haem group as its prosthetic group. It catalyses the decomposition of hydrogen peroxide into water and oxygen: \(2H_2O_2 \rightarrow 2H_2O + O_2\). Carbonic anhydrase contains zinc, RuBisCO uses Mg²⁺, and succinate dehydrogenase uses FAD as its cofactor.
Q 20
Biotechnology
Which organism was used by Eli Lilly to produce human insulin (Humulin) for the first time?
✅ Correct Answer
B — Bacterium (Escherichia coli)
💡 Explanation
In 1982, Eli Lilly became the first company to market recombinant human insulin, commercially called Humulin. The human insulin gene was cloned and expressed in E. coli bacteria. The A and B chains were separately produced and then combined in vitro to form functional insulin, replacing animal-derived insulin and reducing allergic reactions.
Q 21
Biomolecules
Inhibition of succinic dehydrogenase enzyme by malonate is a classical example of:
✅ Correct Answer
A — Competitive inhibition
💡 Explanation
Malonate is structurally similar to succinate (both are dicarboxylic acids). It competes with succinate for the active site of succinic dehydrogenase, blocking the normal substrate from binding. This is the textbook definition of competitive inhibition — the inhibitor can be overcome by increasing substrate concentration. This inhibition is reversible.
Q 22
Plant Anatomy
Bulliform cells in monocots are responsible for:
✅ Correct Answer
C — Inward curling of leaves in monocots
💡 Explanation
Bulliform cells (motor cells) are large, colourless, bubble-like epidermal cells found on the upper surface of monocot leaves (e.g., grasses). During water stress, they lose turgor and collapse, causing the leaf to curl inward, which reduces the exposed surface area and thereby decreases water loss through transpiration — an adaptation to drought.
Q 23
Plant Physiology
Formation of interfascicular cambium from fully developed parenchyma cells is an example of:
✅ Correct Answer
A — Dedifferentiation
💡 Explanation
Dedifferentiation is the process by which already differentiated, mature cells regain the capacity to divide (meristematic activity). Mature parenchyma cells of the medullary rays become interfascicular cambium to enable secondary growth. This is different from redifferentiation, which is when dedifferentiated cells differentiate again into new structures.
Q 24
Photosynthesis
How many molecules of ATP and NADPH are required for fixation of one CO₂ molecule in the Calvin cycle?
✅ Correct Answer
B — 3 ATP and 2 NADPH per CO₂
💡 Explanation
For each CO₂ fixed, the Calvin cycle consumes 3 ATP and 2 NADPH. For 1 glucose (6 CO₂ fixed), the total is 18 ATP + 12 NADPH. The ATP is used in the phosphoglycerate kinase and RuBP regeneration steps, while NADPH reduces 1,3-bisphosphoglycerate to glyceraldehyde-3-phosphate (G3P).
Q 25
Plant Tissue Culture
The capacity to generate a whole plant from any cell of the plant is called:
✅ Correct Answer
C — Totipotency
💡 Explanation
Totipotency is the inherent potential of a single plant cell to divide, differentiate, and develop into a complete plant organism. First demonstrated by F.C. Steward using carrot phloem cells. It forms the theoretical basis of plant tissue culture and micropropagation. Every cell carries the full genome — totipotency is its complete expression.
Q 26
Cell Division
Spindle fibres fully attach to the kinetochores of chromosomes during which phase?
✅ Correct Answer
D — Metaphase
💡 Explanation
During metaphase, all chromosomes are maximally condensed, aligned at the equatorial plate (metaphase plate), and spindle fibres from both poles are fully attached to kinetochores. This is why metaphase is the ideal stage for karyotyping. Separation of chromatids happens in anaphase, after spindle tension pulls them apart.
Q 27
Biomolecules
Lecithin, found in living tissues, is an example of:
✅ Correct Answer
D — Phospholipids
💡 Explanation
Lecithin (phosphatidylcholine) is a phospholipid composed of glycerol, two fatty acids, a phosphate group, and a choline head group. It is a major structural component of all cell membranes, forms the fluid bilayer described in the Singer-Nicolson model, and is also found in high quantities in egg yolk and brain tissue.
Q 28
Molecular Biology
A transcription unit in DNA is primarily defined by which three regions?
✅ Correct Answer
B — Promoter, Structural gene, Terminator
💡 Explanation
A transcription unit has three essential components: (1) Promoter — where RNA polymerase binds to initiate transcription; (2) Structural gene — the coding sequence transcribed into RNA; (3) Terminator — signals where transcription ends. An operon (like Lac operon) additionally has an operator, but that is a regulatory element, not a basic transcription unit component.
Q 29
Biotechnology
Statement I: Bt toxins are insect group-specific and coded by cry genes. Statement II: Bt toxin becomes active due to the acidic pH of the insect gut.
✅ Correct Answer
A — Statement I is true; Statement II is false
💡 Explanation
Bacillus thuringiensis produces protoxin crystals (Cry proteins) encoded by cry genes — these are indeed group-specific (e.g., cryIAc targets bollworms, cryIIAb targets armyworms). However, Statement II is wrong: Bt toxin is activated by the alkaline pH of the insect midgut (not acidic), which solubilises the crystal and releases the active toxin that binds to gut epithelium.
Q 30
Plant Physiology
Spraying the sugarcane crop with which plant growth regulator increases stem length?
✅ Correct Answer
D — Gibberellin
💡 Explanation
Gibberellins (especially GA₃) dramatically promote internode elongation in stem. In sugarcane, spraying with gibberellin increases the length of internodes, which increases the total cane biomass and sugar yield by up to 20 tonnes per acre. They also break dormancy, promote seed germination, and induce flowering in long-day plants.
Q 31
Chemical Coordination
Which of the following is NOT a steroid hormone?
✅ Correct Answer
B — Glucagon
💡 Explanation
Glucagon is a peptide hormone (29 amino acids) secreted by α-cells of the pancreas — it raises blood glucose by stimulating glycogenolysis. Progesterone, cortisol, and testosterone are all steroid hormones derived from cholesterol. Steroids are lipid-soluble and act via nuclear receptors, while peptide hormones act via cell surface receptors.
Q 32
Evolution
The flippers of penguins and dolphins are an example of:
✅ Correct Answer
A — Convergent evolution
💡 Explanation
Penguins (birds) and dolphins (mammals) evolved flippers from entirely different ancestral forelimb structures, yet they look and function similarly due to the same aquatic environment. This is convergent evolution — unrelated lineages independently evolving similar traits. These structures are analogous (similar function, different origin), not homologous.
Q 33
Genetics
Which of the following factors will NOT affect Hardy-Weinberg equilibrium?
✅ Correct Answer
B — Constant gene pool
💡 Explanation
Hardy-Weinberg equilibrium is maintained when: there is random mating, no mutation, no gene flow, no genetic drift, no natural selection, and a large population. A constant gene pool is the assumption of the equilibrium itself — it is the condition, not a disrupting force. Factors that disturb H-W equilibrium include mutation, migration, genetic drift, natural selection, and non-random mating.
Q 34
Human Physiology
Which factors favour the formation of oxyhaemoglobin in the alveoli?
✅ Correct Answer
D — High pO₂ and lesser H⁺ concentration
💡 Explanation
In the alveoli, conditions favour oxyhaemoglobin formation: high pO₂ (~104 mmHg), low pCO₂ (~40 mmHg), low H⁺ concentration (alkaline pH ~7.4), and lower temperature. High CO₂, high H⁺ (Bohr effect), and high temperature shift the oxygen-dissociation curve to the right, favouring O₂ release in tissues — not loading in lungs.
Q 35
Biotechnology
The "Ti plasmid" of Agrobacterium tumefaciens stands for:
✅ Correct Answer
A — Tumour-inducing plasmid
💡 Explanation
The Ti (Tumour-inducing) plasmid of Agrobacterium tumefaciens contains T-DNA (transfer DNA), which integrates into the host plant genome and causes crown gall disease. Scientists disarmed the Ti plasmid (removed tumour-causing genes) and use it as a natural vector to deliver foreign genes of interest into plant cells — making it the most important tool in plant genetic engineering.
Q 36
Plant Kingdom
In mosses, the first stage of the gametophyte is the protonema stage. The protonema develops directly from spores produced in a capsule. Evaluate these statements:
✅ Correct Answer
A — Both correct and R correctly explains A
💡 Explanation
In mosses, spores germinate to form the protonema, which is the first and filamentous stage of the gametophyte. This is directly and correctly explained by the reason: spores are produced in the sporophyte capsule, and upon germination form protonema. The protonema then differentiates into the leafy gametophore — the second stage of the gametophyte.
Q 37
Plant Reproduction
In angiosperms, the haploid, diploid, and triploid structures of a fertilised embryo sac sequentially are:
✅ Correct Answer
C — Synergids (n), Zygote (2n), Primary endosperm nucleus (3n)
💡 Explanation
After double fertilisation: Synergids are haploid (n) — they guide the pollen tube. The Zygote is diploid (2n) — formed by fusion of male gamete with the egg cell. The Primary Endosperm Nucleus (PEN) is triploid (3n) — formed by fusion of the second male gamete with two polar nuclei, developing into endosperm.
Q 38
Plant Physiology
Which hormone promotes internode/petiole elongation in deep water rice under submerged conditions?
✅ Correct Answer
C — Ethylene
💡 Explanation
Ethylene (the only gaseous plant hormone) promotes rapid internode elongation in deep water rice when submerged — this keeps the leaves above the water surface, allowing photosynthesis and gas exchange. This is a unique case where ethylene promotes growth rather than its typical roles of fruit ripening, leaf abscission, and senescence.
Q 39
Ecology
Among 'The Evil Quartet', which is the most important cause of species extinction?
✅ Correct Answer
A — Habitat loss and fragmentation
💡 Explanation
The "Evil Quartet" are the four main drivers of biodiversity loss: habitat loss/fragmentation, over-exploitation, alien species invasion, and co-extinctions. Of these, habitat destruction is the single biggest cause globally — tropical deforestation, wetland destruction, and coral reef degradation collectively threaten more species than all other factors combined.
Q 40
Photosynthesis
Which micronutrient is required for the splitting of water (photolysis) during photosynthesis?
✅ Correct Answer
A — Manganese (Mn)
💡 Explanation
Manganese (Mn) is an essential component of the Oxygen-Evolving Complex (OEC), also called the water-splitting complex, in Photosystem II. It undergoes cyclical oxidation states (Mn²⁺ → Mn⁴⁺) to catalyse the reaction: \(2H_2O \rightarrow 4H^+ + 4e^- + O_2\). Magnesium is the central atom of chlorophyll, while copper is found in plastocyanin (electron carrier).
Q 41
Plant Morphology
Axile placentation is observed in which plants?
✅ Correct Answer
D — China rose, Petunia and Lemon
💡 Explanation
In axile placentation, the placenta is on the central axis where the septa (partitions) of a multilocular (many-chambered) ovary meet. Classic examples: China rose (Hibiscus), Petunia, Lemon, Tomato. Marginal placentation: Pea; Parietal: Mustard, Cucumber; Free central: Dianthus, Primrose; Basal: Marigold.
Q 42
Photosynthesis
The reaction centre in Photosystem II (PS II) has an absorption maxima at:
✅ Correct Answer
A — 680 nm (P680)
💡 Explanation
The reaction centre of PS II is P680 (absorbs light at 680 nm) and the reaction centre of PS I is P700 (absorbs light at 700 nm). When P680 absorbs light, it ejects an electron that passes through the electron transport chain. Photolysis of water occurs at PS II to replenish the lost electron, releasing O₂ as a byproduct.
Q 43
Cell Biology
In eukaryotes, DNA replication occurs in which phase of the cell cycle?
✅ Correct Answer
B — S phase (Synthesis phase)
💡 Explanation
The S (Synthesis) phase is when DNA replication (semi-conservative) takes place. The DNA content doubles from 2C to 4C without any change in chromosome number. G₁ is the growth phase before DNA replication, G₂ is the preparation phase after replication, and M phase is mitosis itself — no new DNA is made during M phase.
Q 44
Plant Kingdom
Identify the heterosporous pteridophytes from the following:
✅ Correct Answer
B — Selaginella and Salvinia
💡 Explanation
Heterosporous pteridophytes produce two distinct types of spores: microspores (small, develop into male gametophyte) and megaspores (large, develop into female gametophyte). Selaginella and Salvinia are classic heterosporous pteridophytes. Lycopodium, Psilotum, Equisetum are homosporous. Heterospory is considered an evolutionary precursor to seed habit.
Q 45
Photosynthesis
ATP and NADPH₂ required to synthesize one complete glucose molecule in the Calvin cycle are:
✅ Correct Answer
B — 18 ATP and 12 NADPH₂
💡 Explanation
For complete synthesis of 1 glucose (C₆H₁₂O₆), 6 CO₂ must be fixed: \[ 6\,CO_2 + 18\,ATP + 12\,NADPH \rightarrow C_6H_{12}O_6 + 18\,ADP + 12\,NADP^+ \] The 18 ATP are used in the phosphorylation steps (PGA → BPGA and RuBP regeneration), while the 12 NADPH provide reducing power to convert BPGA to G3P.
Q 46
Molecular Biology
What is the role of RNA Polymerase III in eukaryotes?
✅ Correct Answer
B — tRNA, 5S rRNA and snRNA synthesis
💡 Explanation
Eukaryotes have three RNA polymerases: RNA Pol I transcribes rRNA (28S, 18S, 5.8S) in the nucleolus; RNA Pol II transcribes mRNA precursors (hnRNA) and some snRNAs; RNA Pol III transcribes small stable RNAs — tRNA, 5S rRNA, and snRNAs. These three polymerases are inhibited by α-amanitin at different concentrations.
Q 47
Genetics
The first use of recombination frequency for gene mapping was by:
✅ Correct Answer
C — Alfred Sturtevant
💡 Explanation
Alfred Sturtevant (a student of Morgan) created the world's first genetic linkage map in 1913 using Drosophila. He realised that recombination frequency between two genes is proportional to their distance apart on the chromosome — 1% recombination = 1 centimorgan (cM). Morgan discovered linkage; Sturtevant quantified it.
Q 48
Cell Biology
Which components are required for chemiosmosis to produce ATP?
✅ Correct Answer
A — Membrane + Proton pump + Proton gradient + ATP synthase
💡 Explanation
Chemiosmosis (Mitchell's hypothesis) requires: (1) a selectively permeable membrane; (2) a proton pump (ETC complexes) that builds the gradient; (3) a proton gradient (H⁺ concentration difference across the membrane); (4) ATP synthase (Complex V) through which protons flow back, releasing energy used to synthesise ATP from ADP + Pi.
Q 49
Human Health
HIV replicates primarily in which type of blood cells?
✅ Correct Answer
A — T Helper (TH) cells / CD4⁺ lymphocytes
💡 Explanation
HIV (Human Immunodeficiency Virus) selectively infects and destroys CD4⁺ T-helper lymphocytes — the master regulators of immune response. HIV uses reverse transcriptase to convert its RNA genome into DNA, which integrates into the host cell's DNA (provirus). As CD4⁺ count drops below 200 cells/μL, the patient is diagnosed with AIDS, leaving them vulnerable to opportunistic infections.
Q 50
Molecular Biology
Statement I: Prokaryotic DNA is positively charged and associated with negatively charged proteins. Statement II: Eukaryotic DNA wraps around positively charged histones.
✅ Correct Answer
D — Statement I is false; Statement II is true
💡 Explanation
Statement I is wrong: DNA in all organisms is negatively charged (due to phosphate groups). Prokaryotic DNA is associated with positively charged polyamines and basic proteins (not negatively charged ones). Statement II is correct: In eukaryotes, DNA wraps around positively charged histone octamers (rich in Lys and Arg) to form nucleosomes, enabling efficient packaging of DNA in the nucleus.