Biology — NEET Complete Yearwise Papers MCQs with Solutions
Free NEET Complete Yearwise Papers Biology MCQs with step-by-step solutions (100 questions). Part of NEET 2024 Paper. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Biology · medium · theory
Identify the set of correct statements:
A. The flowers of Vallisneria are colorful and produce nectar.
B. The flowers of waterlily are not pollinated by water.
C. In most of water-pollinated species, the pollen grains are protected from wetting.
D. Pollen grains of some hydrophytes are long and ribbon like.
E. In some hydrophytes, the pollen grains are carried passively inside water.
Choose the correct answer from the options given below:
A. A, C, D and E only
B. B, C, D and E only ✓ Correct
C. C, D and E only
D. A, B, C and D only
Solution: A is wrong: wind and water pollinated flowers are NOT colourful and do not produce nectar, and Vallisneria is water pollinated.
The rest are correct. Water lily flowers emerge above the water and are pollinated by insects or wind. In most water-pollinated species the pollen is protected from wetting by a mucilaginous covering, and in seagrasses the pollen grains are long and ribbon-like and are carried passively inside the water.
Q2 — Biology · medium · theory
The type of conservation in which the threatened species are taken out from their natural habitat and placed in a special setting where they can be protected and given special care is called:
A. Semi-conservative method
B. Sustainable development
C. In-situ conservation
D. Ex-situ conservation ✓ Correct
Solution: This is ex-situ ("off site") conservation. Threatened animals and plants are taken out of their natural habitat and kept in a special setting where they are protected and given special care — zoological parks, botanical gardens and wildlife safari parks all serve this purpose.
In-situ conservation, by contrast, protects the species where it already lives.
Q3 — Biology · medium · theory
Inhibition of succinic dehydrogenase enzyme by malonate is a classical example of:
A. Competitive inhibition ✓ Correct
B. Enzyme activation
C. Cofactor inhibition
D. Feedback inhibition
Solution: Malonate closely resembles the substrate succinate, so it competes with succinate for the substrate-binding site of succinic dehydrogenase. With the site occupied the substrate cannot bind and enzyme action declines — the definition of competitive inhibition.
Q4 — Biology · medium · theory
Identify the part of the seed from the given figure which is destined to form the root when the seed germinates.
A. C ✓ Correct
B. D
C. A
D. B
Solution: The primary root, or radicle, is the first organ to appear when a seed germinates.
In the figure the labels from the top are A — coleoptile, B — plumule, C — radicle and D — coleorhiza, so the radicle is C.
Q5 — Biology · medium · theory
Bulliform cells are responsible for
A. Increased photosynthesis in monocots.
B. Providing large spaces for storage of sugars.
C. Inward curling of leaves in monocots. ✓ Correct
D. Protecting the plant from salt stress.
Solution: In grasses, certain adaxial epidermal cells along the veins modify into large, empty, colourless cells called bulliform cells. When they have absorbed water and are turgid the leaf surface is exposed; when they become flaccid under water stress they make the leaves curl inwards, which cuts water loss.
Q6 — Biology · medium · theory
Which of the following are required for the dark reaction of photosynthesis?
A. Light
B. Chlorophyll
C. CO₂
D. ATP
E. NADPH
Choose the correct answer from the options given below:
A. C, D and E only ✓ Correct
B. D and E only
C. A, B and C only
D. B, C and D only
Solution: The dark reaction (Calvin cycle) needs CO₂ for the carboxylation step, and ATP and NADPH — supplied by the light reaction — for the reduction and regeneration steps.
Light and chlorophyll are needed for the LIGHT reaction, not directly for the dark reaction.
Q7 — Biology · medium · theory
Formation of interfascicular cambium from fully developed parenchyma cells is an example for
A. Dedifferentiation ✓ Correct
B. Maturation
C. Differentiation
D. Redifferentiation
Solution: Living differentiated cells that have lost the capacity to divide can regain it under certain conditions. That phenomenon is called dedifferentiation — and the formation of interfascicular cambium and cork cambium from fully differentiated parenchyma is the standard example.
Q8 — Biology · medium · theory
Hind II always cuts DNA molecules at a particular point called recognition sequence and it consists of:
A. 4 bp
B. 10 bp
C. 8 bp
D. 6 bp ✓ Correct
Solution: The recognition sequence of Hind II is six base pairs long, and it cuts to give blunt ends. It was the first restriction endonuclease to be characterised.
Q9 — Biology · medium · theory
Tropical regions show the greatest level of species richness because
A. Tropical latitudes have remained relatively undisturbed for millions of years, hence more time was available for species diversification.
B. Tropical environments are more seasonal.
C. More solar energy is available in tropics.
D. Constant environments promote niche specialization.
E. Tropical environments are constant and predictable.
Choose the correct answer from the options given below:
A. A, B and E only
B. A, B and D only
C. A, C, D and E only ✓ Correct
D. A and B only
Solution: Unlike temperate regions, which were repeatedly disturbed by glaciations, the tropics have remained relatively undisturbed for millions of years, giving a long evolutionary time for diversification (A). Tropical environments are LESS seasonal — more constant and predictable (so B is wrong but E is right), which promotes niche specialisation (D). More solar energy is available in the tropics, raising productivity and hence diversity (C).
Hence A, C, D and E.
Q10 — Biology · medium · theory
Which one of the following is not a criterion for classification of fungi?
A. Mode of spore formation
B. Fruiting body
C. Morphology of mycelium
D. Mode of nutrition ✓ Correct
Solution: The kingdom Fungi is divided into classes on the basis of the morphology of the mycelium, the mode of spore formation and the fruiting body.
Mode of nutrition is not used — all fungi are heterotrophic, so it cannot separate one class from another.
Q11 — Biology · medium · theory
How many molecules of ATP and NADPH are required for every molecule of CO₂ fixed in the Calvin cycle?
A. 3 molecules of ATP and 3 molecules of NADPH
B. 3 molecules of ATP and 2 molecules of NADPH ✓ Correct
C. 2 molecules of ATP and 3 molecules of NADPH
D. 2 molecules of ATP and 2 molecules of NADPH
Solution: For every CO₂ molecule entering the Calvin cycle, 3 ATP and 2 NADPH are required.
The reduction step uses 2 ATP for phosphorylation and 2 NADPH for reduction, and the regeneration step uses 1 more ATP to reform RuBP.
Q12 — Biology · medium · theory
These are regarded as major causes of biodiversity loss:
A. Over exploitation
B. Co-extinction
C. Mutation
D. Habitat loss and fragmentation
E. Migration
Choose the correct option:
A. A, B and E only
B. A, B and D only ✓ Correct
C. A, C and D only
D. A, B, C and D only
Solution: There are four major causes of biodiversity loss, nicknamed "The Evil Quartet": habitat loss and fragmentation, over-exploitation, alien species invasions, and co-extinctions.
Of the options listed, that gives A, B and D. Mutation and migration are not among them.
Q13 — Biology · medium · theory
The capacity to generate a whole plant from any cell of the plant is called:
A. Differentiation
B. Somatic hybridization
C. Totipotency ✓ Correct
D. Micropropagation
Solution: Totipotency is the capacity of a single plant cell to generate a whole plant. It is the basis of tissue culture and of micropropagation — but micropropagation is the technique, not the capacity itself.
Q14 — Biology · medium · theory
The equation of Verhulst-Pearl logistic growth is $\dfrac{dN}{dt} = rN\left[\dfrac{K-N}{K}\right]$. From this equation, $K$ indicates:
A. Carrying capacity ✓ Correct
B. Population density
C. Intrinsic rate of natural increase
D. Biotic potential
Solution: In the logistic growth equation, $N$ is the population density at time $t$, $r$ is the intrinsic rate of natural increase, and $K$ is the CARRYING CAPACITY — the population size the habitat can support, where the sigmoid curve levels off.
Q15 — Biology · medium · theory
Spindle fibers attach to kinetochores of chromosomes during
A. Anaphase
B. Telophase
C. Prophase
D. Metaphase ✓ Correct
Solution: The key features of metaphase are that spindle fibres attach to the kinetochores of the chromosomes, and the chromosomes are moved to the spindle equator and aligned along the metaphase plate by fibres running to both poles.
Q16 — Biology · medium · theory
Identify the type of flowers based on the position of calyx, corolla and androecium with respect to the ovary from the given figures (a) and (b).
A. (a) Perigynous; (b) Epigynous
B. (a) Perigynous; (b) Perigynous ✓ Correct
C. (a) Epigynous; (b) Hypogynous
D. (a) Hypogynous; (b) Epigynous
Solution: When the gynoecium sits in the centre and the other floral parts are on the rim of the thalamus at almost the same level, the flower is perigynous and the ovary is half inferior — as in plum, rose and peach.
Both figures show this arrangement, so both are perigynous.
Q17 — Biology · medium · theory
Match List I with List II.
List I
(A) Rhizopus
(B) Ustilago
(C) Puccinia
(D) Agaricus
List II
(I) Mushroom
(II) Smut fungus
(III) Bread mould
(IV) Rust fungus
Choose the correct answer from the options given below:
A. A-III, B-II, C-I, D-IV
B. A-IV, B-III, C-II, D-I
C. A-III, B-II, C-IV, D-I ✓ Correct
D. A-I, B-III, C-II, D-IV
Solution: Rhizopus is the bread mould, and belongs to Phycomycetes (III).
The other three are common Basidiomycetes: Ustilago is the smut fungus (II), Puccinia the rust fungus (IV), and Agaricus the mushroom (I).
Q18 — Biology · medium · theory
In a plant, black seed colour ($BB/Bb$) is dominant over white seed colour ($bb$). In order to find out the genotype of the black seed plant, with which of the following genotypes will you cross it?
A. $Bb$
B. $BB/Bb$
C. $BB$
D. $bb$ ✓ Correct
Solution: This is a test cross: an organism showing the dominant phenotype, whose genotype is to be determined, is crossed with the homozygous RECESSIVE parent ($bb$) rather than self-crossed.
If any white-seeded offspring appear, the black parent must be $Bb$; if all offspring are black, it is $BB$.
Q19 — Biology · medium · theory
A pink flowered snapdragon plant was crossed with a red flowered snapdragon plant. What type of phenotype/s is/are expected in the progeny?
A. Only pink flowered plants
B. Red, Pink as well as white flowered plants
C. Only red flowered plants
D. Red flowered as well as pink flowered plants ✓ Correct
Solution: Snapdragon flower colour shows incomplete dominance: $RR$ is red, $Rr$ is pink and $rr$ is white.
Crossing pink ($Rr$) with red ($RR$) gives $RR$, $Rr$, $RR$, $Rr$ — that is 50% red and 50% pink, with no white.
Q20 — Biology · medium · theory
Match List I with List II.
List I
(A) Two or more alternative forms of a gene
(B) Cross of F1 with homozygous recessive parent
(C) Cross of F1 progeny with any of the parents
(D) Number of chromosome sets in plant
List II
(I) Back cross
(II) Ploidy
(III) Allele
(IV) Test cross
Choose the correct answer from the options given below:
A. A-III, B-IV, C-I, D-II ✓ Correct
B. A-IV, B-III, C-II, D-I
C. A-I, B-II, C-III, D-IV
D. A-II, B-I, C-III, D-IV
Solution: Two or more alternative forms of a gene — allele (III).
Cross of F1 progeny with the homozygous recessive parent — test cross (IV).
Cross of F1 progeny with ANY of the parents — back cross (I).
Number of chromosome sets in a plant — ploidy (II).
Q21 — Biology · medium · theory
Lecithin, a small molecular weight organic compound found in living tissues, is an example of:
A. Glycerides
B. Carbohydrates
C. Amino acids
D. Phospholipids ✓ Correct
Solution: Lecithin, also called phosphatidyl choline, is a phospholipid — a glycerol backbone carrying two fatty acids and a phosphate-choline head group. It is a major component of biological membranes.
Q22 — Biology · medium · theory
Match List I with List II.
List I
(A) Clostridium butylicum
(B) Saccharomyces cerevisiae
(C) Trichoderma polysporum
(D) Streptococcus sp.
List II
(I) Ethanol
(II) Streptokinase
(III) Butyric acid
(IV) Cyclosporin-A
Choose the correct answer from the options given below:
A. A-III, B-I, C-IV, D-II ✓ Correct
B. A-IV, B-I, C-III, D-II
C. A-III, B-I, C-II, D-IV
D. A-II, B-IV, C-III, D-I
Solution: Clostridium butylicum — butyric acid (III).
Saccharomyces cerevisiae (brewer's yeast) — ethanol (I).
Trichoderma polysporum — cyclosporin-A, the immunosuppressant (IV).
Streptococcus sp. — streptokinase, the clot buster (II).
Q23 — Biology · medium · theory
In the given figure, which component has thin outer walls and highly thickened inner walls?
A. A
B. B
C. C ✓ Correct
D. D
Solution: The structure with unevenly thickened walls is the pair of guard cells, labelled C.
Their outer walls (away from the stomatal pore) are thin while the inner walls (towards the pore) are highly thickened — which is what makes the guard cells bow outwards and open the stoma when they become turgid.
Q24 — Biology · medium · theory
Which of the following is an example of an actinomorphic flower?
A. Pisum
B. Sesbania
C. Datura ✓ Correct
D. Cassia
Solution: A flower is actinomorphic (radially symmetrical) when it can be divided into two equal radial halves by any radial plane through the centre — as in mustard, datura and chilli.
Pisum, Sesbania and Cassia are all zygomorphic (bilaterally symmetrical).
Q25 — Biology · medium · theory
A transcription unit in DNA is defined primarily by the three regions in DNA, and these are, with respect to the upstream and downstream end:
A. Inducer, Repressor, Structural gene
B. Promotor, Structural gene, Terminator ✓ Correct
C. Repressor, Operator gene, Structural gene
D. Structural gene, Transposons, Operator gene
Solution: A transcription unit is defined by three regions: a promoter at the upstream end, the structural gene in between, and a terminator at the downstream end.
Q26 — Biology · medium · theory
What is the fate of a piece of DNA carrying only the gene of interest which is transferred into an alien organism?
A. The piece of DNA would be able to multiply itself independently in the progeny cells of the organisms.
B. It may get integrated into the genome of the recipient.
C. It may multiply and be inherited along with the host DNA.
D. The alien piece of DNA is not an integrated part of chromosome.
E. It shows ability to replicate.
Choose the correct answer from the options given below:
A. B and C only ✓ Correct
B. A and E only
C. A and B only
D. D and E only
Solution: A bare piece of DNA carrying only the gene of interest has no origin of replication, so it cannot multiply on its own in the progeny cells.
What it CAN do is integrate itself into the genome of the recipient, after which it multiplies and is inherited along with the host DNA — statements B and C.
Q27 — Biology · medium · theory
Auxin is used by gardeners to prepare weed free lawns. But no damage is caused to grass as auxin:
A. does not affect mature monocotyledonous plants. ✓ Correct
B. can help in cell division in grasses, to produce growth.
C. promotes apical dominance.
D. promotes abscission of mature leaves only.
Solution: Auxins are widely used as herbicides. 2,4-D, used to kill dicotyledonous weeds, does not affect mature monocotyledonous plants — and grass is a monocot, so the lawn survives while the weeds die.
Q28 — Biology · medium · theory
The cofactor of the enzyme carboxypeptidase is:
A. Flavin
B. Haem
C. Zinc ✓ Correct
D. Niacin
Solution: Metal ions are the common cofactors, and zinc is the prosthetic metal ion of carboxypeptidase — it is held by the enzyme through coordinate bonds and takes part directly in catalysis.
Q29 — Biology · medium · theory
The lactose present in the growth medium of bacteria is transported into the cell by the action of
A. Permease ✓ Correct
B. Polymerase
C. Beta-galactosidase
D. Acetylase
Solution: In the absence of a preferred carbon source such as glucose, lactose supplied in the medium is transported into the bacterial cell through the action of permease (the product of the lac operon's $y$ gene).
Beta-galactosidase then hydrolyses the lactose once it is inside.
Q30 — Biology · medium · theory
Which one of the following can be explained on the basis of Mendel's Law of Dominance?
A. Out of one pair of factors one is dominant and the other is recessive.
B. Alleles do not show any expression and both the characters appear as such in $F_2$ generation.
C. Factors occur in pair in normal diploid plants.
D. The discrete unit controlling a particular character is called factor.
E. The expression of only one of the parental characters is found in a monohybrid cross.
Choose the correct answer from the options given below:
A. B, C and D only
B. A, B, C, D and E
C. A, B and C only
D. A, C, D and E only ✓ Correct
Solution: The Law of Dominance says characters are controlled by discrete units called factors (D), which occur in pairs in a diploid (C); when the pair is dissimilar one is dominant and the other recessive (A), so only one parental character is expressed in a monohybrid cross (E).
B is wrong — alleles DO show expression in the $F_2$ generation.