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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 2. Practise online on Prepizo — no login needed.

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Questions with solutions

Q1 — Biology · medium · theory
The regions with high level of species richness, high degree of endemism and a loss of 70% of the species and habitat are identified as:
A. Natural Reserves
B. Sacred Groves
C. Biodiversity Hotspots  ✓ Correct
D. Biogeographical Regions
Solution: Biodiversity hotspots are regions with very high levels of species richness and a high degree of endemism, which have also lost most of their original species and habitat. India has four: the Western Ghats and Sri Lanka, the Indo-Burma region, the Himalaya, and the Sundaland.
Q2 — Biology · medium · theory
Which of the following simple tissues are commonly found in the fruit walls of nuts and pulp of pear?
A. Sclereids  ✓ Correct
B. Fibres
C. Parenchyma
D. Collenchyma
Solution: Sclereids are spherical, oval or cylindrical sclerenchymatous cells with very thick walls and narrow cavities. They are commonly found in the hard parts of the plant — the fruit walls of nuts, the pulp of fruits like pear and sapota, the seed coats of legumes and the leaves of tea.
Q3 — Biology · medium · theory
In a chromosome, there is a specific DNA sequence, responsible for initiating replication. It is called as:
A. Recognition sequence
B. Cloning site
C. Restriction site
D. ori site  ✓ Correct
Solution: The origin of replication (ori) is the sequence from which replication starts, and any piece of DNA linked to it can be made to replicate within the host cell. A recognition sequence is the palindromic stretch read by a restriction endonuclease, and the cloning site (or restriction site) is where the enzyme cuts the vector.
Q4 — Biology · medium · theory
Given below are two statements: Statement I: When many alleles of a single gene govern a character, it is called polygenic inheritance. Statement II: In polygenic inheritance, the effect of each allele is additive. In the light of the above statements, choose the correct answer from the options given below.
A. Statement I is true but Statement II is false
B. Statement I is false but Statement II is true  ✓ Correct
C. Both Statement I and Statement II are true
D. Both Statement I and Statement II are false
Solution: Statement I is incorrect: polygenic inheritance is a trait governed by MORE THAN ONE GENE, not by many alleles of a single gene. Statement II is correct — in a polygenic trait the phenotype reflects the contribution of each allele, so the effect of each allele is additive.
Q5 — Biology · medium · theory
Which of the following are required for the light reaction of photosynthesis? A. CO₂ B. O₂ C. H₂O D. Chlorophyll E. Light Choose the correct answer from the options given below:
A. A, C, D and E only
B. C, D and E only  ✓ Correct
C. A and B only
D. A, C and E only
Solution: The light reaction needs water (which is split by photolysis), chlorophyll (to trap the light) and light itself. CO₂ is used in the SECOND stage, the dark reaction. Oxygen is a PRODUCT of the light reaction, not a requirement.
Q6 — Biology · medium · theory
Match List-I with List-II: List-I (A) Fleming (B) Robert Brown (C) George Palade (D) Camillo Golgi List-II (I) Disc shaped sacs or cisternae near cell nucleus (II) Chromatin (III) Ribosomes (IV) Nucleus Choose the correct answer from the options given below:
A. A-II, B-IV, C-III, D-I  ✓ Correct
B. A-II, B-III, C-I, D-IV
C. A-I, B-II, C-III, D-IV
D. A-IV, B-II, C-III, D-I
Solution: Fleming gave the name chromatin to the nuclear material stained by basic dyes (II). Robert Brown first described the nucleus as a cell organelle, in 1831 (IV). George Palade first observed ribosomes as dense particles under the electron microscope (III). Camillo Golgi first observed the densely stained, disc-shaped sacs or cisternae near the nucleus that now bear his name (I).
Q7 — Biology · medium · theory
Match List-I with List-II: List-I (Type of Inheritance) (A) Incomplete dominance (B) Co-dominance (C) Pleiotropy (D) Polygenic inheritance List-II (Example) (I) Blood groups in Human (II) Flower colour in Antirrhinum (III) Skin colour in human (IV) Phenylketonuria Choose the correct answer from the options given below:
A. A-III, B-IV, C-II, D-I
B. A-II, B-I, C-IV, D-III  ✓ Correct
C. A-II, B-III, C-I, D-IV
D. A-IV, B-I, C-III, D-II
Solution: Incomplete dominance — flower colour in Antirrhinum, where the heterozygote is pink (II). Co-dominance — human blood groups, where I^A and I^B are both expressed (I). Pleiotropy — phenylketonuria, where a single gene affects several traits (IV). Polygenic inheritance — human skin colour (III).
Q8 — Biology · medium · theory
Which part of the ovule stores reserve food materials?
A. Nucellus  ✓ Correct
B. Integument
C. Placenta
D. Funicle
Solution: The nucellus is the mass of cells enclosed within the integuments, and it has abundant food reserves. The integuments are the protective layers themselves, the placenta is the region of the ovary the ovule is attached to, and the funicle is the stalk of that attachment.
Q9 — Biology · medium · theory
Which one of the following is not found in Gymnosperms?
A. Sieve cells
B. Albuminous cells
C. Tracheids
D. Vessels  ✓ Correct
Solution: Gymnosperms lack VESSELS in their xylem — water is conducted by tracheids alone. Their phloem has sieve cells with albuminous cells (rather than the sieve tubes and companion cells of angiosperms), so the other three are all present.
Q10 — Biology · medium · theory
Which one of the following is not included under in-situ conservation?
A. Wild-life sanctuary
B. Botanical garden  ✓ Correct
C. Biosphere reserve
D. National park
Solution: A botanical garden keeps plants outside their natural habitat under human care, which makes it EX-SITU conservation. Wildlife sanctuaries, biosphere reserves and national parks all protect species where they already live, so they are in-situ.
Q11 — Biology · medium · theory
Given below are two statements: Statement I: The Indian Government has set up GEAC, which will make decisions regarding the validity of GM research. Statement II: Biopiracy is the term used to refer to the use of bio-resources by native people. In the light of the above statements, choose the correct answer from the options given below:
A. Statement I is true but Statement II is false  ✓ Correct
B. Statement I is false but Statement II is true
C. Both Statement I and Statement II are true
D. Both Statement I and Statement II are false
Solution: Statement I is true — GEAC (the Genetic Engineering Approval Committee) makes decisions on the validity of GM research and the safety of introducing GM organisms. Statement II is false: biopiracy is the use of bio-resources by MULTINATIONAL COMPANIES and other organisations without proper authorisation from the countries and people concerned — not use by native people themselves.
Q12 — Biology · medium · theory
Pollen grains remain preserved as fossils due to the presence of:
A. Epidermal layer
B. Tapetum
C. Exine layer  ✓ Correct
D. Intine layer
Solution: The exine is the hard outer layer of the pollen grain, made of sporopollenin — the most resistant organic material known. It withstands high temperatures and strong acids and alkalis, which is why pollen grains survive so well as fossils.
Q13 — Biology · medium · theory
Identify the incorrect pair:
A. Sphenopsida − Adiantum  ✓ Correct
B. Pteropsida − Dryopteris
C. Psilopsida − Psilotum
D. Lycopsida − Selaginella
Solution: Adiantum (the maidenhair fern) belongs to the class PTEROPSIDA, not Sphenopsida. Equisetum is the standard example of Sphenopsida. The other three pairings are correct.
Q14 — Biology · medium · theory
Which of the following is the correct match?
A. Gymnosperms : Cedrus, Pinus, Sequoia  ✓ Correct
B. Angiosperms : Wolffia, Eucalyptus, Sequoia
C. Bryophytes : Polytrichum, Polysiphonia, Sphagnum
D. Pteridophytes : Equisetum, Ginkgo, Adiantum
Solution: Cedrus, Pinus and Sequoia are all gymnosperms — that grouping is correct. The others each contain an intruder: Sequoia is a gymnosperm, not an angiosperm; Polysiphonia is a red alga, not a bryophyte; and Ginkgo is a gymnosperm, not a pteridophyte.
Q15 — Biology · medium · theory
Given below are two statements regarding RNA polymerase in prokaryotes: Statement I: In prokaryotes, RNA polymerase is capable of catalysing the process of elongation during transcription. Statement II: RNA polymerase associates transiently with 'Rho' factor to initiate transcription. In the light of the above statements, choose the correct answer from the options given below:
A. Statement I is true but Statement II is false  ✓ Correct
B. Statement I is false but Statement II is true
C. Both Statement I and Statement II are true
D. Both Statement I and Statement II are false
Solution: Statement I is true — on its own, RNA polymerase can only catalyse elongation. Statement II names the wrong factor: the polymerase associates transiently with the SIGMA ($\sigma$) factor to INITIATE transcription. The rho ($\rho$) factor is the one involved in TERMINATION.
Q16 — Biology · medium · theory
Which of the following is a nucleotide?
A. Uridine
B. Adenylic acid  ✓ Correct
C. Guanine
D. Guanosine
Solution: A nucleotide is a nitrogenous base + sugar + PHOSPHATE. Adenylic acid (adenosine monophosphate) has all three. Uridine and guanosine are nucleosides (base + sugar, no phosphate), and guanine is just a nitrogenous base.
Q17 — Biology · medium · theory
Match List-I with List-II: List-I (A) Vexillary aestivation (B) Epipetalous stamens (C) Epiphyllous stamens (D) Perigynous flower List-II (I) Brinjal (II) Peach (III) Pea (IV) Lily Choose the correct answer from the options given below:
A. A-III, B-I, C-IV, D-II  ✓ Correct
B. A-III, B-IV, C-I, D-II
C. A-III, B-II, C-I, D-IV
D. A-II, B-I, C-IV, D-III
Solution: Vexillary (papilionaceous) aestivation — pea (III). Epipetalous stamens, attached to the petals — brinjal (I). Epiphyllous stamens, attached to the perianth — lily (IV). Perigynous flower — peach (II).
Q18 — Biology · medium · theory
Match List-I with List-II: List-I (A) China rose (B) Mustard (C) Primrose (D) Marigold List-II (I) Free central (II) Basal (III) Axile (IV) Parietal Choose the correct answer from the options given below:
A. A-IV, B-III, C-II, D-I
B. A-II, B-III, C-IV, D-I
C. A-III, B-IV, C-I, D-II  ✓ Correct
D. A-III, B-IV, C-II, D-I
Solution: China rose — axile placentation (III). Mustard — parietal placentation (IV). Primrose — free central placentation (I). Marigold — basal placentation (II).
Q19 — Biology · medium · theory
Which of the following helps in maintenance of the pressure gradient in sieve tubes?
A. Albuminous cells
B. Sieve cells
C. Phloem parenchyma
D. Companion cells  ✓ Correct
Solution: Companion cells are closely associated with the sieve tube elements through pit fields and plasmodesmata, and it is they that help maintain the pressure gradient in the sieve tubes — which drives translocation.
Q20 — Biology · medium · theory
Mesosome in a cell is a:
A. Membrane bound vesicular structure
B. Chain of many ribosomes attached to a single mRNA
C. Special structure formed by extension of plasma membrane  ✓ Correct
D. Medium sized chromosome
Solution: A mesosome is a special membranous structure formed by the extension of the plasma membrane into the bacterial cell, in the form of vesicles, tubules and lamellae. It helps in cell wall formation, DNA replication and distribution to daughter cells, respiration and secretion.
Q21 — Biology · medium · theory
Match List-I with List-II: List-I (A) Abscisic acid (B) Ethylene (C) Gibberellin (D) Cytokinin List-II (I) Promotes female flowers in cucumber (II) Helps seeds to withstand desiccation (III) Helps in nutrient mobilization (IV) Promotes bolting in beet, cabbage etc. Choose the correct answer from the options given below:
A. A-II, B-III, C-IV, D-I
B. A-III, B-II, C-I, D-IV
C. A-II, B-I, C-IV, D-III  ✓ Correct
D. A-II, B-I, C-III, D-IV
Solution: Abscisic acid — helps seeds to withstand desiccation (II). Ethylene — promotes female flowers in cucumber (I). Gibberellin — promotes bolting in beet, cabbage and similar plants (IV). Cytokinin — helps in nutrient mobilisation (III).
Q22 — Biology · medium · theory
Match List-I with List-II: List-I (A) Genetically engineered Human Insulin (B) GM Cotton (C) ADA Deficiency (D) ELISA List-II (I) Gene therapy (II) E. coli (III) Antigen-antibody interaction (IV) Bacillus thuringiensis Choose the correct answer from the options given below:
A. A-III, B-II, C-IV, D-I
B. A-II, B-I, C-IV, D-III
C. A-IV, B-III, C-I, D-II
D. A-II, B-IV, C-I, D-III  ✓ Correct
Solution: Genetically engineered human insulin — produced using E. coli as the host (II). GM cotton — created using genes from Bacillus thuringiensis (IV). ADA deficiency — treated by gene therapy (I). ELISA — based on antigen-antibody interaction (III).
Q23 — Biology · medium · theory
Match List-I with List-II: List-I (A) ETS Complex I (B) ETS Complex II (C) ETS Complex III (D) ETS Complex IV List-II (I) NADH Dehydrogenase (II) Cytochrome bC₁ (III) Cytochrome C Oxidase (IV) Succinate Dehydrogenase Choose the correct answer from the options given below:
A. A-IV, B-I, C-III, D-II
B. A-I, B-IV, C-II, D-III  ✓ Correct
C. A-III, B-I, C-IV, D-II
D. A-I, B-II, C-IV, D-III
Solution: ETS Complex I — NADH dehydrogenase (I). ETS Complex II — succinate dehydrogenase (IV). ETS Complex III — the cytochrome bc₁ complex (II). ETS Complex IV — the cytochrome c oxidase complex (III).
Q24 — Biology · medium · theory
Cryopreservation technique is used for:
A. Protection of environment
B. Protection of Biodiversity hotspots
C. Preservation of gametes in viable and fertile condition for a long period  ✓ Correct
D. In-situ conservation
Solution: Cryopreservation is a form of ex-situ conservation in which gametes of threatened species are preserved in viable and fertile condition for long periods, using liquid nitrogen at very low temperature.
Q25 — Biology · medium · theory
Which of the following are correct about cellular respiration? A. Cellular respiration is the breaking of C−C bonds of complex organic molecules by oxidation. B. The entire cellular respiration takes place in Mitochondria. C. Fermentation takes place under anaerobic condition in germinating seeds. D. The fate of pyruvate formed during glycolysis depends on the type of organism also. E. Water is formed during respiration as a result of O₂ accepting electrons and getting reduced. Choose the correct answer from the options given below:
A. A, C, D, E only  ✓ Correct
B. A, B, E only
C. A, B, C, E only
D. B, C, D, E only
Solution: B is the wrong one: the ENTIRE process does not take place in the mitochondria — glycolysis, the first stage, occurs in the cytoplasm. A, C, D and E are all correct.
Q26 — Biology · medium · theory
Given below are two statements: Statement I: In eukaryotes there are three RNA polymerases in the nucleus in addition to the RNA polymerase found in the organelles. Statement II: All the three RNA polymerases in eukaryotic nucleus have different roles. In the light of the above statements, choose the correct answer from the options given below:
A. Statement I is correct but Statement II is incorrect
B. Statement I is incorrect but Statement II is correct
C. Both Statement I and Statement II are correct  ✓ Correct
D. Both Statement I and Statement II are incorrect
Solution: Both are correct. Eukaryotes have three RNA polymerases in the nucleus, over and above the one found in the organelles. And there is a clear division of labour: RNA polymerase I transcribes rRNAs (28S, 18S and 5.8S), RNA polymerase II the precursor of mRNA, and RNA polymerase III tRNA, 5S rRNA and snRNAs.
Q27 — Biology · medium · theory
Match List-I with List-II: List-I (A) Histones (B) Nucleosome (C) Euchromatin (D) Heterochromatin List-II (I) Loosely packed chromatin (II) Densely packed chromatin (III) Positively charged basic proteins (IV) DNA wrapped around histone octamer Choose the correct answer from the options given below:
A. A-IV, B-III, C-II, D-I
B. A-III, B-I, C-IV, D-II
C. A-II, B-III, C-IV, D-I
D. A-III, B-IV, C-I, D-II  ✓ Correct
Solution: Histones — positively charged basic proteins, rich in lysine and arginine (III). Nucleosome — DNA wrapped around a histone octamer (IV). Euchromatin — loosely packed, transcriptionally active chromatin (I). Heterochromatin — densely packed, transcriptionally inactive chromatin (II).
Q28 — Biology · medium · theory
Given below are two statements: Statement I: Failure of segregation of chromatids during cell cycle resulting in the gain or loss of whole set of chromosome in an organism is known as aneuploidy. Statement II: Failure of cytokinesis after anaphase stage of cell division results in the gain or loss of a chromosome is called polyploidy. In the light of the above statements, choose the correct answer from the options given below:
A. Statement I is true but Statement II is false
B. Statement I is false but Statement II is true
C. Both Statement I and Statement II are true
D. Both Statement I and Statement II are false  ✓ Correct
Solution: Both statements have their definitions crossed over. Failure of segregation of chromatids causes the gain or loss of a single CHROMOSOME (not a whole set) — that is aneuploidy. Failure of cytokinesis after TELOPHASE (not anaphase) results in an increase in a WHOLE SET of chromosomes — that is polyploidy.
Q29 — Biology · medium · theory
Recombination between homologous chromosomes is completed by the end of
A. Diakinesis
B. Zygotene
C. Diplotene
D. Pachytene  ✓ Correct
Solution: Crossing over between the non-sister chromatids of homologous chromosomes takes place during pachytene, catalysed by recombinase at the recombination nodules, and recombination is completed by the end of that stage.
Q30 — Biology · medium · theory
Match List-I with List-II: List-I (A) Metacentric chromosome (B) Sub-metacentric chromosome (C) Acrocentric chromosome (D) Telocentric chromosome List-II (I) Chromosome has a terminal centromere (II) Middle centromere forming two equal arms of chromosome (III) Centromere is slightly away from the middle of chromosome resulting into two unequal arms (IV) Centromere is situated close to its end forming one extremely short and one very long arm Choose the correct answer from the options given below:
A. A-II, B-I, C-IV, D-III
B. A-IV, B-I, C-II, D-III
C. A-I, B-II, C-III, D-IV
D. A-II, B-III, C-IV, D-I  ✓ Correct
Solution: Metacentric — the centromere sits in the middle, giving two equal arms (II). Sub-metacentric — the centromere is slightly away from the middle, giving two unequal arms (III). Acrocentric — the centromere is close to one end, giving one extremely short and one very long arm (IV). Telocentric — the centromere is terminal (I).