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NEET Genetics and Evolution MCQs with answers

431 questions on Genetics and Evolution for NEET 2027, 5 of them from previous NEET papers. Below are 15 to try; tap a question to see its answer and explanation.

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15 sample questions

easyNEET 2024

1. Central dogma states:

  1. RNA→DNA→Protein
  2. DNA→RNA→Protein
  3. Protein→DNA→RNA
  4. DNA→Protein→RNA
Show answer and explanation

Answer: (B) DNA→RNA→Protein

Transcription: DNA→RNA. Translation: RNA→Protein. Reverse transcription: RNA→DNA (retroviruses).

Quick tip: DNA→RNA→Protein. HIV uses reverse transcriptase for RNA→DNA.

NCERT Class 12, Chapter 6, page 110

mediumNEET 2024

2. DNA replication is semi-conservative because:

  1. Both strands new
  2. One old + one new strand
  3. Two forks
  4. Polymerase conserves energy
Show answer and explanation

Answer: (B) One old + one new strand

Meselson-Stahl (1958): each daughter DNA has one parental + one new strand.

Quick tip: Semi = half old, half new. Proved by Meselson-Stahl using ¹⁵N/¹⁴N.

NCERT Class 12, Chapter 6, page 108

mediumNEET 2023

3. ABO blood group is example of:

  1. Dominance
  2. Codominance and multiple allelism
  3. Incomplete dominance
  4. Epistasis
Show answer and explanation

Answer: (B) Codominance and multiple allelism

3 alleles (Iᴬ,Iᴮ,i) = multiple allelism. AB group = codominance.

Quick tip: ABO=codominance+multiple allelism. Sickle cell=codominance. 4 o'clock plant=incomplete dominance.

NCERT Class 12, Chapter 5, page 98

hardNEET 2022

4. Hardy-Weinberg equilibrium requires:

  1. Mutations
  2. Natural selection
  3. Large population, random mating
  4. Genetic drift
Show answer and explanation

Answer: (C) Large population, random mating

5 conditions: large population, random mating, no mutation, no selection, no gene flow.

Quick tip: p²+2pq+q²=1. Five conditions for equilibrium. Violations = evolution occurring.

NCERT Class 12, Chapter 7, page 137

easy

5. In humans, the sex of the child is determined by:

  1. The type of sperm that fertilises the egg
  2. The type of egg produced by the mother
  3. The number of autosomes in the egg
  4. Environmental temperature during development
Show answer and explanation

Answer: (A) The type of sperm that fertilises the egg

All eggs carry an X chromosome. Sperm carry either X or Y; an X-bearing sperm gives a girl (XX) and a Y-bearing sperm gives a boy (XY).

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6. A normal woman whose father was colour-blind marries a colour-blind man. What is the probability that their first child will be a colour-blind daughter?

  1. 25%
  2. 0%
  3. 50%
  4. 75%
Show answer and explanation

Answer: (A) 25%

The woman receives an X chromosome carrying the colour-blind allele from her father, making her a carrier (XᶜX). The man is colour-blind (XᶜY). The possible offspring are XᶜXᶜ (colour-blind female), XᶜX (carrier female), XᶜY (colour-blind male), and XY (normal male). The probability of a child being both female and colour-blind is 1/4 or 25%.

hard

7. In a human population that is in Hardy-Weinberg equilibrium, the frequency of an autosomal recessive condition is 1 in 10,000. What is the expected frequency of carrier (heterozygous) individuals in this population?

  1. 0.0099
  2. 0.0001
  3. 0.0198
  4. 0.0200
Show answer and explanation

Answer: (C) 0.0198

Given q² = 1/10000 = 0.0001, so q = 0.01. Since p + q = 1, p = 1 - 0.01 = 0.99. Frequency of carriers (heterozygotes) = 2pq = 2 × 0.99 × 0.01 = 0.0198.

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8. Out of 64 codons, how many code for amino acids?

  1. 61
  2. 64
  3. 60
  4. 20
Show answer and explanation

Answer: (A) 61

There are 64 codons; 61 code for amino acids and 3 (UAA, UAG, UGA) are stop codons that do not code for any amino acid.

easy

9. In a typical dihybrid cross in garden pea showing independent assortment, what proportion of the F₂ progeny is expected to be homozygous for both dominant traits?

  1. 2/16
  2. 1/16
  3. 4/16
  4. 9/16
Show answer and explanation

Answer: (B) 1/16

In the F₂ generation of a dihybrid cross (RrYy × RrYy), the genotype homozygous dominant for both characters is RRYY. Out of 16 possible combinations, only 1 individual has the RRYY genotype, giving a frequency of 1/16.

hard

10. The saying 'ontogeny recapitulates phylogeny' refers to:

  1. Lamarck's use and disuse
  2. Darwin's theory of natural selection
  3. Haeckel's idea that an organism's development repeats its evolutionary history
  4. de Vries's mutation theory
Show answer and explanation

Answer: (C) Haeckel's idea that an organism's development repeats its evolutionary history

Haeckel proposed that embryonic development (ontogeny) repeats evolutionary history (phylogeny); von Baer disagreed with this interpretation.

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11. The coding region of an mRNA, from the start codon AUG up to and including the stop codon, is 432 nucleotides long. The polypeptide translated from it has how many amino acids?

  1. 144
  2. 432
  3. 145
  4. 143
Show answer and explanation

Answer: (D) 143

432 nucleotides = 144 codons. The stop codon codes for no amino acid, so 144 − 1 = 143 amino acids.

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12. Assuming independent assortment, find the number of different phenotypes in the F₂ of a cross between two Aa individuals (complete dominance).

  1. 3
  2. 2
  3. 4
  4. 1
Show answer and explanation

Answer: (B) 2

With 1 heterozygous gene pair: gametes = 2¹ = 2, F₂ genotypes = 3¹ = 3, F₂ phenotypes = 2¹ = 2.

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13. Which of the following is an example of evolution observable in the short term due to human activity?

  1. Evolution of the horse
  2. Resistance of insects to DDT
  3. Origin of mammals from reptiles
  4. Adaptive radiation of Darwin's finches
Show answer and explanation

Answer: (B) Resistance of insects to DDT

Insect resistance to pesticides such as DDT is evolution under selection by human activity, observable within a few years.

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14. The coding region of an mRNA, from the start codon AUG up to and including the stop codon, is 660 nucleotides long. The polypeptide translated from it has how many amino acids?

  1. 660
  2. 221
  3. 220
  4. 219
Show answer and explanation

Answer: (D) 219

660 nucleotides = 220 codons. The stop codon codes for no amino acid, so 220 − 1 = 219 amino acids.

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15. Sickle-cell anaemia results from substitution of glutamic acid by valine at the sixth position of the beta globin chain, caused by a change of:

  1. GAG to GUG
  2. GUG to GAG
  3. AUG to UAG
  4. GAA to GAC
Show answer and explanation

Answer: (A) GAG to GUG

A single base substitution in the beta globin gene changes the codon GAG to GUG, replacing glutamic acid with valine. The disorder is autosomal recessive.

NCERT Class 12

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