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NEET Biology · Genetics and Evolution

Mechanism of evolution and the Hardy-Weinberg principle

Taught in NCERT Class 12 Biology, Chapter 6: Evolution.

5 verified questions: Easy 0 · Medium 4 · NEET level 1

Sample questions

Sample 1Medium

A bar chart shows the genotypes of a population in Hardy–Weinberg equilibrium for one gene with a dominant allele A and a recessive allele a. The bar for individuals showing the recessive phenotype (aa) reads 16%. What is the frequency of heterozygotes (Aa)?

  1. A

    0.24

  2. B

    0.36

  3. C

    0.16

  4. D

    0.48

Show the answer
The answer is D.

q² = 0.16, so q = 0.4 and p = 1 − 0.4 = 0.6. The heterozygote frequency is 2pq = 2 × 0.6 × 0.4 = 0.48.

  • A.

    This is pq; the heterozygote frequency is 2pq.

  • B.

    This is p², the frequency of AA homozygotes.

  • C.

    This is q², the frequency of aa individuals given in the chart.

NCERT: Class 12 Biology, Chapter 6, 6.7

Sample 2NEET level

In a large, randomly mating population in Hardy-Weinberg equilibrium, 16% of individuals show a recessive phenotype controlled by one gene with two alleles. What is the frequency of heterozygous individuals?

  1. A

    0.24

  2. B

    0.36

  3. C

    0.48

  4. D

    0.16

Show the answer
The answer is C.

The recessive homozygotes are q² = 0.16, so q = 0.4 and p = 1 − 0.4 = 0.6. Heterozygotes are 2pq = 2 × 0.6 × 0.4 = 0.48.

  • A.

    This is pq, without the factor of 2.

  • B.

    This is p², the frequency of dominant homozygotes.

  • D.

    This is q², the frequency of the recessive phenotype given in the question.

NCERT: Class 12 Biology, Chapter 6 (Evolution), Hardy-Weinberg principle

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