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In a population of 2000 individuals, the allele frequency of a recessive allele "r" is 0.4 and the allele frequency of the dominant allele "R" is 0.6. Assuming Hardy–Weinberg equilibrium, how many individuals are expected to be homozygous recessive (rr)?

hard 3 marks Class 12 · Biology · Principles of Inheritance and Variation
Answer
320

The correct answer is 320 — Round if necessary; here it is an integer.

Step-by-step solution

  1. Under Hardy–Weinberg, genotype frequency for rr is q² where q is the frequency of r.
  2. Compute q²: 0.4² = 0.16.
  3. Multiply by population size: 0.16 × 2000 = 320.
  4. Thus, 320 individuals are expected to be homozygous recessive.
  5. Round if necessary; here it is an integer.
Common mistake: Using allele frequency directly instead of genotype frequency

Concept tested: Hardy–Weinberg equilibrium

Published · How this solution was prepared and checked

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