In a village of 4000 individuals, the frequency of the recessive allele ‘a’ for a genetic disorder is 0.2. Assuming Hardy–Weinberg equilibrium, calculate the expected number of individuals who are homozygous recessive (aa).
Answer
Number of aa individuals = 4000 × (0.2)^2 = 160.
The correct answer is Number of aa individuals = 4000 × (0.2)^2 = 160. — Thus, 160 individuals are expected to be homozygous recessive.
Step-by-step solution
- Identify the recessive allele frequency q = 0.2.
- Calculate the genotype frequency of aa as q^2 = (0.2)^2 = 0.04.
- Multiply the genotype frequency by the total population: 0.04 × 4000 = 160.
- Thus, 160 individuals are expected to be homozygous recessive.
Common mistake: Using the dominant allele frequency p instead of q or forgetting to square q
Concept tested: Hardy–Weinberg genotype frequencies
Published · How this solution was prepared and checked
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