Hardy-weinberg Principle

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In a population at Hardy-Weinberg equilibrium, if the frequency of the recessive allele is 0.2, what is the expected frequency of homozygous recessive individuals?

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The Hardy-Weinberg principle states that in a large, randomly mating population free of evolutionary influences, allele and genotype frequencies remain constant, following the relationship p^2 + 2pq + q^2 = 1. Here p and q are the frequencies of the dominant and recessive alleles, with q^2 representing the homozygous recessive genotype frequency. Given the recessive allele frequency q = 0.2, the homozygous recessive frequency is q^2 = 0.2 multiplied by 0.2 = 0.04. The option 0.16 corresponds to p^2 if p were 0.4, or a misread value, and does not match q^2. The value 0.32 actually equals 2pq, the heterozygous frequency (2 multiplied by 0.8 multiplied by 0.2 = 0.32), not the homozygous recessive class. The figure 0.20 is simply q itself, the allele frequency, not the genotype frequency, so squaring is required. As stated in NCERT Class 12, Chapter 7 (Evolution), the principle quantifies allele and genotype frequencies in non-evolving populations. As a numerical sanity check, p = 0.8 gives p^2 = 0.64, 2pq = 0.32, and q^2 = 0.04, and these sum to 1.0, confirming 0.04 as correct.

Hardy-weinberg Principle

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