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In human populations, 4% of the individuals are homozygous recessive to a specific RFLP marker. What percentage of individuals do you expect to be heterozygous for this trait ?
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- 4%
- 8%
- 16%
- 32%
Correct Option: D
According to the Hardy–Weinberg’s principle, in the simplest case of a single locus with two alleles: the dominant allele is denoted A and the recessive a and their frequencies are denoted by p and q; freq(A) = p; freq(a) = q; p + q = 1. If the genotype frequencies are in Hardy–Weinberg proportions resulting from random mating, then we will have freq(AA) = p2 for the AA homozygotes in the population, freq(aa) = q2 for the aa homozygotes, and freq(Aa) = 2pq for the heterozygotes.
So, q2 = 0.04, q = 0.2, p = 1-0.2 = 0.8, p2 + q2 + 2pq = 1, 2pq = 1-(p2 + q2) = 0.32 = 32%
