In a tropical human population in Hardy-Weinberg equilibrium for an autosomal locus determining presence/absence of pigment in the skin,the frequency of albinism (aa)is 1 in 10,000.The frequency of heterozygotes is approximately:

Question

Quizplus · Accepted Answer

In Hardy-Weinberg equilibrium, the frequency of homozygous recessive individuals (aa) is given as p^2 where p is the frequency of the dominant allele (A) and q is the frequency of the recessive allele (a). Thus, p^2 = 1/10,000, or p = sqrt(1/10,000) = 0.01.

The frequency of heterozygous individuals (Aa) can be calculated using the equation 2pq, where p and q are the frequencies of the dominant and recessive alleles, respectively. Thus, 2pq = 2(0.01)(0.99) = 0.198, or about 1 in 50 individuals in the population are heterozygous for this autosomal locus. Therefore, the best answer is B) 1 in 50.

In a Tropical Human Population in Hardy-Weinberg Equilibrium for an Autosomal