In a Hardy-Weinberg population with two alleles that show complete dominance, the frequency of the recessive allele is 0.4. What percentage of the population has a homozygous genotype? A) 36 B) 4 C) 40 D) 16 E) 52

Question

Thaddeus Bray

Biology

6 January, 12:47

In a Hardy-Weinberg population with two alleles that show complete dominance, the frequency of the recessive allele is 0.4. What percentage of the population has a homozygous genotype? A) 36 B) 4 C) 40 D) 16 E) 52

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Franklin Williamson · Answer 1

The correct answer is E) 52

When there is Hardy-Weinberg equilibrium like in this case of a single locus with two alleles denoted A and a with frequencies f (A) = p and f (a) = q, the expected genotype frequencies under random mating are f (AA) = p² for the AA homozygotes, f (aa) = q² for the aa homozygotes, and f (Aa) = 2pq for the heterozygotes. Let's put that A is dominant and a is recessive allele. In Hardy-Weinberg equilibrium we have:

p²+2*p*q+q² = 1 p+q=1 p=1-q

f (a) = 0.4=q q²=0.4²=0.16=16%

p = 1-0.4=0.6 p²=0.6²=0.36=36%

Percentage of a homozygous genotype in the population is 16%+36%=52%