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

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%

idk

explanation:

"do a testcross with a homozygous recessive plant."