Eight percent of a population shows a recessive trait. What would be the frequency of the heterozygotes for that trait if the population is in hardy=w

Population that show recessive trait should be only homozygous recessive. If the population in Hardy Weinberg equilibrium, then the recessive trait gene frequency would be:

homozygous recessive = aa = 8%=0.08

a = √0.08=0.283

a+A=1

A = 1-a = 1 - 0.283 = 0.717

The percentage of heterozygote in population would be:

heterozygote = 2 Aa = (2) (0.717) (0.283) = 0.4058 = 40.58%