The presence of freckles is due to a dominant allele. four percent of the individuals in a particular population lack freckles. use the hardy-weinberg formula to calculate the percentage of individuals in this population who are homozygous dominant for freckles

Let us assign the recessive allele for freckle to be f and the dominant F. According to the statement, we deduce that ff = 0.04.

The frequency of the recessive f allele is, therefore, is 0.2 (square root of 0.04 ).

Hardy-Weinburg p + q = 1

P + 0.2 = 1; p = 1- 0.2 = 0.8

Therefore using the Hardy-Weinburg equation of a population in equilibrium

P2+2pq+q2=1

Heterozygous individuals are;

2pq = 2*0.8*0.2 = 0.32

This is 32% of the population

the answer is c on e2020, bud.