If the frequency of males affected with an X-linked recessive trait in the human population is 0.20. What will be the frequency of affected females?

The correct option is A 0.04
We know that in humans, sex is determined by the XY system of sex determination, where females are homogametic (XX) and males are heterogametic (XY). As females have two X-chromosomes for a recessive trait to appear in them both the chromosomes must carry the mutated allele, whereas in males mutation in a single chromosome is enough.

Based on the above scenario two inferences can be are drawn

1. Since the males have only one X-chromosome the frequency of X-linked allele in the population is equal to the frequency of affected males
2. As females have two X-chromosomes the probability of any female having the mutated allele in question on both the chromosome will be $q^2$ (where q is the frequency of the allele)

So in the given question, the frequency of affected males is 0.20, which means q = 0.20
They asked for the frequency of affected females = $q^2$
= ${(0.20)}^2$
= 0.04

The frequency of affected females in the population is 0.04, which means 4% of the females are showing the phenotype.