Question

The frequency of one form of recessive x-linked color blindness is 5% among male individuals of a population. What is the expected frequency of this form of color blindness among females in the same population? What fraction of females would be heterozygous carriers?

Answer 1

Answer:

A: wild-type allele / a: color blind allele

Because color blindness is recessive and X-linked your assumption p=F(a)=4% is correct as men do only have one copy of the allele. Subsequently F(A)=q=1−p=0.96 is also correct. Therefore:

a) F(Aa)=2pq=7.68% is correct and b) is wrong, a is the color blind allele and F(a)=0.04 therefore it's p2=0.042=0.0016=0.16%.

92% color blind among females seems a bit high :).