# Hardy-Weinberg Color-blind

In a city, 4% of male population have color blindness. How many of the female are (a) color blind carrier, (b) color blind? Suppose the city holds Hardy Weinberg equilibrium.

My progress: 4% of male have color blind => $p=F(cb~allele)=0.04$ and therefore $q=F(not~cb)=0.96$. Since HW equilibrium stand, we get the allele frequency among female is the same as among male. Then (a) $2pq=2*0.04*0.96=0.0768=7.68\%$ and (b) $q^2=0.9216=92.16\%$.

Am I correct?

• en.wikipedia.org/wiki/… Commented Apr 8, 2015 at 11:16
• Are you given any other data, total population, sex population, genotype of colorblind allele ? Commented Apr 8, 2015 at 15:46
• @Macedon93 you don't need more data... HWE assumes organisms are diploid, only sexual reproduction occurs, generations are non overlapping, random mating, infinite population size, allele frequencies are equal in the sexes and there is no migration, mutation or selection. Color blindness is a trait known to be X-linked and recessive. Commented Apr 8, 2015 at 17:02

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 $p^2=0.04^2=0.0016=0.16\%$.