# 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/… – MySky Apr 8 '15 at 11:16
• Are you given any other data, total population, sex population, genotype of colorblind allele ? – Macedon93 Apr 8 '15 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. – cagliari2005 Apr 8 '15 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 $p^2=0.04^2=0.0016=0.16\%$.