Question

The frequency of one form of X-linked color blindness is 5% among males. What is the expected frequency of this form of color blindness in females? What fraction of females would be heterozygous carriers?

Answer 1

The Hardy Weinberg equation is as follows:

p² + 2pq + q² = 1

p= frequency of the "A" allele= normal allele

q = frequency of the "a" allele in the population= Frequency of the male with the trait

p2 = the frequency of the homozygous genotype AA,

q2 = the frequency of the homozygous genotype aa,

2pq = the frequency of the heterozygous genotype Aa= carrier

Both the male and female can be color blind as the defective gene is carried on X chromosome.

q= 5/100=0.05

a) Frequency of females with color blindness= q²= (0.05)²= 0.05 *0.05= 0.0025

b) p+q=1

p= 1-0.05= 0.95

Frequency of carrier females= 2pq= 2* 0.95*0.05= 0.0095- 0.095*100=9.5%

9.5/100= 1/10.53

There is a 50% chance of a woman.

Fraction of women carrier= 1/10.53 * ½=1/21.06 =1/21