Question

Color blindness is an X-linked recessive trait in humans. Consider a population in which the X-linked gene that determines colorblindness is in Hardy-Weinberg equilibrium, so assume that the allele frequencies are the same in males and females. If 15% of males in this population have the allele for color blindness, then:

(A)what percent of females in the population have normal color vision, but are carriers of the colorblind allele?

(B)what percent of females in the population is color blind?

(C)what percent of females in the population have normal color vision and are non-carriers of the colorblind allele?

(D)what percent of males in the population have normal color vision?

Answer 1

Answer:

The frequency of X-linked recessive disease in males is equal to the X-linked recessive frequency as males have only one X-chromosome.

Diseased frequency in males = 15% = 0.15

Recessive allele frequency = q = 0.15

Dominant allele frequency = p = 1-q = 0.85

A). Percent of females in the population have normal color vision, but are carriers of the colorblind allele = 2pq*100 = 2*0.85*0.15*100 = 25.5%

B). Percent of females in the population is color blind = qq*100 = 0.15*0.15*100 = 2.25%

C). Percent of females in the population have normal color vision and are non-carriers of the colorblind allele = pp*100 = 0.85 * 0.85 *100 = 72.25%

D). Percent of males in the population have normal color vision = p*100= 0.85*100= 85%