In a population of 800 males 120 are found to have color bli

In a population of 800 males, 120 are found to have color blindness, an X-linked recessive trait. Give your answer to 2 decimal places. For example, 0.12345 = .12 Calculate the frequency of males with the recessive phenotype. Calculate the frequency of males with the dominant phenotype. Calculate the approximate frequency of homozygous dominant females. Calculate the approximate frequency of female. Calculate the approximate frequency of homozygous recessive females.

Solution

Answer:

Total population = 800

color blind male frequency = colorblind allele frequency = q (becacuse, males have only one X-chromosome)

The frequency of males with the recessive phenotyp = q= 120/800 = 0.15 = 15%

p + q = 1

p = 1-q= 1-0.15 = 0.85 or 85%

The frequency of males with the dominant phenotyp = p= 0.85 or 85%

The approximate frequency of homozygous dominant females = 0.85 * 0.85 = 0.72

The approximate frequency of female heterozygote = 2 * 0.85 * 0.15 = 0.26

The approximate frequency of homozygous recessive females = 0.15 * 0.15 =0.02