Question

Genetic

The frequency of a harmless, X-linked recessive allele is 0.1 in the males of a population. Showing your calculations, determine the genotypic and phenotypic frequencies, by gender, of a population at equilibrium

Answer 1

The males are hemizygous for X chromsomes (which means they have only one x chromosome), where as the females are homozygous for X chromosomes(which means they have two x chromsomes).

The q or receissieve allele frequency = 0.1, so the domianant allele frequency i.e. p = 1 -q = 1-0.1 = 0.9

p = 0.9; q=0.1

Genotypic frequencies:

Males with recessieve allele frequency = 0.1

Males with dominant allele frequency = 0.9

Female with dominant allele frequency = pp = 0.9*0.9 = 0.81

Females with recesseive allele freequency = qq = 0.1*0.1 = 0.01

Females with heterozygous condition = 2pq = 2*0.9*0.1 = 0.09

The phenotypic frequencies:

As it is harm less it can not express in males and females, so, it won't effect any phenotypic differences.

Males with normal phenotype = 1;

Female with normal phenotype = 1.