Question

Assume that a trait is X-linked and recessive.  Nine percent of the females in a given population express the phenotype caused by this gene.  A) What percentage of males would be expected to express this trait? B) What percentage of the females would be heterozygous for the gene?

Answer 1

For X linked recessive trait,

Normal = XX, XY, XcX (carrier)

Affected = XcXc, XcY

XcXc = 9% = 0.09

Xc = √0.09 = 0.3

As per HWE, X + Xc = 1

X = 1 - 0.3 = 0.7

A. Affected males = XcY = 0.3 = 30%

B. Heterozygous females = XcX = 2 × 0.3 × 0.7 = 0.42 = 42%

According to Hardy weinberg equilibrium, the sum of allele frequencies and genotypic frequencies remain constant across generations in a population if no mutation, migration, natural selection and genetic drift are acting on the population. The population should be large in size and must show random mating.

Please give a good rating.