Question

Suppose that the X‑linked recessive condition hemophilia A affects 1 in 5000 males in the United States. Assume that the only two alleles present are normal and disease‑causing and that the population is in Hardy–Weinberg equilibrium. If the population of females in the United States is 160 million, how many females in the country do you expect to have the disease?

Answer 1

We have two alleles, XA and Xa and the genotypic frequencies are given by p^2 + 2pq + q^2 = 1

In this equation XA is p and Xa is q. Therefore p^2 is the frequency of the genotype XAXA, etc.

For X-linked conditions, females have 2 alleles for the gene, but males only have one. So calculate the genotypic frequencies for males and females separately. Since the population is in Hardy-Weinberg equilibruim, the allelic frequency is the same for both sexes.

Since males only have one X chromosome, the frequency of X linked in male is equal to the allelic frequency. So frequency just equals 1/5000 or 0.0002

so q^2 = q for males in X linked

So q = 0.0002

for females, she would need to have both alleles, so her frequency of the homozygous genotype is q^2 = 0.0002^2 = 0.00000004

number of affected females would be q^2 x number of females in the population

so 0.00000004 x 160 mil = 6.4 females would have hemophilia.

Step-by-step explanation: