Question

Cystic fibrosis is an autosomal recessive disorder affecting 1 in 2,500 newborns. At equilibrium, what % of individuals is expected to be homozygous for the wt (non-mutant) allele?

Answer 1

The percentage of individuals expected to be homozygous for the wild-type allele in cystic fibrosis is 4%.

Step-by-step explanation:

The percentage of individuals expected to be homozygous for the wild-type (non-mutant) allele of cystic fibrosis can be determined using the Hardy-Weinberg equilibrium equation. Cystic fibrosis (CF) is an autosomal recessive disorder. In this case, the frequency of the wild-type allele (q) can be calculated by taking the square root of the frequency of affected individuals (p), which is 1 in 2,500. This gives us q = 1/50. To find the percentage of homozygous wild-type individuals, we square q and multiply by 100, which gives us 4%.