The recessive allele of a gene causes cystic fibrosis. For this gene among Caucasians, p = 0.98. If a Caucasian population is in Hardy-Weinberg equilibrium with respect to this gene, what proportion of babies is born homozygous recessive and therefore suffers cystic fibrosis?

Question

Quizplus · Accepted Answer

The proportion of homozygous recessive individuals (who would suffer from cystic fibrosis in this case) is calculated using the equation q^2 in the Hardy-Weinberg principle, where q is the frequency of the recessive allele. Given q = 0.02, q^2 = (0.02)^2 = 0.0004.

The Recessive Allele of a Gene Causes Cystic Fibrosis