The incidence of cystic fibrosis, a recessive genetic disorder in the Caucasian population of United States, is 1 in every 2,500 individuals. Find the number of heterozygous carriers.
(p + q = 1, p2 + 2pq + q2 = 1)
As hardy Weinberg theorem formula is given below freq. of recessive allele/ q2 applying the values = 1/2500 q2= 0.0004 now take the square root so q=0.02 now q=p-1 q=0.98 As the frequency. of heterozygous individuals = 2pq putting values we get the answer = 0.04 so correct option is C hope it helps
