Correct option is D
The Hardy-Weinberg equation states:
p2 +2pq + q 2 = 1 and p + q = 1
where:
- p2 = Frequency of homozygous dominant (normal) individuals
- 2pq2pq2pq = Frequency of heterozygous carriers
- q2 = Frequency of homozygous recessive individuals (affected by cystic fibrosis)
- ppp and qqq are the allele frequencies of the dominant and recessive alleles, respectively.
Cystic fibrosis - f(aa) = 0.002
q2= 0.002
q = 0.04
p = 1 - 0.04
p = 0.96
(carrier ) 2pq =
=
= 7.68 answer

