Correct option is A
The Hardy-Weinberg equation states:
where:
- p2= Frequency of homozygous dominant (normal) individuals
- 2pq2pq2pq= Frequency of heterozygous carriers
- q2= Frequency of homozygous recessive individuals (affected)
- ppp andqq q are the allele frequencies of the dominant and recessive alleles, respectively.
Diseased - f(aa) = 0.002
q2= 0.002
q = 0.04
p = 1 - 0.04
p = 0.96
Answer is 7.68%

