Cystic fibrosis is caused by a recessive allele. Roughly one out of every 500 individuals (0.20%) have this disease. Using the Hardy-Weinberg equation, the percentage of individuals who are carriers of the recessive allele for the disease is:​

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
• $2pq$ = Frequency of heterozygous carriers
• q² = Frequency of homozygous recessive individuals (affected by cystic fibrosis)
• $p$ and $q$ 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 = $$2 \times 0.9553 \times 0.0447$$

= $$0.0768 \times 100$$

=  7.68  answer