- 32%
Explanation:
According to the Hardy Weinberg law, the allele and genotype frequencies in a population remain constant in the absence of factors responsible for evolution. It states that the sum of all genotype frequencies can be represented as the binomial expansion of the square of the sum of p and q.
This sum is equal to one : ( p + q )2 = p2 + 2pq + q2 = 1.
"p" is the frequency of the dominant allele, and "q" is the frequency of the recessive allele. The "2pq" in the equation shows the frequency of heterozygotes in the population.
According to question, if it is a recessive disease then "p" is the frequency of normal dominant allele, "q" will be the frequency of disesease causing recessive allele and "2pq" will be the frequency of carrier individual.
In the question the frequency of disease causing recessive allele is given i.e q = 80% or o.8.
If p + q = 1 then, p = q - 1 => 1 - 0.8 = 0.2 or 20%.
Thus, the frequency of carrier will be = 2 × p × q => 2 × 0.8 × 0.2 = 0.32 or 32%.