Question

In a random mating population, the frequency of a disease-causing recessive allele is 80%. What would be the frequency of carrier individuals in the population?
A. 64%
B. 32%
C. 16%
D. 100%

Answer 1

Hint: Hardy Weinberg principle states that allele and genotype frequencies in a population will remain constant from generation to generation in the absence of other evolutionary influences. A carrier is an individual who carries and is capable of passing on a genetic mutation associated with a disease and may or may not display disease symptoms. Carriers are associated with diseases inherited as recessive traits.

Complete answer: The Hardy–Weinberg principle may be applied in two ways - either a population is assumed to be in Hardy–Weinberg proportions, in which the genotype frequencies can be calculated, or if the genotype frequencies of all three genotypes are known, they can be tested for deviations that are statistically significant.
According to the question,
The genotype frequency of recessive allele is (q) = (80% or 0.8).
We know from Hardy Weinberg principle that,
\[p^2\] + 2pq + \[q^2\] = 1
where,
\[p^2\] = dominant allele
\[q^2\] = recessive allele
2pq= carrier allele
Also, p + q = 1
Where,
p= dominant allele
q= recessive allele
The frequency of dominant allele is
$p = 1 - q$
$p = 1 - 0.8$
$p = 0.2$
Therefore, the frequency of the carrier individual= $2pq$ $= 2 X 0.8 X 0.2 = 0.32 or 32%$
So, the correct option is B, i.e., 32%

Note: The influences that can change Hardy Weinberg principle are genetic drift, mate choice, assortative mating, natural selection, sexual selection, mutation, gene flow, meiotic drive, genetic hitchhiking, population bottleneck, founder effect and inbreeding.