In a certain group of African people, 4% are born with sickle-cell disease, an autosomal recessive disorder. Heterozygous individuals not only don't have sickle-cell disease, but also are resistant to malaria. If this group is in Hardy-Weinberg equilibrium, what percentage of the population is heterozygous and resistant to malaria? A) 45% B) 18% C) 32% D) 80%

the correct answer is C) 32%

Explanation:

Sickle-cell anaemia is an autosomal recessive genetic disorder. Individuals with the homozygous recessive have sickle-shaped blood cells. Whereas, individuals with heterozygous are only carrier of sickle cell trait. The carrier individuals are resistant to malarial parasite and do not have malaria.

As per the question, 4% of an African population is born with sickle-cell disease, then the percentage of the population is heterozygous and resistant to malaria will be:

Hardy-Weinberg formula is equilibrium,

p² + 2pq + q² = 1

p = frequency of the dominant allele in the population

q = frequency of the recessive allele in the population

p2 = percentage of homozygous dominant individuals

q2 = percentage of homozygous recessive individuals

2pq = percentage of heterozygous individuals

Given, homozygous recessive for this gene (q2) is 4% which is 0.04, the square root (q) is 0.2 (20%) then p should be 1-0.2 = 0.8 (20%).

Thus, the frequency of heterozygous individuals = 2pq.

2 (0.8 x 0.2) = 0.32 (32%).