Answer:
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%).
