Question

You may assume the Sickle Cell locus is in Hardy Weinberg equilibrium. Also, the term carrier implies that the person has a particular allele in his/her genotype, but is not affected with the condition. If the proportion of an African population that are susceptible to Malaria is 0.688, what proportion of the population are carriers for the Sickle Cell allele (HbS)

Answer 1

The frequency of the population that are carriers is 0.284 or 28.4%

Step-by-step explanation:

According to Hardy-Weinberg equilibrium equation, p² + q² + 2pq = 1

Also, the sum of the allele frequencies for all the alleles at the locus, p + q = 1.

Where where p is the frequency of the dominant allele and q is the frequency of the recessive allele in the population.

p² = dominant h0m0zygous genotype frequency

2pq = heterozygous genotype frequency

q² = recessive h0m0zygous genotype frequency

From the given data; the dominant h0m0zygous genotype, i.e., the proportion of an African population that are susceptible to Malaria, p² = 0.688

p = √0.6880 = 0.829

But p + q = 1; q = 1 - p

Since p = 0.829, q = 1 - 0.829

q = 0.171

Frequency carriers for Sickle Cell allele, i.e., the heterozygous genotype = 2pq

Frequency of carriers = 2 × 0.829 × 0.171 = 0.284

Therefore, the frequency of the population that are carriers is 0.284 or 28.4%