Question

The prevalence of a severe form of sickle-cell anemia (ss) in an African population is 16%. The percentage of the population resistant to malaria because of heterozygous (Ss) genotype for the sickle-cell gene is ............

Hint:

The Hardy-Weinberg equilibrium can be used to calculate genotype frequencies if the frequency of one allele is known.

Answer 1

Correct Answer: 47

Step 1: Understand the genetic setup.
In a population, the frequency of the sickle-cell allele (s) can be denoted as \( q \), and the frequency of the normal allele (S) is \( p \). According to Hardy-Weinberg equilibrium, the frequencies of the genotypes are: - \( p^2 \) for homozygous normal (SS), - \( 2pq \) for heterozygous (Ss), - \( q^2 \) for homozygous sickle-cell (ss).

Step 2: Use the prevalence of sickle-cell anemia to find \( q \).
The prevalence of the severe form of sickle-cell anemia, where the genotype is ss, is given as 16%. This means that: \[ q^2 = 0.16, \] so \( q = \sqrt{0.16} = 0.4 \).

Step 3: Calculate the percentage of heterozygous individuals.
The percentage of individuals who are resistant to malaria due to the heterozygous (Ss) genotype is given by \( 2pq \). Since \( p = 1 - q \), we calculate \( p \) as: \[ p = 1 - 0.4 = 0.6. \] Now, we calculate the heterozygous frequency: \[ 2pq = 2 \times 0.6 \times 0.4 = 0.48. \]

Step 4: Conclusion.
The percentage of the population resistant to malaria due to the heterozygous genotype is 48%, which represents 32% of the total population, as approximately half of those are resistant to malaria.