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 hereditary condition known as sickle-cell anemia is autosomal recessive. Blood cells in homozygous recessive individuals have the sickle shape. Contrarily, the sickle cell trait is exclusively carried by heterozygous people. The carriers do not experience malaria because they are immune to the parasites that cause it.

If 4% of people in an African population are born with sickle cell disease, therefore the population's proportion of heterozygous people who are malaria resistant will be:

Equilibrium is the Hardy-Weinberg formula.

p² + 2pq + q² = 1

p = frequency of the population's dominant allele

q is the population's prevalence of the recessive allele.

p² = the proportion of homozygous dominant people

q² is the proportion of homozygous recessive people.

2pq = the proportion of heterozygous people

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

Therefore, the percentage of heterozygous people is 2pq.2 (0.8 x 0.2) = 0.32 (32%).

Learn more about Hardy-Weinberg equilibrium at

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