Respuesta :
A whopping 32% of people have the heterozygous malaria resistance gene.
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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