Answer: If this population were in equilibrium and if the sickle-cell allele is recessive, the proportion of the population susceptible to sickle-cell anemia under typical conditions should be 0.20
Explanation: Hardy-Weinberg law provides an equation to relate genotype frequencies and allele frequencies in a randomly mating population. The equation is;
p² + 2pq + q² = 1
For 2 alleles such as A and a, where
p² = homozygous dominant
q² = homozygous recessive and
2pq = heterozygous
From the question, it is said that the sickle-cell allele (SS) constitutes 20% (that is, 20/100) of the hemoglobin alleles in the human gene pool and it is also said to be the homozygous recessive allele.
Therefore, q² = 20/100 = 0.20
