Again,if this population were in equilibrium,and if the sickle-cell allele is recessive,what proportion of the population should be susceptible to sickle-cell disease under typical conditions?
A)0.04
B)0.16
C)0.20
D)0.32
E)0.80

Question

Quizplus · Accepted Answer

The proportion of the population susceptible to sickle-cell disease (homozygous recessive) can be calculated using the Hardy-Weinberg principle, which states that the frequency of the genotype in a population can be represented as p^2 + 2pq + q^2 = 1, where p and q represent the frequency of the dominant and recessive alleles, respectively. If q^2 represents those with sickle-cell disease and q^2 = 0.04, it means 4% of the population is homozygous recessive and thus susceptible to sickle-cell disease under equilibrium conditions.

Again,if This Population Were in Equilibrium,and If the Sickle-Cell Allele