If 20% of the individuals in a population have sickle-cell anemia, according to the Hardy Weinberg equation, what proportion of the population should be sickle-cell carriers?

According to the Hardy Weinberg equation, the proportion of individuals who carry a sickle-cell allele (but do not have sickle-cell anemia) can be calculated using the equation: 2pq = proportion of heterozygotes Where p = frequency of the sickle-cell allele And q = frequency of the normal allele Since sickle-cell anemia has a frequency of 0.2 (or 20%), the frequency of the sickle-cell allele (p) can be calculated as the square root of 0.2 (or 0.447). The frequency of the normal allele (q) can be calculated as 1 - p, which equals 0.553. Plugging these values into the Hardy Weinberg equation: 2pq = 2(0.447)(0.553) = 0.494 Therefore, \approx imately 49.4% of the population should be sickle-cell carriers.