Question

A mutation in one of the hemoglobin genes causes sickle cell anemia. The sickle cell allele, S, severely reduces fitness in people who are homozygotes, SS. In contrast, people with at least one normal hemoglobin allele, A, do not suffer the effects of sickle cell anemia, even if the individual is a heterozygote, AS. Interestingly, though, in areas with a high rate of malaria, heterozygotes that carry the sickle cell allele have a higher fitness than do individuals that are homozygous for the A allele. A single copy of the sickle cell allele provides some resistance against malaria. In areas with malaria, homozygotes of the sickle cell allele, S, still have very low fitness. Suppose that a population of 2000 people that is found in an area with a high prevalence of malaria contains 300 AA homozygotes, 1600 AS heterozygotes, and 100 SS homozygotes. From the number of individuals observed in each genotype, calculate the number of individuals you would expect to find in each genotype if the population is in Hardy-Weinberg equilibrium. expected number of AA individuals: expected number of AS individuals: expected number of SS individuals: Compare the observed number of individuals to the expected number of individuals in each genotype. Do the data seem to support the hypothesis that the population is in Hardy-Weindberg equilibrium? Yes, the expected number of individuals in each genotype are similar to the observed number of individuals. No, the expected number of individuals in each genotype are very different from the observed number of individuals.

Answer 1

p=A q=S Count the number of allele A and S Number of allele A = (300 X 2) + 1600 = 2200 Number of allele S = (100 x 2) + 1600 = 1800 Divide the number of each allele by the number of total alleles Frequency of allele A = 2200/4000 = 0.55 Use the equation p + q = 1 to calculate the frequency of S allele q = 1-p = 1-0.55 = 0.45 Frequency of allele S is 0.45 Now, calculate the frequency of p2 (AA) and q? (AS) Frequency of AA = 0.552 = 0.30 Frequency of SS = 0.452 = 0.20 Calculate the frequency of heterozygous genotype using 2pq 2pq = 2 x 0.55 x 0.45 = 0.49 Now, to calculate the number of individuals you must multiply the total number of individuals by the frequency of each genotype Expected Individuals AA = 0.30 x 2000 = 600 Expected Individuals AS = 0.49 x 2000 = 980 Expected individuals SS = 0.20 x 2000 = 400 In this case, we are not observing the same number of observed and expected individuals. However, both shows a similar pattern. The heterozygous individuals are more numerous in both observed and expected. The SS individuals represents the less number in both. So, we can say that Yes, the expected number of individuals in each genotype are similar to the observed number of individuals