A population of humans in Africa was studied for sickle cell

A population of humans in Africa was studied for sickle cell anemia. Recall that there are two alleles for the hemoglobin beta chain. H_b^S codes for the abnormal hemoglobin which causes sickle shaped red blood cells in homozygous individuals, i.e. sickle cell anemia. H_b^A codes for the \'normal\' hemoglobin. The population was sampled across two generations and both times the population size was 200. In generation 1 the frequency of H_b^S was 0.6. In generation 2 the frequency of H_b^S homozygotes is 0.0 and the frequency of heterozygotes is 0.6. What are the allelic frequencies of adults in generation 1? If there had been a random assortment of alleles, what genotypic frequencies would you expect in generation 2? What are the actual genotypic frequencies in generation 2 for all three genotypes? Taking into account Hardy-Weinberg expectations, what can you conclude about this population? Please explain.

Solution

a) Frequency of Hb^s   = 0.6 (Generation 1)

p + q = 1 (Sum of Allelic frequencies is equal to 1)

q = Hb^s = 0.6

p = 1 - q

p = 1 - 0.6

p = Hb^A = 0.4

b) Hardy Weinberg equilibrium equation is p2 + 2pq + q2

q² = Hb^S/ Hb^S= 0.6 x 0.6 = 0.36

p² = Hb^A / Hb^A = 0.4 x 0.4 = 0.16

2pq = Hb^A / Hb^S = 2 x 0.4 x 0.6 = 0.48

c) p2 (Hb^A/ Hb^A) = ?, q2 (Hb^S/ Hb^S ) = 0, 2pq (Hb^A Hb^S) = 0.6

p2 + 2pq + q2 = 1

p2 + 0.6 + 0 = 1

p2 = Hb^A/ Hb^A = 0.4

d) Hardy - Weinberg equation is that is the sum of allelic frequencies is equal to one and the sum of genotypic frequencies is equal to one. The expected genotype frequency of generation 2 is different from observed genotype frequency so the population does not obey Hardy - Weinberg equation.