Sickle cell anemia is a recessive genetic disease that only appears in homozygous recessive (ss) individuals. It has also been discovered that homozygous carriers of the sickle cell gene (Ss) are more resistant to malaria. Assuming a population in Africa is under Hardy-Weinberg equilibrium, and 9% of the population has sickle cell anemia, what percentage of the population is more resistant to malaria? Right your answer as a whole number percentage. Do not include decimals.
You know that 9% of the population is homozygous recessive. In other words, q2 = 0.09. And you want to find the percentage that is heterozygous, or 2pq. Therefore,
p + q = 1
p + 0.3 = 1
p = 1 0.3 = 0.7
2pq = 2(0.7)(0.3) = 0.42 or 42%
Now just check to make sure your answer is right.
p2 + 2pq + q2 = 1
0.72 + 0.42 + 0.09=1