how do you do this problem? Can u plz explain me how? thnks?

Question

In a certain African human population, 4% of the population is born with a severe form of sickle-cell anemia. What percentage of this population enjoys the selective advantage of the sickle-cell gene by being more resistant to malaria?

Answer 1

use p squared+2pq+q squared
.04 is p squared because p is homozygous dominant which is sickle cell disease. square root is .2
since allele frequencies add to make 1 q is .8
2pq is heterozygotes which is malaria immune non sickle cell.
.8x.2x2=.32
32%

Answer 2

You have to use the Hardy-Weinberg ideas.

p^2 + 2pq = q^2 = 1

q^2 = sickle cell anemia = 0.04
q = 0.2

p + q = 1 says Hardy-W.
p = 0.8

Sickle cell trait is the heterozygous form of this trait. One allele for normal hemoglobin and one allele for sickle hemoglobin.

2pq = 2 * 0.8 * 0.2 = 0.32
32% of the population have the advantageous heterozygous condition.

Answer 3

I think the idea is that the gene has a certain frequency in the population. 4% of matings result in a child that gets a copy of the affected from both their mother and their father, and so has the severe form of anemia.

The Hardy-Weinberg law predicts genotype frequencies from allele frequencies under certain conditions, in which case:

frequency(AA) = p^2
frequency(Aa) = 2pq
frequency(aa) = q^2

AA are the 4 percent that have severe anemia (two copies of the gene).
Aa are the ones that have protection.

p^2 = 0.4 => p = 0.2, q = 0.8

Therefore frequency(Aa) = 2 x 0.2 x 0.8 = 0.32 = 32 percent