hardy-weinberg question? tricky stuff!?

Question

Heterozygosity in sickle-cell anemia provides resistence against malaria. If 9% of an African population is born with a severe form of sickle-cell anemia (recessive), what percentage of the more resistant allele is present in that population?

Answer 1

AA = p(squared)
Aa = 2pq
aa = q(squared)
p + q = 1

First you need to interpret "9% have the disease". This means that 9% of the population have both recessive alleles, which means the frequency of aa is 0.09. You are asked to calculate the percentage (or frequency) of the resistant (or dominant allele), which is denoted A. So by using the above equations:

aa = 0.09, which can also be expressed as a(squared) = 0.09

a = square root of a(squared) = square root of 0.09 = 0.3

The recessive allele (a in this case) is always donated by q, and the dominant allele (A in this case) is always denoted by p. So therefore; a = q.

p + q = 1
p + 0.3 = 1
p = 1 - 0.3
p = 0.7

Since p = A, the frequency of the dominant allele (A) equals 0.7

Answer 2

What do you mean by "more resistant allele"?

Answer 3

AA = p²
Aa = 2pq
aa = q² = .09

q = 0.3