You have sampled a population in which you know that the percentage of the homozygous dominant genotype (AA) is 16%.
Using that 16%, calculate the following - the frequency of the "A" allele. (p)
p2 + 2pq + q2 = 1 and p + q = 1
a
1
b
0.4
c
0.36
d
0.16
b. 0.4
First, we know that the homozygous dominant genotype (AA) makes up 16% of the population, so p^2 = 0.16.
From p^2 + 2pq + q^2 = 1, we can substitute in the value of p^2 we found:
0.16 + 2pq + q^2 = 1
Since p + q = 1, we can rewrite q as (1-p). Substituting this into the equation:
0.16 + 2p(1-p) + (1-p)^2 = 1
0.16 + 2p - 2p^2 + 1 - 2p + p^2 = 1
-0.84 + p^2 = 0
Solving for p:
p^2 = 0.84
p = √0.84
p = 0.4
Therefore, the frequency of the "A" allele (p) is 0.4.