Answer:
p² = 0.36 or 36%
Step-by-step explanation:
Using the Hardy-Weinberg equation to solve this problem:
The basic equation is as follows:
p + q = 1.0 where p and q are the allele frequency and it is always equal to 1.
There are different variations for calculating the homozygous and heterozygous variations if deeper detail, but for this question we will stick to the basic equation for simplification purposes:
If the homozygous recessive is 16% then q²= to 0.16
This means that q is equal to 0.4
If q is equal to 0.4 then p is equal to 1.0-0.4
p is equal to 0.6
This makes p²=.36 or 36%