Answer:
p = 0.36
q = 0.64
p² = 0.13
2pq = 0.46
q² = 0.41
Explanation:
We need to use the Hardy-Weinberg equations:
p + q = 1
p² + 2pq + q² = 1
p: the frequency of the dominant allele
q: the frequency of the recessive allele
p²: the frequency of homozygous dominant
2pq: the frequency of heterozygous
q²: the frequency of homozygous recessive
Here, we know that 328/800 people are homozygous recessive, which means that q² = 328/800 = 0.41.
Then, q = √0.41 = 0.64, and p = 1 - q = 1 - 0.64 = 0.36.
Now, we have p² = (0.36)² = 0.1296 ≈ 0.13 and 2pq = 2 * 0.36 * 0.64 = 0.4608 ≈ 0.46.
The answers are:
p = 0.36
q = 0.64
p² = 0.13
2pq = 0.46
q² = 0.41
