Question

Scenario #1

Answer 1

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