Question

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if the frequency of ( p = 0.8 ), what is the frequency of the heterozygous genotype?
0.16
0.32
1.6
3.2

Explanation:

Step1: Find the frequency of q

According to the Hardy - Weinberg equilibrium \(p + q=1\). Given \(p = 0.8\), then \(q=1 - p\).
\(q=1 - 0.8=0.2\)

Step2: Calculate the frequency of the heterozygous genotype

The formula for the frequency of the heterozygous genotype (\(2pq\)) in the Hardy - Weinberg equilibrium.
Substitute \(p = 0.8\) and \(q = 0.2\) into \(2pq\).
\(2pq=2\times0.8\times0.2\)
\(2pq = 0.32\)

Answer:

0.32