Question

A population consists of 100 individuals of the following genotypes: 60 AA, 20 Aa, 20 aa. Is this population in Hardy-Weinberg equilibrium? Show work for possible partial credit.

Answer 1

No, this is not in the Hardy Weinberg equilibrium.

Explanation -

So, 'A' is the dominant allele and 'a' is the corresponding recessive allele.
Now, let 'p' be the frequency of the dominant allele and 'q' be the frequency of the recessive allele,
Also, 'p²' is the frequency of the homozygous dominant individuals (AA)
'q²' is the frequency of the homozygous recessive individuals (aa)
and '2pq' is the frequency of heterozygous individuals (Aa).

Here, p² = 60/100 = 0.6

So, p = square root of 0.6 = 0.77

Also, q² = 20/100 = 0.2

So, q = square root of 0.2 = 0.45

Now, according to the Hardy Weinberg equilibrium, we know that: p + q = 1.

But here, p + q = 0.77 + 0.45 = 1.22

This contradicts the Hardy Weinberg equilibrium.

Thus, the given data is not in Hardy Weinberg equilibrium.

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