In a population of Mendel's garden peas, the frequency of the dominant A (yellow flower) allele is 80%. Let p represent the frequency of the A allele and q represent the frequency of the a allele. Assuming that the population is in Hardy-Weinberg equilibrium, what are the genotype frequencies?

If the frequency of the dominant allele in the pea population is 0.8, the genotype frequencies in the population would be 0.64 AA, 0.32 Aa, and 0.04 aa.

Explanation:

Hardy-Weinberg equation tells us that:

p2 + 2pq + q2

p2 is the frequency of AA plants

2pq is the frequency of Aa plants, and

q2 the frequency of aa individuals.

A allele in the population is 80%, the frequency is 0.8.

p represent the frequency of A allele. p is 0.8

Therefore calculating for q ( frequency of a allele).

p + q = 1.0

q = 1.0 − p

q = 1.0 − 0.8

q = 0.2

p = 0.8, q = 0.2.

Now we can calculate the predicted frequencies of the different genotypes, remembering that p2 is the frequency of the AA genotype, 2pq is the frequency of the Aa genotype, and q2 the frequency of aa genotype.

p2 = 0.8 x 0.8 = 0.64

2pq = 2 x 0.8 x 0.2 = 0.32

q2 = 0.2 x 0.2 = 0.04.

Thus, we would expect to see genotype frequencies of 0.64 AA, 0.32 Aa, and 0.04 aa in the pea plants.

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