Consider a plant population with the genotype frequencies: A/A – 0.5 A/a – 0.25

Question

Consider a plant population with the genotype frequencies: A/A – 0.5 A/a – 0.25 a/a – 0.25

a. Are these frequencies consistent with Hardy-Weinberg equilibrium? Show how you would determine this.

Does this population appear to be in Hardy-Weinberg equilibrium? Yes / No

b. The A allele is incompletely dominant, so each genotype has a similar but distinct seed shape phenotype. These differences offer varying levels of protection against rodent foraging. The relative fitnesses of the 3 genotypes are A/A = 1.0, A/a = 0.8, and a/a = 0.6. If you started with seeds with the genotype frequencies: AA – 0.36, Aa – 0.48 and aa – 0.16, what will the genotype frequencies be for the plants that grow from the seeds that avoid being eaten?

What are the allele frequencies for those plants?

Explanation / Answer

a. Yes. These frequencies are consistent with Hardy-Weinberg equilibrium.

Hardy-Weinberg equation is p2 + 2pq + q2 = 1

where p2 = Frequency of homozygous genotype AA = 0.5 given

q2 = Frequency of homozygous genotype aa = 0.25 given.

2pq = Frequency of heterozygous genotype Aa = 0.25 given.

So applying value in the equation we get,

0.5 + 0.25 + 0.25 = 1

1 = 1.

b. I am not sure about this answer.

May be genotype frequency AA-0.64, Aa-0.32 and aa-0.44.

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