In a population at hardy-weinberg equilibrium with two alleles, a and a, 264 of the 1200 individuals display the recessive trait (aa). what is the estimated frequency of the a allele?

Assuming that this population is in Hardy-Weinberg equilibrium, the estimated frequency of the a allele is 0.47. And the estimated frequency of the A allele is 0.53.

The Hardy-Weinberg equilibrium theory states that the allelic frequencies in a locus are represented as p and q. Assuming a diallelic gene,

•p ⇒ frequency of the dominant allele,

•q ⇒ frequency of the recessive allele.

•p² (H0m0zyg0us dominant genotypic frequency),

•2pq (Heter0zyg0us genotypic frequency),

•q² (H0m0zyg0us recessive genotypic frequency).

If a population is in H-W equilibrium, it gets the same allelic and genotypic frequencies generation after generation.

The addition of the allelic frequencies equals 1

p + q = 1.

The sum of genotypic frequencies equals 1

p² + 2pq + q² = 1

According to this framework, to get the allelic frequency, first we should take the genotypic frequency of the recessive trait, F(aa).

F(aa) = q² = 264 / 1200 = 0.22

Now, we can get the allelic frequency f(a) = q

f(a) = q = √q² = √0.22 = 0.469 ≅ 0.47

Fynally, we can get f(A) = p by clearing the following equation,

p + q = 1

p = 1 - q

p = 1 - 0.47

p = 0.53

The estimated frequency of the a allele is 0.47.

The estimated frequency of the A allele is 0.53.

You can learn more about the Hardy-Weinberg equilibrium at

link

meiosis is the first to occur after the nuclei of opposite of opposite mating types fuse

your answer is going to be physical weathering