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With a population in Hardy-Weinberg equilibrium, the A allele has a frequency of 0.60, and the frequency of the a allele is 0.40. What is the frequency of the heterozygote

Answer :

MARIANAEGARCIAPEREDOGO

Following the Hardy-Weinberg equilibrium theory, the frequency of the heter0zyg0us genotype is 2pq. In the exposed example, 2pq = 0.48.

Hardy-Winberg equilibrium

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

Assuming a diallelic gene,

→  The allelic frequencies are

  • p is the frequency of the dominant allele,
  • q is the frequency of the recessive allele.

→  The genotypic frequencies after one generation are

  • 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

If the allele A has a frequency of 0.6, and the allele B has a frequency of 0.4, then the frequency of the heter0zyg0us genotype is

2pq = 2 x 0.6 x 0.4 = 0.48

You can learn more about the Hardy-Weinberg equilibrium at

https://brainly.com/question/3406634

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