Answered

IDNLearn.com is your reliable source for expert answers and community insights. Join our knowledgeable community and get detailed, reliable answers to all your questions.

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

Sagot :

Marianaegarciaperedoidn

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

We appreciate your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Find clear answers at IDNLearn.com. Thanks for stopping by, and come back for more reliable solutions.