Following the Hardy-Weinberg equilibrium theory, the frequency of the heter0zyg0us genotype is 2pq. In the exposed example, 2pq = 0.48.
<h3>
Hardy-Winberg equilibrium</h3>
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 =<u> 0.48</u>
You can learn more about the Hardy-Weinberg equilibrium at
brainly.com/question/3406634
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Explanation:
