The frequency of two alleles in a gene pool is 0.1 (A) and .9 (a). What is the percentage in the population of heterozygous in individuals? of homozygous recessives?

Answers

Answer 1
Answer:

The frequency of two alleles in a gene pool is 0.1 (A) and .9 (a). The percentage in the population of heterozygous in individuals is 18%, and the percentage of homozygous recessives is 81%.

The frequency of two alleles in a gene pool is 0.1 (A) and .9 (a). Hence,

p = 0.1 and q = 0.9

The equation for the Hardy-Weinberg principle is:

p² + 2pq + q² = 1

p² + 2pq represents the frequency of heterozygous individuals and q² represents the frequency of homozygous recessives

Substituting the values:

p² + 2pq + q² = 11 × 0.01 + 2 × 0.1 × 0.9 + 0.81 = 1 = 0.01 + 0.18 + 0.81= 1

Therefore, the percentage of heterozygous in individuals is:

p² + 2pq = 0.18 × 100% = 18%

And the percentage of homozygous recessives is:

q² = 0.81 × 100% = 81%

Thus, the percentage in the population of heterozygous individuals is 18%, and the percentage of homozygous recessives is 81%.

Learn more about Hardy-Weinberg principle here:

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Answer 2
Answer: hey its p(A) = 0.1 and q(a) = 0.9
The percentage in the population of heterozygous individuals:
2pq = 2 x 0.1 x 0.9 = 0.18 = 18%


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