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Nine percent of a population is homozygous recessive (aa) at a certain locus. Assuming that the population is in Hardy-Weinburg equilibrium, which of the following is closest to the frequency of the recessive allele (a)?a. 0.09b. 0.27c. 0.30d. 0.90e. 0.97

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Answer:

The frequency of the recessive allele is C: 0.3.

Step-by-step explanation:

According to Hardy-Weinberg, the allelic frequencies in a locus are represented as p and q, referring to the allelic dominant or recessive forms. The genotypic frequencies after one generation are (Homozygous dominant), 2pq (Heterozygous), (Homozygous recessive). Populations in H-W equilibrium will get the same allelic frequencies generation after generation. The sum of these allelic frequencies equals 1, this is p + q = 1.

In the exposed example, the genotypic frequency for homozygous recessive individuals () in the population is aa = 9% = 0.09

This is the proportion or frequency of the recessive genotype in a population, denoted as (aa) that equals .

Allelic frequency for the trait is a, wich equals q. And as we know the genotypic frequency, aa or q², we need to calculate the sqare root of this value: √a² = √0.09 = 0.3

This is the proportion of a specific allele, a, respect to the pool of alleles. Results are expressed in proportions, so the sum of every allelic frequency for one locus equals one.

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