Question

You discover a population of mustard plants and find it to be at Hardy–Weinberg equilibrium with respect to the R locus. Suppose that there are two alleles at this locus and the frequency of one of those alleles (the r-6 allele) is 0.2. Of the individuals that carry at least one r-6 allele, what fraction are heterozygotes?

Answer 1

Answer:

The fraction of heterozygous individuals in the population is 32/100 that equals 0.32 which is the genotipic proportion for these endividuals.

Explanation:

According to Hardy-Weinberg, the allelic frequencies in a locus are represented as p and q, referring to the alleles. The genotypic frequencies after one generation are p² (Homozygous for allele p), 2pq (Heterozygous), q² (Homozygous for the allele q). 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 r-6 allelic frequency is 0,2. This means that if r-6=0.2, then the other allele frequency (R) is=0.8, and the sum of both the allelic frequencies equals one. This is:

p + q = 1

r-6 + R = 1  

0.2 + 0.8 = 1

Then, the genotypic proportion for the homozygous individuals RR is 0.8 ² = 0.64

The genotypic proportion for the homozygous individuals r-6r-6 is 0.2² = 0.04

And the genotypic proportion for heterozygous individuals Rr-6 is 2xRxr-6 = 2 x 0.8 x 0.2 = 0.32