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? A) ¹/₉ B) ¹/₅ C) ¹/₃ D) ⁴/₅ E) ⁸/₉

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The frequency of the r-6 allele is 0.2, so the frequency of the R allele is 0.8. The frequency of the RR homozygotes is 0.8^2 = 0.64. The frequency of the rr homozygotes is 0.2^2 = 0.04. The frequency of the Rr heterozygotes is 2 * 0.8 * 0.2 = 0.32. The fraction of individuals that carry at least one r-6 allele that are heterozygotes is 0.32 / (0.32 + 0.04) = 0.8889, which is closest to 8/9.

You Discover a Population of Mustard Plants and Find It