Can a person have more than two alleles for a single gene?

73. Coloration in the peppered moth (Biston betularia) is determined by a single gene with two alleles showing?

• 73. Coloration in the peppered moth (Biston betularia) is determined by a single gene with two alleles showing complete dominance. Dark moths are homozygous dominant or heterozygous for the gene, light moths are homozygous recessive. In a sample of 100 moths, you determine that 64 of the moths are dark. According to the Hardy-Weinberg rule, the expected frequency of the dominant allele is ____. A. 0.4 can someone explain how .4 is the answer here

• Answer:

  Since you already have the answer, I'm figuring you're not doing homework and I'll help. So, you have two alleles, let's call them "A" for the dominant dark allele and "a" for the recessive allele. This means you should have only 4 possibly genotypes: AA Aa aA aa However, you should only have 2 possible phenotypes: dark (AA, Aa, aA) light (aa) Because any genotype with a dominant "A" allele will always be dark. Only the double recessive "aa" genotype will show up light. So, if we assign the proportion of "A" alleles to the letter "p" and the proportion of "a" alleles to the letter "q", we know that: p + q = 1 Because that's accounting for all of the alleles in all the moths. But, to calculate what "p" and "q" are, we need to look at the proportion of dark and light moths in the sample and--assuming Hardy-Weinberg equilibrium assumptions are met--we should get the following proportions of each genotype: AA = p * p or p^2 Aa = p * q or pq aA = p * q or pq aa = q * q or q ^ 2. Because we can only see the phenotypes, we'll add those proportions together dark (AA, Aa, aA) = p^2 + pq + pq = p^2 + 2pq light (aa) = q^2 We have the observed numbers for the phenotypes: dark = 64/100 = .64 light = (100 - 64)/100 = 36/100 = .36 And setting the estimates to the expected proportions: dark (AA, Aa, aA) .64 = p^2 + 2pq light (aa) .36 = q^2 We pick the easier of the two to solve: light (aa) .36 = q^2 sqrt(.36) = sqrt(q^2) .6 = q So, the frequency of the "a" or light allele is .6. We know that p + q = 1 (or all of them) so: p + q = 1 p = 1 - q p = 1 - (.6) p = .4 Plugging it back into the equation above: dark (AA, Aa, aA) .64 = p^2 + 2pq .64 = (.4)^2 + 2(.4)(.6) .64 = (.16) + 2 (.24) .64 = .16 + .48 .64 = .64 It checks out. Sort of a long, slowish, explanation, but I hope I helped you figure out why... Good luck!