Assume that skin pigmentation in humans is a quantitative trait controlled by tw

Question

Assume that skin pigmentation in humans is a quantitative trait controlled by two loci, A and B. A and B make an equal and additive contribution to pigment production while a and b make no contribution.

A. How many skin-color phenotypes are possible?
B. Identify the genotype(s) possible for a black man, for a white woman and for their children.
C. Assume that a person with intermediate pigmentation is heterozygous at both pigment loci. What genotypes and phenotypes would you expect among the progeny with a black individual?
D. Could two parent of intermediate color produce black children? Explain.

Explanation / Answer

OK...so you are working with two loci: A and B. Each of these exists as as a dominant and a recessive allele. So you have A and a, B and b. You know only A and B make a pigment contribution. That means you can pretty much discount the recessive alleles. So, our options for the dominant alleles are A- -- (one dominant A, all the rest are recessives), A-B- (one dominant allele of each) and -- B- (only one dominant B allele, all the rest being recessive. That gives us the number of main phenotypes (since only the dominant alleles matter for phenotype). A. See above :) B. It may seem tedious, but the only way to do this is to sit down and write them out. For the sake of convenience, let's say the A allele gives us white skin and the B allele gives us black skin. So you know a couple of things: 1. A white individual will have at least one dominant A allele and no dominant B allele. 2. A black individual will have at least one dominant B allele and no dominant A allele. That gives you two genotypes for each. I'll let you write them out. Then to figure what their children might have depends on which combination they have. Just do a Punnett square for each. Any way I cut it I get four possible genotypes for the children. C. The intermediates are heterozygotes for both (AaBb) and you cross them with a black individual (one of your two possibilities from part B). Just do a Punnett square. There should be six possibilities. D. If we assume that intermediate individuals are heterozygous for both (AaBb) as in the previous case, just do a Punnett square and see if any of your outcomes have one dominant B allele and only recessive a alleles (--B-). If so, then yes. If you assume intermediates are homozygous dominant (AABB) then obviously no combination is going to give you recessive genes you would need (aa). Hope that helps!! Good luck.

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