A worked example

# A worked example - B loci were in linkage disequilibrium in...

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A worked example: gamete frequencies in 1000 observations: 580 AB's, 140 Ab's, 60 aB's and 280 ab's. Thus f(A) allele = (520+140)/1000 = 0.66 so f(a) = .34. f(B) = (520+60)/1000 = 0.58 so f(b) = 0.42. At random we expect the following gamete frequencies: f(AB) should be .66(.58)1000 = 383. f(Ab) should be .66(.42)1000 = 277. f(aB) should be .34(.58)1000 = 197 and f(ab) should be .34(.42)1000 = 143. These numbers of expected gametes are clearly different from the observed gametes. We can thus calculate the linkage disequilibrium as d = [.52(.28)] - [.14(.06)] = 0.1372. This tells us that the A and B alleles are in linkage disequilibrium. This disequilibrium will be broken up by recombination and the rate of breakup will be determined by the rate of recombination (see figure 8.2, pg. 202). Now let's say that the A locus was under selection with A alleles favored. If the A and
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Unformatted text preview: B loci were in linkage disequilibrium in the coupling state what would happen to the B alleles? They too would be selected for, but not because they were under selection. This is a very important phenomenon in population and evolutionary genetics called hitchhiking . It demonstrates a very important distinction we must make about selection and phenotype: we need to distinguish between selection "of" and selection "for" If the A allele is favored, there is selection for the A allele and selection of the B allele due to its linkage to the A allele (i.e., linkage between the A and B loci). Now consider the situation where the nose length is the result of interactions between the two loci. In the first case the interaction is additive, in the second case there is epistasis...
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## This note was uploaded on 11/05/2011 for the course BIOLOGY MCB2010 taught by Professor Jessicadigirolamo during the Fall '10 term at Broward College.

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