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Two-locus selection. p 1 ’ =p 1 2 +p 1 p 2 +p 1 p 3 +(1-r)p 1 p 4 +rp 2 p 3 p 2 ’ =p 2 2 +p 1 p 2 +p 2 p 4 +rp 1 p 4 +(1-r)p 2 p 3 p 3 ’ =p 3 2 +p 3 p.

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Presentation on theme: "Two-locus selection. p 1 ’ =p 1 2 +p 1 p 2 +p 1 p 3 +(1-r)p 1 p 4 +rp 2 p 3 p 2 ’ =p 2 2 +p 1 p 2 +p 2 p 4 +rp 1 p 4 +(1-r)p 2 p 3 p 3 ’ =p 3 2 +p 3 p."— Presentation transcript:

1 Two-locus selection

2 p 1 ’ =p 1 2 +p 1 p 2 +p 1 p 3 +(1-r)p 1 p 4 +rp 2 p 3 p 2 ’ =p 2 2 +p 1 p 2 +p 2 p 4 +rp 1 p 4 +(1-r)p 2 p 3 p 3 ’ =p 3 2 +p 3 p 4 +p 1 p 3 +rp 1 p 4 +(1-r)p 2 p 3 p 4 ’ =p 4 2 +p 3 p 4 +p 2 p 4 +(1-r)p 1 p 4 +rp 2 p 3 r is probabilities of cross-over (coefficient of recombination). Usually 0  r  0.5. If r=0.5 then loci are called unlinked (or independent). If r=0 then population transform to one loci population with four alleles. AB Ab aB ab p 1 p 2 p 3 p 4

3 p 1 ’ =p 1 2 +p 1 p 2 +p 1 p 3 +p 1 p 4 - rD p 2 ’ =p 2 2 +p 1 p 2 +p 2 p 4 +p 1 p 4 + rD p 3 ’ =p 3 2 +p 3 p 4 +p 1 p 3 +p 1 p 4 + rD p 4 ’ =p 4 2 +p 3 p 4 +p 2 p 4 +p 1 p 4 - rD AB Ab aB ab p 1 p 2 p 3 p 4 The miniumum and the maximum for D

4 Selection coefficients Haploid genotypes AB, Ab, aB, ab Zygote genotypes ABAB, ABAb, ABaB ABab AbAB, AbAb, AbaB Abab aBAB, aBAb, aBaB aBab abAB, abAb, abaB abab Fitness coefficients BB Bb bb AA w 11 w 12 w 22 Aa w 13 w 14 w 24 aa w 33 w 34 w 44

5 p 1 ’ =( w 11 p 1 2 +w 12 p 1 p 2 +w 13 p 1 p 3 +w 14 p 1 p 4 –w 14 rD)/W p 2 ’ =( w 22 p 2 2 +w 12 p 1 p 2 +w 24 p 2 p 4 +w 33 p 2 p 3 +w 14 rD)/W p 3 ’ =( w 33 p 3 2 +w 34 p 3 p 4 +w 13 p 1 p 3 +w 33 p 2 p 3 +w 14 rD)/W p 4 ’ =( w 44 p 4 2 +w 34 p 3 p 4 +w 24 p 2 p 4 +w 14 p 1 p 4 –w 14 rD)/W Measure of disequilibria D= p 1 p 4 -p 2 p 3 p 1 ’ =p 1 2 +p 1 p 2 +p 1 p 3 +p 1 p 4 - rD p 2 ’ =p 2 2 +p 1 p 2 +p 2 p 4 +p 1 p 4 + rD p 3 ’ =p 3 2 +p 3 p 4 +p 1 p 3 +p 1 p 4 + rD p 4 ’ =p 4 2 +p 3 p 4 +p 2 p 4 +p 1 p 4 - rD

6 One Selected Locus, One Neutral Locus Assume that the A locus is unselected (neutral), while the B ocus experiences selection with fitnesses for BB, Bb, and bb equal to 1, 1+ h s, and 1+s, respectively.

7 p 1 ’ =( w 11 p 1 2 +w 12 p 1 p 2 +w 13 p 1 p 3 +w 14 p 1 p 4 –w 14 rD)/W p 2 ’ =( w 22 p 2 2 +w 12 p 1 p 2 +w 24 p 2 p 4 +w 33 p 2 p 3 +w 14 rD)/W p 3 ’ =( w 33 p 3 2 +w 34 p 3 p 4 +w 13 p 1 p 3 +w 33 p 2 p 3 +w 14 rD)/W p 4 ’ =( w 44 p 4 2 +w 34 p 3 p 4 +w 24 p 2 p 4 +w 14 p 1 p 4 –w 14 rD)/W Haploid genotypes AB, Ab, aB, ab Fitness coefficients BB Bb bb AA w 11 w 12 w 22 Aa w 13 w 14 w 24 aa w 33 w 34 w 44 p 1 ’ =( p 1 2 + (1+hs) p 1 p 2 +p 1 p 3 + (1+hs) p 1 p 4 – (1+hs) rD)/W p 2 ’ =( (1+s) p 2 2 +p 1 p 2 + (1+s) p 2 p 4 +p 2 p 3 + (1+hs) rD)/W p 3 ’ =( p 3 2 + (1+hs) p 3 p 4 +p 1 p 3 +p 2 p 3 + (1+hs) rD)/W p 4 ’ =( (1+s) p 4 2 + (1+hs) p 3 p 4 + (1+s) p 2 p 4 + (1+hs) p 1 p 4 – (1+hs) rD)/W

8 p 1 ’ =( p 1 2 + (1+hs) p 1 p 2 +p 1 p 3 + (1+hs) p 1 p 4 – (1+hs) rD)/W p 2 ’ =( (1+s) p 2 2 +p 1 p 2 + (1+s) p 2 p 4 +p 2 p 3 + (1+hs) rD)/W p 3 ’ =( p 3 2 + (1+hs) p 3 p 4 +p 1 p 3 +p 2 p 3 + (1+hs) rD)/W p 4 ’ =( (1+s) p 4 2 + (1+hs) p 3 p 4 + (1+s) p 2 p 4 + (1+hs) p 1 p 4 – (1+hs) rD)/W Evolutionary operator

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11 Only when D=0 will the frequency of a remain constant over time, as expected from the one-locus model. Otherwise, the a allele will change in frequency because of its association (linkage disequilibrium) with another selected locus: Genetic Hitchhiking. When s is positive (b favored) and D is positive (AB and ab are more common than expected), the a allele will increase in frequency over time. When s is positive (b favored) and D is negative (Ab and aB are more common than expected), the a allele will decrease in frequency over time. If recombination is loose (r > s), disequilibrium decays rapidly and the a allele will change little in frequency. If recombination is tight (r < s), the extent of hitchhiking can b

12 Result 2 Linkage disequilibrium does not always decay to zero but can be maintained indefinitely within a population. In fact, very special conditions are required for linkage equilibrium (D=0) to be maintained when both loci are polymorphic and under selection. Result 3 Continuous cycling can occur (Hastings 1981): "apparent cases of directional selection may be due to stable cycling". Result 4 Even when the double heterozygote has the highest fitness, there may be no stable equilibrium maintaining polymorphisms at both loci! Claim 1 If there is pure directional selection favoring a genotype, say A 2 B 2 (such that w i,j+1 > w i,j and w i+1,j > w i,j ), it has been claimed but not proven that selection will lead eventually to the fixation of A 2 B 2. Nordborg (pers. comm.) ran simulations with millions of fitness sets, which all confirmed this claim, but it has not been shown analytically. More results are known for special cases of the two-locus model: Special Case 1: r=0 When r=0, the system is dynamically the same as a one-locus four allele model. Mean fitness never decreases. Special Case 2: Additive fitnesses If the fitnesses of the A and B loci add together to give two-locus fitnesses, then: There is only one equilibrium with both loci polymorphc for r>0, which has x 1 = p A1 p B1, x 2 = p A1 p B2, x 3 = p A2 p B1, and x 4 = p A2 p B2, where p A1 and p B1 equal the allele frequencies found at equilibrium in the one-locus model with heterozygote advantage. This "internal" equilibrium is globally stable if there is heterozygote advantage at both loci. D=0 at equilibrium. Mean fitness never decreases (Ewens 1969). Away from equilibrium, selection WILL generate linkage disequilibrium within a population that is initially in linkage equilibrium. Special Case 3: Multiplicative fitnesses If the fitnesses of the A and B loci multiply together to give two-locus fitnesses, then: The same internal polymorphism exists with x 1 = p A1 p B1, x 2 = p A1 p B2, x 3 = p A2 p B1, and x 4 = p A2 p B2. This "internal" equilibrium is only stable if there is heterozygote advantage at both loci AND if r is large enough. For tight linkage, the population goes to one of two stable equilibria at which D ≠ 0


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