L567 lecture 3 - L567 lecture 3 L567 lecture 3:...

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L567 lecture 3 L567 lecture 3: Mutation-selection balance. The breeder’s equation Population Genetics revisited. Last week we derived p t + 1 = π ( πΩ 11 + θΩ 12 29 Now, we want to solve for the change in p , delta p (note: we drop the subscripts for p t to give just p ) p = p t +1 - p t = p ( pW 11 + qW 12 ) W - p D p = p ( pW 11 + qW 12 ) W - 1 With some algebra, you can show that p = pq [ p ( W 11 - W 12 ) + q ( W 12 - W 22 )] W There is an equilibrium when p = 0 , where are the equilibria in terms of p and q? What happens when W 11 =W 12 =W 22 ? Is that an equilibrium? If so, what kind? Solving for the case where p = 0 and solving for p, we get 5 p (called p hat, where the hat is the indication of an equilibrium point) 5 p = 12 - 22 2 12 - 11 - 22 when will 5 p be on the interval between zero and one. In other words, when will there be a genetic polymorphism? 1
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L567 lecture 3 Finally, the main point. Natural selection can cause gradual change in gene frequencies (and hence gradual phenotypic change), even when selection is weak. This simple model helped to solve the conflict between the Mendelians and the Biometricians. Tweak number 2 Remember that p = pq [ p ( W 11 - W 12 ) + q ( W 12 - W 22 )] W Now let, W 11 =1 12 =1- ησ 22 =1- σ where h is the dominance coefficient, and s (little s) is the selection coefficient. If h =1, which allele is dominant? If h=0, which allele is dominant? What happens when h =0.5 (we call this co-dominance)? If we assume that s is small, then Wbar is approximately equal to one. (Convince yourself of this.) 2
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L567 lecture 3 Under this assumption, we get p pq [ p (1- (1- hs ))+ q ((1- hs )- (1- s ))] 1 D p pqs [ ph + q (1- h )] So, very clearly, delta p depends on allele frequencies, the strength of selection, and dominance.
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This note was uploaded on 07/01/2011 for the course L 567 taught by Professor Curtis during the Fall '10 term at Indiana State University .

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L567 lecture 3 - L567 lecture 3 L567 lecture 3:...

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