Lecture 11 - QG3 & Sexual selection

Lecture 11 - QG3 & Sexual selection - 1 Lecture 11,...

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1 Lecture 11, con't from last time. .. V. Epistasis The definition of epistasis is when the effects of an allele at one locus depend on the genotype at some other locus, i.e., effects are non-additive across loci. Simplest example: haploid organism, two loci (A, B). "1" alleles add 1 unit of phenotype, so: G ij = α i + α j + ε If ε = 0, then the effects are additive and there is no epistasis. E.g., G 11 = 1 + 1 = 2; G 12 =G 21 = 1 + 0 = 1; G 22 = 0 + 0 = 0. If ε > 0, epistasis is positive, the "whole is greater than the sum of the parts". This is also called "synergistic" epistasis If ε <0, epistasis is negative and epistasis results in "diminishing returns" Diploid organism, two loci. Can write an analogous model, but it is more complicated because it includes effects of dominance. We will skip the formal diploid model. Example 1 (say, bristles/mm 2 in flies): A1A1 A1A2 A2A2 B1B1 5 4 3 B1B2 4 3 2 B2B2 3 2 1 genotypic effect a A = a B = 1; d A = d B = 0 (no dominance, effects additive across loci) Example 2 A1A1 A1A2 A2A2 B1B1 5 5 3 B1B2 4 4 2 B2B2 3 3 1 genotypic effect a A = a B = 1; d A = 1, d B = 0 (dominance at A, effects additive across loci)
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2 Example 3 A1A1 A1A2 A2A2 B1B1 0 4 3 B1B2 4 3 2 B2B2 3 2 0 Now, genotypic effects depend on genotype at other locus, i.e., epistasis IF B1B1, a A = -1.5, d A = 2.5; if B2B2, a A = 1.5, d A = 0.5 (direction defined w/ respect to A1A1 genotype) NOTE: multiplicative selection on a natural (i.e., additive) scale implies additivity on a log scale. Convince yourself using the following scenario: two loci (A, B); two alleles/locus (A1,A2; B1, B2). 1 alleles are completely dominant and 22 homozygotes have fitness 1- s . Therefore, the A2A2/B2B2 homozygote has fitness (1- s ) 2 . Log transforming such that log(1- s ) = (1- s* ), single-locus 22 homozygotes have fitness 1-s* and the double 22 het has fitness 2(1-s*), because log(x y ) = ylog(x) Therefore, if w = log(w), then w AB = w A + w B + ε , where ε is the "coefficient of epistasis" and represents the deviation from additivity of log(fitness) Interactions between loci do not mean that we cannot calculate breeding values (A ijkl. .. ), it just means that the breeding value will have to take account of allele frequencies at other loci. It follows that the genotypic component of variance, V G , will have a component due to interactions between alleles at different loci, so V G = V A + V D + V I Lecture 11.1 - Quantitative Genetics, con't VI. Genotype x Environment Interaction (G x E) Not all genotypes have the same phenotype in every environment Figure, GxE e.g., the "Thrifty Gene(otype)": Alleles that confer susceptibility to diabetes and obesity in the modern environment were beneficial in the ancestral environment of uncertain resources. Thrifty genotypes can efficiently turn food into fat during periods of plentiful resources.
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3 So, Phenotypic value breaks down into: P = G + E + GxE, conclusions about V P follow VII. Cov(G,E) Recall the variance of a sum, Var(X + Y) = Var(X) + Var(Y) + 2Cov(X,Y)
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Lecture 11 - QG3 &amp; Sexual selection - 1 Lecture 11,...

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