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EVE100_107-127_2pp

# EVE100_107-127_2pp - Modes of selection 107 Fut 13.11 Fut...

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1/13/10 1 107 Modes of selection 108 Neither directional nor stabilizing selection maintain genetic variation Fut 13.11 Fut 13.12 For overview see Darimont et al, Proc. Natl Acad. Sci 106:952 (Jan 20 2009) Human predators outpace other agents of trait change in the wild

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1/13/10 2 109 An example of diversifying or disruptive selection in the black-bellied seedcracker ( Pyrenestes ostrinus ) Fut 12.14 110 Measuring fitness - an example Absolute fitness; asexual population (all female) 60 eggs per female viability = 0.05 Fitness = 60(0.05) = 3 Relative fitness Genotype B = 4/4 = 1 Genotype A = 3/4 =0.75 Mean relative fitness of a population depends on genotype frequencies Example: if freq A = 0.2 and freq B = 0.8 then mean fitness = 0.2(.75) + 0.8(1.0) = 0.95 Fut 12.2
1/13/10 3 111 The selection coefficent Genotype A B Relative fitness 0.75 1 Relative fitness 1-s 1 Selection coefficient = 0.25 112 Model of directional selection Assume three genotypes, and variation in viability. We can calculate the genotype frequencies after NS Genotype A 1 A 1 A 1 A 2 A 2 A 2 Freq before sel. p 2 2pq q 2 Rel. fitness w 11 w 12 w 22 Freq after sel. p 2 w 11 2pq w 12 q 2 w 22 Sum after sel p 2 w 11 + 2pqw 12 + q 2 w 22 which is the mean fitness = w Bar We can now calculate the allele frequencies after NS A 1 = [p 2 w 11 + (1/2)2pq w 12 ]/w Bar = [p 2 w 11 + pq w 12 ]/w Bar = (p[pw 11 + qw 12 ])/w Bar = p l A 2 = [q 2 w 22 + (1/2)2pq w 12 ]/w Bar = [q 2 w 22 + pq w 12 ]/w Bar = (q[qw 22 + pw 12 ])/w Bar = q l

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1/13/10 4 113 Genotype A 1 A 1 A 1 A 2 A 2 A 2 Freq before sel. p 2 2pq q 2 Rel. fitness 1 1 1-s Rel freq after sel. p 2 2pq q 2 (1-s) There are fewer individuals after selection!
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