. viability selection sexual selection fecundity selection zygote survival to adult . courtship sexual selection fertilization sexual selection gamete production fecundity selection

Fitness = individual’s genetic contribution to the next generation (zygotezygote); differential survival and/or reproduction absolute fitness, Wij = #offspring, lifespan, etc relative fitness, wij = contribution relative to other genotypes selection differential, sij = strength of selection against a genotype A1A1 A1A2 A2A2 Pr(survival) Wij 80% 40% 20% wij 1.0 0.5 0.25 sij 0 0.5 0.75

A numerical example Find the new genotype and allele frequencies A1A1 A1A2 A2A2 genotype freq. 0.25 0.50 0.25 wij 1.0 0.75 0.25 Survival after selection 0.25(1) 0.5(0.75) 0.25(0.25) 0.25 0.375 0.0625 But what is the sum of these? 0.6875 To make them sum to one (for a new frequency) you must divide by 0.6875 What is 0.6875? It is the mean fitness. (p2w11 +2pqw12+q2w22) New genotype frequencies 0.363 0.546 0.091 What are the new allele frequencies? A1 ~ 0.64 (0.5) A2 ~ 0.36 (0.5) w ’ = 0.363(1) + 0.546(0.75) + 0.091(0.25) = 0.7954

How does selection change genotype and allele frequencies? A1A1 A1A2 A2A2 geno. freq. p2 2pq q2 relative fitness, wij w11 w12 w22 geno. freq. p2w11 2pqw12 q2w22 after selection w w w average relative fitness, w = p2w11 + 2pqw12 + q2w22

Patterns of selection -- Fitness arrays A1A1 A1A2 A2A2 w11 w12 w22 deleterious recessive 1 1 1-s recessive lethal 1 1 0 deleterious dominant 1 1 1+s deleterious dominant 1 1-s 1-s deleterious intermediate 1 1-hs 1-s deleterious recessive 1 1+s 1+s heterozygote advantage 1-s 1 1-t heteroz. disadvantage 1+s 1 1+t

Selection against a recessive allele A1A1 A1A2 A2A2 initial g.f. P2 2pq q2 rel. fitness 1 1 1-s w = p2(1) + 2pq(1) + q2(1-s) = 1 – q2s g.f. > selection p2(1) 2pq(1) q2(1-s) 1-q2s 1-q2s 1-q2s

A numerical example A1A1 A1A2 A2A2 gen. freq. 0.25 0.50 0.25 wij 1.0 1.0 0.4 gen. freq. 0.25(1) 0.5(1) 0.25(0.4) > selection 0.85 0.85 0.85 0.294 0.588 0.118 f’(A1) ~ 0.59 (0.5) f’(A2) ~ 0.42 (0.5) w ’ = 0.294(1) + 0.588(1) + 0.25(0.4) = 0.982

what is the new frequency of A2 ? q’ = Q’ + H’ = + = q’ = 1 2 pq 1-sq2 q2(1-s) 1-sq2 recall that p = 1 - q and q = 1 - p q2(1-s) + pq 1-sq2 q2 – sq2 + q – q2 1-sq2 q(1-sq) 1-sq2

change in the frequency of a lethal recessive in Tribolium castaneum

change in the frequency of a deleterious recessive 2

what is the new frequency of A2 ? q’ = Q’ + H’ = + = q’ = 1 2 pq 1-sq2 q2(1-s) 1-sq2 recall that p = 1 - q and q = 1 - p q2(1-s) + pq 1-sq2 q2 – sq2 + q – q2 1-sq2 q(1-sq) 1-sq2

how much has the frequency of A2 changed after one generation of selection ? Dq = q’ - q = - q = Dq = q(1-sq) 1-sq2 q – sq2 – q + sq3 1-sq2 -sq2(1 – q) 1-sq2

selection against a deleterious recessive allele Dq q

Selection against a dominant allele A1A1 A1A2 A2A2 initial g.f. P2 2pq q2 rel. fitness 1 1-s 1-s w = p2(1) + 2pq(1-s) + q2(1-s) = 1 – sq(2p-q) g.f. > selection p2(1) 2pq(1-s) q2(1-s) 1-sq(2p-q) 1-sq(2p-q) 1-sq(2p-q)

change in the frequency of a deleterious dominant

Selection against a dominant allele Dq q

Selection favoring heterozygotes A1A1 A1A2 A2A2 initial g.f. P2 2pq q2 rel. fitness 1-s 1 1-t w = p2(1-s) + 2pq(1) + q2(1-t) = 1 – p2s - q2t g.f. > selection p2(1-s) 2pq(1) q2(1-t) 1 – p2s - q2t 1 – p2s - q2t 1 – p2s - q2t

q-tq2 1-sp2-tq2 p-sp2 1-sp2-tq2 Dq = - q and Dp = - p at equilibrium, Dq = 0 and Dp = 0 q-tq2 w = q p-sp2 = p w 1 – tq 1 - sp = w w tq = sp = s(1-q) s s + t > q = > t s + t p =

heterozygote advantage Dq

heterozygote advantage at phosphoglucose isomerase in Colias butterflies

glycolysis

enzyme kinetics of phosphoglucose isomerase in Colias

. . . . expected heterozygosity deviation from July .15 .10 .05 -.05 -.05 -.10 3 11 17 23 July

Selection against heterozygotes A1A1 A1A2 A2A2 initial g.f. P2 2pq q2 rel. fitness 1+s 1 1+t w = p2(1+s) + 2pq(1) + q2(1+t) = 1 + p2s + q2t g.f. > selection p2(1+s) 2pq(1) q2(1+t) 1+p2s+q2t 1+p2s+q2t 1+p2s+q2t

at equilibrium, Dq = 0 and Dp = 0 > s s + t t s + t > q = and p =

heterozygote disadvantage

heterozygote disadvantage: translocation heterozygotes in Drosophila

simple models of selection w11 > w12 < w22 w11 > w12 > w22 unstable polymorphism fix A1 relative fitness of A1A1 w11 < w12 > w22 w11 < w12 < w22 stable polymorphism fix A2 relative fitness of A2A2

relative fitness enables different traits and populations to be compared selection can act at many stages in the life cycle; opportunity for opposing selection at different stages directional selection fixes one allele and eliminates all others from the population heterozygote advantage can maintain a balanced polymorphism, but cannot explain the high levels of genetic variation found in natural populations heterozygote disadvantage produces an unstable polymorphism; which allele is fixed depends on chance