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zygote

. zygote. viability selection. 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

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zygote

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

  2. 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

  3. 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 0.25(1) 0.25(0.25) 0.5(0.75) 0.25 0.375 0.0625 w ’ = 0.363(1) + 0.546(0.75) + 0.091(0.25) = 0.7954 Survival after selection 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)

  4. 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

  5. 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

  6. 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

  7. 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

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

  9. change in the frequency of a lethal recessive in Tribolium castaneum

  10. change in the frequency of a deleterious recessive 2

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

  12. 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

  13. selection against a deleterious recessive allele Dq q

  14. 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)

  15. change in the frequency of a deleterious dominant

  16. Selection against a dominant allele Dq q

  17. 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

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

  19. heterozygote advantage Dq

  20. heterozygote advantage at phosphoglucose isomerase in Colias butterflies

  21. glycolysis

  22. enzyme kinetics of phosphoglucose isomerase in Colias

  23. .15 . . .10 . . .05 deviation from expected heterozygosity 0 -.05 -.10 23 3 11 17 July

  24. 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

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

  26. heterozygote disadvantage

  27. heterozygote disadvantage: translocation heterozygotes in Drosophila

  28. 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

  29. 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

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