zygote viability selection survival to adult courtship sexual
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); 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 Pr(survival) Wij wij sij A 1 A 1 A 1 A 2 A 2 A 2 80% 1. 0 0 40% 0. 5 20% 0. 25 0. 75
A numerical example Find the new genotype and allele frequencies A 1 A 1 A 2 A 2 genotype freq. wij 0. 25 1. 0 0. 50 0. 75 0. 25 Survival after selection 0. 25(1) 0. 5(0. 75) 0. 25 0. 375 0. 25(0. 25) 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. (p 2 w 11 +2 pqw 12+q 2 w 22) New genotype frequencies 0. 363 What are the new allele frequencies? 0. 546 0. 091 A 1 ~ 0. 64 (0. 5) A 2 ~ 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? A 1 A 1 A 1 A 2 A 2 A 2 p 2 2 pq q 2 relative fitness, wij w 11 w 12 w 22 geno. freq. after selection p 2 w 11 w 2 pqw 12 w q 2 w 22 w geno. freq. average relative fitness, w = p 2 w 11 + 2 pqw 12 + q 2 w 22
Patterns of selection -- Fitness arrays A 1 A 1 w 11 deleterious recessive lethal deleterious dominant deleterious intermediate deleterious recessive heterozygote advantage heteroz. disadvantage 1 1 1 1 -s 1+s A 1 A 2 w 12 1 1 -s 1 -hs 1+s 1 1 A 2 A 2 w 22 1 -s 0 1+s 1 -s 1+s 1 -t 1+t
Selection against a recessive allele initial g. f. rel. fitness A 1 A 1 A 1 A 2 A 2 A 2 P 2 1 2 pq 1 q 2 1 -s w = p 2(1) + 2 pq(1) + q 2(1 -s) = 1 – q 2 s g. f. > selection p 2(1) 1 -q 2 s 2 pq(1) 1 -q 2 s q 2(1 -s) 1 -q 2 s
A numerical example A 1 A 1 A 1 A 2 A 2 A 2 gen. freq. 0. 25 0. 50 0. 25 wij 1. 0 0. 4 gen. freq. > selection 0. 25(1) 0. 85 0. 294 f’(A 1) ~ 0. 59 (0. 5) 0. 5(1) 0. 85 0. 588 0. 25(0. 4) 0. 85 0. 118 f’(A 2) ~ 0. 42 (0. 5) w ’ = 0. 294(1) + 0. 588(1) + 0. 25(0. 4) = 0. 982
what is the new frequency of A 2 ? q’ = Q’ + 1 2 H’ pq q 2(1 -s) = + 2 1 -sq q’ = q 2(1 -s) + pq 1 -sq 2 = q 2 – sq 2 + q – q 2 1 -sq 2 = q(1 -sq) 1 -sq 2 recall that p = 1 - q and q = 1 - p
change in the frequency of a lethal recessive in Tribolium castaneum
2 change in the frequency of a deleterious recessive
what is the new frequency of A 2 ? q’ = Q’ + 1 2 H’ pq q 2(1 -s) = + 2 1 -sq q’ = q 2(1 -s) + pq 1 -sq 2 = q 2 – sq 2 + q – q 2 1 -sq 2 = q(1 -sq) 1 -sq 2 recall that p = 1 - q and q = 1 - p
how much has the frequency of A 2 changed after one generation of selection ? Dq Dq = q’ - q = q(1 -sq) 1 -sq 2 = q – sq 2 – q + sq 3 1 -sq 2 = -sq 2(1 – q) 1 -sq 2 - q
Dq selection against a deleterious recessive allele q
Selection against a dominant allele initial g. f. rel. fitness A 1 A 1 A 1 A 2 A 2 A 2 P 2 1 2 pq 1 -s q 2 1 -s w = p 2(1) + 2 pq(1 -s) + q 2(1 -s) = 1 – sq(2 p-q) g. f. > selection p 2(1) 1 -sq(2 p-q) 2 pq(1 -s) 1 -sq(2 p-q) q 2(1 -s) 1 -sq(2 p-q)
change in the frequency of a deleterious dominant
Dq Selection against a dominant allele q
Selection favoring heterozygotes initial g. f. rel. fitness A 1 A 1 A 1 A 2 A 2 A 2 P 2 1 -s 2 pq 1 q 2 1 -t w = p 2(1 -s) + 2 pq(1) + q 2(1 -t) = 1 – p 2 s - q 2 t g. f. > selection p 2(1 -s) 2 pq(1) 1 – p 2 s - q 2 t q 2(1 -t) 1 – p 2 s - q 2 t
q-tq 2 Dq = 1 -sp 2 -tq 2 - q p-sp 2 and Dp = - p 2 2 1 -sp -tq at equilibrium, Dq = 0 and Dp = 0 q-tq 2 w p-sp 2 w =q 1 – tq w = 1 - sp w tq = sp = s(1 -q) t p= s+t > > s q= s+t =p
Dq heterozygote advantage
heterozygote advantage at phosphoglucose isomerase in Colias butterflies
glycolysis
enzyme kinetics of phosphoglucose isomerase in Colias
deviation from expected heterozygosity. 15. 10 . 05 0 -. 10 . 3 . . 11 17 July . -. 05 23
Selection against heterozygotes initial g. f. rel. fitness A 1 A 1 A 1 A 2 A 2 A 2 P 2 1+s 2 pq 1 q 2 1+t w = p 2(1+s) + 2 pq(1) + q 2(1+t) = 1 + p 2 s + q 2 t g. f. > selection p 2(1+s) 1+p 2 s+q 2 t 2 pq(1) 1+p 2 s+q 2 t q 2(1+t) 1+p 2 s+q 2 t
at equilibrium, Dq = 0 and Dp = 0 and t p= s+t > > s q = s+t
heterozygote disadvantage
heterozygote disadvantage: translocation heterozygotes in Drosophila
relative fitness of A 1 A 1 simple models of selection w 11 > w 12 > w 22 w 11 > w 12 < w 22 fix A 1 unstable polymorphism w 11 < w 12 > w 22 w 11 < w 12 < w 22 stable polymorphism fix A 2 relative fitness of A 2 A 2
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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