1. . viability selection sexual selection fecundity selection zygote
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 % 40% 20%
wij
sij

3. A numerical example Find the new genotype and allele frequencies
A1A1 A1A2 A2A2
genotype freq
wij
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
What is ?
It is the mean fitness. (p2w11 +2pqw12+q2w22)
New genotype frequencies
What are the new allele frequencies?
A1 ~ (0.5) A2 ~ (0.5)
w ’ = (1) (0.75) (0.25) =

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 s
recessive lethal
deleterious dominant s
deleterious dominant s 1-s
deleterious intermediate hs 1-s
deleterious recessive s 1+s
heterozygote advantage 1-s t
heteroz. disadvantage 1+s t

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

7. A numerical example
A1A1 A1A2 A2A2
gen. freq
wij
gen. freq (1) (1) (0.4)
> selection
f’(A1) ~ (0.5) f’(A2) ~ (0.5)
w ’ = (1) (1) (0.4) = 0.982

8. 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

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

10. change in the frequency of a deleterious recessive2

12. 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

13. 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

14. selection against a deleterious recessive alleleDq q

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

16. change in the frequency of a
deleterious dominant

18. Selection against a dominant alleleDq q

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

20. 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 sp = w w
tq = sp = s(1-q) s
s + t
>
q =
> t
s + t
p =

21. heterozygote advantageDq

22. heterozygote advantage at phosphoglucose
isomerase in Colias butterflies

23. glycolysis

24. enzyme kinetics of phosphoglucose isomerase in Colias

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

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

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

28. heterozygote disadvantage

29. heterozygote disadvantage: translocation
heterozygotes in Drosophila

30. 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

31. 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