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
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
A1A1
A1A2
A2A2
80%
1.0
0
40%
0.5
0.5
20%
0.25
0.75
A numerical example
Find the new genotype and allele frequencies
A1A1
A1A2
A2A2
genotype freq.
wij
0.25
1.0
0.50
0.75
0.25
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. (p2w11 +2pqw12+q2w22)
New genotype frequencies
0.363
What are the new allele frequencies?
0.546
0.091
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
p2
2pq
q2
relative fitness, wij
w11
w12
w22
geno. freq.
after selection
p2w11
w
2pqw12
w
q2w22
w
geno. freq.
average relative fitness, w = p2w11 + 2pqw12 + q2w22
Patterns of selection -- Fitness arrays
A1A1
w11
deleterious recessive
recessive lethal
deleterious dominant
deleterious dominant
deleterious intermediate
deleterious recessive
heterozygote advantage
heteroz. disadvantage
1
1
1
1
1
1
1-s
1+s
A1A2
w12
1
1
1
1-s
1-hs
1+s
1
1
A2A2
w22
1-s
0
1+s
1-s
1-s
1+s
1-t
1+t
Selection against a recessive allele
initial g.f.
rel. fitness
A1A1
A1A2
A2A2
P2
1
2pq
1
q2
1-s
w = p2(1) + 2pq(1) + q2(1-s)
= 1 – q2s
g.f. > selection
p2(1)
1-q2s
2pq(1)
1-q2s
q2(1-s)
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.
> selection
0.25(1)
0.85
0.294
f’(A1) ~ 0.59 (0.5)
0.5(1)
0.85
0.588
0.25(0.4)
0.85
0.118
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’ +
1
2
H’
pq
q2(1-s)
=
+
2
1-sq2
1-sq
q’
=
q2(1-s) + pq
1-sq2
=
q2 – sq2 + q – q2
1-sq2
=
q(1-sq)
1-sq2
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 A2 ?
q’
= Q’ +
1
2
H’
pq
q2(1-s)
=
+
2
1-sq2
1-sq
q’
=
q2(1-s) + pq
1-sq2
=
q2 – sq2 + q – q2
1-sq2
=
q(1-sq)
1-sq2
recall that p = 1 - q
and q = 1 - p
how much has the frequency of A2 changed
after one generation of selection ?
Dq
Dq
= q’ - q
=
q(1-sq)
1-sq2
=
q – sq2 – q + sq3
1-sq2
=
-sq2(1 – q)
1-sq2
- q
Dq
selection against a deleterious recessive allele
q
Selection against a dominant allele
initial g.f.
rel. fitness
A1A1
A1A2
A2A2
P2
1
2pq
1-s
q2
1-s
w = p2(1) + 2pq(1-s) + q2(1-s)
= 1 – sq(2p-q)
g.f. > selection
p2(1)
1-sq(2p-q)
2pq(1-s)
1-sq(2p-q)
q2(1-s)
1-sq(2p-q)
change in the frequency of a
deleterious dominant
Selection against a dominant allele
q
Selection favoring heterozygotes
initial g.f.
rel. fitness
A1A1
A1A2
A2A2
P2
1-s
2pq
1
q2
1-t
w = p2(1-s) + 2pq(1) + q2(1-t)
= 1 – p2s - q2t
g.f. > selection
p2(1-s)
2pq(1)
1 – p2s - q2t 1 – p2s - q2t
q2(1-t)
1 – p2s - q2t
q-tq2
Dq = 1-sp2-tq2 - q
p-sp2
and Dp =
- p
2
2
1-sp -tq
at equilibrium, Dq = 0 and Dp = 0
q-tq2
w
p-sp2
w
=q
1 – tq
w
=
1 - sp
w
tq = sp = s(1-q)
s
q=
s+t
t
p= s+t
=p
heterozygote advantage
Change in allele frequency
0.10
0.05
0.00
-0.05
-0.10
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
Allele frequency, q
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
A1A1
A1A2
A2A2
P2
1+s
2pq
1
q2
1+t
w = p2(1+s) + 2pq(1) + q2(1+t)
= 1 + p2s + q2t
g.f. > selection
p2(1+s)
1+p2s+q2t
2pq(1)
1+p2s+q2t
q2(1+t)
1+p2s+q2t
at equilibrium, Dq = 0 and Dp = 0
>
s
q =
s+t
and
t
p=
s+t
heterozygote disadvantage
Change in allele frequency
0.10
0.05
0.00
-0.05
-0.10
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
Allele frequency, q
heterozygote disadvantage: translocation
heterozygotes in Drosophila
relative fitness of A1A1
simple models of selection
w11 > w12 > w22
w11 > w12 < w22
fix A1
unstable
polymorphism
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