1. Life cycle with under Hardy-Weinberg assumptions
Life cycle with evolutionary forces

2. in Medium Ground Finches
Selection on beak size
in Medium Ground Finches

3. The peppered moth, Biston betularia The rare dark form increased
after pollution killed lichens, making tree trunks dark
But the dark form decreased after pollution was reduced
How do we quantify this change?
10% disadvantage??, 20% disadvantage??

4. Kinds of natural selection

5. A model of natural selection
Genotype-specific survivorship = lii
Genotype-specific fecundity = mii
Offspring
Fitness = genotype-specific survivorship and fecundity
wii = lii • mii
Selection coefficient = s = (1-fitness); e.g. saa = 1-waa
Genotype AA Aa aa
Adults
Survivors
Absolute fitness (After/before) 80/100= /200= /100= 0.5
Relative fitness, wii (= Absolute/best)
Selection coefficient (sii = 1-wii)

6. Selection changes allele frequencies
The frequency of A allele, f(A), in the next generation
weighted fitness of “A” - carrying individuals
weighted fitness of entire population
pt+1 =
Average fitness = w = (=wbar) weighted fitness of all individuals

8. Genetic variation and rate of evolution
Dp = pt+1 - pt and Dp = 0 when pt+1 = pt (at ‘equilibrium’)
p2WAA + pqWAa
Denominator =
“Wbar”
Dp = _________________________
- pt
p2WAA + 2pqWAa + q2Waa
p(pWAA + qWAa)
This simplifies to:
Dp = _________________________
- pt
Wbar pq
pq[p(WAA - WAa) + q(WAa - Waa)]
Dp = ________________________________
0.25
Wbar
Dp = 0 when p or q = 0 (when either allele is fixed)
Dp is maximal when pq is maximal (at intermediate frequencies = genetic variation
Simple version of Fisher’s fundamental theorem of natural selection
1) The rate of evolution is propostional to the genetic variaiton in the population p

9. Fitness and selection coefficients
Fitness = 1 - (selection coefficient)
AA Aa aa
A dominant (fitness)
selection coefficient
a dominant (fitness)
selection coefficient
co-dominance (fitness)
selection coefficient
Overdominance (fitness)
Selection coefficients s= t=0.1

10. h = 0 A1 dominant, A2 recessive
h=1 A1 recessive, A2 dominant
0<h<1 incomplete dominance
h<0 overdominance
h>0 underdominance

11. AA Aa aa
A dominant
h=0 1 1-hs 1-s
s =
Incomplete- or AA Aa aa
co-dominance
h= hs 1-s
s =
A is recessive AA Aa aa
a is dominant
h=1 1 1-hs 1-s
s =

12. Allele frequency change
AA Aa aa
A dominant (fitness)
A dominant
A dominant, a recessive
Adaptive topography
A recessive, a dominant
Allele frequency change
A dominant, a recessive
A recessive, a dominant

13. Equilibrium allele frequency due to balancing selection
AA Aa aa
Fitness
Selection s= t=0.1
AA Aa aa
Fitness
Selection s= t=0.1
pequil = 0.5
pequil = 0.33
Equilibrium allele frequency due to balancing selection
pequil = t/(s+t) = 0.1/0.2 = 0.5
pequil = t/(s+t) = 0.1/0.3 = 0.33

14. Equilibrium and Stability Analyses
Dp = 0 when pt+1 = pt at equilibrium
p(pWAA + qWAa)
Dp = 0 when _________________________
- pt = 0
Wbar
WAA = 1.0, WAa = 0.8, Waa = 0.9
WAA = 0.9, WAa = 1.0, Waa = 0.8
Plot of Dp vs. p with
overdominance shows
a stable polymorphic
equilibrium
Plot of Dp vs. p with
underdominance shows
two stable equilibria,
but neither are polymorphic

16. u
A model of mutation
A a v
u = probability of mutation
(1-u) = probability of NO mutation
v = probability of reverse mutation
We want pt+1 from pt (p in next generation, given p)
pt+1 = pt(probability of not mutating) +
qt(probability of mutating from a to A)
pt+1 = pt(1-u) + qt(v)
Dp = 0 when pt+1 = pt at equilibrium
Dp = pt+1 - p = 0 = pt(1-u) + q(v) - pt
pt(1-u) + q(v) = pt pequilib. = v / (u + v)
qequilib. = u / (u + v)
pt+1 = pt(1-u) + qt(v)

17. Mutation and selection Mutation: increases “a” allele (A -> a)
Rate of change = p • u, which is the same as (1-q) • u
Dq positive : “a” allele increases
Population genetic fight:
mutation introduces bad allele
selection eliminates bad allele
AA Aa aa
1 1 1-s
Selection against ‘a’ ~ -sq2(1-q) = Dq
Dq is negative
See pg 215 (4th Ed.)
Together: Dq = (1-q) • u sq2(1-q)
Increase due to mutation Decrease due to selection
when does Dq = 0? At ‘mutation selection equilibrium’
(1-q) • u = sq2(1-q) math magic -> -> qequilib = √ u/s

18. Cystic fibrosis

19. Does Cystic fibrosis fit a mutation selection balance?
f(CFTR) allele = 0.02
Mutation: A -> a try mutation rate of 10-5
Selection: aa homozygotes are lethal (s = 1.0)
qequilib = √( u/s)
qequilibrium = √( /1) = = expected
= observed
conclusion: too rare compared to observed
Try a higher mutation rate of 10-4
qequilibrium = √(0.0001/1) = 0.01 = expected
still too rare compared to observed
Actual mutation rate = 6.7 x 10-7
We need some other evolutionary force to explain the data
Did CFTR help with typhoid resistance??

20. CFTR mutants are resistant to typhus bacteria
Do CFTR mutants have a benefit and a cost?

21. Genetic load and substitutions
Substitution of new allele
AA Aa aa
1 1+s 1+2s
Requires elimination of old allele
Lots of genetic deaths
Rates of molecular evolution are too fast to be explained by positive selection
Too much genetic load
Substitution load
Load = (Woptimal - Wbar) / Woptimal
Kimura: The majority of substitutions at the molecular level are neutral in evolution
Different proteins evolve at different rates due to different levels of functional constraint •
Observation: molecular clocks
are approximately linear • • •
Fibrino
peptides • •
Percent
sequence
divergence • • •
Hemo
globin • • • • • • • • • • • • •
Cytochrome c •
Date of divergence
(millions of years)

22. Genetic load and polymorphism
Observation from allozyme data:
About 1/4 of loci are polymorphic
About 3000 loci in fruitflies
Heterozygosity ~ 0.3 on average
Is variation maintained by selection?
AA Aa aa
1-s 1 1-t
This requires lots of genetic deaths
Homozygotes eliminated
Genetic load
Segregation load
Load = (Woptimal - Wbar) / Woptimal
For 3000 loci with 10% selection (s and t = 5% each) this load could be huge
Too much genetic death to allow balancing selection
No problem if alleles are neutral
No genetic load with neutrality
SS SF FF SF FF SS SF SF
well
bands
Geno
types
SS SF FF SF SS SF FF SF
Monomorphic
enzyme
Dimeric
Kimura and Ohta:
Protein polymorphism as a phase
Of neutral evolution

23. CCR5-32 allele and potential selection on Humans with AIDS
-32 allele common,
but HIV/AIDS rare.
Selection on 32 allele is weak
-32 allele rare,
but HIV/AIDS common
Selection is strong

27. Patterns of seasonal and geographic variation in chromosome
Inversions in Drosophila pseudoobscura.
Theodosius Dobzhansky

28. Opposing seasonal fluctuations of alternative chromosome inversion
types in one population of Drosophila pseudoobscura.
Selection may maintain variation by favoring alternative inversions
in alternative niches (temporal/thermal niches).

29. Environmental heterogeneity
Multiple niche polymorphism
AA Aa aa
Niche
Niche
___________________________
Average
Relative fitness
“Marginal” overdominance: the average (marginal) fitness
for the heterozygotes is higher than for either homozygote
This can maintain polymorphism
Key question: what is the relative proportion of the two niches?
if one niche is more common than the other, a simple
average fitness is not correct; need a weighted average

30. http://sciencelife. uchospitals

31. Estimation of viabilities
Prion protein gene PRNP
ViabilityMM = P’MM(PMV) / P’MV(PMM)
= 0.133(0.514) / 0.767(0.221)
= 0.303
ViabilityVV = P’VV(PMV) / P’MV(PVV)
= 0.100(0.514) / 0.767/(0.264)
= 0.254
Selection is primarily in females, so viabilities are ½ shown:
Equilibrium frequency = qV = sMM / (sMM + sVV)
= [( )/2] / [( )/2 + ( )/2]
qV = (observed = )

32. Estimating selection by deviation from HWE
F = fixation index = 1 – (Hobs after selection) / (2pqobs after selection)
Multiplicative selection will not deviate from HWE
AA Aa aa
1 (1-s) (1-s)2

33. Intertidal microhabitats have different thermal profiles
Heat coma for barnacles
Cool Site
Hot site
High Exposed (X)
Under Algal cover (A)
Low Intertidal (L)

34. MPI, GPI and sugar metabolism MPI and GPI share
ATP
ATP
ADP
ADP
MPI and GPI share
a product in the forward direction
a substrate in the reverse direction

35. Mpi genotypes show significant habitat association
Year 1
Year 2
Cool sites
Hot sites
Damariscotta River, Maine
Schmidt and Rand (1999) Evolution

36. Gpi genotypes show no significant habitat association
Year 1
Year 2
Cool sites
Hot sites

37. High Exposed
Artificial shade
Algae
Low-Exposed

38. Marginal fitnesses and the Average excess of an allele
Average fitness of an allele =
• freq. of pairing with an ‘A’ allele, giving fitness WAA
+ freq. of pairing with an ‘a’ allele giving fitness WAa .
p2WAA + pqWAa
pt+1 = _________________________
p2WAA + 2pqWAa + q2Waa
p(pWAA + qWAa)
pt+1 = _________________________
p2WAA + 2pqWAa + q2Waa
WA•
pt+1 = p _______
WA• = (pWAA + qWAa)
Wbar
p increases if WA•/Wbar > 1
The main point: this identifies the selection on a specific allele,
averaged across all other alleles in the population with which
that allele can pair. See pg for treatment with 3 alleles