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

in Medium Ground Finches Selection on beak size in Medium Ground Finches http://www.greglasley.net/Images/Medium-Ground-Finch-0002.jpg

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 http://www.ibri.org/Books/Pun_Evolution/Chapter3/fig3-03.jpg How do we quantify this change? 10% disadvantage??, 20% disadvantage??

Kinds of natural selection

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 100 200 100 Survivors 80 160 50 Absolute fitness (After/before) 80/100= 0.8 160/200= 0.8 50/100= 0.5 Relative fitness, wii (= Absolute/best) 1.0 1.0 0.625 Selection coefficient (sii = 1-wii) 0.0 0.2 0.375

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

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 0 1 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

Fitness and selection coefficients Fitness = 1 - (selection coefficient) AA Aa aa A dominant (fitness) 1 1 0.9 selection coefficient 0 0 0.1 a dominant (fitness) 0.9 1.0 1.0 selection coefficient 0.1 0.0 0.0 co-dominance (fitness) 1 0.95 0.90 selection coefficient 0 0.05 0.10 Overdominance (fitness) 0.8 1.0 0.9 Selection coefficients s=0.2 0.0 t=0.1

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

AA Aa aa A dominant 1 1 0.8 h=0 1 1-hs 1-s s = 0 0 0.2 Incomplete- or AA Aa aa co-dominance 1 0.9 0.8 h=0.5 1 1-hs 1-s s = 0 0.1 0.2 A is recessive AA Aa aa a is dominant 1 0.8 0.8 h=1 1 1-hs 1-s s = 0 0.1 0.2

Allele frequency change AA Aa aa A dominant (fitness) 1.0 1.0 0.8 A dominant 1.0 0.8 0.8 A dominant, a recessive Adaptive topography A recessive, a dominant Allele frequency change A dominant, a recessive A recessive, a dominant

Equilibrium allele frequency due to balancing selection AA Aa aa Fitness 0.9 1.0 0.9 Selection s=0.1 0.0 t=0.1 AA Aa aa Fitness 0.8 1.0 0.9 Selection s=0.2 0.0 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

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

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)

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

Cystic fibrosis http://www.hgu.mrc.ac.uk/Research/Devgen/Cysfib/cfgifs/fig2.gif http://www.science.ca/images/scientists/s1-tsui.jpg

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 = √(0.00001/1) = 0.003 = expected 0.02 = 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??

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

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)

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

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

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

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).

Environmental heterogeneity Multiple niche polymorphism AA Aa aa Niche 1 1.0 0.9 0.5 Niche 2 0.5 0.9 1.0 ___________________________ Average 0.75 0.9 0.75 Relative fitness 0.83 1.0 0.83 “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

http://sciencelife. uchospitals http://sciencelife.uchospitals.edu/2009/11/19/evolution-via-cannibalism-the-case-of-kuru/ http://open.jorum.ac.uk/xmlui/bitstream/handle/123456789/940/Items/S250_4_section3.html

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) = [(1-0.303)/2] / [(1-0.303)/2 + (1-0.254)/2] qV = 0.483 (observed = 0.484)

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

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

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

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

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

High Exposed Artificial shade Algae Low-Exposed

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 213-14 for treatment with 3 alleles