MOLECULAR EVOLUTIONFocus is on long-term evolution leading to differences between species.Differences between species are determined by the same factors that determinedifferences between individuals within a species: mutation, selection, and drift. But thereare some critical differences:1. Time periodPopulation genetics: ≤ 4Ne generations (≤ time since most recent common ancestor ofall copies of a gene in a species)Evolutionary genetics: > 4Ne generations2. Selection causes hitchhiking within species; we only discussed this in terms of theAdhF/S difference.3. Method of studyRates of sequence evolutionMolecular evolution is generally studied by following procedure:(1) Sequence one copy of a gene from each of a number of different species oforganisms(2) Align sequences.(3) Calculate the proportion of sites that differ, for each pair of species.d = pairwise sequence difference in differences per bp(4) Correct for multiple hits, especially if d ≥ 0.10sequence divergence = K = -(3/4) ln[1-(4/3)d] in substitutions perbp(5) Calculate rate of sequence divergence if desiredE = K/2T in substitutions per bp per year• A species is a population of organisms.• During speciation, one species splits into two populations that evolveindependently of each other.• To study evolution, we usually sequence one gene from each species.• Many mutations occur in the subsequent evolution of both species, but mostare eliminated. The sequences only show differences that were fixed in oneor the other species.K = frequency of base pair substitutions that occurred along both evolutionarypaths; i.e. the number of mutations that occurred AND were fixed in one or theother species instead of being lost.We can state this with a remarkably simple equation:E = MFrate of molecular Evolution = total Mutation rate X Fixation probabilityM = 2Nuwhereu = mutations per site (bp) per gameteF = mutations fixed/total mutations (substitutions per mutation)2N = number of gametes in population or number of copies of the gene inpopulationK = number (per site) of mutations that occurred AND were fixed in one or theother species instead of being lost.E = 2NuFIf we want to express this in units of time, then we have to incorporategeneration time by assigning units:2N = gametes per year 2N u F gametes X mutations X substitutions year site X gamete mutatione.g.u = 5 × 10-9 mutations per site per gamete2N = 106 gametesF = 10-7 substitutions per mutationE = 5 × 10-10 substitutions per site per yearHaploids, organelles, asexuals: E = NuFThe rate of neutral substitution equals the mutation rate.Neutral mutations:Fn = 1/2N for a new mutationEn = 2Nu(1/2N) = uE = u !!This remarkably simple result is also remarkably important. It means that The mutation rate can be estimated from the substitution rate for neutralmutations.• The mutation rate equals the pseudogene substitution rate becausepseudogene substitutions are neutral.• Synonymous substitution rates in functional protein-coding genes are aboutequal to pseudogene rates in eukaryotes, which shows that they are alsoneutral or effectively neutral (average Ne|s| << 1).Directional selection reduces F and thus EF(new mutation in diploids) = (1 – e-2Nes/N) / (1 – e-4Nes)The three classes of mutations with different levels of polymorphism also havedifferent rates of substitution:(1) neutral s = 0 H ≈ 4Neu F = 1/2N E = u(2) detrimental s < 0 H < 4Neu F < 1/2N E < u(3) advantageous s > 0 H < 4Neu F > 1/2N E > uNote that we are ignoring those subject to balancing selection, as they are rare.The great majority of mutations are either neutral or detrimental, so on average,F < 1/2N and E < u.Emphasize: this is average over a large number of sites. This is actually what we observe:For large sample of genes in mammals:Synonymous rate 3.51 X 10-9 substitutions per site per yearNonsynonymous rate 0.74 X 10-9e.g. comparisons of globin genes in cow and goat K (mean ± std. error)β-globin pseudogenes 9.1 ± 0.9β- and γ-globin exon synonymous 8.6 ± 2.5β- and γ-globin intron 8.1 ± 0.7β- and γ-globin 5’-flanking 5.3 ± 1.2Note that this is evidence of natural selection which is predominantlypurifying,eliminating detrimental mutations.MEASURING THE STRENGTH OF PURIFYING SELECTIONCalculate ratio nonsynonymous substitutions/synonymous substitutions =Kn/Ks or Ka/Ks.0.74 X 10-9 = 0.213.51 X 10-9neutral mutations + detrimental mutations Kn/Ks < 1neutral mutations only Kn/Ks = 1neutral mutations + advantageous mutations Kn/Ks > 1DETECTING POSITIVE SELECTION (FOR ADVANTAGEOUS MUTATIONS)Selection for advantageous mutations: Kn/Ks > 1We can use this to detect positive selection for advantageous mutations. Usecomputer to isolate specific sites and calculate Kn/Ks for each site. Then find iffind some sites have Kn/Ks > 1, these probably had one or more advantageousmutations fixed in fairly recent time.Making Phylogenetic TreesIf we have DNA sequences from the genes of three or more species, we can usethem to recover their evolutionary history by making a phylogenetic tree.Can do same using AA sequences of protein encoded by gene.Saw at beginning of course in the tree of life.Need one gene that is present in all organisms: usually use gene for smallsubunit of ribosomal RNA.Not good for some organisms, so usually use protein-coding genes foreukaryotes.Done by computer.Aligns all sequences.Calculates pairwise d.Corrects for multiple hits to get pairwise K..Uses pairwise K values to make trees.Give e.g. of three species from our bdelloid rotifers plus an outgroup(monogonont rotifer).Makes tree, using any one of a number of different algorithms to make tree thatis simplest or most probable given the data.Ari1/2 WPr1/1 Flt2/1 Ari1/2 - WPr 0.19394 -FlT2/1 0.17876 0.11301 -B.quadrid 0.56295 0.52794 0.53827 -Old way: use morphological differences/similarities.Sequence data have some advantages:• Use neutral or nearly neutral sites, avoids problems due to convergent orparallel evolution.• Hard to know what morphological traits are informative in many organisms.• Can make phylogenetic trees of genes even without seeing the organisms!Ari WPr FlT BqUses:• Reconstruct evolutionary historye.g. tree of life shown at beginning of coursePotential problems.If I sequence a gene from a cow, a langur monkey, and a rhesus monkey, whichtree will I