Presentation on theme: "Coalescence DNA Replication DNA Coalescence"— Presentation transcript:
1 Coalescence DNA Replication DNA Coalescence A coalescent event occurs when two lineages of DNA molecules merge back into a single DNA molecule at some time in the past.
2 COALESCENCE OF n COPIES OF HOMOLOGOUS DNA Gene Tree (all copies of homologous DNA coalesce to a common ancestral molecule)
3 Coalescence in an Ideal Population of N with Ploidy Level x Each act of reproduction is equally likely to involve any of the N individuals, with each reproductive event being an independent eventUnder these conditions, the probability that two gametes are drawn from the same parental individual is 1/NWith ploidy level x, the probability of identity by descent/coalescence from the previous generation is (1/x)(1/N) = 1/(xN)In practice, real populations are not ideal, so pretend the population is ideal but with an “inbreeding effective size” of an idealized population of size Nef; Therefore, the prob. of coalescence in one generation is 1/(xNef)
4 Sample Two Genes at Random The probability of coalescence exactly t generations ago is the probability of no coalescence for the first t-1 generations in the past followed by a coalescent event at generation t:
5 Sample Two Genes at Random The average time to coalescence is:The variance of time to coalescence of two genes (ct) is the average or expectation of (t-xNef)2 :
8 Sample n Genes at Random Once the first coalescent event has occurred, we now have n-1 gene lineages, and therefore we simply repeat all the calculations with n-1 rather than n. In general, the expected time and variance between the k–1 coalescent event and the kth event is:
9 Sample n Genes at Random The average times to the first and last coalescence are:2xNef/[n(n-1)] and 2xNef(1-1/n)Let n = 10 and x=2, then the time span covered by coalescent events is expected to range from Nef to 3.6Nef.Let n = 100, then the time span covered by coalescent events is expected to range from Nef to 3.96Nef.These equations imply that you do not need large samples to cover deep (old) coalescent events, but if you want to sample recent coalescent events, large sample sizes are critical.For n large, the expected coalescent time for all genes is 2xNefdrift determines how rapidly genes coalesce to a common ancestral molecule.
10 Sample n Genes at Random The variance of time to coalescence of n genes is:Note that in both the 2- and n-sample cases, the mean coalescent times are proportional to Nef and the variances are proportional to Nef2.The Standard Molecular Clock is a Poisson Clock in Which the Mean = Variance.The Coalescent is a noisy evolutionary process with much inherent variation that cannot be eliminated by large n’s; it is innate to the evolutionary process itself and is called “evolutionary stochasticity.”
12 Fixation (Coalescence) Times in 105 Replicates of the Same Evolutionary Process Problem: No Replication With Most Real Data Sets. Only 1 Realization.
13 Evolutionary Stochasticity Using the standard molecular clock and an estimator of of per year, the time to coalescence of all mtDNA to a common ancestral molecule has been estimated to be 290,000 years ago (Stoneking et al. 1986). This figure of 290,000 however is subject to much error because of evolutionary stochasticity. When evolutionary stochasticity is taken into account (ignoring sampling error, measurement error, and the considerable ambiguity in ), the 95% confidence interval around 290,000 is 152,000 years to 473,000 years (Templeton 1993) -- a span of over 300,000 years!
14 Coalescence of a mtDNA in an Ideal Population of N♀ haploids Each act of reproduction is equally likely to involve any of the N♀ individuals, with each reproductive event being an independent eventUnder these conditions, the probability that two gametes are drawn from the same parental individual is 1/N♀Under haploidy, the probability of identity by descent/coalescence from the previous generation is (1)(1/N♀) = 1/(N♀)In practice, real populations are not ideal, so pretend the population is ideal but with an “inbreeding effective size” of an idealized population of size Nef♀; Therefore, the prob. of coalescence in one generation is 1/(Nef♀)
15 Expected Coalescence Times for a Large Sample of Genes Mitochondrial DNA2Nef♀=Nef (if Nef♀=1/2Nef)Y-Chromosomal DNA2Nef♂=Nef (if Nef ♂=1/2Nef)X-Linked DNA3NefAutosomal DNA4Nef
16 Estimated Coalescence Times for 24 Human Loci 9Uniparental Haploid DNA Regions8X-Linked Loci76Autosomal LociTMRCA (In Millions of Years)54321FIXCCR5Y-DNAmtDNAMAOXq13.3G6PDAPLXAMELXMC1RECPEDNMS205HFEMX1HS571B2TNFSF5RRM2P4PDHA1Hb-BetaFUT6CYP1A2LactaseFUT2MSN/ALAS2Locus
22 Gene Vs. Allele (Haplotype) Tree haplotypes refer to alternative states of a homologous DNA region regardless of whether or not thatregion corresponds to a gene locus; alleles are alternative states of a homologous DNA regionthat corresponds to a gene locus.
23 Gene Trees vs. Haplotype Trees Gene trees are genealogies of genes. They describe how different copies at a homologous gene locus are “related” by ordering coalescent events.The only branches in the gene tree that we can observe from sequence data are those marked by a mutation. All branches in the gene tree that are caused by DNA replication without mutation are not observable. Therefore, the tree observable from sequence data retains only those branches in the gene tree associated with a mutational change. This lower resolution tree is called an allele or haplotype tree.The allele or haplotype tree is the gene tree in which all branches not marked by a mutational event are collapsed together.
25 Haplotype trees are not new in population genetics; they have been around in the form of inversion trees since the 1930’s.The Inversion Tree Is Not Always The Same As A Tree of Species Or Populations, In This Case Because of:Transpecific PolymorphismEstimated by Maximum Parsimony, minimizing the total number of mutations in the network.
26 Haplotype Trees Can Coalesce Both Within And Between Species The human MHC region fits this pattern; it takes 35 million years to coalesce, so humans and monkeys share polymorphic clades.
27 Ebersberger et al. (2007) Estimated Trees From 23,210 DNA Sequences In Apes & Rhesus Monkey: Below Are The Numbers That Significantly Resolved the “Species Tree”NEVER equate a haplotype tree to a tree of populations -- this was done (and continues to be done) for mt-Eve!
28 Haplotype Trees ≠Species or Population Trees NEVER equate a haplotype tree to a tree of populations -- this was done (and continues to be done) for mt-Eve!
29 It is dangerous to equate a haplotype tree to a species tree It is dangerous to equate a haplotype tree to a species tree. It is NEVER justified to equate a haplotype tree to a tree of populations within a species because the problem of lineage sorting is greater and the time between events is shorter. Moreover, a population tree need not exist at all.
30 Homoplasy & The Infinite Sites Model Homoplasy is the phenomenon of independent mutations (& many gene conversion events) yielding the same genetic state.Homoplasy represents a major difficulty when trying to reconstruct evolutionary trees, whether they are haplotype trees or the more traditional species trees of evolutionary biology.It is common in coalescent theory (and molecular evolution in general) to assume the infinite sites model in which each mutation occurs at a new nucleotide site.Under this model, there is no homoplasy because no nucleotide site can ever mutate more than once. Each mutation creates a new haplotype.
33 E. g., Apoprotein E Gene Region 0.0.51.1.184.108.40.206.220.127.116.11.5Exon123478569AB1*No recombination has been detected in this region.
34 The Apo-protein E Haplotype Tree Homoplasy creates loops, makes it difficult to resolve a tree. 4 x 4 x 15 =240different ways of breaking loops, all being maximum parsimonious.
35 The Apo-protein E Haplotype Tree Use a Finite Sites mutation model that allows homoplasy. Can show that probability of homoplasy between two nodes increasing with increasing number of observed mutational differences. Therefore, allocate homoplasies to longer branches. Called “Statistical Parsimony” because you can use models to calculate the probability of violating parsimony for a given branch length.
36 The Apo-protein E Statistical Parsimony Haplotype Tree Homoplasy is still common, as shown by circled mutations.The Apo-protein E Statistical Parsimony Haplotype TreeThere are only 2 trees under statistical parsimony & use of other coalescent criteria-- e.g. rare-rare haplotype connections much less likely than common-rare.Circles show homoplasy still common.In this case, most of the homoplasy is associated with Alu sequences, a common repeat type in the human genome that is known to cause local gene conversion, which mimics the effects of parallel mutations.
37 Estimated Times To Common Ancestor (Method of Takahata et al. 2001) Dhc Nuc.Diff.Between Humans& ChimpsDh Nuc.Diff.Within HumansTMRCA = 12Dh/Dhc6 Million Years Ago
38 The Apo-protein E Haplotype Coalescent 3.2The Apo-protein E Haplotype Coalescent2.439371.6Years(x 105)4075116324407319980.85229B3084036495147136736241522290754531063701234
39 Estimate the distribution of the age of the haplotype or clade as a Gamma Distribution (Kimura, 1970) with mean T=4N (or N for mtDNA) and Variance T2/(1+k) (Tajima, 1983) where k is the average pairwise divergence among present day haplotypes derived from the haplotype being aged, measured as the number of nucleotide differences. NOTE: VARIANCE INCREASES WITH INCREASING T AND DECREASING k!
40 The Apo-protein E Haplotype Coalescent 3.239372.41.6Years(x 105)4075244011637319980.8f(t)5229B3084036367349514716245451522290737013106234Years (x 105)
41 Because of Deviations From The Infinite Sites Model, Corrections Must Also be Made in How We Count the Number of Mutations That Occurred in The Coalescent Process.
42 The Basic Idea of Coalescence Is That Any Two Copies of Homologous DNA Will Coalesce Back To An Ancestral Molecule Either Within Or Between SpeciesTimet
43 Mutations Can Accumulate in the Two DNA Lineages During This Time, t, to Coalescence. We Quantify This Mutational Accumulation Through A Molecule Genetic DistanceTimetX MutationsY Mutations
44 Molecule Genetic Distance = X + Y Molecule Genetic Distance = X + Y. If = the neutral substitution rate, then the Expected Value of X = t and the Expected Value of Y = t, So the Expected Value of the Genetic Distance = 2tComplication: Only Under The Infinite Sites Model Are X+Y Directly Observable; Otherwise X+Y ≥ The Observed Number of Differences.TimetX MutationsY MutationsUse Models of DNA Mutation To Correct For Undercounting
45 Molecule Genetic Distance = X + Y = 2 t THE JUKES-CANTOR GENETIC DISTANCE Consider a single nucleotide site that has a probability of mutating per unit time (only neutral mutations are allowed). This model assumes that when a nucleotide site mutates it is equally likely to mutate to any of the three other nucleotide states. Suppose further that mutation is such a rare occurrence that in any time unit it is only likely for at most one DNA lineage to mutate and not both. Finally, let pt be the probability that the nucleotide site is in the same state in the two DNA molecules being compared given they coalesced t time units ago. Note that pt refers to identity by state and is observable from the current sequences. Then,
46 Molecule Genetic Distance = X + Y = 2 t THE JUKES-CANTOR GENETIC DISTANCE Approximating the above by a differential equation yields:extract 2t from the equation given above:
47 Molecule Genetic Distance = X + Y = 2 t THE JUKES-CANTOR GENETIC DISTANCE The above equation refers to only a single nucleotide, so pt is either 0 and 1. Hence, this equation will not yield biologically meaningful results when applied to just a single nucleotide. Therefore, Jukes and Cantor (1969) assumed that the same set of assumptions is valid for all the nucleotides in the sequenced portion of the two molecules being compared. Defining as the observed number of nucleotides that are different divided by the total number of nucleotides being compared, Jukes and Cantor noted that pt is estimated by 1-. Hence, substituting 1- for pt yields:
48 Molecule Genetic Distance = X + Y = 2 t THE KIMURA 2-PARAMETER GENETIC DISTANCE The Jukes and Cantor genetic distance model assumes neutrality and that mutations occur with equal probability to all 3 alternative nucleotide states. However, for some DNA, there can be a strong transition bias (e.g., mtDNA):where is the rate of transition substitutions, and 2is the rate of transversion substitutions. The total rate of substitution (mutation)
49 Molecule Genetic Distance = X + Y = 2 t THE KIMURA 2-PARAMETER GENETIC DISTANCE Kimura (J. Mol. Evol. 16: , 1980) showed thatGENETIC DISTANCE = Dt = 2()t = -1/2ln(1-2P-Q) - 1/4ln(1-2Q)where P is the observed proportion of homologous nucleotide sites that differ by a transition, and Q is the observed proportion of homologous nucleotide sites that differ by a transversion.Note that if (no transition bias), then we expect P = Q/2, so = P+Q = 3/2Q, or Q = 2/3. This yields the Jukes and Cantor distance, which is therefore a special case of the Kimura Distance.If (large transition bias), as t gets large, P converges to 1/4 regardless of time, while Q is still sensitive to time. Therefore, for large times and with molecules showing an extreme transition bias, the distances depend increasingly only on the transversions. Therefore, you can get a big discrepancy between these two distances when a transition bias exists and when t is large enough.
50 Molecule Genetic Distance = X + Y = 2 t You can have up to a 12 parameter model for just a single nucleotide (a parameter for each arrowhead). You can add many more parameters if you consider more than 1 nucleotide at a time.If distances are small (Dt ≤ 0.05), most alternatives give about the same value, so people mostly use Jukes and Cantor, the simplest distance. Above 0.05, you need to investigate the properties of your data set more carefully. ModelTest can help you do this (I emphasize help because ModelTest gives some statistical criteria for evaluating 56 different models -- but conflicts frequently arise across criteria, so judgment is still needed).LOOK AT YOUR DATA!
51 Recombination Can Create Complex Networks Which Destroy the “Treeness” of the Relationships Among Haplotypes.
52 (Templeton et al.,AMJHG 66: 69-83, 2000) LD in the human LPL geneRecombination is notUniformly distributed in thehuman genome, but rather isConcentrated into “hotspots” thatSeparate regions of low to noRecombination.HaplotypeTrees can beEstimated for theseTwo regions, but notFor the entire LPL region.Region of Overlap of the Inferred Intervals Of All 26 Recombination and Gene Conversion Events Not Likely to Be Artifacts.Significant |D’|Non-significant |D’|Too Few Observationsfor any |D’| to be significant(Templeton et al.,AMJHG 66: 69-83, 2000)
53 Because of the random mating equation: Dt=D0(1-r)t Linkage Disequilibrium Is Often Interpreted As An Indicator of the Amount of Recombination. This Is Justifiable When Recombination Is Common Relative To Mutation However, in regions of little to no recombination, the pattern of disequilibrium is determined primarily by the historical conditions that existed at the time of mutation, that is the Haplotype Tree.
54 Apoprotein E Gene Region These Two Sites Are in Strong Disequilibrium in All SamplesNote, African-Americans Have More D Than Europeans & EA Because of Admixture: Not All D Reflects LinkageThese Two Sites Show No Significant Disequilibrium in Any SampleApoprotein E Gene Region0.0.51.1.18.104.22.168.22.214.171.124.5Exon123478569AB1*
55 The Apo-protein E Haplotype Tree All Four Gametes Exist Because of Homoplasy, Not RecombinationThese haplotypes Are T at Site 832 & C At Site 3937These haplotypes Are G at Site 832 & T At Site 3937These mutations areWell separated in timeAnd show little D21143015221575290753612656062417201862441The Apo-protein E Haplotype Tree56029370173560495183211282319624545403647153611163These mutations are close in timeAnd show much Disequilibrium2519987536122440639378321998624515560125603856031065229B30849511331273673407510560166246242449519560157522