Presentation on theme: "Coalescence DNA Replication DNA Coalescence A coalescent event occurs when two lineages of DNA molecules merge back into a single DNA molecule at some."— Presentation transcript:
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.
Gene Tree (all copies of homologous DNA coalesce to a common ancestral molecule) COALESCENCE OF n COPIES OF HOMOLOGOUS DNA
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 event Under these conditions, the probability that two gametes are drawn from the same parental individual is 1/N With 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 N ef ; Therefore, the prob. of coalescence in one generation is 1/(xN ef )
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:
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-xN ef ) 2 :
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 k th event is:
Sample n Genes at Random The average times to the first and last coalescence are: 2xN ef /[n(n-1)] and 2xN ef (1-1/n) Let n = 10 and x=2, then the time span covered by coalescent events is expected to range from 0.0444N ef to 3.6N ef. Let n = 100, then the time span covered by coalescent events is expected to range from 0.0004N ef to 3.96N ef. 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 2xN ef
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 N ef and the variances are proportional to N ef 2. 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.”
Fixation (Coalescence) Times in 105 Replicates of the Same Evolutionary Process Problem: No Replication With Most Real Data Sets. Only 1 Realization.
Evolutionary Stochasticity Using the standard molecular clock and an estimator of of 10 -8 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!
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 event Under 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 N ef ♀ ; Therefore, the prob. of coalescence in one generation is 1/(N ef ♀ )
Expected Coalescence Times for a Large Sample of Genes Mitochondrial DNA2N ef ♀ =N ef ( if N ef ♀ = 1 / 2 N ef ) Y-Chromosomal DNA2N ef ♂ =N ef ( if N ef ♂ = 1 / 2 N ef ) X-Linked DNA3N ef Autosomal DNA4N ef
Estimated Coalescence Times for 24 Human Loci 0 1 2 3 4 5 6 7 8 9 Y-DNA mtDNA MAO FIX MSN/ALAS2 Xq13.3 G6PD HS571B2 APLX AMELX TNFSF5 RRM2P4 PDHA1 MC1R ECP EDN MS205 HFE Hb-Beta CYP1A2 FUT6 Lactase CCR5 FUT2 MX1 Locus TMRCA (In Millions of Years) Uniparental Haploid DNA Regions X-Linked Loci Autosomal Loci
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.
The Inversion Tree Is Not Always The Same As A Tree of Species Or Populations, In This Case Because of: Transpecific Polymorphism Haplotype trees are not new in population genetics; they have been around in the form of inversion trees since the 1930’s.
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.
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”
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.
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.
E. g., Apoprotein E Gene Region 0.0.51.1.220.127.116.11.18.104.22.168.5 E x o n 1 E x o n 2 E x o n 3 E x o n 4 7 3 3 0 8 4 7 1 5 4 5 5 6 0 6 2 4 8 3 2 1 1 6 3 1 5 2 2 1 5 7 5 1 9 9 8 2 4 4 0 2 9 0 7 3 1 0 6 3 6 7 3 3 9 3 7 4 0 3 6 4 0 7 5 4 9 5 1 5 2 2 9 A 5 2 2 9 B 5 3 6 1 3 7 0 1* No recombination has been detected in this region.
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.
The Apo- protein E Statistical Parsimony Haplotype Tree 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. Homoplasy is still common, as shown by circled mutations.
Estimated Times To Common Ancestor (Method of Takahata et al. 2001) D h Nuc.Diff. Within Humans D hc Nuc.Diff. Between Humans & Chimps 6 Million Years Ago TMRCA = 12D h /D hc
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 T 2 /(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!
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.
The Basic Idea of Coalescence Is That Any Two Copies of Homologous DNA Will Coalesce Back To An Ancestral Molecule Either Within Or Between Species t Time
Mutations Can Accumulate in the Two DNA Lineages During This Time, t, to Coalescence. We Quantify This Mutational Accumulation Through A Molecule Genetic Distance t Time X Mutations Y Mutations
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 = 2 t t Time X Mutations Y Mutations Complication: Only Under The Infinite Sites Model Are X+Y Directly Observable; Otherwise X+Y ≥ The Observed Number of Differences. Use Models of DNA Mutation To Correct For Undercounting
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 p t 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 p t refers to identity by state and is observable from the current sequences. Then,
Molecule Genetic Distance = X + Y = 2 t THE JUKES-CANTOR GENETIC DISTANCE Approximating the above by a differential equation yields: extract 2 t from the equation given above:
Molecule Genetic Distance = X + Y = 2 t THE JUKES-CANTOR GENETIC DISTANCE The above equation refers to only a single nucleotide, so p t 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 p t is estimated by 1- . Hence, substituting 1- for p t yields:
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 2 is the rate of transversion substitutions. The total rate of substitution (mutation)
Molecule Genetic Distance = X + Y = 2 t THE KIMURA 2-PARAMETER GENETIC DISTANCE Kimura (J. Mol. Evol. 16: 111-120, 1980) showed that GENETIC DISTANCE = D t = 2( )t = - 1 / 2 ln(1-2P-Q) - 1 / 4 ln(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 / 2 Q, 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.
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 (D t ≤ 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!
Recombination Can Create Complex Networks Which Destroy the “Treeness” of the Relationships Among Haplotypes.
(Templeton et al.,AMJHG 66: 69-83, 2000) Region of Overlap of the Inferred Intervals Of All 26 Recombination and Gene Conversion Events Not Likely to Be Artifacts. LD in the human LPL gene Recombination is not Uniformly distributed in the human genome, but rather is Concentrated into “hotspots” that Separate regions of low to no Recombination. Significant |D ’ | Non-significant |D ’ | Too Few Observations for any |D ’ | to be significant Haplotype Trees can be Estimated for these Two regions, but not For the entire LPL region.
Because of the random mating equation: D t =D 0 (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.
Apoprotein E Gene Region 0.0.51.1.22.214.171.124.126.96.36.199.5 E x o n 1 E x o n 2 E x o n 3 E x o n 4 7 3 3 0 8 4 7 1 5 4 5 5 6 0 6 2 4 8 3 2 1 1 6 3 1 5 2 2 1 5 7 5 1 9 9 8 2 4 4 0 2 9 0 7 3 1 0 6 3 6 7 3 3 9 3 7 4 0 3 6 4 0 7 5 4 9 5 1 5 2 2 9 A 5 2 2 9 B 5 3 6 1 3 7 0 1* These Two Sites Are in Strong Disequilibrium in All Samples These Two Sites Show No Significant Disequilibrium in Any Sample Note, African- Americans Have More D Than Europeans & EA Because of Admixture: Not All D Reflects Linkage
The Apo- protein E Haplotype Tree 560 1575 624 1522 5361 4951 832 24401998 3937 5229B 4075 1163 4036 73 471 14 11 19 172018 23 15 12 25 13 10 16 24 2 22 6 7 5 1 1575 560 624 21 26 4 3 31 3106 28 545 27 3673 308 29 3701 8 30 2907 9 These haplotypes Are T at Site 832 & C At Site 3937 These haplotypes Are G at Site 832 & T At Site 3937 These mutations are Well separated in time And show little D These mutations are close in time And show much Disequilibrium All Four Gametes Exist Because of Homoplasy, Not Recombination