Coalescence

DNAReplication

DNACoalescence

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 Nef; Therefore, the prob. of coalescence in one generation is 1/(xNef)

Sample Two Genes at RandomThe 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:

€

Prob.(Coalesce at t) = 1−1

xN ef

⎛

⎝ ⎜

⎞

⎠ ⎟

t −1

1

xN ef

⎛

⎝ ⎜

⎞

⎠ ⎟

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 : €

Expected(Time to Coalesce) = tt =1

∞

∑ 1−1

xN ef

⎛

⎝ ⎜

⎞

⎠ ⎟

t −1

1

xN ef

⎛

⎝ ⎜

⎞

⎠ ⎟= xN ef

€

ct2 = t − xN ef( )

2

t =1

∞

∑ 1−1

xN ef

⎛

⎝ ⎜

⎞

⎠ ⎟

t −1

1

xN ef

⎛

⎝ ⎜

⎞

⎠ ⎟= xN ef (xN ef −1) = x2 N ef

2 − xN ef

Sample n Genes at Random

€

Number of pairs of genes =n

2

⎛

⎝ ⎜

⎞

⎠ ⎟=

n!

(n − 2)!2!=

n(n −1)

2

€

Prob.(coalescence in the previous gen.) =n

2

⎛

⎝ ⎜

⎞

⎠ ⎟1

xN=

n(n −1)

2xN

€

Prob.(no coalescence in the previous gen.) =1 -n(n −1)

2xN

Sample n Genes at Random

€

Prob.(first coalescence in t generations) = 1 -n(n −1)

2xN

⎛

⎝ ⎜

⎞

⎠ ⎟t -1

n(n −1)

2xN

€

E(time to first coalescence) = tt =1

∞

∑ 1 -n(n −1)

2xN

⎛

⎝ ⎜

⎞

⎠ ⎟t -1

n(n −1)

2xN=

2xN

n(n −1)

€

12 = t −

n(n −1)

4 N

⎛

⎝ ⎜

⎞

⎠ ⎟2

1−n(n −1)

2xN

⎛

⎝ ⎜

⎞

⎠ ⎟

t =1

∞

∑t −1

n(n −1)

2xN=

2xN

n(n −1)

2xN

n(n −1)−1

⎛

⎝ ⎜

⎞

⎠ ⎟

Sample n Genes at RandomOnce 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:

€

E(time between k −1 and k coalescent events) =2xN

(n − k +1)(n − k)

€

k2 =

2xN

(n − k +1)(n − k)

2xN

(n − k +1)(n − k)−1

⎛

⎝ ⎜

⎞

⎠ ⎟

€

E(time to coalescence of all n genes) =2xN

(n − k +1)(n − k)k =1

n−1

∑ = 2xN 1− 1n( )

Sample n Genes at RandomThe 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 0.0444Nef to 3.6Nef.•Let n = 100, then the time span covered by coalescent events is expected to range from 0.0004Nef 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 2xNef

Sample n Genes at RandomThe 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 Nef

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.”

€

2xN

(n − k +1)(n − k)

2xN

(n − k +1)(n − k)−1

⎛

⎝ ⎜

⎞

⎠ ⎟

k=1

n−1

∑ ≈ 4 x 2N 2 1

(i)2(i −1)2i= 2

n

∑

Buri’s Experi-ment on Genetic

Drift

GenerationNumber ofPopulationsFixed for bw

Number ofPopulations

Fixed for bw75

1 0 0

2 0 0

3 0 0

4 0 1

5 0 2

6 1 3

7 3 3

8 5 59 5 6

10 7 811 11 1012 12 1713 12 1814 14 2115 18 2316 23 2517 26 2618 27 2819 30 28

0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32

Number of bw75 Alleles

Generation of Fixation

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 Nef♀; Therefore, the prob. of coalescence in one generation is 1/(Nef♀)

Expected Coalescence Times for a Large Sample of Genes

Mitochondrial DNA 2Nef♀=Nef (if Nef♀=1/2Nef)

Y-Chromosomal DNA 2Nef♂=Nef (if Nef ♂=1/2Nef)

X-Linked DNA 3Nef

Autosomal DNA 4Nef

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

TM

RC

A (

In M

illio

ns o

f Y

ear

s)

Uniparental Haploid DNA Regions

X-Linked Loci

Autosomal Loci

Coalescence With Mutation

Mutation Creates

Variation and

Destroys Identity

by Descent

Coalescence Before Mutation… …

€

Prob.(coalescence before mutation) = Prob.(identity by descent)

= 1−1

xN ef

⎛

⎝ ⎜

⎞

⎠ ⎟

t −1

1

xN ef

⎛

⎝ ⎜

⎞

⎠ ⎟(1− μ )2t

=

Prob. of no

coalescence

for t -1 gen.

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟×

Prob. of

coalescence

at gen. t

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟×

Prob. of no

mutation in

2t DNA

replications

⎛

⎝

⎜ ⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ⎟

Mutation Before Coalescence… …

Mutation

€

Prob.(mutation before coalescence) =

1−1

xNef

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

t

2μ(1− μ)2t−1

Mutation and Coalescence: Genetic Diversity

= Expected Heterozygosity (where xNef)

€

Prob.(mutation before coalescence| mutation or coalescence)

=2μ (1− μ )2t −1 1− 1

xNef( )t

2μ (1− μ )2t −1 1− 1xNef( )

t+ 1

xNef(1− μ )2t 1− 1

xNef( )t −1 =

2xN ef μ − 2μ

2xN ef μ − 3μ +1

€

2xNef μ − 2μ

2xNef μ − 3μ +1≈

2xNef μ

2xNef μ +1=

θ

θ +1

Gene Vs. Allele (Haplotype) Tree

Gene Trees vs. Haplotype TreesGene 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.

Unrooted Haplotype Tree

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”

Haplotype Trees ≠Species or Population Trees

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.

Homoplasy & The Infinite Sites Model

QuickTime™ and aTIFF (Uncompressed) decompressor

are needed to see this picture.

QuickTime™ and aTIFF (LZW) decompressor

are needed to see this picture.

Homoplasy & The Infinite Sites Model

Motif Number of

Nucleotides in

Motif

Number of

Polymorphic

Nucleotides

Percent Polymorphic

CG

198 19 9.6%

polymerase arrest

sites with motif

TG(A/G)(A/G)GA

264 8 3.0%

Mononucleotide

Runs ≥ 5

Nucleotides

456 15 3.3%

All Other Sites 8,777 46 0.5%

The distribution of polymorphic nucleotide sites in a 9.7 kb region of the human Lipoprotein Lipase gene over nucleotides associated with three known mutagenic motifs and all remaining nucleotide positions.

E. g., Apoprotein E Gene Region0. 0.5 1. 1.5 2. 2.5 3. 3.5 4. 4.5 5. 5.5

Exon

1

Exon

2

Exon

3

Exon

4

73 308

471

545

560

624

832

1163

1522

1575

1998

2440

2907

3106

3673

3937

4036

4075

4951

5229

A52

29B

5361

3701*

No recombination has been detected in this region.

The Apo-protein E Haplotype

Tree

0

2

3 8

6

14

725

28

11

19

10

26

53611998 2440

2440

3937

3937

0

13

12

0

21

0

560

560

29

30

14

2907

0

27

9560

56016

0

2224

832

Chimpanzee (Outgroup)

17

0

23

20

1998 5

0

18

15

313106

0

The Apo-protein E Haplotype Tree

A C T OC TC

A

T

T

A. Maximum Parsimony

B. Statistical Parsimony

560 624 15751575

T C C560 624 1575

A T C560 624 1575

624

C T624

T T C560 624 1575

A

T

A C T OC TC

A

T

T560 624 15751575

T C C560 624 1575

A T C560 624 1575

624

C T624

T T C560 624 1575

A

T

OR OR OR

OR

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)

Dh Nuc.Diff.Within Humans

Dhc Nuc.Diff.Between Humans

& Chimps

6 Million Years Ago

TMRCA = 12Dh/Dhc

The Apo-protein E

Haplotype Coalescent

3937

4075

5229B

624

308

3673

545

2440 1163

1522

3701

2907

4714951

73

3106

4036

1998

3.2

2.4

1.6

0.8

0

Years(x 105)

9 16 6 27 2 28 1 14 29 30 12 13 17 20 5 31

2 3 4

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!

The Apo-protein E Haplotype Coalescent

3937

4075

5229B

624

308

3673

545

2440 1163

15223701

2907

4714951

73

3106

4036

1998

3.2

2.4

1.6

0.8

0

Years(x 105)

9 16 6 27 2 28 1 14 29 30 12 13 17 20 5 31

2 3 4

Years (x 105)

f(t)

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 M

utations Y M

utat

ions

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 = 2t

t

Time

X M

utations Y M

utat

ions

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 tTHE 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,

€

pt+1 = pt(1− μ )2 +(1− pt )2μ / 3 ≈ (1− 2μ )pt + 2μ (1− pt )/ 3

Molecule Genetic Distance = X + Y = 2 tTHE JUKES-CANTOR GENETIC DISTANCE

€

pt +1 ≈ (1− 2μ) pt + 2μ(1− pt ) /3

€

Δp = pt+1 − pt = −2μpt + 2μ (1− pt )/ 3 = − 83 μpt + 2

3 μ

Approximating the above by a differential equation yields:

€

dpt

dt= − 8

3 μpt + 23 μ

€

pt = 1+ 3e−8μt / 3( ) / 4

extract 2t from the equation given above:

€

pt = 14 + 3

4 e−8μt / 3

34 e−8μt / 3 = pt − 1

4

− 83 μt = l n 4

3 pt − 13( )

2μt = − 34 l n 4

3 pt − 13( ) = DJC

Molecule Genetic Distance = X + Y = 2 tTHE JUKES-CANTOR GENETIC DISTANCE

€

DJC = − 34 l n 4

3 pt − 13( )

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:

€

2μt = − 34 l n 1− 4

3 π( ) ≡ DJC

Molecule Genetic Distance = X + Y = 2 tTHE 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):

Pyrimidines T C

A GPurines

α

α

β β

β β

where α is the rate of transition substitutions, and 2β is the rate of transversion substitutions. The total rate of substitution (mutation) α β

Molecule Genetic Distance = X + Y = 2 tTHE KIMURA 2-PARAMETER GENETIC DISTANCE

Kimura (J. Mol. Evol. 16: 111-120, 1980) showed that

GENETIC DISTANCE = Dt = 2(α β)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 α β (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 α β (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.

Pyrimidines T C

A GPurines

α

α

β β

β β

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!

Recombination Can Create Complex Networks Which Destroy the “Treeness” of the Relationships Among Haplotypes.

(Templeton et al.,AMJHG 66: 69-83, 2000)

0

2

4

6

8

10

12

14

16

18

0 1000 2000 3000 4000 5000 6000 7000 8000 9000 10000

Number of Recombination Events

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 geneRecombination is notUniformly distributed in thehuman genome, but rather isConcentrated into “hotspots” thatSeparate regions of low to noRecombination.

Significant |D’|

Non-significant |D’|

Too Few Observationsfor any |D’| to be significant

HaplotypeTrees can beEstimated for theseTwo regions, but notFor the entire LPL region.

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.

Apoprotein E Gene Region0. 0.5 1. 1.5 2. 2.5 3. 3.5 4. 4.5 5. 5.5

Exon

1

Exon

2

Exon

3

Exon

4

73 308

471

545

560

624

832

1163

1522

1575

1998

2440

2907

3106

3673

3937

4036

4075

4951

5229

A52

29B

5361

3701*

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

560

560

560560

560

1575

624

624624

624

1522

5361

5361

5361

4951

4951

4951

832

83224401998 1998

3937

5229B

4075

1163 4036

73

471

14

1119

17 20 18

23

1512

25

13

10 16

24

2

22

67 5

1

1575

560

560

624

624

21

26

4

3

31

3106

28545

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 areWell separated in timeAnd show little D

These mutations are close in timeAnd show much Disequilibrium

All Four Gametes Exist Because of Homoplasy, Not Recombination