Transcript Although Strength of Genetic Drift is Proportional to 1/2N, Drift Can

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.
COALESCENCE OF
n COPIES OF
HOMOLOGOUS
DNA
Gene Tree
(all copies of
homologous
DNA coalesce to
a common
ancestral
molecule)
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 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:
t 1

1   1 
Prob.(Coalesce at t)  1
 

 xN ef  xNef 
Sample Two Genes at Random
The average time to coalescence is:
t 1


 1
1
Expected(Time to Coalesce )   t1
 
xNef  xNef
t 1 


 xNef

The variance of time to coalescence of two genes (ct) is the
average or expectation of (t-xNef)2 :
t 1


 1 
1
2
2
 ct  t  xNef  1
 
 xNef (xNef  1)  x 2 N ef2  xNef
 xNef  xNef 
t 1

Sample n Genes at Random
n 
n!
n(n  1)
Number of pairs of genes =  

2
2  (n  2)! 2!
Prob.(coalescence in the previous gen.)
n  1
n(n  1)
=   
2xN
2 xN
Prob.(no coalescence in the previous gen.)
n(n 1)
=1 2xN
Sample n Genes at Random
 n(n  1) t -1 n(n  1)
Prob.(first coalescence in t generations) = 1 

2xN  2xN
 n(n 1) t -1 n(n 1)
2xN
E(time to first coalescence) =  t1 

2xN  2xN
n(n 1)
t 1 

t 1
2



 n(n 1)  n(n 1)
n(n
1)
2xN  2xN
2
1  t 

1
 1


4 N  
2xN  2xN
n(n 1) n(n 1) 
t 1 

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:
2xN
E(time between k  1 and k coalescent events) =
(n  k 1)(n  k)


2xN
2xN
 
 1

(n  k 1)(n  k) (n  k 1)(n  k) 
2
k
n1
E(time to coalescence of all n genes) = 
2xN
 2xN1
k 1 (n  k 1)(n  k)

1
n
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 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 Random
The variance of time to coalescence of n genes is:
n


2xN
2xN
1
2 2
 (n  k  1)(n  k) (n  k  1)(n  k) 1 4 x N  (i)2 (i 1)2
k1
i 2
n1
•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.”
Generation
Buri’s
Experiment on
Genetic
Drift
Number of
Populations
Fixed for bw
Number of
Populations
Fixed for bw 75
1
0
0
2
0
0
3
0
0
4
0
1
5
0
2
6
1
3
7
3
3
8
9
10
11
12
13
14
15
16
17
18
19
5
5
7
11
12
12
14
18
23
26
27
30
5
6
8
10
17
18
21
23
25
26
28
28
0
2
4
6
8 10 12 14 16 18 20 22 24 26 28 30 32
Number of bw75 Alleles
Fixation (Coalescence) Times in 105
Replicates of the Same Evolutionary Process
Generation of Fixation
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
0
Locus
MX1
FUT2
CCR5
Lactase
FUT6
CYP1A2
Hb-Beta
HFE
MS205
EDN
ECP
MC1R
PDHA1
RRM2P4
TNFSF5
AMELX
APLX
HS571B2
5
G6PD
7
Xq13.3
9
MSN/ALAS2
FIX
MAO
mtDNA
Y-DNA
TMRCA (In Millions of Years)
Estimated Coalescence Times
for 24 Human Loci
Uniparental Haploid DNA Regions
8
X-Linked Loci
6
Autosomal Loci
4
3
2
1
Coalescence
With Mutation
Mutation
Creates
Variation
and
Destroys
Identity
by
Descent
Coalescence Before Mutation
Prob.(coalescence before mutation )  Prob.(identity by descent)
…
…
t 1


 1 
1
= 1
 
(1  )2t
 xNef  xNef 
Prob. of no
Prob. of no   Prob. of  

 
 mutation in
 coalescence  coalescence 

 
  2t DNA
for t -1 gen.   at gen. t  
replications






Mutation Before Coalescence
…
…
Mutation
Prob.(mutation before coalescence ) 

1 
2t1
1
2

(1

)


 xN 

ef 
t

Mutation and Coalescence:
Genetic Diversity
Prob.(mutation before coalescence | mutation or coalescence )
2 (1  )
1
2t 1

2 (1  )
1
2t 1
1
xN ef

t
1
xN ef
1
xN ef

t

(1  )2t 1
1
xN ef
2xN ef   2

t 1
2xN ef   3 1

2xNef   2
2xNef 



2xNef   3  1 2xNef   1   1
= Expected Heterozygosity (where xNef)
Gene Vs. Allele (Haplotype) Tree
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.
Unrooted Haplotype Tree
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 Polymorphism
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 a
TIFF (Uncompressed) decompressor
are needed to see this picture.
QuickTime™ and a
TIFF (LZW) decompressor
are needed to see this picture.
Homoplasy & The Infinite Sites Model
The distributio n of polymorphi c nucleotide sites i n a 9.7 kb region of the human Lipoprotein
Lipase gene over nucleotides associated with thre e known mutagenic motifs and all remainin g
nucleotide positions.
E. g., Apoprotein E Gene Region
Exon 4
Exon 3
Exon 2
Exon 1
5361
5229B
5229 A
4951
4075
4036
3937
3701*
3673
3106
2907
2440
1998
1575
1522
1163
832
624
560
545
471
30 8
73
No recombination has been detected in this region.
5.5
5.
4.5
4.
3.5
3.
2.5
2.
1.5
1.
0.5
0.
The Apoprotein E
Haplotype
Tree
21
14
24
560
26
22
0
0
9
4
560
1
2907
560
0
30
10
560
16
29
17
19
0
28
15
0
25
1998
7
5361
2
2440
3
2440
11
12
3937
6
3937
8
832
0
1998
23
0
3106
5
0
18
0
13
20
27
Chimpanzee (Outgroup)
31
A. Maximum Parsimony
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.
TCC
624
C
T
560 624 1575
AC T
T
560 624 1575
1575
C
TTC
560 624 1575
T
T
A
A
O
OR
C
624
OR
ATC
T
560 624 1575
OR
B. Statistical Parsimony
TCC
C
624
T
560 624 1575
ACT
560 624 1575
T
1575
C
TTC
560 624 1575
T
T
A
A
O
C
OR
624
T
ATC
560 624 1575
Homoplasy is still common, as shown by circled mutations.
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.
The Apoprotein E
Statistical
Parsimony
Haplotype Tree
Estimated Times To Common Ancestor
(Method of Takahata et al. 2001)
Dhc Nuc.Diff.
Between Humans
& Chimps
Dh Nuc.Diff.
Within Humans
TMRCA = 12Dh/Dhc
6 Million Years Ago
3.2
The Apoprotein E
Haplotype
Coalescent
2.4
3937
1.6
4075
2440
1163
73
1998
5229B
308
4036
4951
3673
624
545
1522
471
2907
3106
3701
9 16 6 27 2 28 1 14 29 30 12 13 17 20 5 31
2
0.8
3
4
0
Years
(x 105)
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
3.2
2.4
3937
1.6 Years
(x 105)
4075
f(t)
1163
2440
73
1998
0.8
5229B
308
4036
4951
3673
624
9 16
2
545
6
27
2 28
471
1522
2907
3701
1
3
Years (x 105)
14
29 30
3106
12 13
17
4
20
5
31
0
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
Time
t
Mutations Can Accumulate in the Two DNA Lineages During
This Time, t, to Coalescence. We Quantify This Mutational
Accumulation Through A Molecule Genetic Distance
Time
t
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
Time
t
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 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  )2  (1 pt )2 / 3  (1 2 )pt  2 (1 pt )/ 3
Molecule Genetic Distance = X + Y = 2 t
THE JUKES-CANTOR GENETIC DISTANCE
pt +1  (1 2)pt  2(1 pt )/3
p  pt1  pt  2pt  2 (1 pt )/ 3   83 pt  23 

Approximating
the above by a differential equation yields:
dpt
  83 pt  23 
dt
extract 2t
from the
equation
given above:

pt  1 3e8t / 3/ 4
pt  14  43 e8 t / 3
3
4

e8 t / 3  pt  14
 83 t  n 43 pt  13 
2t   43 n43 pt  13   DJC
Molecule Genetic Distance = X + Y = 2 t
THE JUKES-CANTOR GENETIC DISTANCE
DJC   43 n43 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:
2t   34 n1 43    DJC
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):

Pyrimidines
T
C



Purines

A

G
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 = 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/ Q, or Q = 2/ . This yields the Jukes and Cantor distance, which is therefore a
2
3
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

Pyrimidines
T
C



Purines

A

G
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!
Recombination Can Create Complex Networks Which Destroy the “Treeness”
of the Relationships Among Haplotypes.
Recombination is not
Uniformly distributed in the
human genome, but rather is
Concentrated into “hotspots” that
Separate regions of low to no
Recombination.
Region of Overlap of the
Inferred Intervals Of All 26
Recombination and Gene
Conversion Events Not
Likely to Be Artifacts.
Number of Recombination Events
18
16
LD in the human LPL gene
Haplotype
Trees can be
Estimated for these
Two regions, but not
For the entire LPL region.
14
12
10
8
6
Significant |D’|
4
2
Non-significant |D’|
0
0
1000
2000
3000
4000
5000
(Templeton et al.,AMJHG 66: 69-83, 2000)
6000
7000
8000
9000
10000
Too Few Observations
for any |D’| to be
significant
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.
Note, AfricanAmericans Have More
D Than Europeans &
EA Because of
Admixture: Not All D
Reflects Linkage
0.
0.5
Apoprotein E Gene Region
1.
1.5
2.
2.5
3.
3.5
4.
4.5
5.
5.5
Exon 4
Exon 3
Exon 2
Exon 1
These Two Sites Show No Significant Disequilibrium in Any Sample
5361
5229B
5229 A
4951
4075
4036
3937
3701*
3673
3106
2907
2440
1998
1575
1522
1163
832
624
560
545
471
30 8
73
These Two Sites Are in Strong Disequilibrium in All Samples
All Four Gametes Exist Because of
Homoplasy, Not Recombination
These mutations are
Well separated in time
And show little D
21
2907
560
545
1998
7
5361
2440
2
6
4036
3937
4951
471
5361
832
1998
560
5229B
3673
5
624
15
3106
4951
13
31
4075
560
10
624
4951
560
12
8
308
24
18
23
560
3
20
73
1163
560
27
17
4 560 1
29 3701 832
11
19
624
28
25
5361
624
624
The Apoprotein E
Haplotype
Tree
1522
30
1575
26
These haplotypes
Are T at Site 832 &
C At Site 3937
14
624
9
560
These haplotypes Are G at
Site 832 & T At Site 3937
1575
22
16
These mutations
are close in time
And show much
Disequilibrium