1. Allele
frequency
Time

2. Allele frequency

3. Ewens Watterson test
Empirical sample of size N gives:
1) estimate of homozygosity = ∑xi2
2) Number of different alleles, n
Apply these values to the expressions
Compare fobs with fequilib by simulation (bootstrap resampling)

7. Allele
frequency
Time

8. Single Nucleotide Polymorphisms and site frequency
Ingman et al. 2000, Nature

9. Nine polymorphic sites in human Y-chromosomes
N=24, n = 9 haplotypes, S = 9 segregating sites
π = , q =

10. Allele (Site) frequency distribution
Excess of low frequency alleles
Excess of intermediate frequency alleles
Neutral (Ewens) frequency spectrum
Ewens (1972): expression for allele frequencies given
sample size, number of observed alleles
mutation - drift equilibrium (expected)
Watterson (1977):
compare Expected (Ewens) and observed
homozygosity H=Sxi2
Slatkin (1994) Exact test of Ewens-Watterson test
Nielsen (2005) Ann. Rev. Genet. 39:197
Sequence 20 alleles, record frequency of each SNP

11. Tajima’s Test p - q is negative p - q is positive π-ø < 0
Joe
Mary
Scott
Anne
David
= Average number of differences between all pairs of sequences.
Affected by site (allele) frequency
 = Variation based on the number of segregating sites
Independent of site frequency
1/i
π-ø < 0
π-ø > 0
p - q is negative
p - q is positive

12. External branches ~= 4Nem = q Selection affects # external branches
Human mtDNA tree (Ingman et al. 2000, Nature)
Neutral, ancestral
African
Dtaj = -0.72,
P >0.10
Admixture
Africa + E. Asia
Dtaj = +0.08,
P >0.10
Founder event
New World
Dtaj = -2.45,
P < 0.01
Tajima (1989): p - q
Fu & Li (1993):
External branches ~= 4Nem = q
Selection affects # external branches
Both are sensitive to
demographic changes and admixture

13. Human MHC – Human Leukocye antigen (HLA) complex
Defficieny of homozygosity
(excess heterozygosity)
Positive Tajima’s D
Balancing selection favoring
heterozygotes, better able
to mount diverse immune
responses

14. Ewens-Watterson tests vs. Tajima’s D test
What’s the difference
Allele frequency vs. sequence difference
Ewens Watterson test uses alleles of ‘state’ (Fast, Slow, etc)
Tajima’s D uses nucleotide divergence among alleles – more power
Garrigan and Hedrick 2003

15. Tajima’s D, Selection and Demography
Polymorphism and divergence of mutations
Tajima’s D, Selection and Demography
Allele frequency
Time
Selective force qπ =/>/< qS Tajima’s D value
Neutrality = 0
Deleterious mutations < negative
Beneficial (sweep) < negative
Balancing selection > positive
Tajima’s D value Selective force Demographic force
D < 0 Selective sweep Bottleneck; expansion
Deleterious Bottleneck, expansion
D > 0 Balancing selection Population mixture

16. Tajima’s D, Selection and Demography
Tajima’s D value Selective force Demographic force
D < 0 Selective sweep Bottleneck; expansion
Deleterious Bottleneck, expansion
D > 0 Balancing selection Population mixture
How do you use the genome to tell the difference?
Compare multiple loci:
selection acts on individual genes
Demographic forces act on all genes
Polymorphism and divergence of mutations
Allele frequency
Time

17. Polymorphism and divergence
polymorphism = difference within species
divergence (fixed difference) = difference between species { A T A A C G A C
Species 1
Species 2 A A C A { G T G G T G G T A G T A
Fixed
difference
Polymorphic
in species 1
Polymorphic
in both species

18. Hudson Kreitman Aguadé 1987
Multilocus approach:
Jody Hey’s web page
Maximum Likelihood aaproach
Wright & Charlesworth (2004)
Demography affects entire genome
Selection acts on single (few) loci

19. HKA Test The “footprint” of balancing selection at Adh in Drosophila
Kreitman and Hudson (1991) Genetics 127:565-82
polymorphism
Adh locus
Adh-dup
Fast/Slow
polymorphism
HKA Test
Adjacent silent sites
in linkage disequilibrium
Fast F
Fast F
Adh
Adh-dup 16 20
Fast F
Fast F
Slow S 50 13
Slow S
Slow S
Slow S
Distant sites in Linkage Equilibrium
P < 0.02

20. Selective sweep in the domestication of maize from teosinte
Adapted from R.-L. Wang, et al., 1999.

21. The effect of recombination on levels of polymorphism
Marker loci # • • • • • • • * • • • • • •
0.012
0.010
0.008
0.006
0.004
0.002
0.000 • • • # •
Rate of
re-
combi-
nation
DNA polymorphism • • • • • • • • •
Aquadro, Begun
& Kindahl, 1994 *
Physical position along 3rd chromosome
0.1
0.05
0.000 • * • •
DNA
divergence # •
Locus 1 33 1 • • • 26 6
Begun & Aquadro 1992
P < 0.05
Rate of recombination

22. * = beneficial mutation
Reduced polymorphism due to selective sweeps (adaptive mutations and hitchhiking) *
* = beneficial mutation *
No recombination:Polymorphism removed *
Free recombination: Locally reduced variation
Reduced polymorphism due to background selection eliminating deleterious mutations X
X = deleterious mutation
“mutation-free” chromosomes {
No recombination:Polymorphism removed
Free recombination: little effect

23. Recombination rate Supports hitchhiking model (weakly)
419 genes, 24 alleles per gene
DNA polymorphism
Tajima’s D
Recombination rate
Supports hitchhiking
model (weakly)
DNA divergence
Recombination rate

24. Gene trees and evolutionary hypotheses
Mitochondrial “eve”-type estimations require
rate constancy between species and neutrality within species
What is a neutral references locus?
Rand et al Genetics 138:

25. Testing for selection in mtDNA
The genetic code and DNA “phenotypes”

26. Polymorphism in mtDNA and the MK test
N S
Population or family
Sister species
or unrelated
strain B.
dN/dS
‘within’
‘between’
Polymorphism and divergence of mutations
Allele frequency
Time
Type Polymorphic Fixed
of within between
mutation populations? species?
Neutral Yes Yes
Beneficial No Yes
Deleterious Yes No
Balanced Yes Yes & No
dN/dS
‘within’
‘between’
= Neutrality Index (NI)
NI < 1.0 implies positive selection
NI > 1.0 implies negative selection
(opposite of simple dN/dS)
Rand & Kann (1996) MBE
Rand (2008) PLoS Biology

27. Polymorphism and Divergence at Silent and Replacement Sites
Rand&Kann 1996
MBE 13:
=> mildly deleterious => advantageous

29. N.I. is very sensitive to selection
Sawyer&Hartl (1992)
Akashi (1995)
Nachman (1998)
Weinreich & Rand (2000)
Kimura (1983)
Negative Neutral Positive
Stopped 3/8/11

30. MK tests for mtDNA have excess amino acid polymorphism
NI > 1 (Negative selection)
Rand & Kann 1996, 1998; Weinreich&Rand 2000

31. Drift, Draft, and apparent positive selection
Sweeps remove variation, but fix common variants
ND (complex I) genes have lower constraint (more variation)
Sweeps and draft may fix an excess of mildly deleterious variants
Meiklejohn, Montooth, Rand 2007 Trends in Genetics.

32. Fay & Wu (2002): A frequency twist to the MK test
The distribution of site frequencies is important
Type of mutation Rare? Intermediate? Common?
Yes Yes yes
No More likely
Yes No No
No Yes No

33. 419 genes, 24 alleles sequenced/gene, compared to D. simulans

35. Human Chimp How does DNA evolve?
1 ATGCCCCAACTAAATACTACCGTATGGCCCACCATAATTACCCCCATACT 50
||||||||||||||||| ||||||| |||||||||||||||||||||||
1 atgccccaactaaataccgccgtatgacccaccataattacccccatact 50
51 CCTTACACTATTCCTCATCACCCAACTAAAAATATTAAACACAAACTACC 100
||| |||||||| ||| |||||||||||||||||||||| |||| ||||
51 cctgacactatttctcgtcacccaactaaaaatattaaattcaaattacc 100
101 ACCTACCTCCCTCACCAAAGCCCATAAAAATAAAAAATTATAACAAACCC 150
| ||||| ||||||||||| ||||||||||||||||| || || ||||||
101 atctacccccctcaccaaaacccataaaaataaaaaactacaataaaccc 150
151 TGAGAACCAAAATGAACGAAAATCTGTTCGCTTCATTCATTGCCCCCACA 200
||||||||||||||||||||||||| |||||||||||| ||||||||||
151 tgagaaccaaaatgaacgaaaatctattcgcttcattcgctgcccccaca 200
201 ATCC 204
||||
201 atcc 204

36. Measuring DNA Evolution
Align sequences between species
Determine length of sequences, L
Count number of differences
Divergence = proportion of differences
D = p-distance = (number of differences) / (length of sequence)
Rate of divergence
 = (sequence divergence) / (age of common ancestor)
 = D / time
Rate of substitution
 = D / 2 x time
time
Example: 5 differences in 100
D = 0.05, t = 6 million years
Divergence = 0.05/6x106
Divergence = 8.3 x 10-9

37. Jukes Cantor One parameter model
= rate of substitution
PA(t) = ¼ + ¾ e-4at = probability that A remains A at time t
PNN = ¼ + ¾ e-8at = probability that two sequences have the same nucleotide at N
D = proportion of different nucleotides = 1 - PNN
Dhat = 3/4(1-e-8t)
K = - ¾ ln (1-4/3p)
where p = proportion of nucleotide differences (# diffs./total bp)

38. Kimura two-parameter modelb
a = rate of transition substitution
b = rate of transversion substitution
PAA(t) = ¼ + ¾ e-4bt + ½ e-2(a+b)t
= probability that A remains A at time t
K = ½ ln(1/[1- 2P-Q]) + ¼ ln(1/[1-2Q])
where P = proportion of transitional differences
Q = proportion of transversional differences

40. Comparison of models
P-distance
Jukes Cantor
Kimura 2-parameter
Tamura-Nei
Etc…

41. Molecular clocks
Approximately constant
Divergence of proteins
K = •f0
Rate of substitution =
Mutation rate x proportion of
neutral mutations
“Saturation” due to multiple
Hits in DNA evolution

42. Anatomy of a phylogenetic tree
Terminal (external) nodes
Taxa =
OTUs =
Operational
taxonomic units
Taxon1
Taxon2
Taxon3
Taxon4
Taxon5
Taxon6
Polytomy
Non-dichotomous
splitting
External branch
Internal branch
Internal nodes
Root

43. Relative rate test KAC = KBC KOC is shared Tajima test
(m1-m2)2 / (m1+m2)
Chi square, df=1
Species O m1 m2
Species A
Species B
Species C

45. DNA test of neutrality Antigen binding sites: dN/dS > 1
“positive” selection
Neutral prediction:
amino acid (nonsynonymous) substitution rate (dN) should be lower than silent (synonymous) substitution rate (dS)
True for most genes
Follows from functional constraint argument
Different for Major Histocompatibility Complec (MHC) loci
Antigen recognition sequence shows dN > dS
Rest of molecule shows dN > dS, as expected
Amino acid mutations are favored in antigen recognition region
Promotes diversity, better recognition of foreign peptides
Rest of molecule: dN/dS < 1
Negative (purifying) selection

46. The coalescent: the genealogy of alleles as descendants
from a single common ancestral DNA strands
Last parents ‘picked’ =
most recent common ancestor
MRCA
Same parents ‘picked’ =
coalescence
Randomly ‘pick’ parents
Sampled lineages
Rosenberg & Nordborg (2003)
More lineages picking parents, faster coalescence
More parents to pick from = larger population size, slower coalescence

47. * * * * Drift vs. coalescence Drift Coalescence
1 allele drifts to fixation in 5 generations, going forward
5 alleles coalesce (*) to a common ancestral allele 5 generations ago

48. Expected time that n alleles persist
E(Tn) = 4N/n(n-1)

49. Felsenstein’s box of bugs model
Size of box ~ Ne
number of bugs ~ number of alleles

50. Time to next coalescent event
Ne = effective population size
n = number of alleles in sample
T = 4Ne/n(n-1)
T2 = 4Ne/2(1) = 2Ne
T3 = 4Ne/3(2) = 2Ne/3
T4 = 4Ne/4(3) = 2Ne/6
T5 = 4Ne/5(4) = 2Ne/10

51. Generating random coalescent trees
i = 6 alleles
Generate i-1 random numbers (x) , 0< x < 1
Rank them, and assign coalescent times
Ti = -2ln(1-x) /i(i-1)
Hedrick, pg 352:
x6 = 0.22, T6 =
X5 = 0.57, T5 =
Units of 2N generations back in time.

53. Mutation on genealogy
Allele frequency in the sample of 5 alleles:
F(A2) = 1/5, f(A1) = 2/5, f(A3) = 2/5

58. Linkage and Linkage Disequilibrium
AB = 25%
Ab = 25%
aB = 25%
ab = 25%
f(Ab) = f(A) x f(b) A B B b a b A a
AB = 50%
Ab = 0%
aB = 0%
ab = 50% A B
f(Ab) ≠ f(A) x f(b) B b a b A a

60. A locus and B locus are in Linkage Disequilibrium
D = f(AB) x f(ab) - f(Ab) x f(aB) Maximum with no recombination
D -> 0 with free recombination (linkage equilibrium)
When allele frequencies are intermediate: f(A) = f(a) = f(B) = f(b) = 0.5,
and maximal LD occurs so that no recombinants are present:
f(AB) = f(ab) = 0.5, so D = 0.5 x 0.5 – 0.0 x 0.0 = 0.25
When allele frequencies are skewed: f(A) = 0.9, f(a) = 0.1; f(B) = 0.9, f(b) = 0.1
and maximal LD occurs so that no recombinants are present, D is less than 0.25:
f(AB) = 0.9, and f(ab) = 0.1, so D = 0.9 x 0.1 – 0.0 x 0.0 = 0.09

61. Linkage disequilibrium (LD) decays with distance and time
AB = (1-r)/2
Ab = r/2
aB = r/2
ab = (1-r)/2 A B a b
r =
Rate of
recombination

62. Mutations on a genealogy
Mutation events are proportional to branch length
Mutations on external branches define “singletons”
Mutations on internal branches define “non-singletons”

63. Heterozygosity on Exernal vs. Internal Branches
Fu and Li test:
Heterozygosity on Exernal vs. Internal Branches
Total length of external branches = 4Ne
Total mutations on external branches = 4Ne•u
Total length of ALL branches = 4Ne • a
Total mutations on ALL branches =
4Ne•a•u
Total mutations on internal branches =
4Ne•u•a - 4Ne•u = 4Neu(a-1)
Fu & Li statistic ~ Difference between expected
number of external and internal mutations
A measure of heterozygosity contributed by
singleton vs. heterozygosity contributed by
non-singletons
1/i
E(external) - E(internal)/(a-1)
G =
√(Variance (E(external) - E(internal)/(a-1)))