Natural Selection and Haplotype Trees 2N - NSLC · Maynard Smith (1970) ... the Operation of Natural Selection a) ... default is the Jukes-Cantor Model. For example, these numbers

1

Natural Selection andHaplotype Trees

Recall From the Neutral Theory:

• Rate of Neutral Evolution Over Long Periods ofTime (interspecific) = 1/2N×(2Νµ) = µ =(theneutral mutation rate)

• Amount of Intraspecific Polymorphism (asmeasured by expected heterozygosity) = θ/(1+θ) ≅ θ = 4Nµ

• Do not know µ, & µ can vary from locus to locusor even base-pair to base-pair, but the Ratio ofIntraspecific Polymorphism to InterspecificDivergence should be proportional to θ/µ = 4Nµ/µ = 4Ν for all loci and base pairs under neutrality

Maynard Smith (1970)Suggested This Ratio As a Test

of Neutrality• Problem Then: Interspecific Divergence Data

Came from Amino Acid Sequencing, andIntraspecific Polymorphism Data Came FromProtein Electrophoresis: The methods were non-comparable, so the Ratio was highly suspect fortechnical reasons

• This problem disappears with Haplotype Treesthat span the intra- and interspecific domains.

Inversion Treesfor Drosophila

AnticipatedThis Approach:

E.g., TheHawaiian

DrosophilaInversion Tree

2

Inversion EvolutionIn Hawaii

• Much Speciation Associated with Inter-IslandFounder Events

• Most Founders Are Probably Single GravidFemales Inseminated By Multiple Males

• If a Female Mated with n Males, Then The FounderEvents Have 2+n X-Chromosomes to 2+2nAutosomes.

• Templeton (1987) Used This to Test Ohta’s “NearlyNeutral” Theory That There Will Be EffectivelyNeutral Evolutionary Rates Whenever |s| <1/(4N).This implies that X–chromosomes should evolvemore rapidly than Autosomes for HawaiianDrosophila.

Can Test This Null Hypothesis With aContingency Chi-Square Test

N=a+b+c+d+e+fnIII=c+fnII=b+enI=a+dSum

nB=d+e+ffedPolymorphicWithin Species

nA=a+b+ccbaFixed BetweenSpecies

SumMutationType III

MutationType II

MutationType I

Under the null hypothesis, the expected number in any cell = (rowsum)(column sum)/N. For Example, the expected number in

Mutation Type I, Tree Position Fixed = nAnI/N. After calculatingthe expected numbers, perform an ordinary chi-square test with

degrees of freedom = (no. of rows-1)(no. of columns-1).

Tree Position:

Basic Idea of The Contingency Chi-Square Test

Choose Tree Positions That Should Be Sensitive Tothe Operation of Natural Selectiona) Fixed differences represent mutations that arose

within a species, increased in frequency untilfixation, and then have persisted in time in theentire species – this demonstrates a certaindegree of evolutionary success

b) Polymorphisms represent mtuations that arosewithin a species but have not gone to fixation –less demonstrated evolutionary success

Choose Mutational Types That Should Be Sensitiveto the Operation of Natural Selection

Inversion EvolutionIn Hawaii

X chrom Chrom 2 Chrom 3 Chrom 4 Chrom 5Fixed between species 59 11 14 32 11Polymorphic within 13 9 18 35 11

χ2 = 23.19, df=4

Control: The Mainland repleta Group of Drosophila

X chrom Chrom 2 Chrom 3 Chrom 4 Chrom 5Fixed between species 12 79 16 2 8Polymorphic within 7 86 15 5 5

χ2 =3.62, df=4

3

Haplotype Trees Allow SimpleTests Of Hypotheses About

Natural Selection Both Withinand Between Species

Redundancy of the Genetic Code

The McDonald-Kreitman (1989)Test For Protein Coding DNA

Example: Eanes et al [1993. PNAS 90:7475] surveyed Drosophila simulans & D.melanogaster) using DNA sequencing to find levels of polymorphism at the G6PDlocus.

Amino acidreplacement Synonymous

polymorphic 2 36fixed 21 26

Use Fisher’s Exact Test (too few samples for standard χ2 contingency tests) to rejecthomogeneity (neutrality) at p<0.001.

Uses Same Tree Positions as Earlier Inversion Work: polymorphic within speciesor fixed between them.Uses the mutational categories (for protein coding DNA) of nonsynonymous(amino acid replacement) and synonymous (silent). It is expected that replacementmutations should be more sensitive to selection than silent mutations.

Positions in the Haplotype Tree• Tip Branch: A branch leading to only one haplotype. (Recent

and with no known descendants, unproven evolutionarily orevolutionary dead-end.)

• Interior Branch: A branch leading from one internal node toanother. (Successful evolutionarily because an internal nodehas two or more descendant haplotypes and have persisted forlonger periods of time on average than tip mutations.)

• Interspecific Branch: A branch leading from a part of the treedefined by haplotypes from one species to a haplotype or cladeof haplotypes in another species or common ancestral species.(Successful evolutionarily because it has gone to fixationwithin a species and has persisted for long periods of time.)

• To Make These Categories Mutually Exclusive, Can SeparateTips and Interiors Only For Intraspecific Portions of theHaplotype Tree.

4

Evolution ofCytochrome Oxidase II

712TipC-term(activesite)

212Interior253Fixed

Probability = 0.004

Probability = 0.27

660Fixed

310Interior28TipN-term

(trans-mem-brane)

Replace-mentSilentTree

PositionRegion

Evolution ofCytochrome Oxidase II

712TipC-term(activesite)

212Interior253Fixed

Probability = 0.004

Probability = 0.27

660Fixed

310Interior28TipN-term

(trans-mem-brane)

Replace-mentSilentTree

PositionRegion

There is natural selectionagainst amino acid changes

in the C-terminal half.

Sometimes, small sample sizesrequire pooling.

Note, the McDonald-Kreitman testpools together tips and intraspecific

interiors versus fixed.This pooling does not make sense.

A better pooling is tips versusintraspecific interiors and fixed

(called “young” vs. “old”)

Nucleotide Polymorphism at the RpII215 Gene inDrosophila subobscura: Weak Selection on

Synonymous Mutations (Lloparta & Aguadé,Genetics 2000)

Synonymous codons are divided into preferred and unpreferred toReflect codon usage in this gene in the genus Drosophila.

5

Neutrality Implies That TheRatio of Fixed Mutations to

Polymorphic Mutations ShouldBe Constant Across All Loci

The HKA Test (Hudson,Kreitman & Aguade,

Genetics 1987) Tests ForHomogeneity Of This RatioAcross ≥ 2 Loci With a Chi-

Square Test Statistic

Can Examine Ratio of NonsynonymousSubstitutions Per Nonsynonymous Site(dN) to Synonymous Substitutions Per

Synonymous Site (dS) on Each Branch of aHaploytpe Tree to Examine Heterogeneity

in Natural Selection:• Under Neutrality, Expect dN/dS = 1

• Under Purifying Selection (favors maintaining thecurrent structure): Expect dN/dS < 1

• Under Directional Selection (favors change): ExpectdN/dS > 1

One Potential Problem With This Approach is theCalculation of the Total Number of

Nonsynonymous and Synonymous Substitutions(the denominators in dN and dS).

This requires a mutation model, and often thedefault is the Jukes-Cantor Model.

For example, these numbers could be stronglyaltered by a strong transition bias. It is essential to

fit the mutational model to the data first.DO NOT USE ANY PROGRAM THAT DOESNOT REQUIRE YOU TO FIT A MUTATION

MODEL!

E.g., Evolution of Abalone Sperm LysinLysin Is Involved in Fertilization and Gamete Recognition

ω = dN/dS

6

E.g., Evolution of Abalone Sperm Lysin

• M0, only one ω on all branches• M1, mixture of conserved sites (ω = 0) and

neutral sites (ω = 1)• M2, M1 plus some selected sites with ω > 1• M3, a mixture of three site classes with ω0, ω1,

and ω2• M7, ω’s are distributed as a Beta pdf• M8, M7 plus selected sites with ω > 1• Use Bayesian procedures to identify specific sites

with high probability of ω > 1

Yang et al. (Mol. Biol. Evol. 2000) Considered Several Models

E.g., Evolution of Abalone Sperm Lysin

E.g., Evolution of Abalone Sperm LysinE.g.,

Evolution ofAbalone

Sperm Lysin

Sites Under PositiveSelection Are

Involved in GameteRecognition of Egg

By the Sperm

7

E.g., Evolutionof the Pyrin

LocusThis locus codes forAn Activator of AnInactivating FactorInvolved in PeritonealFluid Balance (and Other Areas of theBody).


LocusMost BranchesHave dN/dS < 1,

Implying That Mostof the Time

Selection Maintainsa Conserved Amino

Acid Sequence


LocusMost BranchesHave dN/dS < 1,

Implying That Mostof the Time

Selection Maintainsa Conserved Amino

Acid SequenceBut Some Branches

Have dN/dS > 1,Implying Episodes

of SelectivelyDriven Change

Substitutions On the Human/ApeLineage Are Functionally Important.Many Mutations Are Known at ThisLocus That Lead to An AutosomalRecessive Disease Called Familial

Mediterranean Fever. Virtually All ofThese Mutations Are Reversals of the

Amino Acid Substitutions On theBranch Leading to Humans and Apes.

8

In General, Many Human Genetic DiseasesAre Caused By Mutations That Affect

Sites Subject to Directional Selection Or toPurfiying Selection.

Therefore, Evolutionary Analyses AreNow Standard As A Way Of IdentifyingThose Mutations Most Likely to Cause

Disease In Humans.These Studies Also Show That The Impactof Amino Acid Replacements Are Context

Dependent

These Tests Are Also Being Used to Estimate WhichLoci Are Selected in Specific Speciation Events. E.g.,

Evolution of the ASPM Locus In HumansLoss of function mutations in this gene cause severe

reductions in the cerebral cortical size of affected humans

dN/dS ratios in the primate ASPM haplotype tree:

E.g., Evolution of the ASPM Locus

dN/dS

9

Studies Are Now Underway To Identify Those GenesSubjected to Positive Selection In The Lineage LeadingTo Humans. Results As Of 2004 (Vallender & Lahn,

2004):

8Other1Skin Color3Brain Size & Anatomy3Neurotransmission4Dietary Adaptation7Sensory Systems14Reproduction23Host-Pathogen Interactions

No. of GenesFunction of Gene

Coalescent Tests of Neutrality

• Can Compare Observed Statistics toExpected Values Under NeutralCoalescence (Incorporates EvolutionaryStochasticity)

• Can Compare Haplotype Tree Topologiesto Expected Topologies Under NeutralCoalescence

Comparing Observed Statisticsto Expected Values

Expected neutral heterozygosity = θ/(1+θ) ≅ θ

!

"=

2# ij

j=2

n

$i=1

j

$

n(n %1)

The Observed expected heterozygosityGiven a sample of n DNA sequencesWith πij being the number of nucleotideDifferences between sequence i and j is:

!

" =S

1

ii=1

n#1

$

Under a neutral coalescent model and theInfinite sites model, the expected hetero-zygosity is also equal to:

where S is the observed number of segregating sitesin the sample of n sequences

Evolutionary Process(Neutral, Constant Sized

Population)

Current Population

Sample

Statistics

ΠΘ

10

Π versus Θ The Tajima (1989) Test:

• Var(Π−Θ) is determined under the infinitesites model

• If D > 0 significantly, this impliesbalancing selection

• If D < 0 significantly, this impliesdirectional selection

!

D ="#$

Var("#$)

Problems with The Tajima (1989) Test:

• Var(Π−Θ) is determined under the infinite sitesmodel, and it is not easy computationally to dealwith deviations from this model

• The null hypothesis is a neutral model in aconstant sized population. Demographic changesas well as selection can cause D ≠ 0

• Fay & Wu’s H test (Genetics 155:1405-1413,2000) is a variant of D that eliminatesconfoundment with demography, but still dependsupon the infinite sites model.

Coalescent Tests of Neutrality

• Can Compare Observed Statistics toExpected Values Under NeutralCoalescence (Incorporates EvolutionaryStochasticity)

• Can Compare Haplotype TreeTopologies to Expected TopologiesUnder Neutral Coalescence

11

Neutral GeneticDrift, Stable

Population Size

Neutral GeneticDrift,

ExpandingPopulation Size

NegativeSelection

Positive(Directional)Selection orBottleneck

Positive(Diversifying)Selection orSubdivision

Distribution of Recombinants In LPL

Recombination Hotspot

Haplotype Network in 5’ Region ofLPL

5'-1 5'-2 5'-823J

8J14J

44N

3

17

13 7 8

4

5 16 9 10 2

14

36J

12

6

15

188

49N

84R

175'-3

5'-4

17

5'-5

5'-6

5'-7

16

4

4

6

32J

9 10


3'-9

56N

53N

3'-4

64J 43J

59N

3'-12 19J

16J

30J

54N

41N 42N

40J

37J

59 44

50 38 54

50

64

42

65

46

61

58

65

66

63

40

53 6724J34J

4136 69

T-1

36J38 39 41 43 44 46 47 60 5764

T-4

69

3'-10

3'-11

45J

53

59

75R

53

36

55

58

61

38J

51

55

40

45

3756

3'-8

8J 14J

T-3

48

52

29J44

55

35J

41

T-2

49

68

12J

3'-6

36

46

58 55

49

81R

39J

49N

50N

78R

46N

503'7

62

4450

26J

61

32J

40

36

41 58 67

4920J

77R

67N 48N

42

41

62

59

3'-161N

60N

56

59 45 633'-2

9N44N

38 38

63

56 56

3'-351N63

633'-5

53 53

28J

4259 45

55

44

40 53

42

53

42

12


Population Size



NegativeSelection



Page and Holmes 1998

Notice That The Ratio of The Branch LengthsBetween Current Haplotypes And Deep

Interior Nodes Is Quite Different In TheseVarious Topologies. This Is A Robust Result

Under Coalescent Theory.

Just As Ratios Are Frequently Used To Measure“Shape” In Morphology, Ratios Can Be Used ToMeasure The “Shape” Of An Evolutionary Tree

1. Calculate the nearest neighbor distance,dij, for each haplotype i within tree j

2. Calculate the average branch length intree j of all branches that are not part of anearest neighbor comparison (which haveto be branches between interior nodes), aj

3. Calculate the statistics dij/aj for all i and j

For Example, 5’ vs. 3’ LPL

5'-1 5'-2 5'-823J

8J14J

44N

3

17

13 7 8

4

5 16 9 10 2

14

36J

12

6

15

188

49N

84R

175'-3

5'-4

17

5'-5

5'-6

5'-7

16

4

4

6

32J

9 10

3'-9

56N

53N

3'-4

64J 43J

59N

3'-12 19J

16J

30J

54N

41N 42N

40J

37J

59 44

50 38 54

50

64

42

65

46

61

58

65

66

63

40

53 6724J34J

4136 69

T-1

36J38 39 41 43 44 46 47 60 5764

T-4

69

3'-10

3'-11

45J

53

59

75R

53

36

55

58

61

38J

51

55

40

45

3756

3'-8

8J 14J

T-3

48

52

29J44

55

35J

41

T-2

49

68

12J

3'-6

36

46

58 55

49

81R

39J

49N

50N

78R

46N

503'7

62

4450

26J

61

32J

40

36

41 58 67

4920J

77R

67N 48N

42

41

62

59

3'-161N

60N

56

59 45 633'-2

9N44N

38 38

63

56 56

3'-351N63

633'-5

53 53

28J

4259 45

55

44

40 53

42

53

42

vs.

5' Distribution

3' Distribution

0

10

20

30

40

dij / a j

0 0.5 1 1.5 2

exact two sided p value = 0.0000 based upon 10,000 random permutations


Population Size



NegativeSelection



Page and Holmes 1998

Shape Is Still Confounded With Demography,But Selection Is Often Locus Specific, So A

Test For Deviation From Neutrality Is To TestFor Shape Homogeneity Among Two Or MoreTrees. This Is Done With A Non-Parametric

Exact Test

13

The Tree Topologies Indicate Either ASelective Sweep in the 5’ End and/or

Balancing Selection in the 3’ EndDNA region

contrasted to 3’LPL

Kolmogorov-Smirnov

p-Value

Permutation Test

p-Value

PDHA1 0.0000 0.0074

Xq13.3 0.0000 0.0000

!-Globin 0.0000 0.6343

MX1 0.0000 0.0000

MC1R 0.0002 0.0313

EDN 0.0000 0.0000

ECP 0.0000 0.0000

MS205 0.0000 0.0000

Homogeneity of the 5’ LPL With These Other Genes Cannot Be Rejected

These Results Indicate That 3’ LPL Is An Extreme Outlier. Therefore,Had Balancing Selection At the 3’ End.

The Tree Topologies Indicate Either ASelective Sweep in the 5’ End and/or

Balancing Selection in the 3’ EndDNA region

contrasted to 3’LPL

Kolmogorov-Smirnov

p-Value

Permutation Test

p-Value

PDHA1 0.0000 0.0074

Xq13.3 0.0000 0.0000

!-Globin 0.0000 0.6343

MX1 0.0000 0.0000

MC1R 0.0002 0.0313

EDN 0.0000 0.0000

ECP 0.0000 0.0000

MS205 0.0000 0.0000

Homogeneity of the 5’ LPL With These Other Genes Cannot Be Rejected

These Results Do Not Indicate A Selective Sweep At The 5’ End, ButWe Can Test That Hypothesis With Another Non-Parametric Test

Using Recombination


3'-9

56N

53N

3'-4

64J 43J

59N

3'-12 19J

16J

30J

54N

41N 42N

40J

37J

59 44

50 38 54

50

64

42

65

46

61

58

65

66

63

40

53 6724J34J

4136 69

T-1

36J38 39 41 43 44 46 47 60 5764

T-4

69

3'-10

3'-11

45J

53

59

75R

53

36

55

58

61

38J

51

55

40

45

3756

3'-8

8J 14J

T-3

48

52

29J44

55

35J

41

T-2

49

68

12J

3'-6

36

46

58 55

49

81R

39J

49N

50N

78R

46N

503'7

62

4450

26J

61

32J

40

36

41 58 67

4920J

77R

67N 48N

42

41

62

59

3'-161N

60N

56

59 45 633'-2

9N44N

38 38

63

56 56

3'-351N63

633'-5

53 53

28J

4259 45

55

44

40 53

42

53

42

12 Recombination Events OccurredBetween T-1 Haplotypes With T–2,

3, or 4 Haplotypes

In All 12 Cases, the 5’ End Was Of The T-1Type. Under Neutrality, This Has A Probability

of (1/2)12 = 0.002.

Therefore, the 5’ End Experienced A SelectiveSweep Enhanced By Recombination

14

Peeled Haplotype Network in of LPL(the portion of the combined 5’/3’ tree that never

experienced recombination)

52N

1JNR 2JNR

56N

61N

9N

53N

60N

4JN

79R

64J

70R

43J

27J

11J

28J

59N

23J

31J 19J

16J

10J

30J

40J

37J

56 59 45

29

63

38

30

50

3050

64

26

42

26

42

56

29 31 33 565953 13

65

46

29

7 8

26

53

25

65 19 61

58

65

66

8

29

8

8 25

63

40

53

67

34J26

4

63

5

16

66

41

29 36 69

17J

9

10

19

2 17

T-1

36J

5

12

16

31

35

38

39

41

43

44

46

47

60

57

64

T-4

5155

T-3

T-240

454968

32374852

29J

19

44

MaximumParsimony (andother techniques)allow you toinfer the state ofextinct ancestralstates.

Ancestral ReconstructionRevealed That Only 8Amino Acid Changes

Are Responsible For TheDiversified Functions ofThis Important Enzyme

Family.

Evolution of Isocitrate Dehydrogenases

TheseEvolutionaryEvents Took

Place 3.5 BillionYears Ago!

(Zhu, Golding & Dean, Sci. 307: 1279-1282, 2005)

Uses NADPUses NAD

15

Why Did NADP Use EvolveIndependently Several Times?


Ancestral reconstructionshows that this change inphenotype was due to 5

specific amino acidsubstitutions. Engineered

these changes in E. coliNADP Idh to make it NADIdh, and then competed iso-genic strains in chemostats


Uses NADPUses NAD

NAD Idh is much less fit than NADP Idh on acetate, yetmore fit on glucose as an energy source.


Ancestral reconstructionshows that

A switch to NADP occurredmultiple times5 specific amino acidsubstitutions are theproximal cause for thisswitchAdaptations to acetate (&similar energy-poorresources) were the ultimatecauses for these switches


Documents

Natural Selection and Haplotype Trees 2N - NSLC · Maynard Smith (1970) ... the Operation of Natural Selection a) ... default is the Jukes-Cantor Model. For example, these numbers