JamesDegnanUniversityofCanterbury11/1/11

(JointworkwithElizabethAllmanandJohnRhodes)

Thanksto

1.BackgroundA.genetreesvs.speciestreesB.coalescenceandincompletelineagesorting

2.Rootedgenetreeprobabilitiesaspolynomials

3.Unrootedgenetreeprobabilities

Populationgenetics:traditionallyusedtoanalyzesinglepopulations.

Phylogenetics:Whatisthebestwaytoinferrelationshipsbetweenpopulations/species?

Graphic by Mark A. Klinger, Carnegie Museum of Natural History, Pittsburgh

Past

Present

Present

Past

Incompletelineagesorting

ABCDThegenetreeisarandomvariable.Thegenetreedistributionisparameterizedbythespeciestreetopologyandinternalbranchlengths.

Usingtransformedbranchlengths,

genetreeprobabilitiescanbewrittenaslinearcombinationsofmonomials

wherenisthenumberoftips.€

X1 = e−t1 , X2 = e− t2 , etc.

€

X1α1X2

α2Xn−2αn−2

€

x

€

y

Probability:

€

19 p2,2(x)p3,2(y) = 1

6 XY − 16 XY

3

History Probabilityh1:(1,2,3) h2:(1,3,3)h3:(2,2,3)h4:(2,3,3)h5:(3,3,3)

Total

€

(1− X)(1−Y )

€

13 (1− X)Y

€

13 X(1− 3

2Y + 12Y

3)€

€

16 XY − 1

6 XY3

€

118 XY

3

€

1− 23 X − 2

3Y + 13 XY + 1

18 XY3

  Giventhesetofgenetreeprobabilities,canthespeciestreeberecovered?

  Inmanycases,thehighestprobabilitygenetreehasthesametopologyasthespeciestree,butnotalways.

  Themostlikelytripleforanysetofthreetaxaisarootedtripleonthespeciestree,sothespeciestreecanberecoveredbymarginalizinggenetreestotheirrootedtriples.

  Toinferrootedgenetrees,youneedmolecularclockoroutgroup;otherwiseonlyunrootedgenetreescanbeinferred

  Speciestreemethodsusingthecoalescentcurrentlyassumerootedtrees–canspeciestreesbeinferredusingunrootedgenetrees?

  Theprobabilityofanunrootedgenetreeisthesumoftheprobabilitiesofallgenetreeswiththesameunrootedtopology

  Pr[]=P[(((AB)C)D)] +P[(((AB)D)C)]

+P[(((CD)A)B)]+P[(((CD)B)A)]+P[((AB)(CD))]

AB D

C

  Probabilitiesofunrootedgenetreesarelinearcombinationsofprobabilitiesofrootedgenetrees.

  Expressionsforprobabilitiesofunrootedgenetreesareoftensimplerthanrootedgenetreeprobabilitiesforthesamenumberofspecies.

UnrootedGeneTreesProbability

AB

AC

B

A BD

CD

D

C€

1− 23 X

€

13 X

€

13 X

UnrootedGeneTreesProbability

AB

AC

B

A BD

CD

D

C€

1− 23 XY

€

13 XY

€

13 XY

  Ifthespeciestreetopologyisknown,theunrootedgenetreedistributiononlyhasinformationaboutoneinternaledge(orsumofedges).

  Ifthespeciestreeisunknown,theunrootedgenetreedistributiononlyidentifiestheunrootedspeciestreetopology.Thefollowingspeciestreesinducethesameunrootedgenetreedistribution:

  15unrootedtopologies  3rootedspeciestreesshapes

Caterpillar Balanced Pseudocaterpillar

Whathappenswithfivetaxa?

•  15unrootedtopologies•  3rootedspeciestreesshapes

Caterpillar Balanced Pseudocaterpillar

Givenadistributionof155‐taxonunrootedgenetreeprobabilities,whatinformationcanwerecoveraboutthespeciestree?(1)  Unrootedspeciestreetopology?(2)  Rootedspeciestreetopology?(3)  Branchlengths?

A

B C

D

E

€

T1

probability

A

AB

B

CC

D

D

E

E

€

T1

€

T2

probability

A

B

C

DE

€

T3

A

A

A

A

A

B

B

BB

B

CC

C

C

D

D

DD

D

E

E

E

E

E

C

€

T1

€

T2

€

T4

€

T5

€

T7

probability

A

B

C

DE

€

T3

€

T13

€

T6

€

T9

€

T12

5others

(1)Puttingunrootedgenetreesthataretiedinprobabilityintoequivalenceclasses,thesizeoftheseclassesdependsontheunlabeled,rootedspeciestreetopologyonfivetaxa:

Caterpillarclasssizes:1,1,1,2,2,2,6Balancedclasssizes:1,2,2,4,6Pseudocaterpillarsizes:1,2,2,2,8

(2)Theunlabeled,rootedspeciestreetopologycanthereforebedeterminedfromtheclasssizes.

(3)Fromthethefour‐taxonresults,theunrootedspeciestreetopologycanbeidentifiedbydeterminingthemostprobablequartetforeachsubsetoffourspecies

Thereforeweknow(forfivetaxa):(i)labeled,unrootedspeciestreetopology(ii)unlabeled,rootedspeciestreetopology

(4)Thelabeled,rootedspeciestreetopologycanbedeterminedbyfurtherconsideringinvariantsorinequalitiesinunrootedgenetreeprobabilities.

Example:Giventheunrootedspeciestreeandgiventhatthespeciestreeisbalanced,Therootedspeciestreeisoneof:

A

BC

D

E

€

T1

A B C D E A B C D E

€

R1

€

R2

andimplydifferentinvariantsandinequalitiesfortheunrootedgenetreeprobabilities.

Under,isinthe6‐elementclass,and

Under,isinthe4‐elementclass,and

€

R1

€

R2

€

R1

€

T7

€

Pr(T7) < Pr(T5)

€

R2

€

T7

€

Pr(T7) > Pr(T5)

  Similarargumentscanbeusedtoidentifythecaterpillarandpseudocaterpillar.Thusfive‐taxonrootedspeciestreesareidentifiablefromfive‐taxonunrootedgenetreeprobabilities.

  Theresultsgeneralizeimmediatelytolargertrees:theunrootedgenetreedistributionforeachsubset

offivetaxacanbeobtainedbysummingovertreesthatdisplaythefive‐taxontree.

  Thusallrootedquintetsonthespeciestreeareidentifiable.Therootedspeciestreetopologycanbeconstructedfromtherootedquintets.

  Allbranchlengthsonfive‐taxonspeciestreescanberecovered.Example,(((a,b):x,c):y,(d,e):z)

Theorem.

(i)Theunrootedgenetreedistributiondeterminestherootedspeciestreeandbranchlengthswhenthereare5ormoretaxa.

(ii)Theunrootedgenetreedistributiongivenafour‐taxonspeciestreedeterminestheunrootedspeciestree,butnottherootedspecies.

(((AB)C)(DE))

((((AB)C)D)E)

(((AB)(CD))E)

‐‐Agenetreedistributionhasonlyn‐2parameters(speciestreebranchlengths)

‐‐Butthereare(2n‐3)!!genetreeprobabilities

‐‐Therearemanytiesingenetreeprobabilitiesamongstdifferentgenetrees

‐‐Thereareotherlinearconstraints.Forthespeciestree(((AB)C)D),wehave

Pr[((AB)(CD))]–Pr[(((AB)D)C)]–Pr[(((AD)B)C)]=0

‐‐Howmanylinearconstraintsarethere?Howdoesthenumberoflinearconstraintsdependonthenumberoftaxaandspeciestreetopology?

‐‐Whataresomenonlinearconstraints?

Probabilitiesofcoalescenthistories,andthereforeofgenetrees,arepolynomialsinthetransformedbranchlengths.Anexamplepolynomialconstraint: