Substitution Models and theSubstitution Models and the Phylogenetic Assumptions

Vivek Jayaswal Lars S. Jermiin

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Why do we need substitution models?Why do we need substitution models?

1 2 3 4 5 N1 2 3 4 5 … N Human A T C G A … CChimp A G C A A … C

Root

Gorilla ………........... Orangutan ………...........

I1 I2

G ill O t Chi HGorilla Orangutan Chimp Human

4 44

Rootj

)|()|()|()|()|()|( 4321

4

1

4

1

4

1vOPvOPuOPuOPjvPjuPfL

u vjji ∑ ∑∑

= ==

=

I1P(u|j)

I2P(v|j)P(u|j) ( |j)

GorillaP(O1|u)

OrangutanP(O2|u)

ChimpP(O3|v)

HumanP(O4|u)P(O1|u) P(O2|u) ( 3| ) ( 4| )

TopicsTopics

• how correction for multiple substitution are done• how correction for multiple substitution are done• some of the phylogenetic assumptions• how we may evaluate phylogenetic assumptions• an example involving bacterial DNAp g

Substitutions at a Single SiteSubstitutions at a Single Site

S S S C S

AG

Single Substitution

ATG T

Multiple Substitution

TG

Coincidental Substitution

A

A—G

AA—G

G—T

A

A—G A—T

GG

Parallel Substitution

GG

Convergent Substitution

AA

Back Substitution

A

GG

A—G A—G

A

GG

A—G A—TT—G

A

AA

A—GG—A

Note —

A A A

NoteEvery substitution overwrites the evidence of a past state, which leads to an erosion of the historical signal

Modeling Nucleotide SubstitutionsModeling Nucleotide Substitutions

Consider the evolution at a given site in terms of conditionalConsider the evolution at a given site in terms of conditionalrates-of-change from nucleotide i to nucleotide j

− α Aj∑ α AC α AG α AT⎡⎢

⎤⎥

A T

R =

j≠A

α CA − α Cjj≠C∑ α CG α CT

⎢⎢⎢⎢

⎥⎥⎥⎥R =

α GA α GC − α Gjj≠G∑ α GT

α α α α∑

⎢⎢⎢⎢⎢

⎥⎥⎥⎥⎥α TA α TC α TG − α Tj

j≠T∑

⎣⎢⎢ ⎦

⎥⎥CG

Note —Here αij is the conditional rate-of-change from nucleotide i to nucleotide j in R the ‘rate matrix’ and R is the most general Markov model for DNAR — the rate matrix — and R is the most general Markov model for DNA

Modelling a Site in one SequenceModelling a Site in one Sequence

Consider Markov process X that results in nucleotide iConsider Markov process, X, that results in nucleotide ibeing converted to nucleotide j over time t —

Pij t( )= P X t( )= j | X 0( )= i[ ] t j

If the rates of change are constant then P(t) eRt

Time X

constant, then P(t) = eRtt = 0

Ancestori

P t( )= I + Rt +Rt( )2

+Rt( )3

+

In matrix notation, this isP t( )= I + Rt +

2!+

3!+L

=Rt( )k∞

∑ k!k=0∑

Modelling a Site in two SequencesModelling a Site in two Sequences

Consider two Markov processes X and Y and the followingConsider two Markov processes, X and Y and the following scenario —

t

Fij t( )= P X t( )= i, Y t( )= j | X 0( )=Y 0( )[ ] Time XY

t = 0

Ancestor

F t( )= PX t( )( )T F 0( )PY t( )In matrix notation, this is

Take-home Message #1

• The substitution model, R, is an integral part of , , g pthe transition function; it is used to estimate the probability of the present states given R and tprobability of the present states, given R and t

Rate matrix revisitedRate matrix revisited

− r∑ r r r⎡ ⎤

Ri =

rAjj≠A∑ rAC rAG rAT

rCA − rCjj≠C∑ rCG rCT

∑

⎡⎢⎢⎢⎢⎢

⎤⎥⎥⎥⎥⎥i

rGA rGC − rGjj≠G∑ rGT

rTA rTC rTG − rTjj≠T∑

⎣

⎢⎢⎢⎢⎢ ⎦

⎥⎥⎥⎥⎥j≠T⎣⎢ ⎦⎥

• Typically simplified forms of this rate matrix are used

• These matrices belong to the GTR-family of models and can be represented as

⎥⎥⎥⎤

⎢⎢⎢⎡

⎥⎥⎥⎤

⎢⎢⎢⎡

−−

=C

A

CTCGAC

ATAGAC

ππ

Rrrrrrr

⎥⎥

⎦⎢⎢

⎣⎥⎥

⎦⎢⎢

⎣ −−

T

G

GTCTAT

GTCGAG

ππ

rrrrrr

Commonly-used Markov Models

Assumptions

Commonly-used Markov Models

J k & C t (1969) Assumptions−3α α α αα −3α α α

⎡⎢⎢

⎤⎥⎥

Jukes & Cantor (1969)

1 One rate (α)R =α α −3α αα α α −3α⎣

⎢⎢⎢

⎦

⎥⎥⎥

1. One rate (α) 2. Uniform nucleotide content

−(2β +α ) β α β⎡⎢

⎤⎥

Kimura (1980)

R =β −(2β +α ) β αα β −(2β +α ) ββ α β −(2β +α )⎣

⎢⎢⎢⎢

⎦

⎥⎥⎥⎥

1. Two rates (α and β) 2. Uniform nucleotide content

β β ( β )⎣ ⎦

(β ) β β⎡ ⎤

Hasegawa, Kishino & Yano (1985)

R =

−(βπY +απG ) βπC απG βπT

βπ A −(βπ R +απT ) βπG απT

απ A βπC −(βπY +απ A ) βπT

⎡⎢⎢⎢⎢

⎤⎥⎥⎥⎥

1. Two rates (α and β)2. Non-uniform nucleotide content

βπ A απC βπG −(βπ R +απC )⎣⎢

⎦⎥

The Phylogenetic AssumptionsThe Phylogenetic Assumptions

• Given an alignment of nucleotides phylogenetic methods commonlyGiven an alignment of nucleotides, phylogenetic methods commonly assume that the sites have evolved under

• stationary and reversible conditions• homogeneous conditions• independent and identical conditions

Phylogenetic AssumptionsPhylogenetic Assumptions

Consider the following scenarioConsider the following scenario…

21

F1 = fA fC fG fT[ ] F2 = fA fC fG fT[ ]

0− αAjj≠A∑ αAC αAG αAT⎡

⎢⎢

⎤⎥⎥

− αAjj≠A∑ αAC αAG αAT⎡

⎢⎢

⎤⎥⎥

R1 =

j

αCA − αCjj≠C∑ αCG αCT

αGA αGC − αGj∑ αGT

⎢⎢⎢⎢⎢

⎥⎥⎥⎥⎥

R2 =

j

αCA − αCjj≠C∑ αCG αCT

αGA αGC − αGj∑ αGT

⎢⎢⎢⎢⎢

⎥⎥⎥⎥⎥GA GC Gj

j≠G∑ GT

αTA αTC αTG − αTjj≠T∑

⎣

⎢⎢⎢⎢ ⎦

⎥⎥⎥⎥

αGA αGC αGjj≠G∑ αGT

αTA αTC αTG − αTjj≠T∑

⎣

⎢⎢⎢⎢ ⎦

⎥⎥⎥⎥

F0 = fA fC fG fT[ ]Π1 = π A πC πG πT[ ] Π2 = π A πC πG πT[ ]F0 fA fC fG fT[ ]Π1 π A πC πG πT[ ] Π2 π A πC πG πT[ ]

The Stationary ConditionThe Stationary Condition

21

0

Π1 = π A πC πG πT[ ] Π2 = π A πC πG πT[ ]F0 = fA fC fG fT[ ]0 fA fC fG fT[ ]

Note —The stationary condition is met if F0 = Π1 = Π2, implying that the marginal distributions of R and R are the same even though R and R may differ!distributions of R1 and R2 are the same, even though R1 and R2 may differ!

The Reversible ConditionThe Reversible Condition

21

0

R1 Π1 R2 Π2

NotesNotes —If the process is stationary, then Π1 = Π2

Moreover if πR = πR for all i and j then the process is reversibleMoreover, if πiRij = πjRji for all i and j, then the process is reversible

The Homogeneous ConditionThe Homogeneous Condition

21

0R1 R2

Notes —The homogeneous condition is met for the Markov processes, R1 and R2, if R1 = R2

If the homogeneous condition is met, then Π1 = Π2 — however, non-stationary, and therefore non-reversible conditions may still prevail (i e if F ≠ Π = Π )and therefore non-reversible, conditions may still prevail (i.e., if F0 ≠ Π1 = Π2)

IID conditionIID condition12

t

time XY

Seq_1 ACGTGTCCATGATTA...

time XY

Rx Rx …t = 0

Ancestor

Seq_2 ACCTGCCCAAGATAA...

NNotes —For computational reasons, it is convenient to assume that —

Sites in the sequence have evolved independentlySites in the sequence have evolved independentlySites in the sequence have evolved under the same model (Rx = Ry)

Rate Heterogeneity Across SitesRate Heterogeneity Across SitesRNA-coding GenesRNA coding Genes

Take-home Message #2

• Phylogenetic analyses require the users to make y g y qcertain assumptions about the data before these are investigated in detailare investigated in detail

QuestionsQuestions…

• Are these phylogenetic assumptions realistic? p y g p• How can we assess whether the phylogenetic

ti t b th d t ?assumptions are met by the data?

AnswerAnswer…

• Inspecting the data — before or after inferring theInspecting the data before or after inferring the phylogeny — increases the chance of finding out what might have taken place during the evolutionwhat might have taken place during the evolution

Considering the IID ConditionConsidering the IID Condition

• Visual inspection of the alignment might show• Visual inspection of the alignment might show whether some regions evolved faster

• Assume rate-heterogeneity across some sites, and then use phylogenetic methods that accountand then use phylogenetic methods that account for this

Hierarchical Likelihood-Ratio TestHierarchical Likelihood-Ratio Test

Consider the following decision treeConsider the following decision tree…

Yes No Uniform nucleotide content?

One conditional rate?YesNo NoYes

Two or six conditional rates?YesNo YesNo rates?

Source: Posada & Crandall (1998) Bioinformatics 14, 817-818.

Take-home Message #3

• The likelihood-ratio test can be used identify a suitable substitution model for a given data set g

Testing assumptions prior to modellingTesting assumptions prior to modelling

• n × F(t) expected divergence matrix• n × F(t) — expected divergence matrix• N(t) — observed divergence matrix

Examples:

294 0 0 00 372 0 0

⎡⎢⎢

⎤⎥⎥

244 31 8 1128 321 14 9

⎡⎢⎢

⎤⎥⎥N(0) =

0 372 0 00 0 829 00 0 0 655⎣

⎢⎢⎢

⎦

⎥⎥⎥

N(t) =28 321 14 911 13 801 414 10 3 628⎣

⎢⎢⎢

⎦

⎥⎥⎥

⎣ ⎦ ⎣ ⎦

Matched-pairs Tests of SymmetryMatched-pairs Tests of Symmetry

Seq 1 AGACTAGGTCTTGTATAGACTAATGTTCACAGTTTTTTAACTTTGTCAATGGASeq 1 AGACTAGGTCTTGTATAGACTAATGTTCACAGTTTTTTAACTTTGTCAATGGA...Seq 2 AGACGAGGTCGTGTATGGCCTCGTGAGCACGGGTTGTTCACTCCGCCAACGGT...

A C G T Σ2

A 5 4 7 1 17

A C G T Σ2

A 5 4 7 1 17 ( )2Test of Symmetry

C 0 7 2 0 9

G 0 1 5 0 6

C 0 7 2 0 9

G 0 1 5 0 6XBowker

2 =xij − x ji( )xij + x jii< j

∑G 0 1 5 0 6

T 1 4 5 8 18Σ

G 0 1 5 0 6

T 1 4 5 8 18Σ T t f M i l S tΣ1 6 16 19 9Σ1 6 16 19 9

Test of Marginal SymmetryXStuart

2 = DV−1DT ,Dij = xi• − x•i ,

V = covariance matrix of D

Note:These tests statistics are asymptotically χ2-distributed on ν degrees of freedom V covariance matrix of D

Sources: Bowker (1948). JASA 43, 572-574; Stuart (1955). Biometrica 42, 412-416.

χ distributed on ν degrees of freedom

Bacterial 16S rDNA SequencesBacterial 16S rDNA Sequences

Ribosomal RNA from five bacteria was compared using theRibosomal RNA from five bacteria was compared using the matched-pairs test of homogeneity

Probabilities Thermotoga Bacillus Deinococcus Thermus

Aquifex 1 32 10 01 1 79 10 11 3 45 10 10 5 09 10 01Aquifex 1.32 10–01 1.79 10–11 3.45 10-10 5.09 10-01

Thermotoga 2.64 10–12 6.69 10–12 4.15 10–01

Bacillus 9.95 10–01 3.64 10–09

Deinococcus 5 99 10–11Deinococcus 5.99 10

Note —It is highly unlike that these data have evolved under homogeneous conditions, implying that it would be unwise to use a time-reversible Markov model

Source: Ababneh et al. (2006). Bioinformatics 22, 1225-1231.

implying that it would be unwise to use a time-reversible Markov model

Phylogeny of Bacterial Ribosomal RNAPhylogeny of Bacterial Ribosomal RNA

Markov model: GTR

Markov model: General

Source: Ababneh et al. (2006). Bioinformatics 22, 1225-1231.

Take home Message #4Take-home Message #4

• It is important to consider the substitution models carefully when using them in phylogenetic studies y g p y g

Suggested LiteratureSuggested Literature

RDM Page, EC Holmes (1998), Molecular Evolution.• Chapter 5 (Sections 5.2, 5.3) — important reading

W-H Li (1997), Molecular Evolution.• Chapter 3 (pp. 59-78) — contains descriptions that are better than

those in Page & Holmes 1998 — important reading D Posada, KA Crandall, 1998. MODELTEST: testing the model of DNA

b tit ti Bi i f ti 14 817 818 f l disubstitution. Bioinformatics 14, 817-818 — useful readingLS Jermiin et al. (2008). Phylogenetic model Evaluation. Pp 331-363. In

Bioinformatics Volume I: Data Sequences Analysis andBioinformatics - Volume I: Data, Sequences Analysis and Evolution (Ed. Keith J), Humana Press, Totowa, NJ. [2008] —important readingp g