Intro to Comp Genomics

Lecture 11: Using models for sequence evolution

Comparing everything

Our intuition:

Feature X similar among a group of species -> Feature X is important

Feature X can be:

• Sequence• Gene expression (human brain vs chimp brain?)• Genic structure (Exon/intron)• Protein complexes • Protein networks• TF-DNA interaction

Two main difficulties:

Species have common ancestry – a lot of stuff may be similar just because it did not diverge yet

Species are related through phylogenetic trees – similarity should be following a tree structure

Modeling multiple genome sequences

Genome1

Genome2

AGCAACAAGTAAGGGAAACTACCCAGAAAA….AGCCACATGTAACGGTAATAACGCAGAAAA….AGCAACAAGAAAGGGTTACTACGCAGAAAA….

Alignment

Statistics

Genome3

ACGT

A C G T

AC

GT

s

s

s s

sMarkov process

Markov process

Unobserved ancestral

Unobserved ancestral

Tree models

H2

S3

S2 S1

H1

For a locus j:

Extant Species Sj1,..,n

Ancestral species Hj1,..(n-1)

Tree T: Parents relation pa Si , pa Hi

(pa S1 = H1 ,pa S3 = H2

The root: H2)

Val(X) = {A,C,G,T}

An evolutionary model = a joint distribution Pr(h,s)

Locus independence: ),Pr(),Pr( jjj hshs

Tree models

ATree: T, Species S1,..,n

Parents relation pa Si

Markov assumption still in effect..but branching complicates it

C

CC

A

ii paxxiii Qtxx ,)exp()pa|Pr(

)Pr( rooth

We need a little more:

)pa|Pr()Pr(),Pr( ! iirootiroot xxhhs

The model:

)|Pr()|Pr()|Pr()|Pr()Pr()Pr( 111223212 hshshshhhs In the triplet:

Tree models

Toy model:

Triplet phylogeny

Substitution probability on all of the branches:

96.001.002.001.0

01.096.001.002.0

02.001.096.001.0

01.002.001.096.0

)|Pr( yx

Uniform background probability: P(x) = 0.25

H2

S3

S2 S1

H1

Tree models

Marginal probability of Xi (any r.v.) :

ii xXsh

iii shPxXx|,

),()Pr()Pr(

)|Pr(/),(),|Pr(|

sshPsxhxhh

i

i

Given partial observations s:

“ancestral inference”

h

shPs )|,()|Pr(

The Total probability of the data:

likelihood of the model given the data

H2

S3

S2 S1

H1

Tree models

?

A

C A

?

xhh

i

i

shPsxh|

),()|Pr(

Given partial observations s:

)),,Pr(( ACA

The Total probability of the data:

)),,(|Pr( 1 ACAAh

?

?

96.001.002.001.0

01.096.001.002.0

02.001.096.001.0

01.002.001.096.0

)|Pr( yx

Intuition – maximum parsimony

?

A

C A

?

“Parsimony” ~ minimal change

The “small” parsimony problem: Find ancestral sequences that minimize the number of substitutions along the tree branches

What is the minimal number of substitutions?

(All branches are equal, all substitutions are equal)

(The “big” parsimony problem: Find the tree topology that gives minimal parsimony score given a set of loci)

C

C 2 substitutions

A

A 1 substitution

Algorithm (Following Fitch 1971):

Up(i):if(extant) { up_set[i] = Si; return}up(right(i)), up(left(i))up_set[i] = up_set[right[i]] ∩ up_set[left[i]]if(up_set[i] = 0)

D += 1 up_set[i] = up_set[right[i]] + up_set[left[i]]

Down(i):down_set[i] = up_set[sibling[i]] ∩ down_set[par(i)]if(down_set[i] = 0) {

down_set[i] = up_set[sibling[i]] + down_set[par(i)]}

Algorithm:D=0up(root);down_set[root] = 0;down(right(root));down(left(root));

Intuition – maximum parsimony?

S3

S2 S1

? up_set[4]

up_set[5]

Algorithm (Following Fitch 1971):

Up(i):if(extant) { up_set[i] = Si; return}up(right(i)), up(left(i))up_set[i] = up_set[right[i]] ∩ up_set[left[i]]if(up_set[i] = 0)

D += 1 up_set[i] = up_set[right[i]] + up_set[left[i]]

Down(i):down_set[i] = up_set[sib[i]] ∩ down_set[par(i)]if(down_set[i] = 0) {

down_set[i] = up_set[i]}down(left(i)), down(right(i))

Algorithm:D=0up(root);down_set[root] = 0;down(right(root));down(left(root));

Intuition – maximum parsimony?

S3

S2 S1

? down_set[4]

down_set[5]

up_set[3]

Set[i] = up_set[i] ∩ down_set[i]

Algorithm (Following Felsenstein 1981):

Up(i):if(extant) { up[i][a] = (a==Si ? 1: 0); return}up(r(i)), up(l(i))iter on a

up[i][a] = b,c Pr(Xl(i)=b|Xi=a) up[l(i)][b] Pr(Xr(i)=c|Xi=a) up[r(i)][c]

Down(i):

down[i][a]= b,c Pr(Xsib(i)=b|Xpar(i)=c) up[sib(i)][b] Pr(Xi=a|Xpar(i)=c) down[par(i)][c]

down(r(i)), down(l(i))Algorithm:

up(root);LL = 0;foreach a {

L += log(Pr(root=a)up[root][a])down[root][a]=Pr(root=a)

}down(r(root));down(l(root));

Probabilistic inference?

S3

S2 S1

? up[4]

up[5]

Felsentstein

Algorithm (Following Felsenstein 1981):

Up(i):if(extant) { up[i][a] = (a==Si ? 1: 0); return}up(r(i)), up(l(i))iter on a

up[i][a] = b,c Pr(Xl(i)=b|Xi=a) up[l(i)][b] Pr(Xr(i)=c|Xi=a) up[r(i)][c]

Down(i):

down[i][a]= b,c Pr(Xsib(i)=b|Xpar(i)=c) up[sib(i)][b] Pr(Xi=a|Xpar(i)=c) down[par(i)][c]

down(r(i)), down(l(i))Algorithm:

up(root);LL = 0;foreach a {

L += log(Pr(root=a)up[root][a])down[root][a]=Pr(root=a)

}down(r(root));down(l(root));

?

S3

S2 S1

? down[4]

down5]

up[3]

P(hi|s) = up[i][c]*down[i][c]/

(jup[i][j]down[i][j])

Probabilistic inference

Felsentstein

Inference as message passing

s

s

s s

s

s

sYou are P(H|our data)

You are P(H|our data)

I am P(H|all data)

DATA

Inference as message passing

A CC

C

DATA

96.001.002.001.0

01.096.001.002.0

02.001.096.001.0

01.002.001.096.0

)|Pr( yx

Up:(0.01)2,(0.96)2,(0.01)2,(0.02)2

Down:(0.25),(0.25),(0.25),(0.25)

Up:(0.01)(0.96),(0.01)0.96),(0.01)(0.02),(0.02)(0.01)

Learning: Branch decomposition

3

2 1

5

4

5

3

5

4

1

4

2

4

Can we learn each branch independently?

We know how to compute the ML model given two observed species

We have P(S|D) for each species, can we substitute it for the statistics:

))),|(),|(((maxarg )( sXEsXEL ipai

A G C TAGCT

loci

ipaiii sXPsXPXXE ),|(),|()( )(pa

)|Pr(maxarg)|(maxarg DDL

Transition posteriors: not independent!

A CA

C

DATA

96.001.002.001.0

01.096.001.002.0

02.001.096.001.0

01.002.001.096.0

)|Pr( yxDown:(0.25),(0.25),(0.25),(0.25)

Up:(0.01)(0.96),(0.01)0.96),(0.01)(0.02),(0.02)(0.01)

Up:(0.01)(0.96),(0.01)0.96),(0.01)(0.02),(0.02)(0.01)

Learning: Second attempt

3

2 1

5

4 5

3

5

4

1

4

2

4

Can we learn each branch independently?

Given P(Spai->Si|D) for each species, can we substitute it for the statistics?

)),|((maxarg

)|Pr()|Pr(sib

]][[]][[pa]][[sib1

)|,(

)(

)()(

)(

iiipa

ipaiipai

loci ciii

iipa

DXXEL

aXbXaXcX

bXupaXdowncXupZ

SbXaXE

i

Expectation-Maximization

3

2 1

5

4 5

3

5

4

1

4

2

4

)),|((maxarg )(1

iiipaki SXXELL

i

)|Pr()|Pr(sib

]][[]][[pa]][[sib1

)|,(

)()(

)(

aXbXaXcX

bXupaXdowncXupZ

SbXaXE

ipaiipai

loci ciii

iipa

Continuous time

),0[)|Pr(),;,( txXAXAtsxP st

ji

jitP

hthtPhPtP

tP

tP

ijt

ijk

kjik

jij

ij

0

1)(lim

0,)()()(

1)(

0)(

0

Conditions on transitions:

Theorem:

ijij

tij

iiii

tii

qt

tPP

qt

tPP

)(lim)0('

)(1lim)0('

0

0exists (may be infinite)

exists and finite

Think of time steps that are smaller and smaller

Markov

Kolmogorov

Rates and transition probabilitiesThe process’s rate matrix:

nininnn

ni

i

ni

i

qqqq

qqqq

qqqq

Q

..

..........

..........

..

..

210

1121110

0020100

Transitions differential equations (backward form):

)(]1)([)()(

)()()()()(

tPsPtPsP

tPtPsPtPtsP

ijiiik

kjik

ijk

kjikijij

)()()('0 tPqtPqtPs ijiiik

kjikij

)exp()()()(' QttPtQPtP

Matrix exponentialThe differential equation:

)exp()()()(' QttPtQPtP

Series solution:

)exp(!

1

!))'(exp(

!

1)exp(

0

1

0

0

QtQtQi

QtQi

iQt

tQi

Qt

i

i

ii

i

i

i

i

i

1-path 2-path 3-path 4-path 5-path

Summing over different path lengths:

Computing the matrix exponential

t

t

t

ii

i

i

i

i

i

i

i

ne

e

e

t

tti

ti

Q

tQi

Qt

00

00

00

)exp(

)exp()!

1()(

!

1

!

1)exp(

2

1

00

0

Computing the matrix exponential

i

i

itQi

Qt

0 !

1)exp(

Series methods: just take the first k summandsreasonable when ||A||<=1if the terms are converging, you are ok

can do scaling/squaring:

Eigenvalues/decomposition:good when the matrix is symmetricproblems when having similar eigenvalues

Multiple methods with other types of B (e.g., triangular)

m

m

QQ ee

0!

1iQ

i

1SSeB

Learning a rate matrix

3

2 1

5

4 5

3

5

4

1

4

2

4

What if we wish to learn a single rate matrix Q?

)exp()Pr( )( iiipa QtXX

,..),,|Pr(maxarg 21,..,, 121ttQs

nttQ

Learning is easy for a single, fixed length branch. Given (inferred) statistics nk for multiple branch lengths, we must optimize a non linear likelihood function

kijn

ijkjik QttQsL ))(exp(),|( ,,

Learning: Sharing the rate matrix

3

2 1

5

4 5

3

5

4

1

4

2

4

kijn

ijkjik QttQsL ))(exp(),|( ,,

Use generic optimization methods: (BFGS)

?))log(exp(

?),|(

u

Qt

t

tQsLL

k

Protein genes: codes and structure

1 2 3 codons

Intron/exons

Domains

Conformation

Degenerate code

Recombination easier?

Epistasis: fitness correlation between two remote loci

5’ utr3’ utr

The classical analysis paradigm

BLAT/BLAST

Target sequence

Genbank

Matching sequences CLUSTALW

ACGTACAGAACGT--CAGAACGTTCAGAACGTACGGA

Alignment

PhylogeneticModeling

Analysis: rate, Ka/Ks…

Clustalw and multiple alignment

• ClustalW is the semi-standard multiple alignment algorithm when sequences are relatively diverged and there are many of them

ClustalW

Compute pairwise sequence distances (using pairwise alignment)

Build a guide-tree: approximating the phylogenetic relations between the sequences

“Progressive” alignment on the guide tree

S2S1

S4S3

S5 Dist(s1,s2) = best pair align

DistanceMatrix

NeighborJoining

Guide tree is based on pairwise analysis!

From the leafs upwards:Align two children given their “profiles”Several heuristics for gaps

• Other methods are used to multi-align entire genomes, especially when one well annotated model genome is compared to several similar species. Think of using one genome as a “scaffold” for all other sequences.

Nucleotide substitution models

• For nucleotides, fewer parameters are needed:

A

C T

G

A

C T

G

Jukes-Kantor (JK) Kimura

• But this is ignoring all we know on the properties of amino-acids!

Simple phylogenetic modeling: PAM/BLOSSOM62

• Given a multiple alignment (of protein coding DNA) we can convert the DNA to proteins. We can then try to model the phylogenetic relations between the proteins using a fixed rate matrix Q, some phylogeney T and branch lengths ti

When modeling hundreds/thousands amino acid sequences, we cannot learn from the data the rate matrix (20x20 parameters!) AND the branch lengths AND the phylogeny.

Based on surveys of high quality aligned proteins, Margaret Dayhoff and colleuges generated the famous PAM (Point Accepted mutations): PAM1 is for 1% substitution probability.

Using conserved aligned blocks, Henikoff and Henikoff generated the BLOSUM family of matrices. Henikoff approach improved analysis of distantly related proteins, and is based on more sequence (lots of conserved blocks), but filtering away highly conserved positions (BLOSUM62 filter anything that is more than 62% conserved)

Universal amino-acid substitution rates?

Jordan et al., Nature 2005

“We compared sets of orthologous proteins encoded by triplets of closely related genomes from 15 taxa representing all three domains of life (Bacteria, Archaea and Eukaryota), and used phylogenies to polarize amino acid substitutions. Cys, Met, His, Ser and Phe accrue in at least 14 taxa, whereas Pro, Ala, Glu and Gly are consistently lost. The same nine amino acids are currently accrued or lost in human proteins, as shown by analysis of non-synonymous single-nucleotide polymorphisms. All amino acids with declining frequencies are thought to be among the first incorporated into the genetic code; conversely, all amino acids with increasing frequencies, except Ser, were probably recruited late. Thus, expansion of initially under-represented amino acids, which began over 3,400 million years ago, apparently continues to this day. “

You task

• Get aligned chromosome 17 for human, chimp, orangutan, rhesus, marmoset

• Use EM on the known phylogeny to estimate a substitution model from the data (P(x|pax))

• Partition the genome into two parts according to overall conservation (define the criterion yourself). Then train independently two models and compare them.

• Optional: can your models be explained using a single rate matrix and different branch lengths?