genome evolution. amos tanay 2009 genome evolution lecture 7: selection in protein coding genes

Genome Evolution. Amos Tanay 2009

Genome evolution

Lecture 7:

Selection in protein coding genes


Protein genes: codes and structure

1 2 3

codons

Introns/exons

Domains

Conformation

Degenerate code

Recombination easier?

Epistasis: fitness correlation between two remote loci

5’ utr3’ utr


Identifying protein coding genes

From mRNAs

Spliced ESTs : short low quality fragments that are easier to get

Using computational methods. Limited accuracy

Using conservation or mapping from other genomes


Questions on protein function and evolution

Identification:• Identify protein coding genes

– Not yet resolved for new species, but with new technology this question may be less important than before

Structure/Function:• Define functional domains

– Highly important for understanding protein function• Which parts of the proteins are “important” (e.g., catalytic?)

– Difficult since structural modeling is hard and context dependent

Evolution • Identify places and times where a new protein feature emerged

– Positive selection• Understand mutation/selection through codon degeneracy• Understanding processes of duplication and diversitification


The classical analysis paradigm

BLAT/BLAST

Target sequence

Genbank

Matching sequences CLUSTALW

ACGTACAGAACGT--CAGAACGTTCAGAACGTACGGA

Alignment

PhylogeneticModeling

Analysis: rate, Ka/Ks…


Basics: rates of substitution

We observe two sequences

The divergence D is the fraction of different amino acids

We want to find the rate of replacement (or substitution)

ttt DDD )2)(1(1 D

tDdtdD 22/ tt eD 21

tK 2 )ˆ1ln( DK

])ˆ1/[(ˆ)ˆ( LDDKVar

So if we computed the divergence D of two sequence, we can estimate the rate

Where L is the sequence length.

Note that for small D’s, K~D, but for larger values, K takes into account multiple substitutions

And:

)ˆ2/(ˆˆ tK


Basics: rates of substitution - nucleotides

With nucleotides, we cannot ignore mutations that eliminate divergence

)1()31( )()()1( tAtAtA PPP

Probability to have the same value after two branches of length t:

So we can estimate the rate given the observed divergence d:

A

C T

G

A

C T

G

Jukes-Kantor (JK) KimurattAtAtAtA ePPPP 4)()()()1( 4

3

4

14

tNN eP 8

4

3

4

1

)3/ˆ41ln(4

3ˆ)1(4

3 8 dked t


Using universal matrices: 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)

S. Henikoff


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. “


Learning a rate matrix

3

2 1

5

4)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(),|( ,,

Use generic optimization methods: (BFGS)

?))log(exp(

?),|(

u

Qt

t

tQsLL

k


Molecular clocks and lineage acceleration

• How universal is the rate of the evolutionary process?• Mutations may depend on the number of cell division and thus in the

length of generation• Mutations depends on the genomic machinery to prevent them (• Mutations may also depend on the environment• The molecular clock (MC) hypothesis state that evolution is working

in a similar rate for all lineages

A B C

O

Relative rate test:

KOA – KOB = 0 ?

Test: KCA – KCB


Different molecular clocks

Kim et al., 2006 PLoS genet in apes and primates

Cytochrom C: 5 substiutions per 100 residues per 100 million years

Hemoglobin: 20 substiutions per 100 residues per 100 million years

Fibrinopeptiedes: 80 substiutions per 100 residues per 100 million years


Analysis: rate variation

• If our ML model include rate variation, we can use the inferred rates to annotate the protein

• Same can be done by constructing a conservation profile, even if the model is simplistic.

• Shown here are example from Tal Pupko’s work on the Rate4Site and ConSurf programs


Synonymous vs. non synonymous mutations

• Degenerate positions of codon are evolving more rapidly – free from selection on the coding sequence

• This provide us with a powerful “internal control” – we are comparing two different types of evolutionary events at the same loci, so all sources of variation in the mutational process are not affecting us.

Given aligned proteins, we can count:

MA – number of non-synonymous changes

Ms – number of synonymous changes

We then want to estimate:• Ka – rate of non-synonymous mutations (per syn site)• Ks – rate of synonymous mutations (per syn site)• Estimate V(Ka), V(Ks)

• Comparing Ka and Ks can provide evolutionary insights:– Ka/Ks<<1: negative selection may be purging protein modifying mutations

– Ks/Ka>>1: positive selection may help acquiring a new function

– (statistics using, e.g., T-test)

Non-SynSyn

ChangeMAMS

No Change

NA-MANS-MS

Chi-square test

Average number of sites


From dN/dS to Ka and Ks

Consider the divergence of synonymous and non synonymous sites separately.

As discussed before, we can estimate the rates:

)3

)(41ln(

4

3

SS N

dSK )

3

)(41ln(

4

3

AA N

dNK

A more realistic approach should consider the genetic code and other effect

A codon model is defining a rate matrix over nucleotide triplets

We can use various parameterizations, for example:

][][ kQ jij

For transitions For non-synonymous

We learn the ML parameters. Small indicate selection


Codon bias

• Different codons appears in significantly different frequencies, which is not expected assuming neutrality

• Bias is measured in several ways, most popular is the codon adaptation index:

• Possible sources of bias:

– Selection for translational efficiency given different tRNA abudnances• Highly expressed genes tend to have stronger codon adaptation indices

– Sequence context mutational effects• E.g. CpGs are highly mutable

– Selection for low insertion/deletion potential

• Weak selection for codon bias should be stronger for genomes with larger effective population size. In some cases this is true

L

i

iLi X

XCAI

1

max,1

Codon frequency divided by the frequency of the synonymous codon with maximal frequency


Positive selection in humans vs chimp

Nielsen et al., 2005 PloS Biol

Testis genes: P<0.0001

Immunity genes, Gematogenesis, Olfaction P<1e-5Inhibition of apoptosis P<0.005Sensory perception P<0.02

Kn vs Ks Looking at trends for families of genes

Significantly enriched functions/tissues Example


Mcdonald-Kreitman test

rM

sM

sr MMM

Possible neutral replacement mutations

Possible neutral synonymous mutations

Deleterious mutations

s

r

sb

rb

MM

MTMT )3/()3/(

Expected ratio of replacement to synonymous fixed mutations

s

r

sw

rw

MM

MTMT )3/()3/(

Expected ratio of replacement to synonymous polymorphic mutations

Outgroup

Tw

Tb

RFn

SFn

RPn

SPn

Replacement

Synonymous

Fixed Poly

M. Kreitman


Mcdonald-Krietman test - example• Works by comparing Ka/Ks divergence between species and Ka/Ks diversity among

species populations• Negative selection should make the divergence Ka/Ks smaller than the diversity

Ka/Ks• Positive selection should drive the opposite effect

Busstamente et al, Nature 2005

humanchimp


Reminder: the coalescent

Present 10

2)( 5

NTE

6

2)( 4

NTE

3

2)( 3

NTE

NTE 2)( 2

Past

1 2 3 54

Theorem: The amount of time during which there are k lineages, tk has approximately an exponential distribution with mean 2N * (2/(k(k-1)))

In the infinite sites model, mutations occur at distinct sites.

Back in time

)4(,2

2 NkNk

Coalescent:N

kk

2

1

2

)1(

mutation:


Infinite sites model

n

j

n

jtot

n

jjtot

jjjjTE

jtT

22

2

)1(

12

)1(

2)(

Theorem: Let u be the mutation rate for a locus under consideration, and set =4Nu. Under the infinite sites model, the expected number of segregating sites is:

Proof: Let tj be the amount of time in the coalescent during which there are j lineages. We showed earlier that tj has approximately an exponential distribution with mean 2/(j(j-1)). The total amount of time in the tree for a sample size n is:

1

1

1)(

n

i iSE

Mutations occur at rate 2Nu:

)(2)( totn TNuESE

ii N 4


Infinite sites model

Theorem: =4Nu. Under the infinite sites model, the number of segregating sites Sn has

Proof: Let sj be the number of segregating sites created when there were j lineages. While there are j lineages, we may get mutations at rate 2Nuj, and coalescence at rate j(j-1)/2. Mutations occur before coalescence with probability:

1

12

21

1

11)(

n

i

n

in ii

SV

14

4

2/)1(2

2

jNu

Nu

jjNuj

Nuj

,..2,1,01

1

1)Pr(

kj

j

jks

k

j

2

2

2

22

2

2

2

)1(1)1(

)1(

)1(

)1(

1

1)(

jjj

j

j

j

jp

psVar j

k successes:

It’s a shifted geometric distribution:


The HKA test (Hudson, Kreitman, Aguade)

humanchimp

humanchimp

humanchimp

Slow

Fast

PurifyingSelection

Bi

Ai SS ,

iD

Number of segregating sites in locus i and population A and B (Polymorphism)

L loci are sequenced in populations A and B Each locus is supposed to be behaving as an infinite site locus

Number of difference between two random gametes from A and B (Divergence)

Our null hypothesis is of neutral evolution for T’ generations with population sizes 2N and 2Nf, but starting from a single ancestral population of size 2N(1+f)/2

We do not know:

1

1

1)(

An

ii

Ai i

SE

Nfi ,,

We want to allow different loci to have different mutation rates

1

1

1)(

Bn

ii

Bi i

fSE

1

12

2 1)()(

An

ii

Ai

Ai i

SESV

1

12

2 1)()()(

Bn

ii

Bi

Bi i

fSESV

What is the expected divergence?

ii N 4



B

Bi

Ai SS ,

iD



1

1

1)(

An

ii

Ai i

SE

1

1

1)(

Bn

ii

Bi i

fSE

1

12

2 1)()(

An

ii

Ai

Ai i

SESV

1

12

2 1)()()(

Bn

ii

Bi

Bi i

fSESV

A

NTT 2/' )2/)1(( fTED ii

DivergenceIs a Poisson variable

Coalescent in ancestral population

2)2/)1((2/)1()( ffTDVar iiii

Variance of Poisson variable

Variance of S with n=2



Bi

Ai SS ,

iD



1

1

1)(

An

ii

Ai i

SE

1

1

1)(

Bn

ii

Bi i

fSE

1

12

2 1)()(

An

ii

Ai

Ai i

SESV

1

12

2 1)()()(

Bn

ii

Bi

Bi i

fSESV

)2/)1(( fTED ii 2)2/)1(()( fEDDVar iii

There are L+2 parameters that we should estimate. This can be done by solving the equations:

L

iiA

L

i

Ai nCS

11

ˆ)(

L

iiB

L

i

Bi fnCS

11

ˆ)(

Find

L

ii

1

Find f

L

ii

L

ii fTD

11

ˆ)2/)ˆ1(ˆ( Find T

)()()2/)ˆ1(ˆˆBAii

Bi

Ai nCnCfTDSS Find i



Bi

Ai SS ,

iD



1

1

1)(

An

ii

Ai i

SE

1

1

1)(

Bn

ii

Bi i

fSE

1

12

2 1)()(

An

ii

Ai

Ai i

SESV

1

12

2 1)()()(

Bn

ii

Bi

Bi i

fSESV

)2/)1(( fTED ii 2)2/)1(()( fEDDVar iii

The goodness of fit can be expressed as:

L

iiii

L

i

Bi

Bi

Bi

L

i

Ai

Ai

Ai

DraVDED

SraVSES

SraVSESX

1

2

1

2

1

22

)(ˆ/))(ˆ(

)(ˆ/))(ˆ(

)(ˆ/))(ˆ(

Significance is best tested using simulations (although we can assume normality and independence and use chi square with 3L-(L+2) degrees of freedom)


Example: Drosophila chromosome 4

humanchimp

humanchimp

humanchimp

Slow

Fast

PurifyingSelection

Collecting data from a target locus: Size, Polymorphism, divergence

Collecting data from a control neutral locus: Size, Polymorphism, divergence

ciD5’ Adh

Nucs11063326

Polymorphic030

Divergence5478

This data is unlikely given our model, why?

Population dynamics? (small turns bigger?)

Selective sweeps?

Mutational effects (gene conversion?)


Impact of local recombination rate

Presgraves 2005

In drosophila, we find strong correlation between recombination rate and the level of polymorphism:

High recombination regions have high polymorphismLow recombination region have low polymorphism

One possible reason is that high recombination makes mutation rate higher

If this was the reason, we should have observe correlation of recombination and divergence

But divergence and recombination are not correlated

The explanation may therefore be more efficient purifying selection in low recombination regions

We indeed see high dN/dS in high recombination – more efficient fixation of beneficial mutations


Compensatory mutations in proteins?

PDB structuresHomology modelling

3-Alignments

Pairs of interacting residues

Rat Mouse Human

Choi et al, Nat Genet 2005

Find pairs of mutations in interacting residues (DRIP)Coupled: occurring in the same lineageUncoupled: occurring in different lineages


Codon volatility

• Volatility is the number/fraction of adjacent non-synonymous codons

• Genes under positive selection may have increased volatility

• Think about the distance from the stationary codon distribution

• No need to align!!

Plotkin et al, Nature 2004


Using extensive polymorphisms and haplotype data, the most recent examples of positive selection:

Sabeti et al, Nature 2007

the analysis reveals more than 300 strong candidate regions. Focusing on the strongest 22 regions, we develop a heuristic for scrutinizing these regions to identify candidate targets of selection. In a complementary analysis, we identify 26 non-synonymous, coding, single nucleotide polymorphisms showing regional evidence of positive selection. Examination of these candidates highlights three cases in which two genes in a common biological process have apparently undergone positive selection in the same population:LARGE and DMD, both related to infection by the Lassa virus3, in West Africa;SLC24A5 and SLC45A2, both involved in skin pigmentation4, 5, in Europe; and EDAR and EDA2R, both involved in development of hair follicles6, in Asia.


Time resolution of different positive selection methods

Sabeti et al, Science 2005

genome evolution. amos tanay 2009 genome evolution lecture 7: selection in protein coding genes

Documents

amos tanay

genome evolution lecture

henikoff slide

protein genes

evolution identification

protein coding genes

protein function

kaks slide