Gibbs sampling - DTU Health Tech · 2011. 10. 27. · Gibbs sampling A special kind of Monte Carlo...

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Department of Systems Biology Technical University of Denmark 1

Gibbs sampling

Massimo Andreatta Center for Biological Sequence Analysis

Technical University of Denmark [email protected]

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Monte Carlo simulations

MC methods use repeated random sampling to numerically approximate solutions to problems

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A simple example: computing π with sampling

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Ac = πr2

r

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As = 2r( )2

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Ac = πr2

r

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As = 2r( )2

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AcAs

=πr2

4r2=π4

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π = 4 AcAs

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π = 4 AcAs

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X



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π = 4 AcAs

X

X

Throw darts randomly

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hit circlehit square

=hit

hit +miss=AcAs

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X

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π = 4 AcAs

hit=0for N iterations x = random(-1,1) y = random(-1,1) dist=sqrt(x2+y2)

if (dist

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X

hit=0for N iterations x = random(-1,1) y = random(-1,1) dist=sqrt(x2+y2)

if (dist

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-  More iterations more accurate estimate -  After 1,000,000 iterations I got pi ≈ 3,14182...

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Gibbs sampling

A special kind of Monte Carlo method (Markov Chain Monte Carlo, or MCMC) - estimates a distribution by sampling from it - the samples are taken with pseudo-random steps - stepping to the next state only depends on the current state (memory-less chain)

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Gibbs sampling

Stochastic search

Z

f(Z)

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Gibbs sampling

Stochastic search

Zi = current state of the system P = probability of accepting the move T = a scalar lowered during the search

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P =min 1,exp dET

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dE = f (Zi) − f (Zi−1)

Z

f(Z)

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Gibbs sampling - down to biology

Sequence alignment


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P =min 1,exp dET

SLFIGLKGDIRESTVDGEEEVQLIAAVPGK VFRLKGGAPIKGVTF SFSCIAIGIITLYLG IDQVTIAGAKLRSLNWIQKETLVTFKNPHAKKQDV KMLLDNINTPEGIIP ELLEFHYYLSSKLNK LNKFISPKSVAGRFAESLHNPYPDYHWLRT

€

dE = f (Zi) − f (Zi−1)

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Gibbs sampling - down to biology

Sequence alignment


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P =min 1,exp dET

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dE = Ei − Ei−1

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E = Cp,ap,a∑ log

pp,aqa


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dE = f (Zi) − f (Zi−1)

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Gibbs sampling - sequence alignment

State transition


move to state +1

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dE = Ei − Ei−1

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E = Cp,ap,a∑ log

pp,aqa

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move to state +1

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P =min 1,exp dET

Accept or reject the move?

Note that the probability of going to the new state only depends on the previous state


State transition

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dE = Ei − Ei−1

€

E = Cp,ap,a∑ log

pp,aqa

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Ei = 2.52


move to state +1


Numerical example - 1

€

Ei−1 = 2.44

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P = min 1,exp 0.080.2

= min 1 , 1.49[ ] =1 Accept move with

Prob = 100%

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T = 0.2

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move to state +1


Numerical example - 2

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P = min 1,exp −0.090.2

= min 1 , 0.638[ ] = 0.638

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T = 0.2

Accept move with Prob = 63.8%

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Ei = 2.35

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Ei−1 = 2.44

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Now, one thing at a time

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What is the MC temperature?

iteration

T

it’s a scalar decreased during the simulation

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What is the MC temperature?

iteration

T

it’s a scalar decreased during the simulation E.g. same dE=-0.3 but at different temperatures

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P(t1) =min 1,expdEt1

=min 1,exp

−0.30.4

= 0.47

t1=0.4

t3=0.02

€

P(t3) =min 1,exp−0.30.02

≈ 0

t2=0.1

€

P(t2) =min 1,exp−0.30.1

= 0.05

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Z

f(Z)

Move freely around states when the system is “warm”, then cool it off to force it into a state of high fitness

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Why sampling? SLFIGLKGDIRESTVDGEEEVQLIAAVPGK VFRLKGGAPIKGVTF SFSCIAIGIITLYLG IDQVTIAGAKLRSLNWIQKETLVTFKNPHAKKQDV KMLLDNINTPEGIIP ELLEFHYYLSSKLNK LNKFISPKSVAGRFAESLHNPYPDYHWLRT

............

DFAAQVDYPSTGLY

50 sequences 12 amino acids long

try all possible combinations with a 9-mer overlap

450 ~ 1030 possible combinations

...computationally unfeasible

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move to state +1

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P =min 1,exp dET


Single sequence move

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dE = Ei − Ei−1

€

E = Cp,ap,a∑ log

pp,aqa

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move to state +1

€

P =min 1,exp dET


Phase shift move

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dE = Ei − Ei−1

€

E = Cp,ap,a∑ log

pp,aqa

shift all sequences

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A sketch for the alignment algorithm

•  Start from a random alignment •  Set initial temperature •  For N iterations

•  pick a random sequence •  suggest a shift move •  accept or reject the move depending on •  every Psh moves, attempt a phase shift move •  decrease temperature

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P =min 1,exp dET

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Does it work?

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Gibbs sequence alignment - performance

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Aligning scoring matrices

More Gibbs sampling

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Alignment of scoring matrices

4 networks trained on HLA*DRB1-0401

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Combined logo

Equally valid solutions, but with different core registers

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The PSSM-align algorithm

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Individual PSSM

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LIndividual PSSM

1. Extend matrix with BG frequencies

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LAll individual PSSMs



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1.  Extend matrix with BG frequencies

2.  Apply random shift


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2.  Apply random shift 3.  Do Gibbs sampling for

many iterations core

Maximize combined Information Content of the core


€

P =min 1,exp dET

Accept moves with probability:

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2.  Apply random shift 3.  Do Gibbs sampling for

many iterations core

Avg matrix

Offset2

-300

-803

Maximize combined Information Content of the core


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after alignment before alignment

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And more Gibbs sampling

Clustering peptide data

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Gibbs clustering

--ELLEFHYYLSSKLNK----------LNKFISPKSVAGRFAESLHNPYPDYHWLRT-------NKVKSLRILNTRRKL-------MMGMFNMLSTVLGVS----AKSSPAYPSVLGQTI--------RHLIFCHSKKKCDELAAK-

----SLFIGLKGDIRESTV----DGEEEVQLIAAVPGK----------VFRLKGGAPIKGVTF---SFSCIAIGIITLYLG-------IDQVTIAGAKLRSLN--WIQKETLVTFKNPHAKKQDV-------KMLLDNINTPEGIIP

Cluster 2

Cluster 1 SLFIGLKGDIRESTVDGEEEVQLIAAVPGKVFRLKGGAPIKGVTFSFSCIAIGIITLYLGIDQVTIAGAKLRSLNWIQKETLVTFKNPHAKKQDVKMLLDNINTPEGIIPELLEFHYYLSSKLNKLNKFISPKSVAGRFAESLHNPYPDYHWLRTNKVKSLRILNTRRKLMMGMFNMLSTVLGVSAKSSPAYPSVLGQTIRHLIFCHSKKKCDELAAK

Multiple motifs!

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Gibbs clustering - the algorithm

FIGLKGDIREEEVQLIAARLKGGAPIKSCIAIGIITQVTIAGAKLQKETLVTFKLLDNINTPELEFHYYLSSKFISPKSVALHNPYPDYHVKSLRILNTGMFNMLSTVSSPAYPSVLLIFCHSKKK

1. List of peptides

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1. List of peptides 2. create N random groups

-----QVTIAGAKL----------QKETLVTFK----------LEFHYYLSS----------GMFNMLSTV----------SSPAYPSVL-----

-----SLFIGLKGD----------SFSCIAIGI----------KMLLDNINT----------KYVHGTWRS----------NKVKSLRIL-----

-----LHNPYPDYH----------LIFCHSKKK----------RLKGGAPIK----------KFISPKSVA----------EEEVQLIAA-----

g1 g2 gN


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-----QVTIAGAKL----------QKETLVTFK----------LEFHYYLSS----------GMFNMLSTV----------SSPAYPSVL-----



g1 g2 gN

GMFNMLSTV

3 Move sequence


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4b. Remove peptide from its group I



-----QVTIAGAKL----------QKETLVTFK----------LEFHYYLSS----------SSPAYPSVL-----



g1 g2 gN

GMFNMLSTV

3 Move sequence


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5b. Score peptide to a new random group R and in its original group I

€

dE = SR − SI






g1 g2 gN

GMFNMLSTV

3 Move sequence


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P =min 1,exp dET

6b. Accept or reject move



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dE = SR − SI





-----LHNPYPDYH----------LIFCHSKKK----------RLKGGAPIK----------KFISPKSVA----------EEEVQLIAA----------GMFNMLSTV-----

g1 g2 gN

GMFNMLSTV

3 Move sequence


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P =min 1,exp dET

6b. Accept or reject move



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dE = SR − SI





-----LHNPYPDYH----------LIFCHSKKK----------RLKGGAPIK----------KFISPKSVA----------EEEVQLIAA----------GMFNMLSTV-----

g1 g2 gN

GMFNMLSTV

3 Move sequence


And iterate many times, gradually decreasing T

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Two MHC class I alleles: HLA-A*0101 and HLA-B*4402

Does it work ?

Mixture of 100 binders for the two alleles

ATDKAAAAY A*0101EVDQTKIQY A*0101AETGSQGVY B*4402ITDITKYLY A*0101AEMKTDAAT B*4402FEIKSAKKF B*4402LSEMLNKEY A*0101GELDRWEKI B*4402LTDSSTLLV A*0101FTIDFKLKY A*0101TTTIKPVSY A*0101EEKAFSPEV B*4402AENLWVPVY B*4402

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A0101 B4402

G 1

G 2

Mixed

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97 3

3 97

A0101 B4402

G 1

G 2

Resolved

Mixed

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Five MHC class I alleles A0101 A0201 A0301 B0702 B4402

G 0

G 1

G 2

G 3

G 4

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Five MHC class I alleles A0101

0 1 76 1 0

2 4 0 0 95

5 87 5 1 0

93 2 19 0 2

0 6 0 98 3

A0201 A0301 B0702 B4402

G 0

G 1

G 2

G 3

G 4

HLA-B4402 94%

HLA-B0702 92%

HLA-A0201 89%

HLA-A0101 80%

HLA-A0301 97%

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HLA-A*02:01 sub-motifs

= 10 nM = 4 hours

= 10 nM = 1.5 hours

666 peptide binders (aff < 500 nM)

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Splitting with Gibbs clustering

= 10 nM = 3.5 hours

= 10 nM = 2.25 hours

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Gibbs clustering

And what if we don’t know a priori the number of clusters?

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How many clusters?

We could run the algorithm with different number of clusters k and choose the k with highest information content

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How many clusters?


What’s going on ?

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How many clusters?


What’s going on ?

smaller groups tend to have higher information content

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How many clusters?

Let’s look back at the Energy function

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E = Cp,ap,a∑ log

pp,aqa

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How many clusters?


€

E = Cp,ap,a∑ log

pp,aqa

This is equivalent to scoring each sequence S to its matrix

€

E = logpp,aqap,a

∑S∑

L

20

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How many clusters?


€

E = Cp,ap,a∑ log

pp,aqa


€

E = logpp,aqap,a

∑S∑

What is the problem? Overfitting. S was also used to calculate the log-odds matrix!

The contribution of S on the matrix will be larger if the cluster is small.

L

20

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How many clusters?


€

E = Cp,ap,a∑ log

pp,aqa


€

E = logpp,aqap,a

∑S∑



L

20

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How many clusters?

€

E = logpp,aqap,a

∑S∑



€

E = logpp,aS−

qap,a∑

S∑

Before scoring S, remove it and update the matrix

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How many clusters?

€

E = logpp,aqap,a

∑S∑

€

E = logpp,aS−

qap,a∑

S∑

Is this so important..?

YQAFRTKVHSPRTLNAWVYALTVVWLLLSSIGIPAYAVAKCNLNHTPYDINQMLLLMMTLPSIKELENEYYFIENATFFIFAEMLASIDL...

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How many clusters?

€

E = logpp,aqap,a

∑S∑

€

E = logpp,aS−

qap,a∑

S∑

Is this so important..? YES

YQAFRTKVHSPRTLNAWVYALTVVWLLLSSIGIPAYAVAKCNLNHTPYDINQMLLLMMTLPSIKELENEYYFIENATFFIFAEMLASIDL...

SCORE Num of sequences in the cluster

100 20 3

w/o removing

5.52 10.42 26.78

removing 4.11 2.57 0.05

Score YALTVVWLL to a matrix, including vs. excluding YALTVVWLL in the matrix construction

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How many clusters?

Quality of clustering is not only determined by information content of individual clusters (intra-cluster distance), but also by the ability of different groups to discriminate (inter-cluster distance)

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How many clusters?

Quality of clustering is not only determined by information content of individual clusters (intra-cluster distance), but also by the ability of different groups to discriminate (inter-cluster distance)

€

E = logpp,aS−

qap,a∑

S∑

€

E = logpp,aS−

qp,aS

p,a∑

S∑

position and cluster-specific background (the background is calculated on all groups not containing S, it accounts for inter-cluster distance)

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How many clusters?

€

E = logpp,aS−

qp,aS

p,a∑

S∑ −λn

One last thing and we are ready.

A parameter λ to modulate the ‘tightness’ of the clustering (n is the number of clusters)

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How many clusters?

€

E = logpp,aS−

qp,aS

p,a∑

S∑ −λn

One last thing and we are ready.

A parameter λ to modulate the ‘tightness’ of the clustering (n is the number of clusters) position and cluster-specific background

(the background is calculated on all groups not containing S, it accounts for inter-cluster distance)

frequencies are calculated by removing the sequence being scored S

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How many clusters?

!

!

!!

!!

! ! !

!! !

3.0

3.4

3.8

4.2

2 alleles ! lambda=0.02

Groups

KLD

sum

2 4 6 8 10 12

!

!

!

!

! ! ! !!

! ! !

3.2

3.4

3.6

3.8

4.0


Groups

KLD

sum

2 4 6 8 10 12

!

!

!

!

!

!!

!! !

!!

3.4

3.8

4.2


Groups

KLD

sum

2 4 6 8 10 12

!

! !

!

!

!

!

!! !

!

!

3.4

3.8

4.2


Groups

KLD

sum

2 4 6 8 10 12

!

!

!

! !

!

!

!

!

!

!

!

3.3

3.5

3.7

3.9


Groups

KLD

sum

2 4 6 8 10 12

!

!

!

!

! !

!!

!!

! !

2.8

3.2

3.6


Groups

KLD

sum

2 4 6 8 10 12

!

!

!

!

!

! !

! !

! !!

2.8

3.2

3.6


Groups

KLD

sum

2 4 6 8 10 12

!

!

!

!

!

!

!

! !

!!

!

3.0

3.4

3.8


Groups

KLD

sum

2 4 6 8 10 12

Binders for 2 to 9 MHC class I alleles

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How many clusters?

2 3 4 5 6 7 8

34

56

78

9

10 random allele combinations

Alleles

Nu

mb

er

of

clu

ste

rs

!

!

!

! !

! !

! !

!

!

!

!

!

! !

!

!

!

!

!

Lambda penalty

0

0.02

0.04

!

!

2 3 4 5 6 7 8

24

68

10

Lambda = 0.000

Alleles

Nu

mb

er

of

clu

ste

rs

!

!

2 3 4 5 6 7 8

24

68

10

Lambda = 0.020

Alleles

Nu

mb

er

of

clu

ste

rs

!

!

! !

!

2 3 4 5 6 7 8

23

45

67

89

Lambda = 0.040

Alleles

Nu

mb

er

of

clu

ste

rs

CEN

TER FO

R B

IOLO

GIC

AL SEQ

UEN

CE A

NA

LYSIS


In conclusion

•  Sampling methods can solve problems where the search space is too large to be exhaustively explored

•  Gibbs sampling can detect even weak motifs in a sequence alignment (e.g. MHC class II)

•  More than 1,000 papers in PubMed using Gibbs sampling methods

•  Transcription start-sites •  Receptor binding sites •  Acceptor:Donor sites •  ...

Gibbs sampling - DTU Health Tech · 2011. 10. 27. · Gibbs sampling A special kind of Monte Carlo...

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