ScheduleDay 1: Molecular Evolution9.00-10.15 Lecture: Models of Sequence Evolution and Statistical Alignment10.30-12.00 Practical: Molecular Evolution (Phylogenies – PHYLIP+) 2.00-3.30 Lecture: Molecular Evolution & Comparative Genomics3.30-5.00 Student Activity: Prepare projects Day 2: Population Biology and Mapping9.00-10.00 Lecture: Population Genetics and Gene Genealogies10.00-11.00 Exercise: Jukes-Cantor and Rate Matrix11.00-12.00 Lecture: Inferring Recombination Histories 2.00-3.30 Practical: DNA Sequence Analysis (PAML Phase +)3.30-5.00 Student Activity: Prepare projects Day 3: Integrative Genomics (IG)9.00-10.15 Lecture: High Throughput Data, the structure of IG, GF10.30-12.00 Practical: Statistical Alignment & Footprinting2.00-3.30 Lecture (L): Grammars and RNA Prediction3.30-5.00 Student Activity: Prepare projects Day 4: Integrative Genomics (IG)9.00-10.00 Lecture: Networks and other concepts 10.00-11.00 Exercise: Stochastic Context Free Grammars11.00-12.00 Lecture: Concepts, Data Analysis and Functional Studies2.00-3.30 Practical: Detecting Recombinations 3.30-5.00 Student Activity: Prepare projects Day 5: Project Discussion/Presentation9.00-10.00 Project 1 – Population Genomics: 1000 genomes10.00-11.00 Practical – Integrative Data Analysis – Mapping11.00-12.00 Project 2 – Comparative Genomics: Signals2.00-3.00 Project 3 – Integrative Genomics: Basic data types3.00-4.00 Exercise: Statistical Alignment4.00-5.00 Project 4 – Comparative Biology: Networks
The Data & its growth.1976/79 The first viral genome –MS2/X174
1995 The first prokaryotic genome – H. influenzae
1996 The first unicellular eukaryotic genome - Yeast
1997 The first multicellular eukaryotic genome – C.elegans
2000 Arabidopsis thaliana, Drosophila
2001 The human genome
2002 Mouse Genome
2005+ Dog, Marsupial, Rat, Chicken, 12 Drosophilas
1.5.08: Known
>10000 viral genomes
2000 prokaryotic genomes
80 Archeobacterial genomes
A general increase in data involving higher structures and dynamics of biological systems
The Human Genome (Harding & Sanger)
*50.000
*20
globin(chromosome 11)
6*104 bp
3*109 bp
*103
Exon 2Exon 1 Exon 3
5’ flanking 3’ flanking3*103 bp
Myoglobin globin
ATTGCCATGTCGATAATTGGACTATTTGGA
30 bp
aa aa aa aa aa aa aa aa aa aa
DNA:
Protein:
ACGTC
Central Problems: History cannot be observed, only end products.
Even if History could be observed, the underlying process couldn’t !!
ACGCC
AGGCC
AGGCT
AGGCT
AGGTT
ACGTC
ACGCC
AGGCC
AGGCTAGGGC
AGGCT
AGGTT AGTGC
Some DefinitionsState space – a set often corresponding of possible observations ie {A,C,G,T}.
A random variable, X can take values in the state space with probabilities ie P{X=A} = ¼. The value taken often indicated by small letters - x
Stochastic Process is a set of time labeled stochastic variables Xt
ie P{X0=A, X1=C, .., X5=G} =.00122
Time can be discrete or continuous, in our context it will almost always mean natural numbers, N {0,1,2,3,4..}, or an interval on the real line, R.
Time Homogeneity – the process is the same for all t.
Markov Property: ie
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P{X i X i−1,..., X0} = P{X i X i−1}
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P{X i,X i−1,...,X0} = P{X0}P{X1 X0}...P{X i X i−1}
2) Processes in different positions of the molecule are independent, so the probability for the whole alignment will be the product of the probabilities of the individual patterns.
Simplifying Assumptions IData: s1=TCGGTA,s2=TGGTT
1) Only substitutions. s1 TCGGTA s1 TCGGA s2 TGGT-T s2 TGGTT
TGGTTTCGGTA
a - unknown
Biological setup
TT
a1a2
a3a4
a5
G G T T
C G G A
Probability of Data
TGGTT)(TCGGTA)(*)( →→=∑ PPPP
TGGTT)(TCGGA)(*)( →→=∑ PPPP
)1s()1s(*)(5
1iiiiii
aii
iaPaPaPP →→∏= ∑
=
Simplifying Assumptions II
3) The evolutionary process is the same in all positions
4) Time reversibility: Virtually all models of sequence evolution are time reversible. I.e. π i Pi,j(t) = πj Pj,i(t), where πi is the stationary distribution of i and Pt(i->j) the probability that state i has changed into state j after t time. This implies that
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P(a)a
∑ * Pl1(ai → s1i)Pl1
(ai → s2i) = P(s1i) * Pl1 +l2(s1i → s2i)
=a
s1i s2i
l2+l1l1 l2 s2is1i
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P = ∏i=1
5P(ai)
a∑ * P(ai → s1i)P(ai → s2i)
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P = ∏i=1
5P(s1i)P(s1i → s2i)
Simplifying assumptions III
6) The rate matrix, Q, for the continuous time Markov Chain is the same at all times (and often all positions). However, it is possible to let the rate of events, ri, vary from site to site, then the term for passed time, t, will be substituted by ri*t.
5) The nucleotide at any position evolves following a continuous time Markov Chain.
T O A C G TF A -(qA,C+qA,G+qA,T) qA,C qA,G qA,T
R C qC,A -(qC,A+qC,G+qC,T) qC, G qC ,T O G qG,A qG,C -(qG,A+qG,C+qG,T) qG,T
M T qT,A qT,C qT,G -(qT,A+qT,C+qT,G)
Pi,j(t) continuous time markov chain on the state space {A,C,G,T}.
Q - rate matrix:
t1 t2CC
A
ijji q
P=>−
)(lim ,
0 iiii q
P−=
−>−
1)(lim ,
0
i. P(0) = I
Q and P(t)What is the probability of going from i (C?) to j (G?) in time t with rate matrix Q?
vi. QE=0 Eij=1 (all i,j) vii. PE=E viii. If AB=BA, then eA+B=eAeB.
ii. P() close to I+Q for small
iii. P'(0) = Q.
iv. lim P(t) has the equilibrium frequencies of the 4 nucleotides in each row
v. Waiting time in state j, Tj, P(Tj > t) = eqjj
t
.......!3)(
!2)(
!)()exp()(
32
0
++++=== ∑∞
=
tQtQtQIitQtQtP
i
i
Expected number of events at equilibrium
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t −qiiπ inucleotides
∑
Jukes-Cantor (JC69): Total SymmetryRate-matrix, R: T O A C G T
F A R C O G M T
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P = P(s1)∏i=1
5P(s1i → s2i) = ( 1
4)5P(T → T)P(C → G)P(G → G)P(G → T)P(A → T)
= (14
)5( 14
)5(1+ 3e−4 a )2(1− e−4a )3
Stationary Distribution: (1,1,1,1)/4.
Transition prob. after time t, a = *t:
P(equal) = ¼(1 + 3e-4*a ) ~ 1 - 3a P(specific difference) = ¼(1 - e-4*a ) ~ 3a
Principle of Inference: LikelihoodLikelihood function L() – the probability of data as function of parameters: L(Q,D)
If the data is a series of independent experiments L() will become a product of Likelihoods of each experiment, l() will become the sum of LogLikelihoods of each experiment
In Likelihood analysis parameter is not viewed as a random variable.
increases.data as (D)ˆ:yConsistenc trueQ→Q
LogLikelihood Function – l(): ln(L(Q,D))
LikelihoodLogLikelihood
From Q to P for Jukes-Cantor
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−−
−−
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−−
−−
3111131111311113
33
33
α
αααααααααααααααα
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−−
−−
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−−
−−
−
3111131111311113
4
3111131111311113
1i
i
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−3α α α αα −3α α αα α −3α αα α α −3α
⎡
⎣
⎢ ⎢ ⎢ ⎢
⎤
⎦
⎥ ⎥ ⎥ ⎥i= 0
∞
∑
i
t i /i!=1/4[I − (−4αt)i
−3 1 1 11 −3 1 11 1 −3 11 1 1 −3
⎡
⎣
⎢ ⎢ ⎢ ⎢
⎤
⎦
⎥ ⎥ ⎥ ⎥i=1
∞
∑ /i!] =
1/4[I +
3 −1 −1 −1−1 3 −1 −1−1 −1 3 −1−1 −1 −1 3
⎡
⎣
⎢ ⎢ ⎢ ⎢
⎤
⎦
⎥ ⎥ ⎥ ⎥e−4αt ]
Exponentiation/Powering of Matrices
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Qi = BΛB−1BΛB−1...BΛB−1 = BΛiB−1then
€
Q = BΛB−1
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Λ=
l1 0 0 00 λ2 0 00 0 λ3 00 0 0 λ4
⎛
⎝
⎜ ⎜ ⎜ ⎜
⎞
⎠
⎟ ⎟ ⎟ ⎟
If where
€
(tQ)i
i!i= 0
∞
∑ = (tBΛB−1)i
i!= B[ (tΛ)i
i!i= 0
∞
∑i= 0
∞
∑ ]B−1 = B
exp tλ1 0 0 00 exp tλ 2 0 00 0 exp tλ 3 00 0 0 exp tλ 4
⎛
⎝
⎜ ⎜ ⎜ ⎜
⎞
⎠
⎟ ⎟ ⎟ ⎟B−1and
Finding Λ: det (Q-lI)=0
By eigen values:
Numerically:
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(tQ)i
i!i= 0
∞
∑ ~ (tQ)i
i!i= 0
k
∑ where k ~6-10
JC69:
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P(t) =
1 1/4 0 11 1/4 0 −11 −1/4 1 01 −1/4 −1 0
⎛
⎝
⎜ ⎜ ⎜ ⎜
⎞
⎠
⎟ ⎟ ⎟ ⎟
1 0 0 00 exp− 4tα 0 00 0 exp− 4tα 00 0 0 exp− 4tα
⎛
⎝
⎜ ⎜ ⎜ ⎜
⎞
⎠
⎟ ⎟ ⎟ ⎟
1/4 1/4 1/4 1/41/8 1/8 −1/8 −1/80 0 1 −11 −1 0 0
⎛
⎝
⎜ ⎜ ⎜ ⎜
⎞
⎠
⎟ ⎟ ⎟ ⎟
Finding : (Q-liI)bi=0
Kimura 2-parameter model - K80 TO A C G T
F A - R C O G M T a = *t b = *t
Q:
P(t)start
)21(25. )(24 bab ee +−− ++
)1(25. 4be−−
)1(25. 4be−−
)21(25. )(24 bab ee +−− −+
Unequal base composition: (Felsenstein, 1981 F81)
Qi,j = C*πj i unequal j
Felsenstein81 & Hasegawa, Kishino & Yano 85
Tv/Tr & compostion bias (Hasegawa, Kishino & Yano, 1985 HKY85)
()*C*πj i- >j a transition Qi,j = C*πj i- >j a transversion
Rates to frequent nucleotides are high - (π =(πA , πC , πG , πT)
Tv/Tr = (πT πC +πA πG )/[(πT+πC )(πA+ πG )]A
G
T
C
Tv/Tr = () (πT πC +πA πG )/[(πT+πC )(πA+ πG )]
Measuring Selection ThrSerACGTCA
Certain events have functional consequences and will be selected out. The strength and localization of this selection is of great interest.
ThrProPro ACGCCA
-
ArgSer AGGCCG
-
The selection criteria could in principle be anything, but the selection against amino acid changes is without comparison the most important
ThrSer ACGCCG
ThrSer ACTCTG
AlaSer GCTCTG
AlaSer GCACTG
The Genetic Code
i.
3 classes of sites:
4
2-2
1-1-1-1
Problems:
i. Not all fit into those categories.
ii. Change in on site can change the status of another.
4 (3rd) 1-1-1-1 (3rd)
ii. TA (2nd)
Possible events if the genetic code remade from Li,1997
Substitutions Number Percent
Total in all codons 549 100
Synonymous 134 25
Nonsynonymous 415 75
Missense 392 71
Nonsense 23 4
Possible number of substitutions: 61 (codons)*3 (positions)*3 (alternative nucleotides).
Kimura’s 2 parameter model & Li’s Model.
Selection on the 3 kinds of sites (a,b)(?,?)
1-1-1-1 (f*,f*)
2-2 (,f*)
4 (, )
Rates:start
Probabilities:
)21(25. )(24 bab ee +−− ++
)1(25. 4be−−
)1(25. 4be−−
)21(25. )(24 bab ee +−− −+
Sites Total Conserved Transitions Transversions1-1-1-1 274 246 (.8978) 12(.0438) 16(.0584)2-2 77 51 (.6623) 21(.2727) 5(.0649)4 78 47 (.6026) 16(.2051) 15(.1923)
alpha-globin from rabbit and mouse.Ser Thr Glu Met Cys Leu Met Gly GlyTCA ACT GAG ATG TGT TTA ATG GGG GGA * * * * * * * **TCG ACA GGG ATA TAT CTA ATG GGT ATASer Thr Gly Ile Tyr Leu Met Gly Ile
Z(t,t) = .50[1+exp(-2t) - 2exp(-t(+)] transition Y(t,t) = .25[1-exp(-2t )] transversionX(t,t) = .25[1+exp(-2t) + 2exp(-t(+)] identity
L(observations,a,b,f)= C(429,274,77,78)* {X(a*f,b*f)246*Y(a*f,b*f)12*Z(a*f,b*f)16}* {X(a,b*f)51*Y(a,b*f)21*Z(a,b*f)5}*{X(a,b)47*Y(a,b)16*Z(a,b)15}
where a = at and b = bt.
Estimated Parameters: a = 0.3003 b = 0.1871 2*b = 0.3742 (a + 2*b) = 0.6745 f = 0.1663
Transitions Transversions1-1-1-1 a*f = 0.0500 2*b*f = 0.06222-2 a = 0.3004 2*b*f = 0.06224 a = 0.3004 2*b = 0.3741
Expected number of: replacement substitutions 35.49 synonymous 75.93Replacement sites : 246 + (0.3742/0.6744)*77 = 314.72Silent sites : 429 - 314.72 = 114.28 K s = .6644 Ka = .1127
Approaches to Sequence Analysis
s2 s3 s4s1
statistics
GT-CAT
GTTGGT
GT-CA-
CT-CA-
Parsimony, similarity, optimisation.
Data {GTCAT,GTTGGT,GTCA,CTCA}
Actual Practice: 2 phase analysis.
Ideal Practice: 1 phase analysis.
1. TKF91 - The combined
substitution/indel process.
2. Acceleration of Basic
Algorithm
3. Many Sequence Algorithm
4. MCMC Approaches
Thorne-Kishino-Felsenstein (1991) Process
l (birth rate) m (death rate)
A # C G
# ##
#
T= 0
T = t#
s2
s1
s1 s2
rs1 s22. Time reversible:
1. P(s) = (1-lm)(lm)l A #A* .. * T
#T l =length(s)
# - - -
# # # #
*
l & m into Alignment BlocksA. Amino Acids Ignored:
e-mt[1-l](l)k-1
# - - - # # # # k
# - - - -- # # # # k
=[1-e(lm)t]/[mle(lm)t]
pk(t) p’k(t)
[1-l-m](l)k
p’0(t)= m(t)
* - - - -* # # # # k[1-l](l)k
p’’k(t)
B. Amino Acids Considered:
T - - -R Q S W Pt(T-->R)*Q*..*W*p4(t) 4
T - - - -- R Q S W R *Q*..*W*p’4(t) 4
# - - ... -# # # ... #
Differential Equations for p-functions
# - - - ... -- # # # ... #
* - - - ... -* # # # ... #
Initial Conditions: pk(0)= pk’’(0)= p’k (0)= 0 k>1 p1(0)= p0’’(0)= 1. p’0 (0)= 0
pk = t*[l*(k-1) pk-1 + m*k*pk+1 - (l+m)*k*pk]
p’k=t*[l*(k-1) p’k-1+m*(k+1)*p’k+1-(l+m)*k*p’k+m*pk+1]
p’’k=t*[l*k*p’’k-1+m*(k+1)*p’’k+1- [(k+1)l+km]*p’’k]
Basic Pairwise Recursion (O(length3))
Survives: Dies:
i-1j-2
i
j
i-1 i
j-1 j
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……………………
1… j (j) cases
……………………
j
i-1 i
j
ii-1
j-1
])[2(*'*)21( 111 jspssP ji −− →
0… j (j+1) cases
…………………………………………
……………………
i
j
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P(s1i → s2 j )
€
(s2[ j])
€
f (s1[i],s2[ j −1])
€
p2
€
P(s1i−1 → s2 j−2)
e-mt[1-l](l)k-1, where
=[1-e(lm)t]/[mle(lm)t]
Basic Pairwise Recursion (O(length3))
(i,j)
i
j
i-1
j-1
(i-1,j)
(i-1,j-1)
survive
death
(i-1,j-k)
…………..
…………..…………..
Initial condition:
p’’=s2[1:j]
Accelleration of Pairwise Algorithm(From Hein,Wiuf,Knudsen,Moeller & Wiebling 2000)
Corner Cutting ~100-1000
Better Numerical Search ~10-100Ex.: good start guess, 28 evaluations, 3 iterations
Simpler Recursion ~3-10
Faster Computers ~250
1991-->2000 ~106
-globin (141) and -globin (146)(From Hein,Wiuf,Knudsen,Moeller & Wiebling 2000)
430.108 : -log(-globin) 327.320 : -log(-globin --> -globin) 747.428 : -log(-globin, -globin) = -log(l(sumalign))
l*t: 0.0371805 +/- 0.0135899m*t: 0.0374396 +/- 0.0136846s*t: 0.91701 +/- 0.119556
E(Length) E(Insertions,Deletions) E(Substitutions) 143.499 5.37255 131.59
Maximum contributing alignment:
V-LSPADKTNVKAAWGKVGAHAGEYGAEALERMFLSFPTTKTYFPHF-DLS--H---GSAQVKGHGKKVADALTVHLTPEEKSAVTALWGKV--NVDEVGGEALGRLLVVYPWTQRFFESFGDLSTPDAVMGNPKVKAHGKKVLGAFS
NAVAHVDDMPNALSALSDLHAHKLRVDPVNFKLLSHCLLVTLAAHLPAEFTPAVHASLDKFLASVSTVLTSKYRDGLAHLDNLKGTFATLSELHCDKLHVDPENFRLLGNVLVCVLAHHFGKEFTPPVQAAYQKVVAGVANALAHKYH
Ratio l(maxalign)/l(sumalign) = 0.00565064
The invasion of the immortal link
VLSPADNAL.....DLHAHKR 141 AA long
???????????????????? k AA long
2 107 years
2 108 years
2 109 year s
*########### …. ### 141 AA long
*########### …. ###
*########### …. ###
109 years
Algorithm for alignment on star tree (O(length6))(Steel & Hein, 2001)
* (lm)*######
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P(S) =(1−λμ
)[P*(S)+λμ
P#(Tail∑ )P(S−Tail)]
a
s1 s2
s3
*ACGC *TT GT
*ACG GT
Binary Tree Problem
The problem would be simpler if:
s1
s2
s3
s4
a1 a2
ACCT
GTT
TGA
ACG
A Markov chain generating ancestral alignments can solve the problem!!
a1 a2* *# ## -- ## #- #
i. The ancestral sequences & their alignment was known.
ii. The alignment of ancestral alignment columns to leaf sequences was known
How to sum over all possible ancestral sequences and their alignments?:
- # # E # # - E ** l lm (1 l)e-m lm (1 l)(1 e-
m) (1 lm) (1 l)
## l lm (1 l)e-m lm (1 l)(1 e-
m) (1 lm) (1 l) _# l lm (1 l)e-m lm (1 l)(1 e-
m) (1 lm) (1 l)
#- l
€
1−λβe−μ
1−e−μ
€
λβe−μ
1−e−μ
€
(μ−λ)β1−e−μ
Generating Ancestral Alignments
a1 *a2 *
- #l
# # lm (1 l)e-
m
E E (1 lm) (1 l)
The Basic Recursion
S E
”Remove 1st step” - recursion:
”Remove last step” - recursion:
Last/First step removal are inequivalent, but have the same complexities. First step algorithm is the simplest.
Sequence Recursion: First Step Removal
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ε∑
i∈Sα∑ P'(kSi,H α)
H ∈Cα∑ P(α → ε)Pε(Si)
P(Sk): Epifixes (S[k+1:l]) starting in given MC starts in .
P(Sk) = E
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( p 'k
j : H ( j ) = 0
∏ ( tj
) πsj
[ i ( j ) : k ( j )])( pk
j : H ( j ) = 1
∏ ( tj
) πsj
[ i ( j ) + 1 : k ( j )])F(kSi,H)
Where P’(kS i,H) =
Human alpha hemoglobin;Human beta hemoglobin;Human myoglobinBean leghemoglobin
Probability of data e -1560.138
Probability of data and alignment e-1593.223
Probability of alignment given data 4.279 * 10-15 = e-33.085
Ratio of insertion-deletions to substitutions: 0.0334
Maximum likelihood phylogeny and alignmentGerton Lunter
Istvan Miklos
Alexei Drummond
Yun Song
Metropolis-Hastings Statistical AlignmentLunter, Drummond, Miklos, Jensen & Hein, 2005