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Schedule Day 1: Molecular Evolution 9.00-10.15 Lecture: Models of Sequence Evolution and Statistical Alignment 10.30-12.00 Practical: Molecular Evolution (Phylogenies – PHYLIP+) 2.00-3.30 Lecture: Molecular Evolution & Comparative Genomics 3.30-5.00 Student Activity: Prepare projects Day 2: Population Biology and Mapping 9.00-10.00 Lecture: Population Genetics and Gene Genealogies 10.00-11.00 Exercise: Jukes-Cantor and Rate Matrix 11.00-12.00 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, G F 10.30-12.00 Practical: Statistical Alignment & Footprinting 2.00-3.30 Lecture (L): Grammars and RNA Prediction 3.30-5.00 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 Grammars 11.00-12.00 Lecture: Concepts, Data Analysis and Functional Studies 2.00-3.30 Practical: Detecting Recombinations 3.30-5.00 Student Activity: Prepare projects Day 5: Project Discussion/Presentation 9.00-10.00 Project 1 – Population Genomics: 1000 genomes 10.00-11.00 Practical – Integrative Data Analysis – Mapping 11.00-12.00 Project 2 – Comparative Genomics: Signals 2.00-3.00 Project 3 – Integrative Genomics: Basic data types 3.00-4.00 Exercise: Statistical Alignment 4.00-5.00 Project 4 – Comparative Biology: Networks
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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
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The Human Genome (Harding & Sanger) *50.000 *20 globin (chromosome 11) 6*10 4 bp 3*10 9 bp *10 3 Exon 2 Exon 1 Exon 3 5’ flanking 3’ flanking 3*10 3 bp Myoglobin globin ATTGCCATGTCGATAATTGGACTATTTGG A 30 bp aa DNA: Protein :
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ACGTC Central Problems: History cannot be observed, only end products. Even if History could be observed, the underlying process couldn’t !! ACGCC AGGCC AGGCT AGGTT ACGTC ACGCC AGGCC AGGCT AGGGC AGGCT AGGTT AGTGC
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Some Definitions State 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 X t ie P{X 0 =A, X 1 =C,.., X 5 =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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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 I Data: s1=TCGGTA,s2=TGGTT 1) Only substitutions. s1 TCGGTA s1 TCGGA s2 TGGT-T s2 TGGTT TGGTT TCGGTA a - unknown Biological setup T T a1a1 a2a2 a3a3 a4a4 a5a5 G G T T C G G A Probability of Data
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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 P i,j (t) = π j P j,i (t), where π i is the stationary distribution of i and P t (i->j) the probability that state i has changed into state j after t time. This implies that = a s1 i s2 i l 2 +l 1 l1l1 l2l2 s2 i s1 i
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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, r i, vary from site to site, then the term for passed time, t, will be substituted by r i *t. 5) The nucleotide at any position evolves following a continuous time Markov Chain. T O A C G T F A -(q A,C +q A,G +q A,T ) q A,C q A,G q A,T R C q C,A -(q C,A +q C,G +q C,T ) q C, G q C,T O G q G,A q G,C -(q G,A +q G,C +q G,T ) q G,T M T q T,A q T,C q T,G -(q T,A +q T,C +q T,G ) P i,j (t) continuous time markov chain on the state space {A,C,G,T}. Q - rate matrix: t1t1 t2t2 C C A
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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 E ij =1 (all i,j) vii. PE=E viii. If AB=BA, then e A+B =e A e B. 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, T j, P(T j > t) = e q jj t Expected number of events at equilibrium
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Jukes-Cantor (JC69): Total Symmetry Rate-matrix, R: T O A C G T F A R C O G M T 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
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Principle of Inference: Likelihood Likelihood function L() – the probability of data as function of parameters: L( ,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. LogLikelihood Function – l(): ln(L( ,D)) Likelihood LogLikelihood
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From Q to P for Jukes-Cantor
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Exponentiation/Powering of Matrices then If where and Finding : det (Q- I)=0 By eigen values: Numerically: where k ~6-10 JC69: Finding : (Q- I)b i =0
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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
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Unequal base composition: (Felsenstein, 1981 F81) Q i,j = C*π j i unequal j Felsenstein81 & Hasegawa, Kishino & Yano 85 Tv/Tr & compostion bias (Hasegawa, Kishino & Yano, 1985 HKY85) ( )*C*π j i- >j a transition Q i,j = C*π j i- >j a transversion Rates to frequent nucleotides are high - (π =(π A, π C, π G, π T ) Tv/Tr = (π T π C +π A π G ) /[(π T +π C )(π A + π G )] A G T C Tv/Tr = ( ) (π T π C +π A π G ) /[(π T +π C )(π A + π G )]
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Measuring Selection ThrSer ACGTCA Certain events have functional consequences and will be selected out. The strength and localization of this selection is of great interest. Pro ThrPro 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
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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 (3 rd ) 1-1-1-1 (3 rd ) ii. T A (2 nd )
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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).
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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:
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Sites Total Conserved Transitions Transversions 1-1-1-1 274 246 (.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 Gly TCA ACT GAG ATG TGT TTA ATG GGG GGA * * * * * * * ** TCG ACA GGG ATA TAT CTA ATG GGT ATA Ser Thr Gly Ile Tyr Leu Met Gly Ile Z( t, t) =.50[1+exp(-2 t) - 2exp(-t( + )] transition Y( t, t) =.25[1-exp(-2 t )] transversion X( t, t) =.25[1+exp(-2 t) + 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 Transversions 1-1-1-1 a*f = 0.0500 2*b*f = 0.0622 2-2 a = 0.3004 2*b*f = 0.0622 4 a = 0.3004 2*b = 0.3741 Expected number of: replacement substitutions 35.49 synonymous 75.93 Replacement sites : 246 + (0.3742/0.6744)*77 = 314.72 Silent sites : 429 - 314.72 = 114.28 K s =.6644 K a =.1127
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Approaches to Sequence Analysis s2s2 s3s3 s4s4 s1s1 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
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Thorne-Kishino-Felsenstein (1991) Process (birth rate) (death rate) A # C G ### # T= 0 T = t # s2 s1 s2 r s1 s2 2. Time reversible: 1. P(s) = (1- )( ) l A #A *.. * T #T l =length(s) # - - - # # *
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& into Alignment Blocks A. Amino Acids Ignored: e - t [1- ]( ) k-1 # - - - # # k # - - - - - # # # # k =[1-e ( )t ]/[ e ( )t ] p k (t) p’ k (t) [1- - ]( ) k p’ 0 (t)= (t) * - - - - * # # # # k [1- ]( ) k p’’ k (t) B. Amino Acids Considered: T - - - R Q S W P t (T-->R)* Q *..* W *p 4 (t) 4 T - - - - - R Q S W R * Q *..* W *p’ 4 (t) 4
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# - -... - # # #... # Differential Equations for p-functions # - - -... - - # # #... # * - - -... - * # # #... # Initial Conditions: p k (0)= p k ’’(0)= p’ k (0)= 0 k>1 p 1 (0)= p 0 ’’(0)= 1. p’ 0 (0)= 0 p k = t*[ *(k-1) p k-1 + *k*p k+1 - ( )*k*p k ] p’ k = t*[ *(k-1) p’ k-1 + *(k+1)*p’ k+1 -( )*k*p’ k + *p k+1 ] p’’ k = t*[ *k*p’’ k-1 + *(k+1)*p’’ k+1 - [(k+1) +k ]*p’’ k ]
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Basic Pairwise Recursion (O(length 3 )) Survives: Dies: i-1 j-2 i j i-1 i j-1 j …………………… 1… j (j) cases …………………… j i-1i j i j-1 0… j (j+1) cases …………………… i j e - t [1- ]( ) k-1, where =[1-e ( )t ]/[ e ( )t ]
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Basic Pairwise Recursion (O(length 3 )) (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]
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Accelleration of Pairwise Algorithm (From Hein,Wiuf,Knudsen,Moeller & Wiebling 2000) Corner Cutting ~100-1000 Better Numerical Search ~10-100 Ex.: good start guess, 28 evaluations, 3 iterations Simpler Recursion ~3-10 Faster Computers ~250 1991-->2000 ~10 6
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-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)) *t: 0.0371805 +/- 0.0135899 *t: 0.0374396 +/- 0.0136846 s*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---GSAQVKGHGKKVADALT VHLTPEEKSAVTALWGKV--NVDEVGGEALGRLLVVYPWTQRFFESFGDLSTPDAVMGNPKVKAHGKKVLGAFS NAVAHVDDMPNALSALSDLHAHKLRVDPVNFKLLSHCLLVTLAAHLPAEFTPAVHASLDKFLASVSTVLTSKYR DGLAHLDNLKGTFATLSELHCDKLHVDPENFRLLGNVLVCVLAHHFGKEFTPPVQAAYQKVVAGVANALAHKYH Ratio l(maxalign)/l(sumalign) = 0.00565064
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The invasion of the immortal link VLSPADNAL.....DLHAHKR 141 AA long ???????????????????? k AA long 2 10 7 years 2 10 8 years 2 10 9 years * ########### …. ### 141 AA long * ########### …. ### 10 9 years
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Algorithm for alignment on star tree (O(length 6 )) (Steel & Hein, 2001) * ( ) *###### a s1 s2 s3 *ACGC*TT GT *ACG GT
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Binary Tree Problem The problem would be simpler if: s1 s2 s3 s4 a1a2 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?:
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- # # E # # - E * * e - e - # # e - e - _ # e - e - # - Generating Ancestral Alignments a1 * a2 * - # # e - E
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The Basic Recursion SE ”Remove 1 st step” - recursion: ”Remove last step” - recursion: Last/First step removal are inequivalent, but have the same complexities. First step algorithm is the simplest.
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Sequence Recursion: First Step Removal P (S k ): Epifixes (S[k+1:l]) starting in given MC starts in . P (S k ) = E F( k S i,H) Where P’( k S i,H =
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Human alpha hemoglobin; Human beta hemoglobin; Human myoglobin Bean 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 alignment Gerton Lunter Istvan Miklos Alexei Drummond Yun Song
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Metropolis-Hastings Statistical Alignment Lunter, Drummond, Miklos, Jensen & Hein, 2005
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