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Multiple Alignment by profile HMM training and Phylogenetic Trees Elze de Groot & Anastacia Berdnikova Elze de Groot & Anastasia Berdnikova 1 Topics Multiple alignment with known HMM HMM training from unaligned sequences Avoiding local maxima – Simulated annealing – Noise injection – Stochastic sampling traceback algorithm Model surgery Phylogenetic trees Elze de Groot & Anastasia Berdnikova 2 Multiple alignment with known profile HMM Multiple alignment and model known -> align large number of other family members Calculating Viterbi alignment for every sequence Residues in same match state are aligned in columns That´s a difference between profile HMM and traditional multiple alignment Elze de Groot & Anastasia Berdnikova 3 Example Model estimated from an alignment Elze de Groot & Anastasia Berdnikova 4 Example continued The most probable paths and alignment Elze de Groot & Anastasia Berdnikova 5 Profile HMM training from unaligned sequences Algorithm: Elze de Groot & Anastasia Berdnikova 6 Initial Model Choose length of model - M is number of match states - set M to be the average length Choose initial models carefully Randomness in choice of initial model Elze de Groot & Anastasia Berdnikova 7 Parameter Estimation Use forward and backward variables to reestimate emission and transition probability parameters Baum-Welch re-estimation can be replaced by viterbi alternative Elze de Groot & Anastasia Berdnikova 8 Forward Algorithm Elze de Groot & Anastasia Berdnikova 9 Backward algorithm Elze de Groot & Anastasia Berdnikova 10 Baum-Welch re-estimation equations Expected emission counts from sequence x 1 EM k ( a ) f M k (i )bM k (i ) P( x) i| xi a 1 EI k (a) f I k (i )bI k (i ) P( x) i| xi a Elze de Groot & Anastasia Berdnikova 11 Baum-Welch re-estimation equations Expected transition counts from sequence x 1 AX k M k 1 f X k (i )a X k M k 1 eM k 1 ( xi 1)bM k 1 (i 1) P( x) i 1 AX k I k f X k (i )a X k I i eI k ( xi 1)bI k (i 1) P( x) i 1 AX k Dk 1 f X k (i )a X k Dk 1 bDk 1 (i 1) P( x) i Elze de Groot & Anastasia Berdnikova 12 Avoiding local maxima Baum-Welch guaranteed to find local maxima Not guaranteed it is anywhere near global optimum or biologically reasonable solution Reason: models are long -> many options to get wrong solution Elze de Groot & Anastasia Berdnikova 13 Avoiding local maxima Use stochastic search algorithm Commonly used: Simulated annealing Elze de Groot & Anastasia Berdnikova 14 Simulated annealing Some compounds only cristallise if they are slowly annealed from high to low temperature Optimisation problem: minimise function ´energy´ E(x) Maximising function same as minimising negative value of function Elze de Groot & Anastasia Berdnikova 15 Simulated annealing (2) ´temperature´ T Probability of ´state´ x is given by Gibbs distribution 1 1 P(x) exp E(x) Z T 1 Z exp E ( x) dx T x usually multidimensional so impossible to calculate Z Partition function: Elze de Groot & Anastasia Berdnikova 16 Simulated annealing (3) T0, all configurations except with lowest energy are prob 0 (system is ´frozen´) T, All configuration have same prob (system is ´molten´) With crystallisation: minimum can be found by sampling this distribution at high temperature first and then decreasing temperatures Elze de Groot & Anastasia Berdnikova 17 Simulated annealing for HMM Natural energy function negative log of likelihood –logP(data|) 1/ T 1 P ( data | ) 1 1 exp log P(data | ) P(data | )1/ T 1/ T Z P ( data | ´) d ´ T Z Non-trivial, the two methods I´m going to mention are approximations Elze de Groot & Anastasia Berdnikova 18 Noise injection Adding noise to counts estimated in forward-backward procedure and let size of noise decrease slowly In Krogh et al.[1994] the noise was generated by a random walk in the initial model Elze de Groot & Anastasia Berdnikova 19 Simulated annealing Viterbi estimation If there are N sequences, there´s an exact translation from the N paths 1,…, N to the parameters of the model Treat the paths as fundamental parameters in which to maximise the likelihood Simulated annealing done in these variables instead of the model parameters Elze de Groot & Anastasia Berdnikova 20 Simulated annealing Viterbi estimation P( , x | )1/ T Prob( ) 1/ T P ( ´, x | ) ´ Denominator is Z, the partition function -> sum over all paths Can be obtained by modified forward algorithm using exponentiated transmission and emission parameters Elze de Groot & Anastasia Berdnikova 21 Simulated annealing Viterbi estimation Exponentiated transmission parameter – âij = aij1/T Exponentiated emission parameter – êj(x) = ej(x)1/T Used in place of unmodified probability parameters in forward algorithm Z is result of forward algorithm Elze de Groot & Anastasia Berdnikova 22 Simulated annealing Viterbi estimation Algorithm: Stochastic sampling traceback algorithm for HMMs Initialisation: πL+1 = End. Recursion: for L+1 ≥ i ≥ 1, Prob i 1 | i f i 1, i1 â i1 , i / k f i 1,k âi , i Elze de Groot & Anastasia Berdnikova 23 Simulated annealing Viterbi vs Viterbi Key difference: Viterbi selects highest probable path for each sequence Simulated annealing samples each path according to the likelihood of the path Elze de Groot & Anastasia Berdnikova 24 Model Surgery During training a model two things can happen: (a) some match states are redundant and should be absorbed in insert state (b) one or more insert states aborb too much sequence, in which case they should be expanded Elze de Groot & Anastasia Berdnikova 25 Model Surgery How much is a certain transition used by training sequences Usage of match state is sum of counts for all letters in state Elze de Groot & Anastasia Berdnikova 26 Model surgery If match state is used by less than ½ sequences -> delete module If more than ½ of sequences use the transitions into an insert state, this is expanded to new modules Elze de Groot & Anastasia Berdnikova 27 Model surgery – Example SAM I tried a sequence in SAM with and without model surgery Same 7 sequences as in example before Parameters <cutinsert 0.25> <cutmatch 0.5> -> delete any match state used by fewer than half the sequences, and insert match states for any insert node used by greater than one quarter of the sequences Elze de Groot & Anastasia Berdnikova 28 Model surgery – Example SAM Without model surgery >seq1 FPHFD.....L...S.....-HGSAQ >seq2 FESFG.....D...LstpdaVMGNPK >seq3 FDRFKhlkteA...E.....MKASED >seq4 FTQFA.....G...Kdles.IKGTAP >seq5 FPKFK.....G...LttadqLKKSAD >seq6 FSFLK.....GtseV.....PQNNPE >seq7 FGFSG.....A...-.....--SDPG With model surgery >seq1 FPHF.DLS-..-..--HGSAQ >seq2 FESF.GDLStpD..AVMGNPK >seq3 FDRF.KHLK..TeaEMKASED >seq4 FTQFaGKDL..E..SIKGTAP >seq5 FPKF.KGLTtaD..QLKKSAD >seq6 FSFL.KGTS..E..VPQNNPE >seq7 FGFS.G---..-..--ASDPG Elze de Groot & Anastasia Berdnikova 29 Building phylogenetic trees Elze de Groot & Anastasia Berdnikova 30 Overview The tree of life – description Background on trees Elze de Groot & Anastasia Berdnikova 31 Multiple alignment and trees Alignment of sequences should take account of their evolutionary relationship. [Sankoff, Morel & Cedergren, 1973] Several progressive alignment algorithms use a ‘guide tree’ (to guide the clustering process). We begin to build trees. Elze de Groot & Anastasia Berdnikova 32 The tree of life The similarity of molecular mechanisms of the organisms that have been studied strongly suggests that all organisms on Earth had a common ancestor. Thus any sets of species is related, and this relationship is called a phylogeny. Usually the relationship can be represented by a phylogenetic tree. Elze de Groot & Anastasia Berdnikova 33 Zuckerkandl & Pauling’s paper [1962] showed that molecular sequences provide sets of morphological characters that can carry a large amount of information. An assumption: the sequencies we want to analyze on the phylogeny matter have descended from some common ancestral gene in a common ancestral species. Gene duplication exists => we have to check the assumption carefully. Elze de Groot & Anastasia Berdnikova 34 Gene duplication and speciation By another mechanism, gene duplication, two sequences can also be separated and diverge from the common ancestor. Genes which diverged because of speciation are called orthologues. Genes which diverged by gene duplication are called paralogues. Elze de Groot & Anastasia Berdnikova 35 A tree of orthologues: alpha haemoglobins HBA_ACCGE, HBA_AEGMO, HBA_AILFU, HBA_AILME, HBA_ALCAA, HBA_ALLMI, HBA_AMBME, HBA_ANAPL (SWISS-PROT). Elze de Groot & Anastasia Berdnikova 36 A tree of paralogues: HBAT_HUMAN, HBAZ_HUMAN, HBA_HUMAN, HBB_HUMAN, HBD_HUMAN, HBE_HUMAN, HBG_HUMAN, MYG_HUMAN (SWISS-PROT). Elze de Groot & Anastasia Berdnikova 37 Background on trees All trees will be assumed to be binary (an edge that branches splits into two daughter edges). Each edge of the tree has a certain amount of evolutionary divergence associated to it. We adopt the general term ‘length’, which will be represented by lengthes of edges on figures. A true biological phylogeny has a ‘root’, or ultimate ancestor of all sequences. Elze de Groot & Anastasia Berdnikova 38 Rooted and unrooted tree Elze de Groot & Anastasia Berdnikova 39 A tree with a given labelling will be called a labelled branching pattern. We refer to this as the tree topology and denote it by T. Lengths of the edges: ti with a suitable numbering scheme for the is. Elze de Groot & Anastasia Berdnikova 40 Counting and labelling Rooted tree: – n leaves, plus (n-1) branch nodes in addition to leaves -> we have 2n-1 nodes in all, and 2n-2 edges. – leaves – 1..n, branch nodes – n+1 .. 2n-1, (2n-1)th node is root. Elze de Groot & Anastasia Berdnikova 41 Counting and labelling Unrooted tree: – n leaves, 2n-2 nodes and 2n-3 edges. – a root can be added at any of its edges => we can get 2n-3 rooted trees. Elze de Groot & Anastasia Berdnikova 42 Number of rooted and unrooted trees A root can be added at any edge, producing 2n-3 rooted trees from unrooted tree => there are (2n-3) times as many rooted trees as unrooted trees, for a given number n of leaves. Elze de Groot & Anastasia Berdnikova 43 Instead of the root, we can add an extra edge or ‘branch’ with a distinct label in its leaf. Elze de Groot & Anastasia Berdnikova 44 ● There are three such trees with (2n-3)=5 leaves – they are distinct labelled branching patterns. ● There are then five ways of adding a further branch labelled with a distinct label (‘5’), giving in all 3x5=15 unrooted trees with five leaves. ● The number of unrooted trees with n leaves is equal to 3*5*...*(2n-5) = (2n-5)!! So, we have (2n-3)!! rooted trees with n leaves. Elze de Groot & Anastasia Berdnikova 45 Building phylogenetic trees Questions? Elze de Groot & Anastasia Berdnikova 46 Exercise 7.2 The trees with three and four leaves in Figure 7.3 all have the same unlabelled branching pattern. For both rooted and unrooted trees, how many leaves do there have to be to obtain more than one unlabelled branching pattern? Find a recurrence relation for the number of rooted trees. (Hint: consider the trees formed by joining two trees at their root). Elze de Groot & Anastasia Berdnikova 47 Exercise 7.2 Elze de Groot & Anastasia Berdnikova 48 Exercise 7.3 All trees considered so far have been binary, but one can envisage ternary trees that, in their rooted form, have three branches descending from a branch node. If there are m branch nodes in an unrooted ternary tree, how many leaves are there and how many edges? Elze de Groot & Anastasia Berdnikova 49 Exercise 7.4 Consider next a composite unrooted tree with m ternary branch nodes and n binary branch nodes. How many leaves are there, and how many edges? Let Nm,n denote the number of distinct labelled branching patterns of this tree. Extend the counting argument for binary trees to show that Nm,n = (3m+2n-1)N m,n-1 + (n+1)N m-1,n+1 (Hint: the first term after the ‘=’ counts the number of ways that a new edge can be added to an existing edge, thereby creating an additional binary node; the second term corresponds to edges added at binary nodes, thereby producing ternary nodes.) Elze de Groot & Anastasia Berdnikova 50