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CpG is a pair of nucleotides C and G, appearing successively, in this order, along one DNA strand. CpG islands are particular short subsequences in them the couple CpG is more frequent. These CpG islands are known to appear in biologically more significant parts of the genome.

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Two problems involving CpG islands: 1.Given a short genome sequence, decide if it comes from a CpG islands or not. 2.Given a long DNA sequence, locate all the CpG islands in it.

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We like to show a Markov chain graphically as a collection of ‘states’, each of which corresponds to a particular residue, with arrows between the states.

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A Markov chain is a triplet (Q,{p( x 1 = s )},A), where : Q is a finite set of states. Each state corresponds to a symbol in the alphabet . Initial state probabilities. A is the state transition probabilities, denoted by a st for each s,t Q For each s,t Q the transition probability is: a st P ( x i = t | x i-1 = s ) Formal definition :

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For any probabilistic model of sequences we can write the probability of the sequence as : P(x) = P (x L, x L-1,…..,x 1 ) By applying P(X,Y) = P(X|Y) P(Y) many times:

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Begin and end states ( and ) can be added to a Markov chain. P(x 1 = s) = a s P( | x L = t) = a t

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INPUT : A short DNA sequence X = (x 1, …., x L ) (where = { A,C,G,T }). QUESTION : Decide whether X is a CpG island. We can use two Markov chain models to solve this problem : one for dealing with CpG islands (the ‘+’ model) and the other for dealing with non CpG island (the ‘-’ model).

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c + st is the number of times letter t followed letter s in the CpG chain. The transition probabilities in each model are derived from a collection of human gene sequences, containing 48 putative CpG islands. a + st denote the transition probability of s,t inside a CpG island.

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Each row sums to one The table is asymmetric

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The higher this score, the more likely it is that X is a CpG island. Therefore we can compute a logaritmic likelihood score for a sequence X by :

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CpG islands sequences shown in dark grey and non-CpG sequences in light grey.

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INPUT : A long DNA sequence X = (x 1, …., x L ) (where = { A,C,G,T }). QUESTION : Locate the CpG islands along X.

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Extract a sliding window X k = (x k+1, …, x k+l ) (where l << L and 1 k L – l) from the sequence. Calculate Score(X k ) for each one of the resulting subsequences. Subsequences that receive positive scores are potential CpG islands. A naive approach main disadvantage : We have no information about the lengths of the islands.

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Combine the two markov chains of the previous problem into a unified model, with a small probability of switching from one chain to the other at each transition point. A better solution :

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In the new model : A +, C +, G + and T + emit A,C,G and T respectively in CpG island regions, and A-,C-,G- and T- corresponding in non-islands regions. State: A +, C +, G +, T +, A-, C-, G-, T- Emitted Symbol: A C G T A C G T

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A Hidden Markov Model (HMM) is a triplet M = ( , Q, ), where: is an alphabet of symbols. Q is a finite set of states, capable of emitting symbols from the alphabet . is a set of probabilities, comprised of: - State transition probabilities - Emission probabilities

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State transition probabilities : The state sequence = ( 1,…, L ) is called a path. the probability of a state depends only on the previous state: a kl = P ( i = l | i-1 = k ) Emission probabilities : The symbol sequence is X = (x 1,….,x L ) Emission probability is the probability that symbol b is seen when in state k : e k (b) = P (x i = b | i = k )

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Where for our convenience we denote : 0 = begin and L+1 = end. The probability that the sequence X was generated by the model M given the path is therefore:

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Example : modeling a dishonest casino dealer

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INPUT : A hidden Markov model M = ( , Q, ) and a sequence X , for which the generating path = ( 1,…, L ) is unknown. QUESTION : Find the most probable generating path for X In general there may be many state sequences that could give rise to any particular sequence of symbols.

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In our example : (C+,G+,C+,G+), (C-,G-,C-,G-) and (C+,G-,C+,G-) would all generate the symbol sequence CGCG. However, they do so with very different probabilities. We are looking for a path such that P(X, ) is maximized. = argmax {P(X, ) }

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This algorithm is based on dynamic programming. The most probable path * can be found recursively. Suppose the probability v k (i) of the most probable path ending in state k with observation i is known for all states k. then these probabilities can be calculated for observation x i+1 as :

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The algorithm : Time complexity: O(L|Q| 2 ) Space complexity: O(L|Q|)

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The values of v for the sequence CGCG. Example 1 : CpG islands

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Example 2 : The casino

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INPUT : A hidden Markov model M = ( , Q, ) and a sequence X , for which the generating path = ( 1,…, L ) is unknown. QUESTION : What the probability that observation x i came from state k given the observed sequence? i.e. P( i = k|X). This is the posterior probability of state k at time i when the emitted sequence X is known.

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To answer this question we need two algorithms, both assume that X is known and i = k : Forward algorithm - f k (i) = P(x 1 … x i, i = k) The probability of emitting the prefix ( x 1, …,x i ). Backward algorithm - b k (i) = P(x i+1 … x L | i = k) The probability of the suffix ( x i+1, …,x L ).

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In Markov chains, the probability of a sequence was calculated by the equation : What is the probability P(x) for an HMM ? In order to find P( i = k|X). We need to know p(x).

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many different state paths can generate the same sequence x The probability of x is sum of the probabilities for all possible paths. In HMM… The number of possible paths increases exponentially with the length of the sequence Enumerating all paths is not practical

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one approach : We can use probability of Viterbi path * as an approximation to P(x). In fact, the full probability can itself be calculated by a dynamic programming (like Viterbi), replacing the maximization steps with sums.

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The algorithm :

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Calculate the probability of the suffix (x i+1,…,x L ). b k (i) = P(x i+1 …x L | i = k) The recursion start at the end of the sequence.

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Back to the posterior problem… Now we can calculate P( i = k|X). Where P(x) is the result of the forward (or backward) algorithm.

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In Viterbi, Forward and Backward algorithms : Complexity - Time complexity: O(L|Q| 2 ) - Space complexity: O(L|Q|) working in log space to avoid underflow errors when implemented on computer. Some comments …

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The casino example The posterior probability of being in a fair die throughout the sequence Loaded die used in blue areas Number of the roll Prob. Of fair die

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Uses for Posterior Decoding - Two alternative forms of decoding : 1. When many different paths have almost the same probability as the most probable one, then we may want to consider other possible paths as well. ** = argmax k {P( i = k|X)} * ** may not be a legitimate path, if some transitions are not permitted.

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2.When we ’ re not interested in the state sequence itself, but in some other property derived from it. For example : g(k) = 1 if k {A+,C+,G+,T+}, 0 otherwise. G(i | x) is posterior probability that base i is in a CpG island.

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The most difficult problem faced when using HMMs is specifying the model : Design the structure : which states and the connection between them. The assignment of the transition and the emission probabilities a kl and e k (b).

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Given training sequences X 1, …,X n * of length L 1, …,L n respectively, which were all generated from the HMM M = ( , Q, ). however, the values of the probabilities in , are unknown. We want to construct an HMM that will best characterize X 1, …,X n. We need to assign values to that will maximize the probability of X 1, …,X n.

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since the sequences where generated independently, Probability of X 1, …,X n with is: Using the logarithmic score our goal is to find * such that:

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The sequences X 1, …,X n are usually called the training sequences We shell examine two cases for parameter estimation: Estimation when the state sequence is known. Estimation when the state sequence is unknown.

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Estimation when the state sequence is known When all the paths are known, we can count the number of times each particular transition or emission is used in the set of training sequence. A kl - number of transitions from the state k to l in all the state sequences E k (b) – the number of times that an emission of the symbol b occurred in state k in all the state sequences.

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Laplace`s correction In order to avoid zero probability we will use : Where usually r kl and r k (b) Are usually equal 1.

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Estimation when the state sequence is unknown In the case that the state sequences are not known, the problem of finding the optimal set of parameters * is known to be NP-complete. The Baum-Welch algorthim is a heuristic algorithm for finding a solution to the problem Baum-Welch algorithm, which is a special case of the EM technique (Expectation and maximization).

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Initialization: Assign arbitrary values to . Expectation: calculates A kl and E k (b) as the expected number of times each transition is used, given the training sequences. Probability that a kl is used at position i in sequence x is:

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Sum over all positions and training sequences to get A kl Similary,we can find the expected number of times that letter b appears in state k.

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Maximization: update the values of a kl and e k (b) according to the equations: This process is iterated until the improvement of Score(X 1,…,X n | ) is less then a given parameter .

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Baum-Welch converges to local maximum of the target function Score(X 1,…,X n | ) Main problem : may exist several local maximum. Solution : 1.run the algorithm several times, each time with different initial values for . 2.Start with values that are meaningful.

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References Hidden markov models / ron shamir http://www.math.tau.ac.il/~rshamir/ algmb/98/scribe/html/lec06/node1.html http://www.math.tau.ac.il/~rshamir/ algmb/98/scribe/html/lec06/node1.html Biological Sequence Analysis, Durbin et el,chapter 3

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