1. Professor of Computer Science and Mathematics
Computational Biology Lecture #8: Hidden Markov Models, Motifs & Patterns
Bud Mishra
Professor of Computer Science and Mathematics
11 ¦ 12 ¦ 2001
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©Bud Mishra, 2001

2. Hidden Markov Models (HMM)
Defined by an alphabet S,
A set of (hidden) states Q,
A matrix of state transition probabilities A,
and a matrix of emission probabilities E.
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©Bud Mishra, 2001

3. States ©Bud Mishra, 2001 S = An alphabet of symbols
Q = A set of states that emit symbols from the alphabet S
A = (akl) = |Q| £ |Q| matrix of state transition probabilities
E = (eK(B)) = |Q| £ |S| matrix of emission probabilities
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©Bud Mishra, 2001

4. A Simple Exaple ©Bud Mishra, 2001 Fair Bet Casino:
S = {H, T} ! A two sided coin
Q = {F, B} ! The casino may use a fair coin
Pr(H) = Pr(T) = ½
or a biased coin
Pr(H) = ¾, Pr(T) = ¼.
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©Bud Mishra, 2001

5. Fair Bet Casino Problem
0.9
0.9
0.1 B F
Pr[H]= ½
Pr[T]= ½
Pr[H]= ¾
Pr[T]= ¼
0.1
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©Bud Mishra, 2001

6. Transition and Emission
aFF = aBB = 0.9
aFB = aBF = 0.1
The casino switches the coin with probability = 0.9
eF(H) = ½ eF(T) = ½
eB(H) = ¾ eB(T) = ¼
Given a sequence of coin tosses find out when the dealer used a fair coin ad when the dealer used a biased coin.
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©Bud Mishra, 2001

7. HMM in Genomics ©Bud Mishra, 2001 CG-Islands:
Subsequences around genes where CG appear relatively frequently.
Methylation is suppressed in CG islands.
C within the CG nucleotide is typically methylated.
Methyl C has a tendency to mutate into a T.
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©Bud Mishra, 2001

8. A Path in the HMM ©Bud Mishra, 2001 p = p1 pi2 L pn
= a sequence of states 2 Q* in the hidden Markov model M
x 2 S* = sequence generated by the path p, determined by the model M
P(x| p) = P(p1)[ Õi=1n P(xi | pi) P(pi | pi+1) ]
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©Bud Mishra, 2001

9. A Path in the HMM ©Bud Mishra, 2001
P(x| p) = [Õi=1n P(xi | pi) P(pi | pi+1) ] P(p1)
P(xi | pi) = epi(xi)
P(pi | pi+1) = api, pi+1
p0 = Initial state “begin”
pn+1 = Final state “end”
P(x| p)
= ap0, p1 ep1(x1) ap1, p2 ep2(x2)L epn(xn) apn, pn+1
= ap0, p1 Õi=1n epi(xi) api, pi+1
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©Bud Mishra, 2001

10. Decoding Problem ©Bud Mishra, 2001 p* = arg maxp P(x|p)
For a given sequence x, and a given path p,
The model (Markovian) defines the probability P(x | p)
The dealer knows p and x
The player knows x but not p
“The path of x is hidden.”
Decoding Problem:
Find an optimal path p* for x such that P(x|p) is maximized.
p* = arg maxp P(x|p)
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©Bud Mishra, 2001

11. Dynamic Programming Approach
Principle of Optimality:
Optimal path for the (i+1)-prefix of x
x1 L xi+1
uses a path for an i-prefix of x that is optimal among the paths ending in an (unknown) state pi = k 2 Q
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©Bud Mishra, 2001

12. Dynamic Programming Approach
sk(i) = The probability of the most probable path for the i-prefix ending in state k.
8k 2 Q 81 5 i 5 n
sl(i+1) = el(xi+1). maxk2 Q [sk(i) . akl]
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©Bud Mishra, 2001

13. Dynamic Programming ©Bud Mishra, 2001 i=0
sbegin(0) = 1, sk(0) =0, 8k ¹ begin
0 < i 5 n
sl(i+1) = el(xi+1) ¢ maxk2 Q [ sk(i) ¢ akl ]
i= n+1
P(x | p*) = maxk2 Q sk(n) ak, end
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©Bud Mishra, 2001

14. + maxk2 Q [ Sk(i) + log akl ]
Viterbi Algorithm
Dynamic Programming with log-score function
Sl(i) = log sl(i)
Space complexity = O(n |Q|)
Time complexity = O(n |Q|)
Sl(i+1) = log el(xi+1)
+ maxk2 Q [ Sk(i) + log akl ]
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©Bud Mishra, 2001

15. Estimating the ith State
P(pi = k | x) = Given a sequence x 2 S*, the probability that the HMM was in state k at instant i.
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©Bud Mishra, 2001

16. Forward Estimate: ©Bud Mishra, 2001
fk(i) = P(x1 L xi, pi = k) = Probability of emitting the prefix x1 L xi and reaching the state pi = k
fk(i) = ek(xi) ¢ ål 2 Q fl(i-1) alk
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©Bud Mishra, 2001

17. Backward Estimate: ©Bud Mishra, 2001
bk(i) = P(xi+1 L xn, pi = k) = Probability of being at the state pi = k and emitting the suffix xi+1 L xn .
bk(i) = ål 2 Q ek(xi+1) ¢ bl(i+1) akl
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©Bud Mishra, 2001

18. Applying Bayes’ Rule ©Bud Mishra, 2001 P(pi = k | x) = (1/P(x)
£ P(x1 … xi, pi = k)
£ P(xi+1…xn, pi = k)
= [fk(i) bk(i)] / P(x)
P(x) = åp P(x | p)
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©Bud Mishra, 2001

19. Sequence Motifs ©Bud Mishra, 2001
A Sequence of patterns of biological significance.
Examples:
DNA: Protein binding sites
(e.g. promoters, regulatory sequences)
Protein: sequences corresponding to conserved pieces of structure
(e.g. Local features, At various scales: blocks, domains & families)
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©Bud Mishra, 2001

20. MEME Algorithm ©Bud Mishra, 2001 Description of a motif:
Uses EM (Expectation Minimization) algorithm to find multiple motifs in a set of sequences.
Description of a motif:
W = (Fixed) width of a motif
P = (plc)l2 S, c 2 1..W
= Matrix of probabilities that letter l occurs at position c = |S| £ W matrix
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21. Example ©Bud Mishra, 2001 DNA motif of width
S = { A, T, C, G}
r = motif ) Wr = 3, Pr = 4 £ 3 stochastic matrix
Pr =
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©Bud Mishra, 2001

22. Computational Problem
Given:
A set of sequences, G
A width parameter W
Find:
Motifs of width W common to sequences G and present their probabilistic descriptions.
Note that motif start sites in each sequence are unknown (hidden).
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©Bud Mishra, 2001

23. Basic EM Approach ©Bud Mishra, 2001 G = Training sequences.
|G| = Total number of sequences = m; Minimum length of a sequence in G = l.
Z = m £ l matrix of probabilities
zij = Probability that the motif starts in position j in sequence i.
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©Bud Mishra, 2001

24. EM Algorithm ©Bud Mishra, 2001 Set initial values for P do
Re-estimate Z from P
Re-estimate P from Z
until change in P < e
return P
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©Bud Mishra, 2001

25. EM Algorithm ©Bud Mishra, 2001
Maximize the likelihood of a motif in the training sequence:
si 2 G ith sequence
Iij= {1, if motif starts at posn. j in seq. i
{0, otherwise.
lk = the char. at posn. j+k-1 in seq. Si
Pr (Si | Iij = 1, r) = Õk=1W rl{k, k}.
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©Bud Mishra, 2001

26. Example ©Bud Mishra, 2001 Si = AGGCTGTAGACAC Pr(Si = TGT | Ii5 =1, r)
= rT,1 £ rG,2 £ rT,3
= 0.2 £ 0.1 £ 0.1
= 2 £ 10-3
Pr =
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©Bud Mishra, 2001

27. Estimating Z ©Bud Mishra, 2001 zij = Pr(Iij = 1 | r, Si)
= Estimates the starting position in Si 2 G.
zij = Pr ( Iij =1 | r, Si)
= Pr( Si, Iij = 1 | r)/ Pr(Si | r)
= Pr( Si | Iij = 1, r) Pr(Iij = 1)/ åk Pr( Si | Iik = 1, r) Pr(Iik = 1)
= Pr( Si | Iij = 1, r) / åk Pr( Si | Iik = 1, r)
Follows from an application of the Bayes’ rule and the assumption that “it is equally likely that the motif will start in any position.”
8jk Pr(Iij = 1) = Pr(Iik=1)
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©Bud Mishra, 2001

28. Example ©Bud Mishra, 2001 Si = AGGCTGTAGACAC z’i1 = 6 £ 10-3
Pr =
0.1 £ 0.1 £ 0.6
z’i1 = 6 £ 10-3
z’i2 = 3 £ 10-3
z’i3 = 6 £ 10-3 …
NORMALIZE
0.3 £ 0.1 £ 0.1
0.3 £ 0.2 £ 0.1
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©Bud Mishra, 2001

29. Estimating Pr nck = ås(i) 2 G;{j | s(i, j+k-1) = c} zij
Given Z, estimate the probability that the character c occurs at the kth position of a motif.
nck = ås(i) 2 G;{j | s(i, j+k-1) = c} zij
Expected number of occurrences of the character c at the k^{th} position of a motif \rho (assuming that the motif “start position” is known.)
pck = (nck + 1)/ åd (ndk + 1)
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30. Example ©Bud Mishra, 2001 s1 : A C A G C A s2 : A G G C A G
s3 : T C A G T C
z1,1
z1,3
z2,1
z3,3
pA,1 = (z11 +z13+ z21 + z33 +1)/
(z11 + z12 + L+ z33 + z34 +4)
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31. Meme ©Bud Mishra, 2001 Uses the basic EM approach
Try many starting points.
Allow multiple occurrences of a motif per sequence
Allows multiple motifs to be learned simultaneously.
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©Bud Mishra, 2001

32. Meme ©Bud Mishra, 2001 Initial set of possible motifs:
Take every distinct subsequences of length W in the training set
Derive an initial matrix P
pck = { a if c occurs in position k in the subsequence
{(1-a)/ (|S| - 1)
Otherwise
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©Bud Mishra, 2001

33. Example ©Bud Mishra, 2001 W = 3, r = T A T, a = 0.5
Choose the motif model with the highest likelihood.
Run EM to convergence
Pr =
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©Bud Mishra, 2001