1. Variants of HMMs

2. Higher-order HMMs How do we model “memory” larger than one time point?
P(i+1 = l | i = k) akl
P(i+1 = l | i = k, i -1 = j) ajkl …
A second order HMM with K states is equivalent to a first order HMM with K2 states
aHHT
state HH
state HT
aHT(prev = H)
aHT(prev = T)
aHTH
state H
state T
aTHH
aHTT
aTHT
state TH
state TT
aTH(prev = H)
aTH(prev = T)
aTTH

3. Modeling the Duration of States
1-p
Length distribution of region X:
E[lX] = 1/(1-p)
Geometric distribution, with mean 1/(1-p)
This is a significant disadvantage of HMMs
Several solutions exist for modeling different length distributions X Y p q
1-q

4. Sol’n 1: Chain several statesp
1-p X X X Y q
1-q
Disadvantage: Still very inflexible
lX = C + geometric with mean 1/(1-p)

5. Sol’n 2: Negative binomial distributionp p p
1 – p
1 – p
1 – p X X X Y ……
Duration in X: m turns, where
During first m – 1 turns, exactly n – 1 arrows to next state are followed
During mth turn, an arrow to next state is followed
m – m – 1
P(lX = m) = n – 1 (1 – p)n-1+1p(m-1)-(n-1) = n – 1 (1 – p)npm-n

6. Example: genes in prokaryotes
EasyGene:
Prokaryotic
gene-finder
Larsen TS, Krogh A
Negative binomial with n = 3

7. Solution 3: Duration modeling
Upon entering a state:
Choose duration d, according to probability distribution
Generate d letters according to emission probs
Take a transition to next state according to transition probs
Disadvantage: Increase in complexity:
Time: O(D2) -- Why?
Space: O(D)
where D = maximum duration of state X

8. Connection Between Alignment and HMMs

9. A state model for alignment
(+1,+1)
Alignments correspond 1-to-1 with sequences of states M, I, J I
(+1, 0) J
(0, +1)
-AGGCTATCACCTGACCTCCAGGCCGA--TGCCC---
TAG-CTATCAC--GACCGC-GGTCGATTTGCCCGACC
IMMJMMMMMMMJJMMMMMMJMMMMMMMIIMMMMMIII

10. Let’s score the transitions
s(xi, yj) M
(+1,+1)
Alignments correspond 1-to-1 with sequences of states M, I, J
s(xi, yj)
s(xi, yj) -d -d I
(+1, 0) J
(0, +1) -e -e -e -e
-AGGCTATCACCTGACCTCCAGGCCGA--TGCCC---
TAG-CTATCAC--GACCGC-GGTCGATTTGCCCGACC
IMMJMMMMMMMJJMMMMMMJMMMMMMMIIMMMMMIII

11. How do we find optimal alignment according to this model?
Dynamic Programming:
M(i, j): Optimal alignment of x1…xi to y1…yj ending in M
I(i, j): Optimal alignment of x1…xi to y1…yj ending in I
J(i, j): Optimal alignment of x1…xi to y1…yj ending in J
The score is additive, therefore we can apply DP recurrence formulas

12. Needleman Wunsch with affine gaps – state version
Initialization:
M(0,0) = 0; M(i,0) = M(0,j) = -, for i, j > 0
I(i,0) = d + ie; J(0,j) = d + je
Iteration:
M(i – 1, j – 1)
M(i, j) = s(xi, yj) + max I(i – 1, j – 1)
J(i – 1, j – 1)
e + I(i – 1, j)
I(i, j) = max e + J(i, j – 1)
d + M(i – 1, j – 1)
J(i, j) = max e + J(i, j – 1)
Termination:
Optimal alignment given by max { M(m, n), I(m, n), J(m, n) }

13. Probabilistic interpretation of an alignment
An alignment is a hypothesis that the two sequences are related by evolution
Goal:
Produce the most likely alignment
Assert the likelihood that the sequences are indeed related

14. A Pair HMM for alignments
BEGIN M
P(xi, yj) I
P(xi) J
P(yj)
1 – 2 – 
1 – 2 –    
1 – 2 –    M I J   
END

15. A Pair HMM for unaligned sequences
Model R
1 - 
1 -  I
P(xi) J
P(yj)
BEGIN
END
BEGIN
END
1 - 
1 -   
P(x, y | R) = (1 – )m P(x1)…P(xm) (1 – )n P(y1)…P(yn)
= 2(1 – )m+n i P(xi) j P(yj)

16. To compare ALIGNMENT vs. RANDOM hypothesis
1 – 2 – 
Every pair of letters contributes:
(1 – 2 – ) P(xi, yj) when matched
 P(xi) P(yj) when gapped
(1 – )2 P(xi) P(yj) in random model
Focus on comparison of
P(xi, yj) vs. P(xi) P(yj) M
P(xi, yj)
1 – 2 – 
1 – 2 –     I
P(xi) J
P(yj)   
1 - 
1 - 
BEGIN I
P(xi)
END
BEGIN J
P(yj)
END
1 - 
1 -   

17. To compare ALIGNMENT vs. RANDOM hypothesis
Idea:
We will divide alignment score by the random score,
and take logarithms
Let
P(xi, yj) (1 – 2 – )
s(xi, yj) = log ––––––––––– log –––––––––––
P(xi) P(yj) (1 – )2
(1 – 2 – ) P(xi)
d = – log ––––––––––––––––––––
(1 – ) (1 – 2 – ) P(xi)
 P(xi)
e = – log –––––––––––
(1 – ) P(xi)
Every letter b in random model contributes
(1 – ) P(b)

18. The meaning of alignment scores
Because , , are small, and ,  are very small,
P(xi, yj) (1 – 2 – ) P(xi, yj)
s(xi, yj) = log ––––––––– + log ––––––––––  log –––––––– + log(1 – 2)
P(xi) P(yj) (1 – ) P(xi) P(yj)
(1 –  – ) – 
d = – log ––––––––––––––––––  – log  ––––––
(1 – ) (1 – 2 – ) – 2 
e = – log –––––––  – log 
(1 – )

19. The meaning of alignment scores
The Viterbi algorithm for Pair HMMs corresponds exactly to the Needleman-Wunsch algorithm with affine gaps
However, now we need to score alignment with parameters that add up to probability distributions
 1/mean length of next gap
 1/mean arrival time of next gap
affine gaps decouple arrival time with length
 1/mean length of aligned sequences (set to ~0)
 1/mean length of unaligned sequences (set to ~0)

20. The meaning of alignment scores
Match/mismatch scores:
P(xi, yj)
s(a, b)  log ––––––––––– (let’s ignore log(1 – 2) for the moment – assume no gaps)
P(xi) P(yj)
Example:
Say DNA regions between human and mouse have average conservation of 50%
Then P(A,A) = P(C,C) = P(G,G) = P(T,T) = 1/ (so they sum to ½)
P(A,C) = P(A,G) =……= P(T,G) = 1/ (24 mismatches, sum to ½)
Say P(A) = P(C) = P(G) = P(T) = ¼
log [ (1/8) / (1/4 * 1/4) ] = log 2 = 1, for match
Then, s(a, b) = log [ (1/24) / (1/4 * 1/4) ] = log 16/24 =
Cutoff similarity that scores 0: s*1 – (1 – s)*0.585 = 0
According to this model, a 37.5%-conserved sequence with no gaps would score on average
0.375 * 1 – * = 0
Why?
37.5% is between the 50% conservation model, and the random 25% conservation model !

21. Substitution matrices
A more meaningful way to assign match/mismatch scores
For protein sequences, different substitutions have dramatically different frequencies!
PAM Matrices:
Start from a curated set of very similar protein sequences
Construct ancestral sequences (using parsimony)
Calculate Aab: frequency of letters a and b interchanging
Calculate Bab = P(b|a) = Aab/(c≤d Acd)
Adjust matrix B so that a,b qa qb Bab = 0.01
PAM(1)
Let PAM(N) = [PAM(1)]N -- Common PAM(250)

22. Substitution Matrices
BLOSUM matrices:
Start from BLOCKS database (curated, gap-free alignments)
Cluster sequences according to > X% identity
Calculate Aab: # of aligned a-b in distinct clusters, correcting by 1/mn, where m, n are the two cluster sizes
Estimate
P(a) = (b Aab)/(c≤d Acd); P(a, b) = Aab/(c≤d Acd)

23. BLOSUM matrices
BLOSUM 50
BLOSUM 62
(The two are scaled differently)