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Hidden Markov Models

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1

2

2

1

1

1

1

…

2

2

2

2

…

K

…

…

…

…

x1

K

K

K

K

x2

x3

xK

…

Hidden Markov Models

VITERBI

Initialization:

V0(0) = 1

Vk(0) = 0, for all k > 0

Iteration:

Vl(i) = el(xi) maxkVk(i-1) akl

Termination:

P(x, *) = maxkVk(N)

- FORWARD
- Initialization:
- f0(0) = 1
- fk(0) = 0, for all k > 0
- Iteration:
- fl(i) = el(xi) k fk(i-1) akl
- Termination:
- P(x) = k fk(N)

BACKWARD

Initialization:

bk(N) = 1, for all k

Iteration:

bl(i) = k el(xi+1) akl bk(i+1)

Termination:

P(x) = k a0k ek(x1) bk(1)

P(i = k | x) =

P(i = k , x)/P(x) =

P(x1, …, xi, i = k, xi+1, … xn) / P(x) =

P(x1, …, xi, i = k) P(xi+1, … xn | i = k) / P(x) =

fk(i) bk(i) / P(x)

We can now calculate

fk(i) bk(i)

P(i = k | x) = –––––––

P(x)

Then, we can ask

What is the most likely state at position i of sequence x:

Define ^ by Posterior Decoding:

^i = argmaxkP(i = k | x)

- For each state,
- Posterior Decoding gives us a curve of likelihood of state for each position
- That is sometimes more informative than Viterbi path *

- Posterior Decoding may give an invalid sequence of states (of prob 0)
- Why?

x1 x2 x3 …………………………………………… xN

- P(i = k | x) = P( | x) 1(i = k)
= {:[i] = k}P( | x)

State 1

P(i=l|x)

l

k

1() = 1, if is true

0, otherwise

Variants of 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

aHTT

aTHH

aTHT

state TH

state TT

aTH(prev = H)

aTH(prev = T)

aTTH

- P(i+1 = l | i = k, i -1 = j)
- Vlk(i) = maxj{ Vkj(i – 1) + … }
- Time? Space?

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

Example: exon lengths in genes

p

1-p

X

Y

X

X

q

1-q

Disadvantage: Still very inflexible

lX = C + geometric with mean 1/(1-p)

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 – 1 m – 1

P(lX = m) = n – 1 (1 – p)n-1+1p(m-1)-(n-1) = n – 1 (1 – p)npm-n

p

p

p

1 – p

1 – p

1 – p

Y

X(n)

X(1)

X(2)

……

- EasyGene:
Prokaryotic

gene-finder

Larsen TS, Krogh A

- Negative binomial with n = 3

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 of Viterbi:

Time: O(D)

Space: O(1)

where D = maximum duration of state

F

d<Df

xi…xi+d-1

Pf

Warning, Rabiner’s tutorial claims O(D2) & O(D) increases

emissions

emissions

Recall original iteration:

Vl(i) = maxk Vk(i – 1) akl el(xi)

New iteration:

Vl(i) = maxk maxd=1…DlVk(i – d) Pl(d) akl j=i-d+1…iel(xj)

F

L

d<Df

d<Dl

Pl

Pf

transitions

xi…xi + d – 1

xj…xj + d – 1

Precompute cumulative values

Proteins, Pair HMMs, and Alignment

M

(+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

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

-AGGCTATCACCTGACCTCCAGGCCGA--TGCCC---

TAG-CTATCAC--GACCGC-GGTCGATTTGCCCGACC

IMMJMMMMMMMJJMMMMMMJMMMMMMMIIMMMMMIII

Dynamic Programming:

M(i, j):Optimal alignment of x1…xi to y1…yjending 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…yjending in J

The score is additive, therefore we can apply DP recurrence formulas

Initialization:

M(0,0) = 0;

M(i, 0) = M(0, j) = -, for i, j > 0

I(i,0) = d + ie;J(0, j) = d + je

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

d + M(i – 1, j)

e + J(i, j – 1)

J(i, j) = max

d + M(i, j – 1)

Termination:

Optimal alignment given by max { M(m, n), I(m, n), J(m, n) }

Brief introduction to the evolution of proteins

Protein sequence and structure

Protein classification

Phylogeny trees

Substitution matrices

The Protein Folding Problem

- What determines structure?
- Energy
- Kinematics

- How can we determine structure?
- Experimental methods
- Computational predictions

- The primary structure of a protein is the amino acid sequence

- Twenty different amino acids have distinct shapes and properties

A useful mnemonic for the hydrophobic amino acids is "FAMILY VW"

- helices and sheets are stabilized by hydrogen bonds between backbone oxygen and hydrogen atoms

- Actin is ancient and abundant
- Most abundant protein in cells
- 1-2 actin genes in bacteria, yeasts, amoebas
- Humans: 6 actin genes
- -actin in muscles; -actin, -actin in non-muscle cells
- ~4 amino acids different between each version
MUSCLE ACTIN Amino Acid Sequence

1 EEEQTALVCD NGSGLVKAGF AGDDAPRAVF PSIVRPRHQG VMVGMGQKDS YVGDEAQSKR

61 GILTLKYPIE HGIITNWDDM EKIWHHTFYN ELRVAPEEHP VLLTEAPLNP KANREKMTQI

121 MFETFNVPAM YVAIQAVLSL YASGRTTGIV LDSGDGVSHN VPIYEGYALP HAIMRLDLAG

181 RDLTDYLMKI LTERGYSFVT TAEREIVRDI KEKLCYVALD FEQEMATAAS SSSLEKSYEL

241 PDGQVITIGN ERFRGPETMF QPSFIGMESS GVHETTYNSI MKCDIDIRKD LYANNVLSGG

301 TTMYPGIADR MQKEITALAP STMKIKIIAP PERKYSVWIG GSILASLSTF QQMWITKQEY

361 DESGPSIVHR KCF

- Proteins evolve by both duplication and species divergence

New PDB structures

Mutation rates between amino acids have dramatic differences!

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) = (bAab)/(c≤dAcd); P(a, b) = Aab/(c≤dAcd)

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

Model M

1 – 2

This model generates two sequences simultaneously

Match/Mismatch state M:

P(x, y) reflects substitution frequencies between pairs of amino acids

Insertion states I, J:

P(x), P(y) reflect frequencies of each amino acid

: set so that 1/2 is avg. length before next gap

:set so that 1/(1 – ) is avg. length of a gap

M

P(xi, yj)

1 –

1 –

I

P(xi)

J

P(yj)

optional

Model R

Two sequences are independently generated from one another

P(x, y | R) = P(x1)…P(xm) P(y1)…P(yn) = i P(xi) j P(yj)

1

1

J

P(yj)

I

P(xi)

1 – 2

Every pair of letters contributes:

M

- (1 – 2) P(xi, yj) when matched
- P(xi) P(yj) when gapped
R

- P(xi) P(yj) in random model
Focus on comparison of

P(xi, yj) vs. P(xi) P(yj)

M

P(xi, yj)

1 –

1 –

I

P(xi)

J

P(yj)

1

1

J

P(yj)

I

P(xi)

1 – 2

Every pair of letters contributes:

M

- (1 – 2) P(xi, yj) when matched
- P(xi) P(yj) when gapped
- Some extra term for gap opening
R

- Some extra term for gap opening
- 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

(1 – )

––––––––

(1 – 2)

I

P(xi)

J

P(yj)

Equivalent!

1

1

J

P(yj)

I

P(xi)

Idea:

We will divide M alignment score by R score, and take logarithms

Let

P(xi, yj)

s(xi, yj) = log ––––––––– + log (1 – 2)

P(xi) P(yj)

(1 – ) P(xi)

d = – log –––––––––––––

(1 – 2) P(xi)

P(xi)

e = – log ––––––

P(xi)

=Defn substitution score

=Defn gap initiation penalty

=Defn gap extension penalty

- The Viterbi algorithm for Pair HMMs corresponds exactly to global alignment DP with affine gaps
VM(i, j) = max { VM(i – 1, j – 1), VI( i – 1, j – 1) – d, Vj( i – 1, j – 1) } + s(xi, yj)

VI(i, j) = max { VM(i – 1, j) – d, VI( i – 1, j) – e }

VJ(i, j) = max { VM(i – 1, j) – d, VI( i – 1, j) – e }

- s(.,.) (1 – 2) ~how often a pair of letters substitute one another
- 1/mean length of next gap
- (1 – ) / (1 – 2) 1/mean arrival time of next gap

Match/mismatch scores:

P(xi, yj)

s(a, b) log –––––––––– (ignore log(1 – 2) for the moment)

P(xi) P(yj)

Example:

Genes between human and mouse genes have average conservation of 80%

- Let’s calculate this way the substitution score for a match:
P(a, a) + P(c, c) + P(g, g) + P(t, t) = 0.8 P(x, x) = 0.2

P(a) = P(c) = P(g) = P(t) = 0.25

s(x, x) = log [ 0.2 / 0.252 ] = 1.163

- …and for a mismatch:
P(a, c) +…+P(t, g) = 0.2 P(x, yx) = 0.2/12 = 0.0167

s(x, y x) = log[ 0.0167 / 0.252 ] = -1.322

- What ratio matches/(matches + mism.) gives score 0?
x(#match) – y(#mism) = 0

1.163 (#match) – 1.322 (#mism) = 0

#match = 1.137(#mism)

matches = 53.2%