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# Hidden Markov model

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Hidden Markov model. BioE 480 Sept 16, 2004. In general, we have Bayes theorem: P(X|Y) = P(Y|X)P(X)/P(Y) Event X: the die is loaded, Event Y: 3 sixes. - PowerPoint PPT Presentation
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Hidden Markov model BioE 480 Sept 16, 2004
Transcript Hidden Markov model

BioE 480

Sept 16, 2004 • In general, we have Bayes theorem:

P(X|Y) = P(Y|X)P(X)/P(Y)• Event X: the die is loaded, Event Y: 3 sixes.

• Example: Assume we know that on average extracellular proteins have a slightly different a.a. composition than intracellular ones. Eg. More cysteines. How do we use this information to predict a new protein sequence x=x1x2…xn whether it is intracellular or extracellular.– We first split the training examples from Swiss-Prot into intracellular and

extracellular proteins, leaving aside those unclassifiable.– We then estimate a set of frequencies for intraceullar proteins and a set of

extracellular frequencies.– Also estimate the probability that any new sequence is extracelluar, pext and

intracellular pint , called prior probabilites, because they are best guesses about a sequence before we actually see the sequence itself.

intaq

extaq • We now have:

• Because we assume that every sequence must be either extracellular or intracelluar, we have:

• By Bayes’ theorem,

• This is the number we want: the posterior probability that a sequence is extracellular.– It is our best guess after we have seen the data.

• More complicated: transmembrane proteins have both intra and extra cellular components.

i xi

extx ii

qxPqextxP int)int|(,)|(

int)|()|()( int xPpextxPpxP ext

i xi

extx

exti

extx

ext

ii

i

qpqp

qpxextP

intint)|( • Random Model R: For two sequences x and y, of lengths n and m. If xi is the i th symbol in x, and yi the i th symbol in y. Assume that letter a occurs independently with some frequency qa.

– The probability of the two sequences x and y is just the product of the probabilities of each amino acid:

P(x,y|R) = qxi qyi

• An alternative model: Match Model M: Aligned pairs of residues occur with a joint probability Pab. Its value can be thought of as the probability that the resdiues a and b have each independently been derived from some unknown original residue c in their common ancester.

– c might be the same as a and/or b.– The probability of the whole alignment is:

P(x,y|M) = pxiyi

• The ratio of these two likelihoods is the odds ratio:

P(x,y|M) / P(x,y|R) = pxiyi / (qxi qyi )= pxiyi / qxi qyi

• To make this additive, we take the logarithm of this ratio, the log-odd ratio.

S = s(xi, yi), where s(a, b) = log (pab / qa qb ), Here s(a, b) is the log likelihood ratio of the residue pair (a,b) occurring as an aligned pair, as opposed to an unaligned pair.

A biologist may write down and ad hoc substitution matrix based on intuition, but it actually implies the “target frequencies” pab . Any substitution matrix is making a statement about the probability of observing ab pairs in real alignment.

• How to develop an evolutionary model?

– Parameterized by probability of residue A mutated to residue B: PAB

– Statistical modeling: these parameters cannot be assigned, rather, they have to be estimated from data.

• Suppose we know sequences s and s’ are related: find PAB that maximizes:

– Maximum likelihood: maximize data likelihood under model.

– Results:

) using model |',( ABPssP

. oflength theis column, ain aligned are B andA count when # is )',( Here ,)',(

snssABnAB

AB n

ssnP • Substitution matrix can be obtained when alignment of sequences are compiled.

• Different matrix for different evolutionary time t :

• How do we estimate it?

– The probability of given a residue A and it is substituted by B within evolutionary

distance t :

– Ignore directionality of time:

– Assume that the distribution of amino acid (a.a.) does not change during evolution:

Can be estimated from:

• relative frequency of pair (A,B) in the known alignment of s and s’, and

• relative frequency qA of residue A.

– Substitution matrix over a longer time scale:

)|,()( tBAPtPAB

),|( tABP

)|(),|()|,( tAPtABPtBAP

AA qtABPtBAPqAPtAP ),|()|,( and ,)()|(

Aq

tBAPtABPM

)|,(),|(

kAB MMktABPM ' and )),|((' Regular Expression

• Widely used in many programs, especially those on Unix: awk, grep, sed, and perl.

– Used for searching text files for a pattern. Eg. Search for all files that containing “C.elegans” or “Caenorhabditis elegans” with the regular expression:

% grep “C[\.a-z]* elegans” *– This matches any line containing a C followed by any number of lower-

case leters or “.”, then a space, and then “elegans”.

– Another example, PROSITE.

– Difficulty: need to be very broad and complex, because protein spelling is much more free than English spelling. • Example: ( use DNA because of the smaller number of letters than a.a.)

ACA---ATG

TCAACTATC

ACAC--AGC

AGA---ATC

ACCG--ATC• A regular expression for this is:

[AT][CG][AC][ACGT]*A[TG][GC]

Problem: It does not distinguish:

TGCT--AGG Highly implausible, exceptional character in each position

ACAC--ATC Consensus sequence • Alternative: score sequences by how well they fit the alignment.

– Eg. A proabability of 4/5=0.8 for A in the first position, 1/5=0.2 for a T; etc.

– After the third position in the alignment, 3 out of 5 sequences have “insertions” of varying lengths, so we say the probability of making an insertion is 3/5 and thus 2/5 for not makng one.

– A diagram: This is a hidden Markov model! ACA---ATGTCAACTATCACAC--AGCAGA---ATCACCG--ATC

A 0.8C 0.0G 0.0T 0.2

A 0.0C 0.8G 0.2T 0.0

A ?C ?G ?T ?

A ?C ?G ?T ?

A 1.0C 0.0G 0.0T 0.0

A 0.0C 0.0G 0.2T 0.8

A 0.0C 0.8G 0.2T 0.2

1.0 1.0 1.0

??

?

?

? Hidden Markov Model• A box is called a “(Match) state”:

– one state for each term in the regular expression.

– Probabilities: counting in the multiple alignment how many times each event occurs.

• “Insertion”: A state above the other states.

– Probabilities of NTs: counting all occurrences of the four NTs in this region of the alignment: A 1/5; C 2/5; G 1/5, and T 1/5.

– Probabilities of transitions:

• After sequences 2, 3 and 5 have made one insertion each, there are two more (from sequence 2)

• Total number of transitions back to the main line is 3: there are three sequences that have insertions. All will come back to the main states.

• Therefore, probability of making a transition to the next state: 3/5

• Probability of making a transition to itself: 2/5 --- keep inserting. Scoring Sequences

• Consensus sequences: ACACATC.– Probability of the 1st A: 4/5.– This is multiplied by the probability of the transition from the first state to

the second, which is 1.– ….

• How do we score the exceptional sequence TGCT--AGG?

– This is 2000 times smaller.– We can now get a score for each sequence to measure how well it fits the

motif.

21074801801160406080180180 -......... P(ACACATC)

2100023.0 P • For the other four original sequences:

• The probability depends very strongly on the length of the sequence.

– Probability: Not a good number as score.

– Use log-odds ratio: ln( observed/random), here the random model (null model) is that the sequences are random strings of NTs:

• the probability of a sequence of length L is: 0.25 L • The log-odds score is:

• Other null model: not 0.25, but background NT compositions.– When a sequence fits a HMM very well: high log-odds score– When it fits a null model better: negative score.

25.0log)(log25.0

)(log sequencefor oddslog LSP

SPS

L • The second sequence has raw score as low as the exceptional score

– because it has three inserts.

– But the log-odds score is much higher than the exceptional seq.

– Excellent discrimination.

– But, high log-odds may not be a “hit”: there will always be random hits when searching a database. Need to look at E-value and P-value.

• If the alignment had no gaps or insertions:

– No insert state.

– All probabilities associated with the arrows (the transition probabilities) = 1. Can all be ignored.

– HMM works then exactly as a weight matrix of log-odds scores. What is hidden

• Come back to the occasional dishonest casino: they use a fair die most of the time, with a probability of 0.05 it switching to loaded die, and with a probability of 0.1 of switch back.

• The switch between die is a Markov process (it only depends on the previous state).

• The observation of the sequence of rolls is hidden Markov process because the casino wouldn’t tell you in which role they were using loaded die. Profile HMMs• Profile HMMs: allows position-dependent gap penalties.

– Obtained from a multiple alignment.– Can be used to search a database for other members of the family just like a standard profile.

• Structure of the Model:

– Main states (bottom): model the columns of the alignment, are the main states.• Probabilities are calculated by the frequency of the a.a. or NT.

– Insert states (diamond): model highly variable regions in the alignment• Often the probabilities are a fixed distribution, eg, by composition – Delete states (circle): silent or null state. Do not match any residues, they are there so it is possible to jump over one or more columns:

• For modeling when just a few of the sequences have a “-” at a position.

• Example: • Insertion region (white): an alignment of this region is highly uncertain.• Shaded region: columns that correspond to main states in the HMM model.

– Probabilities: For each non-insert column, we make a main state and set the probabilities equal to the amino acid frequencies.

– Transition probabilities: count how many sequences use the various transitions, like before.

• Delete states: Two transitions from a main state to a delete state, shown with dashed lines:– from “begin” to the first delete state– from main state 12 to delete state 13.– Both correspond to dashes in the alignment:

• Only one sequence has gaps, the probability of these delete transitions is 1/30.– The 4th sequence continues deletion to the end:

• Probability from delete 13 to 14 is 1, and from delete 14 to the end is also 1.  Pseudo-counts• Dangerous to estimate a probability distribution from just a few observed amino

acids.

– If there are two sequences, with Leu at a position:

• P for Leu =1, but P = 0 for all other residues at this position

• But we know that often Val substitutes Leu.

• The probability of the whole sequence are easily become 0 if a single Leu is substituted by a Val.

• Or , the log-odds is minus infinity.

• How to avoid “over-fitting” (strong conclusions drawn from very little evidence)?

• Use pseudocounts:

– Pretend to have more counts than those from the data.

– A. Add 1 to all the counts:

• Leu: 3/23, other a.a.: 1/23 • Adding 1 to all counts is as assuming a priori all a.a. are equally likely.

• Another approach: use background composition as pseudocounts.

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