lect 06 HMM alignment

lect 06 HMM alignment - Bioc 2808 Lecture 5 Sequence...

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Click to edit Master subtitle style 7/29/10 Bioc 2808 Lecture 5 Sequence alignment and Instructor Dr. Junwen Wang
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7/29/10 Today’s topics Profile-HMM (PHMM) and protein sequence alignment Generalized HMM (GHMM) and gene prediction
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7/29/10 33 HMM …… H M M M C …… Observed layer Hidden Layer
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7/29/10 44 Insertion states Deletion states No of matching states = average sequence length in the family PFAM Database - of Protein families ( http://pfam.wustl.edu) Profile HMM(PHMM)
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7/29/10 From multiple alignment to profile HMM Good model of these proteins must reflect: highly conserved positions in the alignment NF. ....A- DF. ....SY NYrqsanS- NFapistAY DFvlamrSF
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7/29/10 Exercise 1. Build profile HMM from multiple NF. ....A- DF. ....SY NYrqsanS- NFapistAY DFvlamrSF Three kinds of states: match insert silent S t a r t
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7/29/10 BME100 Profile HMM from multiple alignment NF. ....A- DF. ....SY NYrqsanS- NFapistAY DFvlamrSF P(N)=0.6 P(D)=0.4 P(R)=0.1 3 P(Q)=0.0 7 P(A)=0.2 • • • Three states: match insert silent P(F)=0.8 P(Y)=0.2 1.0 0.8 0.6 P(S)=0 .6 P(A)=0 .4 P(Y)=0 .67 P(F)=0. 33 0. 4 - 0. 2 0. 4 0. 6 0. 0 0. 0 0. 0 0. 0 0. 0 0. 0 1. 0
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7/29/10 Exercise 2. build another pHMM Three states: match insert silent S t a r t AC - - - AG TC ACT AC AC C - - AC AG - - - AC AC G - - AC
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7/29/10 Score a sequence with the HMM Once we have an HMM for a group of proteins, we are often interested in how well a new sequence fits the model. We want to compute a probability for our sequence with respect to the model.
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7/29/10 Exercise 3. finding the alignment for a sequence P(N)=0 .6 P(D)=0 .4 - P(F)=0. 8 P(Y)=0 .2 P(S)=0 .6 P(A)=0 .4 P(Y)=0 .67 P(F)=0. 33 P(R)=0 .13 P(Q)=0 .07 P(A)=0 .2 • • • 1. 0 0. 4 0. 6 0. 6 0. 8 0. 2 0. 4 0. 0 0. 0 0. 0 0. 0 Given a sequence “DYAF”, can you find the most probable alignment based on following pHMM? 0. 0 0. 0 1. 0
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7/29/10 BME100 One sequence many paths A protein sequence can be represented by many paths through the HMM. P(N)=0 .6 P(D)=0 .4 P(F)=0. 8 P(Y)=0 .2 P(S)=0 .6 P(A)=0 .4 P(Y)=0 .67 P(F)=0. 33 P(R)=0 .13 P(Q)=0 .07 P(A)=0 .2 • • • 1. 0 0. 4 0. 6 0. 6 0. 8 0. 1 0. 4 - 0. 0 0. 0 0. 1 0. 0 DYAF 0. 0 0. 0 1. 0
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7/29/10 Most probable path: Viterbi algorithm P(N) =0.6 P(D) =0.4 - P(F) =0.8 P(Y) =0.2 P(S) =0.6 P(A) =0.4 P(Y) =0.6 7 P(F) P(R) =0.1 3 P(Q) =0.0 7 P(A) =0.2 • • • 1. 0 0.
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lect 06 HMM alignment - Bioc 2808 Lecture 5 Sequence...

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