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Sequence Alignment Cont’d

Sequence Alignment Cont’d

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Sequence Alignment Cont’d

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  1. Sequence AlignmentCont’d

  2. …… query …… scan DB query Indexing-based local alignment (BLAST- Basic Local Alignment Search Tool) • SEED Construct a dictionary of all the words in query (database) & search database (query) in linear time for word matches • EXTEND Initiate fast local alignment procedures to the left and right of each word match • REPORT Retain all alignments that score above a threshold and report them

  3. Indexing-based local alignment—Extensions A C G A A G T A A G G T C C A G T Example: k = 4 The matching word GGTC initiates an alignment Extension to the left and right with no gaps until alignment falls < C below best so far Output: GTAAGGTCC GTTAGGTCC C C C T T C C T G G A T T G C G A

  4. Indexing-based local alignment—Extensions A C G A A G T A A G G T C C A G T Gapped extensions until threshold • Extensions with gaps until score < C below best score so far Reference: Zhang, Bergman, Miller, RECOMB ‘98 Output: GTAAGGTCCAGT GTTAGGTC-AGT C T G A T C C T G G A T T G C G A

  5. Sensitivity-Speed Tradeoff X% Sens. Speed Kent WJ, Genome Research 2002

  6. Sensitivity-Speed Tradeoff Methods to improve sensitivity/speed • Using pairs of words • Using inexact words • Patterns—non consecutive positions ……ATAACGGACGACTGATTACACTGATTCTTAC…… ……GGCACGGACCAGTGACTACTCTGATTCCCAG…… ……ATAACGGACGACTGATTACACTGATTCTTAC…… ……GGCGCCGACGAGTGATTACACAGATTGCCAG…… TTTGATTACACAGAT T G TT CAC G

  7. Measured improvement Kent WJ, Genome Research 2002

  8. Non-consecutive words—Patterns Patterns increase the likelihood of at least one match within a long conserved region Consecutive Positions Non-Consecutive Positions 6 common 5 common 7 common 3 common On a 100-long 70% conserved region: ConsecutiveNon-consecutive Expected # hits: 1.07 0.97 Prob[at least one hit]: 0.30 0.47

  9. Advantage of Patterns 11 positions 11 positions 10 positions

  10. Multiple patterns TTTGATTACACAGAT T G TT CAC G T G T C CAG TTGATT A G • K different patterns • Construct K distinct dictionaries, one for each pattern • Takes K times longer to scan • Patterns can complement one another • Computational problem: • Given: a model (prob distribution) for homology between two regions • Find: best set of K patterns that maximizes Prob(at least one match) How long does it take to search the query? Buhler et al. RECOMB 2003 Sun & Buhler RECOMB 2004

  11. Variants of BLAST • NCBI BLAST: search the universe http://www.ncbi.nlm.nih.gov/BLAST/ • MEGABLAST: http://genopole.toulouse.inra.fr/blast/megablast.html • Optimized to align very similar sequences • Works best when k = 4i  16 • Linear gap penalty • WU-BLAST: (Wash U BLAST) http://blast.wustl.edu/ • Very good optimizations • Good set of features & command line arguments • BLAT http://genome.ucsc.edu/cgi-bin/hgBlat • Faster, less sensitive than BLAST • Good for aligning huge numbers of queries • CHAOS http://www.cs.berkeley.edu/~brudno/chaos • Uses inexact k-mers, sensitive • PatternHunter http://www.bioinformaticssolutions.com/products/ph/index.php • Uses patterns instead of k-mers • BlastZ http://www.psc.edu/general/software/packages/blastz/ • Uses patterns, good for finding genes • Typhon http://typhon.stanford.edu • Uses multiple alignments to improve sensitivity/speed tradeoff

  12. Example Query: Human atoh enhancer, 179 letters [1.5 min] Result: 57 blast hits • gi|7677270|gb|AF218259.1|AF218259 Homo sapiens ATOH1 enhanc... 355 1e-95 • gi|22779500|gb|AC091158.11| Mus musculus Strain C57BL6/J ch... 264 4e-68 • gi|7677269|gb|AF218258.1|AF218258 Mus musculus Atoh1 enhanc... 256 9e-66 • gi|28875397|gb|AF467292.1| Gallus gallus CATH1 (CATH1) gene... 78 5e-12 • gi|27550980|emb|AL807792.6| Zebrafish DNA sequence from clo... 54 7e-05 • gi|22002129|gb|AC092389.4| Oryza sativa chromosome 10 BAC O... 44 0.068 • gi|22094122|ref|NM_013676.1| Mus musculus suppressor of Ty ... 42 0.27 • gi|13938031|gb|BC007132.1| Mus musculus, Similar to suppres... 42 0.27 gi|7677269|gb|AF218258.1|AF218258 Mus musculus Atoh1 enhancer sequence Length = 1517 Score = 256 bits (129), Expect = 9e-66 Identities = 167/177 (94%), Gaps = 2/177 (1%) Strand = Plus / Plus Query: 3 tgacaatagagggtctggcagaggctcctggccgcggtgcggagcgtctggagcggagca 62 ||||||||||||| ||||||||||||||||||| |||||||||||||||||||||||||| Sbjct: 1144 tgacaatagaggggctggcagaggctcctggccccggtgcggagcgtctggagcggagca 1203 Query: 63 cgcgctgtcagctggtgagcgcactctcctttcaggcagctccccggggagctgtgcggc 122 |||||||||||||||||||||||||| ||||||||| |||||||||||||||| ||||| Sbjct: 1204 cgcgctgtcagctggtgagcgcactc-gctttcaggccgctccccggggagctgagcggc 1262 Query: 123 cacatttaacaccatcatcacccctccccggcctcctcaacctcggcctcctcctcg 179 ||||||||||||| || ||| |||||||||||||||||||| ||||||||||||||| Sbjct: 1263 cacatttaacaccgtcgtca-ccctccccggcctcctcaacatcggcctcctcctcg 1318 http://www.ncbi.nlm.nih.gov/BLAST/

  13. The Four-Russian Algorithmbrief overviewA (not so useful) speedup of Dynamic Programming[Arlazarov, Dinic, Kronrod, Faradzev 1970]

  14. Main Observation xl’ xl Within a rectangle of the DP matrix, values of D depend only on the values of A, B, C, and substrings xl...l’, yr…r’ Definition: A t-block is a t  t square of the DP matrix Idea: Divide matrix in t-blocks, Precompute t-blocks Speedup: O(t) B A yr C yr’ D t

  15. The Four-Russian Algorithm Main structure of the algorithm: • Divide NN DP matrix into KK log2N-blocks that overlap by 1 column & 1 row • For i = 1……K • For j = 1……K • Compute Di,j as a function of Ai,j, Bi,j, Ci,j, x[li…l’i], y[rj…r’j] Time: O(N2 / log2N) times the cost of step 4 t t t

  16. The Four-Russian Algorithm t t t

  17. 1 2 2 1 1 1 1 … 2 2 2 2 … K … … … … x1 K K K K x2 x3 xK … Hidden Markov Models

  18. Outline for our next topic • Hidden Markov models – the theory • Probabilistic interpretation of alignments using HMMs Later in the course: • Applications of HMMs to biological sequence modeling and discovery of features such as genes

  19. Example: The Dishonest Casino A casino has two dice: • Fair die P(1) = P(2) = P(3) = P(5) = P(6) = 1/6 • Loaded die P(1) = P(2) = P(3) = P(5) = 1/10 P(6) = 1/2 Casino player switches back-&-forth between fair and loaded die once every 20 turns Game: • You bet $1 • You roll (always with a fair die) • Casino player rolls (maybe with fair die, maybe with loaded die) • Highest number wins $2

  20. Question # 1 – Evaluation GIVEN A sequence of rolls by the casino player 1245526462146146136136661664661636616366163616515615115146123562344 QUESTION How likely is this sequence, given our model of how the casino works? This is the EVALUATION problem in HMMs Prob = 1.3 x 10-35

  21. Question # 2 – Decoding GIVEN A sequence of rolls by the casino player 1245526462146146136136661664661636616366163616515615115146123562344 QUESTION What portion of the sequence was generated with the fair die, and what portion with the loaded die? This is the DECODING question in HMMs FAIR LOADED FAIR

  22. Question # 3 – Learning GIVEN A sequence of rolls by the casino player 1245526462146146136136661664661636616366163616515615115146123562344 QUESTION How “loaded” is the loaded die? How “fair” is the fair die? How often does the casino player change from fair to loaded, and back? This is the LEARNING question in HMMs Prob(6) = 64%

  23. The dishonest casino model 0.05 0.95 0.95 FAIR LOADED P(1|F) = 1/6 P(2|F) = 1/6 P(3|F) = 1/6 P(4|F) = 1/6 P(5|F) = 1/6 P(6|F) = 1/6 P(1|L) = 1/10 P(2|L) = 1/10 P(3|L) = 1/10 P(4|L) = 1/10 P(5|L) = 1/10 P(6|L) = 1/2 0.05

  24. Definition of a hidden Markov model Definition: A hidden Markov model (HMM) • Alphabet = { b1, b2, …, bM } • Set of states Q = { 1, ..., K } • Transition probabilities between any two states aij = transition prob from state i to state j ai1 + … + aiK = 1, for all states i = 1…K • Start probabilities a0i a01 + … + a0K = 1 • Emission probabilities within each state ei(b) = P( xi = b | i = k) ei(b1) + … + ei(bM) = 1, for all states i = 1…K 1 2 K …

  25. A HMM is memory-less At each time step t, the only thing that affects future states is the current state t P(t+1 = k | “whatever happened so far”) = P(t+1 = k | 1, 2, …, t, x1, x2, …, xt) = P(t+1 = k | t) 1 2 K …

  26. 1 1 1 1 … 2 2 2 2 … … … … … K K K K … A parse of a sequence Given a sequence x = x1……xN, A parse of x is a sequence of states  = 1, ……, N 1 2 2 K x1 x2 x3 xK

  27. 1 1 1 1 … 2 2 2 2 … … … … … K K K K … Likelihood of a parse A compact way to write a01 a12……aN-1N e1(x1)……eN(xN) Number all parameters aij and ei(b); n params Example: a0Fair : 1; a0Loaded : 2; … eLoaded(6) = 18 Then, count in x and  the # of times each parameter j = 1, …, n occurs F(j, x, ) = # parameter j occurs in (x, ) (call F(.,.,.) the feature counts) Then, P(x, ) = j=1…n jF(j, x, ) = = exp[j=1…nlog(j)F(j, x, )] 1 Given a sequence x = x1……xN and a parse  = 1, ……, N, To find how likely is the parse: (given our HMM) P(x, ) = P(x1, …, xN, 1, ……, N) = P(xN, N | N-1) P(xN-1, N-1 | N-2)……P(x2, 2 | 1) P(x1, 1) = P(xN | N)P(N | N-1) ……P(x2 | 2)P(2 | 1)P(x1 | 1)P(1) = a01 a12……aN-1Ne1(x1)……eN(xN) 2 2 K x1 x2 x3 xK

  28. Example: the dishonest casino Let the sequence of rolls be: x = 1, 2, 1, 5, 6, 2, 1, 5, 2, 4 Then, what is the likelihood of  = Fair, Fair, Fair, Fair, Fair, Fair, Fair, Fair, Fair, Fair? (say initial probs a0Fair = ½, aoLoaded = ½) ½  P(1 | Fair) P(Fair | Fair) P(2 | Fair) P(Fair | Fair) … P(4 | Fair) = ½  (1/6)10  (0.95)9 = .00000000521158647211 ~= 0.5  10-9

  29. Example: the dishonest casino So, the likelihood the die is fair in this run is just 0.521  10-9 OK, but what is the likelihood of = Loaded, Loaded, Loaded, Loaded, Loaded, Loaded, Loaded, Loaded, Loaded, Loaded? ½  P(1 | Loaded) P(Loaded, Loaded) … P(4 | Loaded) = ½  (1/10)9  (1/2)1 (0.95)9 = .00000000015756235243 ~= 0.16  10-9 Therefore, it somewhat more likely that all the rolls are done with the fair die, than that they are all done with the loaded die

  30. Example: the dishonest casino Let the sequence of rolls be: x = 1, 6, 6, 5, 6, 2, 6, 6, 3, 6 Now, what is the likelihood  = F, F, …, F? ½  (1/6)10  (0.95)9 = 0.5  10-9, same as before What is the likelihood  = L, L, …, L? ½  (1/10)4  (1/2)6 (0.95)9 = .00000049238235134735 ~= 0.5  10-7 So, it is 100 times more likely the die is loaded

  31. The three main questions on HMMs • Evaluation GIVEN a HMM M, and a sequence x, FIND Prob[ x | M ] • Decoding GIVEN a HMM M, and a sequence x, FIND the sequence  of states that maximizes P[ x,  | M ] • Learning GIVEN a HMM M, with unspecified transition/emission probs., and a sequence x, FIND parameters  = (ei(.), aij) that maximize P[ x |  ]

  32. Let’s not be confused by notation P[ x | M ]: The probability that sequence x was generated by the model The model is: architecture (#states, etc) + parameters  = aij, ei(.) So, P[x | M] is the same with P[ x |  ], and P[ x ], when the architecture, and the parameters, respectively, are implied Similarly, P[ x,  | M ], P[ x,  |  ] and P[ x,  ] are the same when the architecture, and the parameters, are implied In the LEARNING problem we always write P[ x |  ] to emphasize that we are seeking the * that maximizes P[ x |  ]

  33. Problem 1: Decoding Find the best parse of a sequence

  34. 1 2 2 1 1 1 1 … 2 2 2 2 … K … … … … x1 K K K K … x2 x3 xK Decoding GIVEN x = x1x2……xN We want to find  = 1, ……, N, such that P[ x,  ] is maximized * = argmax P[ x,  ] We can use dynamic programming! Let Vk(i) = max{1… i-1} P[x1…xi-1, 1, …, i-1, xi, i = k] = Probability of most likely sequence of states ending at state i = k

  35. Decoding – main idea Given that for all states k, and for a fixed position i, Vk(i) = max{1… i-1} P[x1…xi-1, 1, …, i-1, xi, i = k] What is Vl(i+1)? From definition, Vl(i+1) = max{1… i}P[ x1…xi, 1, …, i, xi+1, i+1 = l ] = max{1… i}P(xi+1, i+1 = l | x1…xi,1,…, i) P[x1…xi, 1,…, i] = max{1… i}P(xi+1, i+1 = l | i ) P[x1…xi-1, 1, …, i-1, xi, i] = maxk [P(xi+1, i+1 = l | i = k) max{1… i-1}P[x1…xi-1,1,…,i-1, xi,i=k]] = el(xi+1)maxk aklVk(i)

  36. The Viterbi Algorithm Input: x = x1……xN Initialization: V0(0) = 1 (0 is the imaginary first position) Vk(0) = 0, for all k > 0 Iteration: Vj(i) = ej(xi)  maxk akj Vk(i – 1) Ptrj(i) = argmaxk akj Vk(i – 1) Termination: P(x, *) = maxk Vk(N) Traceback: N* = argmaxk Vk(N) i-1* = Ptri (i)

  37. The Viterbi Algorithm x1 x2 x3 ………………………………………..xN Similar to “aligning” a set of states to a sequence Time: O(K2N) Space: O(KN) State 1 2 Vj(i) K

  38. Viterbi Algorithm – a practical detail Underflows are a significant problem P[ x1,…., xi, 1, …, i ] = a01 a12……ai e1(x1)……ei(xi) These numbers become extremely small – underflow Solution: Take the logs of all values Vl(i) = logek(xi) + maxk [ Vk(i-1) + log akl ]

  39. Example Let x be a long sequence with a portion of ~ 1/6 6’s, followed by a portion of ~ ½ 6’s… x = 123456123456…123456626364656…1626364656 Then, it is not hard to show that optimal parse is (exercise): FFF…………………...FLLL………………………...L 6 characters “123456” parsed as F, contribute .956(1/6)6 = 1.610-5 parsed as L, contribute .956(1/2)1(1/10)5 = 0.410-5 “162636” parsed as F, contribute .956(1/6)6 = 1.610-5 parsed as L, contribute .956(1/2)3(1/10)3 = 9.010-5