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. Sequence Alignment I Lecture #2 This class has been edited from Nir Friedman’s lecture which is available at www.cs.huji.ac.il/~nir. Changes made by Dan Geiger, then Shlomo Moran. www.cs.huji.ac.il Background Readings: The second chapter (pages 12-45) in the text book, Biological Sequence Analysis, Durbin et al., 2001.

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2 Sequence Comparison Much of bioinformatics involves sequences u DNA sequences u RNA sequences u Protein sequences We can think of these sequences as strings of letters u DNA & RNA: alphabet ∑ of 4 letters u Protein: alphabet ∑ of 20 letters

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3 Sequence Comparison (cont) u Finding similarity between sequences is important for many biological questions For example: u Find similar proteins Allows to predict function & structure u Locate similar subsequences in DNA Allows to identify (e.g) regulatory elements u Locate DNA sequences that might overlap Helps in sequence assembly g1g1 g2g2

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4 Sequence Alignment Input: two sequences over the same alphabet Output: an alignment of the two sequences Example: u GCGCATGGATTGAGCGA u TGCGCCATTGATGACCA A possible alignment: -GCGC-ATGGATTGAGCGA TGCGCCATTGAT-GACC-A

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5 Alignments -GCGC-ATGGATTGAGCGA TGCGCCATTGAT-GACC-A Three elements: u Perfect matches u Mismatches u Insertions & deletions (indel)

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6 Choosing Alignments There are many possible alignments For example, compare: -GCGC-ATGGATTGAGCGA TGCGCCATTGAT-GACC-A to ------GCGCATGGATTGAGCGA TGCGCC----ATTGATGACCA-- Which one is better?

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7 Scoring Alignments Intuition: u Similar sequences evolved from a common ancestor u Evolution changed the sequences from this ancestral sequence by mutations: Replacements: one letter replaced by another Deletion: deletion of a letter Insertion: insertion of a letter u Scoring of sequence similarity should examine how many and which operations took place

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8 Simple Scoring Rule Score each position independently: u Match: +1 u Mismatch: -1 u Indel -2 Score of an alignment is sum of position scores

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9 Example Example: -GCGC-ATGGATTGAGCGA TGCGCCATTGAT-GACC-A Score: (+1x13) + (-1x2) + (-2x4) = 3 ------GCGCATGGATTGAGCGA TGCGCC----ATTGATGACCA-- Score: (+1x5) + (-1x6) + (-2x11) = -23

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10 More General Scores u The choice of +1,-1, and -2 scores is quite arbitrary u Depending on the context, some changes are more plausible than others Exchange of an amino-acid by one with similar properties (size, charge, etc.) vs. Exchange of an amino-acid by one with opposite properties u Probabilistic interpretation: (e.g.) How likely is one alignment versus another ?

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11 Additive Scoring Rules u We define a scoring function by specifying a function (x,y) is the score of replacing x by y (x,-) is the score of deleting x (-,x) is the score of inserting x u The score of an alignment is the sum of position scores

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12 The Optimal Score The optimal (maximal) score between two sequences is the maximal score of all alignments of these sequences, namely, u Computing the maximal score or actually finding an alignment that yields the maximal score are closely related tasks with similar algorithms. u We now address these problems.

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13 Computing Optimal Score u How can we compute the optimal score ? If | s | = n and | t | = m, the number A( m,n ) of possible “legal” alignments is large! Exercise: Show that u The additive form of the score allows us to perform dynamic programming to compute optimal score efficiently.

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14 Recursive Argument Suppose we have two sequences: s[1..n+1] and t[1..m+1] The best alignment must be one of three cases: 1. Last match is ( s[n+1],t[m +1] ) 2. Last match is ( s[n +1], - ) 3. Last match is ( -, t[m +1] )

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15 Recursive Argument Suppose we have two sequences: s[1..n+1] and t[1..m+1] The best alignment must be one of three cases: 1. Last match is ( s[n+1],t[m +1] ) 2. Last match is ( s[n +1], - ) 3. Last match is ( -, t[m +1] )

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16 Recursive Argument Suppose we have two sequences: s[1..n+1] and t[1..m+1] The best alignment must be one of three cases: 1. Last match is ( s[n+1],t[m +1] ) 2. Last match is ( s[n +1], - ) 3. Last match is ( -, t[m +1] )

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17 Recursive Argument Define the notation: Using our recursive argument, we get the following recurrence for V : V[i,j]V[i+1,j] V[i,j+1]V[i+1,j+1]

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18 Recursive Argument u Of course, we also need to handle the base cases in the recursion: AA - We fill the matrix using the recurrence rule: S T versus

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19 Dynamic Programming Algorithm We continue to fill the matrix using the recurrence rule S T

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20 Dynamic Programming Algorithm V[0,0]V[0,1] V[1,0]V[1,1] +1 -2 -A A- -2 (A- versus -A) versus S T

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21 Dynamic Programming Algorithm S T

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22 Dynamic Programming Algorithm Conclusion: d( AAAC, AGC ) = -1 S T

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23 Reconstructing the Best Alignment u To reconstruct the best alignment, we record which case(s) in the recursive rule maximized the score S T

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24 Reconstructing the Best Alignment u We now trace back a path that corresponds to the best alignment AAAC AG-C S T

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25 Reconstructing the Best Alignment u Sometimes, more than one alignment has the best score S T AAAC A-GC AAAC -AGC AAAC AG-C

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26 Time Complexity Space: O(mn) Time: O(mn) Filling the matrix O(mn) Backtrace O(m+n) S T

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27 Space Complexity In real-life applications, n and m can be very large u The space requirements of O(mn) can be too demanding If m = n = 1000, we need 1MB space If m = n = 10000, we need 100MB space u We can afford to perform extra computation to save space Looping over million operations takes less than seconds on modern workstations u Can we trade space with time?

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28 Why Do We Need So Much Space? Compute V(i,j), column by column, storing only two columns in memory (or line by line if lines are shorter). 0 -2 -4 -6 -8 -2 1 -3 -5 -4 0 -2 -4 -6 -3 -2 0 A 1 G 2 C 3 0 A 1 A 2 A 3 C 4 Note however that u This “trick” fails when we need to reconstruct the optimizing sequence. Trace back information requires O(mn) memory bytes. To compute V[n,m]=d(s[1..n],t[1..m]), we need only O(min(n,m)) space:

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29 Space Efficient Version: Outline u If n=1 align s[1,1] and t[1,m] Else, find position (n/2, j) at which some best alignment crosses a midpoint s t u Construct alignments A=s[1,n/2] vs t[1,j] B=s[n/2+1,n] vs t[j+1,m] Return AB Input: Sequences s[1,n] and t[1,m] to be aligned. Idea: perform divide and conquer

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30 Finding the Midpoint The score of the best alignment that goes through j equals: d(s[1,n/2],t[1,j]) + d(s[n/2+1,n],t[j+1,m]) Thus, we need to compute these two quantities for all values of j s t

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31 Finding the Midpoint (Algorithm) Define u F[i,j] = d(s[1,i],t[1,j]) u B[i,j] = d(s[i+1,n],t[j+1,m]) F[i,j] + B[i,j] = score of best alignment through (i,j) We compute F[i,j] as we did before We compute B[i,j] in exactly the same manner, going “backward” from B[n,m] u Requires linear space complexity

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32 Time Complexity Analysis Time to find a mid-point: cnm ( c - a constant) u Size of recursive sub-problems is (n/2,j) and (n/2,m-j-1), hence T(n,m) = cnm + T(n/2,j) + T(n/2,m-j-1) Lemma: T(n,m) 2cnm Proof (by induction): T(n,m) cnm + 2c(n/2)j + 2c(n/2)(m-j-1) 2cnm. Thus, time complexity is linear in size of the problem At worst, twice the cost of the regular solution.

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33 Local Alignment Consider now a different question: Can we find similar substrings of s and t Formally, given s[1..n] and t[1..m] find i,j,k, and l such that d(s[i..j],t[k..l]) is maximal

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34 Local Alignment u As before, we use dynamic programming We now want to set V[i,j] to record the best alignment of a suffix of s[1..i] and a suffix of t[1..j] u How should we change the recurrence rule? Same as before but with an option to start afresh u The result is called the Smith-Waterman algorithm

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35 Local Alignment New option: u We can start a new match instead of extending a previous alignment Alignment of empty suffixes

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36 Local Alignment Example s = TAATA t = TACTAA S T

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37 Local Alignment Example s = TAATA t = TACTAA S T

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38 Local Alignment Example s = TAATA t = TACTAA S T

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39 Local Alignment Example s = TAATA t = TACTAA S T

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40 Local Alignment Example s = TAATA t = TACTAA S T

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41 Variants of Sequence Alignment We have seen two variants of sequence alignment: u Global alignment u Local alignment Other variants in the book and in tutorial time: 1. Finding best overlap 2. Using an affine cost d(g) = -d –(g-1)e for gaps of length g. The –d is for introducing a gap and –e for continuing the gap. We used d=e=2. We could use smaller e. These variants are based on the same basic idea of dynamic programming.

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42 Remark: Edit Distance Instead of speaking about the score of an alignment, one often talks about an edit distance between two sequences, defined to be the “cost” of the “cheapest” set of edit operations needed to transform one sequence into the other. Cheapest operation is “no change” Next cheapest operation is “replace” The most expensive operation is “add space”. Our goal is now to minimize the cost of operations, which is exactly what we actually did.

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43 Where do scoring rules come from ? We have defined an additive scoring function by specifying a function ( , ) such that (x,y) is the score of replacing x by y (x,-) is the score of deleting x (-,x) is the score of inserting x But how do we come up with the “correct” score ? Answer: By encoding experience of what are similar sequences for the task at hand.

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44 Why use probability to define and/or interpret a scoring function ? Similarity is probabilistic in nature because biological changes like mutation, recombination, and selection, are not deterministic. We could answer questions such as: How probable two sequences are similar? Is the similarity found significant or random? How to change a similarity score when, say, mutation rate of a specific area on the chromosome becomes known ?

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45 A Probabilistic Model u For now, we will focus on alignment without indels. u For now, we assume each position (nucleotide /amino-acid) is independent of other positions. u We consider two options: M: the sequences are Matched (related) R: the sequences are Random (unrelated)

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46 Unrelated Sequences u Our random model of unrelated sequences is simple Each position is sampled independently from a distribution over the alphabet We assume there is a distribution q( ) that describes the probability of letters in such positions. u Then:

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47 Related Sequences We assume that each pair of aligned positions (s[i],t[i]) evolved from a common ancestor Let p(a,b) be a distribution over pairs of letters. p(a,b) is the probability that some ancestral letter evolved into this particular pair of letters.

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48 Odd-Ratio Test for Alignment If Q > 1, then the two strings s and t are more likely to be related (M) than unrelated (R). If Q < 1, then the two strings s and t are more likely to be unrelated (R) than related (M).

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49 Score(s[i],t[i]) Log Odd-Ratio Test for Alignment Taking logarithm of Q yields If log Q > 0, then s and t are more likely to be related. If log Q < 0, then they are more likely to be unrelated. How can we relate this quantity to a score function ?

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50 Probabilistic Interpretation of Scores u We define the scoring function via u Then, the score of an alignment is the log-ratio between the two models: Score > 0 Model is more likely Score < 0 Random is more likely

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51 Modeling Assumptions u It is important to note that this interpretation depends on our modeling assumption!! u For example, if we assume that the letter in each position depends on the letter in the preceding position, then the likelihood ratio will have a different form.

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52 Constructing Scoring Rules The formula suggests how to construct a scoring rule: Estimate p(·,·) and q(·) from the data Compute (a,b) based on the estimated p(·,·) and q(·) u How to estimate these parameters is the subject matter of parameter estimation in Statistics.

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. Sequence Alignment I Lecture #2 This class has been edited from Nir Friedman’s lecture. Changes made by Dan Geiger, then Shlomo Moran. Background Readings:

. Sequence Alignment I Lecture #2 This class has been edited from Nir Friedman’s lecture. Changes made by Dan Geiger, then Shlomo Moran. Background Readings:

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