Lecture 4: September 19

Transcription

1 CSCI1810: Computational Molecular Biology Fall 2017 Lecture 4: September 19 Lecturer: Sorin Istrail Scribe: Cyrus Cousins Note: LaTeX template courtesy of UC Berkeley EECS dept. Disclaimer: These notes have not been subjected to the usual scrutiny reserved for formal publications. They may be distributed outside this class only with the permission of the Instructor. 4.1 Time Complexity of Global lignment In complexity analysis, the question we are asking is, how long does it take to run an algorithm? Four our purposes, assignment, comparisons, and arithmetic all take one unit of time. The time complexity of the Needleman-Wunsch algorithm is a function of the input sequences x and y. In this case, the same number of operations are required for any input sequences, so long as they are held constant, so all that matter are the lengths of these sequences, m and n. In the end, we will be performing asymptotic complexity analysis, which eliminates constant factors, so we re not too concerned with getting the exact number of each operation correct. Instead, we re concerned with how the number of operations changes as a function of m and n. Recall from the previous lecture, the pseudocode for the Needleman-Wunsch algorithm is as follows: 1: function Global lignment(x Σ m, y Σ n 2: S 0,0 0 3: for i {1, 2,..., m} do 4: S i,0 S i 1,0 + δ(x i,- 5: end for 6: for j {1, 2,..., n} do 7: S 0,j S 0,j 1 + δ(-, y i 8: for i {1, 2,..., m} do S i 1,j 1 + δ(x i, y i 9: S i,j max S i 1,j + δ(x i,- S i,j 1 + δ(-, y i 10: end for 11: end for 12: return S m,n 13: end function Line 2 requires 1 operation. Line 4 requires 3 operations, and is repeated m times. Line 7 requires 3 operations, and is repeated n times. Line 9 requires 9 operations, and is repeated mn times. Line 12 is a single operation. In order to simplify the analysis of this expression, we use what is called asymptotic notation, wherein constant factors do not matter. Specifically, we will use big-oh notation, which is used to take an upper 4-1

2 4-2 Lecture 4: September 19 bound on the asymptotic performance of an algorithm. By asymptotic performance, we mean that we we want to describe how an algorithm performs as the size of the input tends to infinity. Formally, we say that if m lim f(n, n,m g(n, m < then f(n, m O(n, m. Informally, this means that as the problem size increases (in our case, x and y get larger, the ratio between f(n, m and g(n, m converges to a constant. So in the end, we get that the time complexity of global alignment is as follows: 9mn + 3m + 3n + 2 O(9mn + 3n m = O(mn Usually, we can simply drop all but the highest order terms in an expression to obtain the simplest form of the big-oh complexity class. More formally, we can show this as follows: 9mn + 3n m lim n,m mn ( = lim 9 n,m = lim n,m m + 3 n + 1 mn ( ( lim + lim n,m m n,m n + ( lim n,m 1 mn = 9 Because this limit exists and is finite, we may conclude that 9mn + 3n m O(mn. 4.2 Local lignment Defining Local lignment So far, we have examined the problem of calculating the optimal alignment between strings x and y. We now study a related problem; that of local alignment. What if we have 2 strings, and we know that the majority of these strings differ greatly, but we want to identify the sections of each string that most strongly align to one another. motivating biological example would be if we had 2 distantly related genomes, and we wanted to identify genes that were highly conserved between the two. In this case, we can t expect the strings to align well globally, but if we can find two regions that are highly conserved between the strings, then we can expect these regions to align well. Before formalizing the above intuition and defining local alignment, we need a few definitions. prefix a of a string b is a string that may be obtained by removing characters from the end of b. Similarly, a suffix c of a string b may be obtained by removing characters from the beginning of b. Then, a substring

3 Lecture 4: September Prefixes, Suffixes, and Substrings of CTG: Prefixes: C CT CTG CTG Suffixes: G TG CTG CTG CTG Substrings: CTG CTG CT C CTG CT C CTG CT C TG T G Figure 4.1: Prefixes, Suffixes, and Subsequences or subsequence 1 a of a string b may be obtained by removing characters from either end of b (but not from the middle. See Figure 4.1 for some examples of prefixes, suffixes, and substrings. t this time, you should convince yourself that the set of all suffixes of prefixes of x is equivalent to the set of all prefixes of suffixes of x, and both are equivalent to the set of all substrings of x. We are now ready to define local alignment. The optimal local alignment of two strings x and y is simply the optimal global alignment of any x, y such that x is a substring of x and y is a substring of y. This definition is deceptively simple. Note that it encapsulates our intuition, allowing us to identify regions of two strings that align well, even when the remainder of the strings aligns poorly. Furthermore, we shall soon see an efficient algorithm (Smith-Waterman exists to calculate optimal local alignments that is asymptotically equivalent to the Needleman-Wunsch algorithm for global alignment Naïvely Computing Optimal Local lignments With the definition provided above, we have all the tools we need to calculate local alignments. We simply enumerate over all pairs of substrings of x and y, and evaluate the global alignment score of each, and in the end take the maximum over all such scores. So we could just apply our global alignment algorithm for every pair of subsequences. But how many pairs of subsequences are there, and how long would this take? simple counting argument 2 tells us that there are no more than ( ( n unique subsequences of a string of length n, and thus n+1 ( m pairs of substrings of x and y. Enumerating them, and running a O(nm algorithm on each pair thus runs in O(n 3 m 3 time. This is asymptotically inferior to the Smith-Waterman algorithm, and is prohibitively expensive for aligning long sequences. In the next section, we come up with a more sophisticated algorithm that efficiently computes local alignments. 1 Note that some authors define subsequence such that characters to be removed from the middle of a sequence; This definition is not of interest to us in this context. The term substring is generally less ambiguous. 2 Either we may select the empty subsequence, or we may select two distinct values a, b in {0, 1,..., n}, and take the substring x a, x a+1,..., x b.

4 4-4 Lecture 4: September Computing Optimal Local lignments Recall that the optimal substructure property of global alignment ensures that S i,j is the score of the optimal alignment of the first i characters of x and the first j characters of y. In other words, by taking the maximum over S, we can calculate the optimal alignment score over all prefixes of x and y. Recall that substrings are suffixes of prefixes, and note that we re already halfway to a solution. In order to handle suffixes, we introduce another trick. s we align the strings x and y, we need to consider that we can start the alignment at any character of x and any character of y, as this is equivalent to choosing suffixes. Now, suppose we have suffixes x b, y b such that x = x a x b and y = y a y b, where denotes concatenation. If the alignment score of x b and y b can be improved by extending the suffixes, then we should do so, but if it can not, then we should not. We can represent this resetting by adding a 0 term to the maximum in our recurrence relationship. By doing this, we ensure that we start a new prefix when doing so yields a higher alignment score than not doing so. Combining these two techniques, by adding a 0 term to the maximum of the recurrence relationship, and identifying the highest scoring cell in the entire matrix, we identify the optimal suffix of the optimal prefix, or equivalently the optimal substring. Recall that the recurrence relationship for local alignment is as follows: V 0,0 = 0 0 V V i,j = max i 1,j 1 + δ(x i, y j V i 1,j + δ(x i, - V i,j 1 + δ(-, y i We now translate this recurrence into pseudocode to calculate the optimal local alignment score between strings x and y. Local Pairwise lignment Given: 1. Sequence x Σ m. 2. Sequence y Σ n. 3. Similarity matrix δ. Produce: The optimal global alignment similarity score over any strings x, y where x is a substring of x and y a substring of y. 1: function Local lignment(x Σ m, y Σ n 2: for i {0, 1, 2,..., m} do 3: V i,0 0 4: end for 5: for j {1, 2,..., n} do 6: V 0,j 0 7: end for 8: for j {1, 2,..., n} do

5 Lecture 4: September : for i {1, 2,..., m} do 0 V 10: V i,j max i 1,j 1 + δ(x i, y i V i 1,j + δ(x i, V i,j 1 + δ(, y i 11: end for 12: end for 13: return max V i,j i {0, 1,..., m} j {0, 1,..., n} 14: end function Note that this algorithm differs from local alignment in only a few ways. V i,0 and V 0,j are now initialized to 0 (this follows from the assumption that δ(x,- and δ(-, x are negative, which is a bit simpler, the maximum on line now includes a 0, requiring one additional comparison, and on line 13, we need to take the maximum score over the entire matrix, rather than just returning V m,n Local lignment and Scoring Functions So far, when discussing global alignment we have mostly been considering the unit cost function. In global alignment, the actual cost values of a scoring matrix can be adjusted by adding a to each match/mismatch and a 2 to each gap (you should convince yourself that any alignment over x, y of lengths m, n changes score by exactly a(m+n 2, and thus that the relative scores between alignments don t change. In local alignment, this is not the case: notably, if all entries of δ are positive, the optimal local alignment will be the optimal global alignment (as the 0 term of the maximum is never taken, and scores increase monotonically as i, j increase, thus the maximum score will occur at cell V i,j, and if all entries of δ are negative, the optimal local alignment will be the empty alignment. Furthermore, we also require that gap costs are nonpositive, otherwise the initialization as given above is incorrect. reasonable way to handle this situation is to set δ i,j = 1 for all gaps and mismatches, and 1 for matches.