SE701:Global Sequence Alignment

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http://www.cs.princeton.edu/courses/archive/fall06/cos126/assignments/sequence.html

The problem. Through your research you have found the following sequence of a gene in a previously unstudied organism.

   A A C A G T T A C C

What is the function of the protein that this gene encodes? You could begin a series of uninformed experiments in the lab to determine what role this gene plays. However, there is a good chance that it is a variant of a known gene in a previously studied organism. Since biologists and computer scientists have laboriously determined (and published) the genetic sequence of many organisms (including humans), you would like to leverage this information to your advantage. We'll compare the above genetic sequence with one which has already been sequenced and whose function is well understood.

   T A A G G T C A

If the two genetic sequences are similar enough, we might expect them to have similar functions. We would like a way to quantify "similar enough."

Edit-distance. In this assignment we will measure the similarity of two genetic sequences by their edit distance, a concept first introduced in the context of coding theory, but which is now widely used in spell checking, speech recognition, plagiarism detection, file revisioning, and computational linguistics. We align the two sequences, but we are permitted to insert gaps in either sequence (e.g., to make them have the same length). We pay a penalty for each gap that we insert and also for each pair of characters that mismatch in the final alignment. Intuitively, these penalties model the relative likeliness of point mutations arising from deletion/insertion and substitution. We produce a numerical score according to the following simple rule, which is widely used in biological applications:

   Penalty 	Cost
   Per gap 	  2  
   Per mismatch 	1
   Per match 	0

As an example, two possible alignments of AACAGTTACC and TAAGGTCA are:

    Sequence 1   	 A  	 A  	 C  	 A  	 G  	 T  	 T  	 A  	 C  	 C 
    Sequence 2   	T 	A 	A 	G 	G 	T 	C 	A 	- 	-
    Penalty   	1 	0 	1 	1 	0 	0 	1 	0 	2 	2
   	        	
    Sequence 1   	 A  	 A  	 C  	 A  	 G  	 T  	 T  	 A  	 C  	 C 
    Sequence 2   	T 	A 	- 	A 	G 	G 	T 	- 	C 	A
    Penalty   	1 	0 	2 	0 	0 	1 	0 	2 	0 	1

The first alignment has a score of 8, while the second one has a score of 7. The edit-distance is the score of the best possible alignment between the two genetic sequences over all possible alignments. In this example, the second alignment is in fact optimal, so the edit-distance between the two strings is 7. Computing the edit-distance is a nontrivial computational problem because we must find the best alignment among exponentially many possibilities. For example, if both strings are 100 characters long, then there are more than 10^75 possible alignments.

A recursive solution. We will calculate the edit-distance between the two original strings x and y by solving many edit-distance problems on the suffixes of the two strings. We use the notation x[i] to refer to character i of the string. We also use the notation x[i..M] to refer to the suffix of x consisting of the characters x[i], x[i+1], ..., x[M-1]. Finally, we use the notation opt[i][j] to denote the edit distance of x[i..M] and y[j..N]. For example, consider the two strings x = "AACAGTTACC" and y = "TAAGGTCA" of length M = 10 and N = 8, respectively. Then, x[2] is 'C', x[2..M] is "CAGTTACC", and y[8..N] is the empty string. The edit distance of x and y is opt[0][0].

Now we describe a recursive scheme for computing the edit distance of x[i..M] and y[j..N]. Consider the first pair of characters in an optimal alignment of x[i..M] with y[j..N]. There are three possibilities:

  1. The optimal alignment matches x[i] up with y[j]. In this case, we pay a penalty of either 0 or 1, depending on whether x[i] equals y[j], plus we still need to align x[i+1..M] with y[j+1..N]. What is the best way to do this? This subproblem is exactly the same as the original sequence alignment problem, except that the two inputs are each suffixes of the original inputs. Using our notation, this quantity is opt[i+1][j+1].
  2. The optimal alignment matches the x[i] up with a gap. In this case, we pay a penalty of 2 for a gap and still need to align x[i+1..M] with y[j..N]. This subproblem is identical to the original sequence alignment problem, except that the first input is a proper suffix of the original input.
  3. The optimal alignment matches the y[j] up with a gap. In this case, we pay a penalty of 2 for a gap and still need to align x[i..M] with y[j+1..N]. This subproblem is identical to the original sequence alignment problem, except that the second input is a proper suffix of the original input. 

The key observation is that all of the resulting subproblems are sequence alignment problem on suffixes of the original inputs. To summarize, we can compute opt[i][j] by taking the minimum of three quantities:

   opt[i][j] = min { opt[i+1][j+1] + 0/1, opt[i+1][j] + 2, opt[i][j+1] + 2 }

This equation works assuming i < M and j < N. Aligning an empty string with another string of length k requires inserting k gaps, for a total cost of 2k. Thus, in general we should set opt[M][j] = 2(N-j) and opt[i][N] = 2(M-i). For our example, the final matrix is:

          |  0  1  2  3  4  5  6  7  8
      x\y |  T  A  A  G  G  T  C  A  -
   -----------------------------------
    0  A  |  7  8 10 12 13 15 16 18 20
    1  A  |  6  6  8 10 11 13 14 16 18
    2  C  |  6  5  6  8  9 11 12 14 16
    3  A  |  7  5  4  6  7  9 11 12 14
    4  G  |  9  7  5  4  5  7  9 10 12
    5  T  |  8  8  6  4  4  5  7  8 10
    6  T  |  9  8  7  5  3  3  5  6  8
    7  A  | 11  9  7  6  4  2  3  4  6
    8  C  | 13 11  9  7  5  3  1  3  4
    9  C  | 14 12 10  8  6  4  2  1  2
   10  -  | 16 14 12 10  8  6  4  2  0

By examining opt[0][0], we conclude that the edit distance of x and y is 7.

A dynamic programming approach. A direct implementation of the above recursive scheme will work, but it is spectacularly inefficient. If both input strings have N characters, then the number of recursive calls will exceed 2^N. To overcome this performance bug, we use dynamic programming. (Read the first section of Section 9.6 for an introduction to this technique.) Dynamic programming is a powerful algorithmic paradigm, first introduced by Bellman in the context of operations research, and then applied to the alignment of biological sequences by Needleman and Wunsch. Dynamic programming now plays the leading role in many computational problems, including control theory, financial engineering, and bioinformatics, including BLAST (the sequence alignment program almost universally used by molecular biologist in their experimental work). The key idea of dynamic programming is to break up a large computational problem into smaller subproblems, store the answers to those smaller subproblems, and, eventually, use the stored answers to solve the original problem. This avoids recomputing the same quantity over and over again. Instead of using recursion, use a nested loop that calculates opt[i][j] in the right order so that opt[i+1][j+1], opt[i][j+1], and opt[i+1][j+1] are all computed before we try to compute opt[i][j].

Recovering the alignment itself. The above procedure describes how to compute the edit distance between two strings. We now outline how to recover the optimal alignment itself. The key idea is to retrace the steps of the dynamic programming algorithm backwards, re-discovering the path of choices (highlighted in red in the table above) from opt[0][0] to opt[M][N]. To determine the choice that led to opt[i][j], we consider the three possibilities:

  1. The optimal alignment matches x[i] up with y[j]. In this case, we must have opt[i][j] = opt[i+1][j+1] if x[i] equals y[j], or opt[i][j] = opt[i+1][j+1] + 1 otherwise.
  2. The optimal alignment matches x[i] up with a gap. In this case, we must have opt[i][j] = opt[i+1][j] + 2.
  3. The optimal alignment matches y[j] up with a gap. In this case, we must have opt[i][j] = opt[i][j+1] + 2. 

Depending on which of the three cases apply, we move down, right, or diagonally towards opt[M][N], printing out x[i] aligned with y[j] (case 1), x[i] aligned with a gap (case 2), or y[j] aligned with a gap (case 3). In the example above, we know that the first T aligns with the first A because opt[0][0] = opt[1][1] + 1, but opt[0][0] ≠ opt[1][0] + 2 and opt[0][0] ≠ opt[0][1] + 2. The optimal alignment is the second candidate alignment in the edit-distance section.