Computational Molecular Biology

Transcription

1 Computational Molecular Biology Erwin M. Bakker Lecture 2 Materials used from R. Shamir [2] and H.J. Hoogeboom [4]. 1 Molecular Biology Sequences DNA A, T, C, G RNA A, U, C, G Protein A, R, D, N, C E, Q, G, H, I L, K, M, F, P S, T, W, Y, V 2 1

2 Central Dogma of Molecular Biology From Sequence to Function A, T, C, G DNA Transcription and Splicing A, U, C, G RNA Translation A, R, D, N, C E, Q, G, H, I L, K, M, F, P S, T, W, Y, V 3 Protein Biological Motivation for Pairwise Sequence Alignment Storing, retrieving and comparing molecular biology sequences in databases Comparing two or more sequences for similarities Reconstructing long sequences of DNA from overlapping sequence fragments Physical and genetic mapping from probe data Detecting and exploring frequently occurring patterns in sequences Detecting informative areas in protein and DNA sequences Etc. 4 2

3 Sequence Similarity DNA Resemblance Common ancestral origins Evolution from mutations Local modifications Insertion Deletion Substitution Reciprocity: If a sequence S becomes equal to a sequence T after an insertion, then T becomes equal to S after a deletion. 5 => insertion/deletion is called indel. Sequence Distance Definition: The edit distance d(s,t) between two sequences S and T is equal to the minimal number of edit operations needed to transform one sequence into the other. Edit operations: Insertion Deletion Substitution 6 3

4 Sequence Distance Example: Edit Distance d(s,t) Given two sequences S: ACTGGAC, and T: ACCGGA S: ACTGGAC -> (delete C ) ACTGGA -> (substitute C for T ) T: ACCGGA The minimal number of edit operations necessary to transform string S into string T is equal to 2, i.e., d(s,t) = 2 7 Alignment S: GTAGTACAGCTCAGTTGGGATCACAGGCTTCT T: GTAGAACGGCTTCAGTTGTCACAGCGTTC S : T : GTAGTACAGCT CAGTTGGGATCACAGGCTTCT GTAGAACGGCTTCAGTTG TCACAGCGTTC Definition An alignment of two sequences S and T is obtained by inserting spaces at the beginning, end or into S and T, such that: 8 The resulting two sequences S and T contain the same number of characters, i.e., S = T. The characters of S and T are placed in a 1 to 1 correspondence. i.e., each i th character S i of S corresponds to the i th character T i of T, where i ϵ {1,, S }. 4

5 Some Alignment Examples A G T C A s A G A C A DNA sequences. A G T C A d A G - C A A G T - C A A G T A C A i K S Q E T K S Q E T K S Q E - T s d i K V Q E T K - Q E T K S Q E V T Protein sequences: Peptide or amino acid sequences. 9 s = substitution; d = deletion; i = insertion; Scoring an Alignment S: GTAGTACAGCT CAGTTGGGATCACAGGCTTCT T: GTAGAACGGCTTCAGTTG TCACAGCGTTC By Defining Distances Match = 0, substitution = 1, indel = 2 => distance(s,t) = 14 (4x1 + 5x2) 10 Match = 0, d(a,t) = d(g,c)=1, d(a,g)=d(g,t)=1.5, indel = 2 => distance(s,t) = 14.5 By Defining Similarity Match = 1, substitution = 0, indel = -1.5 => similarity(s,t) = 15.5 ( = 16.5) Substitution/scoring matrix s(i,, indel equals to s(i,-), or s(-, 5

6 Scoring an Alignment 11 Models for Alignment (1970) (1981) 12 Ends Free Alignment 6

7 Models for Alignment Global Alignment Input: Two sequences S and T of almost equal length. Question: What is the maximum similarity between S and T? Find the best alignment of S and T. Local Alignment Input: Two sequences S and T. Question: What is the maximum similarity between a subsequence of S and a subsequence of T? Find the most similar subsequences. Ends free alignment Input: Two sequences S and T. Question: Find a best alignment between subsequences of S and T when at least one of these subsequences is a prefix of the original sequence and one is a suffix. (alignment between the endpoints of the original sequences) 13 Models for Alignment Definition: A gap is a maximal contiguous run of spaces in a single sequence within a given alignment. The length of a gap is the number of indel operations on it. A gap penalty function is a function that measures the cost of a gap as a function of its length. Alignment with Gap Penalty Input: Two sequences S and T. Question: Find a best alignment between the two sequences using the gap penalty function. 14 7

8 Global Alignment Input: S: ACGCTTTG T: CATGTAT Alignments: S : T : S : T : AC GCTTTG CATG TAT ACGCTTTG CATG TAT Homework: How many are there? S : T : ACGCTTTG CATGTAT 15 Global Alignment Input: Two sequences of length n,m ϵ N S = S 1,,S n and T = T 1,,T m Question: What is the maximum similarity between S and T? Find an optimal alignment of S and T. 16 8

9 Global Alignment Needleman-Wunsch (1970) Lemma: Let A(i, be the optimal alignment score of S 1 i and T 1 j, where 0 i n, and 0 j m, then: A( i,0) A(0, i k0 ( S, ) j k0 k (, T ) k A( i 1, j 1) ( Si, T j ) A( i, max A( i 1, ( Si, ) A( i, j 1) (, T ) j, for 1 i n and 1 j m Where σ(a,b) equals the score of the alignment of character a with character b (including spaces - ). 17 Global Alignment Needleman-Wunsch (1970) Lemma: Let A(i, be the optimal alignment score of S 1 i and T 1 j, where 0 i n, and 0 j m, then: A( i,0) A(0, i k0 ( Sk, ) A(i-1,j-1) A(i-1, j k0 (, T ) k A( i 1, j 1) ( Si, T j ) A( i, max A( i 1, ( Si, ) A( i, j 1) (, T ) j A(i, j-1), for 1 i n and 1 j m A(i, Where σ(a,b) equals the score of the alignment of character a with character b (including spaces - ). 18 9

10 Global Alignment A( i,0) A(0, Proof: k0 Initial condition: Matching the first i elements of S with 0 elements of T is done by matching the first i elements of S with i spaces in T, the cost clearly equals A(i, 0) as defined. Similarly, a cost equal to A(0, follows from matching the first j elements of T with j spaces in S. i k0 j ( S, ) k (, T ) k For example: S: ACGCTTTGTCTCTGTG S: ACGCTTTG T: CATGTATG T: CATGTATGTACTGTAC position i position j 19 Global Alignment A( i 1, j 1) ( Si, T j ) A( i, max A( i 1, ( Si, ) Proof (cont d): A( i, j 1) (, T j ) Consider an optimal alignment S 1 i and T 1 j. Then there are 3 cases: 1. S i aligned with T i, then the score is equal to σ(s i, T j ) + score for the optimal alignment of S 1 i-1 with T 1 j-1. This clearly equals A(i-1,j-1) + σ(s i,t j ). 2. S i aligned with - in T, then the score is equal to σ(s i, - ) + score for the optimal alignment of S 1 i-1 with T 1 j. This clearly equals A(i-1, + σ(s i, - ). 3. T j aligned with - in S, then the score is equal to σ( - T j ) + score for the optimal alignment of S 1 i with T 1 j-1. This clearly equals A(i,j-1) + σ( -,T j )

11 Global Alignment A(i-1,j-1) A(i, j-1) A(i-1, A(i, Algorithm to compute A(i, for all i, j Initialize: A(0,.), and A(.,0) for i=1 to n do { for j=1 to m do // Note only 2 rows are necessary here { Calculate A(i, using A(i-1,j-1), A(i,j-1), A(i-1, } } 21 Recursion 22 11

12 Recursion (explosion) a a a 0 1 a a, n n1 n n1 1 Calculate a 6 Answer = a 6 8 Fibonacci 3 3 a a a 4 a a a a 3 a a 2 a 1 a 2 a 1 a 1 a 0 1 a 2 0 a 1 a 1 a 0 a 1 a 0 a 1 a 0 a 1 a 0 23 Recursion Dynamic Programming a a a a a, n n1 n n1 1 a 1 a 2 a 3 a 4 a 5 a 6 a 7 a 8 a 9 a Dynamic Programming Memorization Store intermediate results in a table Determine intermediate results bottom-up 24 12

13 Global Alignment A(i-1,j-1) A(i, j-1) A(i-1, A(i, Algorithm to compute A(i, for all i, j Initialize: A(0,.), and A(.,0) for i=1 to n do { for j=1 to m do // Note only 2 rows are necessary here { Calculate A(i, using A(i-1,j-1), A(i,j-1), A(i-1, } } 25 Global Alignment Dynamic Programming T S 1 5 T 1..3 = S Optimal alignment score 26 13

14 Global Alignment Dynamic Programming T S 1 5 T 1..3 = S S 1 6 T 1..5 = 27 Global Alignment Needleman-Wunsch (1970) Lemma: Let A(i, be the optimal alignment score of S 1 i and T 1 j, where 0 i n, and 0 j m, then: A( i,0) A(0, i k0 ( S, ) j k0 k (, T ) k A( i 1, j 1) ( Si, T j ) A( i, max A( i 1, ( Si, ) A( i, j 1) (, T ) j, for 1 i n and 1 j m Where σ(a,b) equals the score of the alignment of character a with character b (including spaces - )

15 Global Alignment Dynamic Programming Calculate the initial conditions T T 1..3 = S 29 Global Alignment Dynamic Programming Sequence T Sequence S Assume gap cost: -1 Match t with t gives: = -2 = 1 = -3 Sequence S Sequence T 15

16 Global Alignment Sequence T Sequence S Assume gap cost: -1 Match t with t gives: = -2 = 1 = -3 Sequence S Sequence T Global Alignment Maintain trace while calculating the optimal alignment scores. T S 32 16

17 Sequence S Global Alignment Dynamic Programming - c Sequence T a a c t g g Sequence T c t t t Sequence S 33 Trace back to obtain an optimal global alignment. Note that, here three such optimal global alignments exist. Homework: Implement yourself in C++. Global Alignment (Eddy S.R. 2004a) 34 17

18 Global Alignment Complexity Time Complexity: O(nm) If only the value A(S,T) is required: Space Complexity: O(n+m) Can you do better? If the alignment has to be constructed: Space Complexity: O(nm) 35 Global Alignment Complexity Time Complexity: O(nm) If only the value A(S,T) is required: Space Complexity: O(min(n,m)) better If the alignment has to be constructed: Space Complexity: O(nm) 36 18

19 Alignment Graph Definition: Given two sequences S and T of lengths n and m respectively. An alignment graph is a directed graph G=(V,E) on (n+1) x (m+1) nodes, each labeled with a distinct pair (i,, where 0 i n, and 0 j m, with the following weighted edges: ( (i,, (i+1, ) with weight σ(s i+1,-) ( (i,, (i,j+1) ) with weight σ(-, T j+1 ) ( (i,, (i+1,j+1) ) with weight σ(s i+1,t j+1 ) 37 Alignment Graph (i, σ(-, T j+1 ) (i,j+1) (0,0) σ(s i+1,-) σ(s i+1,t j+1 ) (i+1, (i+1,j+1) 38 (n,m) 19

20 Alignment Graph A path P from node (0,0) to (n,m) in the alignment graph G corresponds to an alignment of sequence S with sequence T. Where the weight of the path P equals the alignment score. The global alignment problem thus translates to the following graph problem: Find the heaviest path P from node (0,0) to node (n,m) in G. 39 Global Alignment in Linear Space D.S. Hirschberg, 1977 D.S. Hirschberg, Algorithms for the longest common subsequence problem. J.ACM, 24: , A(i, denoted the score of an optimal alignment of the first i characters of S against the first j characters in T. Let A r (i, denote an optimal alignment of the last i characters of S against the last j characters in T. Lemma: A(n,m) = max 0 k m { A(n/2,k) + A r (n/2, m-k) } 40 20

21 Global Alignment in Linear Space D.S. Hirschberg, 1977 Lemma: A(n,m) = max 0 k m { A(n/2,k) + A r (n/2, m-k) } Sequence T (0,0) A k row n/2 of A r row n/2 of A Sequence S n/2 A r 41 (n,m) Global Alignment in Linear Space D.S. Hirschberg, 1977 The Algorithm 1. Compute A(S,T) while saving the n/2-th row. We denote A(S,T) as the Forward Matrix F. 2. Compute A r (S,T) while saving the n/2-th row. We denote A r (S,T) as the Backward Matrix B. 3. Find the column k * so that the crossing point (n/2,k * ) satisfies F(n/2, k*) + B(n/2, m-k*) = F(n,m). (Here the stored rows are used.) 4. Now that k * is found, recursively partition the problem to two sub problems: i. Find the path from (0,0) to (n/2,k * ). ii. Find the path from (n,m) to (n/2, m-k * )

22 Global Alignment in Linear Space D.S. Hirschberg, 1977 Lemma: The time complexity T of Hirschberg s algorithm is O(nm) Proof: Let T*(n,m) be the time to find only the value of an n x m problem. Let T(n,m) be the time to find the path (solution) of an n x m problem using Hirschberg s algorithm. T(n,m) = 2T * (n,m) + T(n/2,k * ) + T(n/2,m-k * ) = 2T * (n,m) + 2T*(n/2,k * ) + 2T * (n/2,m-k * ) + = 2T*(n,m) + 2T*(n/2,m) + 2T*(n/4,m) + 2T*(n/8,m) + Since T * (n,m) cnm for some c, and 1+1/2+1/4+1/8+ <=2, it follows that T(n,m) 4T*(n,m) <= 4cnm. Hence the time complexity is still O(nm) 43 Global Alignment in Linear Space D.S. Hirschberg, 1977 Lemma: The space complexity of Hirshberg s algorithm is O( min(n,m) ) Proof: Each dynamic programming computation requires storing one additional row (the middle on) for determining k*, which can be discarded once the middle point k* is found. If n<m we can store the middle column instead. Therefore the space complexity is O( min(n,m) ) The answer needs space complexity O(n+m)

23 Local Alignment Example S: G G T C T G A G T: A A A C G A Match = 2, indel/substitution = -1 Best local alignment S: G G T C T G A G T: A A A C _ G A 45 Local Alignment Motivation (1/2) Coding and non-coding regions of DNA Mutations in non-coding regions (introns) are expected to be more likely than mutations in coding regions (exons). As mutations in exons will have a direct impact on the organism. Therefore a best match between two stretches of DNA from different species is most likely between 2 exons (i.e., subsequences)

24 Local Alignment Motivation (2/2) Protein Domains Different kind of proteins and proteins of different species often show local similarities, so called homeoboxes (most probably functional subunits). 47 Local Alignment Local Alignment Problem Given two sequences S and T, find subsequences s of S and t of T whose similarity is maximal over all pairs of subsequences of S and T. Note that, a subsequence here is a contiguous subsequence

25 Local Alignment Local Suffix Alignment Definition Given sequences S and T, and indices i and j, the local suffix alignment problem is finding a (possibly empty) suffix s of S 1 i and a (possibly empty) suffix t of T 1 j such that the score of the alignment of s and t is maximal over all alignments of suffixes of S 1 i and T 1 j. S T s t i j Remark: The solution of the local alignment problem is the same as the maximal solution to the local suffix alignment problem over all i and j. 49 Local Alignment Algorithm Let A(i, the value of the optimal local suffix alignment for a given pair i,j of indices Let the weights be limited to σ(x,y) 0, if x, y match, and σ(x,y) 0, if x, y do not match or one of them is equal to a space Note: the maximal A(i, over all i, j is the value we are looking for

26 Local Alignment Algorithm Algorithm Sketch Compute the local suffix alignment (for all i and of S i = S 1...i and T j = T 1 j. By using the global alignment algorithm where the prefixes of S and T whose alignments are 0 are discarded, i.e., subsequences may start from indices 1. Search the results and find the indices i* and j* of S and T respectively, after which the similarity (obtained by local suffix alignment) only decreases. 51 Local Alignment Algorithm Let A(i, be the optimal local suffix alignment score of S 1 i and T 1 j, where 0 i n, and 0 j m, then: i, j : A( i,0) 0, A(0, 0 0 A( i 1, j 1) ( S, ) i T j A( i, max A( i 1, ( Si, ) A( i, j 1) (, T j ), for 1 i n and 1 j m Compute i* and j* such that A(i*,j*) = max 1 i n, 1 j m A(i,. This value is the optimal local alignment score

27 Local Alignment Obtain optimal local alignment sequences by backtracking 53 Local Alignment Obtain optimal local alignment sequences by backtracking 54 27

28 Local Alignment Complexity Lemma: Local alignment can be solved in linear space The optimal local alignment identifies the subsequences s and t whose global alignment is optimal over all pairs of subsequences. Hirschberg s method for global alignment can then be used to find the actual alignment of subsequences s and t: Using the recursion i* and j* can be calculated using a row or column only. Hence the end points (i*,j*) can be computed in linear space. Finding the start positions can be done using reverse dynamic programming starting in (i*,j*). 55 Local Alignment Complexity Time complexity O(nm) Space complexity O(min(n,m)) Note: answer requires space O(n+m) 56 28

29 End-Space Free Alignment Example A T C G G C T A C C G A G T A C T A C G A G C T A A T C A C T A A T C G A G C T A C T 57 End-Space Free Alignment Motivation Shotgun Sequence Assembly A large number of partially overlapping sequences coming from many copies of one original but unknown DNA sequence R has to be searched for pairs of overlapping subsequences in order to reconstruct the original DNA sequence. Two subsequences from different parts of R will have a low global alignment score as well as a low end-space free alignment score. Two overlapping subsequences from the same part of R will still have a low global alignment score but a high end-space free alignment score

30 End-Space Free Alignment Example 59 End-Space Free Alignment End-Space Free Alignment Problem Input: Two sequences S and T. Question: Find a best alignment between subsequences of S and T when at least one of these subsequences is a prefix of the original sequence and one (not necessarily the other, i.e, complete overlap is possible) is a suffix. Hereby costs of indels at the end or beginning of the sequences are not counted

31 End-Space Free Algorithm Initial Conditions Set the initial conditions to allow zero weight to leading indel operations in (at most) one of the sequences. Compute Optimal Value Fill the table with the values of A(i, (as before). Then search for the maximal value in either of the ending rows, thus allowing (at most) one sequence to end before the other, with zero weight for all indel operations from there on. This value is the best value. Determine Sequence The aligned sequence is tracked from cell (0,0) in the table until the end of one sequence (bottom row /right most column). From there on, all indel operations until cell (n,m) are not counted in the total value (though they are present in the table). 61 End-Space Free Alignment Define the End-Space Free Alignment score as A(S,T), then: i, j : A( i 1, j 1) ( Si, T j ) A( i, max A( i 1, ( Si, ) A( i, j 1) (, T ) 62 A( i,0) 0, A(0, 0 * Search for i such that : * Search for j such that : Define alignment score : j * A( i, m) max * A( n, j ) max * A( n, j ) A( S, T ) max * A( i, m), for 1 i n and 1 1in, m n,1 jn A( i, A( i, j m 31

32 End-Space Free Alignment S ends before T, 63 End-Space Free Alignment j* j* is where S ends before T i* is where T ends before S i* 64 32

33 Gap Penalty Definition: A gap is a maximal, consecutive run of spaces in a single sequence of a given alignment. Definition: The length of a gap is the number of indel operations. 65 Gap Penalty Motivation DNA Sequences Insertion or deletion of an entire subsequence often occurs as a single mutational event. A set of these events can create many gaps of varying sizes. Protein Sequences Two protein sequences may be similar except for some subunits that exist in the one but not the other

34 Gap Penalty Motivation cdna matching DNA transcribes to pre-mrna the complement of the gene s DNA (with introns and exons). After splicing mrna (only transcribed exons) results. When mrna is captured from the cell, so called cdna can be transcribed which has to be matched with the DNA in order to find the gene from which it originally resulted. cdna does not contain the gaps that the original DNA exhibits because of the intron regions. 67 Gap Penalty Constant Gap Penalty g(k) = k*g Affine Gap Penalty g(k) = k*g + s Convex Gap Penalty each additional gap contributes less to the gap than the previous space. General Gap Penalty g(k) arbitrary 68 34

35 Matrix B gap in T Matrix A no gap Matrix C gap in S Algorithm for Sequence Alignment with Affine Gap Penalty Model Using 3 Matrices for tracking the gap penalties: B: S i T -----j A: S i T j C: S -----i T j V taken the max over the three matrices. 69 Algorithm for Sequence Alignment with Affine Gap Penalty Model B gap in T A no gap C gap in S <- new gap, from previous match <- new gap in T, from gap in S Still a time complexity of 70 for a 35

36 References [1] R. Shamir, Algorithms in Molecular Biology (2009 version), [2] R. Shamir, Pairwise Alignment, Scribe from Lecture, [3] A.P. Gultyaev, Lecture Notes, [4] H.J. Hoogeboom, Lecture Slides,