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Sequence Alignment ILecture #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, and finally Benny Chor.
Background Readings: Gusfield, chapter 11. Durbin et. al., chapter 2.
2
Sequence Comparison
Much of bioinformatics involves sequences DNA sequences RNA sequences Protein sequences
We can think of these sequences as strings of letters
DNA & RNA: alphabet ∑of 4 letters (A,C,T/U,G) Protein: alphabet ∑ of 20 letters (A,R,N,D,C,Q,…)
3
Sequence Comparison (cont)
Finding similarity between sequences is important for many biological questions
For example: Find similar proteins
Allows to predict function & structure Locate similar subsequences in DNA
Allows to identify (e.g) regulatory elements Locate DNA sequences that might overlap
Helps in sequence assembly
g1
g2
4
Sequence Alignment
Input: two sequences over the same alphabet
Output: an alignment of the two sequences
Example: GCGCATGGATTGAGCGA TGCGCCATTGATGACCA
A possible alignment:
-GCGC-ATGGATTGAGCGA
TGCGCCATTGAT-GACC-A
5
Alignments
-GCGC-ATGGATTGAGCGA
TGCGCCATTGAT-GACC-A
Three “components”: Perfect matches Mismatches Insertions & deletions (indel)
Formal definition of alignment:
6
Choosing Alignments
There are many (how many?) possible alignments
For example, compare:
-GCGC-ATGGATTGAGCGA
TGCGCCATTGAT-GACC-A
to
------GCGCATGGATTGAGCGA
TGCGCC----ATTGATGACCA--
Which one is better?
7
Scoring Alignments
Motivation: Similar (“homologous”) sequences evolved from a
common ancestor In the course of evolution, the sequences changed
from the ancestral sequence by random mutations: Replacements: one letter changed to another Deletion: deletion of a letter Insertion: insertion of a letter
Scoring of sequence similarity should reflect how many and which operations took place
8
A Naive Scoring Rule
Each position scored independently, using: Match: +1 Mismatch : -1 Indel -2
Score of an alignment is sum of position scores
9
Example
-GCGC-ATGGATTGAGCGA
TGCGCCATTGAT-GACC-A
Score: (+1x13) + (-1x2) + (-2x4) = 3
------GCGCATGGATTGAGCGA
TGCGCC----ATTGATGACCA--
Score: (+1x5) + (-1x6) + (-2x11) = -23
according to this scoring, first alignment is better
10
More General Scores
The choice of +1,-1, and -2 scores is quite arbitrary 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 very
different properties Probabilistic interpretation: (e.g.) How likely is one
alignment versus another ?
11
Additive Scoring Rules
We define a scoring function by specifying
(x,y) is the score of replacing x by y (x,-) is the score of deleting x (-,x) is the score of inserting x
The score of an alignment is defined as the
sum of position scores
: ( { }) ( { })
12
The Optimal Score
The optimal (maximal) score between two sequences is the maximal score of all alignments of these sequences, namely,
Computing the maximal score or actually finding an alignment that yields the maximal score are closely related tasks with similar algorithms.
We now address these problems.
1 21 2 alignment of s &sV(s ,s ) max score(alignment)
13
Computing Optimal Score
How can we compute the optimal score ? If |s| = n and |t| = m, the number A(m,n) of
possible alignments is large!
Exercise: Show that
So it is not a good idea to go over all alignments The additive form of the score allows us to apply
dynamic programming to compute optimal score efficiently.
2
( , )m n m n
A m nm m
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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] )
(s[1..n 1], t[1..m 1]) (s[1.., n], t[1..m])
(s[n 1], t[m 1])
V V
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] )
(s[1..n 1], t[1..m 1]) (s[1.., n], t[1..m 1])
(s[n 1], )
V V
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] )
(s[1..n 1], t[1..m 1]) (s[1.., n 1], t[1..m]V )
( , t[m 1])
V
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Useful Notation
(ab)use of notation:
V[i,j] = value of optimal alignment between
i prefix of s and j prefix of t.
[i, j] (s[1..i],t[1.V )V .j]
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Recursive Argument
(ab)use of notation:
V[i,j] = value of optimal alignment between
i prefix of s and j prefix of t. Using our recursive argument, we get the following
recurrence for V:
[i, j] (s[i 1], t[ j 1])
[i 1, j 1] max [i, j 1] (s[i 1], )
V
V V
V[i 1, j] ( , t[ j 1])
[i, j] (s[1..i], t[1.V )V .j]
V[i,j] V[i+1,j]
V[i,j+1] V[i+1,j+1]
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Recursive Argument
Of course, we also need to handle the base cases in the recursion (boundary of matrix):
V[0,0] 0
V[i 1,0] V[i,0] (s[i 1], )
V[0, j 1] V[0, j] ( , t[ j 1])
0 A 1
G 2
C 3
0 0 -2 -4 -6
A 1 -2
A 2 -4
A 3 -6
C 4 -8
AA- -
We fill the “interior of matrix” using our recurrence rule
ST
vs.
20
Dynamic Programming Algorithm
We continue to fill the matrix using the recurrence rule
0
A 1
G 2
C 3
0 0 -2 -4 -6
A 1 -2
A 2 -4
A 3 -6
C 4 -8
ST
21
Dynamic Programming Algorithm
0
A 1
G 2
C 3
0 0 -2 -4 -6
A 1 -2 1
A 2 -4
A 3 -6
C 4 -8
V[0,0] V[0,1]
V[1,0] V[1,1]
+1-2 -A A-
-2 (A- versus -A)
versus
ST
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Dynamic Programming Algorithm
0
A 1
G 2
C 3
0 0 -2 -4 -6
A 1 -2 1 -1 -3
A 2 -4 -1 0
A 3 -6 -3
C 4 -8 -5
ST
(hey, what is the scoring function (x,y) ? )
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Dynamic Programming Algorithm
0
A 1
G 2
C 3
0 0 -2 -4 -6
A 1 -2 1 -1 -3
A 2 -4 -1 0 -2
A 3 -6 -3 -2 -1
C 4 -8 -5 -4 -1
Conclusion: V(AAAC,AGC) = -1
ST
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Reconstructing the Best Alignment
To reconstruct the best alignment, we record which case(s) in the recursive rule maximized the score
0A1
G2
C3
0 0 -2 -4 -6
A 1 -2 1 -1 -3
A 2 -4 -1 0 -2
A 3 -6 -3 -2 -1
C 4 -8 -5 -4 -1
ST
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Reconstructing the Best Alignment
We now trace back a path that corresponds to the best alignment
0A1
G2
C3
0 0 -2 -4 -6
A 1 -2 1 -1 -3
A 2 -4 -1 0 -2
A 3 -6 -3 -2 -1
C 4 -8 -5 -4 -1
AAACAG-C
ST
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Reconstructing the Best Alignment
More than one alignment could have the best score
(sometimes, even exponentially many)
0A1
G2
C3
0 0 -2 -4 -6
A 1 -2 1 -1 -3
A 2 -4 -1 0 -2
A 3 -6 -3 -2 -1
C 4 -8 -5 -4 -1
ST
AAACA-GC
AAAC-AGC
AAACAG-C
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Time Complexity
Space: O(mn)Time: O(mn) Filling the matrix O(mn) Backtrace O(m+n)
0A1
G2
C3
0 0 -2 -4 -6
A 1 -2 1 -1 -3
A 2 -4 -1 0 -2
A 3 -6 -3 -2 -1
C 4 -8 -5 -4 -1
ST
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Space Complexity
In real-life applications, n and m can be very large The space requirements of O(mn) can be fairly
demanding If m = n = 10,000, we need 100MB space If m = n = 100,000, we need 10GB space
We can afford to perform extra computation to save space Looping over million operations takes less than
seconds on modern workstations
Can we trade space with time?
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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
-1
-3
-5
-4
-1
0
-2
-4
-6
-3
-2
-1
-1
0A1
G2
C3
0
A 1
A 2
A 3
C 4
Note however that This “trick” fails if we want to
reconstruct the optimal alignment. Trace back information seemingly
requires keeping all back pointers, O(mn) memory.
Will get back to this.
To compute just the value V[n,m]=V(s[1..n],t[1..m]), we need only O(min(n,m)) space:
30
Local Alignment
The alignment version we studies so far is called
global alignment: We align the whole sequence s
to the whole sequence t.
Global alignment is appropriate when s,t are highly
similar (examples?), but makes little sense if they
are highly dissimilar. For example, when s (“the query”)
is very short, but t (“the database”) is very long.
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Local Alignment
When s and t are not necessarily similar, we may want to consider a different question:
Find similar subsequences of s and t Formally, given s[1..n] and t[1..m] find i,j,k, and l
such that V(s[i..j],t[k..l]) is maximal
This version is called local alignment.
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Local Alignment
As before, we use dynamic programming We now want to setV[i,j] to record the maximum
value over all alignments of a suffix of s[1..i] and a suffix of t[1..j] In other words, we look for a suffix of a prefix. How should we change the recurrence rule?
Same as before but with an option to start afresh
The result is called the Smith-Waterman algorithm, after its inventors (1981).
33
Local Alignment
New option: We can start a new match instead of extending a
previous alignment
V[i, j] (s[i 1], t[ j 1])
V[i 1, j 1] max V[i, j 1] (s[i 1], )
V[i 1, j] ( , t[ j 1])
0
Alignment of empty suffixesV[0,0] 0
V[i 1,0] max( , V[i,0] (s[i 1], ))
V[0, j 1] max( , V[0, j] ( , t[ j 1 )0 ]
0
)
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Local Alignment Example
0
A 1
T 2
C 3
T 4
A 5
A 6
0 0 0 0 0 0 0 0
T 1 0
A 2 0
A 3 0
T 4 0
A 5 0
s = TAATAt = TACTAA
ST
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Local Alignment Example
0
T 1
A 2
C 3
T 4
A 5
A 6
0 0 0 0 0 0 0 0
T 1 0 1 0 0 1 0 0
A 2 0 0 2 0 0 2 1
A 3 0
T 4 0
A 5 0
s = TAATAt = TACTAA
ST
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Local Alignment Example
0T1
A2
C3
T4
A5
A6
0 0 0 0 0 0 0 0
T 1 0 1 0 0 1 0 0
A 2 0 0 2 0 0 2 1
A 3 0 0 1 1 0 1 3
T 4 0 0 0 0 2 0 1
A 5 0 0 1 0 0 3 1
s = TAATAt = TACTAA
ST
What score should we take?1 (lower right corner); 3 (max); or something else?
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Local Alignment Example
0T1
A2
C3
T4
A5
A6
0 0 0 0 0 0 0 0
T 1 0 1 0 0 1 0 0
A 2 0 0 2 0 0 2 1
A 3 0 0 1 1 0 1 3
T 4 0 0 0 0 2 0 1
A 5 0 0 1 0 0 3 1
s = TAATAt = TACTAA
ST
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Local Alignment Example
0T1
A2
C3
T4
A5
A6
0 0 0 0 0 0 0 0
T 1 0 1 0 0 1 0 0
A 2 0 0 2 0 0 2 1
A 3 0 0 1 1 0 1 3
T 4 0 0 0 0 2 0 1
A 5 0 0 1 0 0 3 1
s = TAATAt = TACTAA
ST
