Notes on Sequence Alignment - Bioinformatics | CAP 5510, Study notes of Computer Science

Download Notes on Sequence Alignment - Bioinformatics | CAP 5510 and more Study notes Computer Science in PDF only on Docsity! 8/20/2005 Su-Shing Chen, CISE 1 CAP 5510-5 Sequence Alignment BIOINFORMATICS Su-Shing Chen CISE 8/20/2005 Su-Shing Chen, CISE 2 Sequence Alignment
Through the generations of a population, a sequence changes in mutation.
Simple mutations are switching from one nucleotide to another (substitution), insertion and deletion.
In natural selection or random drift, mutations spread to an entire species.
At a site, in a population, there is no single nucleotide type – polymorphism. 8/20/2005 Su-Shing Chen, CISE 5 Edit Distance
The sum of substitution, deletion and insertion from one aligned to another.
CGGGTA - - TCCAA
CCC – TAGGTCCCA
0 s s d 0 0 I I 0 0 0 s 0 = 6
A C D E H G H K K L M N P Q R S T V W Y
A C C G H G - - K L M N P Q R S T V W Y
0 0 s s 0 0 d d 0 0 0 0 0 0 0 0 0 0 0 0 = 4 8/20/2005 Su-Shing Chen, CISE 6 Alignment Algorithms for Two Sequences
A simple scoring scheme = (the number of matches) – (the number of mismatches and indels).
Scoring schemes are similar to the simple scheme based on statistical analysis of existing data.
Protein sequence scorings are more complicated. 8/20/2005 Su-Shing Chen, CISE 7 Optimal Scores
Best scores indicate relatedness under evolution models.
Two sequences related share common patterns useful to find functional domains.
Comparing scores among several different species for the phylogenetic tree. 8/20/2005 Su-Shing Chen, CISE 10 Discrete Random Variables
A discrete random variable X is a numerical quantity taking some discrete set of values randomly in an experiment.
The probability distribution of X is the set of values that X can take and their probabilities.
Fair coin tossing twice: X = 0, 1, 2 probabilities 0.25, 0.50 and 0.25. 8/20/2005 Su-Shing Chen, CISE 11 Probability, Distribution, and Mean
Probability of Y = y: PY(y) = g(y), y = y1, y2, …, g(y) is a mathematical function of y.
Distribution function FY(y) = Σ ψ<=y PY(ψ)
The mean (expected value) of a random variable is the sum of all possible values y the random variable Y can take with its probability:
µ = Ε(Y) = Σ yPY(y) 8/20/2005 Su-Shing Chen, CISE 12 Variance of Discrete Random Variable
The variance of Y is a measure of the dispersion of the probability distribution of Y around its mean.
σ2 = Σ (y-µ)2 PY(y) 8/20/2005 Su-Shing Chen, CISE 15 Position 1 2 3 4 5 0.33 0.34 0.19 0.20 0.21 0.22 0.27 0.23 0.24 0.21 0.31 0.18 0.34 0.30 0.25 0.14 0.21 0.24 0.26 0.33 1 1 1 1 1 A G C T N U C L E O T I D C 8/20/2005 Su-Shing Chen, CISE 16 Statistical Tests for Significant Similarity
G G A G A C T G T A G A C A G C T A A T G C T A T T
G A A C G C C C T A G C C A C G A G C C C T T A T C
0 1 0 1 1 0 1 1 0 0 0 1 0 0 1 1 1 1 1 1 1 1 0 0 0 1
Exactly matched subsequences with length Y and Ymax (logest subsequence) = ymax (observed) = 3
To assess whether the observed ymax = 3 is significant?
What is the probability distribution of Ymax, when the null hypothesis that there is no significant similarity between the two sequences (one is generated independently of the other)?
Exact matching is not wise: Evolution usually has some differences – Well matching subsequences. 8/20/2005 Su-Shing Chen, CISE 17 Well Matching Subsequences
Allow k mismatches in subsequences.
The length Y is the number of trials up to but not including the k+1-th failure.
The probability distribution of Y is the generalized geometric distribution:
FY(y) = Σ j=k y (j k) p j-k(1-p)k+1,
y=k,k+1,k+2,…
P(Y<=y-1) =Σ j=k y-1 (j k) p j-k(1-p)k+1,
y=k+1,k+2,… 8/20/2005 Su-Shing Chen, CISE 20 Some Local Scores
Default - give scores –2 for residue matched with a gap, -1 with a different residue, and +1 with the same.
PAM
BLOSUM 8/20/2005 Su-Shing Chen, CISE 21 Dynamic Programming Algorithms
Given a scoring scheme of substitution matrix and linear gap penalty.
Find optimal gapped, global alignments.
X=X1X2…Xm and Y=Y1Y2…Yn
C(m,n)=all possible global alignments 8/20/2005 Su-Shing Chen, CISE 22 How many alignments are there?
G(m,n) = # groups of alignments with the same combination of aligned residue pairs ignoring the indels.
G(m,n) < C(m,n).
G(m,n) = Σk(m k) (n k)=(m+n n).
G(n,n) = (2n n)~22n/sq-root of πn
G(1000,1000)~10 600. 8/20/2005 Su-Shing Chen, CISE 25 Algorithm (2)
B = (m+1)x(n+1) matrix.
B(m,n) = solution=highest scoring alignment.
X1,i = X1X2…Xi
Y1,j = Y1Y2…Yj
Xi Xi -
Yj - Yj
B(i,j) = B(i-1,j-1)+s(i,j),
B(i,j) = B(i-1,j)-d,
B(i,j) = B(i,j-1)-d. 8/20/2005 Su-Shing Chen, CISE 26 Algorithm (3)
B(i,j) = max{B(i-1,j-1)+s(i,j), B(i-1,j)- d, B(i,j-1)-d}.
Run time O(mn).
X=GAATCT, Y=CATT.
GAATCT.
C-AT-T, CA-T-T, -CAT-T. 8/20/2005 Su-Shing Chen, CISE 27 B(i,j)=max[B(i-1,j-1)+s(i,j),B(i-1,j)-2,B(i,j-1)-2] B(6,4)=-2B(6,3)=-5B(6,2)=-8B(6,1)=-9B(6.0)=-12T B(5,4)=-2B(5,3)=-3B(5,2)=-6B(5,1)=-7B(5,0)=-10C B(4,5)=0B(4,4)=-1B(4,3)=-4B(4,1)=-7B(4,0)=-8T B(3,4)=-3B(3,3)=-1B(3,2)=-2B(3,1)=-5B(3,0)=-6A B(2,4)=-4B(2,3)=-2B(2,2)=0B(2,1)=-3B(2,0)=-4A B(1,4)=-7B(1,3)=-5B(1,2)=-3B(1,1)=-1B(1,0)=-2G B(0,4)=-8B(0,3)=-6B(0,2)=-4B(0,1)=-2B(0,0)=0- TTAC- 8/20/2005 Su-Shing Chen, CISE 30 Smith-Waterman Algorithm
L(i,j) = max{0, B(Xh,i,Yk,j )}.
If B(Xh,i,Yk,j ) is negative, it is good to remove the part of negative score.
Use dynamic programming.
Initializing L(i,0)=0=L(0,j). 8/20/2005 Su-Shing Chen, CISE 31 Smith-Waterman Algorithm (2)
L(i,j) = max{0,L(i-1,j-1)+s(i,j), L(i- 1,j)-d, L(i,j-1)-d}.
X2 X3 – X4 X5
Y5 Y6 Y7 Y8 Y9