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Algorithms for Regulatory Motif Discovery

This document explores advanced algorithms for motif discovery in biological sequences, including the use of probabilistic models like positional weight matrices and sequence logos. It details both straightforward and complex string matching techniques, such as exact string matching using comparison algorithms and the Knuth-Morris-Pratt and Karp-Rabin algorithms. Furthermore, the text discusses running times for various algorithms, providing insight into the efficiency of string searches and motif detection in genomic data. Emphasis is placed on the applicability of these methods in computational biology.

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Algorithms for Regulatory Motif Discovery

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  1. Algorithms for Regulatory Motif Discovery Xiaohui Xie University of California, Irvine

  2. Positional weight matrix representation Lambda cI/cro binding sites Sequence Logo

  3. Probabilistic model for motif discovery • Input: a set of sequence: {S1, S2 …,SN}, each of which is of length w, i.e. |Si|=w for all i. • Two models: • Motif model: P(y|) • Background model: P(y|) • Hidden variables: {z1, z2 …,zN}, where zi =1 if Si is a motif site and 0 otherwise. • P(Si|zi=1) = P(Si|) and P(Si|zi=0) = P(Si|) • P(zi=1|Si) = P(Si|zi=1)P(zi=1)/[P(Si|zi=1)P(zi=1)+P(Si|zi=0)P(zi=0)]

  4. Probabilistic model for motif discovery • Input: a set of sequences: {S1, S2 …,SN}, each of which is of length w, i.e. |Si|=w for all i. • Parameter estimation problem: • Find the  and  that maximize P(S1, S2 …,SN| , ) • If zi =1 for all i, then ij = cij/N where cij is the number of letter j occurring at position i in the set of sequences.

  5. String matching

  6. Exact String Matching • Input: Two strings T[1…n] and P[1…m], containing symbols from alphabet . Example:  = {A,C,G,T} T[1…12] = “CAGTACATCGAT” P[1..3] = “AGT” • Goal: find all “shifts” 0≤s ≤n-m such that T[s+1…s+m] = P

  7. Simple Algorithm fors← 0ton-m Match ← 1 for j ← 1 tom if T[s+j]≠P[j] then Match ← 0 exit loop if Match=1 then outputs

  8. Analysis • Running time of the simple algorithm: • Worst-case: O(nm) • Average-case (random text): O(n) (expectation) • Ts = time spend on checking shift s (the number of comparisons until 1st mismatch) • E[Ts] < 2 (why) • E[SsTs] = SsE[Ts] = O(n)

  9. Worst-case • Is it possible to achieve O(n) for any input ? • Knuth-Morris-Pratt’77: deterministic • Karp-Rabin’81: randomized • Digression: what is the difference between • Algorithm that is fast on a random input (as seen on the previous slide) • Randomized algorithm (as in the rest of this lecture)

  10. Karp-Rabin Algorithm • Idea: semi-numerical approach: • Consider all m-mers: • T[1…m], T[2…m+1], …, T[m-n+1…n] • Map each T[s+1…s+m] into a number ts • Map the pattern P[1…m] into a number p • Report the m-mers that map to the same value as p • Problem: how to map all m-mers in O(n) time ?

  11. Implementation • Attempt I: • Assume S={0,1} (for {A,C,G,T} convert: A:00, C:01, G:10, T:11) • Think about each T[s+1…s+m] as a number in binary representation, i.e., • ts=T[s+1]2m-1+T[s+2]2m-2+…+T[s+m]20 • Find a fast way of computing ts+1 given ts • Output all s such that ts is equal to the number p represented by P

  12. Magic formula • How to transform ts=T[s+1]2m-1+T[s+2]2m-2+…+T[s+m]20 into ts+1=T[s+2]2m-1+T[s+3]2m-2+…+T[s+m+1]20 ? • Three steps: • Subtract T[s+1]2m-1 • Multiply by 2 (i.e., shift the bits by one position) • Add T[s+m+1]20 • Therefore: ts+1= (ts- T[s+1]2m-1)*2 + T[s+m+1]20

  13. Algorithm ts+1= (ts- T[s+1]2m-1)*2 + T[s+m+1]20 • Can compute ts+1 from ts using 3 arithmetic operations • Therefore, we can compute all t0,t1,…,tn-m using O(n) arithmetic operations • We can compute a number corresponding to P using O(m) arithmetic operations • Are we done ?

  14. Problem • To get O(n) time, we would need to perform each arithmetic operation in O(1) time • However, the arguments are m-bit long ! • If m large, it is unreasonable to assume that operations on such big numbers can be done in O(1) time • We need to reduce the number range to something more managable

  15. Hashing • We will instead compute • t’s=T[s+1]2m-1+T[s+2]2m-2+…+T[s+m]20 mod q where q is an “appropriate” prime number • One can still compute t’s+1 from t’s : t’s+1= (t’s- T[s+1]2m-1)*2+T[s+m+1]20 mod q • If q is not large, we can compute all t’s (and p’) in O(n) time

  16. Problem • Unfortunately, we can have false positives, i.e., T[s+1…s+m]P but ts mod q = p mod q • Our approach: • Use a random q • Show that the probability of a false positive is small → randomized algorithm

  17. Aligning two sequences • Longest common substring (LCS) problem (no gaps): • Input: Two strings T[1…n] and P[1…m], n≥m • Goal: Largest k such that T[i+1…i+k] = P[j+1…j+k] for some i,j • How can we solve this problem efficiently ? • Hint: How can we check if LCS has length k ?

  18. Checking for a common m-mer

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