1. Fa05CSE 182 CSE182-L4: Scoring matrices, Dictionary Matching

2. Fa05CSE 182 Class Mailing List fa05_182@cs.ucsd.edu To subscribe, send email to –fa05_182-subscribe@cs.ucsd.edufa05_182-subscribe@cs.ucsd.edu You can subscribe from the course web page Use the list for all course related queries, discussions,…

3. Fa05CSE 182 Silly Quiz Name a famous Bioinformatics Researcher Name a famous Bioinformatics Researcher who is a woman

4. Fa05CSE 182 Scoring DNA DNA has structure.

5. Fa05CSE 182 DNA scoring matrices So far, we considered a simple match/mismatch criterion. The nucleotides can be grouped into Purines (A,G) and Pyrimidines. Nucleotide substitutions within a group (transitions) are more likely than those across a group (transversions)

6. Fa05CSE 182 Scoring proteins –Scoring protein sequence alignments is a much more complex task than scoring DNA Not all substitutions are equal –Problem was first worked on by Pauling and collaborators –In the 1970s, Margaret Dayhoff created the first similarity matrices. “One size does not fit all” Homologous proteins which are evolutionarily close should be scored differently than proteins that are evolutionarily distant Different proteins might evolve at different rates and we need to normalize for that

7. Fa05CSE 182 PAM 1 distance Two sequences are 1 PAM apart if they differ in 1 % of the residues. PAM 1 (a,b) = Pr[residue b substitutes residue a, when the sequences are 1 PAM apart] 1% mismatch

8. Fa05CSE 182 PAM1 matrix Align many proteins that are very similar –Is this a problem? PAM1 distance is the probability of a substitution when 1% of the residues have changed Estimate the frequency P b|a of residue a being substituted by residue b. S(a,b) = log 10 (P ab /P a P b ) = log10(P b|a /P b )

9. Fa05CSE 182 PAM 1

10. Fa05CSE 182 PAM distance Two sequences are 1 PAM apart when they differ in 1% of the residues. When are 2 sequences 2 PAMs apart? 1 PAM 2 PAM

11. Fa05CSE 182 Higher PAMs PAM 2 (a,b) = ∑ c PAM 1 (a,c). PAM 1 (c,b) PAM 2 = PAM 1 * PAM 1 (Matrix multiplication) PAM 250 –= PAM 1 *PAM 249 –= PAM 1 250

12. Fa05CSE 182 Note: This is not the score matrix: What happens as you keep increasing the power?

13. Fa05CSE 182 Scoring using PAM matrices Suppose we know that two sequences are 250 PAMs apart. S(a,b) = log 10 (P ab /P a P b )= log 10 (P b|a /P b ) = log 10 (PAM 250 (a,b)/P b )

14. Fa05CSE 182 BLOSUM series of Matrices Henikoff & Henikoff: Sequence substitutions in evolutionarily distant proteins do not seem to follow the PAM distributions A more direct method based on hand-curated multiple alignments of distantly related proteins from the BLOCKS database. BLOSUM60 Merge all proteins that have greater than 60%. Then, compute the substitution probability. –In practice BLOSUM62 seems to work very well.

15. Fa05CSE 182 PAM vs. BLOSUM What is the correspondence? PAM1 Blosum1 PAM2 Blosum2 Blosum62 PAM250 Blosum100

16. Fa05CSE 182 Dictionary Matching, R.E. matching, and position specific scoring

17. Fa05CSE 182 Dictionary Matching Q: Given k words (s i has length l i ), and a database of size n, find all matches to these words in the database string. How fast can this be done? 1:POTATO 2:POTASSIUM 3:TASTE P O T A S T P O T A T O dictionary database

18. Fa05CSE 182 Dict. Matching & string matching How fast can you do it, if you only had one word of length m? –Trivial algorithm O(nm) time –Pre-processing O(m), Search O(n) time. Dictionary matching –Trivial algorithm (l 1 +l 2 +l 3 …)n –Using a keyword tree, l p n (l p is the length of the longest pattern) –Aho-Corasick: O(n) after preprocessing O(l 1 +l 2..) We will consider the most general case

19. Fa05CSE 182 Direct Algorithm P O P O P O T A S T P O T A T O P O T A T O Observations: When we mismatch, we (should) know something about where the next match will be. When there is a mismatch, we (should) know something about other patterns in the dictionary as well.

20. Fa05CSE 182 PO T A TO T UISM SET A The Trie Automaton Construct an automaton A from the dictionary –A[v,x] describes the transition from node v to a node w upon reading x. –A[u,’T’] = v, and A[u,’S’] = w –Special root node r –Some nodes are terminal, and labeled with the index of the dictionary word. 1:POTATO 2:POTASSIUM 3:TASTE 1 2 3 w vu S r

21. Fa05CSE 182 An O(l p n) algorithm for keyword matching Start with the first position in the db, and the root node. If successful transition –Increment current pointer –Move to a new node –If terminal node “success” Else –Retract ‘current’ pointer –Increment ‘start’ pointer –Move to root & repeat

22. Fa05CSE 182 Illustration: PO T A TO T UISM SET A P O T A S T P O T A T O l c v S 1

23. Fa05CSE 182 Idea for improving the time P O T A S T P O T A T O Suppose we have partially matched pattern i (indicated by l, and c), but fail subsequently. If some other pattern j is to match –Then prefix(pattern j) = suffix [ first c-l characters of pattern(i)) l c 1:POTATO 2:POTASSIUM 3:TASTE P O T A S S I U M T A S T E Pattern i Pattern j

24. Fa05CSE 182 Improving speed of dictionary matching Every node v corresponds to a string s v that is a prefix of some pattern. Define F[v] to be the node u such that s u is the longest suffix of s v If we fail to match at v, we should jump to F[v], and commence matching from there Let lp[v] = |s u | PO T A TO T UISM SET A 1 2345 6 7 8 910 11 S

25. Fa05CSE 182 An O(n) alg. For keyword matching Start with the first position in the db, and the root node. If successful transition –Increment current pointer –Move to a new node –If terminal node “success” Else (if at root) –Increment ‘current’ pointer –Mv ‘start’ pointer –Move to root Else –Move ‘start’ pointer forward –Move to failure node

26. Fa05CSE 182 Illustration P O T A S T P O T A T O POTATO T UISM SET A lc v S 1

27. Fa05CSE 182 Time analysis In each step, either c is incremented, or l is incremented Neither pointer is ever decremented (lp[v] < c-l). l and c do not exceed n Total time <= 2n P O T A S T P O T A T O lc

28. Fa05CSE 182 Blast: Putting it all together Input: Query of length m, database of size n Select word-size, scoring matrix, gap penalties, E-value cutoff

29. Fa05CSE 182 Blast Steps 1.Generate an automaton of all query keywords. 2.Scan database using a “Dictionary Matching” algorithm (O(n) time). Identify all hits. 3.Extend each hit using a variant of “local alignment” algorithm. Use the scoring matrix and gap penalties. 4.For each alignment with score S, compute the bit-score, E- value, and the P-value. Sort according to increasing E-value until the cut-off is reached. 5.Output results.

30. Fa05CSE 182 Protein Sequence Analysis What can you do if BLAST does not return a hit? –Sometimes, homology (evolutionary similarity) exists at very low levels of sequence similarity. A: Accept hits at higher P-value. –This increases the probability that the sequence similarity is a chance event. –How can we get around this paradox? –Reformulated Q: suppose two sequences B,C have the same level of sequence similarity to sequence A. If A& B are related in function, can we assume that A& C are? If not, how can we distinguish?