Fa05 CSE 182

CSE182-L4: Scoring matrices, Dictionary Matching

Fa05 CSE 182

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Silly Quiz

• Name a famous Bioinformatics Researcher

• Name a famous Bioinformatics Researcher who is a woman

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Scoring DNA

• DNA has structure.

QuickTime™ and aTIFF (LZW) decompressor

are needed to see this picture.

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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)

QuickTime™ and aTIFF (LZW) decompressor

are needed to see this picture.

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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

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PAM 1 distance

• Two sequences are 1 PAM apart if they differ in 1 % of the residues.

• PAM1(a,b) = Pr[residue b substitutes residue a, when the sequences are 1 PAM apart]

1% mismatch

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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 Pb|a of residue a being substituted by residue b.

• S(a,b) = log10(Pab/PaPb) = log10(Pb|a/Pb)

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PAM 1

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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

1 PAM

2 PAM

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Higher PAMs

• PAM2(a,b) = ∑c PAM1(a,c). PAM1 (c,b)

• PAM2 = PAM1 * PAM1 (Matrix multiplication)

• PAM250

– = PAM1*PAM249

– = PAM1250

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

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Scoring using PAM matrices

• Suppose we know that two sequences are 250 PAMs apart.

• S(a,b) = log10(Pab/PaPb)= log10(Pb|a/Pb) = log10(PAM250(a,b)/Pb)

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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.

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PAM vs. BLOSUM

• What is the correspondence?

• PAM1 Blosum1• PAM2 Blosum2

• Blosum62

• PAM250 Blosum100

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Dictionary Matching, R.E. matching, and position specific scoring

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Dictionary Matching

• Q: Given k words (si has length li), and a database of size n, find all matches to these words in the database string.

• How fast can this be done?

1:POTATO2:POTASSIUM3:TASTE

P O T A S T P O T A T O

dictionary

database

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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 (l1+l2+l3…)n

– Using a keyword tree, lpn (lp is the length of the longest pattern)

– Aho-Corasick: O(n) after preprocessing O(l1+l2..)

• We will consider the most general case

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Direct Algorithm

P O P O P O T A S T P O T A T OP O T A T OP O T A T OP O T A T OP 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.

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P O T A T O

T UIS M

S ETA

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:POTATO2:POTASSIUM3:TASTE

1

2

3

w

vu

S

r

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An O(lpn) 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

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Illustration:

P O T A T O

T UIS M

S ETA

P O T A S T P O T A T Ol c

v

S

1

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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:POTATO2:POTASSIUM3:TASTE

P O T A S S I U MT A S T E

Pattern i

Pattern j

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Improving speed of dictionary matching

• Every node v corresponds to a string sv that is a prefix of some pattern.

• Define F[v] to be the node u such that su is the longest suffix of sv

• If we fail to match at v, we should jump to F[v], and commence matching from there

• Let lp[v] = |su|

P O T A T O

T UIS M

S ETA

1 2 3 4 5

67

89 10

11S

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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

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Illustration

P O T A S T P O T A T O

P O T A T O

T UIS M

S ETA

lc

v S

1

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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 Ol c

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Blast: Putting it all together

• Input: Query of length m, database of size n

• Select word-size, scoring matrix, gap penalties, E-value cutoff

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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.

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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?