Designing Spaced Seeds

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Designing Spaced Seeds
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Project/Exam deadlines

• May 2
• Send email to me with a title of your project
• May 9
• Each student/group gives a 10 min. presentation
  on their proposed project.
• Show preliminary computations. What is the test
  plan? What is the data like, and how much is
  there.
• Last week of classes
• A 20 min. presentation from each group
• A written report on the project
• A take home exam, due electronically on the date
  of the final exam

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Accuracy

• Consider a 64bp sequence that is 70 similar to
  the query.
• Pr(an 11 mer matches) 0.3
• Pr(A spaced seed 11101001.. Matches) 0.466
• This non-intuitive result leads to selection of
  spaced words that are an order of magnitude
  faster for identical specificity and sensitivity
• Implemented in PATTERNHUNTER

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How to compute a spaced seed

• No good algorithm is known.
• Iterate over all (M choose W) seeds.
• Use a computation to decide Pr(match)
• Choose the seed that maximizes probability.

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Prob. Computation for Spaced Seeds

• Given a specific seed Q(M,W), compute the
  probability of a hit in a sequence of length L.
• We can assume that there is a probability p of
  match.
• The match mismatch string is a binary string with
  probability p of 1

1
L
1 1 1 0 1 1 1 0 1 1 1 0 1 1 1 1 1 1 0 1 1 1 1 0 0
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Prob. Computation for Spaced Seeds

• Given a specific seed Q(M,W), compute the
  probability of a hit in a sequence of length L.
• Q is a binary string of length M, with W 1s
• We try to match the binary match string S which
  is a random binary string with probability p of
  success.

M
1
L
1100.11..0

• PQ Prob. (Q matches random S at some location)
• How can we compute PQ?

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Computing F(i,b)

• For a specific string b, define
• F(i,b) Prob. (Q matches a random string S of
  length i, s.t. S ends in b)

i
1
b

• Why is it sufficient to compute f(I,b) for all I,
  b?
• PQ f(L,?)

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Computing f(i,b)

• Define B1 as the set of all strings b that match
  a suffix of Q

b

• We have two possibilities
• b ? B1 b is consistent with a suffix of Q.
• b ? B0 B-B1

110001
Q
110001
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Computing f(i,b)

• Case b ? B0
• f(i,b) f(i-1,bgtgt1)

b
Q

• Case b ? B1 and b M
• f(i,b) 1

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Computing f(i,b)

• Case b ? B1, bltM
• f(i,b) pf(i-1,1b) (1-p) pf(i-1,0b)
• Note that if b ? B1 , then 1b ? B1
• However, it is possible that 0b ? B1
• We want to iterate only over b ? B1
• Find smallest j s.t. 0bgtgtj? B1
• f(i,b) pf(i-1,1b) (1-p)f(i-j,0bgtgtj)

Q
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Efficiency

• B1 M2M-W
• The iteration proceeds for all i, and all b?B1,
  and each comparison needs O(M) steps
• O(M2M-W L M) O(M22M-W L)

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More efficient algorithm for spaced seed design

• Due to Buhler, Keich, and Sun
• Consider seed ? (weight w, span s).
• Let Q? be the set of all possible 2s-w strings
  matching ?.

1 1 0 0 0 1 0 1
1 1 0 0 0 1 0 1 1 1 0 0 0 1 1 1 1 1 0 0 1 1 0 1 1
1 0 0 1 1 1 1
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Trie construction

• Our goal is to make an automaton that accepts all
  strings which contain a string from Q.
• Make a trie T? from Q?.
• T? is a DFA that precisely accepts Q?
• Can we convert T? to an DFA that accepts all
  strings that matches a string from Q? as a
  suffix?

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Failure links

• Use of failure links allows us to traverse any
  string till Q is reached.
• Note the DFA has special structure. Does it
  help? No failure links when outgoing edge is 0.
  Therefore, we fail only when we see a 1.

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Substring automaton

• We started with a Trie that only accepts Q?
• Next, we use failure links to accept any string
  with a suffix from Q?.
• Finally, make every accepting state an absorbing
  one, to accept all strings containing a string
  from Q? as suffix.

0,1
1
0
0
1
1
1
1
Ex ? 1001
0
1
1
1
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Computing sensitivity of ?

• Compute the probability that a random string of
  length l will match ??
• Equivalently What is the probability that a
  random string of length l that starts at the
  begin node will end in an accepting state of A?.
• Case 1 Each bit of S is 1 with probability p
• P(q,t)Probability that we reach q after reading
  the first t bits.

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Complexity

• Size of the Automaton W2M-W
• What is the in-degree?
• Claimed complexity ?(W2M-WL)
• O(M2/W) faster then the previous algorithm

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Generalizing the match string

• The match string may have a different
  distribution
• Errors do not fall independently at random
• Instead of independent bernoulli trials, we can
  have a higher order markovian process generating
  the match string.
• The algorithm of Keich et al. Cannot deal with
  this extension, but it is natural in Mandala

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Experimental Results with Mandala

• 428 human/mouse genomic aligned regions.
• Repeat mask the alignments and separate into
  coding/non-coding regions.
• A total of 1136000 similarities (alignments) were
  pulled. These are used to check for sensitivity
  (accuracy) of filters.

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Effect of Span

• Solid line 0-th order model
• Dashed line 5-th order model.
• W11 throughout larger span implies more gaps,
  span11 implies ungapped (BLASTN) seed

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Accuracy of different seeds

• Non-coding
• Coding

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Model order

• Non-Coding solid line
• coding dashed line

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What about multiple keywords

• All of the analysis is for ungapped alignments.
• With indels, multiple words might be more
  sensitive.
• Mandala works for multiple keywords also.
• Can we make the algorithm more efficient?
• In particular, there is an explosion of states in
  making a deterministic automaton? Can we match a
  non-deterministic automaton?

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Regular Expressions

• Concise representation of a set of strings over
  alphabet ?.
• Described by a string over
• R is a r.e. if and only if

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Regular Expression

• Q Let ?A,C,E
• Is (AC)EEC a regular expression?
• (AC)?
• AC..E?

• Q When is a string s in a regular expression?
• R (AC)EEC
• Is CEEC in R?
• AEC?
• ACEE?

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Regular Expression Automata

• Every R.E can be expressed by an automaton (a
  directed graph) with the following properties
• The automaton has a start and end node
• Each edge is labeled with a symbol from ?, or ?

• Suppose R is described by automaton A
• S ? R if and only if there is a path from start
  to end in A, labeled with s.

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Examples Regular Expression Automata

• (AC)EEC

C
A
E
E
start
end
C
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Constructing automata from R.E

• R ?
• R ?, ? ? ?
• R R1 R2
• R R1 R2
• R R1

?
?
?
?
?
?
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Regular Expression Matching

• Given a database D, and a regular expression R,
  is a substring of D in R?

• Is there a string Dl..c that is accepted by the
  automaton of R?

• Simpler Q Is D1..c accepted by the automaton
  of R?

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Alg. For matching R.E.

• If D1..c is accepted by the automaton RA
• There is a path labeled D1Dc that goes from
  START to END in RA

?
D1
D2
Dc
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Alg. For matching R.E.

• If D1..c is accepted by the automaton RA
• There is a path labeled D1Dc that goes from
  START to END in RA
• There is a path labeled D1..Dc-1 from START
  to node u, and a path labeled Dc from u to the
  END

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D.P. to match regular expression

• Define
• Au,? Automaton node reached from u after
  reading ?
• Eps(u) set of all nodes reachable from node u
  using epsilon transitions.
• Nc subset of nodes reachable from START node
  after reading D1..c
• Q when is v ? Nc?

u
?
v
?
u
Eps(u)
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D.P. to match regular expression

• Q when is v ? Nc?
• A If for some u ? Nc-1, w Au,Dc,
• v ? w Eps(w)

w
u
D1 .. Dc-1
?
Dc
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Algorithm
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The final step

• We have answered the question
• Is D1..c accepted by R?
• Yes, if END ? Nc
• We need to answer
• Is Dl..c (for some l, and some c) accepted by R

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