(Regulatory-) Motif Finding

1
(Regulatory-) Motif Finding
2
Finding Regulatory Motifs
. . .

• Given a collection of genes with common
  expression,
• Find the TF-binding motif in common

3
Expectation Maximization

• Initialize parameters ? (M, B), ?
• Try different values of ? from N-1/2 up to 1/(2K)
• Repeat
• Expectation
• Maximization
• Until change in ? (M, B), ? falls below ?
• Report results for several good ?

?
1 ?
M1
M1
MK
B
A C G T








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Gibbs Sampling in Motif Finding
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Gibbs Sampling

• Given
• x1, , xN,
• motif length K,
• background B,
• Find
• Model M
• Locations a1,, aN in x1, , xN
• Maximizing log-odds likelihood ratio

6
Gibbs Sampling

• AlignACE first statistical motif finder
• BioProspector improved version of AlignACE
• Algorithm (sketch)
• Initialization
• Select random locations in sequences x1, , xN
• Compute an initial model M from these locations
• Sampling Iterations
• Remove one sequence xi
• Recalculate model
• Pick a new location of motif in xi according to
  probability the location is a motif occurrence

7
Gibbs Sampling

• Initialization
• Select random locations a1,, aN in x1, , xN
• For these locations, compute M

• That is, Mkj is the number of occurrences of
  letter j in motif position k, over the total

8
Gibbs Sampling

• Predictive Update
• Select a sequence x xi
• Remove xi, recompute model

M

• where ?j are pseudocounts to avoid 0s,
• and B ?j ?j

9
Gibbs Sampling

• Sampling
• For every K-long word xj,,xjk-1 in x
• Qj Prob word motif M(1,xj)??M(k,xjk-1)
• Pi Prob word background B(xj)??B(xjk-1)
• Let
• Sample a random new position ai according to the
  probabilities A1,, Ax-k1.

Prob
0
x
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Gibbs Sampling

• Running Gibbs Sampling
• Initialize
• Run until convergence
• Repeat 1,2 several times, report common motifs

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Advantages / Disadvantages

• Very similar to EM
• Advantages
• Easier to implement
• Less dependent on initial parameters
• More versatile, easier to enhance with heuristics
• Disadvantages
• More dependent on all sequences to exhibit the
  motif
• Less systematic search of initial parameter space

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Repeats, and a Better Background Model

• Repeat DNA can be confused as motif
• Especially low-complexity CACACA AAAAA, etc.
• Solution
• more elaborate background model
• 0th order B pA, pC, pG, pT
• 1st order B P(AA), P(AC), , P(TT)
• Kth order B P(X b1bK) X, bi?A,C,G,T
• Has been applied to EM and Gibbs (up to 3rd
  order)

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Phylogenetic Footprinting(Slides by Martin Tompa)
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Phylogenetic Footprinting(Tagle et al. 1988)

• Functional sequences evolve slower than
  nonfunctional ones
• Consider a set of orthologous sequences from
  different species
• Identify unusually well conserved regions

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Substring Parsimony Problem

• Given
• phylogenetic tree T,
• set of orthologous sequences at leaves of T,
• length k of motif
• threshold d
• Problem
• Find each set S of k-mers, one k-mer from each
  leaf, such that the parsimony score of S in T
  is at most d.
• This problem is NP-hard.

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Small Example
Size of motif sought k 4
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Solution
Parsimony score 1 mutation
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CLUSTALW multiple sequence alignment (rbcS
gene) Cotton ACGGTT-TCCATTGGATGA---AATGAGATAAGAT-
--CACTGTGC---TTCTTCCACGTG--GCAGGTTGCCAAAGATA------
-AGGCTTTACCATT Pea GTTTTT-TCAGTTAGCTTA---GTGGGCATC
TTA----CACGTGGC---ATTATTATCCTA--TT-GGTGGCTAATGATA-
------AGG--TTAGCACA Tobacco TAGGAT-GAGATAAGATTA---
CTGAGGTGCTTTA---CACGTGGC---ACCTCCATTGTG--GT-GACTTA
AATGAAGA-------ATGGCTTAGCACC Ice-plant TCCCAT-ACAT
TGACATAT---ATGGCCCGCCTGCGGCAACAAAAA---AACTAAAGGATA
--GCTAGTTGCTACTACAATTC--CCATAACTCACCACC Turnip ATT
CAT-ATAAATAGAAGG---TCCGCGAACATTG--AAATGTAGATCATGCG
TCAGAATT--GTCCTCTCTTAATAGGA-------A-------GGAGC Wh
eat TATGAT-AAAATGAAATAT---TTTGCCCAGCCA-----ACTCAGT
CGCATCCTCGGACAA--TTTGTTATCAAGGAACTCAC--CCAAAAACAAG
CAAA Duckweed TCGGAT-GGGGGGGCATGAACACTTGCAATCATT--
---TCATGACTCATTTCTGAACATGT-GCCCTTGGCAACGTGTAGACTGC
CAACATTAATTAAA Larch TAACAT-ATGATATAACAC---CGGGCAC
ACATTCCTAAACAAAGAGTGATTTCAAATATATCGTTAATTACGACTAAC
AAAA--TGAAAGTACAAGACC Cotton CAAGAAAAGTTTCCACCCTC
------TTTGTGGTCATAATG-GTT-GTAATGTC-ATCTGATTT----AG
GATCCAACGTCACCCTTTCTCCCA-----A Pea C---AAAACTTTTCA
ATCT-------TGTGTGGTTAATATG-ACT-GCAAAGTTTATCATTTTC-
---ACAATCCAACAA-ACTGGTTCT---------A Tobacco AAAAAT
AATTTTCCAACCTTT---CATGTGTGGATATTAAG-ATTTGTATAATGTA
TCAAGAACC-ACATAATCCAATGGTTAGCTTTATTCCAAGATGA Ice-p
lant ATCACACATTCTTCCATTTCATCCCCTTTTTCTTGGATGAG-ATA
AGATATGGGTTCCTGCCAC----GTGGCACCATACCATGGTTTGTTA-AC
GATAA Turnip CAAAAGCATTGGCTCAAGTTG-----AGACGAGTAAC
CATACACATTCATACGTTTTCTTACAAG-ATAAGATAAGATAATGTTATT
TCT---------A Wheat GCTAGAAAAAGGTTGTGTGGCAGCCACCTA
ATGACATGAAGGACT-GAAATTTCCAGCACACACA-A-TGTATCCGACGG
CAATGCTTCTTC-------- Duckweed ATATAATATTAGAAAAAAAT
C-----TCCCATAGTATTTAGTATTTACCAAAAGTCACACGACCA-CTAG
ACTCCAATTTACCCAAATCACTAACCAATT Larch TTCTCGTATAAGG
CCACCA-------TTGGTAGACACGTAGTATGCTAAATATGCACCACACA
CA-CTATCAGATATGGTAGTGGGATCTG--ACGGTCA Cotton ACC
AATCTCT---AAATGTT----GTGAGCT---TAG-GCCAAATTT-TATGA
CTATA--TAT----AGGGGATTGCACC----AAGGCAGTG-ACACTA Pe
a GGCAGTGGCC---AACTAC--------------------CACAATTT-
TAAGACCATAA-TAT----TGGAAATAGAA------AAATCAAT--ACAT
TA Tobacco GGGGGTTGTT---GATTTTT----GTCCGTTAGATAT-G
CGAAATATGTAAAACCTTAT-CAT----TATATATAGAG------TGGTG
GGCA-ACGATG Ice-plant GGCTCTTAATCAAAAGTTTTAGGTGTGA
ATTTAGTTT-GATGAGTTTTAAGGTCCTTAT-TATA---TATAGGAAGGG
GG----TGCTATGGA-GCAAGG Turnip CACCTTTCTTTAATCCTGTG
GCAGTTAACGACGATATCATGAAATCTTGATCCTTCGAT-CATTAGGGCT
TCATACCTCT----TGCGCTTCTCACTATA Wheat CACTGATCCGGAG
AAGATAAGGAAACGAGGCAACCAGCGAACGTGAGCCATCCCAACCA-CAT
CTGTACCAAAGAAACGG----GGCTATATATACCGTG Duckweed TTA
GGTTGAATGGAAAATAG---AACGCAATAATGTCCGACATATTTCCTATA
TTTCCG-TTTTTCGAGAGAAGGCCTGTGTACCGATAAGGATGTAATC La
rch CGCTTCTCCTCTGGAGTTATCCGATTGTAATCCTTGCAGTCCAATT
TCTCTGGTCTGGC-CCA----ACCTTAGAGATTG----GGGCTTATA-TC
TATA Cotton T-TAAGGGATCAGTGAGAC-TCTTTTGTATAACTGT
AGCAT--ATAGTAC Pea TATAAAGCAAGTTTTAGTA-CAAGCTTTGCA
ATTCAACCAC--A-AGAAC Tobacco CATAGACCATCTTGGAAGT-TT
AAAGGGAAAAAAGGAAAAG--GGAGAAA Ice-plant TCCTCATCAAA
AGGGAAGTGTTTTTTCTCTAACTATATTACTAAGAGTAC Larch TCTT
CTTCACAC---AATCCATTTGTGTAGAGCCGCTGGAAGGTAAATCA Tur
nip TATAGATAACCA---AAGCAATAGACAGACAAGTAAGTTAAG-AGA
AAAG Wheat GTGACCCGGCAATGGGGTCCTCAACTGTAGCCGGCATCC
TCCTCTCCTCC Duckweed CATGGGGCGACG---CAGTGTGTGGAGGA
GCAGGCTCAGTCTCCTTCTCG
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An Exact Algorithm(generalizing Sankoff and
Rousseau 1975)
Wu s best parsimony score for subtree rooted
at node u, if u is labeled with string s.
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Running Time
Wu s ? min ( Wv t d(s, t) )
v child t

of u
Average sequence length
O(k?42k ) time per node
Number of species
Total time O(n k (42k l ))
Motif length
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Limits of Motif Finders
0
???
gene

• Given upstream regions of coregulated genes
• Increasing length makes motif finding harder
  random motifs clutter the true ones
• Decreasing length makes motif finding harder
  true motif missing in some sequences

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Limits of Motif Finders

• A (k,d)-motif is a k-long motif with d random
  differences per copy
• Motif Challenge problem
• Find a (15,4) motif in N sequences of length L
• CONSENSUS, MEME, AlignACE, most other programs
  fail for N 20, L 1000

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Example Application Motifs in Yeast

• Group
• Tavazoie et al. 1999, G. Churchs lab, Harvard
• Data
• Microarrays on 6,220 mRNAs from yeast Affymetrix
  chips (Cho et al.)
• 15 time points across two cell cycles

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Processing of Data

• Selection of 3,000 genes
• Genes with most variable expression were selected
• Clustering according to common expression
• K-means clustering
• 30 clusters, 50-190 genes/cluster
• Clusters correlate well with known function
• AlignACE motif finding
• 600-long upstream regions
• 50 regions/trial

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Motifs in Periodic Clusters
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Motifs in Non-periodic Clusters
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Rapid Global Alignments

• How to align genomic sequences in (more or less)
  linear time

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(No Transcript)
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(No Transcript)
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Motivation

• Genomic sequences are very long
• Human genome 3 x 109 long
• Mouse genome 2.7 x 109 long
• Aligning genomic regions is useful for revealing
  common gene structure
• Useful to compare regions gt 1,000,000-long

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Main Idea

• Genomic regions of interest contain ordered
  islands of similarity, such as genes
• Find local alignments
• Chain an optimal subset of them
• Refine/complete the alignment
• Systems that use this idea to various degrees
• MUMmer, GLASS, DIALIGN, CHAOS, AVID, LAGAN, TBA,
  others

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Saving cells in DP

1. Find local alignments
2. Chain -O(NlogN) L.I.S.
3. Restricted DP

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Methods to CHAIN Local Alignments
Sparse Dynamic Programming O(N log N)
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The Problem Find a Chain of Local Alignments
(x,y) ? (x,y) requires x lt x y lt y
Each local alignment has a weight FIND the chain
with highest total weight
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Quadratic Time Solution

• Build Directed Acyclic Graph (DAG)
• Nodes local alignments (xa,xb) ? (ya,yb)
  score
• Directed edges local alignments that can be
  chained
• edge ( (xa, xb, ya, yb) , (xc, xd, yc, yd) )
• xa lt xb lt xc lt xd
• ya lt yb lt yc lt yd
• Each local alignment
• is a node vi with
• alignment score si

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Quadratic Time Solution

• Dynamic programming
• Initialization
• Find each node va s.t. there is no edge (u,v0)
• Set score of V(a) to be sa
• Iteration
• For each vi, optimal path ending in vi has total
  score
• V(i) max ( weight(vj, vi) V(j) )
• Termination
• Optimal global chain
• j argmax ( V(j) ) trace chain from vj
• Worst case time quadratic

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Sparse Dynamic Programming

• Back to the LCS problem
• Given two sequences
• x x1, , xm
• y y1, , yn
• Find the longest common subsequence
• Quadratic solution with DP
• How about when hits xi yj are sparse?

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Sparse Dynamic Programming
15 3 24 16 20 4 24 3 11 18










4
20
24
3
11
15
11
4
18
20

• Imagine a situation where the number of hits is
  much smaller than O(nm) maybe O(n) instead

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Sparse Dynamic Programming L.I.S.

• Longest Increasing Subsequence
• Given a sequence over an ordered alphabet
• x x1, , xm
• Find a subsequence
• s s1, , sk
• s1 lt s2 lt lt sk

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Sparse LCS expressed as LIS
x

• Create a sequence w
• Every matching point x-to-y, (i, j), is inserted
  into a sequence as follows
• For each position j of x, from smallest to
  largest, insert in z the points (i, j), in
  decreasing column i order
• The 11 example points are inserted in the order
  given
• Any two points (ya, xa), (yb, xb) can be chained
  iff
• a is before b in w, and
• ya lt yb

15 3 24 16 20 4 24 3 11 18
6
4
2 7
1 8
10

9
5
11
3
4
20
24
3
11
15
11
4
18
20
y
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Sparse LCS expressed as LIS
x

• Create a sequence w
• w (4,2) (3,3) (10,5) (2,5) (8,6) (1,6) (3,7)
  (4,8) (7,9) (5,9) (9,10)
• Consider now ws elements as ordered
  lexicographically, where
• (ya, xa) lt (yb, xb) if ya lt yb
• Claim An increasing subsequence of w is a common
  subsequence of x and y

15 3 24 16 20 4 24 3 11 18
6
4
2 7
1 8
10

9
5
11
3
4
20
24
3
11
15
11
4
18
20
y
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Sparse Dynamic Programming for LIS
x

• Algorithm
• initialize empty array L
• / at each point, lj will contain the last
  element of the longest j-long increasing
  subsequence that ends with the smallest wi /
• for i 1 to w
• binary search for wi in L, to find lj lt wi
  lj1
• replace lj1 with wi
• keep a backptr lj ? wi
• Thats it!!!

15 3 24 16 20 4 24 3 11 18
6
4
2 7
1 8
10

9
5
11
3
4
20
24
3
11
15
11
4
18
20
y
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Sparse Dynamic Programming for LIS
x

• Example
• w (4,2) (3,3) (10,5) (2,5) (8,6) (1,6) (3,7)
  (4,8) (7,9) (5,9) (9,10)
• L
• (4,2)
• (3,3)
• (3,3) (10,5)
• (2,5) (10,5)
• (2,5) (8,6)
• (1,6) (8,6)
• (1,6) (3,7)
• (1,6) (3,7) (4,8)
• (1,6) (3,7) (4,8) (7,9)
• (1,6) (3,7) (4,8) (5,9)
• (1,6) (3,7) (4,8) (5,9) (9,10)

15 3 24 16 20 4 24 3 11 18
6
4
2 7
1 8
10

9
5
11
3
4
20
24
3
11
15
11
4
18
20
y
Longest common subsequence s 4, 24, 3, 11, 18
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Sparse DP for rectangle chaining

• 1,, N rectangles
• (hj, lj) y-coordinates of rectangle j
• w(j) weight of rectangle j
• V(j) optimal score of chain ending in j
• L list of triplets (lj, V(j), j)
• L is sorted by lj
• L is implemented as a balanced binary tree

h
l
y
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Sparse DP for rectangle chaining

• Go through rectangle x-coordinates, from lowest
  to highest
• When on the leftmost end of i
• j rectangle in L, with largest lj lt hi
• V(i) w(i) V(j)
• When on the rightmost end of i
• j rectangle in L, with largest lj ? li
• If V(i) gt V(j)
• INSERT (li, V(i), i) in L
• REMOVE all (lk, V(k), k) with V(k) ? V(i) lk ?
  li

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Example
x
2
1 5
5
6
2 6
9
10
3 3
11
12
4 4
14
15
5 2
16
y
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Time Analysis

• Sorting the x-coords takes O(N log N)
• Going through x-coords N steps
• Each of N steps requires O(log N) time
• Searching L takes log N
• Inserting to L takes log N
• All deletions are consecutive, so log N per
  deletion
• Each element is deleted at most once N log N for
  all deletions
• Recall that INSERT, DELETE, SUCCESSOR, take O(log
  N) time in a balanced binary search tree