ch04 motifs

Finding Regulatory Motifs

in DNA Sequences

1. Implanting Patterns in Random Text

2. Gene Regulation

3. Regulatory Motifs

4. The Gold Bug Problem

5. The Motif Finding Problem

6. Brute Force Motif Finding

7. The Median String Problem

8. Search Trees

9. Branch-and-Bound Motif Search

10. Branch-and-Bound Median String Search

11. Consensus and Pattern Branching: Greedy Motif Search

12. Planted Motif Search: Exhaustive Motif Search

Outline

Random Sample

atgaccgggatactgataccgtatttggcctaggcgtacacattagataaacgtatgaagtacgttagactcggcgccgccg

acccctattttttgagcagatttagtgacctggaaaaaaaatttgagtacaaaacttttccgaatactgggcataaggtaca

tgagtatccctgggatgacttttgggaacactatagtgctctcccgatttttgaatatgtaggatcattcgccagggtccga

gctgagaattggatgaccttgtaagtgttttccacgcaatcgcgaaccaacgcggacccaaaggcaagaccgataaaggaga

tcccttttgcggtaatgtgccgggaggctggttacgtagggaagccctaacggacttaatggcccacttagtccacttatag

gtcaatcatgttcttgtgaatggatttttaactgagggcatagaccgcttggcgcacccaaattcagtgtgggcgagcgcaa

cggttttggcccttgttagaggcccccgtactgatggaaactttcaattatgagagagctaatctatcgcgtgcgtgttcat

aacttgagttggtttcgaaaatgctctggggcacatacaagaggagtcttccttatcagttaatgctgtatgacactatgta

ttggcccattggctaaaagcccaacttgacaaatggaagatagaatccttgcatttcaacgtatgccgaaccgaaagggaag

ctggtgagcaacgacagattcttacgtgcattagctcgcttccggggatctaatagcacgaagcttctgggtactgatagca

Implanting Motif AAAAAAAGGGGGGG

atgaccgggatactgatAAAAAAAAGGGGGGGggcgtacacattagataaacgtatgaagtacgttagactcggcgccgccg

acccctattttttgagcagatttagtgacctggaaaaaaaatttgagtacaaaacttttccgaataAAAAAAAAGGGGGGGa

tgagtatccctgggatgacttAAAAAAAAGGGGGGGtgctctcccgatttttgaatatgtaggatcattcgccagggtccga

gctgagaattggatgAAAAAAAAGGGGGGGtccacgcaatcgcgaaccaacgcggacccaaaggcaagaccgataaaggaga

tcccttttgcggtaatgtgccgggaggctggttacgtagggaagccctaacggacttaatAAAAAAAAGGGGGGGcttatag

gtcaatcatgttcttgtgaatggatttAAAAAAAAGGGGGGGgaccgcttggcgcacccaaattcagtgtgggcgagcgcaa

cggttttggcccttgttagaggcccccgtAAAAAAAAGGGGGGGcaattatgagagagctaatctatcgcgtgcgtgttcat

aacttgagttAAAAAAAAGGGGGGGctggggcacatacaagaggagtcttccttatcagttaatgctgtatgacactatgta

ttggcccattggctaaaagcccaacttgacaaatggaagatagaatccttgcatAAAAAAAAGGGGGGGaccgaaagggaag

ctggtgagcaacgacagattcttacgtgcattagctcgcttccggggatctaatagcacgaagcttAAAAAAAAGGGGGGGa

Where is the Implanted Motif?

atgaccgggatactgataaaaaaaagggggggggcgtacacattagataaacgtatgaagtacgttagactcggcgccgccg

acccctattttttgagcagatttagtgacctggaaaaaaaatttgagtacaaaacttttccgaataaaaaaaaaggggggga

tgagtatccctgggatgacttaaaaaaaagggggggtgctctcccgatttttgaatatgtaggatcattcgccagggtccga

gctgagaattggatgaaaaaaaagggggggtccacgcaatcgcgaaccaacgcggacccaaaggcaagaccgataaaggaga

tcccttttgcggtaatgtgccgggaggctggttacgtagggaagccctaacggacttaataaaaaaaagggggggcttatag

gtcaatcatgttcttgtgaatggatttaaaaaaaaggggggggaccgcttggcgcacccaaattcagtgtgggcgagcgcaa

cggttttggcccttgttagaggcccccgtaaaaaaaagggggggcaattatgagagagctaatctatcgcgtgcgtgttcat

aacttgagttaaaaaaaagggggggctggggcacatacaagaggagtcttccttatcagttaatgctgtatgacactatgta

ttggcccattggctaaaagcccaacttgacaaatggaagatagaatccttgcataaaaaaaagggggggaccgaaagggaag

ctggtgagcaacgacagattcttacgtgcattagctcgcttccggggatctaatagcacgaagcttaaaaaaaaggggggga

ImplantingAAAAAAGGGGGGG with 4 Mutations

atgaccgggatactgatAgAAgAAAGGttGGGggcgtacacattagataaacgtatgaagtacgttagactcggcgccgccg

acccctattttttgagcagatttagtgacctggaaaaaaaatttgagtacaaaacttttccgaatacAAtAAAAcGGcGGGa

tgagtatccctgggatgacttAAAAtAAtGGaGtGGtgctctcccgatttttgaatatgtaggatcattcgccagggtccga

gctgagaattggatgcAAAAAAAGGGattGtccacgcaatcgcgaaccaacgcggacccaaaggcaagaccgataaaggaga

tcccttttgcggtaatgtgccgggaggctggttacgtagggaagccctaacggacttaatAtAAtAAAGGaaGGGcttatag

gtcaatcatgttcttgtgaatggatttAAcAAtAAGGGctGGgaccgcttggcgcacccaaattcagtgtgggcgagcgcaa

cggttttggcccttgttagaggcccccgtAtAAAcAAGGaGGGccaattatgagagagctaatctatcgcgtgcgtgttcat

aacttgagttAAAAAAtAGGGaGccctggggcacatacaagaggagtcttccttatcagttaatgctgtatgacactatgta

ttggcccattggctaaaagcccaacttgacaaatggaagatagaatccttgcatActAAAAAGGaGcGGaccgaaagggaag

ctggtgagcaacgacagattcttacgtgcattagctcgcttccggggatctaatagcacgaagcttActAAAAAGGaGcGGa

Now Where is the Motif?

atgaccgggatactgatagaagaaaggttgggggcgtacacattagataaacgtatgaagtacgttagactcggcgccgccg

acccctattttttgagcagatttagtgacctggaaaaaaaatttgagtacaaaacttttccgaatacaataaaacggcggga

tgagtatccctgggatgacttaaaataatggagtggtgctctcccgatttttgaatatgtaggatcattcgccagggtccga

gctgagaattggatgcaaaaaaagggattgtccacgcaatcgcgaaccaacgcggacccaaaggcaagaccgataaaggaga

tcccttttgcggtaatgtgccgggaggctggttacgtagggaagccctaacggacttaatataataaaggaagggcttatag

gtcaatcatgttcttgtgaatggatttaacaataagggctgggaccgcttggcgcacccaaattcagtgtgggcgagcgcaa

cggttttggcccttgttagaggcccccgtataaacaaggagggccaattatgagagagctaatctatcgcgtgcgtgttcat

aacttgagttaaaaaatagggagccctggggcacatacaagaggagtcttccttatcagttaatgctgtatgacactatgta

ttggcccattggctaaaagcccaacttgacaaatggaagatagaatccttgcatactaaaaaggagcggaccgaaagggaag

ctggtgagcaacgacagattcttacgtgcattagctcgcttccggggatctaatagcacgaagcttactaaaaaggagcgga

Why Finding the Hidden Motif is Difficult

atgaccgggatactgatAgAAgAAAGGttGGGggcgtacacattagataaacgtatgaagtacgttagactcggcgccgccg

acccctattttttgagcagatttagtgacctggaaaaaaaatttgagtacaaaacttttccgaatacAAtAAAAcGGcGGGa

tgagtatccctgggatgacttAAAAtAAtGGaGtGGtgctctcccgatttttgaatatgtaggatcattcgccagggtccga

gctgagaattggatgcAAAAAAAGGGattGtccacgcaatcgcgaaccaacgcggacccaaaggcaagaccgataaaggaga

tcccttttgcggtaatgtgccgggaggctggttacgtagggaagccctaacggacttaatAtAAtAAAGGaaGGGcttatag

gtcaatcatgttcttgtgaatggatttAAcAAtAAGGGctGGgaccgcttggcgcacccaaattcagtgtgggcgagcgcaa

cggttttggcccttgttagaggcccccgtAtAAAcAAGGaGGGccaattatgagagagctaatctatcgcgtgcgtgttcat

aacttgagttAAAAAAtAGGGaGccctggggcacatacaagaggagtcttccttatcagttaatgctgtatgacactatgta

ttggcccattggctaaaagcccaacttgacaaatggaagatagaatccttgcatActAAAAAGGaGcGGaccgaaagggaag

ctggtgagcaacgacagattcttacgtgcattagctcgcttccggggatctaatagcacgaagcttActAAAAAGGaGcGGa

AgAAgAAAGGttGGG

cAAtAAAAcGGcGGG..|..|||.|..|||

Challenge Problem

• Find a motif in a sample of 20 “random” sequences (e.g. 600

nucleotides long).

• Each sequence contains an implanted pattern of length 15.

• Each pattern appears with 4 mismatches.

• More generally, an (n, k) motif is a pattern of length n which

appears with k mismatches within a DNA sequence.

• So our challenge problem is to find a (15,4) motif in a group

of 20 sequences.

Why (15,4)-motif is hard to find?

• Goal: recover original pattern P from its (unknown!) instances:

P1 , P2 , … , P20

• Problem: Although P and Pi are similar for each i (4 mutations for a (15,4) motif), given two different instances Pi and Pj, they may differ twice as much (4 + 4 = 8 mutations for a (15,4) motif).

• Conclusions:

1. Pairwise similiarities are misleading.

2. Multiple similarities are difficult to find.

Combinatorial Gene Regulation

• A microarray experiment showed

that when gene X is knocked out,

20 other genes are not expressed

• Motivating Question: How can

one gene have such drastic

effects?

DNA Microarray

Regulatory Proteins

• Answer: Gene X encodes regulatory protein, a.k.a. a

transcription factor (TF)

• The 20 unexpressed genes rely on gene X’s TF to induce

transcription

• A single TF may regulate multiple genes

Regulatory Regions

• Every gene contains a regulatory region (RR) typically stretching 100-1000 bp upstream of the transcriptional start site.

• Located within the RR are the Transcription Factor Binding Sites(TFBS), also known as motifs, which are specific for a given transcription factor.

• TFs influence gene expression by binding to a specific TFBS.

• A TFBS can be located anywhere within the regulatory region.

• TFBS may vary slightly across different regulatory regions since non-essential bases could mutate.

Transcription Factors and Motifs: Example

geneATCCCG

geneTTCCGG

geneATCCCG

geneATGCCG

geneATGCCC

Transcription Factors and Motifs: Illustration

http://www.cs.uiuc.edu/homes/sinhas/work.html

Motif Logos

• Motifs can mutate on

unimportant bases.

• The five motifs in five different

genes have mutations in position

3 and 5.

• Representations called motif

logos illustrate the conserved

and variable regions of a motif.

• At right is an example of a

motif logo.

TGGGGGA

TGAGAGA

TGGGGGA

TGAGAGA

TGAGGGA

Motif Logos: An Additional Example

(http://www-lmmb.ncifcrf.gov/~toms/sequencelogo.html)

Identifying Motifs

• Recall that a TFBS is represented by a motif.

• Therefore finding similar motifs in multiple genes’ regulatory regions suggests a regulatory relationship among those genes.

Identifying Motifs: Complications

• We do not know the motif sequence in advance.

• We do not know where the motif is located relative to the genes’ start.

• As we have seen, a motif can differ slightly from one gene to the next.

• How do we discern a motif with a real pattern from “random” motifs that don’t represent real correlation between genes?

Detour: A Motif Finding Analogy

• The Motif Finding Problem is similar to the problem posed by

Edgar Allan Poe (1809 – 1849) in his short story “The Gold

Bug.”

The Gold Bug Problem

• “Here Legrand, having re-heated the parchment, submitted it to

my inspection. The following characters were rudely traced, in a

red tint, between the death's head and the goat:”

53++!305))6*;4826)4+.)4+);806*;48!8`60))85;]8*:+*8!83(88)5*!;

46(;88*96*?;8)*+(;485);5*!2:*+(;4956*2(5*-4)8`8*; 4069285);)6

!8)4++;1(+9;48081;8:8+1;48!85;4)485!528806*81(+9;48;(88;4(+?3

4;48)4+;161;:188;+?;

• Legrand’s Goal: Decipher the message on the parchment.

Gold Bug Problem: Assumptions

• The encrypted message is in English

• Each symbol corresponds to one letter in the English alphabet

• Conversely, no letter corresponds to more than one symbol

• No punctuation marks are encoded

Gold Bug Problem: Naïve Approach

• Count the frequency of each symbol in the encrypted message

• Find the frequency of each letter in the alphabet in the English language

• Compare the frequencies of the previous steps, try to find a correlation and map the symbols to a letter in the alphabet

Gold Bug Problem: Symbol Frequencies

• Gold Bug Message:

• English Language:

e t a o i n s r h l d c u m f p g w y b v k x j q z

Most frequent Least frequent

Symbol 8 ; 4 ) + * 5 6 ( ! 1 0 2 9 3 : ? ` - ] .

Frequency 34 25 19 16 15 14 12 11 9 8 7 6 5 5 4 4 3 2 1 1 1

Gold Bug Problem: Symbol Frequencies

• Result of using symbol frequencies:

sfiilfcsoorntaeuroaikoaiotecrntaeleyrcooestvenpinelefheeosnlt

arhteenmrnwteonihtaesotsnlupnihtamsrnuhsnbaoeyentacrmuesotorl

eoaiitdhimtaecedtepeidtaelestaoaeslsueecrnedhimtaetheetahiwfa

taeoaitdrdtpdeetiwt

• The result does not make sense.

• Therefore, we must use some other method to decode the

message.

Gold Bug Problem: l-tuple count

• A better approach is to examine the frequencies of l-tuples,

which are subsequences of 2 symbols, 3 symbols, etc.

• “The” is the most frequent 3-tuple in English and “;48” is the

most frequent 3-tuple in the encrypted text.

• We make inferences of unknown symbols by examining other

frequent l-tuples.

Gold Bug Problem: l-tuple count

• Mapping “the” to “;48” and substituting all occurrences of the

symbols:

53++!305))6*the26)h+.)h+)te06*the!e`60))e5t]e*:+*e!e3(ee)5*!t

h6(tee*96*?te)*+(the5)t5*!2:*+(th956*2(5*h)e`e*th0692e5)t)6!e

)h++t1(+9the0e1te:e+1the!e5th)he5!52ee06*e1(+9thet(eeth(+?3ht

he)h+t161t:1eet+?t

The Gold Bug Message Decoding: Second Attempt

• Make inferences:




he)h+t161t:1eet+?t






he)h+t161t:1eet+?t

• “thet(ee” most likely means “the tree”

• Infer “(“ = “r”






he)h+t161t:1eet+?t

• “thet(ee” most likely means “the tree”

• Infer “(“ = “r”

• “th(+?3h” becomes “thr+?3h”

• Can we guess “+” and “?”?

Gold Bug Problem: Solution

• Using inferences like these to figure out all the mappings, the

final message is:

AGOODGLASSINTHEBISHOPSHOSTELINTHEDEVILSSEATWENYONEDEGRE

ESANDTHIRTEENMINUTESNORTHEASTANDBYNORTHMAINBRANCHSEVENT

HLIMBEASTSIDESHOOTFROMTHELEFTEYEOFTHEDEATHSHEADABEELINE

FROMTHETREETHROUGHTHESHOTFIFTYFEETOUT

Gold Bug Problem: Solution

• Using inferences like these to figure out all the mappings, the

final message is:

AGOODGLASSINTHEBISHOPSHOSTELINTHEDEVILSSEATWENYONEDEGRE

ESANDTHIRTEENMINUTESNORTHEASTANDBYNORTHMAINBRANCHSEVENT

HLIMBEASTSIDESHOOTFROMTHELEFTEYEOFTHEDEATHSHEADABEELINE

FROMTHETREETHROUGHTHESHOTFIFTYFEETOUT

• Punctuation is important:A GOOD GLASS IN THE BISHOP’S HOSTEL IN THE DEVIL’S SEA,

TWENTY ONE DEGREES AND THIRTEEN MINUTES NORTHEAST AND BY NORTH,

MAIN BRANCH SEVENTH LIMB, EAST SIDE, SHOOT FROM THE LEFT EYE OF

THE DEATH’S HEAD A BEE LINE FROM THE TREE THROUGH THE SHOT,

FIFTY FEET OUT.

Gold Bug Problem: Prerequisites

• There are two prerequisites that we need to solve the gold bug problem:

1. We need to know the relative frequencies of single letters, as well as the frequencies of 2-tuples and 3-tuples in English.

2. We also need to know all the words in the English language.

Gold Bug Problem and Motif Finding: Similarities

Motif Finding:

• Nucleotides in motifs encode for a message in the “genetic” language.

• In order to solve the problem, we analyze the frequencies of patterns in the nucleotide sequences

• Knowledge of established regulatory motifs makes the Motif Finding problem simpler.

Gold Bug Problem:

• Symbols in “The Gold Bug” encode for a message in English.

• In order to solve the problem, we analyze the frequencies of patterns in the text written in English

• Knowledge of the words in the English language helps

Gold Bug Problem and Motif Finding: Differences

• Motif Finding is more difficult than the Gold Bug problem:

1. We don’t have the complete dictionary of motifs

2. The “genetic” language does not have a standard

“grammar”

3. Only a small fraction of nucleotide sequences encode for

motifs; the size of the data is enormous

The Motif Finding Problem: Informal Statement

• Given a random sample of DNA sequences:

cctgatagacgctatctggctatccacgtacgtaggtcctctgtgcgaatctatgcgtttccaaccat

agtactggtgtacatttgatacgtacgtacaccggcaacctgaaacaaacgctcagaaccagaagtgc

aaacgtacgtgcaccctctttcttcgtggctctggccaacgagggctgatgtataagacgaaaatttt

agcctccgatgtaagtcatagctgtaactattacctgccacccctattacatcttacgtacgtataca

ctgttatacaacgcgtcatggcggggtatgcgttttggtcgtcgtacgctcgatcgttaacgtacgtc

• Find the pattern that is implanted in each of the individual sequences, namely, the motif

The Motif Finding Problem: Additional Info

• The hidden sequence is of length 8.

• The pattern is not necessarily the same in each array because

random mutations (substitutions) may occur in the sequences,

as we have seen.

The Motif Finding Problem

• The patterns revealed with no mutations:

cctgatagacgctatctggctatccacgtacgtaggtcctctgtgcgaatctatgcgtttccaaccat


aaacgtacgtgcaccctctttcttcgtggctctggccaacgagggctgatgtataagacgaaaatttt


ctgttatacaacgcgtcatggcggggtatgcgttttggtcgtcgtacgctcgatcgttaacgtacgtc

acgtacgt

Consensus String


• The patterns with 2-point mutations:

cctgatagacgctatctggctatccaGgtacTtaggtcctctgtgcgaatctatgcgtttccaaccat

agtactggtgtacatttgatCcAtacgtacaccggcaacctgaaacaaacgctcagaaccagaagtgc

aaacgtTAgtgcaccctctttcttcgtggctctggccaacgagggctgatgtataagacgaaaatttt

agcctccgatgtaagtcatagctgtaactattacctgccacccctattacatcttacgtCcAtataca

ctgttatacaacgcgtcatggcggggtatgcgttttggtcgtcgtacgctcgatcgttaCcgtacgGc


• The patterns with 2-point mutations:






• Can we still find the motif now?

Defining Motifs

• To define a motif, lets say we know where the motif starts in the sequence.

• The motif start positions in their sequences can be represented as s = (s1,s2,s3,…,st).

Motifs: Profiles and Consensus

• Line up the patterns by their start indexes

s = (s1, s2, …, st)

• Construct profile matrix with frequencies of each nucleotide in columns

• Consensus nucleotide in each position has the highest score in column

a G g t a c T tC c A t a c g t

Alignment a c g t T A g ta c g t C c A tC c g t a c g G

_________________

A 3 0 1 0 3 1 1 0Profile C 2 4 0 0 1 4 0 0

G 0 1 4 0 0 0 3 1T 0 0 0 5 1 0 1 4

_________________

Consensus A C G T A C G T

Consensus String

• Think of the consensus string as an “ancestor” motif, from

which mutated motifs emerged

• The distance between a real motif and the consensus sequence

is generally less than the distance between two real motifs

Consensus String

Evaluating Motifs

• We have a guess about the consensus sequence, but how

“good” is this consensus?

• We need to introduce a scoring function to compare different

consensus strings.

• Keep in mind that we really want to choose the starting

positions, but since the consensus is obtained from an array of

starting positions, we will determine how to compare

consensus strings and then work backward to choosing starting

positions.

Parameters: Definitions

• t - number of sample DNA sequences

• n - length of each DNA sequence

• DNA - sample of DNA sequences (stored as a t x n array)

• l - length of the motif (l-mer)

• si - starting position of an l-mer in sequence i

• s = (s1, s2,… st) - array of motif’s starting positions

Parameters: Example






l = 8 (length of the motif)

t=5

DNA

s1 = 26 s2 = 21 s3= 3 s4 = 56 s5 = 60s

n = 69

(starting positions)

Scoring Motifs

• Given starting positions s = (s1, … st ) and DNA:

score(s, DNA) =

• Here count(k, i) represents the frequency of nucleotide k in the motif starting at si

• At right is an example for a given array of starting positions

l

t

l

i GCTAk

ikcount

1 },,,{

),(max

Profile

a G g t a c T t

C c A t a c g t

a c g t T A g t

a c g t C c A t

C c g t a c g G

_________________

A 3 0 1 0 3 1 1 0

C 2 4 0 0 1 4 0 0

G 0 1 4 0 0 0 3 1

T 0 0 0 5 1 0 1 4

_________________

Consensus a c g t a c g t

Score 3+4+4+5+3+4+3+4=30

Finding the Best Profile Matrix

• If starting positions s = (s1, s2,… st ) are given, finding the

consensus is easy even with mutations in the sequences

because we can simply construct the profile matrix in order to

find the consensus string.

• But…the starting positions s are usually not given. How can

we find the “best” profile matrix?

The Motif Finding Problem: Formal Statement

• Goal: Given a set of DNA sequences, find a set of tl-mers, one from each sequence, that maximizes the consensus score

• Input: A t x n matrix of DNA, and l, the length of the pattern to find

• Output: An array of t starting positions s = (s1, s2, … st ) maximizing score(s, DNA)

The Motif Finding Problem: Brute Force Method

• Compute the scores for each possible combination of starting positions s.

• The best score will determine the best profile and the consensus pattern in DNA.

• The goal is to maximize Score(s, DNA) by varying the starting positions si, where:

si = [1, …, n-l+1]

i = [1, …, t]

BruteForceMotifSearch: Pseudocode

1. BruteForceMotifSearch(DNA, t, n, l )

2. bestScore 0

3. for each s=(s1,s2 , . . ., st ) from (1,1 . . . 1) to (n-l+1, . . ., n-l+1)

4. if score (s, DNA) >bestScore

5. bestScore score (s, DNA)

6. bestMotif (s1,s2 , . . . , st)

7. return bestMotif

BruteForceMotifSearch: Running Time

• Varying (n – l + 1) positions in each of t sequences, we’re

looking at (n – l + 1)t sets of starting positions

• For each set of starting positions, the scoring function

makes l operations, so the algorithm’s complexity is

l (n – l + 1)t = O(l nt)

• That means that for t = 8, n = 1000,l = 10 we must perform

approximately 1020 computations – the algorithm will take

billions of years to complete on such a problem instance

BruteForceMotifSearch: Running Time

• Varying (n – l + 1) positions in each of t sequences, we’re

looking at (n – l + 1)t sets of starting positions

• For each set of starting positions, the scoring function

makes l operations, so the algorithm’s complexity is

l (n – l + 1)t = O (l nt )

• That means that for t = 8, n = 1000,l = 10 we must perform

approximately 1020 computations – the algorithm will take

billions of years to complete on such a problem instance

• Conclusion: We need to view the problem in a new light

Changing Gears: The Median String Problem

• Given a set of t DNA sequences, find a pattern that appears in

all t sequences with the minimum number of mutations.

• This pattern will be the motif.

• Key Difference: Rather than varying the starting positions and

trying to find a consensus string representing a motif, we will

instead vary all possible motifs directly.

Hamming Distance

• Given two nucleotide strings v and w, dH(v,w) is the number of

nucleotide pairs that do not match when v and w are aligned.

• For example:

dH(AAAAAA, ACAAAC) = 2

Total Distance: Definition

• For each DNA sequence i, compute all dH(v, x), where x is an l-mer with starting position si (1 < si < n – l + 1)

• Find minimum of dH(v, x) among all l-mers in sequence i

• TotalDistance(v, DNA) is the sum of the minimum Hamming

distances for each DNA sequencei

• TotalDistance(v, DNA) = minsdH(v, s), where s is the set of

starting positions s1, s2,… st

Total Distance: Example

• Given v = “acgtacgt” and s the starting points below:

acgtacgt

cctgatagacgctatctggctatccacgtacAtaggtcctctgtgcgaatctatgcgtttccaaccat

acgtacgt


acgtacgt

aaaAgtCcgtgcaccctctttcttcgtggctctggccaacgagggctgatgtataagacgaaaatttt

acgtacgt


acgtacgt

ctgttatacaacgcgtcatggcggggtatgcgttttggtcgtcgtacgctcgatcgttaacgtaGgtc

v is the sequence in red, x is the sequence in blue



acgtacgt


acgtacgt


acgtacgt


acgtacgt


acgtacgt



dH(v, x) = 1



acgtacgt


acgtacgt


acgtacgt


acgtacgt


acgtacgt



dH(v, x) = 1

dH(v, x) = 0



acgtacgt


acgtacgt


acgtacgt


acgtacgt


acgtacgt



dH(v, x) = 2

dH(v, x) = 1

dH(v, x) = 0



acgtacgt


acgtacgt


acgtacgt


acgtacgt


acgtacgt



dH(v, x) = 2

dH(v, x) = 1

dH(v, x) = 0

dH(v, x) = 0



acgtacgt


acgtacgt


acgtacgt


acgtacgt


acgtacgt



dH(v, x) = 2

dH(v, x) = 1

dH(v, x) = 0

dH(v, x) = 0

dH(v, x) = 1



acgtacgt


acgtacgt


acgtacgt


acgtacgt


acgtacgt



• TotalDistance(v, DNA) = 1+0+2+0+1 = 4

dH(v, x) = 2

dH(v, x) = 1

dH(v, x) = 0

dH(v, x) = 0

dH(v, x) = 1

The Median String Problem: Formulation

• Goal: Given a set of DNA sequences, find a median string.

• Input: A t x n matrix DNA, and l, the length of the pattern to

find.

• Output: A (median) string v of l nucleotides that minimizes

TotalDistance(v, DNA) over all strings of that length.

The Median String Problem: Formulation

• Goal: Given a set of DNA sequences, find a median string.

• Input: A t x n matrix DNA, and l, the length of the pattern to

find.

• Output: A (median) string v of l nucleotides that minimizes

TotalDistance(v, DNA) over all strings of that length.

• Note: This implies the natural brute force algorithm of

calculating TotalDistance(v, DNA) for all strings v (next slide)

MedianStringSearch: Pseudocode

1. MedianStringSearch (DNA, t, n, l )

2. bestWord AAA…A

3. bestDistance ∞

4. for each l-mer v from AAA…A to TTT…T

5. if TotalDistance (v, DNA) < bestDistance

6. bestDistance TotalDistance (v, DNA)

7. bestWord v

8. return bestWord

Motif Finding Problem = Median String Problem

• The Motif Finding is a maximization problem while Median

String is a minimization problem.

• However, the Motif Finding problem and Median String

problem are computationally equivalent.

• We need to show that minimizing TotalDistance is equivalent

to maximizing Score.

Motif Finding Problem == Median String Problem

• At any column iScorei + TotalDistancei = t

• Because there are l columns

Score + TotalDistance = l * t

• Rearranging:

Score = l * t - TotalDistance

• l * t is constant, so the minimization

of the right side is equivalent to the

maximization of the left side

l

t

a G g t a c T t

C c A t a c g t

Alignment a c g t T A g t

a c g t C c A t

C c g t a c g G

_________________

A 3 0 1 0 3 1 1 0

Profile C 2 4 0 0 1 4 0 0

G 0 1 4 0 0 0 3 1

T 0 0 0 5 1 0 1 4

_________________

Consensus a c g t a c g t

Score 3+4+4+5+3+4+3+4

TotalDistance 2+1+1+0+2+1+2+1

Sum 5 5 5 5 5 5 5 5

Motif Finding Problem vs. Median String Problem

• Why bother reformulating the Motif Finding problem into the

Median String problem?

• The Motif Finding Problem needs to examine all possible choices for s. Recall that this is (n — l + 1)t possibilities!!!

• The Median String Problem needs to examine all 4l

combinations for v. This number is typically smaller, although if l is large using brute force will still be

infeasible.

Median String: Improving the Running Time

1. MedianStringSearch (DNA, t, n, l )

2. bestWord AAA…A

3. bestDistance ∞

4. for each l-mer v from AAA…A to TTT…T

5. if TotalDistance (v, DNA) < bestDistance

6. bestDistance TotalDistance (v, DNA)

7. bestWord v

8. return bestWord

Structuring the Search

• For the Median String Problem we need to consider all 4l

possible l-mers:

aa… aa

aa… ac

aa… ag

aa… at

.

.

tt… tt

How to organize this search?

l

Alternative Representation of the Search Space

• Let A = 1, C = 2, G = 3, T = 4

• Then the sequences from AA…A to TT…T become:

11…11

11…12

11…13

11…14

.

.

44…44

• Notice that the sequences above simply list all numbers as if we

were counting on base 4 without using 0 as a digit.

l

First Try: Linked List

• Suppose l = 2

aa ac ag at ca cc cg ct ga gc gg gt ta tc tg tt

• We need to visit all the predecessors of a sequence before

visiting the sequence itself.

• Unfortunately this isn’t very efficient.

Start

Better Structure: Search Tree

• Instead, let’s try grouping the sequences by their prefixes.



• Instead, let’s try grouping the sequences by their prefixes


a- c- g- t-


• Instead, let’s try grouping the sequences by their prefixes


a- c- g- t-

--

root

Analyzing Search Trees

• Characteristics of search trees:

• The sequences are contained in its leaves

• The parent of a node is the prefix of its children

• How can we move through the tree?

Moving through the Search Trees

• Four common moves in a search tree that we are about to

explore:

1. Move to the next leaf

2. Visit all the leaves

3. Visit the next node

4. Bypass the children of a node

Move 1: Visit the Next Leaf

1. NextLeaf( a, L, k ) // a : the array of digits2. for i L to 1 // L: length of the array3. if ai <k // k : max digit value4. aiai + 15. return a6. ai 17. return a

• Given a current leaf a , we need to compute the “next” leaf:

Note: In the case of the nucleotide alphabet, we are using k = 4

Move 1: Visit the Next Leaf

• The algorithm is common addition base-k:

1. Increment the least significant digit

2. “Carry the one” to the next digit position when the digit is at maximal value

NextLeaf: Example

• Moving to the next leaf:

1- 2- 3- 4-

11 12 13 14 21 22 23 24 31 32 33 34 41 42 43 44

--Current Location

NextLeaf: Example


1- 2- 3- 4-

11 12 13 14 21 22 23 24 31 32 33 34 41 42 43 44

\

--Next Location

NextLeaf: Example


1- 2- 3- 4-

11 12 13 14 21 22 23 24 31 32 33 34 41 42 43 44

--Next Location

NextLeaf: Example


• Etc.

1- 2- 3- 4-

11 12 13 14 21 22 23 24 31 32 33 34 41 42 43 44

--Next Location

Move 2: Visit All Leaves

• Printing all l-mers in ascending order:

1. AllLeaves(L, k) // L: length of the sequence

2. a (1,...,1) // k : max digit value

3. while forever // a: array of digits

4. output a

5. a NextLeaf(a, L, k)

6. if a = (1,...,1)

7. return

Visit All Leaves: Example

• Moving through all the leaves in order:

1- 2- 3- 4-

11 12 13 14 21 22 23 24 31 32 33 34 41 42 43 44

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

--

Order of steps

Depth First Search

• So we can search through the leaves.

• How about searching through all vertices of the tree?

• We will do this with a depth firstsearch.

• Specifically we need an algorithm to visit the next vertex.

Move 3: Visit the Next Vertex

1. NextVertex(a, i, L, k ) // a: the array of digits2. if i < L // i : prefix length3. ai+1 1 // L: max length4. return (a, i +1 ) // k: max digit value5. else6. for j l to 17. if aj < k8. aj aj +19. return (a, j )10. return (a,0)

Next Vertex: Example

• Moving to the next vertex:

1- 2- 3- 4-

11 12 13 14 21 22 23 24 31 32 33 34 41 42 43 44

--Current Location



1- 2- 3- 4-

11 12 13 14 21 22 23 24 31 32 33 34 41 42 43 44

--



1- 2- 3- 4-

11 12 13 14 21 22 23 24 31 32 33 34 41 42 43 44

--

Location after 5

next vertex moves

Move 4: Bypass

• Given a prefix (internal vertex), find the next vertex after

skipping all the prefix’s children.

1. ByPass(a, i, L, k) // a: array of digits

2. for j i to 1 // i : prefix length

3. if aj < k // L: maximum length

4. ajaj +1 // k : max digit value

5. return (a, j )

6. return (a, 0)

Bypass: Example

• Bypassing the descendants of “2-”:

1- 2- 3- 4-

11 12 13 14 21 22 23 24 31 32 33 34 41 42 43 44

--Current Location

Bypass: Example

• Bypassing the descendants of “2-”:

1- 2- 3- 4-

11 12 13 14 21 22 23 24 31 32 33 34 41 42 43 44

--Next Location

Revisiting Brute Force Search

• Now that we have method for navigating the tree, lets look again

at BruteForceMotifSearch.

Brute Force Search Revisited

1. BruteForceMotifSearchAgain(DNA, t, n, l)

2. s (1,1,…, 1)


4. while forever

5. s NextLeaf (s, t, n — l +1 )6. if score (s,DNA) >bestScore


8. bestMotif (s1,s2 , . . . , st)

9. return bestMotif

Can We Streamline the Algorithm?

• Sets of s = (s1, s2, …, st) may have a weak profile for the first ipositions (s1, s2, …, si)

• Every row of the alignment matrix may add at most l to score

• Optimistic View: all subsequent (t-i) positions (si+1, …st) add

(t – i ) * l to score(s, i, DNA)

• So if score(s, i, DNA) + (t – i ) * l < bestScore, it makes no sense to search through vertices of the current subtree

• In this case, we can use ByPass() to skip frivolous branches.

• This leads to what is called in general a “branch and bound” method for motif finding.

Branch and Bound Algorithm for Motif Search

• Since each level of the tree goes

deeper into search, discarding a

prefix discards all following

branches.

• This saves us from looking at

(n – l + 1)t-i leaves.

• We use NextVertex() and

ByPass() to navigate the tree.

Branch and Bound Motif Search: Pseudocode

1. BranchAndBoundMotifSearch(DNA, t, n,l )

2. s (1,…,1)

3. bestScore 0

4. i 1

5. while i > 0

6. if i <t

7. optimisticScore score(s, i, DNA) + (t – i ) * l

8. if optimisticScore < bestScore

9. (s, i) ByPass(s, i, n – l +1)

10. else

11. (s, i) NextVertex(s, i, n – l +1)

12. else

13. if score(s, DNA) > bestScore

14. bestScore score(s)

15. bestMotif (s1, s2, s3, …, st )

16. (s, i) NextVertex(s, i, t, n – l + 1)

17. return bestMotif

Median String Search Improvements

• Recall the computational differences between motif search and

median string search.

• The Motif Finding Problem needs to examine all (n – l +1)t

combinations for s.

• The Median String Problem needs to examine 4l

combinations of v. This number is usually relatively small.

• We want to use the median string algorithm with the Branch

and Bound trick!

Median String Search with Branch and Bound

• Note that if the total distance for a prefix is greater than that

for the best word so far:

TotalDistance (prefix, DNA) >BestDistance

then there is no use exploring the remaining part of the word.

• We can eliminate that branch and BYPASS exploring that

branch further.

Median String Search with Branch and Bound

1. BranchAndBoundMedianStringSearch(DNA, t, n, l )2. s (1,…,1)3. bestDistance ∞4. i 15. while i > 06. if i < l7. prefix string corresponding to the first i nucleotides of s8. optimisticDistance TotalDistance(prefix, DNA)9. if optimisticDistance > bestDistance10. (s, i ) ByPass(s, i, l, 4)11. else12. (s, i )NextVertex(s, i, l, 4)13. else 14. word nucleotide string corresponding to s15. if TotalDistance(s, DNA ) < bestDistance16. bestDistance TotalDistance(word, DNA)17. bestWord word18. (s, i ) NextVertex(s, i, l, 4)19. return bestWord

Improving the Bounds

• Given an l-mer w, divided into two parts at point i:

• u : prefix w1, …, wi,

• v : suffix wi+1, ..., wl

• Find minimum distance for u in a sequence.

• No instances of u in the sequence have distance less than the minimum distance.

• Note: this doesn’t tell us anything about whether u is part of any motif. We only get a minimum distance for prefix u.


• Repeating the process for the suffix v gives us a

minimum distance for v.

• Since u and v are two substrings of w, and included in

motif w, we can assume that the minimum distance of

u plus minimum distance of v can only be less than

the minimum distance for w.


• If d(prefix) + d(suffix) > bestDistance:

• Motif w (prefix.suffix) cannot give a better (lower) score

than d(prefix) + d(suffix)

• In this case, we can ByPass()

Better Bounded Median String Search

1. ImprovedBranchAndBoundMedianString(DNA, t, n, l )2. s= (1, 1, …, 1)3. bestDistance = ∞4. i = 15. while i > 06. if i < l7. prefix = nucleotide string corresponding to (s1, s2, s3, …, si )8. optimisticPrefixDistance = TotalDistance (prefix, DNA )9. if optimisticPrefixDistance <bestSubstring [i ]10. bestSubstring[i ] = optimisticPrefixDistance11. if (l – i <i )12. optimisticSuffixDistance = bestsubstring [l -i ] 13. else14. optimisticSuffixDistance = 0;15. if optimisticPrefixDistance + optimisticSuffixDistance > bestDistance16. (s, i ) = ByPass(s, i, l, 4)17. else18. (s, i ) = NextVertex(s, i, l, 4)19. else20. word = nucleotide string corresponding to (s1,s2, s3, …, st)21. if TotalDistance(word, DNA ) < bestDistance22. bestDistance = TotalDistance(word, DNA)23. bestWord = word24. (s,i ) = NextVertex(s, i, l, 4)25. return bestWord

Notes on Motif Finding

• Exhaustive Search and Median String are both exact

algorithms.

• They always find the optimal solution, though they may be too

slow to perform practical tasks.

• Many algorithms sacrifice optimal solution for speed.

CONSENSUS: Greedy Motif Search

• Finds two closest l -mers in sequences 1 and 2 and forms

2 x l alignment matrix with score(s, 2, DNA).

• At each of the following t-2 iterations CONSENSUS finds a “best” l -mer in sequence i from the perspective of the already

constructed (i-1) x l alignment matrix for the first (i-1)

sequences.

• In other words, it finds an l - mer in sequence i maximizing

score(s, i, DNA)

under the assumption that the first (i-1) l -mers have been

already chosen.

• CONSENSUS sacrifices optimal solution for speed: in fact the bulk of the time is actually spent locating the first 2 l -mers.

Some Motif Finding Programs

• CONSENSUS

Hertz, Stromo (1989)

• GibbsDNA

Lawrence et al (1993)

• MEME

Bailey, Elkan (1995)

• RandomProjections

Buhler, Tompa (2002)

• MULTIPROFILER Keich,

Pevzner (2002)

• MITRA

Eskin, Pevzner (2002)

• Pattern Branching

Price et al.,

lo (2003)

ch04 motifs

Documents

implanted motif

bound motif search10

greedy motif search12

brute force motif finding7

median string problem8

median string search11

search trees9

regulatory motifs4