1. CSE 5290: Algorithms for Bioinformatics Fall 2011
Suprakash Datta
Office: CSEB 3043
Phone: ext 77875
Course page:
11/13/2018
CSE 5290, Fall 2011

2. Biological (genomic) data
ATATTGAATTTTCAAAAATTCTTACTTTTTTTTTGGATGGACGCAAAGAAGTTTAATAATCATATTACATGGCATTACCACCATATATATCCATATCTAATCTTACTTATATGTTGTGGAAATGTAAAGAGCCCCATTATCTTAGCCTAAAAAAACCTTCTCTTTGGAACTTTCTAATACGCTTAACTGCTCATTGCTATATTGAAGTACGGATTAGAAGCCGCCGAGCGGGCGACAGCCCTCCGACGGAAGACTCTCCTCTGCGTCCTCGTCTTCACCGGTCGCGTTCCTGAAACGCAGATGTGCCTCGCGCCGCACTGCTCCGAACAATAAAGATTCTACAATACTCTTTTATGGTTATGAAGAGGAAAAATTGGCAGTAACCTGGCCCCACAAACCTTCAAATTAACGAATCAAATTAACAACCATAGGATGAATGCGATTAGTTTTTTAGCCTTATTTCTGGGGTAATTAATCAGCGAAGCGATGATTTTTGATCTATTAACAGATATATAAATGGAAGCTGCATAACCACTTTAACTAATACTTTCAACATTTTCAGTTTGTATTACTTCTTATTCAAATGTCATAAAAGTATCAACAAAAAATTTAATATACCTCTATACTTTAACGTCAAGGAGAAAAAACTATAATGACTAAATCTCATTCAGAAGAAGTGATTGTACCTGAGTTCAACTAGCGCAAAGGAATTACCAAGACCATTGGCCGAAAAGTGCCCGAGCATAATTAAGAAATTTATAAGCGCTTATGATGCTAAACCGGTTTGTTGCTAGATCGCCTGGTAGAGTCAATCTAATTGGTGAACATATTGATTATTGTGACTTCTCGGTTTTACCTTTAGCTATTGATTGATATGCTTTGCGCCGTCAAAGTTTTGAACGAGAAAAATCCATCCATTACCTTAATAAATGCTGATCCCAAATTTGCTCAAAGGAATCGATTTGCCGTTGGACGGTTCTTATGTCACAATTGATCCTTCTGTGTCGGACTGGTCTAATTACTTTAAATGTGGTCTCCATGTTGCACTCTTTTCTAAAGAAACTTGCACCGGAAAGGTTTGCCAGTGCTCCTCTGGCCGGGCTGCAAGTCTTCTGTGAGGGTGATGTACCATGGCAGTGATTGTCTTCTTCGGCCGCATTCATTTGTGCCGTTGCTTTAGCTGTTGTTAAAGCGAATATGGGCCCTGGTTATCATATCCAAGCAAAATTTAATGCGTATTACGGTCGTTGCAGAACATTATGTTGGTGTTAACAATGGCGGTATGGATCAGGCTGCCTCTGTTTGGTGAGGAAGATCATGCTCTATACGTTGAGTTCAAACCGCAGTTGAAGGCTACTCCGTTTAAATTTCCGCAATTAAAAAACCATGAATAGCTTTGTTATTGCGAACACCCTTGTTGTATCTAACAAGTTTGAAACCGCCCCAACCAACTATAATTTAAGAGTGGTAGAAGTCACCAGCTGCAAATGTTTTAGCTGCCACGTACGGTGTTGTTTTACTTTCTGGAAAAGAAGGATCGAGCACGAATAAAGGTAATCTAAGAGTTCATGAACGTTTATTATGCCAGATATCACAACATTTCCACACCCTGGAACGGCGATATTGAATCCGGCATCGAACGGTTAACAAAGGCTAGTACTAGTTGAAGAGTCTCTCGCCAATAAGAAACAGGGCTTTAGTGTTGACGATGTCGCACAATCCTTGAATTGTTCTCGCGAAATTCACAAGAGACTACTTAACAACATCTCCAGTGAGATTTCAAGTCTTAAAGCTATATCAGAGGGCTAAGCATGTGTATTCTGAATTTAAGAGTCTTGAAGGCTGTGAAATTAATGACTACAGCGAGCTTTACTGCCGACGAAGACTTTTTCAAGCAATTTGGTGCCTTGATGCGAGTCTCAAGCTTCTTGCGATAAACTTTACGAATGTTCTTGTCCAGAGATTGACAAAATTTGTTCCATTGCTTTGTCAAATGGATCATGGTTCCCGTTTGACCGGAGCTGGCTGGGGTGGTTGTACTGTTCACTTGGTTCCAGGGGGCCCAAATGGCAACATAGAAAAGGTAAGAAGCCCTTGCCAATGAGTTCTACAAGGTCAAGTACCCTAAGATCACTGATGCTGAGCTAGAAAATGCTATCATCGTCTCTAAACCAATTGGGCAGCTGTCTATATGAATTATAAGTATACTTCTTTTTTTTACTTTGTTCAGAACAACTTCTCATTTTTTTCTACTCATAACTAGCATCACAAAATACGCAATAATAACGAGTAGTAACACTTTTATAGTTCATACATGCTTCAACTACTTAATAAATGATTGTATGATAGTTTTCAATGTAAGAGATTTCGATTATCCACAAACTTTAAAACACAGGGACAAAATTCTTGATATGCTTTCAACCGCTGCGTTTTGGACCTATTCTTGACATGATATGACTACCATTTTGTTATTGTACGTGGGGCAGTTGACGTCTTATCATATGTCAAAGTCATTTGCGAAGCTTGGCAAGTTGCCAACTGACGAGATGCAGTAAAAAGAGATTGCCGTCTTGAAACTTTTTGTCCTTTTTTTTTTCCGGGGACTCTACGAACCCTTTGTCCTACTGATTAATTTTGTACTGAATTTGGACAATTCAGATTTTAGTAGACAAGCGCGAGGAGGAAAAGAAATGACAAAAATTCCGATGGACAAGAAGATAGGAAAAAAAAAAAGCTTTCACCGATTTCCTAGACCGGAAAAAAGTCGTATGACATCAGAATGAAATTTTCAAGTTAGACAAGGACAAAATCAGGACAAATTGTAAAGATATAATAAACTATTTGATTCAGCGCCAATTTGCCCTTTTCCATTCCATTAAATCTCTGTTCTCTCTTACTTATATGATGATTAGGTATCATCTGTATAAAACTCCTTTCTTAATTTCACTCTAAAGCATCCCATAGAGAAGATCTTTCGGTTCGAAGACATTCCTACGCATAATAAGAATAGGAGGGAATAATGCCAGACAATCTATCATTACATTAGCGGCTCTTCAAAAAGATTGAACTCTCGCCAACTTATGGAATCTTCCAATGAGACCTTTGCGCCAAATAATGTGGATTTGGAAAAAGTATAAGTCATCTCAGAGTAATATAACTACCGAAGTTTATGAGGCATCGAGCTTTGAAGAAAAAGTAAGCTCAGAAAAACCTCAATAGCTCATTCTGGAAGAAAATCTATTATGAATATGTGGTCGTTGACAAATCAATCTTGGGTGTTTCTATTCTGGATTCATTTATGTACACAGGACTTGAAGCCCGTCGAAAAAGAAAGGCGGGTTTGGTCCTGGTACAATTATTGTTACTTCTGGCTTGCTGAATGTTTCAATATCCACTTGGCAAATTGCAGCTACAGGTCTACAACTGGGTCTAAATTGGTGGCAGTGTTGGATAACAATTTGGATTGGGTACGGTTTCGTGTGCTTTTGTTGTTTTGGCCTCTAGAGTTGGATCTGCTTATCATTTGTCATTCCCTATATCATCTAGAGCATCATTCGGTATTTTCT
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3. Annotated data 11/13/2018 CSE 5290, Fall 2011
ATATTGAATTTTCAAAAATTCTTACTTTTTTTTTGGATGGACGCAAAGAAGTTTAATAATCATATTACATGGCATTACCACCATATATATCCATATCTAATCTTACTTATATGTTGTGGAAATGTAAAGAGCCCCATTATCTTAGCCTAAAAAAACCTTCTCTTTGGAACTTTCTAATACGCTTAACTGCTCATTGCTATATTGAAGTACGGATTAGAAGCCGCCGAGCGGGCGACAGCCCTCCGACGGAAGACTCTCCTCTGCGTCCTCGTCTTCACCGGTCGCGTTCCTGAAACGCAGATGTGCCTCGCGCCGCACTGCTCCGAACAATAAAGATTCTACAATACTCTTTTATGGTTATGAAGAGGAAAAATTGGCAGTAACCTGGCCCCACAAACCTTCAAATTAACGAATCAAATTAACAACCATAGGATGAATGCGATTAGTTTTTTAGCCTTATTTCTGGGGTAATTAATCAGCGAAGCGATGATTTTTGATCTATTAACAGATATATAAATGGAAGCTGCATAACCACTTTAACTAATACTTTCAACATTTTCAGTTTGTATTACTTCTTATTCAAATGTCATAAAAGTATCAACAAAAAATTTAATATACCTCTATACTTTAACGTCAAGGAGAAAAAACTATAATGACTAAATCTCATTCAGAAGAAGTGATTGTACCTGAGTTCACTAGCGCAAAGGAATTACCAAGACCATTGGCCGAAAAGTGCCGAGCATAATTAAGAAATTTATAAGCGCTTATGATGCTAAACCGAGTTGTATGTATTTGGCCTTATGTAGCTCGCGCCCGTTCGAGATAAGGATGTTTCTAGAAATCCGTAAAGATATAGAGATGTACACACATCTACATTTGTAACTCTATTTATAGTTAGAAACTTGTCCTCGAGGTCTCTCTATAAACCTTTTTGTACGTCCATAAATGTGGAAATCTACCGATCTTTTTGTCTCCGTATATAGGGAAACAGCGTTTTCGCTATACCTGGGTACAAACAGAGTTTTGTAGCTCCACGTTTCGCTGTCTCGTTCCGTGGAGCCCTGGGGGTCCTTAGACATATACTCTTTTTACATAGTTGGATGGGGGTCTGTACCTAGTTAGCTCTAGCTCGAGACAGCGATAGAGAATTTTGTATACTTGTCCGTTTACGTGTACCCGGCGATCTGTCTATGTCTATGGATAGCCACGTTTATGTCGTTTTGTAGGTCGTTGTATATCGATATATAGAGCGCGGATAATTAGGTAGGTCGACCGCGCTGTGGCTCTATCTCTAGTTATTTGTAGGTCGATGTGTAGATGTAATTCTAGCTGGACATCCATACCTACCTGTGTTTGTAGGTATTTCCATAAAACCACAGCGATGTTTGTAGAAAACGCGCGCTACCCCTACACCGCTATATACATAATATATCTCTGTACAAAGATGTATAGAGATAAAGACACAGTTCGAAACCTATCGACTTGGACAAACAGTTGTTTATTTTTAAGTCGCTCGACCGAACTAGTTACACCGAGATCGATTTGTTTCTCTATACACCTCTCTCTGTGGAGAAACAGAGCGAGAAGTAGATTTCGAGAAGCCACCGGGACAATTACAGAAAGCGGTAGATTTACATACAAAGAAGGAGACTTATCGATACACATAGAGGTATCGATAACGATGTATACCTACATCCAGCTCCATACCTAAAGGTAGAAAGACATGTGTCGACATGTTTACGTTTAGATATGGACCTAGATATGTCTACGGACACGAGACTTACACGTCTAGATAGTTGTGTATTTCTCTCTGGACAGATCTGTAAAGGTACGTCTACATGTCGATATCGGTCTGTCGGTAGATAAAAATCTATATTTAGACCGATAACTAGGTCTCGATGTCGTTCTCTAACGATGGACCTGTAGACCGAAAAAGAACTTTTTGTTTTCCACAAGTCTAGACTTTTTTGGGTCTAGACACTTGTGGAATTCGAGATAGGGCTCTCCCTCTGGCTCTATAGATCGACGGGTATCGCTATCGAGACGTGGGTCGAGATGGGTATCTCGCTATACATGGATTTCCAACCTTGTAGGTGTCTCTCGAAGGCGGTAGGGACACAAAATAGCTGTAGCTACAACTACGTATCGATACATAAAGAGCTACAAATTGGGCAGCTGTCTATATGAATTATAAGTATACTTCTTTTTTTTACTTTGTTCAGAACAACTTCTCATTTTTTTCTACTCATAACTAGCATCACAAAATACGCAATAATAACGAGTAGTAACACTTTTATAGTTCATACATGCTTCAACTACTTAATAAATGATTGTATGATAGTTTTCAATGTAAGAGATTTCGATTATCCACAAACTTTAAAACACAGGGACAAAATTCTTGATATGCTTTCAACCGCTGCGTTTTGGACCTATTCTTGACATGATATGACTACCATTTTGTTATTGTACGTGGGGCAGTTGACGTCTTATCATATGTCAAAGTCATTTGCGAAGCTTGGCAAGTTGCCAACTGACGAGATGCAGTAAAAAGAGATTGCCGTCTTGAAACTTTTTGTCCTTTTTTTTTTCCGGGGACTCTACGAACCCTTTGTCCTACTGATTAATTTTGTACTGAATTTGGACAATTCAGATTTTAGTAGACAAGCGCGAGGAGGAAAAGAAATGACAAAAATTCCGATGGACAAGAAGATAGGAAAAAAAAAAAGCTTTCACCGATTTCCTAGACCGGAAAAAAGTCGTATGACATCAGAATGAAATTTTCAAGTTAGACAAGGACAAAATCAGGACAAATTGTAAAGATATAATAAACTATTTGATTCAGCGCCAATTTGCCCTTTTCCATTCCATTAAATCTCTGTTCTCTCTTACTTATATGATGATTAGGTATCATCTGTATAAAACTCCTTTCTTAATTTCACTCTAAAGCATACCGCCAGGTACGTACGTATACAGAAATACATGTATCTGTGGATATCCGTACATCGAGCCACATATCCCTTTAACTGGCGAAATATACTTATACCGAAAATTAGAGGGAACGCGGTATATGTACGACCGACACAATGAAACTAGATTGCGTAATTTCTAGTGTAAACAAATATGGCTATCTAAATGTCTCTAGGTACATCGAAAGAAAGTTACATATATTTAAATCGATAACTACGTAGATGGGTTTCTAGTTGTAGAGCGACAAATCTCGAAAGCTCTTTTTGGAGAGGTAGATATATAGTATATATCGCTGTCGAAGTATACAAATATCTACTTCGATAACTAACCAACGGTATCGGTCTAGAAAAGTCTCGCCAGGTCCGTAAACAAAGAGGTACATAACGAGACCGGTGGGTGTTTCGGTATACACTTGTGGGTATCGAGACATGTATGTTTGTGTGTAACTATATCCAAGGTCTTTGTGGACTTGTAGAGGTGTATCTGTCGATATTTACGTC
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4. What do we do with genomes?
Inter-species comparison
Intra-species comparison
Diseases from a genomic standpoint, drug/therapy design
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5. Importance of algorithms
– Compare human vs. mouse (blocks of 1,000 nucleotides)
• 3,000,000*3,000,000 comparisons, each 1,000*1,000 operations (w/dynamic progr.)
• At 1 trillion operations per second, it would take 104 days
– Search all regulatory motifs of length 20 in the human genome:
• 426 years
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6. Clustering flow cytometry data
1 million vectors
Each of length 25 (real numbers)
Need quick output!
Results should be biologically meaningful!
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7. R: introduction
Why R?
Lots of available libraries (statistics, machine learning,…..)
Very good visualization capability
Free
Multiplatform
Easy to publish code
Biologists use it!
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8. R - contd Grew out of a popular statistics package
Used extensively by statisticians and computational biologists
Lots of resources (see class web page)
Some similarities with MatLab
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9. R – strengths and weaknesses
Allows very quick testing of ideas
Libraries available for most purposes
Allows integration with C code
Weaknesses
Not as efficient as MatLab on matrix operations
Not very good at handling large data sets
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10. Next:
Ch 3 of text: Some techniques that biologists use for data gathering
Things to do…
Read Ch 1 and 2 on your own.
Get familiar with R
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11. Analyzing a Genome How to analyze a genome in four easy steps.
Cut it
Use enzymes to cut the DNA in to small fragments.
Copy it
Copy it many times to make it easier to see and detect.
Read it
Use special chemical techniques to read the small fragments.
Assemble it
Take all the fragments and put them back together. This is hard!!!
Bioinformatics takes over
What can we learn from the sequenced DNA.
Compare interspecies and intraspecies.
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12. 8.1 Copying DNA
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13. Why we need so many copies
Biologists needed to find a way to read DNA codes.
How do you read base pairs that are angstroms in size?
It is not possible to directly look at it due to DNA’s small size.
Need to use chemical techniques to detect what you are looking for.
To read something so small, you need a lot of it, so that you can actually detect the chemistry.
Need a way to make many copies of the base pairs, and a method for reading the pairs.
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14. Polymerase Chain Reaction (PCR)
Used to massively replicate DNA sequences.
How it works:
Separate the two strands with low heat
Add some base pairs, primer sequences, and DNA Polymerase
Creates double stranded DNA from a single strand.
Primer sequences create a seed from which double stranded DNA grows.
Now you have two copies.
Repeat. Amount of DNA grows exponentially.
1→2→4→8→16→32→64→128→256…
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15. Polymerase Chain Reaction
Problem: Modern instrumentation cannot easily detect single molecules of DNA, making amplification a prerequisite for further analysis
Solution: PCR doubles the number of DNA fragments at every iteration
1… … … …
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16. Denaturation
Raise temperature to 94oC to separate the duplex form of DNA into single strands
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17. Design primers
To perform PCR, a 10-20bp sequence on either side of the sequence to be amplified must be known because DNA polymerase requires a primer to synthesize a new strand of DNA
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18. Annealing
Anneal primers at 50-65oC
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19. Annealing
Anneal primers at 50-65oC
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20. Extension
Extend primers: raise temp to 72oC, allowing Taq polymerase to attach at each priming site and extend a new DNA strand
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21. Extension
Extend primers: raise temp to 72oC, allowing Taq polymerase to attach at each priming site and extend a new DNA strand
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22. Repeat
Repeat the Denature, Anneal, Extension steps at their respective temperatures…
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23. Polymerase Chain Reaction
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24. Cloning DNA DNA Cloning Use Polymerase Chain Reaction (PCR)
Insert the fragment into the genome of a living organism and watch it multiply.
Once you have enough, remove the organism, keep the DNA.
Use Polymerase Chain Reaction (PCR)
Vector DNA
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25. 8.2 Cutting and Pasting DNA
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26. Restriction Enzymes Discovered in the early 1970’s
Used as a defense mechanism by bacteria to break down the DNA of attacking viruses.
They cut the DNA into small fragments.
Can also be used to cut the DNA of organisms.
This allows the DNA sequence to be in a more manageable bite-size pieces.
It is then possible using standard purification techniques to single out certain fragments and duplicate them to macroscopic quantities.
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27. Cutting DNA Restriction Enzymes cut DNA
Only cut at special sequences
DNA contains thousands of these sites.
Applying different Restriction Enzymes creates fragments of varying size.
Restriction Enzyme “A” Cutting Sites
Restriction Enzyme “B” Cutting Sites
“A” and “B” fragments overlap
Restriction Enzyme “A” & Restriction Enzyme “B” Cutting Sites
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28. Pasting DNA
Two pieces of DNA can be fused together by adding chemical bonds
Hybridization – complementary base-pairing
Ligation – fixing bonds with single strands
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29. 8.3 Measuring DNA Length
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30. Electrophoresis
A copolymer of mannose and galactose, agarose, when melted and recooled, forms a gel with pores sizes dependent upon the concentration of agarose
The phosphate backbone of DNA is highly negatively charged, therefore DNA will migrate in an electric field
The size of DNA fragments can then be determined by comparing their migration in the gel to known size standards.
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31. Reading DNA Electrophoresis
Reading is done mostly by using this technique. This is based on separation of molecules by their size (and in 2D gel by size and charge).
DNA or RNA molecules are charged in aqueous solution and move to a definite direction by the action of an electric field.
The DNA molecules are either labeled with radioisotopes or tagged with fluorescent dyes. In the latter, a laser beam can trace the dyes and send information to a computer.
Given a DNA molecule it is then possible to obtain all fragments from it that end in either A, or T, or G, or C and these can be sorted in a gel experiment.
Another route to sequencing is direct sequencing using gene chips.
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32. Assembling Genomes Must take the fragments and put them back together
Not as easy as it sounds.
SCS Problem (Shortest Common Superstring)
Some of the fragments will overlap
Fit overlapping sequences together to get the shortest possible sequence that includes all fragment sequences
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33. Assembling Genomes DNA fragments contain sequencing errors
Two complements of DNA
Need to take into account both directions of DNA
Repeat problem
50% of human DNA is just repeats
If you have repeating DNA, how do you know where it goes?
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34. 8.4 Probing DNA
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35. DNA probes
Oligonucleotides: single-stranded DNA nucleotides long
Oligonucleotides used to find complementary DNA segments.
Made by working backwards---AA sequence----mRNA---cDNA.
Made with automated DNA synthesizers and tagged with a radioactive isotope.
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36. DNA Hybridization
Single-stranded DNA will naturally bind to complementary strands.
Hybridization is used to locate genes, regulate gene expression, and determine the degree of similarity between DNA from different sources.
Hybridization is also referred to as annealing or renaturation.
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37. Create a Hybridization Reaction
1. Hybridization is binding two genetic sequences. The binding occurs because of the hydrogen bonds [pink] between base pairs.
2. When using hybridization, DNA must first be denatured, usually by using use heat or chemical. T C T A G C G T C A T G T
TAGGC T
ATCCGACAATGACGCC
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38. Create a Hybridization Reaction - 2
Once DNA has been denatured, a single-stranded radioactive probe [light blue] can be used to see if the denatured DNA contains a sequence complementary to probe.
Sequences of varying homology stick to the DNA even if the fit is poor.
ACTGC
ACTGC
ATCCGACAATGACGCC
Great Homology
ACTGC
ATCCGACAATGACGCC
Less Homology
ATTCC
ATCCGACAATGACGCC
ACCCC
Low Homology
ATCCGACAATGACGCC
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39. Labeling technique for DNA arrays
RNA samples are labeled using fluorescent nucleotides (left) or
radioactive nucleotides (right), and hybridized to arrays. For fluorescent
labeling, two or more samples labeled with differently colored fluorescent
markers are hybridized to an array. Level of RNA for each gene in the
sample is measured as intensity of fluorescence or radioactivity binding
to the specific spot. With fluorescence labeling, relative levels of expressed
genes in two samples can be directly compared with a single array.
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40. DNA Arrays--Technical Foundations
An array works by exploiting the ability of a given mRNA molecule to hybridize to the DNA template.
Using an array containing many DNA samples in an experiment, the expression levels of hundreds or thousands genes within a cell by measuring the amount of mRNA bound to each site on the array.
With the aid of a computer, the amount of mRNA bound to the spots on the microarray is precisely measured, generating a profile of gene expression in the cell.
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41. An experiment on a microarray
In this schematic:   GREEN represents Control DNA
RED represents Sample DNA
  YELLOW represents a combination of Control and Sample DNA
  BLACK represents areas where neither the Control nor Sample DNA
  Each color in an array represents either healthy (control) or diseased (sample) tissue. The location and intensity of a color tell us whether the gene, or mutation, is present in the control and/or sample DNA.
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42. DNA Microarray
Millions of DNA strands
build up on each location.
Tagged probes become hybridized to the DNA chip’s microarray.
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43. DNA Microarray
Affymetrix
Microarray is a tool for analyzing gene expression that consists of a glass slide.
Each blue spot indicates the location of a PCR product. On a real microarray, each spot is about 100um in diameter.
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44. Photolithography
Light directed oligonucleotide synthesis.
A solid support is derivatized with a covalent linker molecule terminated with a photolabile protecting group.
Light is directed through a mask to deprotect and activate selected sites, and protected nucleotides couple to the activated sites.
The process is repeated, activating different set of sites and coupling different based allowing arbitrary DNA probes to be constructed at each site.
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45. Affymetrix GeneChip® Arrays
A combination of photolithography and combinatorial chemistry to manufacture
GeneChip® Arrays. With a minimum number of steps, Affymetrix produces arrays with thousands of different probes packed at extremely high density. Enable to obtain high quality, genome-wide data using small sample volumes.
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46. Affymetrix GeneChip® Arrays
Data from an experiment showing the
expression of thousands of genes on
a single GeneChip® probe array.
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47. Next DNA Mapping and Brute Force Algorithms How is DNA sequenced?
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48. Full Restriction Digest
• Cutting DNA at each restriction site creates
multiple restriction fragments:
• Is it possible to reconstruct the order of
the fragments from the sizes of the fragments
{3,5,5,9} ?
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49. Full Restriction Digest: Multiple Solutions
• Alternative ordering of restriction fragments: vs
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50. Measuring Length of Restriction Fragments
Restriction enzymes break DNA into restriction fragments.
Gel electrophoresis is a process for separating DNA by size and measuring sizes of restriction fragments
Can separate DNA fragments that differ in length in only 1 nucleotide for fragments up to 500 nucleotides long
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51. Partial Restriction Digest
The sample of DNA is exposed to the restriction enzyme for only a limited amount of time to prevent it from being cut at all restriction sites
This experiment generates the set of all possible restriction fragments between every 2 (not necessarily consecutive) cuts
This set of fragment sizes is used to determine the positions of the restriction sites in the DNA sequence
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52. Partial Digest Example
Partial Digest results in the following 10 restriction fragments:
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53. Multiset of Restriction Fragments
We assume that multiplicity of a fragment can be detected, i.e., the number of restriction fragments of the same length can be determined (e.g., by observing twice as much fluorescence intensity for a double fragment than for a single fragment)
Multiset: {3, 5, 5, 8, 9, 14, 14, 17, 19, 22}
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54. Partial Digest Fundamentals
the set of n integers representing the location of all cuts in the restriction map, including the start and end X:
the total number of cuts n:
the multiset of integers representing lengths of each of the nC2 fragments produced from a partial digest
DX:
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55. One More Partial Digest Example2 4 7 10 5 8 3 6
Representation of DX = {2, 2, 3, 3, 4, 5, 6, 7, 8, 10} as a two dimensional table, with elements of
X = {0, 2, 4, 7, 10}
along both the top and left side. The elements at (i, j) in the table is xj – xi for 1 ≤ i < j ≤ n.
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56. Partial Digest Problem: Formulation
Goal: Given all pairwise distances between points on a line, reconstruct the positions of those points
Input: The multiset of pairwise distances L, containing n(n-1)/2 integers
Output: A set X, of n integers, such that DX = L
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57. Partial Digest: Multiple Solutions
It is not always possible to uniquely reconstruct a set X based only on DX.
For example, the set
X = {0, 2, 5}
and (X + 10) = {10, 12, 15}
both produce DX={2, 3, 5} as their partial digest set.
The sets {0,1,2,5,7,9,12} and {0,1,5,7,8,10,12} present a less trivial example of non-uniqueness. They both digest into:
{1, 1, 2, 2, 2, 3, 3, 4, 4, 5, 5, 5, 6, 7, 7, 7, 8, 9, 10, 11, 12}
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58. Homometric Sets 1 2 5 7 9 12 4 6 8 11 3 10 1 5 7 8 10 12 4 6 9 11 2 3
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59. Brute Force Algorithms
Also known as exhaustive search algorithms; examine every possible variant to find a solution
Efficient in rare cases; usually impractical
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60. Partial Digest: Brute Force
Find the restriction fragment of maximum length M. M is the length of the DNA sequence.
For every possible set
X={0, x2, … ,xn-1, M}
compute the corresponding DX
If DX is equal to the experimental partial digest L, then X is the correct restriction map
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61. BruteForcePDP BruteForcePDP(L, n): M <- maximum element in L
for every set of n – 2 integers 0 < x2 < … xn-1 < M
X <- {0,x2,…,xn-1,M}
Form DX from X
if DX = L
return X
output “no solution”
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62. Efficiency of BruteForcePDP
BruteForcePDP takes O(M n-2) time since it must examine all possible sets of positions.
One way to improve the algorithm is to limit the values of xi to only those values which occur in L.
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63. AnotherBruteForcePDP
AnotherBruteForcePDP(L, n)
M <- maximum element in L
for every set of n – 2 integers 0 < x2 < … xn-1 < M
X <- { 0,x2,…,xn-1,M }
Form DX from X
if DX = L
return X
output “no solution”
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64. AnotherBruteForcePDP
AnotherBruteForcePDP(L, n)
M <- maximum element in L
for every set of n – 2 integers 0 < x2 < … xn-1 < M from L
X <- { 0,x2,…,xn-1,M }
Form DX from X
if DX = L
return X
output “no solution”
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65. Efficiency of AnotherBruteForcePDP
It’s more efficient, but still slow
If L = {2, 998, 1000} (n = 3, M = 1000), BruteForcePDP will be extremely slow, but AnotherBruteForcePDP will be quite fast
Fewer sets are examined, but runtime is still exponential: O(n2n-4)
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66. Defining D(y, X) Before describing PartialDigest, first define D(y, X)
as the multiset of all distances between point y and all other points in the set X
D(y, X) = {|y – x1|, |y – x2|, …, |y – xn|}
for X = {x1, x2, …, xn}
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67. PartialDigest Algorithm
PartialDigest(L):
width <- Maximum element in L
DELETE(width, L)
X <- {0, width}
PLACE(L, X)
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68. PartialDigest Algorithm (cont’d)
PLACE(L, X)
if L is empty
output X
return
y <- maximum element in L
Delete(y,L)
if D(y, X ) Í L
Add y to X and remove lengths D(y, X) from L
PLACE(L,X )
Remove y from X and add lengths D(y, X) to L
if D(width-y, X ) Í L
Add width-y to X and remove lengths D(width-y, X) from L
Remove width-y from X and add lengths D(width-y, X ) to L
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69. An Example
L = { 2, 2, 3, 3, 4, 5, 6, 7, 8, 10 }
X = { 0 }
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70. An Example
L = { 2, 2, 3, 3, 4, 5, 6, 7, 8, 10 }
X = { 0 }
Remove 10 from L and insert it into X. We know this must be
the length of the DNA sequence because it is the largest
fragment.
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71. An Example
L = { 2, 2, 3, 3, 4, 5, 6, 7, 8, 10 }
X = { 0, 10 }
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72. An Example
L = { 2, 2, 3, 3, 4, 5, 6, 7, 8, 10 }
X = { 0, 10 }
Take 8 from L and make y = 2 or 8. But since the two cases are symmetric, we can assume y = 2.
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73. An Example
L = { 2, 2, 3, 3, 4, 5, 6, 7, 8, 10 }
X = { 0, 10 }
We find that the distances from y=2 to other elements in X are D(y, X) = {8, 2}, so we remove {8, 2} from L and add 2 to X.
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74. An Example
L = { 2, 2, 3, 3, 4, 5, 6, 7, 8, 10 }
X = { 0, 2, 10 }
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75. An Example
L = { 2, 2, 3, 3, 4, 5, 6, 7, 8, 10 }
X = { 0, 2, 10 }
Take 7 from L and make y = 7 or y = 10 – 7 = 3. We will
explore y = 7 first, so D(y, X ) = {7, 5, 3}.
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76. An Example
L = { 2, 2, 3, 3, 4, 5, 6, 7, 8, 10 }
X = { 0, 2, 10 }
For y = 7 first, D(y, X ) = {7, 5, 3}. Therefore we
remove {7, 5 ,3} from L and add 7 to X.
D(y, X) = {7, 5, 3} = {½7 – 0½, ½7 – 2½, ½7 – 10½}
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77. An Example
L = { 2, 2, 3, 3, 4, 5, 6, 7, 8, 10 }
X = { 0, 2, 7, 10 }
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78. An Example
L = { 2, 2, 3, 3, 4, 5, 6, 7, 8, 10 }
X = { 0, 2, 7, 10 }
Take 6 from L and make y = 6. Unfortunately
D(y, X) = {6, 4, 1 ,4}, which is not a subset of L. Therefore we won’t explore this branch. 6
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79. An Example
L = { 2, 2, 3, 3, 4, 5, 6, 7, 8, 10 }
X = { 0, 2, 7, 10 }
This time make y = 4. D(y, X) = {4, 2, 3 ,6}, which is a
subset of L so we will explore this branch. We remove
{4, 2, 3 ,6} from L and add 4 to X.
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80. An Example
L = { 2, 2, 3, 3, 4, 5, 6, 7, 8, 10 }
X = { 0, 2, 4, 7, 10 }
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81. An Example
L = { 2, 2, 3, 3, 4, 5, 6, 7, 8, 10 }
X = { 0, 2, 4, 7, 10 }
L is now empty, so we have a solution, which is X.
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82. An Example
L = { 2, 2, 3, 3, 4, 5, 6, 7, 8, 10 }
X = { 0, 2, 7, 10 }
To find other solutions, we backtrack.
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83. An Example
L = { 2, 2, 3, 3, 4, 5, 6, 7, 8, 10 }
X = { 0, 2, 10 }
More backtrack.
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84. An Example
L = { 2, 2, 3, 3, 4, 5, 6, 7, 8, 10 }
X = { 0, 2, 10 }
This time we will explore y = 3. D(y, X) = {3, 1, 7}, which is
not a subset of L, so we won’t explore this branch.
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85. An Example
L = { 2, 2, 3, 3, 4, 5, 6, 7, 8, 10 }
X = { 0, 10 }
We backtracked back to the root. Therefore we have found all the solutions.
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86. Analyzing PartialDigest Algorithm
Still exponential in worst case, but is very fast on average
Informally, let T(n) be time PartialDigest takes to place n cuts
No branching case: T(n) < T(n-1) + O(n)
Quadratic
Branching case: T(n) < 2T(n-1) + O(n)
Exponential
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87. Double Digest Mapping One with only first enzyme
Double Digest is yet another experimentally method to construct restriction maps
Use two restriction enzymes; three full digests:
One with only first enzyme
One with only second enzyme
One with both enzymes
Computationally, Double Digest problem is more complex than Partial Digest problem
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88. Next: Finding Regulatory Motifs in DNA sequences 11/13/2018
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89. Combinatorial Gene Regulation
A microarray experiment showed that when gene X is knocked out, 20 other genes are not expressed
How can one gene have such drastic effects?
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90. Regulatory Proteins
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
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91. Regulatory Regions
Every gene contains a regulatory region (RR) typically stretching bp upstream of the transcriptional start site
Located within the RR are the Transcription Factor Binding Sites (TFBS), also known as motifs, specific for a given transcription factor
TFs influence gene expression by binding to a specific location in the respective gene’s regulatory region - TFBS
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92. Transcription Factor Binding Sites
A TFBS can be located anywhere within the Regulatory Region.
TFBS may vary slightly across different regulatory regions since non-essential bases could mutate
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93. Motifs and Transcriptional Start Sites
ATCCCG
gene
TTCCGG
gene
ATCCCG
gene
ATGCCG
gene
ATGCCC
gene
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94. Transcription Factors and Motifs
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95. Motif Logo TGGGGGA TGAGAGA TGAGGGA
Motifs can mutate on non important 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
TGGGGGA TGAGAGA TGAGGGA
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96. Identifying Motifs Genes are turned on or off by regulatory proteins
These proteins bind to upstream regulatory regions of genes to either attract or block an RNA polymerase
Regulatory protein (TF) binds to a short DNA sequence called a motif (TFBS)
So finding the same motif in multiple genes’ regulatory regions suggests a regulatory relationship amongst those genes
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97. Identifying Motifs: Complications
We do not know the motif sequence
We do not know where it is located relative to the genes start
Motifs can differ slightly from one gene to the next
How to discern it from “random” motifs?
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98. Motifs: summary Short strings Transcription factors bind to motifs
Regulate downstream genes
Challenge: must find patterns without knowing them
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99. How can we do this? Statistics? Brute force search 11/13/2018
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100. Statistical approaches
Deciphering encoded English text.
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101. Motif Finding and The Gold Bug Problem: Similarities
Nucleotides in motifs encode for a message in the “genetic” language. Symbols in “The Gold Bug” encode for a message in English
In order to solve the problem, we analyze the frequencies of patterns in DNA/Gold Bug message.
Knowledge of established regulatory motifs makes the Motif Finding problem simpler. Knowledge of the words in the English dictionary helps to solve
the Gold Bug problem.
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102. Similarities (cont’d)
Motif Finding:
In order to solve the problem, we analyze the frequencies of patterns in the nucleotide sequences
Gold Bug Problem:
In order to solve the problem, we analyze the frequencies of patterns in the text written in English
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103. Similarities (cont’d)
Motif Finding:
Knowledge of established motifs reduces the complexity of the problem
Gold Bug Problem:
Knowledge of the words in the dictionary is highly desirable
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104. Motif Finding and The Gold Bug Problem: Differences
Motif Finding is harder than Gold Bug problem:
We don’t have the complete dictionary of motifs
The “genetic” language does not have a standard “grammar”
Only a small fraction of nucleotide sequences encode for motifs; the size of data is enormous
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105. Challenge Problem Find a motif in a sample of
- 20 “random” sequences (e.g. 600 nt long)
- each sequence containing an implanted
pattern of length 15,
- each pattern appearing with 4 mismatches
as (15,4)-motif.
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106. Exhaustive searches
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107. The Motif Finding Problem
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
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108. The Motif Finding Problem (cont’d)
Additional information:
The hidden sequence is of length 8
The pattern is not exactly the same in each array because random point mutations may occur in the sequences
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109. The Motif Finding Problem (cont’d)
The patterns revealed with no mutations:
cctgatagacgctatctggctatccacgtacgtaggtcctctgtgcgaatctatgcgtttccaaccat
agtactggtgtacatttgatacgtacgtacaccggcaacctgaaacaaacgctcagaaccagaagtgc
aaacgtacgtgcaccctctttcttcgtggctctggccaacgagggctgatgtataagacgaaaatttt
agcctccgatgtaagtcatagctgtaactattacctgccacccctattacatcttacgtacgtataca
ctgttatacaacgcgtcatggcggggtatgcgttttggtcgtcgtacgctcgatcgttaacgtacgtc
acgtacgt
Consensus String
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110. The Motif Finding Problem (cont’d)
The patterns with 2 point mutations:
cctgatagacgctatctggctatccaGgtacTtaggtcctctgtgcgaatctatgcgtttccaaccat
agtactggtgtacatttgatCcAtacgtacaccggcaacctgaaacaaacgctcagaaccagaagtgc
aaacgtTAgtgcaccctctttcttcgtggctctggccaacgagggctgatgtataagacgaaaatttt
agcctccgatgtaagtcatagctgtaactattacctgccacccctattacatcttacgtCcAtataca
ctgttatacaacgcgtcatggcggggtatgcgttttggtcgtcgtacgctcgatcgttaCcgtacgGc
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111. The Motif Finding Problem (cont’d)
The patterns with 2 point mutations:
cctgatagacgctatctggctatccaGgtacTtaggtcctctgtgcgaatctatgcgtttccaaccat
agtactggtgtacatttgatCcAtacgtacaccggcaacctgaaacaaacgctcagaaccagaagtgc
aaacgtTAgtgcaccctctttcttcgtggctctggccaacgagggctgatgtataagacgaaaatttt
agcctccgatgtaagtcatagctgtaactattacctgccacccctattacatcttacgtCcAtataca
ctgttatacaacgcgtcatggcggggtatgcgttttggtcgtcgtacgctcgatcgttaCcgtacgGc
Can we still find the motif, now that we have 2 mutations?
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112. 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)
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113. Motifs: Profiles and Consensus
Line up the patterns by their start indexes
s = (s1, s2, …, st)
Construct matrix profile 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 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
Profile C G T
Consensus A C G T A C G T
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114. Consensus
Think of consensus as an “ancestor” motif, from which mutated motifs emerged
The distance between a real motif and the consensus sequence is generally less than that for two real motifs
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115. Consensus (cont’d)
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116. Evaluating Motifs
We have a guess about the consensus sequence, but how “good” is this consensus?
Need to introduce a scoring function to compare different guesses and choose the “best” one.
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117. Defining Some Terms t - number of sample DNA sequences
n - length of each DNA sequence
DNA - sample of DNA sequences (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
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118. Parameters l = 8 DNA t=5 s s1 = 26 s2 = 21 s3= 3 s4 = 56 s5 = 60
cctgatagacgctatctggctatccaGgtacTtaggtcctctgtgcgaatctatgcgtttccaaccat
agtactggtgtacatttgatCcAtacgtacaccggcaacctgaaacaaacgctcagaaccagaagtgc
aaacgtTAgtgcaccctctttcttcgtggctctggccaacgagggctgatgtataagacgaaaatttt
agcctccgatgtaagtcatagctgtaactattacctgccacccctattacatcttacgtCcAtataca
ctgttatacaacgcgtcatggcggggtatgcgttttggtcgtcgtacgctcgatcgttaCcgtacgGc
l = 8
DNA
t=5
n = 69 s
s1 = s2 = s3= s4 = s5 = 60
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119. Scoring Motifs Given s = (s1, … st) and DNA: Score(s,DNA) = l t
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 C G T
Consensus a c g t a c g t
Score =30 t
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120. The Motif Finding Problem
If starting positions s=(s1, s2,… st) are given, finding consensus is easy even with mutations in the sequences because we can simply construct the profile to find the motif (consensus)
But… the starting positions s are usually not given. How can we find the “best” profile matrix?
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121. The Motif Finding Problem: Formulation
Goal: Given a set of DNA sequences, find a set of l-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)
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122. The Motif Finding Problem: Brute Force Solution
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]
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123. BruteForceMotifSearch
BruteForceMotifSearch(DNA, t, n, l)
bestScore  0
for each s=(s1,s2 , . . ., st) from (1, ) to (n-l+1, . . ., n-l+1)
if (Score(s,DNA) > bestScore)
bestScore  score(s, DNA)
bestMotif  (s1,s2 , , st)
return bestMotif
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124. Running Time of BruteForceMotifSearch
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 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 – it will take billions of years
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125. 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
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126. Hamming Distance Hamming distance:
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
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127. Total Distance: Definition
(Intuition: let v be a (candidate) motif)
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,s) is the sum of the minimum Hamming distances for each DNA sequence i
TotalDistance(v,DNA) = mins dH(v, s), where s is the set of starting positions s1, s2,… st
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128. 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 string v of l nucleotides that minimizes TotalDistance(v,DNA) over all strings of that length
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129. Median String Search Algorithm
MedianStringSearch (DNA, t, n, l)
bestWord  AAA…A
bestDistance  ∞
for each l-mer s from AAA…A to TTT…T
if TotalDistance(s,DNA) < bestDistance
bestDistanceTotalDistance(s,DNA)
bestWord  s
return bestWord
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130. 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
Need to show that minimizing TotalDistance is equivalent to maximizing Score
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131. We are looking for the same thing
At any column i Scorei + TotalDistancei = t
Because there are l columns
Score + TotalDistance = l * t
Rearranging:
Score = l * t - TotalDistance
l * t is constant the minimization of the right side is equivalent to the maximization of the left side
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
Profile C G T
Consensus a c g t a c g t
Score
TotalDistance
Sum t
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132. 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 the combinations for s. That is (n - l + 1)t combinations!!!
The Median String Problem needs to examine all 4l combinations for v. This number is relatively smaller
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133. Motif Finding: Improving the Running Time
Recall the BruteForceMotifSearch:
BruteForceMotifSearch(DNA, t, n, l)
bestScore  0
for each s=(s1,s2 , . . ., st) from (1, ) to (n-l+1, . . ., n-l+1)
if (Score(s,DNA) > bestScore)
bestScore  Score(s, DNA)
bestMotif  (s1,s2 , , st)
return bestMotif
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134. Structuring the Search
How can we perform the line
for each s=(s1,s2 , . . ., st) from (1, ) to (n-l+1, . . ., n-l+1) ?
We need a method for efficiently structuring and navigating the many possible motifs
This is not very different than exploring all t-digit numbers
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135. Median String: Improving the Running Time
MedianStringSearch (DNA, t, n, l)
bestWord  AAA…A
bestDistance  ∞
for each l-mer s from AAA…A to TTT…T if TotalDistance(s,DNA) < bestDistance
bestDistanceTotalDistance(s,DNA)
bestWord  s
return bestWord
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136. 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
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137. 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
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138. aa ac ag at ca cc cg ct ga gc gg gt ta tc tg tt
Linked List
Suppose l = 2
aa ac ag at ca cc cg ct ga gc gg gt ta tc tg tt
Need to visit all the predecessors of a sequence before visiting the sequence itself
Start
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139. aa ac ag at ca cc cg ct ga gc gg gt ta tc tg tt
Linked List (cont’d)
Linked list is not the most efficient data structure for motif finding
Let’s try grouping the sequences by their prefixes
aa ac ag at ca cc cg ct ga gc gg gt ta tc tg tt
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140. aa ac ag at ca cc cg ct ga gc gg gt ta tc tg tt
Search Tree
a c g t-
aa ac ag at ca cc cg ct ga gc gg gt ta tc tg tt
root --
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141. Analyzing Search Trees
Characteristics of the 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?
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142. Visit the Next Leaf NextLeaf( a,L, k ) // a : the array of digits
Given a current leaf a , we need to compute the “next” leaf:
NextLeaf( a,L, k ) // a : the array of digits
for i  L to // L: length of the array
if ai < k // k : max digit value
ai  ai + 1
return a
ai  1
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143. NextLeaf (cont’d) The algorithm is common addition in radix k:
Increment the least significant digit
“Carry the one” to the next digit position when the digit is at maximal value
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144. NextLeaf: Example Moving to the next leaf: Current Location --
Current Location
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145. NextLeaf: Example (cont’d)
Moving to the next leaf: --
Next Location
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146. Visit All Leaves Printing all permutations in ascending order:
AllLeaves(L,k) // L: length of the sequence
a  (1,...,1) // k : max digit value
while forever // a : array of digits
output a
a  NextLeaf(a,L,k)
if a = (1,...,1)
return
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147. Tree Search So we can search leaves
How about searching all vertices of the tree?
We can do this with a depth first search
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148. Visit the Next Vertex NextVertex(a,i,L,k) // a : the array of digits
if i < L // i : prefix length
a i+1  // L: max length
return ( a,i+1) // k : max digit value
else
for j  l to 1
if aj < k
aj  aj +1
return( a,j )
return(a,0)
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149. Example Moving to the next vertex: Current Location 1- 2- 3- 4-
Current Location --
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CSE 5290, Fall 2011

150. Example Moving to the next vertices:
Location after 5 next vertex moves --
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151. Bypass Move
Given a prefix (internal vertex), find next vertex after skipping all its children
Bypass(a,i,L,k) // a: array of digits
for j  i to // i : prefix length
if aj < k // L: maximum length
aj  aj +1 // k : max digit value
return(a,j)
return(a,0)
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CSE 5290, Fall 2011

152. Bypass Move: Example Bypassing the descendants of “2-”:
Current Location --
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CSE 5290, Fall 2011

153. Example Bypassing the descendants of “2-”: Next Location 1- 2- 3- 4-
Next Location --
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CSE 5290, Fall 2011

154. Revisiting Brute Force Search
Now that we have a method for navigating the tree, lets look again at BruteForceMotifSearch
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CSE 5290, Fall 2011

155. Brute Force Search Again
BruteForceMotifSearchAgain(DNA, t, n, l)
s  (1,1,…, 1)
bestScore  Score(s,DNA)
while forever
s  NextLeaf (s, t, n- l +1)
if (Score(s,DNA) > bestScore)
bestScore  Score(s, DNA)
bestMotif  (s1,s2 , , st)
return bestMotif
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CSE 5290, Fall 2011

156. Can We Do Better?
Sets of s=(s1, s2, …,st) may have a weak profile for the first i positions (s1, s2, …,si)
Every row of alignment may add at most l to Score
Optimism: if all subsequent (t-i) positions (si+1, …st) add
(t – i ) * l to Score(s,i,DNA)
If Score(s,i,DNA) + (t – i ) * l < BestScore, it makes no sense to search in vertices of the current subtree
Use ByPass()
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CSE 5290, Fall 2011

157. 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
Use NextVertex() and ByPass() to navigate the tree
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CSE 5290, Fall 2011

158. Pseudocode for Branch and Bound Motif Search
BranchAndBoundMotifSearch(DNA,t,n,l)
s  (1,…,1)
bestScore  0
i  1
while i > 0
if i < t
optimisticScore  Score(s, i, DNA) +(t – i ) * l
if optimisticScore < bestScore
(s, i)  Bypass(s,i, n-l +1)
else
(s, i)  NextVertex(s, i, n-l +1)
if Score(s,DNA) > bestScore
bestScore  Score(s)
bestMotif  (s1, s2, s3, …, st)
(s,i)  NextVertex(s,i,t,n-l + 1)
return bestMotif
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CSE 5290, Fall 2011

159. 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 relatively small
We want to use median string algorithm with the Branch and Bound improvement.
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CSE 5290, Fall 2011

160. Branch and Bound Applied to Median String Search
Note that if the total distance for a prefix is greater than that for the best word so far:
TotalDistance (prefix, DNA) > BestDistance
there is no use exploring the remaining part of the word
We can eliminate that branch and BYPASS exploring that branch further
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CSE 5290, Fall 2011

161. Bounded Median String Search
BranchAndBoundMedianStringSearch(DNA,t,n,l )
s  (1,…,1)
bestDistance  ∞
i  1
while i > 0
if i < l
prefix  string corresponding to the first i nucleotides of s
optimisticDistance  TotalDistance(prefix,DNA)
if optimisticDistance > bestDistance
(s, i )  Bypass(s,i, l, 4)
else
(s, i )  NextVertex(s, i, l, 4)
word  nucleotide string corresponding to s
if TotalDistance(s,DNA) < bestDistance
bestDistance  TotalDistance(word, DNA)
bestWord  word
(s,i )  NextVertex(s,i,l, 4)
return bestWord
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CSE 5290, Fall 2011

162. 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
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CSE 5290, Fall 2011

163. Improving the Bounds (cont’d)
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
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CSE 5290, Fall 2011

164. Better Bounds
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165. Better Bounds (cont’d)
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()
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CSE 5290, Fall 2011

166. More on the Motif Problem
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
Examples: Local search, Stochastic sampling…
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CSE 5290, Fall 2011