Graph Algorithms 8.6-8.10

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# Graph Algorithms 8.6-8.10 - PowerPoint PPT Presentation

Graph Algorithms 8.6-8.10. CS 6030 – Bioinformatics Summer II 2012 Jason Eric Johnson. Sequencing by Hybridization. DNA Array gives all strings of length l How do we find the order? Spectrum( s,l ) String s of length n Spectrum is multiset of n-l+1 l- mers in s.

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## Graph Algorithms 8.6-8.10

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### Graph Algorithms 8.6-8.10

CS 6030 – Bioinformatics

Summer II 2012

Jason Eric Johnson

Sequencing by Hybridization
• DNA Array gives all strings of length l
• How do we find the order?
• Spectrum(s,l)
• String s of length n
• Spectrum is multiset of n-l+1 l-mers in s
Sequencing by Hybridization
• s = TATGGTGC
• l = 3
• Spectrum(s,l) = {TAT,ATG,TGG,GGT,GTG,TGC}
• Problem:
• Input: Set S of all l-mers from s
• Output: String s s.t. Spectrum(s,l) = S
Sequencing by Hybridization
• Special case of Shortest Superstring Problem
• SBH is linear-time
• SSP (NP-Complete) is more general
• In SSP, no guaranteed overlap
• In SBH, we know the length of the target sequence
Sequencing by Hybridization
• There is a problem with DNA Arrays
• No good way to distinguish a match from a highly stable mismatch
• Mismatch could give strong hybridization signal
• Need longer probes to deal with mutations
SBH: Hamiltonian Path Approach
• Two l-mers overlap if overlap(p,q) = l – 1
• Last l-1 letters of p are same as first l-1 of q
• Make each l-mer in Spectrum(s,l) a node
• Construct directed graph(s) that connect every p and q with a directed edge
• 1 to 1 correspondence between paths that visit each vertex exactly once (Hamiltonian Paths) and DNA fragments with Spectrum(s,l)
SBH: Hamiltonian Path Approach

S = { ATG AGG TGC TCC GTC GGT GCA CAG }

H

ATG

AGG

TGC

TCC

GTC

GCA

CAG

GGT

ATG

C

A

G

G

T

C

C

Path visited every VERTEX once

A more complicated graph:

S = { ATG TGG TGC GTG GGC GCA GCG CGT }

SBH: Hamiltonian Path Approach
S = { ATG TGG TGC GTG GGC GCA GCG CGT }

Path 1:

SBH: Hamiltonian Path Approach

ATGCGTGGCA

Path 2:

ATGGCGTGCA

SBH: Hamiltonian Path Approach
• Problem is that there is no efficient algorithm
• As overlap graph gets larger, this is not a useful technique since the Hamiltonian Path problem is NP-Complete
SBH: Eulerian Path Approach
• This leads to simple linear-time algorithm for sequence reconstruction
• Construct graph whose edges correspond to l-mers
• Find path(s) that visit each edge exactly once

CG

GT

TG

CA

AT

GC

Path visited every EDGE once

GG

SBH: Eulerian Path Approach

S = { ATG, TGC, GTG, GGC, GCA, GCG, CGT }

Vertices correspond to ( l – 1 ) – mers : { AT, TG, GC, GG, GT, CA, CG }

Edges correspond to l – mers from S

S = { AT, TG, GC, GG, GT, CA, CG } corresponds to two different paths:SBH: Eulerian Path Approach

CG

CG

GT

GT

TG

AT

TG

GC

AT

GC

CA

CA

GG

GG

ATGGCGTGCA

ATGCGTGGCA

SBH: Eulerian Path Approach
• If for every vertex the number of incoming edges is equal to the number of outgoing edges, the graph is balanced
• Theorem: A connected graph is Eulerian if and only if each of its vertices is balanced
• Theorem: A connected graph has an Eulerian path if and only if it contains at most two semi-balanced vertices and all other vertices are balanced
Some Difficulties with SBH
• Fidelity of Hybridization: difficult to detect differences between probes hybridized with perfect matches and 1 or 2 mismatches
• Array Size: Effect of low fidelity can be decreased with longer l-mers, but array size increases exponentially in l. Array size is limited with current technology.
• Practicality: SBH is still impractical. As DNA microarray technology improves, SBH may become practical in the future
• Practicality again: Although SBH is still impractical, it spearheaded expression analysis and SNP analysis techniques
Fragment Assembly
• Now that we have our reads sequenced, we need to assemble them into the entire DNA sequence
Fragment Assembly
• We have some problems:
• Errors in reads (1% to 3%)
• Which strand did the read come from?
• Did the read come from the target DNA sequence or its Watson-Crick complement?
• Repeats in DNA (this is the major problem)
• See page 278 for puzzle example
Fragment Assembly
• Very difficult to put it all together if repeats are longer than read length
• Could solve this by increasing read length, but the technology isn’t there yet
Fragment Assembly
• One approach is to break the sequence into about 30,000 Bacterial Artificial Chromosomes
• Sequence each BAC individually
• Put them all together
• Used and shown effective (if cumbersome) by the Human Genome Project
Fragment Assembly
• Another option (used in mouse genome assembly) is the Weber-Meyers approach
• Pairs reads that are separated by a fixed-size gap
• Gap size L is chosen to be longer than most repeats
• Unlikely both reads lie in large repeat
• Read that is in unique portion of DNA tells us which copy of a repeat the mate is in
Fragment Assembly
• Most algorithms consist of these steps:
• Overlap
• Layout:
• Find order of reads along DNA
• Consensus:
• Derive DNA sequence from layout
Overlap
• Find the best match between the suffix of one read and the prefix of another
• Due to sequencing errors, need to use dynamic programming to find the optimal overlap alignment
• Apply a filtration method to filter out pairs of fragments that do not share a significantly long common substring

T GA

TACA

| ||

||

TAGA

TAGT

• Sort all k-mers in reads (k ~ 24)
• Find pairs of reads sharing a k-mer
• Extend to full alignment – throw away if not >95% similar

TAGATTACACAGATTAC

|||||||||||||||||

TAGATTACACAGATTAC

• A k-mer that appears N times, initiates N2 comparisons
• For an Alu that appears 106 times  1012 comparisons – too much
• Solution:

Discard all k-mers that appear more than

t Coverage, (t ~ 10)

Create local multiple alignments from the overlapping reads

TAGATTACACAGATTACTGA

TAGATTACACAGATTACTGA

TAG TTACACAGATTATTGA

TAGATTACACAGATTACTGA

TAGATTACACAGATTACTGA

TAGATTACACAGATTACTGA

TAG TTACACAGATTATTGA

TAGATTACACAGATTACTGA

Layout
• Repeats are a major challenge
• Do two aligned fragments really overlap, or are they from two copies of a repeat?
• Solution: repeat masking – hide the repeats!!!
• Masking results in high rate of misassembly (up to 20%)
• Misassembly means alot more work at the finishing step
Consensus
• A consensus sequence is derived from a profile of the assembled fragments
• A sufficient number of reads is required to ensure a statistically significant consensus
Derive Consensus Sequence

TAGATTACACAGATTACTGA TTGATGGCGTAA CTA

Derive multiple alignment from pairwise read alignments

TAGATTACACAGATTACTGACTTGATGGCGTAAACTA

TAG TTACACAGATTATTGACTTCATGGCGTAA CTA

TAGATTACACAGATTACTGACTTGATGGCGTAA CTA

TAGATTACACAGATTACTGACTTGATGGGGTAA CTA

TAGATTACACAGATTACTGACTTGATGGCGTAA CTA

Derive each consensus base by weighted voting

Protein Sequencing and Identification
• Protein can be digested into peptides by proteases (such as trypsin)
• Can then sequence the fragments individually and re-assemble
• Mass spectrometry allows us to find proteins involved in cell death, for example
Protein Sequencing and Identification
• Tandem mass spectrometer breaks peptides into smaller fragments
• These fragments have electrical charge
• Fragments are spun around in an magnetic field until they hit a detector
• Larger masses are harder to spin than smaller ones, so mass can be determined by the amount of energy required to fling fragments around
Protein Sequencing and Identification
• The problem we encounter is how to reconstruct the amino acid sequence of the peptide from the masses of the broken pieces
References
• Generated from:
• An Introduction to Bioinformatics Algorithms, Neil C. Jones, Pavel A. Pevzner, A Bradford Book, The MIT Press, Cambridge, Mass., London, England, 2004
• Slides 4, 8-10, 13, 14, 16, 23-29 from http://bix.ucsd.edu/bioalgorithms/slides.php#Ch8