Bioalgo 2012-02-graphs
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The document discusses graph algorithms and their applications in bioinformatics. It introduces graph theory concepts like the Hamiltonian cycle problem and describes how it relates to finding knight tours on a chessboard. It also discusses how the traveling salesman problem and Hamiltonian cycle problem are connected and how graph algorithms can be used for problems in DNA sequencing and fragment assembly.
Bioalgo 2012-02-graphs
- 2. Outline 1. Introduction to Graph Theory 2. The Hamiltonian & Eulerian Cycle Problems 3. Basic Biological Applications of Graph Theory 4. DNA Sequencing 5. Shortest Superstring & Traveling Salesman Problems 6. Sequencing by Hybridization 7. Fragment Assembly & Repeats in DNA 8. Fragment Assembly Algorithms
- 3. Section 1: Introduction to Graph Theory
- 4. Knight Tours • Knight Tour Problem: Given an 8 x 8 chessboard, is it possible to find a path for a knight that visits every square exactly once and returns to its starting square? • Note: In chess, a knight may move only by jumping two spaces in one direction, followed by a jump one space in a perpendicular direction. http://www.chess-poster.com/english/laws_of_chess.htm
- 5. 9th Century: Knight Tours Discovered
- 6. • 1759: Berlin Academy of Sciences proposes a 4000 francs prize for the solution of the more general problem of finding a knight tour on an N x N chessboard. • 1766: The problem is solved by Leonhard Euler (pronounced “Oiler”). • The prize was never awarded since Euler was Director of Mathematics at Berlin Academy and was deemed ineligible. 18th Century: N x N Knight Tour Problem Leonhard Euler http://commons.wikimedia.org/wiki/File:Leonhard_Euler_by_Handmann.png
- 7. • A graph is a collection (V, E) of two sets: • V is simply a set of objects, which we call the vertices of G. • E is a set of pairs of vertices which we call the edges of G. Introduction to Graph Theory
- 8. • A graph is a collection (V, E) of two sets: • V is simply a set of objects, which we call the vertices of G. • E is a set of pairs of vertices which we call the edges of G. • Simpler: Think of G as a network: Introduction to Graph Theory http://uh.edu/engines/epi2467.htm
- 9. • A graph is a collection (V, E) of two sets: • V is simply a set of objects, which we call the vertices of G. • E is a set of pairs of vertices which we call the edges of G. • Simpler: Think of G as a network: • Nodes = vertices Introduction to Graph Theory http://uh.edu/engines/epi2467.htm Vertex
- 10. • A graph is a collection (V, E) of two sets: • V is simply a set of objects, which we call the vertices of G. • E is a set of pairs of vertices which we call the edges of G. • Simpler: Think of G as a network: • Nodes = vertices • Edges = segments connecting the nodes Introduction to Graph Theory http://uh.edu/engines/epi2467.htm Vertex Edge
- 11. Section 2: The Hamiltonian & Eulerian Cycle Problems
- 12. • Input: A graph G = (V, E) • Output: A Hamiltonian cycle in G, which is a cycle that visits every vertex exactly once. Hamiltonian Cycle Problem
- 13. • Input: A graph G = (V, E) • Output: A Hamiltonian cycle in G, which is a cycle that visits every vertex exactly once. • Example: In 1857, William Rowan Hamilton asked whether the graph to the right has such a cycle. Hamiltonian Cycle Problem
- 14. • Input: A graph G = (V, E) • Output: A Hamiltonian cycle in G, which is a cycle that visits every vertex exactly once. • Example: In 1857, William Rowan Hamilton asked whether the graph to the right has such a cycle. • Do you see a Hamiltonian cycle? Hamiltonian Cycle Problem
- 15. • Input: A graph G = (V, E) • Output: A Hamiltonian cycle in G, which is a cycle that visits every vertex exactly once. • Example: In 1857, William Rowan Hamilton asked whether the graph to the right has such a cycle. • Do you see a Hamiltonian cycle? Hamiltonian Cycle Problem
- 16. • Input: A graph G = (V, E) • Output: A Hamiltonian cycle in G, which is a cycle that visits every vertex exactly once. • Example: In 1857, William Rowan Hamilton asked whether the graph to the right has such a cycle. • Do you see a Hamiltonian cycle? Hamiltonian Cycle Problem
- 17. • Input: A graph G = (V, E) • Output: A Hamiltonian cycle in G, which is a cycle that visits every vertex exactly once. • Example: In 1857, William Rowan Hamilton asked whether the graph to the right has such a cycle. • Do you see a Hamiltonian cycle? Hamiltonian Cycle Problem
- 18. • Input: A graph G = (V, E) • Output: A Hamiltonian cycle in G, which is a cycle that visits every vertex exactly once. • Example: In 1857, William Rowan Hamilton asked whether the graph to the right has such a cycle. • Do you see a Hamiltonian cycle? Hamiltonian Cycle Problem
- 19. • Input: A graph G = (V, E) • Output: A Hamiltonian cycle in G, which is a cycle that visits every vertex exactly once. • Example: In 1857, William Rowan Hamilton asked whether the graph to the right has such a cycle. • Do you see a Hamiltonian cycle? Hamiltonian Cycle Problem
- 20. • Input: A graph G = (V, E) • Output: A Hamiltonian cycle in G, which is a cycle that visits every vertex exactly once. • Example: In 1857, William Rowan Hamilton asked whether the graph to the right has such a cycle. • Do you see a Hamiltonian cycle? Hamiltonian Cycle Problem
- 21. • Input: A graph G = (V, E) • Output: A Hamiltonian cycle in G, which is a cycle that visits every vertex exactly once. • Example: In 1857, William Rowan Hamilton asked whether the graph to the right has such a cycle. • Do you see a Hamiltonian cycle? Hamiltonian Cycle Problem
- 22. • Input: A graph G = (V, E) • Output: A Hamiltonian cycle in G, which is a cycle that visits every vertex exactly once. • Example: In 1857, William Rowan Hamilton asked whether the graph to the right has such a cycle. • Do you see a Hamiltonian cycle? Hamiltonian Cycle Problem
- 23. • Input: A graph G = (V, E) • Output: A Hamiltonian cycle in G, which is a cycle that visits every vertex exactly once. • Example: In 1857, William Rowan Hamilton asked whether the graph to the right has such a cycle. • Do you see a Hamiltonian cycle? Hamiltonian Cycle Problem
- 24. • Input: A graph G = (V, E) • Output: A Hamiltonian cycle in G, which is a cycle that visits every vertex exactly once. • Example: In 1857, William Rowan Hamilton asked whether the graph to the right has such a cycle. • Do you see a Hamiltonian cycle? Hamiltonian Cycle Problem
- 25. • Input: A graph G = (V, E) • Output: A Hamiltonian cycle in G, which is a cycle that visits every vertex exactly once. • Example: In 1857, William Rowan Hamilton asked whether the graph to the right has such a cycle. • Do you see a Hamiltonian cycle? Hamiltonian Cycle Problem
- 26. • Input: A graph G = (V, E) • Output: A Hamiltonian cycle in G, which is a cycle that visits every vertex exactly once. • Example: In 1857, William Rowan Hamilton asked whether the graph to the right has such a cycle. • Do you see a Hamiltonian cycle? Hamiltonian Cycle Problem
- 27. • Input: A graph G = (V, E) • Output: A Hamiltonian cycle in G, which is a cycle that visits every vertex exactly once. • Example: In 1857, William Rowan Hamilton asked whether the graph to the right has such a cycle. • Do you see a Hamiltonian cycle? Hamiltonian Cycle Problem
- 28. • Input: A graph G = (V, E) • Output: A Hamiltonian cycle in G, which is a cycle that visits every vertex exactly once. • Example: In 1857, William Rowan Hamilton asked whether the graph to the right has such a cycle. • Do you see a Hamiltonian cycle? Hamiltonian Cycle Problem
- 29. • Input: A graph G = (V, E) • Output: A Hamiltonian cycle in G, which is a cycle that visits every vertex exactly once. • Example: In 1857, William Rowan Hamilton asked whether the graph to the right has such a cycle. • Do you see a Hamiltonian cycle? Hamiltonian Cycle Problem
- 30. • Input: A graph G = (V, E) • Output: A Hamiltonian cycle in G, which is a cycle that visits every vertex exactly once. • Example: In 1857, William Rowan Hamilton asked whether the graph to the right has such a cycle. • Do you see a Hamiltonian cycle? Hamiltonian Cycle Problem
- 31. • Input: A graph G = (V, E) • Output: A Hamiltonian cycle in G, which is a cycle that visits every vertex exactly once. • Example: In 1857, William Rowan Hamilton asked whether the graph to the right has such a cycle. • Do you see a Hamiltonian cycle? Hamiltonian Cycle Problem
- 32. • Input: A graph G = (V, E) • Output: A Hamiltonian cycle in G, which is a cycle that visits every vertex exactly once. • Example: In 1857, William Rowan Hamilton asked whether the graph to the right has such a cycle. • Do you see a Hamiltonian cycle? Hamiltonian Cycle Problem
- 33. • Input: A graph G = (V, E) • Output: A Hamiltonian cycle in G, which is a cycle that visits every vertex exactly once. • Example: In 1857, William Rowan Hamilton asked whether the graph to the right has such a cycle. • Do you see a Hamiltonian cycle? Hamiltonian Cycle Problem
- 34. • Input: A graph G = (V, E) • Output: A Hamiltonian cycle in G, which is a cycle that visits every vertex exactly once. • Example: In 1857, William Rowan Hamilton asked whether the graph to the right has such a cycle. • Do you see a Hamiltonian cycle? Hamiltonian Cycle Problem
- 35. • Let us form a graph G = (V, E) as follows: • V = the squares of a chessboard • E = the set of edges (v, w) where v and w are squares on the chessboard and a knight can jump from v to w in a single move. • Hence, a knight tour is just a Hamiltonian Cycle in this graph! Knight Tours Revisited
- 36. • Theorem: The Hamiltonian Cycle Problem is NP-Complete. • This result explains why knight tours were so difficult to find; there is no known quick method to find them! Hamiltonian Cycle Problem
- 37. • Recall the Traveling Salesman Problem (TSP): • n cities • Cost of traveling from i to j is given by c(i, j) • Goal: Find the tour of all the cities of lowest total cost. • Example at right: One busy salesman! • So we might like to think of the Hamiltonian Cycle Problem as a TSP with all costs = 1, where we have some edges missing (there doesn’t always exist a flight between all pairs of cities). Hamiltonian Cycle Problem as TSP http://www.ima.umn.edu/public-lecture/tsp/index.html
- 38. • The city of Konigsberg, Prussia (today: Kaliningrad, Russia) was made up of both banks of a river, as well as two islands. • The riverbanks and the islands were connected with bridges, as follows: • The residents wanted to know if they could take a walk from anywhere in the city, cross each bridge exactly once, and wind up where they started. The Bridges of Konigsberg http://www.math.uwaterloo.ca/navigation/ideas/Zeno/zenocando.shtml
- 39. • 1735: Enter Euler...his idea: compress each land area down to a single point, and each bridge down to a segment connecting two points. The Bridges of Konigsberg
- 40. • 1735: Enter Euler...his idea: compress each land area down to a single point, and each bridge down to a segment connecting two points. • This is just a graph! The Bridges of Konigsberg http://www.math.uwaterloo.ca/navigation/ideas/Zeno/zenocando.shtml
- 41. • 1735: Enter Euler...his idea: compress each land area down to a single point, and each bridge down to a segment connecting two points. • This is just a graph! • What we are looking for, then, is a cycle in this graph which covers each edge exactly once. The Bridges of Konigsberg http://www.math.uwaterloo.ca/navigation/ideas/Zeno/zenocando.shtml
- 42. • 1735: Enter Euler...his idea: compress each land area down to a single point, and each bridge down to a segment connecting two points. • This is just a graph! • What we are looking for, then, is a cycle in this graph which covers each edge exactly once. • Using this setup, Euler showed that such a cycle cannot exist. The Bridges of Konigsberg http://www.math.uwaterloo.ca/navigation/ideas/Zeno/zenocando.shtml
- 43. Eulerian Cycle Problem • Input: A graph G = (V, E). • Output: A cycle in G that touches every edge in E (called an Eulerian cycle), if one exists. • Example: At right is a demonstration of an Eulerian cycle. http://mathworld.wolfram.com/EulerianCycle.html
- 44. Eulerian Cycle Problem • Theorem: The Eulerian Cycle Problem can be solved in linear time. • So whereas finding a Hamiltonian cycle quickly becomes intractable for an arbitrary graph, finding an Eulerian cycle is relatively much easier. • Keep this fact in mind, as it will become essential.
- 45. Section 3: Basic Biological Applications of Graph Theory
- 46. Modeling Hydrocarbons with Graphs • Arthur Cayley studied chemical structures of hydrocarbons in the mid-1800s. • He used trees (acyclic connected graphs) to enumerate structural isomers. Hydrocarbon StructureArthur Cayley http://www.scientific-web.com/en/Mathematics/Biographies/ArthurCayley01.html
- 47. T4 Bacteriophages: Life Finds a Way • Normally, the T4 bacteriophage kills bacteria • However, if T4 is mutated (e.g., an important gene is deleted) it gets disabled and loses the ability to kill bacteria • Suppose a bacterium is infected with two different disabled mutants– would the bacterium still survive? • Amazingly, a pair of disabled viruses can still kill a bacterium. • How is this possible? T4 Bacteriophage
- 48. Benzer’s Experiment • Seymour Benzer’s Idea: Infect bacteria with pairs of mutant T4 bacteriophage (virus). • Each T4 mutant has an unknown interval deleted from its genome. • If the two intervals overlap: T4 pair is missing part of its genome and is disabled—bacteria survive. • If the two intervals do not overlap: T4 pair has its entire genome and is enabled – bacteria are killed. http://commons.wikimedia.org/wiki/File:Seymour_Benzer.gif Seymour Benzer
- 50. Benzer’s Experiment and Graph Theory • We construct an interval graph: • Each T4 mutant forms a vertex. • Place an edge between mutant pairs where bacteria survived (i.e., the deleted intervals in the pair of mutants overlap) • As the next slides show, the interval graph structure reveals whether DNA is linear or branched.
- 51. Interval Graph: Linear Genomes
- 52. Interval Graph: Linear Genomes
- 53. Interval Graph: Linear Genomes
- 54. Interval Graph: Linear Genomes
- 55. Interval Graph: Linear Genomes
- 56. Interval Graph: Linear Genomes
- 57. Interval Graph: Linear Genomes
- 58. Interval Graph: Linear Genomes
- 59. Interval Graph: Linear Genomes
- 60. Interval Graph: Linear Genomes
- 61. Interval Graph: Linear Genomes
- 62. Interval Graph: Branched Genomes
- 63. Interval Graph: Branched Genomes
- 64. Interval Graph: Branched Genomes
- 65. Interval Graph: Branched Genomes
- 66. Interval Graph: Branched Genomes
- 67. Interval Graph: Branched Genomes
- 68. Interval Graph: Branched Genomes
- 69. Interval Graph: Branched Genomes
- 70. Interval Graph: Branched Genomes
- 71. Interval Graph: Branched Genomes
- 72. Interval Graph: Branched Genomes
- 73. Linear Genome Branched Genome Linear vs. Branched Genomes: Interval Graphs • Simply by comparing the structure of the two interval graphs, Benzer showed that genomes cannot be branched!
- 75. • Sanger Method (1977): Labeled ddNTPs terminate DNA copying at random points. • Both methods generate labeled fragments of varying lengths that are further electrophoresed. • Gilbert Method (1977): Chemical method to cleave DNA at specific points (G, G+A, T+C, C). DNA Sequencing: History Frederick Sanger Walter Gilbert
- 76. Sanger Method: Generating Read 1. Start at primer (restriction site). 2. Grow DNA chain. 3. Include ddNTPs. 4. Stop reaction at all possible points. 5. Separate products by length, using gel electrophoresis.
- 77. Sanger Method: Sequencing • Shear DNA into millions of small fragments. • Read 500 – 700 nucleotides at a time from the small fragments.
- 78. Fragment Assembly • Computational Challenge: assemble individual short fragments (“reads”) into a single genomic sequence (“superstring”). • Until late 1990s the so called “shotgun fragment assembly” of the human genome was viewed as an intractable problem, because it required so much work for a large genome. • Our computational challenge leads to the formal problem at the beginning of the next section.
- 79. Section 5: Shortest Superstring & Traveling Salesman Problems
- 80. Shortest Superstring Problem (SSP) • Problem: Given a set of strings, find a shortest string that contains all of them. • Input: Strings s1, s2,…., sn • Output: A “superstring” s that contains all strings s1, s2,…., sn as substrings, such that the length of s is minimized.
- 81. SSP: Example
- 82. SSP: Example
- 83. SSP: Example
- 84. SSP: Example • So our greedy guess of concatenating all the strings together turns out to be substantially suboptimal (length 24 vs. 10).
- 85. SSP: Example • So our greedy guess of concatenating all the strings together turns out to be substantially suboptimal (length 24 vs. 10). • Note: The strings here are just the integers from 1 to 8 in base-2 notation.
- 86. SSP: Issues • Complexity: NP-complete (in a few slides). • Also, this formulation does not take into account the possibility of sequencing errors, and it is difficult to adapt to handle that consideration.
- 87. • Given strings si and sj , define overlap(si , sj ) as the length of the longest prefix of sj that matches a suffix of si . The Overlap Function
- 88. • Given strings si and sj , define overlap(si , sj ) as the length of the longest prefix of sj that matches a suffix of si . • Example: • s1 = aaaggcatcaaatctaaaggcatcaaa • s2 = aagcatcaaatctaaaggcatcaaa The Overlap Function
- 89. • Given strings si and sj , define overlap(si , sj ) as the length of the longest prefix of sj that matches a suffix of si . • Example: • s1 = aaaggcatcaaatctaaaggcatcaaa • s2 = aagcatcaaatctaaaggcatcaaa aaaggcatcaaatctaaaggcatcaaa aaaggcatcaaatctaaaggcatcaaa The Overlap Function
- 90. • Given strings si and sj , define overlap(si , sj ) as the length of the longest prefix of sj that matches a suffix of si . • Example: • s1 = aaaggcatcaaatctaaaggcatcaaa • s2 = aagcatcaaatctaaaggcatcaaa aaaggcatcaaatctaaaggcatcaaa aaaggcatcaaatctaaaggcatcaaa • Therefore, overlap(s1 , s2 ) = 12. The Overlap Function
- 91. Why is SSP an NP-Complete Problem? • Construct a graph G as follows: • The n vertices represent the n strings s1, s2,…., sn. • For every pair of vertices si and sj , insert an edge of length overlap( si, sj ) connecting the vertices. • Then finding the shortest superstring will correspond to finding the shortest Hamiltonian path in G. • But this is the Traveling Salesman Problem (TSP), which we know to be NP-complete. • Hence SSP must also be NP-Complete! • Note: We also need to show that any TSP can be formulated as a SSP (not difficult).
- 92. Reducing SSP to TSP: Example 1 • Take our previous set of strings S = {000, 001, 010, 011, 100, 101, 110, 111}.
- 93. Reducing SSP to TSP: Example 1 • Take our previous set of strings S = {000, 001, 010, 011, 100, 101, 110, 111}. • Then the graph for S is given at right.
- 94. Reducing SSP to TSP: Example 1 • Take our previous set of strings S = {000, 001, 010, 011, 100, 101, 110, 111}. • Then the graph for S is given at right. • One minimal Hamiltonian path gives our previous superstring, 0001110100.
- 95. Reducing SSP to TSP: Example 1 • Take our previous set of strings S = {000, 001, 010, 011, 100, 101, 110, 111}. • Then the graph for S is given at right. • One minimal Hamiltonian path gives our previous superstring, 0001110100. • Check that this works!
- 96. Reducing SSP to TSP: Example 2 • S = {ATC, CCA, CAG, TCC, AGT}
- 97. Reducing SSP to TSP: Example 2 ATC CCA TCC AGT CAG 2 2 22 1 1 1 0 1 1 • S = {ATC, CCA, CAG, TCC, AGT} • The graph is provided at right.
- 98. Reducing SSP to TSP: Example 2 ATC CCA TCC AGT CAG 2 2 22 1 1 1 0 1 1 • S = {ATC, CCA, CAG, TCC, AGT} • The graph is provided at right. • A minimal Hamiltonian path gives as shortest superstring ATCCAGT.
- 99. Reducing SSP to TSP: Example 2 ATC CCA TCC AGT CAG 2 2 22 1 1 1 0 1 1 ATC • S = {ATC, CCA, CAG, TCC, AGT} • The graph is provided at right. • A minimal Hamiltonian path gives as shortest superstring ATCCAGT.
- 100. Reducing SSP to TSP: Example 2 ATC CCA TCC AGT CAG 2 2 22 1 1 1 0 1 1 ATCC • S = {ATC, CCA, CAG, TCC, AGT} • The graph is provided at right. • A minimal Hamiltonian path gives as shortest superstring ATCCAGT.
- 101. Reducing SSP to TSP: Example 2 ATC CCA TCC AGT CAG 2 2 22 1 1 1 0 1 1 ATCCA • S = {ATC, CCA, CAG, TCC, AGT} • The graph is provided at right. • A minimal Hamiltonian path gives as shortest superstring ATCCAGT.
- 102. Reducing SSP to TSP: Example 2 ATC CCA TCC AGT CAG 2 2 22 1 1 1 0 1 1 ATCCAG • S = {ATC, CCA, CAG, TCC, AGT} • The graph is provided at right. • A minimal Hamiltonian path gives as shortest superstring ATCCAGT.
- 103. Reducing SSP to TSP: Example 2 ATC CCA TCC AGT CAG 2 2 22 1 1 1 0 1 1 • S = {ATC, CCA, CAG, TCC, AGT} • The graph is provided at right. • A minimal Hamiltonian path gives as shortest superstring ATCCAGT. ATCCAGT
- 105. • 1988: SBH is suggested as an an alternative sequencing method. Nobody believes it will ever work. • 1991: Light directed polymer synthesis is developed by Steve Fodor and colleagues. • 1994: Affymetrix develops the first 64-kb DNA microarray. First microarray prototype (1989) First commercial DNA microarray prototype w/16,000 features (1994) 500,000 features per chip (2002) Sequencing by Hybridization (SBH): History
- 106. • Attach all possible DNA probes of length l to a flat surface, each probe at a distinct known location. This set of probes is called a DNA array. • Apply a solution containing fluorescently labeled DNA fragment to the array. • The DNA fragment hybridizes with those probes that are complementary to substrings of length l of the fragment. How SBH Works Hybridization of a DNA Probe http://members.cox.net/amgough/Fanconi-genetics-PGD.htm
- 107. How SBH Works • Using a spectroscopic detector, determine which probes hybridize to the DNA fragment to obtain the l–mer composition of the target DNA fragment. • Reconstruct the sequence of the target DNA fragment from the l-mer composition. DNA Microarray http://www.wormbook.org/chapters/www_germlinegenomics/germlinegenomics.html
- 108. How SBH Works: Example • Say our DNA fragment hybridizes to indicate that it contains the following substrings: GCAA, CAAA, ATAG, TAGG, ACGC, GGCA. • Then the most logical explanation is that our fragment is the shortest superstring containing these strings! • Here the superstring is: ATAGGCAAACGC DNA Microarray Interpreted
- 109. l-mer Composition • Spectrum( s, l ): The unordered multiset of all l-mers in a string s of length n. • The order of individual elements in Spectrum( s, l ) does not matter.
- 110. l-mer Composition • Spectrum( s, l ): The unordered multiset of all l-mers in a string s of length n. • The order of individual elements in Spectrum( s, l ) does not matter. • For s = TATGGTGC all of the following are equivalent representations of Spectrum( s, 3):
- 111. l-mer Composition • Spectrum( s, l ): The unordered multiset of all l-mers in a string s of length n. • The order of individual elements in Spectrum( s, l ) does not matter. • For s = TATGGTGC all of the following are equivalent representations of Spectrum( s, 3): {TAT, ATG, TGG, GGT, GTG, TGC}
- 112. l-mer Composition • Spectrum( s, l ): The unordered multiset of all l-mers in a string s of length n. • The order of individual elements in Spectrum( s, l ) does not matter. • For s = TATGGTGC all of the following are equivalent representations of Spectrum( s, 3): {TAT, ATG, TGG, GGT, GTG, TGC} {ATG, GGT, GTG, TAT, TGC, TGG}
- 113. l-mer Composition • Spectrum( s, l ): The unordered multiset of all l-mers in a string s of length n. • The order of individual elements in Spectrum( s, l ) does not matter. • For s = TATGGTGC all of the following are equivalent representations of Spectrum( s, 3): {TAT, ATG, TGG, GGT, GTG, TGC} {ATG, GGT, GTG, TAT, TGC, TGG} {TGG, TGC, TAT, GTG, GGT, ATG}
- 114. l-mer Composition • Spectrum( s, l ): The unordered multiset of all l-mers in a string s of length n. • The order of individual elements in Spectrum( s, l ) does not matter. • For s = TATGGTGC all of the following are equivalent representations of Spectrum( s, 3): {TAT, ATG, TGG, GGT, GTG, TGC} {ATG, GGT, GTG, TAT, TGC, TGG} {TGG, TGC, TAT, GTG, GGT, ATG} • Which ordering do we choose?
- 115. l-mer Composition • Spectrum( s, l ): The unordered multiset of all l-mers in a string s of length n. • The order of individual elements in Spectrum( s, l ) does not matter. • For s = TATGGTGC all of the following are equivalent representations of Spectrum( s, 3): {TAT, ATG, TGG, GGT, GTG, TGC} {ATG, GGT, GTG, TAT, TGC, TGG} {TGG, TGC, TAT, GTG, GGT, ATG} • Which ordering do we choose? Typically the one that is lexicographic, meaning in alphabetical order (think of a phonebook).
- 116. • Different sequences may share a common spectrum. • Example: Different Sequences, Same Spectrum Spectrum GTATCT, 2 Spectrum GTCTAT, 2 AT, CT, GT, TA, TC
- 117. The SBH Problem • Problem: Reconstruct a string from its l-mer composition • Input: A set S, representing all l-mers from an (unknown) string s. • Output: A string s such that Spectrum( s, l ) = S • Note: As we have seen, there may be more than one correct answer. Determining which DNA sequence is actually correct is another matter.
- 118. SBH: Hamiltonian Path Approach • Create a graph G as follows: • Create one vertex for each member of S. • Connect vertex v to vertex w with a directed edge (arrow) if the last l – 1 elements of v match the first l – 1 elements of w. • Then a Hamiltonian path in this graph will correspond to a string s such that Spectrum( s, l )!
- 119. SBH: Hamiltonian Path Approach • Example: S = {ATG TGG TGC GTG GGC GCA GCG CGT}
- 120. SBH: Hamiltonian Path Approach • Example: S = {ATG TGG TGC GTG GGC GCA GCG CGT}
- 121. SBH: Hamiltonian Path Approach • Example: S = {ATG TGG TGC GTG GGC GCA GCG CGT}
- 122. SBH: Hamiltonian Path Approach • Example: S = {ATG TGG TGC GTG GGC GCA GCG CGT}
- 123. SBH: Hamiltonian Path Approach • Example: S = {ATG TGG TGC GTG GGC GCA GCG CGT}
- 124. SBH: Hamiltonian Path Approach • Example: S = {ATG TGG TGC GTG GGC GCA GCG CGT}
- 125. SBH: Hamiltonian Path Approach • Example: S = {ATG TGG TGC GTG GGC GCA GCG CGT}
- 126. SBH: Hamiltonian Path Approach • Example: S = {ATG TGG TGC GTG GGC GCA GCG CGT}
- 127. SBH: Hamiltonian Path Approach • Example: S = {ATG TGG TGC GTG GGC GCA GCG CGT}
- 128. SBH: Hamiltonian Path Approach • Example: S = {ATG TGG TGC GTG GGC GCA GCG CGT}
- 129. SBH: Hamiltonian Path Approach • Example: S = {ATG TGG TGC GTG GGC GCA GCG CGT}
- 130. SBH: Hamiltonian Path Approach • Example: S = {ATG TGG TGC GTG GGC GCA GCG CGT}
- 131. SBH: Hamiltonian Path Approach • Example: S = {ATG TGG TGC GTG GGC GCA GCG CGT}
- 132. SBH: Hamiltonian Path Approach • Example: S = {ATG TGG TGC GTG GGC GCA GCG CGT} • There are actually two Hamiltonian paths in this graph:
- 133. SBH: Hamiltonian Path Approach • Example: S = {ATG TGG TGC GTG GGC GCA GCG CGT} • There are actually two Hamiltonian paths in this graph: • Path 1:
- 134. SBH: Hamiltonian Path Approach • Example: S = {ATG TGG TGC GTG GGC GCA GCG CGT} • There are actually two Hamiltonian paths in this graph: • Path 1: Gives the string S =
- 135. SBH: Hamiltonian Path Approach • Example: S = {ATG TGG TGC GTG GGC GCA GCG CGT} • There are actually two Hamiltonian paths in this graph: • Path 1: Gives the string S = ATG
- 136. SBH: Hamiltonian Path Approach • Example: S = {ATG TGG TGC GTG GGC GCA GCG CGT} • There are actually two Hamiltonian paths in this graph: • Path 1: Gives the string S = ATGC
- 137. SBH: Hamiltonian Path Approach • Example: S = {ATG TGG TGC GTG GGC GCA GCG CGT} • There are actually two Hamiltonian paths in this graph: • Path 1: Gives the string S = ATGCG
- 138. SBH: Hamiltonian Path Approach • Example: S = {ATG TGG TGC GTG GGC GCA GCG CGT} • There are actually two Hamiltonian paths in this graph: • Path 1: Gives the string S = ATGCGT
- 139. SBH: Hamiltonian Path Approach • Example: S = {ATG TGG TGC GTG GGC GCA GCG CGT} • There are actually two Hamiltonian paths in this graph: • Path 1: Gives the string S = ATGCGTG
- 140. SBH: Hamiltonian Path Approach • Example: S = {ATG TGG TGC GTG GGC GCA GCG CGT} • There are actually two Hamiltonian paths in this graph: • Path 1: Gives the string S = ATGCGTGG
- 141. SBH: Hamiltonian Path Approach • Example: S = {ATG TGG TGC GTG GGC GCA GCG CGT} • There are actually two Hamiltonian paths in this graph: • Path 1: Gives the string S = ATGCGTGGC
- 142. SBH: Hamiltonian Path Approach • Example: S = {ATG TGG TGC GTG GGC GCA GCG CGT} • There are actually two Hamiltonian paths in this graph: • Path 1: Gives the string S = ATGCGTGGCA
- 143. SBH: Hamiltonian Path Approach • Example: S = {ATG TGG TGC GTG GGC GCA GCG CGT} • There are actually two Hamiltonian paths in this graph: • Path 1: Gives the string S = ATGCGTGGCA • Path 2:
- 144. SBH: Hamiltonian Path Approach • Example: S = {ATG TGG TGC GTG GGC GCA GCG CGT} • There are actually two Hamiltonian paths in this graph: • Path 1: Gives the string S = ATGCGTGGCA • Path 2: Gives the string S =
- 145. SBH: Hamiltonian Path Approach • Example: S = {ATG TGG TGC GTG GGC GCA GCG CGT} • There are actually two Hamiltonian paths in this graph: • Path 1: Gives the string S = ATGCGTGGCA • Path 2: Gives the string S = ATG
- 146. SBH: Hamiltonian Path Approach • Example: S = {ATG TGG TGC GTG GGC GCA GCG CGT} • There are actually two Hamiltonian paths in this graph: • Path 1: Gives the string S = ATGCGTGGCA • Path 2: Gives the string S = ATGG
- 147. SBH: Hamiltonian Path Approach • Example: S = {ATG TGG TGC GTG GGC GCA GCG CGT} • There are actually two Hamiltonian paths in this graph: • Path 1: Gives the string S = ATGCGTGGCA • Path 2: Gives the string S = ATGGC
- 148. SBH: Hamiltonian Path Approach • Example: S = {ATG TGG TGC GTG GGC GCA GCG CGT} • There are actually two Hamiltonian paths in this graph: • Path 1: Gives the string S = ATGCGTGGCA • Path 2: Gives the string S = ATGGCG
- 149. SBH: Hamiltonian Path Approach • Example: S = {ATG TGG TGC GTG GGC GCA GCG CGT} • There are actually two Hamiltonian paths in this graph: • Path 1: Gives the string S = ATGCGTGGCA • Path 2: Gives the string S = ATGGCGT
- 150. SBH: Hamiltonian Path Approach • Example: S = {ATG TGG TGC GTG GGC GCA GCG CGT} • There are actually two Hamiltonian paths in this graph: • Path 1: Gives the string S = ATGCGTGGCA • Path 2: Gives the string S = ATGGCGTG
- 151. SBH: Hamiltonian Path Approach • Example: S = {ATG TGG TGC GTG GGC GCA GCG CGT} • There are actually two Hamiltonian paths in this graph: • Path 1: Gives the string S = ATGCGTGGCA • Path 2: Gives the string S = ATGGCGTGC
- 152. SBH: Hamiltonian Path Approach • Example: S = {ATG TGG TGC GTG GGC GCA GCG CGT} • There are actually two Hamiltonian paths in this graph: • Path 1: Gives the string S = ATGCGTGGCA • Path 2: Gives the string S = ATGGCGTGCA
- 153. SBH: A Lost Cause? • At this point, we should be concerned about using a Hamiltonian path to solve SBH. • After all, recall that SSP was an NP-Complete problem, and we have seen that an instance of SBH is an instance of SSP. • However, note that SBH is actually a specific case of SSP, so there is still hope for an efficient algorithm for SBH: • We are considering a spectrum of only l-mers, and not strings of any other length. • Also, we only are connecting two l-mers with an edge if and only if the overlap between them is l – 1, whereas before we connected l-mers if there was any overlap at all. • Note: SBH is not NP-Complete since SBH reduces to SSP, but not vice-versa.
- 154. SBH: Eulerian Path Approach • So instead, let us consider a completely different graph G: • Vertices = the set of (l – 1)-mers which are substrings of some l-mer from our set S. • v is connected to w with a directed edge if the final l – 2 elements of v agree with the first l – 2 elements of w, and the union of v and w is in S. • Example: S = {ATG, TGG, TGC, GTG, GGC, GCA, GCG, CGT}.
- 155. SBH: Eulerian Path Approach • So instead, let us consider a completely different graph G: • Vertices = the set of (l – 1)-mers which are substrings of some l-mer from our set S. • v is connected to w with a directed edge if the final l – 2 elements of v agree with the first l – 2 elements of w, and the union of v and w is in S. • Example: S = {ATG, TGG, TGC, GTG, GGC, GCA, GCG, CGT}. • V = {AT}. AT GT CG CAGCTG GG
- 156. SBH: Eulerian Path Approach • So instead, let us consider a completely different graph G: • Vertices = the set of (l – 1)-mers which are substrings of some l-mer from our set S. • v is connected to w with a directed edge if the final l – 2 elements of v agree with the first l – 2 elements of w, and the union of v and w is in S. • Example: S = {ATG, TGG, TGC, GTG, GGC, GCA, GCG, CGT}. • V = {AT, TG}. AT GT CG CAGCTG GG
- 157. SBH: Eulerian Path Approach • So instead, let us consider a completely different graph G: • Vertices = the set of (l – 1)-mers which are substrings of some l-mer from our set S. • v is connected to w with a directed edge if the final l – 2 elements of v agree with the first l – 2 elements of w, and the union of v and w is in S. • Example: S = {ATG, TGG, TGC, GTG, GGC, GCA, GCG, CGT}. • V = {AT, TG, GG}. AT GT CG CAGCTG GG
- 158. SBH: Eulerian Path Approach • So instead, let us consider a completely different graph G: • Vertices = the set of (l – 1)-mers which are substrings of some l-mer from our set S. • v is connected to w with a directed edge if the final l – 2 elements of v agree with the first l – 2 elements of w, and the union of v and w is in S. • Example: S = {ATG, TGG, TGC, GTG, GGC, GCA, GCG, CGT}. • V = {AT, TG, GG, GC}. AT GT CG CAGCTG GG
- 159. SBH: Eulerian Path Approach • So instead, let us consider a completely different graph G: • Vertices = the set of (l – 1)-mers which are substrings of some l-mer from our set S. • v is connected to w with a directed edge if the final l – 2 elements of v agree with the first l – 2 elements of w, and the union of v and w is in S. • Example: S = {ATG, TGG, TGC, GTG, GGC, GCA, GCG, CGT}. • V = {AT, TG, GG, GC, GT}. AT GT CG CAGCTG GG
- 160. SBH: Eulerian Path Approach • So instead, let us consider a completely different graph G: • Vertices = the set of (l – 1)-mers which are substrings of some l-mer from our set S. • v is connected to w with a directed edge if the final l – 2 elements of v agree with the first l – 2 elements of w, and the union of v and w is in S. • Example: S = {ATG, TGG, TGC, GTG, GGC, GCA, GCG, CGT}. • V = {AT, TG, GG, GC, GT, CA}. AT GT CG CAGCTG GG
- 161. SBH: Eulerian Path Approach • So instead, let us consider a completely different graph G: • Vertices = the set of (l – 1)-mers which are substrings of some l-mer from our set S. • v is connected to w with a directed edge if the final l – 2 elements of v agree with the first l – 2 elements of w, and the union of v and w is in S. • Example: S = {ATG, TGG, TGC, GTG, GGC, GCA, GCG, CGT}. • V = {AT, TG, GG, GC, GT, CA, CG}. AT GT CG CAGCTG GG
- 162. SBH: Eulerian Path Approach • So instead, let us consider a completely different graph G: • Vertices = the set of (l – 1)-mers which are substrings of some l-mer from our set S. • v is connected to w with a directed edge if the final l – 2 elements of v agree with the first l – 2 elements of w, and the union of v and w is in S. • Example: S = {ATG, TGG, TGC, GTG, GGC, GCA, GCG, CGT}. • V = {AT, TG, GG, GC, GT, CA, CG}. • E = shown at right. AT GT CG CAGCTG GG
- 163. SBH: Eulerian Path Approach • So instead, let us consider a completely different graph G: • Vertices = the set of (l – 1)-mers which are substrings of some l-mer from our set S. • v is connected to w with a directed edge if the final l – 2 elements of v agree with the first l – 2 elements of w, and the union of v and w is in S. • Example: S = {ATG, TGG, TGC, GTG, GGC, GCA, GCG, CGT}. • V = {AT, TG, GG, GC, GT, CA, CG}. • E = shown at right. AT GT CG CAGCTG GG
- 164. SBH: Eulerian Path Approach • So instead, let us consider a completely different graph G: • Vertices = the set of (l – 1)-mers which are substrings of some l-mer from our set S. • v is connected to w with a directed edge if the final l – 2 elements of v agree with the first l – 2 elements of w, and the union of v and w is in S. • Example: S = {ATG, TGG, TGC, GTG, GGC, GCA, GCG, CGT}. • V = {AT, TG, GG, GC, GT, CA, CG}. • E = shown at right. AT GT CG CAGCTG GG
- 165. SBH: Eulerian Path Approach • So instead, let us consider a completely different graph G: • Vertices = the set of (l – 1)-mers which are substrings of some l-mer from our set S. • v is connected to w with a directed edge if the final l – 2 elements of v agree with the first l – 2 elements of w, and the union of v and w is in S. • Example: S = {ATG, TGG, TGC, GTG, GGC, GCA, GCG, CGT}. • V = {AT, TG, GG, GC, GT, CA, CG}. • E = shown at right. AT GT CG CAGCTG GG
- 166. SBH: Eulerian Path Approach • So instead, let us consider a completely different graph G: • Vertices = the set of (l – 1)-mers which are substrings of some l-mer from our set S. • v is connected to w with a directed edge if the final l – 2 elements of v agree with the first l – 2 elements of w, and the union of v and w is in S. • Example: S = {ATG, TGG, TGC, GTG, GGC, GCA, GCG, CGT}. • V = {AT, TG, GG, GC, GT, CA, CG}. • E = shown at right. AT GT CG CAGCTG GG
- 167. SBH: Eulerian Path Approach • So instead, let us consider a completely different graph G: • Vertices = the set of (l – 1)-mers which are substrings of some l-mer from our set S. • v is connected to w with a directed edge if the final l – 2 elements of v agree with the first l – 2 elements of w, and the union of v and w is in S. • Example: S = {ATG, TGG, TGC, GTG, GGC, GCA, GCG, CGT}. • V = {AT, TG, GG, GC, GT, CA, CG}. • E = shown at right. AT GT CG CAGCTG GG
- 168. SBH: Eulerian Path Approach • So instead, let us consider a completely different graph G: • Vertices = the set of (l – 1)-mers which are substrings of some l-mer from our set S. • v is connected to w with a directed edge if the final l – 2 elements of v agree with the first l – 2 elements of w, and the union of v and w is in S. • Example: S = {ATG, TGG, TGC, GTG, GGC, GCA, GCG, CGT}. • V = {AT, TG, GG, GC, GT, CA, CG}. • E = shown at right. AT GT CG CAGCTG GG
- 169. SBH: Eulerian Path Approach • So instead, let us consider a completely different graph G: • Vertices = the set of (l – 1)-mers which are substrings of some l-mer from our set S. • v is connected to w with a directed edge if the final l – 2 elements of v agree with the first l – 2 elements of w, and the union of v and w is in S. • Example: S = {ATG, TGG, TGC, GTG, GGC, GCA, GCG, CGT}. • V = {AT, TG, GG, GC, GT, CA, CG}. • E = shown at right. AT GT CG CAGCTG GG
- 170. SBH: Eulerian Path Approach • So instead, let us consider a completely different graph G: • Vertices = the set of (l – 1)-mers which are substrings of some l-mer from our set S. • v is connected to w with a directed edge if the final l – 2 elements of v agree with the first l – 2 elements of w, and the union of v and w is in S. • Example: S = {ATG, TGG, TGC, GTG, GGC, GCA, GCG, CGT}. • V = {AT, TG, GG, GC, GT, CA, CG}. • E = shown at right. AT GT CG CAGCTG GG
- 171. SBH: Eulerian Path Approach • So instead, let us consider a completely different graph G: • Vertices = the set of (l – 1)-mers which are substrings of some l-mer from our set S. • v is connected to w with a directed edge if the final l – 2 elements of v agree with the first l – 2 elements of w, and the union of v and w is in S. • Example: S = {ATG, TGG, TGC, GTG, GGC, GCA, GCG, CGT}. • V = {AT, TG, GG, GC, GT, CA, CG}. • E = shown at right. AT GT CG CAGCTG GG
- 172. SBH: Eulerian Path Approach • Key Point: A sequence reconstruction will actually correspond to an Eulerian path in this graph. • Recall that an Eulerian path is “easy” to find (one can always be found in linear time)…so we have found a simple solution to SBH! • In our example, two solutions: AT GT CG CAGCTG GG
- 173. SBH: Eulerian Path Approach • Key Point: A sequence reconstruction will actually correspond to an Eulerian path in this graph. • Recall that an Eulerian path is “easy” to find (one can always be found in linear time)…so we have found a simple solution to SBH! • In our example, two solutions: 1. ATG AT GT CG CAGCTG GG
- 174. SBH: Eulerian Path Approach • Key Point: A sequence reconstruction will actually correspond to an Eulerian path in this graph. • Recall that an Eulerian path is “easy” to find (one can always be found in linear time)…so we have found a simple solution to SBH! • In our example, two solutions: 1. ATGG AT GT CG CAGCTG GG
- 175. SBH: Eulerian Path Approach • Key Point: A sequence reconstruction will actually correspond to an Eulerian path in this graph. • Recall that an Eulerian path is “easy” to find (one can always be found in linear time)…so we have found a simple solution to SBH! • In our example, two solutions: 1. ATGGC AT GT CG CAGCTG GG
- 176. SBH: Eulerian Path Approach • Key Point: A sequence reconstruction will actually correspond to an Eulerian path in this graph. • Recall that an Eulerian path is “easy” to find (one can always be found in linear time)…so we have found a simple solution to SBH! • In our example, two solutions: 1. ATGGCG AT GT CG CAGCTG GG
- 177. SBH: Eulerian Path Approach • Key Point: A sequence reconstruction will actually correspond to an Eulerian path in this graph. • Recall that an Eulerian path is “easy” to find (one can always be found in linear time)…so we have found a simple solution to SBH! • In our example, two solutions: 1. ATGGCGT AT GT CG CAGCTG GG
- 178. SBH: Eulerian Path Approach • Key Point: A sequence reconstruction will actually correspond to an Eulerian path in this graph. • Recall that an Eulerian path is “easy” to find (one can always be found in linear time)…so we have found a simple solution to SBH! • In our example, two solutions: 1. ATGGCGTG AT GT CG CAGCTG GG
- 179. SBH: Eulerian Path Approach • Key Point: A sequence reconstruction will actually correspond to an Eulerian path in this graph. • Recall that an Eulerian path is “easy” to find (one can always be found in linear time)…so we have found a simple solution to SBH! • In our example, two solutions: 1. ATGGCGTGC AT GT CG CAGCTG GG
- 180. SBH: Eulerian Path Approach • Key Point: A sequence reconstruction will actually correspond to an Eulerian path in this graph. • Recall that an Eulerian path is “easy” to find (one can always be found in linear time)…so we have found a simple solution to SBH! • In our example, two solutions: 1. ATGGCGTGCA AT GT CG CAGCTG GG
- 181. SBH: Eulerian Path Approach • Key Point: A sequence reconstruction will actually correspond to an Eulerian path in this graph. • Recall that an Eulerian path is “easy” to find (one can always be found in linear time)…so we have found a simple solution to SBH! • In our example, two solutions: 1. ATGGCGTGCA AT GT CG CAGCTG GG
- 182. SBH: Eulerian Path Approach • Key Point: A sequence reconstruction will actually correspond to an Eulerian path in this graph. • Recall that an Eulerian path is “easy” to find (one can always be found in linear time)…so we have found a simple solution to SBH! • In our example, two solutions: 1. ATGGCGTGCA 2. ATG AT GT CG CAGCTG GG
- 183. SBH: Eulerian Path Approach • Key Point: A sequence reconstruction will actually correspond to an Eulerian path in this graph. • Recall that an Eulerian path is “easy” to find (one can always be found in linear time)…so we have found a simple solution to SBH! • In our example, two solutions: 1. ATGGCGTGCA 2. ATGC AT GT CG CAGCTG GG
- 184. SBH: Eulerian Path Approach • Key Point: A sequence reconstruction will actually correspond to an Eulerian path in this graph. • Recall that an Eulerian path is “easy” to find (one can always be found in linear time)…so we have found a simple solution to SBH! • In our example, two solutions: 1. ATGGCGTGCA 2. ATGCG AT GT CG CAGCTG GG
- 185. SBH: Eulerian Path Approach • Key Point: A sequence reconstruction will actually correspond to an Eulerian path in this graph. • Recall that an Eulerian path is “easy” to find (one can always be found in linear time)…so we have found a simple solution to SBH! • In our example, two solutions: 1. ATGGCGTGCA 2. ATGCGT AT GT CG CAGCTG GG
- 186. SBH: Eulerian Path Approach • Key Point: A sequence reconstruction will actually correspond to an Eulerian path in this graph. • Recall that an Eulerian path is “easy” to find (one can always be found in linear time)…so we have found a simple solution to SBH! • In our example, two solutions: 1. ATGGCGTGCA 2. ATGCGTG AT GT CG CAGCTG GG
- 187. SBH: Eulerian Path Approach • Key Point: A sequence reconstruction will actually correspond to an Eulerian path in this graph. • Recall that an Eulerian path is “easy” to find (one can always be found in linear time)…so we have found a simple solution to SBH! • In our example, two solutions: 1. ATGGCGTGCA 2. ATGCGTGG AT GT CG CAGCTG GG
- 188. SBH: Eulerian Path Approach • Key Point: A sequence reconstruction will actually correspond to an Eulerian path in this graph. • Recall that an Eulerian path is “easy” to find (one can always be found in linear time)…so we have found a simple solution to SBH! • In our example, two solutions: 1. ATGGCGTGCA 2. ATGCGTGGC AT GT CG CAGCTG GG
- 189. SBH: Eulerian Path Approach • Key Point: A sequence reconstruction will actually correspond to an Eulerian path in this graph. • Recall that an Eulerian path is “easy” to find (one can always be found in linear time)…so we have found a simple solution to SBH! • In our example, two solutions: 1. ATGGCGTGCA 2. ATGCGTGGCA AT GT CG CAGCTG GG
- 190. But…How Do We Know an Eulerian Path Exists? • A graph is balanced if for every vertex the number of incoming edges equals to the number of outgoing edges. We write this for vertex v as: in(v)=out(v) • Theorem: A connected graph is Eulerian (i.e. contains an Eulerian cycle) if and only if each of its vertices is balanced. • We will prove this by demonstrating the following: 1. Every Eulerian graph is balanced. 2. Every balanced graph is Eulerian.
- 191. Every Eulerian Graph is Balanced • Suppose we have an Eulerian graph G. Call C the Eulerian cycle of G, and let v be any vertex of G. • For every edge e entering v, we can pair e with an edge leaving v, which is simply the edge in our cycle C that follows e. • Therefore it directly follows that in(v)=out(v) as needed, and since our choice of v was arbitrary, this relation must hold for all vertices in G, so we are finished with the first part.
- 192. Every Balanced Graph is Eulerian • Next, suppose that we have a balanced graph G. • We will actually construct an Eulerian cycle in G. • Start with an arbitrary vertex v and form a path in G without repeated edges until we reach a “dead end,” meaning a vertex with no unused edges leaving it. • G is balanced, so every time we enter a vertex w that isn’t v during the course of our path, we can find an edge leaving w. So our dead end is v and we have a cycle.
- 193. Every Balanced Graph is Eulerian • We have two simple cases for our cycle, which we call C: 1. C is an Eulerian cycle G is Eulerian DONE. 2. C is not an Eulerian cycle. • So we can assume that C is not an Eulerian cycle, which means that C contains vertices which have untraversed edges. • Let w be such a vertex, and start a new path from w. Once again, we must obtain a cycle, say C’.
- 194. Every Balanced Graph is Eulerian • Combine our cycles C and C’ into a bigger cycle C* by swapping edges at w (see figure). • Once again, we test C*: 1. C* is an Eulerian cycle G is Eulerian DONE. 2. C* is not an Eulerian cycle. • If C* is not Eulerian, we iterate our procedure. Because G has a finite number of edges, we must eventually reach a point where our current cycle is Eulerian (Case 1 above). DONE.
- 195. • A vertex v is semi-balanced if either in(v) = out(v) + 1 or in(v) = out(v) – 1 . • 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. • If G has no semi-balanced vertices, DONE. • If G has two semi-balanced vertices, connect them with a new edge e, so that the graph G + e is balanced and must be Eulerian. Remove e from the Eulerian cycle in G + e to obtain an Eulerian path in G. • Think: Why can G not have just one semi-balanced vertex? Euler’s Theorem: Extension
- 196. • Fidelity of Hybridization: It is difficult to detect differences between probes hybridized with perfect matches and those with one mismatch. • Array Size: The 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. Some Difficulties with SBH
- 197. • Practicality Again: Although SBH is still impractical, it spearheaded expression analysis and SNP analysis techniques. • Practicality Again and Again: In 2007 Solexa (now Illumina) developed a new DNA sequencing approach that generates so many short l-mers that they essentially mimic a universal DNA array. Some Difficulties with SBH
- 198. Section 7: Fragment Assembly & Repeats in DNA
- 201. DNA Shake DNA fragments Traditional DNA Sequencing
- 202. DNA Shake DNA fragments Traditional DNA Sequencing
- 203. DNA Shake DNA fragments Vector Circular genome (bacterium, plasmid) Traditional DNA Sequencing
- 204. + DNA Shake DNA fragments Vector Circular genome (bacterium, plasmid) Traditional DNA Sequencing
- 205. + DNA Shake DNA fragments Vector Circular genome (bacterium, plasmid) Traditional DNA Sequencing
- 206. + = DNA Shake DNA fragments Vector Circular genome (bacterium, plasmid) Traditional DNA Sequencing
- 207. + = DNA Shake DNA fragments Vector Circular genome (bacterium, plasmid) Traditional DNA Sequencing
- 208. + = DNA Shake DNA fragments Vector Circular genome (bacterium, plasmid) Known location (restriction site) Traditional DNA Sequencing
- 209. Different Types of Vectors Vector Size of Insert (bp) Plasmid 2,000 - 10,000 Cosmid 40,000 BAC (Bacterial Artificial Chromosome) 70,000 - 300,000 YAC (Yeast Artificial Chromosome) > 300,000 Not used much recently
- 211. Electrophoresis Diagrams: Hard to Read
- 212. Reading an Electropherogram • Reading an Electropherogram requires four processes: 1. Filtering 2. Smoothening 3. Correction for length compressions 4. A method for calling the nucleotides – PHRED
- 214. Shotgun Sequencing Cut many times at random (hence shotgun) Genomic Segment
- 215. Shotgun Sequencing Cut many times at random (hence shotgun) Genomic Segment
- 216. Shotgun Sequencing Cut many times at random (hence shotgun) Genomic Segment
- 217. Shotgun Sequencing Cut many times at random (hence shotgun) Genomic Segment Get one or two reads from each segment
- 218. Shotgun Sequencing Cut many times at random (hence shotgun) Genomic Segment Get one or two reads from each segment ~500 bp ~500 bp
- 219. Fragment Assembly • Cover region with ~7-fold redundancy. • Overlap reads and extend to reconstruct the original genomic region. Reads
- 220. Read Coverage • Length of genomic segment: L • Number of reads: n • Length of each read: l • Define the coverage as: C = n l / L • Question: How much coverage is enough? • Lander-Waterman Model: Assuming uniform distribution of reads, C = 10 results in 1 gap in coverage per million nucleotides. C
- 221. • Repeats: A major problem for fragment assembly. • More than 50% of human genome are repeats: • Over 1 million Alu repeats (about 300 bp). • About 200,000 LINE repeats (1000 bp and longer). Repeat Repeat Repeat Challenges in Fragment Assembly
- 222. • A Triazzle ® puzzle has only 16 pieces and looks simple. • BUT… there are many repeats! • The repeats make it very difficult to solve. • This repetition is what makes fragment assembly is so difficult. DNA Assembly Analogy: Triazzle http://www.triazzle.com/
- 223. Repeat Type Explanation • Low-Complexity DNA (e.g. ATATATATACATA…) • Microsatellite repeats (a1…ak)N where k ~ 3-6 (e.g. CAGCAGTAGCAGCACCAG) • Gene Families genes duplicate & then diverge • Segmental duplications ~very long, very similar copies Repeat Classification
- 224. Repeat Classification Repeat Type Explanation •SINE Transposon Short Interspersed Nuclear Elements (e.g., Alu: ~300 bp long, 106 copies) •LINE Transposon Long Interspersed Nuclear Elements ~500 - 5,000 bp long, 200,000 copies •LTR retroposons Long Terminal Repeats (~700 bp)
- 226. Assembly Method: Overlap-Layout-Consensus • Assemblers: ARACHNE, PHRAP, CAP, TIGR, CELERA
- 227. Assembly Method: Overlap-Layout-Consensus • Assemblers: ARACHNE, PHRAP, CAP, TIGR, CELERA • Three steps:
- 228. Assembly Method: Overlap-Layout-Consensus • Assemblers: ARACHNE, PHRAP, CAP, TIGR, CELERA • Three steps: 1. Overlap: Find potentially overlapping reads. Overlap
- 229. Assembly Method: Overlap-Layout-Consensus • Assemblers: ARACHNE, PHRAP, CAP, TIGR, CELERA • Three steps: 1. Overlap: Find potentially overlapping reads. 2. Layout: Merge reads into contigs and contigs into supercontigs. Layout Overlap
- 230. Assembly Method: Overlap-Layout-Consensus • Assemblers: ARACHNE, PHRAP, CAP, TIGR, CELERA • Three steps: 1. Overlap: Find potentially overlapping reads. 2. Layout: Merge reads into contigs and contigs into supercontigs. 3. Consensus: Derive the DNA sequence and correct any read errors. Consensus ..ACGATTACAATAGGTT.. Layout Overlap
- 231. Step 1: Overlap • Find the best match between the suffix of one read and the prefix of another. • Due to sequencing errors, we 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.
- 232. TAGATTACACAGATTAC TAGATTACACAGATTAC ||||||||||||||||| T GA TAGA | || TACA TAGT || Step 1: Overlap • 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.
- 233. • A k-mer that appears N times initiates N2 comparisons. • For an Alu that appears 106 times, we will have 1012 comparisons – this is too many. • Solution: Discard all k-mers that appear more than t Coverage, (t ~ 10) Step 1: Overlap
- 234. • We next create local multiple alignments from the overlapping reads. TAGATTACACAGATTACTGA TAGATTACACAGATTACTGA TAG TTACACAGATTATTGA TAGATTACACAGATTACTGA TAGATTACACAGATTACTGA TAGATTACACAGATTACTGA TAG TTACACAGATTATTGA TAGATTACACAGATTACTGA Step 2: Layout
- 235. Step 2: 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!
- 236. Step 2: 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 a high rate of misassembly (~20 %).
- 237. Step 2: 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 a high rate of misassembly (~20 %). • Misassembly means a lot more work at the finishing step.
- 238. • Repeats shorter than read length are OK. • Repeats with more base pair differences than the sequencing error rate are OK. • To make a smaller portion of the genome appear repetitive, try to: • Increase read length • Decrease sequencing error rate Step 2: Layout
- 239. Step 3: Consensus • A consensus sequence is derived from a profile of the assembled fragments. • A sufficient number of reads are required to ensure a statistically significant consensus. • Reading errors are corrected.
- 240. • Derive multiple alignment from pairwise read alignments. • Derive each consensus base by weighted voting. TAGATTACACAGATTACTGA TTGATGGCGTAA CTA TAGATTACACAGATTACTGACTTGATGGCGTAAACTA TAG TTACACAGATTATTGACTTCATGGCGTAA CTA TAGATTACACAGATTACTGACTTGATGGCGTAA CTA TAGATTACACAGATTACTGACTTGATGGGGTAA CTA TAGATTACACAGATTACTGACTTGATGGCGTAA CTA Step 3: Consensus Multiple Alignment Consensus String
- 241. • Each vertex represents a read from the original sequence. • Vertices are connected by an edge if they overlap. Overlap Graph: Hamiltonian Approach
- 242. Repeat Repeat Repeat • Each vertex represents a read from the original sequence. • Vertices are connected by an edge if they overlap. Overlap Graph: Hamiltonian Approach
- 243. Repeat Repeat Repeat • Each vertex represents a read from the original sequence. • Vertices are connected by an edge if they overlap. Overlap Graph: Hamiltonian Approach
- 244. Repeat Repeat Repeat • Each vertex represents a read from the original sequence. • Vertices are connected by an edge if they overlap. Overlap Graph: Hamiltonian Approach
- 245. Repeat Repeat Repeat • Each vertex represents a read from the original sequence. • Vertices are connected by an edge if they overlap. Overlap Graph: Hamiltonian Approach
- 246. Repeat Repeat Repeat • Each vertex represents a read from the original sequence. • Vertices are connected by an edge if they overlap. Overlap Graph: Hamiltonian Approach
- 247. Repeat Repeat Repeat • Each vertex represents a read from the original sequence. • Vertices are connected by an edge if they overlap. Overlap Graph: Hamiltonian Approach
- 248. Repeat Repeat Repeat • Each vertex represents a read from the original sequence. • Vertices are connected by an edge if they overlap. Overlap Graph: Hamiltonian Approach
- 249. Repeat Repeat Repeat • A Hamiltonian path in this graph provides a candidate assembly. • Each vertex represents a read from the original sequence. • Vertices are connected by an edge if they overlap. Overlap Graph: Hamiltonian Approach
- 250. • So finding an alignment corresponds to finding a Hamiltonian path in the overlap graph. • Recall that the Hamiltonian path/cycle problem is NP- Complete: no efficient algorithms are known. Overlap Graph: Hamiltonian Approach
- 251. • So finding an alignment corresponds to finding a Hamiltonian path in the overlap graph. • Recall that the Hamiltonian path/cycle problem is NP- Complete: no efficient algorithms are known. Overlap Graph: Hamiltonian Approach
- 252. • So finding an alignment corresponds to finding a Hamiltonian path in the overlap graph. • Recall that the Hamiltonian path/cycle problem is NP- Complete: no efficient algorithms are known. Overlap Graph: Hamiltonian Approach
- 253. • So finding an alignment corresponds to finding a Hamiltonian path in the overlap graph. • Recall that the Hamiltonian path/cycle problem is NP- Complete: no efficient algorithms are known. Overlap Graph: Hamiltonian Approach
- 254. • So finding an alignment corresponds to finding a Hamiltonian path in the overlap graph. • Recall that the Hamiltonian path/cycle problem is NP- Complete: no efficient algorithms are known. Overlap Graph: Hamiltonian Approach
- 255. • So finding an alignment corresponds to finding a Hamiltonian path in the overlap graph. • Recall that the Hamiltonian path/cycle problem is NP- Complete: no efficient algorithms are known. Overlap Graph: Hamiltonian Approach
- 256. • So finding an alignment corresponds to finding a Hamiltonian path in the overlap graph. • Recall that the Hamiltonian path/cycle problem is NP- Complete: no efficient algorithms are known. Overlap Graph: Hamiltonian Approach
- 257. • So finding an alignment corresponds to finding a Hamiltonian path in the overlap graph. • Recall that the Hamiltonian path/cycle problem is NP- Complete: no efficient algorithms are known. Overlap Graph: Hamiltonian Approach
- 258. • So finding an alignment corresponds to finding a Hamiltonian path in the overlap graph. • Recall that the Hamiltonian path/cycle problem is NP- Complete: no efficient algorithms are known. Overlap Graph: Hamiltonian Approach
- 259. • So finding an alignment corresponds to finding a Hamiltonian path in the overlap graph. • Recall that the Hamiltonian path/cycle problem is NP- Complete: no efficient algorithms are known. Overlap Graph: Hamiltonian Approach
- 260. • So finding an alignment corresponds to finding a Hamiltonian path in the overlap graph. • Recall that the Hamiltonian path/cycle problem is NP- Complete: no efficient algorithms are known. Overlap Graph: Hamiltonian Approach
- 261. • So finding an alignment corresponds to finding a Hamiltonian path in the overlap graph. • Recall that the Hamiltonian path/cycle problem is NP- Complete: no efficient algorithms are known. Overlap Graph: Hamiltonian Approach
- 262. • So finding an alignment corresponds to finding a Hamiltonian path in the overlap graph. • Recall that the Hamiltonian path/cycle problem is NP- Complete: no efficient algorithms are known. Overlap Graph: Hamiltonian Approach
- 263. • So finding an alignment corresponds to finding a Hamiltonian path in the overlap graph. • Recall that the Hamiltonian path/cycle problem is NP- Complete: no efficient algorithms are known. Overlap Graph: Hamiltonian Approach
- 264. • So finding an alignment corresponds to finding a Hamiltonian path in the overlap graph. • Recall that the Hamiltonian path/cycle problem is NP- Complete: no efficient algorithms are known. Overlap Graph: Hamiltonian Approach
- 265. • So finding an alignment corresponds to finding a Hamiltonian path in the overlap graph. • Recall that the Hamiltonian path/cycle problem is NP- Complete: no efficient algorithms are known. Overlap Graph: Hamiltonian Approach
- 266. • So finding an alignment corresponds to finding a Hamiltonian path in the overlap graph. • Recall that the Hamiltonian path/cycle problem is NP- Complete: no efficient algorithms are known. • Note: Finding a Hamiltonian path only looks easy because we know the optimal alignment before constructing overlap graph. Overlap Graph: Hamiltonian Approach
- 267. • The “overlap-layout-consensus” technique implicitly solves the Hamiltonian path problem and has a high rate of mis- assembly. • Can we adapt the Eulerian Path approach borrowed from the SBH problem? • Fragment assembly without repeat masking can be done in linear time with greater accuracy. EULER Approach to Fragment Assembly
- 268. Repeat Repeat Repeat Repeat Graph: Eulerian Approach • Gluing each repeat edge together gives a clear progression of the path through the entire sequence.
- 269. Repeat Repeat Repeat • Gluing each repeat edge together gives a clear progression of the path through the entire sequence. Repeat Graph: Eulerian Approach
- 270. Repeat Repeat Repeat Repeat Graph: Eulerian Approach • Gluing each repeat edge together gives a clear progression of the path through the entire sequence.
- 271. Repeat Repeat Repeat Repeat Graph: Eulerian Approach • Gluing each repeat edge together gives a clear progression of the path through the entire sequence. • In the repeat graph, an alignment corresponds to an Eulerian path…linear time reduction!
- 272. Repeat1 Repeat1Repeat2 Repeat2 • The repeat graph can be easily constructed with any number of repeats. Repeat Graph: Eulerian Approach
- 273. Repeat1 Repeat1Repeat2 Repeat2 Repeat Graph: Eulerian Approach • The repeat graph can be easily constructed with any number of repeats.
- 274. Repeat1 Repeat1Repeat2 Repeat2 Repeat Graph: Eulerian Approach • The repeat graph can be easily constructed with any number of repeats.
- 275. • Problem: In previous slides, we have constructed the repeat graph while already knowing the genome structure. • How do we construct the repeat graph just from fragments? • Solution: Break the reads into smaller pieces. ? Making Repeat Graph From Reads Only
- 276. Repeat Sequences: Emulating a DNA Chip • A virtual DNA chip allows one to solve the fragment assembly problem using our SBH algorithm.
- 277. Construction of Repeat Graph • Construction of repeat graph from k-mers: emulates an SBH experiment with a huge (virtual) DNA chip. • Breaking reads into k-mers: Transforms sequencing data into virtual DNA chip data.
- 278. • Error correction in reads: “Consensus first” approach to fragment assembly. • Makes reads (almost) error-free BEFORE the assembly even starts. • Uses reads and mate-pairs to simplify the repeat graph (Eulerian Superpath Problem). Construction of Repeat Graph
- 279. • If an error exists in one of the 20-mer reads, the error will be perpetuated among all of the smaller pieces broken from that read. • However, that error will not be present in the other instances of the 20-mer read. • So it is possible to eliminate most point mutation errors before reconstructing the original sequence. Minimizing Errors
- 280. • Graph theory has a wide range of applications throughout bioinformatics, including sequencing, motif finding, protein networks, and many more. Graph Theory in Bioinformatics
- 281. • Simons, Robert W. Advanced Molecular Genetics Course, UCLA (2002). http://www.mimg.ucla.edu/bobs/C159/Presentations/Benzer.pdf • Batzoglou, S. Computational Genomics Course, Stanford University (2004). http://www.stanford.edu/class/cs262/handouts.html References