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Physical Mapping of DNA

BIO/CS 471 – Algorithms for Bioinformatics. Physical Mapping of DNA. Landmarks on the genome. Identify the order and/or location of sequence landmarks on the DNA. Restriction Enzyme Digests. Hybridization Mapping. Clones. BamH1 – GGATCC. y. x. z. w. Probes.

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Physical Mapping of DNA

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  1. BIO/CS 471 – Algorithms for Bioinformatics Physical Mapping of DNA

  2. Landmarks on the genome • Identify the order and/or location of sequence landmarks on the DNA Restriction Enzyme Digests Hybridization Mapping Clones BamH1 – GGATCC y x z w Probes Physical Mapping

  3. Producing a map of the genome m u a e j i x f w z d u z d m Physical Mapping

  4. Restriction Fragment Mapping 3 8 6 10 A: 4 5 11 7 B: 3 1 5 2 6 3 7 A + B: Physical Mapping

  5. Set Partitioning • The Set Partition Problem • Input: X = {x1, x2, x3, … xn} • Output: Partition of X into Y and Z such thatY = Z • This problem is NP Complete • Suppose we have X = {3, 9, 6, 5, 1} • Can we recast the problem as a double digest problem? Physical Mapping

  6. Reduction from Set Partition • X = {3, 9, 6, 5, 1} 1 3 5 6 9 12 12 1 3 5 6 9 9 3 5 6 1 12 12 9 3 5 6 1 Physical Mapping

  7. NP-Completeness • Suppose we could solve the Double Digest Problem in polynomial time… Instance ofSet Partition Instance ofDouble Digest Convert* Solve in polynomial time *polynomial time conversion Physical Mapping

  8. Hybridization Mapping • The sequence of the clones remains unknown • The relative order of the probes is identified • The sequence of the probes is known in advance Clones y x z w Probes Physical Mapping

  9. Interval graph representation a b c d e Becomes a graph coloring problem, which is (you guessed it) NP-Complete b a c e d Physical Mapping

  10. Simplifying Assumptions • Probes are unique • Hybridize only once along the target DNA • There are no errors • Every probe hybridizes at every possible position on every possible clone Physical Mapping

  11. Consecutive Ones Problem (C1P) Rearrange the columns such that all the ones in every row are together: Probes Clones: Physical Mapping

  12. An algorithm for C1P • Separate the rows (clones) into components • Permute the components • Merge the permuted components S1 = {1, 2, 4, 5, 7, 9} S2 = {2, 3, 4, 5, 6, 7, 8, 9} Physical Mapping

  13. Partitioning clones into components • Component graph Gc • Nodes correspond to clones • Connect li and lj iff: Physical Mapping

  14. Component Graph l2 l3 a b l1 l4 g l8 l5 d Here, connected components are labeled with greek letters. l6 l7 Physical Mapping

  15. Assembling a component • Consider only row 1 of the following: • Placing all of the ones together, we can place columns 2, 7, and 8 in any order l2 l1 l3 {2, 7, 8} {2, 7, 8} {2, 7, 8} l1 … 0 1 1 1 0 … Physical Mapping

  16. Row 2 • Because of the way we have constructed the component, l2 will have some columns with 1’s where l1 has 1’s, and some where l2 does not. • Shall we place the new 1’s to the right or left? • Doesn’t matter because the reverse permutation is the same answer. Physical Mapping

  17. Adding row 2 • Placing column 5 to the left partially resolves the {2, 7, 8} columns S1 = {2, 7, 8} S2 = {2, 5, 7} {5} {2, 7} {2, 7} {8} l1 … 0 0 1 1 1 0 … l2 … 0 1 1 1 0 0 … Physical Mapping

  18. Additional Rows • Select a new row k from the component such that edges (i, j) and (i, k) exist for two already added rows i, and k. • Look at the relationship between i and k, and between i and j to determine if k goes on the same side or the opposite side of j as i. k i j Physical Mapping

  19. Definitions • Let • Place k on the same side as i if • Else, place on the opposite side k i j Physical Mapping

  20. Placing Rows Place k on the same side as i if • Or: i l2 j i l1 l3 k So we place l3 on the opposite side of l2 (relative to l1), which is the right. Physical Mapping

  21. Placing rows • Repeat for every row in the component {5} {2} {7} {8} {1,4} {1,4} l1 … 0 0 1 1 1 0 0 0 … l2 … 0 1 1 1 0 0 0 0 … l3 … 0 0 0 1 1 1 1 0 … Physical Mapping

  22. Joining Components • New graph: Gm – the merge graph (directed) • Nodes are connected components of Gc • Edge (, ) iff every set in  is a subset of a set in  l2 a l3   b l1 Physical Mapping

  23. Constructing Gm l2 l3 a b l1 l4 g l8 l5 d l6 l7 a b d g Physical Mapping

  24. All rows in g will share the same disjoint/subset relationship with each row of a Different components  disjoint or subset Same component  shares a 1 column That column matches in a row of a, then subset, else disjoint Properties of Gm a b d g Physical Mapping

  25. Ordering components • Vertices without incoming edges: freeze their columns • Process the rest in topological order. •  is a singleton and a subset a b d g Physical Mapping

  26. Ordering Components (2) • Find the leftmost column with a 1 • In the current assembly, find the rows that contain all ones for that column • Merge the columns Physical Mapping

  27. Ordering Components (3) Physical Mapping

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