1. Genome Rearrangements …and YOU!! Presented by: Kevin Gaittens

2. Overview Bio background Definitions and Set-up Reality-Desire Good Components Bad Components Fin

3. Biological Bakground Comparing entire genomes across species Need “distance” measure Interested in larger differences than just single insertions/deletions etc. Genome Rearrangements – chromosome piece (gene) being moved or copied to another location or transferring to another chromosome altogether

4. Definitions Block – section of genome possibly containing more than one gene; one unit Homologous – when two blocks contain the same genes. Homologous blocks have the same number label Reversal – reversing a series of blocks and also their orientations; distance is measured in number of reversals

5. Example of Reversal 3 4 1 2 5 3 2 1 4 5 Red – right orientation Black – left orientation

6. Goals Want shortest number of reversals to transform one genome to another –Parsimony assumption – assume Nature changes optimally Desire polynomial time solution Oriented has a poly-time solution, unoriented NP-hard

7. Example 1 2 3 4 5 5 2 1 3 4 Add circle if orientation changes

8. One solution 1 2 3 4 5 1 2 5 4 3 1 2 5 3 45 2 1 3 4

9. Breakpoints Act as a minimum Happens in the case of: –first/last label in original not the first/last label in the target –OR 2 labels are consecutive in original, but not in target –OR consecutive in original and target but duel orientation is different between blocks …5 4… and …5 4… –NOTE: If a pair of labels is an exact reversal in the target, there is NO breakpoint …4 5… and …5 4… do not have a breakpoint

10. Breakpoints for Last Example 1 2 3 4 5 Goal reminder: 5 2 1 3 4 1 is different than first of target No breakpoint between 1 and 2 since exact reversal in target 2 and 3 not consecutive in target 3 and 4 match, thus no breakpoint 5 is different from last in target 4 and 5 are not consecutive in target

11. Mathy Stuff :o) Let L be finite set of labels L 0 = U { a, a } for all a in L | x | -> remove arrows Ex: | a | = | a | = a

12. Cont’d Oriented permutation over L is a mapping α: [1..n] -> L 0 such that for any a ε L, there is exactly one i ε [1..n] with |α(i)| =a Basically, permutation “picks” an orientation for each label. If a is picked, then a will not be

13. Example n = 4 L = {1, 2, 3, 4} α = ( 2, 1, 4, 3 ) So α(3) = 4

14. Identity Permutation Special case Permutation I such that I(i) = i for all i between 1 and n For n = 3, I = ( 1 2 3)

15. Reversals Let i and j be two indices with 1 ≤ i, j ≤ n [i,j] indicates a reversal affecting elements α(i) through α(j)

16. Example Given α = ( 2, 3, 4, 1) α[2,3] = ( 2, 4, 3, 1) Note: similar to boxing scheme used earlier

17. More Math! In general: Α[i, j](k) = α(i + j – k) if i ≤ k ≤ j α(k) otherwise α(k) means reversal of orientation of α(k)

18. Sorting by Reversals Is the main goal Given 2 permutations α and β, seek minimum number of reversals to transform α into β Αp 1 p 2 p 3 …p t = β where p 1, p 2,…, p t are reversals t is called the reversal distance of α with respect to β and denoted by d β (α)

19. Sorting con’t Look for reversals that “make progress” towards β d β (αp) < d β (α) or d β (αp) = d β (α) - 1

20. Breakpoints Add labels L and R to α to get “extended version” One example of a α is: (L, 2, 3, 1, 6, 5, 4, R) If B is identity, then breakpoints at…

21. Breakpoints none at 5 4, reverse pair 4 5 is in β L 2 3 1 6 5 4 R L 1 2 3 4 5 6 R 2 is not the first block of β 2 and 3 are consecutive, but the orientations are different than what they need and are not a complete reversal 3 and 1 are not consecutive in β 1 and 6 are not consecutive in β 6 and 5 are consecutive, but not a complete reversal (orientation of 6 prevents it) 4 is not the final block in β

22. Breakpoints con’t Can remove at most 2 breakpoints with each reversal Thus, b(α) – b(αp) ≤ 2 This also means that b(α)/2 ≤ d(α) This is a lower bound for d(α)

23. Bps cont’d b(α)/2 is lower bound However, this is rarely achievable Want a better lower bound Look to something called reality-desire diagram

24. Reality-Desire Happens when 2 labels are adjacent, but do not “want” to be adjacent Reality – neighbor a certain label has in α Desire – neighbor the label has in β

25. Diagram Oriented labels can be viewed as a battery Positive terminal at tip of arrow Negative at tail - a +

26. Example ααpααp Desire Reality

27. Example Extended α: L 3 2 1 4 5 R Replace labels by terminals & reality edges: L -3 +3 +2 -2 +1 -1 -4 +4 +5 -5 R Add desire edges

28. Diagram To create diagram of reality-desire: –Arrange all terminal nodes around a circle with L and R at the top –L to the left of R and all other nodes following α counterclockwise –Reality edges will be along circumference –Desire edges will be the chords

29. Diagram of Reality-Desire Happens where not breakpoint

30. Interpretation Number of cycles in RD(α) is c β (α) and is number of connected parts c β (β) has no breakpoints Notice c β (β)=n+1 –Why?

31. Effects of a Reversal Let (s,t) and (u,v) be two reality edges characterizing a reversal p with (s,t) preceding in the permutation α. Then RD(αp) differs from RD(α) by: 1. Reality edges (s,t) and (u,v) are replaced by (s,u) and (t,v) 2. Desire edges remain unchanged 3. The section of the circle going from node t to node u, including these extremities, in counterclockwise direction, is reversed.

32. Our Example Reversing (-1,-4) and (+4, +5)

33. Definitions Let e and f be two reality edges belonging to the same cycle in RD(α) If orientations induced by e and f coincide, they are convergent –Walk counterclockwise from start of e (passing through desire edges) until you reach the beginning of f. If the end of f is still counterclockwise, then converge Divergent otherwise

34. Walking Convergent Still counterclockwise (+3,+2) to (-1,-4)

35. How Reversals Affect Cycles If e and f belong to different cycles, c(αp)=c(α) -1

36. If e and f belong to the same cycles and converge c(αp)=c(α)

37. If e and f belong to the same cycles and diverge c(αp)=c(α) +1

38. Summary If e and f: belong to different cycles, c(αp)=c(α) -1 belong to same cycle & converge, c(αp)=c(α) belong to same cycle & diverge, c(αp)=c(α)+1

39. Lower Bound Since number of cycles changes by at most 1 per reversal, can get a new lower bound for reversals Suppose αp 1 p 2..p t =β --c β (αp 1 p 2...p t )=c β (β)=n+1 c β (αp 1 ) – c β (α) ≤ 1 c β (αp 1 p 2 ) – c β (αp 1 ) ≤ 1 … c β (αp 1...p t ) – c β (αp 1...p t-1 ) ≤ 1

40. Lower Bound Add to get n+1 – c β (α) ≤ t If p 1,p 2,...,p t is an optimal sorting, then t=d β (α) n+1 – c β (α) ≤ d β (α) Very good lower bound

41. Good/Bad Cycles A cycle is “good” if it has two divergent reality edges If not, it is considered “bad” Good cycles have at least two desire edges that cross –Not all cycles that have crossing edges are good Call cycles “proper” if they have at least four edges

42. Good/Bad cont’d If we only have good cycles, lower bound d(α) ≥ n+1 – c(α) is an equality How could it be possible for it to be an equality if there are a few bad cycles mixed in to start?

43. Interleave Twisting another cycle while breaking another is only possible if the two cycles are such that some desire edge from one of the cycles crosses some desire edge from the other These two cycles “interleave” in this case

44. Interleave

45. Interleaving Graph Important to verify which cycles interleave with which other cycles Take as nodes the proper cycles of RD(α) Two nodes adjacent iff the cycles interleave Connected components are classified as good or bad If a component contains all bad cycles, it is bad. Otherwise, it is said to be good

46. RD to Interleave Gray filled-in circles are good cycles

47. Choosing a Reversal

48. C is the only good cycle Let e = (L, +3), f=(-3,-4), g=(-1,+2) f & g converge, so not a good choice

49. e and g e and g diverge and produce 2 good components with 1 cycle each

50. e and f e and f produce a single good component with two cycles

51. Reversal Choosing cont’d A reversal characterized by two divergent edges of the same cycle is a sorting reversal iff its application does not lead to the creation of bad components So reversing e & f or e & g are both acceptable

52. Bad Components Good components can be sorted as in previous slide First step in dealing with bad components is to classify them Component Y “separates” components X and Z if all chords in RD(α) that link a terminal in X to one in Z cross a desire edge of Y

53. E separates F and D What are some other separations?

54. Definitions Hurdle – bad component that does not separate two bad components Nonhurdle – bad component that separates two bad components

55. Definitions cont’d X protects nonhurdle Y if removal of X would cause Y to become a hurdle –If anytime Y separates 2 bad components, X is one of them Superhurdle – hurdle that protects a nonhurdle Simple hurdle – does not protect a nonhurdle F protects E

56. Classification

57. Formula for Reversal Distance d(α) = n + 1 – c(α) + h(α) + f(α) h(α) = number of hurdles f(α) = 0 or 1 1 if α is a fortress A nonhurdle will become a hurdle at some point

58. Fortress A fortress is a permutation where there are an odd number of hurdles and all of them are super hurdles. They require an extra reversal since a nonhurdle will become a hurdle at some point

59. Definitions X and Y are “opposite” hurdles when we find the same number of hurdles when walking around the circle counterclockwise from X to Y as we do clockwise. Note: only when even number hurdles

60. Hurdle Cutting Reverse edges in same component Used only with simple hurdles

61. Final Algorithm While α not B: If there is a good component in RD(α) then pick two divergent edges in this component ensuring that it does not create a bad component Else if h(α) is even then return merging of two opposite hurdles else if there is a simple hurdle return a reversal cutting this hurdle else //fortress return merging of any two hurdles

62. Fortress Handling Fortress, so choose any 2 hurdles and merge C is good C A B

63. Complexity Construction RD(α) takes linear Finding the cycles is O(n) For each cycle, determine good/bad –This is O(n) per cycle, so O(n 2 ) total Determining interleaving can be done in O(n 2 ) Counting hurdles etc. can be done linearly with the other knowledge

64. Complexity cont’d Figuring out a Sorting Reversal for good components is the worst since need ensure we don’t create bad components Since reversal is identified with a pair of edges, O(n 2 ) reversals. For each one, O(n 2 ) time checking the resulting permutation. O(n 4 ) total We need to do this d β (α) times so O(n 5 ) all together

65. Final Slide, Huzzah! Found accurate distance measure for genome movements Found a poly-time solution for solving the problem Played with fun graphs