1. Genome Rearrangements Unoriented Blocks

2. Quick Review Looking at evolutionary change through reversals Find the shortest possible series of reversals that transform gene A into gene B It has been shown that this results in an NP-Hard problem

3. Oriented Blocks 12345521341234552134 1234512543125345213412345125431253452134

4. Unoriented Blocks Orientation of the blocks in the genomes is unknown 21375486123456782137548612345678

5. Definitions unoriented permutation - a mapping from {1,2,…,n} to a set L of n labels. reversal – reverses the order of a segment of consecutive labels.

6. Definitions (cont.) reversal distance – if p 1,p 2,…p t is a shortest series of reversals such that αp 1 p 2 …p t = β, t is the reversal distance of α with respect to β, denoted by d β (α)

7. Example 1 21375486123456782137548612345678 Assign labels 1 through 8 to the blocks in the lower chromosome Transfer the labels to the upper chromosome giving equal labels to homologous blocks We obtain a starting permutation in the upper chromosome and our goal is to sort it into the lower one, the identity Figure below shows two chromosomes with homologous blocks

8. Example 1 (cont.) 21375486123754861234578612345768123456782137548612375486123457861234576812345678

9. Best Solution? How do we know that this is the shortest series of reversals? To decide what the reversal distance should be, we look at the breakpoints

10. Breakpoints A breakpoint of an unoriented permutation α is a pair of labels adjacent in α but not in the target. In the case of the identity, this means adjacent labels that are not consecutive.

11. Example 2 Assume the identity is the target… Breakpoints with oriented blocks: L52134R Breakpoints with unoriented blocks: L52134R

12. Example 2 (cont.) L21375486RL21375486R b(α) denotes the number of breakpoints of α a reversal can remove at most two breakpoints hence: d(α) > ( b(α) / 2 ) where d(α) is the reversal distance using this rule, we see that d(α) > 4 for the above example

13. Strips L45321R If we have two adjacent labels that do not make a breakpoint, they must be of the form: …x(x+1) or …x(x-1)

14. Strips (cont.) strip – a sequence of consecutive labels surrounded by breakpoints but with no internal breakpoints Two types of strips: increasing decreasing

15. Special Rules A single label surrounded by breakpoints is said to be a strip that is both increasing and decreasing L and R are always considered part of an increasing strip, even if they are by themselves L and R are considered a single element for the purpose of defining strips. If 0, 1, … is a strip and …, n, n+1 is a strip, we consider these two sequences as a single strip. They are linked by the common element L = R.

16. Example 3 L12873564RL12873564R Strips increasing: (R,L,1,2) (5,6) decreasing: (8,7) both:(3) (4)

17. Theorem 1 If label k belongs to a decreasing strip and k - 1 belongs to an increasing strip, then there is a reversal that removes at least one breakpoint L4523176R kk-1

18. Proof Labels k – 1 and k must belong to different strips, since only single elements are said to be both increasing and decreasing. The above statement implies that each one is the last element in its strip (each is followed by a breakpoint).

19. Proof (cont.) Two possible schemes: … (k - 1) … k… … k … (k - 1) … Performing a reversal on the area between the breakpoints brings k and k-1 together, reducing the number of breakpoints by at least one.

20. Example 4 L4523176RL4523176RL4567132RL4567132RL4523176RL4523176RL4567132RL4567132R k k-1

21. Observations All permutations have at least one increasing strip (L or R) All permutations do not necessarily have a decreasing strip If there is a decreasing strip, the previous proof shows that there is a breakpoint- removing reversal

22. Theorem 2 If label k belongs to a decreasing strip and k + 1 belongs to an increasing strip, then there is a reversal that removes at least one breakpoint. L5423167R kk+1

23. Proof Two possible schemes: (k + 1) …k … k …(k + 1) … Performing a reversal on the area between the breakpoints brings k and k+1 together, reducing the number of breakpoints by at least one.

24. Example 5 L5423167RL5423167RL1324567RL1324567RL5423167RL5423167RL1324567RL1324567R k+1k+1 k

25. The Result The two proofs just explained show that, as long as we have decreasing strips, we can always reduce the number of breakpoints. Notice that this also applies to single-element strips What about when there are no decreasing strips?

26. Theorem 3 Let α be a permutation with a decreasing strip. If all reversals that remove breakpoints from α leave no decreasing strips, then there is a reversal that removes two breakpoints from α.

27. Proof Let k be the smallest label involved in a decreasing strip. p is the reversal uniting k and k - 1 k – 1 must be to the left of k, otherwise p leaves a decreasing strip. … (k – 1) … k …

28. Proof (cont.) Let ℓ be the largest label involved in a decreasing strip. σ is the reversal uniting ℓ and ℓ + 1 ℓ + 1 must be to the right of ℓ, otherwise σ leaves a decreasing strip … ℓ …(ℓ + 1) …

29. Proof (cont.) Observe that k must be inside the interval reversed by σ, otherwise σ would leave k ’s decreasing strip intact. Likewise, ℓ must belong to the interval of p … (k – 1) ℓ …k(ℓ + 1) …

30. Proof (cont.) … (k – 1)ℓ…k(ℓ + 1) … We can see that p = σ must be true The reversal removes two breakpoints because k is united with k – 1 and ℓ is united with ℓ + 1

31. Example 6 L78354612RL78354612R Reversals that remove breakpoints L78354612RL78354612R L78345612RL78345612R k-1 ℓ ℓ + 1 k

32. Sorting a Permutation We can use an algorithm that sorts a permutation using at most 2 * d(α) reversals (that is, twice as many reversals as the minimum possible) Algorithm assumes that the target is the identity (1,2,3,4….)

33. General Idea A main loop looks at the current permutation and selects the best possible reversal to apply Update the current permutation and report the reversal applied The loop stops when the current permutation is the identity

34. Choosing the Reversals If there is a decreasing strip, look for a reversal that reduces the number of breakpoints and leaves a decreasing strip. If no such reversal exists, there is a reversal that encompasses all the decreasing strips and removes two breakpoints. If there are no decreasing strips, select a reversal that cuts two breakpoints.

35. Sorting Algorithm Algorithm: Sorting Unoriented Permutation input: permutation α output: series of reversals that sort α list  empty while α != I do if α has a decreasing strip then k  smallest label in a decreasing strip p  reversal that cuts after k and after k-1 if αp has no decreasing strip then ℓ  largest label in a decreasing strip p  reversal that cuts before ℓ and before ℓ+1 else p  reversal that cuts the first two breakpoints α  αp list  list+p return list L 1 2. 8 7. 3. 5 6. 4. R list  empty k  3 p  (8 7 3) αp = L 1 2 3. 7 8. 5 6. 4. R α  αp list  (8 7 3) k  4 p  (7 8 5 6 4) αp = L 1 2 3 4. 6 5. 8 7. R α  αp list  (8 7 3), (7 8 5 6 4) k  5 p  (6 5) αp = L 1 2 3 4 5 6. 8 7. R α  αp list  (8 7 3), (7 8 5 6 4), (6 5) k  7 p  (8 7) αp = L 1 2 3 4 5 6 7 8 R α  αp list  (8 7 3), (7 8 5 6 4), (6 5), (8 7)

36. Another Example list  empty k  1 p  (2 1) αp = L 1 2 3. 7. 5 4. 8. 6. R α  αp list  (2 1) k  4 p  (7 5 4) αp = L 1 2 3 4 5. 7 8. 6. R α  αp list  (2 1), (7 5 4) k  6 p  (7 8 6) αp = L 1 2 3 4 5 6. 8 7. R α  αp list  (2 1), (7 5 4), (7 8 6) k  7 p  (8 7) αp = L 1 2 3 4 5 6 7 8 R list  (2 1), (7 5 4), (7 8 6), (8 7) L. 2 1. 3. 7. 5 4. 8. 6. R

37. But is it Optimal? It has been shown: d(α) > ( b(α) / 2 ) For the previous example: b(α) = 7 d(α) >= 4 Although the algorithm produces the optimal result in this instance, it is not guaranteed to do so. The algorithm may produce a list containing more reversals than are actually necessary to solve the problem.

38. Theorem 4 The number of iterations in algorithm Sorting Unoriented Permutation is less than or equal to the number of breakpoints in the initial permutation

39. Proof Must prove that, on average, each iteration removes at least one breakpoint. We can see this is true because the only time we remove 0 breakpoints, is immediately after we have removed 2, keeping the average of 1 breakpoint per iteration intact.