Genome Rearrangements

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

• 1 2 3 4 5
• 5 2 1 3 4

1 2 3 4 5 1 2 5 4 3 1 2 5 3 4 5 2 1 3 4
4
Unoriented Blocks

• Orientation of the blocks in the genomes is
  unknown

2 1 3 7 5 4 8 6 1 2 3 4 5 6 7 8
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 p1,p2,pt is a shortest
  series of reversals such that ap1p2pt ß
  , t is the reversal distance of a with
  respect to ß, denoted by dß(a)

7
Example 1

Figure below shows two chromosomes with
homologous blocks
2 1 3 7 5 4 8 6 1 2 3 4 5 6 7 8

• 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

8
Example 1 (cont.)

• 2 1 3 7 5 4 8 6
• 1 2 3 7 5 4 8 6
• 1 2 3 4 5 7 8 6
• 1 2 3 4 5 7 6 8
• 1 2 3 4 5 6 7 8

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 a is a
  pair of labels adjacent in a 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
• L 5 2 1 3 4 R
• Breakpoints with unoriented blocks
• L 5 2 1 3 4 R

12
Example 2 (cont.)

• L 2 1 3 7 5 4 8 6 R

• b(a) denotes the number of breakpoints of a
• a reversal can remove at most two breakpoints
  hence d(a) gt ( b(a) / 2 ) where d(a) is
  the reversal distance
• using this rule, we see that d(a) gt 4 for the
  above example

13
Strips

• L 4 5 3 2 1 R
• If we have two adjacent labels that do not make a
  breakpoint, they must be of the form
• x(x1)
• 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, n1 is a strip, we consider these
  two sequences as a single strip. They are linked
  by the common element L R.

16
Example 3

• L 1 2 8 7 3 5 6 4 R
• 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
• L 4 5 2 3 1 7 6 R

k
k-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

• L 4 5 2 3 1 7 6 R
• L 4 5 2 3 1 7 6 R
• L 4 5 6 7 1 3 2 R
• L 4 5 6 7 1 3 2 R

k-1
k
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.
• L 5 4 2 3 1 6 7 R

k
k1
23
Proof

• Two possible schemes
• (k 1) k
• k (k 1)
• Performing a reversal on the area between the
  breakpoints brings k and k1 together, reducing
  the number of breakpoints by at least one.

24
Example 5

• L 5 4 2 3 1 6 7 R
• L 5 4 2 3 1 6 7 R
• L 1 3 2 4 5 6 7 R
• L 1 3 2 4 5 6 7 R

k
k1
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 a be a permutation with a decreasing strip.
  If all reversals that remove breakpoints from a
  leave no decreasing strips, then there is a
  reversal that removes two breakpoints from a.

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.
• s is the reversal uniting l and l 1
• l 1 must be to the right of l, otherwise s
  leaves a decreasing strip
• l (l 1)

29
Proof (cont.)

• Observe that k must be inside the interval
  reversed by s, otherwise s 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 s 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

• L 7 8 3 5 4 6 1 2 R
• Reversals that remove breakpoints
• L 7 8 3 5 4 6 1 2 R
• L 7 8 3 4 5 6 1 2 R

k-1
l 1
l
k
32
Sorting a Permutation

• We can use an algorithm that sorts a permutation
  using at most 2 d(a) 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 Reversal s

• 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
L 1 2 . 8 7 . 3 . 5 6 . 4 .
R list ? empty k ? 3 p ? (8 7 3) ap L 1 2
3 . 7 8 . 5 6 . 4 . R a ? ap list ? (8 7
3) k ? 4 p ? (7 8 5 6 4) ap L 1 2 3 4
. 6 5 . 8 7 . R a ? ap list ? (8 7 3), (7 8
5 6 4) k ? 5 p ? (6 5) ap L 1 2 3 4
5 6 . 8 7 . R a ? ap list ? (8 7 3), (7 8 5 6
4), (6 5) k ? 7 p ? (8 7) ap L 1 2 3 4
5 6 7 8 R a ? ap list ? (8 7 3), (7 8 5
6 4), (6 5), (8 7)

• Algorithm Sorting Unoriented Permutation
• input permutation a
• output series of reversals that sort a
• list ? empty
• while a ! I do
• if a has a decreasing strip then
• k ? smallest label in a decreasing strip
• p ? reversal that cuts after k and after
  k-1
• if ap has no decreasing strip then
• l ? largest label in a decreasing
  strip
• p ? reversal that cuts before l and
  before l1
• else
• p ? reversal that cuts the first two
  breakpoints
• a ? ap
• list ? listp
• return list

36
Another Example
L . 2 1 . 3 . 7 . 5 4 . 8 . 6 . R

• list ? empty
• k ? 1
• p ? (2 1)
• ap L 1 2 3 . 7 . 5 4 . 8 . 6 . R
• a ? ap
• list ? (2 1)
• k ? 4
• p ? (7 5 4)
• ap L 1 2 3 4 5 . 7 8 . 6 . R
• a ? ap
• list ? (2 1), (7 5 4)
• k ? 6
• p ? (7 8 6)
• ap L 1 2 3 4 5 6 . 8 7 . R
• a ? ap
• list ? (2 1) , (7 5 4) , (7 8 6)

k ? 7 p ? (8 7) ap L 1 2 3 4 5 6
7 8 R list ? (2 1), (7 5 4), (7 8 6), (8 7)
37
But is it Optimal?

• It has been shown
• d(a) gt ( b(a) / 2 )
• For the previous example
• b(a) 7
• d(a) gt 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.