1. Lecture 3: Genome Rearrangements and Duplications

2. Breakpoint graph 1-dimensional construction
Transform p = < 2, -4, -3, 5, -8, -7, -6, 1 > into g = < 1, 2, 3, 4, 5, 6, 7, 8 > by reversals.
Vertices: i ® ia ib i ® ib ia and 0b, 9a
Edges: match the ends of consecutive blocks in p, g
Superimpose matchings

3. Breakpoint graph Breakpoints
Each reversal goes between 2 breakpoints, so d ³ # breakpoints / 2 = 6/2 = 3.
Theorem (Hannenhalli-Pevzner 1995): d = n + 1 – c + h + f where c = # cycles; h,f are rather complicated, but can be computed from graph in polynomial time.
Here, d = – = 4
Breakpoints are not independent. Breakpoint graph shows dependencies between the breakpoints.

4. Breakpoint graph Þ rearrangement scenario

5. Oriented and Unoriented Cycles
Proper reversal acts on black edges: c(ρ π) – c (π) = 1 C F
Unoriented Cycles E
No proper reversal acting on an unoriented cycle
These are “impediments” in sorting by reversals.

6. Interleaving Edges
Interleaving edges are grey edges that cross each other
Example: Edges (0,1) and (18, 19) are interleaving
Cycles are interleaving if they have an interleaving edge
These 2 grey edges interleave

7. Interleaving Graphs
An Interleaving Graph is defined on the set of cycles in the Breakpoint graph and are connected by edges where cycles are interleaved A A B B D C C E E F F B D A E F C

8. Interleaving Graphs
Label oriented cycles. Component oriented if contains oriented cycle. A A B B D C C E E F F B D A E F C

9. Interleaving Graphs
Remove oriented components from interleaving graph. A A B C D E E F C D B A E F

10. Hurdles
Hurdle: Minimal or maximal unoriented component under containment partial order. A A E E A E
h(π) = 1

11. Reversal Distance with Hurdles
Hurdles are obstacles in the genome rearrangement problem
They cause a higher number of required reversals for a permutation to transform into the identity permutation 3 2 1 3 -1 -2 1 -3 -2
c(π) = 2, h(π) = 1 1 2 3
Every hurdle can be transformed into oriented cycles by reversal on arbitrary cycle in hurdle.

12. Reversal Distance with Hurdles
Hurdles are obstacles in the genome rearrangement problem
They cause a higher number of required reversals for a permutation to transform into the identity permutation
Let h(π) be the number of hurdles in permutation π
Taking into account of hurdles, the following formula gives a tighter bound on reversal distance:
d(π) ≥ n+1 – c(π) + h(π)
Every hurdle can be transformed into oriented cycles by reversal on arbitrary cycle in hurdle.
** Doing so, might cause problems with overlapping hurdles

13. Superhurdles “Protect” non-hurdles
Deletion of superhurdles creates another hurdle

14. Superhurdles “Protect” non-hurdles
Deletion of superhurdles creates another hurdle
Superhurdle
Superhurdle

15. Superhurdles “Protect” non-hurdles
Deletion of superhurdles creates another hurdle
Hurdle
Hurdle

16. Fortresses
A permutation π with an odd number of hurdles, all of which are superhurdles
Theorem (Hannenhalli-Pevzner 1995): d(π) = n + 1 – c(π) + h(π) + f
where c = # cycles; h = # hurdles
f = 1 if π is fortress.

17. GRIMM-Synteny on X chromosome 2-dimensional breakpoint graph

18. GRIMM-Synteny on X chromosome 2-dimensional breakpoint graph

19. Coming Next Other rearrangement operations
Duplications
Rearrangements and Phylogeny
Multiple Genomic Distance Problem: Given permutations 1, …, k find a permutation  such that
k=1, k d(1, ) is minimal.

20. Other Types of Rearrangements
So far:
Discussed reversals.
Also: translocations, fissions, fusions (modeled as reversals in concatenate of chromosomes
( ) (–6 – –2)
( –2) (–6 –1 4 10)

21. Other Types of Rearrangements
Transpositions
Duplication Transposition
Duplications are very frequent in cancer genomes.

22. Duplications HARD!!! (NP-hard?) What problem to solve?
Given G  {1, .., n}N
(“permutation with duplicates”)
Find reversals 1, 2, …, t, duplications 1, …, s, and permutation  such that
 (1, …, t, 1, …, s) i = G and s + t is minimal
???
HARD!!!
(NP-hard?)

23. Duplications (2) What problem to solve?
Given: G  {1, .., n}N (“permutation with duplicates”) ,
H =  G for some permutation 
Find: Reversals 1, 2, …, t such that 1 …t G = H and t is minimal
Signed reversal distance with duplicates
NP-hard (Chen, et al. 2005)
If 1-1 mapping of repeated elements (orthologs) in G to H then problem reduces to reversal distance.

24. Duplications (3) Solution when at most two duplicates per gene
What problem to solve?
Given: P {1, .., n}N
(permutation with duplicates)
Find: Permutation  and reversals 1, 2, …, s, duplications 1, … t such that
1, …, s1, …, t  = P
and t minimal.
Solution when at most two duplicates per gene
El-Mabrouk and Sankoff (2002)

25. Whole Genome Duplication
Genome is doubled – extra copy of each element.
Subsequently undergoes reversals.
Genome Halving Problem. Given a duplicated genome P, recover the ancestral pre-duplicated genome R minimizing the reversal distance from the perfect duplicated genome R © R to the duplicated genome P. (El-Mabrouk and Sankoff )

26. Whole Genome Duplication
Genome is doubled – extra copy of each element.
Subsequently undergoes reversals.
If copies of each element labeled uniquely, then problem reduces to reversal distance problem.

27. Reversal Distance and Duplications
Let d(G,H) = reversal distance b/w G and H
Problem of computing d(P, R  R) is unsolved
minR d(P, R  R) solvable in polynomial time

28. Breakpoint Graph p g G( p,g ) 0h 2t 2h 4h 4t 3h 3t 5t 5h 8h 8t 7h 7t2 -4 -3 5 -8 -7 -6 1 9 0h 2t 2h 4h 4t 3h 3t 5t 5h 8h 8t 7h 7t 6h 6t 1t 1h 9t g 1 2 3 4 5 6 7 8 9 0h 1t 1h 2t 2h 3t 3h 4t 4h 5t 5h 6t 6h 7t 7h 8t 8h 9t G(
p,g ) 2 -4 -3 5 -8 -7 -6 1 9 0b 2a 2b 4b 4a 3b 3a 5a 5b 8b 8a 7b 7a 6b 6a 1a 1b 9a

29. Genome Halving: Exhaustive
Doubled genome with 2n genes
Compute reversal distance on all 2n labeling of genes.

30. Genome Halving
Weak Genome Halving Problem. For a given duplicated genome P, find a perfect duplicated genome R © R and a labeling of gene copies that maximizes the number of black-gray cycles c(G) in the breakpoint graph G(P,R © R) of the labeled genomes P and R  R.
(Alekseyev and Pevzner 2006)
Theorem (Hannenhalli-Pevzner 1995): d(π) = n + 1 – c(π) + h(π) + f
where c = # cycles; h = # hurdles
f = 1 if π is fortress.

31. Contracted Breakpoint Graph
Breakpoint graph construction p 2 -4 -3 5 -8 -7 -6 1 9 0h 2t 2h 4h 4t 3h 3t 5t 5h 8h 8t 7h 7t 6h 6t 1t 1h 9t g 1 2 3 4 5 6 7 8 9 0h 1t 1h 2t 2h 3t 3h 4t 4h 5t 5h 6t 6h 7t 7h 8t 8h 9t G(
p,g ) 2 -4 -3 5 -8 -7 -6 1 9 0h 2t 2h 4h 4t 3h 3t 5h 5t 8h 8t 7h 7t 6h 6t 1t 1h 9t
Implicit were obverse edges (xt, xh)
 is black-obverse alternating path
 is gray-observe alternating path

32. Contracted Breakpoint Graph
With duplicates, pair of vertices with same label.
Contract these identical vertices

33. Contracted Breakpoint Graph
P = −a−b+g+d+f+g+e−a+c−f−c−b−d−e
R = −a−b−d−g+f−c−e
G’(P,R © R)
Each gray edge is pair of parallel edges

34. Cycle Decompositions Genomes P and Q
G(P,Q) breakpoint graph for some labeling
Black-gray cycle decomposition
???
G’(P,Q) contracted breakpoint graph
Induced black-gray cycle decomposition
Labeling Problem. Given a black-gray cycle decomposition of the contracted breakpoint graph G′(P,Q) of duplicated genomes P and Q, find labeling of P and Q that induces this cycle decomposition.

35. Maximal black-gray cycle decomposition
P = −a−b+g+d+f+g+e−a+c−f−c−b−d−e
R = −a−b−d−g+f−c−e
G’(P,R © R)
BG graph
corresponding to G’
Maximal black-gray cycle decomposition

36. P as black-observe cycle
Cycle Decomposition
P = −a−b+g+d+f+g+e−a+c−f−c−b−d−e
R = −a−b−d−g+f−c−e
P as black-observe cycle

37. Genome Halving Algorithm: Outline
Input: Doubled genome P
Construct BO (black-obverse) graph for P by gluing identical edges
Introduce gray edges “optimally” to create BOG (black-observe-gray) graph G’ with single gray-observe cycle (!!!)
R = gray-observe cycle in G’
Find maximal black-gray cycle decomposition of G’
Q = R  R