1. A Unifying View of Genome Rearrangement
Anne Bergeron, Julia Mixtacki and Jens Stoye
Presentation by Colleen Maquiling January 2017

2. Plan Introduction Graphs Genome Adjacency Graph Sorting with DCJ
Conclusion

3. A B Distance 1 2 3 4 5 6 … Introduction
“Given two genomes A and B, what is the shortest sequence of rearrangement operations that transforms A into B?”
Distance A B …

4. Double cut and join Introduction
Inversions Translocations Fissions Fusions
Transpositions
Double cut and join

5. Graphs
cycles
{p}
p q
{p,q}
paths

6. Graphs: Double Cut and Join
The double cut and join (DCJ) operation acts on two vertices u and v of a graph with vertices of degree one or two in the following ways:

7. Graphs: DCJ (Paths)
Translocation
u={p,q}
v={r,s}

8. Graphs: DCJ (Paths)
Translocation
u={p,q}
v={r}

9. Graphs: DCJ (Paths)
Path fusion/fission
u={q}
w={q,r}
v={r}

10. Graphs: DCJ (Path/Cycle)
inversion
integration
excisions

11. Graphs: DCJ (Path/Cycle)
inversion
linearization
circularization

12. Graphs: DCJ (Cycles)
inversion
integration
excisions

13. Graphs: Lemma 1
The application of a single DCJ operation changes the number of circular or linear components by at most one.

14. Genomes Adjacencies: {ah, bt}, {ah, bh}, {at, bt}, {at, bh}
A gene is an oriented sequence of DNA that starts with a tail and ends with a head. b a
{ah}
{at}
{bh}
{bt}
Adjacencies:
{ah, bt}, {ah, bh}, {at, bt}, {at, bh}
Telomeres: {ah}, {at}…

15. Genomes: Graph Representation
A = {{at}, {ah, ct}, {ch, dh}, {dt}, {bh, et}, {eh, bt}, {ft}, {fh, gt}, {gh}} 7 genes: a, b, c, d, e, f, g

16. Genomes: Sorting & Distance Problem
Given two genomes A and B who have the same genes (not in the same order), find a shortest sequence of DCJ operations that transforms A into B. The length of such a sequence is called the DCJ distance between A and B, denoted by dDCJ (A, B).

17. Genomes: Sorting Example
A = {{at}, {ah, ct}, {ch, dh}, {dt}, {bh, et}, {eh, bt}, {ft}, {fh, gt}, {gh}}
B = {{ah, bt}, {bh, at}, {ct}, {ch, dt}, {dh}, {et}, {eh}, {fh, gt}, {gh, ft}}
ah bt
bh at ct
ch dt dh et eh
fh gt
gh ft

18. Genomes: Sorting Example
A = {{at}, {ah, ct}, {ch, dh}, {dt}, {bh, et}, {eh, bt}, {ft}, {fh, gt}, {gh}}integration {{at}, {ah, bt}, {ch, dh}, {dt}, {bh, et}, {eh, ct}, {ft}, {fh, gt}, {gh}} excision {{et}, {ah, bt}, {ch, dh}, {dt}, {bh, at}, {eh, ct}, {ft}, {fh, gt}, {gh}} inversion {{et}, {ah, bt}, {ch, dt}, {dh}, {bh, at}, {eh, ct}, {ft}, {fh, gt}, {gh}} path fission {{et}, {ah, bt}, {ch, dt}, {dh}, {bh, at}, {eh}, {ct}, {ft}, {fh, gt}, {gh}} circulization
B = {{ah, bt}, {bh, at}, {ct}, {ch, dt}, {dh}, {et}, {eh}, {fh, gt}, {gh, ft}}

19. Adjacency Graph
The adjacency graph AG(A,B) is a graph whose set of vertices are the adjacencies and telomeres of A and B.
For each u ∈ A and v ∈ B there are |u ∩ v| edges between u and v.

20. Adjacency Graph: Example
A = {{at}, {ah, ct}, {ch, dh}, {dt}, {bh, et}, {eh, bt}, {ft}, {fh, gt}, {gh}}
B = {{ah, bt}, {bh, at}, {ct}, {ch, dt}, {dh}, {et}, {eh}, {fh, gt}, {gh, ft}}

21. Sorting by DCJ : Lemma 2
Let A and E be two genomes defined on the same set of N genes, then we have A=E if and only if N=C+I/2 where C is the number of cycles and I the number of odd paths in AG(A,E).

22. N=C+I/2 Sorting by DCJ : Lemma 2
A = {{at}, {ah, ct}, {ch, dh}, {dt}, {bh, et}, {eh, bt}, {ft}, {fh, gt}, {gh}} at
ah ct
ch dh dt
bh et
eh bt ft
fh gt gh at
ah ct
ch dh dt
bh et
eh bt ft
fh gt gh
E = {{at}, {ah, ct}, {ch, dh}, {dt}, {bh, et}, {eh, bt}, {ft}, {fh, gt}, {gh}}
N=C+I/2

23. Sorting by DCJ : Lemma 3
The application of a single DCJ operation changes the number of odd paths in the adjacency graph by –2, 0, or 2.
odd
odd
odd
odd -2
even
even -2
even

24. Sorting by DCJ : Lemma 3 even even even even +2 odd odd even odd even
even
even
even
even +2
odd
odd
even
odd
even
odd
odd
even
odd

25. Sorting by DCJ : Lemma 4
Let A and B be two genomes defined on the same set of N genes, then we have dDCJ(A,B) ≥ N−(C+I/2) where C is the number of cycles and I the number of odd paths in AG(A,B).
Lemma 1: The application of a single DCJ operation changes the number of circular or linear components by at most one.
Lemma 2: Let A and B be two genomes defined on the same set of N genes, then we have A=B if and only if N=C+I/2 where C is the number of cycles and I the number of odd paths in AG(A,B).
Lemma 3: The application of a single DCJ operation changes the number of odd paths in the adjacency graph by –2, 0, or 2.

26. Sorting by DCJ : Algorithm
A = {{at}, {ah, ct}, {ch, dh}, {dt}, {bh, et}, {eh, bt}, {ft}, {fh, gt}, {gh}}
2 odd paths
1 cycle
B = {{ah, bt}, {bh, at}, {ct}, {ch, dt}, {dh}, {et}, {eh}, {fh, gt}, {gh, ft}}
ah bt
bh at ct
ch dt dh et eh
fh gt
gh ft
B = {{ah, bt}, {bh, at}, {ct}, {ch, dt}, {dh}, {et}, {eh}, {fh, gt}, {gh, ft}}

27. Sorting by DCJ : Algorithm (Adjacencies)
u=ah, ct
v=eh, bt
u=ah bt
x=ah bt A
v=ct, eh
x=ah, bt ct eh B ct eh

28. Sorting by DCJ : Algorithm (Telomeres)
ch dt dh
fh gt
gh ft
ah bt
bh at et ct eh
v=ct, eh A B
p=ct eh
v=ct, eh
v=ct
p=ct eh A B

29. Sorting by DCJ : Algorithm Result
A = {{ah, bt}, {bh, at}, {ct}, {ch, dt}, {dh}, {et}, {eh}, {fh, gt}, {gh, ft}}
ah bt
bh at ct
ch dt dh et eh
fh gt
gh ft
B = {{ah, bt}, {bh, at}, {ct}, {ch, dt}, {dh}, {et}, {eh}, {fh, gt}, {gh, ft}}
dDCJ(A,B)=N−(C+I/2)=7-(1+2/2)=5

30. Sorting by DCJ : Theorem 1
Let A and B be two genomes defined on the same set of N genes, then we have dDCJ(A,B) = N−(C+I/2) where C is the number of cycles and I the number of odd paths in AG(A,B).

31. Conclusion
Transposition
dDCJ(A,B)
Efficient algorithm