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

Plan Introduction Graphs Genome Adjacency Graph Sorting with DCJ Conclusion

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 1 2 3 4 5 6 …

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

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

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:

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

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

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

Graphs: DCJ (Path/Cycle) inversion integration excisions

Graphs: DCJ (Path/Cycle) inversion linearization circularization

Graphs: DCJ (Cycles) inversion integration excisions

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

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}…

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

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).

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

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}}

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.

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}}

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).

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

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

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

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.

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}}

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

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

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

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).

Conclusion Transposition dDCJ(A,B) Efficient algorithm