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Genome Rearrangement
By
Ghada Badr
Part II
Genome Models

Genomes can be modeled by permutations:
each gene
can be assigned a unique number and is exactly found
once in the genome.
Signed
Permutation: Each gene may be assigned + or sign to indicate the strand it resides on.
Unsigned
Permutation: If the corresponding strand is
unknown.
2
Genome Rearrangement
Our problem:
Given a set of genomes and a set of possible
evolutionary events (operations), find a shortest set
of events transforming those genomes into one
another.
What are the Rearrangement events (Operation)?
3
Rearrangement Operations
Rearrangement operations affect gene order
and gene content. There are various types:
In case of single-chromosome genome:
• Inversions
• Transpositions
• Reverse transpositions
• Gene Duplications
• Gene loss
In case of multiple-chromosomes genomes we add:
• Translocations
• fusions
• fissions
4
Rearrangement Problems
Our problem:
Given a set of genomes and a set of possible
evolutionary events (operations), find a shortest set
of events transforming those genomes into one
another.
Any set of operations yields a distance between
genomes, by counting the minimum number of
operations needed to transform one genome
into the other.
5
Rearrangement Problems
Our problem:
Given a set of genomes and a set of possible
evolutionary events (operations), find a shortest set
of events transforming those genomes into one
another.
Two classical problems
• Computing the distance d()
• Computing one optimal sorting sequence of
events.
6
Rearrangement Operations
Can we have a unifying framework in which circular and
linear chromosomes can coexist throughout evolving
genomes?
Can we have a unifying view of Genome Rearrangements?
(Bergeron 2006)
A Double Cut and Join Operation DCJ was introduced.
7
Rearrangement Operations - DCJ
•
Double Cut-and-Join DCJ was first proposed by
Yancopoulos et. al. (2005).
•
Allows to model all the classical operations (inversions,
translocations, fissions, fusions, transposition, and block
interchanges) with a single operation.
•
This general model accounts for the genomic evidence of
the coexistence of both linear and circular chromosomes
in many genomes.
•
Both the DCJ sorting and distance problems can be
solved in O(n) time by Bergeron et. al. (2006)
8
Rearrangement Operations - DCJ
•
A gene a is an oriented sequence of DNA that starts with
a tail at and ends with a head ah.
•
Two consecutive genes do not necessarily have the
same orientation, thus adjacency of two consecutive
genes a and b, can be of four different types:
{ah,bt},{ah,bh},{at,bt},{at,bh}
 ,  ,  , 
An extremity that is not adjacent to any other gene is
called telomeres by a singleton set {ah} or {at}.
•
•
We can use adjacencies to represent both genomes with
multiple or uni-chromosomes.
9
Rearrangement Operations - DCJ
•
A genome is a set of adjacencies and telomeres
such that the tail or head of any gene appears in
exactly one adjacency or telomere.
Example
Genome A: chr1: a c -d
chr2: b e
chr3: f g
Replace each gene by two extremities
at ah ct ch dh dt
 bt bh et eh
ft fh gt gh
Adjacencies: {ah, ct}{ch, dh} {bh, et} {fh, gt}
Telomere:{at} {dt} {bt} {eh}{ft}{gh}
A = {{at}{ah, bt}{bh, ct}{ch, dt}{dh} {et} {eh,ft} {fh,gt} {gh}}
10
Rearrangement Operations - DCJ
•
DCJ operations:
a) {p,q}{r,s}
{p,r}{s,q} or { p,s} {q,r}
11
Rearrangement Operations - DCJ
•
DCJ operations:
b) {p,q}{r}
{p,r}{q} or{p}{q,r}
12
Rearrangement Operations - DCJ
•
DCJ operations:
c) {q} {r}
{q,r}
13
Rearrangement Operations - DCJ
•
DCJ operations:
Example:
Genome A: chr1: a c -d
chr2: b e
chr3: f g
Adjacencies and telomeres are:
{ah, ct}{ch, dh} {bh, et} {fh, gt} {at} {dt} {bt} {eh}{ft}{gh}
{ah,ct}{fh, gt} -->{ah,fh}{ct,gt}  Genome A: chr1: a -f
chr2: b e
chr3: d -c g
{ah,ct}{fh, gt} -->{ah,gt}{ct,fh} 
Genome A: chr1: a g
chr2: b e
chr3: f c -d
14
DCJ sorting and Distance problems
Problem: Given two genomes A and B defined on the
same set of genes, 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, dcj(A,B).
15
DCJ sorting and Distance problems
Example:
Genome A: chr1: a c -d
chr2: b e
chr3: f g
Replace each gene by two extremities
at ah ct ch dh dt
 bt bh et eh
ft fh gt gh
Genome B: chr 1: a b c d 
chr 2: e f g
at ah bt bh ct ch dt dh
et eh ft fh gt gh
Get adjacencies and telomeres for each genome:
A= {{ah, ct}{ch, dh} {bh, et} {fh, gt} {at} {dt} {bt} {eh}{ft}{gh}}
B = {{at}{ah, bt}{bh, ct}{ch, dt}{dh} {et} {eh,ft} {fh,gt} {gh}}
16
DCJ sorting and Distance problems
Greedy Algorithm to sort by DCJ:
{ah, ct}{ch, dh} {bh, et} {fh, gt} {at} {dt} {bt} {eh}{ft}{gh}
{ah, bt}{ch, dh} {bh, et} {fh, gt} {at} {dt} {ct} {eh}{ft}{gh}
Genome A: chr1: a c -d
chr2: b e
chr3: f g
Genome A: chr1:
chr2:
chr3:
{ah, bt} {ch, dh} {bh, ct} {fh, gt} {at} {dt} {et} {eh}{ft}{gh} Genome A: chr1:
chr2:
chr3:
{ah, bt} {ch, dt} {bh, ct} {fh, gt} {at} {dh} {et} {eh} {ft}{gh} Genome A: chr1:
chr2:
chr3:
{at}{ah, bt}{bh, ct}{ch, dt}{dh} {et} {eh,ft} {fh,gt} {gh}
a b e
c -d
f g
a b c -d
e
f g
a b c d
e
f g
Genome B: chr1: a b c d
chr2: e f g17
DCJ sorting and Distance problems
Optimal and O(n) time.
18
DCJ sorting and Distance problems
Adjacency Graph (bipartite graph):
{ah, ct}{ch, dh} {bh, et} {fh, gt} {at} {dt} {bt} {eh}{ft}{gh}
{at}{ah, bt}{bh, ct}{ch, dt}{dh} {et} {eh,ft} {fh,gt} {gh}
Vertices: adjacencies and telomeres
Edges: connect an edge from A to B between adjacencies
or telomers that have common elements.
Graph can be easily constructed in O(n) time and space
19
DCJ sorting and Distance problems
Adjacency Graph (bipartite graph): IF SORTED
{at}{ah, bt}{bh, ct}{ch, dt}{dh} {et} {eh,ft} {fh,gt} {gh}
{at}{ah, bt}{bh, ct}{ch, dt}{dh} {et} {eh,ft} {fh,gt} {gh}
In each iteration: the algorithm increments C by one or
I by two
When sorted: n = C + I/2
dcj(A,B)  n
20
DCJ sorting and Distance problems
Adjacency Graph (bipartite graph):
{ah, ct}{ch, dh} {bh, et} {fh, gt} {at} {dt} {bt} {eh}{ft}{gh}
{at}{ah, bt}{bh, ct}{ch, dt}{dh} {et} {eh,ft} {fh,gt} {gh}
1 cycle
4 odd paths
1 even path
dcj(A,B) = n - (cycles + oddPath/2)
= 7-1-4/2 = 4
21
Genome Rearrangement and phylogeny

Genome rearrangements events are rare, these changes
of gene orders enable biologists to reconstruct histories far
back in time.

Extend the notion of genome rearrangement distance to
the optimal positioning of Steiner points in the appropriate
space of a given distance metric.

Two phylogenetic versions of the Steiner Problem (the
first inside the other):


Inner problem: optimizing internal nodes of a given tree, where n
leaves are labeled.
Outer problem: optimizing over all trees with n leaves.
22
Genome Rearrangement and phylogeny

We will discuss the inner problem defined as follows:
Given a fixed phylogeny (tree) T, together with a set of K
permutations (genome), each of size n corresponding to the terminal
(leaf) nodes.
Find a set of permutations corresponding to the internal nodes such
that the total weight w(T) is minimized, where w(T) is defined as:
w(T) = ∑ d(x,y) for all (x,y) in T
Here d(.,.) is the genome rearrangement distance metric defined on
pairs of permutations.
23
Genome Rearrangement and phylogeny

Consider a heuristic for the problem of computing the
internal nodes, where T is a star on three vertices.

We will study a more basic problem, the median problem.

Divide the problem on an arbitrary binary tree into a number
of overlapping median problems and apply the median
algorithm iteratively to search for a heuristic solution to the
original problem.

internal nodes retain biological meaning, and edges
represent transitions between states of genome.
24
Median Problem

The median-based method for phylogeny reconstruction
was first proposed by Sankoff and Blanchette (1998).

The idea is to build the global solution by aggregating
local solutions for the simplest problem: Find a Steiner
point M of three genomes.

After an initialization step, the algorithm iterates over a
tree, repeatedly resetting the permutations of internal
nodes to the medians of their three neighbors. Continue
till a convergence occurs.
25
Median Problem

The median of three or simply the median problem:
Find a permutation such that the sum of distances is
minimized between  and each of the starting permutation
 = {}.

Find a permutation M that minimizes the median score
S(), where:
S() = d1, M + d2,M + d3,M
26
Median Problem
Constructing phylogeny from medians
27
Median Problem

The median problem: Find a permutation such that the
sum of distances is minimized between  and each of the
starting permutation  = {}.

What are the distance measures that we can use?

Distances: breakpoint, reversal …

A breakpoint median has no straightforward biological
interpretation and they are not unique.

Breakpoint medians score poorly compared to reversal
medians.
28
Reversal Median

Reversal median Problem: Find a solution to the
median problem using the reversal distance.

Find a permutation such that the sum of reversal
distances is minimized between  and each of the
starting genomes.

The reversal median is NP-hard problem.

Why?
29
Reversal Median
Reversal graph for n = 3
Vertices: all permutations of n = 3.
Edges: connect an edge between 1 and 2 if reversal
distance d(1, 2) = 1.
30
Reversal Median
Reversal graph for n = 3


distance d(i, k) = shortest path between v1 and v2.
Finding the median is equivalent to finding the minimum
Steiner tree for the graph.
31
Reversal Median
Reversal graph for n = 3



The graph is huge |V| = n!.2n
A feasible graph-search algorithm is not possible!
What technique we can use to develop an algorithm for
this kind of problems?
32
Reversal Median

We will study a branch-and-bound algorithm by Adam
Siepel 2001.

This algorithm depends only on the availability of a
rapidly computable distance metric.
33
Reversal Median

The median score S() of a set of equally sized
permutations  = {}, separated by distances d1,2,
d1,3, and d2,3, obeys these bounds:
d1,2 + d1,3+ d2,3
2

S()
 min { (d
1,2+d2,3),(d1,2+d1,3),
(d2,3+d1,3)}
34
Reversal Median
•
Assume that is in the shortest path between  and the
median M, and is separated from  by distances d1,,
d1,, and d2,, the median score S()
d2, + d3,+ d2,3
d1, +
S() d
+min{(d2,+d3,),(d3,+d2,3), (d2,3+d2,)}
1,
2
35
Reversal Median
Algorithm (sketch):

Establish upper and lower bounds using a rapid reversal distance algorithm,
Mmin and Mmax.

Start with one of the three permutations, say .

Assume the median is M = .

Push the corresponding vertex v in a priority stack s for the best scoring
vertices.

While s is not empty




Pop the most promising vertex v from s.
If best score of v  Mmax then stop
Generate all possible vertices  that can be obtained from v by single reversal.
For each possible unmarked 






Calculate bound for the previous equation min, max.
If max= Mmin then M =  and stop. (median is found)
Add  to stack s only if max< Mmax (pruning)
update Mmax= max if max< Mmax .
End for loop.
End while loop.
36
Reversal Median
Algorithm (sketch):

Establish upper and lower bounds using a rapid reversal distance algorithm,
Mmin and Mmax.

Start with one of the three permutations, say .

Assume the median is M = .

Push the corresponding vertex v in a priority stack s for the best scoring
vertices.

While s is not empty




Pop the most promising vertex v from s.
If best score of v  Mmax then stop
Generate all possible vertices  that can be obtained from v by single reversal.
For each possible unmarked 






Calculate bound for the previous equation min, max.
If max= Mmin then M =  and stop. (median is found)
Add  to stack s only if max< Mmax (pruning)
update Mmax= max if max< Mmax .
End for loop.
End while loop.
37
Reversal Median
Algorithm (sketch):

Establish upper and lower bounds using a rapid reversal distance algorithm,
Mmin and Mmax.

Start with one of the three permutations, say .

Assume the median is M = .

Push the corresponding vertex v in a priority stack s for the best scoring
vertices.

While s is not empty




Pop the most promising vertex v from s.
If best score of v  Mmax then stop
Generate all possible vertices  that can be obtained from v by single reversal.
For each possible unmarked 






Calculate bound for the previous equation min, max.
If max= Mmin then M =  and stop. (median is found)
Add  to stack s only if max< Mmax (pruning)
update Mmax= max if max< Mmax .
End for loop.
End while loop.
38
Reversal Median
Algorithm (sketch):

Establish upper and lower bounds using a rapid reversal distance algorithm,
Mmin and Mmax.

Start with one of the three permutations, say .

Assume the median is M = .

Push the corresponding vertex v in a priority stack s for the best scoring
vertices.

While s is not empty




Pop the most promising vertex v from s.
If best score of v  Mmax then stop
Generate all possible vertices  that can be obtained from v by single reversal.
For each possible unmarked 






Calculate bound for the previous equation min, max.
If max= Mmin then M =  and stop. (median is found)
Add  to stack s only if max< Mmax (pruning)
update Mmax= max if max< Mmax .
End for loop.
End while loop.
39
Reversal Median
Algorithm (sketch):

Establish upper and lower bounds using a rapid reversal distance algorithm,
Mmin and Mmax.

Start with one of the three permutations, say .

Assume the median is M = .

Push the corresponding vertex v in a priority stack s for the best scoring
vertices.

While s is not empty




Pop the most promising vertex v from s.
If best score of v  Mmax then stop
Generate all possible vertices  that can be obtained from v by single reversal.
For each possible unmarked 






Calculate bound for the previous equation min, max.
If max= Mmin then M =  and stop. (median is found)
Add  to stack s only if max< Mmax (pruning)
update Mmax= max if max< Mmax .
End for loop.
End while loop.
40
Reversal Median
Algorithm (sketch):

Establish upper and lower bounds using a rapid reversal distance algorithm,
Mmin and Mmax.

Start with one of the three permutations, say .

Assume the median is M = .

Push the corresponding vertex v in a priority stack s for the best scoring
vertices.

While s is not empty




Pop the most promising vertex v from s.
If best score of v  Mmax then stop
Generate all possible vertices  that can be obtained from v by single reversal.
For each possible unmarked 






Calculate bound for the previous equation min, max.
If max= Mmin then M =  and stop. (median is found)
Add  to stack s only if max< Mmax (pruning)
update Mmax= max if max< Mmax .
End for loop.
End while loop.
O(n3d) with d = min{d1,2 + d1,3+ d2,3}
With faster average running time
41
Conclusions

Described Double Cut and Join DCJ operation:
A unifying view of genome rearrangements.

Presented a branch and bound median-based
approach for building phylogeny using reversal
distance.

Many other problems in genome
rearrangement as “Genome halving problem”
42
Genome Halving
a
f c
g
g a
b
d
e
f
b
d e
a
f c
b
g
d
e
c
43
Genome Halving
Duplication
a
f c
g
g a
b
d
e
f
b
d e
a
f c
b
g
d
e
c
44
Genome Halving
b
c
d e
g
a
f
b
c
d e
g a
f
a
f c
g
g a
b
d
e
f
b
d e
a
f c
b
g
d
e
c
45
Genome Halving
b
c
d e
g
a
f
b
c
d e
g a
f
a
f c
g
g a
b
d
e
f
b
d e
a
f c
b
g
d
e
c
46
Genome Halving
b
c
d e
g
a
f
b
c
d e
g
a
f
b
c
d e
g a
f
a
f c
g
g a
b
d
e
f
b
d e
a
f c
b
g
d
e
c
47
Genome Halving
b
c
d e
g
a
f
a
f c
b
g
d
e
48
Genome Halving
b
c
d e
g
a
f
b
c
d e
g
a
f
b
c
d e
g a
f
a
f c
b
g
d
e
49
Genome Halving
b
c
d e
g
a
f
b
c
d e
g
a
f
b
c
d e
g a
f
a
f c
g
g a
b
d
e
f
b
d e
a
f c
b
g
d
e
c
50
Genome Halving
b
c
d e
g
a
f
b
c
d e
g a
f
a
f c
g
g a
b
d
e
f
b
d e
c
51
Genome Halving
b
c
d e
g
a
f
b
c
d e
g a
f
a
f c
g
g a
b
d
e
f
b
d e
c
52
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1.
2.
3.
4.
5.
6.
7.
8.
9.
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Yancopoulos S., Attie O., Friedberg R. Efficient sorting of genomic
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