Genome Rearrangements

5. Lecture WS 2003/04

Bioinformatics III 1

Genome Rearrangements

Compare to other areas in bioinformatics we still know very little about

the rearrangement events that produced the existing varieties of

genomic architectures ...

Some material of this lecture borrowed from:Nipun Mehra, www.stanford.edu/class/cs374/Notes/lec17.pptwww.sna.csie.ndhu.edu.tw/~lung/seminar/20020502.ppt

Bafna V., and P.A. Pevzner. "Sorting by reversals: genome rearrangements in plant organelles and evolutionary history of X Chromosome."

Hannenhalli S., and P.A. Pevzner. "Transforming cabbage into turnip: polynomial algorithm for sorting signed permutations by reversals.“

“Computational Molecular Biology” book by P.A. Pevzner, MIT press, chapter 10

http://www.stanford.edu/class/cs374/Notes/lec17.ppt



Processes of Evolution

- Substitution

- Insertion

- Deletion

- Translocation

- Inversion/ Reversal

- Duplication



What is a reversal = inversion ?

Break and Invert

A T G C C T G T A C T A

T A C G G A C A T G A T

A T G T A C A G G C T A

T A C A T G T C C G A T

• Purines (A, G) and Pyrimidines (C, T) switch strands

•Many organisms have highly similar genes but very different

gene orders.•Very prominent in Prokaryotes, Mitochondrial DNA and

Mamallian X-chromosome.



Types of Genome Rearrangements

Two genomes may have many genes in common, but the genes may be

arranged in a different sequence or be moved between chromosomes. Such

differences in gene orders are the results of rearrangement events that are

common in molecular evolution.

For example, in unichromosomal genomes, the most common rearrangement

events are reversals, in which a contiguous interval of genes is put into the

reverse order.

For multichromosomal genomes, the most common rearrangement events are

reversals, translocations, fissions, and fusions.

The pairwise genome rearrangement problem is to find an optimal scenario

transforming one genome to another via these rearrangement events.



Representation of a genome

We consider a unichromosomal genome to bef a sequence of n genes. The

genes are represented by numbers 1, 2, ..., n.

The two orientations of gene i are represented by i and -i.

A genome is represented as a signed permutation of the numbers 1, 2, ..., n.

For example, a unichromosomal genome with n = 5 genes is 5 -3 4 2 -1



Multichromosal Genome

A multichromosomal genome consists of n genes spread over m

chromosomes. We represent it as a signed permutation of 1, 2, ..., n, with

delimiters "$" or ";" inserted between the chromosomes. For example, a genome

with 12 genes spread over 3 chromosomes is

7 -2 8 3 $ 5 9 -6 -1 12 $11 4 10 $ The order of the chromosomes and the direction of the chromosomes do not

matter in the multichromosomal algorithms. Thus, we could represent this same

genome by flipping the first chromosome (reverse the order of its entries and

negate them) and then moving the last chromosome to the beginning:

11 4 10 $ -3 -8 2 -7 $ 5 9 -6 -1 12 $



Unichromosomal genomes: sorting by reversal

A reversal in a signed permutation is an operation that takes an interval in a

permutation, reverses the order of the numbers, and changes all their signs. For

example,

5 1 3 2 -9 7 -4 6 8

5 1 -7 9 -2 -3 -4 6 8

The reversal distance between two genomes is the minimum number of

reversals it takes to get from one genome to the other.

For a given pair of genomes, the reversal distance is unique, but there are

usually many possible reversal scenarios with this distance.

However, it is possible that this mathematical notion of reversal distance can

underestimate the actual number of steps that occurred biologically.



Multichromosomal genomes: rearrangement operations

We treat four elementary rearrangement events in multichromosomal genomes:

reversals, translocations, fusions, and fissions.

Reversal: An interval within a single chromosome may be reversed in the

same fashion as a reversal acts in the unichromosomal case:

7 -2 8 3 $ 7 -2 8 3 $

5 9 -6 -1 12 $ 5 9 -12 1 6 $

11 4 10 $ 11 4 10 $

Note: When the programs are run in unichromosomal mode, the genomes

3 1 2 and -2 -1 -3are considered different (one reversal apart, distance=1), while in

multichromosomal mode, those same genomes are considered equivalent

(distance=0) because we have simply flipped an entire chromosome, which

gives an equivalent genome in the multichromosomal mode.



Two chromosomes "A B" and "C D" may be rearranged into "A D" and "C B".

(The letters A, B, C, D stand for sequences of genes.)

Because flipping chromosomes does not alter a genome (only its

representation is altered), "A -C" and "-B D" is another possible translocation.

(-B means to reverse the order of the genes in sequence B and negate each

one.)

For example, a translocation on chromosomes 1 and 3 is

7 -2 8 3 $ 7 -2 8 -4 -11 $

5 9 -6 -1 12 $ 5 9 -6 -1 12 $

11 4 10 $ -3 10 $

Translocation



Fussion & Fission

Fusion: Two chromosomes may be fused together into a single chromosome.

Due to chromosome flippings, there are four distinct fusions between each pair of

chromosomes. Here is one of the fusions between chromosomes 1 and 3:

7 -2 8 3 $ 7 -2 8 3 -10 -4 -11 $

5 9 -6 -1 12 $ 5 9 -6 -1 12 $

11 4 10 $

Fission: A chromosome may be broken into two chromosomes between any pair

of genes:

7 -2 8 3 $ 7 -2 8 3 $

5 9 -6 -1 12 $ 5 9 $

11 4 10 $ -6 -1 12 $

11 4 10 $



Signed and unsigned genomes

Most comparative mapping techniques determine the physical locations and

relative order of genes in each chromosome, but do not determine which of two

orientations each gene has.

Current sequencing methods do provide the orientations. It turns out that the

genome rearrangement problem (uni- and multichromosomal) for unsigned

permutations is NP-hard, but the same problems for signed data can be done in

polynomial time.

Fortunately, with many genomes currently being sequenced, it is likely that

many comparative maps (corresponding to unsigned permutations) will soon be

replaced by sequencing data (corresponding to signed permutations).




For example, to turn the unsigned genome

1 2 3 4 5

into the unsigned genome

1 4 3 2 5

requires one unsigned reversal. An assignment of signs may be designed in

the source and destination genomes that give a signed reversal scenario

requiring this same number of steps. Here, we get

1 2 3 4 5

1 -4 -3 -2 5

which also takes one step. Note that there may be other sign assignments

taking this minimum number of steps.




It is possible that correctly signed data would have increased the number of

steps:

1 2 3 4 5

1 -4 -3 -2 5

1 -4 3 -2 5

If the data collection method did not determine signs, it is impossible to know

mathematically whether the one step or two step scenario is more biologically

accurate; the mathematical problem the genome rearrangement programs solve

is to find the signs giving the minimum possible distance.



X-Alignments

• Implication: The reversals took place equidistant from the center of chromosome.• Those along the diagonal are orthologs between species.• Those along anti-diagonal are duplicates separated by inversion, within species.

• The “X” Factor discovered by Eisen et al • Alignment of whole genomes of prokaryotes like bacteria revealed X-like

patterns in dot plots – called X-alignments.



A biological model case

8 7 6 5 4 3 2 1 11 10 9

4 3 2 8 7 1 5 6 11 10 9

cabbage

turnip

Palmer and Herbon found that the mitochondrial genomes in cabbage and

turnip had very similar gene sequences, but with fairly different gene orders.

How to design a „transformation“ of cabbage into turnip?

Mitochondrial DNA of cabbage and turnip are composed of five conserved

blocks of genes that are shuffled in cabbage as compared to turnip. Every

conserved block has a direction that is shown by a + or – sign.



Inversion, Transposition and inverted Transposition

inversion

transposition

inverted transposition



Oriented/Unoriented Blocks

Remember that the unoriented case results in an NP-Hard problem, whereas

the oriented case can be solved in polynomial time.

2 1 3 7 5 4 8 6

1 2 3 4 5 6 7 8

8 7 6 5 4 3 2 1 11 10 9

4 3 2 8 7 1 5 6 11 10 9

UNORIENTED BLOCKS

ORIENTED BLOCKS

Polynomial Time

NP-Hard



Sorting by Reversals

8 7 6 5 4 3 2 1 11 10 9

8 7 6 5 4 3 2 1 11 10 9

8 2 3 4 5 6 7 1 11 10 9

4 3 2 8 7 1 5 6 11 10 9

8 2 3 4 5 1 7 6 11 10 9

4 3 2 8 5 1 7 6 11 10 9

4 3 2 8 7 1 5 6 11 10 9

4 3 2 8 7 1 5 6 11 10 9

Cabbage

Turnip



Permutation () : an ordered arrangement of

the set { 1,2,…,n}

Reversal () :a rearrangement that inverts a

block in {3 4 7 6 1 5 2 } (3,6) ={3 4 5 1 6 7 2}

Signed Permutation (): a permutation

where the elements are oriented

a reversal switches element orientation

{+3 -4 +7 -6 +1 -5 +2 } (3,6) ={+3 -4 +5 -1 +6 -7 +2}



easy to do by eye ...

8 7 6 5 4 3 2 1 11 10 9

8 7 6 5 4 3 2 1 11 10 9

8 2 3 4 5 6 7 1 11 10 9

4 3 2 8 7 1 5 6 11 10 9

8 2 3 4 5 1 7 6 11 10 9

4 3 2 8 5 1 7 6 11 10 9

4 3 2 8 7 1 5 6 11 10 9

4 3 2 8 7 1 5 6 11 10 9

1

12

123

12….t=

= t …. 21



Formal Approach: Sorting by Reversals

The order of genes in 2 organisms is represented by permutations = 12 ... n and = 12 ... n. A reversal of an interval [i,j] is the permutation

1 2 ... i-1 i i+1 ... j-1 j j+1 ... n

1 2 ... i-1 j j-1 ... i+1 i j+1 ... n

(i,j) has the effect of reversing the order of ii+1 ... j and transforming

1 ... i-1i ... j j+1 ... n into •(i,j) = 1 ... i-1j ... ij+1 ... n .

Given permutations and , the reversal distance problem is to find a series of

reversals 12 ... t such that •1•2 ... t = and t is minimal.

t is called the reversal distance between and .



Breakpoint Graph

Sort a permutation is a hard problem.

Breakpoints were introduced by Watterson et al. (1982) and by Nadeau and Taylor

(1984) and correlations were noticed between the reversal distance and the

number of breakpoints.

Let i j if |i – j| = 1. Extend a permutation = 12 ... n by adding 0 = 0 and

n+1 = n + 1. We call a pair of elements (i,i+1), 0 i n, of an adjacency

if i i+1, and a breakpoint if i i+1.

2 3 1 4 6 5 7

0 2 3 1 4 6 5 7 8

adjacencies

breakpoints

As the identity permutation has no

breakpoints, sorting by reversals

corresponds to eliminating breakpoints.

An observation that every reversal can

eliminate at most 2 breakpoints implies that

the reversal distance d() b() / 2 where

b() is the number of breakpoints in .

However, this is a clear overestimate.



Breakpoint Graph

The breakpoint graph of a permutation is an edge-colored graph G() with n +

2 vertices {0, 1 ... n, n+1} {0, 1, ..., n, n+1}. We join vertices i and i+1 by a

black edge for 0 i n. We join vertices i and j by a gray edge if i j.

Black path

0 2 3 1 4 6 5 7

Grey path

0 2 3 1 4 6 5 7

Superposition of black and grey paths formsthe breakpoint graph:

A breakpoint graph is obtained by a super-position of a black pathtraversing the vertices0, 1, ..., n, n+1 in the order given by the permutation and a graypath traversing the verticesin the order given by theidentity permutation.



Cycle decomposition

A cycle in an edge-colored graph G is called alternating if the colors of every two

consecutive edges of this cycle are distinct. In the following, cycles will mean

alternating cycles.

Cycle decomposition ofthe breakpoint graph:

0 2 3 1 4 6 5 7

0 2 3 1 4 6 5 7

0 2 3 1 4 6 5 7

0 2 3 1 4 6 5 7

A vertex v in a graph G is called balanced if the

number of black edges incident to v equals the

number of grey edges incident to v.

A balanced graph is a graph in which every

vertex is balanced. G() is a balanced graph.

Therefore, there exists a cycle decomposition

of G() into edge-disjoint alternating cycles

(every edge in the graph belongs to exactly one

cycle in the decomposition). Cycles in an edge

decomposition may be self-intersecting. The

previous breakpoint graph can be decomposed

into 4 cycles, one of which is self-intersecting.



Cycle decomposition

What is the decomposition of the breakpoint graph into a maximum number c()

of edge-disjoint alternating cycles? Here, c() = 4.

Cycle decompositions play an important role in estimating reversal distances.

When a reversal is applied to a permutation, the number of cycles in a maximum

decomposition can change by at most one (while the number of breakpoints

can change by two).

Bafna&Pevzner (1996) proved the bound:

d() n + 1 - c()

Which is much tighter than the bound in terms of breakpoints d() b() / 2.

For many biological problems, d() = n + 1 - c().

Therefore, the reversal distance problem reduces to the problem of finding

the maximal cycle decomposition.



Effects of reversals on cycles

(A) For reversals acting on two

cycles, (b – c) = 1.

(B) For reversals acting on an

unoriented cycle, (b – c) = 0.

(C) For reversals acting on an

oriented cycle, (b – c) = -1

Hannenvalli, Pevzner, Journal of the ACM 46, 1 (1999)



Effect of reversals on gray edges

(a) A proper reversal on an oriented gray edge.

(b) A nonproper reversal on an unoriented gray edge.




Transform signed into unsigned permutation

(a) Optimal sorting of a

permutation (3 5 8 6 4 7 9 2 1 10

11) by 5 reversals.

(b) Breakpoint graph of this

permutation: black edges

connect adjacent vertices that

are not consecutive, gray edges

connect consecutive vertices that

are not adjacent.

(c) Transformation of a signed

permutation into an unsigned

permutation and the breakpoint

graph G(); (d) Interleaving graph

H with two oriented and

one unoriented unoriented

component.





The Problems

Minimum Sorting by Reversals (MinSortRv):

Given a permutation , what is the shortest sequence (12….t ) of reversals

that sorts ? (Distance: d())

Complexity remains open. (NP-Hard)

Minimum Signed Sorting by Reversals (SignedSortRv):

Given a signed permutation , what is the shortest sequence (12….t ) of

reversals that sorts ?

Solvable in polynomial time.



KS93 -Kececioglu and Sanko“Exact and approximation algorithms for the inversion distance between two chromosomes", 4th CPM- studied MinSortRv- introduced notion of breakpoints- 2 approximation algorithm

BP93 -Bafna and Pevzner“Genome Rearrangements and Sorting by Reversals", 34th FOCS- breakpoint graph and cycle decomposition- introduced signed sorting SignedSortRv- 3/2 approx algorithm for SignedSortRv- 7/4 approx algorithm for MinSortRv

HP95 - Hannenhali and Pevzner “Transforming Cabbage into Turnip”, 27th STOC- SignedSortRv resolved- O(n4) algorithm- introduced hurdles and fortresses- d() = b() - c() + h() + f()

Important developments



KS93- Breakpoints

Extend to include element 0 (L) on the left and element n+1 (R) on the right.

A breakpoint occurs between two adjacent elements that do not differ by 1

Example: = { 3 5 6 7 2 1 4 8 } has 5 breakpoints, (b() = 5).

R 3 5 6 7 2 1 4 8 L

Breakpoints partition sequence into strips that are increasing or decreasing.

Reversals add or remove breakpoints. Sorted permutation has 0 breakpoints.

i-reversal (i = 0,1, 2): a reversal that decreases number of breakpoints by i.

Theorem (KS): Let contain a decreasing strip. Then has a 1- or 2-reversal.

If every reversal that removes a breakpoint of results in a permutation with no

decreasing strips, then has a 2-reversal.



Algorithm KS()

i 0

while contains a breakpoint

i i+1

the reversal that removes the most

breakpoints, resolving ties in favor of

reversals that leave a decreasing strip

return

Optimal reversal distance is at least b()/2

KS returns a solution that is at most 2*optimal = b()



BP93 – Breakpoint Graph

Vertices: elements of (plus 0 (L) and n+1 (R) )

2 3 1 6 5 4L R

+-6 6

THE DIAGRAM OF REALITY AND DESIRE



Construction of a diagram of reality and desire

3 2 1 4 5L R

L -3 +3 +2 -2 +1 -1 -4 +4 +5 -5 R

L -3 +3 +2 -2 +1 -1 -4 +4 +5 -5 R

Reality edges

L -3 +3 +2 -2 +1 -1 -4 +4 +5 -5 R

Desire edges

1 2 3 4 5L RDesire

Reality



L -3 +3 +2 -2 +1 -1 -4 +4 +5 -5 R

L R-3

+3

+2

-2

+1 -1 -4

+4

+5

-5



c() = number of cycles in a maximum cycle decomposition

Observation: reversals affect c().

Example:

{L [+1 -1] –2 +2 +3 -3 R}- removes 2 breakpoints and 1 cycle.

effect of reversals on c()

L +1 -1 -2 +2 +3 -3 R L -1 +1 -2 +2 +3 -3 R



d() >= b() - c()

Cycles of length 4 are eliminated by 2-reversals.

Let c4() = number of 4-cycles.

(c() - c4()) : Cycles of length > 4 include at least three breakpoints

d() >= b() – c4() - (c() - c4()) / 3



Algorithm BP()

while contains a breakpoint

if has no decreasing strips

if a 4-cycle C remains

Find cycle C’ that crosses C

0-reversal on C’, 2-reversal on C

else

Regular 0-reversal

else

Regular greedy choice

Algorithm BP produces a solution that is at most (3*optimal)/2



A FB C

D

E Interleaving Graph

HP95 – Hurdles and Fortress



HurdlesA hurdle is a bad component that does not separate any other two bad

components. Separation is an important concept, in that a reversal through

reality edges in different components A and C will result in every component B,

that separates A and C being twisted. A bad component becomes good when

twisted.

Bad Components

Non-Hurdles Hurdles

SimpleHurdles

SuperHurdles

B

A FCD

E



Fortress

A permutation a is called fortress f() when its reality and desire diagram

contains an odd number of hurdles and all of them are super hurdles.

Fortresses are permutations that require

one extra reversal to sort, due to their

special structure

A smallest possible fortress.



Algorithm HP()If there is a good component in RD() then

pick two divergent edges e,f in this component,

making sure the corresponding reversal does

not create any bad components

Return the reversal characterized by e and f

Else

if h() is even then

Return merging of two opposite hurdles

else

if h() is odd and there is a simple hurdle

return a reversal cutting this hurdle

else // fortress

return merging of any two hurdles

d() b() - c() + h() + f() h(): number of hurdles

f(): 0/1, according to being a fortress or not



Hurdles

(a) Unoriented component U separates U‘ and U‘‘ by virtue of the edge (0, 1)

(b) Hurdle U does not separate U‘ and U‘‘.




Effects of reversals on cycles


Reversal on a cycle C (i) deletes vertex C from the interleaving graph; (ii) changes

the orientation of vertices in V(C); (iii) complements the subgraph induced by V(C).



Merging hurdles




Hannenvalli-Pevzner algorithm




Improvements of Hannenhalli-Pevzner algorithm

Several websites offer programs to sort permutations by reversals. At their roots

is the Hannenhalli-Pevzner algorithm for sorting signed permutations by

reversals. Successive authors improved the algorithm.

• By the Hannenhalli&Pevzner algorithm, the distance computation is

performed in time O(n4).

• improvements in the algorithm developed by Haim Kaplan, Ron Shamir and

Robert E. Tarian bring the time to compute distance down to O(n2).

• GRAPPA is written by a multitude of authors. It reduces the distance

computation time to O(n) using improvements by David A. Bader, Bernard

M.E. Moret and Mi Yan.

The main purpose of GRAPPA is to construct phylogenetic trees for multiple

signed unichromosomal genomes; the distance computation on which we are

focused here is but a mere subroutine in that context.



Algorithm by Kaplan, Shamir, Tarjan

The algorithm has three main stages:

1. Pre-process the permutation. This pre-processing contains 3 sub stages:

1a. Unsign the permutation, e.g., p will be unsigned to the permutation 0,

(7,8), (4,3), (1,2), (5,6), (12,11), (9,10), 13.

1b. Define the Overlap graph of the permutation

1c. Find the connected components of the overlap graph

2. Clear the hurdles. A hurdle is a problematic connected component of the

overlap graph. In this stage each reversal merges two hurdles in distinct

connected components into one non-hurdle component.

3. Generate a sequence of safe reversals. A safe reversal is defined as a

reversal that reduces b-c (the number of breakpoints minus the number of

cycles) without creating new hurdles.



Multichromosomal genomes: more tricky

Word problems and insertions/deletions

So far we did not consider "word problems" in which some genes are repeated,

1 2 -1 3 4

nor did we allow gaps in the numbering (as may arise from insertion/deletion),

1 3 -9 -7 5

Distinguish between microrearrangements (e.g. intrachromosomal

rearrangements with a span < 1 Mb) and macrorearrangements (e.g.

intrachromosomal rearrangements of larger span as well as interchromosomal

rearrangements). The existing rearrangement algorithms do not distinguish

between these two types of rearrangements.

First identify conserved synteny blocks (segments that can be converted into

conserved segments by microrearrangements).



Genome Rearrangements: Synteny

(a) Human and mouse synteny blocks of

conserved gene order. Every block

corresponds to a rectangle, with a diagonal

showing whether the arrangements of

anchors in human and mouse (within the

synteny block) are the same or reversed.

(b) Combining anchors into clusters by the

GRIMM-Synteny algorithm at G = 100 kb. The

edges in the anchor graph connect the closest

ends of the anchors. The anchors are color-

coded by the resulting clusters. At G = 1 Mb,

this forms a single cluster, which in turn forms

a synteny block (the lower right block in the

human 18/mouse 17 rectangle in a).

Pevzner, Tesler, Genome Res 13, 37 (2003)



From Anchors to Breakpoint Graphs

X-chromosome: from local similarities, to

synteny blocks, to breakpoint graph, to

rearrangement scenario. (a) Dot-plot of

anchors. Anchors are enlarged for

visibility. (b) Clusters of anchors. (c)

Rectified clusters. (d) Synteny blocks. (e)

Synteny blocks (symbolic representation

as genome rearrangement units). (f) 2D

breakpoint graph superimposed on

synteny blocks. The projections of the 2D

graph onto the human and mouse axes

form the conventional breakpoint graphs.

(g) 2D breakpoint graph. The four cycles

in the breakpoint graph are shown by

different colors. (h) A most parsimonious

rearrangement scenario for human and

mouse X-chromosomes. Pevzner, Tesler, Genome Res 13, 37 (2003)



Genome Rearrangements

Construction of the breakpoint

graph from synteny blocks.

(a) Solid path through human.

(b) Dotted path through mouse.

(c) Superposition of paths.

(d) Remove blocks to obtain

cycles.




Multichromosomal breakpoint graph

Multichromosomal breakpoint

graph of the whole human and

mouse genomes. The

conventional chromosome

order and orientation are not

suitable for such graphs; an

optimal chromosome order and

orientation were determined by

the algorithm in Tesler (2002).

Three "null chromosomes," N1,

N2, N3, were added to mouse

to equalize the number of

chromosomes in the two

genomes.




Summary

Computational studies on genome rearrangements assume that evolution took

the path of shortest reversal distance.

Due to algorithmic improvements, the computational costs could be

significantly reduced.

It may be very advantageous to simultaneously analyze more than 2 genomes

at the same time!

Documents

Genome Rearrangements