Genome Rearrangement

ByGhada Badr

Part I

Genome, chromosome, gene, gene order

QuickTime™ and aTIFF (Uncompressed) decompressor

are needed to see this picture.

• The entire complement of genetic material carried by an individual is called the genome.• Each genome contains one or more DNA molecules, one per chromosome

• A gene is a segment of DNA sequence with a specific function

• Genes can be ordered by their DNA sequence location. • DNA consists of two complementary strands twisted around each other to form a right-handed double helix. • A sign (+/-) is usually used to indicate on which strand a gene is located.

F5’ 3’3’ 5’ QuickTime™ and a

TIFF (Uncompressed) decompressorare needed to see this picture.

Gene order: A -B C D -E F

A B C D FE

The DNA molecule (chromosome) may be circular or linear

• The genome is structurally specific to each species, and it changes only slowly over time. Therefore genome comparison among different species can provide us with much evidence about evolution.

• Genome rearrangements are an important aspect of the evolution of species. Even when the gene content of two genomes is almost identical, gene order can be quite different.

A -B C D F-E

B -E F-D CA

Genome 1

Genome 2

Gene order analysis on a set of organisms is a powerful technique for genomic comparison phylogenetic inference.

General Definition for the problem:Given a set of genomes and a set of

possible evolutionary events (operations), find a shortest set of events transforming (sorting) those genomes into one another.

What genome means and what events are, makes the diversity of the problem.

Since these events are rare, scenarios minimizing their number are more likely close to reality.Many models have been proposed.

Genes (or blocks of contiguous genes) are a good example of homologous markers, segments of genomes, that can be found in several species.

The simplest possible model is:

The order of genes in each genome is known,

All the genomes share the same set of genes,

All genomes contain a single copy of each gene, and

All genomes consist of a single chromosome.

Genome Models

Genomes can be modeled by each gene can be assigned a unique number and is exactly found once in the genome.

Genome Models

permutations:

Signed Permutation: Each gene may be assigned + or - sign to indicate the strand it resides on.

Unsigned Permutation: If the corresponding strand is unknown.

Genes (markers) are represented by integers: 1, 2, . . . . , n, with +,- sign to indicate the strand they lie on.

The order and orientation of genes of one genome in relation to the other is represented by a signed permutation .

= (2n-1n) of size n over {-n, ... , -1, 1, ... , n}, such that for each i from 1 to n, either i or -i is mandatory represented, but not both.

Permutaions

The identity permutation n = (1, 2, 3, . . . . , n).

When multiple genomes with the same gene content are compared, one of them is chosen as a base (reference), i.e, represented as n, and all other identical genes are given the same integer values.

Permutaions

Identity permutation:

In order to sort a permutation this means that we want to apply some operations on to change it to n.

If (1 = 2) We say that is sorted with respect to .

If (1 2) We say that is unsorted with respect to .

Permutaions

Sorted/unsorted permutation:

Permutaions

Fruit Fly

Mosquito

Silkworm

Locust

Centipede

Example: Mitochondrial Genomes of 6 Arthropoda

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

1 2 3 4 5 6 8 7 9 -10 11 12 13 14 15 16 17

1 2 3 4 5 6 7 8 9 10 11 12 14 13 15 16 17

1 2 3 5 4 6 7 8 9 10 11 12 13 14 15 16 17

1 3 4 5 6 7 8 9 10 11 -2 12 13 14 15 16 17

1 3 4 5 6 7 8 9 10 11 -2 12 16 13 14 15 17

1= (1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12 , 13 , 14 , 15 , 16 , 17)

2= (1 , 2 , 3 , 4 , 5 , 6 , 8 , 7 , 9 ,-10 , 11 , 12 , 13 , 14 , 15 , 16 , 17)

3= (1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12 , 14 , 13 , 15 , 16 , 17)

4= (1 , 2 , 3 , 5 , 4 , 6 , 7 , 8 , 9 , 10 , 11 , 12 , 13 , 14 , 15 , 16 , 17)

5= (1 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , -2 , 12 , 13 , 14 , 15 , 16 , 17)

6= (1 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , -2 , 12 , 16 , 13 , 14 , 15 , 17)

Permutaions

Fruit Fly

Mosquito

Silkworm

Locust

Centipede

1= (1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12 , 13 , 14 , 15 , 16 , 17)

2= (1 , 2 , 3 , 4 , 5 , 6 , 8 , 7 , 9 ,-10 , 11 , 12 , 13 , 14 , 15 , 16 , 17)

3= (1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12 , 14 , 13 , 15 , 16 , 17)

4= (1 , 2 , 3 , 5 , 4 , 6 , 7 , 8 , 9 , 10 , 11 , 12 , 13 , 14 , 15 , 16 , 17)

5= (1 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , -2 , 12 , 13 , 14 , 15 , 16 , 17)

6= (1 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , -2 , 12 , 16 , 13 , 14 , 15 , 17)

is linear when it represents a linear chromosome, or circular when it represents a circular chromosome.

When = (2n-1n) is circular: ’ = (-nn-121) all permutations obtained by shifts on or ’

shift( , i) = (n-i+1n-i+2n-1n1n-

i are all equivalent. Example: (-3,2,1,-4) & (-1,-2,3,4)

Permutaions

Linear and circular permutation:

Permutaions

For a given permutation = (2n-

1n), there is a point between each pair of consecutive values i and i+1 in .

If is linear: there are two additional points, one before and one after n.

If is circular: there is one additional point between nand 1.

Pts() = n+1 if linear, and pts() = n if circular.

Points in permutations

Permutaions

For a given = (2n-1n)

If is linear: a linear extension of is ’= (0, 2n-1n, n+1)

If is circular: a linear extension of is ’= (0, 2n-1n-1, n)

Linear extension of a permutation:

Now: we want to compare our genomes.

Permutaions

Example: = (4,8,9,7,6,5,1,3,2) ’= (0,4,8,9,7,6,5,1,3,2,10)

’= (0.4.8.9.7.6.5.1.3.2.10)

Then Pts() = 10

Problem: Given two genomes, How do we measure their similarity and/or distance?

Permutations - similarity/distance

A Related Problem: Given two permutations,How do we measure their similarity and/ordistance?

A distance measure should be a metric on the set of genomes.

A Metric d on a set S (d: S S R) satisfies the following three axioms:

1. Positivity: for all s, t in S, d(s,t) 0, and d(s,t)=0 iff s = t.

2. Symmetry: for all s, t in S, d(s,t) = d(t,s).3. Triangular inequality: for all s, t, u in S, d(s,u) d(s,t) + d(t,u).

Measures of similarity between permutations that are used in computational biology are numerous in literature.

First measures used are (will be useful later on):

Breakpoints (Introduced by Sankoff and Blanchette (1997))

Common intervals

Permutations-distance - Breakpoints

When analyze with respect to , each point in can be an adjacency or a breakpoint.

A point (pair of consecutive values) (i, i+1) in is an adjacency between and : when either (i, i+1) or (-I+1, -i) are consecutive in .

If is linear: we have adjacency before if is also the first value in , and an adjacency after n, if n is also last value in .

If is circular: we assume that n is also last value in and (n, 1) is an adjacency if is also the first value in .

Breakpoint distance counts the lost adjacencies between genomes.

The breakpoint distance between and is:

brp() = pts() - adj() where: pts() is the number of points in . adj() is the number of adjacencies.

• If is sorted ( = ): has only adjacencies and no breakpoints (brp() = 0).

• If is unsorted ( ): has at least one breakpoint (brp() 0).

Back to our Example: = (4,8,9,7,6,5,1,3,2) ’= (0,4,8,9,7,6,5,1,3,2,10)

’= (0.4.8.9.7.6.5.1.3.2.10) Then Pts() = 10, brp()?Adjacencies?n= (0.1.2.3.4.5.6.7.8.9.10)

(8,9) (7,6) (6,5) (3,2) adj() = 4 brp() = pts() - adj() = 10 - 4 = 6

Breakpoint distance is based on the notion of conserved adjacencies and can be defined on a set of more than two genomes.

It is easy to compute.

It always fails to capture more global relations between genomes.

The first generalization of adjacencies is the notion of common intervals.

Common intervals: subsets of genes that appear consecutively together in two or more genomes, where genes are the same in each interval but may be not in the same order or orientation.

Permutations-distance - Common Intervals

1= (1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12 , 13 , 14 , 15 , 16 , 17) 2= (1 , 2 , 3 , 4 , 5 , 6 , 8 , 7 , 9 ,-10 , 11 , 12 , 13 , 14 , 15 , 16 , 17) 3= (1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12 , 14 , 13 , 15 , 16 , 17) 4= (1 , 2 , 3 , 5 , 4 , 6 , 7 , 8 , 9 , 10 , 11 , 12 , 13 , 14 , 15 , 16 , 17) 5= (1 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , -2 , 12 , 13 , 14 , 15 , 16 , 17) 6= (1 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , -2 , 12 , 16 , 13 , 14 , 15 , 17)

Example (circular chromosomes)

If compare the first 4 species: they share 6 adjacencies{1,2}, {2,3},{11.12},{15,16},{16,17},{17,1}If compare all 6 species: they share only 1 adjacency{17,1}

Common intervals: subsets of genes that appear consecutively together in two or more genomes, where genes are the same in each interval but may be not in the same order or orientation.

1= (1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12 , 13 , 14 , 15 , 16 , 17) 2= (1 , 2 , 3 , 4 , 5 , 6 , 8 , 7 , 9 ,-10 , 11 , 12 , 13 , 14 , 15 , 16 , 17) 3= (1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12 , 14 , 13 , 15 , 16 , 17) 4= (1 , 2 , 3 , 5 , 4 , 6 , 7 , 8 , 9 , 10 , 11 , 12 , 13 , 14 , 15 , 16 , 17) 5= (1 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , -2 , 12 , 13 , 14 , 15 , 16 , 17) 6= (1 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , -2 , 12 , 16 , 13 , 14 , 15 , 17)

Example (circular chromosomes)

The six permutations are very similar.

The genes in the interval [1,12] are all the same, as genes in the intervals [3,6], [6,9],[9,11], and [12,17].

We can use common intervals as a measure of similarity between species.

Disadvantage: All these measures do not reflect rearrangement operations or explain what happened to the genome over time.

Back to our original 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.

Rearrangement operations (events)

What are the Rearrangement events (Operation)?These events (Operation) could be applied to a single gene or to a group of genes, intervals.

Rearrangement operations

Fruit Fly

Mosquito

Silkworm

Locust

Centipede

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

Rearrangement operations affect gene orderand 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

Rearrangement Operations

Rearrangement Operations - Single Chro.

Inversion

Fruit Fly

Mosquito

Silkworm

Locust

Centipede

An inversion.Example: Mitochondrial Genomes of 6 Arthropoda

Transposition

Fruit Fly

Mosquito

Silkworm

Locust

Centipede

A transposition

Reverse Transposition

Fruit Fly

Mosquito

Silkworm

Locust

Centipede

A reverse transposition

Rearrangement Operations - Multiple Chro.

Translocation

Fusion

Fission

Fusion

Fission

Fusion

Fission

Fusion

Fission

Fusion

Fission

Fusion

Fission

58[Source: Linda Ashworth, LLNL]

DOE Human Genome Program Report

From 24 chromosomes

To 21 chromosomes

Rearrangement Problems

Any set of operations yields a distance between genomes, by counting the minimum number of operations needed to transform one genome into the other.

Rearrangement Problems

• Computing the distance d()• Computing one optimal sorting sequence of events.

Two classical problems

• Given a permutation , calculate reversal distance d() and find one optimal sequence of reversals sorting .

Assumption:

Only reversals are allowed.

No duplication in genes.

Genomes are unichromosomal.

Reversal Distance - Sorting by Reversals

QuickTime™ and aTIFF (LZW) decompressor

A reversal is represented as a set of genes appearing together in the given genome.

This approach is symmetric

Vertices: all permutations of n = 3.Edges: connect an edge between 1 and 2 if reversal distance

d(1, 2) = 1.

Reversal graph for n = 3

Reversal distance d(i, k) = length of shortest path between vi and vk.

The graph is huge |V| = n!.2n

A feasible graph-search algorithm is not possible!

The classical approach for solving these two problems in polynomial time was developed by Hannenhalli and Pevzner. (1995)

The reversal distance can be computed in O(n) time by Bader et. al. (2000)

The fastest algorithm to find an optimal sorting sequence is < O(n2) by Tannier et. al. (2007)

Most approaches are based on a special structure called the breakpoint graph.

Breakpoint Graph: edges are black or gray.

Given = (n-1n)

If is linear: we add the values 0, and n+1, the represents the extremities of the chromosome obtaining:

= (0, n-1n, n+1)

If is circular: assume n = n and add only the value 0, obtaining:

= (0, n-1n-1, n)

Black edge: Links each pair of consecutive value in

by a horizontal (a point in ).

Gray edges: Link the extremities of black edges such that the values will be in order.

Graph: collection of cycles, where black and gray edges alternate.

Trivial cycle: one black and one gray edge (adjacency)

Long Cycle: four or more edges ( 2 breakpoints)

When sorted

Linear and circular permutations are different in breakpoint graph construction.

Same analyses.

Linear Circular

= (-3 , 2 , 1 , -4)

When sorted

If is sorted:

• Only adjacencies, no breakpoints.

• Breakpoint graph is a collection of trivial cycles.

• # cycles in sorted graph cyc() = pts()

= (-3 , 2 , 1 , -4) sorted

If is unsorted:

• At least one breakpoint, at least one long cycle.

• # cycles cyc() is at most = pts() - 1

= (-3 , 2 , 1 , -4) sorted

Observation: To sort a permutation , we would like to increase the number of cycles in its breakpoint graph.

The effects of a reversal over a breakpoint graph .

Split reversal

Neutral reversal

Joint reversal

cyccyc cyccyc

cyccyc

The effects of a reversal over a breakpoint graph .

QuickTime™ and aTIFF (LZW) decompressorare needed to see this picture.QuickTime™ and aTIFF (LZW) decompressorare needed to see this picture.

Observation: To sort , we must maximize the number of split reversals in the sorting sequence s.

If s has only split reversals: what will be the reversal distance d(Hint: in terms of pts() and cyc())

dpts() - cyc()

Are we done?

dpts() - cyc()

A split reversal does not always exist.

For example, if all black edges in the graph have the same direction.

In this case, we need to add some joint and/or neutral reversals in the sorting sequence s.

• It is always possible to calculate the number of non-split reversals in a sorting sequence.

• It will be the number of non-split reversals to sort some hard components in the graph with no orientation, unoriented components.

• Unoriented components can be a hurdle hrdor more hardly a fortress frt() in the breakpoint graph.

• Hardles are very rare, and fortresses are even more rare in permutations that represent real genomes.

• In practice, split reversals are sufficient to sort the permutation.

Can we choose any split reversal? only safe reversals.

Safe reversal: a split reversal not producing hurdles.

Unsafe reversal Safe reversal

There is always a safe reversal for any oriented .

The final formula for the reversal distance d is:

dpts() - cyc() + hrd() + frt()

Where: • frt() = 1, if is a fortress, and 0 otherwise. • pts() = n+1, if is linear, and n if is circular.

Algorithm: Get optimal sorting sequence s that sorts Input: A signed permutation . Output: An optimal sequence of reversals sorting .1. Construct the breakpoint graph of .2. S [empty]3. If frt() = 1 then4. choose a reversal to eliminate the fortress5. 6 s s . [concatenate the reversal to s] 7. End if8. While there is hurdles in do9. choose a reversal to eliminate the hurdle10. s s . [concatenate the reversal to s]12. End while13. While is not sorted do14. choose a safe split reversal to 15. 6 s s . [concatenate the reversal to s]17. End while18. return s

ComplexityO(n5)

Tools: GRIMM & GRAPPA

We can have more than one optimal solution

conclusions

• Represented linear and circular genomes as permutations in our simple model.

• Described first measures for similarity between permutation were breakpoint and common intervals --> has no biological interpretation.

• Used genome rearrangement events to describe similarity/distances between genomes --> has more biological meaning.

• Described in details one distance measure (reversal distance) and events (reversals) to sort genomes.

Thank you

Questions?

Next Lecture?

Genome Rearrangement

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