Genome Rearrangement By Ghada Badr Part I Genome, chromosome, gene, gene order
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 Genome, chromosome, gene, gene order
A gene is a segment of DNA sequence with a specific function Genome, chromosome, gene, gene order
A C D F 5 3 3 5 B E Gene order:A -B C D EF 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. Genome, chromosome, gene, gene order
A B C D E F H I K J The DNA molecule (chromosome) may be circular or linear Genome Rearrangement A -B C D -E F B -E F -D A C
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 -E F Genome 1 B -E F -D A C Genome 2 Genome Rearrangement Gene order analysis on a set of organisms is a powerful technique for genomic comparison phylogenetic inference. Genome Rearrangement 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. Genome Models 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. 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. Permutaions Genes (markers) are represented by integers:
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 . = ( 2 n-1 n) 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 Identity permutation:
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 Sorted/unsorted 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 Example: Mitochondrial Genomes of 6 Arthropoda
1= (1 , 2 ,3 , 4,5,6,7,8,9, 10 , 11 , 12 , 13 , 14 , 15 , 16 , 17) Fruit Fly Mosquito Silkworm Locust Tick Centipede 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 Example: Mitochondrial Genomes of 6 Arthropoda
1= (1 , 2 ,3 , 4,5,6,7,8,9, 10 , 11 , 12 , 13 , 14 , 15 , 16 , 17) Fruit Fly Mosquito Silkworm Locust Tick Centipede 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 Linear and circular permutation:
is linear when it represents a linear chromosome, or circular when it represents a circular chromosome. When = ( 2 n-1 n) is circular: = (-n n-1 2 1) all permutations obtained by shifts on or shift( , i) = (n-i+1 n-i+2n-1 n1 n-i are all equivalent. Example: (-3,2,1,-4) & (-1,-2,3,4) Permutaions Points in permutations
For a given permutation = ( 2 n-1 n), 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. Permutaions Linear extension of a permutation:
For a given = ( 2 n-1 n) If is linear: a linear extension of is = (0, 2 n-1 n, n+1) If is circular: a linear extension of is = (0, 2 n-1 n-1, n) Permutaions Example: = (4,8,9,7,6,5,1,3,2)
= (0,4,8,9,7,6,5,1,3,2,10) = ( ) Then Pts() = 10 Now: we want to compare our genomes. Permutations - similarity/distance
Problem: Given two genomes, How do we measure their similarity and/or distance? A Related Problem: Given two permutations, How do we measure their similarity and/or distance? Permutations - similarity/distance
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: Positivity: for all s, t in S, d(s,t) 0, and d(s,t)=0 iff s = t. Symmetry: for all s, t in S, d(s,t) = d(t,s). Triangular inequality: for all s, t, u in S, d(s,u) d(s,t) + d(t,u). Permutations - similarity/distance
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 . Permutations-distance - Breakpoints
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 onebreakpoint (brp() 0). Breakpoint distance counts the lost adjacencies between genomes. The breakpoint distance between and is: Permutations-distance - Breakpoints
Back to our Example: = (4,8,9,7,6,5,1,3,2) = (0,4,8,9,7,6,5,1,3,2,10) = ( ) Then Pts() = 10, brp()? Adjacencies? n= ( ) (8,9) (7,6) (6,5) (3,2) adj() = 4 brp() = pts() - adj() = = 6 Permutations-distance - Breakpoints
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. Permutations-distance - 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. Example (circular chromosomes) 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) 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} Permutations-distance - 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. Example (circular chromosomes) 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) 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]. Permutations-distance - Common Intervals
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. Rearrangement operations (events)
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. 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
Example: Mitochondrial Genomesof 6 Arthropoda Fruit Fly Mosquito Silkworm Locust Tick Centipede 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 Rearrangement Operations - Single Chro.
Inversion Rearrangement Operations - Single Chro.
Inversion Rearrangement Operations - Single Chro.
Inversion Rearrangement Operations - Single Chro.
Example: Mitochondrial Genomesof 6 Arthropoda An inversion. Fruit Fly Mosquito Silkworm Locust Tick Centipede Rearrangement Operations - Single Chro.
Transposition Rearrangement Operations - Single Chro.
Transposition Rearrangement Operations - Single Chro.
Transposition Rearrangement Operations - Single Chro.
Example: Mitochondrial Genomesof 6 Arthropoda Fruit Fly Mosquito Silkworm Locust Tick Centipede A transposition Rearrangement Operations - Single Chro.
Reverse Transposition Rearrangement Operations - Single Chro.
Reverse Transposition Rearrangement Operations - Single Chro.
Reverse Transposition Rearrangement Operations - Single Chro.
Example: Mitochondrial Genomesof 6 Arthropoda Fruit Fly Mosquito Silkworm Locust Tick Centipede A reverse transposition Rearrangement Operations - Multiple Chro.
Translocation Rearrangement Operations - Multiple Chro.
Translocation Rearrangement Operations - Multiple Chro.
Translocation Rearrangement Operations - Multiple Chro.
Translocation Rearrangement Operations - Multiple Chro.
Translocation Rearrangement Operations - Multiple Chro.
Translocation Rearrangement Operations - Multiple Chro.
Fusion Fission Rearrangement Operations - Multiple Chro.
Fusion Fission Rearrangement Operations - Multiple Chro.
Fusion Fission Rearrangement Operations - Multiple Chro.
Fusion Fission Rearrangement Operations - Multiple Chro.
Fusion Fission Rearrangement Operations - Multiple Chro.
Fusion Fission Rearrangement Operations - Multiple Chro.
From 24 chromosomes To 21 chromosomes [Source: Linda Ashworth, LLNL] DOE Human Genome Program Report Rearrangement Problems
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. 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
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. Two classical problems Computing the distance d() Computing one optimal sorting sequence of events. Reversal Distance - Sorting by Reversals
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
A reversal is represented as a set of genes appearing together in the given genome. Reversal Distance - Sorting by Reversals Reversal Distance - Sorting by Reversals Reversal Distance - Sorting by Reversals Reversal Distance - Sorting by Reversals Reversal Distance - Sorting by Reversals Reversal Distance - Sorting by Reversals
This approach is symmetric Reversal Distance - Sorting by Reversals
Reversal graph for n = 3 Vertices: all permutations of n = 3. Edges: connect an edge between 1 and 2 ifreversal distance d(1, 2) = 1. Reversal Distance - Sorting by Reversals
Reversal graph for n = 3 Reversal distance d(i, k) = length of shortest path between vi and vk. Reversal Distance - Sorting by Reversals
Reversal graph for n = 3 The graph is huge |V| = n!.2n A feasible graph-search algorithm is not possible! Reversal Distance - Sorting by Reversals
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. Reversal Distance - Sorting by Reversals
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) Reversal Distance - Sorting by Reversals
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) Reversal Distance - Sorting by Reversals
+5 When sorted Reversal Distance - Sorting by Reversals
+5 When sorted Reversal Distance - Sorting by Reversals
+5 When sorted Reversal Distance - Sorting by Reversals
+5 When sorted Reversal Distance - Sorting by Reversals
+5 When sorted Reversal Distance - Sorting by Reversals
+5 When sorted Reversal Distance - Sorting by Reversals
+5 When sorted Reversal Distance - Sorting by Reversals
+5 When sorted Reversal Distance - Sorting by Reversals
= (-3 , 2 , 1 , -4) Linear Circular Linear and circular permutations are different in breakpoint graph construction. Same analyses. Reversal Distance - Sorting by Reversals
+5 When sorted Reversal Distance - Sorting by Reversals
+5 When sorted Reversal Distance - Sorting by Reversals
= (-3 , 2 , 1 , -4) sorted If is sorted: Only adjacencies, no breakpoints. Breakpoint graph is a collection of trivial cycles. # cycles in sorted graph cyc() = pts() Reversal Distance - Sorting by Reversals
= (-3 , 2 , 1 , -4) sorted If is unsorted: At least one breakpoint, at least one long cycle. # cycles cyc() is at most = pts() - 1 Observation: To sort a permutation , we would like to increase the number of cycles in its breakpoint graph. Reversal Distance - Sorting by Reversals
The effects of a reversal over a breakpoint graph . Split reversal Joint reversal cyc( ) cyc() cyc( ) cyc() Neutral reversal cyc( ) cyc() Reversal Distance - Sorting by Reversals
The effects of a reversal over a breakpoint graph . Reversal Distance - Sorting by Reversals
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()) d()pts() - cyc() Are we done? Reversal Distance - Sorting by Reversals
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. d()pts() - cyc() Reversal Distance - Sorting by Reversals
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 hrd()or 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. Reversal Distance - Sorting by Reversals
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 . Reversal Distance - Sorting by Reversals
The final formula for the reversal distance d() is: d()pts() - cyc() + hrd() + frt() Where: frt() = 1, if is a fortress, and 0 otherwise. pts() = n+1, if is linear, and n if is circular. Reversal Distance - Sorting by Reversals
Algorithm: Get optimal sorting sequence s that sorts Input:A signed permutation . Output: An optimal sequence of reversals sorting . Construct the breakpoint graph of . S [empty] If frt() = 1then choose a reversal to eliminate the fortress s s . [concatenate the reversal to s] End if Whilethere is hurdles in do choose a reversal to eliminate the hurdle End while While is not sorted do choose a safe split reversal to return s Reversal Distance - Sorting by Reversals
Algorithm: Get optimal sorting sequence s that sorts Input:A signed permutation . Output: An optimal sequence of reversals sorting . Construct the breakpoint graph of . S [empty] If frt() = 1then choose a reversal to eliminate the fortress s s . [concatenate the reversal to s] End if Whilethere is hurdles in do choose a reversal to eliminate the hurdle End while While is not sorted do choose a safe split reversal to return s Reversal Distance - Sorting by Reversals
Algorithm: Get optimal sorting sequence s that sorts Input:A signed permutation . Output: An optimal sequence of reversals sorting . Construct the breakpoint graph of . S [empty] If frt() = 1then choose a reversal to eliminate the fortress s s . [concatenate the reversal to s] End if Whilethere is hurdles in do choose a reversal to eliminate the hurdle End while While is not sorted do choose a safe split reversal to return s Reversal Distance - Sorting by Reversals
Algorithm: Get optimal sorting sequence s that sorts Input:A signed permutation . Output: An optimal sequence of reversals sorting . Construct the breakpoint graph of . S [empty] If frt() = 1then choose a reversal to eliminate the fortress s s . [concatenate the reversal to s] End if Whilethere is hurdles in do choose a reversal to eliminate the hurdle End while While is not sorted do choose a safe split reversal to return s ComplexityO(n5) Tools: GRIMM & GRAPPA Reversal Distance - Sorting by Reversals
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?