 Research
 Open Access
Genome rearrangements with duplications
 Martin Bader^{1}Email author
https://doi.org/10.1186/1471210511S1S27
© Bader; licensee BioMed Central Ltd. 2010
 Published: 18 January 2010
Abstract
Background
Finding sequences of evolutionary operations that transform one genome into another is a classical problem in comparative genomics. While most of the genome rearrangement algorithms assume that there is exactly one copy of each gene in both genomes, this does not reflect the biological reality very well  most of the studied genomes contain duplicated gene content, which has to be removed before applying those algorithms. However, dealing with unequal gene content is a very challenging task, and only few algorithms allow operations like duplications and deletions, especially if the duplicated or deleted segments are of arbitrary size.
Results
We recently proposed a heuristic algorithm for sorting unichromosomal genomes by reversals, block interchanges, tandem duplications, and deletions. In this paper, we extend this approach to multichromosomal genomes. We are now able to sort a multichromosomal ancestral genome into a genome of a descendant by a large set of different operations, including tandem duplications and deletions of segments of arbitrary size.
Conclusion
Experimental results show that our algorithm finds sorting sequences that have a weight close to the true evolutionary distance.
Keywords
 Tandem Duplication
 Synteny Block
 Black Edge
 Breakpoint Graph
 Sorting Sequence
Background
During evolution, genomes are subject to genome rearrangements, which are large scale mutations that can alter the ordering and orientation (strandedness) of the genes on the chromosomes or even change the genome content by inserting, deleting, or duplicating genes. Because these events are rare compared to point mutations, they can give us valuable information about ancient events in the evolutionary history of organisms. For this reason, one is interested in the most "plausible" genome rearrangement scenario between two genomes. More precisely, given two genomes, one wants to find an optimal sequence of rearrangements that transforms this genome into the other. In the classical approach, each gene has exactly one copy in each genome, and only operations that do not change the genome content are considered. A breakthrough in the research of these "classical operations" was Hannenhalli and Pevzner's algorithm for sorting by reversals [1], and due to further research, we are now able to sort multichromosomal genomes by reversals, translocations, fusions, and fissions in polynomial time [2, 3]. If one also considers transpositions, the problem gets more complicated, and there are only approximation algorithms known [4–6]. To simplify the existing algorithms, Yancopoulos et al. invented the double cut and join operator (DCJ), which can simulate reversals, translocations, fusions, fissions, and block interchanges (a more generalized form of transpositions), resulting in a simple and efficient algorithm [7].
However, restricting the genes to be unique in each genome does not reflect the biological reality very well, as in most genomes that have been studied, there are some genes that are present in two or more copies. This holds especially for the genomes of plants, and one of the most prominent genomes is the one of the flowering plant Arabidopsis thaliana, where large segments of the genome have been duplicated (see e.g. [8]). There are various evolutionary events that can change the content of the genome, like duplications of single genes, horizontal gene transfer, or tandem duplications. For a nice overview in the context of comparative genomics, see [9]. In a pioneering work, Sankoff tackled the challenge of genomes with duplicated genes with his "exemplar model" [10], where the following problem was examined. Given two genomes with duplicated genes, identify in both genomes the "true exemplars" of each gene and remove all other genes, such that the rearrangement distance between these modified genomes is minimized. This approach was later extended to the "matching model", where one searches for a maximum matching between the copies of each gene such that the genome rearrangement distance according to this matching is minimized [11]. However, both approaches have been proven to be NPhard for the breakpoint distance and the reversal distance [11–13]. Note that these approaches do not construct the evolutionary events that changed the genome contents, i.e. the set of operations is still restricted to the classical set of operations. The first approach that explicitly constructed duplication events was done by ElMabrouk [14], where one searches for a hypothetical ancestor with unique gene content, such that the reversal and duplication distance from this ancestor to a given descendant (with duplicated genes) is minimized. This work has been further extended during the last years (see e.g. [13, 15]). Still, the duplications were technically limited to have the length of one element, and therefore the algorithms can only be applied if no large segmental duplication happened during evolution. One idea to overcome this problem was to simulate duplications by insertions, as it has been done in [16–18]. Recently, Yancopoulos and Friedberg provided a mathematical model (but without algorithm) of a genome rearrangement distance for genomes with unequal gene content [19], combining the DCJ operator with arbitrary but lengthweighted insertions and deletions. To the best of our knowledge, the first work that allowed duplications of arbitrary segments was done by OzeryFlato and Shamir [20], who demonstrated that a simple greedy algorithm can find biologically realistic sorting scenarios for most karyotypes in the Mitelman Database of Chromosome Abberations in Cancer [21]. Further simplifications of the model led to an algorithm with provable approximation ratio of 3 [22] (note that the algorithm performs much better in practice). Recently, we proposed a heuristic algorithm for sorting a unichromosomal genome by DCJs, tandem duplications, and deletions of arbitrary segments [23]. In this work, we extend this approach to multichromosomal genomes. We are now able to sort an ancestral multichromosomal genome by a large set of operations, including duplications and deletions of arbitrary size. As a constraint, two chromosomes in the ancestral genome must be either disjoint or identical. Although this restriction seems to be very limiting, this is often fulfilled in practice. A possible application is to examine the evolution of a cancer karyotype out of a diploid set of healthy chromosomes.
Methods
Preliminaries
A chromosome is a string over the alphabet Σ = {1, ..., n}, where each element may have a positive or negative orientation (indicated by or ). We get the inverse of an element (indicated by ) by inverting its orientation. The reflection of a chromosome is the chromosome . It is considered to be equivalent to π^{ i }. A genome π= {π^{ i }, ..., π^{ m }} is a multiset of chromosomes. The multiplicity mult(π, x) of an element x is the number of its occurrences (with arbitrary orientation) in π. A segment is a consecutive sequence of elements of a chromosome. Each element x can also be represented by the ordered set of its extremities x_{ t }(the tail ) and x_{ h }(the head ), where the ordering of the extremities is x_{ t }x_{ h }if x has positive orientation, and x_{ h }x_{ t }otherwise. For example, the chromosome can also be written as (1_{ t }1_{ h }2_{ t }2_{ h }). The two extremities belonging to the same element are called coelements. We say the tail/head x_{ t/h }of an element x is a telomere if there is a chromosome in π beginning or ending with x_{t /h}. The value t(π, x_{ t }) determines how often x_{ t }is a telomere in π, the value t(π, x_{ h }) is defined analogously. Two consecutive extremities in a chromosome π^{ i }which are no coelements form an inner adjacency w.r.t. another genome ρ if they are also consecutive in a chromosome of ρ, otherwise they form an inner breakpoint. An extremity which is a telomere in π forms a telomere adjacency w.r.t. another genome ρ if it is also a telomere in ρ, otherwise it forms a telomere breakpoint. For example, if ρ = {(1_{ t }1_{ h }2_{ t }2_{ h }3_{ t }3_{ h })} and π = {(1_{ t }1_{ h }2_{ t }2_{ h }), (1_{ t }1_{ h }3_{ t }3_{ h })}, then 1_{ t }and 3_{ h }form telomere adjacencies, while 2_{ h }forms a telomere breakpoint. Applying an operation op to a genome π yields the genome op(π). A genome rearrangement problem is defined as follows. Given two genomes ρ and π and a set of possible operations, where each operation is assigned a weight, find a sequence of minimum weight (i.e. the sum of the weights of the operations is minimized) that transforms ρ into π. This minimum weight will be denoted by d(ρ, π). In our algorithm, we will consider all operations listed in Subsection "Operations". For simplification, we will assume that two chromosomes in ρ are either disjoint (i.e. they have no element in common) or identical. Furthermore, each element in Σ must appear in at least one chromosome of ρ. Note that these restrictions only hold for the genome ρ, not for π.
Operations
The following set of operations will be considered by our algorithm. A reversal inverts the order of the elements of a segment. Additionally, the orientation of each element in the segment is inverted. A transposition cuts a segment out of a chromosome, and reinserts it at another position in the same chromosome. If we additionally apply a reversal on this segment, we speak of an inverted transposition. A fusion concatenates two chromosomes. Both chromosomes π^{ i }and π^{ j }can be inverted before the operation, i.e. we can replace π^{ i }or π^{ j }by its reflection. A fission splits a chromosome into two chromosomes. A translocation splits two chromosomes π^{ i }and π^{ j }into π^{i, t}, π^{i, h}and π^{j, t}, π^{j, h}, and then concatenates π^{i, t}with π^{j, h}and π^{j, t}with π^{i, h}. Again, both chromosomes can be inverted before the operation. A tandem duplication inserts an identical copy of a segment immediately after this segment in a chromosome. A transposition duplication inserts an identical copy of a segment at an arbitrary position in the genome. A deletion cuts a segment out of a chromosome. A chromosome duplication creates an identical copy of a chromosome. A chromosome deletion deletes a chromosome.
All operations have weight 1, except for (inverted) transpositions and transposition duplications, which have weight 2. These weights are chosen rather by mathematical than biological reasons, but still result in biologically realistic scenarios (see also Section "Discussion"). To simplify the analysis of the effects of the operations in Section "A lower bound", we will there use the double cut and join operator (DCJ), which has been introduced in [7]. A DCJ cuts a genome at two positions, and rejoins the cut ends in two new pairs. It is a wellknown fact that reversals, translocations, fusions, and fission can be described by one DCJ operation, while transpositions can be described by two DCJ operations. Duplications and deletions cannot be described by DCJ operations and therefore must be examined separately.
The breakpoint graph
A lower bound
In other words, for each telomere x_{ t }(or x_{ h }) in ρ, we count how often we must create this telomere in π such that t(ρ, x_{ t }) ≤ t(π, x_{ t }) (or t(ρ, x_{ h }) ≤ t(π, x_{ h })). Additionally, we add the amount of telomeres x_{ t }and x_{ h }in π that have to be removed, i.e. they are not telomeres in ρ.
Lemma 1. If π = ρ, then C(ρ, π) is maximized, and L(ρ, π) and T(ρ, π) are minimized.
Proof. If π = ρ, the set of gray edges and the set of black edges in the breakpoint graph are equal. Thus, removing black edges does not increase the number of components, and adding black edges can only decrease the number of components. Therefore, changing π cannot increase C(ρ, π). L(ρ, π) and T(, π) are both positive numbers, and if π = ρ, they are equal to 0 and therefore minimized. □
In [23], we have shown that all operations can either change the number of components by 1, or change the number of loops by 2. These observations still hold for our slightly modified breakpoint graph. We will now examine how an operation can change T(ρ, π).
Inverse reversal, translocation, transposition
These operations can be simulated by one or two inverse DCJs (which is equivalent to a normal DCJ), thus it is sufficient to examine the effects of a DCJ (note that transpositions, which require two DCJs, have weight 2). A DCJ can only change T(ρ, π) if one of its edges is the end of a chromosome. Then, a telomere x_{t /h}is removed and a new telomere y_{t /h}is created. This decreases T (ρ, π) by 1 or 2 if x_{t /h}is not a telomere in π and y_{t /h}is a telomere in π, otherwise the operation does not decrease T(ρ, π). However, in the first case, the DCJ did not cut any black edge, as we neither draw black edges for telomeres in ρ nor for telomeres in π.
Inverse fusion (fission)
Splitting a chromosome will only decrease T(ρ, π) if both new telomeres are also telomeres in ρ. In this case, no black edge in the breakpoint graph is removed, i.e. C(ρ, π) and L(ρ, π) remain unchanged. T(ρ, π) is decreased by at most 2.
Inverse fission (fusion)
Concatenating two chromosomes can decrease T(ρ, π) by at most 2. This operation never removes a black edge, thus C(ρ, π) cannot be increased and L(ρ, π) cannot be decreased.
Inverse tandem duplication
This operation does never change the set of telomeres in π, and therefore cannot change T(ρ, π).
Inverse transposition duplication
This operation can decrease T(ρ, π) only if the duplicated segment is at a chromosome end, and the new chromosome end (after deleting the segment) is a telomere in ρ. In this case, no black edge with multiplicity 1 is removed, therefore C(ρ, π) and L(ρ, π) remain unchanged. The decrement of T(ρ, π) is ≤ 2.
Inverse deletion (insertion)
This operation can only change T(ρ, π) if we insert a segment at a chromosome end. In this case, no black edge is removed, i.e. C(ρ, π) cannot be increased and L(ρ, π) cannot be decreased. T(ρ, π) is decreased by at most 2.
Inverse chromosome duplication
This operation can decrease T(ρ, π) by at most 2 (if the telomeres of the removed chromosome are not telomeres in ρ). Only black edges with multiplicity ≥ 2 are removed, thus C(ρ, π) and L(ρ, π) remain unchanged.
Inverse chromosome deletion (chromosome insertion)
This operation can decrease T(ρ, π) by at most 2 (if the telomeres of the new chromosome are also telomeres in ρ). In the breakpoint graph, no black edges are removed, i.e. C(ρ, π) cannot be increased and L(ρ, π) cannot be decreased.
where c(ρ) = n + (number of different chromosomes in ρ), and L_{ i }(ρ, π) is the number of vertices with a loop in component C_{ i }in the breakpoint graph of ρ and π.
Proof. An operation that decreases T(ρ, π) will neither increase C(ρ, π) nor decrease L(ρ, π), therefore we can separate every sequence into operations that decrease T(ρ, π) and operations that decrease . Each operation decreases T(ρ, π) by at most 2, so we need at least operations of the first kind. Furthermore, if ρ = π, then C(ρ, π) = c(ρ) and therefore CL(ρ, π) = 0. As each operation decreases CL(ρ, π) by at most one (the proof in [23] still holds), we need at least CL(ρ, π) operations of the second kind. Therefore, any sorting sequence must have at least lb(ρ, π) operations. □
Corollary 1. lb(ρ, ρ) = 0.
For example, if ρ = and π = , then lb(ρ, π) = 0 and τ(ρ, π) = 2.
Lemma 2. If ρ = π, then τ(ρ, π) = 0. Otherwise, τ(ρ, π) > 0.
Proof. It is clear that τ(ρ, π) = 0 if ρ = π. In order to minimize τ(ρ, π), it is necessary to minimize m(ρ, π) and to maximize ia(ρ, π) and ta(ρ, π). ia(ρ, ρ) and ta(ρ, ρ) are independent of π and therefore fixed. Each inner adjacency is weighted by 2. We can interpret this as if each of the adjacent extremities is weighted by 1. Therefore, we can say that each element in π can account at most 2 to ia(ρ, π) + ta(ρ, π), and this value is reached if there is an adjacency on both sides of the element. Thus, the contribution to τ(ρ, π) of all occurrences of an element x in π is minimized if mult(ρ, x) = mult(π, x) and no extremity of x is part of any breakpoint. Every additional occurrence of x may increase ia(ρ, π) + ta(ρ, π) by 2, but also increases m(ρ, π) by 4 and therefore increases τ(ρ, π) by at least 2. This means that τ(ρ, π) is minimized if each element has the same multiplicity in ρ and π, and the breakpoint graph contains no breakpoints. This is the case if and only if ρ and π are identical. □
The algorithm
The algorithm uses a greedy strategy to sort π into ρ by inverse operations. For better readability, we will simply write "operation" instead of "inverse operation" in this section. First, we define Δlb(op) = lb(ρ, π)  lb(ρ, op(π)) and Δτ(op) = τ(ρ, π)  τ(ρ, op(π)). The score σ of an operation op is defined as the tuple σ(op) = (Δlb(op), Δτ(op)). The comparison operator between two scores is defined by σ(op_{1}) >σ(op_{2}) if Δlb(op_{1}) > Δlb(op_{2}) ∨ Δ(lb(op_{1}) = Δlb(op_{2}) ∧ Δτ (op_{1}) > Δτ (op_{2})). In each step, we search for operations that decrease the lower bound, and apply the one with the greatest score. If no such operation exists, we use additional heuristics to find operations that do not change the lower bound but have a positive score (i.e. σ (op) >(0, 0)). There is still the possibility that we even do not find such an operation. In this case, we use a fallback algorithm that is guaranteed to terminate.
Operations that decrease the lower bound
Finding operations that increase C(ρ, π) can be done by finding 1bridges and 2bridges in the breakpoint graph and verifying additional preconditions, as shown in [23]. The only difference is that now a DCJ can cut only one black edge. This is the case when the other cutting point contains a telomere in ρ or π. Thus, we also have to consider DCJs that act on a 1bridge and a telomere. Such a DCJ can be interpreted as inverse reversal, translocation, fission, or transposition. In the last case, we have to find a third cutting point in the same chromosome such that the resulting inverse transposition still increases C(ρ, π). Also finding operations that decrease L(ρ, π) is straightforward and can be done as in [23]. The remaining task is to find operations that decrease T(ρ, π). For this, we create a list of telomeres in π that are not telomeres in ρ, and another list of inner breakpoints in π where at least one of the adjacent elements is a telomere in ρ. Operations that decrease T(ρ, π) must act on one or two points of these lists, depending on the operation type. Creating the lists can be done by a linear scan over π, therefore all operations that decrease T(ρ, π) can be found in quadratic time. The only exceptions are inverse deletions and inverse chromosome deletions, which may add segments of arbitrary content. Practical tests have shown that it is better to only allow the insertion of segments without any breakpoints. This does not only lead to better sorting sequences, but also keeps the time complexity of finding the operations in O(n^{2}).
Additional operations
If there is no operation that decreases lb(ρ, π), we may still find operations that do not change the lower bound but decrease τ(ρ, π). Searching for all these operations would exceed our computing capacity, so we just search for the following subset of these operations that can be found easily.

There are inverse tandem duplications and transposition duplications that do not change σ(ρ, π), but decrement τ(ρ, π). We therefore search for identical consecutive segments that are maximal, i.e. they cannot be extended in any direction, and check the effect on σ(ρ, π) and τ(ρ, π) if we remove one of them. This operation corresponds either to an inverse tandem duplication, or to an inverse transposition duplication.

Depending on the telomeres of a chromosome, the lower bound can remain unchanged during an inverse chromosome duplication, but τ(ρ, π) can decrease. We therefore search for identical chromosomes and check the score of removing one of them.

Inserting a segment of consecutive elements x with mult(ρ, x) >mult(π, x) decreases τ(ρ, π). We search for such segments of maximal length and try to insert them by an inverse deletion. Note that this is not always possible as this operation can increase the lower bound by merging two components.

Creating inner or telomere adjacencies never increases the lower bound, but decreases τ(ρ, π). We therefore search for DCJs and inverse fissions that create new adjacencies without splitting old ones.
The fallback algorithm
It is possible that there is neither an operation that decreases lb(ρ, π), nor an operation that decreases τ(ρ, π), so the main algorithm gets stuck. However, this case cannot occur if all elements have the same multiplicity in ρ and in π.
Lemma 3. If ρ ≠ π and mult(ρ, x) = mult(π, x) holds for all elements x, then there is an operation with positive score.
Proof. When the preconditions are fulfilled, there must be at least one breakpoint in π. We have to distinguish three cases. (1) This is a telomere breakpoint. W.l.o.g. a chromosome in π ends with x_{ h }, but x_{ h }is not a telomere in ρ. Then, mult(ρ, x) = mult(ρ, x + 1) (as they are in the same chromosome), and therefore there must be another breakpoint including (x + 1)_{ t }. An operation that creates an adjacency between x_{ h }and (x + 1)_{ t }will not decrease the lower bound, but decrease τ (ρ, π) by at least 2. (2) The breakpoint is an inner breakpoint between two extremities that are telomeres in ρ. In this case, the score of cutting the chromosome at this breakpoint is (1, 2), because both extremities become telomeres (i.e. T(ρ, π) increases by 2), and we create two telomere adjacencies. (3) The breakpoint is an inner breakpoint, and at least one of the adjacent extremities is not a telomere in ρ. W.l.o.g., the breakpoint is of the form (x_{ h }, y_{ h }), and x_{ h }is not a telomere in ρ. Then, mult(ρ, x) = mult(ρ, x + 1), thus there must be another breakpoint including (x + 1)_{ t }. An operation that creates an adjacency between x_{ h }and (x + 1)_{ t }will not increase the lower bound, but decrease τ(ρ, π) by at least 2. □
The fallback algorithm will first ensure that the precondition of the lemma holds. For each chromosome ρ^{ i }in ρ, we determine the element x with the most occurrences in π. We then create maximal segments of consecutive elements ... such that each element z in the segment belongs to ρ^{ i }and mult(π, z) <mult(π, x), and add this segment by an inverse deletion to π. Note that this can be done without creating new breakpoints. This step is repeated until all elements belonging to the same component in ρ have the same multiplicity in π. We then transform ρ into ρ' by applying chromosome duplications and chromosome deletions on ρ such that for each element x, mult(ρ', x) = mult(π, x). Now, we apply our normal algorithm to sort π into ρ'. To ensure that the precondition of Lemma 3 is always fulfilled, we forbid operations that create or delete elements, i.e. any kind of duplication or deletion. Due to Lemma 3, the algorithm will find a sorting sequence that transforms π into ρ'. As last step, we have to undo the operations that transformed ρ into ρ'.
Results
We tested our algorithm on artificial data and on cancer karyotypes from the "Mitelman Database of Chromosome Aberrations in Cancer" [21].
Artificial data
Cancer karyotypes
In principle, our algorithm correctly identified all operations. The triple translocation t(3; 7; 9)(q 23; q 32; q 22) and the new chromosome +i(7)(q 10) are not allowed operations in our model. Our algorithm replaced the triple translocation by two translocations, and the new chromosome by a tandem duplication with a subsequent fission, which are the best possible explanations within our model.
Discussion
In the last sections, we have shown that our algorithm will terminate in any case, and finds rearrangement scenarios of reasonable quality. However, the weights of the operations are chosen due to a mathematical model and do not reflect the biological reality. This leaves room for further investigations. For example, the algorithm could be improved by giving unwanted operations a larger weight or completely omit them. While adapting the theoretical model to other weights seems to be the obvious way to improve the algorithm, it might also be worth to examine how robust the results are w.r.t. the chosen weights. In other words, does the optimal rearrangement scenario change when we use other weights? Some observations in the genome can be explained at best by a specific operation (e.g. a duplicated chromosome is most likely caused by a single chromosomal duplication), no matter how this operation is weighted. Such observations are predominant in closely related genomes, and the corresponding operations can be reconstructed even with a wrong weighting scheme. In more diverged genomes, there are often different possible rearrangement scenarios, and the weighting scheme matters. Thus, further investigations should examine what the "critical distance" between two genomes is, i.e. up to which distance the optimal rearrangement scenario is mostly robust w.r.t. the weighting scheme.
Conclusion
We presented an algorithm to sort multichromosomal genomes by a wide range of different operations. Although our results are promising, this algorithm should be seen as a single step towards an algorithm that produces biologically reliable results. While one direction of further research should investigate the chosen weighting scheme (see Section "Discussion"), other possible improvements are closer lower bounds or better heuristics.
Declarations
Acknowledgements
MB wants to thank Enno Ohlebusch for valuable comments.
This article has been published as part of BMC Bioinformatics Volume 11 Supplement 1, 2010: Selected articles from the Eighth AsiaPacific Bioinformatics Conference (APBC 2010). The full contents of the supplement are available online at http://www.biomedcentral.com/14712105/11?issue=S1.
Authors’ Affiliations
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