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Evolution through segmental duplications and losses: a Super-Reconciliation approach.
Algorithms for Molecular Biology ( IF 1.5 ) Pub Date : 2020-05-26 , DOI: 10.1186/s13015-020-00171-4
Mattéo Delabre 1 , Nadia El-Mabrouk 1 , Katharina T Huber 2 , Manuel Lafond 3 , Vincent Moulton 2 , Emmanuel Noutahi 1 , Miguel Sautie Castellanos 1
Affiliation  

The classical gene and species tree reconciliation, used to infer the history of gene gain and loss explaining the evolution of gene families, assumes an independent evolution for each family. While this assumption is reasonable for genes that are far apart in the genome, it is not appropriate for genes grouped into syntenic blocks, which are more plausibly the result of a concerted evolution. Here, we introduce the Super-Reconciliation problem which consists in inferring a history of segmental duplication and loss events (involving a set of neighboring genes) leading to a set of present-day syntenies from a single ancestral one. In other words, we extend the traditional Duplication-Loss reconciliation problem of a single gene tree, to a set of trees, accounting for segmental duplications and losses. Existency of a Super-Reconciliation depends on individual gene tree consistency. In addition, ignoring rearrangements implies that existency also depends on gene order consistency. We first show that the problem of reconstructing a most parsimonious Super-Reconciliation, if any, is NP-hard and give an exact exponential-time algorithm to solve it. Alternatively, we show that accounting for rearrangements in the evolutionary model, but still only minimizing segmental duplication and loss events, leads to an exact polynomial-time algorithm. We finally assess time efficiency of the former exponential time algorithm for the Duplication-Loss model on simulated datasets, and give a proof of concept on the opioid receptor genes.

中文翻译:

通过分段重复和损失进行演变:一种超和解方法。

经典的基因和物种树和解用于推断基因得失的历史,解释了基因家族的进化,并假设每个家族都有独立的进化。尽管此假设对于基因组中相距较远的基因是合理的,但不适用于分组为同义模块的基因,这更可能是协调进化的结果。在这里,我们介绍了超和解问题,该问题包括推断一段复制和丢失事件(涉及一组邻近基因)的历史,从而导致单个祖先导致一组当今的同义。换句话说,我们将传统的单基因树的重复-丢失对账问题扩展到一组树,以解决分段重复和丢失问题。超和解的存在取决于单个基因树的一致性。另外,忽略重排意味着存在也取决于基因顺序的一致性。我们首先表明,重构最简约的超和解(如果有)的问题是NP难的,并给出了精确的指数时间算法来解决。或者,我们表明,考虑到进化模型中的重排,但仍仅使分段重复和丢失事件最小化,可以得出精确的多项式时间算法。最后,我们在模拟数据集上评估了复制-损失模型的前指数时间算法的时间效率,并给出了关于阿片受体基因的概念证明。忽略重排意味着存在也取决于基因顺序的一致性。我们首先表明,重构最简约的超和解(如果有)的问题是NP难的,并给出了精确的指数时间算法来解决。或者,我们表明,考虑到进化模型中的重排,但仍仅使分段重复和丢失事件最小化,可以得出精确的多项式时间算法。最后,我们在模拟数据集上评估了复制-损失模型的前指数时间算法的时间效率,并给出了关于阿片受体基因的概念证明。忽略重排意味着存在也取决于基因顺序的一致性。我们首先表明,重构最简约的超和解(如果有)的问题是NP难的,并给出了精确的指数时间算法来解决。或者,我们表明,考虑到进化模型中的重排,但仍仅使分段重复和丢失事件最小化,可以得出精确的多项式时间算法。我们最终在模拟数据集上评估了复制-损失模型的前指数时间算法的时间效率,并给出了关于阿片受体基因的概念证明。我们表明,在进化模型中考虑重排,但仍仅使分段重复和丢失事件最小化,导致了精确的多项式时间算法。最后,我们在模拟数据集上评估了复制-损失模型的前指数时间算法的时间效率,并给出了关于阿片受体基因的概念证明。我们表明,在进化模型中考虑重排,但仍仅使分段重复和丢失事件最小化,导致了精确的多项式时间算法。最后,我们在模拟数据集上评估了复制-损失模型的前指数时间算法的时间效率,并给出了关于阿片受体基因的概念证明。
更新日期:2020-05-26
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