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arXiv 1808.07098q-bio.PEmath.CO
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最多两个替换数据的最大简约树的唯一性:遗传学经典Buneman定理的推广

On the uniqueness of the maximum parsimony tree for data with up to two substitutions: an extension of the classic Buneman theorem in phylogenetics

Mareike Fischer

摘要:

进化遗传学的主要目标之一是在给定基本物种集的数据时重建正确的进化树。这些数据通常以DNA、RNA或蛋白质比对的形式出现,它们由各种字符(也通常称为位点)组成。然而,通常,基于最大简约性等标准的树重建方法可能无法为给定的数据集提供唯一的树,或者更糟糕的是,重建“错误”的树(即与生成数据的树不同的树)。另一方面,长期以来已知的是,如果比对由对应于特定树的边缘的所有字符组成,即它们都需要在该树上实现精确的$k=1$替换,则该树将通过最大简约方法恢复。这是基于数学遗传学中的布尼曼定理。这是本手稿的目标,以扩展这一经典的结果如下:我们证明,如果一个对齐的所有字符,需要完全$k=2$取代一个特定的树,这棵树将永远是唯一的最大简约树(我们还表明,这可以推广到字符,需要在最多$k=2$取代)。特别是,这也证明了一个猜想的基础上,最近发表的观察Goloboff等。肯定的特殊情况下,$k=2$。

英文摘要:

One of the main aims of phylogenetics is the reconstruction of the correct evolutionary tree when data concerning the underlying species set are given. These data typically come in the form of DNA, RNA or protein alignments, which consist of various characters (also often referred to as sites). Often, however, tree reconstruction methods based on criteria like maximum parsimony may fail to provide a unique tree for a given dataset, or, even worse, reconstruct the `wrong' tree (i.e. a tree that differs from the one that generated the data). On the other hand it has long been known that if the alignment consists of all the characters that correspond to edges of a particular tree, i.e. they all require exactly $k=1$ substitution to be realized on that tree, then this tree will be recovered by maximum parsimony methods. This is based on Buneman's theorem in mathematical phylogenetics. It is the goal of the present manuscript to extend this classic result as follows: We prove that if an alignment consists of all characters that require exactly $k=2$ substitutions on a particular tree, this tree will always be the unique maximum parsimony tree (and we also show that this can be generalized to characters which require at most $k=2$ substitutions). In particular, this also proves a conjecture based on a recently published observation by Goloboff et al. affirmatively for the special case of $k=2$.

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