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解决一个系统发育问题需要多少个核苷酸?一种适用于现有序列的新统计方法的应用。

How many nucleotides are required to resolve a phylogenetic problem? The use of a new statistical method applicable to available sequences.

作者信息

Lecointre G, Philippe H, Vân Lê H L, Le Guyader H

机构信息

Laboratoire d'Ichtyologie générale et appliquée (CNRS GDR 1005), Muséum National d'Histoire Naturelle, Paris, France.

出版信息

Mol Phylogenet Evol. 1994 Dec;3(4):292-309. doi: 10.1006/mpev.1994.1037.

Abstract

The evolution of bootstrap proportions (BP) with sequence length was studied using a 28S ribosomal RNA data set. For different sequence lengths, informative sites were jackknifed several times. Bootstrapping was subsequently performed on each of these subsamples. For each node, BPs so obtained were plotted against sequence length, showing the evolution of the robustness with increasing number of informative sites. For robust nodes (BP of 100%), the pattern of BPs is unvarying and is described by a simple function BP = 100 (1-e-b(x-x')), where x is the number of informative sites and b and x' are two parameters estimated using a nonlinear regression procedure. When a node has a BP < 100% and the pattern of BPs fits this function, it is possible to estimate the number of informative sites required to obtain a given average BP. The method also identifies nonrobust nodes (nonascending clusters of BP dots), for which it seems to be more cost effective and fruitful to turn to other species and/or genes rather than to continue sequencing longer gene lengths from the same species to reach a BP of 95%.

摘要

利用一个28S核糖体RNA数据集研究了自展比例(BP)随序列长度的演变。对于不同的序列长度,对信息位点进行了多次刀切法抽样。随后对这些子样本中的每一个进行自展分析。对于每个节点,将如此获得的BP值相对于序列长度作图,显示出随着信息位点数量增加稳健性的演变。对于稳健节点(BP为100%),BP的模式是不变的,并且由一个简单函数BP = 100(1 - e^(-b(x - x')))描述,其中x是信息位点的数量,b和x'是使用非线性回归程序估计的两个参数。当一个节点的BP < 100%且BP的模式符合此函数时,就有可能估计获得给定平均BP所需的信息位点数量。该方法还能识别不稳健的节点(BP点的非上升簇),对于这些节点,转向其他物种和/或基因似乎比继续从同一物种测序更长的基因长度以达到95%的BP更具成本效益且富有成效。

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