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对于均方根偏差为6埃的蛋白质结构进行随机预测的概率是多少?

What is the probability of a chance prediction of a protein structure with an rmsd of 6 A?

作者信息

Reva B A, Finkelstein A V, Skolnick J

机构信息

Institute of Mathematical Problems of Biology, Russian Academy of Sciences, Pushchino, Moscow Region, Russian Federation.

出版信息

Fold Des. 1998;3(2):141-7. doi: 10.1016/s1359-0278(98)00019-4.

Abstract

BACKGROUND

The root mean square deviation (rmsd) between corresponding atoms of two protein chains is a commonly used measure of similarity between two protein structures. The smaller the rmsd is between two structures, the more similar are these two structures. In protein structure prediction, one needs the rmsd between predicted and experimental structures for which a prediction can be considered to be successful. Success is obvious only when the rmsd is as small as that for closely homologous proteins (< 3 A). To estimate the quality of the prediction in the more general case, one has to compare the native structure not only with the predicted one but also with randomly chosen protein-like folds. One can ask: how many such structures must be considered to find a structure with a given rmsd from the native structure?

RESULTS

We calculated the rmsd values between native structures of 142 proteins and all compact structures obtained in the threading of these protein chains over 364 non-homologous structures. The rmsd distributions have a Gaussian form, with the average rmsd approximately proportional to the radius of gyration.

CONCLUSIONS

We estimated the number of protein-like structures required to obtain a structure within an rmsd of 6 A to be 10(4)-10(5) for chains of 60-80 residues and 10(11)-10(12) structures for chains of 160-200 residues. The probability of obtaining a 6 A rmsd by chance is so remote that when such structures are obtained from a prediction algorithm, it should be considered quite successful.

摘要

背景

两条蛋白质链对应原子之间的均方根偏差(rmsd)是衡量两个蛋白质结构相似性的常用指标。两个结构之间的rmsd越小,这两个结构就越相似。在蛋白质结构预测中,需要预测结构与实验结构之间的rmsd,当rmsd小到与紧密同源蛋白质的rmsd相当时(< 3 Å),预测才被认为是成功的。在更一般的情况下,为了评估预测的质量,不仅要将天然结构与预测结构进行比较,还要与随机选择的类蛋白质折叠结构进行比较。人们可能会问:必须考虑多少个这样的结构才能找到一个与天然结构具有给定rmsd的结构?

结果

我们计算了142种蛋白质的天然结构与这些蛋白质链在364个非同源结构上穿线得到的所有紧密结构之间的rmsd值。rmsd分布呈高斯形式,平均rmsd大致与回转半径成正比。

结论

我们估计,对于60 - 80个残基的链,要获得一个rmsd在6 Å以内的结构,需要10⁴ - 10⁵个类蛋白质结构;对于160 - 200个残基的链,则需要10¹¹ - 10¹²个结构。偶然获得6 Å rmsd的概率非常低,因此当通过预测算法获得这样的结构时,应该认为是相当成功的。

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