The Ames test: The two-fold rule revisited

Neal F. Cariello , Walter W. Piegorsch
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引用次数: 70

Abstract

Mutagenicity in the Ames assay is evaluated by comparing the number of revertants observed in treated cultures to those in untreated cultures. Often, some form of the ‘2-fold rule’ is employed, whereby a compound is judged mutagenic if a 2-fold or greater increase is seen in a treated culture. In order to understand the underpinnings of this approach, we study some of its statistical properties. We assume that the number of revertants on any plate from a given two-group experiment follows a Poisson distribution and we address the following questions: (1) what is the false-positive error probability of observing at least a doubling of the number of colonies from the control to the treatment group?; (2) if a given mean number of colonies is postulated for a control group, what number of colonies above the observed control mean provides a false-positive rate of 5%? We also present results for question 1 in the case where the number of revertants follows a negative binomial distribution.

艾姆斯测试:重新审视双重规则
在Ames试验中,通过比较处理过的培养物与未处理过的培养物中观察到的回变物的数量来评估致突变性。通常,采用某种形式的“2倍规则”,即如果在处理过的培养物中看到2倍或更大的增加,则判断该化合物具有诱变性。为了理解这种方法的基础,我们研究了它的一些统计特性。我们假设在给定的两组实验中,任一平板上的回复菌数量遵循泊松分布,并解决以下问题:(1)观察到从对照组到处理组的菌落数量至少增加一倍的假阳性误差概率是多少?(2)若假设一个对照组的平均菌落数,高于观察到的控制平均值的菌落数是多少,则假阳性率为5%?我们还提供了问题1的结果,在这种情况下,回复者的数量遵循负二项分布。
本文章由计算机程序翻译,如有差异,请以英文原文为准。
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