You have been screening compounds with the Ames test to determine if they are mutagenic. If a compound is mutagenic, more colonies will grow in the presence of the mutagen. You have obtained the following data from triplicate samples:
Number of mutant colonies
Control: 55, 59, 61 Mean = 58.33333 SD= 3.05505
Sample X: 52, 43, 42 Mean = 45.66667 SD= 5.507571
p - value = 0.025273
Question: Would you recommend that Sample X be included as a food additive? Why
(hint: What is the control for the Ames test?)
To determine if Sample X should be included as a food additive, we need to compare the mutant colony formation in the presence of Sample X to the control.
The control for the Ames test is a substance known to be non-mutagenic. In this case, the control yielded a mean of 58.33333 mutant colonies with a standard deviation of 3.05505.
Sample X, on the other hand, resulted in a mean of 45.66667 mutant colonies with a standard deviation of 5.507571.
To assess the statistical significance of the difference between the control and Sample X, we can calculate the p-value. The p-value represents the probability of obtaining the observed difference (or a more extreme difference) by chance alone.
In this case, the calculated p-value is 0.025273, which is less than the commonly used significance level of 0.05. This suggests that the difference between the control and Sample X is statistically significant.
Since the number of mutant colonies is greater in the presence of Sample X compared to the control, it implies that Sample X has mutagenic properties. Therefore, based on the data and statistical analysis, it is not recommended to include Sample X as a food additive, as it may pose mutagenic risks.