Question

Examining Genetic Alleles in Fast-Twitch Muscles Exercise 7.24 discusses a study investigating the \(A C T N 3\) genotypes \(R R, R X,\) and \(X X .\) The same study also examines the \(A C T N 3\) genetic alleles \(R\) and \(X,\) also associated with fast-twitch muscles. Of the 436 people in this sample, 244 were classified \(R\) and 192 were classified \(X .\) Does the sample provide evidence that the two options are not equally likely? (a) Conduct the test using a chi-square goodnessof-fit test. Include all details of the test. (b) Conduct the test using a test for a proportion, using \(H_{0}: p=0.5\) where \(p\) represents the proportion of the population classified \(R .\) Include all details of the test. (c) Compare the p-values and conclusions of the two methods.

Short answer

The calculation results in a chi-square goodness-of-fit test of approximately 0.07 and a test for a proportion of approximately 0.0062. The results suggest that the two alleles are not equally likely.

Step-by-step solution

Chi-square goodness-of-fit test

The null hypothesis \(H_0\) is that the two alleles are equally likely, so we expect \(E = n/2 = 436/2 = 218\) for both alleles. The observed values are \(O_R = 244\) and \(O_X = 192\). We calculate the chi-square statistic \(\chi^2 = \Sigma (O-E)^2/E = (244-218)^2/218 + (192 - 218)^2/218 = 3.2752\). The degree of freedom is \(df = k - 1 = 2-1 = 1\) where k is the number of classifications.

Find P-value for Chi-square test

Using chi-square distribution table or chi-square calculator with \(\chi^2 = 3.2752\) and \(df = 1\), we find that the P-value is around 0.07.

Conducting test for a proportion

In this test, conduct a one sample z-test for a proportion. The null hypothesis \(H_0: p = 0.5\), indicating that R and X are equally likely. Let p represent the observed proportion of individuals classified as R, \(p = R/n = 244/436 = 0.5596\). The standard error (SE) of the proportion is \(\sqrt{ p(1-p)/n } = \sqrt{ 0.5 * (1 - 0.5)/436 } = 0.02385\). The Z-score is \((p-0.5)/SE = 0.0596/0.02385=2.5\).

Find P-value for the proportion test

Using a z-table or z-score calculator, you can find the P-value associated with a Z = 2.5 which is approximately 0.0062

Compare p-values and draw conclusion

P-value for chi-square test is 0.07, and for proportion test it's 0.0062. Both p-values are less than a typical significance level (\(\alpha = 0.05\)), so we reject the null hypothesis in both tests. Therefore, there is sufficient evidence to reject the null hypothesis that the proportions of R and X are equal.

Interpret results

In conclusion, this sample provides evidence that the two alleles are not equally likely in the population.

Key concepts

Genetic Alleles

When examining genetic alleles, like the ACTN3 gene variants associated with fast-twitch muscle fibers, it's important to understand that alleles are different forms of a gene. In the exercise, the focus is on the R and X alleles of the ACTN3 gene.

Each individual inherits two alleles for each gene, one from each parent. The combination of these alleles determines the trait – in this case, the predisposition for fast-twitch muscles. The expected frequency of these alleles within a population can be modeled and compared with observed data to provide insights into genetic distribution and diversity.

In our example, a sample of 436 individuals showed a division between the R allele and the X allele. The next step, after obtaining such observed frequencies, is often to apply a statistical test, like the chi-square goodness-of-fit test, to determine if the observed distribution significantly differs from what was expected.

Test for a Proportion

A test for a proportion is employed when we want to analyze whether the observed proportion of a sample fits a specified population proportion. This type of hypothesis test evaluates how likely it is to observe our sample data given an assumption about the population proportion.

In the context of our genetics exercise, the null hypothesis is that 50% of the population has the R allele (H_0: p = 0.5). The sample proportion, however, can be different from this assumed value. We use statistical tools to determine if this difference is due to random sample variability or if it's statistically significant. The test computes a Z-score that measures how many standard errors the observed proportion is away from the assumed population proportion. A large absolute value of the Z-score suggests that the observed proportion is unlikely to have occurred by chance alone and may prompt us to reject our null hypothesis.

Statistical Significance

Statistical significance is a key concept that helps to determine if the results of a statistical test reflect a genuine effect or if they could have occurred by random chance. It's evaluated using a p-value, which is the probability of observing results as extreme as those in the study, assuming the null hypothesis is true.

In the context of the sample problem, two p-values were obtained from the chi-square goodness-of-fit test and the test for the proportion. Comparing these p-values to a chosen significance level, typically α = 0.05, allows us to decide whether to reject the null hypothesis or not. A p-value less than α indicates that the observed results are unlikely to have happened due to chance, and thus we deem them statistically significant.

Null Hypothesis

The null hypothesis, denoted as H_0, is a statement used in statistics that suggests there is no effect or no difference. It serves as a starting point for statistical significance testing. For example, our null hypothesis might claim that two genetic alleles occur with equal probability in a population.

In the exercise on genetic alleles, the null hypothesis posits that the proportion of the R allele is 0.5, meaning that it is equally likely to occur as the X allele. The goal of the statistical tests is to challenge this presumption and to determine if the evidence from our sample is strong enough to reject the null hypothesis. If we do reject it, we're saying that our sample provides enough evidence to support the existence of a difference in the proportion of alleles, steering us away from the initial assumption of equivalence.