- We calculate the Chi-squared statistic using the formula \(\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}\).
- Substituting our data, we find \(\chi^2 = \frac{(2-1)^2}{1} + \frac{(28-20)^2}{20} + \frac{(80-89)^2}{89} \approx 5.11\).
- Comparing this to the critical value of \(3.84\) for \(df = 1\) at \(\alpha = 0.05\), we find \(\chi^2 > 3.84\).
- Therefore, the population is \(\boxed{\text{not in Hardy-Weinberg equilibrium}}\).
Explanation
- Analyze the data and identify the approach
To determine if the population is in Hardy-Weinberg equilibrium, we compare the observed genotype counts to the expected counts provided in the data. The data presents three categories of genotypes: CC, Cc, and a third listed as CC, which we interpret as cc to represent the standard genetic model.
The observed counts (\(O_i\)) are: \(O_{CC} = 2\), \(O_{Cc} = 28\), and \(O_{cc} = 80\).
The expected counts (\(E_i\)) are: \(E_{CC} = 1\), \(E_{Cc} = 20\), and \(E_{cc} = 89\).
The total number of individuals is \(N = 1 + 20 + 89 = 110\).
- Calculate the Chi-squared statistic
We use the Chi-squared (\(\chi^2\)) goodness-of-fit test to see if the differences between observed and expected values are statistically significant. The formula for the test statistic is:
\[\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}\]
We calculate the contribution of each genotype:
- For CC: \(\frac{(2 - 1)^2}{1} = \frac{1^2}{1} = 1\)
- For Cc: \(\frac{(28 - 20)^2}{20} = \frac{8^2}{20} = \frac{64}{20} = 3.2\)
- For cc: \(\frac{(80 - 89)^2}{89} = \frac{(-9)^2}{89} = \frac{81}{89} \approx 0.9101\)
Summing these gives the total \(\chi^2\) statistic:
\[\chi^2 = 1 + 3.2 + 0.9101 = 5.1101\]
- Compare with critical value and conclude
For this test, the degrees of freedom (\(df\)) is given by the number of categories minus 1 minus the number of parameters estimated from the data. With 3 categories and 1 parameter estimated, \(df = 3 - 1 - 1 = 1\).
The critical value for a Chi-squared distribution with \(df = 1\) at the standard significance level of \(\alpha = 0.05\) is \(3.84\).
Since our calculated \(\chi^2\) value of approximately \(5.11\) is greater than the critical value of \(3.84\), we reject the null hypothesis. This indicates that the observed genotype frequencies deviate significantly from the expected values under Hardy-Weinberg equilibrium. Therefore, the population is not in Hardy-Weinberg equilibrium.
Examples
In genetics, the Hardy-Weinberg equilibrium is a fundamental principle used to determine if evolutionary forces like natural selection, mutation, or non-random mating are acting on a population. By applying the Chi-squared test, researchers can quantify whether the observed distribution of traits in a group of organisms deviates from what would be expected by random genetic combinations, helping them identify if a population's genetic structure is stable or evolving over time. For instance, if a researcher observes \(\chi^2 = 5.11\) with a critical value of \(3.84\), they can mathematically justify the conclusion that the population is changing.