Here is the question:

If gene frequencies are in equilibrium, the genotypes, AA, Aa, and aa occur in a population with frequencies (1-0)^2, 20 (1-0), and 0^2. Plato et al.published the following data on haptoglobin type in a sample of 190 people:

Hp1-1 = 10
Hp1-2 = 68
Hp2-2 = 112

Test the goodness of fit of the data to the genetic model of this problem. Use Pearsons chi-square test. You may use the following information: the MLE estimator for 0 is 0^mle = 2X^3 + X^2 / 2n = 0.7684.

To test the goodness of fit of the data to the genetic model, we will use the Pearson's chi-square test. The chi-square test compares the observed frequencies with the expected frequencies under the assumed genetic model.

To conduct the chi-square test, we need to calculate the expected frequencies for each genotype. Given the gene frequencies, we can calculate the expected frequencies as follows:

Expected frequency for AA genotype = (1-0)^2 * total sample size
Expected frequency for Aa genotype = 2 * 0 * (1-0) * total sample size
Expected frequency for aa genotype = 0^2 * total sample size

Given the sample size of 190 people, we can calculate the expected frequencies:

Expected frequency for AA genotype = (1-0)^2 * 190 = 190
Expected frequency for Aa genotype = 2 * 0 * (1-0) * 190 = 0
Expected frequency for aa genotype = 0^2 * 190 = 0

Now, we can set up our null hypothesis (H0) and alternative hypothesis (H1) for the chi-square test:

H0: The observed frequencies are consistent with the expected frequencies under the genetic model.
H1: The observed frequencies are not consistent with the expected frequencies under the genetic model.

Next, we calculate the chi-square statistic:

chi-square = sum[(observed frequency - expected frequency)^2 / expected frequency]

Using the observed frequencies and expected frequencies, we can calculate the chi-square statistic:

chi-square = [(10 - 190)^2 / 190] + [(68 - 0)^2 / 0] + [(112 - 0)^2 / 0] = infinitely large

Now we need to compare the calculated chi-square statistic with the critical chi-square value at a given significance level (usually chosen as 0.05). If the calculated chi-square value exceeds the critical chi-square value, we reject the null hypothesis.

However, in this case, the observed frequencies in the Aa and aa genotypes are zero, which creates problems in calculating the chi-square statistic. This violates the assumptions of the chi-square test, as it requires non-zero observed and expected frequencies for all categories.

To address this, we can perform an exact test, such as Fisher's exact test, which can handle zero frequencies. In Fisher's exact test, we can directly calculate the p-value based on the observed frequencies.