Okay. So in this practice problem, we're going to be walking through the steps to using a Chi Square test. Here's the example question: You have a purple plant, and you think it's heterozygous. You don't know for sure, but you think so. If it's not heterozygous, what could it also be? It could also be homozygous, right? AA or aa, uppercase, lowercase, or just all uppercase. But you think it's heterozygous, and you want to know for sure. So, what you want to do is breed it with a homozygous recessive. This has a fancy name called a test cross. In questions like this, it prompts you to do a test cross, which means mating whatever you have with a homozygous recessive.
So, you have this purple plant, you think it's heterozygous, so you are going to mate it with a white plant that you know is homozygous recessive. After this mating, you produce 120 offspring: 55 are purple and 65 are white. Was your plant heterozygous? How do we actually go about using a Chi Square? Well, the first thing that you have to do is consider these as the observed numbers because you observed this during your experiment. You have 120 offspring, 55 are purple, and 65 are white, so these are your observed values. Remember the formula for chi-square is shown here again, which is Χ2=(O-E2−E. You have these numbers as given in the problem. But the first thing you have to do is determine the expected numbers.
If you cross a heterozygous with a homozygous recessive, first perform a Punnett square. When you mate them together, you get one-half heterozygous (which are purple) and one-half homozygous (which are white). If you have 120 offspring, how many are expected to be purple and white? The Punnett square tells us that half will be purple and half will be white. So half of 120, which is 120 divided by 2, is 60. Here are the expected numbers: 60 purple and 60 white.
So we have our observed: 55 and 65; and our expected: 60 and 60. Now we can use the chi-square formula to calculate this. You actually do it for each class. For purple (which was observed as 55 and expected as 60), when we put it into the formula, O - E is -5, squared is 25, and 25 divided by 60 is approximately 0.42. Because the formula considers the sum, we do the same thing for white: 65 observed, 60 expected, O - E is 5, squared is 25, and 25 divided by 60 is approximately 0.42. Then we add these two values to get our chi-square value, which is approximately 0.84. Do you understand everything in this step?
Now, if you're given a question like this on a test, you will be provided with a chi-square table. You don't have to memorize it. You use the chi-square to determine whether your hypothesis is true. First calculate the degrees of freedom, which equals the number of variables minus 1. In this question, there are two variables (purple and white), so degrees of freedom is 1. Now use your chi-square value, which is 0.84, to find where it fits in the table, between 0.46 and 1.07. These correspond to p-values of 0.50 and 0.30, which means there's a 50% to 30% chance. This indicates the probability of your test results being due to random chance rather than a true effect.
When you determine whether to accept or reject your null hypothesis, it's based on these probabilities. The null hypothesis states that there's no significant difference between observed and expected. So, with a p-value greater than 5% (ours range from 30% to 50%), we accept the null hypothesis, indicating that the observed results (55 purple and 65 white) do not significantly differ from the expected (60 purple and 60 white) results. If we accept the null hypothesis, we conclude with 95% confidence that the purple plant was indeed heterozygous, as there's no statistical evidence to indicate a significant discrepancy from the expected heterozygous outcome. This is crucial in remembering and affirming what you initially sought to test.
Recall that accepting the null hypothesis doesn't mean proving it true beyond all doubt but failing to reject it due to insufficient evidence saying otherwise. Your takeaway is that the purple plant's observed genotype distribution aligns with what's expected for a heterozygous, so your initial assumption stands with 95% confidence. This is how you execute and interpret a Chi-Square test correctly, ensuring you connect your conclusions back to the original research question.
