Experimental Exercise 2: Using Chi-square Analysis to Understand Epistatic Inher
ID: 150681 • Letter: E
Question
Experimental Exercise 2: Using Chi-square Analysis to Understand Epistatic Inheritance Patterns in Maize 5. Use Chi -square to test which of 2 possible ratios is the most probable ratio of progeny kernels. Use this information to determine the cross type and parental genotype. Cross type does not refer to the type of epistasis!
Colorless aleurone Ratios Degrees Chi Accept freedom valuerejeet Colored aleurone Ear code of square Cross type purple red yellow white tested E 257 1109 Parental genotype(s) for ear E (5 pts)Explanation / Answer
12:3:1 Complete dominance at both gene pairs; however, when one gene is dominant, it hides the phenotype of the other gene Dominant epistasis
15:1 Complete dominance at both gene pairs; however, when either gene is dominant, it hides the effects of the other gene Duplicate dominant epistasis
2 test is simply a way of quantifying the various deviations expected by chance if a hypothesis is true. For example, consider a simple hypothesis that a certain plant is a heterozygote (monohybrid) of genotype A/a. To test this hypothesis, we would make a testcross to a/a and predict a 1:1 ratio of A/a and a/a in the progeny. Even if the hypothesis is true, we do not always expect an exact 1:1 ratio. We can model this experiment with a barrel full of equal numbers of red and blue marbles. If we blindly removed samples of 100 marbles, on the basis of chance we would expect samples to show small deviations such as 52 red: 48 blue quite commonly and larger deviations such as 60 red:40 blue less commonly. The 2 test allows us to calculate the probability of such chance deviations from expectations if the hypothesis is true. But, if all levels of deviation are expected with different probabilities even if the hypothesis is true, how can we ever reject a hypothesis? It has become a general scientific convention that a probability value of less than 5 percent is to be taken as the criterion for rejecting the hypothesis. The hypothesis might still be true, but we have to make a decision somewhere, and the 5 percent level is the conventional decision line. The logic is that, although results this far from expectations are expected 5 percent of the time even when the hypothesis is true, we will mistakenly reject the hypothesis in only 5% of cases and we are willing to take this chance of error.
Let’s consider an example taken from gene interaction. We cross two pure lines of plants, one with yellow petals and one with red. The F1 are all orange. When the F1 is selfed to give an F2, we find the following resu
What hypothesis can we invent to explain the results? There are at least two possibilities:
Hypothesis 1. Incomplete dominance
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Hypothesis 2. Recessive epistasis of r (red) on Y (orange) and y (yellow)
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The statistic 2 is always calculated from actual numbers, not from percentages, proportions, or fractions. Sample size is therefore very important in the 2 test, as it is in most considerations of chance phenomena. Samples to be tested generally consist of several classes. The letter O is used to represent the observed number in a class, and E represents the expected number for the same class based on the predictions of the hypothesis. The general formula for calculating 2 is as follows:
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To convert the 2 value into a probability, we use Table 4-1, which shows 2 values for different degrees of freedom (df). For any total number of progeny, if the number of individuals in two of the three phenotypic classes is known, then the size of the third class is automatically determined. Hence, there are only 2 degrees of freedom in the distribution of individuals among the three classes. Generally, the number of degrees of freedom (shown as the different rows of Table 4-1) is the number of classes minus 1. In this case, it is 31=2. Looking along the 2-df line, we find that the 2 value places the probability at less than 0.025, or 2.5 percent. This means that, if the hypothesis is true, then deviations from expectations this large or larger are expected approximately 2.5 percent of the time. As mentioned earlier, by convention the 5 percent level is used as the cutoff line. When values of less than 5 percent are obtained, the hypothesis is rejected as being too unlikely. Hence the incomplete dominance hypothesis must be rejected.
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