Therefore, to distinguish one-gene segregation from two-gene segregation we need a statistical test to distinguish 3: 1 from 15: 1. Intuitively, we know that in order to get statistical significance, we need to look at a sufficient number of individuals For a chi-square test you start with a specific hypothesis that gives a precise expectation. The test is then applied to the actual experimental results and will give the probability of obtaining the results under the hypothesis. The test is useful for ruling out hypotheses that would be very unlikely to give the actual results Say we look at 16 flies in the F2 and observe 14 not paralyzed and 2 paralyzed flies Inder the hypothesis of two genes we expect 15 not paralyzed flies and 1 paralyzed fly We calculate the value x2 using the formula below. Where o is the number of individuals observed in each class and e is the number of individuals expected for each class (OE)21212 0.067+1=1067 E 15 (all classes) degrees of freedom(df)=number of classes -1 From the table using 1 df, 0.05< p<0.5 The convention we use is that p s0.05 constitutes a deviation from expectation that is significant enough to reject the hypothesis. Therefore, on the basis of this sample of 16 flies we can't rule out the hypothesis that two genes are required Say we look at 64 F2 flies and find that 12 are paralyzed. For the hypothesis of two genes the expectation is that 4 would be paralyzed. The x2 for this data 2 =107+16=171 From the table p<0.005 so we reject the two-gene hypothesi Lets use this data to test the hypothesis of one gene segregation which would be expected to give 16 paralyzed flies from 64 F2 flies. x2=48+16=031=13 From the table using 1 df, 0.5< p<0.5. thus the data still fits the hypothesis of one gene segregationTherefore, to distinguish one-gene segregation from two-gene segregation we need a statistical test to distinguish 3 : 1 from 15 : 1. Intuitively, we know that in order to get statistical significance, we need to look at a sufficient number of individuals. For a chi-square test you start with a specific hypothesis that gives a precise expectation. The test is then applied to the actual experimental results and will give the probability of obtaining the results under the hypothesis. The test is useful for ruling out hypotheses that would be very unlikely to give the actual results. Say we look at 16 flies in the F2 and observe 14 not paralyzed and 2 paralyzed flies. Under the hypothesis of two genes we expect 15 not paralyzed flies and 1 paralyzed fly. We calculate the value χ2 using the formula below. Where O is the number of individuals observed in each class and E is the number of individuals expected for each class. 12 Σ (O–E)2 12 χ2 = = + = 0.067 + 1 = 1.067 E 15 1 (all classes) degrees of freedom (df) = number of classes – 1 From the table using 1 df, 0.05 < p < 0.5 The convention we use is that p ≤ 0.05 constitutes a deviation from expectation that is significant enough to reject the hypothesis. Therefore, on the basis of this sample of 16 flies we can’t rule out the hypothesis that two genes are required. Say we look at 64 F2 flies and find that 12 are paralyzed. For the hypothesis of two genes the expectation is that 4 would be paralyzed. The χ2 for this data: 82 82 χ2 = + = 1.07 + 16 = 17.1 60 4 From the table p < 0.005 so we reject the two-gene hypothesis. Let’s use this data to test the hypothesis of one gene segregation which would be expected to give 16 paralyzed flies from 64 F2 flies, 42 42 χ2 = + = 0.33 + 1 = 1.33 48 16 From the table using 1 df, 0.5 < p < 0.5. Thus the data still fits the hypothesis of one￾gene segregation