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Unformatted text preview: Lecture 4 Lecture 4 Lecture 4 Lecture 4 Lecture 4 From the last lecture, we followed gene segregation in a cross of a true breeding shibire fly with a wild type fly. Shibire x wild type ↓ F 1 : all not paralyzed ↓ F 2 : 3 not paralyzed : 1 paralyzed This is the segregation pattern expected for a single gene. But in an actual experiment how do we know that the phenotypic ratio is really 3 : 1 ? There is no logical way to prove that we have a 3 :1 ratio. Nevertheless, we can think of an alternative hypothesis then show that the alternative hypothesis does not fit the data. Usually, we then adopt the simplest hypothesis that still fits the data. A possible alternative hypothesis is that recessive mutations in two different genes are needed to get a paralyzed fly. In this case a true breeding paralyzed fly would have genotype: a / a , b / b Whereas wild type would have genotype: A / A , B / B F 1 : A / a B / b not paralyzed F 2 : p( a / a and b / b ) = ( 1/ 4 ) 2 = 1 / 16 p( a / a and B / – ) = 1 / 4 x 3 / 4 = 3 / 16 p( b / b and A / – ) = 3 / 16 p( A / – and B / – ) = the rest = 9 / 16 This is the classic ratio for two gene segregation 9 : 3 : 3 : 1 paralyzed For our hypothesis we should see a phenotypic ratio of 15 not paralyzed : 1 paralyzed. 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 chi-square test chi-square test chi-square test 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 F 2 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 χ 2 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. 1 2 Σ (O–E) 2 1 2 χ 2 = = + = 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 F 2 flies and find that 12 are paralyzed. For the hypothesis of two genes the expectation is that 4 would be paralyzed. The χ 2 for this data: 8 2 8 2 χ 2 = + = 1.07 + 16 = 17.1 60 4 From the table p < 0.005 so we reject the two-gene hypothesis....
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- Fall '04
- Genetics, Zygosity, Sex linkage, 1.067 15 1 degrees