In other words the genes are linked 29 the chi square

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Unformatted text preview: btained from a breeding experiment to determine the total number of offspring, the number of different classes of offspring and the number of offspring observed in each class 2. Calculate the number of offspring that would be expected for each class if the null hypothesis were correct (percentage predicted x total number of offspring) 30 The Chi Square Test 3. χ2 = Σ (Number observed-Number expected)2 Number expected 4. Determine the degrees of freedom (df) : number of classes -1 5. Use the chi-square value and number of degrees of freedom to determine a p value : the probability that a deviation from the predicted numbers at least as large as that observed in the experiment will occur 31 by chance (Table 5.1 in text) The Chi Square Test 6. Evaluate the significance of the p value. The convention is that a 0.05 p value is the boundary between accepting and rejecting the null hypothesis. Very small p values indicate a high degree of significant difference 0.05 = 5% chance that the observation is due to chance alone (in other words, a 95% probability that the observation is not due to chance. It is a REAL difference) 32 chromatids from Ps F1 Are genes A and B linked in testcross AB/ab x ab/ab ? F2 progeny AaBb aabb Aabb aaBb observed 17 14 8 11 50 expected no linkage Null hypothesis: 2 gene pairs are assorting independently Hypothesis predicts # of parental phenotypes = # of recombinant phenotypes (1/2 parental class, 1/2 recombinant) Expected values would be 25 for each class phenotype! 33 parentals Are genes A and B linked in testcross AB/ab x ab/ab ? progeny AaBb aabb Aabb aaBb observed 17 14 8 11 50 expected …if unlinked, then independent assortment: Ab AB ab aB F1 AaBb test cross aabb ab Ab ab R AB ab P ab ab P aB ab R 34 Are genes A and B linked in testcross AB/ab x ab/ab ? progeny AaBb parental class aabb Aabb recombinant aaBb class observed expected 17 25 31 14 8 25 19 11...
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