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Lecture5Oct5b - Lab this week Homework 4 is due in lab Data...

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Lab this week Homework 4 is due in lab. Data analysis Fly Lab #2 PV92 Crime lab
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CHI SQUARE ANALYSIS
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Outcome of a Dihybrid Cross +/+; +/+ p/p; e/e X +/p; +/e +/p; +/e X + ; + +; e p; + p; e P F1 Gametes
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Outcome of a Dihybrid Cross Female + ; + + ; e p ; + p ; e Male + ; + +/+; +/+ +/+ ; e/+ p/+ ; +/+ p/+ ; e/+ + ; e +/+; +/ e +/+; e/e p/+; +/e p/+; e/e p ; + +/p; +/+ +/p; e/+ p/p; +/+ p/p; e/+ p ; e +/p; +/e +/p; e/e p/p; +/e p/p; e/e Phenotypes dark bodies and pink eyes dark bodies wild type pink eyes Genotypes
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Example of Null (flies) If e and p are autosomal recessive and inherited according to Mendel’s laws of equal segregation and independent assortment, the observed ratio of progeny resulting from a dihybrid cross should not differ from the expected ratio of 9 wild-type for both traits:3 with dark bodies:3 with pink eyes:1 with a dark body and pink eyes.
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Analysis of Data Expected: 56.25 wild-type for both traits 18.75 dark bodies 18.75 pink eyes 6.25 mutant for both traits Observed: 61 wild-type for both traits 20 dark bodies 17 pink eyes 2 mutant for both traits Individual Data 100 flies
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Chi-square Analysis Formula: X 2 = Sum ((observed-expected) 2 /expected) Determine the number of degrees of freedom (number of classes of progeny expected-1) Use Table of Critical Values to determine the p-value If the p-value is less than or equal to 0.05, the null- hypothesis can be rejected.
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Chi-square analysis Wild-type: (61-56.25) 2 /56.25= 0.401 Dark bodies: (20-18.75) 2 /18.75=
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