Lecture06-14Apr11

Lecture06-14Apr11 - Chiasmata = Crossovers Alfred...

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BIS 101-002 14 Apr 08 Allele and Gene Interactions
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BIS 101 14 Apr 11 Midterm I Review Session: Wednesday April 20, 6:10-8:00 pm Giedt 1003 Genetic Linkage, Genetic Mapping
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Epistasis Analysis Starting Point: Mutations in 2 genes that function in the same pathway and have different single mutants phenotypes. gene A encodes Enzyme A gene B encodes Enzyme B Yellow Hair Brown Hair Black Hair Enzyme A Enzyme B A_;B_ Determining the order that genes function in a pathway by determining the phenotype of double mutants Double mutant phenotype: aa;bb = Single mutant phenotypes: A_ = black hair aa = B_ = black hair bb =
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Recessive Epistasis (9:3:4) 9: B_;E_ 3: bb;E_ 3:B_;ee + 1:bb;ee Two genes, in dependent pathway.
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Dominant Epistasis (12:3:1) Two genes in dependent pathway.
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Molecular Model for Suppression (one example:dominant suppressor)
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Molecular Model for Synthetic Lethality (one example)
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Example of Redundancy fgf24 -/- wild-type fgf24 + ;fgf8 + fgf24;fgf8 + fgf24 + ;fgf8 fgf24;fgf8
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Penetrance and Expressivity
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Incomplete Penetrance of a Dominant allele
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Unformatted text preview: Chiasmata = Crossovers Alfred Sturtevant QuickTime and a decompressor are needed to see this picture. QuickTime and a decompressor are needed to see this picture. Barbara McClintock QuickTime and a decompressor are needed to see this picture. Maize QuickTime and a decompressor are needed to see this picture. Chiasmata Two Factor Cross Three Factor Cross Chi Square analysis In biological applications, a probability of 5% is usually adopted as the standard. There is only a 5% chance that, given your hypothesis, you would land outside +/- 2 SD. Accept Hypothesis Reject Hypothesis Dihybrid cross (9:3:3:1) Dihybrid cross YyRr X YyRr 9:3:3:1 F 1 : All round yellow: RrYy F 2 : Phenotype Observed Expected obs-exp round yellow 315 312.7 2.2 round green 108 104.3 3.7 wrinkled yellow 101 104.3 -3.3 wrinkled green 32 34.7 -2.7 total: 556 556 X 2 = 0.015 + 0.13 + 0.1 + 0.21 = 0.455 df = 3 P= 0.9 :. Accept hypothesis....
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Lecture06-14Apr11 - Chiasmata = Crossovers Alfred...

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