Lecture 16 & 17

Lecture 16 & 17 - Lectures 16 & 17: Mapping by...

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Keywords: Key Figures: Two point cross  Fig. 5.10 Three point cross   Single Crossover    Double Crossover    Interference Solved Problems beginning on pg. 127: S1, S2, S3A Practice problems (end of chapt. 5): C10, C11, C12, E5, E6, E8, E9, E11,  E12, E14, E15, E17, E20, E21, E23, E24
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y+ w+ m+ Recombination frequency, RF = % of recombinant types y and w; recombinant types 1.1% w and m; recombinant types 32.8% 1% RF = 1 map unit (m.u.) = 1 centimorgan (cM) y+ and m+ are 1.1 mu apart w+ and m+ are 32.8 mu apart From before:
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Two point cross: a cross involving two genes -determines the map distance between them y - w 1.1 cM w - m 32.8 cM y - m 34.3 cM y m The first genetic map. A. Sturtevant (as a sophomore in Morgan’s lab) You can build a genetic map:
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Lecture 16 & 17 - Lectures 16 & 17: Mapping by...

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