exercise 11b - Vivian Mai-Tran Lab Analysis 11b 1. ng of...

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Vivian Mai-Tran Lab Analysis 11b 1. ng of vector DNA in each μL of each Ligation Reaction Tube- 20 ng of vector DNA / 10μL = 2 ng/μL in each ligation reaction tube Total volume and ng/μL of vector DNA present in each transformation reaction tube- 50 μL bacteria + 3 μL ligation reaction = 53 μL for each transformation reaction tube (2 ng/μL)(3 μL/ 53 μL) = 0.1132 ng/μL of vector DNA in each transformation reaction tube Volume and nanogram of pBluescript DNA in each transformation reaction tube- 947 μL of LB broth + 53 μL of ligation reaction = 1000μL for each reaction tube (20 ng)(3/10) = 6 ng of pBluescript DNA present in each transformation reaction tube Nanograms of pBluescript in each dilution and 100 μL transformation reaction- Test 1:10 - (0.1132 ng/μL)(100 μL of T Trf) = 11.32 ng or 0.01132 μg Test 1:100 - (0.1132 ng/μL)(10 μL of T Trf) = 1.132 ng or 0.001132 μg Control 1 - (0.1132 ng/μL)(10 μL of C1 Trf) = 1.132 ng or 0.001132 μg Control 2 - (0.1132 ng/μL)(10 μL of C2 Trf) = 1.132 ng or 0.001132 μg
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This note was uploaded on 02/22/2010 for the course BIO 50100 taught by Professor Brand during the Fall '09 term at University of Texas at Austin.

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