LAB 6 - LAB 6 NOTE: These labs are known for high failure...

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LAB 6 NOTE: These labs are known for high failure rates. Therefore, if I show an example using data, a graph, etc… yours may be totally different depending upon mine and yours success with the lab. So don’t freak if your data is completely different from mine. And therefore the analysis of results sometimes may need to be answered differently since you base your analysis off your data and not mine. DISCLAMIER: DO NOT CHEAT! Analysis of Results 1. LB+ lawn LB- lawn LB/AMP+ 10 LB/AMP- 0 2. Compare and contrast the number of colonies on each of the following pairs of plates. What does each pair tell you about the experiment? a. LB+ and LB- both have lawns since there was no antibiotic in the agar. b. LB/AMP- and LB/AMP+ have ampicillin. The negative did not contain colonies because they were not ampicillin resistant. For the positive there were colonies because some cells took up plasmid and became ampicillin resistant. c. LB/AMP+ contained 10 colonies because some cells were transformed. LB+ did not have colonies, only a lawn due to no ampicillin. 3. …excess DNA may actually interfere with the transformation process. a. Determine the total mass of pAMP used. Note: the (m) used here is supposed to be the special symbol but I can’t find it on insert symbols. Volume X concentration = 10 ml X .005 mg/ml = .05 mg pAMP used. b. Calculate the total volume of cell suspension prepared. 250 ml CaCl2 + 10 ml pAMP + 250 ml LB = total volume of 510 ml. c. Now calculate the fraction of the total cell suspension that was spread on the plate. Number of ml 100 / total volume 510 ml = fraction = 20%.
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LAB 6 - LAB 6 NOTE: These labs are known for high failure...

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