10-12-06 BIMM100 Assign 3p

10-12-06 BIMM100 Assign 3p - BIMM 100 FALL 2006 Assignment...

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BIMM 100 FALL 2006 Assignment 3 1. (a) Figure 1a shows the growth curve for E. coli bacteria grown in a glucose-containing broth (culture A). Figure 1b shows the growth curve for E. coli grown in a broth that contains equal parts of glucose and lactose (culture B). Describe what is happening at the different time points and explain why the two curves are different. Figure 1(a) Figure 1(b) (b) Which gene is primarily responsible for the difference in the two curves? Describe an experiment to prove that this gene’s expression is different in the two bacteria cultures, and also that its expression is different in phase I and phase II of culture B. (c) IPTG is a chemical that mimics the action of lactose by binding to the lac repressor. The enzyme adenylate cyclase synthesizes cAMP from ATP. Explain how the bacteria growth curve will change if adenylate cyclase is overexpressed in E. coli and grown in culture A with IPTG added. What about culture B with IPTG added? How will it change the results of your
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10-12-06 BIMM100 Assign 3p - BIMM 100 FALL 2006 Assignment...

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