Question

You have E. coli growing in the following flasks: Flask A-contains glucose and nitrate, incubated aerobically Flask B-contains glucose and nitrate, incubated anaerobically Flask C-contains only glucose, incubated anaerobically Predict the following after 24 hours of incubation at 37oC.

A. Which flask will have the most number of cells?

B. Which flask will have the least number of cells?

C. Explain why you chose the flasks you did for each condition. Give specific features of metabolism to support your answer.

An engineered organism was created to catabolize substrate A to products B and C for an energy yield of -∆G 25kJ/mol.

A. If this reaction is coupled to another reaction that is a positive + ∆G, what is the maximum ∆G that would still allow the reaction to move forward?

B. If the concentration of products (B and C) is building up, what can you add or remove to make the reaction continue forward? Give two possible solutions [A]<[C][D]

Answer 1

A. The flask A will have the most number of the E.coli cells.

B. The flask C will have the least number of the E.coli cells.

C. E.coli is a facultative bacteria, which means that it can grow in both aerobic and anaerobic conditions. However, it gorws better in the awrobic conditions. Moreover, E.coli derives energy from the glucose, but nitrogen is also necessary for its growth. The flask A have both these conditions, and hence, it will have the most number of cells.

The flask C only contains glucose (no nitrogen) and is kept in anerobic environment. Thus, it will have least number of cells.