Escherichia coli bacteria can use the enzyme lactase to digest lactose and use it as an energy source. However, it is more efficient for bacteria to use glucose as an energy source if it is present. Which of the methods of regulating enzyme activity would be most energy efficient for a bacterium that has plenty of glucose?

A. The bacterium should produce an irreversible, allosteric inhibitor of lactase.
B. The bacterium should produce a reversible, competitive inhibitor of lactase.
C. The bacterium should produce a reversible, allosteric inhibitor of lactase.
D. The bacterium should produce an irreversible, competitive inhibitor of lactase.
E. The bacterium should stop production of lactase.
F. The bacterium should use feedback inhibition to stop the metabolic pathway that lactase starts.

E. coli bacteria can degrade lactose, but lactose is not preferred as source of fuel or energy to glucose. If glucose is present, E. coli would preferably employ it over lactose as Glucose needs little process and minimal energy to degrade when compared to lactose. Although, if lactose is the major sugar that is present, the E. coli will have no option than to employ it as it's source of fuel or energy.

The formation of lactase enzyme utilizes energy, which cannot be utilised in the presence of high level glucose.