Questions: Microbial Substrate Utilization and Metabolic Induction

Question 1

E. coli is placed in a medium containing both glucose and lactose in equal amounts. What growth pattern would you observe over time?

• A. E. coli immediately metabolizes both sugars simultaneously, maximizing growth rate from the start
• B. E. coli grows exponentially on glucose first (catabolite repression prevents lac operon induction), pauses briefly to induce the lac operon after glucose is exhausted, then resumes growth on lactose
• C. E. coli preferentially metabolizes lactose because it provides more ATP per mole than glucose
• D. E. coli enters a prolonged lag phase because competing regulatory signals from glucose and lactose create a stalemate

Answer: (B)

This is the classic diauxic growth curve. While glucose is present, high glucose levels suppress adenylate cyclase, keeping cAMP low. Low cAMP means the CAP protein cannot activate the lac promoter, so the lac operon is not induced even if lactose is present and the repressor has been released by allolactose. Only after glucose is depleted does cAMP rise, allowing CAP to activate the lac promoter. The brief pause between the two exponential phases represents the lag time needed to synthesize the lac enzymes de novo. Options A and C reflect the misconception that bacteria simultaneously optimize all substrates.

Question 2

The lac operon is fully induced when lactose is present in the growth medium and the lac repressor has been released from the operator. Why might the operon still fail to be transcribed at high levels?

• A. Because allolactose degrades faster than it can accumulate, preventing repressor release
• B. Because glucose may still be present, keeping cAMP levels low and preventing CAP from activating the lac promoter — both repressor removal AND CAP activation are required for full induction
• C. Because the lac repressor requires glucose as a co-repressor to remain inactive
• D. Because lactose permease cannot import lactose when glucose is being metabolized simultaneously

Answer: (B)

This question targets the two-part logic of lac operon regulation. Repressor removal (by allolactose) and CAP activation (by cAMP-CAP binding the promoter) are independent requirements — both must be satisfied for full transcription. When glucose is present, adenylate cyclase is inhibited and cAMP levels drop. Even if allolactose has released the repressor, without cAMP the CAP protein cannot bind the promoter and RNA polymerase transcribes the operon inefficiently. This dual control ensures that lactose enzymes are only produced when lactose is present AND glucose is absent — an elegant priority system.

Question 3

Substrate induction controls enzyme concentration (Vmax) rather than the activity of existing enzyme molecules, making it a regulatory mechanism that operates at a slower timescale than allosteric control but allows a much larger range of adjustments.

• T. True
• F. False

Answer: True

Allosteric regulation modulates the activity of enzymes already present — it changes kcat or apparent Km in seconds. Substrate induction changes how much enzyme is synthesized — controlling Vmax — which takes minutes (for transcription, translation, and protein folding). This is a coarser but potentially much larger-magnitude response: allosteric regulation might reduce activity 2-10 fold, while induction can change enzyme concentration from near zero to thousands of molecules per cell. The two mechanisms complement each other: allosteric regulation responds to rapid fluctuations; induction adjusts capacity for sustained changes in available substrates.

Question 4

To avoid wasting energy, bacteria constitutively produce most their catabolic enzymes at low baseline levels so they are ready to exploit any available substrate immediately.

• T. True
• F. False

Answer: False

This is the opposite of microbial metabolic strategy. Inducible enzyme systems exist precisely because producing unnecessary enzymes is energetically costly — ribosomes, amino acids, and ATP are consumed to make proteins that provide no current benefit. The adaptive solution is to make catabolic enzymes ONLY when their substrate is present (substrate induction) and to prioritize the best substrate (catabolite repression). A bacterium maintaining constitutive expression of dozens of catabolic enzyme sets would be at a severe fitness disadvantage relative to one that produces each set on demand.

Question 5

Why is catabolite repression adaptive for bacteria? What problem would arise without it?

Model answer: Catabolite repression creates a hierarchy of substrate preference — glucose first, then less preferred sugars — by linking cAMP levels (and thus CAP-activated transcription) inversely to glucose availability. This is adaptive because glucose is the most efficiently metabolized carbon source for E. coli: it yields the most ATP per unit metabolic investment. Without catabolite repression, a bacterium in a glucose-plus-lactose environment would wastefully produce lac enzymes (and enzymes for other alternative sugars), consuming energy and ribosomes to make proteins that provide no net benefit while glucose is available. It might also dilute its metabolic capacity by running multiple catabolic pathways simultaneously at partial efficiency. Catabolite repression solves this by ensuring the best available substrate is fully exploited before resources are committed to alternatives.

The diauxic growth curve is the observable consequence of catabolite repression — not a deficiency in bacterial flexibility, but an evolved metabolic economy that prioritizes the most valuable substrate. The two-phase growth is the experimentally visible signature of substrate hierarchy.