Classifying an Inhibitor from Two Velocity Points
A purified dehydrogenase is assayed in 1.0 mL at 25 degrees C. Without inhibitor, Vmax = 100 micromol/min and Km = 5.0 mM for its substrate. A candidate drug is added to 10 micromolar (well below enzyme concentration effects), and the team measures only two points: at [S] = 5.0 mM, v = 33 micromol/min; at [S] = 10 mM, v = 50 micromol/min. Literature binding data give a Ki of 10 micromolar against the free enzyme. Plasma substrate concentration in patients is about 5.0 mM, and the substrate's solubility limit in the assay buffer and in plasma is 80 mM. The team must classify the inhibition, judge whether 'just add more substrate' is a real strategy, and predict the signature of a noncompetitive inhibitor with the same Ki.
1. Show that the two velocity points are consistent with competitive inhibition and extract the apparent Km in the presence of drug.
Read the worked answer
Uninhibited reference
- v = Vmax[S]/(Km + [S]) = 100 x 5.0/(5.0 + 5.0) = 50 micromol/min
- So the uninhibited enzyme is already at half-maximal velocity at 5.0 mM.
Inhibited data
- At [S] = 10 mM, v = 50, which is exactly Vmax/2 (Vmax is unchanged only if inhibition is competitive).
- Half-maximal velocity occurs when [S] = Km_app, so Km_app = 10 mM (a 2-fold increase).
- Cross-check at [S] = 5.0 mM: v = 100 x 5.0/(10 + 5.0) = 33.3, matching the measured 33.
Consistency with the Ki
- Km_app = Km(1 + [I]/Ki) = 5.0(1 + 10/10) = 10 mM. Agreement is quantitative.
Judgment The distinguishing feature is that velocity still rises above the inhibited plateau as [S] increases (33 -> 50). A pure noncompetitive inhibitor would have capped velocity regardless of [S]. Competitive inhibition: same Vmax, larger Km.
2. Evaluate the proposal to overcome inhibition by raising substrate concentration in patients.
Read the worked answer
What would be required
- To reach 90% of Vmax with Km_app = 10 mM, you need [S] = 9 x Km_app = 90 mM.
Constraint check
- Solubility limit is 80 mM, below 90 mM. At 80 mM the fraction of Vmax is 80/(10 + 80) = 88.9% - close but not 90%, and uncomfortably near precipitation.
- At the actual plasma level of 5.0 mM, velocity is 33% of Vmax; merely doubling substrate to 10 mM gets only 50%.
Verdict Substrate escalation has severely diminishing returns: the dose-response saturates hyperbolically, so each increment buys less. It is not a practical route to near-full activity, and precipitation risk plus off-target effects argue against it. A better strategy is to lower [I] (dose reduction, better pharmacokinetics) or find an inhibitor that does not compete at the substrate site.
3. Predict the velocities a noncompetitive inhibitor with the same Ki and [I] would produce at [S] = 5.0 and 10 mM, and state the diagnostic signature.
Read the worked answer
Noncompetitive (pure) case
- Vmax_app = Vmax/(1 + [I]/Ki) = 100/2 = 50 micromol/min
- Km unchanged at 5.0 mM
Velocities
- [S] = 5.0 mM: v = 50 x 5.0/(5.0 + 5.0) = 25
- [S] = 10 mM: v = 50 x 10/(5.0 + 10) = 33.3
Contrast with the observed drug
- Observed drug: 33 at 5 mM and 50 at 10 mM (velocity climbs toward the full Vmax of 100).
- Noncompetitive: 25 at 5 mM and 33 at 10 mM (ceiling is 50 no matter how much substrate is added).
Diagnostic signature A double-reciprocal (Lineweaver-Burk) plot gives parallel lines: same x-intercept (-1/Km), raised y-intercept (1/Vmax_app). Competitive inhibition instead gives lines that intersect on the y-axis (same Vmax, shifted x-intercept). Conclusion: because substrate can partially rescue the observed inhibition, it is competitive, not noncompetitive.