Enzyme Inhibition Data: When the Kinetic Pattern and the Protocol Disagree
A postgraduate student studies a purified dehydrogenase that converts substrate S to product P. Steady-state assays at 25 °C, with S varied from 0.25 to 20 mM, give a clean Michaelis–Menten fit: Km = 2.0 mM and Vmax = 120 micromol per min per mg. She repeats the series with 5 micromolar of inhibitor I1 and separately with 5 micromolar of inhibitor I2. I1 gives Km = 6.0 mM and Vmax = 120. I2 gives Km = 2.0 mM and Vmax = 48. Her report concludes that I1 is competitive, I2 is non-competitive, and recommends 'raising substrate concentration to restore full activity in the presence of I2'. A reviewer notes two loose ends: I1 was added 15 minutes before the substrate in her protocol, and enzyme pre-incubated with I1 and then dialysed against buffer recovered only 55% of its original activity. The reviewer also asks whether I2 could be an allosteric regulator rather than a dead-end inhibitor.
1. Is the classification of I1 as a competitive inhibitor supported by the data? Use the constants to quantify the inhibition and explain what the dialysis result does to the conclusion.
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The pattern fits competition: Km rises from 2.0 to 6.0 mM (alpha = 3) while Vmax is unchanged at 120. A competitive inhibitor raises apparent Km by the factor (1 + [I]/Ki), so Ki = [I]/(alpha − 1) = 5/2 = 2.5 micromolar. Substrate can out-compete it, so the ceiling is unaffected.
The dialysis result contradicts that: a reversible competitive inhibitor should wash out and restore close to 100% of activity. Recovering only 55% after pre-incubation and dialysis means a large fraction of the enzyme was inactivated irreversibly, or by very slow, tight binding. Two explanations survive: either 20 mM substrate is not enough to displace a very tight inhibitor, so the measured Vmax is underestimated; or I1 is a time-dependent inactivator and the 'competitive' shape is an artefact of the pre-incubation step.
Decisive test: pre-incubate enzyme with I1 for 0, 5, 15 and 30 minutes, then dilute 100-fold into assay. Reversible inhibition vanishes on dilution; inactivation does not. Extend substrate to at least 10 × Km before claiming Vmax is unchanged.
2. I2 leaves Km unchanged and lowers Vmax to 48. What inhibition model does this fit, and what evidence would separate a dead-end inhibitor from an allosteric regulator?
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Quantification: alpha-prime = 120/48 = 2.5, so on a pure non-competitive model Ki = 5/(2.5 − 1) = 3.3 micromolar, with the inhibitor binding E and ES about equally and lowering the catalytic constant rather than blocking substrate binding.
Why the data cannot decide: classical non-competitive inhibition is exactly what an allosteric inhibitor looks like when the substrate curve stays hyperbolic. Both models predict unchanged Km and reduced Vmax in this range.
Discriminating evidence:
- Shape of the v–[S] curve - allosteric regulators commonly produce sigmoidal kinetics (Hill coefficient above 1) under at least one condition; a dead-end inhibitor does not.
- Completeness of inhibition at saturating [I] - partial inhibition points to a separate site with its own residual activity.
- Direct binding - isothermal titration calorimetry or equilibrium dialysis showing a second, non-catalytic site with a dissociation constant consistent with the kinetics.
- Reversal by a known activator without changing [S], or site-directed mutagenesis of the putative allosteric pocket.
One clean binding experiment settles what kinetic fitting cannot.
3. Evaluate the recommendation to raise substrate concentration to overcome I2. Write the relevant rate expressions and state when substrate escalation would and would not work.
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The recommendation is wrong for I2.
Competitive inhibition (I1): v = Vmax[S] / (Km(1 + [I]/Ki) + [S]). As [S] tends to infinity the rate tends to Vmax, so substrate escalation does restore the ceiling.
Non-competitive inhibition (I2): v = (Vmax/(1 + [I]/Ki)) × [S] / (Km + [S]). The ceiling itself is divided by (1 + [I]/Ki), so at saturating [S] the rate still only reaches Vmax/2.5 = 48. No amount of substrate recovers it; adding more only wastes material and risks substrate inhibition or precipitation.
What actually works: lower or remove the inhibitor (dialysis, if the binding is reversible and the dialysis experiment is repeated properly), or change inhibitor class. Note the exception - if I2 turns out to be an allosteric regulator whose effect is reversed by an activator or a conformational change, the pharmacology is different. That is one more reason to establish reversibility and binding site before designing around the inhibitor.