Two Substrate Concentrations, One Inhibitor, One Wrong Colleague
A biochemist assays a purified hydrolase at 37 C, pH 7.4, using two substrate concentrations. Without inhibitor, the initial velocities at [S] = 2 mM and [S] = 8 mM are 30 and 60 micromol/min respectively. The identical two reactions are then run with a fixed concentration of compound X added, yielding 15 and 30 micromol/min. Total enzyme concentration, assay volume, temperature, and detection method are the same in all four runs. Substrate is present in vast excess over enzyme, so free [S] can be treated as constant, and no substrate inhibition or product inhibition is observed. A colleague glances at the numbers and says compound X must be competitive, 'because you can always out-compete a competitive inhibitor simply by adding more substrate.' There is one afternoon of instrument time left.
1. Estimate Km and Vmax from the uninhibited data, showing the algebra rather than guessing.
Read the worked answer
Use v = Vmax[S] / (Km + [S]) twice and divide.
Run 1: 30 = Vmax(2) / (Km + 2) --> Vmax = 15(Km + 2) Run 2: 60 = Vmax(8) / (Km + 8) --> Vmax = 7.5(Km + 8)
Set equal: 15Km + 30 = 7.5Km + 60 --> 7.5Km = 30 --> Km = 4 mM Then Vmax = 15(4 + 2) = 90 micromol/min.
Check: 90(2)/(6) = 30 and 90(8)/(12) = 60. Both match.
A useful sanity check on the shape: at [S] = 2 mM you are at Km/2, so v should be one-third of Vmax (30 of 90) - and it is. At 8 mM you are at 2 x Km, so v should be two-thirds of Vmax (60 of 90) - also true.
2. Determine the inhibition type from the inhibited data and evaluate the colleague's claim.
Read the worked answer
The inhibited velocities are exactly half the uninhibited velocities at both substrate concentrations (15/30 = 0.50; 30/60 = 0.50).
Fit them: with inhibitor, 15 = V'max(2)/(K'm + 2) and 30 = V'max(8)/(K'm + 8). The same algebra gives K'm = 4 mM and V'max = 45 micromol/min.
- Km unchanged (4 vs 4 mM)
- Vmax halved (45 vs 90)
That is the signature of pure noncompetitive inhibition (a mixed inhibitor with alpha = alpha' = 1): the inhibitor binds free enzyme and enzyme-substrate complex equally well and simply removes a fixed fraction of active enzyme from circulation.
The colleague is wrong. His rule is true only for competitive inhibition, where inhibitor and substrate compete for the same site and the fraction inhibited shrinks as [S] rises. Here the ratio v(inhibited)/v(uninhibited) is 0.50 at 2 mM, at 8 mM, and would be 0.50 at any [S] - the curves converge on different asymptotes (90 vs 45), so they never meet. No amount of substrate rescues the rate.
3. Predict the inhibited velocity at [S] = 16 mM and use that number to advise whether raising substrate is a workable strategy against compound X.
Read the worked answer
With inhibitor: v = 45(16) / (4 + 16) = 720 / 20 = 36 micromol/min. Without inhibitor at the same [S]: v = 90(16) / 20 = 72 micromol/min.
So quadrupling [S] from 8 to 16 mM raised the inhibited rate only from 30 to 36, and the inhibited/uninhibited ratio stayed at 0.50.
Advice: raising substrate is not a workable strategy. The inhibited velocity is ceiling-bound at V'max = 45, which is only half of the uninhibited Vmax; even at saturating [S] the best achievable is 50% of the uninhibited maximum. Worse, pushing [S] to 16 mM buys a 20% gain for a doubling of substrate cost, and the assay is already near saturation (8 mM is 2 x Km).
What would distinguish the mechanisms on paper: if compound X were competitive, you would have seen the same Vmax in both arms and a Km shifted from 4 mM to something larger (Km_app = Km(1 + [I]/Ki)), with the two curves converging at high [S]. That pattern is absent. If instrument time is short, the highest-value follow-up is a Ki determination (vary [I] at fixed, saturating [S]) rather than more substrate titration.