The inhibitor that lowered both numbers
An undergraduate assays a liver enzyme with substrate S and a compound X sold as a 'competitive inhibitor'. Each 1.0 mL reaction contains 0.02 µg enzyme, 50 mM buffer pH 7.4, and 1% DMSO (used to dissolve X). Initial rates at 25 °C are measured at five substrate concentrations with no inhibitor and with 5 µM X: [S] = 1, 2, 4, 8, 16 mM; v without X = 2.0, 3.3, 5.0, 6.7, 8.0 µmol min⁻¹; v with X = 1.8, 2.7, 3.8, 4.6, 5.2 µmol min⁻¹. A control lacking X but containing 1% DMSO matches the no-inhibitor line exactly. A classmate insists the data must be wrong because 'a competitive inhibitor cannot change Vmax', and proposes simply adding more substrate to rescue the rate. X has never been tested against this enzyme, which is a homodimer, and no structural information on X binding is available.
1. Classify the inhibition using a quantitative analysis of the data. Do not rely on the trend alone.
Read the worked answer
Take reciprocals and fit each line:
| 1/[S] (mM⁻¹) | 1/v no X | 1/v with X |
|---|---|---|
| 1.000 | 0.500 | 0.556 |
| 0.500 | 0.303 | 0.370 |
| 0.250 | 0.200 | 0.263 |
| 0.125 | 0.149 | 0.217 |
| 0.0625 | 0.125 | 0.192 |
No X: slope = (0.125 − 0.500)/(0.0625 − 1.000) = 0.400; intercept = 0.100. Hence Vmax = 10.0 µmol min⁻¹ and Km = 0.400 × 10.0 = 4.0 mM.
With X: slope = (0.192 − 0.556)/(0.0625 − 1.000) = 0.388; intercept = 0.168. Hence Vmax′ ≈ 5.95 µmol min⁻¹ and Km′ ≈ 0.388 × 5.95 = 2.31 mM.
The slopes are equal within reading error, but the intercept rises. Parallel double-reciprocal lines mean the inhibitor lowers Vmax and Km by the same factor (≈0.60 here). That is uncompetitive inhibition: X binds the ES complex, not free E. It is not competitive (which would leave Vmax unchanged and raise Km, giving lines crossing on the x-axis) and not pure non-competitive (which would leave Km unchanged).
2. The classmate proposes raising [S] to 200 mM to 'outcompete' X. Predict the outcome quantitatively and explain at the molecular level why the strategy fails.
Read the worked answer
For uncompetitive inhibition the inhibited rate is v_i = Vmax′[S]/(Km′ + [S]) = 5.95[S]/(2.31 + [S]).
At [S] = 200 mM: v_i = 5.95 × 200/202.31 = 5.88 µmol min⁻¹, v_0 = 10.0 × 200/204.0 = 9.80 µmol min⁻¹. Ratio = 0.60: the enzyme is still 40% inhibited, exactly as at saturating substrate.
More generally the fractional inhibition is (Vmax/Vmax′) × (Km′ + [S])/(Km + [S]); as [S] → ∞ this tends to Vmax′/Vmax = 0.60. At very low [S] the ratio tends to 1, so raising [S] makes the inhibition worse, not better.
Mechanistically, X binds only the ES complex. Adding substrate increases the concentration of the very species X binds, so the ESX pool grows. Substrate and X are not competing for the same site, so mass action cannot displace X; only lowering [I] or removing X restores the rate.
3. Design one experiment that would distinguish uncompetitive inhibition from mixed inhibition, and state the expected result for each.
Read the worked answer
Vary [X] over at least four concentrations at each of five or more [S] values, and fit the whole family of double-reciprocal lines globally.
- Uncompetitive: lines remain parallel at every [X]; only the intercept (1/Vmax_app) changes. In the general modifier scheme this means α = 1 (X does not bind free E) and α′ > 1.
- Mixed: both slope and intercept change with [X]; lines intersect at a point to the left of the y-axis (or, if α = α′, on the x-axis for pure non-competitive).
An independent binding test is stronger: equilibrate X with enzyme in the absence and presence of a saturating substrate analogue. Uncompetitive X shows no detectable binding to free E but binds in the presence of substrate; mixed X binds free E as well. Include the 1% DMSO control in every arm, because DMSO itself can perturb Km.