Tyrosinase inhibition with a drifting pH
During a biochemistry lab, a student measures mushroom tyrosinase activity with L-DOPA as substrate at 25 °C in 50 mM phosphate buffer, nominal pH 6.8. A 1.0 mL reaction contains 0.20 µg enzyme and substrate concentrations from 0.05 to 2.0 mM. Without inhibitor, the fitted parameters are Vmax = 0.80 µmol·min⁻¹ and Km = 0.25 mM. With 0.10 mM of inhibitor X added to the same enzyme amount, Vmax remains 0.80 µmol·min⁻¹ but Km rises to 0.50 mM. The student notices that the measured pH falls to 6.2 by the end of each 3-minute run because the buffer was prepared at 10 mM, not 50 mM. A separate control shows that at pH 6.2 the enzyme's catalytic constant is 30% lower, while substrate binding is unchanged. The student must decide whether inhibitor X is competitive, calculate its Ki, and determine how the pH drift affects the apparent kinetic constants if the entire run is used for fitting.
1. Classify inhibitor X and calculate Ki from the given Km values. State any assumption needed for the calculation.
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Classification. Vmax is unchanged while Km rises, so inhibitor X is competitive: it binds the free enzyme and is overcome by higher [S].
Calculation. For competitive inhibition, Km_app = Km(1 + [I]/Ki). Substitute: 0.50 = 0.25(1 + 0.10/Ki). Divide by 0.25: 2 = 1 + 0.10/Ki, so 0.10/Ki = 1 and Ki = 0.10 mM.
Assumption. The calculation assumes rapid equilibrium, one binding site, and that the added inhibitor concentration is approximately the free concentration (no significant depletion by binding). If the inhibitor were tight-binding, this estimate would need a Morrison correction.
2. If the student fits the entire 3-minute run despite the pH falling to 6.2, predict the direction of error in apparent Vmax and Km. Explain.
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At pH 6.2 the catalytic constant is 30% lower, so the enzyme's turnover is slower; fitting the whole run will underestimate Vmax. The apparent Vmax should be lower than the true pH 6.8 value by roughly the time-weighted loss of activity (not exactly 30% unless the entire run is at 6.2).
For Km, the answer is more subtle. In Michaelis-Menten terms, Km = (k-1 + kcat)/k1. If substrate binding (k1, k-1) is unchanged and only kcat falls, Km can decrease slightly, not increase. If kcat is much smaller than k-1, the change is negligible. Thus the pH drift cannot explain the observed doubling of Km; the inhibitor effect on Km is real and not an artifact of pH. The main artifact is a lower apparent Vmax.
3. Give one specific experimental change that would best isolate the inhibitor's effect from the pH drift, and explain why it controls the confound.
Read the worked answer
Use a 50 mM phosphate buffer (or a pH-stat) to hold pH at 6.8 throughout the assay, and measure initial rates only during the first 10–20 s before any drift. Matching the no-inhibitor control to the same final pH (6.2) would separate pH effects from inhibitor effects, but holding pH constant is stronger because it prevents the confound rather than correcting for it. Also include a no-enzyme blank and pre-incubate enzyme with inhibitor at 25 °C. The key is that both inhibited and uninhibited runs must experience the same pH history; otherwise the apparent Km and Vmax differences mix inhibitor and pH effects.