The idea this lesson is working on — Students expect enough substrate to overcome any inhibitor. The book derives the rate equation, then the lab floods two inhibited samples: one climbs toward the uninhibited curve and the other stays under a ceiling of 33. Ask: what would have to be true of the inhibitor for substrate to push it out? The idea to reach is that the binding site decides which constant moves.
Km and Vmax say different things — Michaelis and Menten’s 1913 paper on invertase is often cited with a constant near 0.0167 M; they assumed rapid equilibrium binding. Briggs and Haldane’s 1925 steady-state treatment gives the same form with Km = (k₋₁ + kcat)/k₁, which equals the dissociation constant only when kcat is small. The factor of ten per 5.7 kJ/mol is RT ln 10 at 298 K.
Inhibitors distinguished by their effect — Pure non-competitive inhibition, binding enzyme and complex equally, lowers Vmax and leaves Km unchanged; most real cases are mixed. Fomepizole, a competitive inhibitor of alcohol dehydrogenase, has largely replaced ethanol as the antidote for methanol and ethylene glycol poisoning.