Enzyme-rate models without assuming a binding location
A class receives fitted rate models for a harmless hypothetical enzyme system; no experiment or chemical treatment is being prescribed. Both curves follow the simple Michaelis-Menten form v = Vmax times S divided by Km plus S over the stated measurement range. Without inhibitor, Vmax is 60 rate units and Km is 2 concentration units. With inhibitor, the fitted Vmax remains 60 while the fitted apparent Km rises to 8. The students are told to treat these as ideal reversible-model fits with comparable enzyme amounts and stable assay conditions. One student says that any inhibition must lower Vmax. Another says the fitted pattern proves the inhibitor occupies a physically identical pocket to the substrate. The class must calculate rates and distinguish a kinetic model from a direct structural observation.
1. Calculate the rates at S = 2 and identify the model pattern.
Read the worked answer
Without inhibitor, v = 60 times 2/(2+2) = 30 units. With inhibitor, v = 60 times 2/(8+2) = 12 units. The preserved limiting Vmax with increased apparent Km is consistent with simple reversible competitive kinetics under the stated model; it does not require Vmax to fall.
2. What happens at high substrate, and what cannot be established from these fits alone?
Read the worked answer
As S becomes much larger than the relevant Km, both fitted rates approach 60. Raising substrate can therefore reduce inhibition in this model. The fitted curves establish a kinetic relationship, not a unique atomic binding location: mutually exclusive substrate and inhibitor binding can arise through different structural mechanisms. Direct structural or additional mechanistic evidence would be required to identify a pocket.