Repair an enzyme inventory before claiming a steady rate

Try: all free is not a steady state

Begin with the reference packet and propose 100% free enzyme. The ten nM enzyme units are conserved, but association forms ES at 20 nM/s while no ES is yet available to release product. The proposal is a possible initial composition, not a stationary one. Move the free share until ES formation equals its two removal paths. Predict before opening the worked partition.

Understand the cycle

E denotes free enzyme, S free substrate, ES their bound intermediate and P product. The original teaching cycle is E + S ⇌ ES → E + P. Binding removes free enzyme temporarily; dissociation and product release both regenerate it. One enzyme unit remains present throughout. This is a mechanism sketch, not a structural drawing of a named enzyme.

The association constant is stipulated as 2 µM⁻¹ s⁻¹, while dissociation and product release each have constant 1 s⁻¹. Free substrate is externally held. We omit product reversal, enzyme degradation and other binding states. The exercise does not simulate a closed batch or diagnose a real enzyme from a curve.

Calculate a small case

At S = 1 µM, association flux is 2E and total ES removal is 2ES, both in nM/s when enzyme concentrations are nM. Stationarity requires E = ES. Conservation then gives E + ES = 10 nM, so each is 5 nM. Association is 10 nM/s, dissociation 5 nM/s and product release 5 nM/s. Net ES change is zero even while product is formed.

Stationarity is not equilibrium. The held substrate supplies a continuing flow, and the irreversible product step is a deliberate approximation. It cannot establish an equilibrium product ratio. Away from stationarity, the displayed product flux is still a conditional instantaneous value; it is not a justified constant rate over an interval.

Experiment: substrate and enzyme are different controls

At S = 4 µM with total enzyme 10 nM, the stationary free share is 20%: E = 2 and ES = 8 nM. Four times substrate gives only 8/5 = 1.6 times the reference product flux. Explain the finite bound-enzyme inventory. Now double enzyme at the original substrate: E = ES = 10 nM and product flux doubles. Compare the fixed graph scale before concluding that the same fraction means the same amount.

Try low substrate and zero substrate. A proposal containing ES with no free substrate can release product momentarily while ES drains; it is not the stationary endpoint. Explain the sign of both enzyme derivatives and why their sum remains zero.

Apply and continue

For a new total enzyme of 12 nM at S = 2 µM, derive the free/bound partition using both balance and conservation. Then state what measurements would be needed to test the model: calibrated initial product formation, known enzyme inventory, controlled substrate and conditions, and uncertainty. Matching this public calculation does not validate a real mechanism or independent practical skill.

The AP Biology enzyme investigation compares stipulated rate curves and ambiguity. This activity instead constructs the free/bound inventory that those reduced curves leave implicit. Return to pre-equilibrium validity to contrast a rapid equilibrium approximation with stationary flux balance.

OpenStax Chemistry 2e, Catalysis provides the conceptual context of regenerated catalysts and biological enzymes. The constants, cases, diagram and calculations here are original teaching constructions, not measured G6PD data. The derived binding calculation supplies optional explanatory depth; it is not a claim that all enzyme-kinetics algebra is an AP exam requirement.