Enzyme catalysis, kinetics and inhibition

An enzyme changes a reaction’s route and rate

Enzymes are biological macromolecular catalysts. Most are proteins; catalytic RNA molecules are also included. Catalysis can provide an alternative reaction route with different activation barriers, increasing the approach to equilibrium without changing that reaction’s equilibrium under the same conditions.

Substrate identity, enzyme state and conditions belong to a rate description. An enzyme may require restoration of its active state after a cycle; catalytic turnover does not justify claiming that every enzyme is chemically unchanged in every exceptional reaction.

A thermodynamically favorable transformation can remain slow. Conversely, a fast catalytic route does not make an unfavorable net transformation thermodynamically favorable. Coupled processes and maintained reservoirs need a fuller reaction account.

A rate equation defines its parameters and conditions

For an invented single-substrate initial-rate model, use v=VS/(K_m+S), with S free substrate concentration and V the limiting rate at the specified active enzyme amount and fixed conditions. Stipulate an applicable quasi-steady range, negligible product, a supplied substrate reservoir and unchanged assay volume. This is not the initial transient or a complete cellular pathway model.

Assign V=12.0 μmol min⁻¹ and K_m=2.00 mmol L⁻¹. At S=0.200, 2.00 and 20.0 mmol L⁻¹, rates are respectively 1.09090909, 6.00 and 10.9090909 μmol min⁻¹. At S=K_m the equation gives half the limiting rate. At finite positive S the uninhibited rate is below V.

In this example assign active enzyme amount 0.200 μmol and one catalytic centre per active enzyme entity. Then k_cat=V/n_active=60.0 min⁻¹=1.00 s⁻¹. Amount/time divided by active amount is a first-order turnover quantity. A preparation’s total protein mass would not automatically supply that active-site inventory.

K_m has concentration units and belongs to the specified Michaelis–Menten function. It is not generally an equilibrium binding K_d. A saturating curve can be compatible with several mechanisms; additional evidence is needed to identify elementary steps.

A competitive rate pattern can preserve the limiting rate

Use an operational competitive model v=VS/(αK_m+S), where α=1+I/K_ic and I is the maintained free inhibitor concentration. Assign I=6.00 μmol L⁻¹ and K_ic=3.00 μmol L⁻¹, so α=3.00. The inhibitor ratio is dimensionless even though substrate and inhibitor are expressed using different concentration scales.

With the previous V and K_m, apparent K_m becomes 6.00 mmol L⁻¹ while V remains 12.0 μmol min⁻¹. At S=2.00 mmol L⁻¹, v=12×2/(6+2)=3.00 μmol min⁻¹.

At S=20.0 mmol L⁻¹ the inhibited rate is 9.23076923, still below the uninhibited 10.9090909 μmol min⁻¹. The shared high-substrate asymptote does not mean the inhibition is exactly absent at a finite concentration. The operational pattern does not alone prove a unique physical binding site, and tight-binding depletion would require another model.

Mixed inhibition changes two apparent parameters

Stipulate v=VS/(αK_m+α′S). Rearranging gives apparent limiting rate V_app=V/α′ and apparent K_m,app=αK_m/α′. Assign α=3.00 and α′=2.00. Then V_app=6.00 μmol min⁻¹, K_m,app=3.00 mmol L⁻¹ and the rate at S=2.00 mmol L⁻¹ is 2.40 μmol min⁻¹.

For the special pure noncompetitive case α=α′=3.00, V_app=4.00 while K_m,app remains 2.00. For the stipulated uncompetitive case α=1, α′=3.00, V_app=4.00 and K_m,app=0.6666667 mmol L⁻¹. These distinct parameter pairs come from the defined function.

“Noncompetitive” has been used inconsistently, so state the actual case and rate equation. In multi-substrate systems the operational classification can also depend on which substrate is varied and the conditions of the others. None of these assigned functions alone locates binding groups or counts sites.

Faster equilibration can leave the same final composition

Consider a separate closed ideal first-order toy A ⇌ B with forward rate coefficient k_f=0.400 min⁻¹, reverse k_r=0.100 min⁻¹, total concentration 10.0 mmol L⁻¹ and initial B=0. Give A and B the same standard-state convention. At equilibrium k_f c_A=k_r c_B, so c_B/c_A=4 and equilibrium A/B concentrations are 2.00/8.00 mmol L⁻¹.

The time relation is c_B(t)=8.00[1−exp(−0.500t)] mmol L⁻¹ with t numerically in minutes. At t=2.00 min, c_B=5.05696447 mmol L⁻¹. A stipulated catalytic change that multiplies both coefficients by ten reaches that same composition at 0.200 min, while retaining the same equilibrium ratio four.

Changing only the forward coefficient by ten would give a ratio forty and a different equilibrium; it would not represent this same simple catalytic scaling under unchanged thermodynamic conditions. The toy excludes enzyme-binding sequestration and external energy input. It distinguishes rate information from equilibrium without claiming to model a real enzyme.

A local assay does not determine system-level flux

An initial-rate assay can characterize an operational response under its stated substrate, product and enzyme conditions. A biological network also has transport, competing reactions, changing pools and regulatory interactions. A half-response concentration or catalytic turnover therefore does not automatically identify the step controlling the entire network.

Use the mathematical model only within its declared conditions, check the active catalytic-centre inventory and distinguish measured quantities from assigned teaching inputs. The original curves here are calculated rate patterns, not enzyme data or cellular performance measurements.

Worked examples

Try it yourself

1. Double the active enzyme amount with the same conditions and maintained substrate. Find limiting V, K_m and rates at the three specified S values.

   Show the answer and reasoning

   V=24.0 μmol min⁻¹; K_m=2.00 mmol L⁻¹; rates 2.18181818, 12.0 and 21.8181818 μmol min⁻¹.

   The assigned model makes V proportional to active centres; k_cat and the substrate half-response remain unchanged.

2. At S=20.0 mmol L⁻¹, compare the competitive α=3 rate with the uninhibited rate. Is the inhibition exactly gone?

   Show the answer and reasoning

   9.23076923 versus 10.9090909 μmol min⁻¹; it remains present at that finite S.

   Use both denominators. Equal limiting asymptotes do not imply equal finite-concentration rates.

3. Give V_app, K_m,app and rate at S=2.00 for pure noncompetitive α=α′=3 and uncompetitive α=1, α′=3.

   Show the answer and reasoning

   Pure noncompetitive: 4.00 μmol min⁻¹, 2.00 mmol L⁻¹, rate 2.00. Uncompetitive: 4.00, 0.6666667, rate 3.00 in corresponding units.

   Substitute into V_app=V/α′ and K_m,app=αK_m/α′, then check with the original denominator.

4. Multiply both A⇌B rate coefficients by ten. Find equilibrium B and time to the earlier two-minute composition. Would scaling only k_f preserve K?

   Show the answer and reasoning

   Equilibrium B remains 8.00 mmol L⁻¹; the same composition occurs at 0.200 min. Scaling only k_f changes the ratio to forty.

   The common scaling preserves k_f/k_r but speeds its relaxation. A one-sided coefficient change is a different model.

What to carry forward

• A catalyst changes approach to equilibrium under unchanged thermodynamic conditions.
• A limiting rate and active-centre amount define the assigned turnover.
• K_m is not generally a binding dissociation constant.
• Specify the operational inhibition equation and substrate conditions.
• A local rate model does not establish network control.