Enzyme Kinetics Simulator

Worked example: competitive vs. noncompetitive, side by side

Same baseline enzyme in both cases, Vmax = 100 µM/min, Km = 25 µM, [S] = 25 µM, inhibitor at [I] = 50 µM with Ki = 50 µM (all tool defaults), so r = [I]/Ki = 1.

Both inhibitors are present at the same concentration and affinity (same r), but they change completely different things: competitive inhibition doubles the apparent Km and leaves Vmax alone, while noncompetitive inhibition halves Vmax and leaves Km alone. At this particular [S] (equal to the uninhibited Km), that difference in mechanism is why competitive gives a higher velocity (33.33) than noncompetitive (25.00) even though both use the same inhibitor concentration and Ki.

Set the simulator to these exact values (Vmax=100, Km=25, [S]=25, I=50, Ki=50) and switch between inhibition types to reproduce every number above.

Catalytic efficiency: k_cat and k_cat/K_m

V_max and K_m describe a whole reaction mixture, but V_max depends on how much enzyme you happened to use. k_cat (the turnover number) fixes that: it's V_max divided by total enzyme concentration, so it's a property of one enzyme molecule at saturation, how many substrate molecules it converts per second when it's never waiting around for more substrate to bind. k_cat/K_m (the specificity constant) goes a step further, and it's usually the single number a question is really asking for when it says "how efficient is this enzyme": it combines how fast the enzyme is (k_cat) with how well it grabs onto substrate in the first place (1/K_m) into one value, and it's what actually matters at the low, non-saturating substrate concentrations most enzymes see inside a cell.

This uses whichever V_max and K_m are set above (the uninhibited, intrinsic values, not the apparent ones under inhibition, since k_cat/K_m describes the enzyme's own chemistry, not a particular inhibitor).

Diffusion-limited ("catalytically perfect") enzymes, catalase and triose phosphate isomerase are the two classic textbook examples, reach k_cat/K_m around 10⁸–10⁹ M⁻¹s⁻¹, the physical speed limit set by how fast enzyme and substrate can diffuse into contact with each other in solution. At that point, essentially every collision between enzyme and substrate leads to product, the enzyme's chemistry is no longer the bottleneck, diffusion is. Most real enzymes fall well short of this, in the 10⁴–10⁷ M⁻¹s⁻¹ range.

One more use for k_cat/K_m: it's exactly how "which substrate does this enzyme prefer" gets answered quantitatively. Given two possible substrates, whichever one has the higher k_cat/K_m is the one the enzyme actually processes faster when both are present at low, competing concentrations, regardless of which one has the lower K_m alone.

FAQ

How can I tell inhibition type from a Lineweaver-Burk plot?
Same y-intercept (shared 1/Vmax) = competitive. Same x-intercept (shared −1/Km) = noncompetitive. Parallel lines, no intersection = uncompetitive.

Why doesn't competitive inhibition change Vmax?
The inhibitor competes for the same active site as substrate. At high enough [S], substrate outcompetes the inhibitor often enough to still reach the same maximum velocity. It just takes more substrate to get there, which is why apparent Km rises while Vmax stays put.

Can real inhibition be a mix of these types?
Yes, competitive, uncompetitive, and pure noncompetitive are special cases of the general "mixed inhibition" model, where the inhibitor's affinity for free enzyme and for the ES complex can differ. This tool's α and α′ factors are exactly that framework; pure noncompetitive is just the case where both affinities are equal.

Why do some textbooks separate "noncompetitive" from "mixed"?
Strictly, pure noncompetitive requires equal affinity for E and ES (Ki = Ki′). When the affinities differ, both Km and Vmax change, and some textbooks reserve "mixed inhibition" for that broader case instead of calling it noncompetitive.

What's the actual difference between Km and kcat/Km?
Km alone only tells you binding affinity, a low Km means tight binding, but says nothing about how fast the enzyme actually converts that bound substrate into product. kcat/Km folds both binding (1/Km) and turnover speed (kcat) into a single efficiency number, which is why it (not Km alone) is what you compare to judge which of two enzymes, or two substrates, is genuinely better.

Does competitive inhibition change kcat/Km?
Yes, even though it leaves Vmax (and therefore kcat) completely unchanged. Competitive inhibition raises the apparent Km, and since kcat stays fixed while Km(app) goes up, the apparent kcat/Km falls, the enzyme's intrinsic chemistry hasn't changed, but its practical efficiency at that inhibitor concentration has.

Practice problems

1. Vmax = 100 µM/min, Km = 25 µM, [S] = 50 µM. With uncompetitive inhibition at [I] = 100 µM and Ki = 50 µM, find v.

2. Vmax = 100 µM/min, Km = 25 µM, [S] = 100 µM. With competitive inhibition at [I] = 50 µM and Ki = 25 µM, find v.

3. An enzyme has Vmax = 60 µM/min and Km = 15 µM in an assay with [E]total = 2 nM. Find kcat and kcat/Km, and say roughly how efficient this enzyme is compared to a diffusion-limited one.

3.33×10⁷ M⁻¹s⁻¹ is a genuinely fast, well-optimized enzyme, but it's still roughly 3–30× below the 10⁸–10⁹ diffusion limit, so there's still real room for tighter binding (lower Km) or faster turnover (higher kcat) before collisions with substrate become the bottleneck instead of the enzyme's own chemistry.