Competitive Enzyme Inhibition Calculator
Work out how much a competitive inhibitor slows an enzyme, the apparent Km it creates and the substrate that undoes it.
What it does
Many drugs work by blocking an enzyme, and the commonest way to block is to pretend to be the substrate. The inhibitor has a shape close enough to the real thing to fit into the active site, and the two then compete for the same place. That is competitive inhibition, and it is the model behind the action of a good share of the drugs that are prescribed today.
The biochemistry text behind this page sets out the kinetics. When the substrate and the inhibitor bind to the same site they do it in a mutually exclusive way, and the rate with inhibitor present becomes the maximum rate times the substrate concentration, divided by the Michaelis constant multiplied by one plus the inhibitor concentration over the inhibition constant, plus the substrate concentration.
Two features of that equation matter more than the rest. The maximum rate is unchanged, because at high substrate concentration every enzyme molecule still ends up working, just more slowly. The Michaelis constant is multiplied by one plus the inhibitor concentration over Ki, and that multiplied value is the apparent Km, which is what a Lineweaver-Burk plot shows as a steeper line with the same intercept.
This page runs the whole calculation from a maximum rate, a Michaelis constant, an inhibition constant, a substrate concentration and an inhibitor concentration. It returns the rate with and without the inhibitor, the apparent Km, the percentage inhibition, the inhibitor concentration that would halve the rate at that substrate level, and the substrate concentration that would undo the inhibition completely.
The limits are honest. The model assumes a single binding site, fast reversible equilibrium and no time dependence of the inhibitor, and it says nothing about partial inhibition, allostery or the way a real enzyme is used in a cell. A measured Km that does not fit the model is a sign that the mechanism is something else.
How to use
- Enter the maximum rate Vmax in micromoles per minute.
- Enter the Michaelis constant Km in millimoles, taken from a fit of the uninhibited reaction.
- Enter the inhibition constant Ki in millimoles. The lower it is, the tighter the inhibitor binds.
- Enter the substrate concentration you are interested in, and the inhibitor concentration you have added.
- Read the two rates to see the size of the slowdown, and the percent inhibition for a single number.
- Read the apparent Km to see how much extra substrate the enzyme now needs, and the substrate that restores the original rate.
Formula
- v
- Initial reaction rate with the inhibitor present, in micromoles per minute.
- Vmax
- Maximum rate of the reaction with the enzyme saturated by substrate.
- S
- Substrate concentration, in millimoles.
- I
- Inhibitor concentration, in millimoles.
- Ki
- Inhibition constant, the inhibitor concentration that binds half of the free enzyme.
The inhibitor concentration that halves the rate at a given substrate level is Ki times one plus S over Km, and the substrate concentration that restores the uninhibited rate is S times one plus I over Ki. Both follow from rearranging the same equation, and both are worth knowing when you are designing an experiment or reading one.
Worked example
Example: a 0.2 mM inhibitor against an enzyme with Km 0.5 mM
Defaults: a maximum rate of 100 umol per min, Km 0.5 mM, Ki 0.2 mM, 0.5 mM of substrate and 0.2 mM of inhibitor.
- Without the inhibitor the rate is Vmax times S over Km plus S, which is 100 x 0.5 / 1 = 50.0000 umol per min.
- The factor that multiplies Km is one plus I over Ki, which is 1 + 0.2 / 0.2 = 2.
- So the apparent Km becomes 0.5 x 2 = 1.0000 mM.
- The inhibited rate is 100 x 0.5 over 1.0 + 0.5, which is 33.3333 umol per min.
- That is a slowdown of (50 - 33.3333) / 50 = 33.33 percent inhibition.
- Half inhibition at this substrate level needs I = Ki x (1 + S / Km) = 0.2 x 2 = 0.4000 mM of inhibitor.
- The tool shows 50.0000, 33.3333, 1.0000, 33.33 %, 0.4000 and 1.0000 mM, which is the substrate that restores the original rate.
Reference table
| Inhibitor | Apparent Km | Rate | Inhibition |
|---|---|---|---|
| 0 mM | 0.5000 mM | 50.0000 umol per min | 0.00 % |
| 0.1 mM | 0.7500 mM | 40.0000 umol per min | 20.00 % |
| 0.2 mM | 1.0000 mM | 33.3333 umol per min | 33.33 % |
| 0.4 mM | 1.5000 mM | 25.0000 umol per min | 50.00 % |
| 1.0 mM | 3.0000 mM | 14.2857 umol per min | 71.43 % |
Limits
- The model assumes a single binding site and a fast reversible equilibrium, which many real inhibitors break.
- It is the model for competitive inhibition only. Non-competitive and uncompetitive inhibition change Vmax as well, and the numbers here do not describe them.
- The Ki and Km values are fitted numbers that carry their own error, and the calculated inhibition inherits that error.
- It ignores enzyme concentration, so it says nothing about how much enzyme the pathway needs, and nothing about reversibility over time.
Common mistakes
- Expecting the maximum rate to drop with a competitive inhibitor. It does not, because extra substrate outcompetes the inhibitor.
- Reading the apparent Km as a change in the real Km. The enzyme has not changed, the inhibitor has made more substrate necessary.
- Mixing units. If Km is in millimoles and Ki in micromoles the factor one plus I over Ki comes out a hundred times too large.
- Forgetting that a low Ki means a powerful inhibitor, so a Ki much smaller than Km stops the enzyme even at low substrate.
- Using the percent inhibition at one substrate level as a general property of the inhibitor, when it falls as the substrate rises.
Try it yourself
Quick checks. Answers sit tucked under each question.
What happens if the inhibitor concentration is doubled to 0.4 mM?
The apparent Km rises to 1.5000 mM, the rate falls to 25.0000 umol per min and the inhibition reaches 50.00 percent. This is the inhibitor level that gives half inhibition at 0.5 mM of substrate, which matches the Ki times one plus S over Km rule.
How much substrate is needed to get back to 50 umol per min?
1.0000 mM, which is double the original 0.5 mM. The substrate has to grow by the same factor of one plus I over Ki, which is 2 here, that the apparent Km grew by, and the rate then returns to its old value.
What does an inhibitor with a Ki of 0.01 mM do to this enzyme?
It is a far tighter binder than the substrate, because the Ki is twenty times smaller than the Km. At 0.2 mM of inhibitor the factor one plus I over Ki becomes 21, the apparent Km rises to 10.5 mM and the rate at 0.5 mM of substrate drops to about 4.5 umol per min, so the enzyme is almost stopped.
Questions
What is competitive inhibition?
It is inhibition in which the inhibitor and the substrate bind to the same site on the enzyme, so they compete. The substrate and the inhibitor are said to bind in a mutually exclusive way.
What happens to Vmax and Km?
Vmax is unchanged, because a high substrate concentration still drives the reaction at full speed. The apparent Km rises to Km times one plus I over Ki, which is why the Lineweaver-Burk plot keeps the same intercept and gets a steeper slope.
What is the difference between Km and Ki?
Km is the substrate concentration that gives half of Vmax. Ki is the inhibitor concentration that binds half of the free enzyme, so it measures how tightly the inhibitor holds on. The course notes behind this page describe Ki as the same kind of number for inhibitors.
How do I find the inhibitor concentration that halves the reaction?
It is Ki times one plus S over Km. At a substrate level of 0.5 mM and a Km of 0.5 mM that is twice the Ki, so 0.4000 mM of inhibitor gives the half rate.
Can competitive inhibition be overcome?
Yes, and that is what makes it interesting in a cell. Raising the substrate concentration outcompetes the inhibitor, and the reaction speed comes back, which is why a competitive inhibitor slows an enzyme rather than switching it off.
How do I tell competitive inhibition from other kinds?
By the shape of the double reciprocal plot. A competitive inhibitor leaves the intercept at one over Vmax and moves the x-intercept closer to zero, while non-competitive and uncompetitive inhibition change the intercept as well.
Sources
- Biology LibreTexts, fundamentals of biochemistry 6.4: enzyme inhibition, with the competitive inhibition equation
- Chemistry LibreTexts, biochemistry 5.4: enzyme inhibition and the increased apparent Km
- University of California Davis lecture notes: enzyme inhibition kinetics and how to calculate Ki