Library / Solution chemistry, equilibria and kinetics / SCH-45
Solution chemistry, equilibria and kinetics
Initial rate with a competitive inhibitor
Initial rate with a reversible competitive inhibitor at free concentration I and binding constant Ki, which raises the apparent Km but leaves Vmax unchanged.
- Ref
- SCH-45
- Rev
- v1
- Field
- Solution chemistry, equilibria and kinetics
- Result
- vi [mol/(L·s)]
A competitive inhibitor binds to the same site as the substrate, so it raises the apparent Michaelis constant by the factor 1 + I/Ki and leaves Vmax unchanged: vi = Vmax S/[Km(1 + I/Ki) + S]. Enough substrate can outcompete the inhibitor and return the rate to Vmax. I must be the free inhibitor concentration, which can differ from the total when the inhibitor binds tightly. At I = 0 the formula reduces to the uninhibited rate.
Initial rate with a reversible competitive inhibitor at free concentration I and binding constant Ki, which raises the apparent Km but leaves Vmax unchanged. At I = 0 it becomes the uninhibited Michaelis–Menten rate.
Calculator
- Uninhibited rate, Vmax S/(Km + S)0.0015 mol/(L·s)
- Apparent Km, Km(1 + I/Ki)0.003 mol/L
- Inhibition, 100(1 − vi/v₀)33.33 %
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- The inhibitor slows the reaction from 0.0015 to 0.001 mol/(L·s), 33.3 % inhibition, by raising Km to 0.003 mol/L.
- Raising the substrate concentration reduces the inhibition, which distinguishes competitive inhibition from the non-competitive kind, where Vmax itself falls.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (mol/(L·s)) | Local step | Change in result |
|---|---|---|---|---|---|---|
| Vmax Maximum rate | 1e-8 to 10 mol/(L·s) | 5e-9 | 5 | 5 | +10% | +10 % |
| Km Michaelis constant | 1e-6 to 1 mol/L | 1.998e-6 | 0.001998 | 0.001996 | +10% | −4.76 % |
| S Substrate concentration | 1e-6 to 1 mol/L | 6.664e-7 | 0.001994 | 0.001993 | +10% | +4.76 % |
| Ki Inhibitor binding constant | 1e-8 to 1 mol/L | 3e-8 | 0.001499 | 0.001499 | +10% | +3.13 % |
| I Free inhibitor concentration | 0 to 1 mol/L | 5.976e-6 | 0.0015 | 0.001494 | +10% | −3.23 % |
Sweep of Free inhibitor concentration
| Free inhibitor concentration (mol/L) | Inhibited rate (mol/(L·s)) |
|---|---|
| 0 | 0.0015 |
| 0.125 | 4.651e-5 |
| 0.25 | 2.362e-5 |
| 0.375 | 1.583e-5 |
| 0.5 | 1.19e-5 |
| 0.625 | 9.539e-6 |
| 0.75 | 7.958e-6 |
| 0.875 | 6.826e-6 |
| 1 | 5.976e-6 |
Sweep one variable
Variables and how each one contributes
Inhibited rate
What the equation gives you. Everything else feeds it.
Maximum rate
The maximum rate, in mol/(L·s). Raises the rate in direct proportion. Raising it by 10 % (to 0.0022) moves the result by +10 %, from 0.001 to 0.0011 mol/(L·s).
Substrate concentration
The substrate concentration, in mol/L. Raises the rate, and overcomes more of the inhibition. Raising it by 10 % (to 0.0033) moves the result by +4.76 %, from 0.001 to 0.001048 mol/(L·s).
Michaelis constant
The Michaelis constant without inhibitor, in mol/L. A larger Km lowers the rate. Raising it by 10 % (to 0.0011) moves the result by −4.76 %, from 0.001 to 9.524×10−4 mol/(L·s).
Free inhibitor concentration
The free inhibitor concentration, in mol/L. A larger inhibitor concentration lowers the rate. Raising it by 10 % (to 0.0022) moves the result by −3.23 %, from 0.001 to 9.677×10−4 mol/(L·s).
Inhibitor binding constant
The inhibitor binding constant, in mol/L. A larger Ki means weaker inhibition, so the rate rises. Raising it by 10 % (to 0.0011) moves the result by +3.13 %, from 0.001 to 0.001031 mol/(L·s).
Worked example with the values above
- Apparent Km: Km(1 + I/Ki) = 0.001 × (1 + 0.002/0.001) = 0.003 mol/L.
- Denominator: 0.003 + 0.003 = 0.006.
- vi = 0.002 × 0.003 / 0.006 = 0.001 mol/(L·s).
Assumptions and limits
- Irreversible first-order steps for the consecutive scheme, and a single substrate with a steady-state intermediate for the enzyme.
- Constant temperature, pH and volume, with negligible product and substrate depletion for initial rates.
Common mistakes
- Using the formulas after substantial substrate depletion, where the rate no longer is the initial rate.
- Mixing up k₁ and k₂ in the consecutive formulas.
- Using the total inhibitor concentration in place of the free one.
Statistics and uncertainty
Fitted enzyme constants, Km and Vmax, carry uncertainties of 10 % to 30 % from the data scatter, and a wide range of substrate concentrations is needed to define both.
Where it is used
- Following intermediates in sequential reactions and decay chains.
- Describing enzyme saturation and inhibition.
Maths you need
- Integrated rate law: concentration as a function of time
- Steady state: the intermediate’s concentration stays constant
- Michaelis constant: Km, the substrate concentration at half of Vmax