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Library / Solution chemistry, equilibria and kinetics / SCH-44

Solution chemistry, equilibria and kinetics

Michaelis–Menten initial rate

Initial rate of a single-substrate enzyme reaction from the substrate concentration S, the maximum rate Vmax and the Michaelis constant Km.

Ref
SCH-44
Rev
v1
Field
Solution chemistry, equilibria and kinetics
Result
v0 [mol/(L·s)]

With a steady-state enzyme–substrate complex, the initial rate of a single-substrate reaction is v₀ = Vmax S/(Km + S). At low S the rate is proportional to S, and at high S it saturates at Vmax. At S = Km the rate is half of Vmax, which defines Km. It applies for initial rates, before the substrate or product changes appreciably, and for a single substrate.

Initial rate of a single-substrate enzyme reaction from the substrate concentration S, the maximum rate Vmax and the Michaelis constant Km. It is half of Vmax when S = Km. Steady-state intermediate and negligible depletion.

Calculator

Solve for
mol/(L·s)
mol/(L·s)
1e-810
mol/L
1e-61
mol/L
1e-61
  • Fraction of Vmax, S/(Km + S)0.75
  • Substrate for 90 % of Vmax, 9Km0.009 mol/L
  • Rate at S = Km, Vmax/20.001 mol/(L·s)

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.

Initial rate (mol/(L·s)) current value 0.0015 mol/(L·s)
VmaxMaximum rateKey driver
7.5e-97.5
SSubstrate concentration
2e-60.002
KmMichaelis constant
5.98e-60.002
  • At 0.003 mol/L of substrate the enzyme runs at 75 % of its maximum rate, 0.0015 mol/(L·s).
  • Because the returns fall as S rises, going from 10 Km to 100 Km only raises the rate from 91 % to 99 % of Vmax.
Chart data as tables

Ranked variables

VariableTypical rangeLowest resultHighest resultSwing (mol/(L·s))Local stepChange in result
Vmax Maximum rate1e-8 to 10 mol/(L·s)7.5e-97.57.5+10%+10 %
S Substrate concentration1e-6 to 1 mol/L1.998e-60.0019980.001996+10%+2.33 %
Km Michaelis constant1e-6 to 1 mol/L5.982e-60.0019990.001993+10%−2.44 %

Sweep of Substrate concentration

Substrate concentration (mol/L)Initial rate (mol/(L·s))
1e-61.998e-6
5.623e-61.118e-5
3.162e-56.131e-5
1.778e-43.02e-4
0.0010.001
0.0056230.001698
0.031620.001939
0.17780.001989
10.001998

Sweep one variable

Initial rate (mol/(L·s))Hover, touch, or use the arrow keys on the chart
05e−40.0010.00150.002 1e−61e−51e−40.0010.010.11 Substrate concentration (mol/L), log scale now 0.0015

Variables and how each one contributes

v0

Initial rate

Result · mol/(L·s)

What the equation gives you. Everything else feeds it.

Vmax

Maximum rate

mol/(L·s) · typical 1e-8 to 10

The maximum rate at saturating substrate, in mol/(L·s). Raises the rate in direct proportion. Raising it by 10 % (to 0.0022) moves the result by +10 %, from 0.0015 to 0.00165 mol/(L·s).

S

Substrate concentration

mol/L · typical 1e-6 to 1

The initial free substrate concentration, in mol/L. Raises the rate, with falling returns as saturation sets in. Raising it by 10 % (to 0.0033) moves the result by +2.33 %, from 0.0015 to 0.001535 mol/(L·s).

Km

Michaelis constant

mol/L · typical 1e-6 to 1

The Michaelis constant, in mol/L. A larger Km means weaker binding, so the rate falls. Raising it by 10 % (to 0.0011) moves the result by −2.44 %, from 0.0015 to 0.001463 mol/(L·s).

Worked example with the values above

  1. Denominator: Km + S = 0.001 + 0.003 = 0.004 mol/L.
  2. Saturation fraction: S/(Km + S) = 0.75.
  3. v₀ = 0.002 × 0.75 = 0.0015 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

Related formulas

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