Substrate for a Target Fraction of Vmax (Michaelis-Menten Inverse)
The assay-design inverse of Michaelis-Menten.
Example
You enter
- Km (substrate at half Vmax) 25
- Target velocity (% of Vmax) 90
You get
- Required substrate [S] 225
- As a multiple of Km 9 x Km
No published worked example is on file for this calculation. It is checked against a hand-worked derivation of its method and an independent reference test.
Details, formula, and sources
ScopeEnter Km and the fraction of Vmax you want, get the substrate concentration to use. [S] = Km x f/(1 - f). Half of Vmax takes exactly one Km (the definition of Km), but 90% takes 9 x Km and 99% takes 99 x Km -- the runaway excess that shows why an enzyme approaches Vmax but never reaches it. Feeds straight back into the forward calculation. Steady state, substrate in excess of enzyme; the assay conditions govern.
Formula or table[S] = Km x f/(1 - f) with f = target_velocity/Vmax; at f = 0.5, [S] = Km; fold_km = f/(1 - f).
Source and editionThe Michaelis-Menten equation inverted for the substrate concentration at a target fraction of Vmax, by name.
Free accessStandard biochemistry enzyme-kinetics relation solved for the substrate; the assay conditions govern.
What governsEstimate. AHJ and licensed professional govern.
API field nameskm, target_percent, substrate, fold_km
- Inverse of the forward tile [S] = Km f/(1 - f); feeding the result into michaelis-menten with the same Km returns the target fraction of VmaxMichaelis-Menten
- Steady state substrate far in excess of enzyme; single substrate, no cooperativity or inhibitionMichaelis-Menten