- Maximum velocity Vmax
- 100µmol/min
- Michaelis constant Km
- 2mM
- Substrate concentration [S]
- 5mM
71.4286µmol/min
Open with these values71.4286µmol/min
Result: 71.4286 µmol/minThe reaction velocity is v = Vmax × [S] ÷ (Km + [S]). At [S] = Km the enzyme runs at exactly half of Vmax; far above Km it approaches Vmax and more substrate stops helping. Km and [S] must share one concentration unit — the velocity comes out in the unit of Vmax.
71.4286µmol/min
Open with these values50.0000µmol/min
Open with these values150.0000µmol/min
Open with these valuesv = (Vmax × [S]) ÷ (Km + [S])
An enzyme has no single speed. It runs at whatever rate the substrate on offer allows, and this page gives that rate for one particular offer. The equation splits cleanly in two: Vmax sets the scale, and the ratio of [S] to Km decides what fraction of that scale you get. The defaults show it. With Km = 2 mM and [S] = 5 mM the substrate sits at 2.5 times Km, so the enzyme runs at 2.5 ÷ 3.5 = 0.7143 of full speed — against a Vmax of 100 µmol/min, 71.4286 µmol/min. Any other pair with the same ratio yields the same fraction, which is why Km and [S] only ever have to match each other and never Vmax. What the result does not contain is the two numbers it was built from. Km and Vmax are measured, and measured afresh for every enzyme, substrate, temperature and pH; the page stores neither, because there is no single correct value to store. You bring them, usually from a fitted curve of your own, and the velocity is only as good as that fit. The equation then assumes a steady state: one substrate, a complex forming and breaking down at a constant level, nothing inhibiting. Two substrates, an allosteric enzyme or a competitive inhibitor all bend the same curve, and the arithmetic here is then the wrong tool rather than an imprecise one.
Km is the substrate concentration at which the reaction runs at exactly half of Vmax, so it carries the unit of [S]. A low Km means half speed is reached at little substrate, usually read as tight binding.
Only the ratio of Km and [S] enters the formula, so both have to use the same concentration unit. The velocity then comes out in whatever unit Vmax was given in.
At 500 times Km the velocity is 99.8004 against a Vmax of 100 — the curve only approaches Vmax. Adding substrate there changes almost nothing.
It holds for a single substrate, an enzyme-substrate complex forming and breaking down at a constant level, and no inhibitors. Multiple substrates, allosteric regulation or inhibition need extended models.
A low Km means the enzyme works fast.
Km is a concentration, not a rate: it says at what substrate level half of Vmax is reached. The speed itself is set by Vmax.
Doubling the substrate doubles the velocity.
At [S] = Km the enzyme is already at half of Vmax, and at 500 times Km it reaches 99.8004 out of 100. The curve flattens as it approaches Vmax.
Vmax in µmol/min forces Km and [S] into mM.
Only Km and [S] have to match each other. Enter both in µM and the velocity still comes out in the unit of Vmax.
| Vmax, Km, [S] | Substrate against Km | Velocity (µmol/min) |
|---|---|---|
| 100, 2, 0 | no substrate | 0.0000 |
| 100, 2, 2 | equal to Km | 50.0000 |
| 50, 1, 1 | equal to Km | 25.0000 |
| 100, 2, 5 | 2.5 times Km | 71.4286 |
| 200, 5, 15 | 3 times Km | 150.0000 |
| 10, 0.5, 10 | 20 times Km | 9.5238 |
| 100, 2, 1000 | 500 times Km | 99.8004 |
Multiply Vmax by the substrate concentration and divide by the sum of Km and that concentration: v = (Vmax × [S]) ÷ (Km + [S]). With Vmax = 100 µmol/min, Km = 2 mM and [S] = 5 mM the velocity is 500 ÷ 7 = 71.4286 µmol/min.
Km is the substrate concentration at which the reaction runs at exactly half of Vmax. A low Km means the enzyme reaches half speed at little substrate, usually read as tight binding; a high Km means more substrate is needed.
Vmax is the maximum rate the enzyme reaches when every active site is occupied. Beyond that point more substrate no longer speeds the reaction up, and the velocity only approaches Vmax.
Km and the substrate concentration must use the same concentration unit, because only their ratio enters the formula. The velocity then comes out in whatever unit Vmax was given in — here µmol/min against mM.
A single substrate, a steady state in which the enzyme-substrate complex forms and breaks down at a constant level, and no inhibitors. Multiple substrates, allosteric regulation or inhibition need extended models.
Information, not professional advice.
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