Michaelis-Menten Kinetics

Model enzyme kinetics with the Michaelis-Menten equation. Compare inhibition effects.

Worked example — Michaelis-Menten Kinetics

Vmax (apparent)	100.00
Km (apparent)	5.00
Plot type V vs [S] Lineweaver-Burk	5.0[S]V

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Saturation is asymptotic, and brutally slow 🖖

The curve approaches V_max but never arrives, and the arithmetic of how slowly is worth having in front of you. Since v/V_max = [S]/(K_m + [S]), reaching half speed costs exactly one K_m of substrate. Reaching 90% costs nine. Reaching 99% costs ninety-nine. Set [S] to K_m in the tool and read 0.5·V_max; multiply [S] by a hundred and you still have not touched the ceiling. That is why V_max can never be measured directly — it is read off a fit, never off a reading, and it is the reason the reciprocal plot in the next block was invented at all.

Straightening the curve to read it 🖖

The tool's Lineweaver-Burk plot flips the equation into 1/v = (K_m/V_max)(1/[S]) + 1/V_max, turning the bending saturation curve into a straight line. Before curve-fitting software existed, this let scientists extend the line with a ruler and read the constants off directly: the y-intercept gives 1/V_max and the x-intercept gives -1/K_m. The catch is that low [S] points, with their huge reciprocals, dominate the fit and exaggerate measurement error.

The same curve hides everywhere 🖖

This rectangular hyperbola is not unique to enzymes. The identical shape appears as the Langmuir adsorption isotherm for gas molecules sticking to a surface, as Monod's equation for microbial growth versus nutrient supply, and as the Hill equation for oxygen loading onto hemoglobin. Any system where a fixed number of binding sites gradually saturates produces this shape, so fitting V_max and K_m really means finding a maximum capacity and a half-saturation point.

Problem solved in full

An enzyme with V max = 100 and K m = 5 5 steps
An enzyme with V_max = 100 and K_m = 5. Show what K_m actually measures, and then work out how much substrate it takes to get from 10% of full speed to 90%.
1. The rate law is a saturating curve: proportional to substrate when there is little, flat when there is plenty. K_m is the only thing setting where the transition happens.
  
  $v = \frac{V _{m a x} [ S ]}{K _{m} + [ S ]}$
2. Put [S] equal to K_m and the expression collapses. K_m is the half-saturation concentration — not a rate, not a binding energy, and not a property you can read off the flat part of the curve.
  
  $[S] = K_{m} ⟹ v = \frac{V _{m a x} K _{m}}{2 K _{m}} = \frac{V _{m a x}}{2} = 50$
3. Invert the rate law to ask the reverse question: what concentration gives a chosen fraction of V_max?
  
  $v = f V_{m a x} ⟺ [S] = K_{m} \frac{f}{1 - f}$
4. Ten per cent needs one ninth of K_m; ninety per cent needs nine times it.
  
  $f = 0.1 : [S] = \frac{K _{m}}{9} = 0.556; f = 0.9 : [S] = 9 K_{m} = 45$
5. So the middle 80% of the curve spans an eighty-one-fold range in substrate.
  
  $\frac{9 K _{m}}{K _{m} /9} = 81 ⟹ an 81-fold range for the middle 80%$
Answer

The tool prints V_max = 100.00 and K_m = 5.00. The 81 is the number that matters in a laboratory. Saturation is approached so slowly that you cannot measure V_max by adding more substrate — reaching 99% would take 99 K_m, and reaching it exactly takes infinity — which is why V_max is always obtained by fitting, never by observation. It is also why the Lineweaver–Burk plot was invented and why it is now discouraged: taking reciprocals makes the crowded, noisy low-concentration points dominate the fit. Switch the plot type and watch which points spread out.

$K_{m} is the half-saturation concentration; 10% \to 90% costs 81 \times [S]$

References (2)

The 1913 paper, in translation, with the original constant: K. A. Johnson and R. S. Goody, "The Original Michaelis Constant: Translation of the 1913 Michaelis–Menten Paper." Biochemistry 50(39), 8264–8269, 2011.
Steady-state kinetics, and why V_max is fitted rather than measured: A. Cornish-Bowden, Fundamentals of Enzyme Kinetics, 4th ed., ch. 2–3. Wiley-Blackwell, 2012. ISBN 978-3-527-33074-4.

By the 42Math team · Updated 2026-06-30

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Example problems

Typical - Vmax 100.00 with Km 5.00 — and at a substrate concentration equal to Km the rate is exactly half, 50.00.
High affinity - A tighter enzyme: Km drops to 0.50, so half-maximal rate arrives at a tenth of the substrate.
Competitive inh. - A competitive inhibitor raises apparent Km to 13.33 and leaves Vmax untouched at 100.00.
Non-competitive inh. - Non-competitive inhibition does the reverse — Vmax falls to 37.50 while Km stays at 5.00.