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.

Problems 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.

$K_{m} is the half-saturation concentration; 10% \to 90% costs 81 \times [S]$
A competitive inhibitor, and the substrate that beats it 7 steps
The same enzyme meets a competitive inhibitor at 5 units, with Kᵢ = 3. Work out what moves and what does not, and then find how much extra substrate it takes to get the original rate back.

Open this problem in the tool
1. A competitive inhibitor binds the empty active site, so the enzyme now has two things it can be doing instead of working. Its dissociation constant says how tightly: small Kᵢ means a grip that is hard to break.
  
  $E + I ⇌ E I, K_{i} = \frac{[ E ] [ I ]}{[ E I ]} = 3$
2. Follow that through the steady-state derivation and only one thing changes. The free enzyme is split between working and blocked, and the split is a single number, the same at every substrate concentration.
  
  $α = 1 + \frac{[ I ]}{K _{i}} = 1 + \frac{5}{3} = \frac{8}{3} = 2.667$
3. That factor lands on Kₘ and nowhere else. The tool reports an apparent Kₘ of 13.33 against the 5.00 it had before, and an apparent Vₘₐₓ still sitting at 100.00.
  
  $v = \frac{V _{m a x} [ S ]}{α K _{m} + [ S ]} ⟹ K_{m}^{app} = α K_{m} = 2.667 \times 5 = 13.33, V_{m a x}^{app} = 100.00$
4. Vₘₐₓ survives because substrate and inhibitor are competing for the same site, and substrate can always be made to win by sheer numbers. Enough of it and the inhibitor never gets there first.
5. At a fixed substrate concentration the cost is real. Where the enzyme ran at half speed before, it now runs at 27.27, so the inhibitor has taken away nearly half the output without touching the enzyme’s ceiling.
  
  $[S] = 5 : v = \frac{100 \times 5}{13.33 + 5} = 27.27 against \frac{100 \times 5}{5 + 5} = 50.00$
6. So how much more substrate buys the rate back? Set the inhibited rate equal to the old one and solve. The substrate required goes up by exactly the same factor the apparent Kₘ did, which is the most useful line in the whole calculation: measure how far you have to push the substrate, and you have measured the inhibitor.
  
  $\frac{V _{m a x} [ S ]}{α K _{m} + [ S ]} = \frac{V _{m a x}}{2} ⟹ [S] = α K_{m} = 13.33 = α \times 5$
7. And this is how the two kinds of inhibition are told apart on a graph rather than by assertion. Switch the tool to the non-competitive preset at the same 5 units and the same Kᵢ = 3, and the numbers trade places: Vₘₐₓ falls to 37.50 and Kₘ stays at 5.00. That inhibitor binds somewhere other than the active site, so more substrate cannot argue with it.
  
  $non-competitive: V_{m a x}^{app} = \frac{V _{m a x}}{α} = \frac{100}{2.667} = 37.50, K_{m}^{app} = 5.00$
Answer

Kₘ rises from 5.00 to 13.33, Vₘₐₓ does not move, and 2.667× the substrate restores the original rate. One factor, α = 1 + [I]/Kᵢ, does both jobs: it multiplies the apparent Kₘ, and it is the ratio by which substrate has to rise to win the rate back, from any starting concentration and not just this one. What the inhibitor costs is substrate, never the ceiling.

$K_{m}^{app} = α K_{m} = 13.33, V_{m a x} unmoved; [S] \to α [S] restores the rate$

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 · Reviewed against published sources. Corrections welcome.