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Home/Chemistry/Michaelis–Menten Kinetics

Michaelis–Menten Kinetics

v vs [S] saturation and Lineweaver–Burk line from slope Km/Vmax and intercept 1/Vmax.

Parameters

10
2

Measured values

Km (model)2.00
Vmax (model)10.0
v at [S]=Km5.000

About this model

Michaelis–Menten enzyme kinetics relate initial rate v to substrate concentration [S] by v = V_max [S] / (K_m + [S]), saturating at V_max when enzyme is fully bound. The Lineweaver–Burk plot graphs 1/v versus 1/[S] as a straight line with slope K_m/V_max and intercept 1/V_max. The simulator shows both the hyperbolic v([S]) saturation curve and the double-reciprocal line side by side. Assumptions are the standard quasi-steady-state single-substrate mechanism without product inhibition, cooperativity, or allosteric regulation. You vary K_m and V_max to see how substrate affinity and catalytic capacity reshape the saturation curve and the Lineweaver–Burk slope and intercepts.

Who it's for: Biochemistry and physical chemistry courses on enzyme kinetics and linearization of rate laws.

Key terms

  • Michaelis–Menten
  • Km
  • Vmax
  • Lineweaver–Burk
  • Enzyme kinetics
  • Saturation

How it works

Saturation kinetics: v = Vmax[S]/(Km + [S]). The Lineweaver–Burk plot (1/v vs 1/[S]) linearizes the curve so that slope = Km/Vmax and the y-intercept is 1/Vmax.

Key equations

Equilibrium E + S ⇌ ES → E + P gives the same form when steady-state applies to ES.

Frequently asked questions

What does K_m mean physically?
K_m is the substrate concentration at which v = V_max/2. Under Michaelis–Menten assumptions it reflects the substrate level needed to half-saturate the enzyme; it is not universally equal to a simple dissociation constant when catalytic steps are fast.
Why use a Lineweaver–Burk plot if it distorts errors?
Historically it linearizes the hyperbola so slope and intercept give K_m and V_max by eye. It overweight low-[S] points, so modern fitting prefers nonlinear regression on v([S]); the plot remains useful pedagogically and for spotting inhibition patterns.
Does saturation mean the reaction stops speeding up because of equilibrium?
Saturation means free enzyme is depleted: nearly all enzyme exists as ES, so adding more substrate cannot increase the ES concentration further. The catalytic step still turns over at its limiting rate V_max; equilibrium of the overall reaction is a separate thermodynamic question.