Michaelis–Menten calculator: Km, Vmax, kcat and kcat/Km

$$v=\frac{V_{\max }[\mathrm{S}]}{K_{\mathrm{m}}+[\mathrm{S}]}$$

The calculator fits this equation to every replicate rate as its own point by ordinary (unweighted) nonlinear least squares, using the Levenberg–Marquardt method. Starting values come from the data ($V_{\max }$ from the largest rate, $K_{\mathrm{m}}$ from the $[\mathrm{S}]$ where the rate is half of it). Standard errors are the square roots of the diagonal of the covariance matrix, $s^2\cdot (J^{\mathsf{T}}J{)}^{-1}$, where $J$ is the Jacobian of the model at the solution and $s^2=\mathrm{SS}/(n-2)$ is the residual variance; 95% confidence intervals are the estimate $\pm t(0.975,n-2)\times \mathrm{SE}$. $R^2=1-{\mathrm{SS}}_{\mathrm{residual}}/{\mathrm{SS}}_{\mathrm{total}}$.

kcat and kcat/Km. The turnover number is $k_{\mathrm{cat}}=V_{\max }/[\mathrm{E}]$, with $[\mathrm{E}]$ the molar concentration of active sites; the rate must be in molar units per time (µM/min, nM/s…), and $k_{\mathrm{cat}}$ is reported in ${\mathrm{s}}^{-1}$. The specificity constant $k_{\mathrm{cat}}/K_{\mathrm{m}}$ is in ${\mathrm{M}}^{-1}{\mathrm{s}}^{-1}$. Its standard error comes from the delta method:

$$\begin{array}{rl}& \mathrm{Var}(V_{\max }/K_{\mathrm{m}})\\ & =\frac{\mathrm{Var}(V_{\max })}{K_{\mathrm{m}}^2}\\ & +\frac{V_{\max }^2\mathrm{Var}(K_{\mathrm{m}})}{K_{\mathrm{m}}^4}\\ & -\frac{2V_{\max }\mathrm{Cov}(V_{\max },K_{\mathrm{m}})}{K_{\mathrm{m}}^3}\end{array}$$

with $[\mathrm{E}]$ treated as exact. Michaelis and Menten’s own 1913 analysis in fact estimated $V_{\max }/K_{\mathrm{m}}$, the quantity that equals $k_{\mathrm{cat}}/K_{\mathrm{m}}\times [\mathrm{E}]$ (Johnson and Goody 2011).

Substrate inhibition. When the rate falls at high substrate, the model

$$v=\frac{V_{\max }[\mathrm{S}]}{K_{\mathrm{m}}+[\mathrm{S}](1+[\mathrm{S}]/K_{\mathrm{i}})}$$

adds a dissociation constant $K_{\mathrm{i}}$ for a second, unproductive substrate binding. It is compared with plain Michaelis–Menten by the extra sum-of-squares F test,

$$F=\frac{({\mathrm{SS}}_{\mathrm{MM}}-{\mathrm{SS}}_{\mathrm{SI}})/({\mathrm{df}}_{\mathrm{MM}}-{\mathrm{df}}_{\mathrm{SI}})}{{\mathrm{SS}}_{\mathrm{SI}}/{\mathrm{df}}_{\mathrm{SI}}}$$

In the substrate-inhibition model $K_{\mathrm{m}}$ is no longer exactly the $[\mathrm{S}]$ at half of $V_{\max }$.

Why not the Lineweaver–Burk line? Lineweaver and Burk (1934) plotted $1/v$ against $1/[\mathrm{S}]$, which turns the hyperbola into a straight line with slope $K_{\mathrm{m}}/V_{\max }$ and intercept $1/V_{\max }$. Taking reciprocals distorts the errors: a small error on the slowest rate (lowest $[\mathrm{S}]$) becomes a large error in $1/v$, and those points, furthest from the axis, then dominate a least-squares line. Dowd and Riggs (1965) compared the linear transformations and found the double-reciprocal plot gave the least reliable estimates. Fit the hyperbola directly and use the double-reciprocal plot only to display the data, for example the pattern of lines with an inhibitor.

Assumptions. Initial rates (less than about 10% of substrate converted), $[\mathrm{E}]$ much lower than $[\mathrm{S}]$, a single substrate (or the others saturating), and errors of similar size across the curve. Spread $[\mathrm{S}]$ from about $0.2\times K_{\mathrm{m}}$ to $5\times K_{\mathrm{m}}$ or higher; if the highest $[\mathrm{S}]$ is below about $2\times K_{\mathrm{m}}$, $V_{\max }$ and $K_{\mathrm{m}}$ are strongly correlated and poorly determined.

Sources: Johnson KA, Goody RS. The original Michaelis constant: translation of the 1913 Michaelis–Menten paper. Biochemistry 2011;50:8264–8269, doi:10.1021/bi201284u. Lineweaver H, Burk D. The determination of enzyme dissociation constants. J Am Chem Soc 1934;56:658–666, doi:10.1021/ja01318a036. Dowd JE, Riggs DS. A comparison of estimates of Michaelis–Menten kinetic constants from various linear transformations. J Biol Chem 1965;240:863–869.

The built-in example is a synthetic data set generated from $V_{\max }=1.2\mathrm{\mu }\mathrm{M}/\min$ and $K_{\mathrm{m}}=18\mathrm{\mu }\mathrm{M}$, with eight substrate concentrations (2.5 to 320 µM, two-fold steps) measured in triplicate with about 4% random noise. The enzyme concentration is 10 nM.

1. Fit the hyperbola to all 24 rates (22 degrees of freedom): $V_{\max }$ = 1.153 ± 0.012 µM/min (95% CI 1.128 to 1.178) and $K_{\mathrm{m}}$ = 16.74 ± 0.67 µM (95% CI 15.35 to 18.13), $R^2=0.9948$. Both intervals are close to the true values; the highest $[\mathrm{S}]$ is $19\times K_{\mathrm{m}}$, so the plateau is well defined.
2. Turnover number. $V_{\max }$ in M/s is $1.153\times {10}^{-6}/60=1.922\times {10}^{-8}\mathrm{M}/\mathrm{s}$; divided by $[\mathrm{E}]=10\times {10}^{-9}\mathrm{M}$ this gives $k_{\mathrm{cat}}$ = 1.92 ± 0.02 s⁻¹ (115 min⁻¹).
3. Catalytic efficiency. $k_{\mathrm{cat}}/K_{\mathrm{m}}=1.922/(16.74\times {10}^{-6}\mathrm{M})$ = 1.15 × 10⁵ M⁻¹ s⁻¹, with a standard error of $3.8\times {10}^3{\mathrm{M}}^{-1}{\mathrm{s}}^{-1}$ (3.3%) from the delta method. The relative error of the ratio is smaller than the sum of the two relative errors because $K_{\mathrm{m}}$ and $V_{\max }$ are positively correlated.
4. Compare with Lineweaver–Burk. A straight line through $1/v$ against $1/[\mathrm{S}]$ for the same 24 points gives $V_{\max }=1.167\mathrm{\mu }\mathrm{M}/\min$ and $K_{\mathrm{m}}=17.53\mathrm{\mu }\mathrm{M}$ with $R^2=0.9991$. The high R² does not make it better: it reflects the transformed scale, which the lowest-$[\mathrm{S}]$ points dominate. With noisier data at low $[\mathrm{S}]$ the two estimates part further.
5. Substrate inhibition? Switching to the substrate-inhibition model on these data drives $K_{\mathrm{i}}$ to very large values, and the calculator reports that no substrate inhibition is detectable, as expected for data generated without it.