I am studying a biological system using ODEs. As part of my work I need to use the '$k_\text{cat}$' version of the Michaelis-Menten reaction, which is:

$$\frac{V_\text{max}\cdot [\ce{S}]}{(K_\mathrm m+[\ce{S}])} = \frac{k_\text{cat} \cdot [\ce{E}_\text{free}] \cdot [\ce{S}]}{(K_\mathrm m+[\ce{S}])}$$

I have found in a publication a value of a $V_\text{max}$ for one of my reactions:

$$V_\text{max}=\mathrm{\frac{11.4\;pmol}{min^{-1}{\;pmol^{-1}\;enzyme}} = \frac{0.19\;nmol}{s^{-1}\;nmol^{-1}\;enzyme}}$$

My question is, since this $V_\text{max}$ value is quoted in terms of 1 unit of enzyme (i.e. 0.19 nmol is the maximum one nmol of enzyme can catalyze in a second), can I use this $V_\text{max}$ value as a $k_\text{cat}$?

here is the abstract I found the information from :

Retinoic acid (RA) is a critical signaling molecule that performs multiple functions required to maintain cellular viability. It is also used in the treatment of some cancers. Enzymes in the CYP26 family are thought to be responsible for the elimination of RA, and CYP26A1 appears to serve themost critical functions in this family. In spite of its importance, CYP26A1 has neither been heterologously expressed nor characterized kinetically. We expressed the rCYP26A1 in baculovirus-infected insect cells and purified the hexahistidine tagged protein to homogeneity. Heme incorporation was determined by carbon monoxide difference spectrum and a type 1 spectrum was observed with RA binding to CYP26A1.We found that RA is a tight binding ligand of CYP26A1 with low nM binding affinity. CYP26A1 oxidized RA efficiently (depletion Km 9.4 3.3 nM and Vmax 11.3 4.3 pmoles min1 pmole P4501 ) when supplemented with P450 oxidoreductase and NADPH but was independent of cytochrome b5. 4-Hydroxy-RA (4-OH-RA) was the major metabolite produced by rCYP26A1 but two other primary products were also formed. 4-OH-RA was further metabolized by CYP26A1 to more polar metabolites and this sequential metabolism of RA occurred in part without 4-OH-RA leaving the active site of CYP26A1. The high efficiency of CYP26A1 in eliminating both RA and its potentially active metabolites supports the major role of this enzyme in regulating RA clearance in vivo. These results provide a biochemical framework for CYP26A1 function and offer insight into the role of CYP26A1 as a drug target as well as in fetal development and cell cycle regulation.


1 Answer 1


$V_{max}=k_{cat}[E_0]$ where $[E_0]$ is the total enzyme concentration and not just the free enzyme. See my answer here.

You cannot use the value of $V_{max}$ as $k_{cat}$ because while the latter is the property of the enzyme (actually the enzyme-substrate pair), the former is the property of the chemical sample i.e. it is a system parameter.

Obviously, the $V_{max}$ will be same as $k_{cat}$ for a unit-concentration of total enzyme (unit concentration does not mean units of enzyme. It means, for whatever you are using as the units of concentration in your equation, lets say unit-x, the concentration of the enzyme should be 1 unit-x)

  • $\begingroup$ Hi WYSIWYG, would you know of a way to convert my Vmax value into a usable Kcat value then? Thanks $\endgroup$ Sep 11, 2015 at 10:02
  • $\begingroup$ @user3059024 Can you provide a link to that paper? I need to know the context. $\endgroup$
    Sep 11, 2015 at 10:06
  • $\begingroup$ Sure, here it is ac.els-cdn.com/S0006295208007235/… $\endgroup$ Sep 11, 2015 at 16:16
  • 1
    $\begingroup$ Technically speaking you need to make a few more assumptions to use $V_{max}=k_{cat} [E_0]$. If a single enzyme molecule has two active sites that both have equal activity, then you probably would want to use $V_{max} = 2 k_{cat} [E_0]$, for example. And of course the standard restrictions about substrate concentration being in vast excess relative to enzyme would also apply... $\endgroup$
    – Curt F.
    Sep 12, 2015 at 1:17
  • 2
    $\begingroup$ @CurtF. Well yes.. you can always extend the model to include more complex kinetics.. The standard restrictions are the part of the MM model and actually it really doesn't explicitly state that. The model just assumes either equilibrium or QSSA which usually holds when the substrate is in excess. $\endgroup$
    Sep 12, 2015 at 4:07

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