Vmax depends upon and is, ideally, directly proportional to the concentration of the enzyme. Specific activity divides Vmax by the amount of protein present (typically in mg, as in µmol/min/mg). Specific activity can be used to describe the level of activity of any enzyme-containing preparation, without requiring that the enzyme be purified.

kcat divides Vmax (in concentration units) by the enzyme concentration to get s-1. Purified enzyme is normally used for measuring this parameter.

You can express the enzyme activity in whatever way suits your needs, but you should be explicit about how the calculation is made, the source of the enzyme, what protein concentration was used, and the specific activity of the preparation so that others can repeat the work.

If you keep the enzyme concentration constant throughout all experiments, and its activity remains constant throughout, then, for the sake of your calculations, you can leave Vmax in the form of moles of product formed per unit of time or, dividing by the volume of the reaction, as concentration of product formed per unit of time.

Hello Adam B Shapiro

First of all, thank you very much for your detailed reply. I am actually struggling to reproduce the kinetic parameters of a study for my own research. They created an in silico model with a series of reactions and metabolites. For example, the Vmax value for the enzyme CYP3A4 is computed as 700 µmol/h. However, when I look at the paper they are referencing, I can't obtain the same value.

In the referenced paper, the Vmax is given as 2.7 nmol/min/nmol enzyme (supp-4). The supplementary material (supp 1) indicates that the enzyme concentration was 40 pmol/mL, and the final volume was 200 µL. With these values, I am not getting the same Vmax that they used in the computer model.

I am simply multiplying the Vmax value by the number of moles of enzyme, and then by 60 (to convert to per hour). So, I feel like I might be missing a step along the way :(

Using Michaelis-Menten kinetic parameters allows the initial rate of the reaction to be calculated at a constant substrate concentration, and under some conditions can be used to calculate the instantaneous rate at any time point during substrate consumption in vitro. The kinetics of the reaction in vivo are often not suitable for calculation of the rate by M-M kinetics, however. First, the enzyme concentration is usually unknown. Second, for enzyme reactions that metabolize drugs, the drug concentration varies over time due to its ADME properties (which vary between individuals). Third, metabolic enzymes are not single-substrate enzymes, so you also have to consider the concentration of the other substrate and use a more complicated kinetic rate equation to describe the rate. Fourth, products of the reaction can inhibit the reaction if they accumulate. This can be taken into account in the kinetic rate equation, but only if the product concentration and the kinetic parameters for product inhibition are known. Fifth, there may be competition by other substances for the same enzyme. Sixth, expression of metabolic enzymes can be induced by their substrates, causing the enzyme concentration to change over time. There are probably other complicating factors, as well, that I haven't thought of.

I hope some of this is relevant to the calculation in the paper. My point is that the calculation may have been based on things besides the simple Michaelis-Menten rate equation for a single-substrate enzyme.

Back to the calculation:

The value of kcat was 2.7 min-1 (which is extremely slow, by the way). The rate using the simple single-substrate M-M equation is v = kcat[E][S]/{Km + [S]}. The value of Km is not mentioned, but estimated from the figure, it is roughly 10 µM. The enzyme concentration was 40 nM. The volume of the reaction was 200 µL. We need to know the substrate concentration to complete the calculation of the instantaneous rate in this simplistic situation.

Remember that this is for a single-substrate enzyme, which is not the case for metabolic enzymes, so the only way to use this simple equation is to assume that the other substrate is at a saturating concentration ([S]>>Km) and there is no substrate inhibition, no product inhibition, and no competition by other substances.

Hello Adam B Shapiro

What you say about the kinetic laws for metabolic enzymes makes absolute sense. That said, the computational model that I am working on does use MM for the metabolism of the drug sorafenib by the enzyme CYP3A4, probably under the assumptions you just mentioned.

The substrate concentration part is where I am confused, because the concentrations of the drug are varied during the kinetics assay (0.1 - 100 µM).

Also, how did you calculate kcat from the Vmax? I thought kcat would be the Vmax value (2.7) divided by the enzyme concentration.

kcat is, indeed, Vmax/[E], where Vmax is in units of product concentration per unit time. You stated that Vmax was 2.7 nmol/min/nmol enzyme. The volume part of the concentration units is missing, or to put it another way, concentration was multiplied by the volume, which is the same for both the products and the enzyme (nmol/L x L = nmol). So we cancel the nmol units and arrive at kcat = 2.7/min.

As for the Vmax for CYP3A4 being 700 µmol/h, this requires knowing the enzyme concentration and the volume: Vmax = kcat[E]. You said the volume was 200 µL and the enzyme concentration was 40 pmol/mL = 40 nmol/L. The kcat was 2.7/min or 162/h. Assuming the substrate concentration was at saturation ([S]>>Km) for both substrates without substrate inhibition, the enzyme would be operating at Vmax.

Vmax = kcat[E] = (162/h) x (40 nmol/L) = 6480 nmol/L/h

Multiply by the volume: (6480 nmol/L/h) x (2 x 10-4 L) = 1.296 nmol/h

Since this result is nowhere near 700 µmol/h, something is not right. We're off by a factor of 540,000. I suspect the experimental methods given for the enzyme assay for enzyme concentration and volume differ from the figures used for the simulation.

Yes, that is the value I get as well. I got these values from the following section in their supplementary material.

Kinetics of sorafenib N-oxide formation by CYP3A4

Incubation time and protein concentration was optimized for sorafenib N-oxide formation by CYP3A4 by performing a time course up to 60 min and evaluating protein concentrations of 20, 40, and 80 pmol/mL. Reactions (final volume of 200 µL) consisted of CYP3A4 40 pmol/mL, varying concentrations of sorafenib (0.1 - 100 µM), and an incubation time of 30 min.  Reactions were performed in triplicate following the procedure described for the CYP enzyme screen. Reaction rate (velocity) was calculated as nmol sorafenib N-oxide per minute per nmol CYP3A4. Substrate concentration (x-axis) was plotted against CYP3A4 velocity (y-axis) and the apparent kinetic parameters Vmax and Km were estimated by fitting the Michaelis-Menten equation to the data by nonlinear regression analysis as implemented in GraphPad Prism.e

This whole thing with units is very perplexing I must say. My project is related to computational systems biology, where ODE and mass action-based kinetic laws have the quantity, volume, and time units. When I need to create a computational model for drug metabolism, I find a variety of units for Vmax pmol/min/mg protein, pmol/min/pmol enzyme, and sometimes even peak area/min/mg, which are actually specific activity units. I also even see some studies using these units for kcat. So, it is very confusing when different studies use various terms and units interchangeably

Hello again Adam B Shapiro ,

Following up on our previous conversation, I am curious as to whether I am following the correct logic here:

If Vmax is given in pmol/mg/min, then one has to know the enzyme concentration in mg/L, correct? I assume that's how one cancels out the mg unit in the denominator.

If you want to get Vmax in the form of pmol product/min, then multiply by the number of mg of enzyme present in the reaction, not the concentration. If you want to get to pM product/min, the multiply by the enzyme concentration in mg/L.