To fit Michaelis Menten equation in order to get kcat and Km, you need to use 5-10 concentrations of substrate and record the initial velocity of each point. The y axis in Michaelis Menten equation is initial velocity, the x axis is the corresponding concentration of substrate.

The figure you showed is the reaction progress curve. It is produced by the reaction of one concentration of substrate with time. You cannot fit a progress curve to the regular M-M equation. To get kcat and Km by reaction progress curve, you need to fit intergrated Michaelis Menten equation, which is much more difficult than regular one.

Cydi from EZBiolab

Hi there,

The graphics you show is a kinetics from which you can calculate reaction rate when steady state is settled (ie. the curve is linear and the slope is the rate of reaction). The curve shows that steady state take time to settle (around 200 time units). If you want to establish MM equation appplied to your system you have to repeat the experiment with varying substrate concentrations in order to be able to trace the v vs [substrate] relationship.

Thanks for the answers!

Zhang et al. (Analytical Biochemistry 211, 7-15 (1993)) provide an integrated form of the MM equation which i have unsuccessfully tried to fit to my curves:

t = [P]/(kcat*E0)+(Km/(kcat*E0))*ln([P]inf/([P]inf-[P]))

[P] - product concentration

[P]inf - product concentration at infinite time

E0 - total enzyme concentration

Looks like the data from a simple time course is insufficient for the fit, or?

The tricky part is that in this equation, time (t) is dependent while product concentration becomes independent. There is no a simple math solution to make [P] as a function of t. I haven't figure out a way to reliably apply integrated form of the MM equation.

Cydi from EZBiolab

Progress curves can be fit using numerical integration to obtain Km and Vmax with software such as Kintek Global Kinetic Explorer. It should be done simultaneously with progress curves collected at multiple substrate concentration and a single enzyme concentration. The possibility of product inhibition needs to be considered during this analysis.

There is some data missing from the beginning of this progress curve. This lost data can be dealt with in the Kintek program by allowing it to know how much time was skipped.

There is still the question of why the progress curve has the shape that it does. Instead of gradually slowing down as the substrate is consumed, the reaction reaches a steady state rate after an initial period with a faster rate. This looks like burst kinetics, but the enzyme concentration seems too low for that to apply. This situation needs to be understood before attempting progress curve analysis. Perhaps the enzyme is being modified during the reaction. For example, it might be dephosphorylating itself in trans.

Meanwhile, if you can get earlier data points, you may be able to at least make initial rate measurements for the early part of the progress curve in order to get kinetic constants for that part.

Most of the above answers are just educated guesses, BECAUSE YOU DID NOT PUT THE "X" AXIS VARIABLE IN YOUR PLOT!

A plot without variables is meaningless.

1. Assuming the "X variable" is time..

Michaelis Menten model is

v0=Vmax*[S]/(Km+[S])

The variables are v0 vs [S] substrate concentration. where v0 is Initial velocity= usually given as mass of product accumulated over time at the onset of the progress curve, the one that apparently you showed us).

I recommend you to read the book by A. Cornish-Bowden and C. W. Wharton, Enzyme Kinetics: In Focus (Series) Oxford University Press, 1988.

2. Assuming the "X variable" of your plot is other than time... Your questions cannot be given a coherent answer.

Best wishes

Rogelio

Jou's point about Lambert-Beer's law is well worth to discuss further. It is generally believed that Lambert-Beer's law is valid until OD reaches 1. I don't know its theoretical base. Or it is just from experience.

The absorbance in the progress curve shown in the question approaches 1. Its irregular shape may be caused by departure from Lambert-Beer's law of the product. It is necessary to calibrate the absorbance of product to make sure the Lambert-Beer's law is still hold in this range. This should be done for every kinetic experiment measured by absorbance if measurement is over 0.5, I think.

Cydi from EZBiolab

It is of course difficult to tell without access to the original data, but from "eyeballing" the curve I'd say that you have

You can try to fit the data to curves and then decide wether the residuals can be explained by the standard deviation of the y-values (0-hypothesis, data fit the curve) or not (alternative hypothesis, data do not fit the curve). There is a t-test for this (Kreyszig, 7th ed. 1979, pp. 276), but with some loss of precission you can also use a sign test.

@ Rogelio:

If somebody is talking about a kinetic curve, it involves the X axis is actually time. If it is not actually the case, it means the guy uses an inappropriate scientific vocabulary.

OK Dominique Liger

Then the Michaelis-Menten curve is NOT A KINETIC curve, for the X axis is [Susbtrate].

In any case, the above curve cannot be directly related to the Michaelis-Menten equation if the X-axis is time (yet, it is not indicated). Even assuming the curves are Absorbance (of a product) vs. time, the curves need to be converted to velocity vs. [substrate], by derivation, or the integrated form of the MM equation need to be used to fit the data, before you can say that the curve does not conform to MM.

Unfortunately, the accumulation of product does complicate the analysis, because the MM assumes no product is present, and thus ignores backwards reaction rate and product inhibition, so only a gross approximation of Vmax & Km can be obtained. Furthermore, usually, the change in substrate concentration observed in a single curve rarely spans the range to provide a reasonable coverage of substrate concentrations to render a good statistical fit.

For the above reasons, it is customary to measure the initial velocity (the slope of the Product vs. time extrapolated at zero time, multiplied by the inverse of the stoichiometry coefficient) and do repeated measures of velocity at variable starting substrate concentrations.

Best wishes,

Rogelio

If you click on the graph you see that it is a progress curve (absorbance vs time), unfortunately the image is clipped when not active. From the curve-fitting perspective that makes no difference as ideally the progress curve would be a hyperbola too. That this is not the case here I have discussed above.

Dear Engelbert Buxbaum ,

Progress curves could be numerically approximated by an hyperbolae. But they are complex curves of exponential functions, that if you assume continuous functions can be used to describe discrete phenomena, such as chemical reactions (a close approximation, but not an exact description).

You can check this book for a more complete description of the solution to the differential equations that integrate the function of S vs time for a simple enzyme reaction:

Gutfreund, H. (2002). Kinetics for the life sciences: Receptors, transmitters and catalysts. Cambridge: Cambridge University Press.

ISBN: 052148586X 9780521485869

What defines MM behaviour is the relationship between two variables: "velocity vs. [substrate]" when:

A) product accumulation is small.

B) Enzyme concentration is small compared to substrate concentration

C) The system has reached a steady state.

In a progress curve, the velocity is represented by the slope, and the substrate concentration is hidden, as it is the difference S0-Sx where S0 is the starting substrate concentration and Sx is the substrate concentration at time x.

Unfortunately, as I mentioned before, the full curve has to be sampled, product inhibition has to be ignored, and the starting concentration needs to be carefully chosen to allow acceptable coverage of substrate concentrations. Thus, this strategy is rarely used.

Best wishes,

Rogelio

So the curve shown above is definitely a kinetics and its shape (initial burst over the first tens of seconds and then steady state) tends to prove the reaction mechanism goes through several intermediate steps (including a phospoenzyme form, I guess). Still the reaction when in the steady state phase may actually obey the MM equation but it just can't be stated from the analysis of a single kinetics (apart from the fact that measuring a constant velocity is itself in agreement with MM equation).

The size of the burst is around 0.1 absorbance unit (steady state slope extrapolated on Y axis minus baseline), right? What is the concentration of product corresponding to 0.1 absorbance unit, is it comparible with enzyme concentration? If yes, the burst is due to the single turn-over of the enzyme which means the reaction is very slow and at minima occuring via a limiting step which is late in the mechanism (ie. after the step responsible for the increase in absorbance). If the concentration of product formed during the burst is much more than enzyme concentration then the burst is due to a multiple turn over process.

The burst occurs at very short reaction times when the Enzyme concentration is high, relative to substrate, and as Steady State is approached, before it is reached. This usually happens in ms times, unless the enzyme has an unsually low kCAT.

You are overintrerpreting the data. You cannot deduce the Kinetic Mechanism from a single experiment. The existance of a covalent intermediate cannot be deduced from this simple experiment.

In principle, Km and Vmax can both be determined from a single progress curve. This becomes immediately clear if one considers that such a curve contains information about the velocity at an infinite number of concentrations between start concentration [S]0 and the equilibrium concentration [S]∞. However, these measurements are not statistically independent (any error made with [S]0 will affect the entire curve), it is therefore also clear that the precission of the results is less than that obtained from the classical v vs. [S] plot. Nevertheless, progress curve analysis is legitimate and has been reviewed recently in doi:10.1006/meth.2001.1177 (http://ronduggleby.com/publications/Pub120.pdf) and earlier in doi:10.1016/0076-6879(79)63010-0.

As I have already pointed out, however, in this particular case the progress curve is not a simple hyperbola, which makes progress curve analysis a much less promising proposition.

I have also pointed out that the starting absorbance value is very high, which limits the precission of absorbance measurements (some assays show non-linearity already at A > 0.4). Thus, an appropriate blank should be used, so that the baseline is near 0.

@ Rogelio: I'm only making hypotheses which are actually already documented for some other phosphatases (existence of a slow burst phase, phosphoenzyme intermediate) and raising absolutely no conclusion.

About the burst phase I have to strongly disagree with what you wrote: excess of enzyme over substrate is only required if you intend to run a single turn over experiment which usually aims at determining individual rate constantes.

The burst phase finishes when steady state is settled which implies multiple turn over and therefore excess of substrate vs enzyme.

I saw this type of reaction progress curve with a DNA polymerase reaction. My interpretation was that product dissociation/substrate rebinding was rate limiting. When the enzyme reached the end of the single-stranded part of the DNA, it was slow to dissociate and rebind a new substrate strand.

Dear Dominique,

I did not write "excess", I wrote "large", and large enzyme concentration is required because, in order to to observe de burst, you have to be able to quantify the product as it forms INSIDE THE ACTIVE SITE", and ever for large quantum yield fluorophores, this implies nearly µM enzyme concentrations. Substrate concentration can be, and usually is, even higher than the enzyme.

Burst occurs because the product release (or one step thereafter) is rate-limiting, and the enzyme-product complex has to accumulate before the steady-state can be reached. From this initial pre-steady-state curve of reaction progress, it is possible to titrate the concentration of active sites, and the apparent rate constant of substrate-enzyme formation.

For most enzymes, all of these happens in a time span of a few ms and it is very unlikely the above curve can be interpreted in such terms.

Rogelio Rodríguez-Sotres: I have to agree : you did not write "excess". But you did not write "large" either. As you actually wrote in your penultimate message: " The burst occurs at very short reaction times when the Enzyme concentration is high, relative to substrate,... " which I surely misinterprated into [E]0>[S]0. Or maybe the writing was itself confusing...

OK, accept my apology.

Rogelio Rodríguez-Sotres : no problem!