I may be wrong, but I don't think it makes sense to talk about the Km for a mixture of enzymes. I also don't understand why having two enzymes use the same substrate at the same time would result in a Km lower than that for either enzyme, instead of in between the two Kms.

The difference between 0.5, 0.7, and 0.9 mM is so small that it could easily be due to experimental variation. You should make the measurement several times and calculate the standard deviation to see if there is a statistically significant difference.

Dear Chouchene:

Have you used the same substrate concentration in all the assays (enzymes alone and mixed)? If you, then you have positive synergic effect because the Km value decreases which means that the affinity by the substrate is somehow increasing.

Good luck,

Dear Rosa Maria,

Thank you for your response, yes I have used the same substrate concentration in all the assays at the same conditions (Temperature, pH, buffers ...).

Best regards,

I may be wrong, but I don't think it makes sense to talk about the Km for a mixture of enzymes. I also don't understand why having two enzymes use the same substrate at the same time would result in a Km lower than that for either enzyme, instead of in between the two Kms.

The difference between 0.5, 0.7, and 0.9 mM is so small that it could easily be due to experimental variation. You should make the measurement several times and calculate the standard deviation to see if there is a statistically significant difference.

Hi Chouchene, if you are using the same substrate for 2 hydrolytic enzymes that hydrolyze it in the same place then a case for cooperation between them is a bit weak. The Km values are pretty close and if you do the experiments 3-4 times, the SD of the values will probably overlap.

In my opinion, the Km values are not enough to establish a synergistic effect between two enzymes. Maybe the catalytic efficiency (Kcat) can give a better indices about the enzyme catalysis... Also, I will advice you vary the concentrations of the enzymes while doing enzyme mix...the 50/50 mix cannot adequately show the synergy.

Thank you for your responses. About the statistic, the 3 Km determined are significantly differents. ANOVA 5%

Just to echo Adam's concern is that Km values of enzymes are usually reported as 1 enzyme reacting with 1 substrate.

I have never read a paper that mixes 2 different enzymes with 1 substrate and reports 1 Km.

In my field, it is like mixing together mouse thrombin with human thrombin with a single substrate and measuring the Km as it were a single thrombin. That kind of experiment is not realistic and will not be publishable.

Thank you for your response, so how we can prove a synergistic effect between 2 enzymes ?

Please explain what you mean by a synergistic effect between enzymes. Can you give an example of such an effect?

how do you arrange your experiment? do you estimte the initial velocity matehmaticaly from the tangent to the progress curves (concentration of the product versus time)? It is posssible that if you have done only the one-point progress curve, you obtain a wrong velocities,? The mistakes can also come because the concentration of the substrate used simultaneously by two enzymes is lower than you think? Km is the substrate concentration!

the detrermination of Km for two enzymes has not any sense.

I estimate the initial velocity by the concentration of the product versus time (n=3) at 5 different substrate concentrations.

Synergistic effect mean that if we have 2 enzymes who belongs to the same family but from different sources. If the first enzyme give us at the same conditions a concentration of product called A. the second enzyme give a concentration B. If we make the two enzymes in a mix (50/50) we have a concentration of product > A+B (at the same conditions of temperature, time, pH, buffer and substrate concentration).

As usual, Adam's opinion is very valuable in suggesting what is a likely explanation for the preliminary findings. I usually think of synergy being worth considering for enzymes that catalyse adjacent steps in a metabolic pathway i.e. acting on different substrates. This is clearly not the same as envymes competing for the same substrate.

Peter

Chouchenne, The definition of synergy you just gave based on rates of product formation seems more reasonable to me than your original one based on Kms. However, as before, you have to be sure that the so-called synergistic effect is statistically significant, not just experimental variation. Also, there has to be some physical explanation for the effect, if it exists.

For example, imagine (1) that each enzyme (call them A and B) can dimerize, with equilibrium dissociation constants in the same range as the concentrations present in the experiment, (2) the dimer is more active than the monomer, (3) heterodimerization to form AB is possible, and (4) the AB dimer is more active than either the AA or BB dimer.

Really thank you for your responses and your help. In my opinion, the decrease of Km in the mix involve a bettre affinity of enzymes to the substrate that maybe explain why we have an increase in the rate of product formation.

You write: " I estimate the initial velocity by the concentration of the product versus time (n=3) at 5 different substrate concentrations. "

I do not understand HOW you estimate=calculate the initial reaction rate. using one point only? you must do several measurements of the product concentration, and obtain a curve and calculate the tangent to this curve in order to obtain vo.

Which substrate concentration do you give when you use two enzymes?

therefore you can not have any proper Michaelis Menten curve, not having a proper [S] at x axis and not having a proper vo at y axis. thus you can not have any results from such experiment.

I agree with others that the most likely reason for your observation is experimental error. Km-values usually have a much larger standard deviation than the Vmax determined in the same experiment.

So the first thing to do is to make a good Michaelis/Menten-plot.

• For each substrate concentration determine product at at least 5, better 10 time points. Make sure those points are on a straight line (that is, plot them), then determine the slope of the line, which is your velocity (each with a standard deviation).
• Do this for at least 12 different concentrations (doi:10.1006/jtbi.1996.0023), dispersed over the range 0.1 to at least 5, it technically possible (solubility, substrate inhibition) 10 times Km.
• Make a Lineweaver-Burk-plot, just to ensure that the data are linear (if they are not, you need to measure an even wider range of substrate concentrations and use the Hill-equation).
• Do not determine Vmax and Km from the LB-plot, but use nonlinear curve fitting of the untransformed data (if you can, by simplex algorithm (doi:10.1093/comjnl/7.4.308), but Marquardt (http://137.204.42.130/~bittelli/materiale\_lettura\_fisica\_terreno/marquardt\_63.pdf) will do in a pinch). Again, also determine the standard deviation for your parameters (doi:10.1016/0076-6879(92)10009-3).
• The direct plot (http://www.biochemj.org/bj/139/0715/1390715.pdf) may also be used to determine Km and Vmax, in that case the quartils become your error (for a detailed example, see doi:10.1007/978-3-319-19920-7_6).
• Then compare your parameters from different experiments with a t-test, to determine the probability of H0: The values come from the same population.

Mechanistically, morpheeins (https://en.wikipedia.org/wiki/Morpheein) are the only way I can think of to explain your result, if it should proof real.

Hi Mr Buxbaum,

Thank your for your detailed answer.

Best regards

Alaeddine