Dear friends,
I am working on Michaelis–Menten kinetics of an enzyme. The enzyme’s molecular weight is 50 KDa, it elutes as a dimer during gel filtration. The Michaelis–Menten saturation curve of enzyme is very strange, as the specific activity of enzyme decreases beyond a certain substrate concentration (graph attached). This observation is quiet consistent even if I change enzyme reaction parameter beyond optimal. Is this observation can be interrelated as negative cooperativity?
Regards
John is correct, what you are looking at is likely some form of substrate inhibition (depending on the type of mechanism the enzyme uses the standard uncompetitive model fit may or may not be the correct choice for interpretation).
Were you to have cooperative effects, you would see a sigmoidal shape to your reaction profile rather than a hyperbolic shape. This image illustrates that well enough: http://www.mikeblaber.org/oldwine/BCH4053/Lecture26/allost_plot.jpg
Hi there,
It looks like inhibition caused by excess of substrate. If you enzyme is purely michaelian at low substrate concentration and obeys to the scheme E+SES--->E+P, the velocity equation is v0=Vmax*[S0]/(Km+[S0]+[S0]2/Ki) where Vmax and Km are michaelian parameters and Ki is the dissociation constant concerning a second substrate molecule binding the active site in the ES complex to form reversiblely a unproductive ESS complex. Ki will be the concentration of substrate needed to decrease v0 to vmax/2 whereas Km is the concentration required to increase v0 to vmax/2 coming from low S concentration. Coming back to the equation, if working at low substrate concentration, [S0]2/Ki tends to zero as Ki is expected to be much greater than Km so the equation becomes the usual michaelian one whereas at high substrate concentration Km becomes negligible towards [S0] and the equation becomes v0=Vmax/(1+[S0]/Ki) if plotting 1/v0=f([S0]) you get experimental points aligned the slope being 1/(Vm*Ki) and extrapolation on y axis gives 1/vmax. The other solution is to fit your whole data with the global velocity equation using non linear regression with three constants : Km, Vm and Ki...
According to the reaction scheme S acts as the substrate but also as an uncompetitive inhibitor (only binding to ES complex) but the uncompetitive effect of S is only showing up at high concentration whereas the usual uncompetitive inhibitor acts as such whatever its concentration as long as it's not zero. So these are definitely different cases but with common features.
I agree that substrate inhibition is a possibility, but I also wonder about the upturn in the curve at the highest concentration. If this is reproducible, it is not consistent with substrate inhibition (or negative cooperativity). It would be helpful to know more about the assay, such as the nature of the substrate and the method of detection.
Dear Adam,
Yes that upturn in curve at higher concentration of substrate is fairly reproducible. Ill be very thankful if you can help me to comprehend this observation. The detail needed is as follows:
The enzyme is an exoprotease & substrate is most preferred oligopeptide. The assay in use is "modified ninhydrin assay", This assay detect free α-NH2 group at N terminus of peptide or amino acid. The assay produces a colored product which is quantified by colorimetry.
Thanks
A possible explanation for the upturn at the highest substrate concentration is that there is a background level of detection of the substrate in the ninhydrin assay. Have you made a standard curve with the substrate, rather than the product?
Thank you all for these helpful answers.
Dr. John H.T. Luong
I will repeat the experiment at 6 uM (sorry, I mis-typed this before, thanks John) substrate concentration.
Dr. Dominique Liger
Thanks for your insightful information.
Dr. Robert J. Bauer
I will try to fit my data in standard uncompetitive model.
Dr. Adam B Shapiro
I will check my calculation and further evaluate the reproducibility of that upturn in curve at higher concentration of substrate.
Regards
Dear Rahul,
All the answers above seem very precise, but I`d also consider inhibition by the product. There are some proccessive enzymes (i.e. beta-glucanases and cellulases) for which the increase in product concentration inhibits either the substrate binding or the catalysis. This might be the case for your exoprotease: the higher the product concentration, the higher the inhibition by digested protein fragments, which could bound to the substrate binding site with a Ki higher than the substrate itself.
Good luck,
this phenomanon can be caused by many mechanism. but the negative cooperation is not the cause for it. the substrate may act negative regulator for this enzyme (or substrate inhibition mechanism). see my paper: Zhao, Q. (2015). A thermodynamic and theoretical view for enzyme regulation. Biochemistry (Moscow), 80(1), 1-7.
This is not really an answer but a question to help us all understand exactly what we are thinking about here. How exactly do you estimate a rate? Is this based on several time points or do you simply assume that a single measurement at e.g. 10 minutes is sufficient? Many artifacts arise from the latter, because you need to know you are actually comparing linear initial rates.
The plot is rather odd in that there is a sharp and linear decrease in activity and then an upward blip. The likely cause is methodological rather than from a fundamental enzyme property such as substrate inhibition or negative cooperativity
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