while plotting velocity vs substrate graph...initially the velocity increases and then it starts saturating but not completely saturated..due to the solubility limit of substrate I can't increase the substrate conc. further . while fitting the data with graph pad and sigma plot kinetics module....the value of vmax is interpolated very high and errorous (90% error).. though the R-square value is 95%.. what is the mistake I am doing?
If the data are tight enough, it should not be necessary reach saturation in order to measure the kinetic parameters with a Michaelis plot, but there must be enough curvature for the extrapolation to be accurate. The more scatter in the data, the closer to saturation you will have to get for a good measurement. In your case, the limited solubility of the substrate prevents you from reaching saturation, so you will have to rely on getting tight data in the achievable concentration range.
An alternative approach to using a Michaelis plot is to perform progress curve analysis, in which the product versus time curves at several substrate concentrations are fit globally to the integrated Michaelis-Menten equation to get the kinetic constants. It is helpful, when doing this, to have data for many time points.
If the data are tight enough, it should not be necessary reach saturation in order to measure the kinetic parameters with a Michaelis plot, but there must be enough curvature for the extrapolation to be accurate. The more scatter in the data, the closer to saturation you will have to get for a good measurement. In your case, the limited solubility of the substrate prevents you from reaching saturation, so you will have to rely on getting tight data in the achievable concentration range.
An alternative approach to using a Michaelis plot is to perform progress curve analysis, in which the product versus time curves at several substrate concentrations are fit globally to the integrated Michaelis-Menten equation to get the kinetic constants. It is helpful, when doing this, to have data for many time points.
December 23, 015
Answer
Three parameters are required to carry out a correct kinetic study:
1) Temperature
2) Enzyme amount
3) Reaction time
When enzyme activity/time (velocity: v) is determined as function of the substrate concentration, the value must be proportional to the enzyme amount during the considered reaction time at the fixed temperature. This is what we call initial velocity (vi). Obtention of vi for each S concentration is required for correct kinetic study.
It is not requisite to obtain a saturated curve, v =f (S). Just plot 1/v = f (1/[S]).
If the plot is a straight line, the studied enzyme kinetic obeys to the equation of Michaelis-Menten. For precision, it is easy to visualize dispersion of the points on both sides of the plot. The reciprocal plot, so obtained permits to calculate Km and Vm.
In the case, the points (1/v f 1/S) do not fit into a straight line, the studied enzyme kinetics do not fit the equation of Michaelis-Menten; then you have to search another model, as allosteric kinetic, or even, if necessary, you establish a particular model, specific to the studied enzyme.
I hope this comment is helpful
Thanks.
I plotted reciprocal plots for my substrate. now if saturation is not reached then i am finding 1/vmax intercept on y axis becomes negative and that is kind of impossible.
Anyways i will do enzyme optimization one time and check
Thanks
For kinetic study you can first determine optimum enzyme concentration with constant substrate which can also give velocity data.Based on the results, the optimum enzyme concentration can be used further.
January 4, 2016
Dear Puneet
According to your description and with the assumption that the assays you performed were systematically correct, the kinetic of your enzyme is probably allosteric. Use your data for the Hill equation in its log term: log (v/Vm-v) = log S. A straight line is expected.
January 4, 2016
Dear Puneet
According to your description and with the assumption that the assays you performed were systematically correct, the kinetic of your enzyme is probably allosteric. Use your data for the Hill equation in its log term: log (v/Vm-v) = log S. A straight line is expected.
January 4, 2016
Dear Puneet
According to your description and with the assumption that the assays you performed were systematically correct, the kinetic of your enzyme is probably allosteric. Use your data for the Hill equation in its log term: log (v/Vm-v) = log S. A straight line is expected.
What does it means "due to the solubility limit of substrate I can't increase the substrate conc"? What is your avarage ratio of [S] and [E]? Variation of substrate at comparable hight concentrations is crucial for MM kinetics. Considering the nature of KM as a composite value including kinetic constants of enzyme-substrate-complex-formation, enzyme-substrate-complex-decay and kcat , it is essentially required that [S] is much > than [E]. At [S] < [E] the catalyzed reaction is solely controlled by difusion, and v(diff) is in fact a function of 1/[E] (since binding of small S to comparable large E slows down difusion significantly)! Consequently, v(observed) will be close to 0 approaching equilibrium. Due to substrate limitation/product accumulation at low initial [S] this happens immediately. Take into account, that low [S]/[E] will cause significant changes in [S] even at very early stages in the reaction course, this renders impossible to obtain initial velocities in a strictly concentration dependent manner. As a consequence, you data will fail to fit the theoretical MM model. With other words: Trying to fit data which cannot follow the MM model must simply yield poor data match!
if the curvature of the v = f([S]) plot is not enough visible, you will never obtain correct value for Vmax, whatever the plot you will use. About the R-square value, it could be correct for the data (95% is not so nice), but will not give any information about the error on Vmax. For example, if you can obtain only velocity at low substrate concentration, you will obtain a nice straight line, with R-square of 99% but you will never obtained any information about Vmax.
I think that the only way to get correct value of VM is to success to obtain higher substrate concentration. What is your enzyme and your substrate ? Did you tried mixture of solvants to solubilize your substrate ?
Can we get a sigmoidal kinetics when we use polysaccharides as a substrate??
@Joseph Kreit
I am working on a mannase enzyme and my substrate is locust bean gum. Is it possible to get allosteric kinetics with a polysaccharide substrate
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