Hello Abdelmageed, micromoles are a unit of quantity (1 mole = 6.02214129(27)×1023 molecules, Avogadro's number), while the micromolar expresses a concentration, being the number of moles in 1 litre.
The method to calculate the Kcat will vary according to the enzyme, but grossly, it represents the speed at which 1 enzyme converts one substrate (http://www.pearsonhighered.com/mathews/ch11/c11kkkk.htm). Locically, this is usually expressed in moles as nobody works with a single enzyme nor substrate. This will be calculated by prurient a known qty of enzymes in optimal reaction conditions with its substrate. Then the conversion of the substrate will be measured. As I said before, the technic aspect varies depending on the enzyme: a kinase (suppose 1 uMole) can be put in contact with its substrate and the phosphorylated substrates (addition of a phosphate group) is measured. The conversion 1 enzyme = X substrates/minute will be your Kcat.
I hope it helps you!
Hello Abdelmageed, micromoles are a unit of quantity (1 mole = 6.02214129(27)×1023 molecules, Avogadro's number), while the micromolar expresses a concentration, being the number of moles in 1 litre.
The method to calculate the Kcat will vary according to the enzyme, but grossly, it represents the speed at which 1 enzyme converts one substrate (http://www.pearsonhighered.com/mathews/ch11/c11kkkk.htm). Locically, this is usually expressed in moles as nobody works with a single enzyme nor substrate. This will be calculated by prurient a known qty of enzymes in optimal reaction conditions with its substrate. Then the conversion of the substrate will be measured. As I said before, the technic aspect varies depending on the enzyme: a kinase (suppose 1 uMole) can be put in contact with its substrate and the phosphorylated substrates (addition of a phosphate group) is measured. The conversion 1 enzyme = X substrates/minute will be your Kcat.
I hope it helps you!
To calculate Kcat you have to know the concentration of your enzyme you've added to your reaction or assays (Et) then easly can calculate the constant Kcat. It is equal Vmax/Et.
for first times really Graphpad will be great help for you as Thiru said.
You can attempt this problem from a parameter estimation perspective. I am working on modelling biological systems involving enzymes and the way I do it is with parameter estimation as I have a set of ODEs to be solved.
Hi Abdelmageed, You can calculate Km and Vmax using the non-linear regression program HYPER from Dr John Easterby, University of Liverpool (http://homepage.ntlworld.com/john.easterby/software.html ). This is freely downloadable and has the advantage that you get Km/Vmax +/- SD so you can compare separate experiments done on separate days and weight your data. kcat is the turnover no. i.e. the rate constant for s -> P. Therefore it is an apparent first order rate constant (units time-1). A kcat of 10,000 s means the enzyme turns 10,000 molecules of s to P every second. In fact, kcat is the rate of the slowest step in a multi-step mechanism so it might be an electrophilic/nucleophilic attack or a conformational change. Vmax is the maximum value for initial rate v0, which therefore has units of micromoles/min (the International Unit is defined as that amount of enzyme catalysing conversion of 1 micromole S to P per minute). Vmax = kcat [Et] where Et is the enzyme concentration of your assay. If you are working with a pure enzyme calculate the molarity of this in your assy and [VERY IMPORTANT] express it in micromolar. The reason for this is that you already have Vmax in micromolar and, when you divide Vmax by [Et], these units cancel out leaving you with min-1 as units for kcat. This should give you kcat +/- SD. Best if luck
To me micromole is the total amount of substance whereas micromolar is the concentration (amount of substance per volume of solute) of the solution.
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