Hi everyone,
sorry if this is a simple question, or if I get anything wrong here, but I'm very outside of my comfort area when it comes to enzyme activity-related calculations.
So, I need to know the concentration of enzymes that a paper used for their stock solution of Catalase.
Here is a quote from the methods: "Enzyme activities of stock solutions were 3 mM/s for GOX and 998 s-1 for CAT. To obtain a defined, stable oxygen concentration of 2% on cell surface stock solutions were diluted by 1:10,000 for GOX and 1:1,000 for CAT."
For the GOX, I think I can manage to calculate the stock, but the catalase is the issue
So, the Kcat = 998 s-1
The Vmax = 1uM/ min = 0.0166uM/ s
For for the calculation: Kcat = Vmax / E[t] , is it as simple as rearranging it to E[t] = Vmax / Kcat?
That Vmax figure is from the data sheet of the catalase, with 1 unit being equal to 1uM H2O2 processed per minute (not 100% sure this is what is meant by Vmax), hence me dividing by 60 to get the uM per second.
I also know that 1mg of Catalase = 20,000U
The issue is that I don't know how to put this all together. I am currently trying to get more familiar with enzyme kinetics, but this is taking some time. I would very much appreciate if anyone could offer some advice to help speed things up so I can start with my experiments.
If I am missing some information here please let me know.
Best regards,
Ciarán
The value kcat is the number of product molecules formed per molecule of enzyme per second, and is reported in units of s-1. So I don't think that the description of the stock solution in terms of s-1 makes sense , nor can it be converted to an enzyme concentration. Perhaps you mis-transcribed the units in your quotation from the published account, or there was an error in the paper.
For example, let's suppose the authors meant 998 µM/s as the activity of the stock solution, since the GOX stock solution was defined in similar terms (3 mM/s).
998 µmole/(L-s) = 59,880 µmole/(L-min) = 59,880 U
(The definition you gave for a unit of catalase activity is 1 µmole/(L-min).)
The specific activity you mentioned was 20,000 U/mg.
Thus the stock solution contained 59,880 U/(20,000 U/mg) = ~3 mg.
We don't have information about the volume of the stock solution, unfortunately, so we can't calculate its concentration.
What you need for enzyme concentrations are values of enzyme activity numbers (in e.g. umol substrate/min) which you then can relater to a given volume of enzyme preparation. The kcat or specific activity values (mol/(mol s) and umol/(min mg protein), respectively) are intrinsic characteristcs of the enzyme, and you need additional data (like the total activity or the protein amount in your preparation) to get information on enzyme concentration.
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