- The enzyme works at a rate of 2 µM/min • ml
- Total volume of the solution containing the enzyme and its substrate = 600 ml
Dear Ivanildo,
For calculating the quantity of substrate degraded by an enzyme, you should indicate the time (in minutes) of incubation of the substrate with enzyme. If you have this number, you can multiply with the rate of the reaction and the total volume of the reaction. Then you convert the number of µM to grams using the molecular weight of the substrate
Hoping this will be helpful,
Rafik
Dear Ivanildo,
For calculating the quantity of substrate degraded by an enzyme, you should indicate the time (in minutes) of incubation of the substrate with enzyme. If you have this number, you can multiply with the rate of the reaction and the total volume of the reaction. Then you convert the number of µM to grams using the molecular weight of the substrate
Hoping this will be helpful,
Rafik
Did you mean 2 micromoles/(min-ml) or 2 micromolar/(min-ml)? The first one makes sense; the second one doesn't because it has 2 volume units in it [micromoles/(liter-min-ml].
If the amount of time elapsed during the measurement of 2 micromoles/(min-ml) was the same amount of time elapsed during the reaction in the 600 ml volume, then it would be OK to just multiply 2 micromoles/min/ml by the reaction time and the volume to calculate the number of micromoles formed.
However, if the 2 micromoles/(min-ml) measurement was the initial rate of the reaction, but the reaction in the large volume occurred over a longer time than the initial rate measurement, then it is possible that the reaction rate slowed down over time. In that case, the simple multiplication would overestimate the amount of product formed.
The best way to measure the amount of product present after the reaction in the 600 ml volume is to directly measure it by some analytical procedure, such as HPLC or mass spectrometry, using a standard curve for comparison.
Dear Adam Shapiro,
Forgive me for the delay! Really the degradation rate is in micromoles/min/ml. My question is if we multiply directly 2micromoles x 60min x 12-h = 1440 micromoles or if is there some biochemistry formula for this calculation. Thank you in advance for your willingness to solve this question.
Best Regards,
Ivanildo C Medeiros
Dear Prof. Rafik Karaman,
I thought that for the calculations will be necessary to use some formula of enzyme kinetics. According with your answer I can multiply 2 micromoles x 60 min x 12-h x 600 ml = 864000 micromol em 12-h. All rigth? Thank you in advance your kindness in help me.
Best Regards,
Ivanildo C Medeiros
Dear Mirza Arshad,
Thank you very much for you kind attention!
Ivanildo C Medeiros
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