An enzyme acts as a catalyst in the fermentation of A to form product R. An aque
ID: 516436 • Letter: A
Question
An enzyme acts as a catalyst in the fermentation of A to form product R. An aqueous feed stream containing the enzyme and compound A flows into a CSTR at 25 L/min, and the initial concentration of A is 2 mol^- L^-1. Determine the turnover number in min^-1 and expressed to two decimal places if the total enzyme concentration is 2.5 mol^- L^-1. If needed, you may assume that the total enzyme concentration and volumetric flow rates are constant and the reaction follows Michaelis Menten kinetics. The maximum rate of destruction of the substrate is 0.4 mol/(L min). When the substrate concentration is 0.5 mol/L, the rate of destruction of the substrate is 0.2 mol/(L min). i.e. half of the maximum rate. Added thought question (not to be answered here): What does the turnover number physically refer to?Explanation / Answer
Turn over number= Vmax/ET, ET= concentration of Enzyme = 2.5 mol/L, Vmax= 0.4 mol/L.min
Hence turn over number= 0.4/2.5 min=0.16/min
Turn over no signifies the no of conversions of substrate ( A in this case) molecules to R in a single active site of an enzyme. So it siginifes the efficiency of an enzyme