For an enzymatically catalyzed reaction in which the measured values for KM and Vmax are, respectively, 10.0 mM and 0.172 mM min-1, what is the value of the turnover number? The concentration of enzyme used in the reaction is 10.0 μM.

17.2 minutes is the value of the turnover number.

Explanation:

Using Michaelis-Menten equation:

Where  :

= max rate velocity

[S] = substrate concentration

= Michaelis-Menten constant

V = reaction rate

= catalytic rate constant

= initial enzyme concentration

We have :

is the rate is obtained when all enzyme is bonded to the substrate. is termed as the turnover number.

17.2 minutes is the value of the turnover number.