Question

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.

Answer 1

17.2 minutes is the value of the turnover number.

Step-by-step explanation:

Using Michaelis-Menten equation:

`V = V_(max)* ([S])/( (Km + [S]))`

`V_(max )=k_(cat)* E_o`

Where :

`V_(max)` = max rate velocity

[S] = substrate concentration

`K_m` = Michaelis-Menten constant

V = reaction rate

`k_(cat)` = catalytic rate constant

`E_o` = initial enzyme concentration

We have :

`K_m=10.0mM`

`E_o=10.0\mu M=10.0* 0.001 mM`

`V_(max)=0.172 mM/min`

`V_(max)` is the rate is obtained when all enzyme is bonded to the substrate.
`k_(cat)` is termed as the turnover number.

`k_(cat)=(V_(max))/(E_o)=(0.172 mM)/(10.0* 0.001 mM)`

`=17.2 minutes`

17.2 minutes is the value of the turnover number.