Question

For an enzymatically catalyzed reaction in which the measured values for KM and Vmax are, respectively, 6.20 mM and 1.02 mM min-1, what is the value of the turnover number? The concentration of enzyme used in the reaction is 50.0 μM. turnover number =

Answer 1

Michaelis–Menten equation is used typically in enzymatic reactions.

V = Vmax*[S] / (Km + [S])

Where

Vmax = max rate velocity

[S] = substrate concentration

Km = Michaelis–Menten constant

V = reaction rate

Michaelis–Menten equation is used typically in enzymatic reactions.

V = Vmax*[S] / (Km + [S])

Where

Vmax = max rate velocity

[S] = substrate concentration

Km = Michaelis–Menten constant

V = reaction rate

Then, for a turnover number

Kcat = Vmax/[ET]

Vmax = 1.02 mM/min

[ET] = 50 microM = 50*10^-3 mM

Kcat(turnover) = 1.02/(50*10^-3) = 20.4 min