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

Answer:

17.2 minutes is the value of the turnover number.

Explanation:

Using Michaelis-Menten equation:

$V = V_{max}\times \frac{[S]}{ (Km + [S])}$

$V_{max }=k_{cat}\times E_o$

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 :

$K_m=10.0mM$

$E_o=10.0\mu M=10.0\times 0.001 mM$

$V_{max}=0.172 mM/min$

$V_{max}$ is the rate is obtained when all enzyme is bonded to the substrate. $k_{cat}$ is termed as the turnover number.

$k_{cat}=\frac{V_{max}}{E_o}=\frac{0.172 mM}{10.0\times 0.001 mM}$

$=17.2 minutes$

17.2 minutes is the value of the turnover number.