Question

At high (saturating) substrate concentrations, the rate of an enzyme-catalyzed reaction approaches Vmax. How close does the reaction rate actually get to Vmax? Determine how high (i.e. how many times Km) the substrate concentration must be for the reaction rate to be:

a. 98% Vmax (show your work) (2)

b. 99% Vmax (answer only) (1)

c. 99.9% Vmax (answer only) (1)

Answer 1

The rate of an enzymatic reaction can be calculate using the Michaelis-Menten model, which states that the rate is given by:

$rate=\frac{V_{max}[S]}{K_{M}+[S]}$

When the rate is 98% V_max, we have:

$0.98\cdot V_{max}=\frac{V_{max}[S]}{K_{M}+[S]}$

$0.98=\frac{[S]}{K_{M}+[S]}$

Which can be rearranged to calculate the concentration of substrate as a function of K_M.

$0.98\cdot (K_{M}+[S])=[S]$

$0.98K_{M}=0.02[S]$

$[S]=\frac{0.98K_{M}}{0.02}=49K_{M}$

With an analogous calculation, for rate = 99% of Vmax we arrive to:

$[S]=\frac{0.99K_{M}}{0.01}=99K_{M}$

And for rate = 99.9% of Vmax:

$[S]=\frac{0.999K_{M}}{0.001}=999K_{M}$