Enzyme X converts substrate S to product P. When the total concentration of X is 1 μM, Vmax and Km are 1.0 x 10^-3 M/s and 0.10 mM, respectively. What is the value for the kcat/Km?

I'm stuck because my notes indicate that we need to know the concentration of the substrate [S] to apply any of the equations. I know that if [S] >> Km, you can adjust the Michaelis-Menten equation to be (Kcat)(X) = Vmax, but the problem doesn't tell me [S], and when I tried solving for Kcat using the equation above, and then dividing that by the given Km, I didn't get the correct answer. Thanks!

Feb 3, 2018

${k}_{\text{cat}} / {K}_{m} = 10.0$

Explanation:

It depends on some of the simplifying assumptions (or reasons) as shown in the referenced paper.
$X + S \to P$
$v = {V}_{\text{max}} \frac{S}{K} _ m = k ' \left[S\right]$
$k ' = {V}_{\text{max}} / {K}_{m} = 1.00 \times {10}^{- 3} / 0.10 = 0.010$

$v = {k}_{\text{cat}} \left[E\right] \frac{S}{K} _ m = k ' ' \left[E\right] \left[S\right]$
$k ' ' = {k}_{\text{cat}} / {K}_{m}$

$v = k ' \left[S\right]$, so $v = 0.010 \left[S\right]$ and then

$k ' ' \left[E\right] \left[S\right] = 0.010 \left[S\right]$

${k}_{\text{cat}} / {K}_{m} = \frac{0.010 \left[S\right]}{\left[E\right] \left[S\right]} = \frac{0.010}{E}$

$\left[E\right] = 1 u M = 0.001 m M$

${k}_{\text{cat}} / {K}_{m} = \frac{0.010}{0.001} = 10.0$

https://www.ou.edu/OpenEducation/ou-resources/biochemical-methods/lab-11/michaelis-menten-derivation.pdf

Other Values and interactive graph here:
http://www.physiologyweb.com/calculators/michaelis_menten_equation_interactive_graph.html