Question

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!

Answer 1

`k_"cat"/K_m = 10.0`

Explanation:

It depends on some of the simplifying assumptions (or reasons) as shown in the referenced paper.
`X + S -> P`
`v = V_"max"[S]/K_m = k'[S]`
`k' = V_"max"/K_m = 1.00 xx 10^(-3)/0.10 = 0.010`

`v = k_"cat"[E][S]/K_m = k''[E][S]`
`k'' = k_"cat"/K_m`

`v = k'[S]`, so `v = 0.010[S]` and then

`k''[E][S] = 0.010[S]`

`k_"cat"/K_m =( 0.010[S])/([E][S]) = 0.010/[E]`

`[E] = 1uM = 0.001mM`

`k_"cat"/K_m = 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