# Calculating Vmax (or Kcat) given specific activity of an enzyme over length of time incubated with an activator

Is it possible to figure out the $V_\mathrm{max}$ (or $k_\mathrm{cat}$) and/or $K_\mathrm m$ from a plot of enzyme activity expressed as specific activity over time (i.e. $\mathrm{U/mg}$ on the $Y$ and time on the $x$ axis). I've figured out how to express specific activity as turnover but I'm not sure this helps with calculating $V_\mathrm{max}$ or $k_\mathrm{cat}$.

The reaction: p38 phosphorylates MSK1 with Michaelis–Menten kinetics

Experimental conditions: $1\ \mathrm{U/mL}$ of p38 was added at time $t=0\ \mathrm s$ to a solution containing MSK1. They don't say how much MSK1 but express its activity as $\mathrm{U/mg}$ (which is analogous to $\mathrm{(\mu mol/min)/mL}$, otherwise known as specific activity)

The Data:

Time         MSK1 activity
(s)          ((µmol/min)/mg)
__________________________________________________
0               0
300             175
600             350
1200             450
1800             475
2400             500


Calculated turnover:

Time         MSK1 turnover
(s)          (1/(nmol s))
__________________________________________________
18000          300
36000          600
72000         1200
108000         1800
144000         2400
0            0


In actual fact, I though $k_{\rm cat}$ was the turnover and that this was a constant, specific for each enzyme. This data however suggests that as the reaction progresses more MSK1 is produced the turnover number increases, which makes sense and therefore calls into question the two definitions of $k_{\rm cat}$.

## 1 Answer

The enzyme catalysis reaction following MM kinetics can be depicted as:

$$\Large\ce{E + S<=>[k_f][k_r]ES->[k_{cat}]E + P }$$

Note that $V_{max} = k_{cat}.[E_0]$ i.e. conversion rate constant (turnover number) times the total amount of enzyme.

You should note that the enzyme here is p38 and substrate is MSK1 (which in turn is an enzyme). When MSK1 gets phosphorylated, it assumes an active form. MSK1 activity (and specific activity) would reflect the concentration of phosphorylated MSK1 (MSK1-P) i.e. the product. Note that the $k^{^{MSK1-P}}_{cat}$ is its intrinsic property and it does not change; only the $[E^{^{MSK1-P}}_0]$ changes.

I hope your doubt is clear.