How can a Michaelis–Menten formalism be used when enzyme concentration isn't constant?

Question

I understand that $V_\mathrm{max} = k_3[\ce{E}]_0$ in ordinary Michaelis–Menten (MM) kinetics. According to the lecture notes provided by my university (I don't believe they are available online), when the amount of enzyme is treated as a dynamic variable the reaction may be described by

$$\frac{\mathrm d[\ce{P}]}{\mathrm dt} = \frac{V'_\mathrm{max}[\ce{S}][\ce{E}]}{K_\mathrm{M}+[\ce{S}]}.$$

What does $V'_\mathrm{max}$ represent mathematically in this case? And how can such an equation be derived?

I am also told that this equation assumes that the concentration of the enzyme changes slowly compared to the change in $\ce{P}.$ Why?

Answer 1

d[P]/dt = (V'max[S][E])/(Km+[S])

This is a typo or uncommon usage. V'max would have the dimensions of rate divided by concentration, so it would something like a specific activity, but not quite; E would have to be the total enzyme concentration in this context.

The following are correct

$$ d[\mathrm{P}]/dt = k_\mathrm{cat} [\mathrm{ES}] \tag{1}$$

$$ d[\ce{P}]/dt = E_0 \ k_\mathrm{cat} [\ce{S}]/(K_\mathrm{m} +\ce{S}) \tag{2}$$

$$ d[\ce{P}]/dt = V_\mathrm{max} [\ce{S}]/(K_\mathrm{m} +\ce{S}) \tag{3}$$

(1) follows from the reaction scheme. (2) and (3) rely on the steady state approximation. The free enzyme concentration is not expected to show up in any formulation of a rate (because the free enzyme needs to bind to substrate first before it can engage in catalysis.

In ordinary MM kinetics, I understand that Vmax=k3*E0

k3 is an unusual choice for the rate constant of the second step (enzyme substrate complex reacting to yield product). It is often named $k_\mathrm{cat}$, while the rate constants for the binding step are often named $k_1$ and $k_{-1}$.