Finding specific activity of an enzyme from Km values for a reversible enzyme reaction

Question

I have an enzymatic equation in the form:

$$E + S\xleftrightarrow[k_{-1} = 2.6\text{ x }10^{-2} \text{s}^{-1}]{k_1 = 3 \text{ x }10^8 \text{M}^{-1}\text{s}^{-1}} EX \xleftrightarrow[k_{-2} = 3.8\text{ x }10^{-8} \text{M}^{-1}\text{s}^{-1}]{k_2 = 1.45 \text{ x }10^3 \text{s}^{-1}} E + P$$

For the forward reaction, I have calcluated the $K_m$ value to be:

$${K_m}_f = \frac{k_{-1}+k_2}{k_1} = 4.83\text{ x }10^{-6}\text{M}$$

And for the reverse, I calculated $K_m$ to be:

$${K_m}_r = \frac{k_{-1}+k_2}{k_{-2}} = 3.82\text{ x }10^{-6}\text{M}$$

From this info, I need to somehow calculate the specific activity of the enzyme in both the forward and reverse reactions.

The weight of one enzyme unit is 50,000 kDa.

Specific activity is defined as the number of 'Units' per mg of protein.

One enzyme unit is defined as the amount of enzyme that catalyzes the formation of 1 μmol of product per min under optimal assay conditions, which are assumed for this question (saturation of substrate in forward reaction / saturation of product in reverse reaction).

Can anyone point me in the right direction? I have a feeling that this is related to the $V_{max}$ term in the Michaelis-Menten equation:

$$v = \frac{V_{max}\text{[S]}}{K_m + \text{[S]}}$$

In that $K_m = \text{[S] at}\frac{1}{2}V_{max}$.

So, $$[S]_{f} \text{ at } V_{max} = 2(4.83\text{ x }10^{-6}\text{M}) = 9.66\text{ x }10^{-6}\text{M}$$

So, $$[S]_{r} \text{ at } V_{max} = 2(3.82\text{ x }10^{-6}\text{M}) = 7.64\text{ x }10^{-6}\text{M}$$

but I dont know how to relate it to specific activity.

Answer 1

If the enzyme is staturated with substrate, the reaction rate will be $k_2$.

1450 moles per second, per mole of enzyme.

87,000 moles per minute, per mole of enzyme.

87,000,000,000 micromoles per minute, per mole of enzyme.

Based upon "50,000 kDa" (which looks like a typo, 50 kDa sounds more reasonable), one mole of enzyme is 50,000,000 grams or 50,000,000,000 milligrams.

So 87/50 = 1.74 units per milligram.

Then for when product is staturated, use $k_{-1}$ instead of $k_2$.