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.

  • $\begingroup$ v only equals Vmax in the limit of infinite substrate concentration. $\endgroup$
    – DavePhD
    Jan 26 '15 at 18:41
  • $\begingroup$ @DavePHD I suppose that makes my substrate concentration calculations kind of pointless, but do you have any other ideas for what I can do to get these values? $\endgroup$
    – Kestrel
    Jan 26 '15 at 18:56
  • 1
    $\begingroup$ The Michaelis-Menten equation that you have is derived assuming there is no reverse reaction (from product to the enzyme-substate complex). See sys-bio.org/wp-content/uploads/downloads/2012/03/… for the corresponding equation considering reversiblity. Also, I never heard of an assay with saturated product. The way you say "saturation of substrate in forward reaction / saturation of product in reverse reaction" sounds like an equilibrium situation, where there is no net formation of product. $\endgroup$
    – DavePhD
    Jan 26 '15 at 20:01
  • $\begingroup$ @DavePHD This is part of an assignment question. The exact question is: "If the subunit molecular weight of the enzyme is 50,000 kDa, calculate the specific activity if: a) there are conditions of saturating substrate, and b) if there are conditions of saturating product. $\endgroup$
    – Kestrel
    Jan 26 '15 at 20:26

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 $k_2$.

  • $\begingroup$ Thank you so much. If you wouldn't mind, could you explain some of your thought process behind your solution? This one really stumped me. $\endgroup$
    – Kestrel
    Jan 26 '15 at 20:32
  • $\begingroup$ well, if you like math the reasoning would be consider equation 3.2 sys-bio.org/wp-content/uploads/downloads/2012/03/… in the limit of infinite substrate, and it becomes k2 times the enzyme concentation (Et). If you don't like math the reasoning would be if the enzyme is staturated with substrate, only what happens to the enzyme-substrate complex affects the rate. $\endgroup$
    – DavePhD
    Jan 26 '15 at 20:41

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