[Babcock_Hall]

We have been trying to write rate equations for four cases of enzyme inactivation via covalent modification and protection against inactivation by a reversibly bound molecule.  I is the inactivator, and P is the protecting agent, which could be a substrate, product or reversible inhibitor of E.  We are following up the work of Jack Kyte, who was the first person that I know of to derive an equation for inactivation in which the inactivator decays via a first-order process (J. Biological Chemistry 1981).  When the inactivator does not decay and is present in excess, the loss of enzyme activity follows a first-order decay.

I.   stable inactivator; protection is complete under saturating amounts of P
II.  stable inactivator; protection is incomplete at saturating P
III. inactivator decomposes; protection is complete
IV.  inactivator decomposes; protection is incomplete.

I have a Chemdraw scheme if anyone wants to see it.  When one performs inactivation studies with enzymes, it is typical to follow the fraction of enzyme that remains active with time of incubation with I.  Free enzyme E and E•P are catalytically active, and E-I and E-I•P are inactive.  E-I symbolizes a covalently modified form of the enzyme E.  I am assuming that the binding of P to E is rapid and reversible.

The rate constant for reaction of I with E is k_i, and the rate constant for the reaction of I with the E•P complex is k_ip.  K_p is the dissociation constant of P from E.  With some help I have worked out integrated forms of rate equations for these four cases.  The equations look similar to some that have been derived for similar cases.  My question involves the beginning of our derivations.  For case IV I wrote

d{[E-I] + [E-I•P]}/dt = k_i[E] + k_iP[E•P]

I am not sure that it is OK to have a sum on the lefthand side.  I have searched for information on how to write rate equations which include rapid equilibrium steps, but it would seem that I have looked in the wrong places.  Most of the journal articles I have seen do not present derivations.  Any thoughts or advice would be appreciated.

[Babcock_Hall]

Here is a ChemDraw figure in png format.  The top figure covers the first two cases, and the bottom figure covers the last two cases.  When k_iP is zero, protection can be defined as being complete.

[mjc123]

Try writing [E_act] = [E] + [E.P]    [E] = [E_act]α      [E.P] = [E_act](1-α)
K_p = [E.P]/[E][P]    K_p[P] = (1-α)/α     α = 1/(K_p[P] + 1)
d[E_act]/dt = -[E_act][I ]{k_iα + k_ip(1-α)}
If conditions are such that [P] is fairly constant, and thus so is α, this is easy. Bit more complicated if it isn't.

[Babcock_Hall]

I will look over what you have written tonight.  The concentration of P is constant in one experiment.  For case IV the solution I had found was:

ln(fraction of active enzyme) = (-1)[Inactivator]₀(k_app/k')(1 - exp(-k't)). 

I defined k_app = (k_i + k_iP[P]/K_P)/(1 + [P]/K_P)

BTW, In accordance with biochemical convention, I defined K_P as a dissociation constant, but this is an unimportant detail.  The situation in which the inactive form of I, namely I', itself binds to the enzyme would be an interesting question to explore.

[mjc123]

That assumes k' >> k_app[E_act]₀?

[Babcock_Hall]

I don't remember making that assumption.  One often assumes that E, P, and E•P are in rapid equilibrium in these situations.  I'll see if I can upload my derivation later.

[Babcock_Hall]

I may have lost some formatting, but this is my derivation.

[mjc123]

When you write

Replacing [I ] with a function that describes its decay,
d{[E-I] +[E-I¥P]}/dt = {[Etot] Ð ([E-I] + [E-I¥P])} /(1 + [P]/KP){ki + kiP[P]/KP}[I0]exp(-kÕt)
kÕ   the first-order rate constant for breakdown of I to IÕ

You are effectively assuming that decomposition of I is much faster than reaction with enzyme, so the decay of [I ] can be described solely in terms of the former. I agree with the rest of your derivation. (and yes, you lost a bit of formatting!)

[Babcock_Hall]

At this site we cannot upload .doc or .docx files, and when I used a .txt files, I lost the Greek letters.  I can put them back in here, if we need them.  I reread a portion of Jack Kyte's 1981 paper, and he assumed that [E-I] was negligible with respect to [Inactivator]₀.  This is usually true with respect to the relative concentrations of inactivator versus enzyme, although there are a few exceptions.  I am not sure whether or not this addresses the point you were making.

[Babcock_Hall]

We fit some data to our equation, and it did not look bad.  We are assaying a protein tyrosine phosphatase, and arsenate is the protecting agent.

[Babcock_Hall]

This should look better.  It is a pdf file.