Please let me know the rate equation for galactose oxidase. If two molecules are attached in TS. Is it possible that rate is dependent on both the molecules?
The enzymatic reaction with galactose oxidase is second order kinetics. This reaction was studied extensively and some of these studies are shown below.
At concentrations below Km, apparent second-order rate constants for protio-substrate oxidation (kred = 1.5 x 10(4) M-1 s-1) and O2 reduction (kox = 8 x 10(6) M-1 s-1) have been estimated from measurements both by steady-state oxygen electrode and by enzyme-monitored turnover (Biochemistry. 1998 Jun 9;37(23):8426-36. ).
The order of the reaction is determined on the rate-limiting step. In the case two compounds are involved in the rate-limiting step the reaction will be second order, however, if one of them is in large access the reaction will be pseudo first order. If only one compound is involved in the TS the reaction will be first order.
1-Biochemistry. 1998 Jun 9;37(23):8426-36.
Kinetic isotope effects as probes of the mechanism of galactose oxidase.
Whittaker MM1, Ballou DP, Whittaker JW.
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Abstract
Galactose oxidase (GO) is a member of the family of radical-coupled copper oxidases, enzymes containing a free radical coordinated to copper in the active site. In catalysis GO cycles between an oxidized state (comprising Cu(II) with a unique cysteinyl-tyrosine radical) and a reduced state (comprising Cu(I) with the singlet cysteinyl-tyrosine) as it catalyzes the two-electron oxidation of alcohols to aldehydes and the subsequent reduction of O2 to H2O2. A ping-pong mechanism involving radical intermediates has been proposed for GO catalysis. Previous steady-state kinetics studies have demonstrated a KIE of 7-8 that was attributed to substrate oxidation, a process involving the stereospecific abstraction of the pro-S hydrogen from the 6-hydroxymethyl group of galactose. We have used rapid kinetics methods to measure the anaerobic reduction of GO substrate at 4 degreesC and carry out enzyme-monitored turnover experiments using 6-protio and 6-deutero substrates, both in H2O and D2O. At concentrations below Km, the apparent second-order rate constant for protio-substrate oxidation, kred, was 1.59 x 10(4) M-1 s-1, while that for deuterated substrate was 7.50 x 10(2) M-1 s-1, a KIE of 21.2. Steady-state measurements of oxygen consumption at low galactose concentrations reveal an unusually large isotope effect (kH/kD = 22.5 +/- 2) for oxidation of 1-O-methyl-6, 6'-di-[2H]-alpha-d-galactopyranoside, and at high galactose concentrations, where the oxygen half-reaction is rate-limiting in catalysis, a surprisingly large KIE (kH/kD = 8 +/- 1) for the reduction of O2 to H2O2. There is no detectable solvent isotope effect (
If the enzyme indeed follows steady-state Ping Pong bisubstrate kinetics as mentioned in Rafik's first paper, and there is no funny business like substrate inhibition or cooperativity, then the kinetic rate equation for the forward reaction may be written as follows:
V = (Vmax[A][B])/(KmA[B] + KmB[A] + [A][B])
[A] and [B] are the concentrations of the 2 substrates, and A is the first substrate in the reaction. KmA and KmB are the Michaelis constants for substrate A and B, respectively. Now, which substrate is A and which is B?
Thank you Dr. Rafik and Dr. Adam. Rate equation must follows Michaelis equation. However, In one of our model system it follows normal 2nd order kinetics as stated by Dr. Rafik. Is there any discrepancy?