Rapid Equilibrium Random bisubstrate mechanism

Also, after rearranging, we obtain a usual velocity equation for the Rapid Equilibrium Random bisubstrate mechanism ... 

Why product inhibition occurs. The products of reaction are formed at the active site of enzyme and are the substrates for the reverse reaction. Consequently, a product may act as an inhibitor by occupying the same site as the substrate from which it is derived. In the Rapid Equilibrium Random bisubstrate mechanism, most ligand dissociations are very rapid compared to the interconversion of EAB and EPQ. Thus, the levels of EP and EQ are essmtiaUy zero in the absence of added P and Q. In the presence of only one of the products, the reverse reaction can be neglected, as the concentration of the other product is essentially zero during the early part of the reaction. Nevertheless, the forward reaction will be inhibited because finite P (or Q) ties up some of the enzyme. The type of this product inhibition depends on the number and type of enzyme-product complexes that can form. Consequently, product inhibition studies can be very valuable in the diagnostics of kinetic mechanisms (Rudolph, 1979). 

Fromm and Rudolph have discussed the practical limitations on interpreting product inhibition experiments. The table below illustrates the distinctive kinetic patterns observed with bisubstrate enzymes in the absence or presence of abortive complex formation. It should also be noted that the random mechanisms in this table (and in similar tables in other texts) are usually for rapid equilibrium random mechanism schemes. Steady-state random mechanisms will contain squared terms in the product concentrations in the overall rate expression. The presence of these terms would predict nonhnearity in product inhibition studies. This nonlin-earity might not be obvious under standard initial rate protocols, but products that would be competitive in rapid equilibrium systems might appear to be noncompetitive in steady-state random schemes , depending on the relative magnitude of those squared terms. See Abortive Complex... 

In an ordered bisubstrate mechanism one must vary the second substrate, B, and determine V/JSTb, regardless of whether the label is in A or B, since ly AwiU not show an isotope effect. For a random mechanism one must vary both A and B, since one may see different isotope effects on VfK and V/Aa, a distinction that may help to characterize the mechanism. The effects onV/K and V/Aa shouldbe different when one or both substrates are sticky, that is, dissociate more slowly from the enzyme than they react to give products. The substrate with the lower V7A is the sticky one. Larger effects on Vthan on either V7A show that both substrates are sticky smaller ones show that a slow step follows release of the first product. A rapid equilibrium random mechanism will show equal isotope effects on V, V/Ab, and F/Aa, aU larger than unity. 

This example clearly shows that completely randomized steady-state bisubstrate reactions wiU produce extremely complex rate equations which are, in most cases, unmanageable and almost useless for practical purposes. Thus, for example, the rate equation for an Ordered Bi Bi mechanism has 12 terms in the denominator (compare Eq. (9.8)). A completely Random Bi Bi mechanism yields an even more comphcated rate equation with 37 new terms in the denominator. Eor this reason, and in such cases, we shah usuahy revert to simplifying assumptions, usually introducing the rapid equilibrium segments in the mechanism in order to reduce the rate equations to manageable forms. 

---