Inhibitor binding ternary complex

An uncompetitive inhibitor is much like a noncompetitive inhibitor except that an uncompetitive inhibitor binds only the enzyme-substrate complex (Scheme 4.14). The inhibitor-bound ternary complex cannot form product. Uncompetitive inhibitors cause both Vmax and Km to decrease by the same factor (Figure 4.17). Because the slope of a Lineweaver-Burk plot is Km/Vmxi, the slope of the line of an inhibited enzyme is unchanged from the uninhibited enzyme.4... 

The catalytic subunit of cAPK contains two domains connected by a peptide linker. ATP binds in a deep cleft between the two domains. Presently, crystal structures showed cAPK in three different conformations, (1) in a closed conformation in the ternary complex with ATP or other tight-binding ligands and a peptide inhibitor PKI(5-24), (2) in an intermediate conformation in the binary complex with adenosine, and (3) in an open conformation in the binary complex of mammalian cAPK with PKI(5-24). Fig.l shows a superposition of the three protein kinase configurations to visualize the type of conformational movement. 

In such inhibition, the inhibitor and die substrate can simultaneously bind to the enzyme. The nature of the enzyme-inhibitor-substrate binding has resulted in a ternary complex defined as EIS. The Ks and Kt are identical to the corresponding dissociation constants. It is also assumed that the EIS does not react further and is unable to deliver any product P. The rate equation for non-competitive inhibition, unvAX, is influenced ... 

In non-competitive inhibition, the substrate (S) and inhibitor (I) have equal potential to bind to the free enzyme (E). The inhibitor forms a ternary complex with enzyme-substrate (ES) whereas the substrate will form another ternary complex with enzyme-inhibitor (El). Since the non-competitive inhibitor had no effect on the binding of substrate to the enzyme, the Km value remained consistent (or unchanged). There are two different ways for the formation of ESI ternary complex this complex would not form the product and therefore was decreased. Non-competitive inhibitor had no effect on substrate binding or the enzyme-substrate affinity, therefore the apparent rate constant (K ) was unchanged.5 A possible reason for product inhibition was because of the nature of 2-ethoxyethanol,... 

An inhibitor that binds exclusively to the ES complex, or a subsequent species, with little or no affinity for the free enzyme is referred to as uncompetitive. Inhibitors of this modality require the prior formation of the ES complex for binding and inhibition. Hence these inhibitors affect the steps in catalysis subsequent to initial substrate binding that is, they affect the ES —> ES1 step. One might then expect that these inhibitors would exclusively affect the apparent value of Vm and not influence the value of KM. This, however, is incorrect. Recall, as illustrated in Figure 3.1, that the formation of the ESI ternary complex represents a thermodynamic cycle between the ES, El, and ESI states. Hence the augmentation of the affinity of an uncompetitive inhibitor that accompanies ES complex formation must be balanced by an equal augmentation of substrate affinity for the El complex. The result of this is that the apparent values of both Vmax and Ku decrease with increasing concentrations of an uncompetitive inhibitor (Table 3.3). The velocity equation for uncompetitive inhibition is as follows ... 

Noncompetitive inhibitors, conversely, do not affect substrate binding, but produce a ternary complex (enzyme-substrate-inhibitor) which either decomposes slowly, or fails to decompose (i.e., is inactive). Consequently, the primary effect of a noncompetitive inhibitor is to reduce the apparent value of Vmax. 

Competitive inhibition involves (only) the substrate (S) and the inhibitor (I) competing for one type of site on the enzyme (E), in fast, reversible steps, followed by the slow decomposition of the complex ES to form product (P) the complex El is assumed to be inactive. The fact that there is only one type of binding site on the enzyme implies that a ternary complex EIS cannot be formed. 

Consider the standard Uni Uni mechanism (E + A EX E + P). A noncompetitive inhibitor, I, can bind reversibly to either the free enzyme (E) to form an El complex (having a dissociation constant K s), or to the central complex (EX) to form the EXl ternary complex (having a dissociation constant Xu). Both the slope and vertical intercept of the standard double-reciprocal plot (1/v vx. 1/[A]) are affected by the presence of the inhibitor. If the secondary replots of the slopes and the intercepts (thus, slopes or vertical intercepts vx [I]) are linear (See Nonlinear Inhibition), then the values of those dissociation constants can be obtained from these replots. If Kis = Xu, then a plot of 1/v vx 1/[A] at different constant concentrations of the inhibitor will have a common intersection point on the horizontal axis (if not. See Mixed-Type Inhibition). Note that the above analysis assumes that the inhibitor binds in a rapid equilibrium fashion. If steady-state binding conditions are present, then nonlinearity may occur, depending on the magnitude of the [I] and [A] terms in the rate expression. See also Mixed Type Inhibition... 

Substrate binding also induces a conformational change in this enzyme. When both coenzymes and substrate bind the "closed" conformation of the enzyme is formed by a rotation of the catalytic domains of the two subunits relative to the coenzyme-binding do-mains.50 51a Structures of ternary complexes with inhibitors and with substrates have also been established. 

A crystal structure of a ternary complex of horse liver alcohol dehydrogenase with NADH and the inhibitor, dimethyl sulfoxide, first at 4.5 A resolution1365 and a further refinement to 2.9 A resolution,1366 has been published by Eklund et al. The gross structure of the ternary complex is similar to that of the free enzyme structure. Each subunit is divided into a coenzyme-binding domain and a catalytic domain. The subunits are joined together near the... 

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