BIOCHEMISTRY  DR. JAKUBOWSKI
06/12/2014
Learning Goals/Objectives for Chapter 6C: After class and this reading, students will be able to

Reversible uncompetitive inhibition occurs when I binds only to ES and not free E. One can hypothesize that on binding S, a conformational change in E occurs which presents a binding site for I. Inhibition occurs since ESI can not form product. It is a dead end complex which has only one fate, to return to ES. This is illustrated in the chemical equations and molecular cartoon below.
Let us assume for ease of equation derivation that I binds reversibly to ES with a dissociation constant K_{i}i. The second "i" in the subscript "ii" indicates that the intercept of the 1/v vs 1/S LineweaverBurk plot changes while the slope stays constant. Kii is also named K_{i}u, where the subscript "u" stands for the uncompetitive inhibition constant.
A look at the top mechanism shows that in the presence of I, as S increases to infinity, not all of E is converted to ES. That is, there is a finite amount of ESI, even at infinite S. Now remember that Vm = kcatEo if and only if all E is in the form ES . Under these conditions, the apparent Vm, Vmapp is less than the real Vm without inhibitor. In addition, the apparent Km, Kmapp, will change. We can use LaChatelier's principle to understand this. If I binds to ES alone, and not E, it will shift the equilibrium of E + S <==> ES to the right , which would have the affect of decreasing the Kmapp (i.e. it would appear that the affinity of E and S has increased.). The double reciprocal plot (Lineweaver Burk plot) offers a great way to visualize the inhibition. In the presence of I, both Vm and Km decrease. Therefore, 1/Km, the xintercept on the plot, will get more negative, and 1/Vm will get more positive. It turns out that they change to the same extent. Therefore the plots will consist of a series of parallel lines, which is the hallmark of uncompetitive inhibition.
An equation, shown in the diagram above, can be derived which shows the effect of the uncompetitive inhibitor on the velocity of the reaction. The only change is that the S term in the denominator is multiplied by the factor 1+I/Kii. We would like to rearrange this equation to show how Km and Vm are affected by the inhibitor, not S, which obviously isn't. Rearranging the equation as shown above shows that Kmapp = Km/(1+I/Kii) and Vmapp = Vm/(1+I/Kii). This shows that the apparent Km and Vm do decrease as we predicted. Kii is the inhibitor dissociation constant in which the inhibitor affects the intercept of the double reciprocal plot. Note that if I is zero, Km and Vm are unchanged.
Java Applet: Uncompetitive Inhibition
Wolfram Mathematica CDF Player  Uncompetitive Inhibition v vs S (free plugin required)
Interactive SageMath Uncompettive Inhibition v vs S
4/6/14Wolfram Mathematica CDF Player  Uncompetitive Inhibition  Lineweaver Burk(free plugin required)
Interactive SageMath Uncompettive Inhibition Lineweaver Burk
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