Non competitive inhibitors affects the function of enzymes and slows the rate of reaction. Can I say the same for uncompetitive inhibitors ? Because Vmax decreases and it takes longer to form products .

However when I plot the graph both gradients needs to be the same ? So I'm a little confuse here to whether for uncompetitive inhibitors, it does affect the function of enzymes and slow down the rate of reaction .

  • $\begingroup$ A competitive inhibitor does not affect Vmax. An uncompetitive inhibitor decreases the apparent Vmax but has no effect on the specificity constant (Vmax/Km). A mixed inhibitor affects the apparent values of both Vmax and the specificiity constant. A non-competitive inhibitor (A MUCH ABUSED TERM) is best thought of as a special (unlikely) case of mixed inhibitor were both of these kinetic constants are affected to the same extent. See here $\endgroup$ – user1136 Aug 24 '17 at 10:23
  • $\begingroup$ In addition, Lineweaver-Burk (L-B) plots of competitive inhibition intersect on the y-axis. L-B plots of uncompetitive inhibition are parallel, L-B plots of mixed inhibition intersect at a point but not on either axis, and L-B plots of non-competitive inhibition intersect on the (minus) x-axis: there is no change in the Michaelis constant in non-competitive inhibition (as both the 'fundamental' kinetic constants are changed to the same extent -an unlikely scenario?) $\endgroup$ – user1136 Aug 24 '17 at 10:35

I am not really sure to understand completely your question, but both non competitive and uncompetitive inhibitors affect the rate (Vmax) of the reaction by decrease it. In fact in both case, the enzyme (E) takes longer to transform the substrates (S) in to products (or to release it).

To better allow you understand I put below two mechanism of action for uncompetitive inhibition and non competitive inhibition.

Uncompetitive inhibition

Uncompetitive inhibition is unique in that the inhibitor binds to the enzyme-substrate complex. This could imply that the binding site for the inhibitor is accessible only after the enzyme has bound to its substrate. This reduction in the effective concentration of the E-S complex increases the enzyme's apparent affinity for the substrate through Le Chatelier's principle (Km is lowered) and decreases the maximum enzyme activity (Vmax), as it takes longer for the substrate or product to leave the active site.

Non competitive inhibition

Non-competitive inhibition models a system where the inhibitor and the substrate may both be bound to the enzyme at any given time. When both the substrate and the inhibitor are bound, the enzyme-substrate-inhibitor complex cannot form product and can only be converted back to the enzyme-substrate complex or the enzyme-inhibitor complex [...]

This type of inhibition reduces the maximum rate of a chemical reaction without changing the apparent binding affinity of the catalyst for the substrate.

Hope this could help.

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