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 .


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.