Proteins and Enzymes

ID #2329

Just to clarify: Competitive and non-competitive inhibitors do not stay in the active site for very long because they do not have a covalent bond with the active site, correct? So the fact that it's weaker just makes it fall out quicker than a natural substrate because it has a covalent bond (the natural substrate)?


Two answers your question, I'll break down what you wrote into to parts. First, you wrote: "Competitive and non-competitive inhibitors do not stay in the active site..." 
Be careful to keep the binding sites of competitive and non-competitive inhibitors separated. Competitive inhibitors bind to the active site and compete with the natural substrate. Non-competitive inhibitors do not bind to the active site, but instead bind an allosteric site (aka a site in the enzyme specific for the non-competitive inhibitor that is anywhere else other than the active site). 
"...(they don't stay at their sites) for very long because they do not have a covalent bond with the active site, correct? So the fact that it's weaker just makes it fall out quicker than a natural substrate because it has a covalent bond (the natural substrate)?" 
Regardless of which type of inhibitor we're discussing, it's better to think of the competitive inhibitor, non-competitive inhibitor, or the natural substrate association with the enzyme as fluctuating (kind of in a bound, then not bound but still nearby the binding site, then bound, then not bound but still nearby...etc. association) instead of thinking of it as two things being stuck together for an extended period of time. 
The natural substrate will remain associated with the active site for as long as it takes the enzyme to convert it to product. Also, the natural substrate does NOT have to be covalently bonded to the active site (check out the animation on sucrase under lecture 6 on the lecture notes part of the website to see an example of this). 
The competitive inhibitor will remain associated with the enzyme's active site in that fluctuating state because the enzyme can't convert it to product. It stays that way until something else comes along that binds the enzyme during the competitive inhibitor's unbound but still nearby part of the fluctuation. This is how two things (ie. the competitive inhibitor and the natural substrate) can compete for the same space at the same time. This is also why it is important to know (or to be told) how much of each there are in the solution, because a larger amount (or larger concentration) of one means its more likely to be at the active site at any given time. 

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