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    So there's a graph of how the rate of reaction is affected by increasing the substrate concentration (it increases to a certain point, then plateaus as the enzyme's became the limiting factor). If a competitive inhibitor was involved why would the graph plateau LATER? I don't get it surely it should plateau faster seeing as the active sites are more likely to be occupied.
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    Where a competitive inhibitor is involved, the inhibitor is competing with the actual substrate molecules for all the active sites available. Assuming it's a reversible competitive inhibitor, imagine that with equal amounts of inhibitor and substrate, 50% of the enzymes will have proper substrate, and 50% the inhibitor, so the reaction rate will be going half as fast.

    Remember that only the enzyme-substrate complexes can contribute to the rate of reaction, not the enzyme-inhibitor complexes. I think this is where you may be confused; yes all the active sites may be filled, but not necessarily contributing to the reaction.

    In order to minimise the effect of the inhibitor involved, there needs to be WAY more substrate molecules than inhibitor, so that the maximum number of active sites can be filled with the substrate, not the inhibitor, and give the max rate of reaction.
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    (Original post by Qaisha)
    Where a competitive inhibitor is involved, the inhibitor is competing with the actual substrate molecules for all the active sites available. Assuming it's a reversible competitive inhibitor, imagine that with equal amounts of inhibitor and substrate, 50% of the enzymes will have proper substrate, and 50% the inhibitor, so the reaction rate will be going half as fast.

    Remember that only the enzyme-substrate complexes can contribute to the rate of reaction, not the enzyme-inhibitor complexes. I think this is where you may be confused; yes all the active sites may be filled, but not necessarily contributing to the reaction.

    In order to minimise the effect of the inhibitor involved, there needs to be WAY more substrate molecules than inhibitor, so that the maximum number of active sites can be filled with the substrate, not the inhibitor, and give the max rate of reaction.
    Thank you
 
 
 
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