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    why does rate of reaction level off when we have a non-competitive inhibitor;

    Y axis = rate
    x = substrare conc.
    thanks
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    The rate of reaction levels off because non-competitive inhibitors bind irreversibly to the enzyme at a site away from the active site (called the allosteric site). This affects the 3D makeup of the enzyme and changes the shape of its active site, meaning that substrates no longer fit in the active site. The fact that they bind irreversibly to the enzyme means that, no matter how much the concentration of substrate is increased (x axis), the maximum rate of enzyme activity will never be reached - hence the levelling off.
    Hope that helps!
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    couldnt have worded that any better than baconcornflakes, a small thing(very small) to add is that my teacher said the examiners like the word complementary. So in this case, inhibitor binds to enzyme, changes shape of active site, active site no longer complementary to substrate. Basically me repeating bacon
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    But then why does rate not reach and level off at zero if tertiary structure is affected
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    Does it have anything to do with the fact that they still function sligtly or not
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    The non-competitive inhibitor, if there's less of it than there is of enzyme, will only affect some of the enzymes, hence there will still be some enzyme activity.
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    The inhibitor does not have to stay on the enzyme it's attached to, it can affect many more enzymes. Unlike the competitive inhibitor, it doesn't compete with the substrate for the active site, so it can attach itself to the enzyme whenever it wants to
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    (Original post by baconcornflakes)
    The rate of reaction levels off because non-competitive inhibitors bind irreversibly
    They don't bind irreversibly, but the rest of your explaination is fine.



    (Original post by Eloades11)
    Unlike the competitive inhibitor, ir doesn't compete with the substrate for the active site, so it can attach itself to the enzyme whenever it wants to
    It's not binding to the active site, but the binding is still going to have some specificity - it's not just going to bind to the enzyme anywhere.

    EDIT: sorry, I read it as "wherever", not "whenever" :p:
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    (Original post by Markh1000)
    It's not binding to the active site, but the binding is still going to have some specificity - it's not just going to bind to the enzyme anywhere.

    EDIT: sorry, I read it as "wherever", not "whenever" :p:
    haha its fine, I see what you're trying to say
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    (Original post by jsmith6131)
    But then why does rate not reach and level off at zero if tertiary structure is affected
    The y-axis is the rate of reaction, the gradient is zero means that the rate will also be nought, not necessarily need to reach zero
 
 
 
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