Km and Vmax

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coolchix321

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What does Km tell you exactly? Whats the point?
Why does Km increase when there is a competitive inhibitor?
Why does Vmax decrease with a noncompetitive inhibitor?

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Km and Vmax are related to enzyme kinetics in a biological system. Km is the substrate concentration that is required for the reaction to occur at 1/2 Vmax. In other words, it is how much substrate is needed for the reaction to occur at 1/2 its max possible rate. Obviously Vmax is the maximum rate that the reaction can proceed at.

The reason Km increases with a competitive inhibitor is because the inhibitor is directly competing with the substrate for a fixed number of active sites on enzymes. Hence, Km increases. If there is a competitive inhibitor you will need more substrate to get the same 1/2 Vmax (this is why Km increases).

Vmax decreases with a noncompetitive inhibitor because the inhibitor does not bind to the active site. It binds allosterically so it is not competing directly. Its almost as if the substrate cannot win this binding race. So overall the reaction will not occur as fast as it could.
 
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For your first question, the most simple answer is Km tells you binding affinity. In general, if you have a very high Km value, the binding affinity is small. A low Km means binding is strong. Don't memorize here... realize if Km is small, that means that it requires less substrate to get the enzyme to do it's thing. The enzyme loves it, and it binds tightly. A high Km value means you need a lot of substrate to get the enzyme to act on it. Enzyme feels kinda meh about the substrate in this case.

Vmax is simply the rate at which enzyme's do their thing. This is concentration dependent. The more enzymes you have, the higher the Vmax because even more enzyme's are doing their thing.

Now, knowing this information we can apply it to certain inhibitors. Noncompetitive inhibitors bind else where on the enzyme. The active site becomes pretty messed up due to this because the binding of the inhibitor else where on the enzyme changed the enzyme's conformation. Remember the active site and substrate are "lock and key", if you beat the hell out of the lock, the key won't properly fit anymore. If the key doesn't properly fit anymore, the lock necessarily will not unlock at the turn of the key either. Ie. the rate at which the enzyme does its thing (Vmax) will go down.

Now you have a competitive inhibitor. Remember Km is the value which tells us the affinity for the enzyme to the substrate. In competitive inhibition there is something other than the original substrate binding to the enzymes' active site. So, before it was just the enzyme and substrate, it could easily take it up and bind it. Now you have the substrate + the inhibitor. The enzyme is less likely to pick up the actual substrate because now it can bind the inhibitor too, i.e affinity for the substrate goes down... which means Km value goes up (due to reasons explained in first paragraph).

Hope this helps. Sorry if I made a mistake... if something seems off lemme know, no time to proof read lol
 
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Thanks for your help
could you please elaborate on this
i am still somewhat confused..

A low Km means binding is strong. Don't memorize here... realize if Km is small, that means that it requires less substrate to get the enzyme to do it's thing. The enzyme loves it, and it binds tightly. A high Km value means you need a lot of substrate to get the enzyme to act on it. Enzyme feels kinda meh about the substrate in this case.

thanks
 
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Thanks for your help
could you please elaborate on this
i am still somewhat confused..

A low Km means binding is strong. Don't memorize here... realize if Km is small, that means that it requires less substrate to get the enzyme to do it's thing. The enzyme loves it, and it binds tightly. A high Km value means you need a lot of substrate to get the enzyme to act on it. Enzyme feels kinda meh about the substrate in this case.

thanks
Sure, so first check out the file that I've attached... look at the first Km value. You'll notice that it was projected down from Vmax / 2 on the graph. Notice the X-axis is substrate concentration. So the first Km value, to the left, you can see that it requires LESS substrate to reach half Vmax. You can't really use Vmax in these Michaelis Mentin graphs because as you can see it's asymptotic, so we work with half Vmax (in case you were wondering why half Vmax).

The second Km value projected down from Vmax / 2 is higher, and this indicates MORE substrate was necessary to reach half Vmax.

So, if you think about it... if you need less substrate to get the job done, then the enzyme is utilizing that substrate very well (strong binding affinity). If you need more substrate to reach half Vmax, the enzyme doesn't have as strong of a binding affinity for the substrate.

Does that help at all?
 

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I am having trouble understanding a few concepts from acid and bases. For example, how do you determine if a something is a strong base or weak acid or base? I know what the strong acids are... you have to memorize the list... but how can you tell if something is a strong base or weak acid?
Also, during a titration with an amino acid, such as alanine, with a strong base, the COOH is depronated first, which leaves it with a negative charge, and it is still acidic (low pH), but I thought that a positive charge on the Nitrogen is what made it acidic. And once more protons were taken off from the COOH and NH3, then it became basic? Am I missing something here?
 
I realize this forum is extremely old, but I can't help but thank phEight for the great explanation on Vmax and Km. I never really understood it in biochem. Felt like defining Km was kind of redundant when there's already a Vmax... Anyway, I'm studying for the MCAT now, and decided to once again search for a suitable explanation that makes sense. Finally! Thank you!!
 
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Can someone explain why Vmax and Km both decrease in uncompetitive inhibition?

Vmax decreases because the inhibitor is preventing the product from forming and being released from the enzyme. Km decreases because the inhibitor is effectively increasing the enzyme's affinity for its substrate - so much so that, as mentioned, it is not able to release the substrate. Hope this helps.
 
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