Real urgent need!!!! Vmax and Km --> it's not what you think it is btw!!

[Meeee!!!!]

Hey guys,

So, um, I understand that Km increases when you add a competitive inhibitor and that Vmax decreases when you add a non-competitive inhibitor. But...

1. why is it that the Vmax stays the same when you add a competitive inhibitor and ....

2. why does Km stay the same when you add a non-competitive inhibitor



Hoping you guys can helps. Cheers 🙂

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[AnonPanda1]

Km represents two things, the ease at which the enzyme grabs substrates and the concentration at which the reaction is at Vmax/2. Vmax stays the same when you add a competitive inhibitor because if you add enough substrate, it'll outcompete the competitive inhibitor. Then it appears as if it is business as usual.

Km stays the same when you add a non-competitive inhibitor because the enzyme binds to the substrate without change and the inhibitor binds whenever it wants whether the enzyme is free or bound to a substrate. The inhibitor only prevents the reaction.

 

[wizzed101]

1. Because when you add enough substrate the inhibitory effect is no more.

2. That is just one special case and actually doesn't happen most of the time because for a reversible "non-" competitive inhibitor, nothing guarantees that the affinity of the inhibitor for the enzyme alone is the same as the affinity for the enzyme substrate complex.

I think the people at the AAMC want you to study that case because it resembles irreversible enzyme inactivator.

 

[Domepiece]

Hey guys,

So, um, I understand that Km increases when you add a competitive inhibitor and that Vmax decreases when you add a non-competitive inhibitor. But...

1. why is it that the Vmax stays the same when you add a competitive inhibitor and ....

2. why does Km stay the same when you add a non-competitive inhibitor



Hoping you guys can helps. Cheers 🙂

@AnonPanda1 and @wizzed101 have given you great answers. Here is another way to think about it:

1) Km is mathematically defined as the substrate concentration @ 1/2 Vmax. Vmax is the rate of catalysis when enzyme is saturated with substrate. A competitive inhibitor will compete with substrate for binding site access and therefore the addition of more substrate increases the probability of it ending up in the active site over a competitive inhibitor, as panda mentioned. This increases the substrate concentration required for enzyme saturation (and the concentration of substrate required to achieve 1/2 Vmax as well, thus the increased Km). At enzyme saturation, the rate at which the enzyme produces its product has not been altered and Vmax is identical to normal.

2) Km can also be thought of conceptually as measuring the binding affinity of an enzyme for its substrate. Introduction of a non-competitive inhibitor does not influence an enzymes binding affinity for its substrate, thus Km is unaltered. However, it does inhibit the enzyme's ability to catalyze its final product, despite binding site enzyme-substrate interactions, thus the reduced Vmax.

 

[Kpw101]

1.) Competitive inhibitors bind to the active site and can be overcome with increasing substrate concentration since the substrate will then out compete the inhibitor for the active site. It takes more substrate to get to Vmax and 1/2 of Vmax which is why Km increases. (More substrate is needed to get to 1/2 Vmax.)

2.) Km stays the same for noncompetitive inhibitors because they bind to a site other than the active site so they don't affect the Enzyme's affinity for substrate. Be aware though that while Km stays the same for noncompetitive inhibitors, binding to a site other than the active site isn't the only reason Km stays the same. Noncompetitive inhibitors bind equally well to both enzyme alone, and the enzyme substrate complex. Anything that deviates from this equal binding, you start to get into uncompetitive and mixed-type inhibition which does change Km.

 

[Meeee!!!!]

Km represents two things, the ease at which the enzyme grabs substrates and the concentration at which the reaction is at Vmax/2. Vmax stays the same when you add a competitive inhibitor because if you add enough substrate, it'll outcompete the competitive inhibitor. Then it appears as if it is business as usual.

Km stays the same when you add a non-competitive inhibitor because the enzyme binds to the substrate without change and the inhibitor binds whenever it wants whether the enzyme is free or bound to a substrate. The inhibitor only prevents the reaction.

Hey, just wanted some clarification on what you were saying ... you were mentioning how Km also represents "the ease at which the enzyme grabs substrates ", right?

I just want to check my understanding of why Km increases when the affinity of the drug to the enzymes decrease. My reasoning is as follows...

Increasing the Km would means that the concentration of substrate increases (i.e. the amount of drug in the body increases). This would go to the extent that it may be that the amount of drug exceeds the number of enzymes present. Thus, leading to the decrease in the affinity between the drug and the enzymes (due to the amount of drug exceeding the number of enzymes).

So basically what affects the ease at which drug and enzyme bind to each other (i.e. the affinity) is the concentration of the substrate. And Km and affinity are directly proportionate to each other.

Is my reasoning correct?

Please let me know.

Cheerio 🙂

 

[AnonPanda1]

Hey, just wanted some clarification on what you were saying ... you were mentioning how Km also represents "the ease at which the enzyme grabs substrates ", right?

I just want to check my understanding of why Km increases when the affinity of the drug to the enzymes decrease. My reasoning is as follows...

Increasing the Km would means that the concentration of substrate increases (i.e. the amount of drug in the body increases). This would go to the extent that it may be that the amount of drug exceeds the number of enzymes present. Thus, leading to the decrease in the affinity between the drug and the enzymes (due to the amount of drug exceeding the number of enzymes).

So basically what affects the ease at which drug and enzyme bind to each other (i.e. the affinity) is the concentration of the substrate. And Km and affinity are directly proportionate to each other.

Is my reasoning correct?

Please let me know.

Cheerio 🙂

You're extrapolating too much and your logic is backwards. Just know that as the affinity to the drug decreases, Km increases. Don't try to tie in substrate concentration affecting Km. Km is independent of concentration for a certain substrate.

 

[Meeee!!!!]

You're extrapolating too much and your logic is backwards. Just know that as the affinity to the drug decreases, Km increases. Don't try to tie in substrate concentration affecting Km. Km is independent of concentration for a certain substrate.

I really needed that hahaha. Thanks for that AnonPanda 1.

I really should dig too much into it. I will be spending too much time.

Thanks again 🙂

 

[Kpw101]

Hey, just wanted some clarification on what you were saying ... you were mentioning how Km also represents "the ease at which the enzyme grabs substrates ", right?

I just want to check my understanding of why Km increases when the affinity of the drug to the enzymes decrease. My reasoning is as follows...

Increasing the Km would means that the concentration of substrate increases (i.e. the amount of drug in the body increases). This would go to the extent that it may be that the amount of drug exceeds the number of enzymes present. Thus, leading to the decrease in the affinity between the drug and the enzymes (due to the amount of drug exceeding the number of enzymes).

So basically what affects the ease at which drug and enzyme bind to each other (i.e. the affinity) is the concentration of the substrate. And Km and affinity are directly proportionate to each other.

Is my reasoning correct?

Please let me know.

Cheerio 🙂

Okay, enzyme concentration NEVER changes in MM equations! We're not looking at what affects the drug and enzyme affinity, we're looking at the data empirically in the presence of an inhibitor! Try to think of it this way. Km represents the SUBSTRATE concentration at which the reaction rate is half of it's absolute potential maximum. So if Km is low you can deduce that it does not take much substrate to get the enzyme to function at half capacity ------> High(er) affinity! You can guess what happens when Km is higher: It takes more substrate to get to half capacity which means less affinity for substrate. Competitive makes Km higher because it competes for the enzyme site --> less affinity. Noncompetitive does not change Km because it does not compete for the enzyme site. Uncompetitive lowers Km because uncomp. inhibitors only bind to Enzyme-Substrate complexes effectively rendering some enzyme useless: The potential Vmax goes down and Km appears lower in uncomp. inhibition because active sites are taken up and sequestered --> higher affinity. That was a mouthful! Let me know if any of this does not make sense. You really just have to sit down and work it through logic, not memory. I would draw a diagram of cause and effect to really solidify this information. 🙂

 

[Meeee!!!!]

Okay, enzyme concentration NEVER changes in MM equations! We're not looking at what affects the drug and enzyme affinity, we're looking at the data empirically in the presence of an inhibitor! Try to think of it this way. Km represents the SUBSTRATE concentration at which the reaction rate is half of it's absolute potential maximum. So if Km is low you can deduce that it does not take much substrate to get the enzyme to function at half capacity ------> High(er) affinity! You can guess what happens when Km is higher: It takes more substrate to get to half capacity which means less affinity for substrate. Competitive makes Km higher because it competes for the enzyme site --> less affinity. Noncompetitive does not change Km because it does not compete for the enzyme site. Uncompetitive lowers Km because uncomp. inhibitors only bind to Enzyme-Substrate complexes effectively rendering some enzyme useless: The potential Vmax goes down and Km appears lower in uncomp. inhibition because active sites are taken up and sequestered --> higher affinity. That was a mouthful! Let me know if any of this does not make sense. You really just have to sit down and work it through logic, not memory. I would draw a diagram of cause and effect to really solidify this information. 🙂

Thanks a lot. I think it makes a bit more sense. 🙂

 

[AnonPanda1]

I really needed that hahaha. Thanks for that AnonPanda 1.

I really should dig too much into it. I will be spending too much time.

Thanks again 🙂

You seem to enjoy enzyme kinetics a lot. Consider joining a biochemical research lab! You'll be able to go beyond what a textbook can teach and some hands on experience is always fun.