There is an error: for UCI kcat must be lower since Vmax is lowered. What doesn't change is catalytic efficiency (kcat/Km) on account of the lower Km,app. (cf. the corresponding video in this series)
Hello sir, your lectures are so amazing to me and others all over the world. Thanks for everything you did and also doing now. Sir, I have a question on uncompetitive inhibition, You say the turnover number does not affected by inhibitor but you also say turnover number means transformation of substrates into products in a certain period time. In this inhibition there will no products be formed from the ESI complex.
Hello sir, First of all I would like to say that your lectures are amazing. Thanks for all the help. I have a query. In the second part of this topic you explained how in non competitive inhibition the kcat is lowered but not the Km. My question is that once the inhibitor binds to the enzyme it changes the shape of the active site which lowers the Kat. But once the shape of the active site is changed how does the substrate still has a likelihood of binding the active site? The binding of substrate should also decrease!
@@quyennguyentri8993 The inhibition is reversible, meaning the inhibitor has a chance of detaching from the enzyme. Once it detaches, the substrate can bind again.
So with non-competitive inhibition, the inhibitor does not change the shape of the enzyme, because both the substrate and the inhibitor can bind to the enzyme uncompetitively, forming the enzyme substrate-inhibitor complex. However, when the inhibitor binds to the enzyme-substrate mixture - the enzyme cannot catalyse the reaction of the substrate and therefore no products are formed - lowering the kcat value. But the inhibitor can also dissociate away from the enzyme which can enable the enzyme-substrate mixture to catalyse (providing no inhibitor is bound) forming products. I think your explanation is referring to uncompetitive inhibition. I know this post is 3yrs old but thought it may help someone else :).
hi great Video :-) I have a question about the kompetitive Inhibitor ; Why does E(total) not Change , if we have a competive Inhibitor than we won´t have an ES complexe than E(total ) should become lower no?
Hello sir, First of all I would like to say that your lectures are amazing. Thanks for all the help. i want to ask you something. if you have used superpro, I want to ask, what is meant by "K mic" in superpro? The equation is 1-[S]/Kmic. I still don't know what the meaning of the equation is. Thank you so much for the answer.
Hai Sir, I have a query!! In non- competitive inhibition, why doesn't the inhibitor bind directly to the active site of the enzyme. If that even happens, what will the result be??
Question regarding competitiv inhibitors. You mention that the Km value does change, namely it increases, however, the Vmax value remains the same. All good. The issue: When you calculate Km, you use V max / 2. Right? But if the Vmax doesn't change, the Km value would be the same in both cases (when there's a competitiv inhibitor versus when there isn't). But as was just mentioned, the Km does in fact change, which should yield a different Km value but how can it do that if the Vmax remains the same? Hopefully you understand my reasoning. Probably something I just missed... appreciate some help
Vmax/2 is NOT actually equal to Km, it “corresponds” to Km, they form a couple of coordinates (x,y) where x is Km and y is Vm/2. So, since we project the Vm/2 value on the curve to find the Km, that means that the latter "depends" on the slope of the curve, which in not the same between the inhibitive and the non inhibitive case. I hope I got this right myself, and that I used the proper scientifique expressions since English is not my learning language
For uncompetitive inhibition, why do we reach the conclusion for kcat based on the enzyme's behavior when the inhibitor is not binded? Logically, shouldn't we consider the situation when the inhibitor is binded, since we're analyzing the inhibitor's affect?
Oh it stopped prematurely but ok. I think the reason we consider the activity of the active site when it's unbinded by inhibitor is because the active site is unaffected by the inhibitor, so it has normal kcat. But for non-competitive inhibition, the active site is affected by the inhibitor, so some of the active site cannot catalyze reactions anymore. Thus, kcat decreases?
I want to know who is responsible for naming two very different things respectively uncompetitive and non-competitive. ^^
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There is an error: for UCI kcat must be lower since Vmax is lowered. What doesn't change is catalytic efficiency (kcat/Km) on account of the lower Km,app.
(cf. the corresponding video in this series)
You are an angel! seriously!
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Clear and perfect explanation. Thanks
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Hello sir, your lectures are so amazing to me and others all over the world. Thanks for everything you did and also doing now. Sir, I have a question on uncompetitive inhibition, You say the turnover number does not affected by inhibitor but you also say turnover number means transformation of substrates into products in a certain period time. In this inhibition there will no products be formed from the ESI complex.
this guy nails it every time
Hello sir,
First of all I would like to say that your lectures are amazing. Thanks for all the help.
I have a query. In the second part of this topic you explained how in non competitive inhibition the kcat is lowered but not the Km. My question is that once the inhibitor binds to the enzyme it changes the shape of the active site which lowers the Kat. But once the shape of the active site is changed how does the substrate still has a likelihood of binding the active site? The binding of substrate should also decrease!
Have you got the answer yet, after 2 year :3
@@quyennguyentri8993 The inhibition is reversible, meaning the inhibitor has a chance of detaching from the enzyme. Once it detaches, the substrate can bind again.
So with non-competitive inhibition, the inhibitor does not change the shape of the enzyme, because both the substrate and the inhibitor can bind to the enzyme uncompetitively, forming the enzyme substrate-inhibitor complex. However, when the inhibitor binds to the enzyme-substrate mixture - the enzyme cannot catalyse the reaction of the substrate and therefore no products are formed - lowering the kcat value. But the inhibitor can also dissociate away from the enzyme which can enable the enzyme-substrate mixture to catalyse (providing no inhibitor is bound) forming products. I think your explanation is referring to uncompetitive inhibition. I know this post is 3yrs old but thought it may help someone else :).
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hi great Video :-) I have a question about the kompetitive Inhibitor ; Why does E(total) not Change , if we have a competive Inhibitor than we won´t have an ES complexe
than E(total ) should become lower no?
Hello sir,
First of all I would like to say that your lectures are amazing. Thanks for all the help. i want to ask you something. if you have used superpro, I want to ask, what is meant by "K mic" in superpro? The equation is 1-[S]/Kmic. I still don't know what the meaning of the equation is. Thank you so much for the answer.
This man deserves a gold metal.
IMMENSELY BENEFITTED
This is a super great explanation! Thank you very much :) (Btw you only have thumbs up! :) )
Hai Sir, I have a query!! In non- competitive inhibition, why doesn't the inhibitor bind directly to the active site of the enzyme. If that even happens, what will the result be??
Question regarding competitiv inhibitors.
You mention that the Km value does change, namely it increases, however, the Vmax value remains the same. All good.
The issue: When you calculate Km, you use V max / 2. Right? But if the Vmax doesn't change, the Km value would be the same in both cases (when there's a competitiv inhibitor versus when there isn't). But as was just mentioned, the Km does in fact change, which should yield a different Km value but how can it do that if the Vmax remains the same? Hopefully you understand my reasoning.
Probably something I just missed... appreciate some help
Vmax/2 is NOT actually equal to Km, it “corresponds” to Km, they form a couple of coordinates (x,y) where x is Km and y is Vm/2. So, since we project the Vm/2 value on the curve to find the Km, that means that the latter "depends" on the slope of the curve, which in not the same between the inhibitive and the non inhibitive case.
I hope I got this right myself, and that I used the proper scientifique expressions since English is not my learning language
Ur awsomee
I love you sir 😍
For uncompetitive inhibition, why do we reach the conclusion for kcat based on the enzyme's behavior when the inhibitor is not binded? Logically, shouldn't we consider the situation when the inhibitor is binded, since we're analyzing the inhibitor's affect?
Oh it stopped prematurely but ok. I think the reason we consider the activity of the active site when it's unbinded by inhibitor is because the active site is unaffected by the inhibitor, so it has normal kcat. But for non-competitive inhibition, the active site is affected by the inhibitor, so some of the active site cannot catalyze reactions anymore. Thus, kcat decreases?
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I think its more complicated explanation i really have been exhausted