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 :).
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.
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??
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.
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?
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?
![Enzyme Kinetics with Michaelis-Menten Curve | V, [s], Vmax, and Km Relationships](/img/n.gif)
I want to know who is responsible for naming two very different things respectively uncompetitive and non-competitive. ^^
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Thanks! Glad you liked it :)
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Thank you so much, our of the few videos, I've watched, this was the most thorough and easy to understand. I really appreciate your enunciations too!
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 :).
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)
Finally I got it !
Thank you for making this clear !
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Thank you for all your videos 🙏🏻
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Clear and perfect explanation. Thanks
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Sylvia Eunhyoung Kim :-) thank you Sylvia! thanks for watching and subscribing! Appreciate the support
It is my big pleasure. I deeply appreciate your efforts and excellent explanation.
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 Professor
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TheAnGryPOolMaN thank you tons! i appreciate that! :-D
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I learned this in 20 minutes as opposed to my professor's 3-4 hour lecture
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??
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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.
IMMENSELY BENEFITTED
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?
This is a super great explanation! Thank you very much :) (Btw you only have thumbs up! :) )
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
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?
The god
I love u
I think its more complicated explanation i really have been exhausted