You're literally the only person that thoroughly explains why the km value stays the same while the vmax decreases in non-competitive inhibition. I especially appreciate the comparisons to the other types of inhibition as well. Thank you!
I believe that many people prefer listening to your lectures than the lectures of their professors! Your explanations are clearly understandable to help me understand the materials in English during studying in the USA as an international student.
I left my biochem class in the middle of lecture cause I would rather watch your video about the subject. And what that MEANS is, you're such a good teacher thank you
this is by far the best med school teaching channel I've ever seen .. is there any way we can access the whiteboard material ? .. as in a picture or a pdf file
For noncompetitive inhibition Kcat value is lowered because substrate binding efficiency is reduced and yet substrate binding likelihood is unchanged leading to a constant Km value. Isn't that contradictory??
That's not quite right. Kcat is the turnover rate and Km is the affinity which are two different parameters. Kcat is dependent on Vmax which is lowered so Kcat is lower in that case. Km is dependent on Vmax(/2) and the slope of the plot. For a specific slope the Km needn't change in spite of lower Vmax.
@@Parralyzed in the case of uncompetetitive inhibition Kcat is unchanged even when the Vmax is lowered. I.e kcat dosent exactly follow what Vmax does. So lowered Kcat in noncompetivite inhibition can be explained by the fact that it is the inhibitor which reduces the efficiency of the active site to convert the substrate into product by changing its conformation and simply not because the Vmax value is decreased.
Hi! Thanks for being so concise yet thorough. Sorry for the complicated question, but i am lacking surrounding expertise. I have data that cannot be fitted to the Michaelis Menten equation - i tried with Excel solver and Mathematica, both give me ridiculously high Vmax and Km parameters. Also, the velocities are sigmoidal in response to [S], so I think my inhibitor is creating an allosteric effect. How can I derive Vmax and Km using rate vs [S] data and the MWC model? Any help would be appreciated :) Elena
There is some wrong info in this video. Kcat follows what Vmax does. Vmax= Kcat*[E]total [E]total does not change, this is because it is the total amount of enzyme, either bound with inhibitor or free. This means that if Vmax changes, either increases or decreases, Kcat will do the same thing. Competitive inhibition: Km increases and Vmax is unchanged and Kcat is unchanged Uncompetitive Inhibition: Km decreases and Vmax decreases and Kcat decreases Noncompetitive inhibition: Km is unchanged and Vmax decreases and Kcat decreases mixed inhibition: Km is either increased or decreased and Vmax is decreased, and Kcat is decreased Kcat follows Vmax.
[E]total refers to enzyme concentration *excluding* enzyme inactivated by the inhibitor. As far as I can tell, some people do follow the convention in your comment, where [E]total does include inactivated forms, and in that case, yes, uncompetitive inhibitors would give a lower kcat. This isn't the norm though, and what this video states is true according to the standard. As Vmax and total catalytic active sites are lowered proportionally to one another, kcat remains the same.
You're literally the only person that thoroughly explains why the km value stays the same while the vmax decreases in non-competitive inhibition. I especially appreciate the comparisons to the other types of inhibition as well. Thank you!
I believe that many people prefer listening to your lectures than the lectures of their professors! Your explanations are clearly understandable to help me understand the materials in English during studying in the USA as an international student.
Wow, this was the best explanation for this topic. I can't thank you enough.
I left my biochem class in the middle of lecture cause I would rather watch your video about the subject. And what that MEANS is, you're such a good teacher thank you
hehe
A convoluted topic explained so very well, thank you.
Great Explaination. Thanks
this is by far the best med school teaching channel I've ever seen .. is there any way we can access the whiteboard material ? .. as in a picture or a pdf file
For noncompetitive inhibition Kcat value is lowered because substrate binding efficiency is reduced and yet substrate binding likelihood is unchanged leading to a constant Km value. Isn't that contradictory??
That's not quite right. Kcat is the turnover rate and Km is the affinity which are two different parameters. Kcat is dependent on Vmax which is lowered so Kcat is lower in that case. Km is dependent on Vmax(/2) and the slope of the plot. For a specific slope the Km needn't change in spite of lower Vmax.
@@Parralyzed in the case of uncompetetitive inhibition Kcat is unchanged even when the Vmax is lowered. I.e kcat dosent exactly follow what Vmax does. So lowered Kcat in noncompetivite inhibition can be explained by the fact that it is the inhibitor which reduces the efficiency of the active site to convert the substrate into product by changing its conformation and simply not because the Vmax value is decreased.
this is very well explained , easy to follow and understand thanks a lot
PAWA pawa you're welcome :)
Amazing. Thank you.
thanxs for these clear explanations :)
Miia Saadi you're welcome, thanks for watching :)
thanks a lot from Algeria
Hi! Thanks for being so concise yet thorough.
Sorry for the complicated question, but i am lacking surrounding expertise. I have data that cannot be fitted to the Michaelis Menten equation - i tried with Excel solver and Mathematica, both give me ridiculously high Vmax and Km parameters. Also, the velocities are sigmoidal in response to [S], so I think my inhibitor is creating an allosteric effect. How can I derive Vmax and Km using rate vs [S] data and the MWC model?
Any help would be appreciated :)
Elena
There is some wrong info in this video. Kcat follows what Vmax does.
Vmax= Kcat*[E]total
[E]total does not change, this is because it is the total amount of enzyme, either bound with inhibitor or free.
This means that if Vmax changes, either increases or decreases, Kcat will do the same thing.
Competitive inhibition: Km increases and Vmax is unchanged and Kcat is unchanged
Uncompetitive Inhibition: Km decreases and Vmax decreases and Kcat decreases
Noncompetitive inhibition: Km is unchanged and Vmax decreases and Kcat decreases
mixed inhibition: Km is either increased or decreased and Vmax is decreased, and Kcat is decreased
Kcat follows Vmax.
[E]total refers to enzyme concentration *excluding* enzyme inactivated by the inhibitor.
As far as I can tell, some people do follow the convention in your comment, where [E]total does include inactivated forms, and in that case, yes, uncompetitive inhibitors would give a lower kcat.
This isn't the norm though, and what this video states is true according to the standard. As Vmax and total catalytic active sites are lowered proportionally to one another, kcat remains the same.
You’re the best
You are a master
you are awesome
still soooo complicated :( did not got it yet
I thought during uncompetitive inhibition both km and vmax decrease.