Good lecture as always, although I have some points: For the competetive inhibition you write "Km increases + lowers the enzyme's affinity for the substrate". I would say this statement is slightly wrong. The enzyme's affinity for the substrate actually doesn't change - it remains the same. However the Km _appears_ to have been increased, and thus it _looks_ like the affinity for the substrate has decreased. That's why some textbooks say that "Km_app" (and not just "Km") has been increased. Of course the Km_app increases because we need more substrate to outcompete the inhibitor to reach 50% of maximum velocity/activity. Second point: For the uncompetetive inhbition you write that Km decreases, but you don't explain why. While it's obvious that Vmax decrease, because at high [S] a lot of the enzymes are inhibited by the inhibitor, it's not so obvious why the Km decreases. To understand why the apparant Km decreases for uncompetive inhbition, it's important to understand that the enzyme and substrate also bind reversible. So we have a dynamic equilibrium: E + S ES When you add an uncompetetive inhibitor, the inhibitor will only bind to the ES-complex. So the following reaction takes place: ES + I ES-I So basically we're removing ES-complexes from the 1st-mentioned equilibrium, and according to Le Châtelier's principle, this will result in a shift in the equlibrium towards the right, so basically the net result is that more substrate is being used to create ES-complexes. Therefore it appears that the enzyme's affinity for the substrate has been increased (the apparent Km decreases). For the noncompetive inhbition the Km doesn't change, because the inhibitor can bind to both the free enzyme AND the ES-complex. So for the equlibrium: E + S ES both E and ES is getting removed equally so there is no actual 'shift' in the equlibrium, so the enzyme's affinity for the substrate appears to remain unchanged unlike for the uncompetetive inhbition. My own rule to remember the difference in non- and uncompetetive inhibition, is that uncompetetive inhbition seems "unreal" :D (the inhibitor can only bind to the ES-complex and not the enzyme alone) while the noncompetetive inhbition is easier to understand (can both bind to the enzyme and the ES-complex.)
This is so helpful, have EXAM in 2 days and you just saved my life! Thank you so much, really love the way you explain all your lecture. Really appreciate you!:)
Thank you for doing this lectures, with all this well structured and easy to understand informations, my essays feels like walk in the park. Thank you!
For the irreversible inhibitor, if cyanide binds to the active site of Hb oR Mb, how is that classified in terms of competitive, non-competitive and uncompetitive inhibition? How does Km gets affected by this? I am confused.
I don't understand why Km isn't affected with non-competitive inhibition. If Km is the concentration of substrate needed to reach half of Vmax, wouldn't it need to be higher if a chunk of your enzymes have been put out of business but are still nabbing substrate molecules? I would understand if non-competitive inhibition prevented the substrate from attaching to the enzyme in the first place because then they would still be free to go into actually active enzyme molecules, but if a portion of your substrate keeps spending time getting caught up in inactive enzyme molecules, wouldn't the speed of the reaction decrease, not just the maximum possible speed?
Our professors will teach us very well. But I'll go monthly once to college. 😂😂😂 Maximum I'll go for 2-5 classes for entire academic year. So to prepare for exam watching this video 😁😁 sister.
Good lecture as always, although I have some points:
For the competetive inhibition you write "Km increases + lowers the enzyme's affinity for the substrate". I would say this statement is slightly wrong. The enzyme's affinity for the substrate actually doesn't change - it remains the same. However the Km _appears_ to have been increased, and thus it _looks_ like the affinity for the substrate has decreased. That's why some textbooks say that "Km_app" (and not just "Km") has been increased. Of course the Km_app increases because we need more substrate to outcompete the inhibitor to reach 50% of maximum velocity/activity.
Second point: For the uncompetetive inhbition you write that Km decreases, but you don't explain why. While it's obvious that Vmax decrease, because at high [S] a lot of the enzymes are inhibited by the inhibitor, it's not so obvious why the Km decreases.
To understand why the apparant Km decreases for uncompetive inhbition, it's important to understand that the enzyme and substrate also bind reversible. So we have a dynamic equilibrium:
E + S ES
When you add an uncompetetive inhibitor, the inhibitor will only bind to the ES-complex. So the following reaction takes place:
ES + I ES-I
So basically we're removing ES-complexes from the 1st-mentioned equilibrium, and according to Le Châtelier's principle, this will result in a shift in the equlibrium towards the right, so basically the net result is that more substrate is being used to create ES-complexes. Therefore it appears that the enzyme's affinity for the substrate has been increased (the apparent Km decreases).
For the noncompetive inhbition the Km doesn't change, because the inhibitor can bind to both the free enzyme AND the ES-complex. So for the equlibrium:
E + S ES
both E and ES is getting removed equally so there is no actual 'shift' in the equlibrium, so the enzyme's affinity for the substrate appears to remain unchanged unlike for the uncompetetive inhbition.
My own rule to remember the difference in non- and uncompetetive inhibition, is that uncompetetive inhbition seems "unreal" :D (the inhibitor can only bind to the ES-complex and not the enzyme alone) while the noncompetetive inhbition is easier to understand (can both bind to the enzyme and the ES-complex.)
+Christofferr Andersen Point to my self: learn to spell to compet_i_tive and inhibition.
+Christofferr Andersen You answered my questions before I had the chance to ask them. Thank you very much. :)
+Christoffer Andersen As I learnt; If the Km is increased, then the affinity lowers. But this is just what i've learnt.
I've wondered and looked for answers in books without finding a clear explanation, this makes sense, thank you.
Why watch this video if you know it all already? Are you bored
This is so helpful, have EXAM in 2 days and you just saved my life! Thank you so much, really love the way you explain all your lecture. Really appreciate you!:)
Thank you for doing this lectures, with all this well structured and easy to understand informations, my essays feels like walk in the park. Thank you!
Great lecture, on point and helpful . Makes biochem look less gibberish . Thanks!
i loved the lecture and enjoyed it teacher
Your way of teacheaing makes biochemistry piece of the cake
Thank you!
thank you so much for this video, its really helpful and i wish you were my teacher in this subject coz u really make everything much easier
Thank you this is great :)
thanks for the great lecture and i would appreciate if you can explained the phenomena of synergism, antagonism and the difference.
For the irreversible inhibitor, if cyanide binds to the active site of Hb oR Mb, how is that classified in terms of competitive, non-competitive and uncompetitive inhibition? How does Km gets affected by this? I am confused.
love your way of teaching!!! thank you
Very helpful and informative video
I see AK Lectures. I click.
Thank you.
Why did km decreases that mean the substrate affinity is high in uncompetitive inhibition ?
I don't understand why Km isn't affected with non-competitive inhibition. If Km is the concentration of substrate needed to reach half of Vmax, wouldn't it need to be higher if a chunk of your enzymes have been put out of business but are still nabbing substrate molecules? I would understand if non-competitive inhibition prevented the substrate from attaching to the enzyme in the first place because then they would still be free to go into actually active enzyme molecules, but if a portion of your substrate keeps spending time getting caught up in inactive enzyme molecules, wouldn't the speed of the reaction decrease, not just the maximum possible speed?
thank you so much
how we can stop a non competitive inhibitor ??? answer pleaaaase i need it
You the man!
how can be non. competitve inhibitor effect neutralized
Amazing
AMAZING!! Thank you, now if only my prof could teach 1/2 as well!
Our professors will teach us very well.
But I'll go monthly once to college.
😂😂😂 Maximum I'll go for 2-5 classes for entire academic year.
So to prepare for exam watching this video 😁😁 sister.
PROGRESS,
why does the affinity for substrate decrease when Km increases?
+Keenan Kam see my answer above.
Jin Keenan Kam
cool
❤️❤️❤️
i love you
😅