Graffix

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If an inhibitor binds to the substrate would it be considered a non-competitive inhibitor? But then unlike normal non-competitive inhibitors, can increasing substrate concentration reverse the inhibition (in this case it would act more like a competitive inhibitor) ?
 

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If an inhibitor binds to the substrate would it be considered a non-competitive inhibitor? But then unlike normal non-competitive inhibitors, can increasing substrate concentration reverse the inhibition (in this case it would act more like a competitive inhibitor) ?
If the substrate competes for the active site, we have a competitive inhibitor. A great example is methanol and ethanol BOTH compete for alcohol dehydrogenase. Thus if you drank methanol.....we can OVERCOME the methanol by giving IV ethanol. The substrates resemble the active site...The Vmax will be UNCHANGED. In noncompetitive inhibition, a molecule binds to an enzyme somewhere other than the active site. This changes the enzyme's three-dimensional structure so that its active site can still bind substrate with the usual affinity, but is no longer in the optimal arrangement to stabilize the transition state and catalyze the reaction. On the macroscopic scale, noncompetitive inhibition lowers the Vmax. Thus, the enzyme simply cannot catalyze the reaction with the same efficiency as the uninhibited enzyme. Note that noncompetitive inhibition cannot be overcome by raising the substrate concentration like competitive inhibition can. This is a very interesting topic that you will enjoy when you do BioChemistry in Dental School.

Hope this helps.

Dr. Romano
 
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Graffix

Graffix

2+ Year Member
May 12, 2015
140
87
Status
Pre-Dental
If the substrate competes for the active site, we have a competitive inhibitor. A great example is methanol and ethanol BOTH compete for alcohol dehydrogenase. Thus if you drank methanol.....we can OVERCOME the methanol by giving IV ethanol. The substrates resemble the active site...The Vmax will be UNCHANGED. In noncompetitive inhibition, a molecule binds to an enzyme somewhere other than the active site. This changes the enzyme's three-dimensional structure so that its active site can still bind substrate with the usual affinity, but is no longer in the optimal arrangement to stabilize the transition state and catalyze the reaction. On the macroscopic scale, noncompetitive inhibition lowers the Vmax. Thus, the enzyme simply cannot catalyze the reaction with the same efficiency as the uninhibited enzyme. Note that noncompetitive inhibition cannot be overcome by raising the substrate concentration like competitive inhibition can. This is a very interesting topic that you will enjoy when you do BioChemistry in Dental School.

Hope this helps.

Dr. Romano
Hi Dr. Romano!
I will clarify, I understand most of it now I believe. I am still confused as to what happens when the inhibitor binds to the substrate instead of the enzyme.

Km is substrate affinity. Vmax is max velocity of the reaction.

So in the scheme: E + S <=> ES => E + P

If the inhibitor binds to "E" at the active site to make "EI" it is competitively inhibiting the enzyme and fighting space on the active site with "S" . The inhibition can be overcome by increasing the substrate concentration. Vmax would be the same and Km would be changed.

If the inhibitor binds to "E" and "ES" to make "EI" and "EIS" at an allosteric site then it would decommission, for lack of a better name, the enzyme and the Vmax would decrease but Km would not change. This would be a non-competitive inhibitor.

If the inhibitor binds to "ES" to make "ESI" and prevent it from making product. It would be a uncompetitive inhibitor. Km and Vmax both decrease. Also only by increasing number of enzymes?

Then, if "I" binds to "S" ... It will reduce the concentration of substrate? Vmax and Km would be unchanged and it could be overcome by increasing substrate concentration?

Thanks so much for your help!!!