Foggybrain

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Sep 13, 2017
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I know that an uncompetitive inhibitor (UCI) decreases both Km and Vmax, as it binds allosterically and to either E or ES
Kcat is defined as the catalytic turnover rate of the enzyme = # products produced per second per active site
Would you agree that with an UCI, Km decreases, hence the E affinity towards S increases, hence E has a higher turn over rate, hence a higher Kcat should be expected?
I couldn't find the answer online. Thank you for taking the time.
 
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Although this isn’t always true, in the context of MCAT Kcat is usually the same as K2, which is the rate constant from ES complex to product. Since Vmax goes down and it’s dependent on K2 (assuming constant enzyme concentration), I would say Kcat should decrease.
Also, turnover rate is number of substrates CONVERTED TO PRODUCT per one active site per second.
 

Foggybrain

2+ Year Member
Sep 13, 2017
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Although this isn’t always true, in the context of MCAT Kcat is usually the same as K2, which is the rate constant from ES complex to product. Since Vmax goes down and it’s dependent on K2 (assuming constant enzyme concentration), I would say Kcat should decrease.
Also, turnover rate is number of substrates CONVERTED TO PRODUCT per one active site per second.
Thank you for your response. You mentioned that since Vmax decreases, so should Kcat (=K2) as they are directly proportional. Is this purely based on this equation: Vmax = Kcat X [E] ? ([E] being the total enzyme concentration, which is a constant). Thank you for your time.
 
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Yes, exactly. K2 = Kcat, assuming there’s only one active site. If there’s more than one, say n active sites, then Kcat = K2/n because of Kcat’s “per active site” condition. Regardless, Kcat should go down, since K2 goes down.
 
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