laczlacylaci

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Jun 20, 2016
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I understand to identify a competitive inhibitor, the Km increases and the Vmax doesn't change.

I was going through a SB C/P problem (#77) and thought how could Km change based on a constant Vmax?
If Km=1/2Vmax, how does Km decrease/increase while Vmax stays the same?

According to the equation, wouldn't we always expect that with an increase in Vmax, there would be an increase in Km

Probably overthinking again.:yawn:
 

theonlytycrane

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Km is equal to the substrate concentration when velocity = 1/2 vmax. It doesn't equal 1/2 vmax itself!

In the case of a competitive inhibitor, more substrate is needed to get the reaction velocity up to 1/2 vmax. This increased substrate concentration required is the reason that Km is increased.
 
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laczlacylaci

laczlacylaci

2+ Year Member
Jun 20, 2016
264
45
Cheese State
Status
Pre-Medical
Km is equal to the substrate concentration when velocity = 1/2 vmax. It doesn't equal 1/2 vmax itself!

In the case of a competitive inhibitor, more substrate is needed to get the reaction velocity up to 1/2 vmax. This increased substrate concentration required is the reason that Km is increased.
Oh thats right. #taking break this weekend :yawn:
 
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