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#1
06-26-2012
 USMLE Forums Master Steps History: 1+CK+CS Posts: 1,039 Threads: 189 Thanked 557 Times in 318 Posts Reputation: 567
Reaction kinetics question

Consider a reaction that can be catalyzed by one of two enzymes, A and B, with the following kinetics:

Km (M) Vmax (mmol/min)
A 5x10-6 20
B 5x10-4 30

At a concentration of 5x10-4 substrate, the velocity of the reaction catalyzed by enzyme A will be:

a) 10
b) 15
c) 20
d) 25
e) 30

#2
06-26-2012
 USMLE Forums Scout Steps History: 1+CK+CS Posts: 40 Threads: 1 Thanked 50 Times in 16 Posts Reputation: 60

Hi,
I'll go with C) 20. The Vmax will stay the same.
#3
06-28-2012
 USMLE Forums Master Steps History: 1+CK+CS Posts: 1,039 Threads: 189 Thanked 557 Times in 318 Posts Reputation: 567
?

In case of enzyme A if we give 5x10-4 of substrate it's more than Km (5x10-6) but how do we now that velocity of reaction reaches Vmax and just 15 (which is between V for Km - 10, and maximal velocity)? Any ideas?
#4
06-28-2012
 USMLE Forums Addict Steps History: 1+CK+CS Posts: 196 Threads: 9 Thanked 97 Times in 64 Posts Reputation: 107

When you have a substrate concentration a 100 times greater than the Km, V is almost gonna be equal to Vmax.
#5
06-28-2012
 USMLE Forums Master Steps History: 1+CK+CS Posts: 1,039 Threads: 189 Thanked 557 Times in 318 Posts Reputation: 567
:)

Quote:
 Originally Posted by smanthrav When you have a substrate concentration a 100 times greater than the Km, V is almost gonna be equal to Vmax.
Thanks do you think we have to memorize this equation or just remember that 50-100-times increase will move it to Vmax?
#6
06-28-2012
 USMLE Forums Addict Steps History: 1+CK+CS Posts: 196 Threads: 9 Thanked 97 Times in 64 Posts Reputation: 107

Don't think you need to. You know the concept behind it now and that should be enough, I think. I copypasted it...I dont remember the exact equation either.
 The above post was thanked by: Casandra (06-28-2012)

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