madoo
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a simple kaplan q in ch 8
consider a reaction that can be catalyzed by enzyme with the following kinetics
Km(M) 5x0.000001
Vmax(mmol/min) 20
at a concentration 5x0.0001 M substrate the velocity of the reaction catalyzed by this enzyme will be
A:10
B:15
C:20
D:25
E:30
sure 10 is wrong "it is half of the Vmax so the substrate concentration is Km
25, 30 is above Vmax so they are wrong
in the answer they say it is 20
but how they get it
by michaelis-menten , lineweaver-burk equations "equal 19.8"
but it takes too much time to calculate the equation
how about in the exam ??
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| Jackofknives
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no need to calc. 15 is too low. they gave u the options so you can clearly see which answer is correct. in case you didnt notice, the Conc induced was 100 times the Km
the reasoning is like this
v = (Vmax*[S]) / ([S] + Km)
when [S]>>Km
v = Vmax
Edited by Jackofknives on 10/27/07 - 01:53 AM
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| madoo
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well seems logic
thanks
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