1. An enzyme with a Km = 2 x 10-4 M for its substrate and a Vmax of 5.5 x 10-5 m
ID: 600172 • Letter: 1
Question
1. An enzyme with a Km = 2 x 10-4 M for its substrate and a Vmax of 5.5 x 10-5 moles perliter-min was assayed for its catalytic rate with its substrate present at 2 x 10-4 M.
Calculate the anticipated initial rate under these conditions.
2. If the enzyme in problem #1 was incubated with a competitive inhibitor at 2.5 x 10-3 M
concentration and if the inhibitor demonstrated a KI = 2.5 x 10-3 M, what would be the
initial rate in the presence of the inhibitor? All other conditions are the same as those
presented in problem #1.
Explanation / Answer
1.
V=vmax[s]/km +[s]
(5.5*10^-5 moles/lt-min)(2x10^-4M)/(2x10^-4M)+(2*10^-4M)=2.7*10-5 moles/lt-min
2.
v=vmax[s]/km^app+[s] km^app=km(1+ 1/ki)
km^app=(2*10^-4)/(1+(2.5*10^-3/2.5*10^-3) =4*10^-4
v=(5.5*10^-5)(2*10^-4)/(4*10^-4)+(2*10^-4)=1.8*10^-5 moles/l-min
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