An enzyme catalyzes the following chemical reaction: S<-k1 & k2 -> P a). In the
ID: 58905 • Letter: A
Question
An enzyme catalyzes the following chemical reaction:
S<-k1 & k2 -> P
a). In the first experiment, with [Et] at 4 nM, the Vmax is 1.6 M s-1, what is the turn over number (kcat) of the enzyme?
b). In another experiment, with [Et] at 1 nM and [S] at 30 M, the velocity of the reaction V0 = 300 nM s-1, what is the measured Km of the enzyme?
c). Further research showed that the enzyme used in previous experiments is contaminated with a reversible inhibitor. After careful removal of all the inhibitor, the Vmax measured in experiment (a) is increased to 4.8 M s-1 and the measured Km in (b) is now 15 M. What kind of inhibitor contaminated the enzyme?
Explanation / Answer
1) Then Kcat of the enzyme can be calculated as -
Vmax = Kcat [s]
So, Kcat = Vmax / [ Total enzyme concentration i.e Et ] = 1.6 x 10-6 / 4 x 10-9 = 400 S-1
2) We know, V0 = Vmax x [s] / Km + 30
0.3 = 1.6 x 30 / Km + 30
Km = 450 µM
3) since both KM and Vmax get changes then the inhibitor is, of course, uncompetitive.