A research group discovers a new version of happyase, which they call happyase*,
ID: 705950 • Letter: A
Question
A research group discovers a new version of happyase, which they call happyase*, that catalyzes the chemical reaction HAPPY, ?? SAD The researchers begin to characterize the enzyme a In the first experiment, with [Et] at 6 nM, they find bl In another experiment, with [Et] at 3 nM and that the Vmax is 3.08 LM s-1. Based on this experiment, what is the kcat for happyase*? [HAPPY] at 35 LM, the researchers find that Vo = 1180 nM s-. What is the measured Km of happyase* for its substrate HAPPY? Number Number cat C] Further research shows that the purified happyase* used in the first two experiments was actually contaminated with a reversible inhibitor called ANGER. When ANGER is carefully removed from the happyase* preparation, and the two experiments repeated, the measured Vmax in (a) is increased to 7.70 Ms-1, and the measured Km in (b) is now 12.1 M. For the inhibitor ANGER, calculate the values of a and a. Number Number Please scroll d) Based on the information given above, what type of inhibitor is ANGER? down Stion Check AnExplanation / Answer
ans)
Km = Michaelis constant
K cat = it is the turnover number and relates Vmax which is the maximum velocity to Et which is total active site concentration.
(a) From the question we have, Et = 6 nM =0.006 µM
Vmax = 3.08 µM/sec
since, Kcat = Vmax / Et
=3.08 µM/sec / 0.006 µM
Kcat =513 per sec.
(b):
Kcat = 513 per sec, Et = 3 nM
since, Kcat = Vmax / Et
or, Vmax = Et * Kcat
Vmax = 3 * 513= 1539 nM/sec = 0.1539 µM/sec
As, Vo = Vmax (S) / Km +(S)
on cross multiplying the terms
(Vo = 1180 nM/sec = 0.1180 µM/sec and S = 35 µM)
0.1180 µM/sec = 0.1539 µM/sec * (35 µM) / (Km + 35 µM)
( 0.1180 µM/sec) ( Km + 35 µM) = 0.1539 µM/sec * 35 µM
0.1180 µM/sec * Km = 1.26 µM2/sec
Therefore, Km = 10.7 µM
ANGER is to be mixed to know what type of inhibitior it is, as it is affecting both Vmax and Km.
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