The following kinetic data were obtained for an enzyme in the absence of any inh
ID: 62870 • Letter: T
Question
The following kinetic data were obtained for an enzyme in the absence of any inhibitor (1) and in the presence of an inhibitors (2) at 150 nM concentration: Substrate No inhibitor (1) Inhibitor (2) Assume [ET] is the same for all the experiments. Please draw a double recipr2ca11ot using the data above and determine: a). The Vmax and Km for the enzyme. b). The apparent Vmax and Km of the enzyme in presence of the inhibitor. c). What kind of inhibitor is used in the experiment? d). The K, of the inhibitor.Explanation / Answer
a)
Reciprocate the values in the given data:
The Vmax for No inhibitor:
y = 0.1264x + 0.0195
Vmax = 1/ 0.0195
= 51.28 µmole S-1
The Km for No inhibitor:
y = 0.1264x + 0.0195
Km = 0.0195/ 0.1264
= 0.154.
b)
The Vmax for inhibitor 1:
y = 0.4231x + 0.0198
Vmax = 1/ 0.0198
= 50.50 µmole S-1
The Km for No inhibitor:
y = 0.4231x + 0.0198
Km = 0.0198/ 0.4231
= 0.046.
c)
The kind of inhibitor used in the experiment is competitive inhibitor.
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