An enzyme has a Km equal to 1.0 x 10-4 M (molar) for a certain substrate S. When [S] = 1.0 x 10-4 M, 2% of the substrate is transformed into product in 1 minute. Which of the following statements are correct? Hint: use the Michaelis-Menten equation; Remember that a reaction rate can be expressed in molar units / min (M / min). Select one or more than one: a) When [S] = 1.0 x 10-4 M, the initial reaction rate, V0, is 2.0 x 10-7 M / min b) For this enzyme Vmax = 4.0 x 10-7 M / min c) For this enzyme Vmax is equal to 2.002 x 10 -7 M / min. d) When [S] 0 = 1.0 x 10-3 M, the initial speed is 3.64 x 10-7 M / min. e) When [S] = 0.01 M, the initial speed is 4.0 x 10-3 M / min
An enzyme has a Km equal to 1.0 x 10-4 M (molar) for a certain substrate S. When [S] = 1.0 x 10-4 M, 2% of the substrate is transformed into product in 1 minute.
Which of the following statements are correct?
Hint: use the Michaelis-Menten equation; Remember that a reaction rate can be expressed in molar units / min (M / min).
Select one or more than one:
a) When [S] = 1.0 x 10-4 M, the initial reaction rate, V0, is 2.0 x 10-7 M / min
b) For this enzyme Vmax = 4.0 x 10-7 M / min
c) For this enzyme Vmax is equal to 2.002 x 10 -7 M / min.
d) When [S] 0 = 1.0 x 10-3 M, the initial speed is 3.64 x 10-7 M / min.
e) When [S] = 0.01 M, the initial speed is 4.0 x 10-3 M / min
Km = 1.0 x 10-4 M
[S] = 1.0 x 10-4 M
Vo = Vmax[S]/{Km + [S]}
At Km = [S]
Vo = Vmax/2
(a) and (b) If Vo = 2.0 x 10-7 M/min
Vmax = 4.0 x 10-7 M/min
(c) wrong option
(d) Vo = Vmax[S]/{Km + [S]}
Vo = 4.0 x 10-7 M/min x 1.0 x 10-3 M/{1.0 x 10-4 M + 1.0 x 10-3 M}
Vo = 3.64 x 10-7 M/min
(e) Vo = Vmax[S]/{Km + [S]}
Vo = 4.0 x 10-7 M/min x 0.01 M/{1.0 x 10-4 M + 0.01 M}
Vo = 3.96 x 10-7 M/min
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