The following data were obtained in a study of an enzyme known to follow Michaelis-Menten kinetics: V₀
Substrate added
( $\mu$ mol/min)
(mmol/L)
217
0) 8
325
2
433
4
488
6
647
1,000
The K_m for this enzyme is approximately:

Question

The following data were obtained in a study of an enzyme known to follow Michaelis-Menten kinetics: V₀
Substrate added
(

\mu

mol/min)
(mmol/L)
217
0) 8
325
2
433
4
488
6
647
1,000
The K_m for this enzyme is approximately:

Quizplus · Accepted Answer

In Michaelis-Menten kinetics, KM represents the substrate concentration at which the reaction velocity is half of Vmax. To find KM, we need to plot the data in the form of a Michaelis-Menten curve (velocity vs substrate concentration). However, since a graph is not provided, we can estimate KM using the data given.

We can use the formula for Michaelis-Menten kinetics:

V0 = (Vmax [S])/(KM + [S])

At low substrate concentrations, the velocity of the reaction is directly proportional to substrate concentration. This is known as the linear phase of the reaction. From the data given, we can see that at low substrate concentrations (0-2 mmol/L), the velocity is increasing linearly with substrate concentration. Therefore, we can estimate Vmax as the velocity observed at the highest substrate concentration (1000 mmol/min).

Substituting the values for Vmax and [S] into the Michaelis-Menten equation for the highest substrate concentration gives:

1000 = (Vmax [S])/(KM + 1000)

Solving for KM, we get:

KM = [S](Vmax - V0)/V0

Using the values of [S] and V0 from the second highest substrate concentration (6 mmol/L, 647 mmol/min), we get:

KM = (6)(1000-647)/647 = 2.02 mM

Therefore, the best choice is C) 2 mM.

The Following Data Were Obtained in a Study of an Enzyme