Question

Consider the following data for a newly identified enzyme:

[S] (μmol L-1)	Vo (μmol L-1 min-1)
0.1	0.27
2.0	5.0
10.0	20
20.0	40
40.0	64
60.0	77.5
100.0	100
200.0	120
1,000.0	150
2,000.0	155
3,000.0	155
4,000.0	155

Calculate V_o (in μmol L^-1 min^-1)  when [S] is 133 μM

Answer 1

Recall that this plot is used for Michaelis –Menten equation. It helps to calculate the constants much easier (via graph)

From Michaelis–Menten equation.

V = Vmax*[S] / (Km + [S])

Where

Vmax = max rate velocity (M/s)

[S] = substrate concentration (M)

Km = Michaelis–Menten constant [M]

V = reaction rate [M/s]

Now… mathematical manipulation

Inverse:

1 / [ V = Vmax*[S] / (Km + [S]) ]

1/V = (Km + [S]) / (Vmax*[S])

Separate species

1/V = (Km) / (Vmax*[S])  + [S] / (Vmax*[S])

1/V = Km / Vmax*[S]  + 1/ Vmax

Now, note that:

1/V = Km / Vmax*[S]  + 1/ Vmax

1/V = y-axis

Km/Vmax = slope

1/[S] = x-axis

1/Vmax = y-intercept

-1/Km = x-intecept ( when y = 0 )

Now… Plot the your DATA:

x-axis = 1/[S]

y-axis = 1/V

Slope = Km/Vmax =0.3696

y-intercept = 1/Vmax = 0.0078

get V when

1/V = 0.3696*1/133 + 0.0078

V = (0.3696*1/133 + 0.0078)^-1

V = 94.527 μmol L-1 min-1