Use the following information and the equation of the line y = 1.00 x 10 4 x + 1

Question

Use the following information and the equation of the line y = 1.00 x 104x + 135 to find the Vmax, Km, kcat, and Km. Include proper units.

In a test of a certain inhibition enzyme, a scientist used 0, 35.0 nM and 75.0 nM of the inhibitor in a 1.00 mL assay volume, she added 23.78 mg of the enzyme, which as a molecular weight of 58.0 kD. The rates were measured in mM/min and the substrate concentrations were measured in mM.

This equation for a line was the result of a lineweaver-burke plot for the test in which 0 nm of inhibitor was used...y = 1.00 x 104x + 135

Use this information and the equation of the line above to find the Vmax, Km, kcat. Include proper units.

Explanation / Answer

The general equation for the Lineweaver Burke plot can be written as shown below.

1/v0 = (KM/vmax)1/[S]0 + 1/vmax

Compare the given equation with the above equation, then you will get the following conclusions.

At the 0 nM concentration of inhibitor, KM/vmax = 104 and 1/vmax = 135

i.e. vmax = 7.4*10-3 mM/min

And KM = 104 * vmax = 104 * 7.4*10-3 mM/min = 74 mM

i.e. KM = 74 mM

Now, kcat = vmax/[E]0 ......... equation 1

Here, [E]0 = {23.78 * 10-3 g / 58*103 g mol-1} / 1 mL

= 0.41*10-6 mol/mL

= 0.41*10-6*103 mmol/mL

= 0.41*10-3 M (since 1 mmol/mL = 1 M)

[E]0 = 0.41 mM (since 1 M = 103 mM)

Substitute the values of vmax and [E]0 in the equation 1, then you will get

kcat = (7.4*10-3 mM/min) / 0.41 mM

= 18.05*10-3/min

i.e. kcat = 1.805*10-2/min

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