Question

Use the following information and the equation of the line y = 1.00 x 10⁴x + 135 to find the V_max, K_m, k_cat, and K_m. 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 V_max, K_m, k_cat. Include proper units.

Answer 1

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

1/v₀ = (K_M/v_max)1/[S]₀ + 1/v_max

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

At the 0 nM concentration of inhibitor, K_M/v_max = 10⁴ and 1/v_max = 135

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

And K_M = 10⁴ * v_max = 10⁴ * 7.4*10^-3 mM/min = 74 mM

i.e. K_M = ​74 mM

Now, k_cat = v_max/[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

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

= 18.05*10^-3/min

i.e. k_cat = 1.805*10^-2/min