Question

The toxic effects of ingesting methanol can be reduced by administering ethanol. The ethanol acts as a competitive inhibitor of methanol by displacing it from LADH. If an individual has ingested 50 mL of methanol, how much 100 proof whiskey (50% ethanol by volume) must be imbibed to reduce the activity (Vo) of his LADH towards methanol to 3% of its original value? The adult human body contains ~40L of aqueous fluids throughout which ingested alcohols are rapidly and uniformly mixed. Assume the concentration of methanol in the body is 18mM and whiskey is 6.5M ethanol. Assume the KM values of LADH for ethanol and methanol to be 10-3 M and 10-2 M, respectively, and that Ki = KM for ethanol.

Answer 1

We will use the follwoing Michaelis-Menten equation,

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

Given, [MeOH] = 18 mM (0.018 M) and [EtOH] = 6.5 M

Km for MeOH = 10^-2 M and EtOH = 10^-3 M

density = mass/volume

or, mass = density x volume

The amount ingested is

(50 mL meOH) (0.79 g/mL) / (32 g/mol) = 1.23 mol meOH

Total volume = 40 L

molarity = moles/L

Molarity of [meOH] = 1.23 mol / 40 L = 0.031 M meOH

So, V/Vmax = 1 / (1 + KM / [S]) = 1/(1 + 1 x 10^-2/0.031) = 1/3

V*/Vmax = 1 / (1 + KM* / [S]) = 2/300

KM* = 149 * [meOH] = 3

Now you need the alpha that you calculated,

alpha = 1 + [I] / Ki

KM*(meOH) = alpha * KM(meOH) = (1 + [etOH] / 1e-3) * 1e-2 = 75

[etOH] = 6.5 M

We have 40 L of fluid in our body, so the amount of etOH is

(6.5 M etOH / L solution) * (40 L solution) = 260 mol etOH

260 (46 g/mol) / (0.8 g/mL) = 14.95 L EtOH

So for 100 proof (50% ethanol)

(14950 mL etOH) x (200 mL whiskey / 100 mL etOH) = 29.9 L whiskey