The toxic effects of ingesting methanol can be reduced by administering ethanol.

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.

Explanation / Answer

According to Michaelis-Menten equation,

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

Given,

[Methanol] = 18 mM = 18 X 10-3 M and [Ethanol] = 6.5 M

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

density = mass/volume

or, mass = density x volume

The amount of methanol ingested can be calculated as

Denisty of methanol = 0.79 g /mL

so mass of methanol = 50 X 0.79 = 39.5 grams

moles of methanol = mass / molecular weight = 39.5 / 32 = 1.23moles

Total volume = 40 L

now , molarity = moles of solute / volume of solution in L

Molarity of methanol = 1.23 mol / 40 L = 0.0307 M methanol

So, V/Vmax = 1 / (1 + KM / [S]) = 1/(1 + 1 x 10^-2/0.0307) = 1.032

we want to reduce that with a competitive inhibitor by 3% of 1.032 = 3X1.032 / 100 = 3.096 / 100
V*/Vmax = 1 / (1 + KM* / [S]) = 3.096 / 100
which requires an apparent KM, KM*, of
KM* = 31.29 * [meOH] = 31.29 X 0.0307 = 0.96

Now we need the alpha to be calculated by

alpha = 1 + [I] / Ki

Ki = KM for etOH, so what we need is the concentration of ethanol [etOH] that yields an effective KM for methanol of 0.96 :
KM X(meOH) = alpha X KM(meOH) = (1 + [ethanol] / 10^-3) * 10^-2 = 0.96
[etOH] = 6.5 M
We still have 40 L of fluid in our alky, so the amount of etOH is
(6.5 mol etOH / L solution) * (40 L solution) = 260 mol etOH

Let us calcualte the volume of ethanol = 9568 mL
So we need to give 100 proof whisky
(9568 mL etOH) x (200 mL whiskey / 100 mL etOH) = 19.136 L whiskey

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