Question

Ethanol in the body is oxidized to acetaldehyde by liver alcohol dehydrogenase (LADH).

Other alcohols are also oxidized by LADH. Methanol, which is mildly intoxicating, is oxidized by LADH to the highly toxic product formaldehyde.

The toxic effects of methanol (a component of many commercial solvents and "wood alcohol" in moonshine) can be reduced by administering ethanol. The ethanol acts as a competitive inhibitor of methanol by displacing it from LADH. This provides sufficient time for the methanol to be harmlessly excreted by the kidneys.

If an individual has ingested 100 mL of methanol (a lethal dose) how much 100 proof whiskey (50% ethanol by volume) must he or she imbibe to reduce the enzyme activity (the v0) of his or her LADH towards methanol to 5% of its original value?

Assume that the human body contains 40 L of aqueous fluids throughout which the ingested alcohols are rapidly and thoroughly mixed. Also assume that any methanol or whisky ingested does not change the overall amount of fluids in the body. The densities of ethanol and methanol are both 0.79 g/mL. Assume the Km values are of LADH for ethanol and methanol are 10–3 M and 10–2 M, respectively, and that Ki = Km for ethanol.

Answer 1

This is a case of competitive inhibition, and the velocity of the enzyme is defined as:

In the presence and absence of the inhibitor, respectively.

Since we want V_0i to be 5% of V₀, we have:

We know the values of Km (constant for methanol), Ki (constant for ehtanol) and we can calculate the concentration of methanol:

In a total volume of 40 L, this gives a concentration of 0.062M for methanol. We can now calculate the necessary concentration of ethanol:

If we re-arrange this equation:

We need 0.1368 moles of ethanol per liter. If the toatl volume is 40 liters, we need: 5.472 moles = 252 g = 319.1 mL

If the whiskey is 50% ethanol in volume, we need: 638.2 mL of whisky