Question

The metabolism of ethanol in our bodies is catalyzed by the enzyme liver alcohol dehydrogenase (LADH) in the human body. In contrast, methanol is converted to formaldehyde (also catalyzed by LADH), which can cause blindness or even death. An antidote for methanol is ethanol, which acts as a competitive inhibitor for LADH. The excess methanol can then be safely discharged from the body How much absolute (100%) ethanol would a person have to consume after ingesting 50 mL of methanol (a lethal dosage) to reduce the activity of LADH to 3% of the original value? Assume that the total fluid volume in the person's body is 38 liters and that the densities of ethanol and methanol are 0.789 g/mL and 0.791 g/mL, respectively. The KM value for methanol is 1.0 x 10^(-2) M and the KI value for ethanol is 1.0 x 10^(-3) M. State any assumptions.

Answer 1

Equation for rate of the reaction with competitve inhibition is

where

Now given density of ethanol ,

density of methanol,

Net Volume = 38 litres.

Volume of ingested methanol, v = 50ml.

molar mass of ethanol = 46 gms/mol.

molar mass of methanol = 32 gms/mol.

K_M = 10^-2 M and K_I = 10^-3 M

First we'll find [S] = concentration of ingested methanol.

Let V₀ = 3 and V_max = 100, as we need to bring down the activity of LADH to 3% of the maximum activity.

On substituting the values in the equation we get,

thus the volume of ethanol required is 234.17 ml.