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Oxidizing ethanol

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Ethanol in the body is oxidized to acetaldehyde by liver alcohol dehydrogenase (LADH). Other alcohols are also oxidized by LADH. For example, methanol, which is mildly intoxicating, is oxidized by LADH to the quite toxic product formaldehyde. The toxic effects of ingesting methanol (a component of many commercial solvents) can be reduced by administering ethanol. The ethanol acts as a competitive inhibitor of the 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 be imbibed to reduce the activity of the LADH towards methanol to 5% of its original value? The adult human body contains ~40 liters of aqueous fluids throughout which ingested alcohols are rapidly and uniformly mixed. The densities of ethanol and methanol are both 0.79 g cm-3. Assume the Km values of LADH for ethanol and methanol to be 1.0 x 10-3M and 1.0 10-2M, respectively, and that KI = Km for ethanol.

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Solution explains how ethanol is oxidized.

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Hi

This is a great question! You have been given a bunch of information about kinetic parameters for the liver acohol dehydrogenase enzyme (LADH) and you are asked to determine how much ethanol would be required to effectively "outcompete" methanol for the enzyme active site. This was a commonly used treatment for methanol poisoning in the past, although I think there are other methods used sometimes now (without the ethanol "side effects"...)

Obviously, this question requires the use of Michaelis-Menton enzyme kinetics and the associated mathematical expressions for enzyme activity. You have been given the information that methanol is a competitive inhibitor for LADH. Ethanol and methanol both act as substrates for this enzyme. The Michaelis-Menton constant, Km, tells you what substrate concentration ([S]) is required for half of the available enzyme active sites are bound to substrate, which corresponds to the [S] required to achieve ½ the maximum reaction velocity (Vmax). A higher Km value means that enzyme binding/turnover to ...

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