A teenager drank ethanol laced with methanol, a poisonous alcohol. Was rushed to a hospital, the physician pumps his stomach and ask the teenager to drink vodka (40% ethanol) for another 24h. The nurse asks her, why she did that? The Dr. replies, liver alcohol dehydrogenase (ADH) makes methanol a poison, but its Km with ADH is 10mM, while the Km for ethanol is 1mM. The nurse asks, “what about the k2? The Dr. says, I made it irrelevant. Explain why the Dr. did this and why the k2 is irrelevant?
in poisoning of methanol, Doctors prescribes ethanol drinking, because in the liver alcohol dehydrogenase (ADH) enzyme present which metabolises alcohol to aldehydes and acids which are less toxic as compared to alcohols, but in case of methanol drinking ADH metabolises methanol to form formaldehyde and formic acid which are more toxic than the methanol. so to avoid metabolism of methanol in the liver a competative inhibitor of methanol for ADH is added and it is ethanol. it competatively inihibit methanol binding to ADH.
K2 is the rate constant for the convertion of enzyme substrate complex in to enzyme product, that means it applicable when substrate binds to the enzyme, but in this case Dr. inhibiting the binding of methanol to ADH by competative means, So, K2 value does not have any relevence.
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