Calculate the percentage of alcohol in the patient's body.State the chemical reaction that took place in the patient's body.
A 40 year old manager, at a prominent legal firm in L.A., experiences abdominal pain post lunch. He goes to the restroom a couple of times, but continues to experience pain and cramps. He eventually decides to get himself examined. While questioning him, you learn that he consumed two bottles of beer at lunch. The alcohol (ethyl alcohol/ethanol) content in a 12 oz. bottle of beer is about 5%. The man's weight is 150 lbs and he was drinking over an hour period.
Identify the substance produced in the liver that helps catalyze this reaction.
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Let us first determine the cause of problem. Most of the alcohol (ethyl) that we are taking in when we drink is being completely oxidized to carboxylic acid. However, some of it goes into other parts and being excreted. Oxidation of ethanol may not be complete thus can be oxidized into ethanal (an aldehyde) which is toxic to our body at significant concentrations. This oxidation is being catalysed by a cytosolic enzyme alcohol dehydrogenase (ADH). The equation for this reaction is:
`CH_3CH_2OH + NAD^+ -> CH_3CHO + NADH + H^+` .
We can represent the percentage of alcohol in body as:
(mass of ethanol)/(mass of body) x 100
Mass of ethanol:
12 oz of beer is equal to 354.882 mL so 2 bottles of beer would have 709.764 mL. The percentage of alcohol in the beer is said to be 5% and no other descriptions indicated so we assume it is 5grams ethanol per 100 mL.
`709.764 mL * (5 grams ethanol)/(100mL)`
= 35.4882 grams of ethanol.
Mass of the old manager:
`150 lb * (1 kg)/(2.20462 lb)`
= 68.0389 kg = 6803.89 grams
`(35.4882)/(6803.89) * 100`
= 0.52 % -> answer
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