Calculate the percentage of alcohol in the patient's body.
A 40 year old male consumed two 12 oz. bottles of beer at lunch. He is experiencing abdominal pains post lunch. The man weighs 150 lbs. and was drinking over an hour period. The alcohol (ethyl alcohol/ethanol) content in a 12 0z. bottle of beer is about 5%. Calculate the percentage of alcohol in the patient's body, state the chemical reaction that took place in the patient's body, and identify the substance produced in the liver that helps catalyze this reaction.
The equation for calculating blood alcohol content is as follows:
%BAC = (A x 5.14/W x r) - .015 x H, where:
A = liquid oz of alcohol consumed = 2*12* 5% = 1.2
W= weight of the person in lbs = 150 lbs
r= gender alcohol distribution ratio = 0.73 for men
H= hours drinking = 1 hour
Substituting the values into the equation we get:
%BAC = (1.2 x 5.14/150 x 0.73) - .015 x 1 = 0.015%
Therefore, the percentage of alcohol in the individual’s body is 0.015%
The chemical reaction taking place in the patient’s body is below, and it is catalyzed by the cytosolic enzyme alcohol dehydrogenase (ADH).
CH3CH2OH + NAD+ -> CH3CHO + NADH + H+
That is, the highly toxic substance acetaldehyde is produced.