A person with type 1 diabetes ran out of her prescription insulin and has not been able to inject insulin for the past 3 days. The patient is hyperventilating to compensate for her metabolic acidosis. Which of the following reactions explains this partially compensating respiratory alkalosis?
The patient is "blowing off CO2 " to reduce acid. The equation described in answer C is the conversion of carbon dioxide into a soluble form, then into bicarbonate. During an acidosis, the high levels of protons push the reaction described in answer C to the left, to the formation of water and carbon dioxide. As the carbon dioxide is exhaled, and the concentration of carbon dioxide decreases, more carbon dioxide is formed, thereby reducing the pool of free protons and raising the pH. The protonation of ammonia to form ammonium ion takes place in the kidney and not the lungs. Its primary purpose is to alkalinize the urine if it is too acidic. The reaction described in answer B is the dissociation of a proton from b-hydroxybutyrate (a ketone body) to form the anion of b-hydroxybutyrate and a proton. This is the reaction that is occurring to bring about the ketoacidosis and is not the compensatory respiratory alkalosis. Reaction D is the dissociation of water, which cannot buffer the acidosis.
H+NH3=NH4+
CH3CHOHCH2COOH=CH3CHOHCH2COO-+H+
CO2+H2O=H2CO3=H++HCO3-
H2O=H++HO-
Miscellaneous (Bio-Chemistry)