High-AG vs normal-AG metabolic acidosis
Two dogs both present with metabolic acidosis (HCO₃⁻ 12 mEq/L in both). Dog A: Na 145, Cl 110. Dog B: Na 145, Cl 120. Compute the anion gap for each, classify each as high-AG or normal-AG, and pick the differential that fits each.
Hint
AG = Na − (Cl + HCO₃⁻). Dog reference range is 13–25 mEq/L. The anion gap reflects unmeasured anions, mostly proteins plus any organic acids that have accumulated.
Another hint
High AG = organic acid added (lactate, ketones, ethylene glycol metabolites, uremic acids). Normal AG with low HCO₃⁻ = HCO₃⁻ lost and replaced by Cl⁻ (GI loss, RTA, dilution).
Show worked answer
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Dog A: AG = 145 − (110 + 12) = 23 mEq/L. Upper end of normal (reference 13–25), bordering on high.
$$\text{AG}_A = 145 - (110 + 12) = 23 \,mEq/L$$ -
Dog B: AG = 145 − (120 + 12) = 13 mEq/L. Low-normal.
$$\text{AG}_B = 145 - (120 + 12) = 13 \,mEq/L$$ -
Dog A has a borderline-high AG with low HCO₃⁻ and a normal chloride. Pattern fits an organic acid: lactate (sepsis, hypoperfusion), ketones (DKA), uremic acids, ethylene glycol.
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Dog B has a normal AG with hyperchloremia. Pattern fits HCO₃⁻ loss replaced by chloride: small bowel diarrhea, renal tubular acidosis, dilutional acidosis.
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Both dogs have the same HCO₃⁻ but very different mechanisms. The anion gap is the key distinguishing test.
Dog A: AG ≈ 23 (high-normal), high-AG pattern fits an organic acidosis (lactic acidosis from sepsis). Dog B: AG = 13 (normal), hyperchloremic, fits HCO₃⁻ loss from diarrhea.