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Zorluk: OrtaInterpreting Laboratory Results

A 48-year-old man is brought to the emergency department by emergency medical services after being found unresponsive in a park. His medical history is unknown. On physical examination, he is unarousable and his breath has a faint, sweet chemical odor. Vital signs show a temperature of 36.2∘C36.2^\circ\text{C} (97.2∘F97.2^\circ\text{F}), blood pressure of 110/70 mm Hg110/70\text{ mm Hg}, heart rate of 105/min105/\text{min}, and respiratory rate of 22/min22/\text{min}.

Initial laboratory studies are obtained:

Laboratory ParameterResultReference Range
Sodium138 mEq/L138\text{ mEq/L}136−145 mEq/L136-145\text{ mEq/L}
Potassium4.5 mEq/L4.5\text{ mEq/L}3.5−5.0 mEq/L3.5-5.0\text{ mEq/L}
Chloride102 mEq/L102\text{ mEq/L}98−106 mEq/L98-106\text{ mEq/L}
Bicarbonate14 mEq/L14\text{ mEq/L}22−28 mEq/L22-28\text{ mEq/L}
Blood urea nitrogen (BUN)14 mg/dL14\text{ mg/dL}8−20 mg/dL8-20\text{ mg/dL}
Glucose126 mg/dL126\text{ mg/dL}70−110 mg/dL70-110\text{ mg/dL}
Measured osmolality318 mOsm/kg H2O318\text{ mOsm/kg H}_2\text{O}275−295 mOsm/kg H2O275-295\text{ mOsm/kg H}_2\text{O}

Arterial blood gas testing on room air shows a pH of 7.24, pCO2p\text{CO}_2 of 28 mm Hg28\text{ mm Hg}, and pO2p\text{O}_2 of 92 mm Hg92\text{ mm Hg}. A serum ethanol level is undetectable.

What is this patient's serum osmolal gap?

Cevap: 30 mOsm/kg H2O

Cevap

30
The correct answer is 30. The patient's presentation of unresponsiveness, high anion gap metabolic acidosis (anion gap of 22), and an elevated osmolal gap is highly suspicious for a toxic alcohol ingestion (such as methanol or ethylene glycol). To find the osmolal gap, one must first determine the calculated osmolality using the formula: 2[Na+]+Glucose18+BUN2.82[\text{Na}^+] + \frac{\text{Glucose}}{18} + \frac{\text{BUN}}{2.8}. Using the patient's lab values, this is 2(138)+12618+142.8=276+7+5=288 mOsm/kg H2O2(138) + \frac{126}{18} + \frac{14}{2.8} = 276 + 7 + 5 = 288\text{ mOsm/kg H}_2\text{O}. The osmolal gap is then obtained by subtracting the calculated osmolality from the measured osmolality: 318−288=30318 - 288 = 30. A normal osmolal gap is typically ≤10 mOsm/kg H2O\le 10\text{ mOsm/kg H}_2\text{O}. A gap of 30 indicates the presence of unmeasured, osmotically active substances in the blood.

Adım Adım Çözüm

1
Identify the formula for calculated serum osmolality.
Calculated Osmolality=2[Na+]+Glucose18+BUN2.8\text{Calculated Osmolality} = 2[\text{Na}^+] + \frac{\text{Glucose}}{18} + \frac{\text{BUN}}{2.8}
This standard formula converts the major serum solutes (sodium and its accompanying anions, glucose, and urea) into their milliosmole equivalents.
2
Input the patient's laboratory values into the formula.
2(138)+12618+142.82(138) + \frac{126}{18} + \frac{14}{2.8}
Sodium is 138 mEq/L, glucose is 126 mg/dL, and BUN is 14 mg/dL.
3
Calculate the individual components and sum them up.
276+7+5=288 mOsm/kg H2O276 + 7 + 5 = 288\text{ mOsm/kg H}_2\text{O}
Completing the arithmetic provides the expected osmolality based on normal physiological solutes.
4
Determine the osmolal gap.
Osmolal Gap=318−288=30 mOsm/kg H2O\text{Osmolal Gap} = 318 - 288 = 30\text{ mOsm/kg H}_2\text{O}
The osmolal gap is the difference between the directly measured osmolality and the calculated osmolality.

Anahtar Kavram

The serum osmolal gap is calculated by subtracting the calculated serum osmolality from the measured serum osmolality. A normal gap is typically less than 10 mOsm/kg H2O; an elevated gap in the setting of a high anion gap metabolic acidosis strongly suggests toxic alcohol ingestion (e.g., methanol, ethylene glycol).
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