Lab case 420 Interpretation


PH = 7.24 that is moderate acidaemia.

HCO3 = 16 mmol/L, so we have metabolic acidosis.

Next, we need to calculate the anion gap and the compensation.

This patient has high K level. For more accurate interpretation, we will include it in the calculation of AG and we will consider the normal AG level 16.

Accordingly, AG for this patient will be (Na + K) – (Cl + HCO3) = 13.3, So we have NAGMA.

To calculate the compensation, we will use Winter’s formula, that is:

Expected pCO2 = 1.5 x HCO3 + 8 (+/-2) = 32, so we have well compensated NAGMA with no additional respiratory process.

Other abnormal findings:

Na = 128 mmol/L, that is moderate hyponatraemia

K = 6.3 mmol/L, that is moderate hyperKalaemia.

Glucose = 3.4 mmol/L (in non-diabetic patient that is considered normal but on the low side of the scale).

Lactate = 3.4 mmol/L, that is high.


Next, we need to look at the causes of these derangements for this patient. For the differential diagnosis of NAGMA, we use the mnemonic USED CARP.

  • U = Ureteroenterostomy
  • S = Small bowel fistula
  • E = Extra chloride
  • D = Diarrhea
  • C = Carbonic anhydrase inhibitors
  • A = Adrenal insufficiency/ Addison’s disease
  • R = Renal tubular acidosis
  • P = Pancreatic fistula.

Among the causes above Addison’s disease fit (Hyponatraemia, Hyperkalaemia, hypoglycemia) and the high lactate is due to poor tissue perfusion secondary to hypotension.

This patient’s condition improved after she was given hydrocortisone.