Lab case 428 interpretation

Answers:

PH = 7.479, that is mild acidaemia.

HCO3 = 30 mmol/L (that is high). So, we have metabolic alkalosis.

Next we check the respiratory compensation for metabolic alkalosis. For That we use the following equation:  Expected pCO2 = 0.7 x HCO3 + 20 (+/- 5).

Accordingly, expected pCO2 should be: 0.7 x 30 + 20 = 41 mmHg. So we have well compensated metabolic alkalosis.

Other abnormal findings:

Cl = 94 mmol/L, that is hypochloraemia.

Lactate = 2.0 mmol/L, that is mild hyperlactataemia.

So, we have hypochloraemic metabolic alkalosis.

For the causes of metabolic alkalosis we use the mnemonic CLEVER PD.

  • C – contraction (dehydration)
  •  L – liquorice (diuretic), laxative abuse
  •  E – endocrine (Conn’s, Cushing’s)
  •  V – vomiting, GI loss (villous adenoma)
  •  E – excess alkali (antacids)
  •  R – renal (Bartter’s), severe K depletion
  •  P – post hypercapnia
  •  D – diuretics

This child had pyloric stenosis confirmed by U/S.

However, the classic blood gases for pyloric stenosis is Hypochloraemic, hypokalaemic metabolic alkalosis. This patient had K level of 4.6 mmol/L.

The other abnormality is sodium level of 140 which is normal but more to the high side while the chloride level was low. Usually the changed in these 2 ions go in the same direction.

The good practice of always checking the Anion Gap in every blood gases reveal Anion Gap of 16, So this patient has additional HAGMA.

This patient had ketones level of 8.4 mmol/L.

K level was normalised due to the presence of additional metabolic acidosis.

 

Special thanks to Dr Paul Koh for providing the details for this case.