Sun28November0440PM 1

A 70-year-old man underwent emergency surgery for an acute abdomen.

Following surgery, he was noted to have become oliguric.

Investigations revealed the following:

Sodium 121 mmol/L (137-144)

Potassium 6.6 mmol/L (3.5-4.9)

Chloride 92 mmol/L (95-107)

Urea 17.2 mmol/L (2.5-7.5)

Creatinine 250 µmol/L (60-110)

pH 7.16 (7.36-7.44)

Standard bicarbonate 15.6 mmol/L (20-28)

Which of the following is the calculated anion gap for this patient?

(Please select 1 option)

15 mmol/L

10 mmol/L

25 mmol/L

20 mmol/L Correct

5 mmol/L

Explanation

Anion gap is calculated as (Na + K) - (Cl + HCO3).

Therefore in this patient, the calculated value is 20 mmol/L.

The normal anion gap is between 8-16 mmol/l. The excessive value here reflects the presence of other acidic anions, and in this case with the metabolic acidosis, the constituents may be lactate, etc.

Answer Statistics

1

11%

2

9%

3

2%

4

79%

5

1%

Times answered: 266