A 68-year-old man is evaluated in the intensive care unit for decreased urine output. Thre… — USMLE Step 2 CK

A 68-year-old man is evaluated in the intensive care unit for decreased urine output. Three days ago, he was admitted with septic shock secondary to a perforated appendix and underwent an emergent exploratory laparotomy. Since admission, he has received broad-spectrum antibiotics and aggressive intravenous fluid resuscitation. Over the past 24 hours, his total urine output has been $300$ mL. His blood pressure is $108/64$ mmHg and his heart rate is $92$ /min. The patient appears euvolemic on physical examination.

Laboratory studies are as follows:

Serum studies
Sodium: $140$ mEq/L
Potassium: $5.0$ mEq/L
Chloride: $104$ mEq/L
Bicarbonate: $18$ mEq/L
Blood urea nitrogen: $72$ mg/dL
Creatinine: $3.5$ mg/dL (baseline: $1.0$ mg/dL)

Spot urine studies
Sodium: $60$ mEq/L
Creatinine: $50$ mg/dL

Urinalysis reveals muddy brown granular casts.

What is this patient's calculated fractional excretion of sodium ( $FE_{Na}$ ) expressed as a percentage?

Answer: 3 %

Answer

The correct value is 3. This patient's clinical presentation of septic shock, prolonged hypotension, and muddy brown granular casts on urinalysis is highly suggestive of acute tubular necrosis (ATN), an intrinsic form of acute kidney injury. The fractional excretion of sodium (

FE_{Na}

) is a useful calculation to help differentiate prerenal causes of acute kidney injury from intrinsic causes. It is calculated using the formula:

FE_{Na} = \frac{U_{Na} \times S_{Cr}}{S_{Na} \times U_{Cr}} \times 100

. By substituting this patient's laboratory values into the formula, the calculation is:

\frac{60 \times 3.5}{140 \times 50} \times 100 = \frac{210}{7000} \times 100 = 3\%

. Because the renal tubular epithelial cells are damaged in ATN, they lose their ability to appropriately reabsorb sodium from the filtrate. This leads to an inappropriately elevated amount of sodium spilling into the urine, resulting in an

FE_{Na}

> 2\%

Step-by-Step Solution

Identify the standard formula used to calculate the fractional excretion of sodium.

FE_{Na} = \frac{U_{Na} \times S_{Cr}}{S_{Na} \times U_{Cr}} \times 100

This formula measures the percentage of sodium filtered by the kidney that is ultimately excreted in the urine, helping to distinguish prerenal disease from intrinsic kidney injury.

Extract the specific values needed from the patient's laboratory studies.

U_{Na} = 60

mEq/L,

S_{Cr} = 3.5

mg/dL,

S_{Na} = 140

mEq/L, and

U_{Cr} = 50

mg/dL

Accurate identification of the serum and urine values for sodium and creatinine is necessary to perform the calculation.

Calculate the numerator by multiplying urine sodium and serum creatinine.

60 \times 3.5 = 210

This represents the first half of the clearance ratio.

Calculate the denominator by multiplying serum sodium and urine creatinine.

140 \times 50 = 7000

This represents the second half of the clearance ratio.

Divide the numerator by the denominator and multiply by 100 to find the percentage.

(210 / 7000) \times 100 = 0.03 \times 100 = 3\%

This step converts the absolute fractional value into a percentage, as specifically requested by the question stem.

Key Concept

The fractional excretion of sodium (

FE_{Na}

) is calculated as

(U_{Na} \times S_{Cr}) / (S_{Na} \times U_{Cr}) \times 100

. An

FE_{Na}

greater than

2\%

is typically consistent with intrinsic acute kidney injury (such as acute tubular necrosis), whereas a value less than

1\%

suggests prerenal azotemia.

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