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Question 101Question →

A 24-year-old woman with a history of type 1 diabetes mellitus presents to the emergency department with nausea, vomiting, and diffuse abdominal pain. She reports missing her insulin doses for the past two days due to illness. On physical examination, she appears lethargic. Her blood pressure is 105/65 mm Hg, heart rate is 118/min, and respiratory rate is 26/min with deep, rapid breaths. Laboratory studies show the following:

Laboratory testResult
Serum sodium140 mEq/L
Serum potassium5.2 mEq/L
Serum chloride98 mEq/L
Serum bicarbonate12 mEq/L
Blood glucose485 mg/dL
Serum creatinine1.1 mg/dL

What is the patient's calculated serum anion gap? (Provide the numerical value in mEq/L).

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Answer: 30

Answer

30
The serum anion gap is calculated using the standard formula: Na+−(Cl−+HCO3−)Na^+ - (Cl^- + HCO_3^-). In this patient, substituting the given laboratory values yields: 140−(98+12)=140−110=30140 - (98 + 12) = 140 - 110 = 30 mEq/L. A normal anion gap is typically between 8 and 12 mEq/L. This patient has a significantly elevated anion gap of 30 mEq/L, which, in the context of her hyperglycemia and missed insulin doses, is highly consistent with diabetic ketoacidosis, a classic cause of high anion gap metabolic acidosis.

Step-by-Step Solution

1
Identify the standard formula for calculating the serum anion gap.
Anion Gap = Na+−(Cl−+HCO3−)Na^+ - (Cl^- + HCO_3^-)
The anion gap is a calculated value used to differentiate the causes of metabolic acidosis by estimating the unmeasured anions in the serum.
2
Extract the relevant serum electrolyte values from the provided laboratory results.
Na+=140Na^+ = 140 mEq/L, Cl−=98Cl^- = 98 mEq/L, and HCO3−=12HCO_3^- = 12 mEq/L
These three specific electrolyte values are the components required for the standard anion gap calculation. Serum potassium is generally omitted from this formula on the USMLE.
3
Substitute the values into the formula and perform the calculation.
Anion Gap = 140−(98+12)=140−110=30140 - (98 + 12) = 140 - 110 = 30 mEq/L
This mathematical operation yields the final numerical value of the patient's anion gap.

Key Concept

Calculation of the serum anion gap is a fundamental step in evaluating a patient with metabolic acidosis, utilizing the formula: Anion Gap = Na+−(Cl−+HCO3−)Na^+ - (Cl^- + HCO_3^-).
Question 102Question →

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 300300 mL. His blood pressure is 108/64108/64 mmHg and his heart rate is 9292/min. The patient appears euvolemic on physical examination.

Laboratory studies are as follows:

Serum studies
Sodium: 140140 mEq/L
Potassium: 5.05.0 mEq/L
Chloride: 104104 mEq/L
Bicarbonate: 1818 mEq/L
Blood urea nitrogen: 7272 mg/dL
Creatinine: 3.53.5 mg/dL (baseline: 1.01.0 mg/dL)

Spot urine studies
Sodium: 6060 mEq/L
Creatinine: 5050 mg/dL

Urinalysis reveals muddy brown granular casts.

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

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Answer: 3

Answer

3
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 (FENaFE_{Na}) is a useful calculation to help differentiate prerenal causes of acute kidney injury from intrinsic causes. It is calculated using the formula: FENa=UNa×SCrSNa×UCr×100FE_{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: 60×3.5140×50×100=2107000×100=3%\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 FENaFE_{Na} of >2%> 2\%.

Step-by-Step Solution

1
Identify the standard formula used to calculate the fractional excretion of sodium.
FENa=UNa×SCrSNa×UCr×100FE_{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.
2
Extract the specific values needed from the patient's laboratory studies.
UNa=60U_{Na} = 60 mEq/L, SCr=3.5S_{Cr} = 3.5 mg/dL, SNa=140S_{Na} = 140 mEq/L, and UCr=50U_{Cr} = 50 mg/dL
Accurate identification of the serum and urine values for sodium and creatinine is necessary to perform the calculation.
3
Calculate the numerator by multiplying urine sodium and serum creatinine.
60×3.5=21060 \times 3.5 = 210
This represents the first half of the clearance ratio.
4
Calculate the denominator by multiplying serum sodium and urine creatinine.
140×50=7000140 \times 50 = 7000
This represents the second half of the clearance ratio.
5
Divide the numerator by the denominator and multiply by 100 to find the percentage.
(210/7000)×100=0.03×100=3%(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 (FENaFE_{Na}) is calculated as (UNa×SCr)/(SNa×UCr)×100(U_{Na} \times S_{Cr}) / (S_{Na} \times U_{Cr}) \times 100. An FENaFE_{Na} greater than 2%2\% is typically consistent with intrinsic acute kidney injury (such as acute tubular necrosis), whereas a value less than 1%1\% suggests prerenal azotemia.
Question 103Question →

A 42-year-old man presents to the emergency department with a 3-day history of severe nausea, nonbilious vomiting, and diffuse abdominal pain. He has a history of poorly controlled type 1 diabetes mellitus. His temperature is 37.2∘C37.2^\circ\text{C} (99.0∘F99.0^\circ\text{F}), blood pressure is 90/60 mm Hg90/60\text{ mm Hg}, pulse is 118/min118/\text{min}, and respirations are 26/min26/\text{min}. Physical examination reveals dry mucous membranes, decreased skin turgor, and a diffusely tender abdomen without rebound or guarding.

Laboratory studies show the following:
Serum sodium: 142 mEq/L142\text{ mEq/L}
Serum potassium: 3.2 mEq/L3.2\text{ mEq/L}
Serum chloride: 92 mEq/L92\text{ mEq/L}
Serum bicarbonate: 14 mEq/L14\text{ mEq/L}
Blood glucose: 550 mg/dL550\text{ mg/dL}
Serum blood urea nitrogen: 45 mg/dL45\text{ mg/dL}
Serum creatinine: 1.8 mg/dL1.8\text{ mg/dL}

Assuming a normal serum anion gap of 12 mEq/L12\text{ mEq/L} and a normal serum bicarbonate of 24 mEq/L24\text{ mEq/L}, what is this patient's delta-delta ratio (ΔAG/ΔHCO3−\Delta\text{AG} / \Delta\text{HCO}_3^-)?

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Answer: 2.4

Answer

The patient's delta-delta ratio is 2.4.
This patient presents with diabetic ketoacidosis (DKA), which causes a high anion gap metabolic acidosis (HAGMA). He also has a history of severe vomiting, which typically causes a metabolic alkalosis due to loss of gastric hydrochloric acid. To uncover this mixed disorder, the delta-delta ratio is calculated. The measured anion gap is 142−(92+14)=36 mEq/L142 - (92 + 14) = 36\text{ mEq/L}. The change in the anion gap (ΔAG\Delta\text{AG}) is 36−12=24 mEq/L36 - 12 = 24\text{ mEq/L}. The change in bicarbonate (ΔHCO3−\Delta\text{HCO}_3^-) is 24−14=10 mEq/L24 - 14 = 10\text{ mEq/L}. The delta-delta ratio is 24/10=2.424 / 10 = 2.4. In a pure HAGMA, the drop in bicarbonate should roughly equal the rise in the anion gap (ratio of 1.01.0 to 2.02.0). A ratio greater than 2.02.0 confirms that the bicarbonate has not dropped as much as expected, indicating the presence of a concurrent metabolic alkalosis, consistent with his history of vomiting.

Step-by-Step Solution

1
Calculate the patient's measured serum anion gap.
AG=142−(92+14)=36 mEq/L\text{AG} = 142 - (92 + 14) = 36\text{ mEq/L}
Calculating the anion gap is the first step in analyzing any acid-base disorder to determine if unmeasured anions are present.
2
Calculate the change in anion gap (ΔAG\Delta\text{AG}).
ΔAG=36−12=24 mEq/L\Delta\text{AG} = 36 - 12 = 24\text{ mEq/L}
This value represents the excess unmeasured anions (such as ketones) added to the blood.
3
Calculate the change in serum bicarbonate (ΔHCO3−\Delta\text{HCO}_3^-).
ΔHCO3−=24−14=10 mEq/L\Delta\text{HCO}_3^- = 24 - 14 = 10\text{ mEq/L}
This value represents how much bicarbonate was consumed to buffer the excess acid.
4
Calculate the delta-delta ratio.
ΔRatio=2410=2.4\Delta\text{Ratio} = \frac{24}{10} = 2.4
The ratio of ΔAG\Delta\text{AG} to ΔHCO3−\Delta\text{HCO}_3^- reveals whether the change in the anion gap corresponds to the drop in bicarbonate. A ratio greater than 2.0 indicates an underlying concurrent metabolic alkalosis.

Key Concept

The delta-delta ratio is used to identify mixed acid-base disorders in the presence of a high anion gap metabolic acidosis. A ratio greater than 2.0 indicates a concurrent metabolic alkalosis.
Question 104Question →

A 54-year-old man is evaluated in the hospital for decreased urine output. He was admitted 7 days ago for infective endocarditis and has been receiving intravenous vancomycin and gentamicin. Over the past 24 hours, his urine output has dropped to 300 mL. His vital signs are within normal limits and he appears euvolemic on examination. Laboratory studies show the following:

Laboratory ParameterSerumUrine
Sodium140 mEq/L70 mEq/L
Potassium5.2 mEq/L25 mEq/L
Creatinine4.0 mg/dL50 mg/dL
Urea nitrogen (BUN)40 mg/dL-
Osmolality290 mOsm/kg H₂O310 mOsm/kg H₂O

Microscopic urinalysis reveals muddy brown granular casts.

Based on the provided laboratory findings, what is this patient's fractional excretion of sodium (FeNaFe_{Na})? (Express your answer as a percentage)

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Answer: 4

Answer

4.0
The fractional excretion of sodium (FeNaFe_{Na}) measures the percentage of filtered sodium that is ultimately excreted in the urine. It is calculated using the formula: FeNa=UNa×PCrPNa×UCr×100Fe_{Na} = \frac{U_{Na} \times P_{Cr}}{P_{Na} \times U_{Cr}} \times 100. Plugging in the correctly extracted values from the table (UNa=70U_{Na} = 70, PCr=4.0P_{Cr} = 4.0, PNa=140P_{Na} = 140, UCr=50U_{Cr} = 50) yields 2807000×100=4.0%\frac{280}{7000} \times 100 = 4.0\%. A value greater than 2% in the setting of oliguria is highly suggestive of intrinsic acute kidney injury, specifically acute tubular necrosis (ATN). This correlates perfectly with the patient's history of aminoglycoside (gentamicin) use, isotonic urine (osmolality ~300 mOsm/kg), and the pathognomonic finding of muddy brown granular casts on urinalysis.

Step-by-Step Solution

1
Recall the Fractional Excretion of Sodium (FeNa) formula.
FeNa=UNa×PCrPNa×UCr×100Fe_{Na} = \frac{U_{Na} \times P_{Cr}}{P_{Na} \times U_{Cr}} \times 100
This formula normalizes sodium excretion to the glomerular filtration rate, helping differentiate the etiology of acute kidney injury.
2
Identify the correct serum and urine values from the vignette's data table.
UNaU_{Na} = 70 mEq/L, PCrP_{Cr} = 4.0 mg/dL, PNaP_{Na} = 140 mEq/L, UCrU_{Cr} = 50 mg/dL.
Careful extraction prevents using incorrect parameters such as BUN or potassium, which act as distractors.
3
Perform the multiplication for the numerator and denominator.
Numerator = 280, Denominator = 7000.
70×4.0=28070 \times 4.0 = 280 and 140×50=7000140 \times 50 = 7000.
4
Divide the numerator by the denominator and convert to a percentage.
280/7000=0.04280 / 7000 = 0.04, which translates to 4.0%.
This yields the final diagnostic percentage, which is > 2%, consistent with acute tubular necrosis.

Key Concept

The Fractional Excretion of Sodium (FeNa) is calculated to differentiate prerenal azotemia (FeNa < 1%) from intrinsic acute kidney injury (FeNa > 2%).
Question 105Question →

A 35-year-old automobile mechanic is evaluated in the emergency department for altered mental status. His spouse reports finding him confused and stumbling in their unventilated garage. Vital signs show a blood pressure of 132/82 mm Hg, a pulse of 98/min, and a respiratory rate of 26/min. He appears lethargic and is oriented only to person.

Laboratory studies reveal the following:

Laboratory StudyResult
Sodium140 mEq/L
Potassium4.5 mEq/L
Chloride105 mEq/L
Bicarbonate12 mEq/L
Blood urea nitrogen (BUN)14 mg/dL
Creatinine1.2 mg/dL
Glucose90 mg/dL

The patient's measured serum osmolality is 330 mOsm/kg H₂O.

What is this patient's calculated serum osmolar gap (in mOsm/kg H₂O)?

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Answer: 40

Answer

The patient's calculated serum osmolar gap is 40 mOsm/kg H₂O.
The patient presents with altered mental status and a high anion gap metabolic acidosis (Anion Gap = 140−[105+12]=23140 - [105 + 12] = 23), raising high clinical suspicion for toxic alcohol ingestion, such as ethylene glycol or methanol. To further evaluate this, the serum osmolar gap must be calculated. The osmolar gap is the difference between the measured serum osmolality (provided by the laboratory) and the calculated serum osmolality.

Using conventional units (mg/dL), the calculated osmolality is determined by the formula:

Calculated Osmolality=2[Na+]+BUN2.8+Glucose18{\text{Calculated Osmolality}} = 2[\text{Na}^+] + \frac{\text{BUN}}{2.8} + \frac{\text{Glucose}}{18}

Substituting the patient's values:

Calculated Osmolality=2(140)+142.8+9018=280+5+5=290 mOsm/kg H2O{\text{Calculated Osmolality}} = 2(140) + \frac{14}{2.8} + \frac{90}{18} = 280 + 5 + 5 = 290 \text{ mOsm/kg H}_2\text{O}

Next, the osmolar gap is computed:

Osmolar Gap=Measured Osmolality−Calculated Osmolality{\text{Osmolar Gap}} = {\text{Measured Osmolality}} - {\text{Calculated Osmolality}}

Osmolar Gap=330−290=40 mOsm/kg H2O{\text{Osmolar Gap}} = 330 - 290 = 40 \text{ mOsm/kg H}_2\text{O}

A normal osmolar gap is typically less than 10 mOsm/kg H₂O. A significantly elevated osmolar gap in the setting of a high anion gap metabolic acidosis strongly supports the presence of an unmeasured osmotically active substance, aligning with a diagnosis of toxic alcohol poisoning.

Step-by-Step Solution

1
Identify the formula for calculated serum osmolality.
Calculated Osmolality = 2×[Na+]+BUN2.8+Glucose182 \times [\text{Na}^+] + \frac{\text{BUN}}{2.8} + \frac{\text{Glucose}}{18}
This formula converts conventional units (mg/dL) of BUN and glucose into mOsm/kg and accounts for the main physiological contributors to serum osmolality.
2
Insert the patient's laboratory values into the formula to find the calculated osmolality.
Calculated Osmolality = 2×140+142.8+9018=280+5+5=2902 \times 140 + \frac{14}{2.8} + \frac{90}{18} = 280 + 5 + 5 = 290 mOsm/kg H₂O.
Determining the expected baseline osmolality is required before the gap can be derived.
3
Subtract the calculated osmolality from the measured osmolality.
Osmolar gap = 330−290=40330 - 290 = 40 mOsm/kg H₂O.
The osmolar gap quantifies the amount of unmeasured, osmotically active substances in the blood.

Key Concept

Calculation of the serum osmolar gap and its utility in identifying toxic alcohol ingestion.
Question 106Question →

A 30-year-old male electrician is brought to the emergency department after sustaining a high-voltage electrical injury at a construction site. The accident occurred exactly 2 hours prior to his arrival. On physical examination, the patient is awake and alert. Vital signs show a heart rate of 115/min and blood pressure of 105/65 mm Hg. He has deep partial-thickness and full-thickness burns involving his entire right arm and his entire anterior trunk (chest and abdomen). No other areas of the body are burned. His body weight is 75 kg. Two large-bore peripheral intravenous lines are established, and a Foley catheter is placed.

Based on current Advanced Trauma Life Support (ATLS) guidelines for high-voltage electrical burns, what should be the continuous intravenous fluid infusion rate (in mL/hr) of lactated Ringer solution for this patient over the next 6 hours?

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Answer: 675

Answer

675
The correct infusion rate is 675 mL/hr. Electrical burns require aggressive resuscitation at 4 mL/kg/%TBSA4 \text{ mL/kg/\%TBSA}. For a 75 kg patient with a 27% TBSA burn (9% right arm + 18% anterior trunk), the 24-hour requirement is 4×75×27=8100 mL4 \times 75 \times 27 = 8100 \text{ mL}. Half of this volume (4050 mL) must be infused within the first 8 hours of injury. Because 2 hours have already passed, the 4050 mL must be infused over the remaining 6 hours (4050÷6=6754050 \div 6 = 675).

Step-by-Step Solution

1
Determine the Total Body Surface Area (TBSA) burned.
27%
Using the Rule of Nines, the entire right arm is 9% and the entire anterior trunk (chest and abdomen) is 18%. 9%+18%=27%9\% + 18\% = 27\%.
2
Calculate the total 24-hour fluid volume requirement.
8100 mL
Current ATLS guidelines mandate 4 mL/kg/%TBSA4 \text{ mL/kg/\%TBSA} for electrical burns to prevent rhabdomyolysis-induced renal failure. Total volume = 4×75 kg×27%=8100 mL4 \times 75 \text{ kg} \times 27\% = 8100 \text{ mL}.
3
Calculate the fluid volume to be given in the first 8 hours post-injury.
4050 mL
Half (50%) of the total 24-hour calculated volume must be administered during the first 8 hours following the time of injury. 8100 mL÷2=4050 mL8100 \text{ mL} \div 2 = 4050 \text{ mL}.
4
Calculate the hourly infusion rate for the remaining time in the first 8-hour window.
675 mL/hr
The patient arrived 2 hours after the injury, leaving 6 hours to infuse the first half of the fluid. 4050 mL÷6 hours=675 mL/hr4050 \text{ mL} \div 6 \text{ hours} = 675 \text{ mL/hr}.

Key Concept

ATLS electrical burn fluid resuscitation (4 mL/kg/%TBSA) and accounting for pre-hospital time delay.
Question 107Question →

An ambulatory care physician evaluates a 33-year-old woman who reports a one-month history of persistent fatigue and severe muscle weakness. She denies any recent illnesses, diarrhea, vomiting, or medication use. Her vital signs are within normal limits. Physical examination reveals diffusely decreased muscle strength in all extremities with diminished deep tendon reflexes.

Laboratory studies reveal the following:

SerumValue
Sodium138 mEq/L
Potassium2.8 mEq/L
Chloride115 mEq/L
Bicarbonate13 mEq/L
Blood urea nitrogen12 mg/dL
Creatinine0.8 mg/dL
UrineValue
Sodium70 mEq/L
Potassium30 mEq/L
Chloride55 mEq/L
pH6.5
Specific gravity1.010

Based on the laboratory data provided, what is this patient's urine anion gap (in mEq/L)?

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Answer: 45

Answer

The urine anion gap is calculated as 45 mEq/L.
The urine anion gap (UAG) is a critical calculation used in the evaluation of a normal anion gap metabolic acidosis (hyperchloremic acidosis) to distinguish between renal and extrarenal (usually gastrointestinal) etiologies. It provides an indirect estimate of urinary ammonium (NH4+NH_4^+) excretion. The formula is UAG=UNa+UK−UClUAG = U_{Na} + U_{K} - U_{Cl}. In a healthy physiologic response to acidemia, the kidneys increase acid excretion in the form of NH4+NH_4^+. Because ammonium is positively charged, it is accompanied by chloride (Cl−Cl^-) into the urine to maintain electroneutrality. This leads to a high urinary chloride concentration and a negative UAG (typically −20-20 to −50-50 mEq/L), an appropriate compensatory response classically seen in severe diarrhea. Conversely, a positive UAG (as seen in this patient with a value of 45 mEq/L) indicates an inability of the kidneys to appropriately excrete NH4+NH_4^+, which is a hallmark of Renal Tubular Acidosis (RTA). Given this patient's concurrent hypokalemia, inappropriately basic urine pH (6.5), and positive UAG, Distal (Type 1) RTA is the definitive diagnosis.

Step-by-Step Solution

1
Identify the clinical indication and the appropriate diagnostic formula.
The formula for Urine Anion Gap is UAG=UNa+UK−UClUAG = U_{Na} + U_{K} - U_{Cl}.
The UAG helps differentiate between gastrointestinal (negative UAG) and renal (positive UAG) causes of a normal anion gap metabolic acidosis.
2
Extract the necessary variables from the provided urinalysis data panel.
UNaU_{Na} = 70 mEq/L, UKU_{K} = 30 mEq/L, and UClU_{Cl} = 55 mEq/L.
Only urine electrolyte concentrations are utilized in this formula; serum electrolytes serve as distractors for this specific calculation step.
3
Perform the calculation using the extracted values.
UAG=70+30−55=45UAG = 70 + 30 - 55 = 45 mEq/L.
This yields a positive integer, indicating impaired renal ammonium excretion consistent with Distal (Type 1) Renal Tubular Acidosis.

Key Concept

Calculation and clinical utility of the Urine Anion Gap (UAG) in the differential diagnosis of normal anion gap metabolic acidosis.
Estimated Time:1m 30s
Question 108Question →

A patient in the surgical intensive care unit develops oliguria on the second day following an elective partial colectomy. Over the past 8 hours, the urinary catheter has drained only 150 mL of highly concentrated urine. The patient has been receiving maintenance intravenous fluids at a conservative rate due to a known history of heart failure with preserved ejection fraction.

Vital signs show a temperature of 37.2°C (99.0°F), heart rate of 105/min, and blood pressure of 102/65 mmHg. Physical examination reveals dry oral mucosa and flat jugular veins.

Laboratory studies are obtained:

Laboratory ParameterValue
Serum sodium140 mEq/L
Serum blood urea nitrogen48 mg/dL
Serum creatinine2.0 mg/dL
Urine sodium28 mEq/L
Urine creatinine80 mg/dL

Based on the provided laboratory values, calculate the patient's fractional excretion of sodium (FENa). (Enter your answer as a percentage.)

Show answer & explanation

Answer: 0.5

Answer

0.5
The correct answer is derived mathematically using the Fractional Excretion of Sodium (FENa) formula. By multiplying the urine sodium by the serum creatinine (28 × 2.0 = 56) and dividing that product by the multiplication of the serum sodium and urine creatinine (140 × 80 = 11200), the resulting ratio is 0.005. Multiplying this decimal by 100 expresses the final FENa as 0.5%. Physiologically, a value below 1% indicates robust tubular sodium reabsorption, which strongly suggests a prerenal mechanism for the patient's oliguria, consistent with volume depletion.

Step-by-Step Solution

1
Identify the formula used to calculate the fractional excretion of sodium.
FENa=UNa×PCrPNa×UCr×100FENa = \frac{U_{Na} \times P_{Cr}}{P_{Na} \times U_{Cr}} \times 100
This formula relates the clearance of sodium to the clearance of creatinine to determine if the renal tubules are appropriately reabsorbing sodium in the setting of decreased renal perfusion.
2
Extract the specific variables needed from the provided laboratory table.
Urine sodium (UNaU_{Na}) = 28 mEq/L; Plasma (Serum) creatinine (PCrP_{Cr}) = 2.0 mg/dL; Plasma (Serum) sodium (PNaP_{Na}) = 140 mEq/L; Urine creatinine (UCrU_{Cr}) = 80 mg/dL.
Accurate variable identification prevents common inversion errors during calculation.
3
Substitute the values into the formula and calculate the percentage.
FENa=28×2.0140×80×100=5611200×100=0.005×100=0.5FENa = \frac{28 \times 2.0}{140 \times 80} \times 100 = \frac{56}{11200} \times 100 = 0.005 \times 100 = 0.5%
Completing the calculation yields the final value, which is less than 1%, clinically supporting a prerenal etiology for the patient's acute kidney injury.

Key Concept

Calculation of the Fractional Excretion of Sodium (FENa) to evaluate and differentiate etiologies of acute kidney injury.
Question 109Question →

A 48-year-old man with a history of severe alcohol use disorder is brought to the emergency department by paramedics after being found obtunded in a park. His temperature is 36.2∘C36.2^\circ\text{C} (97.2∘F97.2^\circ\text{F}), blood pressure is 110/70 mmHg110/70\text{ mmHg}, pulse is 105/min105/\text{min}, and respiratory rate is 18/min18/\text{min}. The patient is unarousable and responds only to painful stimuli.

Initial laboratory studies reveal the following:

Laboratory ParameterValue
Sodium138 mEq/L138\text{ mEq/L}
Potassium4.2 mEq/L4.2\text{ mEq/L}
Chloride104 mEq/L104\text{ mEq/L}
Bicarbonate22 mEq/L22\text{ mEq/L}
Blood urea nitrogen (BUN)28 mg/dL28\text{ mg/dL}
Creatinine1.1 mg/dL1.1\text{ mg/dL}
Glucose126 mg/dL126\text{ mg/dL}
Ethanol92 mg/dL92\text{ mg/dL}
Measured serum osmolality340 mOsm/kg H2O340\text{ mOsm/kg H}_2\text{O}

Based on the provided laboratory results, what is this patient's serum osmolar gap in mOsm/kg H2O\text{mOsm/kg H}_2\text{O}?

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Answer: 27

Answer

27
The serum osmolar gap is defined as the difference between the measured serum osmolality and the calculated serum osmolality. The standard formula for calculated serum osmolality in the United States (using mg/dL\text{mg/dL} for glucose and BUN) is 2×[Na+]+[Glucose]18+[BUN]2.82 \times [\text{Na}^+] + \frac{[\text{Glucose}]}{18} + \frac{[\text{BUN}]}{2.8}. However, when ethanol is present and measured, its contribution must also be subtracted to isolate the gap caused by other unidentified toxic alcohols (e.g., methanol, ethylene glycol). The conversion factor for ethanol is 4.64.6. Therefore, the complete calculation for this patient is 2×138+12618+282.8+924.6=276+7+10+20=313 mOsm/kg H2O2 \times 138 + \frac{126}{18} + \frac{28}{2.8} + \frac{92}{4.6} = 276 + 7 + 10 + 20 = 313 \text{ mOsm/kg H}_2\text{O}. Subtracting this from the measured osmolality of 340 mOsm/kg H2O340 \text{ mOsm/kg H}_2\text{O} gives a gap of 27 mOsm/kg H2O27 \text{ mOsm/kg H}_2\text{O}.

Step-by-Step Solution

1
Identify the need to calculate the serum osmolar gap.
The patient has altered mental status and a measured serum osmolality of 340 mOsm/kg H2O340 \text{ mOsm/kg H}_2\text{O}, indicating a potential toxic alcohol ingestion.
Calculating the osmolar gap is essential to unmask the presence of unmeasured osmotically active substances (like methanol, ethylene glycol, or isopropyl alcohol) not accounted for by standard chemistry panels.
2
Calculate the expected (calculated) serum osmolality.
Using the expanded formula: 2×138+12618+282.8+924.6=276+7+10+20=313 mOsm/kg H2O2 \times 138 + \frac{126}{18} + \frac{28}{2.8} + \frac{92}{4.6} = 276 + 7 + 10 + 20 = 313 \text{ mOsm/kg H}_2\text{O}.
Because ethanol is present in the patient's blood, its osmotic contribution must be included in the calculation. Failure to include it will result in an artificially elevated osmolar gap.
3
Determine the osmolar gap.
340 (Measured)−313 (Calculated)=27 mOsm/kg H2O340 \text{ (Measured)} - 313 \text{ (Calculated)} = 27 \text{ mOsm/kg H}_2\text{O}.
The osmolar gap is the difference between the measured and calculated osmolality. A normal gap is typically less than 10-15 mOsm/kg H2O10\text{-}15 \text{ mOsm/kg H}_2\text{O}. A gap of 2727 is elevated, suggesting ingestion of an additional toxic alcohol.

Key Concept

Calculation of the serum osmolar gap, including the proper osmotic contribution of ethanol.
Question 110Question →

A 34-year-old female is brought to the emergency department via helicopter after being rescued from a residential house fire. The burn injury occurred at 14:00, and she arrives at the trauma bay at 17:00. Her weight is 90 kg90 \text{ kg}. Physical examination reveals deep partial-thickness and full-thickness burns involving her entire anterior trunk (chest and abdomen), her entire right arm (circumferential), and the anterior surface of her left leg. The remainder of her skin is uninjured. En route to the hospital, paramedics successfully intubated her for airway protection and administered 740 mL740 \text{ mL} of lactated Ringer solution intravenously. According to current Advanced Trauma Life Support (ATLS) guidelines (2 mL/kg/%TBSA2 \text{ mL/kg/\%TBSA} for adult thermal burns), what should be the intravenous fluid infusion rate in mL/hr\text{mL/hr} for the remainder of the first 8-hour post-injury period?

Show answer & explanation

Answer: 500

Answer

500
The correct infusion rate is 500 mL/hr500 \text{ mL/hr}. First, estimate the total body surface area (TBSA) burned using the Rule of Nines: the entire anterior trunk accounts for 18%18\%, the entire right arm is 9%9\%, and the anterior surface of the left leg is 9%9\%. Total TBSA = 18%+9%+9%=36%18\% + 9\% + 9\% = 36\%. Next, apply the ATLS fluid resuscitation guideline for adult thermal burns (2 mL/kg/%TBSA2 \text{ mL/kg/\%TBSA}) to determine the total 24-hour requirement: 2 mL×90 kg×36=6480 mL2 \text{ mL} \times 90 \text{ kg} \times 36 = 6480 \text{ mL}. Half of this total volume, which is 3240 mL3240 \text{ mL}, is required during the first 8 hours post-injury. Because the patient already received 740 mL740 \text{ mL} in transit, the remaining volume needed to complete the first 8-hour requirement is 3240 mL−740 mL=2500 mL3240 \text{ mL} - 740 \text{ mL} = 2500 \text{ mL}. Finally, because the patient arrived at the hospital 3 hours after the burn occurred (14:00 to 17:00), there are exactly 5 hours remaining in the initial 8-hour window. The appropriate infusion rate is calculated as 2500 mL/5 hours2500 \text{ mL} / 5 \text{ hours}, which equals 500 mL/hr500 \text{ mL/hr}.

Step-by-Step Solution

1
Calculate the Total Body Surface Area (TBSA) burned using the Rule of Nines.
Anterior trunk (18%18\%) + entire right arm (9%9\%) + anterior left leg (9%9\%) = 36%36\% TBSA.
Accurate TBSA is required as a primary variable for burn fluid resuscitation formulas.
2
Calculate the total 24-hour fluid requirement using the modern ATLS guideline for thermal burns.
2 mL×90 kg×36%=6480 mL2 \text{ mL} \times 90 \text{ kg} \times 36\% = 6480 \text{ mL}.
This determines the baseline total volume of lactated Ringer solution needed for the first day.
3
Determine the target fluid volume for the first 8 hours post-injury.
6480 mL/2=3240 mL6480 \text{ mL} / 2 = 3240 \text{ mL}.
Standard burn resuscitation protocols dictate that half of the total 24-hour volume must be administered in the first 8 hours from the time of the burn.
4
Adjust for fluids already administered in the prehospital setting.
3240 mL−740 mL=2500 mL3240 \text{ mL} - 740 \text{ mL} = 2500 \text{ mL}.
Accounts for the volume given during transport to prevent dangerous over-resuscitation and compartment syndrome.
5
Calculate the infusion rate for the remaining time in the 8-hour window.
2500 mL/5 hours=500 mL/hr2500 \text{ mL} / 5 \text{ hours} = 500 \text{ mL/hr}.
Because 3 hours elapsed between the injury (14:00) and arrival (17:00), only 5 hours remain in the initial 8-hour post-injury window.

Key Concept

Burn fluid resuscitation calculations require TBSA estimation using the Rule of Nines, application of the ATLS volume guidelines, adjustments for time elapsed since the burn, and deduction of prehospital fluids administered.
Question 111Question →

Emergency medical services bring a 47-year-old mechanic to the hospital after he was found deeply lethargic in his workshop. His coworkers report that he had been working alone on a vehicle's cooling system and seemed disoriented before collapsing. On presentation, his temperature is 36.8∘C36.8^\circ C (98.2∘F98.2^\circ F), blood pressure is 105/65105/65 mmHg, pulse is 112112/min, and respiratory rate is 2424/min. Physical examination reveals an unarousable patient with deep, rapid breathing. Arterial blood gas analysis on room air shows a pH of 7.21, PaCO2PaCO_2 of 24 mmHg, and PaO2PaO_2 of 95 mmHg.

Laboratory evaluation reveals the following:
ParameterValue
Sodium142142 mEq/L
Potassium4.84.8 mEq/L
Chloride104104 mEq/L
Bicarbonate1010 mEq/L
Blood urea nitrogen1414 mg/dL
Creatinine1.61.6 mg/dL
Glucose9090 mg/dL
Measured serum osmolality334334 mOsm/kg H2OH_2O

Based on the provided laboratory results, what is the patient's calculated serum osmolar gap in mOsm/kg H2OH_2O?

Show answer & explanation

Answer: 40

Answer

40
The correct answer is 40. The serum osmolar gap is defined as the difference between the directly measured serum osmolality and the calculated serum osmolality. The formula for calculated serum osmolality in conventional units (mg/dL) is: Calculated Osmolality=2×[Na+]+[Glucose]18+[BUN]2.8Calculated\ Osmolality = 2 \times [Na^+] + \frac{[Glucose]}{18} + \frac{[BUN]}{2.8}. Using the patient's values, this is computed as 2(142)+9018+142.8=284+5+5=2942(142) + \frac{90}{18} + \frac{14}{2.8} = 284 + 5 + 5 = 294 mOsm/kg H2OH_2O. The measured osmolality is 334 mOsm/kg H2OH_2O. Subtracting the calculated value from the measured value yields an osmolar gap of 334−294=40334 - 294 = 40 mOsm/kg H2OH_2O. A normal osmolar gap is typically less than 10 to 15 mOsm/kg H2OH_2O. In a patient with central nervous system depression, high anion gap metabolic acidosis, and exposure to automotive fluids, a markedly elevated osmolar gap strongly suggests intoxication with an unmeasured osmotically active alcohol, such as ethylene glycol (antifreeze).

Step-by-Step Solution

1
Calculate the expected serum osmolality using the standard formula for conventional units.
294294 mOsm/kg H2OH_2O
This establishes the baseline osmolality contributed by the primary endogenous osmoles (sodium, glucose, and urea).
2
Subtract the calculated osmolality from the measured serum osmolality.
334−294=40334 - 294 = 40 mOsm/kg H2OH_2O
An elevated difference (gap >10 mOsm/kg) indicates the presence of unmeasured exogenous osmotically active substances, such as ethylene glycol.

Key Concept

Calculation and interpretation of the serum osmolar gap in toxic alcohol ingestion.
Question 112Question →

A 68-year-old man is evaluated in the hospital for decreased urine output. He was admitted 3 days ago for severe community-acquired pneumonia and has been receiving intravenous antibiotics. His blood pressure is 98/60 mm Hg and heart rate is 102/min. Physical examination shows dry mucous membranes and flat jugular veins. Laboratory studies are obtained to evaluate his acute kidney injury:

Laboratory TestResult
Serum sodium135 mEq/L
Serum creatinine1.5 mg/dL
Blood urea nitrogen38 mg/dL
Urine sodium27 mEq/L
Urine creatinine60 mg/dL

What is the patient's fractional excretion of sodium (FeNa) expressed as a percentage?

Show answer & explanation

Answer: 0.5

Answer

0.5
The correct calculation yields a fractional excretion of sodium (FeNa) of exactly 0.5%. A FeNa of < 1% indicates a prerenal etiology for acute kidney injury. In this patient, signs of volume depletion (hypotension, tachycardia, dry mucous membranes) point toward prerenal azotemia secondary to dehydration, leading the kidneys to avidly reabsorb sodium and water in an attempt to restore intravascular volume.

Step-by-Step Solution

1
Identify the formula for Fractional Excretion of Sodium (FeNa).
FeNa=UNa×PCrPNa×UCr×100FeNa = \frac{U_{Na} \times P_{Cr}}{P_{Na} \times U_{Cr}} \times 100
This formula relates the amount of sodium excreted to the amount of creatinine excreted, thereby correcting for urine concentration.
2
Extract the necessary values from the provided laboratory table.
UNa=27U_{Na} = 27 mEq/L, PCr=1.5P_{Cr} = 1.5 mg/dL, PNa=135P_{Na} = 135 mEq/L, and UCr=60U_{Cr} = 60 mg/dL.
These specific variables are required for the FeNa equation. The Blood Urea Nitrogen (BUN) value of 38 mg/dL is a distractor for this specific calculation.
3
Substitute the values into the FeNa equation.
FeNa=27×1.5135×60×100FeNa = \frac{27 \times 1.5}{135 \times 60} \times 100
Plugging the values into the correct positions in the numerator and denominator ensures mathematical accuracy.
4
Perform the arithmetic calculation.
40.58100×100=0.005×100=0.5\frac{40.5}{8100} \times 100 = 0.005 \times 100 = 0.5
This yields the final FeNa value as a percentage, which can then be interpreted clinically.

Key Concept

Calculation and interpretation of Fractional Excretion of Sodium (FeNa) in the evaluation of Acute Kidney Injury.
Question 113Question →

A 64-year-old woman is brought to the emergency department by her family due to profound weakness and confusion. She has had 5 days of severe, voluminous watery diarrhea. Her medical history includes type 2 diabetes mellitus and hypertension. On examination, blood pressure is 82/50 mm Hg, heart rate is 118/min, and respiratory rate is 24/min. She has dry mucous membranes, flat neck veins, and delayed capillary refill.

Laboratory studies show:
Serum Na+Na^+: 140 mEq/L
Serum K+K^+: 3.2 mEq/L
Serum Cl−Cl^-: 112 mEq/L
Serum HCO3−HCO_3^-: 8 mEq/L
Blood urea nitrogen: 54 mg/dL
Serum creatinine: 2.8 mg/dL (baseline 0.9 mg/dL)
Serum lactate: 5.2 mmol/L (Normal: 0.5-1.0 mmol/L)
Serum albumin: 4.0 g/dL

Assuming a reference anion gap of 12 mEq/L and a normal serum bicarbonate of 24 mEq/L, what is this patient's calculated delta ratio (also known as the Δ/Δ\Delta/\Delta ratio)?

Show answer & explanation

Answer: 0.5

Answer

0.5
The correct answer is 0.5. This patient presents with hypovolemic shock (hypotension, tachycardia, poor perfusion) secondary to severe watery diarrhea, leading to an elevated lactate level. Therefore, she has a mixed acid-base disorder: a high anion gap metabolic acidosis (HAGMA) from lactic acidosis, and a normal anion gap metabolic acidosis (NAGMA) from gastrointestinal bicarbonate losses. The delta ratio is used to uncover this mixed disorder. It is calculated by dividing the excess anion gap (ΔAG=20−12=8\Delta AG = 20 - 12 = 8) by the bicarbonate deficit (ΔHCO3−=24−8=16\Delta HCO_3^- = 24 - 8 = 16). The resulting ratio of 0.5 falls between 0.4 and 0.8, which is the classic diagnostic range for a mixed HAGMA and NAGMA.

Step-by-Step Solution

1
Calculate the serum anion gap.
2020 mEq/L
Determines the presence and magnitude of unmeasured anions contributing to the acidosis.
2
Calculate the ΔAG\Delta AG.
88 mEq/L
Quantifies the excess gap above the normal baseline (20−1220 - 12).
3
Calculate the ΔHCO3−\Delta HCO_3^-.
1616 mEq/L
Quantifies the total deficit of bicarbonate below the normal baseline (24−824 - 8).
4
Divide ΔAG\Delta AG by ΔHCO3−\Delta HCO_3^-.
0.50.5
Provides the delta ratio, which identifies complex mixed acid-base disorders.

Key Concept

In patients with a high anion gap metabolic acidosis (HAGMA), the delta ratio (ΔAG/ΔHCO3−\Delta AG / \Delta HCO_3^-) is calculated to identify concurrent concealed acid-base disorders. A ratio < 0.8 indicates a concurrent normal anion gap metabolic acidosis (NAGMA), which in this patient is due to severe diarrhea.
Question 114Question →

A 28-year-old woman with a history of Sjögren syndrome is evaluated in the clinic for progressive muscle weakness over the past week. She reports no vomiting, diarrhea, or recent illnesses. Her vital signs are within normal limits. Physical examination reveals 3/5 strength in the proximal muscles of her lower extremities. Laboratory studies show:

Serum:
Na +^+: 139 mEq/L
K +^+: 2.8 mEq/L
Cl −^-: 114 mEq/L
HCO 3−_3^-: 14 mEq/L

Arterial blood gas (room air):
pH: 7.28
pCO 2_2: 32 mm Hg

Urine:
Na +^+: 62 mEq/L
K +^+: 30 mEq/L
Cl −^-: 55 mEq/L
pH: 6.0

What is the calculated urine anion gap for this patient in mEq/L?

Show answer & explanation

Answer: 37

Answer

The calculated urine anion gap is 37 mEq/L.
The urine anion gap (UAG) is calculated using the formula: UAG=UNa+UK−UClUAG = U_{Na} + U_{K} - U_{Cl}. Using the patient's urine electrolytes, the calculation is 62+30−55=3762 + 30 - 55 = 37 mEq/L. In the context of normal anion gap metabolic acidosis, a positive UAG indicates impaired renal excretion of ammonium (NH 4+_4^+). This points to a renal etiology, specifically renal tubular acidosis (RTA). This patient's presentation with profound hypokalemia, progressive muscle weakness, and Sjögren syndrome is classic for Type 1 (distal) RTA.

Step-by-Step Solution

1
Identify the patient's primary acid-base disorder.
The patient has a normal anion gap metabolic acidosis (NAGMA).
Serum pH is 7.28 and HCO 3−_3^- is 14 mEq/L, indicating metabolic acidosis. The serum anion gap is calculated as 139 - (114 + 14) = 11 mEq/L, which is within the normal range.
2
Determine the appropriate diagnostic calculation to find the etiology.
Calculate the urine anion gap (UAG).
The UAG helps differentiate between renal (e.g., renal tubular acidosis) and gastrointestinal (e.g., diarrhea) causes of NAGMA by serving as a surrogate marker for unmeasured urinary ammonium (NH 4+_4^+) excretion.
3
Apply the UAG formula using the given laboratory values.
UAG=UNa+UK−UClUAG = U_{Na} + U_{K} - U_{Cl}
Unlike the serum anion gap, potassium is an essential component of the urine anion gap equation due to its significant variation and quantity in the urine.
4
Calculate the final numerical value.
UAG=62+30−55=37UAG = 62 + 30 - 55 = 37 mEq/L.
A positive UAG indicates impaired renal ammonium excretion, which is consistent with Type 1 (distal) renal tubular acidosis, a condition commonly associated with Sjögren syndrome.

Key Concept

Calculation and clinical utility of the urine anion gap in evaluating normal anion gap metabolic acidosis.
Question 115Question →

A 68-year-old man is admitted to the hospital with acute decompensated heart failure. He is started on an intravenous bumetanide infusion. On hospital day 3, his urine output decreases to 300 mL over the last 24 hours. Physical examination shows jugular venous distention, bibasilar crackles, and bilateral lower extremity pitting edema.

Laboratory studies are obtained:
Serum sodium: 140 mEq/L140 \text{ mEq/L}
Serum potassium: 3.5 mEq/L3.5 \text{ mEq/L}
Serum chloride: 98 mEq/L98 \text{ mEq/L}
Serum bicarbonate: 26 mEq/L26 \text{ mEq/L}
Serum blood urea nitrogen (BUN): 60 mg/dL60 \text{ mg/dL}
Serum creatinine: 2.0 mg/dL2.0 \text{ mg/dL}

A spot urine sample is analyzed:
Urine sodium: 105 mEq/L105 \text{ mEq/L}
Urine blood urea nitrogen (BUN): 600 mg/dL600 \text{ mg/dL}
Urine creatinine: 100 mg/dL100 \text{ mg/dL}

Based on these findings, what is the patient's fractional excretion of urea nitrogen (expressed as a percentage)?

Show answer & explanation

Answer: 20

Answer

20
The correct calculation for the fractional excretion of urea nitrogen is 20%20\%. To determine the etiology of acute kidney injury (AKI) in a patient receiving diuretic therapy, the fractional excretion of urea (FeUreaFeUrea) should be used rather than the fractional excretion of sodium (FeNaFeNa). Diuretics impair sodium reabsorption in the tubules, which can falsely elevate the urine sodium concentration and the FeNaFeNa, masking a prerenal state. Urea reabsorption, primarily occurring in the proximal tubule, is less affected by diuretic use. The formula is FeUrea=UrineBUN×SerumCrSerumBUN×UrineCr×100FeUrea = \frac{Urine_{BUN} \times Serum_{Cr}}{Serum_{BUN} \times Urine_{Cr}} \times 100. Plugging in the provided values gives 600×2.060×100×100=20%\frac{600 \times 2.0}{60 \times 100} \times 100 = 20\%. A FeUreaFeUrea of less than 35%35\% is indicative of prerenal AKI, which is consistent with this patient's presentation of low forward cardiac output in the setting of acute decompensated heart failure.

Step-by-Step Solution

1
Identify the indication for calculating FeUreaFeUrea instead of FeNaFeNa.
The patient is on bumetanide (a loop diuretic).
Diuretics increase urinary sodium excretion, making FeNaFeNa unreliable (falsely elevated) for diagnosing prerenal AKI. Urea reabsorption is less affected by diuretics.
2
Select the correct formula for FeUreaFeUrea.
FeUrea=UrineBUN×SerumCrSerumBUN×UrineCr×100FeUrea = \frac{Urine_{BUN} \times Serum_{Cr}}{Serum_{BUN} \times Urine_{Cr}} \times 100
This formula compares the clearance of urea to the clearance of creatinine.
3
Plug the laboratory values into the formula.
FeUrea=600×2.060×100×100FeUrea = \frac{600 \times 2.0}{60 \times 100} \times 100
Extract values: UrineBUN=600Urine_{BUN} = 600, SerumCr=2.0Serum_{Cr} = 2.0, SerumBUN=60Serum_{BUN} = 60, UrineCr=100Urine_{Cr} = 100.
4
Perform the calculation.
12006000×100=0.2×100=20\frac{1200}{6000} \times 100 = 0.2 \times 100 = 20
Basic arithmetic yields the final percentage.

Key Concept

In patients receiving diuretics who develop acute kidney injury, the fractional excretion of urea (FeUreaFeUrea) is a more accurate marker for prerenal etiology than the fractional excretion of sodium (FeNaFeNa). A FeUrea<35%FeUrea < 35\% indicates prerenal AKI.
Question 116Question →

Paramedics transport a 45-year-old male to the hospital after his wife discovered him confused and stumbling inside their garage. The patient reports severe bilateral flank pain and appears visibly intoxicated, though his wife states he has not consumed alcohol. Vital signs include a temperature of 37.1°C (98.8°F), blood pressure of 145/90 mm Hg, heart rate of 98/min, and respiratory rate of 22/min.

Laboratory evaluation reveals:
ParameterResult
Serum sodium142 mEq/L
Serum potassium4.8 mEq/L
Serum chloride102 mEq/L
Serum bicarbonate12 mEq/L
Blood urea nitrogen28 mg/dL
Serum creatinine1.5 mg/dL
Serum glucose108 mg/dL
Serum ethanolUndetectable
Measured serum osmolality335 mOsm/kg H₂O

Based on the provided laboratory data, determine this patient's serum osmolar gap in mOsm/kg H₂O.

Show answer & explanation

Answer: 35

Answer

35
The patient's osmolar gap is 35 mOsm/kg H₂O. When combined with a significant metabolic acidosis (anion gap of 28 mEq/L) and flank pain, this elevated osmolar gap is highly characteristic of ethylene glycol toxicity, likely causing calcium oxalate crystal deposition in the kidneys.

Step-by-Step Solution

1
Recall the formula for calculating serum osmolality in conventional units.
Calculated osmolality = 2×[Na+]+BUN2.8+Glucose182 \times [Na^+] + \frac{BUN}{2.8} + \frac{Glucose}{18}
This formula estimates the baseline contribution of the major extracellular osmoles to the total serum osmolality.
2
Substitute the provided laboratory values into the formula.
2×142+282.8+10818=284+10+6=3002 \times 142 + \frac{28}{2.8} + \frac{108}{18} = 284 + 10 + 6 = 300 mOsm/kg H₂O
This establishes the patient's expected serum osmolality before considering the presence of unmeasured exogenous toxic osmoles.
3
Calculate the serum osmolar gap by finding the difference between measured and calculated osmolality.
335−300=35335 - 300 = 35 mOsm/kg H₂O
An elevated osmolar gap (>10 mOsm/kg H₂O) in the context of an unexplained high anion gap metabolic acidosis strongly suggests toxic alcohol ingestion.

Key Concept

Calculation and interpretation of the serum osmolar gap to evaluate for toxic alcohol ingestion.
Estimated Time:1m 0s
Question 117Question →

A 52-year-old woman is evaluated in the emergency department for severe, watery diarrhea over the past five days. She reports feeling profoundly weak. She has no significant past medical history. Her blood pressure is 105/65 mm Hg, pulse is 112/min, and respiratory rate is 22/min. Physical examination shows dry mucous membranes and delayed capillary refill.

Laboratory results:
- Serum sodium: 138 mEq/L
- Serum potassium: 3.2 mEq/L
- Serum chloride: 114 mEq/L
- Serum bicarbonate: 12 mEq/L
- Arterial blood gas (room air): pH 7.24, PaCO2 24 mm Hg, PaO2 96 mm Hg

Using Winter's formula, calculate the patient's expected PaCO2. (Provide the exact median numerical value in mm Hg, excluding the ±2 range)

Show answer & explanation

Answer: 26

Answer

26
In the setting of a primary metabolic acidosis (as seen in this patient with severe diarrhea and a low serum bicarbonate), the lungs compensate by increasing ventilation to blow off CO2CO_2. Winter's formula, ExpectedPaCO2=(1.5×HCO3−)+8±2Expected PaCO2 = (1.5 \times HCO_3^-) + 8 \pm 2, is used to determine if the respiratory compensation is appropriate. By plugging in the patient's bicarbonate of 12 mEq/L, the calculation is (1.5×12)+8=26(1.5 \times 12) + 8 = 26. The median expected PaCO2 is 26 mm Hg. Since the patient's measured PaCO2 is 24 mm Hg (which falls within the 26±226 \pm 2 range of 24-28 mm Hg), she has an appropriately compensated simple metabolic acidosis.

Step-by-Step Solution

1
Recall Winter's formula for respiratory compensation.
ExpectedPaCO2=(1.5×HCO3−)+8Expected PaCO2 = (1.5 \times HCO_3^-) + 8
This formula predicts the appropriate compensatory decrease in PaCO2 for a given drop in bicarbonate during simple metabolic acidosis.
2
Identify the serum bicarbonate value from the laboratory panel.
HCO3−HCO_3^- = 12 mEq/L
The bicarbonate concentration is the primary variable driving the expected respiratory compensation.
3
Substitute the bicarbonate value into Winter's formula.
ExpectedPaCO2=(1.5×12)+8Expected PaCO2 = (1.5 \times 12) + 8
Setting up the equation allows for the direct calculation of the target PaCO2.
4
Complete the arithmetic calculation.
18+8=2618 + 8 = 26
Yields the exact median expected PaCO2 in mm Hg, before applying the ±2\pm 2 range.

Key Concept

Calculation of expected respiratory compensation (Winter's formula) in metabolic acidosis
Question 118Question →

A 52-year-old woman with a history of type 1 diabetes mellitus is brought to the emergency department after 33 days of intractable vomiting and generalized abdominal pain. She ran out of her prescribed insulin 44 days ago. On physical examination, she appears lethargic with dry mucous membranes and decreased skin turgor. Temperature is 37.2∘37.2^\circ C (99.0∘99.0^\circ F), blood pressure is 95/6095/60 mm Hg, pulse is 118118/min, and respirations are 2626/min and deep.

Laboratory studies reveal:

TestResultReference Range
Serum sodium140140 mEq/L136−145136-145 mEq/L
Serum potassium4.84.8 mEq/L3.5−5.03.5-5.0 mEq/L
Serum chloride9292 mEq/L98−10698-106 mEq/L
Serum bicarbonate1414 mEq/L22−2822-28 mEq/L
Blood urea nitrogen4242 mg/dL7−187-18 mg/dL
Serum creatinine1.61.6 mg/dL0.6−1.20.6-1.2 mg/dL
Blood glucose480480 mg/dL70−11070-110 mg/dL

Assume a normal anion gap of 1212 mEq/L and a normal serum bicarbonate of 2424 mEq/L.

What is the patient's calculated delta ratio (also known as the Δ\Delta AG / Δ\Delta HCO 3−_3^- ratio)?

Show answer & explanation

Answer: 2.2

Answer

The calculated delta ratio is 2.2.
The patient's clinical presentation of diabetic ketoacidosis combined with intractable vomiting strongly suggests a mixed acid-base disorder. Diabetic ketoacidosis causes a high anion gap metabolic acidosis (HAGMA), while vomiting induces a metabolic alkalosis due to the loss of gastric hydrochloric acid. To confirm this mathematically, the delta ratio is calculated. The patient's measured anion gap is 140−(92+14)=34140 - (92 + 14) = 34 mEq/L. The change in anion gap (ΔAG\Delta\text{AG}) is 34−12=2234 - 12 = 22 mEq/L. The change in bicarbonate (ΔHCO3−\Delta\text{HCO}_3^-) is 24−14=1024 - 14 = 10 mEq/L. Dividing the ΔAG\Delta\text{AG} by the ΔHCO3−\Delta\text{HCO}_3^- yields a ratio of 2.22.2. In a pure HAGMA, the drop in bicarbonate should be roughly equivalent to the rise in the anion gap (ratio between 1.01.0 and 2.02.0). A delta ratio >2.0> 2.0 indicates that the bicarbonate level has not fallen as much as expected, which confirms the presence of a concurrent metabolic alkalosis.

Step-by-Step Solution

1
Calculate the patient's anion gap (AG).
AG=140−(92+14)=34\text{AG} = 140 - (92 + 14) = 34 mEq/L
Establishing the initial anion gap is necessary to identify the presence of a high anion gap metabolic acidosis and to compute the delta anion gap.
2
Calculate the change in anion gap (ΔAG\Delta\text{AG}).
ΔAG=34−12=22\Delta\text{AG} = 34 - 12 = 22 mEq/L
The numerator of the delta ratio represents how much the anion gap has increased above the normal baseline.
3
Calculate the change in bicarbonate (ΔHCO3−\Delta\text{HCO}_3^-).
ΔHCO3−=24−14=10\Delta\text{HCO}_3^- = 24 - 14 = 10 mEq/L
The denominator of the delta ratio represents how much the bicarbonate has decreased below the normal baseline due to buffering the excess acid.
4
Divide ΔAG\Delta\text{AG} by ΔHCO3−\Delta\text{HCO}_3^-.
Delta Ratio=2210=2.2\text{Delta Ratio} = \frac{22}{10} = 2.2
The resulting ratio categorizes the nature of the acid-base disorder, determining if a concomitant metabolic alkalosis or normal anion gap metabolic acidosis is present.

Key Concept

In a high anion gap metabolic acidosis, calculating the delta ratio (Δ\Delta AG / Δ\Delta HCO 3−_3^-) helps uncover hidden, concurrent mixed acid-base disorders such as a superimposed metabolic alkalosis.

Alternative Method

Instead of using the delta ratio, clinicians frequently calculate the 'corrected bicarbonate' (or expected bicarbonate) by adding the ΔAG\Delta\text{AG} to the patient's measured bicarbonate: 14+22=3614 + 22 = 36 mEq/L. Because this value is significantly higher than the normal 2424 mEq/L, it leads to the exact same clinical conclusion of a superimposed metabolic alkalosis.
Estimated Time:2m 0s
Question 119Question →

A 45-year-old male is evaluated in the surgical intensive care unit on postoperative day 3 following an extensive small bowel resection for mesenteric ischemia. Over the last 12 hours, the nursing staff reports that his urine output has decreased to 15 mL/hr15 \text{ mL/hr}. He has been receiving maintenance intravenous fluids but has had large volumes of nasogastric tube output. His temperature is 37.2∘C37.2^\circ\text{C} (99.0∘F99.0^\circ\text{F}), blood pressure is 90/55 mmHg90/55 \text{ mmHg}, and pulse is 118/min118/\text{min}. Physical examination reveals flat jugular veins, delayed capillary refill, and dry mucous membranes.

Laboratory studies are obtained:

ParameterResult
Serum sodium140 mEq/L140 \text{ mEq/L}
Serum blood urea nitrogen (BUN)42 mg/dL42 \text{ mg/dL}
Serum creatinine1.4 mg/dL1.4 \text{ mg/dL}
Urine sodium30 mEq/L30 \text{ mEq/L}
Urine creatinine60 mg/dL60 \text{ mg/dL}
Urine osmolality550 mOsm/kg550 \text{ mOsm/kg}

Based on the provided laboratory values, what is this patient's fractional excretion of sodium (FENa)? (Provide your answer as a numerical percentage, e.g., if the calculated value is 1.2%, enter 1.2)

Show answer & explanation

Answer: 0.5

Answer

0.5
The fractional excretion of sodium (FENa) measures the percentage of filtered sodium that is eventually excreted in the urine. It is calculated using the formula: FENa=UNa×PCrPNa×UCr×100\text{FENa} = \frac{U_{Na} \times P_{Cr}}{P_{Na} \times U_{Cr}} \times 100. Plugging in the patient's specific lab values yields 30×1.4140×60×100=428400×100=0.5%\frac{30 \times 1.4}{140 \times 60} \times 100 = \frac{42}{8400} \times 100 = 0.5\%. A FENa of less than 1%1\% indicates appropriate and robust renal conservation of sodium, strongly suggesting a prerenal etiology for the acute kidney injury (AKI). This aligns with the patient's clinical presentation of significant hypovolemia due to poor intake and high nasogastric output.

Step-by-Step Solution

1
Recall the formula for Fractional Excretion of Sodium (FENa).
FENa=UNa×PCrPNa×UCr×100\text{FENa} = \frac{U_{Na} \times P_{Cr}}{P_{Na} \times U_{Cr}} \times 100
FENa evaluates acute kidney injury by comparing the fractional clearance of sodium to the fractional clearance of creatinine, which differentiates prerenal from intrinsic renal causes.
2
Extract the necessary variables from the patient's laboratory results.
Urine Sodium (UNaU_{Na}) = 30 mEq/L30 \text{ mEq/L}, Serum Creatinine (PCrP_{Cr}) = 1.4 mg/dL1.4 \text{ mg/dL}, Serum Sodium (PNaP_{Na}) = 140 mEq/L140 \text{ mEq/L}, Urine Creatinine (UCrU_{Cr}) = 60 mg/dL60 \text{ mg/dL}.
Accurate identification and isolation of the four required variables avoids calculation errors.
3
Input the variables into the FENa formula and perform the calculation.
FENa=30×1.4140×60×100=428400×100=0.005×100=0.5%\text{FENa} = \frac{30 \times 1.4}{140 \times 60} \times 100 = \frac{42}{8400} \times 100 = 0.005 \times 100 = 0.5\%
Solving the equation yields the physiological parameter required to determine the specific etiology of the patient's oliguria.

Key Concept

The fractional excretion of sodium (FENa) is an essential diagnostic calculation used to distinguish between prerenal acute kidney injury (FENa < 1%) and intrinsic acute tubular necrosis (FENa > 2%).
Estimated Time:1m 30s
Question 120Question →

A 35-year-old electrical lineman is brought to the emergency department after sustaining a high-voltage electrical injury from a power transformer. The accident occurred exactly 2 hours prior to his arrival due to extrication difficulties. He is awake but confused. His blood pressure is 140/85 mm Hg and his heart rate is 112/min. A Foley catheter is placed and immediately drains dark, tea-colored urine. He weighs 90 kg. Physical examination reveals deep entry and exit wounds on his upper extremities, as well as associated full-thickness thermal burns covering exactly 20% of his total body surface area (TBSA).

According to the current Advanced Trauma Life Support (ATLS) guidelines for electrical burns, what should be the initial intravenous fluid infusion rate (in mL/hr) of lactated Ringer solution upon his arrival?

Show answer & explanation

Answer: 600

Answer

600
The correct initial fluid rate is 600600 mL/hr. Patients with high-voltage electrical injuries are at significant risk for rhabdomyolysis and subsequent myoglobinuric acute renal failure (indicated by the tea-colored urine). Therefore, the current Advanced Trauma Life Support (ATLS) guidelines mandate a higher fluid resuscitation volume of 44 mL/kg/%TBSA for electrical burns, compared to 22 mL/kg/%TBSA for standard thermal burns. For this patient, the 24-hour fluid requirement is 4×90 kg×20%=72004 \times 90 \text{ kg} \times 20\% = 7200 mL. Half of this total volume (36003600 mL) must be infused within the first 88 hours from the time of injury. Because the patient presented 22 hours post-injury, the remaining time to infuse this volume is 66 hours. The required infusion rate is 36006=600\frac{3600}{6} = 600 mL/hr.

Step-by-Step Solution

1
Identify the appropriate formula for electrical burns.
44 mL/kg/%TBSA
Current ATLS guidelines recommend 44 mL/kg/%TBSA for electrical burns (unlike thermal burns, which use 22 mL) to increase urine output and prevent pigment nephropathy from myoglobinuria.
2
Calculate the total 24-hour fluid requirement.
72007200 mL
Using the formula: 4×90 kg×20%=72004 \times 90 \text{ kg} \times 20\% = 7200 mL.
3
Determine the fluid volume required for the first 8 hours.
36003600 mL
Half of the calculated 24-hour total volume is administered in the first 8 hours post-injury. 72002=3600\frac{7200}{2} = 3600 mL.
4
Adjust the timeline for the delayed presentation.
66 hours remaining
The fluid resuscitation timeline starts from the exact time of the injury. Since the patient arrived 22 hours after the incident, there are 8−2=68 - 2 = 6 hours remaining to infuse the initial half of the fluids.
5
Calculate the initial hourly infusion rate.
600600 mL/hr
Divide the required volume by the remaining hours in the initial 8-hour window. 36006=600\frac{3600}{6} = 600 mL/hr.

Key Concept

Fluid resuscitation in electrical burns requires 44 mL/kg/%TBSA, and the initial infusion rate must be adjusted for any time elapsed since the injury.
Estimated Time:1m 30s
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