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Question 1Question →

A 62-year-old man is admitted to the intensive care unit with septic shock secondary to severe pneumonia. He remains hypotensive despite adequate intravenous fluid resuscitation and requires vasopressor support. Laboratory studies are obtained to evaluate his acid-base status.

Laboratory ParameterValue
Sodium (Na+Na^+)138 mEq/L138 \text{ mEq/L}
Potassium (K+K^+)4.2 mEq/L4.2 \text{ mEq/L}
Chloride (Cl−Cl^-)98 mEq/L98 \text{ mEq/L}
Bicarbonate (HCO3−HCO_3^-)12 mEq/L12 \text{ mEq/L}

Based on the provided laboratory results, what is the patient's calculated serum anion gap?

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Answer: 28

Answer

28
The serum anion gap is calculated using the formula AG=Na+−(Cl−+HCO3−)AG = Na^+ - (Cl^- + HCO_3^-). Substituting the patient's values yields 138−(98+12)=138−110=28 mEq/L138 - (98 + 12) = 138 - 110 = 28 \text{ mEq/L}. An elevated anion gap (normal is typically 8-12 mEq/L) in this patient with septic shock suggests a high anion gap metabolic acidosis, most likely due to lactic acidosis from poor tissue perfusion.

Step-by-Step Solution

1
Identify the formula for serum anion gap.
AG=Na+−(Cl−+HCO3−)AG = Na^+ - (Cl^- + HCO_3^-)
The anion gap is essential for determining the presence and etiology of a metabolic acidosis.
2
Substitute the patient's laboratory values into the formula.
AG=138−(98+12)AG = 138 - (98 + 12)
Sodium is the primary measured extracellular cation, while chloride and bicarbonate are the primary measured extracellular anions.
3
Calculate the sum of the measured anions and subtract from the measured cation.
138−110=28138 - 110 = 28
This yields the concentration of unmeasured anions in the serum.

Key Concept

Calculation of the serum anion gap
Question 2Question →

To evaluate the efficacy of a novel sodium-glucose cotransporter-2 (SGLT2) inhibitor in preventing cardiovascular mortality, a multicenter, double-blind trial enrolls patients with chronic heart failure. Participants are randomly assigned to receive either the SGLT2 inhibitor or a placebo. After a median follow-up of three years, the data are analyzed. Among the 2,0002,000 patients in the intervention group, 240240 experienced a cardiovascular mortality event. In the control group, which also consisted of 2,0002,000 patients, 320320 experienced a cardiovascular mortality event. Based on these findings, what is the number needed to treat to prevent one cardiovascular mortality event over the study period?

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Answer: 25

Answer

25 patients
The Number Needed to Treat (NNT) represents the number of patients who must receive a particular intervention in order to prevent one additional adverse outcome. It is calculated as 1ARR\frac{1}{ARR} (Absolute Risk Reduction). In this study, the control event rate (CER) is 3202,000=0.16\frac{320}{2,000} = 0.16, and the experimental event rate (EER) is 2402,000=0.12\frac{240}{2,000} = 0.12. The ARR is 0.16−0.12=0.040.16 - 0.12 = 0.04. The inverse of 0.040.04 (10.04\frac{1}{0.04}) yields an NNT of 2525. Therefore, 2525 patients need to be treated with the SGLT2 inhibitor to prevent one cardiovascular mortality event.

Step-by-Step Solution

1
Calculate the Control Event Rate (CER).
CER=3202000=0.16CER = \frac{320}{2000} = 0.16
The baseline risk of the event in the unexposed/placebo population is required to determine how much the treatment reduces that risk.
2
Calculate the Experimental Event Rate (EER).
EER=2402000=0.12EER = \frac{240}{2000} = 0.12
The risk of the event in the treated population is needed to compare against the baseline.
3
Calculate the Absolute Risk Reduction (ARR).
ARR=CER−EER=0.16−0.12=0.04ARR = CER - EER = 0.16 - 0.12 = 0.04
ARR represents the absolute difference in event rates between the two groups.
4
Calculate the Number Needed to Treat (NNT).
NNT=1ARR=10.04=25NNT = \frac{1}{ARR} = \frac{1}{0.04} = 25
NNT is the inverse of ARR, indicating how many patients need to receive the intervention to prevent one additional adverse outcome.

Key Concept

Number Needed to Treat (NNT) is calculated as the inverse of the Absolute Risk Reduction (ARR).
Question 3Question →

A randomized, double-blind, placebo-controlled trial evaluates a new calcitonin gene-related peptide (CGRP) antagonist for the prevention of chronic migraines. A total of 1,000 adult patients with a history of chronic migraines are enrolled and randomized in a 1:1 ratio to receive either the CGRP antagonist or a matched placebo. The primary study endpoint is the proportion of patients achieving at least a 50% reduction in monthly migraine days over a 12-week period.

At the conclusion of the 12-week study, the following results are reported:
- Placebo group: 150 out of 500 patients met the primary endpoint.
- Treatment group (CGRP antagonist): 275 out of 500 patients met the primary endpoint.

The findings are determined to be statistically significant (p<0.01p < 0.01). Based on these trial results, what is the number needed to treat (NNT) to achieve the primary endpoint in one additional patient? (Provide the exact whole number).

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Answer: 4

Answer

4
The Number Needed to Treat (NNT) is the number of patients who must receive a particular treatment to achieve one additional desired outcome (or prevent one adverse outcome) compared to the control group. It is calculated as the inverse of the Absolute Risk Reduction (ARR). In this study, the rate of the desired outcome in the treatment group (EER) is 275/500=0.55275 / 500 = 0.55, and the rate in the placebo group (CER) is 150/500=0.30150 / 500 = 0.30. The ARR is the absolute difference between these rates: 0.55−0.30=0.250.55 - 0.30 = 0.25. Therefore, the NNT is 1/0.25=41 / 0.25 = 4. This means 4 patients need to be treated with the CGRP antagonist to achieve a 50% reduction in monthly migraine days in one patient who would not have achieved it with the placebo.

Step-by-Step Solution

1
Calculate the Control Event Rate (CER).
CER = 150 / 500 = 0.30 (or 30%)
To determine the baseline rate of the outcome in the unexposed/placebo population.
2
Calculate the Experimental Event Rate (EER).
EER = 275 / 500 = 0.55 (or 55%)
To determine the rate of the outcome in the population exposed to the new intervention.
3
Calculate the Absolute Risk Reduction (ARR).
ARR = |EER - CER| = |0.55 - 0.30| = 0.25
To quantify the absolute difference in outcome rates between the two groups.
4
Calculate the Number Needed to Treat (NNT).
NNT = 1 / ARR = 1 / 0.25 = 4
NNT is the inverse of the ARR, representing the number of patients who need to receive the intervention to achieve one additional positive outcome compared to the control.

Key Concept

The Number Needed to Treat (NNT) is calculated as 1/ARR1 / ARR (Absolute Risk Reduction). It represents the number of patients that must be treated with the experimental intervention rather than the control to prevent one additional adverse event or achieve one additional beneficial outcome.
Question 4Question →

During a routine follow-up, a 28-year-old female patient complains of new-onset fatigue and profound muscle weakness. She has a notable past medical history of recurrent nephrolithiasis. Her vital signs are within normal limits.

Laboratory testing reveals the following values:

Serum ParameterValue
Sodium (Na+Na^+)138 mEq/L
Potassium (K+K^+)2.8 mEq/L
Chloride (Cl−Cl^-)112 mEq/L
Bicarbonate (HCO3−HCO_3^-)14 mEq/L
Urine ParameterValue
Sodium (Na+Na^+)65 mEq/L
Potassium (K+K^+)25 mEq/L
Chloride (Cl−Cl^-)55 mEq/L
pH6.8

To help differentiate the cause of her normal anion gap metabolic acidosis, the attending physician requests calculation of the urine anion gap.

What is this patient's urine anion gap (in mEq/L)?

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Answer: 35

Answer

35
The patient presents with a normal anion gap metabolic acidosis (NAGMA), likely secondary to type 1 (distal) renal tubular acidosis given her severe hypokalemia, nephrolithiasis, and alkaline urine pH. To confirm the renal etiology, the urine anion gap is calculated using the formula UNa+UK−UClU_{Na} + U_{K} - U_{Cl}. Plugging in the patient's urine electrolyte values yields 65+25−55=3565 + 25 - 55 = 35 mEq/L. A positive urine anion gap indicates impaired renal ammonium excretion, which points toward a renal tubular acidosis rather than gastrointestinal bicarbonate loss (where the UAG would typically be negative).

Step-by-Step Solution

1
Identify the standard formula used for calculating the urine anion gap.
UAG=UNa+UK−UClUAG = U_{Na} + U_{K} - U_{Cl}
The urine anion gap is used to indirectly estimate urinary ammonium excretion when evaluating a normal anion gap metabolic acidosis.
2
Select the correct values from the laboratory data table.
Urine Na+=65Na^+ = 65 mEq/L, Urine K+=25K^+ = 25 mEq/L, Urine Cl−=55Cl^- = 55 mEq/L.
It is crucial to use urine electrolytes, not serum electrolytes, for this specific calculation.
3
Calculate the final result by plugging the values into the formula.
65+25−55=3565 + 25 - 55 = 35
Performing the arithmetic determines whether the gap is positive or negative, which aids in narrowing the differential diagnosis.

Key Concept

The urine anion gap (UAG) is calculated as UNa+UK−UClU_{Na} + U_{K} - U_{Cl} and provides an indirect measurement of urinary ammonium excretion to help distinguish between renal and gastrointestinal causes of a normal anion gap metabolic acidosis.
Question 5Question →

A 7878-year-old woman is evaluated in the hospital for worsening renal function. She was admitted 44 days ago for acute decompensated heart failure and has been receiving aggressive intravenous bumetanide therapy. Over the past 2424 hours, her urine output has decreased to 300 mL300\text{ mL}. On physical examination, her blood pressure is 105/65 mmHg105/65\text{ mmHg} and pulse is 92/min92/\text{min}. Jugular venous pressure is undetectable, and her lungs are clear to auscultation. Laboratory studies are obtained:

TestResult
Serum
Sodium140 mEq/L140\text{ mEq/L}
Blood urea nitrogen50 mg/dL50\text{ mg/dL}
Creatinine2.5 mg/dL2.5\text{ mg/dL} (baseline 1.1 mg/dL1.1\text{ mg/dL})
Urine
Sodium112 mEq/L112\text{ mEq/L}
Urea nitrogen500 mg/dL500\text{ mg/dL}
Creatinine100 mg/dL100\text{ mg/dL}

Based on the laboratory values, calculate the fractional excretion of urea (FEUreaFE_{Urea}) as a percentage to help determine the etiology of her acute kidney injury.

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Answer: 25

Answer

25
The fractional excretion of urea (FEUrea) is calculated using the formula: FEUrea=UUrea×PCrPUrea×UCr×100FE_{Urea} = \frac{U_{Urea} \times P_{Cr}}{P_{Urea} \times U_{Cr}} \times 100. Substituting the patient's laboratory values yields: 500×2.550×100×100=12505000×100=25%\frac{500 \times 2.5}{50 \times 100} \times 100 = \frac{1250}{5000} \times 100 = 25\%. In patients receiving loop or thiazide diuretics, FEUrea is the preferred diagnostic metric over FeNa. Diuretics directly inhibit sodium reabsorption in the loop of Henle or distal convoluted tubule, artificially elevating urinary sodium and falsely increasing FeNa (in this patient, FeNa is calculated to be 2%2\%). However, urea reabsorption occurs primarily in the proximal tubule and is largely unaffected by these diuretics. An FEUrea<35%FE_{Urea} < 35\% strongly suggests prerenal AKI, which perfectly aligns with this patient's clinical picture of over-diuresis (undetectable jugular venous pressure, cleared lungs, and relative hypotension).

Step-by-Step Solution

1
Assess the clinical scenario to choose the correct diagnostic calculation.
Determine that FEUrea should be used instead of FeNa.
The patient is actively receiving a loop diuretic (bumetanide), which artificially increases urinary sodium excretion and renders FeNa unreliable for diagnosing prerenal AKI.
2
Set up the Fractional Excretion of Urea (FEUrea) formula.
FEUrea=UUrea×PCrPUrea×UCr×100FE_{Urea} = \frac{U_{Urea} \times P_{Cr}}{P_{Urea} \times U_{Cr}} \times 100
This formula compares the clearance of urea to the clearance of creatinine to assess proximal tubular reabsorption.
3
Substitute the patient's laboratory values into the formula.
FEUrea=500×2.550×100×100FE_{Urea} = \frac{500 \times 2.5}{50 \times 100} \times 100
Accurate placement of serum and urine values is necessary to perform the calculation.
4
Perform the arithmetic calculation.
FEUrea=12505000×100=0.25×100=25%FE_{Urea} = \frac{1250}{5000} \times 100 = 0.25 \times 100 = 25\%
To obtain the final percentage required by the question format.

Key Concept

In patients receiving loop diuretics, the Fractional Excretion of Urea (FEUrea) is a more reliable indicator than FeNa for distinguishing between prerenal AKI and intrinsic AKI. An FEUrea < 35% indicates a prerenal etiology.
Estimated Time:1m 30s
Question 6Question →

A 24-year-old man is brought to the emergency department by paramedics after being found unresponsive in his garage. An open bottle of antifreeze was found nearby. His temperature is 36.5∘C36.5^\circ\text{C} (97.7∘F97.7^\circ\text{F}), blood pressure is 110/70 mm Hg110/70\text{ mm Hg}, pulse is 112/min112\text{/min}, and respirations are 28/min28\text{/min} and deep. Physical examination reveals a comatose patient with no focal neurologic deficits.

Laboratory studies show the following:
Serum sodium: 138 mEq/L138\text{ mEq/L}
Serum potassium: 5.0 mEq/L5.0\text{ mEq/L}
Serum chloride: 104 mEq/L104\text{ mEq/L}
Serum bicarbonate: 10 mEq/L10\text{ mEq/L}
Blood urea nitrogen: 14 mg/dL14\text{ mg/dL}
Serum creatinine: 1.2 mg/dL1.2\text{ mg/dL}
Serum glucose: 90 mg/dL90\text{ mg/dL}
Measured serum osmolality: 315 mOsm/kg H2O315\text{ mOsm/kg H}_2\text{O}

Arterial blood gas on room air:
pH: 7.207.20
pCO 2_2: 24 mm Hg24\text{ mm Hg}
pO 2_2: 95 mm Hg95\text{ mm Hg}

What is this patient's serum osmolar gap? (Enter the value only)

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Answer: 29

Answer

The patient's serum osmolar gap is 29 mOsm/kg.
The serum osmolar gap is calculated by subtracting the calculated serum osmolality from the measured serum osmolality. The calculated serum osmolality is 2×[Na+]+[Glucose]18+[BUN]2.82 \times [Na^+] + \frac{[Glucose]}{18} + \frac{[BUN]}{2.8}. For this patient, the calculated osmolality is 2×138+9018+142.8=286 mOsm/kg2 \times 138 + \frac{90}{18} + \frac{14}{2.8} = 286 \text{ mOsm/kg}. The osmolar gap is 315−286=29 mOsm/kg315 - 286 = 29 \text{ mOsm/kg}. A normal osmolar gap is typically less than 10 mOsm/kg. An elevated osmolar gap in the setting of a high anion gap metabolic acidosis is highly suggestive of toxic alcohol ingestion (e.g., ethylene glycol or methanol).

Step-by-Step Solution

1
Identify the need to calculate the osmolar gap.
The patient has a high anion gap metabolic acidosis and suspected toxic alcohol ingestion, which requires calculation of the osmolar gap to support the diagnosis.
Ethylene glycol and methanol are unmeasured osmotically active substances that increase the osmolar gap.
2
Calculate the expected serum osmolality.
Expected Osmolality = 2×138+9018+142.8=276+5+5=286 mOsm/kg H2O2 \times 138 + \frac{90}{18} + \frac{14}{2.8} = 276 + 5 + 5 = 286 \text{ mOsm/kg H}_2\text{O}.
The calculated serum osmolality accounts for the primary normal osmotically active solutes in the blood: sodium, glucose, and urea.
3
Calculate the osmolar gap.
Osmolar Gap = Measured Osmolality - Calculated Osmolality = 315−286=29 mOsm/kg315 - 286 = 29 \text{ mOsm/kg}.
The difference between measured and calculated osmolality represents the concentration of unmeasured osmoles (e.g., ethylene glycol).

Key Concept

The serum osmolar gap is used to detect the presence of unmeasured osmotically active substances (such as toxic alcohols) in the blood. It is calculated as the difference between the measured serum osmolality and the calculated serum osmolality.
Estimated Time:1m 30s
Question 7Question →

A 7272-year-old woman is evaluated on hospital day 44 for worsening renal function. She was admitted for acute decompensated heart failure and has been receiving intravenous furosemide. Over the past 4848 hours, her weight has decreased by 4.5 kg4.5\text{ kg}, but her urine output has dropped to 300 mL300\text{ mL} over the last 1212 hours. Her blood pressure is 95/60 mmHg95/60\text{ mmHg} and heart rate is 102/min102\text{/min}. Laboratory studies show:

LabValue
Serum sodium136 mEq/L136\text{ mEq/L}
Serum blood urea nitrogen (BUN)75 mg/dL75\text{ mg/dL}
Serum creatinine2.5 mg/dL2.5\text{ mg/dL} (baseline 1.1 mg/dL1.1\text{ mg/dL})
Urine sodium110 mEq/L110\text{ mEq/L}
Urine urea nitrogen450 mg/dL450\text{ mg/dL}
Urine creatinine100 mg/dL100\text{ mg/dL}

What is this patient's fractional excretion of urea (FEUreaFE_{Urea})? (Express the answer as a percentage).

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Answer: 15

Answer

15
The correct answer is 15%15\%. This patient presents with acute kidney injury (AKI) in the setting of aggressive diuresis for heart failure. Distinguishing between prerenal azotemia (hypovolemia) and intrinsic renal disease (such as acute tubular necrosis) is critical for her ongoing management. While the fractional excretion of sodium (FENaFE_{Na}) is typically less than 1%1\% in prerenal AKI, loop diuretics directly inhibit sodium reabsorption, causing a falsely elevated FENaFE_{Na} (in this patient's case, it would be calculated as roughly 2.02%2.02\%). In patients on active diuretic therapy, the fractional excretion of urea (FEUreaFE_{Urea}) is a much more accurate diagnostic marker because urea reabsorption is closely linked to proximal tubule water reabsorption and is significantly less affected by loop diuretics. An FEUreaFE_{Urea} of less than 35%35\% strongly suggests prerenal azotemia. The calculation is performed as follows: FEUrea=UUrea×PCrPUrea×UCr×100=450×2.575×100×100=11257500×100=15%FE_{Urea} = \frac{U_{Urea} \times P_{Cr}}{P_{Urea} \times U_{Cr}} \times 100 = \frac{450 \times 2.5}{75 \times 100} \times 100 = \frac{1125}{7500} \times 100 = 15\%.

Step-by-Step Solution

1
Evaluate the clinical scenario and the limitation of standard AKI diagnostic indices.
Identify that the patient's active furosemide use falsely elevates sodium excretion, making FE_Na inaccurate (calculated here as 2.02%).
Loop diuretics inhibit sodium reabsorption in the thick ascending limb of the loop of Henle, invalidating FE_Na as a marker for volume status.
2
Select the appropriate diagnostic calculation for patients on diuretics.
Use the Fractional Excretion of Urea (FE_Urea).
Urea reabsorption occurs primarily in the proximal tubule and is less affected by loop diuretics. An FE_Urea < 35% reliably indicates a prerenal state.
3
Apply the FE_Urea formula.
FE_Urea = (Urine Urea_nitrogen * Serum Creatinine) / (Serum BUN * Urine Creatinine) * 100
This formula compares the clearance of urea to the clearance of creatinine to determine the fraction of filtered urea that is ultimately excreted.
4
Substitute the provided laboratory values into the formula.
(450 * 2.5) / (75 * 100) * 100 = 1125 / 7500 * 100
Accurate substitution of variables is required to solve the clearance ratio.
5
Perform the final calculation.
0.15 * 100 = 15%
To express the fractional excretion as a percentage.

Key Concept

In patients receiving active diuretic therapy, the fractional excretion of urea (FE_Urea) is a more reliable indicator of prerenal azotemia than the fractional excretion of sodium (FE_Na).
Question 8Question →

Paramedics bring a 35-year-old male to the emergency department after he was rescued from a residential fire. The patient weighs 75 kg. According to the paramedics, the burn injury occurred exactly 2 hours prior to arrival. Physical examination reveals deep partial-thickness and full-thickness burns covering his entire posterior trunk and both of his entire upper extremities. The remainder of his skin is intact. Two large-bore intravenous lines are established.

Using the traditional Parkland formula (44 mL/kg/% total body surface area), what should be the intravenous fluid administration rate (in mL/hr) prescribed for the remaining time to complete the first 8-hour post-injury resuscitation goal?

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Answer: 900

Answer

900
The traditional Parkland formula is calculated as 44 mL of crystalloid per kg of body weight per percentage of TBSA burned. For this 75 kg patient with 36% TBSA burns (18% for the entire posterior trunk + 9% for each upper extremity), the total 24-hour fluid requirement is 4×75×36=10,8004 \times 75 \times 36 = 10,800 mL. Half of this volume (5,4005,400 mL) must be administered within the first 8 hours from the time of the burn. Because the injury occurred 2 hours prior to arrival, only 6 hours remain in this initial window. Therefore, the 5,4005,400 mL must be infused over 6 hours, requiring a rate of 900900 mL/hr.

Step-by-Step Solution

1
Calculate the burned Total Body Surface Area (TBSA) using the Rule of Nines.
18% (posterior trunk) + 9% (right arm) + 9% (left arm) = 36% TBSA.
The Parkland formula requires an accurate estimation of the total burn surface area.
2
Calculate the total 24-hour fluid resuscitation volume using the traditional Parkland formula.
4 mL/kg × 75 kg × 36% = 10,800 mL.
This establishes the total volume of Lactated Ringer's needed for the first 24 hours after the injury.
3
Determine the fluid volume required for the first 8 hours.
10,800 mL / 2 = 5,400 mL.
Standard burn resuscitation guidelines dictate that 50% of the calculated 24-hour volume is given in the first 8 hours post-injury.
4
Calculate the intravenous fluid rate based on the elapsed time since the injury.
5,400 mL / (8 hours - 2 hours) = 5,400 mL / 6 hours = 900 mL/hr.
The first 8-hour window begins at the time of injury, not at the time of hospital admission. Because 2 hours have already elapsed, the fluid must be infused over the remaining 6 hours.

Key Concept

Burn fluid resuscitation calculations must account for the time elapsed since the injury when setting the infusion rate for the first 8 hours.
Question 9Question →

Paramedics deliver a 68-year-old nursing home resident to the acute care bay after staff found him profoundly lethargic amid multiple episodes of bilious emesis over the preceding 48 hours. Medical history is significant for heart failure with preserved ejection fraction, severe osteoarthritis, and recently diagnosed type 2 diabetes mellitus. On initial assessment, he is obtunded but protects his airway.

Vitals:
ParameterValue
Temperature36.8°C (98.2°F)
Heart rate118/min
Blood pressure88/54 mm Hg
Respiratory rate10/min
SpO293% (room air)
Laboratory data:
TestResult
Sodium142 mEq/L
Potassium3.1 mEq/L
Chloride90 mEq/L
Bicarbonate16 mEq/L
Blood urea nitrogen58 mg/dL
Creatinine2.7 mg/dL
Glucose512 mg/dL
Arterial blood gas (room air):
ParameterValue
pH7.23
PaO274 mm Hg
PaCO240 mm Hg

Assuming a normal reference anion gap of 12 mEq/L and a normal reference serum bicarbonate of 24 mEq/L, what is this patient's delta ratio (ΔAG/ΔHCO3−\Delta AG / \Delta HCO_3^-)?

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Answer: 3

Answer

The patient's calculated delta ratio is 3.0.
This patient presents with a severe triple acid-base disorder: high anion gap metabolic acidosis (likely diabetic ketoacidosis or lactic acidosis from hypoperfusion), metabolic alkalosis (from severe vomiting), and respiratory acidosis (from obtundation causing hypoventilation). The delta ratio compares the rise in the anion gap (ΔAG\Delta AG) to the fall in bicarbonate (ΔHCO3−\Delta HCO_3^-). For every 1 mEq/L increase in unmeasured acids, the bicarbonate should theoretically fall by 1 mEq/L, yielding a ratio of 1.0 (or up to 2.0 in lactic acidosis due to varying intracellular buffering efficiency). A ratio >2.0> 2.0 indicates that the bicarbonate has not fallen as much as expected given the massive acid load. In this patient, the massive acid load (ΔAG=24\Delta AG = 24) only dropped the bicarbonate by 8 mEq/L. This proves the existence of a concurrent metabolic alkalosis (from the loss of gastric HClHCl) generating additional bicarbonate that masks the severity of the acidosis. Additionally, Winter's formula predicts a PaCO2 of 1.5(16)+8=32±21.5(16) + 8 = 32 \pm 2 mm Hg; the measured PaCO2 of 40 mm Hg confirms a concurrent respiratory acidosis.

Step-by-Step Solution

1
Calculate the serum anion gap using the formula AG=Na+−(Cl−+HCO3−)AG = Na^+ - (Cl^- + HCO_3^-).
142−(90+16)=142−106=36142 - (90 + 16) = 142 - 106 = 36 mEq/L.
The first step in evaluating any metabolic acidosis is determining if unmeasured anions are present.
2
Calculate the change in anion gap (ΔAG\Delta AG).
36−12=2436 - 12 = 24 mEq/L.
This determines the exact magnitude of excess unmeasured acid that has accumulated in the blood above the normal baseline.
3
Calculate the change in bicarbonate (ΔHCO3−\Delta HCO_3^-).
24−16=824 - 16 = 8 mEq/L.
This determines how much the serum bicarbonate has dropped from its baseline normal value due to buffering the acid load.
4
Calculate the delta ratio by dividing ΔAG\Delta AG by ΔHCO3−\Delta HCO_3^-.
24/8=3.024 / 8 = 3.0.
Evaluating the ratio between the rise in unmeasured acids and the fall in bicarbonate helps identify concurrent secondary metabolic acid-base disorders.

Key Concept

In patients with a high anion gap metabolic acidosis, the delta ratio (ΔAG/ΔHCO3−\Delta AG / \Delta HCO_3^-) is calculated to uncover hidden concurrent metabolic disorders. A ratio greater than 2.0 indicates a superimposed metabolic alkalosis.
Estimated Time:3m 0s
Question 10Question →

A 55-year-old man presents to the emergency department with a 4-day history of profuse watery diarrhea. He reports orthostatic dizziness, generalized weakness, and a significantly reduced urine output over the past 24 hours. He has no significant past medical history and takes no medications. Vital signs show a temperature of 37.1∘C37.1^\circ\text{C} (98.8∘F98.8^\circ\text{F}), blood pressure of 88/55 mm Hg88/55\text{ mm Hg}, and a heart rate of 122/min122/\text{min}. Physical examination reveals dry mucous membranes, decreased skin turgor, and flat jugular veins.

Laboratory studies are as follows:

Serum
Sodium: 140 mEq/L140\text{ mEq/L}
Potassium: 3.2 mEq/L3.2\text{ mEq/L}
Chloride: 110 mEq/L110\text{ mEq/L}
Bicarbonate: 18 mEq/L18\text{ mEq/L}
Blood urea nitrogen (BUN): 60 mg/dL60\text{ mg/dL}
Creatinine: 2.5 mg/dL2.5\text{ mg/dL}

Urine
Sodium: 14 mEq/L14\text{ mEq/L}
Creatinine: 50 mg/dL50\text{ mg/dL}

What is this patient's fractional excretion of sodium (FeNaFe_{Na}) in percentage?

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Answer: 0.5

Answer

0.5
The fractional excretion of sodium (FeNaFe_{Na}) measures the percentage of filtered sodium that is eventually excreted in the urine. It is widely used to distinguish between prerenal acute kidney injury (AKI) and acute tubular necrosis (ATN). 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 patient's laboratory values yields 14×2.5140×50×100=0.5%\frac{14 \times 2.5}{140 \times 50} \times 100 = 0.5\%. In the setting of oliguria, an FeNa<1%Fe_{Na} < 1\% typically indicates prerenal azotemia, representing an intact renal tubular mechanism trying to conserve sodium and water in response to hypovolemia (consistent with this patient's severe dehydration from diarrhea).

Step-by-Step Solution

1
Identify the values required for the FeNa calculation.
UNa=14U_{Na} = 14, PCr=2.5P_{Cr} = 2.5, PNa=140P_{Na} = 140, UCr=50U_{Cr} = 50
These specific variables are needed to use the FeNaFe_{Na} equation, which helps differentiate prerenal azotemia from intrinsic renal disease.
2
Apply the FeNa formula.
FeNa=14×2.5140×50×100Fe_{Na} = \frac{14 \times 2.5}{140 \times 50} \times 100
The standard formula requires multiplying the ratio of urine-to-plasma sodium by the ratio of plasma-to-urine creatinine.
3
Calculate the final percentage.
0.5%0.5\%
This calculation yields the final percentage, which is less than 1%1\% and highly indicative of a prerenal etiology.

Key Concept

Fractional Excretion of Sodium (FeNa) Calculation
Question 11Question →

Paramedics bring a lethargic 45-year-old male patient to the acute care unit. According to his brother, the patient has consumed heavy amounts of alcohol daily for the past two weeks while completely abstaining from solid foods. Over the last 96 hours, he has suffered from intractable emesis. Physical examination reveals a deeply furrowed tongue, delayed capillary refill, and a loss of skin elasticity.

Laboratory evaluations reveal the following:

ParameterResult
Sodium (Na+Na^+)140 mEq/L
Potassium (K+K^+)3.2 mEq/L
Chloride (Cl−Cl^-)89 mEq/L
Bicarbonate (HCO3−HCO_3^-)18 mEq/L
Blood urea nitrogen42 mg/dL
Serum creatinine1.6 mg/dL
Plasma glucose70 mg/dL

Urinalysis is strongly positive for ketones.

Assume a baseline normal serum anion gap of 12 mEq/L and a normal serum bicarbonate of 24 mEq/L. What is the calculated delta ratio (ΔAGΔHCO3−\frac{\Delta AG}{\Delta HCO_3^-}) for this patient?

Show answer & explanation

Answer: 3.5

Answer

The calculated delta ratio is 3.5.
The patient's anion gap is calculated as 140−(89+18)=33140 - (89 + 18) = 33 mEq/L. The change in anion gap (ΔAG\Delta AG) is 33−12=2133 - 12 = 21 mEq/L. The change in bicarbonate (ΔHCO3−\Delta HCO_3^-) is 24−18=624 - 18 = 6 mEq/L. The delta ratio is 21/6=3.521 / 6 = 3.5. A ratio greater than 2.0 reveals that the drop in bicarbonate is much less than expected for the degree of anion gap elevation, which is characteristic of a concurrent metabolic alkalosis (in this case, from severe volume depletion and intractable emesis superimposed on alcoholic ketoacidosis).

Step-by-Step Solution

1
Calculate the Anion Gap (AG)
3333 mEq/L
To identify the presence and magnitude of a high anion gap metabolic acidosis.
2
Calculate ΔAG\Delta AG
2121 mEq/L
To determine how much the anion gap has increased above the baseline of 12 mEq/L.
3
Calculate ΔHCO3−\Delta HCO_3^-
66 mEq/L
To determine how much the bicarbonate level has decreased from the baseline of 24 mEq/L.
4
Calculate the Delta Ratio
3.53.5
To assess for additional hidden acid-base disorders, such as a concurrent metabolic alkalosis.

Key Concept

The delta ratio (ΔAG/ΔHCO3−\Delta AG / \Delta HCO_3^-) is used in high anion gap metabolic acidosis to identify mixed acid-base disorders; a ratio significantly greater than 2.0 indicates a concurrent metabolic alkalosis.
Question 12Question →

A 42-year-old man is brought to the emergency department via helicopter following a structural fire. The injury occurred exactly 2 hours prior to arrival. He weighs 90 kg90 \text{ kg}. Physical examination reveals deep partial-thickness and full-thickness burns involving his entire anterior trunk, his entire right arm, and the entire anterior surface of his right leg. The remainder of his skin is intact. During transport, paramedics administered 480 mL480 \text{ mL} of intravenous lactated Ringer solution. The trauma team initiates fluid resuscitation using the traditional Parkland formula (4 mL×weight in kg×%TBSA4 \text{ mL} \times \text{weight in kg} \times \% \text{TBSA}). At what rate (in mL/hr\text{mL/hr}) should the lactated Ringer infusion be set for the remainder of the first 8-hour period following the injury?

Show answer & explanation

Answer: 1000

Answer

1000
The patient requires 12,960 mL total in 24 hours (4×90×364 \times 90 \times 36). Half of this (6,480 mL) belongs in the first 8 hours. Deducting the 480 mL given en route leaves 6,000 mL. Because 2 hours have passed since the injury, this 6,000 mL must be given over the remaining 6 hours, yielding a rate of 1000 mL/hr.

Step-by-Step Solution

1
Calculate the Total Body Surface Area (TBSA) affected by the burns.
Entire anterior trunk (18%) + entire right arm (9%) + anterior right leg (9%) = 36% TBSA.
The Rule of Nines is required to accurately estimate burn surface area for fluid calculation.
2
Calculate the total 24-hour fluid requirement using the traditional Parkland formula.
4 mL×90 kg×36=12,960 mL4 \text{ mL} \times 90 \text{ kg} \times 36 = 12,960 \text{ mL}.
The Parkland formula dictates the total volume of lactated Ringer solution needed in the first 24 hours post-injury.
3
Determine the fluid requirement for the first 8 hours.
12,960 mL÷2=6,480 mL12,960 \text{ mL} \div 2 = 6,480 \text{ mL}.
Standard burn resuscitation guidelines state that half of the total 24-hour volume must be given in the first 8 hours from the time of injury.
4
Calculate the remaining volume to be administered.
6,480 mL−480 mL=6,000 mL6,480 \text{ mL} - 480 \text{ mL} = 6,000 \text{ mL}.
Fluids given by EMS en route must be deducted from the total 8-hour requirement.
5
Calculate the required infusion rate for the remaining time.
6,000 mL÷6 hours=1000 mL/hr6,000 \text{ mL} \div 6 \text{ hours} = 1000 \text{ mL/hr}.
The 8-hour window begins at the time of injury. Since 2 hours have elapsed, there are 6 hours left to infuse the remaining 6,000 mL.

Key Concept

Burn fluid resuscitation requires accurate TBSA estimation using the Rule of Nines, application of the appropriate formula, and adjustment of the infusion rate based on the time elapsed since the injury and any pre-hospital fluids already administered.
Question 13Question →

A 54-year-old man is hospitalized for the management of severe pneumonia complicated by sepsis. On hospital day 3, his urine output drops to 200200 mL over a 12-hour period. To evaluate the etiology of his oliguric acute kidney injury, the following laboratory studies are obtained:

Serum sodium: 140140 mEq/L
Serum creatinine: 2.02.0 mg/dL
Urine sodium: 5656 mEq/L
Urine creatinine: 4040 mg/dL

What is this patient's fractional excretion of sodium (FENa)? Give your answer as a percentage.

Show answer & explanation

Answer: 2

Answer

2.0
The correct value is 2.0%. 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×SCrSNa×UCr×100FE_{Na} = \frac{U_{Na} \times S_{Cr}}{S_{Na} \times U_{Cr}} \times 100. By substituting the patient's specific laboratory values into this formula, we get 56×2.0140×40×100=1125600×100=2.0%\frac{56 \times 2.0}{140 \times 40} \times 100 = \frac{112}{5600} \times 100 = 2.0\%. In the context of oliguric acute kidney injury, an FENaFE_{Na} of ≥2%\geq 2\% points toward an intrinsic renal etiology, which aligns perfectly with acute tubular necrosis secondary to this patient's sepsis.

Step-by-Step Solution

1
Identify the proper formula for calculating 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 helps differentiate between prerenal azotemia and intrinsic renal failure (like acute tubular necrosis).
2
Substitute the patient's serum and urine laboratory values into the equation.
FENa=56×2.0140×40×100FE_{Na} = \frac{56 \times 2.0}{140 \times 40} \times 100
Applying the specific clinical data provided in the stem.
3
Perform the mathematical calculation.
1125600×100=0.02×100=2.0\frac{112}{5600} \times 100 = 0.02 \times 100 = 2.0
To obtain the final percentage value.

Key Concept

Calculation of Fractional Excretion of Sodium (FENa) to evaluate acute kidney injury
Question 14Question →

A 38-year-old mechanic presents to the urgent care clinic complaining of severe flank pain, nausea, and blurry vision that started earlier this morning. He admits to accidentally ingesting a few mouthfuls of a liquid from an unmarked container in his garage while working late last night. His temperature is 36.8°C (98.2°F), blood pressure is 142/88 mm Hg, pulse is 98/min, and respiratory rate is 22/min. Cardiopulmonary examination is unremarkable. Initial laboratory tests reveal the following:

* Serum sodium: 138 mEq/L138 \text{ mEq/L}
* Serum potassium: 4.5 mEq/L4.5 \text{ mEq/L}
* Serum chloride: 104 mEq/L104 \text{ mEq/L}
* Serum bicarbonate: 14 mEq/L14 \text{ mEq/L}
* Blood urea nitrogen (BUN): 28 mg/dL28 \text{ mg/dL}
* Serum glucose: 126 mg/dL126 \text{ mg/dL}
* Measured serum osmolality: 315 mOsm/kg H2O315 \text{ mOsm/kg H}_2\text{O}

Based on these findings, what is the patient's calculated serum osmolar gap?

Show answer & explanation

Answer: 22

Answer

22
The calculated serum osmolality is derived using the formula 2×[Na+]+[Glucose]18+[BUN]2.82 \times [\text{Na}^+] + \frac{[\text{Glucose}]}{18} + \frac{[\text{BUN}]}{2.8}. Substituting the patient's values yields 2×138+12618+282.8=276+7+10=293 mOsm/kg H2O2 \times 138 + \frac{126}{18} + \frac{28}{2.8} = 276 + 7 + 10 = 293 \text{ mOsm/kg H}_2\text{O}. The osmolar gap is the measured osmolality minus the calculated osmolality (315−293=22 mOsm/kg H2O315 - 293 = 22 \text{ mOsm/kg H}_2\text{O}). A normal osmolar gap is typically less than 10-12; an elevated gap in this clinical context strongly supports the diagnosis of toxic alcohol ingestion (e.g., methanol or ethylene glycol).

Step-by-Step Solution

1
Determine the formula for calculated serum osmolality.
Calculated Osmolality = 2 × [Na+] + [Glucose]/18 + [BUN]/2.8
This standard formula accounts for the major osmotically active solutes in the extracellular fluid in the United States unit system (mg/dL).
2
Insert the patient's laboratory values into the formula.
Calculated Osmolality = (2 × 138) + (126 / 18) + (28 / 2.8)
To compute the expected osmolality based on native physiological solutes.
3
Perform the arithmetic for each component.
276 + 7 + 10 = 293 mOsm/kg H2O
Resolving the equation yields the total calculated osmolality.
4
Calculate the osmolar gap.
Measured Osmolality (315) - Calculated Osmolality (293) = 22 mOsm/kg H2O
The osmolar gap represents the presence of unmeasured osmotically active substances, such as toxic alcohols (e.g., methanol, ethylene glycol).

Key Concept

Calculation and clinical application of the serum osmolar gap in evaluating suspected toxic alcohol ingestion.
Question 15Question →

A 58-year-old man with a history of chronic alcohol use disorder is brought to the emergency department due to profound weakness and a four-day history of intractable vomiting. He reports being unable to keep any food or fluids down. On physical examination, he appears lethargic and has dry mucous membranes with decreased skin turgor.

Laboratory studies reveal the following:

Serum:
Sodium: 140 mEq/L
Potassium: 3.1 mEq/L
Chloride: 88 mEq/L
Bicarbonate: 14 mEq/L
Blood urea nitrogen: 35 mg/dL
Creatinine: 1.6 mg/dL

Arterial blood gas on room air reveals a pH of 7.32 and a pCO2pCO_2 of 29 mm Hg. The physician suspects a complex mixed acid-base disorder driven by alcoholic ketoacidosis and protracted emesis.

Assuming a baseline normal anion gap of 12 mEq/L and a baseline normal serum bicarbonate of 24 mEq/L, what is the calculated delta-delta ratio (ΔAG/ΔHCO3−\Delta AG / \Delta HCO_3^-) for this patient?

Show answer & explanation

Answer: 2.6

Answer

2.6
In a patient presenting with a high anion gap metabolic acidosis, calculating the delta-delta ratio is essential to evaluate for mixed metabolic disorders. First, the patient's anion gap is calculated as 140−(88+14)=38140 - (88 + 14) = 38 mEq/L. The ΔAG\Delta AG is the patient's calculated AG minus the normal baseline AG (38−12=2638 - 12 = 26). The ΔHCO3−\Delta HCO_3^- is the normal baseline HCO3−HCO_3^- minus the patient's measured HCO3−HCO_3^- (24−14=1024 - 14 = 10). The delta-delta ratio is 26/10=2.626 / 10 = 2.6. Because 2.6 is greater than 2.0, this confirms a concurrent metabolic alkalosis superimposed on the high anion gap metabolic acidosis, which perfectly aligns with the clinical history of alcoholic ketoacidosis paired with profound vomiting.

Step-by-Step Solution

1
Calculate the actual anion gap (AG).
AG=140−(88+14)=38AG = 140 - (88 + 14) = 38 mEq/L.
An elevated anion gap confirms the presence of a high anion gap metabolic acidosis (HAGMA).
2
Calculate the delta anion gap (ΔAG\Delta AG).
ΔAG=38−12=26\Delta AG = 38 - 12 = 26 mEq/L.
This value represents the excess unmeasured anions contributing to the acidosis.
3
Calculate the delta bicarbonate (ΔHCO3−\Delta HCO_3^-).
ΔHCO3−=24−14=10\Delta HCO_3^- = 24 - 14 = 10 mEq/L.
This value represents the apparent fall in bicarbonate due to buffering the excess acid.
4
Calculate the delta-delta ratio.
ΔAG/ΔHCO3−=26/10=2.6\Delta AG / \Delta HCO_3^- = 26 / 10 = 2.6.
A ratio greater than 2.0 indicates that the bicarbonate level is higher than expected, revealing a concurrent metabolic alkalosis (e.g., from vomiting) superimposed on the HAGMA.

Key Concept

The delta-delta ratio (ΔAG/ΔHCO3−\Delta AG / \Delta HCO_3^-) is used in the setting of a high anion gap metabolic acidosis to identify hidden mixed acid-base disorders. A ratio > 2.0 indicates a concurrent metabolic alkalosis, whereas a ratio < 1.0 indicates a concurrent normal anion gap metabolic acidosis.
Estimated Time:1m 30s
Question 16Question →

A 54-year-old man with a history of severe alcohol use disorder is admitted to the intensive care unit with acute necrotizing pancreatitis. Over the past 72 hours, he has experienced intractable vomiting and an inability to tolerate any oral intake. On physical examination, the patient is lethargic. Mucous membranes are extremely dry, and skin turgor is poor. Vital signs show a blood pressure of 88/50 mmHg, a pulse of 118/min, and a respiratory rate of 22/min.

Laboratory studies reveal the following:

VariableValue
Sodium (Na+Na^+)138 mEq/L
Potassium (K+K^+)3.1 mEq/L
Chloride (Cl−Cl^-)92 mEq/L
Bicarbonate (HCO3−HCO_3^-)16 mEq/L
Blood urea nitrogen (BUN)45 mg/dL
Creatinine2.1 mg/dL
Albumin1.6 g/dL
Lactic acid6.2 mmol/L

The physician evaluates the patient's acid-base status. Assuming a baseline normal anion gap of 12 mEq/L and a normal serum bicarbonate of 24 mEq/L, what is the patient's calculated delta ratio (ΔAG/ΔHCO3−\Delta\text{AG}/\Delta\text{HCO}_3^-) after appropriately correcting the anion gap for his hypoalbuminemia?

Show answer & explanation

Answer: 3

Answer

3.0
This patient has an elevated anion gap metabolic acidosis secondary to profound lactic acidosis (from tissue hypoperfusion) and likely starvation ketoacidosis. Concurrently, his intractable vomiting introduces a primary metabolic alkalosis. To accurately quantify this mixed disorder, the delta ratio must be calculated. First, the measured anion gap is calculated as 138−(92+16)=30138 - (92 + 16) = 30 mEq/L. Because albumin is a major circulating unmeasured anion, severe hypoalbuminemia (1.6 g/dL) falsely lowers the measured anion gap. The corrected anion gap is 30+2.5×(4.0−1.6)=3630 + 2.5 \times (4.0 - 1.6) = 36 mEq/L. The ΔAG\Delta\text{AG} is 36−12=2436 - 12 = 24 mEq/L. The change in bicarbonate (ΔHCO3−\Delta\text{HCO}_3^-) is 24−16=824 - 16 = 8 mEq/L. The delta ratio is 24/8=3.024 / 8 = 3.0. A delta ratio greater than 2.0 indicates that the bicarbonate is higher than expected, confirming a concurrent metabolic alkalosis (from vomiting).

Step-by-Step Solution

1
Calculate the measured anion gap (AG).
138 - (92 + 16) = 30 mEq/L.
The standard anion gap is necessary to identify the presence of unmeasured anions (e.g., lactate).
2
Correct the anion gap for hypoalbuminemia.
30 + 2.5 \times (4.0 - 1.6) = 30 + 2.5 \times 2.4 = 30 + 6.0 = 36 mEq/L.
Albumin is a major unmeasured weak acid. In severe hypoalbuminemia, the measured anion gap is falsely lowered and must be corrected to reveal the true extent of the metabolic acidosis.
3
Calculate the change in anion gap (ΔAG\Delta\text{AG}).
36 - 12 = 24 mEq/L.
This value represents the excess unmeasured anions above the baseline normal value of 12 mEq/L.
4
Calculate the change in bicarbonate (ΔHCO3−\Delta\text{HCO}_3^-).
24 - 16 = 8 mEq/L.
This determines how much the serum bicarbonate has fallen from the normal baseline of 24 mEq/L due to buffering.
5
Calculate the delta ratio.
24 / 8 = 3.0.
The delta ratio compares the rise in unmeasured anions to the fall in bicarbonate, helping to identify mixed acid-base disorders.

Key Concept

In patients with high anion gap metabolic acidosis, the measured anion gap must be corrected for hypoalbuminemia before calculating the delta ratio to accurately identify mixed acid-base disorders.
Question 17Question →

A 52-year-old man is brought to the emergency department by ambulance after being found obtunded in his garage. His medical history is significant for major depressive disorder and alcohol use disorder. He is unresponsive to verbal stimuli but localizes to pain. Vital signs are within normal limits except for a respiratory rate of 24/min24/\text{min}. Laboratory evaluation reveals the following:

TestResult
Sodium (Na+Na^+)138 mEq/L138 \text{ mEq/L}
Potassium (K+K^+)4.5 mEq/L4.5 \text{ mEq/L}
Chloride (Cl−Cl^-)100 mEq/L100 \text{ mEq/L}
Bicarbonate (HCO3−HCO_3^-)12 mEq/L12 \text{ mEq/L}
Blood urea nitrogen (BUN)28 mg/dL28 \text{ mg/dL}
Creatinine1.4 mg/dL1.4 \text{ mg/dL}
Glucose144 mg/dL144 \text{ mg/dL}
Serum ethanolUndetectable
Measured serum osmolality320 mOsm/kg320 \text{ mOsm/kg}

Arterial blood gas on room air shows a pH of 7.227.22, PCO2PCO_2 of 26 mm Hg26 \text{ mm Hg}, and PO2PO_2 of 92 mm Hg92 \text{ mm Hg}.

Based on these laboratory results, what is the patient's calculated serum osmolar gap?

Show answer & explanation

Answer: 26

Answer

26
The patient's osmolar gap is 26 mOsm/kg26 \text{ mOsm/kg}. This is calculated by taking the measured osmolality (320 mOsm/kg320 \text{ mOsm/kg}) and subtracting the calculated osmolality (294 mOsm/kg294 \text{ mOsm/kg}). The formula for calculated osmolality is (2×Na+)+(Glucose/18)+(BUN/2.8)(2 \times Na^+) + (\text{Glucose}/18) + (\text{BUN}/2.8). This elevated gap, alongside a high anion gap metabolic acidosis (138−(100+12)=26138 - (100 + 12) = 26), strongly suggests toxic alcohol ingestion, such as methanol or ethylene glycol, given the clinical context of obtundation and negative ethanol levels.

Step-by-Step Solution

1
Calculate the expected serum osmolality.
Calculated osmolality = 294 mOsm/kg294 \text{ mOsm/kg}.
The calculated osmolality requires serum sodium, glucose, and BUN. The formula is 2[Na+]+Glucose18+BUN2.82[Na^+] + \frac{\text{Glucose}}{18} + \frac{\text{BUN}}{2.8}.
2
Calculate the osmolar gap.
Osmolar gap = 26 mOsm/kg26 \text{ mOsm/kg}.
The osmolar gap is the difference between the measured serum osmolality (320 mOsm/kg320 \text{ mOsm/kg}) and the calculated serum osmolality (294 mOsm/kg294 \text{ mOsm/kg}).

Key Concept

The serum osmolar gap is used to detect the presence of unmeasured osmotically active substances (such as toxic alcohols). It is calculated by subtracting the calculated serum osmolality from the measured serum osmolality.
Question 18Question →

A 3434-year-old woman with a history of type 11 diabetes mellitus is brought to the emergency department due to severe nausea, diffuse abdominal pain, and intractable vomiting for the past 4848 hours. She reports missing her last three doses of insulin because she was unable to tolerate any oral intake. On physical examination, her blood pressure is 98/60 mm Hg98/60\text{ mm Hg}, heart rate is 118 /min118\text{ /min}, and respiratory rate is 22 /min22\text{ /min}. Her mucous membranes are remarkably dry, and skin turgor is decreased.

Laboratory testing reveals:

TestResult
Arterial pH\text{pH}7.307.30
pCO2\text{pCO}_229 mm Hg29\text{ mm Hg}
pO2\text{pO}_295 mm Hg95\text{ mm Hg}
Na+\text{Na}^+144 mEq/L144\text{ mEq/L}
K+\text{K}^+3.2 mEq/L3.2\text{ mEq/L}
Cl−\text{Cl}^-94 mEq/L94\text{ mEq/L}
HCO3−\text{HCO}_3^-14 mEq/L14\text{ mEq/L}
Glucose\text{Glucose}550 mg/dL550\text{ mg/dL}
BUN\text{BUN}38 mg/dL38\text{ mg/dL}
Creatinine\text{Creatinine}1.6 mg/dL1.6\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}, calculate the patient's delta ratio (ΔAGΔHCO3−\frac{\Delta\text{AG}}{\Delta\text{HCO}_3^-}) to evaluate for the presence of a mixed acid-base disorder.

Show answer & explanation

Answer: 2.4

Answer

The calculated delta ratio is 2.4.
The patient has a primary high anion gap metabolic acidosis secondary to diabetic ketoacidosis (AG = 3636). In an uncomplicated AGMA, for every 1 mEq/L1\text{ mEq/L} rise in the anion gap, the serum bicarbonate should fall by approximately 1 mEq/L1\text{ mEq/L}, yielding a delta ratio between 1.01.0 and 2.02.0. This patient's delta ratio is 36−1224−14=2.4\frac{36-12}{24-14} = 2.4. A delta ratio greater than 2.02.0 indicates that the bicarbonate has not dropped as much as expected for the amount of acid present. This means there is a concurrent primary metabolic alkalosis adding bicarbonate to the serum, which in this clinical vignette is driven by profound volume contraction and acid loss from intractable vomiting.

Step-by-Step Solution

1
Calculate the measured Anion Gap (AG).
AG=144−(94+14)=36 mEq/L\text{AG} = 144 - (94 + 14) = 36\text{ mEq/L}.
Establishing the presence of an anion gap metabolic acidosis (AGMA) is the first step in acid-base evaluation.
2
Calculate the Delta 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 contributing to the acidosis.
3
Calculate the Delta 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 the amount of bicarbonate consumed to buffer the excess acid.
4
Calculate the Delta Ratio.
ΔAGΔHCO3−=2410=2.4\frac{\Delta\text{AG}}{\Delta\text{HCO}_3^-} = \frac{24}{10} = 2.4.
Comparing the change in anion gap to the change in bicarbonate identifies occult mixed metabolic acid-base disorders.

Key Concept

The delta ratio (ΔAGΔHCO3−\frac{\Delta\text{AG}}{\Delta\text{HCO}_3^-}) is used to diagnose mixed metabolic acid-base disorders in the setting of a high anion gap metabolic acidosis. A ratio >2.0> 2.0 indicates a concurrent metabolic alkalosis, such as from severe vomiting.
Estimated Time:2m 0s
Question 19Question →

A 22-year-old man with type 1 diabetes mellitus presents to the emergency department with abdominal pain, nausea, and persistent vomiting for the past 2 days. He reports missing his last several doses of insulin. His vital signs include a heart rate of 118/min and a blood pressure of 90/60 mm Hg. Laboratory studies show the following:

Laboratory TestResult
Sodium (Na+Na^+)142 mEq/L
Potassium (K+K^+)4.5 mEq/L
Chloride (Cl−Cl^-)92 mEq/L
Bicarbonate (HCO3−HCO_3^-)14 mEq/L
Blood urea nitrogen35 mg/dL
Creatinine1.4 mg/dL
Glucose450 mg/dL

Assuming a normal reference serum anion gap of 12 mEq/L and a normal serum bicarbonate of 24 mEq/L, what is the patient's delta ratio (calculated as the change in anion gap divided by the change in serum bicarbonate)? (Provide the exact calculated decimal value).

Show answer & explanation

Answer: 2.4

Answer

2.4
The delta ratio is used to determine if a mixed acid-base disorder is present in the setting of a high anion gap metabolic acidosis. First, the patient's anion gap is calculated as Na+−(Cl−+HCO3−)=142−(92+14)=36Na^+ - (Cl^- + HCO_3^-) = 142 - (92 + 14) = 36 mEq/L. The change in the anion gap (ΔAG\Delta AG) is the calculated gap minus the normal gap: 36−12=2436 - 12 = 24 mEq/L. The change in bicarbonate (ΔHCO3−\Delta HCO_3^-) is the normal bicarbonate minus the patient's bicarbonate: 24−14=1024 - 14 = 10 mEq/L. The delta ratio is ΔAG÷ΔHCO3−=24÷10=2.4\Delta AG \div \Delta HCO_3^- = 24 \div 10 = 2.4. A ratio greater than 2.0 indicates that the bicarbonate level is higher than would be expected from the buffering of the excess acid alone, implying a concurrent primary metabolic alkalosis (in this case, secondary to the patient's persistent vomiting).

Step-by-Step Solution

1
Calculate the patient's anion gap.
142−(92+14)=36142 - (92 + 14) = 36 mEq/L
The basic anion gap formula is Na+−(Cl−+HCO3−)Na^+ - (Cl^- + HCO_3^-).
2
Calculate the change in the anion gap (ΔAG\Delta AG).
36−12=2436 - 12 = 24 mEq/L
This value represents the excess unmeasured anions above the normal baseline.
3
Calculate the change in serum bicarbonate (ΔHCO3−\Delta HCO_3^-).
24−14=1024 - 14 = 10 mEq/L
This value represents the amount of bicarbonate consumed by buffering the excess acid.
4
Calculate the delta ratio.
24÷10=2.424 \div 10 = 2.4
The delta ratio compares the rise in unmeasured anions to the fall in bicarbonate to help identify mixed acid-base disorders.

Key Concept

Calculation and interpretation of the delta ratio in mixed acid-base disorders
Question 20Question →

Emergency medical services transport a 30-year-old factory worker to the regional trauma center after an industrial flash fire. The primary survey confirms a patent airway, symmetric breath sounds, and stable hemodynamics. Secondary evaluation of the patient's integumentary system reveals partial-thickness and full-thickness burns covering the entire anterior aspect of his trunk as well as the entire anterior aspect of his left leg. The genital region, head, posterior torso, arms, right leg, and posterior left leg are completely spared. Based on the standard adult Rule of Nines, what is the estimated percentage of total body surface area affected by these burns?

Show answer & explanation

Answer: 27

Answer

27
In the initial management of burn patients, accurately estimating the total body surface area (TBSA) involved is crucial for calculating fluid resuscitation requirements. The adult Rule of Nines allocates TBSA as follows: head and neck (9%), each upper extremity (9% total; 4.5% anterior, 4.5% posterior), anterior trunk (18%), posterior trunk (18%), each lower extremity (18% total; 9% anterior, 9% posterior), and the perineum (1%). This patient presents with burns on the complete anterior trunk (18%) and the anterior aspect of the left leg (9%). Adding these two areas together (18 + 9) gives a total estimated burn area of 27%.

Step-by-Step Solution

1
Determine the percentage of total body surface area (TBSA) for the anterior trunk using the Rule of Nines.
18%
The anterior torso (comprising the chest and abdomen) represents 18% of the adult TBSA.
2
Determine the percentage of TBSA for the anterior aspect of one leg.
9%
An entire adult lower extremity represents 18% of TBSA. Therefore, the anterior half of one leg accounts for 9%.
3
Sum the regional percentages to find the total burn area.
27%
Adding the anterior trunk (18%) and the anterior left leg (9%) yields a total of 27%.

Key Concept

The Rule of Nines is a standardized method used to quickly estimate the percentage of total body surface area (TBSA) affected by burns in an adult to guide fluid resuscitation.
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