Tüm alıştırma soruları

127 soru

Soru 61Soru →

A hospital ward team is evaluating a 74-year-old female who was admitted three days ago for an acute exacerbation of chronic systolic heart failure. Since admission, she has received continuous intravenous bumetanide. Over the past 24 hours, her urine output has significantly decreased, and routine morning laboratory studies reveal an acute rise in her serum creatinine from a baseline of 1.1 mg/dL1.1 \text{ mg/dL}.

The following laboratory values are obtained:

ParameterValue
Serum sodium136 mEq/L136 \text{ mEq/L}
Serum potassium3.8 mEq/L3.8 \text{ mEq/L}
Serum chloride98 mEq/L98 \text{ mEq/L}
Blood urea nitrogen (BUN)72 mg/dL72 \text{ mg/dL}
Serum creatinine2.4 mg/dL2.4 \text{ mg/dL}
Urine sodium45 mEq/L45 \text{ mEq/L}
Urine creatinine60 mg/dL60 \text{ mg/dL}
Urine urea nitrogen450 mg/dL450 \text{ mg/dL}

Due to the confounding effect of ongoing loop diuretic therapy, the standard fractional excretion of sodium (FENa) may be misleading. The attending physician asks for the calculation of the most appropriate alternative diagnostic index to differentiate between prerenal azotemia and intrinsic acute tubular necrosis in this specific clinical context.

What is the value of this alternative index (expressed as a percentage)?

Cevabı ve açıklamayı göster

Cevap: 25

Cevap

25
The correct answer is 25. The patient has acute kidney injury in the setting of active loop diuretic therapy (bumetanide), which makes the fractional excretion of sodium (FENa) unreliable. The most appropriate alternative index is the fractional excretion of urea (FEUrea). Using the formula FEUrea=Urine Urea×Serum CrSerum BUN×Urine Cr×100\text{FE}_{\text{Urea}} = \frac{\text{Urine Urea} \times \text{Serum Cr}}{\text{Serum BUN} \times \text{Urine Cr}} \times 100, the calculation is 450×2.472×60×100=25%\frac{450 \times 2.4}{72 \times 60} \times 100 = 25\%.

Adım Adım Çözüm

1
Determine the appropriate diagnostic index for acute kidney injury in a patient on active loop diuretic therapy.
The fractional excretion of urea (FEUrea) should be used instead of the fractional excretion of sodium (FENa).
Loop diuretics inhibit sodium reabsorption in the thick ascending limb, which falsely elevates urinary sodium excretion and FENa, rendering it unreliable for differentiating prerenal azotemia from acute tubular necrosis.
2
Recall the formula for the fractional excretion of urea.
FEUrea=Urine Urea×Serum CrSerum Urea×Urine Cr×100\text{FE}_{\text{Urea}} = \frac{\text{Urine Urea} \times \text{Serum Cr}}{\text{Serum Urea} \times \text{Urine Cr}} \times 100
This formula calculates the clearance of urea relative to the clearance of creatinine.
3
Insert the provided laboratory values into the formula.
FEUrea=450×2.472×60×100\text{FE}_{\text{Urea}} = \frac{450 \times 2.4}{72 \times 60} \times 100
Substituting urine urea nitrogen (450 mg/dL450 \text{ mg/dL}), serum creatinine (2.4 mg/dL2.4 \text{ mg/dL}), BUN/serum urea (72 mg/dL72 \text{ mg/dL}), and urine creatinine (60 mg/dL60 \text{ mg/dL}).
4
Perform the final arithmetic calculation.
10804320×100=0.25×100=25%\frac{1080}{4320} \times 100 = 0.25 \times 100 = 25\%
This yields the final percentage required by the question, correctly pointing to a prerenal etiology (value < 35%).

Anahtar Kavram

In patients receiving diuretic therapy, the fractional excretion of urea (FEUrea) is the most accurate diagnostic index to differentiate prerenal azotemia from acute tubular necrosis. A value less than 35% indicates a prerenal etiology.
Soru 62Soru →

A 5858-year-old man with a history of decompensated alcoholic cirrhosis presents to the emergency department with diffuse abdominal pain and a low-grade fever over the past 2424 hours. Vital signs are temperature 37.9∘C37.9^\circ C (100.2∘F100.2^\circ F), blood pressure 105/65105/65 mmHg, heart rate 9898/min, and respiratory rate 1818/min. Physical examination reveals scleral icterus and a distended, diffusely tender abdomen with a positive fluid wave. To evaluate for spontaneous bacterial peritonitis, a diagnostic paracentesis is performed. The procedure is complicated by a traumatic tap, yielding noticeably bloody fluid. The ascitic fluid laboratory analysis reveals the following:

Ascitic fluid red blood cell (RBC) count: 62,500/μL62,500/\mu L
Ascitic fluid white blood cell (WBC) count: 1,250/μL1,250/\mu L
Ascitic fluid differential: 50%50\% neutrophils, 30%30\% lymphocytes, 20%20\% macrophages

Based on these results, what is the corrected ascitic fluid absolute neutrophil count (in cells/μL\mu L) that should be used to evaluate the patient for spontaneous bacterial peritonitis?

Cevabı ve açıklamayı göster

Cevap: 375

Cevap

The corrected absolute neutrophil count is 375375 cells/μL\mu L.
In the evaluation of spontaneous bacterial peritonitis (SBP), the diagnostic threshold is an absolute neutrophil count (ANC) of ≥250\ge 250 cells/μL\mu L in the ascitic fluid. When a paracentesis is traumatic (bloody), peripheral blood contaminates the fluid, falsely elevating the WBC and neutrophil counts. To correct for this, clinicians must subtract 11 neutrophil for every 250250 red blood cells (RBCs) in the fluid. First, calculate the uncorrected ANC: 1,250 WBCs×0.50=6251,250 \text{ WBCs} \times 0.50 = 625 neutrophils/μL\mu L. Next, calculate the correction factor: 62,500 RBCs/250=25062,500 \text{ RBCs} / 250 = 250 neutrophils/μL\mu L. Finally, subtract the correction from the uncorrected ANC: 625−250=375625 - 250 = 375 cells/μL\mu L. Because the corrected value (375375) is ≥250\ge 250 cells/μL\mu L, the patient has SBP and requires empiric antibiotic therapy.

Adım Adım Çözüm

1
Calculate the uncorrected absolute neutrophil count.
1,250×0.50=6251,250 \times 0.50 = 625 cells/μL\mu L
To establish the baseline number of neutrophils in the sample before accounting for peripheral blood contamination.
2
Calculate the expected neutrophil contamination from the traumatic tap.
62,500/250=25062,500 / 250 = 250 neutrophils/μL\mu L
Peripheral blood artificially elevates the ascitic neutrophil count. The widely accepted correction factor is subtracting 11 neutrophil for every 250250 RBCs present in the fluid.
3
Determine the corrected absolute neutrophil count.
625−250=375625 - 250 = 375 cells/μL\mu L
Subtracting the peripheral contamination from the uncorrected count isolates the true ascitic neutrophil response, determining if the patient meets the diagnostic threshold for spontaneous bacterial peritonitis (≥250\ge 250 cells/μL\mu L).

Anahtar Kavram

Correcting ascitic fluid absolute neutrophil count in the setting of a traumatic paracentesis
Soru 63Soru →

A 42-year-old woman presents to the emergency department complaining of sudden-onset shortness of breath and right-sided pleuritic chest pain. She recently returned from a 14-hour international flight. She has no significant past medical history and takes no medications other than oral contraceptives.

Vital signs:
Temperature: 37.1°C (98.8°F)
Blood pressure: 128/78 mm Hg
Heart rate: 115/min
Respiratory rate: 26/min
Oxygen saturation: 88% on room air

Arterial blood gas analysis on room air reveals:
pH: 7.48
PaCO2PaCO_2: 28 mm Hg
PaO2PaO_2: 55 mm Hg
HCO3−HCO_3^-: 20 mEq/L

Assume the inspired partial pressure of oxygen (PIO2P_IO_2) is 150 mm Hg and the respiratory quotient (RR) is 0.8.

What is this patient's alveolar-arterial (A−aA-a) oxygen gradient in mm Hg?

Cevabı ve açıklamayı göster

Cevap: 60

Cevap

60
The alveolar-arterial (A-a) gradient evaluates the integrity of the alveolar-capillary membrane and helps differentiate causes of hypoxemia. It is calculated by first determining the alveolar oxygen tension (PAO2P_AO_2) using the alveolar gas equation: PAO2=PIO2−(PaCO2/R)P_AO_2 = P_IO_2 - (PaCO_2 / R). Given PIO2=150P_IO_2 = 150 mm Hg, PaCO2=28PaCO_2 = 28 mm Hg, and R=0.8R = 0.8, the PAO2P_AO_2 is 150−(28/0.8)=115150 - (28 / 0.8) = 115 mm Hg. The A-a gradient is then PAO2−PaO2P_AO_2 - PaO_2, which is 115−55=60115 - 55 = 60 mm Hg. A normal A-a gradient is typically less than 10-15 mm Hg (or approximated by Age/4+4Age/4 + 4). An elevated gradient in this patient is highly consistent with a pulmonary embolism, which creates a ventilation-perfusion (V/Q) mismatch.

Adım Adım Çözüm

1
Calculate the alveolar oxygen tension (PAO2P_AO_2)
The expected alveolar oxygen level is determined using the formula: PAO2=PIO2−(PaCO2/R)P_AO_2 = P_IO_2 - (PaCO_2 / R)
This establishes the ideal amount of oxygen present in the alveoli available for gas exchange.
2
Substitute the provided values into the alveolar gas equation
PAO2=150−(28/0.8)=150−35=115P_AO_2 = 150 - (28 / 0.8) = 150 - 35 = 115 mm Hg
Using the given PIO2P_IO_2 of 150 mm Hg and PaCO2PaCO_2 of 28 mm Hg divided by the respiratory quotient of 0.8.
3
Calculate the A-a gradient
Gradient = 115−55=60115 - 55 = 60 mm Hg
Subtract the measured arterial oxygen (PaO2PaO_2) from the calculated alveolar oxygen (PAO2P_AO_2) to find the difference.

Anahtar Kavram

Calculation of the alveolar-arterial (A-a) oxygen gradient to evaluate hypoxemia and ventilation-perfusion mismatch.
Soru 64Soru →

A 72-year-old man is admitted to the hospital with a 3-day history of severe diarrhea and decreased oral intake. His past medical history is significant for heart failure with preserved ejection fraction, for which he takes furosemide and lisinopril daily. On physical examination, his blood pressure is 90/60 mm Hg90/60\text{ mm Hg} and heart rate is 112/min112/\text{min}. Mucous membranes are dry, and capillary refill time is prolonged.

Initial laboratory studies are obtained:

ParameterSerum ValueUrine Value
Sodium135 mEq/L135\text{ mEq/L}108 mEq/L108\text{ mEq/L}
Creatinine2.0 mg/dL2.0\text{ mg/dL}80 mg/dL80\text{ mg/dL}
Urea nitrogen (BUN)60 mg/dL60\text{ mg/dL}360 mg/dL360\text{ mg/dL}

The patient's baseline serum creatinine is 1.0 mg/dL1.0\text{ mg/dL}. The physician suspects prerenal acute kidney injury secondary to volume depletion. Given the patient's active loop diuretic use, the fractional excretion of sodium (FENa) may be artifactually elevated and unreliable.

What is the calculated value (expressed as a percentage) of the alternative fractional excretion index that should be used to confirm the etiology of his kidney injury?

Cevabı ve açıklamayı göster

Cevap: 15

Cevap

15
The Fractional Excretion of Urea (FEUrea) is the most appropriate index here. Unlike sodium, urea reabsorption in the proximal tubule is largely uninhibited by loop and thiazide diuretics. The formula is FEUrea=(Uurea×PCr)/(Purea×UCr)×100FEUrea = (U_{urea} \times P_{Cr}) / (P_{urea} \times U_{Cr}) \times 100. Plugging in the values: (360×2.0)/(60×80)×100=720/4800×100=15%(360 \times 2.0) / (60 \times 80) \times 100 = 720 / 4800 \times 100 = 15\%. An FEUrea <35%<35\% is indicative of a prerenal state.

Adım Adım Çözüm

1
Identify the correct diagnostic index
Fractional Excretion of Urea (FEUrea)
FENa is unreliable in patients on diuretics like furosemide because diuretic-induced natriuresis artificially elevates urine sodium. FEUrea is less affected by diuretic use.
2
Set up the mathematical formula for FEUrea
FEUrea=UrineUrea×SerumCrSerumBUN×UrineCr×100FEUrea = \frac{Urine Urea \times Serum Cr}{Serum BUN \times Urine Cr} \times 100
This formula calculates the percentage of filtered urea that is ultimately excreted in the urine.
3
Input the laboratory parameters and calculate
360×2.060×80×100=15%\frac{360 \times 2.0}{60 \times 80} \times 100 = 15\%
Plugging in the given values yields the final percentage required to confirm prerenal acute kidney injury.

Anahtar Kavram

In patients with acute kidney injury taking diuretics, Fractional Excretion of Urea (FEUrea) is a more reliable indicator of prerenal etiology than Fractional Excretion of Sodium (FENa). An FEUrea < 35% strongly suggests prerenal azotemia.
Tahmini Süre:2m 0s
Soru 65Soru →

A 39-year-old previously healthy woman is brought to the emergency department after her roommates found her lethargic and confused. Empty bottles of an unknown liquid were found in her room. On arrival, she is difficult to rouse and responds poorly to stimuli. Vital signs show a temperature of 36.8∘C36.8^\circ\text{C} (98.2∘F98.2^\circ\text{F}), blood pressure of 115/75 mm Hg115/75\text{ mm Hg}, pulse of 110/min110/\text{min}, and respiratory rate of 28/min28/\text{min} with deep, rapid breaths. Laboratory studies report the following:

ParameterValueReference Range
Sodium138 mEq/L138\text{ mEq/L}136−145 mEq/L136-145\text{ mEq/L}
Potassium4.8 mEq/L4.8\text{ mEq/L}3.5−5.0 mEq/L3.5-5.0\text{ mEq/L}
Chloride102 mEq/L102\text{ mEq/L}98−106 mEq/L98-106\text{ mEq/L}
Bicarbonate8 mEq/L8\text{ mEq/L}22−28 mEq/L22-28\text{ mEq/L}
Blood urea nitrogen (BUN)28 mg/dL28\text{ mg/dL}7−18 mg/dL7-18\text{ mg/dL}
Creatinine1.5 mg/dL1.5\text{ mg/dL}0.6−1.2 mg/dL0.6-1.2\text{ mg/dL}
Glucose108 mg/dL108\text{ mg/dL}70−110 mg/dL70-110\text{ mg/dL}

Additional laboratory analysis:
* Measured serum osmolality: 340 mOsm/kg H2O340\text{ mOsm/kg } H_2O
* Arterial blood gas (room air) pH: 7.157.15
* Arterial PaCO2: 20 mm Hg20\text{ mm Hg}

Based on these laboratory results, what is this patient's osmolar gap? (Express the result in mOsm/kg H2O\text{mOsm/kg } H_2O.)

Cevabı ve açıklamayı göster

Cevap: 48

Cevap

48
The correct osmolar gap is 48 mOsm/kg48\text{ mOsm/kg}. To find this, first determine the calculated serum osmolality using the standard formula: 2×[Na+]+BUN2.8+Glucose182 \times [\text{Na}^+] + \frac{\text{BUN}}{2.8} + \frac{\text{Glucose}}{18}. Substituting the given values yields 2×138+282.8+10818=276+10+6=292 mOsm/kg2 \times 138 + \frac{28}{2.8} + \frac{108}{18} = 276 + 10 + 6 = 292\text{ mOsm/kg}. The osmolar gap is then found by subtracting this calculated value from the measured serum osmolality provided by the laboratory: 340−292=48 mOsm/kg340 - 292 = 48\text{ mOsm/kg}. A gap greater than 10−15 mOsm/kg10-15\text{ mOsm/kg}, combined with an elevated anion gap metabolic acidosis, strongly suggests the presence of unmeasured low-molecular-weight substances, such as methanol or ethylene glycol (antifreeze).

Adım Adım Çözüm

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 accounts for the major osmotically active solutes in the serum (sodium, urea, and glucose) to estimate the baseline osmolality.
2
Substitute the patient's sodium, BUN, and glucose values into the formula.
2×138+282.8+10818=276+10+6=292 mOsm/kg2 \times 138 + \frac{28}{2.8} + \frac{108}{18} = 276 + 10 + 6 = 292\text{ mOsm/kg}
Converting BUN and glucose from mg/dL to mOsm/L requires the specific division constants (2.8 for BUN, 18 for glucose) to determine their exact osmotic contribution.
3
Calculate the osmolar gap.
Measured osmolality (340340) - Calculated osmolality (292292) = 48 mOsm/kg48\text{ mOsm/kg}
The osmolar gap measures the difference between the directly measured osmolality and the calculated baseline, revealing the presence of unmeasured osmotically active substances like toxic alcohols.

Anahtar Kavram

Calculation and interpretation of the osmolar gap in the evaluation of a patient with suspected toxic alcohol ingestion.
Tahmini Süre:1m 30s
Soru 66Soru →

A 32-year-old woman with a history of type 1 diabetes mellitus is brought to the emergency department by her husband due to altered mental status and 3 days of severe nausea, recurrent vomiting, and generalized abdominal pain. She has been unable to retain any oral intake and missed her last three doses of insulin.

On physical examination, her temperature is 37.1°C (98.8°F), blood pressure is 98/60 mm Hg, pulse is 122/min, and respirations are 24/min and deep. Her mucous membranes are remarkably dry, and skin turgor is decreased. Her breath has a fruity odor.

Laboratory studies show:
Serum sodium: 140 mEq/L140 \text{ mEq/L}
Serum potassium: 4.1 mEq/L4.1 \text{ mEq/L}
Serum chloride: 92 mEq/L92 \text{ mEq/L}
Serum bicarbonate: 14 mEq/L14 \text{ mEq/L}
Serum glucose: 540 mg/dL540 \text{ mg/dL}
Serum BUN: 45 mg/dL45 \text{ mg/dL}
Serum creatinine: 1.9 mg/dL1.9 \text{ mg/dL}
Serum ketones: Positive

Arterial blood gas on room air:
pH: 7.28
PaCO2\text{PaCO}_2: 29 mmHg29 \text{ mmHg}
PaO2\text{PaO}_2: 98 mmHg98 \text{ mmHg}

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

Cevabı ve açıklamayı göster

Cevap: 2.2

Cevap

The calculated delta ratio is 2.2.
The correct calculated delta ratio is 2.2. The patient presents with a high anion gap metabolic acidosis secondary to diabetic ketoacidosis (DKA). To identify a superimposed acid-base disorder, the delta ratio is calculated. The measured anion gap is 140 - (92 + 14) = 34 mEq/L. The change in anion gap (ΔAG) is 34 - 12 = 22 mEq/L. The change in bicarbonate (ΔHCO3) is 24 - 14 = 10 mEq/L. The delta ratio is ΔAG / ΔHCO3 = 22 / 10 = 2.2. Because this ratio is greater than 2.0, it indicates that the drop in bicarbonate is less than expected for the corresponding increase in unmeasured anions. This points to a concurrent metabolic alkalosis, which is clinically consistent with her history of severe, recurrent vomiting.

Adım Adım Çözüm

1
Calculate the measured anion gap (AG).
AG = 34 mEq/L
The anion gap is needed to identify the presence of unmeasured anions and to determine the numerator of the delta ratio. AG = Na - (Cl + HCO3) = 140 - (92 + 14) = 34.
2
Calculate the change in anion gap (ΔAG).
ΔAG = 22 mEq/L
The delta ratio uses the change from baseline normal values. ΔAG = Measured AG - Normal AG = 34 - 12 = 22.
3
Calculate the change in bicarbonate (ΔHCO3).
ΔHCO3 = 10 mEq/L
We must determine how much bicarbonate was consumed by the acid load. ΔHCO3 = Normal HCO3 - Measured HCO3 = 24 - 14 = 10.
4
Calculate the final delta ratio.
Delta ratio = 2.2
Divide ΔAG by ΔHCO3. 22 / 10 = 2.2. A ratio > 2.0 indicates a concurrent metabolic alkalosis, consistent with this patient's history of recurrent vomiting.

Anahtar Kavram

The delta ratio (ΔAG / ΔHCO3-) assesses mixed acid-base disorders in the setting of a high anion gap metabolic acidosis. A ratio greater than 2.0 suggests a superimposed metabolic alkalosis, while a ratio less than 1.0 suggests a superimposed non-anion gap metabolic acidosis.
Tahmini Süre:2m 0s
Soru 67Soru →

A 62-year-old woman with a history of severe protein-calorie malnutrition and chronic diarrhea is admitted to the hospital with generalized weakness and lethargy. Vital signs are within normal limits. An arterial blood gas analysis reveals a pHpH of 7.317.31 and a PaCO2PaCO_2 of 32 mmHg32\text{ mmHg}.

Serum laboratory studies report the following:

Laboratory TestResultReference Range
Sodium (Na+Na^+)138 mEq/L138\text{ mEq/L}136−145 mEq/L136-145\text{ mEq/L}
Potassium (K+K^+)3.4 mEq/L3.4\text{ mEq/L}3.5−5.0 mEq/L3.5-5.0\text{ mEq/L}
Chloride (Cl−Cl^-)111 mEq/L111\text{ mEq/L}95−105 mEq/L95-105\text{ mEq/L}
Bicarbonate (HCO3−HCO_3^-)16 mEq/L16\text{ mEq/L}22−28 mEq/L22-28\text{ mEq/L}
Blood urea nitrogen32 mg/dL32\text{ mg/dL}7−18 mg/dL7-18\text{ mg/dL}
Creatinine1.4 mg/dL1.4\text{ mg/dL}0.6−1.2 mg/dL0.6-1.2\text{ mg/dL}
Albumin1.6 g/dL1.6\text{ g/dL}3.5−5.0 g/dL3.5-5.0\text{ g/dL}
Lactate3.8 mmol/L3.8\text{ mmol/L}0.5−1.0 mmol/L0.5-1.0\text{ mmol/L}

What is the patient's albumin-corrected serum anion gap in mEq/L\text{mEq/L}? (Assume a standard reference serum albumin concentration of 4.0 g/dL4.0\text{ g/dL}).

Cevabı ve açıklamayı göster

Cevap: 17

Cevap

The patient's albumin-corrected serum anion gap is 17 mEq/L.
The correct albumin-adjusted anion gap is 17 mEq/L17\text{ mEq/L}. Albumin is a major unmeasured negatively charged protein in the blood. When albumin levels drop significantly, as in severe malnutrition, the baseline 'normal' anion gap shrinks. A standard uncorrected calculation in this patient yields an anion gap of 11 mEq/L11\text{ mEq/L}, which falsely appears normal. Applying the standard correction factor (2.5 mEq/L2.5\text{ mEq/L} added for every 1.0 g/dL1.0\text{ g/dL} deficit of albumin below the norm of 4.0 g/dL4.0\text{ g/dL}) reveals a true anion gap of 17 mEq/L17\text{ mEq/L}. This appropriately unmasks an underlying high anion gap metabolic acidosis, which is highly consistent with the patient's elevated serum lactate of 3.8 mmol/L3.8\text{ mmol/L}.

Adım Adım Çözüm

1
Calculate the uncorrected serum anion gap.
11 mEq/L11\text{ mEq/L}
Using the formula Na+−(Cl−+HCO3−)Na^+ - (Cl^- + HCO_3^-), the calculation is 138−(111+16)=11138 - (111 + 16) = 11. This establishes the baseline apparent anion gap.
2
Determine the patient's albumin deficit.
2.4 g/dL2.4\text{ g/dL}
Subtracting the patient's measured albumin (1.6 g/dL1.6\text{ g/dL}) from the standard reference normal (4.0 g/dL4.0\text{ g/dL}) quantifies the lack of unmeasured anions (4.0−1.6=2.44.0 - 1.6 = 2.4).
3
Calculate the correction value for the anion gap.
6 mEq/L6\text{ mEq/L}
For every 1.0 g/dL1.0\text{ g/dL} decrease in serum albumin, the anion gap is falsely lowered by approximately 2.5 mEq/L2.5\text{ mEq/L}. Multiplying the deficit (2.42.4) by 2.52.5 yields 66.
4
Calculate the final albumin-corrected anion gap.
17 mEq/L17\text{ mEq/L}
Adding the correction value (66) to the uncorrected anion gap (1111) reveals the true anion gap (1717), uncovering an occult high anion gap metabolic acidosis.

Anahtar Kavram

In patients with significant hypoalbuminemia, the calculated serum anion gap must be mathematically corrected (adding 2.5 mEq/L2.5\text{ mEq/L} for every 1.0 g/dL1.0\text{ g/dL} decrease in albumin below 4.0 g/dL4.0\text{ g/dL}) to accurately interpret acid-base status and avoid missing a hidden high anion gap metabolic acidosis.

Alternatif Yöntem

An alternative and clinically common rough estimate is to simply expect the 'normal' anion gap threshold to decrease by 2.52.5 for every 1 g/dL1\text{ g/dL} drop in albumin. In this patient, the expected normal upper limit anion gap would drop from 1212 down to 66. Since the patient's uncorrected measured gap is 1111, it significantly exceeds the newly adjusted threshold of 66, confirming the presence of a high anion gap disorder.
Tahmini Süre:1m 30s
Soru 68Soru →

A 72-year-old woman is evaluated in the hospital for acute kidney injury. She was admitted 6 days ago for urosepsis and has been treated with intravenous ampicillin, gentamicin, and intravenous fluids. Over the past 24 hours, her urine output has decreased to 400 mL. Her blood pressure is 122/76 mm Hg, and heart rate is 84/min. She appears euvolemic, and her mucous membranes are moist. Laboratory studies show:

ParameterResult
Serum chemistry
Sodium140 mEq/L140 \text{ mEq/L}
Potassium4.5 mEq/L4.5 \text{ mEq/L}
Chloride105 mEq/L105 \text{ mEq/L}
Bicarbonate22 mEq/L22 \text{ mEq/L}
Blood urea nitrogen35 mg/dL35 \text{ mg/dL}
Creatinine2.5 mg/dL2.5 \text{ mg/dL}
Urine studies
Sodium56 mEq/L56 \text{ mEq/L}
Creatinine50 mg/dL50 \text{ mg/dL}

Based on these laboratory results, what is the patient's fractional excretion of sodium (FENa) in percent?

Cevabı ve açıklamayı göster

Cevap: 2

Cevap

The patient's fractional excretion of sodium (FENa) is 2%.
The correct calculated value is 2%. The fractional excretion of sodium (FENa) measures the proportion of sodium filtered by the glomeruli that is ultimately excreted in the urine. It is widely utilized to differentiate between prerenal and intrinsic causes of acute kidney injury (AKI). In prerenal AKI, the kidneys avidly reabsorb sodium to restore intravascular volume, leading to a FENa < 1%. Conversely, in intrinsic renal disease such as acute tubular necrosis (ATN), the damaged tubules lose their ability to concentrate urine and reabsorb sodium effectively, resulting in a FENa typically > 2%. In this vignette, the patient developed AKI after exposure to gentamicin, a classic nephrotoxic agent known for causing ATN. Her calculated FENa of 2% corroborates this intrinsic tubular injury mechanism.

Adım Adım Çözüm

1
Identify the standard formula for calculating the Fractional Excretion of Sodium (FENa).
FENa=Urine Na×Serum CrSerum Na×Urine Cr×100\text{FENa} = \frac{\text{Urine Na} \times \text{Serum Cr}}{\text{Serum Na} \times \text{Urine Cr}} \times 100
This equation calculates the percentage of filtered sodium that is eventually excreted in the urine, serving as a critical index to differentiate the etiology of acute kidney injury.
2
Extract the specific necessary laboratory values from the clinical vignette.
Urine Na=56 mEq/L\text{Urine Na} = 56 \text{ mEq/L}, Serum Cr=2.5 mg/dL\text{Serum Cr} = 2.5 \text{ mg/dL}, Serum Na=140 mEq/L\text{Serum Na} = 140 \text{ mEq/L}, and Urine Cr=50 mg/dL\text{Urine Cr} = 50 \text{ mg/dL}.
Accurate variable substitution is required to compute the index correctly.
3
Perform the mathematical calculation.
FENa=56×2.5140×50×100=1407000×100=0.02×100=2%\text{FENa} = \frac{56 \times 2.5}{140 \times 50} \times 100 = \frac{140}{7000} \times 100 = 0.02 \times 100 = 2\%
Solving the equation yields the final physiological percentage used for diagnostic reasoning.

Anahtar Kavram

Fractional Excretion of Sodium (FENa) calculation and interpretation for distinguishing intrinsic acute kidney injury from prerenal disease.
Soru 69Soru →

A 41-year-old man is brought to the emergency department by ambulance after being found obtunded in an alleyway. Medical records indicate a history of severe alcohol use disorder and major depressive disorder. Vital signs are temperature 36.1°C (97.0°F), blood pressure 105/65 mm Hg, pulse 112/min, and respirations 24/min. Physical examination shows a somnolent man who arouses only to painful stimuli. Pupils are equal and reactive. Cardiopulmonary examination is unremarkable.

Laboratory results are as follows:

TestResultReference Range
Sodium138 mEq/L136-145 mEq/L
Potassium4.8 mEq/L3.5-5.0 mEq/L
Chloride100 mEq/L95-105 mEq/L
Bicarbonate14 mEq/L22-28 mEq/L
Blood urea nitrogen14 mg/dL7-18 mg/dL
Creatinine1.3 mg/dL0.6-1.2 mg/dL
Glucose90 mg/dL70-110 mg/dL
Ethanol138 mg/dL0 mg/dL
Serum osmolality (measured)345 mOsm/kg H2O275-295 mOsm/kg H2O

Arterial blood gas analysis on room air reveals:
pH: 7.23
PaCO2: 28 mm Hg
PaO2: 96 mm Hg

Based on the laboratory results provided, calculate the patient's osmolar gap.

Cevabı ve açıklamayı göster

Cevap: 29

Cevap

29
The osmolar gap is the difference between the directly measured serum osmolality and the calculated serum osmolality. When evaluating a patient for toxic alcohol ingestion, any known ethanol must be included in the calculation to isolate the gap caused specifically by unmeasured toxins like methanol or ethylene glycol. The standard formula using US units (mg/dL) is: Calculated Osmolality = 2[Na+]+Glucose18+BUN2.8+Ethanol4.62[\text{Na}^+] + \frac{\text{Glucose}}{18} + \frac{\text{BUN}}{2.8} + \frac{\text{Ethanol}}{4.6}. Using the provided laboratory values, this equals 2(138)+9018+142.8+1384.6=276+5+5+30=316 mOsm/kg H2O2(138) + \frac{90}{18} + \frac{14}{2.8} + \frac{138}{4.6} = 276 + 5 + 5 + 30 = 316 \text{ mOsm/kg H}_2\text{O}. Subtracting this calculated value from the measured osmolality of 345 yields an osmolar gap of 29 mOsm/kg H2O29 \text{ mOsm/kg H}_2\text{O}. An osmolar gap greater than 10-15 is generally considered abnormal and highly suggestive of an unmeasured alcohol.

Adım Adım Çözüm

1
Determine the indication for calculating the osmolar gap.
The patient has an elevated anion gap (138−(100+14)=24138 - (100 + 14) = 24) and a history of substance abuse, raising suspicion for toxic alcohol ingestion.
An unexplained high anion gap metabolic acidosis requires calculation of the osmolar gap to identify hidden osmotically active substances.
2
Calculate the theoretical serum osmolality using the standard formula.
2(138)+9018+142.8+1384.6=276+5+5+30=316 mOsm/kg H2O2(138) + \frac{90}{18} + \frac{14}{2.8} + \frac{138}{4.6} = 276 + 5 + 5 + 30 = 316 \text{ mOsm/kg H}_2\text{O}.
This formula accounts for the primary solutes in the blood, including ethanol, which is known to be present in this patient.
3
Determine the osmolar gap.
Measured osmolality (345) - Calculated osmolality (316) = 29 mOsm/kg H2O29 \text{ mOsm/kg H}_2\text{O}.
The difference between measured and calculated osmolality reveals the concentration of unmeasured osmotically active agents.

Anahtar Kavram

Calculation and interpretation of the osmolar gap in the presence of ethanol.
Soru 70Soru →

A 42-year-old man with known nephrotic syndrome is brought to the emergency department due to altered mental status and fever. His family reports he developed a spreading, painful, erythematous rash on his right lower extremity over the past three days. Vital signs reveal a temperature of 39.1∘C39.1^\circ\text{C} (102.4∘F102.4^\circ\text{F}), heart rate of 118/min118/\text{min}, blood pressure of 88/54 mmHg88/54\text{ mmHg}, and respiratory rate of 26/min26/\text{min}.

Initial laboratory studies show:

TestResult
Sodium (Na+Na^+)140 mEq/L140\text{ mEq/L}
Potassium (K+K^+)4.2 mEq/L4.2\text{ mEq/L}
Chloride (Cl−Cl^-)108 mEq/L108\text{ mEq/L}
Bicarbonate (HCO3−HCO_3^-)16 mEq/L16\text{ mEq/L}
Blood urea nitrogen45 mg/dL45\text{ mg/dL}
Creatinine2.1 mg/dL2.1\text{ mg/dL}
Albumin1.2 g/dL1.2\text{ g/dL}
Lactate4.5 mmol/L4.5\text{ mmol/L}

What is the patient's albumin-corrected serum anion gap in mEq/L? (Assume a normal serum albumin concentration of 4.0 g/dL4.0\text{ g/dL}).

Cevabı ve açıklamayı göster

Cevap: 23

Cevap

23
The patient has a profound high anion gap metabolic acidosis secondary to septic shock and lactic acidosis. However, his underlying nephrotic syndrome has caused severe hypoalbuminemia (1.2 g/dL). Since albumin normally accounts for a large portion of the unmeasured anions in the blood, its depletion artificially lowers the calculated anion gap. The measured gap is 140 - (108 + 16) = 16 mEq/L. To correct for hypoalbuminemia, the deficit (4.0 - 1.2 = 2.8 g/dL) is multiplied by 2.5 mEq/L to get a correction factor of 7 mEq/L. Adding this back to the measured gap (16 + 7) yields a true, corrected anion gap of 23 mEq/L.

Adım Adım Çözüm

1
Calculate the uncorrected, measured serum anion gap.
16 mEq/L
The baseline anion gap evaluates the difference between primary measured cations and anions using the standard formula: Na - (Cl + HCO3).
2
Calculate the patient's albumin deficit.
2.8 g/dL
Because albumin is the primary unmeasured circulating anion, significant deficits must be quantified against the normal baseline of 4.0 g/dL.
3
Calculate the adjustment factor for the anion gap.
7 mEq/L
For every 1 g/dL decrease in serum albumin below 4.0 g/dL, the anion gap is falsely lowered by 2.5 mEq/L. Multiplying the 2.8 deficit by 2.5 yields the necessary correction.
4
Apply the correction factor to the measured anion gap.
23 mEq/L
Adding the 7 mEq/L correction to the 16 mEq/L measured gap reveals the true physiological anion gap.

Anahtar Kavram

Hypoalbuminemia artificially lowers the measured anion gap because albumin acts as a significant unmeasured anion. To correctly interpret acid-base status, the measured gap must be corrected by adding 2.5 mEq/L for every 1 g/dL decrease in serum albumin below the normal level of 4.0 g/dL.
Soru 71Soru →

A tertiary care transplant center is validating a new rapid biomarker assay to detect an invasive opportunistic fungal infection in post-operative patients. According to internal hospital data, the baseline prevalence of this fungal infection among their solid organ transplant recipients is 25%25\%. During the validation phase, the rapid assay correctly identified 90%90\% of patients who had biopsy-proven fungal tissue invasion. The assay also correctly returned a negative result in 90%90\% of transplant recipients who were proven to be free of the infection. A 58-year-old man who recently received a renal allograft is tested as part of a routine post-operative monitoring protocol, and his biomarker assay returns a positive result. Based on the test's performance characteristics in this population, what is the probability (expressed as a percentage) that he actually has the invasive fungal infection?

Cevabı ve açıklamayı göster

Cevap: 75

Cevap

75
The patient's probability of actually having the infection given a positive test result is defined as the Positive Predictive Value (PPV). To calculate PPV, we can construct a 2×22 \times 2 contingency table using a hypothetical cohort of 1,0001,000 solid organ transplant recipients. Based on the 25%25\% baseline prevalence, there are 250250 patients with the disease and 750750 without it. The assay's 90%90\% sensitivity correctly identifies 225225 true positive cases from the diseased group (250×0.90250 \times 0.90). The assay's 90%90\% specificity correctly identifies 675675 true negative cases, which leaves 7575 false positive cases from the disease-free group (750×0.10750 \times 0.10). The PPV is the ratio of true positives to the total number of positive results. Therefore, PPV=225225+75=225300=0.75\text{PPV} = \frac{225}{225 + 75} = \frac{225}{300} = 0.75, which equals 75%75\%.

Adım Adım Çözüm

1
Extract the provided statistical parameters from the clinical vignette
Prevalence = 25%25\%, Sensitivity = 90%90\%, Specificity = 90%90\%
These are the core variables required to calculate the positive predictive value (PPV) of the diagnostic assay.
2
Set up a hypothetical population cohort
Assume a population size of 1,0001,000 patients
Using a hypothetical round number population simplifies the calculation of absolute numbers for a 2×22 \times 2 contingency table.
3
Calculate the total number of patients with and without the infection
Infected = 250250 (1,000×0.251,000 \times 0.25); Uninfected = 750750 (1,000×0.751,000 \times 0.75)
The baseline prevalence splits the total population into true positive and true negative base groups.
4
Calculate True Positives (TP) and False Positives (FP)
TP=250×0.90=225\text{TP} = 250 \times 0.90 = 225; FP=750×(1−0.90)=75\text{FP} = 750 \times (1 - 0.90) = 75
TP represents infected patients correctly identified by sensitivity. FP represents uninfected patients incorrectly identified as positive based on the false positive rate (which is 1−Specificity1 - \text{Specificity}).
5
Calculate the Positive Predictive Value (PPV)
PPV=225225+75=225300=0.75\text{PPV} = \frac{225}{225 + 75} = \frac{225}{300} = 0.75
PPV is the proportion of true positives out of all positive test results, which answers the patient's probability of actually having the disease given a positive result.
6
Convert the resulting PPV probability into a percentage
0.75×100=75%0.75 \times 100 = 75\%
The question specifically requests the probability expressed as a percentage.

Anahtar Kavram

The positive predictive value (PPV) of a test depends heavily on the prevalence of the disease in the tested population, in addition to the test's inherent sensitivity and specificity.
Soru 72Soru →

An international research consortium recently published the findings of a large-scale, phase III, double-blind trial. The study assessed a novel oral small-molecule immunosuppressant against a placebo for maintaining clinical remission in adults with severe ulcerative colitis.

A total of 6,0006,000 participants were enrolled and randomly assigned in a 1:11:1 ratio to receive either the novel drug or the placebo. All participants were followed for a 5252-week maintenance phase. By the end of the study, clinical disease relapse occurred in 435435 patients in the placebo group and 312312 patients in the intervention group.

Based on these trial results, what is the number needed to treat (NNT) to prevent one disease relapse over the 5252-week period? (Calculate the exact value and apply the standard biostatistical rounding rule for NNT).

Cevabı ve açıklamayı göster

Cevap: 25

Cevap

25
The correct answer is derived by calculating the absolute risk reduction (4.1%4.1\%) and finding its reciprocal to get the number needed to treat (24.3924.39). Following standard epidemiological guidelines, NNT values are always rounded up to the next highest integer, yielding 2525.

Adım Adım Çözüm

1
Determine the number of patients in the control and experimental groups.
Control group = 3,0003,000 patients; Experimental group = 3,0003,000 patients.
To calculate event rates, we need the denominator for each group. The total is 6,0006,000 randomized in a 1:11:1 ratio.
2
Calculate the Control Event Rate (CER).
CER=4353000=0.145\text{CER} = \frac{435}{3000} = 0.145
CER is the incidence of the outcome in the placebo group.
3
Calculate the Experimental Event Rate (EER).
EER=3123000=0.104\text{EER} = \frac{312}{3000} = 0.104
EER is the incidence of the outcome in the intervention group.
4
Calculate the Absolute Risk Reduction (ARR).
ARR=CER−EER=0.145−0.104=0.041\text{ARR} = \text{CER} - \text{EER} = 0.145 - 0.104 = 0.041
ARR represents the absolute difference in risk between the two groups.
5
Calculate the Number Needed to Treat (NNT) and apply the correct clinical rounding rule.
NNT=10.041≈24.39\text{NNT} = \frac{1}{0.041} \approx 24.39
Rounded up to the next whole number: 2525
NNT is calculated as the inverse of ARR. Because one cannot treat a fraction of a patient, NNT is always rounded UP to the next whole number to guarantee the prevention of at least one outcome.

Anahtar Kavram

Calculation of Absolute Risk Reduction (ARR) and Number Needed to Treat (NNT) from raw event data, followed by the application of the conservative NNT rounding rule (always rounding up).
Soru 73Soru →

A randomized, double-blind, placebo-controlled trial is conducted to evaluate the efficacy of a novel glucagon-like peptide-1 (GLP-1) receptor agonist for the secondary prevention of major adverse cardiovascular events (MACE). The study enrolls patients with type 2 diabetes mellitus and established cardiovascular disease. Participants are randomized in a 1:1 ratio to receive either the GLP-1 receptor agonist or a placebo, in addition to standard of care.

Over a 3-year follow-up period, the investigators report the following outcomes:

GroupTotal Patients (nn)MACE Occurred
Placebo20002000260260
GLP-1 Receptor Agonist20002000160160

Based on the study results, what is the number needed to treat (NNT) with the novel GLP-1 receptor agonist to prevent one major adverse cardiovascular event over the 3-year follow-up period?

Cevabı ve açıklamayı göster

Cevap: 20

Cevap

20
The number needed to treat (NNT) is an epidemiological measure used to evaluate the clinical impact of an intervention. It is calculated as the inverse of the Absolute Risk Reduction (ARR), using the formula NNT=1ARRNNT = \frac{1}{ARR}. First, determine the event rates for both groups. The Control Event Rate (CER) is 2602000=0.13\frac{260}{2000} = 0.13 (13%13\%), and the Experimental Event Rate (EER) is 1602000=0.08\frac{160}{2000} = 0.08 (8%8\%). The ARR is the absolute difference between these two rates: 0.13−0.08=0.050.13 - 0.08 = 0.05 (5%5\%). The NNT is therefore 10.05=20\frac{1}{0.05} = 20. This indicates that 2020 patients must be treated with the novel GLP-1 receptor agonist over a 3-year period to prevent one major adverse cardiovascular event.

Adım Adım Çözüm

1
Calculate the Control Event Rate (CER).
CER=2602000=0.13CER = \frac{260}{2000} = 0.13 (13%13\%)
To determine the baseline risk of the adverse event in the untreated (placebo) population.
2
Calculate the Experimental Event Rate (EER).
EER=1602000=0.08EER = \frac{160}{2000} = 0.08 (8%8\%)
To determine the risk of the adverse event in the population receiving the intervention.
3
Calculate the Absolute Risk Reduction (ARR).
ARR=CER−EER=0.13−0.08=0.05ARR = CER - EER = 0.13 - 0.08 = 0.05 (5%5\%)
To find the absolute difference in event risk between the control and intervention groups.
4
Calculate the Number Needed to Treat (NNT).
NNT=1ARR=10.05=20NNT = \frac{1}{ARR} = \frac{1}{0.05} = 20
To determine how many patients need to receive the intervention to prevent one additional adverse event.

Anahtar Kavram

The Number Needed to Treat (NNT) is the inverse of the Absolute Risk Reduction (ARR) and represents the number of patients who must receive a specific intervention over a defined time period to prevent one additional adverse outcome.
Soru 74Soru →

A clinical research organization is finalizing the methodology for a randomized, placebo-controlled trial evaluating a novel pharmacologic agent for generalized anxiety disorder. The investigators stipulate that they are willing to accept a 5% risk of concluding the agent is effective when it is truly no different from placebo. Furthermore, formal sample size calculations reveal a 0.14 probability of concluding that the drug has no effect when it actually provides a clinically meaningful benefit to patients.

Based on these parameters, what is the planned statistical power of this trial, expressed as a percentage?

Cevabı ve açıklamayı göster

Cevap: 86

Cevap

The planned statistical power of the trial is 86%.
Statistical power is defined mathematically as 1−β1 - \beta, where β\beta is the probability of committing a Type II error (failing to detect a true difference). In this scenario, the probability of concluding there is no effect when the drug actually provides a benefit is given as 0.14. Therefore, the statistical power is 1−0.14=0.861 - 0.14 = 0.86, which equates to 86%. The 5% risk mentioned in the stem represents the Type I error rate (α\alpha), which is used to set the significance threshold but is not directly subtracted to calculate power.

Adım Adım Çözüm

1
Identify the Type II error rate (β\beta) from the scenario.
β=0.14\beta = 0.14
The scenario defines a 0.14 probability of concluding there is no effect when the drug actually works, which is the definition of a Type II error.
2
Apply the formula for statistical power.
Power=1−β\text{Power} = 1 - \beta
Power and Type II error are complementary probabilities under the alternative hypothesis.
3
Calculate the final value.
1−0.14=0.861 - 0.14 = 0.86, or 86%86\%
To find the percentage, the proportion is multiplied by 100.

Anahtar Kavram

Statistical power is the probability of correctly rejecting the null hypothesis when the alternative hypothesis is true, calculated as 1−β1 - \beta (where β\beta is the Type II error rate).
Tahmini Süre:1m 0s
Soru 75Soru →

An elderly man is admitted due to a two-week history of worsening dyspnea, orthopnea, and increasing lower extremity edema. His past medical history is significant for ischemic cardiomyopathy with a left ventricular ejection fraction of 30%30\%. To manage his fluid overload at home, he recently doubled his daily dose of oral torsemide independently.

Upon admission, his blood pressure is 92/58 mmHg92/58\text{ mmHg}, pulse is 108/min108/\text{min}, and respiratory rate is 22/min22/\text{min}. Jugular venous pressure is visibly elevated. Lung auscultation reveals bilateral basilar crackles, and a third heart sound (S3) is present on cardiac exam.

Laboratory evaluation yields the following results:
Serum chemistry:
Sodium: 132 mEq/L132\text{ mEq/L}
Potassium: 3.1 mEq/L3.1\text{ mEq/L}
Chloride: 94 mEq/L94\text{ mEq/L}
Bicarbonate: 29 mEq/L29\text{ mEq/L}
Blood urea nitrogen (BUN): 75 mg/dL75\text{ mg/dL}
Creatinine: 2.5 mg/dL2.5\text{ mg/dL} (baseline 1.2 mg/dL1.2\text{ mg/dL})

Spot urine chemistry:
Sodium: 66 mEq/L66\text{ mEq/L}
Urea nitrogen: 450 mg/dL450\text{ mg/dL}
Creatinine: 60 mg/dL60\text{ mg/dL}

To evaluate the etiology of the acute kidney injury in the setting of recent loop diuretic use, the medical team decides to calculate a specific fractional excretion index that is less confounded by his medication regimen than sodium excretion.

What is the calculated value of this preferred index, expressed as a percentage?

Cevabı ve açıklamayı göster

Cevap: 25

Cevap

25
The fractional excretion of urea (FEUrea) is the most accurate index to determine the etiology of acute kidney injury in patients actively receiving diuretic therapy. Diuretics increase distal sodium delivery and urinary sodium excretion, often rendering the FENa >1%> 1\% even in severe prerenal states (like heart failure exacerbations). Urea reabsorption occurs primarily in the proximal tubule and is less affected by loop diuretics. The FEUrea is calculated as (Urine Urea Nitrogen ×\times Serum Creatinine) / (BUN ×\times Urine Creatinine) ×\times 100. Plugging in the patient's values: (450×2.5)/(75×60)×100=25%(450 \times 2.5) / (75 \times 60) \times 100 = 25\%. A value <35%< 35\% is strongly indicative of prerenal azotemia, consistent with this patient's cardiorenal syndrome.

Adım Adım Çözüm

1
Identify the clinical problem and the limitation of standard diagnostic indices.
The patient has AKI and is on a loop diuretic, making the Fractional Excretion of Sodium (FENa) unreliable.
Loop diuretics block sodium reabsorption in the loop of Henle, causing natriuresis even when effective arterial blood volume is low, which can falsely elevate the FENa above 1%1\% in prerenal states.
2
Select the appropriate alternative diagnostic index.
The Fractional Excretion of Urea (FEUrea) should be calculated.
Urea handling is linked to proximal tubular fluid reabsorption and is relatively unaffected by loop diuretics, making it a reliable marker for prerenal azotemia in this setting.
3
Recall the formula for FEUrea.
FEUrea=(UUrea×PCrBUN×UCr)×100FE_{Urea} = \left( \frac{U_{Urea} \times P_{Cr}}{BUN \times U_{Cr}} \right) \times 100
This formula normalizes urea clearance to creatinine clearance to determine the fraction of filtered urea that is ultimately excreted.
4
Substitute the patient's laboratory values into the formula.
Numerator: 450×2.5=1125450 \times 2.5 = 1125. Denominator: 75×60=450075 \times 60 = 4500. Ratio: 1125/4500=0.251125 / 4500 = 0.25. As a percentage: 0.25×100=250.25 \times 100 = 25.
Using consistent units (mg/dL) allows the units to cancel out, providing an accurate percentage. A value less than 35%35\% confirms prerenal azotemia.

Anahtar Kavram

In patients with acute kidney injury who are receiving diuretic therapy, the fractional excretion of urea (FEUrea) is preferred over the fractional excretion of sodium (FENa) to distinguish prerenal azotemia from intrinsic acute tubular necrosis.
Soru 76Soru →

A 3636-year-old man is brought to the emergency department by emergency medical services after being found obtunded in his garage. Beside him were several unlabeled containers of antifreeze and an empty bottle of vodka. 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 22/min22/\text{min}. The patient is responsive only to deep pain.

Laboratory evaluation yields the following results:

TestResult
Serum sodium140 mEq/L140\text{ mEq/L}
Serum potassium4.0 mEq/L4.0\text{ mEq/L}
Serum chloride102 mEq/L102\text{ mEq/L}
Serum bicarbonate14 mEq/L14\text{ mEq/L}
Blood urea nitrogen28 mg/dL28\text{ mg/dL}
Serum creatinine1.4 mg/dL1.4\text{ mg/dL}
Serum glucose108 mg/dL108\text{ mg/dL}
Blood ethanol138 mg/dL138\text{ mg/dL}
Measured serum osmolality358 mOsm/kg H2O358\text{ mOsm/kg H}_2\text{O}
Arterial blood gas pH7.247.24
Arterial PCO2P\text{CO}_232 mmHg32\text{ mmHg}

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

(Enter your answer as a whole number).

Cevabı ve açıklamayı göster

Cevap: 32

Cevap

32
The osmolar gap represents the difference between directly measured serum osmolality and the calculated serum osmolality. A normal gap is ≤10 mOsm/kg H2O\le 10\text{ mOsm/kg H}_2\text{O}. An elevated gap in the setting of a high anion gap metabolic acidosis strongly suggests intoxication with an unmeasured osmotically active substance, such as methanol or ethylene glycol. When calculating the expected osmolality, one must include measured ethanol if it is present, using the formula: (2×[Na+])+([Glucose]/18)+([BUN]/2.8)+([Ethanol]/4.6)(2 \times [Na^+]) + ([Glucose]/18) + ([BUN]/2.8) + ([Ethanol]/4.6). Plugging in the patient's values yields: 280+6+10+30=326 mOsm/kg H2O280 + 6 + 10 + 30 = 326\text{ mOsm/kg H}_2\text{O}. Subtracting this from the measured osmolality of 358 mOsm/kg H2O358\text{ mOsm/kg H}_2\text{O} gives an osmolar gap of 32 mOsm/kg H2O32\text{ mOsm/kg H}_2\text{O}, indicating the presence of another toxic alcohol (likely ethylene glycol from antifreeze).

Adım Adım Çözüm

1
Set up the equation for calculated serum osmolality.
Calculated Osmolality=(2×[Na+])+[Glucose]18+[BUN]2.8+[Ethanol]4.6Calculated\ Osmolality = (2 \times [Na^+]) + \frac{[Glucose]}{18} + \frac{[BUN]}{2.8} + \frac{[Ethanol]}{4.6}
Serum osmolality is primarily determined by sodium and its accompanying anions, along with glucose and blood urea nitrogen. In cases of alcohol ingestion, ethanol significantly contributes and must be accounted for to find the true baseline gap.
2
Substitute the patient's laboratory values into the osmolality equation.
Calculated Osmolality=(2×140)+10818+282.8+1384.6Calculated\ Osmolality = (2 \times 140) + \frac{108}{18} + \frac{28}{2.8} + \frac{138}{4.6}
Applying the specific measured variables allows for the conversion of mg/dL into mmol/L (mOsm/kg).
3
Calculate the specific contributions of each component and sum them.
280+6+10+30=326 mOsm/kg H2O280 + 6 + 10 + 30 = 326\text{ mOsm/kg H}_2\text{O}
This yields the theoretical serum osmolality based solely on standard measured solutes plus the co-ingested ethanol.
4
Calculate the osmolar gap by subtracting the calculated osmolality from the measured osmolality.
358−326=32 mOsm/kg H2O358 - 326 = 32\text{ mOsm/kg H}_2\text{O}
The difference reveals the presence of unmeasured osmotically active substances (in this vignette, likely ethylene glycol from the antifreeze), confirming a toxic alcohol ingestion.

Anahtar Kavram

The osmolar gap is calculated by subtracting the calculated serum osmolality (which must incorporate ethanol if present) from the directly measured serum osmolality to identify unmeasured toxic alcohols.
Soru 77Soru →

A 22-year-old man with a history of type 1 diabetes mellitus is brought to the emergency department due to altered mental status, nausea, and diffuse abdominal pain. He has not taken his insulin for the past two days. Vital signs show a heart rate of 118/min118/\text{min}, blood pressure of 105/65 mmHg105/65\text{ mmHg}, and respiratory rate of 28/min28/\text{min} with deep, rapid breaths.

Laboratory results are as follows:

Serum sodium: 142 mEq/L142\text{ mEq/L}
Serum potassium: 4.8 mEq/L4.8\text{ mEq/L}
Serum chloride: 98 mEq/L98\text{ mEq/L}
Serum bicarbonate: 14 mEq/L14\text{ mEq/L}
Blood urea nitrogen: 32 mg/dL32\text{ mg/dL}
Serum creatinine: 1.4 mg/dL1.4\text{ mg/dL}
Serum glucose: 520 mg/dL520\text{ mg/dL}

Arterial blood gas (room air):
pH: 7.217.21
pCO2pCO_2: 24 mmHg24\text{ mmHg}
pO2pO_2: 98 mmHg98\text{ mmHg}

Using Winter's formula, what is the mean expected pCO2pCO_2 (in mmHg) to determine if this patient has an appropriately compensated primary metabolic acidosis?

Cevabı ve açıklamayı göster

Cevap: 29

Cevap

The mean expected pCO2 is 29 mmHg.
In patients with a primary metabolic acidosis, the physiological response is hyperventilation to reduce pCO2pCO_2 and drive the pH back toward normal. Winter's formula predicts this expected respiratory compensation. The formula is: Expected pCO2=(1.5×[HCO3−])+8±2pCO_2 = (1.5 \times [\text{HCO}_3^-]) + 8 \pm 2. Using the patient's serum bicarbonate of 14 mEq/L14\text{ mEq/L}, the mean expected pCO2pCO_2 is (1.5×14)+8=29 mmHg(1.5 \times 14) + 8 = 29\text{ mmHg}. Because the patient's actual measured pCO2pCO_2 is 24 mmHg24\text{ mmHg} (which is lower than the expected range of 27−31 mmHg27-31\text{ mmHg}), he has a concomitant primary respiratory alkalosis, likely secondary to profound anxiety, pain, or early sepsis.

Adım Adım Çözüm

1
Analyze the arterial blood gas to identify the primary acid-base disorder.
The patient has acidemia (pH 7.21) driven by a low serum bicarbonate (14 mEq/L14\text{ mEq/L}), confirming a primary metabolic acidosis.
Establishing the primary disturbance dictates which compensation formula should be applied.
2
Calculate the expected respiratory compensation.
Apply Winter's formula to find the expected pCO2pCO_2 level.
The lungs compensate for metabolic acidosis by increasing ventilation to blow off CO2. Winter's formula determines if this response is appropriate or if a secondary respiratory disorder exists.
3
Perform the mathematical calculation using the patient's bicarbonate level.
Expected pCO2=1.5×14+8=21+8=29 mmHgpCO_2 = 1.5 \times 14 + 8 = 21 + 8 = 29\text{ mmHg}.
Plugging the serum bicarbonate into the formula yields the mean expected pCO2pCO_2.

Anahtar Kavram

In the setting of a primary metabolic acidosis, Winter's formula (pCO2=1.5×[HCO3−]+8±2pCO_2 = 1.5 \times [\text{HCO}_3^-] + 8 \pm 2) must be used to calculate the expected respiratory compensation and identify concomitant mixed acid-base disorders.
Tahmini Süre:1m 0s
Soru 78Soru →

A 27-year-old man with a history of type 1 diabetes mellitus is brought to the emergency department due to altered mental status. His roommate reports that the patient has been suffering from a severe gastrointestinal illness for the past three days, characterized by profuse, intractable vomiting, and has been unable to tolerate food, fluids, or his scheduled insulin doses. On physical examination, the patient is lethargic and tachycardic, with deep, labored respirations. His mucous membranes are markedly parched, and skin turgor is poor.

Laboratory results are as follows:

Laboratory TestResultReference Range
Serum sodium144 mEq/L144 \text{ mEq/L}136−145 mEq/L136 - 145 \text{ mEq/L}
Serum potassium3.4 mEq/L3.4 \text{ mEq/L}3.5−5.0 mEq/L3.5 - 5.0 \text{ mEq/L}
Serum chloride82 mEq/L82 \text{ mEq/L}95−105 mEq/L95 - 105 \text{ mEq/L}
Serum bicarbonate16 mEq/L16 \text{ mEq/L}22−28 mEq/L22 - 28 \text{ mEq/L}
Blood urea nitrogen45 mg/dL45 \text{ mg/dL}7−18 mg/dL7 - 18 \text{ mg/dL}
Serum creatinine2.1 mg/dL2.1 \text{ mg/dL}0.6−1.2 mg/dL0.6 - 1.2 \text{ mg/dL}
Serum glucose680 mg/dL680 \text{ mg/dL}70−110 mg/dL70 - 110 \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 calculated delta ratio (ΔAG/ΔHCO3−\Delta AG / \Delta HCO_3^-)?

Cevabı ve açıklamayı göster

Cevap: 4.25

Cevap

4.25
The patient presents with diabetic ketoacidosis (DKA) complicated by severe vomiting. This clinical scenario typically creates a mixed acid-base disorder: a high anion gap metabolic acidosis (HAGMA) from ketoacid accumulation, and a concurrent metabolic alkalosis from the loss of gastric hydrochloric acid (HCl) due to vomiting. To detect hidden metabolic disorders in the presence of a HAGMA, the delta ratio (ΔAG/ΔHCO3−\Delta AG / \Delta HCO_3^-) is calculated. The ratio assesses the stoichiometric relationship between the rise in the anion gap and the fall in bicarbonate. A pure HAGMA typically yields a ratio between 1.0 and 2.0 (often closer to 1.0 in DKA due to urinary ketone loss). A ratio greater than 2.0 indicates that the bicarbonate level is higher than expected for the degree of acidosis, signifying a concurrent primary metabolic alkalosis. In this patient, the ratio is 4.25, strongly supporting the diagnosis of a mixed HAGMA (from DKA) and metabolic alkalosis (from vomiting).

Adım Adım Çözüm

1
Calculate the serum anion gap (AG).
144−(82+16)=46 mEq/L144 - (82 + 16) = 46 \text{ mEq/L}
The anion gap is necessary to identify the presence and magnitude of a high anion gap metabolic acidosis (HAGMA).
2
Calculate the delta anion gap (ΔAG\Delta AG).
46−12=34 mEq/L46 - 12 = 34 \text{ mEq/L}
The ΔAG\Delta AG represents the excess unmeasured anions contributing to the acidosis above the baseline normal value.
3
Calculate the delta bicarbonate (ΔHCO3−\Delta HCO_3^-).
24−16=8 mEq/L24 - 16 = 8 \text{ mEq/L}
The ΔHCO3−\Delta HCO_3^- represents the amount of bicarbonate that has been consumed to buffer the acid.
4
Divide the ΔAG\Delta AG by the ΔHCO3−\Delta HCO_3^- to determine the delta ratio.
34/8=4.2534 / 8 = 4.25
The delta ratio reveals hidden concurrent metabolic acid-base disorders in the presence of a HAGMA.

Anahtar Kavram

Delta ratio calculation and interpretation in mixed acid-base disorders
Soru 79Soru →

A 68-year-old woman is evaluated in the hospital for decreased urine output. She was admitted 2 days ago for severe gastroenteritis. Her past medical history is significant for hypertension treated with amlodipine. Her blood pressure is 98/60 mm Hg, and her pulse is 112/min. Mucous membranes are dry. Laboratory studies show the following:

TestResult
Serum sodium140 mEq/L
Serum potassium3.8 mEq/L
Serum chloride102 mEq/L
Blood urea nitrogen42 mg/dL
Serum creatinine2.0 mg/dL
Urine sodium14 mEq/L
Urine creatinine40 mg/dL

Based on the laboratory results provided, what is the patient's fractional excretion of sodium (FENa)? Give your answer as a percentage.

Cevabı ve açıklamayı göster

Cevap: 0.5

Cevap

0.5
The fractional excretion of sodium (FENa) evaluates the percentage of sodium filtered by the glomeruli that is eventually excreted in the urine. It is calculated using the formula: FENa=UNa×PCrPNa×UCr×100FENa = \frac{U_{Na} \times P_{Cr}}{P_{Na} \times U_{Cr}} \times 100. Plugging the patient's laboratory values into this formula yields 14×2.0140×40×100=285600×100=0.5%\frac{14 \times 2.0}{140 \times 40} \times 100 = \frac{28}{5600} \times 100 = 0.5\%. A FENa of less than 1% indicates a prerenal etiology for the patient's acute kidney injury, which is consistent with her history of severe gastroenteritis and clinical signs of profound volume depletion.

Adım Adım Çözüm

1
Identify the standard formula for fractional excretion of sodium.
FENa=UNa×PCrPNa×UCr×100FENa = \frac{U_{Na} \times P_{Cr}}{P_{Na} \times U_{Cr}} \times 100
This formula helps differentiate prerenal azotemia from intrinsic acute kidney injury by comparing sodium clearance to creatinine clearance.
2
Extract the specific serum and urine values from the clinical vignette.
UNa=14U_{Na} = 14, PNa=140P_{Na} = 140, UCr=40U_{Cr} = 40, PCr=2.0P_{Cr} = 2.0
Accurate identification of the variables is necessary for the subsequent mathematical calculation.
3
Calculate the ratio and multiply by 100 to find the percentage.
14×2.0140×40×100=285600×100=0.5%\frac{14 \times 2.0}{140 \times 40} \times 100 = \frac{28}{5600} \times 100 = 0.5\%
Completing the mathematical operation yields the final FENa value.

Anahtar Kavram

Calculating the fractional excretion of sodium (FENa) to determine the etiology of acute kidney injury.
Tahmini Süre:1m 0s
Soru 80Soru →

A 55-year-old man is evaluated in the emergency department for progressive abdominal distension and bilateral lower extremity edema over the past 3 weeks. He has a history of chronic hepatitis C infection and alcohol use disorder. On physical examination, his vital signs are stable. The abdomen is distended with prominent shifting dullness and a positive fluid wave. A diagnostic paracentesis is performed, yielding clear, yellow fluid.

Concurrent laboratory studies show:

Serum:
Albumin: 3.1 g/dL
Total protein: 6.5 g/dL
Total bilirubin: 2.5 mg/dL

Ascitic fluid:
Albumin: 1.3 g/dL
Total protein: 2.2 g/dL
Leukocyte count: 210/mm³ (10% neutrophils, 90% mononuclear cells)

To help determine the etiology of the patient's ascites, the physician calculates the serum-ascites albumin gradient (SAAG). What is the patient's SAAG in g/dL?

Cevabı ve açıklamayı göster

Cevap: 1.8

Cevap

1.8
The patient's SAAG is 1.8 g/dL. SAAG is correctly calculated by subtracting the ascitic fluid albumin concentration (1.3 g/dL) from the serum albumin concentration (3.1 g/dL). A SAAG of 1.1 g/dL or greater indicates that the ascites is driven by portal hypertension (such as from hepatic cirrhosis), which aligns with this patient's history of hepatitis C and alcohol use.

Adım Adım Çözüm

1
Identify the formula for calculating SAAG.
SAAG = Serum Albumin - Ascitic Fluid Albumin
The SAAG formula is required to classify the ascites as related to portal hypertension or another etiology.
2
Locate the patient's serum albumin and ascitic fluid albumin levels from the laboratory data.
Serum albumin is 3.1 g/dL, and ascitic fluid albumin is 1.3 g/dL.
Accurate identification of the specific albumin values is necessary, avoiding confusion with total protein levels.
3
Subtract the ascitic fluid albumin from the serum albumin.
3.1 - 1.3 = 1.8
Executing the subtraction provides the final SAAG value.

Anahtar Kavram

The serum-ascites albumin gradient (SAAG) is calculated by subtracting the ascitic fluid albumin from the serum albumin, with a value ≥ 1.1 g/dL indicating portal hypertension.
ÖncekiSayfa 4 / 7Sonraki
Tüm alıştırma soruları — USMLE Step 3 | Examkin