Tüm alıştırma soruları

160 soru

Soru 61Soru →

An 82-year-old man with advanced Alzheimer dementia is brought to the emergency department from a skilled nursing facility due to progressive lethargy over the past 3 days. The facility staff report that he has been refusing meals and liquids. On examination, temperature is 37.1°C (98.8°F), blood pressure is 98/60 mm Hg, pulse is 112/min, and respiratory rate is 16/min. The patient appears cachectic with profoundly dry mucous membranes and poor skin turgor. His current weight is 60 kg (132 lb).

Laboratory studies show:
Serum sodium: 168 mEq/L
Serum potassium: 4.1 mEq/L
Serum chloride: 129 mEq/L
Serum bicarbonate: 24 mEq/L
Blood urea nitrogen: 56 mg/dL
Serum creatinine: 1.8 mg/dL

Based on an estimated total body water of 50% of his body weight, what is this patient's calculated free water deficit in liters?

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Cevap: 6

Cevap

6
The free water deficit is calculated using the formula: Water Deficit=TBW×(Actual Serum Na+140−1)Water\ Deficit = TBW \times \left(\frac{Actual\ Serum\ Na^+}{140} - 1\right). Given a weight of 6060 kg and an estimated TBW fraction of 50%50\%, the TBW is 3030 L. The actual serum sodium is 168168 mEq/L. Substituting these values into the formula yields: 30×(168140−1)=30×(1.2−1)=30×0.2=630 \times \left(\frac{168}{140} - 1\right) = 30 \times (1.2 - 1) = 30 \times 0.2 = 6 liters. This represents the volume of solute-free water the patient has lost to develop this degree of hypernatremia.

Adım Adım Çözüm

1
Calculate the patient's Total Body Water (TBW).
TBW=60×0.50=30TBW = 60 \times 0.50 = 30 L
Free water deficit calculations require an accurate estimate of the patient's baseline TBW, which is typically 50% of lean body weight for elderly men.
2
Set up the free water deficit formula.
Deficit=TBW×(Actual [Na+]Target [Na+]−1)Deficit = TBW \times \left(\frac{Actual\ [Na^+]}{Target\ [Na^+]} - 1\right)
This formula relates the current total body water to the proportional excess of sodium concentration relative to a normal target (140 mEq/L).
3
Substitute the calculated TBW and provided serum sodium values into the formula to find the final deficit.
Deficit=30×(168140−1)=30×(1.2−1)=30×0.2=6Deficit = 30 \times \left(\frac{168}{140} - 1\right) = 30 \times (1.2 - 1) = 30 \times 0.2 = 6 L
This quantifies the exact volume of pure water required to dilute the current total body solute back to a normal serum sodium concentration of 140 mEq/L.

Anahtar Kavram

Free water deficit calculation in hypernatremia
Tahmini Süre:1m 30s
Soru 62Soru →

A 72-year-old man is evaluated in the hospital for decreased urine output. He was admitted 3 days ago for an acute exacerbation of heart failure with reduced ejection fraction. His treatment regimen includes intravenous furosemide, lisinopril, carvedilol, and spironolactone. His weight has decreased by 4 kg (8.8 lb) since admission, and his peripheral edema has significantly improved. However, his urine output over the last 12 hours has been only 250 mL.

Vitals:
Blood pressure: 102/60 mm Hg
Heart rate: 94/min

Laboratory results from today:
Serum:
Sodium: 135 mEq/L
Potassium: 4.8 mEq/L
Chloride: 98 mEq/L
Bicarbonate: 28 mEq/L
Blood urea nitrogen: 80 mg/dL
Creatinine: 2.4 mg/dL

Urine:
Sodium: 75 mEq/L
Creatinine: 120 mg/dL
Urea nitrogen: 600 mg/dL

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

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Cevap: 15

Cevap

15
The correct calculation uses the formula FeUrea=Urine Urea Nitrogen×Serum CreatinineSerum BUN×Urine Creatinine×100FeUrea = \frac{\text{Urine Urea Nitrogen} \times \text{Serum Creatinine}}{\text{Serum BUN} \times \text{Urine Creatinine}} \times 100. Plugging in the provided values gives 600×2.480×120×100\frac{600 \times 2.4}{80 \times 120} \times 100, which simplifies to 14409600×100=15%\frac{1440}{9600} \times 100 = 15\%. In a patient receiving loop diuretics (like furosemide), Fractional Excretion of Sodium (FeNa) is falsely elevated due to diuretic-induced natriuresis. Urea handling in the proximal tubule is unaffected by loop diuretics, making FeUrea a much more reliable indicator of prerenal azotemia. A value below 35%35\% supports the diagnosis of prerenal acute kidney injury due to aggressive over-diuresis.

Adım Adım Çözüm

1
Identify the clinical context and indication for calculating FeUrea.
The patient has AKI while on furosemide, making FeNa unreliable. FeUrea is indicated.
Loop diuretics inhibit Na+ reabsorption, falsely elevating FeNa even in prerenal states. Urea is reabsorbed proximally, independent of loop diuretics.
2
Recall the Fractional Excretion of Urea formula.
FeUrea=UUrea×PCrPBUN×UCr×100FeUrea = \frac{U_{Urea} \times P_{Cr}}{P_{BUN} \times U_{Cr}} \times 100
Fractional excretion compares the clearance of a substance (urea) to the clearance of creatinine.
3
Extract the correct values from the laboratory panel.
UUreaU_{Urea} = 600 mg/dL, PCrP_{Cr} = 2.4 mg/dL, PBUNP_{BUN} = 80 mg/dL, UCrU_{Cr} = 120 mg/dL
Accurate variable selection is required; distractors like Urine Na and Serum Na must be ignored.
4
Execute the calculation.
600×2.480×120×100=15%\frac{600 \times 2.4}{80 \times 120} \times 100 = 15\%
Mathematical derivation of the fractional excretion percentage.

Anahtar Kavram

Fractional Excretion of Urea (FeUrea) is used to differentiate prerenal AKI from acute tubular necrosis in patients on diuretic therapy, where FeNa is often falsely elevated. A FeUrea < 35% indicates a prerenal etiology.
Soru 63Soru →

A 68-year-old man is brought to the emergency department with 3 days of severe, watery diarrhea and poor oral intake. He reports feeling lightheaded upon standing. His temperature is 37.1∘37.1^\circ C (98.8∘98.8^\circ F), blood pressure is 112/72112/72 mm Hg supine and 90/6090/60 mm Hg standing, and pulse is 92/min supine and 116/min standing. Physical examination reveals dry mucous membranes and delayed capillary refill.

Laboratory results are as follows:

Serum studies:
Sodium: 140140 mEq/L
Potassium: 3.53.5 mEq/L
Chloride: 110110 mEq/L
Bicarbonate: 2020 mEq/L
Blood urea nitrogen: 4242 mg/dL
Creatinine: 1.41.4 mg/dL

Urine studies:
Sodium: 2020 mEq/L
Creatinine: 4040 mg/dL

Based on these findings, what is the patient's fractional excretion of sodium?

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Cevap: 0.5

Cevap

0.5
The correct calculation of the Fractional Excretion of Sodium (FENa) requires the formula: FENa=UNa×PCrPNa×UCr×100FENa = \frac{U_{Na} \times P_{Cr}}{P_{Na} \times U_{Cr}} \times 100. Plugging in the patient's values yields (20×1.4)/(140×40)×100(20 \times 1.4) / (140 \times 40) \times 100, which simplifies to 28/5600×100=0.5%28 / 5600 \times 100 = 0.5\%. In the context of this patient's severe diarrhea, poor oral intake, and orthostatic hypotension, a FENa <1%< 1\% accurately supports the diagnosis of a prerenal state (hypovolemia) with an appropriate, intact renal tubular reabsorption of sodium.

Adım Adım Çözüm

1
Identify the formula for Fractional Excretion of Sodium (FENa).
FENa=UNa×PCrPNa×UCr×100FENa = \frac{U_{Na} \times P_{Cr}}{P_{Na} \times U_{Cr}} \times 100
This formula is used to differentiate prerenal acute kidney injury from intrinsic renal disease by measuring the fraction of filtered sodium that is excreted.
2
Substitute the patient's laboratory values into the equation.
UNa=20U_{Na} = 20 mEq/L, PCr=1.4P_{Cr} = 1.4 mg/dL, PNa=140P_{Na} = 140 mEq/L, UCr=40U_{Cr} = 40 mg/dL
Accurate variable placement is required for correct calculation.
3
Calculate the final percentage.
(20×1.4)/(140×40)×100=(28/5600)×100=0.5%(20 \times 1.4) / (140 \times 40) \times 100 = (28 / 5600) \times 100 = 0.5\%
Mathematical simplification yields the final FENa percentage.

Anahtar Kavram

Calculation of Fractional Excretion of Sodium (FENa)
Soru 64Soru →

A 24-year-old woman with a history of type 1 diabetes mellitus presents to the emergency department with a 3-day history of intractable nausea, vomiting, and diffuse abdominal pain. She reports that her insulin pump malfunctioned four days ago. On physical examination, she appears lethargic and has dry mucous membranes. Her pulse is 118/min, blood pressure is 98/60 mm Hg, and respirations are 26/min and deep.

Laboratory studies show:
Serum:
Sodium: 140 mEq/L140 \text{ mEq/L}
Potassium: 4.8 mEq/L4.8 \text{ mEq/L}
Chloride: 88 mEq/L88 \text{ mEq/L}
Bicarbonate: 16 mEq/L16 \text{ mEq/L}
Blood urea nitrogen: 42 mg/dL42 \text{ mg/dL}
Creatinine: 1.6 mg/dL1.6 \text{ mg/dL}
Glucose: 520 mg/dL520 \text{ mg/dL}

Arterial blood gas (room air):
pH: 7.34
pCO2\text{pCO}_2: 31 mm Hg31 \text{ mm Hg}
pO2\text{pO}_2: 95 mm Hg95 \text{ mm Hg}

Assume a normal serum anion gap of 12 mEq/L12 \text{ mEq/L} and a normal serum bicarbonate of 24 mEq/L24 \text{ mEq/L}.

What is this patient's delta-delta ratio (ΔAGΔHCO3−\frac{\Delta \text{AG}}{\Delta \text{HCO}_3^-})?

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Cevap: 3

Cevap

3.0
The correct answer is 3.0. The patient has a calculated anion gap of 36 mEq/L (140 - [88 + 16]). The change in anion gap (Δ\Delta AG) is 24 mEq/L (36 - 12). The change in bicarbonate (ΔHCO3−\Delta\text{HCO}_3^-) is 8 mEq/L (24 - 16). The delta-delta ratio is ΔAG/ΔHCO3−=24/8=3.0\Delta\text{AG} / \Delta\text{HCO}_3^- = 24 / 8 = 3.0. In an uncomplicated high anion gap metabolic acidosis (HAGMA), the drop in bicarbonate should roughly equal the rise in the anion gap, yielding a ratio between 1.0 and 2.0. A ratio greater than 2.0 indicates that the bicarbonate level is higher than expected, signaling a concurrent metabolic alkalosis. This is consistent with the patient's history of intractable vomiting (loss of gastric HCl) in the setting of diabetic ketoacidosis.

Adım Adım Çözüm

1
Calculate the serum anion gap.
36 mEq/L
To identify the presence and magnitude of an unmeasured anion accumulation (high anion gap metabolic acidosis).
2
Calculate the delta anion gap (ΔAG\Delta\text{AG}).
24 mEq/L
To determine how much the anion gap has increased above the normal baseline.
3
Calculate the delta bicarbonate (ΔHCO3−\Delta\text{HCO}_3^-).
8 mEq/L
To determine how much the serum bicarbonate has decreased from the normal baseline.
4
Calculate the delta-delta ratio (ΔAGΔHCO3−\frac{\Delta\text{AG}}{\Delta\text{HCO}_3^-}).
3.0
To evaluate for the presence of a concurrent mixed acid-base disorder.

Anahtar Kavram

Delta-delta ratio in the evaluation of mixed acid-base disorders
Tahmini Süre:1m 30s
Soru 65Soru →

During a routine follow-up at the nephrology clinic, a 42-year-old woman is evaluated for persistent hypokalemia and mild muscle weakness. She has a history of Sjögren syndrome and has had two episodes of calcium phosphate nephrolithiasis in the past five years. She reports no diarrhea, vomiting, or recent dietary changes. Her blood pressure is 118/76 mm Hg. Laboratory studies show:

Serum sodium: 140 mEq/L
Serum potassium: 2.9 mEq/L
Serum chloride: 115 mEq/L
Serum bicarbonate: 15 mEq/L
Blood urea nitrogen: 12 mg/dL
Serum creatinine: 0.8 mg/dL

Urine studies obtained simultaneously show:
Urine sodium: 55 mEq/L
Urine potassium: 25 mEq/L
Urine chloride: 60 mEq/L

What is this patient's urine anion gap in mEq/L? (Provide your answer as a number only)

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Cevap: 20

Cevap

20
The patient has a normal anion gap metabolic acidosis (NAGMA), indicated by a serum anion gap of 10 mEq/L (140−(115+15)140 - (115 + 15)). To evaluate the underlying cause of a NAGMA, the urine anion gap (UAG) is calculated to estimate renal ammonium (NH4+NH_4^+) excretion. The formula is: UAG=(UNa+UK)−UClUAG = (U_{Na} + U_{K}) - U_{Cl}. Using the patient's values: UAG=(55+25)−60=80−60=20UAG = (55 + 25) - 60 = 80 - 60 = 20 mEq/L. In normal physiology and gastrointestinal bicarbonate loss (e.g., diarrhea), the kidneys increase NH4+NH_4^+ production, which is excreted with chloride, resulting in a negative UAG. A positive UAG, as seen in this patient, indicates a failure of the kidneys to excrete adequate acid, which is diagnostic for renal tubular acidosis (RTA). Her history of Sjögren syndrome and calcium phosphate stones further points to Type 1 (distal) RTA.

Adım Adım Çözüm

1
Calculate the serum anion gap.
Serum AG = 140−(115+15)=10140 - (115 + 15) = 10 mEq/L.
Confirms the presence of a normal anion gap metabolic acidosis (NAGMA), prompting the need for a urine anion gap.
2
Set up the urine anion gap formula.
UAG=(UNa+UK)−UClUAG = (U_{Na} + U_{K}) - U_{Cl}
The formula estimates unmeasured urinary ammonium excretion.
3
Insert the provided urine electrolyte values.
UAG=(55+25)−60UAG = (55 + 25) - 60
Accurate substitution of variables is required for the calculation.
4
Perform the final calculation.
80−60=2080 - 60 = 20 mEq/L
Yields the correct numerical answer indicating impaired ammonium excretion.

Anahtar Kavram

The urine anion gap (UAG=UNa+UK−UClUAG = U_{Na} + U_{K} - U_{Cl}) differentiates renal from gastrointestinal causes of normal anion gap metabolic acidosis. A positive value reflects impaired urinary ammonium excretion, characteristic of distal renal tubular acidosis.
Soru 66Soru →

A 65-year-old man is admitted to the hospital for acute decompensated heart failure and is treated with aggressive intravenous furosemide. On hospital day 3, his blood pressure is 105/65 mm Hg and heart rate is 92/min. Over the past 24 hours, his urine output has decreased to 400 mL. His serum creatinine rises from a baseline of 1.1 mg/dL to 2.8 mg/dL.

Laboratory studies are obtained to evaluate his acute kidney injury:

Serum chemistry:
Sodium: 136 mEq/L
Potassium: 3.8 mEq/L
Chloride: 98 mEq/L
Bicarbonate: 28 mEq/L
Blood urea nitrogen (BUN): 84 mg/dL
Creatinine: 2.8 mg/dL

Urine chemistry:
Sodium: 70 mEq/L
Urea nitrogen: 420 mg/dL
Creatinine: 56 mg/dL

Because recent loop diuretic administration impairs tubular sodium reabsorption, the fractional excretion of sodium (FENaFE_{Na}) can be misleadingly elevated. Instead, the fractional excretion of urea (FEUreaFE_{Urea}) is preferred to differentiate between prerenal azotemia and intrinsic acute tubular necrosis in this clinical scenario.

Calculate the patient's FEUreaFE_{Urea}. Express your answer as a percentage.

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Cevap: 25

Cevap

25
The correct calculation for FEUreaFE_{Urea} is [(420×2.8)/(84×56)]×100=25%[(420 \times 2.8) / (84 \times 56)] \times 100 = 25\%. In the evaluation of acute kidney injury, establishing whether the etiology is prerenal (decreased perfusion) or intrinsic (tubular damage) is critical. While the fractional excretion of sodium (FENaFE_{Na}) is standard, loop diuretics like furosemide block the Na-K-2Cl cotransporter in the thick ascending limb of the loop of Henle, forcibly wasting sodium into the urine. This often results in an FENa>1%FE_{Na} > 1\% (in this patient, roughly 2.6%), which falsely points toward intrinsic damage. Because urea reabsorption occurs primarily in the proximal tubule and is largely unaffected by loop diuretics, FEUreaFE_{Urea} remains a highly sensitive and specific metric. An FEUreaFE_{Urea} less than 35% strongly confirms a prerenal state driven by severe heart failure and aggressive diuresis.

Adım Adım Çözüm

1
Identify the correct formula for the Fractional Excretion of Urea (FEUrea).
FEUrea=(UrineUrea×SerumCrSerumBUN×UrineCr)×100FE_{Urea} = \left( \frac{Urine_{Urea} \times Serum_{Cr}}{Serum_{BUN} \times Urine_{Cr}} \right) \times 100
This formula standardizes urea clearance against creatinine clearance to assess proximal tubular handling of urea independent of sodium.
2
Extract the relevant variables from the provided laboratory data.
Urine Urea Nitrogen = 420 mg/dL; Serum Creatinine = 2.8 mg/dL; Serum BUN = 84 mg/dL; Urine Creatinine = 56 mg/dL.
Accurate identification is required to avoid incorrectly mixing serum and urine values during calculation.
3
Calculate the numerator by multiplying Urine Urea by Serum Creatinine.
420×2.8=1176420 \times 2.8 = 1176
This represents the relative amount of urea excreted compared to the serum creatinine level.
4
Calculate the denominator by multiplying Serum BUN by Urine Creatinine.
84×56=470484 \times 56 = 4704
This represents the filtered load based on serum BUN normalized by the urine creatinine level.
5
Divide the numerator by the denominator and multiply by 100.
(1176/4704)×100=0.25×100=25%(1176 / 4704) \times 100 = 0.25 \times 100 = 25\%
This calculation yields the final percentage of filtered urea that is actually excreted by the kidneys.

Anahtar Kavram

Fractional Excretion of Urea (FEUrea) is a more reliable marker than FENa for differentiating prerenal azotemia from intrinsic acute tubular necrosis in patients actively receiving loop diuretics.
Soru 67Soru →

A 45-year-old man with a history of poorly controlled type 2 diabetes mellitus is brought to the emergency department due to confusion, nausea, and abdominal pain for the past 2 days. He has missed his last several doses of insulin. His temperature is 37.1°C (98.8°F), blood pressure is 100/65 mm Hg, pulse is 112/min, and respirations are 24/min. Physical examination shows dry mucous membranes and a fruity breath odor.

Laboratory studies show:

Laboratory TestResult
Sodium142 mEq/L
Potassium4.8 mEq/L
Chloride102 mEq/L
Bicarbonate14 mEq/L
Blood urea nitrogen25 mg/dL
Creatinine1.4 mg/dL
Glucose450 mg/dL

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

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Cevap: 26

Cevap

26
The serum anion gap is calculated using the formula: Anion Gap=Na+−(Cl−+HCO3−)\text{Anion Gap} = \text{Na}^+ - (\text{Cl}^- + \text{HCO}_3^-). Using the patient's laboratory values, the calculation is 142−(102+14)=142−116=26142 - (102 + 14) = 142 - 116 = 26 mEq/L. A normal anion gap is typically between 8 and 12 mEq/L. This patient has an elevated anion gap of 26 mEq/L, consistent with high anion gap metabolic acidosis secondary to diabetic ketoacidosis.

Adım Adım Çözüm

1
Identify the formula for serum anion gap.
Anion Gap=Na+−(Cl−+HCO3−)\text{Anion Gap} = \text{Na}^+ - (\text{Cl}^- + \text{HCO}_3^-)
This is the standard formula used to evaluate the etiology of metabolic acidosis.
2
Extract the necessary serum electrolyte values from the vignette.
Na+=142\text{Na}^+ = 142, Cl−=102\text{Cl}^- = 102, HCO3−=14\text{HCO}_3^- = 14.
These are the three variables required for the calculation.
3
Perform the calculation.
142−(102+14)=142−116=26142 - (102 + 14) = 142 - 116 = 26.
To determine the exact numerical value of the anion gap.

Anahtar Kavram

Calculation of the serum anion gap in the evaluation of metabolic acidosis
Soru 68Soru →

Paramedics are dispatched to the home of a 42-year-old man after he was discovered unresponsive in his garage. According to his husband, the patient has a history of severe depression and recently experienced a job loss. An empty, unlabeled container of sweet-smelling fluid was found nearby. Vital signs upon arrival at the hospital include a temperature of 36.5∘C36.5^\circ\text{C} (97.7∘F97.7^\circ\text{F}), blood pressure of 110/70 mm Hg110/70 \text{ mm Hg}, heart rate of 105/min105/\text{min}, and respiratory rate of 24/min24/\text{min}. The patient is obtunded, responding only by withdrawing from painful stimuli.

Laboratory studies are obtained:

Serum chemistry:
Sodium: 138 mEq/L138 \text{ mEq/L}
Potassium: 4.5 mEq/L4.5 \text{ mEq/L}
Chloride: 102 mEq/L102 \text{ mEq/L}
Bicarbonate: 12 mEq/L12 \text{ mEq/L}
Blood urea nitrogen (BUN): 28 mg/dL28 \text{ mg/dL}
Creatinine: 1.2 mg/dL1.2 \text{ mg/dL}
Glucose: 108 mg/dL108 \text{ mg/dL}

Arterial blood gas (room air):
pH: 7.227.22
pCO2\text{pCO}_2: 26 mm Hg26 \text{ mm Hg}
pO2\text{pO}_2: 95 mm Hg95 \text{ mm Hg}

Additional diagnostics:
Measured serum osmolality: 335 mOsm/kg H2O335 \text{ mOsm/kg H}_2\text{O}

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

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Cevap: 43

Cevap

43
The correct osmolar gap is 43 mOsm/kg H2O. This patient presents with a high anion gap metabolic acidosis (AG = 138 - 102 - 12 = 24 mEq/L) and an obtunded state after suspected toxic alcohol (ethylene glycol) ingestion. To determine the osmolar gap, one must first determine the calculated serum osmolality using the formula: (2 * Na) + (Glucose / 18) + (BUN / 2.8). In this patient, (2 * 138) + (108 / 18) + (28 / 2.8) = 276 + 6 + 10 = 292 mOsm/kg H2O. Subtracting this calculated value from the measured osmolality (335 mOsm/kg H2O) yields an osmolar gap of 43. A normal gap is generally less than 10; an elevated gap in this context confirms the presence of an unmeasured osmotically active substance.

Adım Adım Çözüm

1
Recall the formula for calculating estimated serum osmolality in standard US units.
Calculated Osmolality = (2 \times \text{Na}) + (\text{Glucose} / 18) + (\text{BUN} / 2.8)
This formula estimates the total concentration of the major physiological solutes in the blood.
2
Input the patient's specific laboratory values into the formula.
Calculated Osmolality = (2 \times 138) + (108 / 18) + (28 / 2.8)
Sodium is 138 mEq/L, Glucose is 108 mg/dL, and BUN is 28 mg/dL.
3
Perform the calculation to find the estimated osmolality.
276 + 6 + 10 = 292 \text{ mOsm/kg H}_2\text{O}
This provides the baseline physiological osmolality without any unaccounted external osmoles.
4
Calculate the osmolar gap by subtracting the calculated osmolality from the measured osmolality.
335 - 292 = 43 \text{ mOsm/kg H}_2\text{O}
The difference reveals the presence of unmeasured osmoles, which helps diagnose toxic alcohol ingestion.

Anahtar Kavram

The serum osmolar gap is calculated by subtracting the calculated serum osmolality from the measured serum osmolality. The calculated osmolality is (2×Na)+(Glucose/18)+(BUN/2.8)(2 \times \text{Na}) + (\text{Glucose} / 18) + (\text{BUN} / 2.8). An elevated gap (>10>10) in the context of an anion gap metabolic acidosis strongly suggests toxic alcohol ingestion (e.g., ethylene glycol or methanol).
Soru 69Soru →

A 68-year-old man with a history of type 2 diabetes mellitus is brought to the emergency department due to progressively worsening confusion and profound weakness over the past 4 days. He recently recovered from a mild upper respiratory illness. On examination, he is lethargic but arousable. Vital signs show a blood pressure of 92/58 mm Hg92/58\text{ mm Hg}, a heart rate of 118/min118/\text{min}, and a respiratory rate of 16/min16/\text{min}. His mucous membranes are parched, and skin turgor is significantly decreased.

Initial laboratory evaluation reveals:
Serum sodium: 129 mEq/L129\text{ mEq/L}
Serum potassium: 4.4 mEq/L4.4\text{ mEq/L}
Serum chloride: 95 mEq/L95\text{ mEq/L}
Serum bicarbonate: 22 mEq/L22\text{ mEq/L}
Blood urea nitrogen: 48 mg/dL48\text{ mg/dL}
Serum creatinine: 1.9 mg/dL1.9\text{ mg/dL}
Serum glucose: 600 mg/dL600\text{ mg/dL}

In order to properly select the composition of intravenous fluids for resuscitation, the patient's true sodium status must be evaluated.

What is the patient's corrected serum sodium concentration in mEq/L\text{mEq/L}? (Assume a correction factor of 1.6 mEq/L1.6\text{ mEq/L} for each 100 mg/dL100\text{ mg/dL} increase in glucose above 100 mg/dL100\text{ mg/dL}).

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Cevap: 137

Cevap

137
In the setting of severe hyperglycemia, the high concentration of glucose in the extracellular fluid creates an osmotic gradient that draws water out of cells. This transcellular fluid shift dilutes the serum sodium concentration, leading to a phenomenon known as translocational hyponatremia. To determine the patient's true sodium status, a correction factor must be applied. The standard correction adds 1.6 mEq/L1.6\text{ mEq/L} to the measured sodium for every 100 mg/dL100\text{ mg/dL} increase in glucose above a baseline of 100 mg/dL100\text{ mg/dL}. For a glucose level of 600 mg/dL600\text{ mg/dL}, the excess is 500 mg/dL500\text{ mg/dL}. Multiplying 55 by 1.61.6 yields a correction of 8 mEq/L8\text{ mEq/L}. Adding this to the measured sodium of 129 mEq/L129\text{ mEq/L} results in a corrected serum sodium of 137 mEq/L137\text{ mEq/L}. This indicates that the patient actually has a normal serum sodium concentration, which is critical for choosing the appropriate intravenous fluid for resuscitation.

Adım Adım Çözüm

1
Identify the patient's measured serum sodium and serum glucose levels.
Measured Na = 129 mEq/L129\text{ mEq/L}, Glucose = 600 mg/dL600\text{ mg/dL}.
These are the two variables required to calculate the corrected serum sodium.
2
Calculate the amount of glucose elevation above the normal baseline of 100 mg/dL100\text{ mg/dL}.
600−100=500 mg/dL600 - 100 = 500\text{ mg/dL} above normal.
The correction factor is only applied to the excess glucose driving the osmotic shift.
3
Determine the total sodium correction value by applying the given factor.
Correction = 5×1.6=8 mEq/L5 \times 1.6 = 8\text{ mEq/L}.
For every 100 mg/dL100\text{ mg/dL} of excess glucose, sodium is artificially lowered by 1.6 mEq/L1.6\text{ mEq/L}.
4
Add the correction value to the measured serum sodium.
129+8=137 mEq/L129 + 8 = 137\text{ mEq/L}.
This yields the true serum sodium concentration, guiding appropriate fluid therapy.

Anahtar Kavram

Pseudohyponatremia due to hyperglycemia (translocational hyponatremia)
Soru 70Soru →

Paramedics arrive at the emergency room with a 52-year-old man who was discovered unresponsive in a public park. The patient has a documented medical history of severe alcohol use disorder. Upon initial evaluation, his body temperature is 36.2∘C36.2^\circ\text{C} (97.2∘F97.2^\circ\text{F}), pulse is 102/min102/\text{min}, blood pressure is 110/70 mm Hg110/70\text{ mm Hg}, and respirations are 24/min24/\text{min}. He is unarousable to painful stimuli. Cardiopulmonary examination is unremarkable. Initial laboratory studies show the following:

Laboratory ParameterValue
Serum sodium137 mEq/L137\text{ mEq/L}
Serum potassium4.5 mEq/L4.5\text{ mEq/L}
Serum chloride98 mEq/L98\text{ mEq/L}
Serum bicarbonate12 mEq/L12\text{ mEq/L}
Blood urea nitrogen56 mg/dL56\text{ mg/dL}
Serum creatinine1.8 mg/dL1.8\text{ mg/dL}
Serum glucose162 mg/dL162\text{ mg/dL}
Serum ethanol138 mg/dL138\text{ mg/dL}
Measured serum osmolality368 mOsm/kg H2O368\text{ mOsm/kg H}_2\text{O}

Arterial blood gas (room air) shows a pH of 7.227.22 and a PaCO2\text{PaCO}_2 of 26 mm Hg26\text{ mm Hg}.

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

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Cevap: 35

Cevap

35
The correct osmolar gap is 35. The patient presents with an elevated anion gap metabolic acidosis (Anion Gap = 137 - 98 - 12 = 27). In such cases, evaluating the osmolar gap is crucial to identifying toxic alcohol ingestion. The calculated serum osmolality must account for sodium, glucose, blood urea nitrogen (BUN), and any measured ethanol. The full formula is: 2×[Na+]+[Glucose]18+[BUN]2.8+[Ethanol]4.62 \times [\text{Na}^+] + \frac{[\text{Glucose}]}{18} + \frac{[\text{BUN}]}{2.8} + \frac{[\text{Ethanol}]}{4.6}. Plugging in the values gives 2(137)+16218+562.8+1384.6=274+9+20+30=333 mOsm/kg2(137) + \frac{162}{18} + \frac{56}{2.8} + \frac{138}{4.6} = 274 + 9 + 20 + 30 = 333\text{ mOsm/kg}. The osmolar gap is the measured osmolality minus the calculated osmolality (368−333=35368 - 333 = 35). An osmolar gap greater than 10 to 15 mOsm/kg suggests the presence of unmeasured osmotically active substances, such as methanol or ethylene glycol, which require urgent intervention.

Adım Adım Çözüm

1
Recall the formula for calculated serum osmolality.
Calculated Osmolality = 2×[Na+]+[Glucose]18+[BUN]2.8+[Ethanol]4.62 \times [\text{Na}^+] + \frac{[\text{Glucose}]}{18} + \frac{[\text{BUN}]}{2.8} + \frac{[\text{Ethanol}]}{4.6}
To determine the osmolar gap, the expected (calculated) osmolality must first be determined, accounting for all major osmotically active substances including ethanol.
2
Input the patient's laboratory values into the formula.
Calculated Osm = 2(137)+16218+562.8+1384.62(137) + \frac{162}{18} + \frac{56}{2.8} + \frac{138}{4.6}
Standard conversion factors are used to convert mg/dL to mmol/L (mOsm/kg).
3
Compute the sum of the components.
274+9+20+30=333 mOsm/kg H2O274 + 9 + 20 + 30 = 333\text{ mOsm/kg H}_2\text{O}
This represents the osmolality expected from the known solutes in the patient's blood.
4
Calculate the osmolar gap.
Osmolar gap = Measured Osmolality - Calculated Osmolality = 368−333=35 mOsm/kg H2O368 - 333 = 35\text{ mOsm/kg H}_2\text{O}
The difference reveals the presence of unmeasured osmotically active substances (e.g., toxic alcohols like methanol or ethylene glycol).

Anahtar Kavram

The serum osmolar gap is calculated by subtracting the calculated osmolality from the measured osmolality. When ethanol is present, it must be included in the calculated osmolality formula to accurately assess for the presence of other toxic alcohols.
Tahmini Süre:1m 30s
Soru 71Soru →

A 25-year-old female with a history of type 1 diabetes mellitus is brought to the emergency department by her roommate due to severe lethargy and abdominal pain. Over the preceding 48 hours, she has experienced intractable nonbilious emesis and has consequently withheld her scheduled insulin doses. On physical examination, she appears confused and has markedly dry mucous membranes with decreased skin turgor.

Laboratory results:
- Serum Na+Na^+: 140 mEq/L140 \text{ mEq/L}
- Serum K+K^+: 3.4 mEq/L3.4 \text{ mEq/L}
- Serum Cl−Cl^-: 86 mEq/L86 \text{ mEq/L}
- Serum HCO3−HCO_3^-: 14 mEq/L14 \text{ mEq/L}
- Serum glucose: 545 mg/dL545 \text{ mg/dL}
- Blood urea nitrogen: 38 mg/dL38 \text{ mg/dL}
- Serum creatinine: 1.6 mg/dL1.6 \text{ mg/dL}

Assuming a normal 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-delta ratio?

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Cevap: 2.8

Cevap

2.8
The patient's presentation of hyperglycemia, elevated anion gap, and insulin nonadherence is consistent with diabetic ketoacidosis (DKA), which classically causes a primary high anion gap metabolic acidosis. However, her history of intractable emesis also introduces a primary metabolic alkalosis. The delta-delta ratio (ΔAG/ΔHCO3−\Delta AG / \Delta HCO_3^-) helps diagnose mixed acid-base disorders. First, calculate the Anion Gap: Na+−(Cl−+HCO3−)=140−(86+14)=40 mEq/LNa^+ - (Cl^- + HCO_3^-) = 140 - (86 + 14) = 40 \text{ mEq/L}. Next, calculate the ΔAG\Delta AG: Measured AG - Normal AG = 40−12=2840 - 12 = 28. Then, calculate the ΔHCO3−\Delta HCO_3^-: Normal HCO3−HCO_3^- - Measured HCO3−HCO_3^- = 24−14=1024 - 14 = 10. Finally, the ratio is ΔAG/ΔHCO3−=28/10=2.8\Delta AG / \Delta HCO_3^- = 28 / 10 = 2.8. A ratio between 1.0 and 2.0 typically indicates an isolated high anion gap metabolic acidosis. A ratio greater than 2.0 indicates that the serum bicarbonate is higher than expected for the degree of anion gap elevation, confirming a concurrent primary metabolic alkalosis.

Adım Adım Çözüm

1
Calculate the patient's anion gap (AG).
AG=140−(86+14)=40 mEq/LAG = 140 - (86 + 14) = 40 \text{ mEq/L}.
To identify the presence and severity of a high anion gap metabolic acidosis.
2
Calculate the delta anion gap (ΔAG\Delta AG).
ΔAG=40−12=28 mEq/L\Delta AG = 40 - 12 = 28 \text{ mEq/L}.
To determine the excess amount of unmeasured anions driving the acidosis.
3
Calculate the delta bicarbonate (ΔHCO3−\Delta HCO_3^-).
ΔHCO3−=24−14=10 mEq/L\Delta HCO_3^- = 24 - 14 = 10 \text{ mEq/L}.
To determine the expected drop in bicarbonate due to buffering the excess acid.
4
Calculate the delta-delta ratio.
28/10=2.828 / 10 = 2.8.
To assess for concurrent mixed acid-base disorders by comparing the rise in unmeasured anions to the fall in bicarbonate.

Anahtar Kavram

Delta-Delta Ratio Calculation
Tahmini Süre:1m 0s
Soru 72Soru →

A 60-year-old man is brought to the emergency department by a neighbor who found him lethargic and confused in his apartment. The neighbor notes that the patient has a history of severe alcohol use disorder and has been drinking heavily for the past week. He has also had multiple episodes of non-bilious vomiting over the last three days. Physical examination reveals dry mucous membranes and decreased skin turgor. Laboratory studies are obtained:

Laboratory StudyResult
Serum sodium140 mEq/L140\text{ mEq/L}
Serum potassium3.2 mEq/L3.2\text{ mEq/L}
Serum chloride88 mEq/L88\text{ mEq/L}
Serum bicarbonate16 mEq/L16\text{ mEq/L}
Blood urea nitrogen (BUN)45 mg/dL45\text{ mg/dL}
Serum creatinine1.8 mg/dL1.8\text{ mg/dL}
Blood glucose65 mg/dL65\text{ mg/dL}

Arterial blood gas on room air reveals a pH of 7.307.30 and PaCO 2_2 of 32 mm Hg32\text{ mm Hg}.

Calculate the patient's delta-delta ratio, defined as the change in anion gap divided by the change in serum bicarbonate (ΔAGΔHCO3−\frac{\Delta \text{AG}}{\Delta \text{HCO}_3^-}). Assume a normal anion gap of 12 mEq/L12\text{ mEq/L} (calculated without potassium) and a normal serum bicarbonate of 24 mEq/L24\text{ mEq/L}.

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Cevap: 3

Cevap

3
The patient presents with alcoholic ketoacidosis (AKA) from heavy alcohol use, which generates a high anion gap metabolic acidosis. He also has a concurrent metabolic alkalosis from severe vomiting (loss of gastric HCl). The delta-delta ratio quantifies the relationship between the accumulation of unmeasured anions and the consumption of bicarbonate. An anion gap of 36 mEq/L yields a ΔAG of 24, while the bicarbonate of 16 mEq/L yields a ΔHCO3 of 8. The resulting ratio of 3.0 (which is > 2.0) mathematically confirms the presence of a superimposed metabolic alkalosis, as the rise in unmeasured anions significantly exceeds the drop in bicarbonate.

Adım Adım Çözüm

1
Calculate the patient's current anion gap (AG).
AG = 140 - (88 + 16) = 36 mEq/L
The anion gap is needed to determine the presence of a high anion gap metabolic acidosis (HAGMA) and to calculate the numerator of the delta-delta ratio.
2
Calculate the change in anion gap (ΔAG).
ΔAG = Measured AG - Normal AG = 36 - 12 = 24 mEq/L
This value represents the excess unmeasured anions contributing to the acidosis.
3
Calculate the change in serum bicarbonate (ΔHCO3-).
ΔHCO3- = Normal HCO3- - Measured HCO3- = 24 - 16 = 8 mEq/L
This value represents the amount of bicarbonate buffered by the excess acid.
4
Calculate the delta-delta ratio.
Delta Ratio = ΔAG / ΔHCO3- = 24 / 8 = 3
The delta-delta ratio helps identify concurrent metabolic acid-base disorders.

Anahtar Kavram

The delta-delta ratio is used in the evaluation of high anion gap metabolic acidosis to identify concurrent mixed metabolic acid-base disorders; a ratio greater than 2.0 indicates a superimposed metabolic alkalosis.
Soru 73Soru →

A 38-year-old mechanic is evaluated in the hospital after being found confused and lethargic in his workshop. He has a history of alcohol use disorder. Vital signs show a temperature of 37.1°C (98.8°F), blood pressure of 135/85 mmHg, pulse of 92/min, and respiratory rate of 22/min. Physical examination reveals a disoriented patient with dry mucous membranes.

Initial laboratory evaluation reveals the following:

ParameterValue
Sodium140 mEq/L
Potassium4.1 mEq/L
Chloride98 mEq/L
Bicarbonate14 mEq/L
Blood urea nitrogen20 mg/dL
Serum creatinine1.5 mg/dL
Glucose105 mg/dL

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

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Cevap: 28

Cevap

28
The serum anion gap is calculated by subtracting the primary measured serum anions (chloride and bicarbonate) from the primary measured serum cation (sodium). Using the formula Anion Gap = Na+−(Cl−+HCO3−)Na^+ - (Cl^- + HCO_3^-), the calculation for this patient is 140−(98+14)=140−112=28140 - (98 + 14) = 140 - 112 = 28 mEq/L. A normal anion gap is typically between 8 and 12 mEq/L. This patient has a significantly elevated high anion gap metabolic acidosis (HAGMA), which is consistent with toxic alcohol ingestion (such as methanol or ethylene glycol), given his presentation and history.

Adım Adım Çözüm

1
Identify the formula for the serum anion gap.
Anion Gap = Na+−(Cl−+HCO3−)Na^+ - (Cl^- + HCO_3^-)
The standard serum anion gap is calculated using the primary measured serum cation and the primary measured serum anions.
2
Extract the required laboratory values from the clinical vignette.
Na+Na^+ = 140 mEq/L, Cl−Cl^- = 98 mEq/L, and HCO3−HCO_3^- = 14 mEq/L.
These specific values are required to populate the formula.
3
Calculate the sum of the measured anions.
98+14=11298 + 14 = 112 mEq/L
This determines the total measured negative charge to be subtracted from sodium.
4
Subtract the sum of the anions from the serum sodium concentration.
140−112=28140 - 112 = 28 mEq/L
This yields the final unmeasured anion gap.

Anahtar Kavram

Calculation of the serum anion gap to evaluate metabolic acidosis.
Soru 74Soru →

A 35-year-old man is brought to the emergency department after sustaining injuries in a kitchen grease fire. On examination, he is alert and oriented. His blood pressure is 125/80 mm Hg, heart rate is 98/min, and respiratory rate is 16/min. Physical examination reveals deep partial-thickness burns involving the entire anterior surface of his chest and abdomen, as well as the entire anterior surface of both arms. His airway is patent with no signs of inhalational injury. Using the Rule of Nines, what is the estimated total body surface area (TBSA) percentage affected by these burns?

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Cevap: 27

Cevap

27
Based on the Rule of Nines for adults, the anterior trunk (chest and abdomen) accounts for 18% of the total body surface area. The entire surface of one arm accounts for 9%, meaning the anterior surface of one arm is 4.5%. Since the anterior surfaces of both arms are involved, this contributes an additional 9% (4.5% x 2). Adding these values together yields a total estimated burn area of 27% (18% + 9%).

Adım Adım Çözüm

1
Determine the percentage for the anterior trunk.
18%
According to the Rule of Nines, the entire anterior trunk (chest and abdomen combined) accounts for 18% of total body surface area.
2
Determine the percentage for the anterior surface of both arms.
9%
Each entire arm accounts for 9% of TBSA, making the anterior surface of one arm 4.5%. Since both arms are involved anteriorly, 4.5% + 4.5% = 9%.
3
Calculate the total body surface area affected.
27%
Adding the affected areas: 18% (anterior trunk) + 9% (anterior arms) = 27%.

Anahtar Kavram

The Rule of Nines is a standardized method used to quickly estimate the total body surface area (TBSA) affected by partial-thickness and full-thickness burns in adults, which is essential for guiding fluid resuscitation.
Soru 75Soru →

A 50-year-old construction worker is evacuated by helicopter to the trauma center following a localized industrial explosion. The patient weighs 80 kg80\text{ kg}. Paramedics report the incident occurred exactly 2 hours2\text{ hours} prior to arrival. Primary survey reveals a patent airway with no signs of inhalation injury, clear breath sounds bilaterally, and a heart rate of 115 beats/min115\text{ beats/min}. Secondary survey is notable for deep partial-thickness and full-thickness thermal burns covering the entirety of his posterior trunk (including buttocks), his entire left lower extremity, and his entire right upper extremity. Superficial (first-degree) burns are noted on his face but are not blistering. Large-bore intravenous access is established. The trauma team initiates fluid resuscitation using lactated Ringer solution according to current Advanced Trauma Life Support (ATLS) guidelines (2 mL/kg/%TBSA2\text{ mL/kg/\%TBSA} for adult thermal burns).

What should be the initial rate of the fluid infusion (in mL/hr\text{mL/hr}) programmed on the intravenous pump for the time remaining in the first 8-hour post-injury window?

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Cevap: 600

Cevap

600
The correct calculation requires summing the TBSA (18%18\% posterior trunk + 18%18\% left leg + 9%9\% right arm = 45%45\%), multiplying by 80 kg80\text{ kg} and 2 mL2\text{ mL} (totaling 7,200 mL7,200\text{ mL} for 24 hours), dividing in half for the first 8 hours (3,600 mL3,600\text{ mL}), and dividing by the 66 hours remaining in that window (3,600 mL/6 hr=600 mL/hr3,600\text{ mL} / 6\text{ hr} = 600\text{ mL/hr}).

Adım Adım Çözüm

1
Calculate affected Total Body Surface Area (TBSA)
45%45\%
Using the Rule of Nines: Posterior trunk is 18%18\%, left leg is 18%18\%, and right arm is 9%9\%. Superficial burns are not included in TBSA calculations for fluid resuscitation.
2
Calculate total 24-hour fluid requirement
7,200 mL7,200\text{ mL}
Formula: 2 mL×80 kg×45=7,200 mL2\text{ mL} \times 80\text{ kg} \times 45 = 7,200\text{ mL}.
3
Determine fluid volume for the first 8 hours
3,600 mL3,600\text{ mL}
Half (50%50\%) of the total 24-hour volume is administered during the first 8 hours post-injury.
4
Calculate infusion rate for the remaining time in the first 8-hour window
600 mL/hr600\text{ mL/hr}
The injury occurred 22 hours prior to arrival, leaving 66 hours (8−2=68 - 2 = 6) in the initial window. Rate = 3,600 mL/6 hr=600 mL/hr3,600\text{ mL} / 6\text{ hr} = 600\text{ mL/hr}.

Anahtar Kavram

Calculation of initial burn fluid resuscitation using the Rule of Nines, the ATLS modified formula (2 mL/kg/%TBSA for adult thermal burns), and adjusting the rate for pre-hospital time elapsed since the injury.
Soru 76Soru →

A 3535-year-old man weighing 70 kg70 \text{ kg} is brought to the emergency department 2 hours2 \text{ hours} after being rescued from a structural fire. On examination, he is hemodynamically stable, and his airway is secure with no evidence of inhalational injury. He has sustained deep partial-thickness and full-thickness burns involving his anterior chest, anterior abdomen, entire right arm, and the anterior aspect of his left leg. The remainder of his skin is intact. Intravenous access is obtained. According to current Advanced Trauma Life Support (ATLS) guidelines for adult thermal burns, what should be the initial rate of intravenous lactated Ringer solution administration for this patient? (Enter the value in mL/hr\text{mL/hr})

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Cevap: 420

Cevap

420
The correct initial fluid resuscitation rate is 420 mL/hr420 \text{ mL/hr}. According to the Advanced Trauma Life Support (ATLS) 10th edition guidelines, adult patients with thermal burns should be resuscitated using 2 mL2 \text{ mL} of lactated Ringer solution per kilogram of body weight per percent of affected Total Body Surface Area (TBSA). First, the Rule of Nines is applied: the anterior chest (9%9\%), anterior abdomen (9%9\%), entire right arm (9%9\%), and anterior left leg (9%9\%) yield a 36%36\% TBSA burn. The total 24-hour fluid requirement is 2 mL×70 kg×36=5,040 mL2 \text{ mL} \times 70 \text{ kg} \times 36 = 5,040 \text{ mL}. Half of this volume (2,520 mL2,520 \text{ mL}) must be administered within the first 8 hours8 \text{ hours} from the time of the burn. Because the patient presented 2 hours2 \text{ hours} after the injury, the 2,520 mL2,520 \text{ mL} must be infused over the remaining 6 hours6 \text{ hours}, resulting in a rate of 2,520 mL÷6 hours=420 mL/hr2,520 \text{ mL} \div 6 \text{ hours} = 420 \text{ mL/hr}.

Adım Adım Çözüm

1
Calculate the burned Total Body Surface Area (TBSA).
36%36\%
Using the Rule of Nines, the anterior chest is 9%9\%, anterior abdomen is 9%9\%, entire right arm is 9%9\%, and anterior left leg is 9%9\%.
2
Calculate the total 24-hour fluid requirement using the ATLS 10th edition formula.
5,040 mL5,040 \text{ mL}
The current guideline for adult thermal burns is 2 mL×weight (kg)×TBSA (%)2 \text{ mL} \times \text{weight (kg)} \times \text{TBSA (\%)}. Therefore, 2×70×36=5,040 mL2 \times 70 \times 36 = 5,040 \text{ mL}.
3
Determine the fluid volume required for the first 8-hour block.
2,520 mL2,520 \text{ mL}
Half of the total 24-hour calculated volume is given in the first 8 hours8 \text{ hours} post-injury.
4
Adjust the infusion time based on the delayed presentation.
6 hours6 \text{ hours}
The 8-hour block begins at the time of injury. Since 2 hours2 \text{ hours} have already elapsed, the 2,520 mL2,520 \text{ mL} must be infused over the remaining 6 hours6 \text{ hours}.
5
Calculate the required hourly infusion rate.
420 mL/hr420 \text{ mL/hr}
2,520 mL2,520 \text{ mL} divided by 6 hours6 \text{ hours} equals 420 mL/hr420 \text{ mL/hr}.

Anahtar Kavram

Management of acute burn fluid resuscitation involves applying the Rule of Nines, utilizing the updated ATLS guidelines (2 mL/kg/%TBSA2 \text{ mL/kg/\%TBSA} for adult thermal burns), and dynamically adjusting the infusion rate for the time elapsed since the injury.
Soru 77Soru →

A 32-year-old male utility worker is brought to the emergency department after sustaining a high-voltage electrical injury while repairing a power line. The incident occurred exactly 2 hours prior to his arrival. On presentation, he is awake, alert, and complains of severe pain in his extremities. His temperature is 36.8∘C36.8^\circ \text{C} (98.2∘F98.2^\circ \text{F}), heart rate is 112/min112/\text{min}, and blood pressure is 110/70 mm Hg110/70 \text{ mm Hg}. Physical examination reveals deep, full-thickness burns involving his entire anterior trunk and his entire right lower extremity. There are also deep, charred entry and exit wounds on his right hand and left foot, though these cover a negligible surface area. His body weight is estimated at 75 kg75 \text{ kg}. A Foley catheter is inserted to monitor urine output, and large-bore intravenous access is obtained. According to the Advanced Trauma Life Support (ATLS) guidelines for high-voltage electrical injuries, what is the appropriate initial intravenous fluid infusion rate (in mL/hour\text{mL/hour}) for this patient?

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Cevap: 900

Cevap

900
The appropriate initial fluid infusion rate is 900 mL/hour because high-voltage electrical burns require 4 mL/kg/%TBSA for fluid resuscitation, totaling 10,800 mL for this 75 kg patient with 36% TBSA burns. Half of this total (5,400 mL) must be administered in the first 8 hours post-injury. Because 2 hours have already elapsed, the 5,400 mL must be infused over the remaining 6 hours, yielding a rate of 900 mL/hour.

Adım Adım Çözüm

1
Calculate the Total Body Surface Area (TBSA) burned using the Rule of Nines.
TBSA = 36%
The entire anterior trunk accounts for 18% and the entire right lower extremity accounts for 18% (18+18=3618 + 18 = 36).
2
Select the correct ATLS fluid resuscitation multiplier for the injury mechanism.
4 mL/kg/%TBSA
High-voltage electrical injuries require 4 mL/kg/%TBSA to ensure adequate diuresis and prevent acute kidney injury from rhabdomyolysis, unlike the 2 mL/kg/%TBSA used for standard thermal burns.
3
Calculate the total estimated fluid requirement for the first 24 hours.
10,800 mL
4 mL×75 kg×36%=10,800 mL4 \text{ mL} \times 75 \text{ kg} \times 36\% = 10,800 \text{ mL}.
4
Determine the fluid volume to be administered in the first 8 hours.
5,400 mL
Half of the total 24-hour calculated volume is given in the first 8 hours (10,800÷2=5,400 mL10,800 \div 2 = 5,400 \text{ mL}).
5
Calculate the target hourly infusion rate, accounting for time elapsed since the injury.
900 mL/hour
The patient presented 2 hours post-injury, leaving 6 hours for the first 8-hour window (8−2=68 - 2 = 6). The rate is 5,400 mL÷6 hours=900 mL/hour5,400 \text{ mL} \div 6 \text{ hours} = 900 \text{ mL/hour}.

Anahtar Kavram

High-voltage electrical burns require fluid resuscitation at 4 mL/kg/%TBSA, with half the volume administered over the first 8 hours starting from the time of injury, requiring rate adjustments for delayed presentations.
Tahmini Süre:2m 0s
Soru 78Soru →

A 35-year-old man is brought to the emergency department after a motor vehicle collision. On initial evaluation, he opens his eyes only when the physician loudly calls his name. When asked what happened, he responds with single, random words that do not form coherent sentences. When pressure is applied to his right fingernail bed, he briskly flexes his elbow to pull his right arm away from the stimulus, but does not reach toward the examiner's hand. What is this patient's Glasgow Coma Scale (GCS) score?

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Cevap: 10

Cevap

10
The Glasgow Coma Scale (GCS) evaluates a patient's level of consciousness based on three categories: Eye Opening (E), Verbal Response (V), and Best Motor Response (M). This patient opens his eyes to sound (E3), uses inappropriate words rather than conversational sentences (V3), and withdraws from pain without localizing it (M4). The total score is calculated by adding these components: E3 + V3 + M4 = 10 points. Calculating an accurate GCS score is a critical initial step in the primary survey of any trauma patient to determine the severity of traumatic brain injury and the potential need for definitive airway management.

Adım Adım Çözüm

1
Evaluate Eye Opening (E)
3 points
The patient opens his eyes in response to a verbal stimulus (his name being called).
2
Evaluate Verbal Response (V)
3 points
The patient speaks single, random words but does not engage in a conversation.
3
Evaluate Motor Response (M)
4 points
The patient exhibits normal flexion and withdrawal from a painful stimulus without purposefully localizing the source of pain.
4
Calculate the total score
10 points
The total Glasgow Coma Scale (GCS) score is the sum of the Eye, Verbal, and Motor components (3 + 3 + 4).

Anahtar Kavram

Glasgow Coma Scale (GCS) Scoring
Tahmini Süre:45s
Soru 79Soru →

A 24-year-old man is brought to the emergency department after a high-speed motorcycle collision. He complains of severe pain in his left leg. Physical examination reveals a deformed left knee with an obvious posterior dislocation. The dislocation is promptly reduced in the emergency department. Afterward, palpation of the left pedal pulses reveals they are diminished compared to the right side. The patient remains hemodynamically stable. To evaluate for potential popliteal artery injury, the physician measures systolic blood pressures using a continuous-wave Doppler ultrasound and a sphygmomanometer. The following systolic pressures are recorded:

LocationArterySystolic Pressure (mmHg)
Right ArmBrachial130
Left ArmBrachial140
Right LegDorsalis Pedis135
Right LegPosterior Tibial138
Left LegDorsalis Pedis60
Left LegPosterior Tibial70

What is the Ankle-Brachial Index (ABI) of the injured extremity?

Cevabı ve açıklamayı göster

Cevap: 0.5

Cevap

0.5
The correct Ankle-Brachial Index (ABI) is 0.5. The ABI is a crucial non-invasive test used to screen for peripheral vascular injury, highly relevant after a posterior knee dislocation which carries a high risk of popliteal artery injury. To properly calculate the ABI, you must divide the highest systolic blood pressure at the ankle of the affected leg by the highest systolic blood pressure measured in either arm. In this patient, the highest left ankle pressure is the posterior tibial artery (70 mmHg70\text{ mmHg}). The highest brachial pressure out of both arms is the left arm (140 mmHg140\text{ mmHg}). Therefore, the ABI is 70/140=0.570 / 140 = 0.5. An ABI <0.9< 0.9 in the setting of trauma strongly suggests an arterial injury and requires immediate further evaluation, typically with a CT angiography. Notably, if the patient were hemodynamically unstable with hard signs of vascular injury (e.g., pulsatile bleeding, expanding hematoma, absent pulses), delaying for an ABI would be contraindicated, and the patient would require immediate surgical exploration.

Adım Adım Çözüm

1
Determine the highest ankle systolic pressure on the injured (left) extremity.
70 mmHg
The Ankle-Brachial Index uses the highest pressure measured from either the dorsalis pedis or posterior tibial artery of the affected limb.
2
Determine the highest brachial systolic pressure out of both arms.
140 mmHg
The denominator for the ABI calculation must always be the highest systemic systolic pressure, which represents the true central arterial pressure, regardless of which side is injured.
3
Calculate the ratio by dividing the result of Step 1 by the result of Step 2.
70 / 140 = 0.5
This yields the correct Ankle-Brachial Index. An ABI < 0.9 suggests vascular injury.

Anahtar Kavram

Calculation and interpretation of the Ankle-Brachial Index (ABI) in lower extremity trauma to assess for vascular injury.
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A 45-year-old industrial worker is brought to the emergency department after a chemical vat exploded. He sustained deep partial-thickness burns. Physical examination reveals burns covering his entire anterior trunk (chest and abdomen), his entire right upper extremity (both anterior and posterior surfaces), and his entire right lower extremity (both anterior and posterior surfaces). No other areas of the body are affected.

Based on the Rule of Nines, what is the estimated total body surface area (TBSA) percentage affected by burns in this patient?

Cevabı ve açıklamayı göster

Cevap: 45

Cevap

45
The adult Rule of Nines is a standardized method used to estimate the total body surface area (TBSA) affected by burns. It divides the body into sections representing multiples of 9%. In adults, the anterior trunk accounts for 18%, the posterior trunk accounts for 18%, each entire upper extremity accounts for 9%, each entire lower extremity accounts for 18%, the entire head and neck accounts for 9%, and the perineum accounts for 1%. For this patient, the calculation includes the anterior trunk (18%), the entire right upper extremity (9%), and the entire right lower extremity (18%). Adding these values together (18 + 9 + 18) yields a TBSA of 45%.

Adım Adım Çözüm

1
Determine the percentage for the anterior trunk.
18%
According to the adult Rule of Nines, the anterior portion of the torso (chest and abdomen combined) is assigned 18%.
2
Determine the percentage for the entire right upper extremity.
9%
The entire arm (anterior and posterior surfaces combined) is assigned 9%.
3
Determine the percentage for the entire right lower extremity.
18%
The entire leg (anterior and posterior surfaces combined) is assigned 18%.
4
Sum the regional percentages to find the total body surface area (TBSA).
45%
18% + 9% + 18% = 45% total body surface area affected.

Anahtar Kavram

Estimation of burn size using the Rule of Nines
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