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Question 121Question →

A 34-year-old man is brought to the emergency department after being found unresponsive in his garage. Emergency medical personnel found a half-empty container of commercial antifreeze nearby. His blood pressure is 145/90 mm Hg, pulse is 112/min, and respirations are 24/min. Physical examination reveals deep, rapid breathing.

Laboratory studies show the following:
Serum sodium: 142 mEq/L
Serum potassium: 4.5 mEq/L
Serum chloride: 104 mEq/L
Serum bicarbonate: 12 mEq/L
Blood urea nitrogen: 28 mg/dL
Serum creatinine: 1.4 mg/dL
Serum glucose: 126 mg/dL
Measured serum osmolality: 345 mOsm/kg $H_2O$

Arterial blood gas on room air shows a pH of 7.22 and $pCO_2$ of 26 mm Hg.

Based on the provided laboratory values, what is this patient's serum osmolar gap?

Show answer & explanation

Answer: 44

Answer

The patient's calculated serum osmolar gap is 44 mOsm/kg.

The correct osmolar gap is derived by subtracting the calculated serum osmolality from the measured serum osmolality. Calculated osmolality is found using the formula:

2 \times Sodium + (BUN / 2.8) + (Glucose / 18)

. For this patient,

2(142) + (28 / 2.8) + (126 / 18) = 284 + 10 + 7 = 301

mOsm/kg. Subtracting this expected value from the measured osmolality of 345 mOsm/kg yields a gap of 44 mOsm/kg, strongly suggesting toxic alcohol (ethylene glycol) poisoning.

Step-by-Step Solution

1

Calculate the patient's expected serum osmolality using the standard formula.

Calculated Osmolality =

284 + 10 + 7 = 301

mOsm/kg.

Serum osmolality is primarily determined by sodium and its accompanying anions, urea, and glucose. The formula

(2 \times Na) + (BUN / 2.8) + (Glucose / 18)

converts US units (mg/dL) to SI units (mmol/L) to find the total expected osmoles.

2

Subtract the calculated expected osmolality from the measured osmolality provided in the laboratory results.

345 - 301 = 44

mOsm/kg.

The osmolar gap represents unmeasured osmotically active particles in the blood. A gap greater than 10-15 mOsm/kg is highly suggestive of toxic alcohol ingestion, such as ethylene glycol.

Key Concept

Diagnosis of toxic alcohol ingestion using the osmolar gap, calculated by subtracting the expected osmolality from the measured osmolality.

Question 122Question →

A 35-year-old woman presents for follow-up of generalized weakness and worsening fatigue over the last two weeks. Her medical history is notable for Sjögren syndrome, for which she takes hydroxychloroquine. She appears comfortable and her vital signs are stable. Muscle strength testing demonstrates 4/5 strength in the bilateral quadriceps and hip flexors. Given her symptoms, laboratory testing is obtained:

Serum:
Sodium: 140 mEq/L
Potassium: 2.8 mEq/L
Chloride: 115 mEq/L
Bicarbonate: 15 mEq/L
Blood urea nitrogen: 14 mg/dL
Creatinine: 0.8 mg/dL
Glucose: 95 mg/dL

Urine:
Sodium: 60 mEq/L
Potassium: 25 mEq/L
Chloride: 45 mEq/L
pH: 6.0

To determine the etiology of this patient's metabolic acidosis, the physician decides to calculate the urine anion gap. What is the value of this patient's urine anion gap?

Show answer & explanation

Answer: 40

Answer

40 mEq/L

The patient has a normal anion gap metabolic acidosis (NAGMA), as her serum anion gap is

140 - (115 + 15) = 10

mEq/L (normal: 8-12 mEq/L). In the setting of NAGMA, the urine anion gap (UAG) helps differentiate between renal loss of bicarbonate (e.g., renal tubular acidosis) and extrarenal loss (e.g., diarrhea). The formula for UAG is

U_{Na^+} + U_{K^+} - U_{Cl^-}

. Plugging in this patient's urinary values:

60 \text{ mEq/L} + 25 \text{ mEq/L} - 45 \text{ mEq/L} = 40 \text{ mEq/L}

. A positive UAG indicates impaired renal

NH_4^+

excretion, which is consistent with renal tubular acidosis (Type 1 RTA in this patient with Sjögren syndrome and severe hypokalemia).

Step-by-Step Solution

1

Identify the formula for urine anion gap.

UAG = U_{Na^+} + U_{K^+} - U_{Cl^-}

The urine anion gap is used to indirectly measure urinary ammonium (

NH_4^+

) excretion in a patient with a normal anion gap metabolic acidosis.

2

Extract the appropriate urinary electrolyte values from the laboratory data.

U_{Na^+} = 60

mEq/L,

U_{K^+} = 25

mEq/L,

U_{Cl^-} = 45

mEq/L

These are the measured urinary cations and anions needed for the calculation. Care must be taken not to mistakenly use serum values.

3

Calculate the urine anion gap.

60 + 25 - 45 = 40

mEq/L

Adding the cations and subtracting the chloride yields a positive value of 40 mEq/L, indicating impaired renal ammonium excretion (typical of distal renal tubular acidosis).

Key Concept

The urine anion gap (

U_{Na^+} + U_{K^+} - U_{Cl^-}

) is calculated to differentiate between gastrointestinal and renal causes of normal anion gap metabolic acidosis; a positive value suggests a renal etiology such as renal tubular acidosis.

Question 123Question →

A 56-year-old man with a history of alcohol use disorder and decompensated cirrhosis is admitted to the hospital with worsening lethargy and diffuse abdominal swelling over the past week. His temperature is $37.2^\circ\text{C}$ ( $99.0^\circ\text{F}$ ), blood pressure is $102/64 \text{ mm Hg}$ , pulse is $108/\text{min}$ , and respiratory rate is $22/\text{min}$ . Physical examination reveals scleral icterus, shifting dullness to percussion, and bilateral pitting edema. Laboratory studies are obtained:

Laboratory Parameter	Value
Sodium ( $Na^+$ )	$138 \text{ mEq/L}$
Potassium ( $K^+$ )	$3.8 \text{ mEq/L}$
Chloride ( $Cl^-$ )	$107 \text{ mEq/L}$
Bicarbonate ( $HCO_3^-$ )	$17 \text{ mEq/L}$
Blood urea nitrogen	$32 \text{ mg/dL}$
Creatinine	$1.5 \text{ mg/dL}$
Albumin	$1.6 \text{ g/dL}$

Assuming a normal reference serum albumin of $4.0 \text{ g/dL}$ , what is this patient's albumin-corrected serum anion gap?

Show answer & explanation

Answer: 20

Answer

20

The correct calculation yields

20 \text{ mEq/L}

. The patient's observed anion gap is

14 \text{ mEq/L}

(

138 - [107 + 17]

). Because the patient has severe hypoalbuminemia (

1.6 \text{ g/dL}

), this observed value is misleadingly low. To find the true anion gap, a correction of

2.5 \text{ mEq/L}

is added for every

1.0 \text{ g/dL}

that the albumin is below the reference of

4.0 \text{ g/dL}

. The albumin deficit is

2.4 \text{ g/dL}

(

4.0 - 1.6

). Multiplying

2.4

by

2.5

gives a correction factor of

6.0 \text{ mEq/L}

. Adding this to the observed gap of

14 \text{ mEq/L}

gives a true, corrected anion gap of

20 \text{ mEq/L}

, revealing a hidden high anion gap metabolic acidosis (likely due to lactic acidosis, given his elevated heart rate, low blood pressure, and severe liver disease).

Step-by-Step Solution

1

Calculate the uncorrected serum anion gap.

138 - (107 + 17) = 138 - 124 = 14 \text{ mEq/L}

The standard anion gap estimates unmeasured anions in the blood, primarily driven by the difference between major measured cations and anions.

2

Identify the patient's albumin deficit.

4.0 \text{ g/dL} - 1.6 \text{ g/dL} = 2.4 \text{ g/dL}

Albumin is a negatively charged protein and represents a major portion of the baseline unmeasured anions. Hypoalbuminemia artificially lowers the baseline anion gap.

3

Calculate the albumin correction factor.

2.4 \times 2.5 = 6.0 \text{ mEq/L}

For every

1.0 \text{ g/dL}

drop in serum albumin below the normal

4.0 \text{ g/dL}

, the anion gap is falsely lowered by approximately

2.5 \text{ mEq/L}

.

4

Add the correction factor to the observed anion gap.

14 + 6.0 = 20 \text{ mEq/L}

Adding the lost unmeasured anion equivalent back to the observed gap reveals the true metabolic state, uncovering a high anion gap metabolic acidosis.

Key Concept

In patients with significant hypoalbuminemia (such as those with cirrhosis or nephrotic syndrome), the serum anion gap must be corrected upward (

2.5 \text{ mEq/L}

for every

1 \text{ g/dL}

decrease in albumin below

4.0 \text{ g/dL}

) to accurately identify an underlying high anion gap metabolic acidosis.

Question 124Question →

A 42-year-old woman is brought to the emergency department via ambulance after being rescued from a house fire. Paramedics report the injury occurred exactly 2 hours prior to her arrival. On physical examination, she is in significant pain but has a patent airway with no signs of inhalation injury. She has deep partial-thickness and full-thickness burns involving her entire anterior torso, her entire left arm, and the anterior surface of her left leg. The remainder of her skin is uninjured. Her weight is $75\text{ kg}$ . Large-bore intravenous access is established. Based on the most 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 in $\text{mL/hr}$ ?

Show answer & explanation

Answer: 450

Answer

450

Based on the 10th edition ATLS guidelines, adult thermal burns are resuscitated starting at

2\text{ mL/kg/\%TBSA}

(modified Parkland formula). For this patient (

75\text{ kg}

,

36\%\text{ TBSA}

), the 24-hour requirement is

2 \times 75 \times 36 = 5,400\text{ mL}

. Half of this volume (

2,700\text{ mL}

) is given in the first 8 hours post-injury. Because 2 hours have elapsed since the injury occurred, the

2,700\text{ mL}

must be infused over the remaining 6 hours, resulting in a rate of

450\text{ mL/hr}

.

Step-by-Step Solution

1

Calculate the total percentage of body surface area (%TBSA) burned.

36%

Using the Rule of Nines: anterior torso is 18%, entire left arm is 9%, and anterior left leg is 9%. Sum = 36%.

2

Calculate the total 24-hour fluid resuscitation volume.

5,400 mL

Current ATLS guidelines for adult thermal burns recommend 2 mL x weight (kg) x %TBSA. Therefore, 2 x 75 x 36 = 5,400 mL.

3

Determine the fluid volume required for the first 8 hours post-injury.

2,700 mL

Half of the 24-hour total volume is administered in the first 8 hours post-injury. 5,400 / 2 = 2,700 mL.

4

Calculate the hourly infusion rate.

450 mL/hr

Since 2 hours have elapsed since the injury, the 2,700 mL must be given over the remaining 6 hours (8 - 2 = 6). 2,700 / 6 = 450.

Key Concept

Initial fluid resuscitation for adult thermal burns using the modified Parkland formula (ATLS 10th edition).

Question 125Question →

A $45$ -year-old male construction worker is brought to the emergency department after a high-voltage electrical injury. He accidentally touched a live power line and was thrown to the ground. The incident occurred exactly $3\text{ hours}$ prior to his arrival at the trauma center. No intravenous fluids were administered during transport. The patient weighs $90\text{ kg}$ . On physical examination, there are deep, full-thickness burns covering his entire anterior torso and his entire right upper extremity. The remainder of his skin is unaffected. A Foley catheter is placed, yielding dark red urine. According to the Advanced Trauma Life Support (ATLS) guidelines, at what rate (in $\text{mL/hr}$ ) should intravenous lactated Ringer solution be administered over the remainder of the first 8-hour post-injury window?

Show answer & explanation

Answer: 972

Answer

972 mL/hr

The correct calculation appropriately utilizes the ATLS multiplier for electrical burns (4 mL/kg/%TBSA), divides the 24-hour total in half for the first 8-hour window, and divides that half strictly by the remaining 5 hours since the injury occurred 3 hours prior.

Step-by-Step Solution

1

Calculate the burned body surface area using the Rule of Nines

27% TBSA

Accurate TBSA is required for the fluid resuscitation formula. Anterior torso = 18%, right upper extremity = 9%.

2

Select the appropriate ATLS fluid multiplier for electrical injuries

4 mL/kg/%TBSA

High-voltage electrical burns with suspected myoglobinuria require higher fluid volumes (4 mL) to flush the kidneys compared to standard thermal burns (2 mL).

3

Calculate the total 24-hour fluid volume

9,720 mL

4 mL × 90 kg × 27% = 9,720 mL total for the first 24 hours.

4

Determine the fluid volume for the first 8-hour window

4,860 mL

Standard ATLS guidelines dictate that 50% of the calculated 24-hour fluid volume is given in the first 8 hours post-injury.

5

Calculate the infusion rate over the remaining hours of the first window

972 mL/hr

Because 3 hours have already elapsed since the injury, the 4,860 mL must be infused over the remaining 5 hours (4,860 / 5).

Key Concept

Fluid resuscitation for electrical burns with myoglobinuria utilizes 4 mL/kg/%TBSA, with half administered over the first 8 hours from the time of injury.

Question 126Question →

A 54-year-old man is brought to the emergency department by emergency medical services after his wife found him unresponsive in their garage. She reports that he has a history of major depressive disorder and alcohol use disorder. An open, half-empty bottle of engine antifreeze was found nearby. Vital signs show a temperature of $36.2^\circ\text{C}$ ( $97.2^\circ\text{F}$ ), pulse of $105$ /min, blood pressure of $145/88$ mm Hg, and respirations of $26$ /min. Physical examination reveals a comatose patient with no focal neurological deficits.

Laboratory studies reveal the following:

Parameter	Value
Sodium	$141$ mEq/L
Potassium	$4.2$ mEq/L
Chloride	$104$ mEq/L
Bicarbonate	$8$ mEq/L
Blood urea nitrogen	$42$ mg/dL
Creatinine	$1.8$ mg/dL
Glucose	$126$ mg/dL
Ethanol	Undetectable
Measured serum osmolality	$340$ mOsm/kg $\text{H}_2\text{O}$

Arterial blood gas on room air:
pH: $7.18$
$\text{pCO}_2$ : $20$ mm Hg
$\text{pO}_2$ : $98$ mm Hg

Urinalysis reveals the presence of envelope-shaped crystals.

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

Show answer & explanation

Answer: 36

Answer

36

The patient's osmolar gap is

36

mOsm/kg

\text{H}_2\text{O}

. This is obtained by first finding the calculated serum osmolality using the formula:

2[\text{Na}^+] + \frac{\text{Glucose}}{18} + \frac{\text{BUN}}{2.8}

. Substituting the provided values yields

2(141) + \frac{126}{18} + \frac{42}{2.8} = 282 + 7 + 15 = 304

mOsm/kg

\text{H}_2\text{O}

. Subtracting this calculated value from the measured osmolality (

340

mOsm/kg

\text{H}_2\text{O}

) gives a gap of

36

mOsm/kg

\text{H}_2\text{O}

.

Step-by-Step Solution

1

Determine the formula for calculated serum osmolality.

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

.

This standard calculation is required when assessing a patient for suspected toxic alcohol ingestion.

2

Input the patient's laboratory values into the formula.

\text{Calculated Osmolality} = 2(141) + \frac{126}{18} + \frac{42}{2.8} = 282 + 7 + 15 = 304

mOsm/kg

\text{H}_2\text{O}

.

Dividing glucose by 18 and BUN by 2.8 correctly converts these values from mg/dL to mOsm/L so they can be summed with sodium.

3

Calculate the serum osmolar gap.

\text{Osmolar Gap} = \text{Measured Osmolality} - \text{Calculated Osmolality} = 340 - 304 = 36

mOsm/kg

\text{H}_2\text{O}

.

The difference between measured and calculated osmolality reveals the concentration of unmeasured osmotically active substances (e.g., ethylene glycol).

Key Concept

The serum osmolar gap is calculated by subtracting the calculated serum osmolality (

2[\text{Na}^+] + \frac{\text{Glucose}}{18} + \frac{\text{BUN}}{2.8}

) from the measured serum osmolality. A gap

>10

mOsm/kg

\text{H}_2\text{O}

suggests the presence of unmeasured osmotically active agents such as toxic alcohols (e.g., ethylene glycol, methanol).

Question 127Question →

A 68-year-old man is evaluated in the intensive care unit for oliguria. He was admitted 4 days ago for severe acute pancreatitis complicated by shock, which required aggressive intravenous fluid resuscitation and transient vasopressor support. His blood pressure is now 112/70 mm Hg and his heart rate is 88/min on no vasoactive medications. Urine output over the past 24 hours has been 280 mL. Laboratory studies show the following:

Serum values:
- Sodium: $140$ mEq/L
- Potassium: $4.5$ mEq/L
- Chloride: $105$ mEq/L
- Bicarbonate: $21$ mEq/L
- Blood urea nitrogen: $48$ mg/dL
- Creatinine: $2.8$ mg/dL (baseline $0.9$ mg/dL)

Urine values:
- Sodium: $50$ mEq/L
- Creatinine: $50$ mg/dL

Urinalysis demonstrates the presence of muddy brown granular casts. Based on these findings, what is this patient's calculated fractional excretion of sodium (FENa) in percentage?

Show answer & explanation

Answer: 2

Answer

The patient's fractional excretion of sodium (FENa) is 2%.

The correct calculation yields a fractional excretion of sodium of 2%. This patient has developed acute tubular necrosis (ATN), a form of intrinsic renal failure, as a complication of prolonged hypotension/shock (ischemic ATN). Intrinsic renal failure is characterized by structural damage to the renal tubules (evidenced by muddy brown granular casts). Because the damaged tubules cannot adequately reabsorb sodium, the FENa is typically > 2% in ATN. FENa is determined by the formula:

FENa = \frac{U_{Na} \times P_{Cr}}{P_{Na} \times U_{Cr}} \times 100

. Substituting the values from the vignette gives

\frac{50 \times 2.8}{140 \times 50} \times 100 = 2\%

.

Step-by-Step Solution

1

Extract the four relevant laboratory values required for the FENa calculation.

U_{Na} = 50

,

P_{Cr} = 2.8

,

P_{Na} = 140

, and

U_{Cr} = 50

.

The fractional excretion of sodium relies exclusively on urine and serum concentrations of sodium and creatinine.

2

Apply the standard formula for FENa.

FENa = \frac{U_{Na} \times P_{Cr}}{P_{Na} \times U_{Cr}} \times 100

.

This formula effectively calculates the clearance of sodium divided by the clearance of creatinine, estimating the proportion of filtered sodium that is excreted.

3

Calculate the numerator.

50 \times 2.8 = 140

.

Multiplies urine sodium by serum creatinine.

4

Calculate the denominator.

140 \times 50 = 7000

.

Multiplies serum sodium by urine creatinine.

5

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

\frac{140}{7000} \times 100 = 0.02 \times 100 = 2\%

.

Yields the final percentage indicating the diagnostic category of the acute kidney injury.

Key Concept

Calculation and interpretation of Fractional Excretion of Sodium (FENa) to distinguish intrinsic acute tubular necrosis (ATN) from prerenal azotemia.

Question 128Question →

A 50-year-old man is brought to the emergency department via ambulance exactly 2 hours after being caught in a large brush fire. His estimated weight is $75 \text{ kg}$ . On physical examination, the patient is awake and speaking in full sentences without hoarseness. Airway is patent. He has deep partial-thickness and full-thickness burns involving his entire anterior trunk, his entire right arm, and the anterior surface of his left leg. The remainder of his body surface is unaffected. A large-bore peripheral intravenous line is established. Based on the most current Advanced Trauma Life Support (ATLS) guidelines for adult thermal burns, what is the target intravenous fluid administration rate for the remainder of the first 8 hours post-injury?

Show answer & explanation

Answer: 450

Answer

450

Based on the ATLS 10th edition update, the initial fluid resuscitation for adults with thermal burns is calculated using

2 \text{ mL} \times \text{weight (kg)} \times \text{\%TBSA}

. For a

75 \text{ kg}

man with

36\%

TBSA burns, this equals

5400 \text{ mL}

for the first 24 hours. Half of this volume (

2700 \text{ mL}

) must be given in the first 8 hours after the burn occurred. Because the patient arrived 2 hours post-injury, this

2700 \text{ mL}

must be infused over the remaining 6 hours, resulting in a rate of

450 \text{ mL/hr}

.

Step-by-Step Solution

1

Calculate the total percentage of burned surface area (%TBSA).

Anterior trunk = 18%, Entire right arm = 9%, Anterior left leg = 9%. Total = 18% + 9% + 9% = 36% TBSA.

The Rule of Nines dictates the body is divided into areas representing 9% (or multiples of 9%) to estimate burn extent.

2

Calculate the total fluid requirement for the first 24 hours.

2 \text{ mL} \times 75 \text{ kg} \times 36 = 5400 \text{ mL}

total.

Current ATLS guidelines recommend

2 \text{ mL/kg/\%TBSA}

for adult thermal burns to minimize the risk of over-resuscitation and compartment syndrome.

3

Determine the fluid volume required for the first 8 hours.

5400 \text{ mL} / 2 = 2700 \text{ mL}

.

Half of the total 24-hour fluid requirement must be given within the first 8 hours from the time of injury.

4

Calculate the remaining time for the first 8-hour window.

8 \text{ hours} - 2 \text{ hours elapsed} = 6 \text{ hours remaining}

.

The 8-hour window begins at the time of the injury, not the time of arrival at the hospital.

5

Determine the hourly intravenous fluid rate.

2700 \text{ mL} / 6 \text{ hours} = 450 \text{ mL/hr}

.

This steady rate ensures the patient receives the target 2700 mL over the correct physiological timeframe.

Key Concept

Fluid resuscitation in adult thermal burns (ATLS 10th Edition guidelines)

Estimated Time:1m 30s

Question 129Question →

A 22-year-old woman with a history of bulimia nervosa and type 1 diabetes mellitus is brought to the emergency department by her roommate due to severe lethargy and confusion. The roommate reports that the patient has been intentionally restricting her insulin doses and has been vomiting frequently over the past 3 days. Temperature is 37.0°C (98.6°F), blood pressure is 100/60 mmHg, pulse is 118/min, and respirations are 24/min.

Laboratory studies show:

Parameter	Value
Serum sodium	142 mEq/L
Serum potassium	3.4 mEq/L
Serum chloride	94 mEq/L
Serum bicarbonate	16 mEq/L
Blood glucose	480 mg/dL

Arterial blood gas on room air:

Parameter	Value
pH	7.32
$pCO_2$	32 mmHg
$pO_2$	95 mmHg

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

Show answer & explanation

Answer: 2.5

Answer

The calculated delta ratio is 2.5.

The delta ratio is calculated by dividing the change in anion gap (

\Delta AG

) by the change in serum bicarbonate (

\Delta HCO_3^-

). For this patient, the measured anion gap is

142 - (94 + 16) = 32

. The

\Delta AG

is

32 - 12 = 20

. The

\Delta HCO_3^-

is

24 - 16 = 8

. Therefore, the delta ratio is

\frac{20}{8} = 2.5

. In an uncomplicated high anion gap metabolic acidosis (like pure diabetic ketoacidosis), the ratio is typically between 1.0 and 2.0. A ratio greater than 2.0 indicates that the bicarbonate level is higher than expected for the degree of anion gap elevation, which signifies a concurrent metabolic alkalosis. In this clinical scenario, the metabolic alkalosis is driven by the patient's frequent vomiting, leading to a loss of

H^+

and

Cl^-

.

Step-by-Step Solution

1

Calculate the patient's measured serum anion gap.

AG = 142 - (94 + 16) = 32

mEq/L.

The first step in evaluating a metabolic acidosis is determining the presence and size of an anion gap.

2

Calculate the change in anion gap (

\Delta AG

).

\Delta AG = 32 - 12 = 20

mEq/L.

This value represents the excess unmeasured anions contributing to the acidosis.

3

Calculate the change in serum bicarbonate (

\Delta HCO_3^-

).

\Delta HCO_3^- = 24 - 16 = 8

mEq/L.

This determines how much the bicarbonate has dropped from the normal baseline.

4

Calculate the delta ratio.

Ratio = \frac{20}{8} = 2.5

.

The delta ratio helps identify mixed acid-base disorders, such as a concurrent metabolic alkalosis.

Key Concept

The delta ratio (

\frac{\Delta AG}{\Delta HCO_3^-}

) is used in high anion gap metabolic acidosis to identify concurrent mixed acid-base disorders, with a ratio > 2.0 indicating an additional metabolic alkalosis.

Alternative Method

The delta gap (expected bicarbonate) can also be calculated as: Measured

HCO_3^-

+

\Delta AG

. Here,

16 + 20 = 36

mEq/L. Since the expected normal bicarbonate (36) is significantly higher than the normal baseline of 24, a concurrent metabolic alkalosis is confirmed.

Estimated Time:1m 30s

Question 130Question →

A 50-year-old man is evaluated in the neuro-intensive care unit. He was admitted four days ago after sustaining a severe closed head injury in a high-speed motor vehicle collision. Over the past 24 hours, the nursing staff reports a dramatic increase in his urine output, totaling 7.5 L of pale, dilute urine. He is currently intubated and sedated. His weight on admission was 72 kg, and his current weight is 70 kg. Vital signs show a blood pressure of 102/64 mm Hg and a heart rate of 118/min. Laboratory studies reveal a serum sodium concentration of 160 mEq/L and a serum osmolality of 330 mOsm/kg $H_2O$ . Urine osmolality is 115 mOsm/kg $H_2O$ . The patient is diagnosed with central diabetes insipidus.

What is this patient's estimated free water deficit in liters?

Show answer & explanation

Answer: 6

Answer

6

The patient has developed severe hypernatremia due to massive free water loss from central diabetes insipidus. To safely correct his sodium levels and formulate an intravenous fluid replacement plan, the free water deficit must be calculated. The standard formula is

FWD = TBW \times (\frac{\text{Measured } Na^+}{140} - 1)

. For a 50-year-old adult male, TBW is estimated as 60% of his current body weight (

0.6 \times 70 \text{ kg} = 42 \text{ L}

). Plugging in the serum sodium of 160 mEq/L yields

42 \times (\frac{160}{140} - 1) = 42 \times \frac{1}{7} = 6

liters.

Step-by-Step Solution

1

Calculate Total Body Water (TBW)

42 L

TBW for an adult male is approximately 60% (0.6) of his lean body weight. Current weight (70 kg) should be used.

2

Set up the Free Water Deficit formula

FWD = 42 \times (\frac{160}{140} - 1)

The standard formula requires the TBW and the ratio of the patient's measured serum sodium to the ideal serum sodium (140 mEq/L).

3

Solve the equation mathematically

6

42 \times (\frac{20}{140})

simplifies to

42 \times \frac{1}{7}

, which equals exactly 6.

Key Concept

Calculation of Free Water Deficit in Hypernatremia

Estimated Time:1m 30s

Question 131Question →

A $45$ -year-old man is brought to the burn center via helicopter after an industrial explosion. The incident occurred exactly $3$ hours prior to his arrival. He weighs $100$ kg. Physical examination reveals deep partial-thickness burns over his entire anterior trunk, entire right arm, and entire left arm. His airway is secure, and he is hemodynamically stable. Two large-bore intravenous lines are placed. Based on current Advanced Trauma Life Support (ATLS) guidelines (using $2 \text{ mL/kg} \times \%\text{TBSA}$ for adult thermal burns), at what rate (in $\text{mL/hr}$ ) should the intravenous lactated Ringer's solution be administered for the remainder of the first $8$ -hour post-injury period?

Show answer & explanation

Answer: 720

Answer

720

The correct infusion rate is

720 \text{ mL/hr}

. To determine this, first calculate the TBSA using the Rule of Nines: the entire anterior trunk (

18\%

), right arm (

9\%

), and left arm (

9\%

) sum to

36\%

. Using the modern ATLS fluid resuscitation formula for adults with thermal burns (

2 \text{ mL} \times \text{weight in kg} \times \%\text{TBSA}

), the total 24-hour fluid requirement is

2 \times 100 \times 36 = 7200 \text{ mL}

. Exactly half of this volume (

3600 \text{ mL}

) must be administered in the first 8 hours from the time of the injury. Since the injury occurred

3

hours prior to arrival, there are only

5

hours remaining in this initial 8-hour window. Therefore, the rate is

3600 \text{ mL} / 5 \text{ hours} = 720 \text{ mL/hr}

.

Step-by-Step Solution

1

Calculate the estimated burn size using the Rule of Nines.

36\%

TBSA

Accurate estimation of Total Body Surface Area (TBSA) is required to calculate fluid needs. Anterior trunk =

18\%

, right arm =

9\%

, left arm =

9\%

.

2

Calculate the total 24-hour fluid requirement using the ATLS formula.

7200 \text{ mL}

The formula is

2 \text{ mL} \times \text{weight in kg} \times \%\text{TBSA}

. Therefore,

2 \times 100 \times 36 = 7200 \text{ mL}

.

3

Determine the fluid volume targeted for the first 8 hours post-injury.

3600 \text{ mL}

Half of the total 24-hour calculated volume must be administered during the first 8 hours following the time of the burn.

4

Calculate the infusion rate over the remaining time in the first 8-hour window.

720 \text{ mL/hr}

Because

3

hours have already passed since the injury, the

3600 \text{ mL}

must be delivered over the remaining

5

hours (

3600 / 5 = 720

).

Key Concept

Burn resuscitation fluid calculation requires assessing TBSA, using the

2 \text{ mL/kg/}\text{TBSA}

formula, providing half the total fluid in the first 8 hours, and subtracting any elapsed time from that 8-hour window to determine the current hourly rate.

Estimated Time:1m 30s

Question 132Question →

A 26-year-old man weighing 90 kg is brought to the emergency department 2 hours after being rescued from a house fire. On primary survey, his airway is patent, and he is breathing spontaneously on supplemental oxygen without evidence of inhalation injury. Vital signs include a blood pressure of 110/70 mm Hg, heart rate of 115/min, and respiratory rate of 18/min. Physical examination reveals deep partial-thickness burns with blistering over his entire anterior trunk, entire left arm, and the anterior surface of his left leg. The skin on his face is erythematous and highly sensitive to touch, but no blisters are present. Two large-bore intravenous catheters were placed at the scene, but no intravenous fluids have been administered yet.

Based on current Advanced Trauma Life Support (ATLS) guidelines for adult thermal burns, what is the appropriate targeted intravenous fluid administration rate (in mL/hour) for the remainder of the first 8-hour period following the burn?

Show answer & explanation

Answer: 540

Answer

540

The updated ATLS protocol for adult thermal burns dictates using 2 mL of lactated Ringer solution per kg of body weight per percentage of TBSA. TBSA calculations only include partial-thickness and full-thickness burns (anterior trunk 18% + left arm 9% + anterior left leg 9% = 36%). Superficial face burns are excluded. The 24-hour fluid total is 2 mL x 90 kg x 36% = 6,480 mL. Half of this volume (3,240 mL) is given in the first 8 hours from the time of injury. Since 2 hours have already passed, the 3,240 mL must be infused over the remaining 6 hours, resulting in a target rate of 540 mL/hour.

Step-by-Step Solution

1

Calculate the Total Body Surface Area (TBSA) affected by second- and third-degree burns using the Rule of Nines.

TBSA = 36%

The anterior trunk is 18%, the entire left arm is 9%, and the anterior left leg is 9%. Superficial (first-degree) burns, such as those on the patient's face, do not count toward TBSA for fluid resuscitation.

2

Calculate the total 24-hour resuscitation fluid requirement using the updated ATLS formula for adult thermal burns.

Total 24-hour fluid = 6,480 mL

The current standard is 2 mL x weight in kg x %TBSA. Therefore, 2 x 90 x 36 = 6,480 mL.

3

Determine the fluid volume required for the first 8-hour period post-injury.

First 8-hour volume = 3,240 mL

Exactly half of the total calculated 24-hour fluid must be administered within the first 8 hours following the exact time of the burn.

4

Calculate the appropriate hourly infusion rate based on the time elapsed since the injury.

Infusion rate = 540 mL/hour

The patient arrived 2 hours after the injury, leaving only 6 hours remaining in the initial 8-hour window. Dividing 3,240 mL by 6 hours yields 540 mL/hour.

Key Concept

Burn fluid resuscitation calculations must use the updated ATLS guidelines (2 mL/kg/%TBSA for adults), exclude superficial burns, and account for the time elapsed since the injury.

Alternative Method

An alternative mathematical shortcut is to calculate the initial 8-hour total directly as (1 mL x weight in kg x %TBSA), which yields 1 x 90 x 36 = 3,240 mL, and then divide by the remaining hours.

Estimated Time:2m 0s

Question 133Question →

A 19-year-old college student is brought to the emergency department by his roommates. They report that he ingested a large amount of a sweet-tasting, brightly colored liquid he found in a garage, believing it was a sports drink. On presentation, he is disoriented and slurring his words. His blood pressure is 120/75 mm Hg and his pulse is 100/min.

Laboratory studies:
- Serum sodium: 142 mEq/L
- Serum chloride: 105 mEq/L
- Serum bicarbonate: 14 mEq/L
- Blood urea nitrogen: 14 mg/dL
- Blood glucose: 90 mg/dL
- Measured serum osmolality: 345 mOsm/kg $H_2O$

What is this patient's serum osmolar gap in mOsm/kg $H_2O$ ?

Show answer & explanation

Answer: 51

Answer

51

The osmolar gap is used to detect the presence of unmeasured, osmotically active substances in the blood, such as toxic alcohols (e.g., ethylene glycol, methanol). It is determined by subtracting the calculated serum osmolality from the measured serum osmolality. The formula for calculated osmolality is

(2 \times Na) + \frac{Glucose}{18} + \frac{BUN}{2.8}

. For this patient, the calculated osmolality is

(2 \times 142) + \frac{90}{18} + \frac{14}{2.8} = 284 + 5 + 5 = 294

mOsm/kg

H_2O

. Subtracting this from the measured osmolality of 345 mOsm/kg

H_2O

yields an osmolar gap of 51 mOsm/kg

H_2O

. A normal gap is typically less than 10; an elevated gap in this context strongly suggests ethylene glycol toxicity.

Step-by-Step Solution

1

Recall the formula for calculated serum osmolality

Calculated Osm = (2 \times Na) + \frac{Glucose}{18} + \frac{BUN}{2.8}

This formula estimates the osmotically active particles normally present in the blood, primarily sodium salts, glucose, and urea.

2

Calculate the estimated serum osmolality

Calculated Osm = (2 \times 142) + \frac{90}{18} + \frac{14}{2.8} = 284 + 5 + 5 = 294

mOsm/kg

H_2O

Accurately applying the patient's lab values yields the expected osmolality without the presence of toxic alcohols.

3

Calculate the osmolar gap

Osmolar Gap = Measured Osm - Calculated Osm = 345 - 294 = 51

mOsm/kg

H_2O

The difference between the measured and calculated osmolality represents the concentration of unmeasured osmoles, which in this case points to ethylene glycol ingestion.

Key Concept

Serum Osmolar Gap Calculation

Alternative Method

In systems using SI units (mmol/L) rather than traditional US units, the formula for calculated osmolality does not require division by 18 or 2.8 and is instead:

2 \times Na + Glucose + Urea

.

Estimated Time:1m 30s

Question 134Question →

A 62-year-old man with a history of alcohol-associated cirrhosis is brought to the emergency department due to progressive confusion, lethargy, and fever over the past 2 days. His temperature is $38.6^\circ\text{C}$ ( $101.5^\circ\text{F}$ ), blood pressure is $95/60\text{ mm Hg}$ , heart rate is $112/\text{min}$ , and respiratory rate is $22/\text{min}$ . Physical examination is notable for scleral icterus, spider angiomas, a distended abdomen with a positive fluid wave, and bilateral lower extremity pitting edema.

Laboratory studies show:
Sodium: $136\text{ mEq/L}$
Potassium: $3.8\text{ mEq/L}$
Chloride: $104\text{ mEq/L}$
Bicarbonate: $18\text{ mEq/L}$
BUN: $32\text{ mg/dL}$
Creatinine: $1.5\text{ mg/dL}$
Glucose: $98\text{ mg/dL}$
Albumin: $2.0\text{ g/dL}$
Total bilirubin: $4.5\text{ mg/dL}$

Arterial blood gas on room air reveals a pH of $7.32$ and a $PaCO_2$ of $34\text{ mm Hg}$ .

Assuming a reference normal serum albumin of $4.0\text{ g/dL}$ , what is this patient's albumin-corrected serum anion gap in mEq/L?

Show answer & explanation

Answer: 19

Answer

19

This patient has a metabolic acidosis (pH 7.32, low bicarbonate). To accurately assess the etiology, the serum anion gap must be calculated and corrected for his severe hypoalbuminemia. The uncorrected anion gap is

Na^+ - (Cl^- + HCO_3^-) = 136 - (104 + 18) = 14\text{ mEq/L}

. Because albumin is negatively charged, it constitutes a significant portion of the baseline 'unmeasured anions'. A decrease in albumin falsely lowers the calculated anion gap. The standard correction requires adding

2.5\text{ mEq/L}

to the calculated gap for every

1\text{ g/dL}

decrease in albumin below

4.0\text{ g/dL}

. Here, the deficit is

4.0 - 2.0 = 2.0\text{ g/dL}

. The correction is

2.0 \times 2.5 = 5\text{ mEq/L}

. Adding this to the uncorrected gap yields

14 + 5 = 19\text{ mEq/L}

. This reveals a high anion gap metabolic acidosis, likely secondary to lactic acidosis from sepsis.

Step-by-Step Solution

1

Calculate the uncorrected anion gap.

14\text{ mEq/L}

Using the standard formula:

AG = 136 - (104 + 18) = 14\text{ mEq/L}

.

2

Calculate the albumin deficit.

2.0\text{ g/dL}

Subtract the patient's albumin from normal:

4.0\text{ g/dL} - 2.0\text{ g/dL} = 2.0\text{ g/dL}

.

3

Calculate the albumin-corrected anion gap.

19\text{ mEq/L}

Add the correction factor to the uncorrected gap:

14 + (2.0 \times 2.5) = 14 + 5 = 19\text{ mEq/L}

.

Key Concept

Albumin is a major unmeasured anion; in hypoalbuminemic states, the calculated anion gap is falsely lowered and must be corrected by adding

2.5\text{ mEq/L}

for every

1\text{ g/dL}

decrease in serum albumin.

Question 135Question →

A 68-year-old man is evaluated in the intensive care unit for decreased urine output. He was admitted 4 days ago for septic shock secondary to severe community-acquired pneumonia and initially required norepinephrine for blood pressure support. He has been hemodynamically stable off vasopressors for the past 24 hours. His current blood pressure is 118/76 mm Hg and heart rate is 82/min. Laboratory studies show:

Serum sodium: 140 mEq/L
Serum blood urea nitrogen (BUN): 32 mg/dL
Serum creatinine: 2.0 mg/dL

Urine sodium: 63 mEq/L
Urine creatinine: 30 mg/dL

Urinalysis reveals muddy brown granular casts.

Calculate the patient's fractional excretion of sodium (FeNa). (Express the result as a percentage)

Show answer & explanation

Answer: 3

Answer

3.0

The correct Fractional Excretion of Sodium (FeNa) is calculated using the formula

FeNa = \frac{U_{Na} \times P_{Cr}}{P_{Na} \times U_{Cr}} \times 100

. Plugging in the patient's values yields

\frac{63 \times 2.0}{140 \times 30} \times 100

, which simplifies to

\frac{126}{4200} \times 100 = 3.0\%

. In the clinical context of a patient recovering from septic shock, a FeNa greater than 2% strongly points toward intrinsic acute kidney injury—specifically acute tubular necrosis (ATN). This diagnosis is further corroborated by the presence of muddy brown granular casts on urinalysis.

Step-by-Step Solution

1

Identify the standard formula for calculating FeNa.

FeNa = \frac{U_{Na} \times P_{Cr}}{P_{Na} \times U_{Cr}} \times 100

This formula relates the clearance of sodium to the clearance of creatinine to determine the percentage of filtered sodium that is excreted.

2

Substitute the patient's specific serum and urine chemistries into the formula.

FeNa = \frac{63 \times 2.0}{140 \times 30} \times 100

Applying the clinical laboratory values provided in the vignette.

3

Calculate the products in the numerator and the denominator.

Numerator =

126

, Denominator =

4200

To simplify the mathematical expression before final division.

4

Divide the numerator by the denominator and multiply by 100.

\frac{126}{4200} \times 100 = 0.03 \times 100 = 3.0

To obtain the final percentage value for FeNa.

Key Concept

Fractional Excretion of Sodium (FeNa) Calculation

Estimated Time:1m 30s

Question 136Question →

A 32-year-old male construction worker is brought to the emergency department after a high-voltage electrical injury. He was electrocuted by a live power line exactly 2 hours prior to arrival. On presentation, he is awake and reports severe pain. His vital signs show a blood pressure of 105/65 mm Hg, heart rate of 118/min, and respiratory rate of 20/min. Physical examination reveals deep, charred, full-thickness burns covering his entire right lower extremity, his entire left lower extremity, and his entire right upper extremity. His estimated body weight is 80 kg. A Foley catheter is placed, yielding dark brown urine.

Based on current Advanced Trauma Life Support (ATLS) guidelines for electrical burns, at what rate should the initial intravenous crystalloid infusion be set to deliver the appropriate resuscitation volume for the remainder of the first 8-hour period?

Show answer & explanation

Answer: 1200

Answer

1200

The patient has a high-voltage electrical burn with clinical evidence of myoglobinuria (dark brown urine). Current ATLS guidelines mandate using 4 mL/kg/%TBSA for electrical burns to maintain high urine output and prevent pigment nephropathy. With a 45% TBSA burn (18% for each leg + 9% for the arm) and 80 kg weight, the 24-hour fluid need is 14,400 mL. Half of this total (7,200 mL) must be given in the first 8 hours from the time of injury. Since 2 hours have elapsed prior to arrival, the 7,200 mL must be infused over the remaining 6 hours, resulting in an infusion rate of 1,200 mL/hr.

Step-by-Step Solution

1

Select the correct resuscitation formula based on burn mechanism.

Use the ATLS electrical burn formula: 4 mL/kg/%TBSA.

High-voltage electrical burns cause deep muscle damage and rhabdomyolysis, leading to myoglobinuria (indicated by dark brown urine). Higher fluid volumes (4 mL vs standard 2 mL) are required to flush the kidneys and prevent acute kidney injury.

2

Calculate the Total Body Surface Area (TBSA) burned using the Rule of Nines.

TBSA = 18% (entire right leg) + 18% (entire left leg) + 9% (entire right arm) = 45%.

Accurate TBSA estimation is critical for the Parkland/ATLS fluid calculation.

3

Calculate the total 24-hour resuscitation volume.

4 mL × 80 kg × 45 = 14,400 mL.

This formula provides the estimated fluid requirement for the first 24 hours post-injury.

4

Determine the volume needed for the first 8 hours.

14,400 mL / 2 = 7,200 mL.

ATLS protocols require half of the total 24-hour calculated volume to be administered in the first 8 hours after the time of injury.

5

Calculate the remaining time for the first 8-hour infusion window.

8 hours - 2 hours (time since injury) = 6 hours remaining.

The 8-hour clock starts at the time of the burn, not the time of arrival at the hospital.

6

Calculate the hourly infusion rate.

7,200 mL / 6 hours = 1,200 mL/hr.

This is the continuous rate required to deliver the remaining first-half volume over the correct time period.

Key Concept

Fluid Resuscitation in Electrical Burns

Estimated Time:1m 30s

Question 137Question →

A 52-year-old man weighing $85\text{ kg}$ is brought to the emergency department after being trapped in a residential garage fire. The injury occurred exactly 2 hours prior to his arrival. On examination, he has deep partial-thickness and full-thickness burns involving the entire anterior trunk, the entire right upper extremity, and the anterior surface of the left lower extremity. He has no inhalation injury, and his airway is patent. Intravenous access is established. According to current Advanced Trauma Life Support (ATLS) guidelines for adult thermal burns, what should be the initial intravenous fluid infusion rate in $\text{mL/hr}$ to administer the first half of his calculated resuscitation volume?

Show answer & explanation

Answer: 510

Answer

510

The correct rate is

510\text{ mL/hr}

. First, the total body surface area (TBSA) is calculated using the Rule of Nines: the entire anterior trunk accounts for

18\%

, the entire right upper extremity is

9\%

, and the anterior surface of the left lower extremity is

9\%

, yielding a total of

36\%

TBSA. According to current ATLS guidelines for adult thermal burns, the total 24-hour fluid requirement is calculated as

2\text{ mL} \times 85\text{ kg} \times 36 = 6,120\text{ mL}

. Exactly half of this total (

3,060\text{ mL}

) is mandated to be administered within the first 8 hours from the time of injury. Because 2 hours have already elapsed prior to emergency department arrival, the required

3,060\text{ mL}

must be infused over the remaining 6 hours. Therefore, the hourly rate is

3,060\text{ mL} / 6\text{ hours} = 510\text{ mL/hr}

.

Step-by-Step Solution

1

Calculate the burned Total Body Surface Area (TBSA) using the Rule of Nines.

Anterior trunk (

18\%

) + right arm (

9\%

) + anterior left leg (

9\%

) =

36\%

TBSA.

Accurate estimation of TBSA is required to determine total fluid needs.

2

Calculate the total 24-hour fluid requirement using the current ATLS multiplier for adult thermal burns.

2\text{ mL} \times 85\text{ kg} \times 36\% = 6,120\text{ mL}

.

Current ATLS guidelines recommend

2\text{ mL/kg/\%TBSA}

for adult thermal burns to prevent fluid creep and over-resuscitation.

3

Determine the fluid volume that must be given in the first phase of resuscitation.

6,120\text{ mL} / 2 = 3,060\text{ mL}

.

Half of the calculated 24-hour total must be administered within the first 8 hours from the time of injury.

4

Adjust the remaining time for the first phase based on the pre-hospital delay.

8\text{ hours} - 2\text{ hours elapsed} = 6\text{ hours remaining}

.

The 8-hour window begins at the time of injury, not at the time of arrival to the hospital.

5

Calculate the hourly infusion rate.

3,060\text{ mL} / 6\text{ hours} = 510\text{ mL/hr}

.

This rate ensures the patient receives the correct first-half volume precisely within the 8-hour post-injury window.

Key Concept

Burn fluid resuscitation calculations require applying the Rule of Nines for TBSA, using the

2\text{ mL/kg}

multiplier for adult thermal burns, and subtracting pre-hospital elapsed time from the initial 8-hour administration window.

Question 138Question →

A 76-year-old woman is admitted to the intensive care unit with severe, poorly localized abdominal pain out of proportion to physical examination findings, accompanied by bloody diarrhea. Her medical history is significant for atrial fibrillation. She is hypotensive and tachycardic. Laboratory testing reveals a serum lactate of $8.5$ mmol/L. Arterial blood gas and basic metabolic panel results are as follows:

Laboratory test	Result
Serum sodium	$138$ mEq/L
Serum chloride	$102$ mEq/L
Serum bicarbonate	$14$ mEq/L
Arterial $pH$	$7.26$

The physician determines that the patient has a primary high anion gap metabolic acidosis secondary to suspected acute mesenteric ischemia. According to Winter's formula, what is the exact calculated midpoint of the target expected arterial $pCO_2$ (in mmHg) representing appropriate respiratory compensation?

Show answer & explanation

Answer: 29

Answer

29

In the setting of a primary metabolic acidosis, the lungs compensate by increasing ventilation to blow off

CO_2

, thereby raising the

pH

toward normal. Winter's formula (

Expected\ pCO_2 = 1.5 \times [HCO_3^-] + 8 \pm 2

) is used to determine if this respiratory response is appropriate. For this patient with a serum bicarbonate of

14

mEq/L, the calculated midpoint of the target

pCO_2

is

1.5 \times 14 + 8 = 29

mmHg. If the measured

pCO_2

on an arterial blood gas were significantly higher than this target range (

27-31

mmHg), a concomitant respiratory acidosis would be present; if significantly lower, a concomitant respiratory alkalosis would be present.

Step-by-Step Solution

1

Identify the primary acid-base disorder

Primary metabolic acidosis

The patient has a low

pH

(

7.26

) and a low

HCO_3^-

(

14

mEq/L) along with elevated lactate, indicating a metabolic acidosis.

2

Select the appropriate compensation formula

Winter's formula: Expected

pCO_2 = 1.5 \times [HCO_3^-] + 8

Winter's formula is the standard physiological rule used to predict the expected respiratory compensation (hyperventilation) in the setting of primary metabolic acidosis.

3

Calculate the expected target

pCO_2

1.5 \times 14 + 8 = 29

mmHg

Substituting the patient's measured serum bicarbonate of

14

mEq/L yields the exact midpoint of the expected

pCO_2

range.

Key Concept

Calculating expected respiratory compensation for metabolic acidosis using Winter's formula

Question 139Question →

Helicopter transport arrives at a regional burn center carrying a 30-year-old female who was trapped in a residential fire. She currently weighs $70\text{ kg}$ . The injury occurred exactly $2\text{ hours}$ prior to her arrival. Physical examination reveals full-thickness burns involving her entire anterior trunk, her entire right lower extremity, and her entire left upper extremity. The trauma team confirms this accounts for exactly $45\%$ of her total body surface area (TBSA). Large-bore intravenous access is established. Utilizing the classic Parkland formula ( $4\text{ mL/kg/\%TBSA}$ ), what is the targeted fluid administration rate in $\text{mL/hr}$ for the remainder of the first 8-hour post-burn period?

Show answer & explanation

Answer: 1050

Answer

1050

The correct infusion rate is found by calculating the total 24-hour fluid requirement (

12,600\text{ mL}

), allocating half of that volume to the first 8 hours post-injury (

6,300\text{ mL}

), and dividing that volume by the time remaining in that 8-hour window (

6\text{ hours}

). This yields a target rate of

1,050\text{ mL/hr}

.

Step-by-Step Solution

1

Identify the key variables required for the calculation.

Weight =

70\text{ kg}

, TBSA =

45\%

, Elapsed time =

2\text{ hours}

.

These values are strictly necessary to apply the formula and adjust for the pre-hospital delay.

2

Calculate the total 24-hour fluid volume.

4\text{ mL} \times 70\text{ kg} \times 45 = 12,600\text{ mL}

.

The classic Parkland formula estimates the total fluid deficit over the first 24 hours post-burn.

3

Determine the fluid volume required for the first resuscitation period.

12,600\text{ mL} / 2 = 6,300\text{ mL}

.

Standard burn resuscitation guidelines dictate that half of the total 24-hour volume must be administered in the first 8 hours.

4

Calculate the remaining time in the first 8-hour window.

8\text{ hours} - 2\text{ hours elapsed} = 6\text{ hours remaining}

.

The 8-hour clock begins precisely at the time of injury, not at the time of hospital arrival.

5

Calculate the target hourly infusion rate.

6,300\text{ mL} / 6\text{ hours} = 1,050\text{ mL/hr}

.

The remaining volume must be divided equally over the remaining hours of the first 8-hour period to catch up on fluid deficits safely.

Key Concept

Parkland formula fluid resuscitation and time-delay adjustment

Question 140Question →

A 68-year-old man is admitted to the medical ward following a 4-day history of worsening diarrhea and generalized weakness. He states he has been unable to keep fluids down. His past medical history is significant for hypertension and osteoarthritis. His temperature is 37.0°C (98.6°F), blood pressure is 98/60 mm Hg, and pulse is 108/min. Jugular venous pressure is decreased, and skin turgor is poor. A Foley catheter is placed, yielding a small amount of concentrated urine.

Laboratory evaluation reveals the following:

Parameter	Value
Serum sodium	$140\text{ mEq/L}$
Serum potassium	$3.2\text{ mEq/L}$
Blood urea nitrogen	$45\text{ mg/dL}$
Serum creatinine	$1.5\text{ mg/dL}$
Urine sodium	$14\text{ mEq/L}$
Urine creatinine	$30\text{ mg/dL}$

What is this patient's fractional excretion of sodium?

Show answer & explanation

Answer: 0.5

Answer

0.5

The fractional excretion of sodium (FeNa) measures the percentage of filtered sodium that is excreted in the urine. It is an essential calculation used to differentiate between prerenal and intrinsic etiologies of acute kidney injury. The formula is

FeNa = (U_{Na} / P_{Na}) \times (P_{Cr} / U_{Cr}) \times 100

. Substituting the provided laboratory values yields

(14 / 140) \times (1.5 / 30) \times 100 = 0.1 \times 0.05 \times 100 = 0.5\%

. A value less than

1\%

indicates avid sodium retention by the kidneys, which is characteristic of a prerenal state such as hypovolemia secondary to severe diarrhea.

Step-by-Step Solution

1

Identify the standard equation for fractional excretion of sodium.

FeNa = \frac{U_{Na} \times P_{Cr}}{P_{Na} \times U_{Cr}} \times 100

This formula is used to differentiate the etiology of acute kidney injury.

2

Extract the specific patient values from the clinical vignette.

U_{Na} = 14\text{ mEq/L}

,

P_{Na} = 140\text{ mEq/L}

,

P_{Cr} = 1.5\text{ mg/dL}

,

U_{Cr} = 30\text{ mg/dL}

These are the four variables required to complete the calculation.

3

Substitute the parameters into the equation.

FeNa = \left(\frac{14}{140}\right) \times \left(\frac{1.5}{30}\right) \times 100

To set up the mathematical calculation.

4

Complete the mathematical operation.

FeNa = 0.1 \times 0.05 \times 100 = 0.5\%

To find the final diagnostic value, which supports a prerenal etiology.

Key Concept

Fractional Excretion of Sodium (FeNa) Calculation

Answer

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