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Question 101Question →

A 60-year-old woman is evaluated in the hospital for decreased urine output on her third day of admission for severe gastroenteritis. Her vital signs show a blood pressure of 90/60 mm Hg and a heart rate of 110/min. Physical examination reveals dry mucous membranes and poor skin turgor. Laboratory studies are obtained to determine the etiology of her acute kidney injury:

Laboratory Test	Value
Plasma sodium	$140 \text{ mEq/L}$
Plasma creatinine	$2.0 \text{ mg/dL}$
Urine sodium	$14 \text{ mEq/L}$
Urine creatinine	$100 \text{ mg/dL}$

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

Show answer & explanation

Answer: 0.2

Answer

0.2

The correct calculation yields a fractional excretion of sodium (FENa) of

0.2\%

. FENa measures the percentage of filtered sodium that is excreted in the urine. It is calculated as the ratio of sodium clearance to creatinine clearance. In this patient with a history of severe gastroenteritis, hypotension, and signs of volume depletion, a FENa of less than

1\%

is highly indicative of a prerenal state. The kidneys appropriately reabsorb sodium in response to decreased renal perfusion to restore intravascular volume.

Step-by-Step Solution

1

Identify the relevant diagnostic formula

\text{FENa} = \frac{\text{Urine Na} \times \text{Plasma Cr}}{\text{Plasma Na} \times \text{Urine Cr}} \times 100

To evaluate whether the acute kidney injury is prerenal or intrinsic.

2

Substitute the patient's laboratory values

\text{FENa} = \frac{14 \times 2.0}{140 \times 100} \times 100

Applying the specific plasma and urine values from the vignette.

3

Perform the arithmetic calculation

\frac{28}{14000} \times 100 = 0.002 \times 100 = 0.2

To yield the final percentage value for interpretation.

Key Concept

Calculating the fractional excretion of sodium (FENa) is a crucial step in evaluating acute kidney injury to distinguish between a prerenal etiology (FENa < 1%) and intrinsic renal pathology, such as acute tubular necrosis (FENa > 2%).

Question 102Question →

A $38$ -year-old man is brought to the emergency department by emergency medical services after being found lethargic and confused in his unheated garage during winter. Beside him was an unlabeled jug containing a brightly colored liquid. His blood pressure is $110/70$ mm Hg, pulse is $102$ /min, and respirations are $26$ /min and deep. Physical examination reveals a somnolent patient who localizes to painful stimuli. Cardiopulmonary examination is unremarkable, and he has no focal neurologic deficits. Laboratory results are as follows:

Laboratory Test	Result	Reference Range
Sodium	$142$ mEq/L	$136$ - $145$ mEq/L
Potassium	$5.0$ mEq/L	$3.5$ - $5.0$ mEq/L
Chloride	$104$ mEq/L	$95$ - $105$ mEq/L
Bicarbonate	$10$ mEq/L	$22$ - $28$ mEq/L
Blood urea nitrogen (BUN)	$14$ mg/dL	$7$ - $18$ mg/dL
Creatinine	$1.2$ mg/dL	$0.6$ - $1.2$ mg/dL
Glucose	$198$ mg/dL	$70$ - $110$ mg/dL
Measured serum osmolality	$332$ mOsm/kg H $_2$ O	$275$ - $295$ mOsm/kg H $_2$ O

Based on the provided laboratory results, calculate the patient's osmolar gap. Enter only the numerical value.

Show answer & explanation

Answer: 32

Answer

32

This patient presents with altered mental status and deep respirations (Kussmaul breathing) in the setting of wintertime exposure to a brightly colored, unlabelled fluid, raising strong suspicion for ethylene glycol (antifreeze) toxicity. The laboratory results show a high anion gap metabolic acidosis (Anion gap =

142 - (104 + 10) = 28

). To evaluate for the presence of toxic alcohols, an osmolar gap should be calculated. The formula for calculated serum osmolality is

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

. Using the patient's values, this yields

2(142) + \frac{198}{18} + \frac{14}{2.8} = 284 + 11 + 5 = 300

mOsm/kg. The osmolar gap is the measured osmolality minus the calculated osmolality (

332 - 300 = 32

mOsm/kg). A normal gap is typically

< 10

mOsm/kg; a gap of

32

strongly supports the presence of an unmeasured osmotically active substance, confirming the suspicion of toxic alcohol ingestion.

Step-by-Step Solution

1

Extract the relevant values required for the calculated osmolality formula.

Sodium =

142

, Glucose =

198

, BUN =

14

.

The calculated serum osmolality depends on the primary osmotic drivers in the serum.

2

Apply the calculated osmolality formula:

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

.

2 \times 142 + \frac{198}{18} + \frac{14}{2.8} = 284 + 11 + 5 = 300

mOsm/kg.

This formula estimates the total particle concentration normally present in the blood. The divisors convert mg/dL to mmol/L.

3

Calculate the osmolar gap by subtracting the calculated osmolality from the measured osmolality.

Osmolar gap =

332 - 300 = 32

mOsm/kg.

The difference between measured and calculated osmolality reveals the presence of unmeasured osmotically active substances (e.g., toxic alcohols).

Key Concept

Calculation of the osmolar gap is an essential step in diagnosing toxic alcohol ingestions (such as ethylene glycol or methanol) in patients presenting with altered mental status and a high anion gap metabolic acidosis.

Estimated Time:1m 30s

Question 103Question →

A 34-year-old woman with a history of systemic lupus erythematosus and Sjögren syndrome is evaluated in the emergency department for progressive muscle weakness over the past 3 days. She reports difficulty climbing stairs and getting out of chairs. She has not had diarrhea or vomiting. Her blood pressure is 112/76 mm Hg, and her pulse is 88/min. Physical examination shows 3/5 motor strength in the proximal lower extremities. Laboratory results are as follows:

Serum:
Sodium: $138$ mEq/L
Potassium: $2.4$ mEq/L
Chloride: $115$ mEq/L
Bicarbonate: $13$ mEq/L

Arterial blood gas (room air):
pH: $7.27$
pCO2: $28$ mm Hg

Urine:
Sodium: $42$ mEq/L
Potassium: $35$ mEq/L
Chloride: $51$ mEq/L

To help determine the etiology of this patient's acid-base disturbance, the physician evaluates her urinary electrolytes. What is this patient's urine anion gap in mEq/L?

Show answer & explanation

Answer: 26

Answer

26

The correct calculation for the urine anion gap (UAG) relies on the formula:

UAG = Urine\ Na^+ + Urine\ K^+ - Urine\ Cl^-

. Substituting the provided urine electrolyte values yields:

42\ mEq/L + 35\ mEq/L - 51\ mEq/L = 26\ mEq/L

. In the setting of a normal anion gap metabolic acidosis (confirmed by a serum anion gap of 10), the UAG is used as a surrogate marker for unmeasured urinary ammonium (

NH_4^+

) excretion. A positive UAG (as seen here) indicates impaired ammonium excretion, pointing toward a renal cause such as renal tubular acidosis (RTA). This patient's profound hypokalemia, systemic lupus erythematosus, and Sjögren syndrome strongly suggest type 1 (distal) RTA.

Step-by-Step Solution

1

Analyze the serum electrolytes and arterial blood gas to confirm the primary acid-base disorder.

The patient has a normal anion gap metabolic acidosis. The serum pH is 7.27, bicarbonate is low at 13 mEq/L, and the serum anion gap is 10 mEq/L (Calculated as 138 - (115 + 13)).

Establishing the presence of a normal anion gap metabolic acidosis is the clinical prerequisite for utilizing the urine anion gap.

2

Determine the appropriate formula for calculating the urine anion gap (UAG).

UAG =

Urine\ Na^+ + Urine\ K^+ - Urine\ Cl^-

The UAG indirectly estimates urinary ammonium excretion to differentiate between gastrointestinal bicarbonate loss (negative UAG) and renal tubular acidosis (positive UAG).

3

Substitute the patient's measured urine electrolyte values into the UAG formula and calculate the result.

UAG = 42 + 35 - 51 = 77 - 51 = 26 mEq/L.

Applying the values correctly yields the final diagnostic index necessary to evaluate the etiology of the acidosis.

Key Concept

The urine anion gap is calculated using the formula

(Urine\ Na^+ + Urine\ K^+) - Urine\ Cl^-

and is essential for differentiating between gastrointestinal and renal causes of normal anion gap metabolic acidosis.

Question 104Question →

A clinical study is conducted to evaluate the effect of a novel SGLT2 inhibitor on fasting blood glucose levels in patients with type 2 diabetes. The investigators recruit a total of 120 patients and randomly assign them in a 1:1 ratio to either the intervention group ( $n = 60$ ) or the placebo group ( $n = 60$ ). The primary outcome is the absolute change in fasting blood glucose (measured in mg/dL) from baseline to week 12. The data for this outcome are continuous and demonstrate a normal distribution in both groups. The researchers analyze the primary outcome by comparing the mean change between the two distinct groups using an unpaired Student t-test. What is the degrees of freedom associated with the statistical test for this primary analysis?

Show answer & explanation

Answer: 118

Answer

118

The unpaired Student t-test is utilized to compare the means of two distinct, independent groups when the dependent variable is continuous and normally distributed. The degrees of freedom (

df

) for this test reflects the number of independent values that can vary in the analysis. Because the mean is estimated separately for each of the two groups, two degrees of freedom are lost from the total sample size (

N

). Therefore, the formula is

df = n_1 + n_2 - 2

. Given that each group has 60 patients, the calculation is

60 + 60 - 2 = 118

.

Step-by-Step Solution

1

Identify the statistical test being used and the relevant parameters from the study design.

The test is an unpaired Student t-test. The sample sizes are

n_1 = 60

and

n_2 = 60

.

Determining the correct test is required to know which degrees of freedom formula to apply.

2

Apply the degrees of freedom formula for the identified statistical test.

The formula is

df = n_1 + n_2 - 2

.

In an unpaired t-test, one degree of freedom is lost for each group's mean that is estimated from the sample data.

3

Calculate the final numerical value.

df = 60 + 60 - 2 = 118

.

Substituting the given group sizes yields the exact degrees of freedom.

Key Concept

Degrees of freedom for an unpaired Student t-test

Question 105Question →

An occupational medicine specialist is reviewing health data from a large manufacturing corporation to evaluate the impact of chronic noise exposure on the development of sensorineural hearing loss.

The database includes a 5-year longitudinal cohort of 20,000 factory employees who had normal hearing at baseline. Occupational health records classify employees into two groups: those working in "high-noise" zones (exposed) and those working in administrative or "low-noise" zones (unexposed).

At the end of the 5-year period, audiometric testing yielded the following new-onset sensorineural hearing loss cases:

Exposure Status	Total Employees in Group	New Hearing Loss Cases
High-Noise	5,000	400
Low-Noise	15,000	600

The corporate medical director wants to determine the proportion of all incident cases of sensorineural hearing loss in the entire workforce that could theoretically be prevented if high-noise exposure were completely eliminated.

What is the population attributable risk percent (PAR%) of sensorineural hearing loss due to high-noise exposure in this workforce? (Express your answer as a whole number).

Show answer & explanation

Answer: 20

Answer

20

The correct answer is 20. The Population Attributable Risk Percent (PAR%) indicates the proportion of incident cases in the entire population that are attributable to the exposure. First, calculate the overall incidence (

I_{total}

) which is 1,000 total cases divided by 20,000 total employees (0.05). Next, calculate the incidence in the unexposed group (

I_{unexposed}

) which is 600 cases divided by 15,000 unexposed employees (0.04). The formula for PAR% is

\frac{I_{total} - I_{unexposed}}{I_{total}} \times 100

. Plugging in the values gives

\frac{0.05 - 0.04}{0.05} \times 100 = 20\%

.

Step-by-Step Solution

1

Calculate the total incidence of the outcome in the entire population (

I_{total}

).

I_{total} = 1,000 \text{ cases} / 20,000 \text{ total employees} = 0.05

(or 5%).

PAR% requires knowing the overall incidence of the disease in the population to determine what proportion of all cases is attributable to the exposure.

2

Calculate the incidence of the outcome in the unexposed group (

I_{unexposed}

).

I_{unexposed} = 600 \text{ cases} / 15,000 \text{ unexposed employees} = 0.04

(or 4%).

The unexposed incidence represents the background risk of disease that would exist even if the exposure were entirely eliminated.

3

Calculate the Population Attributable Risk Percent (PAR%) using the formula:

\frac{I_{total} - I_{unexposed}}{I_{total}} \times 100

.

\frac{0.05 - 0.04}{0.05} \times 100 = \frac{0.01}{0.05} \times 100 = 20\%

.

This calculation yields the percentage of all incident cases in the population that can be attributed to the exposure, representing the maximum potential benefit of eliminating the risk factor.

Key Concept

Population Attributable Risk Percent (PAR%) measures the proportion of disease in the total population that is attributable to the exposure. It is calculated as

\frac{I_{total} - I_{unexposed}}{I_{total}} \times 100

.

Alternative Method

PAR% can also be calculated using the prevalence of the exposure in the population (

P_{e}

) and the Relative Risk (

RR

).

P_{e} = \frac{5,000}{20,000} = 0.25

. The exposed incidence is

\frac{400}{5,000} = 0.08

, and unexposed is

\frac{600}{15,000} = 0.04

, making

RR = \frac{0.08}{0.04} = 2.0

. Formula:

PAR\% = \frac{P_{e} \times (RR - 1)}{P_{e} \times (RR - 1) + 1} = \frac{0.25 \times (2 - 1)}{0.25 \times (2 - 1) + 1} = \frac{0.25}{1.25} = 0.20

, or 20%.

Estimated Time:2m 0s

Question 106Question →

An armed forces medical induction center is initiating a screening protocol for latent tuberculosis infection among incoming personnel. The protocol utilizes a highly standardized interferon-gamma release assay. Epidemiological surveillance indicates that the true prevalence of latent tuberculosis in this incoming cohort of $5,000$ recruits is exactly $4\%$ . The laboratory parameters for this specific assay demonstrate a sensitivity of $90\%$ and a specificity of $98\%$ .

If a recruit from this cohort receives a positive result on this assay, what is the probability that the individual actually has a latent tuberculosis infection? (Enter the value as a percentage rounded to the nearest whole number).

Show answer & explanation

Answer: 65

Answer

65

The question asks for the probability that a patient with a positive test result actually has the disease, which is the definition of Positive Predictive Value (PPV). To calculate PPV, one must account for the prevalence of the disease in the specific cohort being tested. Using a 2x2 contingency table for the population of

5,000

recruits, a prevalence of

4\%

equates to

200

individuals with latent tuberculosis and

4,800

healthy individuals. Given a sensitivity of

90\%

, the assay will correctly identify

180

of the

200

diseased individuals (True Positives). Given a specificity of

98\%

, the false positive rate is

2\%

; therefore, the assay will incorrectly flag

96

of the

4,800

healthy individuals (False Positives). The total number of positive test results is the sum of True Positives and False Positives (

180 + 96 = 276

). The PPV is the ratio of True Positives to Total Positives (

180 / 276

), which equals

0.6521

. Converted to a percentage and rounded to the nearest whole number, the correct probability is

65\%

.

Step-by-Step Solution

1

Determine the true disease status distribution in the cohort.

200 recruits have latent TB; 4,800 do not.

Pre-test probability (prevalence) is required to establish the baseline numbers of true positive and true negative candidates in the population.

2

Calculate the number of True Positives.

180 True Positives.

Sensitivity (

90\%

) dictates the proportion of the 200 diseased individuals who will correctly test positive (

200 \times 0.90

).

3

Calculate the number of False Positives.

96 False Positives.

Specificity is

98\%

, meaning the false positive rate is

2\%

. Applying this false positive rate to the 4,800 healthy individuals (

4,800 \times 0.02

) yields 96.

4

Calculate the Positive Predictive Value (PPV).

65.2%, rounded to 65%.

PPV is the ratio of True Positives to all positive test results:

180 / (180 + 96) = 180 / 276 = 0.6521

.

Key Concept

Positive Predictive Value (PPV) depends heavily on the disease prevalence in the tested population, in addition to the test's sensitivity and specificity.

Question 107Question →

A retired teacher with a history of severe alcohol use disorder and biopsy-proven hepatic cirrhosis is admitted to the inpatient medical service for management of new-onset hepatic encephalopathy and tense ascites. On physical examination, the patient appears chronically ill, with scleral icterus and significant peripheral edema.

A comprehensive metabolic panel is obtained to assess the patient's baseline metabolic status. The relevant laboratory results are reported as follows:

Test	Result	Reference Range
Total protein	$4.8$ g/dL	$6.0-8.3$ g/dL
Albumin	$1.5$ g/dL	$3.5-5.0$ g/dL
Total calcium	$7.2$ mg/dL	$8.5-10.5$ mg/dL
Magnesium	$1.9$ mg/dL	$1.7-2.2$ mg/dL
Phosphorus	$3.1$ mg/dL	$2.5-4.5$ mg/dL

Based on the provided laboratory data, what is the patient's corrected serum calcium concentration? (Provide the final numerical value in mg/dL)

Show answer & explanation

Answer: 9.2

Answer

The corrected serum calcium concentration is 9.2 mg/dL.

The patient has marked hypoalbuminemia (

1.5

g/dL) secondary to hepatic cirrhosis. Approximately 40-45% of serum calcium is bound to plasma proteins, primarily albumin. Therefore, a significant drop in albumin leads to a corresponding decrease in the measured total serum calcium without necessarily affecting the physiologically active ionized calcium level (pseudohypocalcemia). To accurately assess the patient's true calcium status when an ionized calcium level is unavailable, the corrected calcium must be calculated using the standard formula:

Corrected\ Calcium = Measured\ Total\ Calcium + [0.8 \times (4.0 - Serum\ Albumin)]

. Substituting the patient's values yields

7.2 + [0.8 \times (4.0 - 1.5)] = 7.2 + 2.0 = 9.2

mg/dL. This corrected value falls well within the normal reference range, indicating that the patient is physiologically eucalcemic and does not require calcium supplementation.

Step-by-Step Solution

1

Identify the patient's measured total serum calcium and serum albumin levels from the laboratory data.

Total calcium = 7.2 mg/dL; Albumin = 1.5 g/dL.

These two values are required to calculate the corrected calcium level in the setting of hypoalbuminemia.

2

Determine the deficit in serum albumin compared to the standard normal physiological reference value of 4.0 g/dL.

4.0 - 1.5 = 2.5 g/dL.

The correction factor depends on how far the patient's albumin has fallen below the normal reference point.

3

Multiply the albumin deficit by the correction factor (0.8 mg/dL) to find the amount of calcium that needs to be mathematically added back.

2.5 \times 0.8 = 2.0 mg/dL.

For every 1.0 g/dL decrease in serum albumin below 4.0 g/dL, measured total serum calcium decreases by approximately 0.8 mg/dL due to lost protein binding sites.

4

Add the calculated correction value to the measured total serum calcium.

7.2 + 2.0 = 9.2 mg/dL.

This yields the corrected total calcium, which better reflects the physiologically active (ionized) calcium status.

Key Concept

Corrected Calcium Calculation in Hypoalbuminemia

Alternative Method

Instead of strictly memorizing the full formula, a quick mental heuristic for the wards is to add 0.8 mg/dL to the measured total calcium for every 1.0 g/dL that the serum albumin is below 4.0 g/dL. In this case, the albumin is 2.5 units below normal (4.0 - 1.5). So,

2.5 \times 0.8 = 2.0

mg/dL must be added to the total measured calcium (

7.2 + 2.0 = 9.2

mg/dL).

Estimated Time:1m 0s

Question 108Question →

A $56$ -year-old male with long-standing Crohn's disease and a history of multiple small bowel resections is brought to the emergency department by his wife. She reports that he has been increasingly lethargic and confused over the past $24$ hours. He has had significantly increased, high-volume watery output from his ileostomy for the last three days. Physical examination reveals dry mucous membranes, poor skin turgor, and a blood pressure of $88/54\text{ mmHg}$ with a heart rate of $118\text{ /min}$ .

Laboratory studies are obtained and reveal the following:

Laboratory Parameter	Result	Reference Range
Sodium, serum	$136\text{ mEq/L}$	$136-145\text{ mEq/L}$
Potassium, serum	$3.4\text{ mEq/L}$	$3.5-5.0\text{ mEq/L}$
Chloride, serum	$106\text{ mEq/L}$	$95-105\text{ mEq/L}$
Bicarbonate, serum	$16\text{ mEq/L}$	$22-28\text{ mEq/L}$
Blood urea nitrogen	$48\text{ mg/dL}$	$7-18\text{ mg/dL}$
Creatinine, serum	$2.1\text{ mg/dL}$	$0.6-1.2\text{ mg/dL}$
Albumin, serum	$1.6\text{ g/dL}$	$3.5-5.0\text{ g/dL}$

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

Show answer & explanation

Answer: 20

Answer

20

The patient is experiencing a high anion gap metabolic acidosis that is visually masked by his profound hypoalbuminemia. Albumin is a major unmeasured anion in the serum. In states of severe malnutrition or protein loss (such as short bowel syndrome), reduced serum albumin artificially shrinks the apparent anion gap. The formula for the uncorrected gap is

\text{Na}^+ - (\text{Cl}^- + \text{HCO}_3^-)

, which gives

136 - (106 + 16) = 14\text{ mEq/L}

. To find the corrected anion gap,

2.5\text{ mEq/L}

is added for every

1.0\text{ g/dL}

the albumin is below the normal reference of

4.0\text{ g/dL}

. The albumin deficit is

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

. Multiplying

2.4

by

2.5

yields a correction factor of

6

. Adding

6

to the uncorrected gap of

14

gives a true anion gap of

20\text{ mEq/L}

. This indicates a significant occult high anion gap metabolic acidosis, likely resulting from acute kidney injury (uremia) and poor tissue perfusion (lactic acidosis) secondary to profound hypovolemia.

Step-by-Step Solution

1

Calculate the standard (uncorrected) anion gap.

136 - (106 + 16) = 136 - 122 = 14\text{ mEq/L}

The basic anion gap must first be derived from the measured primary electrolytes to establish a baseline for correction.

2

Calculate the deficit in serum albumin from the normal expected value.

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

Albumin constitutes the majority of unmeasured anions in the blood. A deficiency in albumin falsely lowers the measured anion gap, masking potential acid-base disorders.

3

Calculate the correction value and add it to the uncorrected anion gap.

Correction =

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

. Corrected AG =

14 + 6 = 20\text{ mEq/L}

.

For every

1.0\text{ g/dL}

decrease in albumin below

4.0\text{ g/dL}

, the anion gap must be increased by

2.5\text{ mEq/L}

to reveal the true metabolic gap.

Key Concept

Patients with severe hypoalbuminemia have a falsely lowered baseline anion gap; an albumin correction factor must be calculated to accurately identify an occult high anion gap metabolic acidosis.

Estimated Time:1m 30s

Question 109Question →

A prospective cohort study is conducted to evaluate the association between the regular use of a novel synthetic oral glucocorticoid and the development of avascular necrosis (AVN) of the femoral head. The study follows $4,000$ patients prescribed the new glucocorticoid and $6,000$ patients prescribed standard nonsteroidal anti-inflammatory drugs (NSAIDs) for similar rheumatologic conditions. Over a $10$ -year follow-up period, $160$ patients in the glucocorticoid group and $60$ patients in the NSAID group develop AVN.

Based on these findings, what percentage of the AVN cases in the glucocorticoid group is theoretically attributable to the use of the new drug?

Show answer & explanation

Answer: 75

Answer

75

The correct answer is

75

. The Attributable Risk Percent (ARP) represents the fraction of disease in the exposed group that is directly attributable to the exposure. It is calculated as

\frac{\text{Incidence in exposed} - \text{Incidence in unexposed}}{\text{Incidence in exposed}}

. In this study, the incidence in the exposed (glucocorticoid) group is

160 / 4,000 = 4\%

. The incidence in the unexposed (NSAID) group is

60 / 6,000 = 1\%

. Therefore,

\text{ARP} = \frac{4\% - 1\%}{4\%} = \frac{3\%}{4\%} = 75\%

. Alternatively, this can be calculated using Relative Risk (RR).

RR = 4\% / 1\% = 4

.

\text{ARP} = \frac{RR - 1}{RR} = \frac{4 - 1}{4} = 75\%

.

Step-by-Step Solution

1

Calculate the incidence of the outcome in the exposed group.

I_e = \frac{160}{4,000} = 0.04

(or

4\%

)

This establishes the total absolute risk of developing the disease when the exposure is present.

2

Calculate the incidence of the outcome in the unexposed group.

I_u = \frac{60}{6,000} = 0.01

(or

1\%

)

This establishes the baseline risk of the disease occurring without the exposure.

3

Calculate the Attributable Risk Percent (ARP).

\text{ARP} = \frac{I_e - I_u}{I_e} \times 100\% = \frac{0.04 - 0.01}{0.04} \times 100\% = 75\%

This formula determines the specific proportion of cases in the exposed group that are a direct result of the exposure.

Key Concept

Attributable Risk Percent (ARP) measures the proportion of disease in an exposed group that is directly attributable to the exposure, calculated as

(I_e - I_u) / I_e

or

(RR - 1) / RR

.

Question 110Question →

A pharmaceutical company is developing a novel Janus kinase (JAK) inhibitor for the treatment of severe, refractory rheumatoid arthritis. Regulatory guidelines stipulate that the manufacturer must conduct two independent, pivotal Phase III randomized controlled trials demonstrating efficacy. The research team designs Trial X with a statistical power of 80% and Trial Y with a statistical power of 90% to detect the predefined minimum clinically important difference. Both trials use a two-sided alpha level of 0.05.

Assuming the new JAK inhibitor is genuinely effective at or above the minimum clinically important difference, what is the probability (expressed as a percentage) that BOTH trials will fail to yield statistically significant results?

Show answer & explanation

Answer: 2

Answer

2

Statistical power (

1 - \beta

) is the probability of correctly rejecting the null hypothesis when a true effect exists. A Type II error (

\beta

) occurs when a study fails to detect a true effect, resulting in a false negative. The probability of Trial X failing is its Type II error rate:

\beta_X = 1 - 0.80 = 0.20

. The probability of Trial Y failing is

\beta_Y = 1 - 0.90 = 0.10

. Because the two pivotal trials are designed to be independent, the joint probability of both trials simultaneously failing to detect the true effect is the product of their individual Type II error rates:

0.20 \times 0.10 = 0.02

. Expressed as a percentage, this is 2%.

Step-by-Step Solution

1

Define statistical power and Type II error

Power =

1 - \beta

, so

\beta

(Type II error rate) =

1 - \text{Power}

To find the probability of a trial failing to detect a true effect, we must calculate its Type II error rate.

2

Calculate Type II error for each trial

Trial X

\beta = 1 - 0.80 = 0.20

; Trial Y

\beta = 1 - 0.90 = 0.10

Each trial has an independent probability of failing to achieve statistical significance despite a true underlying effect.

3

Calculate joint probability

Joint probability =

0.20 \times 0.10 = 0.02

The probability of multiple independent events all occurring is the product of their individual probabilities.

4

Convert to percentage

0.02 \times 100 = 2\%

The question specifically asks for the probability expressed as a percentage.

Key Concept

The probability of failing to detect a true difference in a study is the Type II error rate (

\beta

), calculated as

1 - \text{Power}

. The probability of independent trials both making a Type II error is the product of their individual

\beta

values.

Question 111Question →

A 62-year-old man presents to the clinic to discuss pancreatic cancer screening. He has no current symptoms of abdominal pain, jaundice, or weight loss. Vital signs are within normal limits, and physical examination is unremarkable. His father and paternal uncle both died of pancreatic adenocarcinoma, prompting his enrollment in a high-risk familial surveillance registry. Epidemiological data indicates that the prevalence of early-stage pancreatic adenocarcinoma in this specific surveillance cohort is $10\%$ .

As part of a new clinical protocol, the patient's blood is tested using a novel multi-target serum biomarker panel. The assay has a validated sensitivity of $90\%$ and a specificity of $90\%$ for detecting early-stage disease.

The patient's biomarker panel returns a positive result.

Based on the provided test characteristics and cohort prevalence, what is the post-test probability (in percent) that this patient actually has early-stage pancreatic adenocarcinoma?

Show answer & explanation

Answer: 50

Answer

50

The question asks for the post-test probability of disease following a positive test result, which is the positive predictive value (PPV) of the assay. PPV depends on the test's sensitivity and specificity, as well as the disease prevalence in the tested population.

Using a hypothetical cohort of

1000

patients:
- Number of patients with disease:

1000 \times 0.10 = 100

- Number of patients without disease:

1000 \times 0.90 = 900

Next, calculate the number of positive test results in each group:
- True positives (TP) =

100 \times 0.90 \text{ (sensitivity)} = 90

- False positives (FP) =

900 \times (1 - 0.90 \text{ specificity}) = 90

The total number of positive test results is

90 \text{ (TP)} + 90 \text{ (FP)} = 180

.

PPV =

\frac{\text{TP}}{\text{TP} + \text{FP}} = \frac{90}{180} = 0.50

(or

50\%

).

Therefore, the probability that the patient has early-stage pancreatic adenocarcinoma is

50\%

.

Step-by-Step Solution

1

Determine the disease prevalence in the cohort.

The prevalence is 10%, meaning in a population of 1000, 100 have the disease and 900 do not.

Prevalence establishes the pre-test probability necessary for calculating predictive values.

2

Calculate the true positive (TP) and false positive (FP) counts.

TP = 100 × 0.90 = 90. FP = 900 × 0.10 = 90.

Using absolute numbers derived from sensitivity and specificity clarifies the ratio of true to false positive results.

3

Calculate the Positive Predictive Value (PPV).

PPV = TP / (TP + FP) = 90 / (90 + 90) = 0.50, or 50%.

The PPV directly represents the post-test probability of disease following a positive test result.

Key Concept

Positive predictive value (PPV) and its dependence on disease prevalence.

Question 112Question →

A multicenter clinical trial investigates the efficacy of a newly developed monoclonal antibody targeting proprotein convertase subtilisin/kexin type 9 (PCSK9) in patients with familial hypercholesterolemia. A total of 4,000 patients are enrolled and randomly assigned in equal numbers to receive either the monoclonal antibody or a placebo injection every two weeks. All participants remain on their maximally tolerated statin therapy. Over a 3-year follow-up period, a nonfatal myocardial infarction occurs in 100 of the 2,000 patients in the treatment group, compared to 150 of the 2,000 patients in the placebo group. Based on these findings, what is the number needed to treat (NNT) to prevent one nonfatal myocardial infarction over the 3-year period?

Show answer & explanation

Answer: 40

Answer

40

The Number Needed to Treat (NNT) is calculated as the inverse of the Absolute Risk Reduction (ARR). By determining the incidence of the outcome in both the placebo group (

150/2000 = 7.5\%

) and the treatment group (

100/2000 = 5.0\%

), we find an ARR of

2.5\%

(or

0.025

). Dividing

1

by

0.025

yields exactly

40

, meaning

40

patients must be treated with the PCSK9 inhibitor over a

3

-year period to prevent one nonfatal myocardial infarction.

Step-by-Step Solution

1

Calculate the Control Event Rate (CER) in the placebo group.

CER = \frac{150}{2000} = 0.075

(

7.5\%

)

The CER establishes the baseline risk of the outcome without the experimental intervention.

2

Calculate the Experimental Event Rate (EER) in the treatment group.

EER = \frac{100}{2000} = 0.050

(

5.0\%

)

The EER represents the risk of the outcome when the experimental intervention is applied.

3

Calculate the Absolute Risk Reduction (ARR).

ARR = CER - EER = 0.075 - 0.050 = 0.025

(

2.5\%

)

The ARR quantifies the absolute decrease in the risk of the outcome attributable to the intervention.

4

Calculate the Number Needed to Treat (NNT).

NNT = \frac{1}{ARR} = \frac{1}{0.025} = 40

The NNT indicates how many patients must be treated to prevent one additional adverse outcome.

Key Concept

Number Needed to Treat (NNT)

Question 113Question →

A clinical researcher is evaluating the relationship between smoking status and postoperative pulmonary complications following major abdominal surgery. A cohort of 450 adult patients is recruited. Patients are categorized based on their smoking status prior to surgery into three groups: never smokers, former smokers, and current smokers. Following surgery, patients are monitored for 30 days, and their respiratory outcomes are classified into three mutually exclusive categories: no complications, mild complications (requiring supplemental oxygen only), or severe complications (requiring reintubation or intensive care). The investigator plans to perform a statistical analysis to determine if there is a significant association between smoking status and the severity of postoperative pulmonary complications. Assuming all expected cell counts are greater than 5, the researcher sets up a contingency table for a Chi-square test of independence. What is the degrees of freedom for the test statistic in this analysis?

Show answer & explanation

Answer: 4

Answer

4

The appropriate statistical analysis for comparing two categorical variables (smoking status and complication severity) is the Chi-square test of independence. The degrees of freedom for this test are calculated using the formula

df = (r - 1) \times (c - 1)

, where

r

is the number of rows (categories of the first variable) and

c

is the number of columns (categories of the second variable). With 3 smoking categories and 3 complication categories,

df = (3 - 1) \times (3 - 1) = 2 \times 2 = 4

.

Step-by-Step Solution

1

Identify the appropriate statistical test and its degrees of freedom formula.

The test is a Chi-square test of independence, which uses the formula

df = (r - 1) \times (c - 1)

.

This formula is standard for evaluating contingency tables to determine how many values can vary freely while maintaining fixed marginal totals.

2

Determine the number of categories for each of the two variables.

Smoking status has 3 categories (

r = 3

), and pulmonary complications has 3 categories (

c = 3

).

Mapping the clinical study design to the statistical dimensions is necessary to populate the formula.

3

Calculate the degrees of freedom.

(3 - 1) \times (3 - 1) = 2 \times 2 = 4

.

Applying the variables to the correct statistical formula yields the final value.

Key Concept

Degrees of freedom for a Chi-square test of independence

Question 114Question →

A 58-year-old man with hypertension and stage 3a chronic kidney disease brings a pharmaceutical advertisement to his primary care physician. The advertisement promotes a newly approved medication, "Renalox," highlighting that it "reduces the risk of progressing to end-stage renal disease (ESRD) by 44%!" The patient asks if this medication would be beneficial for him over the next several years.

The physician reviews the phase 3, randomized, double-blind, placebo-controlled trial cited in the advertisement. In this trial, patients with similar baseline characteristics to this patient were randomized 1:1 to receive either Renalox or a placebo. Because participants were followed for variable durations, the study reported event incidence utilizing person-years of follow-up.

The trial results are summarized in the table below:

Outcome	Placebo Arm (N = 2000)	Renalox Arm (N = 2000)
Total Follow-up	8,000 person-years	8,000 person-years
Progression to ESRD	360 events	200 events
All-Cause Mortality	160 events	120 events

Assuming the incidence rate of ESRD remains constant over time, what is the number of patients who need to be treated with Renalox for exactly 5 years to prevent one additional case of progression to ESRD?

Show answer & explanation

Answer: 10

Answer

10

To calculate the Number Needed to Treat (NNT) over a specific timeframe from person-year data, you must first determine the annual absolute rate reduction. The placebo group incidence is

360 \text{ events} / 8000 \text{ person-years} = 0.045 \text{ events/person-year}

. The treatment group incidence is

200 \text{ events} / 8000 \text{ person-years} = 0.025 \text{ events/person-year}

. The absolute difference is

0.020 \text{ events/person-year}

. Since the physician wants to know the NNT for a 5-year period, the cumulative absolute risk reduction over 5 years is

0.020 \times 5 = 0.10

. The NNT is the reciprocal of the absolute risk reduction:

1 / 0.10 = 10

. Therefore, 10 patients must be treated for 5 years to prevent one case of ESRD.

Step-by-Step Solution

1

Select the correct data row for the targeted outcome.

Progression to ESRD: Placebo = 360 events / 8,000 person-years; Renalox = 200 events / 8,000 person-years.

The question specifically asks for the NNT to prevent ESRD, making the 'All-Cause Mortality' row a distractor.

2

Calculate the annual incidence rate for both groups.

Placebo rate = 360 / 8,000 = 0.045 events/person-year. Renalox rate = 200 / 8,000 = 0.025 events/person-year.

Person-year data must be converted into a standardized annual incidence rate to calculate absolute differences.

3

Determine the Absolute Rate Reduction (ARR) per year.

ARR per year = 0.045 - 0.025 = 0.020 events/person-year.

The absolute rate reduction represents the fraction of an event prevented by treating one patient for one year.

4

Scale the ARR to the specific time horizon requested (5 years).

5-year ARR = 0.020 * 5 = 0.10 (or 10%).

Because the clinical question asks for the NNT over a 5-year treatment period, the cumulative absolute risk reduction over that exact timeframe must be determined.

5

Calculate the Number Needed to Treat (NNT).

NNT = 1 / 0.10 = 10.

NNT is the inverse of the absolute risk reduction. Treating 10 patients for 5 years will prevent exactly one progression to ESRD.

Key Concept

Converting person-time incidence rates to time-adjusted absolute risk reductions for Number Needed to Treat (NNT) calculations.

Question 115Question →

A physician is reviewing a pharmaceutical advertisement for a newly approved oral anticoagulant indicated for the prevention of ischemic stroke in patients with non-valvular atrial fibrillation. The advertisement prominently features the headline: "Reduces the risk of ischemic stroke by 25% compared to standard therapy!"

The physician reviews the prescribing information provided in the fine print. According to the data from the pivotal 2-year clinical trial, the incidence of ischemic stroke in the control group receiving standard therapy was 8.0%.

Based on the provided data, what is the number needed to treat (NNT) with the new anticoagulant to prevent one additional ischemic stroke over a 2-year period?

Show answer & explanation

Answer: 50

Answer

50 patients

The number needed to treat (NNT) is the inverse of the absolute risk reduction (ARR). When a drug advertisement provides the relative risk reduction (RRR) and the baseline risk (Control Event Rate, or CER), the ARR must first be derived. The ARR is calculated by multiplying the CER by the RRR (

0.08 \times 0.25 = 0.02

). The NNT is then calculated as 1 / ARR (

1 / 0.02 = 50

). Therefore, 50 patients need to be treated with the new anticoagulant for 2 years to prevent one additional ischemic stroke.

Step-by-Step Solution

1

Identify the given statistical values from the advertisement.

Control Event Rate (CER) = 8.0% (

0.08

) and Relative Risk Reduction (RRR) = 25% (

0.25

).

These values are required to calculate the absolute treatment effect.

2

Calculate the Absolute Risk Reduction (ARR).

ARR = CER

\times

RRR =

0.08 \times 0.25 = 0.02

(or 2.0%).

ARR represents the actual difference in risk between the control and treatment groups, which is necessary for finding the NNT.

3

Calculate the Number Needed to Treat (NNT).

NNT = 1 / ARR = 1 / 0.02 = 50.

NNT is defined as the inverse of the absolute risk reduction.

Key Concept

Calculating Number Needed to Treat (NNT) from Relative Risk Reduction (RRR) and baseline risk.

Estimated Time:1m 0s

Question 116Question →

A clinical researcher is investigating the incidence of electrolyte abnormalities associated with different diuretic regimens in patients with chronic heart failure. The study population is divided into five distinct treatment groups: loop diuretics alone, thiazide diuretics alone, potassium-sparing diuretics alone, sequential nephron blockade (loop plus thiazide diuretics), and combination loop plus potassium-sparing diuretics. At the one-month follow-up, the patients' serum potassium levels are categorized into three distinct states: hypokalemia, normokalemia, and hyperkalemia. The investigator constructs a contingency table to analyze the categorical data and plans to perform a Pearson Chi-square test of independence to determine if there is a significant association between the diuretic regimen and the serum potassium state.

Based on this study design, what is the number of degrees of freedom for this statistical test?

Show answer & explanation

Answer: 8

Answer

8

The Pearson Chi-square test of independence is utilized to determine if there is a statistically significant association between two categorical variables. Data for this test is organized into a contingency table consisting of

R

rows and

C

columns representing the distinct categories of each variable. The formula for calculating the degrees of freedom (

df

) for this inferential test is

df = (R - 1) \times (C - 1)

. In this clinical scenario, the independent categorical variable (diuretic regimen) contains 5 groups (

R = 5

), and the dependent categorical variable (serum potassium state) contains 3 groups (

C = 3

). Therefore, the degrees of freedom is calculated as

(5 - 1) \times (3 - 1) = 4 \times 2 = 8

.

Step-by-Step Solution

1

Determine the number of categories for the first variable.

R = 5

There are five distinct diuretic treatment groups defined in the study design.

2

Determine the number of categories for the second variable.

C = 3

The serum potassium levels are categorized into three distinct, mutually exclusive states.

3

Calculate the degrees of freedom using the standard contingency table formula.

df = 8

Applying the formula

df = (R - 1) \times (C - 1)

, we get

(5 - 1) \times (3 - 1) = 4 \times 2 = 8

.

Key Concept

Degrees of freedom calculation for a Pearson Chi-square test of independence

Question 117Question →

A 52-year-old woman with inadequately controlled rheumatoid arthritis comes to the clinic. Her physician is considering starting her on a newly approved oral Janus kinase (JAK) inhibitor. Before prescribing the medication, the physician reviews the published results of a phase 3, multicenter, randomized, double-blind, placebo-controlled trial that evaluated the safety and efficacy of the drug over a 52-week period.

The trial enrolled a total of $1,500$ patients who were randomized in a 1:1 ratio to receive either the JAK inhibitor or a placebo. The safety data regarding incident herpes zoster infections during the trial are summarized in the table below:

Group	Total Patients	Herpes Zoster Cases
JAK Inhibitor	750	45
Placebo	750	15

Based on these study results, what is the number needed to harm (NNH) for a herpes zoster infection?

Show answer & explanation

Answer: 25

Answer

25

The number needed to harm (NNH) is the number of patients who need to be exposed to a risk factor (or treatment) over a specific period to cause one additional adverse event. It is calculated using the formula:

NNH = \frac{1}{ARI}

, where ARI is the Absolute Risk Increase. The ARI is the difference in adverse event rates between the treatment and control groups. In this trial, the event rate in the treatment group (EER) is

\frac{45}{750} = 0.06

. The event rate in the placebo group (CER) is

\frac{15}{750} = 0.02

. The ARI is

0.06 - 0.02 = 0.04

. Therefore, the

NNH = \frac{1}{0.04} = 25

. This means that for every 25 patients treated with the JAK inhibitor for 52 weeks, one additional patient will develop a herpes zoster infection compared to placebo.

Step-by-Step Solution

1

Calculate the Experimental Event Rate (EER)

EER = 45 / 750 = 0.06

To determine the incidence of the adverse event in the treatment group.

2

Calculate the Control Event Rate (CER)

CER = 15 / 750 = 0.02

To determine the background incidence of the adverse event in the placebo group.

3

Calculate the Absolute Risk Increase (ARI)

ARI = EER - CER = 0.06 - 0.02 = 0.04

To find the absolute difference in risk caused by the medication.

4

Calculate the Number Needed to Harm (NNH)

NNH = 1 / ARI = 1 / 0.04 = 25

To determine how many patients need to be treated to cause one additional adverse event.

Key Concept

Number Needed to Harm (NNH) is the inverse of the Absolute Risk Increase (ARI) and represents the number of patients who need to receive a treatment for one additional adverse event to occur.

Question 118Question →

A hospitalist is reviewing an article from a leading medical journal detailing a phase 3, randomized, double-blind trial. The study evaluated the efficacy and safety of a novel factor XIa inhibitor compared to a placebo for preventing recurrent venous thromboembolism (VTE) in patients with high-risk thrombophilia. A total of 5,200 patients were randomized in a 1:1 ratio to receive either the novel medication or a placebo. The trial summary table is presented below.

Endpoint	Placebo (n=2,600)	Novel Inhibitor (n=2,600)	p-value
Primary Efficacy (Recurrent VTE or VTE-related death)	286 (11.0%)	156 (6.0%)	<0.001
Primary Safety (Major bleeding)	39 (1.5%)	65 (2.5%)	0.012

Based on the study results, what is the number needed to treat (NNT) to prevent one primary efficacy endpoint event? (Provide the exact whole number.)

Show answer & explanation

Answer: 20

Answer

20

The Number Needed to Treat (NNT) is calculated as the inverse of the Absolute Risk Reduction (ARR). The primary efficacy endpoint occurred in 11.0% of the placebo group (CER =

0.11

) and 6.0% of the novel inhibitor group (EER =

0.06

). The ARR is

0.11 - 0.06 = 0.05

. Therefore, the NNT is

\frac{1}{0.05} = 20

. This means 20 patients must be treated with the novel factor XIa inhibitor to prevent one recurrent VTE or VTE-related death.

Step-by-Step Solution

1

Identify the event rates for the primary efficacy endpoint from the provided table.

Control Event Rate (CER) = 11.0% (

0.11

); Experimental Event Rate (EER) = 6.0% (

0.06

).

Calculating the number needed to treat requires the exact incidence rates of the primary outcome in both the control and treatment groups.

2

Calculate the Absolute Risk Reduction (ARR).

ARR = 0.11 - 0.06 = 0.05

ARR represents the absolute difference in the risk of the outcome between the placebo group and the active treatment group.

3

Calculate the Number Needed to Treat (NNT).

NNT = \frac{1}{ARR} = \frac{1}{0.05} = 20

The NNT is defined as the inverse of the Absolute Risk Reduction. It indicates how many patients must be treated to prevent one adverse outcome.

Key Concept

Number Needed to Treat (NNT) Calculation

Question 119Question →

A physician is reviewing a promotional pamphlet for a newly approved oral hypoglycemic agent. The pamphlet includes a safety summary comparing the incidence of severe hypoglycemic episodes between the new medication and placebo over a 52-week clinical trial period.

The safety data excerpt reports the following:
- Novel Agent Group ( $N = 4,000$ ): 180 patients experienced at least one episode of severe hypoglycemia.
- Placebo Group ( $N = 4,000$ ): 112 patients experienced at least one episode of severe hypoglycemia.

Based on the clinical trial data provided in the promotional material, how many patients must be treated with the novel agent instead of placebo for 52 weeks to expect one additional severe hypoglycemic event? (Round your final answer to the nearest whole number).

Show answer & explanation

Answer: 59

Answer

59

The Number Needed to Harm (NNH) is derived from the Absolute Risk Increase (ARI). The ARI is the difference between the Experimental Event Rate (EER) and the Control Event Rate (CER). In this trial, the EER is

180/4000 = 0.045

and the CER is

112/4000 = 0.028

. The difference (ARI) is

0.017

. The NNH is calculated by taking the reciprocal of the ARI (

1/0.017

), which equals

58.82

. Rounding to the nearest whole number yields

59

patients.

Step-by-Step Solution

1

Calculate the Experimental Event Rate (EER).

EER = \frac{180}{4000} = 0.045

(or

4.5\%

)

To determine the absolute risk of the adverse event in the population receiving the new intervention.

2

Calculate the Control Event Rate (CER).

CER = \frac{112}{4000} = 0.028

(or

2.8\%

)

To determine the baseline absolute risk of the adverse event in the control population.

3

Calculate the Absolute Risk Increase (ARI).

ARI = EER - CER = 0.045 - 0.028 = 0.017

(or

1.7\%

)

To find the excess risk of the adverse event directly attributable to the new medication.

4

Calculate the Number Needed to Harm (NNH).

NNH = \frac{1}{ARI} = \frac{1}{0.017} \approx 58.82

NNH indicates how many patients need to be exposed to the risk factor over a specific period to cause one additional harmful event.

5

Round the final value to the nearest whole number.

59

Clinical metrics representing whole patients cannot be expressed as fractions, and standard rounding rules apply.

Key Concept

The Number Needed to Harm (NNH) quantifies the safety of an intervention and is calculated as the inverse of the Absolute Risk Increase (

NNH = 1 / ARI

).

Question 120Question →

An unresponsive 74-year-old woman is brought to the emergency department by emergency medical services after being found on the floor of her apartment by her daughter. The daughter notes that the patient had been increasingly disoriented and complaining of excessive thirst and urination over the past week.

Past medical history: Type 2 diabetes mellitus, hypertension, hyperlipidemia.

Vital signs:
- Temperature: 37.0°C (98.6°F)
- Blood pressure: 92/58 mm Hg
- Heart rate: 118/min
- Respirations: 16/min

Physical examination shows an obtunded patient with dry mucous membranes, decreased skin turgor, and a delayed capillary refill of 4 seconds.

Laboratory studies:

Test	Result	Reference Range
Sodium	130 mEq/L	136-145 mEq/L
Potassium	4.5 mEq/L	3.5-5.0 mEq/L
Chloride	92 mEq/L	98-106 mEq/L
Bicarbonate	24 mEq/L	22-28 mEq/L
Blood urea nitrogen (BUN)	84 mg/dL	7-18 mg/dL
Creatinine	2.1 mg/dL	0.6-1.2 mg/dL
Glucose	900 mg/dL	70-110 mg/dL

What is this patient's calculated serum osmolality in mOsm/kg?

Show answer & explanation

Answer: 340

Answer

340

The patient's calculated serum osmolality is determined using the standard formula:

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

. Substituting the patient's laboratory values yields

2 \times 130 + \frac{900}{18} + \frac{84}{2.8}

, which equals

260 + 50 + 30 = 340

mOsm/kg. This significantly elevated value is consistent with Hyperosmolar Hyperglycemic State (HHS), which often presents with extreme hyperglycemia, high serum osmolality, and profound dehydration.

Step-by-Step Solution

1

Identify the required parameters.

The parameters needed are serum sodium (130 mEq/L), glucose (900 mg/dL), and blood urea nitrogen (84 mg/dL).

Serum osmolality is calculated based on the primary osmotically active solutes in the blood.

2

Convert glucose and BUN from mg/dL to mmol/L.

Glucose:

900 / 18 = 50

mmol/L. BUN:

84 / 2.8 = 30

mmol/L.

The standard osmolality formula requires solutes to be in mmol/L or mEq/L. These conversion factors normalize the mg/dL values.

3

Calculate the serum osmolality using the formula.

2 \times 130 + 50 + 30 = 260 + 50 + 30 = 340

mOsm/kg.

The formula

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

yields the calculated serum osmolality.

Key Concept

The calculated serum osmolality requires converting glucose and BUN to mmol/L. The formula is

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

.

Answer

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