NBRC Mock TMC Exam Review Actual Exam 2026/2027 – Complete Exam-Style Questions with Detailed Rationales | 100% Verified | Pass Guaranteed – A+ Graded

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NBRC Mock TMC Exam Review Actual
Exam 2026/2027 – Complete Exam-Style
Questions with Detailed Rationales | 100%
Verified | Pass Guaranteed – A+ Graded
[SECTION 1: Patient Data Evaluation & Assessment — Questions 1-25]

Q1: A 65-year-old male patient has the following arterial blood gas results: pH 7.28, PaCO2 60
torr, HCO3 28 mEq/L, PaO2 85 torr on room air. Which acid-base disorder is present, and what
is the expected PaCO2 for an uncompensated respiratory acidosis?
A. Metabolic acidosis; Expected PaCO2 25 torr.

B. Metabolic alkalosis; Expected PaCO2 50 torr.

C. Respiratory acidosis; Expected PaCO2 28 torr [CORRECT]

D. Respiratory alkalosis; Expected PaCO2 26 torr.



Correct Answer: C

Rationale: The patient has respiratory acidosis (pH < 7.35, PaCO2 > 45). Using Winter's formula
(Expected PaCO2 = 1.5 x HCO3 + 8 ± 2), we calculate 1.5(28) + 8 = 50. Since the patient's
PaCO2 is 60, which is significantly higher than the expected 50 for a metabolic disorder, this
indicates a superimposed respiratory acidosis. If it were a simple respiratory acidosis, the PaCO2
would be expected to drop the pH by approximately 0.08 per 10 torr rise in PaCO2, which aligns
with the acute respiratory acidosis presentation.


Q2: A 45-year-old female has an ABG: pH 7.48, PaCO2 30 torr, HCO3 22 mEq/L. Which
compensation is occurring?

A. Uncompensated respiratory alkalosis.

B. Compensated metabolic acidosis.

C. Mixed respiratory alkalosis and metabolic acidosis. [CORRECT]

D. Compensated metabolic alkalosis.

,2


Correct Answer: C

Rationale: The patient has a primary respiratory alkalosis (pH > 7.45, PaCO2 < 35). However,
the HCO3 is also low (22), indicating a metabolic acidosis is also present. In pure respiratory
alkalosis, the kidneys would excrete bicarbonate to compensate over time; a low bicarbonate in
an acute setting suggests a mixed disorder (e.g., salicylate toxicity or hypoxia). A pure
respiratory alkalosis would have a normal HCO3 in the acute phase.


Q3: Calculate the A-a gradient for a patient breathing room air (PB is 747 torr) with a PaCO2 of
40 torr and a PaO2 of 60 torr. Assume a respiratory quotient (RQ) of 0.8.
A. 10 mmHg

B. 40 mmHg

C. 15 mmHg

D. 50 mmHg [CORRECT]



Correct Answer: D

Rationale: The A-a gradient formula is PAO2 = (PB - PH2O) x FiO2 - (PaCO2 / RQ). On room
air, FiO2 is 0.21, and PH2O is 47 torr. PAO2 = (747-47) x 0.21 - (40/0.8) = 700 x 0.21 - 50 = 147
- 50 = 97 torr. The A-a gradient is PAO2 - PaO2, so 97 - 60 = 37 mmHg. Self-Correction: Let's
re-calculate carefully. (PB-PH2O) = 747-47=700. 700 * 0.21 = 147. PaCO2/RQ = 40/0.8 = 50.
147-50 = 97. PAO2=97. PaO2=60. Gradient=37.

Correction: My options don't have 37.

Let me re-calculate options.

Maybe the PaCO2 is lower? If PaO2 is 60.

If Gradient is 10, PAO2=70.

If Gradient is 40, PAO2=100.

If Gradient is 15, PAO2=75.
If Gradient is 50, PAO2=110.

Let's check the math again.

(700 * 0.21) = 147.
147 - 50 = 97.

,3


97 - 60 = 37.

Since 37 is not an option, I must adjust the scenario in my head to match one of the options or
check if I missed a variable.

Let's assume PaCO2 is 45. 45/0.8 = 56. 147-56 = 91. 91-60=31.

Let's assume RQ is 1.0 (rare). 40/1 = 40. 147 - 40 = 107. 107 - 60 = 47.

Let's check the options again.

A. 10
B. 40

C. 15
D. 50

Okay, I will generate a question where the math yields one of these numbers to avoid confusion.

New Scenario: PB 747, PaCO2 35, PaO2 95, RQ 0.8.

PAO2 = 700 * 0.21 - (35/0.8) = 147 - 43.75 = 103.25.

Gradient = 103.25 - 95 = 8.25 (approx 10).
Revised Question: PB 747, PaCO2 35, PaO2 55, RQ 0.8.

PAO2 = 147 - 43.75 = 103.25.

Gradient = 103.25 - 55 = 48.25 (approx 50).

I will use this set.



Revised Q3: A patient on room air has a PaO2 of 55 torr, PaCO2 of 35 torr, and a Barometric
Pressure of 747 torr. What is the A-a gradient?

A. 10 mmHg

B. 40 mmHg

C. 15 mmHg
D. 50 mmHg [CORRECT]



Correct Answer: D

, 4


Rationale: First, calculate the Alveolar Oxygen (PAO2): PAO2 = (PB - PH2O) x FiO2 -
(PaCO2/RQ). PAO2 = (747-47) x 0.21 - (35/0.8) = 700 x 0.21 - 43.75 = 147 - 43.75 = 103.25
torr. The A-a gradient is the difference between alveolar and arterial oxygen: 103.25 - 55 = 48.25
torr, which rounds to approximately 50 mmHg. A gradient of 50 mmHg is elevated (normal is <
20 in young, increases with age).



Q4: Which condition is indicated by an elevated A-a gradient in a patient with a normal PaCO2?
A. V/Q mismatch

B. Anemic hypoxia [CORRECT]

C. Hypoventilation
D. Increased FiO2



Correct Answer: A

Rationale: An elevated A-a gradient (PAO2 - PaO2) implies that gas exchange is impaired,
specifically that the blood leaving the alveoli is not fully oxygenated, while the alveoli
themselves are oxygenating normally (since PAO2 is calculated as normal based on FiO2 and
PaCO2). This V/Q mismatch is characteristic of shunts or V/Q defects like PE or pneumonia.
Hypoventilation increases PaCO2 and typically decreases PAO2.



Q5: A 22-year-old male has an ABG of pH 7.50, PaCO2 30 torr, HCO3 24 mEq/L. These values
are consistent with:

A. Metabolic alkalosis with partial respiratory compensation.

B. Acute respiratory alkalosis. [CORRECT]

C. Metabolic acidosis with respiratory compensation.

D. Respiratory acidosis with metabolic compensation.


Correct Answer: B

Rationale: The primary disorder is respiratory alkalosis (low PaCO2, high pH). In acute
respiratory alkalosis, the kidneys have not yet begun to compensate (excrete bicarbonate), so
HCO3 remains normal (22-26). The HCO3 of 24 is within normal limits, confirming an acute
process. If this were chronic, the HCO3 would be decreased due to renal compensation.