CCRN PRACTICE QUESTIONS 3
1. A patient is admitted with unstable angina. He has a long history of hyper-tension and
coronary artery disease. The nurse notes a split S2 on expirationand a single S2 on inspiration
during cardiac auscultation. Blood pressure is 150/88 mm Hg, and heart rate is 88 beats/min.
On the electrocardiogram, there is a normal-appearing P wave in front of each QRS complex,
the PR interval measures 0.2 second consistently, and the QRS complexes measure
0.14 second. They are positive in V5 and V6 and negative in V1 and V2. Thesefindings most
likely indicate which of the following?
a. Left bundle branch block (LBBB)
b. Right bundle branch block (RBBB)
c. Third-degree atrioventricular block
d. Ventricular tachycardia: Correct answer: a
Rationale: Features of LBBB described here are a QRS complex greater than 0.12second in
duration and a QRS complex that is positive in leads V5 and V6 (considerthese left ventricular
leads) and negative in leads V1 and V2 (consider these right ventricular leads). LBBB causes
a paradoxical splitting of S2. This means that it is split on expiration but not on inspiration.
This is paradoxical because it is opposite ofa normal physiologic split of S2, which is split on
inspiration but not split on expiration.Test-Taking Strategy: A P wave precedes each QRS
complex, so eliminate ventric-ular tachycardia, option d. A P wave occurs for each QRS
complex, and the PR interval is consistent, so eliminate third-degree atrioventricular block,
option c. Theproblem is a bundle branch block because it originates above the ventricle (note
normal and consistent P waves), but the QRS complex is wide, indicating that the impulse is
taking longer than normal to travel through the ventricle. Is the block left or right? Remember
that the wide QRS complex will be upright in leads V1 and V2if it is RBBB but upright in
leads V5 and V6 if it is LBBB. Also, in looking at lead V1, ifthere is an associated P wave but
the QRS complex is wide and upright, it is RBBB.If the QRS complex is wide and negative, it
is LBBB. Choose option a.
2. On a pulmonary artery waveform, the dicrotic notch represents closure of
which valve?
a. Aortic valve
b. Pulmonic valve
c. Tricuspid valve
d. Mitral valve: Correct answer: b
Rationale: In a pulmonary artery waveform the three components of the waveformare systole,
dicrotic notch, and diastole. Systole is the pressure generated by the right ventricle so that the
pulmonic valve will be pushed open, the dicrotic notch
is caused by the closure of the pulmonic valve, and diastole is the pressure in the pulmonary
artery during ventricular diastole. The diastolic pressure is a reflection ofthe vascular tone in
the pulmonary vascular bed. If the vessels are constricted or ifthere is back pressure from the
left side of the heart, the diastolic pressure will be high.
Test-Taking Strategy: The waveform from the pulmonary artery cannot reflect what isin front of
,it. Choose the chamber or vessel immediately behind the pulmonary artery.Choose option b.
Also, remember that the dicrotic notch on the arterial waveform represents closure of the
aortic valve.
3. Which of the following statements is true?
a.The diastolic pulmonary artery pressure (PAd) should not be higher than thepulmonary artery
occlusive pressure (PAOP).
b. The PAOP should not be higher than the PAd.
c. The PAd and the PAOP are usually equal.
d. The right atrial pressure (RAP) and the PAOP are usually equal.: Correctanswer: b
Rationale: The PAd is normally 2 to 5 mm Hg higher than the PAOP. PAd may be more than 5
mm Hg higher than the PAOP in patients with pulmonary hypertension.If the PAOP is higher
than the PAd, suspect that there is an occlusion in the catheteror that the catheter is not in the
correct area of the pulmonary vasculature. The RAPis normally lower than the PAOP.
Test-Taking Strategy: Remember that water does not flow uphill. The circuit betweenthe
pulmonary vascular bed and the left atrium is open. The pressure in the pul- monary artery
during diastole is the filling pressure for the left atrium. So the PAd should be slightly higher
than the PAOP.
4. When pulmonary arterial diastolic pressure (PAd) is more than 5 mm Hghigher than
pulmonary artery occlusive pressure (PAOP), it signals which abnormal condition?
a. Right ventricular failure
b. Left ventricular failure
c. Pulmonary hypertension
d. Systemic hypertension: Correct answer: c
Rationale: When the PAd is more than 5 mm Hg higher than the PAOP, it is an indication of
pulmonary hypertension. Possible causes of pulmonary hypertension are passive (e.g., mitral
valve disease) or active (e.g., causes of hypoxemic pul- monary vasoconstriction such as acute
respiratory distress syndrome, chronic ob- structive pulmonary disease, or pulmonary
embolism). Pulmonary embolism causespulmonary hypertension by mechanical obstruction
and by hypoxemic pulmonary vasoconstriction.
Test-Taking Strategy: Consider the pulmonary artery systolic pressure as reflectiveof the right
side of the heart, the PAd as reflective of the pulmonary vascular circuit,and the PAOP as
reflective of the left side of the heart. Therefore the PAd would bereflective of a pulmonary
vascular circuit problem. Left ventricular failure increasesthe PAd, but the PAOP is elevated
also and there is not more than a 5-mm Hg difference between the PAd and the PAOP. Right
ventricular failure increases the RAP. Systemic hypertension increases systemic pressures.
5. Which vasodilator would be best for a patient with a pulmonary artery occlusive pressure
(PAOP) of 24 mm Hg and a systolic vascular resistance(SVR) of 2100 dynes/sec/cm5?
a. Hydralazine (Apresoline)
b. Nitroglycerin (Tridil)
c. Nitroprusside (Nipride)
, d. Morphine sulfate: Correct answer: c
Rationale: Left ventricular preload (as measured by PAOP) and left ventricular afterload (as
measured by SVR) are increased, so venous vasodilation is needed to decrease preload and
arterial vasodilation is needed to decrease afterload. Hy- dralazine dilates arteries only,
morphine sulfate dilates veins only, and nitroglycerindosages must be above 1 mcg/kg/min to
achieve arterial dilating effects. Nitroprus-side is a mixed vasodilator. It dilates arteries and
veins to decrease afterload and preload.
Test-Taking Strategy: Take this step by step. The PAOP indicates increased preload.The SVR
indicated increased afterload. So both preload and afterload need to be decreased. Veins are
pre-heart so dilating veins decreases preload. Arteries are after the heart so dilating arteries
decreases afterload. Now which vasodilator dilatesboth arteries and veins? Of the choices here,
nitroprusside is the best choice.
6. A patient is admitted with acute chest pain and dyspnea. Pulse oximetryindicates an
arterial oxygen saturation (SaO2) of 88%. Readings after inser-tion of the pulmonary artery
catheter included a normal pulmonary artery occlusive pressure (PAOP), an elevated
pulmonary artery, and an elevated right atrial pressure (RAP). The nurse suspects that these
findings are mostindicative of what acute problem?
a. Cardiac tamponade
b. Pulmonary embolism
c. Right ventricular infarction
d. Pericarditis: Correct answer: b
Rationale: Remember that when the pulmonary artery diastolic pressure (PAd) is more than 5
mm Hg greater than the PAOP, pulmonary hypertension is present. Thefindings are consistent
with a problem in the lung that is likely due to a pulmonary embolism.
Pericarditis does not affect hemodynamic parameters unless cardiac tamponade occurs.
Cardiac tamponade would cause the PAd and PAOP to be elevated and within 5 mm Hg of each
other. Elevation and equalization of RAP, PAd, and PAOP isseen in cardiac tamponade and
frequently is referred to as equalization of pressure.Right ventricular infarction would cause an
elevated right atrial pressure with a normal or low PAOP but would not cause hypoxemia.
Test-Taking Strategy: Eliminate answers a and d because they are similar andinvolve the
outside of the heart. Select option b, which is a lung problem.
7. A patient's digital readout of the pulmonary artery pressure suddenly changes from 22/10
mm Hg to 24/2 mm Hg and remains at this pressure. Whichof the following is the most likely
cause of this change?
a. Proximal movement of the catheter
b. Distal movement of the catheter
c. Hypoxemia
d. Pulmonary artery vasodilation: Correct answer: a
Rationale: The sudden decrease in the pulmonary artery diastolic pressure (PAd) most often
indicates that the catheter has flipped back into the right ventricle. Notethat the pulmonary
artery systolic pressure has changed little. Distal migration of thecatheter would cause the PAOP
waveform and pressure. Hypoxemia would increasethe PAd as it causes pulmonary
hypertension. Pulmonary artery vasodilation woulddecrease the systolic and diastolic
1. A patient is admitted with unstable angina. He has a long history of hyper-tension and
coronary artery disease. The nurse notes a split S2 on expirationand a single S2 on inspiration
during cardiac auscultation. Blood pressure is 150/88 mm Hg, and heart rate is 88 beats/min.
On the electrocardiogram, there is a normal-appearing P wave in front of each QRS complex,
the PR interval measures 0.2 second consistently, and the QRS complexes measure
0.14 second. They are positive in V5 and V6 and negative in V1 and V2. Thesefindings most
likely indicate which of the following?
a. Left bundle branch block (LBBB)
b. Right bundle branch block (RBBB)
c. Third-degree atrioventricular block
d. Ventricular tachycardia: Correct answer: a
Rationale: Features of LBBB described here are a QRS complex greater than 0.12second in
duration and a QRS complex that is positive in leads V5 and V6 (considerthese left ventricular
leads) and negative in leads V1 and V2 (consider these right ventricular leads). LBBB causes
a paradoxical splitting of S2. This means that it is split on expiration but not on inspiration.
This is paradoxical because it is opposite ofa normal physiologic split of S2, which is split on
inspiration but not split on expiration.Test-Taking Strategy: A P wave precedes each QRS
complex, so eliminate ventric-ular tachycardia, option d. A P wave occurs for each QRS
complex, and the PR interval is consistent, so eliminate third-degree atrioventricular block,
option c. Theproblem is a bundle branch block because it originates above the ventricle (note
normal and consistent P waves), but the QRS complex is wide, indicating that the impulse is
taking longer than normal to travel through the ventricle. Is the block left or right? Remember
that the wide QRS complex will be upright in leads V1 and V2if it is RBBB but upright in
leads V5 and V6 if it is LBBB. Also, in looking at lead V1, ifthere is an associated P wave but
the QRS complex is wide and upright, it is RBBB.If the QRS complex is wide and negative, it
is LBBB. Choose option a.
2. On a pulmonary artery waveform, the dicrotic notch represents closure of
which valve?
a. Aortic valve
b. Pulmonic valve
c. Tricuspid valve
d. Mitral valve: Correct answer: b
Rationale: In a pulmonary artery waveform the three components of the waveformare systole,
dicrotic notch, and diastole. Systole is the pressure generated by the right ventricle so that the
pulmonic valve will be pushed open, the dicrotic notch
is caused by the closure of the pulmonic valve, and diastole is the pressure in the pulmonary
artery during ventricular diastole. The diastolic pressure is a reflection ofthe vascular tone in
the pulmonary vascular bed. If the vessels are constricted or ifthere is back pressure from the
left side of the heart, the diastolic pressure will be high.
Test-Taking Strategy: The waveform from the pulmonary artery cannot reflect what isin front of
,it. Choose the chamber or vessel immediately behind the pulmonary artery.Choose option b.
Also, remember that the dicrotic notch on the arterial waveform represents closure of the
aortic valve.
3. Which of the following statements is true?
a.The diastolic pulmonary artery pressure (PAd) should not be higher than thepulmonary artery
occlusive pressure (PAOP).
b. The PAOP should not be higher than the PAd.
c. The PAd and the PAOP are usually equal.
d. The right atrial pressure (RAP) and the PAOP are usually equal.: Correctanswer: b
Rationale: The PAd is normally 2 to 5 mm Hg higher than the PAOP. PAd may be more than 5
mm Hg higher than the PAOP in patients with pulmonary hypertension.If the PAOP is higher
than the PAd, suspect that there is an occlusion in the catheteror that the catheter is not in the
correct area of the pulmonary vasculature. The RAPis normally lower than the PAOP.
Test-Taking Strategy: Remember that water does not flow uphill. The circuit betweenthe
pulmonary vascular bed and the left atrium is open. The pressure in the pul- monary artery
during diastole is the filling pressure for the left atrium. So the PAd should be slightly higher
than the PAOP.
4. When pulmonary arterial diastolic pressure (PAd) is more than 5 mm Hghigher than
pulmonary artery occlusive pressure (PAOP), it signals which abnormal condition?
a. Right ventricular failure
b. Left ventricular failure
c. Pulmonary hypertension
d. Systemic hypertension: Correct answer: c
Rationale: When the PAd is more than 5 mm Hg higher than the PAOP, it is an indication of
pulmonary hypertension. Possible causes of pulmonary hypertension are passive (e.g., mitral
valve disease) or active (e.g., causes of hypoxemic pul- monary vasoconstriction such as acute
respiratory distress syndrome, chronic ob- structive pulmonary disease, or pulmonary
embolism). Pulmonary embolism causespulmonary hypertension by mechanical obstruction
and by hypoxemic pulmonary vasoconstriction.
Test-Taking Strategy: Consider the pulmonary artery systolic pressure as reflectiveof the right
side of the heart, the PAd as reflective of the pulmonary vascular circuit,and the PAOP as
reflective of the left side of the heart. Therefore the PAd would bereflective of a pulmonary
vascular circuit problem. Left ventricular failure increasesthe PAd, but the PAOP is elevated
also and there is not more than a 5-mm Hg difference between the PAd and the PAOP. Right
ventricular failure increases the RAP. Systemic hypertension increases systemic pressures.
5. Which vasodilator would be best for a patient with a pulmonary artery occlusive pressure
(PAOP) of 24 mm Hg and a systolic vascular resistance(SVR) of 2100 dynes/sec/cm5?
a. Hydralazine (Apresoline)
b. Nitroglycerin (Tridil)
c. Nitroprusside (Nipride)
, d. Morphine sulfate: Correct answer: c
Rationale: Left ventricular preload (as measured by PAOP) and left ventricular afterload (as
measured by SVR) are increased, so venous vasodilation is needed to decrease preload and
arterial vasodilation is needed to decrease afterload. Hy- dralazine dilates arteries only,
morphine sulfate dilates veins only, and nitroglycerindosages must be above 1 mcg/kg/min to
achieve arterial dilating effects. Nitroprus-side is a mixed vasodilator. It dilates arteries and
veins to decrease afterload and preload.
Test-Taking Strategy: Take this step by step. The PAOP indicates increased preload.The SVR
indicated increased afterload. So both preload and afterload need to be decreased. Veins are
pre-heart so dilating veins decreases preload. Arteries are after the heart so dilating arteries
decreases afterload. Now which vasodilator dilatesboth arteries and veins? Of the choices here,
nitroprusside is the best choice.
6. A patient is admitted with acute chest pain and dyspnea. Pulse oximetryindicates an
arterial oxygen saturation (SaO2) of 88%. Readings after inser-tion of the pulmonary artery
catheter included a normal pulmonary artery occlusive pressure (PAOP), an elevated
pulmonary artery, and an elevated right atrial pressure (RAP). The nurse suspects that these
findings are mostindicative of what acute problem?
a. Cardiac tamponade
b. Pulmonary embolism
c. Right ventricular infarction
d. Pericarditis: Correct answer: b
Rationale: Remember that when the pulmonary artery diastolic pressure (PAd) is more than 5
mm Hg greater than the PAOP, pulmonary hypertension is present. Thefindings are consistent
with a problem in the lung that is likely due to a pulmonary embolism.
Pericarditis does not affect hemodynamic parameters unless cardiac tamponade occurs.
Cardiac tamponade would cause the PAd and PAOP to be elevated and within 5 mm Hg of each
other. Elevation and equalization of RAP, PAd, and PAOP isseen in cardiac tamponade and
frequently is referred to as equalization of pressure.Right ventricular infarction would cause an
elevated right atrial pressure with a normal or low PAOP but would not cause hypoxemia.
Test-Taking Strategy: Eliminate answers a and d because they are similar andinvolve the
outside of the heart. Select option b, which is a lung problem.
7. A patient's digital readout of the pulmonary artery pressure suddenly changes from 22/10
mm Hg to 24/2 mm Hg and remains at this pressure. Whichof the following is the most likely
cause of this change?
a. Proximal movement of the catheter
b. Distal movement of the catheter
c. Hypoxemia
d. Pulmonary artery vasodilation: Correct answer: a
Rationale: The sudden decrease in the pulmonary artery diastolic pressure (PAd) most often
indicates that the catheter has flipped back into the right ventricle. Notethat the pulmonary
artery systolic pressure has changed little. Distal migration of thecatheter would cause the PAOP
waveform and pressure. Hypoxemia would increasethe PAd as it causes pulmonary
hypertension. Pulmonary artery vasodilation woulddecrease the systolic and diastolic
