Cardiovascular and Pulmonary Board Review Questions:

Anatomy, Biochemistry, Physiology

I. Cardiovascular

QNO:1.Correct Answer:
An electrocardiogram of a patient with tachycardia revealed a pattern consistent with a small ventricular posteroseptal infarct resulting from inadequate blood supply. Despite the rapid heart beat, its regularity indicates that the infarct has involved only:
1: The sinoatrial node
2: The atrioventricular node
3: The atrioventricular bundle
4: Both atrioventricular node and bundle
5: A localized region of ventricular myocardium


QNO:2.Correct Answer:
How many layers of plasma membrane must an alveolar oxygen molecule traverse to reach hemoglobin?
1: 2
2: 3
3: 4
4: 5
5: 6


QNO:3.Correct Answer:
If a person suffered a stab injury and air entered the intrapleural space (pneumothorax), the most likely response would be for the
1: lung to expand outward and the chest wall to spring inward
2: lung to expand outward and the chest wall to spring outward
3: lung to collapse inward and the chest wall to collapse inward
4: lung to collapse inward and the chest wall to spring outward
5: lung volume to be unaffected and chest wall to spring outward


QNO:4.Correct Answer:
James Finley, age 18, has the following lipid analyses: immediately after a meal, TG=1500 mg/dL, Chol= 185 mg/dL; after 8-hr fast, TG=1450 mg/dL, Chol=180 mg/dL. (Normals: TG 100-200 mg/dL, Chol < 200 mg/dL). Which of the following lipoprotein fractions would you expect to be elevated?
1: VLDL
2: LDL
3: Chylomicron
4: 1 and 3
5: 1, 2, and 3


QNO:5.Correct Answer:
For the previous question, which of the following is the most likely cause of James's condition?
1: Obesity
2: Lipoprotein lipase deficiency
3: Non-insulin dependent diabetes mellitus
4: LDL receptor deficiency
5: Tangier disease (HDL deficiency)


QNO:7.Correct Answer:
The anatomical basis of conduction in the heart is:
1: Nerve fibers
2: Intercalated discs
3: Adherens junctions
4: Gap junctions
5: Desmosomal junctions


QNO:10.Correct Answer:
A man has received a knife wound in the chest just to the right and slightly below the xiphisternal junction at the level of the 6th costal cartilage. The portion of the heart or great vessels that the knife entered most likely is the:
1: Superior vena cava
2: Right atrium
3: Left ventricle
4: Right ventricle
5: Inferior vena cava


QNO:12.Correct Answer:
Atheromas first develop in the arterial:
1: Endothelium
2: T. intima
3: T. media
4: T. adventitia
5: Elastic laminae


QNO:13.Correct Answer:
Sarah White, a 63-year old homemaker, was found in her car after she ran off the road and hit a tree. She was treated in the ED for multiple contusions, several rib fractures, and a fracture of the left radius. She was transferred to the ICU and still complains of shortness of breath and chest pains. What is the best analysis to monitor during the next 12-24 hr?
1: Plasma LDH concentration
2: Plasma AST (aspartate transaminase) concentration
3: Plasma CK concentration
4: Plasma LDH isoenzyme analysis
5: Plasma CK isoenzyme analysis


QNO:14.Correct Answer:
Cells that commonly form the "foam cells" of an atheroma are derived from the arterial wall and from circulating elements. The latter are most likely:
1: B-cells
2: Neutrophils
3: Monocytes
4: T-cells
5: Platelets


QNO:15.Correct Answer:
In the fetus, there are shunts whereby oxygen-rich blood can bypass the fetal hepatic or pulmonary circulation and be routed to the fetal systemic circulation. In the adult, which of the following structures represents one of these shunts?
1: Foramen lacerum
2: Ligamentum arteriosum
3: Ligamentum teres hepatis
4: Coronary sinus
5: Median umbilical ligament


QNO:16.Correct Answer:
Heart tissue deprived of oxygen due to lung insufficiency will likely
1: be unaffected because heart muscle normally undergoes anaerobic metabolism.
2: become alkalotic because an insufficient supply of protons are produced by the mitochondrial electron transport chain.
3: have an insufficiency of ATP produced by oxidative phosphorylation.
4: use free fatty acids to make glucose for glycolysis in order to maintain ATP levels.
5: shift from anaerobic to aerobic metabolism.


QNO:17.Correct Answer:
A ligature occluding the external carotid artery just distal to the lingual artery would probably stop the flow of blood to the:
1: Sublingual gland
2: Temporalis muscle
3: Thyroid gland
4: Larynx
5: Tongue


QNO:18.Correct Answer:
John Jones, a 58-year old truck driver, appeared in clinic on Monday complaining of lethargy and difficulty in breathing. He indicated this condition had existed over the weekend and was preceded on Friday by a severe bout of "indigestion." Which of the following analyses would be most appropriate for determining whether an MI is contributing to Mr. Jones's condition?
1: Plasma LDH concentration
2: Red cell LDH concentration
3: Plasma CK concentration
4: Plasma LDH isoenzyme analysis
5: Plasma CK isoenzyme analysis



USMLE REVIEW QUESTIONS - PULMONARY PHYSIOLOGY

For clarification or other questions contact Steve Wood, Dept. of Physiology;
wood@brody.med.ecu.edu or telephone 816-2761

Questions start with # 6:

QNO 6.Correct Answer: View Answer
The physiological dead space is:

A: sometimes measured using the arterial PO2.
B: generally smaller than the anatomic dead space.
C: often increased in lung disease.
D: determined primarily by the geometry of the branching airways.


QNO 7.Correct Answer: View Answer
Hypoxic pulmonary vasoconstriction:

A: acts reflexly via the central nervous system.
B: improves matching of ventilation and blood flow in some lung diseases.
C: requires a PO2 of less than 40 Torr in mixed venous blood.
D: is not important in the perinatal period.


QNO 8.Correct Answer: View Answer
Pulmonary vascular resistance increases:

A: at high altitude.
B: during zero gravity (e.g. space flight).
C: with anemia.
D: on exercise.
E: all of the above.


QNO 9.Correct Answer: View Answer
In a subject with normal lungs, which of the following (acting alone) would have the greatest effect on O2 delivery (arterial o2 concentration x cardiac output)?

A: halving alveolar ventilation
B: a 50% right to left shunt
C: doubling inspired PO2
D: hemoglobin reduced to 50% of normal.


QNO 10.Correct Answer: View Answer
A patient is breathing air at sea level and has a respiratory exchange ratio of 1.0. The arterial blood values are: 
          PO2         90 mm Hg 
          PCO2        20 mm Hg 
          pH            7.30
These indicate that the:

A: alveolar-arterial PO2 difference exceeds 20 mm Hg.
B: plasma bicarbonate concentration is low.
C: patient is hyperventilating.
D: patient probably has partially compensated metabolic acidosis.
E: all of the above.


QNO 11.Correct Answer: View Answer
In a patient with anemia and normal lungs:

A: arterial PO2 is reduced.
B: arterial-venous O2 concentration difference is increased.
C: arterial O2 saturation is reduced.
D: PO2 of mixed venous blood is reduced.
E: all of the above.


QNO 12.Correct Answer: View Answer
Features of mild carbon monoxide poisoning include:

A: decreased arterial PO2.
B: decreased hemoglobin-oxygen affinity.
C: increased alveolar ventilation.
D: decreased arterial O2 concentration.
E: all of the above.


QNO 13.Correct Answer: View Answer
The increase in ventilation which occurs immediately following ascent to high altitude:

A: is caused primarily by the reduced work of breathing because of the less dense air.
B: is augmented by the fall in pH of the CSF.
C: is caused by the low PO2 in mixed venous blood.
D: increases still further over the course in the next 1-3 days.
E: all of the above.


QNO 14.Correct Answer: View Answer
In a patient with stable chronic lung disease, elevated PaCO2 and decreased PaO2:

A: CSF pH is near the normal value (7.32).
B: the ventilatory response to CO2 is reduced.
C: changes in PaO2 are more important in determining the level of ventilation than in normal people.
D: arterial [H+] is slightly above normal.
E: all of these.


 Physiology Home Page






















PULMONARY PHYSIOLOGY REVIEW - ANSWERS

Question 6.

The physiological dead space is:

The answer is C. - often increased in lung disease

The PCO2 is used, not the PO2.
Choice B is incorrect because the physiological dead space contains the anatomic dead space, and may be substantially larger if there is inequality of ventilation and blood flow within the lung. This is the reason why the physiological dead space is frequently increased in lung disease (choice C)

Back to Questions




















Question 7.

Hypoxic pulmonary vasoconstriction:

The answer is B. - improves matching of ventilation and blood flow in some lung diseases.

Although the mechanism of hypoxic pulmonary vasoconstriction is not fully understood, we know that central nervous connections are not required because the phenomenon can be demonstrated in isolated lungs. Therefore, choice A is incorrect.

Choice B is correct. Suppose a lobe or lobule of lung is poorly ventilated because of partial bronchial obstruction. The resulting alveolar hypoxia will reduce the blood flow through the mechanism of hypoxic pulmonary vasoconstriction. The result is improvement in the matching of ventilation and blood flow.

Choice C is incorrect. Reducing the PO2 of the blood entering the lung results much less vasoconstriction than reducing the PO2 of alveolar gas.

Choice D is incorrect. Hypoxic pulmonary vasoconstriction is important in the perinatal period. When the newborn baby makes the transition from placental to air breathing, it is important for pulmonary vascular resistance to fall precipitously within a few seconds. As a consequence, pulmonary blood flow dramatically increases from its value of only about 15% of the cardiac output in utero. The increase in pulmonary blood flow is assisted by closure of both the ductus arteriosus and the foramen ovale.

Back to Questions




















Question 8.

Pulmonary vascular resistance increases:

The answer is A. - at high altitude.

Pulmonary vascular resistance increases at high altitude as a result of the global alveolar hypoxia. The exact mechanism is still unknown but is apparently a local effect on the smooth muscle of the pulmonary arterial wall. The increase causes right ventricular hypertrophy with characteristic ECG changes.

Pulmonary vascular resistance during space flight would, if anything decrease as blood flow becomes more uniform. Anemia would decrease viscosity - a term in the resistance formula during exercise, pulmonary capillaries would distend causing resistance to fall.

With exercise, pulmonary arterial pressure tends to rise causing recruitment and distension of pulmonary vessels leading to a fall in pulmonary vascular resistance.

Back to Questions




















Question 9.

In a subject with normal lungs, which of the following (acting alone) would have the greatest effect on O2 delivery (arterial o2 concentration x cardiac output)?

The answer is D. - hemoglobin reduced to 50% of normal.

Halving alveolar ventilation would double alveolar PCO2 and therefore reduce alveolar and arterial PO2 but not enough to halve arterial O2 content. Also, hypoxia would stimulate cardiac output via sympathetic stimulation of heart rate.

A 50% right to left shunt would also lower arterial O2 content but not by 50% since the shunted blood (venous blood) has a significant amount of oxygen (normally about 75% saturated).

Doubling inspired O2 - if you picked this choice …………@#$%!!!!.

Reducing hemoglobin concentration to half normal would reduce arterial O2 content (at any saturation) to half normal, therefore choice D is correct.

Back to Questions




















Question 10.

A patient is breathing air at sea level and has a respiratory exchange ratio of 1.0. The arterial blood values are: 
           PO2        90 mm Hg 
           PCO2       20 mm Hg 
           pH           7.30
These indicate that the:

The answer is E. - all of the above.

The alveolar PO2 is found from the alveolar gas equation:

PAO2 = PIO2 - PACO2/R

We can assume that the inspired PO2 is the sea level normal value of 149 mm Hg. Therefore, neglecting the small correction factor, the alveolar PO2 is 149 - 20/1 or 129 mm Hg. Thus the alveolar-arterial PO2 diffusion is 129 - 90 = 39 mm Hg.

The PCO2 of 20 mm Hg means that the patient is hyperventilating. The combination of the low PCO2 and low pH means that plasma bicarbonate concentration is reduced (Henderson-Hasselbalch equation

pH = pK’ = log [HCO3-]/(PCO2 x a)

The data would fit a metabolic acidosis with partial respiration compensation.

Back to Questions




















Question 11.

In a patient with anemia and normal lungs:

The answer is D. - PO2 of mixed venous blood is reduced.

A patient with anemia and normal lungs typically has a normal arterial PO2. Thus, choice A is incorrect. Because the position of the oxygen dissociation curve is typically normal, the arterial oxygen saturation is also normal and thus, choice C is incorrect.

From the Fick principle:

O2 = x (CaO2 - CO2)
If the oxygen consumption and cardiac output are normal, the arterial-venous O2 concentration difference will also be normal. Thus, choice B is incorrect. As a matter of fact, cardiac output is sometimes reflexly increased in anemia, and if this occurs, the arterial-venous O2 concentration difference will be decreased.

Choice D is correct. Although the arterial PO2 is typically normal, the PO2 of mixed venous blood must fall. This is because the venous oxygen concentration falls to a very low level as the normal amount of oxygen is extracted, and thus the venous PO is abnormally low.

Back to Questions




















Question 12.

Features of mild carbon monoxide poisoning include:

The answer is D. - decreased arterial O2 concentration.

Carbon monoxide poisoning causes some of the hemoglobin in the blood to be combined with CO to give carboxyhemoglobin. As a result, there is a decreased arterial O2 concentration (choice D is correct). However, it is important to realize that this does not decrease the arterial PO2. In this respect, carbon monoxide poisoning is similar to anemia where again the arterial PO2 is typically normal but the arterial oxygen concentration is reduced.

Choice B is not correct. There is an increase in the affinity of the hemoglobin for oxygen as evidenced by the leftward shift of the oxygen dissociation curve. Thus, carbon monoxide poisoning causes a fall in tissue PO2 for two reasons: first less oxygen is carried in the arterial blood, and second the unloading is impeded by the higher hemoglobin affinity.

Choice C is incorrect. Ventilation is unaffected in mild carbon monoxide poisoning; the chemoreceptors respond primarily to the arterial PO2 and, because this is normal, ventilation remains essentially unchanged.

The word “mild” is included in the stem of the question because with very severe carbon monoxide poisoning there may be brain damage which may alter ventilation.

Back to Questions




















Question 13.

The increase in ventilation which occurs immediately following ascent to high altitude:

The answer is D. - increases still further over the course in the next 1-3 days.

The increase in ventilation which occurs immediately following ascent to high altitude is caused by stimulation of the peripheral chemoreceptors by the arterial hypoxemia. Because of the very high blood flow per tissue mass of the carotid chemoreceptors, they essentially respond to arterial PO2 and therefore choice C is incorrect.

The increase in ventilation is partly inhibited by the resulting fall in PCO2 which causes a rise in pH of the CSF. This reduces the stimulation of the central chemoreceptors which works against the increased drive from the peripheral chemoreceptors. However, after a day or so, the pH of the CSF is reduced to some extent by the outward movement of bicarbonate and thus this inhibitory effect on the central chemoreceptors is diminished. Consequently, ventilation increases still further over the course of the 1-3 days.

The increase in ventilation is not caused by the reduced work of breathing the less dense air though it is true that the work of breathing is reduced to some extent at high altitude. If a subject is asked to breathe as fast and as deeply as he can for 15 seconds, it can be shown that the amount of air that is exhaled (measured at BTPS) is greater at high altitude than at sea level, but this does not cause increased ventilation.

Back to Questions




















Question 14.

In a patient with stable chronic lung disease, elevated PaCO2 and decreased PaO2:

The answer is D. - All of these.

Many patients with chronic lung disease have chronic hypoxemia and chronic CO2 retention with an elevated arterial PCO2. Typical values might be an arterial PO2 in the 40s or 50s and an arterial PCO2 in the 50s. Such a patient may have the pH of his CSF near the normal value of about 7.32. Although the pH may have fallen in the initial stages of hypercapnia, the arterial PCO2 is often stable at the high value for so long that the pH of the CSF returns to near normal as a result of an increase in CSF bicarbonate concentration. Therefore, choice A is correct

Such a patient has a reduced ventilatory response to CO2. Indeed this is one reason why there is chronic CO2 retention. Part of the reason for the reduced ventilatory response is the near normal value of the CSF pH. Another reason is that such a patient often has severe airway obstruction with a markedly increased airway resistance. In this case, even if a raised arterial PCO2 causes an increase in the number of impulses to the ventilatory muscles from the respiratory center, they are unable to increase ventilation in the normal way because of the airway obstruction. Thus, choice B is correct.

C is correct. In these patients, the hypoxic ventilatory drive from the peripheral chemoreceptors is often the most important stimulus to ventilation. This has important practical consequences. If such a patient is brought into the emergency room and given oxygen to relieve his arterial hypoxemia, his main stimulus to breathing may be abolished, an he may develop lethal CO2 retention and respiratory acidosis. The best way to manage these patients is to give relatively small increases in oxygen concentration (typically 24 to 28 %) and to monitor the arterial PO2, PCO2 and pH carefully for signs of additional CO2 retention and respiratory acidosis.

Choice D is correct. These patients typically have an almost fully compensated respiratory acidosis, but the arterial hydrogen ion concentration is usually slightly increased.

Back to Questions