Block 3 Pace 6 Nov 2011

1. A gas mixture of several dry gases has a total pressure of 100 mm Hg and a total gas amount of 10 Mol. What is the partial pressure of one of the gases that is in the amount of 2 Mol?

20 mm Hg

30 mm Hg

40 mm Hg

50 mm Hg

60 mm Hg

The partial pressure of a gas in a mixture is directly proportional to the amount of that gas present. In this case, the total gas amount is 10 Mol and the amount of the gas in question is 2 Mol. Therefore, the partial pressure of this gas can be calculated as (2 Mol / 10 Mol) * 100 mm Hg = 20 mm Hg.

Explanation

The partial pressure of a gas in a mixture is directly proportional to the amount of that gas present. In this case, the total gas amount is 10 Mol and the amount of the gas in question is 2 Mol. Therefore, the partial pressure of this gas can be calculated as (2 Mol / 10 Mol) * 100 mm Hg = 20 mm Hg.

2. In a tissue capillary, the interstitial hydrostatic pressure is 2 mm Hg, the capillary hydrostatic pressure is 25 mm Hg and the interstitial oncotic pressure is 7 mm Hg. If the net driving force across the capillary wall is 3 mm Hg favoring filtration, what is the capillary oncotic pressure?

21 mm Hg

23 mm Hg

24 mm Hg

25 mm Hg

27 mm Hg

The correct answer is E. The net driving force for fluid across a capillary wall is calculated by the following:

driving force = (hydrostaticc− (hydrostatici) − (oncoticc - oncotici)

where:

hydrostatici = interstitial hydrostatic pressure

hydrostaticc = capillary hydrostatic pressure

oncotici = interstitial oncotic pressure

oncoticc = capillary oncotic pressure

Substituting the values in the question stem: 3 = (25 - 2) - (x - 7). Simplifying, 3 = 23 - x + 7, therefore = 27.

Explanation

The correct answer is E. The net driving force for fluid across a capillary wall is calculated by the following:

driving force = (hydrostaticc− (hydrostatici) − (oncoticc - oncotici)

where:

hydrostatici = interstitial hydrostatic pressure

hydrostaticc = capillary hydrostatic pressure

oncotici = interstitial oncotic pressure

oncoticc = capillary oncotic pressure

Substituting the values in the question stem: 3 = (25 - 2) - (x - 7). Simplifying, 3 = 23 - x + 7, therefore = 27.

3. The relationship between volume and transmural pressure of a normal lung is illustrated by curve 2. Which curve or line of the figure represents this relationship for the lung of a patient with surfactant deficiency?

1

3

4

not-available-via-ai

Explanation

not-available-via-ai

4. Your patient was a 72 -year-old classic "pink puffer" with a chronic, dry cough and marked weight loss. He was obviously having difficulty breathing. The pathophysiological alteration in his lungs which lead to obstructive features clinically is

Loss of recoil due to alveolar wall destruction

Chronic bronchial inflammation

Irreversible damage to bronchial walls

Replacement of alveolar wall by fibrous tissue

Narrowing of the bronchial lumen

The correct answer is "loss of recoil due to alveolar wall destruction." In this patient, the chronic, dry cough and weight loss are indicative of chronic obstructive pulmonary disease (COPD). COPD is characterized by the destruction of alveolar walls, leading to a loss of elastic recoil in the lungs. This loss of recoil results in air trapping and difficulty breathing, which are the obstructive features seen clinically. Chronic bronchial inflammation and irreversible damage to bronchial walls may contribute to COPD, but the primary pathophysiological alteration is the loss of recoil due to alveolar wall destruction. The other options, such as replacement of alveolar wall by fibrous tissue and narrowing of the bronchial lumen, are not consistent with the given clinical presentation.

Explanation

The correct answer is "loss of recoil due to alveolar wall destruction." In this patient, the chronic, dry cough and weight loss are indicative of chronic obstructive pulmonary disease (COPD). COPD is characterized by the destruction of alveolar walls, leading to a loss of elastic recoil in the lungs. This loss of recoil results in air trapping and difficulty breathing, which are the obstructive features seen clinically. Chronic bronchial inflammation and irreversible damage to bronchial walls may contribute to COPD, but the primary pathophysiological alteration is the loss of recoil due to alveolar wall destruction. The other options, such as replacement of alveolar wall by fibrous tissue and narrowing of the bronchial lumen, are not consistent with the given clinical presentation.

5. A patient suffers from an acute asthma attack. The patient has no history of chronic obstructive pulmonary disease. What change in following respiratory parameters is associated with his asthma attack?

FRC is increased

RV is increased

Vital capacity is decreased

FEV1 is decreased

TLC is decreased

During an acute asthma attack, the airways become inflamed and narrowed, leading to difficulty in breathing. This narrowing of the airways causes a decrease in the forced expiratory volume in one second (FEV1). FEV1 is a measure of the maximum amount of air that can be forcefully exhaled in one second, and it is commonly used to assess lung function. Therefore, a decrease in FEV1 is associated with an asthma attack. The other respiratory parameters listed are not directly affected by an asthma attack.

Explanation

During an acute asthma attack, the airways become inflamed and narrowed, leading to difficulty in breathing. This narrowing of the airways causes a decrease in the forced expiratory volume in one second (FEV1). FEV1 is a measure of the maximum amount of air that can be forcefully exhaled in one second, and it is commonly used to assess lung function. Therefore, a decrease in FEV1 is associated with an asthma attack. The other respiratory parameters listed are not directly affected by an asthma attack.

6. A balloon-tipped catheter is placed into a small branch of the pulmonary artery in a patient. The lumen of the catheter opens distal to the balloon. The pressure measured from the catheter with the balloon deflated is 25/8 mm Hg. When the balloon is inflated, the pressure is 7 mm Hg and non-pulsatile. Which of the following pressures is being approximated when the balloon is inflated?

Left atrial pressure

Left ventricular end diastolic pressure

Left ventricular peak systolic pressure

Pulmonary artery pressure

Right atrial pressure

The correct answer is A. When the balloon is deflated, the catheter simply measures the pulmonary artery
pressure (choice D), which is pulsatile with systolic/diastolic values of 25/8 mm Hg. When the balloon is inflated,
the catheter is "wedged" in a small branch of the pulmonary artery and the pressure that is measured is called
the "pulmonary wedge pressure." Because inflation of the balloon obstructs all blood flow in the artery branch,
the blood vessels distal to the point of obstruction also have no flow. One can think of these distal vessels as
physical extensions of the catheter, as they allow blood pressure to be measured on the other side of the
pulmonary circulation, i.e., in the left atrium. The pulmonary wedge pressure is usually a few mm Hg higher
compared to the left atrial pressure, but the general opinion is that pulmonary wedge pressure is a reflection of
events in the left atrium. It is usually not feasible to measure left atrial pressure directly in the normal human
being because it is difficult to pass a catheter retrograde through the aorta and left ventricle. Therefore, the
pulmonary wedge pressure provides an important clinical estimate of left atrial pressure. Be aware that
pulmonary wedge pressure may also be called pulmonary capillary wedge pressure, pulmonary arterial wedge
pressure, or simply wedge pressure.

In many instances, the pulmonary wedge pressure can provide a reasonable estimate of left ventricular end
diastolic pressure (choice B). However, a notable exception is during mitral stenosis, in which the pressure in
the left atrium (and therefore, the pulmonary wedge pressure) is much higher than the left ventricular end
diastolic pressure because of the high resistance to blood flow through the stenosed valve.

The left ventricular peak systolic pressure (choice C) occurs when the mitral valve is closed, making it
impossible to be approximated using a catheter in the pulmonary artery.

The right atrial pressure (choice E) cannot be measured or approximated from a catheter in the pulmonary
artery.

Explanation

The correct answer is A. When the balloon is deflated, the catheter simply measures the pulmonary artery
pressure (choice D), which is pulsatile with systolic/diastolic values of 25/8 mm Hg. When the balloon is inflated,
the catheter is "wedged" in a small branch of the pulmonary artery and the pressure that is measured is called
the "pulmonary wedge pressure." Because inflation of the balloon obstructs all blood flow in the artery branch,
the blood vessels distal to the point of obstruction also have no flow. One can think of these distal vessels as
physical extensions of the catheter, as they allow blood pressure to be measured on the other side of the
pulmonary circulation, i.e., in the left atrium. The pulmonary wedge pressure is usually a few mm Hg higher
compared to the left atrial pressure, but the general opinion is that pulmonary wedge pressure is a reflection of
events in the left atrium. It is usually not feasible to measure left atrial pressure directly in the normal human
being because it is difficult to pass a catheter retrograde through the aorta and left ventricle. Therefore, the
pulmonary wedge pressure provides an important clinical estimate of left atrial pressure. Be aware that
pulmonary wedge pressure may also be called pulmonary capillary wedge pressure, pulmonary arterial wedge
pressure, or simply wedge pressure.

In many instances, the pulmonary wedge pressure can provide a reasonable estimate of left ventricular end
diastolic pressure (choice B). However, a notable exception is during mitral stenosis, in which the pressure in
the left atrium (and therefore, the pulmonary wedge pressure) is much higher than the left ventricular end
diastolic pressure because of the high resistance to blood flow through the stenosed valve.

The left ventricular peak systolic pressure (choice C) occurs when the mitral valve is closed, making it
impossible to be approximated using a catheter in the pulmonary artery.

The right atrial pressure (choice E) cannot be measured or approximated from a catheter in the pulmonary
artery.

7. Barometric pressure at sea level is about 760 mm Hg. The world's highest summit (Mt. Everest) is 29,035 ft. and the barometric pressure is 247 mm Hg. The inspired PO₂ of a climber who was breathing 50% oxygen at the summit would be about:

65 mm Hg

100 mm Hg

40 mm Hg

150 mm Hg

200 mm Hg

At higher altitudes, the barometric pressure decreases. The barometric pressure at the summit of Mt. Everest is given as 247 mm Hg. The inspired PO2 (partial pressure of oxygen) can be calculated by multiplying the barometric pressure by the fraction of inspired oxygen (FiO2). In this case, the climber is breathing 50% oxygen, so the FiO2 is 0.5. Multiplying 247 mm Hg by 0.5 gives us 123.5 mm Hg, which is the inspired PO2. However, since the question asks for an approximation, the closest option is 100 mm Hg.

Explanation

At higher altitudes, the barometric pressure decreases. The barometric pressure at the summit of Mt. Everest is given as 247 mm Hg. The inspired PO2 (partial pressure of oxygen) can be calculated by multiplying the barometric pressure by the fraction of inspired oxygen (FiO2). In this case, the climber is breathing 50% oxygen, so the FiO2 is 0.5. Multiplying 247 mm Hg by 0.5 gives us 123.5 mm Hg, which is the inspired PO2. However, since the question asks for an approximation, the closest option is 100 mm Hg.

8. Dr. Smolanoff You have been hired by a biotechnology company as a Physician-Consultant to look into a new drug that has been invented. You are told that this new drug will revolutionize the diet industry because it selectively activates uncoupling proteins that perform the same function as thermogenin. Which of the following statements accurately describe what you would expect to see in the mitochondria of cells treated with this drug?

The cells would be capable of making ATP at the same rate as untreated cells.

The membrane potential in the mitochondria would be normal

Co-treatment with oligomycin would restore normal mitochondrial function

Mitochondrial NADH and FADH2 consumption would be inhibited

Mitochondrial oxygen consumption would be much higher than normal

When the drug selectively activates uncoupling proteins that perform the same function as thermogenin, it would lead to an increase in mitochondrial oxygen consumption. This is because uncoupling proteins allow protons to flow back into the mitochondrial matrix without going through ATP synthase, which dissipates the proton gradient and increases oxygen consumption. As a result, the cells treated with this drug would have a higher rate of oxygen consumption compared to untreated cells.

Explanation

When the drug selectively activates uncoupling proteins that perform the same function as thermogenin, it would lead to an increase in mitochondrial oxygen consumption. This is because uncoupling proteins allow protons to flow back into the mitochondrial matrix without going through ATP synthase, which dissipates the proton gradient and increases oxygen consumption. As a result, the cells treated with this drug would have a higher rate of oxygen consumption compared to untreated cells.

9. Dr. Kirera A 65-year-old male patient with history of gastric cancer was admitted to the hospital for operation to remove the tumor. A central venous line was ordered for preoperative nutritional and fluid balance. The venous line was placed via the left subclavian vein and blood returned freely through the catheter. The chest X‐ray film taken immediately after the catheterization revealed a substantial right hemidiaphragmatic elevation. The condition resolved within 2 days after the catheter was readjusted by withdrawing it for about 2cm. Which of the following nerve was likely compressed by the tip of the catheter to elicit this condition?

Right vagus

Right recurrent laryngeal

Right phrenic

Left phrenic

Left recurrent laryngeal

Left vagus

The condition of right hemidiaphragmatic elevation suggests that the catheter tip compressed the right phrenic nerve. The phrenic nerve innervates the diaphragm, which is responsible for breathing. Compression of the right phrenic nerve would result in paralysis or weakness of the right hemidiaphragm, leading to its elevation. Adjusting the catheter by withdrawing it resolved the condition, indicating that the compression on the right phrenic nerve was relieved.

Explanation

The condition of right hemidiaphragmatic elevation suggests that the catheter tip compressed the right phrenic nerve. The phrenic nerve innervates the diaphragm, which is responsible for breathing. Compression of the right phrenic nerve would result in paralysis or weakness of the right hemidiaphragm, leading to its elevation. Adjusting the catheter by withdrawing it resolved the condition, indicating that the compression on the right phrenic nerve was relieved.

10. A patient is suffering from obstructive lung disease. His chest wall and respiratory system compliances are measured as:

Chest wall compliance (C_CW) = 200 ml / cm H₂O
Lung compliance (C_L) = 200 ml /cm H₂O

What is his respiratory system compliance (C_RS) in ml / cm H₂O?

300

50

200

100

400

The respiratory system compliance (CRS) is determined by the sum of the chest wall compliance (CCW) and the lung compliance (CL). In this case, the CCW is 200 ml/cm H2O and the CL is also 200 ml/cm H2O. Therefore, the CRS would be 200 ml/cm H2O + 200 ml/cm H2O = 400 ml/cm H2O.

Explanation

The respiratory system compliance (CRS) is determined by the sum of the chest wall compliance (CCW) and the lung compliance (CL). In this case, the CCW is 200 ml/cm H2O and the CL is also 200 ml/cm H2O. Therefore, the CRS would be 200 ml/cm H2O + 200 ml/cm H2O = 400 ml/cm H2O.

11. A post-operative patient on intravenous fluids develops lesions in the mouth (angular stomatitis). Urinalysis indicates a clinical deficiency of riboflavin. Which of the following TCA enzymes is most likely to be affected?

Citrate synthase

Fumarase

Malate dehydrogenase

Succinate dehydrogenase

Isocitrate dehydrogenase

Succinate dehydrogenase is the correct answer because riboflavin is a cofactor for this enzyme. Riboflavin, also known as vitamin B2, is necessary for the function of succinate dehydrogenase, which is an enzyme involved in the citric acid cycle (TCA cycle). Deficiency of riboflavin can lead to impaired activity of succinate dehydrogenase, resulting in various symptoms including angular stomatitis.

Explanation

Succinate dehydrogenase is the correct answer because riboflavin is a cofactor for this enzyme. Riboflavin, also known as vitamin B2, is necessary for the function of succinate dehydrogenase, which is an enzyme involved in the citric acid cycle (TCA cycle). Deficiency of riboflavin can lead to impaired activity of succinate dehydrogenase, resulting in various symptoms including angular stomatitis.

12. Dr. Beevers A 56-year-old man recently diagnosed with heart failure is started on a therapeutic regimen that includes the drugs furosemide, digoxin, and captopril. What is the mechanism of action of the loop diuretic member of this regimen?

Inhibition of Na+/K+ ATPase

Blockade of L-type Ca2+ channels

Stimulation of guanylyl cyclase

Inhibition of NKCC2

Inhibition of NCC

Loop diuretics, such as furosemide, work by inhibiting the NKCC2 transporter in the thick ascending limb of the loop of Henle in the kidneys. This transporter is responsible for reabsorbing sodium, potassium, and chloride ions from the urine back into the bloodstream. By inhibiting NKCC2, loop diuretics increase the excretion of these ions in the urine, leading to increased urine volume and decreased fluid volume in the body. This mechanism of action is important in treating conditions like heart failure, where reducing fluid volume can help alleviate symptoms.

Explanation

Loop diuretics, such as furosemide, work by inhibiting the NKCC2 transporter in the thick ascending limb of the loop of Henle in the kidneys. This transporter is responsible for reabsorbing sodium, potassium, and chloride ions from the urine back into the bloodstream. By inhibiting NKCC2, loop diuretics increase the excretion of these ions in the urine, leading to increased urine volume and decreased fluid volume in the body. This mechanism of action is important in treating conditions like heart failure, where reducing fluid volume can help alleviate symptoms.

13. A person on a high altitude expedition develops lung edema. Which of following values is most likely increased (in resting condition) as a result of his lung edema?

Arterial PO2

Arterial PCO2

Mixed venous PO2

Mixed venous PCO2

Alveolar- arterial PO2 difference

Lung edema refers to the accumulation of fluid in the lungs, which can impair the exchange of gases. The alveolar-arterial PO2 difference represents the difference in oxygen levels between the alveoli (air sacs in the lungs) and the arterial blood. In a healthy individual, this difference is small because oxygen is efficiently transferred from the alveoli to the blood. However, in the presence of lung edema, the transfer of oxygen is impaired, leading to an increased alveolar-arterial PO2 difference. Therefore, the most likely increased value in this scenario would be the alveolar-arterial PO2 difference.

Explanation

Lung edema refers to the accumulation of fluid in the lungs, which can impair the exchange of gases. The alveolar-arterial PO2 difference represents the difference in oxygen levels between the alveoli (air sacs in the lungs) and the arterial blood. In a healthy individual, this difference is small because oxygen is efficiently transferred from the alveoli to the blood. However, in the presence of lung edema, the transfer of oxygen is impaired, leading to an increased alveolar-arterial PO2 difference. Therefore, the most likely increased value in this scenario would be the alveolar-arterial PO2 difference.

14. At which lung volume or lung capacity is the resistance of alveolar blood vessels (capillaries) is the highest:

Residual volume

Functional residual capacity

60% of total lung capacity

Minimal lung volume

100% of total lung capacity

At 100% of total lung capacity, the resistance of alveolar blood vessels is the highest. This is because at this lung volume or capacity, the alveoli are maximally expanded and the blood vessels are stretched, leading to increased resistance.

Explanation

At 100% of total lung capacity, the resistance of alveolar blood vessels is the highest. This is because at this lung volume or capacity, the alveoli are maximally expanded and the blood vessels are stretched, leading to increased resistance.

15. Dr. Shams A patient inhales as much air as he can (maximal inspiratory level) and then you connect him with a spirometer containing a gas mixture of air and helium. After a homogeneous distribution of helium between spirometer gas and patients' lung volume, you are able to measure the patients' lung volume from helium dilution. Which of the following lung volumes or capacities do you have to know in addition to above measured lung volume to be able to calculate residual volume of this patient?

Tidal volume

Functional residual capacity

Vital capacity

Inspiratory capacity

Expiratory reserve volume

To calculate the residual volume of the patient, you would need to know the vital capacity in addition to the measured lung volume from helium dilution. The residual volume is the amount of air that remains in the lungs after a maximal exhalation, and it is calculated by subtracting the vital capacity from the total lung capacity. Therefore, knowing the vital capacity is necessary to determine the residual volume.

Explanation

To calculate the residual volume of the patient, you would need to know the vital capacity in addition to the measured lung volume from helium dilution. The residual volume is the amount of air that remains in the lungs after a maximal exhalation, and it is calculated by subtracting the vital capacity from the total lung capacity. Therefore, knowing the vital capacity is necessary to determine the residual volume.

16. Shown is a drawing of a section of lung. The arrow is pointing to the nucleus of which cell type?

Dust cell

Type II pneumocyte

Smooth muscle cell

Endothelial cell

Type I pneumocyte

The correct answer is Endothelial cell. The drawing is of a section of lung, and the arrow is pointing to the nucleus of a specific cell type. Endothelial cells are the cells that line the blood vessels and capillaries in the lungs. They play a crucial role in gas exchange and the regulation of blood flow in the lungs.

Explanation

The correct answer is Endothelial cell. The drawing is of a section of lung, and the arrow is pointing to the nucleus of a specific cell type. Endothelial cells are the cells that line the blood vessels and capillaries in the lungs. They play a crucial role in gas exchange and the regulation of blood flow in the lungs.

17. Minute ventilation is controlled by numerous factors including blood gases, mechanical receptors, body temperature, and voluntary effort (input from cortex). Which of the following mechanisms produces the greatest minute ventilation?

Decreased PO2 plus an increased PCO2

Hypotension

Hyperthermia

Voluntary effort

Exercise

Voluntary effort produces the greatest minute ventilation because it involves conscious control and active participation of the individual in increasing their breathing rate and depth. This can result in a significant increase in the amount of air exchanged in the lungs per minute, leading to a higher minute ventilation. The other factors mentioned, such as decreased PO2 plus increased PCO2, hypotension, hyperthermia, and exercise, may also affect minute ventilation to some extent, but voluntary effort has the greatest impact.

Explanation

Voluntary effort produces the greatest minute ventilation because it involves conscious control and active participation of the individual in increasing their breathing rate and depth. This can result in a significant increase in the amount of air exchanged in the lungs per minute, leading to a higher minute ventilation. The other factors mentioned, such as decreased PO2 plus increased PCO2, hypotension, hyperthermia, and exercise, may also affect minute ventilation to some extent, but voluntary effort has the greatest impact.

18. You are given a test tube by your biochemistry professor and asked to identify the purified enzyme it contains. You run a series of biochemical tests and determine that the enzyme requires acetyl CoA for activity, is inhibited by ATP, NADH and succinyl CoA and positively regulated by oxaloacetate. When you hand in the report, what enzyme did you conclude your professor gave you?

Isocitrate dehydrogenase

Fumarase

Citrate synthase

Pyruvate dehydrogenase

Malate dehydrogenase

Based on the information provided, the enzyme in the test tube can be concluded to be Citrate synthase. This is because Citrate synthase requires acetyl CoA for activity, is inhibited by ATP, NADH, and succinyl CoA, and is positively regulated by oxaloacetate. None of the other enzymes listed have all these characteristics.

Explanation

Based on the information provided, the enzyme in the test tube can be concluded to be Citrate synthase. This is because Citrate synthase requires acetyl CoA for activity, is inhibited by ATP, NADH, and succinyl CoA, and is positively regulated by oxaloacetate. None of the other enzymes listed have all these characteristics.

19. A 54-year-old male is seen in clinic with complaints of palpitations and light-headedness. Physical examination is remarkable for a heart rate of greater than 200 beats per minute and a blood pressure of 75/40 mm Hg. What adjustments have probably occurred in the cardiac cycle?

Diastolic time has decreased and systolic time has increased

Diastolic time has decreased but systolic time has decreased more

Systolic time has decreased and diastolic time has increased

Systolic time has decreased but diastolic time has decreased more

Systolic time has decreased but diastolic time has not changed

The correct answer is D. Under normal conditions, one-third of the cardiac cycle is spent in systole and
two-thirds spent in diastole. As heart rate increases dramatically, the time spent in diastole falls precipitously
but the time spent in systole falls only slightly.

A large increase in heart rate must produce a decrease in both diastole and systole (compare with choice A).

The major change with increased heart rate is in diastole, not systole (compare with choice B).

Heart rate cannot increase if diastolic time increases (choice C).

An increase in heart rate must be accompanied by a decrease in diastolic time (compare with choice E).

Explanation

The correct answer is D. Under normal conditions, one-third of the cardiac cycle is spent in systole and
two-thirds spent in diastole. As heart rate increases dramatically, the time spent in diastole falls precipitously
but the time spent in systole falls only slightly.

A large increase in heart rate must produce a decrease in both diastole and systole (compare with choice A).

The major change with increased heart rate is in diastole, not systole (compare with choice B).

Heart rate cannot increase if diastolic time increases (choice C).

An increase in heart rate must be accompanied by a decrease in diastolic time (compare with choice E).

20. A patient is breathing air at sea level and has a respiratory exchange ratio of 0.7. The arterial blood values are: PO₂ 100 mm Hg PCO₂ 28 mm Hg pH 7.60 The alveolar-arterial PO₂ difference of this patient is closest to which of following values in mmHg?

25

20

15

5

10

The alveolar-arterial PO2 difference is a measure of the difference between the partial pressure of oxygen in the alveoli (the air sacs in the lungs) and the arterial blood. A normal value for this difference is around 10-15 mmHg. In this case, the patient's arterial blood PO2 is 100 mmHg, which is within the normal range. Therefore, the alveolar-arterial PO2 difference is likely to be closer to the lower end of the normal range, which is 10 mmHg.

Explanation

The alveolar-arterial PO2 difference is a measure of the difference between the partial pressure of oxygen in the alveoli (the air sacs in the lungs) and the arterial blood. A normal value for this difference is around 10-15 mmHg. In this case, the patient's arterial blood PO2 is 100 mmHg, which is within the normal range. Therefore, the alveolar-arterial PO2 difference is likely to be closer to the lower end of the normal range, which is 10 mmHg.

21. A 65-year-old male visits his family practitioner for a yearly examination. Measurement of his blood pressure reveals a systolic pressure of 190 mm Hg and a diastolic pressure of 100 mm Hg. His heart rate is 74/min and pulse pressure is 90 mm Hg. A decrease in which of the following is the most likely explanation for the high pulse pressure?

Arterial compliance

Cardiac output

Myocardial contractility

Stroke volume

Total peripheral resistance

The correct answer is A. A decrease in arterial compliance indicates that the arterial wall is stiffer (i.e., less
distensible). When the compliance of the arterial system decreases, the rise in arterial pressure becomes
greater for a given stroke volume pumped into the arteries. In the normal young adult, the systolic blood
pressure is about 120 mm Hg and the diastolic blood pressure is about 80 mm Hg. Because the pulse pressure
is the difference between the systolic and diastolic blood pressures, the normal pulse pressure is about 40 mm
Hg in a healthy young adult. However, in older adults the pulse pressure sometimes increases as much as two
times normal because the arteries become hardened by arteriosclerosis.

The cardiac output (choice B) itself has no direct effect on the pulse pressure; however, if a decrease in
cardiac output is associated with a decrease in stroke volume, the pulse pressure would be expected to
decrease.

A decrease in myocardial contractility (choice C) would be expected to decrease stroke volume, and therefore
cause the pulse pressure to decrease.

A decrease in stroke volume (choice D) causes the pulse pressure to decrease because a smaller amount of
blood enters the arterial system with each heartbeat, and the rise and fall of pressure during systole and - 1 -
diastole is decreased.

A decrease in total peripheral resistance (choice E), i.e., vasodilation, does not have a significant effect on the
pulse pressure of the major arteries under normal conditions.

Explanation

The correct answer is A. A decrease in arterial compliance indicates that the arterial wall is stiffer (i.e., less
distensible). When the compliance of the arterial system decreases, the rise in arterial pressure becomes
greater for a given stroke volume pumped into the arteries. In the normal young adult, the systolic blood
pressure is about 120 mm Hg and the diastolic blood pressure is about 80 mm Hg. Because the pulse pressure
is the difference between the systolic and diastolic blood pressures, the normal pulse pressure is about 40 mm
Hg in a healthy young adult. However, in older adults the pulse pressure sometimes increases as much as two
times normal because the arteries become hardened by arteriosclerosis.

The cardiac output (choice B) itself has no direct effect on the pulse pressure; however, if a decrease in
cardiac output is associated with a decrease in stroke volume, the pulse pressure would be expected to
decrease.

A decrease in myocardial contractility (choice C) would be expected to decrease stroke volume, and therefore
cause the pulse pressure to decrease.

A decrease in stroke volume (choice D) causes the pulse pressure to decrease because a smaller amount of
blood enters the arterial system with each heartbeat, and the rise and fall of pressure during systole and - 1 -
diastole is decreased.

A decrease in total peripheral resistance (choice E), i.e., vasodilation, does not have a significant effect on the
pulse pressure of the major arteries under normal conditions.

22. The airway resistance of a person depends on lung volume. At which lung volume or lung capacity is the airway resistance the lowest?

Residual volume

100% of total lung capacity

Functional residual capacity

60% of total lung capacity

Minimal lung volume

At 100% of total lung capacity, the airway resistance is the lowest. This is because at this lung volume, the airways are maximally expanded, allowing for smooth and easy airflow.

Explanation

At 100% of total lung capacity, the airway resistance is the lowest. This is because at this lung volume, the airways are maximally expanded, allowing for smooth and easy airflow.

23. A 40-year-old man was diagnosed with an enlarged left atrium without incidence of mitral valve insufficiency. Which of the following symptoms is the most likely to be associated his condition?

Dysphagia

Dyspnea

Chest pain

Facial edema

Voice hoarseness

Dysphagia refers to difficulty swallowing, which can occur when the left atrium is enlarged and compresses the esophagus. This compression can lead to the sensation of food getting stuck or difficulty in swallowing. It is not directly related to mitral valve insufficiency, which is why the patient in this case does not have that condition. The other symptoms listed are not typically associated with an enlarged left atrium.

Explanation

Dysphagia refers to difficulty swallowing, which can occur when the left atrium is enlarged and compresses the esophagus. This compression can lead to the sensation of food getting stuck or difficulty in swallowing. It is not directly related to mitral valve insufficiency, which is why the patient in this case does not have that condition. The other symptoms listed are not typically associated with an enlarged left atrium.

24. Dr. Bellot A 35 year old woman with a history of dyspnea and chest pain is shown by x-ray to have right ventricular enlargement and is clinically going into right heart failure. Primary pulmonary hypertension is suspected. An intrapulmonary pathological finding characteristic of this condition is:

Bullae formation

Plexiform arteriopathy

Alveolar fibrosis

Absence of pulmonary arterioles

Destruction of alveolar septa

Primary pulmonary hypertension is a condition characterized by increased blood pressure in the pulmonary arteries. Plexiform arteriopathy is a pathological finding commonly seen in primary pulmonary hypertension. It refers to the formation of complex vascular structures within the pulmonary arterioles, which can lead to obstruction of blood flow and further contribute to right ventricular enlargement and right heart failure.

Explanation

Primary pulmonary hypertension is a condition characterized by increased blood pressure in the pulmonary arteries. Plexiform arteriopathy is a pathological finding commonly seen in primary pulmonary hypertension. It refers to the formation of complex vascular structures within the pulmonary arterioles, which can lead to obstruction of blood flow and further contribute to right ventricular enlargement and right heart failure.