Cardiopulmonary Extra Questions

Question 1

The following data were obtained from a patient:

• Mean right atrial pressure = 16 mm Hg
• Right ventricular pressure = 20/3 mm Hg

What is the valve dysfunction? Describe the patient’s heart murmur.

Answer 1

Right atrial pressure is higher than normal, yet right ventricular end diastolic pressure is still normal. This indicates that there is a high resistance to blood flow through the tricuspid valve (tricuspid stenosis). You would expect to hear a presystolic murmur occurring at the time of atrial contraction. In addition, there will be a diastolic decrescendo murmur during the rapid ventricular filling phase (upon mitral valve opening).

Question 2

The normal ranges for systolic/diastolic pulmonary artery pressures are ___/___?

Answer 2

Pulmonary artery systolic pressure = 15 - 30 mm Hg; pulmonary artery diastolic pressure = 4 - 12 mm Hg.

Question 3

A patient has a systolic crescendo-decrescendo murmur and a brachial artery pressure of 110/90 mm Hg. The QRS complex in Lead III has a large negative deflection while there is no net deflection in lead aVR.

What is the valve dysfunction? The patient’s mean electrical axis is?

Answer 3

Aortic stenosis. The patient’s mean electrical axis is minus 60 degrees.

Question 4

A patient has a left bundle branch block. How would this person’s lead I QRS complex differ from normal?

Answer 4

Ventricular depolarization would occur rapidly in the right ventricle since the wave of depolarization can rapidly spread through the right bundle branch as under normal conditions. The right ventricle would depolarize before the left since the wave of depolarization would move slowly through the left ventricle (due to the left bundle branch block). The time to depolarize the both ventricles would be increased, resulting in a wide QRS complex. The QRS complex will be notched, which is indicative of a bundle branch block.

Question 5

The following data were obtained in a 65 kg patient: LV EDV = 200 ml, LV ESV = 140 ml, and heart rate = 50 beats/min. Calculate the patient’s stroke volume, ejection fraction, and cardiac output. Which of these values are abnormal?

Answer 5

Question 6

Increasing the heart rate from 40 to 160 beats per minute alone does not increase cardiac output by a factor of 4. Why?

Answer 6

Increasing the heart rate decreases the duration of diastole, which results in less filling time. As a result, end-diastolic volume will be reduced and stroke volume will decrease (Starling’s law). In addition, there will be a decrease in the duration of systole, resulting in less time for ejection.

Question 7

The mean right atrial pressure in a patient at rest is 2 mm Hg. Is this normal, high or low?

Answer 7

Normal (normal range for right atrial pressure is 0 - 8 mm Hg).

Question 8

A patient has an arterial blood pressure of 175/60 mm Hg. Murmurs are evident during ventricular diastole. What is the valve dysfunction? What would be the net deflection (positive, negative) in lead aVF in this patient?

Answer 8

Aortic insufficiency. The chronic increase in left ventricular volume would result in left ventricular hypertrophy, which would be seen as left axis deviation and a net negative QRS complex in aVF.

Question 9

What is the physiological significance of the low capillary blood flow velocity?

Answer 9

The low capillary blood flow velocity increases the time available for diffusion of solutes between tissue and blood.

Question 10

A 1% albumin solution would have an osmotic pressure _____ that of a 1% NaCl solution.

Answer 10

Both of these solutions would contain 10 g of solute per liter. The high molecular weight of albumin compared to NaCl would cause the osmolality (and therefore osmotic pressure) of the albumin solution to be less than that of the NaCl solution.

Question 11

A patient develops an ectopic pacemaker in the left atrium. How will the P wave and QRS complex in lead I compare to normal?

Answer 11

The patient will have a negative P wave in lead I because the wave of atrial depolarization will be moving from left to right (toward the negative end of lead I). The QRS complex will be normal in lead I since the conduction pathway responsible for ventricular depolarization (i.e., AV

Question 12

Answer 12

The mean electrical axis is - 45 degrees, which is consistent with left ventricular hypertrophy. D.V.’s heart rate is 75 beats per minute.

Question 13

During inspiration, the venous return to the right ventricle is (increased, decreased, unchanged), which results in the force of contraction of the right ventricle being (increased, decreased, unchanged), and in turn causes a(n) (increased, decreased, unchanged) stroke volume.

Answer 13

Increased, Increased, Increased

Question 14

The addition of approximately ____ millosmoles of solute would be required to increase a person’s osmolality from 290 to 310 if the person weighed 100 kg and had a body fat of 15%?

Answer 14

First calculate the total body water:

Total water = 0.85 x 100 x 0.72 = 61.2 liters.

Total solute needed = concentration increase x total water = (20 mosm/liter) x 61.2 liters = 1224 milliosmoles

Question 15

A patient has a diastolic murmur as well as a loud pre-systolic crescendo murmur. Describe the valve dysfunction(s).

Answer 15

Tricuspid or mitral stenosis

Question 16

The following data were obtained from a patient:

• Mean left atrial pressure = 20 mm Hg
• Left ventricular end diastolic pressure = 6 mm Hg

What is the valve dysfunction? Characterize the heart murmur in this patient.

Answer 16

The left ventricular end diastolic pressure is within normal values. The large diastolic pressure difference between the left atrium and left ventricle is characteristic of mitral stenosis. The murmur would be the same as that described in question #6.

Question 17

Left ventricular systole occurs between which heart sounds?

Answer 17

Left ventricular systole is defined as being from the onset of S1 to the onset of S2.

Question 18

How would systolic blood pressure and pulse pressure change during inspiration?

Answer 18

During inspiration, blood is pooled in pulmonary veins because the lower intrathoracic pressure expands the veins. As a result, end diastolic volume in the left ventricle is reduced, resulting in a decreased stroke volume, and therefore lower arterial systolic and pulse pressures.

Question 19

The renal blood flow of a patient’s right kidney = 300 ml/min. Renal artery and renal vein pressures are 95 and 10 mm Hg. Calculate vascular resistance in the right kidney.

Answer 19

Renal Vascular Resistance

= (95-10)/(300/60) = 17 mm Hg/ml/sec.

Multiply by 1330 to get answer in dyne sec/cm5.

Question 20

You plan to expand the plasma volume of a patient by 0.5 liters using IV administration of isotonic saline. The patient weighs 200 lbs and has 17% body fat. You would have to give _____ liters to accomplish this assuming no loss of fluid occurred via the kidneys, GI tract, lungs or skin.

Answer 20

The patient’s weight and percent body fat are not relevant to answer this question. The isotonic saline would stay in the ECF. Since plasma constitutes about one-fourth of the ECF volume you would need 2 liters of isotonic saline to accomplish a 0.5 liter increase in plasma volume.

Question 21

What is the physiological significance of the large capillary surface area?

Answer 21

Although the radius and length of each capillary are small, the large number of capillaries results in the blood flow per capillary being very low. As a result, the capillary surface area in relation to the flow is large. Since the rate of diffusion is directly proportional to the surface area, the large surface area facilitates the exchange of solutes between tissue and blood.

Question 22

A patient has a diastolic decrescendo murmur. The patient’s left ventricular pressure (systolic/diastolic) is 135/8 mm Hg and aortic blood pressure is 135/45 mm Hg. What is the valve dysfunction?

Answer 22

Aortic insufficiency

Question 23

A patient in a nursing home has a nutritional imbalance that causes her to loose both water and solute in excess to intake for a period of 3 days. Her solute loss amounts to 145 mosmoles per day while her water loss amounts to 200 ml per day. Assume this person weighed 100 lbs before these losses occurred and had 17% body fat.

1. After 3 days of nutritional imbalance this person’s total body water would be approximately_____ liters?
2. After 3 days of nutritional imbalance, this person’s body fluid osmolality would be approximately ______ milliosmoles/liter?
3. After the 3 days of nutritional imbalance you would expect this patient’s ratio of ICF to total body water to be _______ that before the nutritional imbalance?

Answer 23

1. Total water = Initial body water - water loss after 3 days

Initial water = (100 / 2.2) x 0.83 x 0.72 = 27.16 liters

Water loss = 0.2 liters/day x 3 days = 0.6 liters

New water = 27.16 - 0.6 = 26.56 liters

2. New mosmolality = (intital mosmoles - mosmoles loss after 3 days) / new water

Inital osmoles = 290 mosmoles/liter x 27.16 liters = 786.4

mosmoles loss = 145 mosmoles/day x 3 days = 435 mosmoles
New mosmolality = (7876-435) / 26.56 liters = 280.17 = 280 mosmoles/liter

3. This person lost solute in excess of water. Accordingly, this constitutes a case of hypotonic contraction in which the ECF contracts and the ICF expands. The shift of water from ECF to ICF would cause the ratio of ICF to total water to increase.

Question 24

A person weighing 55 lbs bets his friend that he can drink 3 liters of water. He rapidly consumes the water and collects the bet. Assume this person has 15% body fat. Answer the following 3 questions assuming this water is absorbed into the body but none is lost via the kidneys, GI tract, lungs, or skin.

What would this cause the persons total body water to increase by?

How would this increase the ECF?

How would this increase the ICF?

Answer 24

1. Total body water would increase by 3 liters since no fluid was lost.
2. ECF volume would increase by one-third of the ingested volume since the water would distribute between ECF and ICF in the ratio 1 to 2. Accordingly, ECF volume would increase by 1 liter.
3. ICF volume would increase by two-thirds of the ingested volume since the water would distribute between ECF and ICF in the ratio 1 to 2. Accordingly, ICF volume would increase by 2 liters.

Question 25

In a normal person aortic blood pressure:

1. is approximately______ mm Hg at the onset of left ventricular ejection.
2. is approximately ____ mm Hg at the time of peak left ventricular systolic pressure.
3. is approximately ____mm Hg at the onset of the second heart sound.

Answer 25

1. 80 mm Hg. Ejection begins when ventricular pressure exceeds aortic pressure. At this time, aortic pressure is at its lowest pressure (diastolic pressure).
2. 120 mm Hg. Because the aortic valve normally offers low resistance to blood flow during ejection, peak systolic pressure and peak aortic pressure are nearly the same (~120 mm Hg).
3. 100 mm Hg. S2 represents closure of the aortic valve which occurs at ~100 mm Hg.

Question 26

Using the ventricular depolarization vector cardiogram shown above, sketch the QRS complex for lead aVF. The mean electrical axis is approximately _______ degrees, which is consistent with ___________________.

Answer 26

The mean electrical axis is approximately -45 to -55 degrees, which is consistent with left ventricular hypertrophy.

Question 27

For a given body weight, an increase in percent body fat would cause plasma volume as percent of body weight to ________?

Answer 27

Total body water decreases as percent body fat increases. Since plasma volume as a fraction of total water remains constant, plasma volume as percent body weight must decrease.

Question 28

Answer 28

7. Resting Mean Arterial Blood Pressure = 80 + (120 - 80)/3 =80+13 = 93mmHg

Exercising Mean Arterial Blood Pressure = 60 + (140 - 60)/3
=60 + 26 = 86mmHg

8. During exercise, dilation of skeletal muscle arterioles decreases TPR. The baroreflex increases cardiac output (CO) to minimize the fall in MAP due to lower TPR. The increased CO is due to increased heart rate as well as increased stroke volume (due to increased inotropic state, increased venous return due to venoconstriction, and lower afterload). The increased stroke volume will increase arterial pulse pressure since more blood is ejected per beat, causing a greater change in arterial blood volume with each beat compared to rest.
9. TPR decreases during exercise due to dilation of arterioles in skeletal muscle. Blood can leave the arterial system at a faster rate when TPR is decreased, which would decrease diastolic pressure. During exercise, heart rate is increased which would increase diastolic pressure. Therefore, there are two opposing influences on arterial diastolic pressure during exercise: lower TPR and increased heart rate. Since the net effect was lower diastolic pressure compared to rest, the effect of TPR on diastolic pressure was greater than the effect of increased heart rate.
10. Resting Pulmonary Vascular Resistance (PVR)
    = (20 mm Hg -5 mm Hg)/(70 b/min x 70 ml/beat /60sec/min)

= 15/82 mm Hg/ml/sec
= 0.18 mm Hg/ml/sec = 239.4 dyne sec/cm5

Question 29

The ratio of left ventricular to right ventricular stroke work is approximately?

Answer 29

The ratio of left to right ventricular stroke work is approximately equal to the ratio of the mean arterial blood pressure to mean pulmonary artery pressure (~ 100 mm Hg/20 mm Hg, or ~ 5). On average, the stroke volume is the same for the right and left ventricles so the difference in stroke work between ventricles is due to the difference in pressure generated by each.

Question 30

The maximum rate of left ventricular blood volume change occurs during the (early, middle, late) period of left ventricular ejection.

Answer 30

Early

Question 31

A patient of yours has a plasma osmolality of 110 mEq/liter. This person weighs 300 lbs and has 30% body fat. The sodium deficit in this patient is approximately_____ mEq if the desired ECF sodium concentration is 140 mEq/liter?

Answer 31

The following formula is generally used to calculate sodium deficit.
Sodium deficit = total body water x (desired - present sodium concentration).

TBW for this patient = (300 / 2.2) x 0.7 x 0.72 = 68.7 liters
Sodium deficit = 68.7 liters x (140-110 mEq/liter) = 2061 mEq.

Question 32

The administration of a diuretic resulted in the person putting out a very dilute urine. Prior to giving the diuretic the patient weighed 220 lbs and had 17% body fat. You estimate that in one day this patient lost a total of 3 liters of water and 480 milliosmoles of solute.

1. At the end of one day of diuresis, this patient’s total body water would be approximately ______ liters?
2. At the end of one day of diuresis, this patient’s total solute would be approximately ______ milliosmoles?
3. At the end of one day of diuresis, this patient’s osmolality would be approximately ______ milliosmoles/liter?

Answer 32

1. Before diuresis:
   Total water = (220 / 2.2) x 0.83 x 0.72 = 59.76 liters

Total Solute = 59.76 x 290 = 17,330 milliosmoles

After diuresis:
Total water = 59.76 - 3 = 56.76 liters

2. Before diuresis:
   Total water = (220 / 2.2) x 0.83 x 0.72 = 59.76 liters

Total Solute = 59.76 x 290 = 17,330 milliosmoles

After diuresis:

Total water = 59.76 - 3 = 56.76 liters
Total solute = 17,330 - 480 = 16,850 millosmoles

3. Before diuresis:
   Total water = (220 / 2.2) x 0.83 x 0.72 = 59.76 liters
   Total Solute = 59.76 x 290 = 17,330 milliosmoles
   After diuresis:
   Total water = 59.76 - 3 = 56.76 liters
   Total solute = 17,330 - 480 = 16,850 millosmoles
   Osmolality = 16,850 / 56.76 liters = 296.86 = 297 millosmoles/liter

Question 33

A patient is given a drug that decreases sodium conductance in atrial and ventricular myocytes. How would this drug affect the P wave and QRS complex?

Answer 33

Decreased Na+ conductance would result in a decreased conduction velocity which would increase the time to completely depolarize the atria and the ventricles. Therefore, the P wave and the QRS complex would be wider.

Question 34

What is the direct effect of anemia on resistance to blood flow?

Answer 34

Since resistance is directly proportional to blood viscosity and blood viscosity is directly proportional to the hematocrit, anemia decreases the resistance to blood flow.

Question 35

A patient has a negative QRS complex in lead I and a systolic crescendo- decrescendo murmur that increases in intensity during inspiration. What is the valve dysfunction?

Answer 35

Pulmonic stenosis. The negative QRS complex in Lead I is consistent with right ventricular hypertrophy. The murmur increases in intensity during inspiration due to a higher end- diastolic volume (as a result of increased venous return to the right atrium). This results in a higher right ventricular stroke volume, which increases the intensity (loudness) of the murmur.

Question 36

A patient has a holosystolic murmur and left ventricular end diastolic volume is approximately 240 ml (normal range is 120 to 140 ml). What is the valve dysfunction?

Answer 36

Mitral insufficiency

Question 37

A 70kg person has 10% body fat.

1. This person’s ECF is approximately how many liters?
2. What is their plasma volume?
3. What percent of the person’s weight is their plasma volume?

Answer 37

1. 15 liters
2. 3.8 liters
3. Plasma as % body weight = 100 x (plasma vol / body weight) = 100 x (3.8 / 70) = 5.4%

Question 38

The osmotic pressure of a 200 mM urea solution would be _____ that of a 100 mM KCl solution?

Answer 38

Both solutions would have an osmolality of 200 milliosmoles/liter (assuming complete dissociation for KCl). Accordingly, these solutions would have the same osmotic pressure.

Question 39

A 70kg person has 10% body fat.

What is this person’s total body water in liters?

Answer 39

45 liters

Question 40

Using the following data, calculate total peripheral resistance in dyne sec/cm5.

MAP=95mmHg,CVP=5mmHg,and CO=6 l/min

Answer 40

TPR = (MAP-CVP)/CO (ml/sec)

= (95-5)/(6000/60) = 90/100 = 0.9 mm Hg/ml/sec.

Multiply this answer by 1330 to get answer in dyne sec/cm5

Question 41

How does an increased afterload shift the position of the Starling curve relating:

• stroke volume and end-diastolic volume
• stroke work and end-diastolic volume

Answer 41

• The curve will be shifted downward with an increase in afterload.
• The position of the curve will not be affected by changes in afterload.

Question 42

A patient has impaired conduction through the right bundle branch. How would this affect the timing of the components of the 2nd heart sound?

Answer 42

In right bundle branch block, depolarization and hence contraction of the right ventricle would be delayed with respect to depolarization and contraction of the left ventricle. Therefore, pulmonic valve closure would be delayed with respect to aortic valve closure to the degree that splitting of S2 could be detected during expiration and inspiration. This condition is known as wide splitting.

Question 43

A patient lost 1 liter of blood due to hemorrhage. Before the hemorrhage the patient weighed 200 lbs, had 17% body fat, and had a hematocrit of 45%. Assume no fluid shifts occurred in response to the fall in blood pressure resulting from the hemorrhage.

1. This hemorrhage would cause plasma volume to decrease by ______ liters?
2. As a result of this hemorrhage, total body solute would decrease by ____ milliosmoles?
3. Following the hemorrhage, the extracellular sodium concentration would be _______ that prior to the hemorrhage?
4. Following the hemorrhage, the ICF volume would be _____ that prior to the hemorrhage?

Answer 43

1. Plasma volume would decrease by 0.55 liters; i.e., 55% of the blood loss.
2. Loss of milliosmoles = 0.55 liters x 290 mosmoles/liter = 159.5 = 160 mosmoles
3. The extracellular sodium concentration would be the same as that prior to hemorrhage. Hemorrhage represents a loss of isotonic fluid from the ECF.
4. Since hemorrhage is the loss of isotonic fluid from the ECF, there are no osmotic forces to drive fluid into or out of the ICF. Accordingly, ICF volume would be the same as that prior to hemorrhage.

Question 44

What changes in the subject’s EKG would you expect to be associated with exercise?

Answer 44

Heart rate is higher because increased SYM activity to the heart increases the rate of firing of the SA node pacemaker. Increased SYM firing also increases AV node calcium conductance, resulting in a decreased PR interval. In ventricular muscle, norepinephrine from SYM nerves increases potassium conductance, which decreases the duration of phase 2 of the action potential, and therefore decreases the duration of the ST segment. In addition, the increased potassium conductance increases the rate of phase 3 repolarization, which decreases the duration of the T wave but increases the height of the T wave.

Question 45

A patient was found to have a resting cardiac index of 5.5 liters/min/meter2. Is this value normal, high or low?

Answer 45

High

Question 46

What are the major factors that affect the development of left ventricular systolic pressure?

Answer 46

The principal factors affecting left ventricular systolic pressure are: 1) the preload or end- diastolic volume, 2) the inotropic state (contractility), and 3) afterload (arterial blood pressure, total peripheral resistance).

Question 47

Answer 47

Question 48

What is the Starling Reserve?

Answer 48

The Starling Reserve represents the maximal extent that stroke volume can be increased by increasing the preload (EDV).

Question 49

What does the area within the left ventricular pressure-volume loop represent?

Answer 49

The area within the left ventricular pressure-volume loop represents the stroke work of the left ventricle (approximately the stroke volume multiplied by the mean arterial blood pressure).

Question 50

What changes in cardiac valves signify the onset and end of left ventricular isovolumic contraction and relaxation?

Answer 50

Isovolumic contraction – the time between mitral valve closure and aortic valve opening.

Isovolumic relaxation – the time between aortic valve closure and mitral valve opening.

Question 51

How does inspiration affect filling of the right heart?

Answer 51

Inspiration increases filling of the right heart while decreasing filling of the left ventricle.

Question 52

Answer 52

Points A and B represent premature ventricular depolarizations which are not preceded by P waves. The wide QRS complex at points A and B indicates that conduction velocity was reduced compared to normal, and the time to completely depolarize the ventricle was increased. The QRS complex at point A results from an ectopic pacemaker located in the left ventricle since the overall direction of depolarization is from left to right (which results in a net negative deflection of this QRS complex). The QRS complex at point B is an ectopic pacemaker located in the right ventricle, resulting in an overall direction of depolarization from right to left (which results in a net positive deflection of this QRS complex).

Question 53

A patient is given a drug that increases the potassium conductance in atrial and ventricular myocytes. What EKG changes would you expect to occur?

Answer 53

An increase in potassium conductance will decrease the duration of the ventricular action and increase the rate of phase 3 repolarization. This will result in a decreased RT interval and increased amplitude of the T wave. Spiked T waves are seen during hyperkalemia as a result of the increased potassium conductance.

Question 54

In the normal heart, right ventricular depolarization is complete before left ventricular depolarization. Why?

Answer 54

Depolarization of the ventricles proceeds from the endocardium to the epicardium. Since the right ventricular wall is thinner than in the left, the time required to completely depolarize the right ventricle is less than the left ventricle.

Question 55

A new cardiovascular drug is found to cause an upward shift of the relationship between stroke work and end-diastolic volume. What is a likely effect of this drug on the heart?

Answer 55

The upward shift in the curve would be due to an increase in the inotropic state - an increase in

Question 56

The following results were obtained from patient RC:

• During left ventricular diastole, blood flow into the left ventricle was 150 ml.
• During left ventricular systole, blood flow into the aorta was 100 ml.

Describe the valve dysfunction. Why are the blood flows different?

Answer 56

Mitral insufficiency. Fifty ml of blood is pumped back into the left atrium during ventricular systole.

Question 57

Increasing the inotropic state when left ventricular end-diastolic volume remains the same will result in:

• (increased, decreased, unchanged) force of contraction and consequently (increased, decreased, unchanged) stroke volume.
• (increased, decreased, unchanged) cardiac output.
• (increased, decreased, unchanged) aortic systolic pressure.

Answer 57

Increased, Increased

Increased

Increased

Question 58

The mean electrical axis of a patient is +60 degrees. In this person, the net QRS deflection is largest in lead ______ and smallest in lead ______.

Answer 58

The net QRS deflection is largest in the lead that is closest to parallel to the mean electrical axis, while the deflection will be smallest in the lead nearly perpendicular to the mean electrical axis. Since the mean electrical axis of +60 degrees is the same as lead II, the net QRS deflection is largest in that lead, and smallest (zero) in lead aVL.

Question 59

A 70kg person has 10% body fat.

1. What is the total millosmoles of solute in this person?
2. An increase in percent body fat without changing their weight will cause their body water to?

Answer 59

1. Total millosmoles = 290 x total water = 290 x 45.4 = 13,166
2. Fat cells contain less water than other cells, an increase in body fat will decrease total body water

Question 60

A 1% NaCl (molecular weight = 58 g) solution would have an osmotic pressure of approximately ____ mm Hg; assume complete dissociation of the NaCl.

Answer 60

Osmolaity of this solution = 10 g/l x (1 mole / 58 g) x 2 = 0.345 osmoles/liter = 345 millosmoles/liter.

Osmotic pressure = 345 mOm x 19.32 mm Hg/mOsm = 6665 mm Hg

Question 61

If left ventricular afterload is increased when preload remains the same, you would predict cardiac output to be (increased, decreased, unchanged).

Answer 61

Increased

Question 62

The IV administration of a hypertonic saline solution to a patient resulted in a plasma osmolality of 310 mosmoles/liter. Prior to the IV administration the patient weighed 250 lbs and had 20% body fat. The IV administration resulted in a water deficit of approximately ______ liters.

Answer 62

Body water prior to IV = (250 / 2.2 ) x 0.8 x 0.72 = 65.45 liters

To raise the osmolality from 290 to 310 would require the new body water = (290/310) x 65.45 = 61.22 liters.

Accordingly, the water deficit = 65.45 - 61.22 = 4.23 liters

Question 63

Decreased arterial compliance associated with aging would (increase, decrease, not affect) arterial systolic pressure and (increase, decrease, not affect) arterial pulse pressure.

Answer 63

Increase, Increase

Question 64

A 60-year-old woman has lived her entire life at an altitude of 12,000 feet. How would her lead V1 QRS complex differ from that of a 60-year-old woman who has lived her entire life at sea level?

Answer 64

The woman living in the mountains would have developed right ventricular hypertrophy (due to hypoxic pulmonary vasoconstriction caused by lower oxygen levels at high altitude). Right ventricular hypertrophy (right axis deviation) would result in a large net positive deflection in the V1 QRS complex. In contrast, the woman residing at sea level would have a net negative deflection of the V1 QRS complex.

Question 65

An infant has an atrial septal defect. Cardiac catheterization reveals that the child’s mean right atrial pressure is 10 mm Hg and mean left atrial pressure is 6 mm Hg. Would the infant’s pulmonary blood flow be greater than or less than systemic blood flow? How would the systemic arterial blood O2 content compare to the pulmonary venous O2 content?

Answer 65

Pulmonary blood flow would be less than systemic blood flow. Some blood is shunted from the right atrium to the left atrium across the atrial septal defect (since right atrial pressure is greater than left atrial pressure). Therefore, the end-diastolic volume and stroke volume of the left ventricle will be greater than that of the right ventricle.

Systemic arterial O2 content will be less than pulmonary venous O2 content. Deoxygenated blood from the right atrium is being mixed with oxygenated blood in the left atrium. As a result, blood O2 content will be lower in the left atrium, left ventricle, and systemic arteries compared to that of the pulmonary veins

Question 66

Patient A weighs 200 lbs and has 15% body fat. Patient B weighs 300 lbs and has 30% body fat. How would the ratio of ECF to ICF volume in patient A compare to that in patient B?

Answer 66

Weight and percent body fat do not influence the distribution of fluid between the ECF and ICF. Accordingly, the ration of ECF to ICF would be the same.

Question 67

A person weighing 55 lbs bets his friend that he can drink 3 liters of water. He rapidly consumes the water and collects the bet. Assume this person has 15% body fat. Answer the following 2 questions assuming this water is absorbed into the body but none is lost via the kidneys, GI tract, lungs, or skin.

1. After equilibration of the ingested water between body fluid compartments, this person’s osmolality would be approximately _______ millosmoles/liter?
2. Assuming this person had a plasma sodium concentration of 140 mEq/liter before water ingestion, the new plasma sodium concentration would be approximately________ mEq/liter?

Answer 67

1. Initial body water = (55/2.2) x 0.72 x 0.85 = 15.3 liters

Intial total milliosmoles = 15.3 lites x 290 milliosmoles/liter = 4437 milliosmoles

New body water = 15.3 + 3 = 18.3 liters

New osmolality = 4437 milliosmoles / 18.3 liters = 242.46 mosmoles/liter

2. ECF volume before = 0.33 x 15 = 4.95 lites
   ECF volume after = 4.95 + 1 = 5.95 liters
   Sodium concentration after = (4.95 / 5.95) x 140 = 116.47 or 116 mEq/liter.

Question 68

The following O2 contents (ml O2 /100 ml blood) were obtained in a patient:

Right atrium = 14, right ventricle = 16, pulmonary artery = 16, brachial artery = 20.

Where is the shunt?

Answer 68

The blood O2 content in the right ventricle is greater than in the right atrium. In this patient, blood from the left ventricle was shunted across a ventricular septal defect to the right ventricle.