Doppler and Hemodynamics
Muhamed Saric
Itzhak Kronzon
1. A 32-year-old woman is referred for evaluation of rheumatic mitral valve stenosis. No mitral regurgitation was noted. The following values were obtained by Doppler echocardiography:
Table 8-1 | ||||||||
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Which of the following statements is TRUE:
A. Mitral valve area can be calculated by dividing 220 into deceleration time.
B. Stroke volume across the mitral valve is 72 mL per beat.
C. Pressure half-time is 355 ms.
D. Mitral valve area is 0.8 cm2.
E. During exertion, her mean gradient is expected to decrease.
View Answer
1. Answer: D. Mitral valve area (MVA) can be calculated using the pressure half-time (PHT) method:
In this question, PHT was not given. However, PHT can be calculated from the stated mitral deceleration time (DT) using the following formula:
Thus, in our patient:
PHT = 0.29 × DT = 0.29 × 910 = 264 ms
MVA = 220/PHT = 220/264 = 0.8 cm2
Alternatively, Eqs. 1 and 2 can be combined into the following one:
In our patient, then:
MVA = 759/DT = 759/910 = 0.8 cm2
Therefore, answer D is correct.
Answer A is incorrect because MVA is calculated by dividing 220 into PHT (Eq. 1) and not DT.
Answer B is incorrect because the stroke volume (SV) across the mitral valve in this patient is 53 mL per beat. Once the MVA is calculated, SV and cardiac output (CO) can be derived using the following formulas:
SV = MVA × VTI
CO = SV × HR
where VTI is the mitral velocity-time integral during diastole, and HR is the heart rate.
In our patient, mitral VTI during diastole was 66 cm and the heart rate was 85 beats/min:
SV = 0.8 cm2 × 66 cm = 53 mL:
CO = 53 mL × 85 beats/min = 4.5 L/min
Answer C is incorrect because, as shown previously, PHT in this patient was 264 ms and not 355 ms.
Answer E is incorrect because the resting gradient of mitral stenosis is expected to increase with augmentation of CO such as during exercise, fever, or pregnancy.
2. A 21-year-old man with dyspnea on exertion and enlarged pulmonary artery on chest x-ray underwent transthoracic echocardiography. The study revealed patent ductus arteriosus and the following:
Table 8-2 | ||||||||||
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Which of the following statements is TRUE:
A. Systemic blood flow (Qs) is 7.8 L/min.
B. The ratio of pulmonic to systemic blood flow (Qp:Qs) is less than 1.
C. Stroke volume entering the lungs is 38 mL per beat.
D. Patient is cyanotic in the lower parts of the body.
E. The ratio of stroke volume through the LVOT and the stroke volume through the RVOT is equal to the Qp:Qs ratio in this patient.
View Answer
2. Answer: E. The patient has a patent ductus arteriosus (PDA) that is an extracardiac shunt resulting from a communication between the descending thoracic aorta (DTA) and the proximal left pulmonary artery. In utero, the blood that reaches the pulmonary artery from the right ventricle cannot enter the collapsed lungs; instead, it is diverted across the ductus arteriosus into the DTA. Soon after birth, the pressure in the pulmonary artery falls below the pressure in DTA and the blood flow in the ductus arteriosus reverses its direction. It now flows from the DTA into the pulmonary artery. High oxygen content of the ductal blood triggers the closure of the ductus arteriosus in most newborns. In rare instances, the communication persists in the postneonatal period giving rise to PDA.
In individuals with PDA, the systemic blood flow (Qs) reaches the right-sided heart through systemic veins and continues through the right ventricular outflow tract (RVOT) into the main pulmonary artery. At that level, Qs is joined by the shunt flow (SF) entering the pulmonary artery through the PDA. The sum of Qs and SF represents the amount of blood flow that enters the pulmonary circulation (Qp).
After passing through the lungs, Qp enters the left-sided heart through the pulmonary veins and exits through the left ventricular outflow tract (LVOT) into the aorta. At the level of the descending aorta, Qp divides into SF, which enters the PDA, and Qs, which continues into the peripheral systemic circulation to ultimately reach the right-sided heart through systemic veins.
Note that in individuals with PDA, the flow across the RVOT represents Qs and the flow across the LVOT represents Qp. Therefore, the answer E is correct.
This is in contrast to atrial and ventricular septal defects where LVOT flow represents Qs and the RVOT flow represents Qp. Since in most individuals with PDA, Qp > Qs, it is the left side of the heart and not the right side of the heart that dilates to accommodate the excess blood flow.
The general echocardiographic formula to calculate volumetric flow (Q) is:
where CSA is the cross-sectional area, VTI is velocity-time integral, and HR is the heart rate.
One can use right and left ventricular outflow tracts to calculate volumetric flow. Since both tracts are assumed to be circular in shape, the CSA can be expressed in the above equations as follows:
where D is the diameter of the outflow tract. Eq. 1, after expressing CSA in terms of Eq. 2, becomes:
Calculations for our patient are summarized in the following table:
Table 8-16 | ||||||||||||||||||||||||||||||||||||
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Answer A is incorrect because the flow rate of 7.8 L/min across the LVOT represents Qp and not Qs in patients with PDA.
Answer B is incorrect because Qp:Qs in this patient is greater than 1 (it is 1.7:1).
Answer C is incorrect because the stroke volume that enters the lungs (97 mL per beat) is the sum of the systemic stroke volume (59 mL per beat) that entered the main pulmonary artery through the RVOT and the shunt flow (38 mL per beat) that came into the pulmonary artery through the PDA.
Answer D is incorrect because Qp is much greater than Qs, the SF is in the left to right direction and the patient is unlikely to be cyanotic. In patients with PDA who develop Eisenmenger physiology, there is a right to left shunt. Such patients are cyanotic in the lower parts of the body because the deoxygenated blood from the pulmonary artery crosses the PDA and enters the DTA past the origins of the aortic arch vessels, which supply a fully oxygenated blood to the head and the arms.
3. A 39-year-old woman was admitted for severe shortness of breath on exertion. On transthoracic echocardiogram, there was mild pulmonic regurgitation. Continuous wave spectral Doppler tracings of the pulmonic regurgitant jet reveal the following:
Table 8-3 | ||||
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Examination of the inferior vena cava by M-mode echocardiography demonstrated the following:
Table 8-4 | ||||
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The following statement is TRUE:
A. Right atrial pressure (RAP) is estimated at 6 mm Hg.
B. Pulmonary artery diastolic pressure is greater than 31 mm Hg.
C. Pulmonary artery diastolic pressure is 36 mm Hg minus the right atrial pressure.
D. Pulmonary artery diastolic pressure cannot be assessed if the pulmonic regurgitation is only mild.
E. Pulmonary artery diastolic pressure is normal.
View Answer
3. Answer: B. This patient with severe shortness of breath has elevated pulmonary artery diastolic pressure (PADP). Using the end-diastolic velocity (V) of the pulmonic regurgitant jet and the 4V2 formula, one can calculate the pressure gradient (&Dgr;P) between the PADP and the end-diastolic right ventricular pressure (RVDP).
In the absence of tricuspid stenosis, RVDP is the same as the right atrial pressure (RAP). Thus the pressure gradient can also be expressed as:
Rearranging Eq. 2, PADP can be calculated in the following manner:
where V is the end-diastolic velocity of the pulmonic regurgitant jet and RAP is right atrial pressure.
RAP can be estimated from the expiratory size of the inferior vena cava (IVC) and the percent decrease in diameter change with inspiration. In our patient, the IVC is dilated (>2.1 cm) and the IVC diameter does not change with inspiration. The estimated RAP is thus greater than 15 mm Hg.
Once RAP is known, we can then calculate PADP:
PADP > 4 × (2 m/s)2 + 15, or greater than 31 mm Hg
Therefore, answer B is correct.
Answer A is incorrect because RAP in this patient is >15 mm Hg as demonstrated earlier.
Answer C is incorrect for two reasons: (1) Pressure gradient between PADP and RVDP is 16 mm Hg and not 36 mm Hg and (2) PADP is calculated by adding RAP to the gradient between PADP and RVDP, and not subtracting from it.
Answer D is incorrect because even in mild pulmonic regurgitation, appropriate spectral Doppler tracings of the regurgitant jet can be obtained.
Answer E is incorrect because normal PADP range is typically between 5 and 16 mm Hg.
4. A 42-year-old man was admitted to the hospital after a 1-month history of intermittent fever and progressive shortness of breath. Blood cultures grew Streptococcus viridans. On transesophageal echocardiogram, perforation of the anterior mitral leaflet and mitral regurgitation were seen. On the color Doppler image, a well-formed flow convergence (PISA) shell was visualized on the ventricular side of the mitral valve in systole. In addition, the following was noted:
Table 8-5 | ||||||||
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The following statement is TRUE:
A. Vena contracta of the mitral regurgitant flow is expected to be less than 0.3 cm.
B. Effective regurgitant orifice area of mitral regurgitation is approximately 0.6 cm2.
C. Instantaneous flow rate across the mitral valve using the PISA method is 70 mL/s.
D. Mitral regurgitation is moderate (2+).
E. Regurgitant volume is 40 mL per beat.
View Answer
4. Answer: B. Severe mitral regurgitation (grades 3+ and 4+) is defined by the following criteria:
Table 8-17 | ||||||||||
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Regurgitant orifice area (ROA) can be calculated using the following formula:
where r is the PISA radius, Valias is the aliasing velocity at which PISA radius is measured, and Vmax is the maximum velocity of the mitral regurgitant (MR) jet on spectral Doppler.
In Eq. 1, the expression 2 × &pgr; × r2 × Valias represents instantaneous flow rate (IFR):
Now Eq. 1 can be expressed as follows:
In our patient, IFR is calculated as follows:
IFR = 2 × 3.14 × (1.0 cm)2 × 45 cm/s = 283 mL/s And ROA as:
ROAMR = 283/500 cm/s = 0.6 cm2
Therefore, answer B is correct.
Answer A is incorrect because the vena contracta in severe mitral regurgitation is ≥0.7 cm.
Answer C is incorrect because the IFR of the MR jet in this patient is 283 mL per second as calculated earlier.
Answer D is incorrect because mitral regurgitation is severe since ROA ≥ 0.4 cm2 (it is 0.6 cm2).
Answer E is incorrect because the regurgitant volume (RegV) in this patient is 79 mL per beat. RV can be calculated as follows:
where VTIMR is the velocity-time integral of the MR jet.
In our patient, RV equals 0.6 cm2 × 140 cm, or 79 mL per beat. This is again consistent with severe mitral regurgitation (RV ≥60 mL per beat).
5. An 84-year-old obese woman with a history of hypertension and chronic renal insufficiency became very short of breath at a rehabilitation facility 2 weeks after elective hip replacement.
Transthoracic echocardiogram revealed normal left ventricular systolic function, no mitral or aortic valve disease, and the following:
Table 8-6 | ||||||||
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The following statement is TRUE:
A. Mean pulmonary artery wedge pressure is markedly elevated.
B. On mitral inflow, E to A ratio is expected to be less than 1.
C. Pulmonary artery systolic pressure is 64 mm Hg.
D. The ratio of peak E wave velocity to the peak medial mitral annular tissue Doppler velocity is expected to be less than 8.
E. Flow propagation velocity of mitral inflow on color M mode is normal for her age.
View Answer
5. Answer: A. The patient presents with shortness of breath due to elevated pulmonary artery wedge pressure (PAWP). In most instances, PAWP elevation is the result of high left atrial pressure (LAP) elevation.
PAWP can be estimated from the following formula:
where E is the peak blood flow velocity of the mitral inflow in cm/s and Vp is the flow propagation velocity of the mitral inflow (in cm/s) obtained by color M-mode. Vp recording of this patient is demonstrated in Figure 8-26.
 Figure 8-26 |
Vp measures the rate at which red blood cells reach the LV apex from the mitral valve level during early diastole. The rate of blood flow from the mitral valve to the LV apex is determined by the rate of LV relaxation during early diastole. Therefore, Vp is an indirect measure of the rate of LV relaxation; the lower the Vp, the slower the LV relaxation and higher the left ventricular diastolic pressure (LVDP) are.
In our patient:
With the value of 26 mm Hg, PAWP is elevated; normal PAWP is ≤12 mm Hg. Therefore, answer A is correct.
Answer B is incorrect because in patients with markedly elevated LAP and PAWP, the peak velocity of the mitral E wave is typically higher than that of the mitral A wave. The patients have either the pseudonormal filling pattern (E/A is between 1.0 and 2.0; E wave deceleration time ≥160 ms) or the restrictive filling pattern (E/A >2 and E wave deceleration time <160 ms).
Answer C is incorrect because the pulmonary artery systolic pressure (PASP) is 64 mm Hg plus the right atrial pressure, or 64 + 15 = 79 mm Hg. In the absence of pulmonic stenosis, PASP is the same as the right ventricular systolic pressure (RVSP). Peak velocity (V) of the tricuspid regurgitant flow can be used to estimate the RV-to-RA pressure gradient (&Dgr;P) at peak systole:
&Dgr;P = 4 × V2 = (4 m/s)2 = 64 mm Hg
By adding RAP to &Dgr;P, RVSP (and, by extension, PASP) can be calculated as follows:
RVSP = PASP = &Dgr;P + RAP = 64 + 15 = 79 mm Hg
Answer D is incorrect because the ratio of mitral E wave to mitral annular tissue Doppler e′ wave is expected to be greater than 15 in patients with markedly elevated LAP and PAWP.
Answer E is incorrect because the normal Vp velocity >55 cm/s in young individuals, and >45 cm/s in middle-aged and elderly individuals.
6. A 44-year-old man with trileaflet aortic valve and dilated aortic root measuring 5.5 cm at the level of sinuses of Valsalva is being evaluated for aortic regurgitation.
The following statement is TRUE:
A. Regurgitant fraction of 65% would indicate that the aortic regurgitation is severe.
B. Like the size of flow convergence (PISA) radius, the size of vena contracta is strongly influenced by Nyquist limit setting.
C. Vena contracta of at least 0.2 cm would indicate that the aortic regurgitation is severe.
D. Regurgitant volume of 30 mL per beat is consistent with severe aortic regurgitation.
E. Vena contracta obtained by two-dimensional echocardiography can be used to calculate regurgitant volume.
View Answer
6. Answer: A. Severe aortic regurgitation (grades 3+ and 4+) is defined by the following criteria:
Table 8-18 | ||||||||||
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 Figure 8-27 |
Therefore, answer A is correct; the regurgitant fraction of 65% indicates a severe aortic regurgitation.
Answer B is incorrect because vena contracta is not strongly influenced by Nyquist limit color Doppler settings. This is in contrast to PISA radius. By changing the color Doppler Nyquist limit, one also automatically changes the velocity filter. The role of the velocity filter is to prevent color encoding of low velocities. By lowering the color Doppler Nyquist limit, one lowers the velocity filter allowing for inclusion of lower velocities and an increase in the color area. Because vena contracta contains predominantly high velocities, altering the Nyquist limit will not change significantly the size of vena contracta diameter. This is in contrast to PISA radius, which becomes progressively larger with lower Nyquist limits.
The impact of changes in color Doppler Nyquist limit on vena contracta is demonstrated in Figure 8-27.
Answer C is incorrect because in severe aortic regurgitation, vena contracta is ≥0.3 cm.
Answer D is incorrect because in severe aortic regurgitation, regurgitant volume is ≥60 mL per beat.
Answer E is incorrect because the diameter of vena contracta obtained by two-dimensional echocardiography should not be used to calculate the regurgitant volume. Instead, the two-dimensional diameter of vena contracta should be used for semiquantitative assessment of the degree of aortic regurgitation.
7. A 62-year-old man with history of treated hypertension, chronic atrial fibrillation, and bicuspid aortic valve had a transthoracic echocardiogram done. The study showed the following:
Table 8-7 | ||||||||
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Systemic blood pressure at the time of the study was 120/70 mm Hg. The following statement is TRUE:
A. Peak-to-peak aortic gradient is 90 mm Hg.
B. Patient is in cardiogenic shock due to left ventricular systolic dysfunction.
C. Mean left atrial pressure is approximately 20 mm Hg.
D. The size of vena contracta is diagnostic of severe mitral regurgitation.
E. Left atrial pressure cannot be estimated by E/e′ method in patients with atrial fibrillation.
View Answer
7. Answer: C. The E/e′ ratio is directly proportional to the left atrial pressure (LAP). The peak velocity of the mitral annular tissue Doppler e′ wave is directly proportional to the rate of LV relaxation during early diastole. The slower the LV relaxation, the higher the left ventricular diastolic pressure (LVDP) is. Once LVDP rises, there is a concomitant rise in the LAP and PAWP rise to allow for better filling of a stiff LV. The higher the LAP, the taller the mitral E wave becomes. In summary, as the LV diastolic dysfunction worsens, the peak velocity of the annular tissue e′ wave gets smaller, the mitral E wave gets higher, and the E/e′ ratio becomes progressively larger reflecting the rising LAP and PAWP.
Table 8-19 | ||||||||||||||||
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The E/e′ ratio can be used to estimate LAP in two ways. One approach is to use it semiquantitatively as shown in table above.
Thus, by E/e′ ratio of 16 alone, our patient has an elevated LAP. The other approach is to estimate LAP numerically using the following equation:
In our patient:
LAP = 1.9 + 1.24 × 16 = 22
An LAP of 22 mm Hg is significantly elevated; normal LAP is <12 mm Hg.
A simplified form of Eq. 1 is:
In our patient, LAP can be estimated by Eq. 2 as 4 + 16, or 20 mm Hg. See Figure 8-28A.
Therefore, answer C is correct.
Answer A is incorrect because the peak-to-peak gradient of aortic stenosis in this patient is 44 mm Hg.
To calculate the peak-to-peak gradient of aortic stenosis, we first need to calculate the peak left ventricular systolic pressure (LVSP), using the following formula:
where &Dgr;PMR is the peak systolic gradient of the mitral regurgitant jet and LAP is the left atrial pressure. After expressing &Dgr;PMR in terms of the peak velocity (V) of the MR jet, Eq. 1 becomes:
In our patient:
LVSP = 4 × (6.0 m/s)2 + 20 = 164 mm Hg
Once LVSP is known, the peak-to-peak aortic gradient (P2P) can be calculated as follows:
where SBP is the systolic blood pressure.
In our patient:
P2P = 164 – 120 = 44 mm Hg
It is important to emphasize that this pressure gradient, which is commonly measured on cardiac catheterization, is not a physiologic one because it represents a pressure difference at separate points in time as demonstrated in Figure 8-28B. P2P is lower than the peak instantaneous gradient obtained by continuous wave Doppler across the aortic valve.
 Figure 8-28A |
 Figure 8-28B |
Answer B is incorrect because left ventricular dP/dt is normal. Patients with cardiogenic shock have low dP/dt values. Normal dP/dt = 1,661 + 323 mm Hg per second.
Answer D is incorrect because in severe mitral regurgitation vena contracta of ≥0.7 cm.
Answer E is incorrect because either Eq. 1 or Eq. 2 is applicable irrespective of the atrial rhythm (normal sinus rhythm, atrial fibrillation, etc.).
8. A 67-year-old man with aortic regurgitation underwent transthoracic echocardiographic examination. There was no mitral stenosis or regurgitation. The following values were obtained:
Table 8-8 | ||||||||||
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Based on the aforementioned data, one can conclude that:
A. Pressure half-time is consistent with severe aortic regurgitation.
B. Aortic valve area can be estimated as 220 divided by pressure half-time.
C. Peak left ventricular systolic pressure (LVSP) is lower than the systolic blood pressure (SBP).
D. Left ventricular end-diastolic pressure (LVEDP) is estimated at 10 mm Hg.
E. Aortic valve area cannot be calculated using continuity equation because there is aortic regurgitation.
View Answer
8. Answer: D. Figure 8-29 shows the continuous wave spectral Doppler tracings of our patient.
 Figure 8-29 |
Using the end-diastolic velocity (V) of the aortic regurgitant jet, one can calculate the pressure gradient (&Dgr;P) between the diastolic blood pressure (DBP) and the LVEDP.
Rearranging Eq. 1, LVEDP can be calculated in the following manner if the DBP is known:
In our patient:
LVEDP = 65 mm Hg – 4 × (3.7 m/s)2 = 10 mm Hg
Therefore, answer D is correct.
Answer A is incorrect because in severe aortic regurgitation pressure half-time is <300 ms.
Answer B is incorrect because the aortic valve (AV) area cannot be calculated by 220 into pressure half-time; that is the formula for calculating the mitral valve area.
Answer C is incorrect because the peak LVSP is always higher than the SBP in patients with aortic stenosis. LVSP becomes progressively higher than SBP as the aortic stenosis becomes more severe. The LVSP-to-SBP pressure gradient is referred to as the peak-to-peak aortic gradient as discussed in answer to Question 7.
Answer E is incorrect because the continuity equation can be used to calculate the AV area in patients with or without aortic regurgitation. The continuity principle states that the stroke volume across the left ventricular outflow tract (LVOT) is the same as the stroke volume across the AV:
Since stroke volume can be expressed as the product of the cross sectional area (CSA) and the flow velocity integral (VTI), Eq. 3 becomes:
In patients with aortic regurgitation, there is an increase in antegrade flow from the left ventricle into the aorta due to augmentation of the true left ventricular stroke volume by the aortic regurgitant volume. However, this increase equally affects the flow through the LVOT and the AV in systole. In Eq. 5, this will be reflected in a proportional increase in VTILVOT and VTIAV.
By continuity equation, aortic valve area (CSAAV) can be calculated as follows:
In aortic regurgitation, there is augmentation of VTILVOT and VTIAV. However, the ratio of the two VTIs remains the same, and therefore the calculated value of CSAAV is not affected by the presence of aortic regurgitation.
9. A 25-year-old woman is being evaluated for percutaneous closure of her secundum atrial septal defect (ASD). Transthoracic echocardiography demonstrated mild tricuspid regurgitation, no pulmonic stenosis, and the following:
Table 8-9 | ||||||||||||||||
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Based on the aforementioned data, one can conclude:
A. Patient should be advised against ASD closure because pulmonary hypertension is present.
B. Pulmonary vascular resistance is approximately 16 Wood units.
C. The ratio of pulmonary to systemic blood flow (Qp:Qs) is approximately 2.5:1.
D. Shunt flow is larger than the pulmonic flow (Qp).
E. Patient is cyanotic.
View Answer
9. Answer: C. Patient has an atrial septal defect (ASD) with a left-to-right shunt. ASD is an intracardiac shunt at the atrial level. Systemic blood flow (Qs) reaches the right atrium through systemic veins. At the level of the right atrium, it is joined by the shunt flow, which enters the right atrium from the left atrium across the ASD. The sum of Qs and the shunt flow then passes through the right ventricular outflow tract (RVOT) into the pulmonary circulation. Therefore, the sum of Qs and the shunt flow represents the pulmonary blood flow (Qp). This Qp reaches the left atrium through the pulmonary veins. At the left atrial level, Qp divides into shunt flow (which traverses ASD to reach the right atrium), and Qs, which enters the left ventricle. Qs then passes through the LVOT into the aorta and eventually reaches the right atrium through systemic veins.
In summary, flow through the LVOT represents Qs while the flow through RVOT represents Qp in patients with ASD.
Shunt calculations for this patient are summarized in the following table:
Table 8-20 | ||||||||||||||||||||||||||||||||||||
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Because Qp:Qs = 2.5:1, answer C is correct.
Answer A is incorrect because the presence of pulmonary hypertension per se does not preclude ASD closure. It is the degree of pulmonary vascular resistance (PVR) that determines whether a patient is a candidate for ASD closure or not, as discussed later.
Answer B is incorrect because the patient’s PVR is essentially normal. Using the Ohm’s law, PVR can be calculated as follows:
where Qp is the pulmonary blood flow (in L/min), and &Dgr;P is the pressure gradient across the pulmonary circulation. &Dgr;P is the difference between the mean pulmonary artery pressure (MPP) and the mean left atrial pressure (LAP). Eq. 1 then becomes:
MPP can be calculated from pulmonary artery systolic pressure (PASP) and the pulmonary artery diastolic pressure (PADP) by using the following equation:
In this patient:
Once MPP is known, we can use Eq. 2 to calculate PVR:
Normal PVR is 1-2 Wood units (80-160 dyne s cm-5). In this patient, PVR is only modestly elevated. In principle, ASD closure should not be performed if PVR is 2/3 or more of the systemic vascular resistance (SVR). Since normal SVR is approximately 13 Wood units (range, 11-16 Wood units, or 900-1,300 dyne s cm-5), PVR ≥9 Wood units usually precludes ASD closure.
Answer D is incorrect because the shunt flow in this patient is 7.2 L/min. Shunt flow is the difference between Qp and Qs. In this patient:
SF = Qp – Qs = 11.9 – 4.7 = 7.2 L/min
Answer E is incorrect because Qp is much larger than Qs, the shunt flow is in the left to right direction, and thus the patient is not expected to be cyanotic.
10. A 35-year-old woman was noted on clinical examination to have a systolic murmur and was referred for transthoracic echocardiography. The examination
revealed perimembranous ventricular septal defect (VSD), mild tricuspid regurgitation, pulmonic stenosis (PS), intact aortic valve, and the following:
revealed perimembranous ventricular septal defect (VSD), mild tricuspid regurgitation, pulmonic stenosis (PS), intact aortic valve, and the following:
Table 8-10 | ||||||||||||
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The following statement is TRUE:
A. Right ventricular systolic pressure (RVSP) is 46 mm Hg.
B. Pulmonary artery systolic pressure (PASP) is 29 mm Hg.
C. RVSP is 84 mm Hg above the right atrial pressure (RAP).
D. PASP is 45 mm Hg higher than the right ventricular systolic pressure.
E. Right ventricular end-diastolic pressure is 28 mm Hg.
View Answer
10. Answer: B. The presence of VSD allows for calculation of the RVSP and, by extension, the PASP if the systolic blood pressure (SBP) is known.
RVSP in a patient with VSD and no left ventricular outflow obstruction can be calculated as follows:
Using the peak systolic velocity (V) across the VSD, peak systolic VSD gradient can be calculated as follows:
By combining Eqs. 1 and 2, RVSP is then calculated as follows:
Thus in this patient,
RVSP = 120 – 4 × (3.0 m/s)2 = 84 mm Hg
When there is no pulmonic stenosis, PASP = RVSP. However, this patient has PS with a PSG of 55 mm Hg across the pulmonic valve. In the presence of PS, the relationship between RVSP and PASP is as follows:
In our patient, PASP = 84 – 55 = 29 mm Hg. Therefore, answer B is correct.
Answer A is incorrect because RVSP in this patient is 84 mm Hg as calculated earlier.
Answer C is incorrect because RAP is not required for RVSP estimation using the VSD method.
Answer D is incorrect because PASP is lower than RVSP due to the presence of PS. RVSP exceeds PASP by 55 mm Hg, which is the peak gradient across the stenosed pulmonic valve.
Answer E is incorrect because the right ventricular end-diastolic pressure (RVEDP) in this patient is 8 mm Hg. If LVEDP is known, RVEDP can be calculated as follows:
Using the end-diastolic velocity (V) across the VSD, the end-diastolic VSD gradient can be calculated as follows:
By combining Eq. 5 and Eq. 6, RVEDP is then calculated as follows:
where V is the end-diastolic velocity across the VSD. In our patient:
RVEDP = 12 – 4 × (1 m/s)2 = 12 – 4 = 8 mm Hg
11. A 21-year-old college student is noted to have fixed splitting of the second heart sound and right bundle branch block. Real-time three-dimensional transesophageal echocardiogram revealed a 1.2-cm secundum atrial septal defect (ASD) that was circular in shape. On color Doppler image, a well-formed hemispheric flow convergence (PISA) shell is seen on the left atrial side of the ASD. The following data were also obtained:
Table 8-11 | ||||||||||||
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The following statement is TRUE:
A. Ratio of pulmonic to systemic flow (Qp:Qs) is 1.8 to 1.0.
B. Shunt flow (SF) across the ASD is approximately 9.0 L/min.
C. The difference between the pulmonic and systemic stroke volume is 180 mL.
D. Systemic stroke volume is 150 mL.
E. Pulmonic blood flow (Qp) is approximately 7.0 L/min.
View Answer
11. Answer: B. The pulmonic flow (Qp) in patients with ASD is the sum of the SF across the ASD and the systemic flow (Qs). SF can be calculated either directly or as the difference between Qp and Qs.
One method for direct calculation of SF is the standard echocardiographic formula for determining flow rate through an orifice:
Flow = CSA × VTI × HR
where CSA is the cross-sectional area of the orifice, VTI is the velocity-time integral at the level of the orifice, and HR is the heart rate.
In the first step, we will calculate the CSA of the ASD whose diameter is 1.2 cm. Since the ASD is circular in shape, then ASD area can be calculated as follows:
CSAASD = (1/2 × ASD diameter)2 × &pgr;
In our patient:
CSAASD = (1/2 × 1.2 cm)2 × 3.14 = 0.36 × 3.14 = 1.13 cm2
Next, we can calculate the stroke volume across the ASD as follows:
ASD shunt stroke volume = CSAASD × VTIASD
In our patient:
ASD shunt stroke volume = 1.13 cm2 × 80 cm = 90 mL per beat.
In the final step, by multiplying the ASD shunt stroke volume by the heart rate, one can calculate the SF across the ASD. In our patient:
ASD shunt flow = 90 mL × 100 beats/min = 9.0 L/min.
Table 8-21 | ||||||||||||||||||||||||||||||||||||||||
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Therefore, the answer B is correct.
Answer A is incorrect because the Qp:Qs in this patient is 2.5:1. In this patient, Qs is calculated at the level of the left ventricular outflow tract (LVOT) using the following formula:
Qs = CSALVOT × VTILVOT × HR
where CSALVOT is the CSA of LVOT, VTILVOT is the velocity-time integral at LVOT level, and HR is the heart rate. In our patient:
Qs = (1/2 × 2.0 cm)2 × &pgr; × 19 cm × 100 beats/min = 60 mL × 100 beats/min = 6.0 L/min
In the next step, we can calculate Qp as follows:
Qp = Qs + ASD shunt flow
In our patient:
Qp = 6.0 L/min + 9.0 L/min = 15.0 L/min
Once Qp and Qs are known, we can calculate Qp:Qs ratio as follows:
Qp:Qs = 15.0 L/min:6.0 L/min = 2.5:1
Answer C is incorrect because the difference between the pulmonic and systemic stroke volumes in this patient is 90 mL/beat. This value represents the ASD shunt stroke volume as calculated earlier.
Answer D is incorrect because the systemic stroke volume in this patient is 60 mL per beat as calculated earlier.
Answer E is incorrect because Qp in this patient is 15.0 L/min as calculated earlier.
Calculations related to this question are summarized in Table 8-21.
12. A 35-year-old woman presents with sudden onset of dyspnea and pulmonary edema. She underwent bedside transthoracic echocardiography, which revealed hyperdynamic left ventricular systolic function, normal aortic valve, and mitral regurgitation. The following data were obtained from the transthoracic echocardiogram:
Table 8-12 | ||||||||||||
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The following statement is TRUE:
A. Peak velocity of the mitral inflow E wave is expected to be low.
B. Left atrial pressure is low.
C. Pulmonary venous flow velocity pattern on spectral Doppler is likely to reveal flow reversal during early diastole.
D. Rate of pressure rise (dP/dt) in the left ventricle is 1,600 mm Hg per second.
E. Left ventricular systolic function is markedly diminished.
View Answer
12. Answer: D. Continuous Doppler spectral tracing of the mitral regurgitant jet can be used to estimate the rate of pressure rise (dP) in the left ventricle over time (dt), a measure of left ventricular systolic function, using the following formula:
where RTI is the relative time interval, measured in seconds, between MR jet velocities of 1 m/s (V1) and 3 m/s (V2). &Dgr;P represents the pressure difference between the left ventricular to LAP gradients at V2 and V1 (Fig. 8-30A).
 Figure 8-30A |
 Figure 8-30B |
This pressure difference can be calculated as follows:
Now, Eq. 1 can be expressed as follows:
In the next step, we will calculate RTI in our patient:
RTI = Time at V2 – Time at V1 = 25 ms – 5 ms = 20 ms
Because in Eq. 2, RTI is expressed in seconds, we have to convert our patient RTI from milliseconds to seconds:
RTI = 20 ms = 0.02 s
Once RTI is known, we can calculate dP/dt in our patient:
dP/dt = 32/0.02 = 1,600 mm Hg/s
Therefore, answer D is correct.
Answer A is incorrect because the peak velocity of mitral E wave in severe mitral regurgitation is expected to be high. Peak velocity across an orifice is directly related to flow across that orifice. Since the flow is the product of stroke volume (SV) and heart rate, peak velocity is then a direct function (f) of SV:
In mitral regurgitation, SV that crosses the mitral valve in diastole is the sum of the systemic SV (SVLVOT) and the regurgitant volume (RegV). Thus, Eq. 5 can be expressed as follows:
The more severe the mitral regurgitation is, the larger the RegV is, and therefore, the higher the peak velocity of the mitral inflow E wave. When native mitral regurgitation is severe (as is the case in this patient as judged by the vena contracta ≥0.7 cm), peak E velocity is expected to be >1.5 m/s. In severe prosthetic mitral regurgitation, the peak E velocity is usually >2.0 m/s.
Answer B is incorrect because LAP in this patient is elevated. The patient presents with severe mitral regurgitation (vena contracta ≥0.7 cm) and pulmonary edema due to elevated LAP.
Using the peak velocity (Vmax) of the mitral regurgitant jet, one can calculate the pressure gradient (&Dgr;P) between the peak left ventricular systolic pressure (LVSP) and the LAP:
In our patient:
&Dgr;P = 4 × (4.0 m/s)2 = 4 × 16 = 64 mm Hg
The sum of this pressure gradient and LAP during systole represents the peak LVSP:
By rearranging Eq. 6, we can solve for LAP:
The LAP calculated by this method represents a value on the CV wave portion of the LAP tracing.
LVSP is not given in the question. In this patient who does not have aortic stenosis or left ventricular outflow obstruction, LVSP is equal to SBP. Thus we can express Eq. 7 as follows:
In our patient, whose SBP was 95 mm Hg and whose &Dgr;P was calculated above at 64 mm Hg, LAP is then calculated as follows:
LAP = 95 mm Hg – 64 mm Hg = 31 mm Hg
This LAP of 31 mm Hg is highly elevated (normal LAP is ≤12 mm Hg).
Answer C is incorrect because in severe mitral regurgitation there may be flow reversal in systolic (S) but not diastolic (D) wave on pulmonary venous flow velocity tracings. An example of S wave reversal due to severe mitral regurgitation is shown in Figure 8-30B.
Answer E is incorrect because dP/dt in this patient is estimated at 1,600 mm Hg/s, which is normal. (Normal dP/dt = 1,661 + 323 mm Hg/s.) The value of 800 mm Hg/s would indicate a markedly diminished LV systolic function as seen in cardiogenic shock, for example.
13. A 29-year-old Bangladeshi woman with rheumatic mitral stenosis is referred to the cardiac catheterization laboratory for percutaneous mitral balloon valvuloplasty. Upon placement of the pigtail catheter in the left ventricle, the following values were obtained:
Table 8-13 | ||||||
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Transesophageal echocardiogram prior to valvuloplasty revealed the absence of both mitral and aortic regurgitation, as well as the following:
Table 8-14 | ||||||||
|---|---|---|---|---|---|---|---|---|
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The following statement is TRUE:
A. Mean left atrial pressure is expected to be lower than the mean left ventricular diastolic pressure.
B. Peak velocity of the mitral inflow E wave is expected to be low.
C. Pressure half-time (PHT) may be unreliable in patients prior to valvuloplasty.
D. Mitral valve area (MVA) is 0.6 cm2.
E. Mean left atrial pressure (LAP) is approximately 28 mm Hg.
View Answer
13. Answer: E. In mitral stenosis, there is a pressure gradient between the left atrium and the left ventricle during diastole. In this patient, the mean diastolic pressure gradient is markedly elevated (21 mm Hg). Mean diastolic pressure gradient of >10 mm Hg is consistent with severe mitral stenosis as shown in the table below.
Table 8-22 | ||||||||||||||||
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In this young patient, left ventricular diastolic pressures are normal. Mean LAP can be calculated as follows:
LAP = Mean mitral gradient in diastole + Early LV diastolic pressure
In our patient:
LAP = 21 mm Hg + 7 mm Hg = 28 mm Hg
Therefore, answer E is correct.
Answer A is incorrect because in mitral stenosis there is an antegrade flow driven by a pressure gradient between the left atrium and the left ventricle in diastole. Therefore, the mean LAP is higher than the mean left ventricular diastolic pressure.
Answer B is incorrect because in mitral stenosis the peak velocity of the mitral E wave is expected to be high. Velocity (V) across an orifice is inversely related to the cross-sectional area (CSA) of the orifice:
For mitral stenosis, CSA equals to the MVA and Eq. 1 becomes
Therefore, the smaller the MVA (i.e., the more severe the mitral stenosis), the higher the peak velocity of the mitral E wave.
Answer C is incorrect because the PHT method may be unreliable immediately after but not before the mitral valvuloplasty. PHT method assumes that the left ventricular pressure and compliance are normal, and therefore that the deceleration slope of the mitral E wave on spectral Doppler tracings in diastole is the function of the MVA alone.
Immediately after valvuloplasty, there is a sudden increase in the mitral orifice area leading to an increase in the stroke volume delivered to the left ventricle in early diastole. Because the left ventricle compliance cannot change acutely, the left ventricular diastolic pressure increases. With the rise in the left ventricular diastolic pressure, the diastolic gradient between the left atrium and the left ventricle decreases and the mitral PHT shortens above and beyond what would be expected by an increase in the MVA alone after valvuloplasty. Therefore, the PHT method may lead to calculation of an erroneously large MVA.
Answer D is incorrect because the MVA by PHT method in this patient is 0.8 cm2:
14. An 81-year-old woman with a systolic heart murmur was referred for an echocardiography. A heavily calcified aortic valve and normal mitral valve were noted on two-dimensional echocardiographic imaging. Doppler echocardiography of the aortic valve revealed:
Table 8-15 | ||||||||
|---|---|---|---|---|---|---|---|---|
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The following statement is TRUE:
A. Aortic valve area (AVA) cannot be calculated because aortic valve velocity-time integral is not stated.
B. Aortic valve stenosis is subvalvular.
C. Aortic valve area is likely to be less than 1 cm2.
D. Left ventricular stroke volume (SV) is 80 mL per beat.
E. Systolic blood pressure is approximately 100 mg Hg above the left ventricular systolic pressure.
View Answer
14. Answer: C. When velocity-time integrals are not available, AVA can be calculated using the following modified continuity equation:
where CSALVOT is the CSA of the LVOT, VLVOT is the peak systolic LVOT velocity, and VAV is the peak systolic AV velocity.
The VLVOT/VAV ratio of the two velocities is referred to as the dimensionless index (DI). Thus Eq. 1 can be expressed as follows:
After expressing the LVOT area in terms of LVOT diameter (D), Eq. 2 becomes:
In our patient:
AVA = 3.14 × (1/2 × 1.9 cm2)2 × (1 m/s/5 m/s)
AVA = 2.84 cm2 × 0.2
AVA = 0.6 cm2
Therefore, answer C is correct.
As a rule, when the dimensionless index (DI) is ≤0.25, the AVA is <1.0 cm2 across the range of LVOT diameters commonly encountered in adults as demonstrated in the following table:
Table 8-23 | ||||||||||||||||||
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Answer A is incorrect because the modified continuity equation using the DI, as explained earlier, can be used to calculate the AVA when velocity-time integrals are unavailable.
Answer B is incorrect because the subvalvular (LVOT) velocity is normal (1.0 m/s).
Answer D is incorrect because the left ventricular SV in this patient is 57 mL per beat. Left ventricular SV can be calculated as follows:
SV = &pgr; × (1/2 × LVOT diameter)2 × VTILVOT
SV = 3.14 × (1/2 × 1.9 cm)2 × 20 cm
SV = 57 mL per beat
Answer E is incorrect because in aortic stenosis, left ventricular peak systolic pressure exceeds the systolic blood pressure. The magnitude of this pressure difference (peak-to-peak gradient) is proportional to the severity of aortic stenosis.
15. This continuous wave spectral Doppler tracing of the tricuspid regurgitant jet comes from an 18-year-old woman with pulmonic valve stenosis (Fig. 8-1). The peak pulmonic valve gradient is 24 mm Hg. Right atrial pressure is estimated at 10 mm Hg. The following is TRUE about this patient:
A. Peak pulmonary artery systolic pressure (PASP) is higher than the right ventricular peak systolic pressure.
B. Right ventricular peak systolic pressure is 64 mm Hg above the pulmonary artery peak systolic pressure.
C. Pulmonary artery peak systolic pressure is 50 mm Hg.
D. Right ventricular peak systolic pressure is 24 mm Hg less than the peak pulmonary artery systolic pressure (PASP).
View Answer
15. Answer: C. Peak RVSP in a patient with or without pulmonic stenosis (PS) can be calculated as follows:
where RAP is the right atrial pressure. Since RV-to-RA systolic gradient can be estimated from the peak systolic velocity of the tricuspid regurgitant (V), Eq. 1 can be expressed as follows:
In the absence of PS, RVSP is equal to PASP. In PS, however, peak RVSP exceeds PASP. The difference between the two pressures represents the peak gradient of PS. Therefore, in patients with PS, PASP is estimated as follows:
In our patient:
RVSP = 4 × (4.0 m/s)2 + 10 = 74 mm Hg
PASP = 74 – 24 = 50 mm Hg
Therefore, answer C is correct.
All calculations are graphically summarized in Figure 8-31 (RVP, right ventricular pressure; RAP, right atrial pressure; PAP, pulmonary artery pressure).
Answer A is incorrect because in the presence of pulmonic valve stenosis, RVSP exceeds PASP as shown in the figure earlier.
Answer B is incorrect because RVSP exceed PASP by 24 mm Hg, the value of the PSG across the pulmonic valve.
Answer D is incorrect because RVSP is 24 mm Hg more than PASP.
Answer E is incorrect because RVSP is 74 mm Hg as calculated earlier.
 Figure 8-31 |
16. An 82-year-old man was referred for evaluation of a systolic ejection murmur. On parasternal long-axis view, the left ventricular outflow tract (LVOT) diameter was measured at 2.0 cm. The above spectral Doppler tracings were obtained in or through the LVOT in the apical 5-chamber view (Fig. 8-2).
The following statement is TRUE:
A. Increased cardiac output alone may explain the elevated gradient across the aortic valve.
B. Marked difference between the subvalvular and valvular velocities in this patient may also be seen in severe aortic regurgitation.
C. Patient has a very severe aortic valve stenosis with a mean gradient of approximately 60 mm Hg.
D. Aortic valve area (AVA) is greater than 1.0 cm2.
E. Patient has hypertrophic obstructive cardiomyopathy (HOCM).
 Figure 8-2 |
View Answer
16. Answer: C. Peak gradient (&Dgr;Pmax) of aortic stenosis can be calculated from the peak systolic velocity (V) across the aortic valve obtained by continuous wave Doppler using the modified Bernoulli equation:
The mean aortic valve gradient (&Dgr;Pmean) is approximately 60% of the peak gradient (&Dgr;Pmax):
In our patient:
&Dgr;Pmax = 4 × (5.0 m/s)2 = 100 mm Hg
&Dgr;Pmean = 0.6 × 100 mm Hg = 60 mm Hg
Therefore, answer C is correct.
Answer A is incorrect because increased cardiac output (as during pregnancy, for instance) leads to a proportional increase in both LVOT and aortic velocities. In this patient, there is a marked difference between the peak systolic LVOT velocity (0.9 m/s) and the peak systolic aortic velocity (5.0 m/s) indicative of aortic stenosis.
Answer B is incorrect because in aortic regurgitation there is a proportional increase in both LVOT and aortic velocities in systole due augmentation of the left ventricular stroke volume by the recirculating regurgitant volume. A wide discrepancy in the peak LVOT and aortic velocities in systole is not expected in severe aortic regurgitation.
Answer D is incorrect because the AVA in this patient is less than 1.0 cm2.
(AVA) in this patient can be estimated using the modified continuity equation:
where CSALVOT is the cross-sectional area of the LVOT, VLVOT is the LVOT peak systolic velocity, and VAV is the peak aortic velocity in systole.
After expressing the LVOT area in terms of LVOT diameter (D), Eq. 3 becomes:
In this patient:
AVA = 3.14 × (1/2 × 2.0)2 × 0.9/5.0 = 0.56 cm2.
Answer E is incorrect because the subvalvular (LVOT) velocity of 1.2 m/s is normal.
17. This continuous wave spectral Doppler tracing from a 21-year-old woman (Fig. 8-3) represents the flow velocity profile in the main pulmonary artery. Based on this tracing, the following is TRUE about this patient:
A. End-diastolic gradient across the pulmonic valve is high.
B. There is severe pulmonic valve stenosis.
C. Pulmonary artery systolic pressure is 9 mm Hg above the right ventricular pressure.
D. Pulmonic valve regurgitation is severe.
E. The velocity profile is diagnostic of patent ductus arteriosus.
 Figure 8-3 |
View Answer
17. Answer: D. The patient has severe pulmonic valve regurgitation, a common long-term complication of tetralogy of Fallot repair.
Because of a large regurgitant orifice, the pressure gradient between pulmonary artery and the right ventricle equalizes rapidly. Equalization is achieved by mid diastole and there is no measurable end-diastolic gradient as demonstrated in Figure 8-32.
This rapid deceleration and premature cessation of the pulmonic regurgitant jet are a characteristic finding of severe pulmonic regurgitation. Therefore, answer D is correct.
Answer A is incorrect because the end-diastolic gradient in severe pulmonic regurgitation is approaching zero.
 Figure 8-32 |
Answer B is incorrect because the peak antegrade velocity across the pulmonic valve in systole is only elevated to about 1.5 m/s (peak systolic gradient = 4 × 1.52 = 9 mm Hg). This is consistent with pulmonic regurgitation alone. During systole, stroke volume is augmented by the recirculating regurgitant volume. This flow augmentation leads to higher systolic velocities across the pulmonic valve based on the fundamental equation of fluid dynamics:
where V is the antegrade velocity across the pulmonic valve, Q is the volumetric flow across the pulmonic valve in systole, SV is the stroke volume, HR is the heart rate, and PVA is the pulmonic valve area. Thus, when the PVA remains constant, any increase in stroke volume leads to elevation in the transvalvular velocity.
Answer C is incorrect, the right ventricular systolic pressure exceeds pulmonary artery by 9 mm Hg (see the figure above).
Answer E is incorrect because in uncomplicated patent ductus arteriosus, antegrade flow occurs during both systole and diastole. In the patient’s tracing there is antegrade flow in systole and retrograde flows in diastole.
18. The tracings (Fig. 8-4) were obtained from an 82-year-old woman with a normal left ventricular ejection fraction of 65%. Figure 8-4A represents blood flow velocity pattern obtained by placing a pulsed Doppler sample volume at the mitral leaflet tips. Figure 8-4B represents tissue Doppler of the lateral mitral annulus. Based on these two tracings, the following is TRUE:
A. The patient has excellent exercise capacity.
B. Abnormal left ventricular relaxation alone explains the mitral inflow pattern.
C. Left atrial pressure is elevated.
D. Patient has normal left ventricular diastolic function.
E. Mitral E wave velocity is expected to increase following the Valsalva maneuver.
 Figure 8-4A |
 Figure 8-4B |
View Answer
18. Answer: C. The tracings were obtained from an elderly woman presenting with acutely decompensated heart failure.
Mean left atrial pressure can be estimated semiquantitatively from the ratio of peak flow velocity of mitral E wave and the peak velocity of mitral annular tissue Doppler e′ wave according to the following chart:
Table 8-24 | ||||||||||||||||
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In our patient, E/lateral e′ is 142/8 or 18. This ratio is consistent with elevated left atrial pressure. Therefore, answer C is correct.
Answer A is incorrect because the patient is likely to have poor exercise capacity with exertional dyspnea given the elevation of left atrial pressure even at rest. With exertion, left atrial pressure is expected to rise even further.
Answer B is incorrect because the patient’s mitral inflow pattern is a combination of abnormal left ventricular relaxation and elevated left atrial pressure. The mitral E/A ratio that is greater than 2 in conjunction with a rapid E wave deceleration time (<160 ms) indicates a restrictive filling pattern. The features of different filling patterns in individuals older than 60 years are summarized in Table 8-25.
Answer D is incorrect because the patient has a restrictive filling pattern. This is an abnormal finding and consistent with severe left ventricular diastolic dysfunction.
Answer E is incorrect because with a Valsalva maneuver the peak velocity of the mitral E wave is expected to decrease. Valsalva maneuver decreases preload and leads to a lower early diastolic pressure gradient between the left atrium and left ventricle. This leads to a lower peak velocity of the mitral E wave and a lower mitral E/A ratio.
19. Figure 8-5A,B were obtained from the same patient at the same heart rate.
The following statement is TRUE:
A. Mitral inflow pattern is diagnostic of restrictive filling.
B. Left ventricular end-diastolic pressure (LVDP) is elevated.
C. The higher the peak velocity of the atrial reversal wave in pulmonary veins, the lower the left ventricular pressure is.
D. The absence of atrial reversal wave in pulmonary vein tracings indicates pulmonary hypertension due to left ventricular dysfunction.
E. Ratio of peak systolic to peak diastolic velocity in pulmonary veins of more than 1 is indicative of elevated left atrial pressure.
 Figure 8-5A |
 Figure 8-5B |
View Answer
19. Answer: B. In sinus rhythm, the left atrium contracts following the P wave on EKG and the blood is propelled both forward into the left ventricle across the mitral valve, as well as backward into the pulmonary veins, which lack valves. The velocity profile of the forward flow is responsible for the mitral inflow A wave, while the retrograde flow into the pulmonary veins is responsible for the atrial reversal (AR) wave.
Table 8-25 | ||||||||||||||||||||||||||||||
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When the LVDP is elevated at the time of atrial contraction, both the peak velocity and the duration of the AR wave are increased. A peak AR velocity of ≥35 cm/s is indicative of elevated LV end-diastolic pressure.
Elevation of LV end-diastolic pressure can also be inferred when the duration of the AR wave is ≥30 ms more than the duration of the mitral inflow A wave. In our patient, peak velocity of AR was 50 cm/s, and AR outlasted mitral A wave by 40 ms (210 – 170 ms); both are indicative of an elevated LVDP. Therefore, answer B is correct.
For further explanation, the reader is referred to Figure 13 in the Canadian Consensus Recommendations for the Measurement and Reporting of Diastolic Dysfunction by Echocardiography (J Am Soc Echocardiogr. 1996;9:736-760).
Answer A is incorrect because a restrictive filling pattern is characterized by a mitral inflow E/A ratio greater than 2; in this patient, peak E wave velocity is barely higher than the peak A wave velocity.
Answer C is incorrect because the higher the peak velocity of the atrial reversal wave in the pulmonary vein spectral tracing, the higher the LVDP.
Answer D is incorrect because with left ventricular dysfunction, there is an increase in the LVDP leading to secondary pulmonary hypertension. Because of LV diastolic pressure elevation, the pulmonary vein atrial reversal wave is likely to be prominent (as explained earlier) rather than absent. Atrial reversal wave is absent in atrial arrhythmias such as atrial fibrillation.
Answer E is incorrect because when left atrial pressure is elevated in older patients, the peak velocity of the systolic wave (S wave) in the pulmonary vein tracings is generally lower than the peak velocity of the diastolic wave (D wave). The higher the left atrial pressure, the lower the S/D ratio is.
20. Upward deflection in respirometry recordings indicates inspiration, while the downward deflection indicates expiration (Fig. 8-6).
The following statement is TRUE:
A. There is no ventricular interdependence.
B. Expiratory increase in diastolic flow reversal in hepatic veins suggests constriction.
C. Abnormal interventricular septal motion is due to right ventricular volume overload.
D. An inspiratory increase in antegrade hepatic vein flow velocities is abnormal.
E. Above M-mode recordings are diagnostic of a large pericardial effusion and tamponade.
 Figure 8-6A |
 Figure 8-6B |
View Answer
20. Answer: B. In constrictive pericarditis, ventricular filling is constrained by an inelastic pericardial sac that envelopes the entire heart except for the cranial portion of the left atrium and the pulmonary veins. This results in (1) ventricular interdependence and (2) a differential impact of negative intrathoracic pressure that develops during inspiration on pulmonary veins and the heart.
Ventricular interdependence refers to diastolic filling of one ventricle at the expense of the other depending on the respiratory phase. In inspiration, the pressure in the intrathoracic systemic vein decreases. This leads to a larger pressure gradient between extra- and intrathoracic systemic veins, which results in improved RV filling. At the same time, the drop in the intrathoracic pressure with inspiration decreases the pulmonary venous pressure. Because of the thickened rigid pericardium, the drop in the intrathoracic pressure cannot be transmitted to the heart; this results in a decreased pressure gradient between pulmonary veins and the left atrium, and decreased LV filling in diastole.
The net effect of inspiration is such that the right ventricle fills at the expense of the left ventricle, and the interventricular septum moves toward the left ventricle. The opposite occurs in expiration. This is illustrated in the M-mode recordings of our patient. The recordings also demonstrate no pericardial effusion.
With inspiration, the drop in intrathoracic pressure enhances forward flow in the hepatic veins in normal individuals; in constrictive pericarditis, there is an exaggeration of this inspiratory forward flow enhancement. During expiration, the rightward shift of the interventricular septum impedes RV filling; the rise in the RV diastolic pressure then leads to an expiratory increase in hepatic vein flow reversal. Therefore, answer B is correct.
Answer A is incorrect because the presence of marked reciprocal changes in the right and left ventricular filling that are phasic with respiration and indicative of ventricular interdependence.
Answer C is incorrect because the abnormal septal motion due to right ventricular overload (as in atrial septal defect or severe tricuspid regurgitation) is characterized by flattening of the interventricular septum with each diastole rather than being phasic with respiration.
Answer D is incorrect because an inspiratory increase in antegrade velocities is a normal finding. During inspiration, the drop in intrathoracic pressures enhances systemic venous return. This increased flow into the right-sided heart elevates antegrade velocities in the hepatic veins.
Answer E is incorrect because the M-mode reveals no echo lucency posterior to the left ventricle that would be diagnostic of a large pericardial effusion. Instead, it shows pericardial thickening. It is important to emphasize, however, that the abnormal interventricular septal motion phasic with respiration is encountered in both tamponade and constrictive pericarditis.
21. A 33-year-old man has had a murmur since childhood. The transthoracic spectral Doppler tracings in Figure 8-7 are obtained from the suprasternal view.
Which of the following statements is TRUE:
A. The pattern of diastolic flow is indicative of severe aortic regurgitation.
B. The tracings are diagnostic of aortic coarctation.
C. Quadricuspid aortic valve is the most common cause of aortic stenosis associated with the above flow velocity pattern.
D. The recordings are obtained from the ascending aorta and represent severe aortic stenosis.
E. Patient’s blood pressure in the legs is markedly higher than that in the arms.
 Figure 8-7 |
View Answer
21. Answer: B. In a normal descending aorta, antegrade flow occurs only in systole and there is a small flow reversal in early diastole as depicted in the following pulsed wave Doppler tracing in Figure 8-33A.
The pulsed-wave Doppler tracings in Figure 8-7 is abnormal as it demonstrates that antegrade flow is throughout the cardiac cycle. In addition, there is a large peak systolic gradient across the coarctation of almost 60 mm Hg. The presence of a holodiastolic antegrade flow in conjunction with a large systolic gradient is indicative of severe aortic coarctation. Therefore, answer B is correct.
Answer A is incorrect because in severe aortic regurgitation, there is a retrograde flow throughout diastole (holodiastolic flow reversal) as demonstrated in Figure 8-33B,C.
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