Echocardiography Board Review: 500 Multiple Choice Questions With Discussion

Chapter 6


1.  101. An intraoperative transesophageal echocardiogram (TEE) revealed mitral regurgitation with the following measurements: regurgitant jet area 4 cm2, PISA radius 0.8 cm at a Nyquist limit of 50 cm/s at a heart rate of 82 beats/min, and arterial blood pressure 80/40 mmHg. This represents:

1.  A. Mild mitral regurgitation (MR)

2.  B. Moderate MR

3.  C. Severe MR

2.  102. For the patient in the above question, if the systolic blood pressure is increased to 145 mmHg, assuming that the effective orifice area is unchanged, then the:

1.  A. MR jet size will double

2.  B. MR jet size will not change

3.  C. MR jet size will more than double

3.  103. For a given regurgitant volume, all of the following result in a reduction in the jet size except:

1.  A. Fast heart rate

2.  B. Doubling the sector angle

3.  C. Increasing the imaging depth

4.  D. Increasing the blood pressure

4.  104. In a patient with severe MR, all of the following factors increase its hemodynamic impact except:

1.  A. Mitral stenosis

2.  B. Left ventricular hypertrophy

3.  C. Compliant left atrium

4.  D. Concomitant aortic regurgitation

5.  105. In a patient undergoing aortic valve replacement (AVR) for aortic stenosis, there was evidence of moderate MR on a preoperative transthoracic echocardiogram. After the AVR, the next step to be taken is:

1.  A. Replace the mitral valve

2.  B. Leave the mitral valve alone

3.  C. Assess for MR with intraoperative TEE, and decide if repair or replacement is needed

4.  D. None of the above

6.  106. A patient with old inferior wall myocardial infarction (MI) has severe MR with a posterolaterally directed jet in the left atrium. The most likely cause of MR in this patient is:

1.  A. Flail posterior leaflet

2.  B. Dilated mitral annulus

3.  C. Tented or apically tethered posterior mitral leaflet

4.  D. Tented or apically tethered anterior mitral leaflet

7.  107. Presence of severe aortic regurgitation (AR) in a patient with mitral stenosis is likely to do the following to the calculated mitral valve area by the pressure half-time method:

1.  A. Overestimate the valve area

2.  B. Underestimate the valve area

3.  C. No effect

8.  108. Presence of atrial septal defect (ASD) in a patient with mitral stenosis is likely to do the following to the calculated mitral valve area by the pressure half-time method:

1.  A. Overestimate the valve area

2.  B. Underestimate the valve area

3.  C. No change

9.  109. In a patient with mitral stenosis, the following diastolic flow measurements were obtained: maximal radius of proximal isovelocity surface area (PISA) 0.8 cm at an aliasing velocity of 50 cm/s, inlet angle 120 degrees, peak inflow velocity 2 m/s. The mitral valve area is:

1.  A. 0.7 cm2

2.  B. 1 cm2

3.  C. 1.2 cm2

4.  D. 1.5 cm2

10. 110. A patient with mitral stenosis without any MR or AR has a stroke volume of 70 cc/beat, a transmitral flow integral of 50 cm, and the mitral valve area is:

1.  A. 0.7

2.  B. 1

3.  C. 1.4

4.  D. None of the above

11. 111. A patient with MR has a transaortic flow of 70 cc/beat by the left ventricular outflow tract (LVOT) method and a transmitral flow of 112 cc/beat by the mitral annular method. The time velocity integral (TVI) of the MR signal by continuous wave Doppler is 60 cm. The effective regurgitant orifice (ERO) area of this patient is:

1.  A. 1.5 cm2

2.  B. 0.7 cm2

3.  C. 0.4 cm2

4.  D. 0.2 cm2

12. 112. For a patient with MR and AR, the following measurements were obtained using echo Doppler: flow across the pulmonary valve 75 cc/beat, flow across the mitral valve 120 cc/beat, flow across the aortic valve 90 cc/beat, TVI of MR signal 90 cm, TVI of AR signal 75 cm. The following statement is accurate in this patient:

1.  A. MR ERO is 0.5 cm2 and AR ERO is 0.2 cm2

2.  B. MR ERO is 1.3 cm2 and AR ERO is 1.2 cm2

3.  C. Cannot be calculated

13. 113. In a patient with isolated AR, the following measurements were obtained: transmitral flow 80 cc/beat, flow across aortic valve 140 cc/beat, TVI of AR signal 100 cm. The AR in this patient is:

1.  A. Mild

2.  B. Moderate

3.  C. Severe

4.  D. Cannot be determined

14. 114. A patient with dilated cardiomyopathy has an end diastolic pulmonary regurgitation (PR) velocity of 2 m/s and the estimated right atrial pressure is 10 mmHg. The following statement is true about this patient:

1.  A. Pulmonary artery (PA) pressure is normal

2.  B. Has mild or moderate pulmonary hypertension

3.  C. Has severe pulmonary hypertension

4.  D. Cannot estimate pulmonary pressure

15. 115. If the patient in question 114 had valvular pulmonary stenosis (PS) with a peak gradient of 36 mmHg, the estimated PA end diastolic pressure would be:

1.  A. 16 mmHg

2.  B. 26 mmHg

3.  C. 36 mmHg

4.  D. 62 mmHg

16. 116. If the patient in question 114 has tricuspid stenosis with a mean diastolic gradient of 8 mmHg across the tricuspid valve, the PA diastolic pressure would be:

1.  A. 26 mmHg

2.  B. 34 mmHg

3.  C. 18 mmHg

4.  D. Cannot be estimated

17. 117. In a patient with valvular PS with right PA branch stenosis, the following measurements were obtained: tricuspid regurgitation (TR) velocity 4 m/s, right atrial (RA) pressure 6 mmHg, systolic velocity across the pulmonary valve 2.5 m/s, velocity across the discrete branch stenosis 2.5 m/s. The systolic pressure in the right pulmonary branch distal to the stenosis is likely to be:

1.  A. 20 mmHg

2.  B. 5 mmHg

3.  C. 70 mmHg

4.  D. Cannot be estimated

18. 118. A 20-year-old patient with a large ventricular septal defect (VSD) underwent PA banding in childhood and was lost to follow-up. A recent echocardiogram revealed the following: peak systolic velocity across the VSD 3 m/s, TR velocity 5 m/s, estimated RA pressure 10 mmHg, cuff blood pressure in the right arm 146/70 mmHg, peak flow velocity across the pulmonary band 4.7 m/s. The following statement is true:

1.  A. This patient has normal PA pressure

2.  B. The patient has severe pulmonary hypertension

3.  C. The patient has features of left ventricular (LV) failure

4.  D. PA pressure cannot be determined

19. 119. The patient has an LVOT velocity of 1 m/s, TVI of 25 cm, LVOT diameter of 2 cm, aortic transvalvular velocity of 1.5 m/s, heart rate of 70 beats/min, and the cardiac output in this patient is:

1.  A. 5.5 L

2.  B. 4.5 L

3.  C. 6.3 L

4.  D. Cannot be determined based on the given data

20. 120. A patient with aortic stenosis has an LVOT diameter of 2 cm, LVOT velocity (V1) of 2.5 m/s, transaortic valve velocity (V2) of 5 m/s, and two-dimensional examination showed moderate systolic anterior motion of the mitral leaflet. Valvular aortic stenosis in this patient is:

1.  A. Mild

2.  B. Moderate

3.  C. Severe

4.  D. Cannot be calculated based on the given data

Answers for chapter 6

1.  101. Answer: C.

Jet area underestimates the severity of MR as the driving pressure is low. The regurgitant flow rate is approximately 200 cc/s. Because of low LV systolic pressure of 80 mmHg, the MR velocity would be in the range of 4 m/s (400 cm/s) assuming an LA pressure of 16 mmHg. Hence, the ERO area would be about 200/400 or 0.5 cm2. Hence, in an intraoperative setting, it is important to bring up the blood pressure before performing MR quantitation.

2.  102. Answer: C.

As the driving pressure across the mitral valve is doubled, the regurgitant volume is doubled, as it is directly proportional to the driving pressure. However, jet size is not only dependent on regurgitant volume but also on the kinetic energy imparted to the jet, which depends on jet velocity and, indirectly, the driving pressure, kinetic energy = 1/2 MV2. In this patient, the jet area would be theoretically about four times larger, that is, 16 cm2 as both regurgitant mass and driving pressure have doubled. However, the constraining effect of the left atrium will make it slightly smaller.

3.  103. Answer: D.

Underestimation of MR can occur due to undersampling in the setting of low frame rate (increasing sector angle and depth) and high heart rates. Increasing blood pressure will increase the driving pressure across the mitral valve and hence the jet size will increase.

4.  104. Answer: C.

A noncompliant left atrium causes a greater rise in left atrial pressure for a given regurgitant volume as it occurs in acute MR. Noncompliant left ventricle and concomitant volume overloads such as aortic regurgitation and anemia increase LV diastolic pressure. In patients with mitral stenosis, the presence of MR increases the transvalvular flow and the gradient.

5.  105. Answer: C.

This patient had evidence of moderate MR on the preoperative echocardiogram. In most patients, just replacing the aortic valve causes MR to regress. The MR, if functional, regresses with AVR, but MR due to structural pathology is unlikely to regress. The functional MR is due to high driving pressure and increased LV end systolic size. With the relief of the high driving pressure, it is likely to regress. But after the AVR, the mitral valve should be assessed intraoperatively to decide if there is a need to address MR. There is no clear consensus on decision-making in patients like these.

6.  106. Answer: C.

In a patient with inferior MI, a posterolateral MR jet can occur due to tented posterior leaflet or flail anterior leaflet. In this patient with inferior MI the likely mechanism of MR is posterolateral displacement of papillary muscle causing apical tethering of posterior mitral leaflet, especially P2 and P3 segments. This jet would be posteriorly directed and originates at the medial commissure.

7.  107. Answer: A.

In the presence of severe AR, the pressure half-time is decreased due to a rise in LV diastolic pressure produced by AR. Hence, in the calculation of mitral valve area by the pressure half-time method, the valve area will be overestimated or mitral stenosis severity will be underestimated. Pressure half-time is decreased due to increase in late left ventricular diastolic pressure, causing a reduction in the LA–LV pressure gradient.

8.  108. Answer: A.

ASD will cause left atrial decompression, and hence would result in rapid reduction in the LA–LV diastolic gradient through diastole, decreasing the pressure half-time. This will cause overestimation of the mitral valve area.

9.  109. Answer: A.

The peak inflow rate has to be corrected for the inlet angle as the shape of the PISA is not hemispheric but two-thirds of a hemisphere. Hence, the peak flow rate is given by the formula 2 × 3.14 × r2 × (angle of inlet/180) × aliasing velocity. By this formula, the flow rate divided by the peak inflow velocity in cm/s gives the mitral valve area in cm2. MVA = {2 × 3.14 × (0.82) × (120/180)}/200 = 0.7cm2.

10. 110. Answer: C.

The transmitral flow volume per beat would be the same as the stroke volume, that is, 70 cc/beat. In the absence of MR, the effective diastolic mitral orifice area would be 70 cc/50 = 1.4 cm2.

11. 111. Answer: B.

The effective regurgitant volume is 42 cm3 (112 – 70). The ERO area is the effective regurgitant volume (cm3)/TVI of MR signal (cm). Hence, 42/60 = 0.7 cm2. The ERO area is 0.7 cm2. Please note the units of each of the measurements: in addition, the flow rate is in cm3/s. Paying attention to these is helpful in formulating the various equations. For example the ERO area can also be obtained by dividing the peak regurgitant flow rate obtained by the PISA method (which is in cm3/s) by the peak regurgitant velocity (which is in cm/s) such that the unit of measurement remains in cm2.

12. 112. Answer. A.

The true forward stroke volume is 75 cc/beat, in the absence of pulmonary regurgitation. Mitral regurgitant volume is 120 – 75 = 45 cc and aortic regurgitant volume is 90 – 75 = 15 cc. Dividing the regurgitant volume by their respective TVIs will yield their effective regurgitant orifice area.

13. 113. Answer. C.

Regurgitant volume is 140–80 = 60 cc/beat and regurgitant fraction is 60/140 = 44%. Effective regurgitant orifice area is regurgitant volume/TVI of aortic signal, that is, 60/100 = 0.6 cm2.The regurgitant fraction in AR depends not only on ERO but diastolic period and driving pressure. Hence, the ERO area is a more reliable index of AR volumetric severity. An ERO area of ≥ 0.4 cm2 is indicative of severe AR.

14. 114. Answer: B.

End-diastolic PR velocity of 2 m/s represents a PA–RV end diastolic gradient of 16 mmHg. Assuming an RV end diastolic pressure of 10 mmHg (same as RA pressure), the PA diastolic pressure will be 26 mmHg, which is in the moderate range.

15. 115. Answer: B.

Systolic gradient across the pulmonary valve does not affect the diastolic pressure gradient based on the simplified Bernoulli equation. The only two determinants of PA end diastolic pressure are the PA–RV end diastolic gradient and the RV end diastolic pressure, which is assumed to be equal to the RA pressure.

16. 116. Answer: C.

In this patient, RV end diastolic pressure equals RA pressure – tricuspid stenosis diastolic gradient (10 – 8 = 2 mmHg). The PA end diastolic pressure will be 16 + 2 = 18 mmHg.

17. 117. Answer: A.

In this patient, the estimated right ventricular systolic pressure (RVSP) is 64 + 6 mmHg = 70 mmHg. Pressure drop across the pulmonary valve is equal to 25 mmHg, resulting in a systolic pressure of 45 mmHg in the main PA. As there is 3–4 cm between the pulmonary valve and the right PA, flow streams would have normalized and would allow us to estimate the pressure drop at the branch stenosis without the limitations of stenosis in series unless there is a substantial pressure recovery. As the pressure drop across the branch stenosis is 25 mmHg, estimated systolic pressure distal to the branch stenosis is 45 – 25 or 20 mmHg.

18. 118. Answer: A.

The RV systolic pressure is 110 mmHg based on TR velocity (5 × 5 × 4 + 10 = 110 mmHg). VSD peak velocity is 3 m/s corresponding to an LV–RV pressure gradient of 36 mmHg. Given the systemic systolic pressure of 146 mmHg (hence an LV systolic pressure of 146 mmHg), the VSD gradient is again concordant with an RV systolic pressure of 110 mmHg. In the absence of PS, this is the pressure in the proximal PA. The pressure gradient across the band is 4.7 × 4.7 × 4 = 88 mmHg. Hence, the PA systolic pressure distal to the band is 110 – 88 = 22 mmHg. Although technically this patient has severe elevation of proximal PA pressure, the PA vascular perfusion pressure is normal, indicating the absence of pulmonary arterial disease, making this patient a candidate for surgical closure of VSD.

19. 119. Answer: A.

The stroke volume equals the cross-sectional area × TVI of LVOT, which is 3.14 × 1 × 1 × 25 = 78 cc. Stroke volume multiplied by heart rate, that is, 78 × 70 = 5.5 L/min equals cardiac output.

20. 120. Answer: D.

In a patient with serial stenosis in close proximity, the continuity equation cannot be applied because of difficulty in obtaining precise subvalvular velocity and cross-sectional area of the flow in the LVOT. In a person without systolic anterior motion the cross-sectional area of subvalvular flow is roughly equal to the cross-sectional area of the LV outflow tract. Subvalvular obstruction will result in flow streams such that the cross-sectional area of flow is less than the anatomic LVOT area.