Doppler

Question 1

Components of the Doppler equation include all the following EXCEPT:

A. The angle between the ultrasound beam and the direction of blood flow must be known for accurate measurement of blood flow.
B. The transmitted ultrasound frequency is an important determinant of the Doppler shift detected.
C. Propagation speed of sound changes relative to the velocity of the red blood cells.
D. The cosine of 0° is 1 and it is assumed in echocardiography that the recorded velocity has been obtained at a near-parallel intercept angle.

Answer 1

C. The Doppler equation relates the change in frequency of transmitted ultrasound as it is backscattered from moving red blood cells:

FD = 2 x FT x [(V x cos 0) ÷ C]

The equation can be solved for velocity:

V = (FD x C) ÷ (2 x FT x cos 0)

FD, Doppler frequency;
FT, transmitted frequency;
V, velocity of red blood cells;
0, angle between the direction of the moving target and the path of the ultrasound beam;
C, velocity of sound in soft tissue;
2, the constant

Question 2

The difference between the transmitted frequency and the reflected frequency is known as the:

A. Bernoulli equation
B. Doppler principle
C. Doppler shift
D. Gorlin equation

Answer 2

C. With Doppler echocardiography high frequency ultrasound (2 to 5 MHz) is reflected off moving red blood cells. The Doppler shift provides information concerning the direction and velocity of blood flow within the heart and great vessels and is expressed in Hertz (Hz).

Question 3

The top normal peak velocity for the aortic valve is:

A. 0.7 m/s
B. 0.9 m/s
C. 1.1 m/s
D. 1.7 m/s

Answer 3

D. The normal maximal velocities in adults are:

Tricuspid: 0.3 to 0.7 m/s
Pulmonary artery: 0.6 to 0.9 m/s
Mitral: 0.6 to 1.3 m/s
LVOT: 0.7 to 1.1 m/s
Aortic valve: 1.0 to 1.7 m/s

Question 4

Minor degrees of tricuspid regurgitation and mitral regurgitation detected by Doppler in structurally normal hearts:

A. Are a common finding
B. Are a rare finding
C. Depend on respiration
D. Vary greatly from one echocardiography laboratory to another

Answer 4

A. Minor degrees of tricuspid regurgitation that may be found by Doppler have a reported prevalence of up to 90% and up to 50% for mitral regurgitation. This condition appears to depend on the patient’s age and physical conditioning.

The peak velocity of tricuspid regurgitation may be used to determine the right ventricular systolic pressure (RVSP) and the systolic pulmonary artery pressure (SPAP).

Question 5

Pulmonary regurgitation as detected by Doppler in structurally normal hearts is:

A. A rare finding
B. A common finding
C. An abnormal finding
D. Dependent upon expiration

Answer 5

B. Pulmonary regurgitation has been reported in up to 90% of patients with normal valve leaflets and structurally normal hearts.

The pulmonary regurgitation end-diastolic velocity may be used to determine the pulmonary artery end-diastolic pressure (PAEDP). The peak velocity of the pulmonary regurgitation may be used to determine the mean pulmonary artery pressure (MPAP).

Question 6

The laminar core of a turbulent jet is called the:

A. Flow convergence region (PISA)
B. Vena contracta
C. Turbulent region
D. Relaminarization

Answer 6

B. Each turbulent jet has four characteristics. The flow convergence region where blood flow “gathers up” proximal to the orifice. The vena contracta is where blood flow becomes uniform as it passes through the orifice. The turbulent region where blood flow becomes chaotic and the relaminarization zone where laminar flow is restored.

In the echocardiography laboratory, the flow convergence region is called the PISA (proximal isovelocity surface area. The PISA may be used to determine the severity of valvular regurgitation.

The vena contracta may be useful in determining the severity of regurgitation (e.g., mitral regurgitation).

The turbulent region appears as a mosaic with color flow Doppler on and is useful when evaluating the severity of mitral regurgitation and tricuspid regurgitation.

Question 7

Pressure recovery may explain discrepancies between the pressure gradient measurements acquired in the cardiac catheterization laboratory and the pressure gradient measurements acquired in the echocardiography laboratory (e.g., aortic stenosis, prosthetic aortic valve). Pressure recovery occurs at the:

A. Flow convergence region (PISA)
B. Vena contracta
C. Turbulent region
D. Relaminarization zone

Answer 7

D. In abnormal flow the maximum pressure difference occurs at the vena contracta. As blood flow becomes laminar distal to the vena contracta and turbulent region pressure recovers.

Pressure recovery can be an explanation for when cardiac catheterization pressure gradients do not match the pressure gradients determined in the echocardiography laboratory especially when comparing gradients in a patient with aortic stenosis or a prosthetic aortic valve.

Question 8

As a valve orifice narrows because of stenosis pressure proximal to the stenosis will:

A. Increase
B. Decrease
C. Increase with inspiration, decrease with expiration
D. Equilibrate

Answer 8

A. With valvular stenosis pressure in the chamber proximal to the stenosis will increase as will the velocity of blood flow across the stenosis.

Question 9

The equation which relates the pressure drop across an area of narrowing is the:

A. Bernoulli equation
B. Continuity equation
C. Doppler equation
D. Velocity ratio equation

Answer 9

A. When a constant volume of blood flow passes through a stenotic site blood flow is accelerated. The resultant decrease in pressure across the stenotic area is related to velocity of the blood flow on the basis of Bernoulli hydrodynamics (equation).

Question 10

The pressure drop between two-chambers may be calculated by the formula:

A. CSA x VTI
B. 220 ÷ pressure half-time
C. 4 x V22
D. Transmitted frequency - received frequency

CSA, cross-sectional area;
VTI, velocity time integral,
V22, velocity across the obstruction

Answer 10

C. The simplified Bernoulli equation 4 x V22 calculates the pressure gradient between two sites. This simplified equation neglects acceleration of flow, viscous friction and the flow velocity proximal to the peak velocity (V12).

V22, velocity across the obstruction;
V12, velocity proximal to the obstruction

Question 11

The simplified Bernoulli equation disregards all of the following factors EXCEPT:

A. Flow acceleration
B. Proximal velocity
C. Velocity at the site of obstruction
D. Viscous friction

Answer 11

C. At most pressure gradients encountered in clinical medicine the contributions from viscous friction and flow acceleration are negligible.
The contributions from the velocity proximal to an obstruction are also negligible. Therefore most gradients can be derived from the determination of the velocity blood flow by the simplified Bernoulli equation:

P = 4 x V22

P, pressure gradient between two-chambers in millimeters of mercury;
V22, velocity across the obstruction in meters per second

Question 12

When evaluating valvular stenosis all of the following are useful
Doppler parameters EXCEPT:

A. Peak velocity
B. Peak instantaneous pressure gradient
C. Mean pressure gradient
D. Chamber dimensions

Answer 12

D. The Bernoulli equation allows for the calculation of the peak
instantaneous pressure gradient (P = 4 x V22). The Bernoulli equation
may be used to determine mean pressure gradient by measuring the velocity at equally spaced points, squaring each velocity, averaging the velocity values and multiplying the average by 4.

The mean pressure gradient may be the most useful parameter when evaluating valvular stenosis. A mean pressure gradient of > 10 mm Hg indicates significant mitral stenosis. A mean pressure gradient of
> 50 mm Hg indicates severe aortic stenosis.

P, pressure gradient between two-chambers in millimeters of mercury;
V22, peak velocity across the obstruction in meters per second

Question 13

Which of the following represent the lengthened Bernoulli equation?

A. CSA x VTI
B. 4 x V22
C. EDV - ESV
D. 4 x V22 - V12

CSA, cross-sectional area;
V2, velocity across the obstruction; EDV, end-diastolic volume;
ESV, end-systolic volume;
V2, peak velocity proximal to the obstruction

Answer 13

D. Generally the proximal velocity to an obstruction can be ignored and
the simplified Bernoulli equation (P = 4 x V22) can be used to calculate
the peak pressure gradient.

If the proximal velocity is > 1.2 m/s then the lengthened Bernoulli equation should be used:

P= 4 x (V22 - V12)

P, pressure gradient between two-chambers in millimeters of mercury;
V22, peak velocity across the obstruction in meters per second;
V12, peak velocity of flow proximal to the obstruction in meters per second

Question 14

The formula that is used to calculate the peak pressure gradient in coarctation of the aorta is:

A. 4 (V22 - V12)
B. 4 (V22)
C. 220 ÷ PHT
D. CSA x VTI

V12, velocity proximal to the obstruction;
V22, velocity across the obstruction in meters per second;
PHT, pressure half-time;
CSA, cross-sectional area;
VTI, velocity time integral

Answer 14

A. To best calculate the peak pressure gradient in aortic coarctation, the lengthened Bernoulli equation should be used. The lengthened Bernoulli equation calculates the velocity proximal to the obstruction (V12), which may be increased in coarctation.

Question 15

A peak velocity of 2 m/s is obtained in a patient with rheumatic mitral stenosis. The peak (maximum) instantaneous pressure gradient is:

A. 2 mm Hg
B. 4 mm Hg
C. 16 mm Hg
D. 26 mm Hg

Answer 15

C. The lengthend Bernoulli equation is:

P= 4 x (V22 - V12)

In mitral valve stenosis the peak velocity on the atrial side of the valve is so low (< 1.2 m/s) that it can be ignored and the pressure drop can be obtained as follows:

P= 4 x V22

P, pressure gradient between two-chambers in millimeters of mercury;
V22, peak velocity across the obstruction in meters per second,
V12, peak velocity of flow proximal to the obstruction in meters
per second

Question 16

In a patient with aortic stenosis the continuous-wave Doppler recordings demonstrate a maximum peak systolic velocity across the aortic valve of 5 m/s. The peak (maximum) instantaneous pressure gradient is:

A. 5 mm Hg
B. 25 mm Hg
C. 100 mm Hg
D. 110 mm Hg

Answer 16

C. The simplified Bernoulli equation may be applied to velocity measurements to make noninvasive estimates of pressure gradients:

P=4 x V22

For this question: P = 4 × 5^2 = 100
mm Hg

In most cases the peak velocity proximal to the aortic valve will be < 1.2 m/s and therefore the simplified Bernoulli equation may be used.

A peak velocity of 5 m/s and a peak instantaneous pressure gradient of 100 mm Hg in a patient with aortic stenosis and no or only mild aortic regurgitation usually indicate severe aortic stenosis.

Question 17

In patients with aortic valve stenosis the pressure gradients measured by Doppler include:

A. Peak (maximum) instantaneous pressure gradient and peak-to-peak gradient
B. Peak (maximum) instantaneous pressure gradient
C. Peak-to-peak pressure gradient
D. Peak-to-mean gradient

Answer 17

B. Doppler measures the peak (maximum) instantaneous pressure gradient and additionally measures the peak-to-peak pressure gradien and mean pressure gradient. The Doppler peak instantaneous gradient is usually greater than the catheterization laboratories peak-to-peak pressure gradient.

The Doppler mean pressure gradient and cardiac catheterization mean pressure gradient should be equal.

Question 18

With aortic valve stenosis and poor global left ventricular systolic function the severity of aortic stenosis by the Doppler pressure gradient may be:

A. Underestimated
B. Overestimated
C. Unaffected
D. Unpredictable

Answer 18

A. Pressure gradients are flow dependent. If there is reduced cardiac output there may be a low peak systolic velocity, peak pressure gradient and mean pressure gradient even in the presence of severe aortic stenosis.

Question 19

A patient with known aortic stenosis presents for evaluation. The ejection fraction is 22%. The peak velocity across the aortic valve as determined by continuous-wave Doppler is 2.3 m/s. The peak instantaneous pressure gradient is 21 mm Hg. The mean pressure gradient is 14 mm Hg. The severity of the aortic stenosis is:

A. Mild
B. Moderate
C. Severe
D. Requires more information

Answer 19

D. Velocities and pressure gradients are flow dependent. Because the patients ejection fraction is markedly reduced the peak velocity, peak pressure gradient and mean pressure gradient should be interpreted with caution. Calculating the aortic valve area by the continuity equation, improving the global left ventricular systolic function with dobutamine or tracing the aortic valve in the short-axis view of the aortic valve to determine aortic valve area are possible solutions to determining severity.

Question 20

With aortic valve stenosis and significant aortic regurgitation the severity of the aortic stenosis by the Doppler pressure gradient may be:

A. Overestimated
B. Underestimated
C. Unaffected
D. Unpredictable

Answer 20

A. Because pressure gradient is flow dependent it may be very high in patients in whom aortic valve flow is high (e.g., those with coexistent aortic regurgitation) and stenosis is only moderate. Likewise, a low gradient may be noted, despite the presence of severe stenosis, when cardiac output is low.

Question 21

Right ventricular systolic pressure may be calculated when the following condition is present:

A. Aortic regurgitation
B. Mitral regurgitation
C. Pulmonary regurgitation
D. Tricuspid regurgitation

Answer 21

D. Utilizing the simplified Bernoulli equation (P = 4 x V22), right ventricular systolic pressure may be calculated when tricuspid regurgitation, ventricular septal defect or patent ductus arteriosus is present. In the absence of right ventricular outflow tract obstruction (e.g., pulmonary stenosis the right ventricular systolic pressure equals the systolic pulmonary artery pressure.

Question 22

Assuming normal intracardiac pressures, the expected continuous-wave Doppler peak velocity of mitral regurgitation would be:

A. 1 m/s
B. 3 m/s
C. 5 m/s
D. 7 m/s

Answer 22

C. The maximum peak systolic velocity of mitral regurgitation represents the peak pressure difference between the left ventricle and left atrium. Since that difference is usually 100 mm Hg, the peak velocity of mitral regurgitation should be approximately 5 m/s. Due to ranges in blood pressure, the expected normal peak velocity range for the peak velocity of mitral regurgitation is 4 to 6 m/s.

Question 23

The continuous-wave Doppler maximum aortic regurgitation velocity reflects the:

A. Maximum instantaneous systolic pressure gradient between the aorta and left ventricle
B. Maximum peak instantaneous diastolic pressure difference between the aorta and the left ventricle
C. Mean diastolic pressure gradient between the aorta and left ventricle
D. Mean systolic pressure gradient between the aorta and the left ventricle

Answer 23

B. The maximum diastolic regurgitant velocity reflects the maximum peak instantaneous pressure difference between the diastolic pressure of the aorta and the diastolic left ventricular pressure. Because the early diastolic pressure gradient between the aorta and the left ventricle is high (approximately 70 mm Hg) the peak aortic regurgitation velocity will range from 3 to 5 m/s.

Question 24

The expected continuous-wave Doppler peak velocity of tricuspid regurgitation assuming normal intracardiac pressures is:

A. 0.5 m/s
B. 1.0 m/s
C. 2.2 m/s
D. 3.3 m/s

Answer 24

C. The peak systolic pressure difference between the right ventricle and right atrium is approximately 20 mm Hg which would be expressed by a peak systolic velocity of 1.7 m/s. As pulmonary pressures increase, the right ventricular systolic pressure increases, the peak systolic pressure gradient between the right ventricle and right atrium increases which will result in an increase in the peak velocity of tricuspid regurgitation.

Question 25

Assuming normal intracardiac pressures, the expected peak velocity of pulmonary regurgitation is: A. 1 m/s B. 2 m/s C. 3 m/s D. 4 m/s

Answer 25

A. The pressure gradient between the pulmonary artery and right ventricle in diastole is approximately 5 mm Hg. The peak velocity of pulmonary regurgitation is expected to be 1 m/s. As the pulmonary pressures increases the diastolic gradient would increase and therefore the peak velocity of pulmonary regurgitation would increase.

Question 26

Assuming normal intracardiac pressures, the expected peak systolic velocity of a ventricular septal defect would be: A. 5 m/s B. 3 m/s C. 1 m/s D. 0.5 m/s

Answer 26

A. The peak systolic pressure gradient between the left ventricle and right ventricle in a ventricular septal defect would be approximately 100 mm Hg which would result in a peak systolic velocity of 5 m/s. Lower velocities could occur if there is hypotension present, there is an increase in pulmonary and right ventricular systolic pressures or if the ventricular septal defect is large (non-restrictive ventricular septal defect).

Question 27

Assuming normal intracardiac pressures, predict the peak systolic velocity for a patent ductus arteriosus. A. 5 m/s B. 3 m/s C. 1 m/s D. 0.5 m/s

Answer 27

A. Because the peak systolic pressure gradient between the aorta and pulmonary artery is approximately 100 mm Hg, a peak velocity of 5 m/s expected. Reasons for a lower peak systolic velocity include systemic hypotension, pulmonary hypertension and/or a large patent ductus arteriosus.

Question 28

Assuming normal intracardiac pressures, predict the peak velocity of atrial septal defect. A. 5 m/s B. 3 m/s C. 1 m/s D. 0.5 m/s

Answer 28

C. Because the peak pressure difference between the left atrium and right atrium is approximately 5 mm Hg, the expected peak velocity of atrial septal defect would be approximately 1 m/s.

Question 29

The tricuspid regurgitation peak velocity is determined to be 3.2 m/s. The inferior vena cava is normal in dimension (< 1.7 cm) and collapsed with a sniff by more than 50%. The right ventricular systolic pressure and systolic pulmonary artery pressure is: A. 3.3 mm Hg B. 11 mm Hg C. 41 mm Hg D. 46 mm Hg

Answer 29

C. RVSP and SPAP (mm Hg) = 4 x (TR peak velocity2) + Right atrial pressure For this question: RVSP and SPAP (mm Hg) = 4 x 3.2^2 + 5 mm Hg = 41 mm Hg The right atrial pressure can be determined by examining the inferior vena cava using the subcostal window. If the inferior vena cava is normal in dimension (<1.7 cm) and collapses by more than 50% with a sniff then the right atrial pressure is assume to be normal (5 mm Hg). If dilated or does not collapse by 50% with a sniff then the right atrial pressure is assumed to be abnormal and 10 mm Hg would be added instead of 5 mm Hg. RVSP, right ventricular systolic pressure; SPAP, systolic pulmonary artery pressure; TR, tricuspid regurgitation

Question 30

The blood pressure in a patient with a ventricular septal defect is 114/77 mm Hg. The peak velocity across the ventricular septal defect as determined with continuous-wave Doppler is 4 m/s. The right ventricular systolic pressure and systolic pulmonary artery pressure is: A. 114 mm Hg B. 64 mm Hg C. 55 mm Hg D. 50 mm Hg

Answer 30

D. RVSP (mm Hg) = BPs - 4 x (VSD peak velocity2) boold For this question: 114 mm Hg - 4 x 4^2 = 50 mm Hg In the absence of right ventricular outflow tract obstruction the right ventricular systolic pressure is equal to the systolic pulmonary artery pressure. RVSP, right ventricular systolic pressure; BPs, systolic blood pressure

Question 31

The blood pressure in a patient with a patent ductus arteriosus is 124/68 mm Hg. The peak velocity across the patent ductus arteriosus as determined by continuous-wave Doppler is 5 m/s. The systolic pulmonary artery pressure is: A. 124 mm Hg B. 100 mm Hg C. 34 mm Hg D. 24 mm Hg

Answer 31

D. SPAP (mm Hg) = BP, - (PDA peak velocity2) For this question: SPAP (mm Hg) = 124 - 4 x 5^2 = 24 mm Hg In the absence of right ventricular outflow tract obstruction the systolic pulmonary artery pressure is equal to the right ventricular systolic pressure. SPAP, systolic pulmonary artery pressure; BP, systolic blood pressure; PDA, patent ductus arteriosus

Question 32

The pulmonary regurgitation is determined to be 2.0 m/s. The inferior vena cava is normal in dimension (< 1.7 cm) and collapses with a sniff by greater than 50%. The pulmonary artery end-diastolic pressure is equal to: A. 2 mm Hg B. 7 mm Hg C. 16 mm Hg D. 21 mm Hg

Answer 32

D. PAEDP (mm Hg) = 4 x (PR end-diastolic velocity2) + Right atrial pressure For this question: PAEDP (mm Hg) = 4 x 2^2 + 5 mm Hg = 21 mm Hg If the inferior vena cava is normal in dimension (< 1.7 cm) and collapses with a sniff by more than 50% as seen in the subcostal window the right atrial pressure is assumed to be normal (5 mm Hg). If the inferior vena cava is dilated or does not respond to a sniff by changing its diameter by > 50% the right atrial pressure is assumed to be abnormal and at least 10 mm Hg should be added. The normal PAEDP is 4 to 12 mm Hg. PAEDP, pulmonary artery end-diastolic pressure; PR, pulmonary regurgitation

Question 33

The peak velocity of pulmonary regurgitation is determined to be 3 m/s. The mean pulmonary artery pressure is: A. 44 mm Hg B. 36 mm Hg C. 9 mm hg D. 3 mm Hg

Answer 33

B. MPAP (mm Hg) = 4 x (PR peak velocity2) For this question: MPAP (mm Hg) = 4 x 3^2 = 36 mm Hg The normal MPAP range is 9 to 18 mm Hg. MPAP, mean pulmonary artery pressure; PR, pulmonary regurgitation

Question 34

The tricuspid regurgitation peak velocity is 3.0 m/s. The right ventricular outflow tract velocity time integral is 20 cm. The pulmonary vascular resistance is: A. Normal B. Increased C. Decreased D. Equal to the peak velocity of the tricuspid regurgitation

Answer 34

A. A simple ratio TRV(m/s)/VTIrvot (cm) > 2 suggests increased pulmonary vascular resistance (PVR). For this question: 3 m/s ÷ 20 cm = .15 TRV, tricuspid regurgitation peak velocity; VTIrvot, velocity time integral of the right ventricular outflow

Question 35

The blood pressure is 120/80 mm Hg. The peak velocity of mitral regurgitation is 5 m/s. The left atrial pressure is: A. 120 mm Hg B. 100 mm Hg C. 20 mm Hg D. 5 mm Hg

Answer 35

C. Left atrial pressure = Systolic blood pressure - 4 x (MR peak velocity2) For this question: 120 mm Hg - 4 x 5^2 = 20 mm Hg MR, mitral regurgitation

Question 36

The formula used to estimate left ventricular end-diastolic pressure (LEDP) from continuous-wave Doppler recording of aortic regurgitation is LVEDP is equal to: A. BPs - Vmax AR B. BPd - Vmax AR C. BPd - 4 x EDV AR D. BPd - 4 x EDV AR2 LVEDP, left ventricular end-diastolic pressure; BP, systolic blood pressure; Vmax, maximum velocity of aortic regurgitation; AR, aortic regurgitation; BPd, diastolic blood pressure; EDV, end-diastolic velocity.

Answer 36

D. LVEDP (mm Hg) = 4 x (EDV AR2). For example if the diastolic blood pressure is 70 mm Hg and the end diastolic velocity of aortic regurgitation is 3 m/s, the LVEDP would be 70 minus 36 mm Hg or 34 mm Hg. LVEDP, left ventricular end-diastolic pressure; BPd, diastolic blood pressure; EDV, end-diastolic velocity; AR, aortic regurgitation

Question 37

The peak velocity across a patent foramen ovale (PFO) is determined to be 1.0 m/s. The right atrial pressure (RAP) is determined to be 5 mm Hg by examination of the characteristics of the inferior vena cava. The left atrial pressure (LAP) is equal to: A. 1 mm Hg B. 4 mm Hg C. 9 mm Hg D. 14 mm Hg

Answer 37

C. LAP (mm Hg) = 4 x (PFO peak velocity2) + RAP For this question: 4 x 1^2 + 5 mm Hg = 9 mm Hg LAP, left atrial pressure; PFO, patent foramen ovale; RAP, right atrial pressure

Question 38

Formulas that may be used to calculate the cross-sectional area of an orifice or vessel through which blood is flowing include all the following EXCEPT: A. 2 x π x r2 B. π x (D ÷ 2)^2 C. 0.785 x D2 D. 1 x (D2 ÷ 4) r, radius; D, diameter

Answer 38

A. To determine blood flow volume using the Doppler technique, the cross-sectional area of the orifice or vessel must be determined. The formula for determining flow volume by Doppler is: V = CSA x VTI For example if the left ventricular outflow tract diameter is 2.0 cm, the cross-sectional area is equal to 3.14 cm2 (.785 x 2^2). To calculate the area of a hemiellipse, use the formula: 2 x π x r V, volume; CSA, cross-sectional area; VTI, velocity time integral; r, radius

Question 39

The left ventricular outflow tract diameter in early ventricular systole as measured in the parasternal long-axis is 2.0 cm. The left ventricular outflow tract time velocity integral is 20 cm. The Doppler stroke volume is: A 2 mL B. 20 mL C. 3.14 cm D. 63 mL

Answer 39

D. Stroke volume is cross-sectional area (CSA cm2) x velocity time integral (VTI cm). To determine cross sectional area use the formula .785 x diameter2. The velocity time integral can be derived by tracing the left ventricular pulsed-wave Doppler of the left ventricular outflow tract. For this question: .785 x 22 x 20 cm = 63 mL The normal stroke volume is 70 to 100 ml.

Question 40

The stroke volume is 63 mL. The heart rate is 100 beats per minutes. The cardiac output is: A. 63 mL B. 6.3 Lpm C. 63000 Lpm D. 6.3 bpm

Answer 40

B. Cardiac output is stroke volume (ml) x heart rate (bpm). The normal cardiac output ranges from 4 to 6 Lpm. For this question: 63 mL x 100 bpm = 6.3 Lpm

Question 41

The use of the continuity equation in patients with aortic stenosis is based on the premise that: A. As the aortic stenosis progresses, V1 increases B. As the aortic stenosis progresses, V2 decreases C. Left ventricular outflow tract flow is greater than flow across the aortic valve D. Flow volume in the left ventricular outflow tract equals the flow volume across the aortic valve

Answer 41

D. Flow across a stenotic or regurgitant orifice is equal to the proximal flow. In aortic stenosis flow across the left ventricular outflow tract is equal to the flow across the aortic valve. This can be written for aortic stenosis as: CSAlvot * VTIlvot = CSAaov * VTIaov To determine aortic valve area the continuity equation is rearranged: AVA (cm2) = (CSAlvot * VTIlvot) ÷ VTIaov

Question 42

The following data is obtained in a patient with aortic stenosis: left ventricular outflow tract diameter is 2.0 cm, peak left ventricular outflow tract velocity integral is 20 cm, the aortic valve time velocity integral is 40 cm. The aortic valve area is: A. 0.3 cm2 B. 0.75 cm2 С. 1.57 сm2 D. 3.14 cm2

Answer 42

C. The aortic valve area may be calculated by the continuity equation: AVA (cm2) = (CSAlvot * VTlvot) ÷ VTIaov (cm) The maximum peak velocity of the LVOT and aortic valve may be substituted for VTI. For this question: (.785 x 2^2 × 20 cm) ÷ 40 cm = 1.57 cm2 AVA, aortic valve area; CSA, cross-sectional area; LVOT, left ventricular outflow tract; VTI, velocity time integral

Question 43

The following data is obtained: left ventricular outflow tract diameter is 2.2 cm, left ventricular outflow tract peak systolic velocity is 1.1 m/s and the peak systolic aortic valve velocity is 5 m/s. The aortic valve area is: A. 0.75 cm2 B. 0.83 cm2 C. 3.14 cm2 D. 100 cm2

Answer 43

B. To determine aortic valve area by echocardiography and cardiac Doppler the continuity equation is used: AVA (cm2) = (CSAlvot * LVOT peak velocity) ÷ Aortic valve peak velocity VTI may be substituted for the peak velocity values. For this question: (.785 x 2.2^2 x 1.1 m/s) ÷ 5 m/s = .83 cm2 AVA, aortic valve area; CSA, cross-sectional area; LVOT, left ventricular outflow tract

Question 44

The following data is obtained in a patient with aortic stenosis: left ventricular outflow tract velocity time integral is 20 cm and the aortic valve velocity time integral is 40 cm. The velocity ratio is: A. 20 B. 40 C. 0.5 D. 800

Answer 44

C. The aortic velocity ratio is determined by the formula: LVOTvtu ÷ Aortic Valvevti For this question: 20 cm ÷ 40 cm = 0.5 A velocity ratio of < 0.25 suggests significant aortic stenosis. The velocity ratio is useful when the left ventricular outflow tract measurement is not possible. The velocity ratio is useful when following patients with aortic valve replacement.

Question 45

The following data is obtained in a patient with a prosthetic mitral valve: left ventricular outflow tract diameter is 2.0 cm, the left ventricular outflow tract velocity time integral is 15 cm and the prosthetic mitral valve velocity time integral is 47 cm. The mitral valve area by the continuity equation is: A. 3.14 cm2 B. 2.0cm2 C. 1.0 cm2 D. 30 cm2

Answer 45

C. The mitral valve area can be determined by the continuity equation: MVA (cm2) = (CSAlvot * VTIlvot) ÷ VTImv For this question: (.785 x 2^2 x 15 cm) ÷ 47 cm = 1 cm2 The continuity equation is recommended when determining the effective orifice area in a mitral valve prosthesis. MVA, mitral valve area; CSA, cross-sectional area; VTI, velocity time integral

Question 46

Determine the mitral effective regurgitant orifice and regurgitant volume using the PISA method: Radius: 1.0 cm Aliasing velocity: 40 cm/s Mitral regurgitation peak velocity: 500 cm/s Mitral regurgitation velocity time integral: 110 cm A. 1 cm2; 50 mL B. 0.50 cm2; 55 mL C. 55 cm2; 50 mL D. 0.40 cm2; 110 mL

Answer 46

B. ERO cm2 =(6.28 x radius2 x aliasing velocity) ÷ MR peak velocity ERO = (6.28 x 1.0^2 x 40 cm/s) ÷ 500 cm/s = 0.50 cm2 RV (ml) = ERO x MRvti 0.50 (cm^2) x 110 cm = 55 mL ERO, effective regurgitant orifice; MR, mitral regurgitation; MRvti, mitral regurgitation velocity time integral

Question 47

All of the following are simplified PISA methods for determining the severity of mitral regurgitation EXCEPT: A. 220 ÷ pressure half-time B. ERO (cm2) = r2 ÷ 2 C. RV (mL) = 2 x r2 x aliasing velocity (cm/s) D. ≥ 0.9 cm PISA radius that is holosystolic indicates significant mitral regurgitation PISA, proximal isovelocity surface area; ERO, effective regurgitant orifice; r, radius of PISA; RV, regurgitant volume

Answer 47

A. Another modified PISA method is: RV (mL) = 6.28 x r2 x aliasing velocity (cm/s) ÷ 3.25 RV, regurgitant volume

Question 48

Determine the mitral regurgitant volume, regurgitant fraction and effective regurgitant orifice using the following information: LVOT diameter: 2.0 cm, LVOTvti: 10 cm, Mitral valve annulus diameter: 3.0 cm, Mitral valve annulus VTI: 15 cm, Mitral regurgitation VTI: 200 cm A. 2 mL; 100%; 2 cm2 B. 74 mL; .70%; 37 cm2 C. 200 mL; 50%;.75 cm2 D. 34 mL; 17%; .17 cm2 LVOT, left ventricular outflow tract; VTI, velocity time integral

Answer 48

B. RV (ml) = .785 x MV diameter2 x MVvti - .785 x LVOT diameter2 & LVOTvti .785 x 3^2 x 15 - .785 x 2^2 x 10 = 74 mL RF (%) = MVrv ÷ MRsv 74 mL ÷ 105 mL = 70% ERO (cm^) = MRrv ÷ MRvti 74 ml ÷ 200 cm = .37 cm2 RV, regurgitant volume; MV, Mitral valve annulus diameter diastole; MVvti, mitral annulus velocity time integral; LVOT, left ventricular outflow tract; LVOIvti, left ventricular outflow tract integral; RF, Regurgitant fraction; MVrv, mitral regurgitation regurgitant volume; MRsv, mitral regurgitation stroke volume; ERO, effective regurgitant orifice; MR, mitral regurgitation velocity time integral

Question 49

Determine the Qp/Qs for an atrial septal defect with the following data: RVOTd = 3.0 cm; RVOTvti = 20 cm; LVOTd = 2.0 cm; LVOTvti = 10 cm A. 2:1 B. 3.3:1 C. 4.5:1 D. 10:1 RVOT, right ventricular outflow tract diameter; RVOTvti, right ventricular outflow tract velocity time integral; LVOTd, left ventricular outflow tract diameter; LVOTvti, left ventricular outflow tract velocity time integral

Answer 49

C. Qp = .785 x RVOTd2 x RVOTvti .785 x 3^2 x 20 cm = 140 mL Qs = .785 x LVOTd2 x LVOTvti .785 x 2^2 x 10 cm = 31 mL Qp/Qs = 140 mL ÷ 31 mL = 4.5 Normal Qp/Qs = 1:1 Significant shunt = > 2:1

Question 50

The mitral valve area can be determined by Doppler with the following formula: A. 220 ÷ pressure half-time B. 220 ÷ deceleration time C. Deceleration time ÷ pressure half-time D. Pressure half-time ÷ 220

Answer 50

A. The formula: 220 ÷ pressure half-time provides an accurate method for determining the mitral valve area in patients with mitral stenosis. Limitations and pitfalls include abnormal myocardial relaxation, significant aortic regurgitation, post-balloon mitral valvuloplasty and rapid heart rates.

Question 51

The pulsed-wave Doppler mitral valve peak E wave velocity is 100 cm/s. The lateral wall mitral annulus tissue Doppler imaging E' wave is 5 cm/s. The diastolic filling pressure is assumed to be: A. Normal B. Increased C. Decreased D. Dependent upon respiration

Answer 51

B. The ratio E/E' may be used to determine filling pressures. In general, if the ratio is < 8, the filling pressures are assumed to be normal. If the ratio is > 15, the filling pressures are assumed to be increased. A recent proposal suggests that an E/E' ratio of > 10 when using the lateral annulus indicates increased filling pressures. An E/E' ratio of >15 obtained from the medial (septal) annulus would indicate increased filling pressures.

Question 52

Predict the tissue Doppler imaging E/E' ratio in a patient with known pseudonormalization of the mitral valve inflow pattern. A. Normal E'/A' ratio B. Increased E'/A' ratio C. Decreased E/A' ratio D. Dependent upon respiration

Answer 52

C. In patients with pseudonormalization of the mitral inflow pattern (Grade II, the tissue Doppler imaging of the mitral annulus will demonstrate E'/A' ratio reversal (an E'/A' ratio of < 1). In addition, the E/E' ratio will be ≥ 10 from the lateral annulus.

Question 53

The S' wave of the mitral valve annulus is determined to be 3 cm/s in peak velocity. This suggests: A. Normal global left ventricular systolic function B. Reduced global left ventricular systolic function C. Hyperdynamic global left ventricular systolic function D. Dependent upon respiration

Answer 53

B. An S' peak velocity of the mitral annulus should be approximately 9 cm/s. A S' peak velocity of < 4 cm/s suggests reduced global left ventricular systolic function.