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55 Cards in this Set

  • Front
  • Back
Pulse duration is affected by:
A. Source of ultrasound
B. Transmission medium
C. Both
D. Neither
Answer: A.
Source of ultrasound. Speed of ultrasound transmission affects only lengths, not
durations.
The pulse repetition frequency (PRF) is affected by:
A. Source of ultrasound
B. Transmission medium
C. Both
D. Neither
Answer: A.
Answer: A.
Source of ultrasound. Speed of ultrasound transmission affects only lengths, not the
durations or frequency.
What happens to the PRF when imaging depth is increased?
A. Increases
B. Decreases
C. Does not change
D. Effect is variable
Answer: B.
It decreases because of an increase in time of flight. PRF¼77 000/depth in cm.
By increasing the PRF, the axial resolution:
A. Increases
B. Decreases
C. Does not change
Answer: C.
The PRF does not affect axial resolution. Axial resolution is determined by spatial
pulse length, which is mainly determined by wavelength (i.e. ultrasound frequency)
and number of cycles in the pulse as transmission speed in biological systems is fairly
fixed.
Imaging at depth affects:
A. Axial resolution
B. Lateral resolution
C. Neither
D. Both
Answer: B.
It decreases because of an increase in time of flight. PRF=77 000/depth in cm.
Reducing the transducer footprint will affect:
A. Lateral resolution
B. Temporal resolution
C. Axial resolution
D. None of the above
87. Answer: A.
It will affect beam width and hence the lateral resolution.
Increasing the transmit power will:
A. Decrease sensitivity
B. Increase lateral resolution
C. Increase penetration
D. None of the above
88. Answer: C.
Penetration increases due to more power. The sensitivity increases, but lateral resolution
decreases due to increasing beam width.
Acoustic impedance equals (rayls):
A. Density in kg/m3speed of sound in m/s
B. Density in kg/m3transducer frequency in MHz
C. Depth in meterstransducer frequency in MHz
D. None of the above
89. Answer: A.
Average soft tissue impedance is 1 630 000 rayls.
Reflection of sound at an interface is affected by:
A. Specific acoustic impedance
B. Transducer frequency
C. Depth
D. None of the above
Answer: A.
The most common cause of coronary sinus dilatation is:
A. Heart failure
B. Persistent left superior vena cava
C. Atrial septal defect
D. None of the above
Answer: A.
Heart failure is the common cause of dilatation of the coronary sinus. Although
persistent left superior vena cava (SVC) causes dilatation of the coronary sinus, it occurs
infrequently. In the absence of heart failure persistent left SVC is the most common
cause of enlarged coronary sinus. Dilatation can occur either due to increased flow in
the coronary sinus or due to increased right atrial pressure. The other causes include
coronary aortic valve fistula and unroofing of the coronary sinus, which causes a left to
right shunt, a variant of atrial septal defect.
92. The following data were obtained from a 72-year-old man with a calcified aortic
valve: left ventricular outflow tract (LVOT) velocity (V1) 0.8 m/s, transaortic
velocity (V2) 4 m/s, LVOT diameter 2 cm. The calculated aortic valve area
(AVA) is:
A. 0.4 cm2
B. 0.6 cm2
C. 0.8 cm2
D. 1 cm2
Answer: B.
The valve area can be calculated with the continuity equation.
A1V1 (LVOT area x LVOT velocity)= (A2V2) (aortic valve area x aortic velocity).
A2 = (A1V1)/V2. A1 = 3.14r2 (r=LVOT diameter/2) = 3.14 x1 x1 = 3.14 cm2.
A2=3.14 x 0.8/4 = 0.6 cm2.
93. The continuity equation is an example of:
A. Law of conservation of mass
B. Law of conservation of energy
C. Law of conservation of momentum
D. None of the above
93. Answer: A.
States that mass cannot be destroyed and hence flow rates at different locations in a flow
stream are the same at a given point in time.
94. Themost practical value for the development of perfluorocarbon bubbles was to improve:
A. Contrast on the right side
B. Stable passage through the transpulmonary bed to improve contrast on the left side
C. Improve contrast visualization in the hepatic bed
D. None of the above
94. Answer: B.
The development of perflorocarbon bubbles increased stable passage through the
pulmonary bed, so that contrast visualization was better on the left side.
95. In a patient with mixed aortic valve disease, the AVA by the Gorlin equation using Fick
cardiac output is likely to be:
A. Less than by the continuity equation
B. More than by the continuity equation
C. The same by both methods
95. Answer: A.
The cardiac output by the Fick method is less than the transaortic flow, which is Fick
cardiac outputþregurgitant volume. Hence the calculation of AVA by Gorlin will
underestimate AVA compared to AVA by the continuity equation
96. In a patient with mixed aortic valve disease, the AVA by the Gorlin equation using
angiographic cardiac output is likely to be:
A. Less than by the continuity equation
B. More than by the continuity equation
C. The same by both method
96. Answer: C.
Angiographic and Doppler cardiac output would be equal.
97. The following measurements were obtained from a mitral regurgitant jet:
Radius of proximal isovelocity surface area¼1 cm, aliasing velocity¼40 cm/s. The
peak regurgitant flow rate equals:
A. 251 cc/s
B. 251 cc/min
C. 125 cc/min
D. 125 cc/s
97. Answer: A.
The regurgitant flow rate is calculated by the formula 2 x 3.14r2 x aliasing velocity. This
formula assumes a hemispherical geometry. Hence it is vital to optimize the aliasing
velocity to maximize the hemisphere of the PISA in all dimensions. Using the formula,
peak flow rate = 2 x 3.14r2 = 2 x 3.14 x 1 x 1 x 40 = 251.2 cc/s.
98. In the patient above the systemic blood pressure is 120/80mmHg in the absence of
aortic stenosis and the left atrial pressure is 20 mmHg. The effective mitral regurgitant
orifice area would be:
A. 0.7 cm2
B. 0.5 cm2
C. 1 cm2
D. Cannot be calculated
Answer: B.
The LA–LV pressure gradient is 100 mmHg, which corresponds to a peak mitral regurgitant
velocity of 5m/s or 500 cm/s. TheEROarea is given by the formula 2x 3.14 x r2 x aliasing
velocity (peak flow rate)/MR velocity. In this patient, peak flow rate = 251 cc/s and
ERO is 251/500 = 0.5 cm2.
99. This effective regurgitant orifice (ERO) area of 0.5 cm2 represents:
A. Mild mitral regurgitation (MR)
B. Moderate MR
C. Severe MR
D. Severity cannot be detected
Answer: C.
This patient has severe MR. The ERO is a fairly stable measure of quantitatingMRas it
represents the defect in the mitral valve coaptation mechanism and is independent of
loading conditions. ERO<0.2 is mild, 0.2–0.4 is moderate and 0.4 cm2 is severe
MR.
100. If the patient in question 99 had a blood pressure of 220/90mmHg with similar
proximal isovelocity surface area (PISA) measurements, the ERO area would:
A. Remain unchanged
B. Be more
C. Be less
Answer: C.
Since the blood pressure is now elevated, the LV–LA pressure gradient is 200 mmHg,
giving rise to an MR jet of 7 m/s. The ERO now is 251/700 =0.3 cm2. If the ERO
were unchanged, the peak flow rate would be increased because of higher driving
pressure and the PISA radius would be increased.
61. Volumetric flow rate decreases with an increase in:
A. Pressure difference
B. Vessel radius
C. Vessel length
D. Blood viscosity
E. Vessel length and blood viscosity
61. Answer: E.
Volume flow rate¼pressure differencepdiameter4/128lengthviscosity.
Hence with an increase in length and viscosity the volume flow rate will decrease.
An increase in driving pressure and radius will increase the flow rate.
62. Which of the following on a color Doppler display is represented in real time?
A. Gray-scale anatomy
B. Flow direction
C. Doppler spectrum
D. Gray-scale anatomy and flow direction
E. All of the above
D
63. Approximately how many pulses are required to obtain one line of color Doppler
information?
A. 1
B. 100
C. 10
D. 10 000
C
64. Multiple focus is not used in color Doppler imaging because:
A. It would not improve the image
B. Doppler transducers cannot focus
C. Frame rates would be too low
D. None of the above
64. Answer: C.
Combination of multiple pulses needed for a scan line, multiple focusing and need for
some width for color flow display box will markedly reduce frame rate.
65. Widening the color box on the display will ————— the frame rate.
A. Increase
B. No change
C. Decrease
D. Cannot be determined
65. Answer: C.
By increasing the number of scan lines per box.
66. The simplified Bernoulli equation is inapplicable under the following circumstances:
A. Serial stenotic lesions
B. Long, tubular lesions
C. Both
D. None of the above
66. Answer: C.
For the simplified Bernoulli equation to work, the lesion has to be a discrete stenosis. In
serial lesions, there may be incomplete recovery of pressure and flow area may be
smaller than the anatomic area before the second lesion is encountered. Hence the
pressure gradient at the first orifice estimated by the simplified Bernoulli equation will
be higher than the actual gradient because of the unmeasured kinetic energy between
two orifices. Hence, the total gradient is not the sum of 4V2 at the two orifices. For
long tubular lesions, viscous forces predominate and Poiseulle’s equation would be
applicable to analyze the pressure–flow relationship
67. The Bernoulli equation is an example of:
A. Law of conservation of mass
B. Law of conservation of energy
C. Law of conservation of momentum
D. None of the above
67. Answer: B.
Describes the relationship between different types of energies as potential (pressure),
kinetic (flow) and viscous forces along a flow stream. Energy can be transformed
from one form to the other, but cannot be destroyed or created.
68. The continuity equation is an example of:
A. Law of conservation of mass
B. Law of conservation of energy
C. Law of conservation of momentum
D. None of the above
68. Answer: A.
Says that mass cannot be destroyed and hence flow rates at different locations in a flow
stream are the same at a given point in time.
69. Effective regurgitant orifice area by the proximal isovelocity surface area (PISA)
method is an example of:
A. Law of conservation of mass
B. Law of conservation of energy
C. Law of conservation of momentum
D. None of the above
A
70. Doppler calculation of aortic valve area is an example of:
A. Law of conservation of mass
B. Law of conservation of energy
C. Law of conservation of momentum
D. None of the above
A
72. Color flow jet area of mitral regurgitation depends upon:
A. Amount of regurgitation alone
B. Driving pressure and the regurgitant volume
C. Presence of aortic regurgitation
D. Degree of mitral stenosis
72. Answer: B.
Driving pressure influences the jet area independent of regurgitant volume as jet area is
proportional to the kinetic energy imparted to the jet, which is proportional to the jet,
volume and also the driving pressure. Increase in driving pressure will also increase the
regurgitant volume for a given regurgitant orifice.
73. Factors influencing mitral regurgitation jet volume also include:
A. Proximity of left atrial wall
B. Heart rate
C. Gain setting
D. Filter setting
E. Left atrial size
F. All of the abo
73. Answer: F.
All of these affect the jet size. Compared to the central jet, a wall-hugging jet is about 50%
smaller for a given volume and a nonwall-hugging eccentric jet may be larger due to theCoanda effect where the jet spreads due to pull towards the wall. Lower gains and higher
filter settings reduce jet size. At a faster heart rate, due to reduced jet sampling the jet size
may be underestimated. Free jet (receiving chamber at least five times the jet size) has a
larger size compared to a contained jet entering a smaller chamber.
74. Amount of mitral regurgitation depends upon:
A. Regurgitant orifice size
B. Driving pressure
C. Duration of systole
D. All of the above
74. Answer: D.
Regurgitant volume is directly proportional to the regurgitant orifice size, driving
pressure and the time over which regurgitation occurs
75. Hemodynamic impact of a given volumetric severity of mitral regurgitation (MR) is
increased by:
A. Nondilated left atrium
B. Left ventricular hypertrophy
C. Presence of concomitant aortic regurgitation
D. All of the above
E. None of the above
75. Answer: D.
Noncompliant left atrium as well as left ventricular hypertrophy will increase the hemodynamic
impact of MR. Presence of aortic regurgitation will add another source of
volume load on the left ventricle. Other factors that may have an adverse impact include
anemia, fever and acuteness of onset.
76. Which feature is consistent with severe MR:
A. Jet size to left atrial area ratio of 0.5
B. The PISA radius of 1.2 cm at an aliasing velocity of 50 cm/s
C. Effective regurgitant orifice area of 0.7 cm2
D. All of the above
E. None of the above
76. Answer: D.
All of the above. Correlates of severe MR include MR jet area of 8 cm2, jet to left
atrial area of >0.4, vena contracta diameter of > 7 mm, effective regurgitant orifice
area of >0.4 cm2 or >40 mm2 and systolic flow reversal in the pulmonary veins. It has
to be kept in mind that wall-hugging jets are smaller for a given regurgitant volume and
the effective orifice area may not be constant during systole
77. When using a fixed-focus probe this parameter cannot be changed by the sonographer:
A. Pulse repetition period
B. Pulse repetition frequency
C. Amplitude
D. Wavelength
77. Answer: D.
The wavelength cannot be changed by the sonographer when using a fixed-focus
probe
41. As frequency increases, backscatter strength:
A. Decreases
B. Increases
C. Does not change
D. Refracts
41. Answer: B.
Smaller wavelengths are more readily reflected compared to longer wavelengths.
42. If an echo arrives 39 ms after a pulse has been emitted, at what depth should the
reflecting object be on the scan line?
A. 3 cm
B. 6 cm
C. 1 cm
D. None of the above
42. Answer: A.
Ultrasound takes 6.5 ms to travel 1 cm in the tissues assuming a speed of 1540 m/s.
39 ms is travel time for 6 cm; hence the object is 3 cm deep.
43. The Doppler shift produced by an object moving at a speed of 1 m/s towards the
transducer emitting ultrasound at 2MHz would be:
A. 2.6 kHz
B. 1.3 kHz
C. 1 MHz
D. 200 Hz
43. Answer: A.
Fd¼(2FoVcos of incident angle)/C where Fd is the Doppler shift, V is the velocity and
C is the speed of sound in the medium. In this example, Fd¼(2200000011)/
1540¼2600 Hz or 2.6 kHz. For each MHz of emitted sound, a target velocity of 1 m/s
will produce a Doppler shift of 1.3 kHz.
44. In the above example, the reflected ultrasound will have a frequency of:
A. 2 002 600 Hz
B. 1 998 700 Hz
C. 1 000 000 Hz
D. 2MHz
44. Answer: A.
As the object is moving directly towards the source of sound, the reflected sound will
have a higher frequency and will equal Fo plus Fd.
45. Reflected ultrasound from an object moving away from the sound source will have a
frequency:
A. Higher than original sound
B. Lower than the original sound
C. Same as the original sound
D. Variable, depending on source of sound and velocity of the moving object
45. Answer: B.
Object moving away will produce a negative Doppler shift.
46. Reflected ultrasound from an object moving perpendicular to the sound source will
have a frequency:
A. Higher than original sound
B. Lower than the original sound
C. Same as the original sound
D. Variable, depending on source of sound and velocity of the moving object
46. Answer: C.
As the cosine of the incident angle of 90d is zero, the Doppler shift is zero (please look at
Doppler equation in question 43). Because of the this angle dependence of the Doppler
shift, the angle between the direction of motion of the object and the ultrasound beam
has to be as close to zero as possible to record the true Doppler shift and hence the true
velocity. Cosine of 0d is 1, cosine of 20d is 0.94 and cosine of 90d is 0. Angle correction
is generally not used for intracardiac flows because of the three-dimensional nature of
intracardiac flows and fallacies of assumed angles in contrast to flow in tubular structures
47. Doppler shift frequency is independent of:
A. Operating frequency
B. Doppler angle
C. Propagation speed
D. Amplitude
47. Answer: D.
Please look up the Doppler equation
48. On a continuous wave Doppler display, amplitude is represented by:
A. Brightness of the signal
B. Vertical extent of the signal
C. Width of the signal
D. None of the above
48. Answer: A.
Amplitude is strength of the returning signal. Vertical extent is the velocity of the object
and horizontal axis is the time axis and gives distribution or timing of the signal in the
cardiac cycle
49. Doppler signals from the myocardium, compared to those from the blood pool, display:
A. Lower velocity
B. Greater amplitude
C. Both of the above
D. None of the above
49. Answer: C.
Myocardium produces stronger or higher amplitude signals that have lower velocities
compared to the blood pool.
50. Doing which of the following modifications to the Doppler processing will allow
myocardial velocities to be recorded selectively compared to blood pool velocities?
A. A band pass filter that allows low velocities
B. A band pass filter that allows high amplitude signals
C. Both
D. Neither
50. Answer: C.
In contrast, blood pool signals are higher velocity and lower amplitude.
51. If the propagation speed is 1.6 mm/ms and the pulse round trip time is 5 ms, the distance
to the reflector is:
A. 8mm
B. 4mm
C. 10mm
D. Cannot be determined
Answer: B.
The distance to the reflector is calculated by the range equation. The formula is
½ (propagation speed (mm/ms)round trip time (ms)). So, solving the equation gives
½(1.6 x 5) = 4 mm.
52. How long after a pulse is sent out by a transducer does an echo from an object at a depth
of 5 cm return?
A. 13 ms
B. 65 ms
C. 5 ms
D. Cannot be determined
Answer: B.
The round trip travel time for 1 cm is 13 ms. Hence for an object at 5 cm the travel time
is 13 ms x 5 = 65 ms.
53. For soft tissues, the attenuation coefficient at 3 MHz is:
A. 1 dB/cm
B. 6 dB/cm
C. 1.5 dB/cm
D. 3 dB/cm
Answer: C.
Attenuation coefficient in soft tissue is equivalent to ½ x frequency (MHz). In the
above question ½ x 3 = 1.5 dB/cm. Multiplying this by the path length (cm) yields the
attenuation (dB).
54. If the density of a medium is 1000 kg/m3 and the propagation speed is 1540 m/s, the
impedance is:
A. 1 540 000 rayls
B. 770 000 rayls
C. 3 080 000 rayls
D. Cannot be determined
54. Answer: A.
Impedance describes the relationship between acoustic pressure and the speed of particle
vibrations in a sound wave. It is equal to the density of a mediumpropagation speed.
Solving the equation gives 1000 x 1540 = 1 540 000 rayls. Impedance is increased if the
density of the medium is increased or the propagation speed is increased.
55. If the propagation speed through medium 2 is greater than the propagation speed
through medium 1 the transmission angle will be ——— the incidence angle.
A. Smaller
B. Larger
C. Equal to
D. Cannot be determined
55. Answer: B.
When the propagation speed in medium 2 is greater than medium 1 the transmission
angle will be greater than the incidence angle.
56. If amplitude is doubled, intensity is:
A. Halved
B. Quadrupled
C. Remains the same
D. Tripled
56. Answer: B.
Intensity is the rate at which energy passes through a unit area. Intensity is equal to
amplitude squared. Hence, if amplitude is doubled, intensity is quadrupled.
57. If both power and area are doubled, intensity is:
A. Doubled
B. Unchanged
C. Halved
D. Tripled
57. Answer: B.
Intensity is given by the equation Power (mW)/Area (cm2). Hence if both power and
area are doubled, intensity will remain the same.
58. Flow resistance in a vessel depends on:
A. Vessel length
B. Vessel radius
C. Blood viscosity
D. All of the above
E. None of the abov
58. Answer: D.
Flow resistance is = 8 x length x viscosity/ (3.14 x radius4).
59. Flow resistance decreases with an increase in:
A. Vessel length
B. Vessel radius
C. Blood viscosity
D. None of the above
59. Answer: B.
Flow resistance decreases with an increase in the vessel radius. Please refer to question
58 for the relationship. Resistance to flow and hence flow rate for a given driving
pressure depends upon radius, length and viscosity.
60. Flow resistance depends most strongly on:
A. Vessel length
B. Vessel radius
C. Blood viscosity
D. All of the above
60. Answer: B.
Flow resistance is inversely related to radius4, hence it is most strongly related to the
vessel radius.