board

24. Which of the following will increase the PRF? A. Reducing depth B. Decreasing transducer frequency C. Reducing sector angle D. Reducing filter

24. Answer: A. Reducing depth reduces time of flight of ultrasound in the body and hence will increase the PRF.

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

69. Answer: A.

11. Increasing depth will change all of the following except: A. Pulse duration B. Pulse repetition period C. Pulse repetition frequency D. Duty factor

11. Answer: A. Pulse duration is the characteristic of the pulse and does not change with depth. Increasing depth will reduce pulse repetition period, frequency and hence the duty factor.

11. Increasing depth will change all of the following except: A. Pulse duration B. Pulse repetition period C. Pulse repetition frequency D. Duty factor

11. Answer: A. Pulse duration is the characteristic of the pulse and does not change with depth. Increasing depthwill reduce pulse repetition period, frequency and hence the duty factor.

12. The two-dimensional images are produced because of this phenomenon when the ultrasound reaches the tissue: A. Refraction B. Backscatter C. Specular reflection D. Transmission

12. Answer: B. Backscatter or diffuse reflection produces most of the clinical images. Specular reflection reaches the transducer only when the incident angle is 908 to the surface, which is not the case in most of the images produced. Refracted and transmitted ultrasounds do not come back to the transducer.

33. The maximum Doppler shift that can be displayed without aliasing with a PRF of 10 kHz is: A. 5 kHz B. 10 kHz C. Depends on depth D. Cannot be determined

33. Answer: A. The Nyquist limit is PRF/2. Hence, a Doppler shift of >5 kHz in this case will cause aliasing. Depth influences the PRF.

34. The PRF is influenced by: A. Transducer frequency B. Depth of imaging C. Both D. Neither

34. Answer: B. The PRF is influenced by pulse duration and time need for ultrasound to travel in tissue. Increasing depth will increase the time spent in the body.

84. What happens to the PRF when imaging depth is increased? A. Increases B. Decreases C. Does not change D. Effect is variable

84. Answer: B. It decreases because of an increase in time of flight. PRF¼77 000/depth in cm.

28. The Nyquist limit can be increased by: A. Increasing the PRF B. Reducing the PRF C. Neither

28. Answer: A. Nyquist limit¼PRF/2.

21. Doppler shift is typically in: A. Ultrasound range B. Infrasound range C. Audible range

21. Answer: C. Doppler shift resulting from moving blood is generally audible. Audible frequency is 20-20 000 Hz.

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

100. 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 251/700¼0.3 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.

9. A sonographer adjusts the ultrasound machine to double the depth of view from 5 cm to 10 cm. If sector angle is reduced to keep the frame rate constant, which of the following has changed? A. Axial resolution B. Temporal resolution C. Lateral resolution D. The wavelength

. 9. Answer: C. Lateral resolution diminishes at depths due to beam divergence. Frame rate determines the temporal resolution. Wavelength is a function of the transducer and is independent of depth and frame rate adjustments

1. The speed of sound in tissues is: A. Roughly 1540 m/s B. Roughly 1540 km/s C. Roughly 1540 cm/s D. Roughly 1540 m/min

1. Answer: A. Roughly 1540 m/s Hence travel time at a depth of 15 cm is roughly 0.1 ms one way (154 000 cm/s or 154 cm/ms or 15 cm per 0.1 ms) or 0.2 ms to and fro. This is independent of transducer frequency and depends only on the medium of transmission.

10. Which of the following properties of a reflected wave is most important in the genesis of a two-dimensional image? A. Amplitude B. Period C. Pulse repetition period D. Pulse duration

10. Answer: A. Amplitude or strength of the reflected beam and its temporal registration, which determines depth registration.

13. Attenuation of ultrasound as it travels to the tissue is increased by: A. Greater depth B. Lower transducer frequency C. Blood rather than soft tissue like muscle D. Bone more than air

13. Answer: A. Attenuation is the loss of ultrasound energy as it travels through the tissue and is caused by absorption and random scatter. It is greater with longer travel path length as it has to go through more tissue. Attenuation is greater at higher frequencies due to smaller wavelength. Attenuation is greatest for air followed by bone, soft tissue and water.

14. The half-intensity depth is a measure of: A. Ultrasound attenuation in tissue B. Half the wall thickness in mm C. Coating on the surface of the transducer D. Half the ultrasound beam width

14. Answer: A. It is a measure of attenuation and reflects the depth at which the ultrasound energy is reduced by half. It is given by the formula: 6 cm/frequency in MHz. For example, for an ultrasound frequency of 3MHz the half-intensity depth is 2 cm, and for 6MHz it is 1 cm.

15. What is the highest pulse repetition frequency (PRF) of a 3MHz pulsed wave transducer imaging at a depth of 7 cm? A. 21 000 Hz B. 2 333 Hz C. 11 000 Hz D. 2.1 million Hz

15. Answer: C. The PRF is independent of transducer frequency and only determined by time of flight, which is the total time taken by ultrasound in the body in both directions. Ultrasound can travel 154 000 cm in a second at a travel speed of 1540 m/s. In other words, at 1 cm depth (2 cm travel distance) the technical limit to the number of pulses that can be sent is 77 000 per second (Hz). Hence the PRF equals 77 000/depth in cm.

16. Examples of continuous wave imaging include: A. Two-dimensional image B. Volumetric scanner-acquired LV image C. Color flow imaging D. Nonimaging Doppler probe (Pedoff)

16. Answer: D. Pedoff is a continuous wave Doppler modality for velocity recording. All other modalities utilize the pulsed wave technique where each of the crystals performs both transmit and receive functions.

17. Which of the following manipulations will increase the frame rate? A. Increase depth B. Increase transmit frequency C. Decrease sector angle D. Increase transmit power

17. Answer: C. Increase in frame rate occurs with reducing sector angle and depth. It is independent of transmit frequency and power.

18. The lateral resolution increases with: A. Decreasing transducer diameter B. Reducing power C. Beam focusing D. Reducing transmit frequency

18. Answer: C. Focusing increases lateral resolution. Increasing transducer diameter and increasing frequency also increase lateral resolution.

19. Axial resolution can be improved by which of the following manipulations? A. Reduce beam diameter B. Beam focusing C. Reduce gain D. Increase transmit frequency

19. Answer: D. Increasing the transmit frequency will reduce the wavelength and hence the spatial pulse length. This will increase the PRF and the axial resolution. Beam diameter and focusing have no effect on axial resolution.

2. The mitral flow measurements in a 62-year-old man are: left ventricular (LV) isovolumic relaxation time (IVRT) 50 ms, E/A ratio 1.5 and E-wave deceleration time 140 ms. This is suggestive of: A. Normal left atrial (LA) pressure B. Abnormal LV relaxation C. High LA pressure D. None of the above

2. Answer: C. High LA pressure. Normal IVRT in adults is 70-100 ms, E/A ratio is about 1 and E-wave deceleration time is 160-250 ms. High LA pressure shortens IVRT and E-wave deceleration time and increases early LV filling. Abnormal LV relaxation has exactly the opposite effect on the mitral flow profile. Very young children may have a pattern mimicking high LA pressure because of superefficient LV relaxation, which promotes early LV filling.

20. Type of sound used in medical imaging is: A. Ultrasound B. Infrasound C. Audible sound

20. Answer: A. Typical frequency is 2-30 MHz: 2-7 MHz for cardiac imaging, 10 MHz for intracardiac echocardiography and 20-30 MHz for intravascular imaging. Ultrasound in the 100-400 MHz range is used for acoustic microscopy. Frequency >20 000 Hz is ultrasound.

22. Duty factor refers to: A. Power the transducer can generate B. Range of frequencies the transducer is capable of C. Physical properties of the damping material D. Fraction of time the transducer is emitting ultrasound

22. Answer: D. It is pulse duration divided by pulse repetition period. Typical value for two-dimensional imaging is 0.1-1% and for Doppler it is 0.5-5%. Example for a 2 MHz transducer: the wavelength in tissue is 0.75mm (period¼0.5 ms); if two periods are in a pulse then pulse duration is 1 ms and if PRF is 1000 Hz (pulse repetition period will be 1 ms or 1000 ms) then the duty factor is 0.1%.

23. Duty factor increases with: A. Increasing gain B. Increasing pulse duration C. Decreasing pulse repetition frequency (PRF) D. Decreasing dynamic range

23. Answer: B. Proportional to pulse duration if the PRP is constant. If pulse duration is constant, decreasing the PRF will reduce the duty factor. Gain and dynamic range have no effect on duty factor.

25. Persistence will have this effect on the image: A. Smoothing of a two-dimensional image B. Better resolution C. Eliminating artifacts D. Spuriously reducing wall thickness

25. Answer: A. Persistence is the process of keeping the prior frames on the display console and will smoothen the image. This reduces random noise and strengthens the signal. However, fast-moving structures can produce artifacts and make the structures look thicker than they are. Some of the other smoothing algorithms include interdigitation and blooming to reduce the spoking appearance produced by the scan lines.

26. Aliasing occurs in this type of imaging: A. Pulsed wave Doppler B. Continuous wave Doppler C. None of the above D. All of the above

26. Answer: A. Aliasing or wrap-around occurs when the Nyquist limit or upper limit of measurable velocity is reached. The Nyquist limit is determined by the PRF. Spectral pulsed wave Doppler and color flow imaging are pulsed wave modalities.

27. The Nyquist limit at a PRF of 1000 Hz is: A. 500 Hz B. 1000 Hz C. 2000 Hz D. Cannot calculate

27. Answer: A. Nyquist limit¼PRF/2.

29. The Nyquist limit can also be increased by: A. Increasing transducer frequency B. Reducing transducer frequency C. Reducing filter D. None of the above

29. Answer: B. Reducing transducer frequency will increase aliasing velocity and reduces range ambiguity. For a given detected Doppler shift, the lower the transducer frequency, the higher is the measured velocity. V in cm/s¼(77Fd in kHz)/Fo in MHz for an incident angle of zero, where Fd is the Doppler shift and Fo is the transmitting frequency.

3. The frame rate increases with: A. Increasing the depth B. Reducing sector angle C. Increasing line density D. Adding color Doppler to B-mode imaging

3. Answer: B. Reducing the sector angle will reduce the time required to complete a frame by reducing the number of scan lines. This increases the temporal resolution. Decreasing depth will increase the frame rate as well. Adding color Doppler will reduce the frame rate.

30. Aliasing can be reduced by: A. Decreasing the depth B. Increasing the PRF C. Reducing the transducer frequency D. Changing to continuous wave Doppler E. All of the above

30. Answer: E. All of the above.

31. What is the purpose of the depth or time gain compensation process adjusted by the echo cardiographer and performed in an ultrasound's receiver? A. Corrects for attenuation and makes the image uniformly bright B. Eliminates image artifacts C. Eliminates aliasing D. None of the above

31. Answer: A. It is postprocessing, which adjusts for loss of ultrasound that occurs at increasing depths.

32. Which of the following increases the Nyquist limit? A. Increasing the depth B. Reducing the sample volume depth C. Increasing the transducer frequency D. None of the above

32. Answer: B. The Nyquist limit is determined by the PRF and PRF¼77 000/depth in cm. Hence decreasing the sample volume depth will increase the PRF, which in turn will increase the Nyquist limit.

35. Two identical structures appear on an ultrasound scan. One is real and the other is an artifact, the artifact being deeper than the real structure. What is this artifact called? A. Shadowing B. Ghosting C. Speed error artifact D. Mirror image

35. Answer: D. Mirror image artifact is a type of artifact where the artifact is always deeper than the real structure and occurs because of the structure or the surface between the two functioning as a mirror.

36. What is determined by the medium through which sound travels? A. Wavelength B. Speed C. Wavelength and speed D. None of the above

36. Answer: C. Speed is determined only by the medium through which sound is traveling. For a given frequency, speed will determine the wavelength: the greater the speed, the longer the wavelength. Period is the time taken for one cycle and is determined by frequency and is independent of the transmission medium.

37. Image quality on an ultrasound scan is dark throughout? What is the first best step to take? A. Increase output power B. Increase receiver gain C. Change to a higher frequency transducer D. Decrease receiver gain

37. Answer: A. The first best action to take is to increase output power. This will brighten the overall image. If the image is still dark, then the receiver gain should be increased.

38. All of the following will improve temporal resolution except: A. Decreasing line density B. Decreasing sector angle C. Increasing frame rate D. Multifocusing

38. Answer: D. Multifocusing will decrease temporal resolution by decreasing the frame rate, whereas all the others will improve temporal resolution by facilitating an increase in the frame rate.

39. Sound travels faster in a medium with which of the following characteristics? A. High density, low stiffness B. Low density, high stiffness C. High density, high stiffness D. Low density, low stiffness

39. Answer: B. Sound travels faster in a medium with low density and high stiffness.

4. The mitral flow measurements in a 1-year-old child are: LV IVRT 50 ms, E/A ratio 2.5 and E-wave deceleration time 120 ms. This is: A. Normal B. Suggestive of abnormal LV relaxation C. Suggestive of high LA pressure D. Is pseudonormal

4. Answer: A. This is normal and results from a very efficient relaxation process, which facilitates early diastolic LV filling. Rapid E-wave deceleration results in physiological S3. And also, as most of the filling occurs in early diastole, children are able to tolerate rapid heart rates and loss of atrial kick without much of a problem. In other words, efficient relaxation mimics high LA pressure in terms of mitral inflow pattern.

40. Which of the following is associated with continuous wave Doppler compared to pulsed wave Doppler? A. Aliasing B. Range specificity C. Ability to record higher velocities D. All of the above

40. Answer: C. Aliasing and range specificity are properties of pulsed wave Doppler. Continuous wave Doppler is associated with range ambiguity. Continuous wave Doppler will also permit recording of higher velocities than pulsed wave Doppler, as it is not limited by the PRF.

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 908 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 08 is 1, cosine of 208 is 0.94 and cosine of 908 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 in question 43.

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.

5. Determination of regurgitant orifice area by the proximal isovelocity surface area (PISA) method is based on: A. Law of conservation of mass B. Law of conservation of energy C. Law of conservation of momentum D. Jet momentum analysis

5. Answer: A. The law of conservation of mass is the basis of the continuity equation. As the flow rate at the PISA surface and the regurgitant orifice are the same, dividing this by the velocity at the regurgitant orifice obtained by continuous wave Doppler gives the effective regurgitant area (regurgitant flow rate in cm3/s divided by flow velocity in cm/s equals effective regurgitant area in cm2).

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

51. 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(1.65)¼4 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

52. 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 ms5¼65 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

53. Answer: C. Attenuation coefficient in soft tissue is equivalent to ½frequency (MHz). In the above question ½3¼1.5 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 10001540¼1 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 above

58. Answer: D. Flow resistance is¼8lengthviscosity/pradius4.

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.

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

62. Answer: 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

63. Answer: 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.

83. The pulse repetition frequency (PRF) is affected by: A. Source of ultrasound B. Transmission medium C. Both D. Neither

83. Answer: A. Source of ultrasound. Speed of ultrasound transmission affects only lengths, not the durations or frequency.

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

70. Answer: A.

71. Calculation of right ventricular systolic pressure from the tricuspid regurgitation velocity signal 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

71. Answer: B.

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 above

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 the Coanda 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.

82. Pulse duration is affected by: A. Source of ultrasound B. Transmission medium C. Both D. Neither

82. Answer: A. Source of ultrasound. Speed of ultrasound transmission affects only lengths, not durations.

85. By increasing the PRF, the axial resolution: A. Increases B. Decreases C. Does not change

85. 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.

86. Imaging at depth affects: A. Axial resolution B. Lateral resolution C. Neither D. Both

86. Answer: B. Lateral resolution drops because of beam divergence and widening.

87. 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.

88. 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.

89. 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.

90. Reflection of sound at an interface is affected by: A. Specific acoustic impedance B. Transducer frequency C. Depth D. None of the above

90. Answer: A.

91. 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

91. 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

92. Answer: B. The valve area can be calculated with the continuity equation. A1V1 (LVOT areaLVOT velocity)¼A2V2 (aortic valve areaaortic velocity). A2¼A1V1/V2. A1¼pr2 (r¼LVOT diameter/2)¼3.1411¼3.14 cm2. A2¼3.140.8/4¼0.6 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 methods

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 2pr2aliasing 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 rate¼2pr2¼23.141140¼251.2 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

98. 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 2pr2aliasing velocity (peak flow rate)/MR velocity. In this patient, peak flow rate¼251 cc/s and ERO is 251/500¼0.5 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

99. 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.