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RDCS Pediatric practice questions

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Question 1Ultrasound Physics, Instrumentation, and Bioeffects/Safety

In soft tissue, how does an increase in ultrasound frequency affect the wavelength?

  1. The wavelength increases.
  2. The wavelength decreases. Correct answer.
  3. The wavelength becomes independent of propagation speed.
  4. The wavelength remains unchanged.

Explanation

Wavelength equals propagation speed divided by frequency. In a given medium the speed is nearly constant, so raising frequency shortens the wavelength.

Question 2Ultrasound Physics, Instrumentation, and Bioeffects/Safety

What is the approximate average propagation speed of sound in soft tissue?

  1. 4080 m/s
  2. 1540 m/s Correct answer.
  3. 330 m/s
  4. 154 m/s

Explanation

Soft tissue averages about 1540 m/s, which is the calibration value used by most clinical scanners.

Question 3Ultrasound Physics, Instrumentation, and Bioeffects/Safety

The piezoelectric effect in an ultrasound transducer refers to the crystal's ability to:

  1. focus the ultrasound beam electronically.
  2. amplify returning echoes before processing.
  3. convert electrical and acoustic energy. Correct answer.
  4. store electrical charge indefinitely.

Explanation

The piezoelectric crystal deforms when voltage is applied and generates voltage when deformed, allowing both transmission and reception.

Question 4Ultrasound Physics, Instrumentation, and Bioeffects/Safety

Attenuation of an ultrasound beam in tissue is best described as:

  1. a progressive increase in beam intensity as depth increases
  2. the production of harmonic frequencies at higher intensities
  3. a decrease in amplitude and intensity with depth Correct answer.
  4. a change in beam direction caused by an interface

Explanation

Attenuation is the combined loss of amplitude and intensity from absorption, reflection, and scattering as sound travels through tissue.

Question 5Ultrasound Physics, Instrumentation, and Bioeffects/Safety

The ALARA principle in pediatric echocardiography means that ultrasound exposure should be:

  1. applied only during continuous wave Doppler.
  2. kept as low as reasonably achievable. Correct answer.
  3. maximized to improve image quality.
  4. limited to adults because children are less sensitive.

Explanation

ALARA stands for as low as reasonably achievable and applies to all patients, especially children, because they may be more sensitive to ultrasound energy.

Question 6Ultrasound Physics, Instrumentation, and Bioeffects/Safety

Which control adjusts the overall brightness of all echoes on the display?

  1. Overall gain Correct answer.
  2. Time gain compensation
  3. Depth
  4. Focus

Explanation

Overall gain amplifies all received echoes equally across the image. Time gain compensation instead compensates selectively for depth-dependent attenuation.

Question 7Ultrasound Physics, Instrumentation, and Bioeffects/Safety

Continuous wave Doppler can measure high velocities without aliasing because it:

  1. automatically shifts the baseline during acquisition.
  2. transmits and receives continuously with separate crystals. Correct answer.
  3. uses one crystal that alternates transmission and reception.
  4. employs a lower pulse repetition frequency.

Explanation

Continuous wave Doppler uses separate transmitting and receiving elements and does not gate returning echoes by time, so there is no Nyquist limit to exceed.

Question 8Ultrasound Physics, Instrumentation, and Bioeffects/Safety

Which change will increase the frame rate on a two-dimensional echocardiogram?

  1. Increasing the number of scan lines
  2. Increasing the depth of the sector
  3. Adding color Doppler to the image
  4. Decreasing the depth of the sector Correct answer.

Explanation

Frame rate is roughly the pulse repetition frequency divided by the number of scan lines per frame. Reducing depth shortens the round-trip time per line and allows more frames per second.

Question 9Ultrasound Physics, Instrumentation, and Bioeffects/Safety

Axial resolution is determined primarily by which factor?

  1. Doppler angle
  2. Beam width at the focus
  3. Dynamic range
  4. Spatial pulse length Correct answer.

Explanation

Axial resolution equals spatial pulse length divided by two. Shorter pulses, achieved with higher frequency and damping, improve it.

Question 10Ultrasound Physics, Instrumentation, and Bioeffects/Safety

Lateral resolution is determined primarily by which factor?

  1. Spatial pulse length
  2. Beam width Correct answer.
  3. Dynamic range
  4. Pulse repetition frequency

Explanation

Lateral resolution is determined by beam width at the depth of interest; a narrower beam gives better lateral resolution.

Question 11Ultrasound Physics, Instrumentation, and Bioeffects/Safety

The Nyquist limit equals:

  1. one-half the pulse repetition frequency. Correct answer.
  2. twice the pulse repetition frequency.
  3. Doppler shift multiplied by the cosine of the angle.
  4. propagation speed divided by frequency.

Explanation

The Nyquist limit is half the pulse repetition frequency. Doppler shifts above this limit alias.

Question 12Ultrasound Physics, Instrumentation, and Bioeffects/Safety

Which adjustment can help resolve aliasing in pulsed wave Doppler?

  1. Decreasing the velocity scale
  2. Increasing the velocity scale Correct answer.
  3. Increasing the sample volume depth
  4. Increasing the transducer frequency

Explanation

Increasing the velocity scale raises the pulse repetition frequency and therefore the Nyquist limit, reducing aliasing. Lower transducer frequency or continuous wave Doppler can also help.

Question 13Ultrasound Physics, Instrumentation, and Bioeffects/Safety

Tissue harmonic imaging primarily improves image quality by:

  1. using higher-frequency harmonics to reduce side lobe artifact. Correct answer.
  2. reducing frame rate to improve axial resolution.
  3. increasing the transmitted fundamental frequency to its maximum.
  4. eliminating the need for time gain compensation.

Explanation

Harmonic imaging listens at twice the transmitted frequency, where the beam is narrower and side lobe clutter is weaker, improving contrast and lateral resolution.

Question 14Ultrasound Physics, Instrumentation, and Bioeffects/Safety

The mechanical index (MI) is most closely related to which potential bioeffect?

  1. Cavitation from peak negative pressure Correct answer.
  2. Aliasing in spectral Doppler
  3. Reverberation artifact
  4. Thermal injury from prolonged exposure

Explanation

The mechanical index estimates nonthermal cavitation risk from peak negative pressure and frequency. It is distinct from the thermal index.

Question 15Ultrasound Physics, Instrumentation, and Bioeffects/Safety

The thermal index (TI) provides an estimate of:

  1. the potential for temperature rise in tissue Correct answer.
  2. the risk of cavitation occurring at the focal zone
  3. the axial resolution achieved by the transducer
  4. the amount of aliasing present in color Doppler

Explanation

The thermal index estimates potential tissue heating from absorbed acoustic energy. TIS, TIB, and TIC apply to soft tissue, bone, and cranial bone respectively.

Question 16Ultrasound Physics, Instrumentation, and Bioeffects/Safety

For accurate velocity measurement by Doppler, the angle between the beam and blood flow should ideally be:

  1. fixed at 45 degrees
  2. between 60 and 90 degrees
  3. near 0 degrees Correct answer.
  4. exactly 90 degrees

Explanation

Doppler shift is proportional to the cosine of the angle; at 90 degrees the shift is zero, and accuracy improves as the angle approaches 0 degrees.

Question 17Ultrasound Physics, Instrumentation, and Bioeffects/Safety

A strong specular reflection occurs at an interface when there is:

  1. a Doppler angle near 90 degrees
  2. a large impedance mismatch Correct answer.
  3. no acoustic impedance difference
  4. a low pulse repetition frequency

Explanation

Specular reflection is strong when acoustic impedance differs markedly between media; a mismatch reflects a large fraction of the beam.

Question 18Ultrasound Physics, Instrumentation, and Bioeffects/Safety

Reverberation artifact appears as:

  1. lateral displacement of a strong reflector
  2. equally spaced echoes at increasing depth Correct answer.
  3. a shadow behind a highly attenuating structure
  4. bright enhancement behind a fluid-filled structure

Explanation

Reverberation occurs when sound bounces between two strong reflectors, creating repeated echoes at regular depth intervals.

Question 19Ultrasound Physics, Instrumentation, and Bioeffects/Safety

Range ambiguity in pulsed wave Doppler occurs when:

  1. echoes from an earlier pulse arrive late Correct answer.
  2. the wall filter is set too high
  3. the sample volume is set too small
  4. the Doppler angle is greater than 60 degrees

Explanation

If the pulse repetition frequency is too high for the depth, echoes from one pulse can return after the next pulse is sent and be assigned to the wrong gate, causing range ambiguity.

Question 20Ultrasound Physics, Instrumentation, and Bioeffects/Safety

Reducing the dynamic range of the ultrasound system will:

  1. decrease the frame rate.
  2. increase the number of gray shades displayed.
  3. increase the Nyquist limit.
  4. make the image more black and white. Correct answer.

Explanation

Dynamic range controls the range of echo strengths displayed as gray shades. Reducing it compresses weak and strong signals into fewer shades, increasing contrast.

Question 21Ultrasound Physics, Instrumentation, and Bioeffects/Safety

The mechanical index is calculated by dividing peak negative pressure (in MPa) by:

  1. the square root of frequency in MHz Correct answer.
  2. two times the propagation speed
  3. the center frequency squared in MHz
  4. the pulse repetition frequency in Hz

Explanation

MI equals peak negative pressure in MPa divided by the square root of center frequency in MHz. Higher pressure or lower frequency raises MI and cavitation risk.

Question 22Ultrasound Physics, Instrumentation, and Bioeffects/Safety

In pediatric echocardiography, which thermal index is most relevant when bone is at or near the focal zone?

  1. MI
  2. TIB Correct answer.
  3. TIC
  4. TIS

Explanation

TIB estimates heating when bone lies in or near the beam focus and is the relevant thermal index for transthoracic pediatric imaging. TIS applies to soft tissue and TIC to transcranial applications.

Question 23Ultrasound Physics, Instrumentation, and Bioeffects/Safety

Side lobes and grating lobes are most likely to produce which artifact?

  1. reduced frame rate from increased line density
  2. aliasing in spectral Doppler
  3. loss of axial resolution from a long pulse
  4. off-axis echoes displayed within the image Correct answer.

Explanation

Side lobes and grating lobes are off-axis beams that can receive echoes from structures outside the main beam, displaying them as clutter within the image.

Question 24Ultrasound Physics, Instrumentation, and Bioeffects/Safety

In color Doppler, aliasing is displayed as:

  1. absence of color signal at the vessel wall
  2. wrap-around to the opposite end of the map Correct answer.
  3. uniformly increased color brightness with depth
  4. a change from red to blue only with flow reversal

Explanation

When the Doppler shift exceeds the Nyquist limit, the color map wraps and high velocities are shown with the opposite color, often bright red to blue or vice versa.

Question 25Ultrasound Physics, Instrumentation, and Bioeffects/Safety

Increasing the color Doppler packet size (ensemble length) generally:

  1. reduces the thermal index to zero during color imaging
  2. eliminates aliasing from all vessels in the image
  3. lowers frame rate but improves velocity accuracy Correct answer.
  4. raises frame rate without reducing velocity accuracy

Explanation

A larger packet uses more pulses per color line, improving velocity estimation but taking more time and lowering frame rate.

Question 26Ultrasound Physics, Instrumentation, and Bioeffects/Safety

When an ultrasound system converts echo return times into distances for display, it assumes an average propagation speed in soft tissue of approximately:

  1. 1,540 m/s Correct answer.
  2. 770 m/s
  3. 4,000 m/s
  4. 3,080 m/s

Explanation

Ultrasound machines assume a constant soft-tissue propagation speed of about 1540 m/s when converting round-trip time into depth. Structures whose true speed differs from this assumption are displayed at a slightly incorrect depth or position. The 1540 m/s value is an average across soft tissues, not a measured value for each patient.

Question 27Ultrasound Physics, Instrumentation, and Bioeffects/Safety

The active element of a diagnostic ultrasound transducer, which converts electrical energy into acoustic energy and back again, is composed of:

  1. Silicone elastomer compound
  2. Tungsten carbide alloy
  3. Lead zirconate titanate Correct answer.
  4. Barium sulfate suspension

Explanation

The active element is a piezoelectric ceramic such as lead zirconate titanate (PZT). An applied voltage changes its thickness and generates a pressure wave, and returning echoes deform it to produce a voltage. Materials that are not piezoelectric cannot perform this bidirectional energy conversion.

Question 28Ultrasound Physics, Instrumentation, and Bioeffects/Safety

The ALARA principle, as applied to diagnostic ultrasound, refers to:

  1. Keeping acoustic output as low as reasonably achievable. Correct answer.
  2. Keeping the number of stored images as small as reasonably achievable.
  3. Keeping the transducer as close to the skin as reasonably achievable.
  4. Keeping the displayed frame rate as high as reasonably achievable.

Explanation

ALARA means the operator uses the lowest acoustic output and exposure time that still yield the needed diagnostic information. Output is reduced after optimization of gain, depth, and processing rather than raising power to compensate. The principle applies to all modes, with extra caution during Doppler and in pediatric patients.

Question 29Ultrasound Physics, Instrumentation, and Bioeffects/Safety

Increasing the transmitted ultrasound frequency while imaging the same structure will generally:

  1. Improve penetration and worsen spatial resolution.
  2. Improve spatial resolution and reduce penetration. Correct answer.
  3. Improve both spatial resolution and penetration.
  4. Reduce both spatial resolution and penetration.

Explanation

Higher frequency shortens the wavelength, which sharpens spatial resolution along and across the beam. Attenuation in soft tissue rises in proportion to frequency (about 0.5 dB/cm/MHz one way), so the beam is weakened sooner and penetration falls. Resolution and penetration therefore always trade off against each other.

Question 30Ultrasound Physics, Instrumentation, and Bioeffects/Safety

Which transducer component reduces the large acoustic impedance mismatch between the piezoelectric element and soft tissue?

  1. The electrical shield
  2. The acoustic lens
  3. The backing block
  4. The matching layer Correct answer.

Explanation

A matching layer has an impedance intermediate between that of the ceramic (roughly 30 MRayl) and soft tissue (about 1.5 MRayl), usually cut to a quarter-wavelength thickness. It increases the fraction of energy transmitted into tissue and received back. Without it, most energy would be reflected at the face of the element.

Question 31Ultrasound Physics, Instrumentation, and Bioeffects/Safety

A 4 MHz transducer operating in soft tissue (propagation speed 1540 m/s) produces a wavelength of approximately:

  1. 0.15 mm
  2. 1.54 mm
  3. 3.85 mm
  4. 0.39 mm Correct answer.

Explanation

Wavelength equals propagation speed divided by frequency: 1540 m/s divided by 4,000,000 Hz gives 0.000385 m, or about 0.39 mm. Because axial resolution is tied to the spatial pulse length, wavelength is a key determinant of image detail. Doubling the frequency would halve this wavelength.

Question 32Ultrasound Physics, Instrumentation, and Bioeffects/Safety

Time gain compensation is used to:

  1. Increase the emitted acoustic power as depth increases.
  2. Suppress high-amplitude echoes returning from slowly moving tissue.
  3. Change the displayed dynamic range of the image.
  4. Amplify echoes from greater depths more than shallow ones. Correct answer.

Explanation

Because attenuation increases with depth, echoes from deep reflectors return weaker than identical reflectors near the transducer. Time gain compensation applies progressively greater receiver gain at later arrival times so that similar interfaces appear equally bright. It affects only the received signal and does not change transmitted output.

Question 33Ultrasound Physics, Instrumentation, and Bioeffects/Safety

Bending of the ultrasound beam as it crosses a curved interface between tissues with different propagation speeds can produce which artifact?

  1. A vertical band of alternating bright and dark echoes.
  2. Ghost echoes appearing at twice the true depth.
  3. Enhancement of echoes deep to a fluid-filled structure.
  4. Displacement of a structure from its true anatomic position. Correct answer.

Explanation

Refraction occurs because the beam changes direction when propagation speed differs across an interface. The system still assumes the beam traveled in a straight line, so returning echoes are mapped to an incorrect location, and edge shadowing may also appear. Curved structures such as vessel walls and the heart borders are common sources.

Question 34Ultrasound Physics, Instrumentation, and Bioeffects/Safety

Which structure is most likely to produce a clean acoustic shadow on a pediatric echocardiogram?

  1. A moderate pericardial effusion
  2. Calcified valve leaflets Correct answer.
  3. The left ventricular blood pool
  4. A thin-walled coronary sinus

Explanation

Shadowing appears deep to structures that strongly reflect or absorb the beam, so almost no sound reaches the tissues beyond them. Calcium, bone, and air are the classic causes. Fluid-filled structures do the opposite and produce enhancement.

Question 35Ultrasound Physics, Instrumentation, and Bioeffects/Safety

The high-pass filter used in Doppler processing is intended to:

  1. Correct the Doppler angle automatically.
  2. Prevent aliasing by doubling the pulse repetition frequency.
  3. Reject low-velocity, high-amplitude signals from moving tissue. Correct answer.
  4. Remove the highest-velocity components of flow from the spectrum.

Explanation

Doppler shifts from moving myocardium and vessel walls are strong but close to zero frequency, so a high-pass threshold removes them from the display. The setting must be balanced, because a filter set too high will also delete genuine low-velocity flow. It does not alter the pulse repetition frequency or angle correction.

Question 36Ultrasound Physics, Instrumentation, and Bioeffects/Safety

In pulsed wave Doppler, increasing the sample volume (gate) length will:

  1. Increase the maximum velocity that can be measured without aliasing.
  2. Reduce the thermal index at the focal zone.
  3. Shorten the pulse and improve axial resolution.
  4. Broaden the spectrum by including a wider range of velocities. Correct answer.

Explanation

A longer range gate spans a larger region along the beam, so echoes from a wider spread of velocities are recorded together. The result is spectral broadening and poorer spatial discrimination of flow. A small gate gives a cleaner, more localized velocity trace.

Question 37Ultrasound Physics, Instrumentation, and Bioeffects/Safety

Attenuation of ultrasound in soft tissue is approximately:

  1. 0.05 dB/cm per MHz, one way
  2. 0.5 dB/cm per MHz, one way Correct answer.
  3. 1540 dB/cm per MHz, one way
  4. 5 dB/cm per MHz, one way

Explanation

The standard rule of thumb is 0.5 dB/cm/MHz measured one way, which doubles to about 1 dB/cm/MHz for the round trip. Attenuation increases linearly with frequency in soft tissue, which is why high-frequency imaging is limited to shallow structures. This relationship underlies the frequency-versus-depth trade-off.

Question 38Ultrasound Physics, Instrumentation, and Bioeffects/Safety

Continuous wave Doppler carries a greater potential for tissue heating than pulsed wave Doppler at comparable output settings mainly because:

  1. Its duty factor approaches 1.0, so temporal-average intensity is high. Correct answer.
  2. It samples at a higher frame rate, depositing more energy per second.
  3. It uses a substantially higher peak negative pressure than pulsed Doppler does.
  4. It focuses the beam more tightly, raising the spatial peak intensity.

Explanation

Thermal bioeffects depend on temporal-average intensity, which is the product of pulse intensity and duty factor. Continuous wave transducers transmit without interruption, giving a duty factor near 1.0, while pulsed systems spend most of the time listening. This is why Doppler exposure deserves particular attention under ALARA.

Question 39Ultrasound Physics, Instrumentation, and Bioeffects/Safety

A 2 MHz transducer interrogates flow of 1 m/s directed straight toward the transducer (0 degrees). The resulting Doppler shift is approximately:

  1. 2.6 kHz Correct answer.
  2. 5.2 kHz
  3. 13 kHz
  4. 0.65 kHz

Explanation

The Doppler equation gives fd = 2 x f0 x v x cos(theta) / c. Substituting 2,000,000 Hz, 100 cm/s, cos 0 degrees = 1, and 154,000 cm/s yields about 2600 Hz. Because the shift scales with transmitted frequency, a higher-frequency probe would produce a proportionally larger shift.

Question 40Ultrasound Physics, Instrumentation, and Bioeffects/Safety

Which intensity descriptor is used in the output display standard because it correlates most closely with temperature rise?

  1. Spatial peak, temporal average (SPTA) Correct answer.
  2. Spatial average, temporal peak (SATP)
  3. Spatial peak, temporal peak (SPTP)
  4. Spatial average, temporal average (SATA)

Explanation

SPTA combines the highest intensity found anywhere in the beam with averaging over the pulse repetition cycle. That combination best predicts the steady temperature rise produced by absorbed acoustic energy. Averaging over space dilutes the peak, and using only the temporal peak ignores duty cycle.

Question 41Normal Pediatric Cardiac Anatomy, Physiology, and Embryology

In a normal fetus near term, which vessel carries the most highly oxygenated blood?

  1. Umbilical vein Correct answer.
  2. Ductus arteriosus
  3. Superior vena cava
  4. Umbilical arteries

Explanation

Blood returning from the placenta in the umbilical vein has the highest oxygen saturation in the fetal circulation, roughly 80%. The umbilical arteries return desaturated blood to the placenta, while the ductus arteriosus and superior vena cava carry mixed or venous blood.

Question 42Normal Pediatric Cardiac Anatomy, Physiology, and Embryology

Which cardiac chamber forms the apex of the normal heart?

  1. Right atrium
  2. Left ventricle Correct answer.
  3. Left atrium
  4. Right ventricle

Explanation

The left ventricle has the thickest wall and extends to form the cardiac apex, which is directed anteriorly, inferiorly, and to the left. The right ventricle forms most of the anterior surface but does not reach the apex in the normal heart.

Question 43Normal Pediatric Cardiac Anatomy, Physiology, and Embryology

The fetal ductus arteriosus directly connects which two structures?

  1. Umbilical vein to inferior vena cava
  2. Right atrium to left atrium
  3. Aorta to superior vena cava
  4. Pulmonary artery to aorta Correct answer.

Explanation

The ductus arteriosus joins the main pulmonary artery near the origin of the left pulmonary artery to the descending thoracic aorta. It allows right ventricular output to bypass the unexpanded fetal lungs.

Question 44Normal Pediatric Cardiac Anatomy, Physiology, and Embryology

Which structure forms the flap that covers the foramen ovale in the fetus?

  1. Septum primum Correct answer.
  2. Septum secundum
  3. Eustachian valve
  4. Thebesian valve

Explanation

The septum primum forms the thin, mobile valve of the foramen ovale that is pushed leftward by right atrial pressure. The septum secundum forms the thicker muscular rim against which the septum primum seals after birth.

Question 45Normal Pediatric Cardiac Anatomy, Physiology, and Embryology

A normal mitral valve consists of how many leaflets?

  1. One
  2. Three
  3. Two Correct answer.
  4. Four

Explanation

The mitral valve is bicuspid, with an anterior and a posterior leaflet suspended by chordae tendineae. The tricuspid valve has three leaflets, and each semilunar valve has three cusps.

Question 46Normal Pediatric Cardiac Anatomy, Physiology, and Embryology

The embryonic truncus arteriosus divides into which pair of great vessels?

  1. Aorta and pulmonary trunk Correct answer.
  2. Coronary sinus and ductus venosus
  3. Right and left atria
  4. Superior and inferior vena cava

Explanation

The spiral aortopulmonary septum divides the truncus arteriosus into the ascending aorta and the main pulmonary trunk. Failure of this septation leaves a single common arterial trunk.

Question 47Normal Pediatric Cardiac Anatomy, Physiology, and Embryology

In normal situs solitus, where does the morphologic right atrium lie relative to the morphologic left atrium?

  1. Anterior and to the left
  2. Posterior and midline
  3. Superior and to the left
  4. On the right side Correct answer.

Explanation

In situs solitus the morphologic right atrium, which receives the caval veins, lies on the right and somewhat anteriorly, while the morphologic left atrium lies posteriorly. Atrial situs is defined by atrial appendage and venous morphology rather than by position alone.

Question 48Normal Pediatric Cardiac Anatomy, Physiology, and Embryology

The fetal ductus venosus shunts blood between which two vessels?

  1. Umbilical artery and iliac artery
  2. Descending thoracic aorta and pulmonary artery
  3. Portal vein and hepatic vein
  4. Umbilical vein and inferior vena cava Correct answer.

Explanation

The ductus venosus carries well-oxygenated blood from the umbilical vein directly into the inferior vena cava, bypassing the hepatic sinusoids. After birth it constricts and becomes the ligamentum venosum.

Question 49Normal Pediatric Cardiac Anatomy, Physiology, and Embryology

Which change is the principal driver of the fall in pulmonary vascular resistance immediately after birth?

  1. Reduction of left atrial pressure
  2. Rise in systemic vascular resistance
  3. Closure of the foramen ovale
  4. Oxygen-induced pulmonary vasodilation Correct answer.

Explanation

Lung expansion and the rise in alveolar oxygen tension relax the pulmonary arterioles, so pulmonary blood flow rises steeply and pulmonary vascular resistance falls. The atrial and ductal events occur in parallel but do not themselves cause pulmonary vasodilation.

Question 50Normal Pediatric Cardiac Anatomy, Physiology, and Embryology

In the normal near-term fetus, which ventricle ejects the larger share of combined ventricular output?

  1. Left ventricle
  2. Only the left ventricle
  3. Right ventricle Correct answer.
  4. Both ventricles equally

Explanation

The fetal right ventricle handles roughly 55 to 60 percent of combined ventricular output because it perfuses the placenta through the ductus arteriosus. After birth, pulmonary vascular resistance falls and the left ventricle becomes the dominant ejecting chamber.