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Iraqi Board Radiology study materials
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Iraqi Board Radiology practice questions

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Question 1Physics, instrumentation and radiation safety

In a diagnostic X-ray tube, what is the primary role of the rotating anode rather than a stationary anode?

  1. To collimate the beam to the selected field size
  2. To spread the electron beam over a larger focal track area so heat is dissipated more effectively Correct answer.
  3. To filter low-energy photons from the emitted beam
  4. To increase the tube's electrical insulation against high voltage

Explanation

The rotating anode spreads bombarding electrons along a circular focal track, so the heat per unit area is far lower than with a stationary focal spot of the same apparent size. This allows higher tube currents and longer exposures while preserving a small effective focal spot for sharp images. Insulation, filtration and collimation are handled by other components.

Question 2Physics, instrumentation and radiation safety

Which interaction between incident X-ray photons and tissue predominates at diagnostic energies and is chiefly responsible for producing subject contrast on a radiograph?

  1. Coherent (Rayleigh) scattering
  2. Photoelectric absorption Correct answer.
  3. Pair production
  4. Compton scattering

Explanation

Photoelectric absorption depends strongly on atomic number and photon energy, so it differentially attenuates bone, soft tissue and fat, generating radiographic contrast. Compton scattering contributes most of the scatter fog at higher energies and degrades contrast. Pair production requires energies above 1.02 MeV and is irrelevant to diagnostic radiography.

Question 3Physics, instrumentation and radiation safety

A radiographer doubles the source-to-image distance while keeping the kilovoltage and milliampere-seconds constant. What happens to the absorbed dose at the image receptor plane?

  1. It doubles
  2. It falls to one quarter Correct answer.
  3. It falls to one half
  4. It remains unchanged

Explanation

Photon fluence from a point source obeys the inverse square law, so doubling the distance reduces the dose rate at the receptor by a factor of four. Compensating for the lower intensity would require a substantial increase in milliampere-seconds. This is why longer distances are used in some examinations to reduce patient skin dose.

Question 4Physics, instrumentation and radiation safety

Which personal dosimeter is designed to record cumulative dose from a single wearing period and is read by heating the detector to release trapped electrons?

  1. Geiger-Muller counter
  2. Scintillation gamma camera
  3. Thermoluminescent dosimeter Correct answer.
  4. Ionisation chamber survey meter

Explanation

In a thermoluminescent dosimeter, radiation traps electrons in metastable states within a crystalline material, and heating releases them as light proportional to absorbed dose. It provides a cumulative record over a wearing period. Survey meters and Geiger counters give real-time readings rather than cumulative personal records.

Question 5Physics, instrumentation and radiation safety

What is the main practical purpose of applying beam collimation to the smallest field that includes the region of interest?

  1. To reduce patient dose and scattered radiation reaching the receptor Correct answer.
  2. To improve the spatial resolution of the receptor itself
  3. To increase the tube output for a given exposure
  4. To shorten the anode warm-up time

Explanation

Tight collimation reduces the volume of tissue irradiated and the amount of scatter generated, lowering patient dose and improving image contrast. It does not change tube output, warm-up behaviour or the intrinsic resolution of the receptor. Collimation is a basic optimisation step in every examination.

Question 6Physics, instrumentation and radiation safety

Which unit is used to express equivalent dose and takes account of the radiation weighting factor?

  1. Sievert Correct answer.
  2. Coulomb per kilogram
  3. Becquerel
  4. Gray

Explanation

The sievert expresses equivalent and effective dose, incorporating radiation and tissue weighting factors to reflect biological risk. The gray expresses absorbed dose, the becquerel expresses activity and coulomb per kilogram expresses exposure. Using the sievert allows dose from different radiation types to be compared on a common risk scale.

Question 7Physics, instrumentation and radiation safety

A computed radiography plate is exposed but not read for several hours. What is the most likely consequence for the resulting image?

  1. Loss of latent image signal with reduced density and increased noise Correct answer.
  2. Complete erasure of the latent image before readout
  3. A shift in the effective kilovoltage of the exposure
  4. Improved contrast because of signal amplification over time

Explanation

The latent image in a photostimulable phosphor plate fades over time, so delayed readout produces a weaker signal, lower density and relatively more noise. This is why plates should be processed promptly after exposure. The kilovoltage is fixed at exposure and cannot change after the event.

Question 8Physics, instrumentation and radiation safety

In a flat-panel detector using amorphous selenium, what is the immediate result of X-ray photons interacting with the photoconductor?

  1. Emission of light photons that are then detected by a photodiode array
  2. Stimulation of a photostimulable phosphor layer
  3. Direct creation of electron-hole pairs that are collected by a charge-collecting electrode array Correct answer.
  4. Ionisation of a gas within a sealed chamber

Explanation

Amorphous selenium is a direct-conversion photoconductor: absorbed photons create electron-hole pairs that drift under an applied bias and are collected by a thin-film transistor array. Light emission followed by photodiode detection describes indirect conversion with a scintillator such as caesium iodide. Gas ionisation and photostimulable phosphors belong to other detector types.

Question 9Physics, instrumentation and radiation safety

Which factor most directly determines the spatial resolution of a digital image reconstructed on a fixed matrix?

  1. The milliampere-seconds product used
  2. The size of the pixel in the reconstructed image Correct answer.
  3. The anode rotation speed
  4. The amount of added aluminium filtration

Explanation

Spatial resolution in a digital image is limited by pixel size, which equals field of view divided by matrix dimensions. Smaller pixels sample finer detail. Milliampere-seconds affects noise, filtration affects beam quality and anode rotation affects heat loading, none of which sets the sampling limit.

Question 10Physics, instrumentation and radiation safety

A fluoroscopy unit operating in pulsed mode at 7.5 pulses per second is compared with continuous fluoroscopy. What is the principal advantage of pulsing?

  1. It eliminates the need for automatic brightness control
  2. It removes the need for anti-scatter grid use
  3. It increases the detector's spatial resolution
  4. It reduces patient and staff dose by limiting the X-ray on-time Correct answer.

Explanation

Pulsed fluoroscopy switches the beam on only during frame acquisition, so the total on-time and hence dose are reduced compared with continuous operation. Automatic brightness control is still needed, and pulsing does not change detector resolution or remove the need for a grid. The dose saving depends on the pulse rate and pulse width.

Question 11Physics, instrumentation and radiation safety

Which statement correctly describes the relationship between tube current and focal spot size selection in a diagnostic X-ray tube?

  1. Tube current is limited only by the generator's kilovoltage setting
  2. Smaller focal spots permit higher tube currents because electrons are more concentrated
  3. Focal spot size has no influence on the maximum permissible tube current
  4. Larger focal spots permit higher tube currents because heat is distributed over a greater area Correct answer.

Explanation

A larger focal spot spreads electron energy over a greater anode area, allowing higher tube currents and longer exposures without exceeding the anode's thermal limits. Smaller focal spots improve geometric sharpness but restrict current. Tube current limits therefore reflect focal spot selection and anode heat capacity.

Question 12Physics, instrumentation and radiation safety

What is the main reason for using a grid with a high grid ratio in a large-patient abdominal radiograph?

  1. To shorten the exposure time required
  2. To increase the field size covered by the collimator
  3. To reduce the patient's entrance surface dose
  4. To remove a greater proportion of scattered radiation before it reaches the receptor Correct answer.

Explanation

Higher grid ratios remove more obliquely travelling scatter, improving contrast in thick body parts where scatter production is high. The trade-off is that grids also absorb some primary radiation, so patient dose and exposure factors must increase. Grids do not enlarge the field or reduce entrance dose.

Question 13Physics, instrumentation and radiation safety

Which of the following best explains why high-frequency generators are preferred over single-phase generators in modern radiography?

  1. They eliminate the need for exposure timing circuits
  2. They require no rectification of the mains supply
  3. They allow the anode to be stationary
  4. They produce a nearly constant potential waveform, giving a more uniform beam and higher effective output Correct answer.

Explanation

High-frequency generators convert the mains supply to a high-frequency waveform and rectify it, producing an almost constant potential across the tube. This yields a beam with higher effective energy for the same kilovoltage and greater output per milliampere-second. Rectification and accurate timing remain necessary, and anode rotation is unrelated.

Question 14Physics, instrumentation and radiation safety

During fluoroscopy, a trainee stands close to the patient's side and the table. Which action most effectively reduces the trainee's occupational dose?

  1. Reducing the collimated field size only
  2. Increasing the fluoroscopic kilovoltage
  3. Increasing the pulse rate to shorten the procedure
  4. Standing further from the patient and using a lead apron Correct answer.

Explanation

Occupational exposure from a patient acting as a scatter source falls sharply with distance, and lead aprons attenuate the remaining scatter. Increasing kilovoltage or pulse rate can raise dose, and collimation alone gives only a modest staff benefit. Distance and shielding are the primary protective measures.

Question 15Physics, instrumentation and radiation safety

A radiograph of the lumbar spine appears excessively noisy despite adequate positioning. Which change is most appropriate to reduce quantum noise while preserving acceptable contrast?

  1. Decrease source-to-image distance
  2. Increase milliampere-seconds and accept a slightly longer exposure Correct answer.
  3. Reduce milliampere-seconds and increase kilovoltage
  4. Reduce the grid ratio to zero

Explanation

Quantum noise reflects too few photons reaching the detector, so increasing milliampere-seconds raises the photon count and reduces noise. A slightly longer exposure is usually acceptable if motion can be controlled. Reducing milliampere-seconds worsens noise, and removing the grid would degrade contrast in a thick region.

Question 16Physics, instrumentation and radiation safety

Which statement about the dose-area product meter used in fluoroscopy is correct?

  1. It measures the tube's anode current during exposure
  2. It estimates the total radiation dose delivered to the patient over the examination Correct answer.
  3. It measures the absorbed dose at a single point on the patient's skin
  4. It records the dose received by the radiologist's hands

Explanation

A dose-area product meter integrates dose and field area over the whole examination, providing a useful estimate of total patient dose and stochastic risk. It is not a point skin-dose measurement, a personal dosimeter or a current meter. Recording dose-area product supports audit and optimisation.

Question 17Physics, instrumentation and radiation safety

A departmental audit finds that repeated radiographs of the pelvis in obese patients are common. Which combination of changes best addresses the underlying physics problem?

  1. Use a smaller focal spot and reduce the exposure time substantially
  2. Increase kilovoltage, use a higher grid ratio and apply automatic exposure control with adequate milliampere-seconds Correct answer.
  3. Increase collimation to a very small field and reduce source-to-image distance
  4. Decrease kilovoltage, remove the grid and reduce milliampere-seconds

Explanation

In obese patients, higher kilovoltage improves penetration, a higher grid ratio controls the increased scatter, and automatic exposure control with sufficient milliampere-seconds ensures adequate photons at the receptor. Together these reduce repeat rates. The other combinations either worsen penetration or increase noise.

Question 18Physics, instrumentation and radiation safety

Which statement best explains why the effective dose from a computed tomography scan is higher than from a single chest radiograph for a comparable anatomical region?

  1. Computed tomography delivers radiation from many projections around the patient, so more tissue is irradiated Correct answer.
  2. Computed tomography uses a much higher tube potential than radiography
  3. Computed tomography images are always reconstructed at a larger pixel size
  4. Computed tomography uses no beam filtration

Explanation

In computed tomography the tube rotates around the patient, so every projection irradiates a large volume of tissue and the contributions accumulate. Tube potentials are broadly similar to radiography, and filtration is routinely used. The geometry and number of projections, not pixel size or kilovoltage alone, explain the higher effective dose.

Question 19Physics, instrumentation and radiation safety

A pregnant patient requires an urgent abdominal radiograph. Which principle should guide the decision according to radiation protection practice?

  1. The examination should be performed at the highest possible kilovoltage to minimise fetal dose
  2. The examination should be delayed until after delivery regardless of clinical urgency
  3. The examination should be refused because any fetal exposure is unacceptable
  4. The examination should proceed if the clinical benefit outweighs the small fetal risk, with dose kept as low as reasonably achievable Correct answer.

Explanation

Justification requires weighing the clinical benefit against the small fetal risk, and optimisation ensures the dose is kept as low as reasonably achievable consistent with the diagnostic requirement. Refusal or unconditional delay could harm both mother and fetus. Simply maximising kilovoltage is not a substitute for proper justification and optimisation.

Question 20Physics, instrumentation and radiation safety

In a digital radiography system, which change would most likely increase the detective quantum efficiency at a fixed exposure?

  1. Increasing the thickness of the scintillator to absorb a greater fraction of incident photons Correct answer.
  2. Increasing the size of the reconstructed pixel matrix without changing field of view
  3. Increasing the amount of scattered radiation reaching the detector
  4. Reducing the tube potential while keeping milliampere-seconds constant

Explanation

Detective quantum efficiency improves when more incident photons are absorbed and converted into signal, so a thicker scintillator raises absorption efficiency and reduces noise relative to dose. Scatter degrades contrast and effective efficiency, while matrix size and tube potential do not directly improve photon absorption.

Question 21Anatomy, embryology and physiology relevant to imaging

During embryonic development, which structure gives rise to the midgut and its associated mesentery?

  1. The vitelline duct remnant
  2. The caudal limb of the primary intestinal loop Correct answer.
  3. The urogenital sinus
  4. The foregut diverticulum

Explanation

The midgut is formed from the primary intestinal loop, which herniates into the umbilical coelom and rotates 270 degrees around the superior mesenteric artery. The caudal limb of this loop forms the distal ileum, caecum, appendix, ascending colon and proximal two-thirds of the transverse colon. The cranial limb forms the jejunum and proximal ileum.

Question 22Anatomy, embryology and physiology relevant to imaging

The caudate lobe of the liver receives its principal venous drainage directly into which vessel?

  1. The right hepatic vein
  2. The inferior vena cava Correct answer.
  3. The portal vein
  4. The middle hepatic vein

Explanation

The caudate lobe (segment I) drains via short hepatic veins directly into the inferior vena cava, independent of the three main hepatic veins. This is why the caudate lobe can hypertrophy in Budd-Chiari syndrome when the main hepatic veins are occluded.

Question 23Anatomy, embryology and physiology relevant to imaging

A patient has a lesion at the level of the C5-C6 intervertebral disc compressing the exiting nerve root. Which myotome is most useful to test for this root?

  1. C8 (finger flexors)
  2. C5 (deltoid and biceps) Correct answer.
  3. C7 (triceps and wrist flexors)
  4. C4 (shoulder shrug)

Explanation

In the cervical spine the nerve root exits above its corresponding vertebra, so a C5-C6 disc herniation typically compresses the C6 root; however, the C5 myotome (deltoid and biceps) is classically tested for the C4-C5 level. For clarity, the C5 root is the one exiting at C4-C5, and deltoid/biceps testing is the standard bedside check for C5.

Question 24Anatomy, embryology and physiology relevant to imaging

Which of the following best describes the normal arrangement of the renal arteries relative to the renal veins at the hilum?

  1. Renal artery lies inferior to the renal vein
  2. Renal artery lies superior to the renal vein
  3. Renal artery lies posterior to the renal vein Correct answer.
  4. Renal artery lies anterior to the renal vein

Explanation

At the renal hilum the renal vein is anterior and the renal artery is posterior, with the renal pelvis most posterior. This relationship is important when planning percutaneous renal access and interpreting contrast-enhanced CT. The left renal vein is longer and crosses anterior to the aorta.

Question 25Anatomy, embryology and physiology relevant to imaging

The blood supply to the head and neck of the pancreas is derived mainly from which artery?

  1. The superior mesenteric artery
  2. The gastroduodenal artery via the superior pancreaticoduodenal arcade Correct answer.
  3. The left gastric artery
  4. The splenic artery

Explanation

The pancreatic head and uncinate process are supplied by the superior pancreaticoduodenal arcade (from the gastroduodenal artery) and the inferior pancreaticoduodenal arcade (from the superior mesenteric artery). The splenic artery supplies the body and tail. This distinction matters in angiography and in planning pancreatic resection.

Question 26Anatomy, embryology and physiology relevant to imaging

A 30-year-old man has a right-sided aortic arch with an aberrant left subclavian artery. Which structure is most likely to be compressed?

  1. The left brachiocephalic vein
  2. The right recurrent laryngeal nerve
  3. The trachea
  4. The oesophagus Correct answer.

Explanation

A right aortic arch with an aberrant left subclavian artery often forms a vascular ring that compresses the oesophagus, causing dysphagia lusoria. The aberrant artery passes behind the oesophagus from a Kommerell diverticulum. Tracheal compression is less typical because the ring is usually incomplete anteriorly.

Question 27Anatomy, embryology and physiology relevant to imaging

The normal fetal circulation includes a shunt that directs blood from the pulmonary trunk to the descending aorta. This structure is the:

  1. Ductus arteriosus Correct answer.
  2. Umbilical vein
  3. Foramen ovale
  4. Ductus venosus

Explanation

The ductus arteriosus connects the pulmonary trunk to the descending aorta, allowing right ventricular output to bypass the non-aerated lungs. It normally closes functionally within hours of birth and anatomically within weeks, forming the ligamentum arteriosum. Persistent patency is common in premature infants.

Question 28Anatomy, embryology and physiology relevant to imaging

Which of the following statements about the blood-brain barrier is most accurate for imaging interpretation?

  1. It prevents all contrast agents from entering the brain
  2. It is absent in the area postrema and pituitary gland Correct answer.
  3. It is composed mainly of astrocytes
  4. It is formed by fenestrated capillary endothelium

Explanation

The blood-brain barrier is formed by tight junctions between non-fenestrated capillary endothelial cells, supported by pericytes and astrocyte end-feet. It is absent at circumventricular organs such as the area postrema, median eminence, pituitary and pineal gland, which is why these structures normally enhance. Gadolinium contrast leaks where the barrier is disrupted.

Question 29Anatomy, embryology and physiology relevant to imaging

The principal arterial supply to the spinal cord in the thoracic region is derived from which vessel?

  1. The posterior inferior cerebellar artery
  2. The artery of Adamkiewicz arising from a posterior intercostal or lumbar artery Correct answer.
  3. The middle meningeal artery
  4. The anterior spinal artery arising from the vertebral arteries

Explanation

The artery of Adamkiewicz (arteria radicularis magna) is the dominant radiculomedullary feeder to the lower thoracic and lumbar cord, usually arising from a left posterior intercostal or lumbar artery. Its injury during aortic surgery can cause anterior spinal artery syndrome. The anterior spinal artery itself is a longitudinal vessel that is supplemented by these radicular feeders.

Question 30Anatomy, embryology and physiology relevant to imaging

Which of the following best describes the normal position of the uterus in a nulliparous woman on MRI?

  1. Anteverted and anteflexed over the bladder Correct answer.
  2. Retroverted and anteflexed
  3. Anteverted and retroflexed
  4. Retroverted and retroflexed over the rectum

Explanation

The normal nulliparous uterus is anteverted (cervix tilted forward on the vagina) and anteflexed (body flexed forward on the cervix), resting over the urinary bladder. Retroversion is a normal variant in about 15-20% of women and is not itself pathological. Position is best assessed on sagittal T2-weighted images.

Question 31Anatomy, embryology and physiology relevant to imaging

A patient has a lesion in the posterior limb of the internal capsule. Which tract is most directly affected?

  1. The inferior cerebellar peduncle
  2. The medial longitudinal fasciculus
  3. The corticospinal tract Correct answer.
  4. The optic radiation

Explanation

The posterior limb of the internal capsule carries the corticospinal and corticobulbar fibres between the motor cortex and the brainstem/spinal cord, along with sensory fibres of the thalamocortical radiation. A small lacunar infarct here can cause a dense contralateral hemiparesis. The optic radiation lies more posteriorly in the retrolentiform part.

Question 32Anatomy, embryology and physiology relevant to imaging

The thyroid gland develops from which embryological structure?

  1. The second branchial arch
  2. The Rathke pouch
  3. The foramen caecum of the tongue Correct answer.
  4. The fourth pharyngeal pouch

Explanation

The thyroid primordium arises as a median endodermal thickening in the floor of the pharynx at the foramen caecum and descends along the thyroglossal duct to its final position. Persistence of the duct can produce a thyroglossal cyst. The ultimobranchial body (fourth/fifth pouch) contributes the parafollicular C cells.

Question 33Anatomy, embryology and physiology relevant to imaging

Which of the following best explains why the right main bronchus is more likely to be aspirated into than the left?

  1. It arises at a sharper angle from the trachea
  2. It is wider, shorter and more vertical Correct answer.
  3. It is narrower and more vertical
  4. It is compressed by the aortic arch

Explanation

The right main bronchus is wider, shorter and more vertical than the left, so aspirated material more often lodges in the right lower lobe. This anatomical asymmetry is a useful clue when interpreting chest radiographs for aspiration. The left bronchus is longer and more horizontal, partly due to the aortic arch crossing it.

Question 34Anatomy, embryology and physiology relevant to imaging

Which of the following statements about the normal cardiac axis on a frontal chest radiograph is correct?

  1. The left heart border is formed mainly by the left atrium
  2. The right heart border is formed mainly by the right atrium Correct answer.
  3. The cardiac apex normally points to the right
  4. The ascending aorta forms the right heart border

Explanation

The right heart border is formed mainly by the right atrium, with the superior vena cava above it. The left border is formed by the aortic knuckle, pulmonary trunk, left atrial appendage and left ventricle. The cardiac apex normally points to the left.

Question 35Anatomy, embryology and physiology relevant to imaging

A patient with a pituitary macroadenoma develops bitemporal hemianopia. Which anatomical structure is compressed?

  1. The optic tract
  2. The optic radiation
  3. The lateral geniculate nucleus
  4. The optic chiasm Correct answer.

Explanation

The optic chiasm lies above the pituitary fossa, and fibres from the nasal retina that cross here subserve the temporal visual fields. A pituitary mass expanding upward compresses the chiasm, producing bitemporal hemianopia. Compression of the optic tract or radiation would produce different field defects.

Question 36Anatomy, embryology and physiology relevant to imaging

The normal spleen lies in the left upper quadrant and is related to which structure on its medial surface?

  1. The stomach Correct answer.
  2. The pancreas
  3. The left kidney
  4. The left adrenal gland

Explanation

The medial (visceral) surface of the spleen has gastric, renal and colic impressions, with the gastric impression being the largest and most medial, related to the stomach. The splenic hilum transmits the splenic vessels and tail of the pancreas. This relationship explains why splenic enlargement can indent the gastric fundus on barium studies.

Question 37Pathology and basic sciences relevant to radiology

Which cellular process is most directly responsible for the increased radiosensitivity observed in cells during the G2/M phase of the cell cycle?

  1. Unrepaired DNA double-strand breaks cannot be repaired before chromosome segregation Correct answer.
  2. Mitotic chromosomes have already completed homologous recombination repair
  3. Repair of sublethal damage is maximal before mitosis begins
  4. Chromatin is condensed and thus more resistant to double-strand breaks

Explanation

Cells are most radiosensitive in late G2 and M phase because chromatin is condensed and there is little time for repair of DNA double-strand breaks before the cell divides. Cells that enter mitosis with unrepaired breaks undergo mitotic catastrophe and die. The other phases, especially late S phase, allow more time for repair and are relatively radioresistant.

Question 38Pathology and basic sciences relevant to radiology

A tissue has an alpha/beta ratio of 10 Gy. Which statement best describes the radiobiological behaviour of this tissue?

  1. It is typical of early-responding tissues and is relatively insensitive to fraction size Correct answer.
  2. It has a low capacity for repair of sublethal damage between fractions
  3. It shows a pronounced shoulder on the survival curve and is spared by fractionation
  4. It is typical of late-responding tissues such as spinal cord and kidney

Explanation

A high alpha/beta ratio (around 10 Gy) is characteristic of early-responding tissues such as skin, mucosa and most tumours. These tissues have a broad shoulder and considerable repair capacity, so they are relatively insensitive to changes in fraction size. Low alpha/beta values (around 2–3 Gy) are seen in late-responding tissues, which are more sensitive to fraction size.

Question 39Pathology and basic sciences relevant to radiology

Which type of cell death is classically described as the principal mode of radiation-induced death in lymphoid tissue and is characterised by internucleosomal DNA fragmentation?

  1. Apoptosis Correct answer.
  2. Necrosis
  3. Autophagy
  4. Mitotic catastrophe

Explanation

Radiation-induced apoptosis is prominent in lymphoid tissues and is characterised by internucleosomal DNA cleavage producing a ladder pattern on electrophoresis, along with chromatin condensation and apoptotic body formation. Necrosis results from severe membrane and metabolic injury. Mitotic catastrophe occurs after cells enter mitosis with unrepaired damage and is common in solid tumours.

Question 40Pathology and basic sciences relevant to radiology

A radiation oncologist wants to compare two treatment plans with different dose distributions. Which radiobiological model is most appropriate for estimating tumour control probability from a non-uniform dose distribution using the linear-quadratic formalism?

  1. Thermoluminescence dosimetry
  2. Nominal standard dose model
  3. Equivalent uniform dose concept with the linear-quadratic model Correct answer.
  4. Linear energy transfer spectrum analysis

Explanation

The equivalent uniform dose (EUD) concept converts a non-uniform dose distribution into a single uniform dose that would produce the same biological effect. When combined with the linear-quadratic model and Poisson statistics, it allows estimation of tumour control probability and normal tissue complication probability. The nominal standard dose model is an older, less biologically grounded approach.

Question 41Pathology and basic sciences relevant to radiology

Which statement correctly describes the relative biological effectiveness (RBE) of high linear energy transfer (LET) radiation?

  1. RBE is lower for neutrons than for X-rays at all dose levels
  2. RBE is independent of dose and dose rate
  3. RBE is highest at low doses because the survival curve has a broad shoulder
  4. RBE decreases as dose per fraction increases because the reference low-LET curve catches up Correct answer.

Explanation

High-LET radiation produces survival curves with little or no shoulder, whereas low-LET radiation has a broad shoulder at low doses. As dose per fraction increases, the low-LET curve becomes steeper and approaches the high-LET curve, so the RBE decreases. RBE is therefore highest at low doses and low dose rates.

Question 42Pathology and basic sciences relevant to radiology

In the context of radiation carcinogenesis, which phenomenon describes the observation that a dose delivered in many small fractions is generally less carcinogenic than the same total dose given as a single exposure?

  1. Dose-rate effect Correct answer.
  2. Oxygen effect
  3. Bystander effect
  4. Adaptive response

Explanation

The dose-rate effect describes the reduced biological effect, including carcinogenic risk, when a given total dose is delivered at low dose rate or in many fractions. This occurs because sublethal damage is repaired between exposures and cells with damage are removed before the next dose. The oxygen effect and bystander effect describe different radiobiological phenomena.

Question 43Pathology and basic sciences relevant to radiology

Which normal tissue is considered a late-responding tissue with a low alpha/beta ratio, making it particularly sensitive to large fraction sizes?

  1. Spinal cord Correct answer.
  2. Testicular seminiferous epithelium
  3. Jejunal mucosa
  4. Bone marrow

Explanation

The spinal cord is a classic late-responding tissue with an alpha/beta ratio of about 2 Gy, so it is sensitive to large dose per fraction and at risk of radiation myelopathy. Jejunal mucosa, bone marrow and seminiferous epithelium are early-responding tissues with high turnover and high alpha/beta ratios.

Question 44Pathology and basic sciences relevant to radiology

A patient with a known p53 mutation is treated with radiotherapy. What is the most likely consequence for tumour response?

  1. Reduced ability to repopulate between fractions
  2. Improved repair of double-strand breaks
  3. Enhanced apoptosis and increased radiosensitivity
  4. Impaired apoptosis and increased radioresistance Correct answer.

Explanation

TP53 is a key regulator of apoptosis and cell cycle arrest after DNA damage. Loss of p53 function impairs radiation-induced apoptosis and allows cells with damaged DNA to continue cycling, contributing to radioresistance. It does not improve DNA repair or reduce repopulation.

Question 45Pathology and basic sciences relevant to radiology

Which molecular pathway is the principal mechanism for repair of radiation-induced DNA double-strand breaks in mammalian cells during the G1 phase of the cell cycle?

  1. Mismatch repair
  2. Base excision repair
  3. Homologous recombination
  4. Non-homologous end joining Correct answer.

Explanation

Non-homologous end joining (NHEJ) is the dominant pathway for repairing DNA double-strand breaks in G1, when no sister chromatid is available as a template. Homologous recombination requires a sister chromatid and operates mainly in S and G2. Mismatch repair and base excision repair address other types of DNA lesions.

Question 46Pathology and basic sciences relevant to radiology

Which statement best describes the oxygen enhancement ratio (OER) for sparsely ionising radiation such as megavoltage X-rays?

  1. The OER is approximately 2.5–3.0 and is maximal at high doses
  2. The OER is approximately 1.0 because oxygen does not affect radiation response
  3. The OER is higher for high-LET radiation than for low-LET radiation
  4. The OER is approximately 2.5–3.0 and is greatest at low doses where the survival curve shoulder is present Correct answer.

Explanation

For sparsely ionising radiation, the oxygen enhancement ratio is about 2.5–3.0. It is greatest at low doses because oxygen interacts with the shoulder region of the survival curve, and it decreases at high doses as the curves converge. High-LET radiation has a lower OER because dense ionisation directly causes lethal damage.

Question 47Pathology and basic sciences relevant to radiology

Which cell survival curve feature is characteristic of high linear energy transfer (LET) radiation such as alpha particles?

  1. A pronounced shoulder indicating extensive sublethal damage repair
  2. A nearly exponential curve with little or no shoulder Correct answer.
  3. A biphasic curve with a resistant tail at low doses
  4. A broad shoulder with a large alpha/beta ratio

Explanation

High-LET radiation produces dense ionisation tracks that cause complex, irreparable DNA damage, so the survival curve is nearly exponential with little or no shoulder. This reflects a low capacity for repair of sublethal damage and a high alpha/beta ratio. Low-LET radiation shows a broad shoulder at low doses.

Question 48Pathology and basic sciences relevant to radiology

Which statement about the bystander effect in radiobiology is correct?

  1. It describes the increased radiosensitivity of cells exposed to high oxygen concentrations
  2. It explains why high-LET radiation has a lower RBE than low-LET radiation
  3. It refers to the repair of sublethal damage between fractions
  4. Biological effects are observed in cells that were not directly irradiated but are in contact with or near irradiated cells Correct answer.

Explanation

The radiation-induced bystander effect describes biological responses such as DNA damage, apoptosis or genomic instability in non-irradiated cells that receive signals from directly irradiated cells or their medium. It is mediated by gap junction communication and soluble factors. It is distinct from the oxygen effect, sublethal damage repair and RBE.

Question 49Pathology and basic sciences relevant to radiology

Which tumour property is most closely associated with a poor response to radiotherapy because of accelerated repopulation during treatment?

  1. High clonogenic fraction with rapid proliferation after a lag period Correct answer.
  2. Low expression of epidermal growth factor receptor
  3. Extensive tumour necrosis
  4. Low mitotic index

Explanation

Accelerated repopulation refers to the increased proliferation of surviving clonogenic tumour cells after a lag period during fractionated radiotherapy. A high clonogenic fraction with rapid proliferation after this lag reduces tumour control and is a rationale for avoiding treatment gaps and considering accelerated fractionation. Low mitotic index and necrosis are not features of accelerated repopulation.

Question 50Pathology and basic sciences relevant to radiology

Which statement correctly describes the relationship between linear energy transfer, relative biological effectiveness and the oxygen enhancement ratio?

  1. As LET increases, both RBE and OER increase
  2. As LET increases, RBE decreases and OER increases
  3. As LET increases, RBE increases and OER decreases Correct answer.
  4. As LET increases, both RBE and OER remain constant

Explanation

Increasing LET produces denser ionisation, more complex DNA damage and a higher RBE, while the dependence on oxygen for fixation of damage falls, so the OER decreases. At very high LET, RBE eventually plateaus or declines due to overkill. This relationship underlies the interest in high-LET particles such as carbon ions.