25 board-style questions for the NMTCB PET specialty exam, covering all four content domains. Try each one, then open the answer for the full explanation. No sign-up needed.
Diagnostic Procedures · Instrumentation & QC · Radiation Protection · Radiopharmaceuticals
1. A patient is injected with FDG at 09:00 for a restaging study. Scanner problems delay the acquisition until 10:45. Compared with the prior study acquired at 60 minutes, the SUVmax of a known lesion will most likely be:
Answer: C — Higher, biasing comparison with the baseline
SUV is decay corrected, so decay itself does not lower it. Most tumours continue to accumulate FDG well past 60 minutes while background clears, so SUVmax rises with uptake time. The comparison with the baseline is therefore biased and the delay must be recorded.
2. A diabetic patient presents for oncology FDG PET with a capillary glucose of 305 mg/dL. Which action is most appropriate?
Answer: C — Reschedule after glycaemic control improves
Hyperglycaemia competes with FDG at the GLUT transporter and lowers lesion conspicuity. Insulin is the trap: it drives FDG into skeletal muscle and myocardium, degrading the study further. Extending uptake or raising activity does not overcome the competitive block.
3. Which feature best distinguishes brown adipose tissue uptake from supraclavicular nodal metastases?
Answer: A — Symmetry, with fat density on the CT
Brown fat is characteristically symmetric and maps to fat attenuation on the CT rather than to a soft tissue nodal mass. SUV overlaps with malignancy and is not discriminating. Unilateral activity should raise rather than lower concern for disease.
4. A cardiac study shows a segment with reduced perfusion and preserved FDG uptake. This finding indicates the segment is:
Answer: D — Viable and may recover after revascularisation
Reduced flow with maintained glucose metabolism is the classic mismatch of hibernating myocardium, which is viable and may recover after revascularisation. A matched reduction in both perfusion and FDG indicates scar.
5. A cardiac sarcoidosis study shows intense, diffuse myocardial FDG uptake. The most likely explanation is that the patient:
Answer: A — Had inadequate myocardial suppression preparation
Diffuse myocardial uptake usually means physiologic glucose use by normal myocytes was not adequately suppressed. Appropriate low-carbohydrate/high-fat dietary preparation and fasting shift normal myocardium toward free-fatty-acid metabolism. Inadequate preparation or carbohydrate exposure can therefore produce diffuse myocardial FDG uptake.
6. A brain FDG study shows hypometabolism of the occipital cortex in addition to temporoparietal regions. This pattern favours:
Answer: B — Dementia with Lewy bodies
Occipital hypometabolism, particularly with relative sparing of the posterior cingulate, favours dementia with Lewy bodies. Alzheimer disease typically spares the occipital cortex. Frontotemporal dementia predominantly affects frontal and anterior temporal regions.
7. A curvilinear band of apparently absent activity is seen at the lung base on the attenuation-corrected images only. The most likely cause is:
Answer: D — Respiratory mismatch between the CT and PET data
CT is acquired in seconds and PET over minutes of breathing, so diaphragm position differs between datasets. The resulting attenuation map error produces a classic cold curvilinear band. That it appears only on corrected images is the key clue.
8. The total energy released when a positron annihilates with an electron is:
Answer: C — 1022 keV, shared equally by two photons
The combined rest mass energy of the positron and electron is 1.022 MeV, released as two photons of 511 keV each. Confusing the total with the per-photon energy is a common error; the scanner detects the two 511 keV photons in coincidence.
9. Photon non-collinearity degrades spatial resolution most severely in scanners with:
Answer: B — A larger detector ring diameter
The two photons depart slightly off 180 degrees, so the positional error grows with the distance travelled to the detectors. A larger bore therefore suffers more blurring from this effect. Crystal speed and energy window affect timing and scatter rejection, not this geometric term.
10. Two photons from separate annihilations reach opposing detectors within the timing window. This event is recorded as a:
Answer: A — Random coincidence
A random, or accidental, coincidence arises from two unrelated annihilations. A scattered coincidence still comes from one annihilation, but a photon has changed direction. Randoms add a broad background and their rate rises roughly with the square of activity.
11. Time-of-flight capability depends chiefly on a crystal property described as:
Answer: C — Short scintillation decay time
TOF needs precise timing, which requires a fast scintillation decay, as in LSO and LYSO at roughly 40 nanoseconds. BGO has excellent stopping power from its high atomic number but decays far too slowly for useful TOF timing.
12. The image-quality benefit of time-of-flight reconstruction is greatest in:
Answer: B — Larger patients
TOF reduces how far noise propagates along each line of response, and the improvement scales with the size of the object. Larger patients, where attenuation and scatter make signal to noise worst, gain the most; small or thin subjects gain least.
13. CT data must be scaled before use as a PET attenuation map because:
Answer: D — CT uses much lower photon energies than 511 keV
Attenuation coefficients depend on photon energy, and the CT beam is far lower in energy than 511 keV. The measured values must be converted to the coefficients that apply at annihilation energy, otherwise correction is systematically wrong.
14. Roughly how many half-value layers of lead are needed to reduce a 511 keV beam to about one eighth of its original intensity?
Answer: D — Three
Each half-value layer halves intensity, so three layers give one eighth. Taking the narrow-beam HVL of lead at 511 keV as roughly 4 mm, that is about 12 mm, far more than the fraction of a millimetre that would suffice for Tc-99m. Shielding design uses larger broad-beam values.
15. Under narrow-beam conditions, the half-value layer of lead at 511 keV is closest to:
Answer: B — 4 mm
Lead attenuates 511 keV photons with a narrow-beam half-value layer of roughly 4 mm, against about 0.25 to 0.3 mm at the 140 keV of Tc-99m. That order of magnitude difference explains why PET shielding is so much heavier.Practical shielding design uses larger broad-beam figures, with AAPM guidance citing roughly 6 mm, because scattered photons are present as well.
16. A technologist wears a standard 0.5 mm lead-equivalent apron while positioning an injected PET patient. The apron:
Answer: A — Provides little useful attenuation at 511 keV
Aprons are designed for diagnostic x-ray energies. At 511 keV, 0.5 mm of lead is a small fraction of a half-value layer, so attenuation is minimal while the weight adds strain. Time and distance are the practical controls in PET.
17. Which practice contributes most to reducing extremity dose during FDG dose preparation?
Answer: C — Using tungsten shields and remote handling
Tungsten shields and automated or remote dispensing keep the hands away from an unshielded source. Gloves stop contamination but not penetrating photons, and abandoning the shield to save time increases dose rather than reducing it.
18. A technologist's annual dose is compared against the 5 rem occupational limit using which quantity?
Answer: B — Total effective dose equivalent
The 5 rem annual limit in 10 CFR Part 20 applies to the total effective dose equivalent, which combines external and internal exposure. Extremity and lens doses are tracked against their own separate limits, which is why PET staff wear both a body badge and a ring.
19. The NRC annual occupational dose limit to the extremities is:
Answer: A — 50 rem
The shallow dose equivalent limit to extremities is 50 rem, or 500 mSv, per year. Ring badges are worn in PET because hand doses during dose drawing and injection can be substantial.
20. A vial is calibrated at 08:00. Approximately what fraction of the F-18 activity remains at 11:40?
Answer: A — One quarter
F-18 has a half-life of 109.8 minutes, so 220 minutes is very close to two half-lives, leaving about one quarter. Treating roughly 110 minutes as one half-life is the practical shortcut used in the department.
21. FDG is retained within the cell because the molecule:
Answer: B — Cannot continue through glycolysis once phosphorylated
Hexokinase phosphorylates FDG to FDG-6-phosphate. Substitution of fluorine at the 2 position blocks further glycolysis, and the charged phosphate prevents the molecule leaving. FDG must cross the membrane to be trapped at all, via GLUT transporters.
22. A rubidium generator is typically replaced every four to six weeks. The property that sets this interval is the half-life of:
Answer: C — Strontium-82, about 25.5 days
Generator life is governed by decay of the parent. Sr-82 has a half-life of about 25.5 days, so useful yield falls after roughly a month. The daughter half-life sets elution timing, not replacement interval, and Ge-68 belongs to a gallium generator.
23. Imaging must begin essentially during the Rb-82 infusion because its half-life is about:
Answer: A — 75 seconds
Rb-82 decays with a half-life of roughly 75 seconds, so acquisition is tied directly to the infusion. The very short half-life keeps patient dose low and allows rest and stress imaging within a single short appointment.
24. A Ga-68 generator remains useful for about a year. Its parent radionuclide is:
Answer: D — Germanium-68
Ge-68 has a half-life of about 271 days and decays to Ga-68, which has a half-life of about 68 minutes. The long parent half-life gives the generator roughly a year of service and makes tracers such as DOTATATE available without a cyclotron.
25. N-13 ammonia is restricted to centres with a cyclotron on site because its half-life is about:
Answer: B — 10 minutes
N-13 decays with a half-life of roughly 9.97 minutes, so it cannot be transported any meaningful distance. This is the main practical reason many departments choose a rubidium generator for perfusion imaging instead.
The full bank has 200 NMTCB PET specialty questions, plus 34 clinical case studies, timed mock exams and flashcards. Every answer is explained the same way.
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