I85PREP
⌘ K
ABR Nuclear Radiology study materials
ABR Nuclear Radiology study resources

ABR Nuclear Radiology practice questions

One question at a time. Build your ABR Nuclear Radiology understanding.

First 50 of 200 questions. Each one lists the choices, the correct answer, and an explanation. The checker at the top of this page saves progress in this browser when it is available.

Continue in the practice tool above to load more questions, 50 at a time.

Question 1Physics, Instrumentation, and Radiation Detection

A radiopharmaceutical has a physical half-life of 6 hours and a biological half-life of 12 hours. What is its effective half-life?

  1. 6 hours
  2. 9 hours
  3. 4 hours Correct answer.
  4. 18 hours

Explanation

Effective half-life combines physical decay and biological clearance through 1/Te = 1/Tp + 1/Tb. Here 1/6 + 1/12 = 3/12, giving Te = 4 hours. The effective half-life is always shorter than either component considered alone.

Question 2Physics, Instrumentation, and Radiation Detection

In a NaI(Tl) gamma camera crystal, what is the principal role of the thallium dopant?

  1. It increases crystal density
  2. It shifts light output into the visible range Correct answer.
  3. It shortens the decay time to picoseconds
  4. It blocks Compton scatter

Explanation

Pure sodium iodide emits light poorly and at wavelengths mismatched to common photocathodes. Thallium activator centers provide luminescent sites that emit at roughly 415 nm, where photomultiplier tubes are efficient. Stopping power comes from the iodine content, not the dopant.

Question 3Physics, Instrumentation, and Radiation Detection

What is the principal function of the photomultiplier tubes in an Anger camera?

  1. To convert light into amplified voltage pulses Correct answer.
  2. To collimate the incoming gamma rays
  3. To convert gamma photons directly into current
  4. To separate radionuclides by decay time

Explanation

The photocathode releases electrons when struck by scintillation light, and the dynode chain multiplies them into a measurable pulse. Pulse amplitude is proportional to the number of light photons collected, which in turn reflects deposited energy. Collimation and gamma absorption occur before the tube.

Question 4Physics, Instrumentation, and Radiation Detection

Positron annihilation in tissue produces two photons with which properties?

  1. 1.022 MeV each, emitted at 180 degrees
  2. 140 keV each, emitted at 180 degrees
  3. 511 keV each, emitted near 180 degrees Correct answer.
  4. 511 keV each, emitted at 90 degrees

Explanation

Annihilation converts the combined rest mass of the positron and electron into two 511 keV photons, 1.022 MeV in total. Conservation of momentum makes them emerge almost back-to-back. Small angular deviations from residual momentum slightly degrade PET spatial resolution.

Question 5Physics, Instrumentation, and Radiation Detection

A technologist sets a 20% energy window centered on the 140 keV photopeak of Tc-99m. What is the approximate lower discriminator setting?

  1. 133 keV
  2. 126 keV Correct answer.
  3. 120 keV
  4. 112 keV

Explanation

A 20% window spans plus or minus 10% of the photopeak energy. Ten percent of 140 keV is 14 keV, so the window runs from 126 to 154 keV. The lower discriminator is therefore about 126 keV.

Question 6Physics, Instrumentation, and Radiation Detection

What is the primary purpose of the collimator on a gamma camera?

  1. To improve the crystal's energy resolution
  2. To restrict photon directions reaching the crystal Correct answer.
  3. To amplify the scintillation light before it is digitized
  4. To reduce the radiation dose to the patient

Explanation

Gamma rays arrive from all directions, so lead septa absorb off-axis photons and define the projection seen by the crystal. Without a collimator the image would be a blurred superposition of activity from the entire field of view. Collimation is achieved at the cost of sensitivity.

Question 7Physics, Instrumentation, and Radiation Detection

A dose calibrator measures the activity of a radiopharmaceutical by which principle?

  1. Scintillation counting in a well crystal
  2. Cerenkov light production in water
  3. Thermoluminescence of doped lithium fluoride
  4. Ionization of gas in a sealed chamber Correct answer.

Explanation

The dose calibrator is a sealed, gas-filled ionization chamber in which radiation creates ion pairs and a measurable current. That current is converted to activity using radionuclide-specific calibration settings. Routine clinical units do not provide energy spectra.

Question 8Physics, Instrumentation, and Radiation Detection

A technologist doubles her distance from an unshielded point source. What happens to the exposure rate?

  1. It halves
  2. It stays the same
  3. It increases fourfold
  4. It falls to one quarter Correct answer.

Explanation

Exposure from a point source falls as the inverse square of distance. Doubling distance divides the exposure rate by 2 squared, which is four. Distance is one of the most effective radiation protection tools.

Question 9Physics, Instrumentation, and Radiation Detection

On a NaI(Tl) gamma spectrum, which feature represents photons that scattered in the patient and then escaped the body?

  1. A broad continuum below the photopeak Correct answer.
  2. A peak at twice the photopeak energy
  3. A peak at the 511 keV annihilation energy
  4. A sharp peak at the full photopeak energy

Explanation

A photon that undergoes Compton scattering deposits only part of its energy before escaping, and the deposited fraction varies from event to event. These events produce a continuum extending from low energy up to the Compton edge rather than a discrete peak. The photopeak arises from full energy absorption.

Question 10Physics, Instrumentation, and Radiation Detection

Why is a 20% energy window centered on the photopeak preferred over a very wide window?

  1. It improves the spatial resolution determined by the collimator
  2. It increases the number of scattered events that are accepted
  3. It rejects most scattered photons while retaining photopeak events Correct answer.
  4. It shortens the physical half-life of the radionuclide

Explanation

Most Compton-scattered photons lose enough energy to fall below a narrow window centered on the photopeak. A 20% window keeps the majority of primary photons while excluding many scatter events, which improves contrast with only a modest loss of counts. Wider windows raise sensitivity but admit more scatter.

Question 11Physics, Instrumentation, and Radiation Detection

How is Tc-99m separated from Mo-99 in a radionuclide generator?

  1. By precipitation of molybdenum with sulfate
  2. By saline elution from an alumina column Correct answer.
  3. By extraction into an organic solvent
  4. By distillation of volatile technetium oxide

Explanation

Mo-99 is loaded as molybdate onto an alumina column, where it binds tightly. Tc-99m forms pertechnetate, which is far less tightly bound and is washed off with 0.9% saline. The daughter can be eluted repeatedly as it regrows from Mo-99 decay.

Question 12Physics, Instrumentation, and Radiation Detection

In a PET block detector, why is the scintillator only partially cut into small elements?

  1. To increase the light output of each crystal element
  2. To permit energy discrimination of 511 keV photons
  3. To reduce the scanner's dead time
  4. To identify which crystal absorbed the photon Correct answer.

Explanation

A single large crystal is partially sawed into small elements, and light is shared among a small number of photomultiplier tubes. Anger-type positioning of the shared light identifies the element that absorbed the photon. This yields fine pixels at much lower cost than one tube per crystal.

Question 13Physics, Instrumentation, and Radiation Detection

In filtered backprojection for SPECT, what is the purpose of the ramp filter?

  1. To smooth the image and reduce statistical noise
  2. To correct for detector nonuniformity
  3. To correct blurring inherent in backprojection Correct answer.
  4. To compensate for photon attenuation in tissue

Explanation

Simple backprojection smears each projection back across the image, producing a characteristic 1/r blur. The ramp filter applies a high-pass weighting in frequency space that cancels this blur and restores edges. Because it amplifies high-frequency noise, it is usually combined with an apodization window.

Question 14Physics, Instrumentation, and Radiation Detection

The Chang method for attenuation correction in SPECT estimates attenuation by using which approach?

  1. Applying a ramp filter before backprojection
  2. Measuring transmission through the patient with an external source
  3. Subtracting a scatter image acquired in a second energy window
  4. Using a uniform attenuation coefficient and computed path lengths Correct answer.

Explanation

Chang correction assumes a constant attenuation coefficient for the body and derives a body contour from emission or transmission data. Each pixel is corrected using its distance to the contour and the assumed coefficient. The correction is often applied iteratively to refine the estimate.

Question 15Physics, Instrumentation, and Radiation Detection

A region of interest records 10,000 counts. What is the approximate percent standard deviation of this measurement?

  1. 1% Correct answer.
  2. 0.1%
  3. 10%
  4. 0.5%

Explanation

Poisson statistics give a standard deviation equal to the square root of the number of counts, here 100. Dividing by 10,000 gives 0.01, or 1%. This equals 1 divided by the square root of N.

Question 16Physics, Instrumentation, and Radiation Detection

A low-energy high-resolution collimator is replaced by a low-energy high-sensitivity collimator. What is the expected effect on image quality?

  1. Worse resolution and increased sensitivity Correct answer.
  2. Improved resolution and decreased sensitivity
  3. Worse resolution and decreased sensitivity
  4. Improved resolution and increased sensitivity

Explanation

Sensitivity and resolution trade off against each other in collimator design. Larger holes or thinner septa admit more photons but permit more off-axis rays, widening the point spread function. A high-sensitivity collimator therefore increases count rate while degrading spatial resolution.

Question 17Physics, Instrumentation, and Radiation Detection

Why is Y-90 imaged by detecting bremsstrahlung rather than a photopeak?

  1. Y-90 emits positrons that annihilate in tissue
  2. Y-90 emits beta particles with no gamma ray Correct answer.
  3. Y-90 gamma emission overlaps the Tc-99m photopeak
  4. Y-90 decays only by isomeric transition

Explanation

Y-90 is essentially a pure beta emitter, and its weak internal pair production is not practical for imaging. As beta particles slow in tissue and surrounding plastic, they produce bremsstrahlung photons over a broad energy range. These are imaged with a medium-energy collimator and a wide window.

Question 18Physics, Instrumentation, and Radiation Detection

How does the random coincidence rate in PET change as activity in the field of view increases?

  1. With the square of activity Correct answer.
  2. Independent of activity
  3. Linearly with activity
  4. Inversely with activity

Explanation

A random coincidence occurs when two unrelated photons from different annihilations fall within the timing window. The chance of such a pairing is proportional to the product of the two singles rates, each of which scales with activity. The random rate therefore grows as the square of activity while trues grow linearly.

Question 19Physics, Instrumentation, and Radiation Detection

Which radionuclide is most appropriately counted with liquid scintillation rather than a NaI well counter?

  1. Tc-99m
  2. I-131
  3. H-3 Correct answer.
  4. F-18

Explanation

Tritium emits a low-energy beta particle with no gamma ray, which is heavily attenuated by the walls of a well counter. Liquid scintillation counting mixes the sample with fluor so the beta energy is converted to light within the cocktail. The gamma emitters listed are more conveniently counted in a well counter.

Question 20Physics, Instrumentation, and Radiation Detection

What is the consequence of inadequate collimator septal thickness for the photon energy being imaged?

  1. A fall in sensitivity across the field of view
  2. Septal penetration that lowers image contrast Correct answer.
  3. Improved spatial resolution at all depths
  4. Reduced acceptance of Compton scatter

Explanation

Septa must absorb photons that would otherwise cross from one hole into an adjacent one. Thin or short septa relative to photon energy allow such penetration, so events are assigned to the wrong position. The result is loss of image contrast rather than improved resolution.

Question 21Physics, Instrumentation, and Radiation Detection

A detector with paralyzable dead time is exposed to steadily increasing activity. What happens to the observed count rate at very high activities?

  1. It increases linearly without limit
  2. It plateaus at about 1/tau
  3. It peaks and then falls toward zero Correct answer.
  4. It decreases steadily even at low activities

Explanation

For a paralyzable system the observed rate is n = m times e to the minus m tau, where m is the true rate. As m rises, n reaches a maximum at m = 1/tau and then declines toward zero because every event extends the dead period. A nonparalyzable system instead plateaus at 1/tau.

Question 22Physics, Instrumentation, and Radiation Detection

A parallel-hole collimator has hole diameter d and hole length L. At a fixed source-to-collimator distance of 10 cm, it is replaced by a collimator with the same hole diameter but twice the length. What happens to collimator spatial resolution?

  1. It improves by exactly a factor of two
  2. It improves but by less than a factor of two Correct answer.
  3. It is unchanged
  4. It worsens by a factor of about two

Explanation

Parallel-hole resolution follows R approximately d(Leff + z)/Leff, where z is the source-to-collimator distance. Doubling Leff changes R by the factor (2Leff + z)/(2(Leff + z)), which lies between 0.5 and 1 whenever z is greater than zero. Resolution therefore improves, but by less than a factor of two.

Question 23Physics, Instrumentation, and Radiation Detection

In PET, how does the noise equivalent count rate behave as administered activity is increased?

  1. It falls monotonically with activity
  2. It is independent of activity
  3. It increases monotonically with activity
  4. It rises, peaks, then declines as randoms grow Correct answer.

Explanation

Noise equivalent count rate is defined as T squared divided by (T + S + kR), so it reflects statistical quality rather than raw counts. At low activity trues dominate and the value rises. At high activity, randoms grow as the square of activity and dead time losses accumulate, so the curve peaks and then falls.

Question 24Physics, Instrumentation, and Radiation Detection

In 2D PET, the attenuation factor applied to a line of response depends on which quantity?

  1. The energy of the emitted photons
  2. The depth of the annihilation in the body
  3. The activity concentration at the site of decay
  4. The total thickness along the response line Correct answer.

Explanation

Both 511 keV photons must escape the patient, so the survival probability is exp(-mu L), where L is the total attenuating path along the line of response. This depends only on total thickness, not on where along the line the annihilation occurred. That property makes measured attenuation correction practical in PET.

Question 25Physics, Instrumentation, and Radiation Detection

A gamma camera has an intrinsic spatial resolution of 3.5 mm, and the collimator resolution at a given source distance is 7.0 mm. What is the approximate system spatial resolution?

  1. 3.5 mm
  2. 10.5 mm
  3. 7.0 mm
  4. 7.8 mm Correct answer.

Explanation

Independent contributions to spatial resolution combine in quadrature, so Rs = square root of (Ri squared + Rc squared). Here that is the square root of (12.25 + 49), which is about 7.8 mm. The system value is always slightly worse than the dominant component.

Question 26Radiopharmacy and Radiopharmaceutical Chemistry

What is the physical half-life of technetium-99m?

  1. Thirteen hours
  2. Sixty-six hours
  3. Six hours Correct answer.
  4. Two hours

Explanation

Tc-99m decays by isomeric transition to Tc-99 with a 6.01-hour half-life. This short half-life permits same-day imaging while limiting absorbed dose.

Question 27Radiopharmacy and Radiopharmaceutical Chemistry

Which positron-emitting radionuclide is used in most clinical 2-[18F]fluoro-2-deoxy-D-glucose PET studies?

  1. Nitrogen-13
  2. Fluorine-18 Correct answer.
  3. Gallium-68
  4. Carbon-11

Explanation

FDG is labeled with fluorine-18, which has a 109.8-minute half-life and decays by positron emission. Its half-life is long enough for transport and typical PET workflows.

Question 28Radiopharmacy and Radiopharmaceutical Chemistry

Which radiopharmaceutical is most commonly used for whole-body bone scintigraphy?

  1. Tc-99m MDP Correct answer.
  2. Tc-99m sulfur colloid
  3. Tc-99m pertechnetate
  4. Tc-99m MAG3

Explanation

Tc-99m MDP accumulates in bone by chemisorption onto hydroxyapatite, with uptake related to osteoblastic activity and blood flow. It is the standard whole-body bone scan agent.

Question 29Radiopharmacy and Radiopharmaceutical Chemistry

A Tc-99m generator contains Mo-99, which decays to Tc-99m. What is the approximate half-life of the parent Mo-99?

  1. 13 hours
  2. 6 hours
  3. 66 hours Correct answer.
  4. 271 days

Explanation

Mo-99 has a half-life of about 66 hours and decays to Tc-99m on the generator column. The daughter Tc-99m has a much shorter 6-hour half-life.

Question 30Radiopharmacy and Radiopharmaceutical Chemistry

Radiochemical purity of a Tc-99m radiopharmaceutical refers to which quantity?

  1. Fraction of activity from the desired radionuclide
  2. Number of particles per milliliter in the dose
  3. Fraction of activity in the intended chemical form Correct answer.
  4. Amount of nonradioactive impurities present

Explanation

Radiochemical purity is the proportion of radioactivity in the desired chemical form, such as Tc-99m MDP rather than free pertechnetate. It is distinct from radionuclidic purity and chemical purity.

Question 31Radiopharmacy and Radiopharmaceutical Chemistry

Iodine-131 is commonly used for radioiodine therapy because it emits which radiation in addition to gamma photons?

  1. Neutrons
  2. Positrons
  3. Alpha particles
  4. Beta particles Correct answer.

Explanation

I-131 emits beta particles that deliver local therapeutic dose and gamma photons that permit imaging. The beta component is principally responsible for tissue ablation.

Question 32Radiopharmacy and Radiopharmaceutical Chemistry

Which parent radionuclide is used in a commercially available gallium-68 generator?

  1. Rubidium-82
  2. Strontium-82
  3. Germanium-68 Correct answer.
  4. Molybdenum-99

Explanation

Ge-68 decays with a 271-day half-life to Ga-68, which has a 68-minute half-life and can be eluted repeatedly from the generator. Sr-82/Rb-82 is a different generator pair.

Question 33Radiopharmacy and Radiopharmaceutical Chemistry

What is the purpose of adding stannous chloride to many Tc-99m radiopharmaceutical kits?

  1. To raise the pH of the final solution
  2. To reduce pertechnetate to a reactive lower state Correct answer.
  3. To bind technetium directly to the vial
  4. To sterilize the kit during reconstitution

Explanation

Stannous chloride is a reducing agent that converts pertechnetate to a lower oxidation state so it can bind the kit ligand. Without adequate stannous ion, labeling efficiency falls.

Question 34Radiopharmacy and Radiopharmaceutical Chemistry

During elution of a Mo-99/Tc-99m generator, which solution is passed through the alumina column?

  1. Normal saline Correct answer.
  2. Dilute hydrochloric acid
  3. Sterile water
  4. Sodium hydroxide

Explanation

Normal saline (0.9% NaCl) elutes sodium pertechnetate from the alumina column while Mo-99 remains bound. Water or extreme pH solutions can impair generator performance.

Question 35Radiopharmacy and Radiopharmaceutical Chemistry

Aluminum breakthrough in a Tc-99m eluate is clinically important primarily because aluminum can do which of the following?

  1. Increase the elution efficiency of the generator
  2. Convert Tc-99m to Mo-99 within the vial
  3. Shorten the physical half-life of Tc-99m
  4. Alter the biodistribution of subsequently prepared radiopharmaceuticals Correct answer.

Explanation

Aluminum breakthrough can cause particle aggregation and alter labeling or biodistribution of Tc-99m radiopharmaceuticals. USP limits aluminum to 10 µg/mL in eluate.

Question 36Radiopharmacy and Radiopharmaceutical Chemistry

Which radiopharmaceutical is most appropriate for measuring effective renal plasma flow because it is cleared by tubular secretion?

  1. Tc-99m sulfur colloid
  2. Tc-99m MAG3 Correct answer.
  3. Tc-99m MDP
  4. Tc-99m DMSA

Explanation

Tc-99m MAG3 is secreted by renal tubular cells and is used to estimate effective renal plasma flow. DMSA binds cortical tubules, while MDP and sulfur colloid are not renal flow agents.

Question 37Radiopharmacy and Radiopharmaceutical Chemistry

What is the principal mechanism by which Tc-99m sestamibi localizes in myocardium?

  1. Extracellular fluid distribution
  2. Capillary blockade after particle trapping
  3. Active transport into sympathetic nerve terminals
  4. Passive diffusion and mitochondrial retention Correct answer.

Explanation

Sestamibi is a lipophilic cationic complex that diffuses into myocytes and is retained by mitochondria in response to membrane potential. This underlies myocardial perfusion uptake.

Question 38Radiopharmacy and Radiopharmaceutical Chemistry

Which agent is used for hepatobiliary imaging because it is taken up by hepatocytes and excreted into bile?

  1. Tc-99m MAA
  2. Tc-99m MDP
  3. Tc-99m mebrofenin Correct answer.
  4. Tc-99m DTPA

Explanation

Tc-99m mebrofenin is an iminodiacetic acid derivative taken up by hepatocytes and excreted into bile. It is used for hepatobiliary imaging and gallbladder evaluation.

Question 39Radiopharmacy and Radiopharmaceutical Chemistry

For sentinel lymph node mapping, which property of Tc-99m sulfur colloid is most important for nodal retention?

  1. Its particle size and phagocytic uptake Correct answer.
  2. Its high-energy gamma emission
  3. Its ability to cross cell membranes
  4. Its rapid renal clearance

Explanation

Tc-99m sulfur colloid particles are phagocytosed by macrophages in lymph nodes, and particle size influences migration and retention. This enables sentinel node localization.

Question 40Radiopharmacy and Radiopharmaceutical Chemistry

FDG PET relies on trapping of the tracer after phosphorylation by which enzyme?

  1. Hexokinase Correct answer.
  2. Phosphofructokinase
  3. Glucose-6-phosphatase
  4. Pyruvate kinase

Explanation

FDG is phosphorylated by hexokinase to FDG-6-phosphate, which is trapped because it is a poor substrate for glucose-6-phosphatase and cannot proceed through glycolysis. The trapped activity reflects glucose uptake.

Question 41Radiopharmacy and Radiopharmaceutical Chemistry

Which radiopharmaceutical is used for PET bone imaging and localizes by chemisorption to hydroxyapatite?

  1. 18F-choline
  2. 18F-florbetapir
  3. 18F-FDG
  4. 18F-sodium fluoride Correct answer.

Explanation

18F-sodium fluoride localizes in bone by chemisorption to hydroxyapatite and is used for PET bone imaging. It is not a glucose, amyloid, or membrane tracer.

Question 42Radiopharmacy and Radiopharmaceutical Chemistry

Radiochemical purity of a Tc-99m labeled kit is typically assessed by which method?

  1. Sterility culture
  2. Gamma camera imaging
  3. Thin-layer chromatography Correct answer.
  4. Dose calibrator linearity

Explanation

Thin-layer chromatography separates the desired labeled compound from free pertechnetate and hydrolyzed reduced technetium, allowing radiochemical purity to be calculated. Gamma imaging and dose calibrator tests do not assess chemical form.

Question 43Radiopharmacy and Radiopharmaceutical Chemistry

Why is 18F-FDG typically formulated in a solution containing sodium chloride and buffered to near-neutral pH?

  1. To prevent phosphorylation by hexokinase
  2. To increase the positron range
  3. To make it suitable for intravenous injection Correct answer.
  4. To convert F-18 to F-19

Explanation

FDG for injection must be isotonic, near-neutral, and sterile to be safe for intravenous administration. These formulation features are part of its quality control.

Question 44Radiopharmacy and Radiopharmaceutical Chemistry

Which chelator is commonly used to attach Ga-68 or Lu-177 to peptides such as DOTATATE?

  1. HYNIC
  2. MAG3
  3. DOTA Correct answer.
  4. DTPA

Explanation

DOTA is a macrocyclic chelator that forms stable complexes with trivalent metals such as Ga-68 and Lu-177. HYNIC is typically used with Tc-99m and requires co-ligands.

Question 45Radiopharmacy and Radiopharmaceutical Chemistry

Which is a pure beta-emitting radiopharmaceutical used for intra-arterial liver tumor therapy?

  1. Ra-223 dichloride
  2. Lu-177 DOTATATE
  3. I-131 sodium iodide
  4. Y-90 microspheres Correct answer.

Explanation

Y-90 microspheres are administered intra-arterially for liver tumors and deliver beta radiation locally after embolization. Y-90 has a 64-hour half-life and essentially no gamma emission.

Question 46Radiopharmacy and Radiopharmaceutical Chemistry

A Mo-99/Tc-99m generator has reached transient equilibrium. After complete elution of Tc-99m, which statement best describes the Tc-99m activity that grows back into the column?

  1. It remains zero until the generator is eluted again
  2. It approaches a plateau set by remaining Mo-99 Correct answer.
  3. It immediately equals the Mo-99 activity at elution
  4. It decays with the 6-hour half-life only

Explanation

After elution, Tc-99m regrows as Mo-99 continues to decay, approaching a transient-equilibrium plateau determined by the remaining Mo-99 activity. The regrowth rate is governed by the 6-hour Tc-99m half-life.

Question 47Radiopharmacy and Radiopharmaceutical Chemistry

A radiopharmaceutical is described as carrier-free. What does this imply about its specific activity?

  1. Specific activity equals the activity concentration of the vial
  2. Specific activity is very high because stable isotope is absent Correct answer.
  3. Specific activity is zero because no carrier is present
  4. Specific activity is fixed by the physical half-life

Explanation

Carrier-free means no stable isotope of the radionuclide has been added, so the specific activity approaches the theoretical maximum. It does not mean zero activity.

Question 48Radiopharmacy and Radiopharmaceutical Chemistry

Which factor most limits the use of carbon-11-labeled radiopharmaceuticals to facilities with an on-site cyclotron?

  1. Its 20-minute half-life allows little transport time Correct answer.
  2. Its decay product is chemically toxic
  3. Its production requires a nuclear reactor
  4. Its gamma emission requires heavy shielding

Explanation

Carbon-11 has a 20.4-minute half-life, so most of the activity decays before long-distance transport; production and use generally require an on-site cyclotron. Carbon-11 is a positron emitter, not a reactor-produced gamma emitter.

Question 49Radiopharmacy and Radiopharmaceutical Chemistry

In radioiodination of a protein with iodine-131 using chloramine-T, what is the principal role of chloramine-T?

  1. It protects the protein from radiolysis
  2. It reduces disulfide bonds in the protein
  3. It oxidizes iodide to a reactive iodinating species Correct answer.
  4. It chelates iodide to the protein backbone

Explanation

Chloramine-T oxidizes iodide to reactive iodine species that undergo electrophilic substitution on tyrosine residues of proteins. It does not chelate iodide or reduce disulfides.

Question 50Radiopharmacy and Radiopharmaceutical Chemistry

A Tc-99m kit is reconstituted with eluate containing a high concentration of oxidizing impurities. Which effect is most likely?

  1. Improved binding of technetium to the chelator
  2. Reduced labeling efficiency and increased free pertechnetate Correct answer.
  3. Increased stannous ion available for reduction
  4. Conversion of Tc-99m to Mo-99

Explanation

Oxidizing impurities consume stannous ion and leave pertechnetate unreduced, lowering labeling efficiency and increasing free pertechnetate. They do not transform Tc-99m into Mo-99.