ASTR BIO Stdy Guide 2019 Flashcards
2019-I1. Which of the following statements concerning the interaction of photons with matter is CORRECT?
A. The probability of the photoelectric effect decreases with the atomic number of the absorber
B. The predominant interaction of 10 keV photons with soft tissue is the Compton process
C. In the Compton process, the energy of the scattered photon is less than that of the incident photon
D. Pair production occurs for photons with energies less than 1.02 MeV
E. There is only partial absorption of the energy of the incident photon in the photoelectric effect
Answer: C
In the Compton process, a photon interacts with an atom causing the ejection of an orbital electron. The incident photon, now with reduced energy, continues along a deflected path.
The probability of the photoelectric effect increases with the atomic number of the absorber Z3/E3 (Answer Choice A).
The predominant interaction of 10 keV photons in soft tissue is the photoelectric effect (Answer Choice B).
Pair production occurs for photons with energies greater than 1.02 MeV and results in the complete conversion of the photon’s energy into the production of a positron and electron (Answer Choice D).
For the photoelectric effect, there is complete absorption of the photon’s energy, resulting in ejection of an electron that possesses kinetic energy equal to the difference between the incident photon’s energy and the electron’s binding energy (Answer Choice E).
2019-I2. Which one of the following is a radiolysis product of water resnsible for the molecular damage caused by the indirect action of ionizing radiation?
A. eaq
B. 1O2
C. OH-
D. OH•
E. O2-
Answer: D
65-75% of the damage caused by indirect action is mediated by the hydroxyl radical, OH•.
Little biological damage is caused by the hydrated electron (eaq; Answer Choice A).
1O2 is produced primarily by photosensitizers and, rarely, by ionizing radiation (Answer Choice B).
Neither OH- nor O2- are primary radiolysis products, although O2- can be produced secondarily by reaction of eaq with O2 (Answer Choices C and E).
2019-I3. The approximate minimum photon energy required to cause ionization is:
A. 10-25 eV
B. 100-250 eV
C. 1-2.5 keV
D. 10-25 keV
E. 100-250 keV
Answer: A
On average, about 25 eV is required to create an ion pair in water, although the minimum energy needed to eject an electron is only 12.6 eV.
2019-I4. Which of the following X-ray interactions with matter is most important for producing high-contrast diagnostic radiographs?
A. Compton process
B. Pair production
C. Photoelectric effect
D. Nuclear disintegration
E. Coherent scattering
Answer: C
The photoelectric effect is the predominant interaction responsible for producing high quality diagnostic radiographs. At relatively low photon energies, the photoelectric effect is the most likely photon interaction and is the desirable type of photon/tissue interaction since there is complete photon absorption with no production of secondary photons.
The other possible tissue interactions at the photon energies used in diagnostic radiology are the Compton effect and coherent scattering. For these interactions, a deflected photon traveling in an altered direction is produced at the site of interaction. If these secondary photons are permitted to reach the film, there would be a reduction in image sharpness and loss of spatial resolution.
Furthermore, with the photoelectric effect, absorption of photons is dependent on the cube of the atomic number of the material. The resultant differential of absorption in tissue allows for the ability to differentiate between bone, soft tissue, and air.
2019-I5. Which of the following pairs of photon energy and predominant atomic interaction at the specified photon energy is correct?
A. 1 keV – pair production
B. 50 keV – triplet production
C. 100 keV – compton process
D. 2 MeV – photoelectric effect
Answer: C
The predominant atomic interaction for 100 keV photons is the Compton process.
Sources provide different answers on minimum energy for triplet production with some stating 2mC2 (1.02 MeV) and some stating 4mC2 (2.04 MeV) The photoelectric effect is predominant for photon energies in the range of 10 keV.
2019-I6. Which of the following statements is correct? High LET radiations:
A. Include 250 kVp X-rays, 200 MeV protons, and 1.1 MV X-rays
B. Produce much higher yields of OH radicals than do either X-rays or y-rays
C. Are components of solar flares but not of cosmic rays
D. Produce less dense ionization tracks than X-rays
E. Produce increased numbers of clustered lesions in DNA than X-rays
Answer: E
High linear energy transfer (LET), or densely ionizing, radiations include particles such as 290 MeV carbon ions, X-particles, and neutrons.
250 kVp X-rays, 200 MeV protons and 1.1 MV X-rays are all low LET, or sparsely ionizing, radiations (Answer Choice A).
Although high LET radiations produce more clustered lesions (multiply damaged sites) in DNA than low LET radiations (Answer Choice E), they actually produce lower yields of OH radicals because of the extensive ion and radical recombination within spurs and blobs (Answer Choice B).
High LET radiations, such as iron or carbon ions, are components of cosmic rays, while solar flares are composed largely of energetic protons (which are low LET; Answer Choice C).
2019-I7. The lifetime of an OH• radical is approximately:
A. 10-15 second
B. 10-9 second
C. 10-1 second
D. 1 second
E. 1 minute
Answer: B
The initial ionization process takes approximately 10-15 second. The primary radicals produced by the ejection of an electron typically have a lifetime of 10-10 second. The resulting hydroxyl radical has a lifetime of approximately 10-9 second. The DNA radicals subsequently produced have a lifetime of approximately 10-5 second.
2019-I8. Regarding pair production and annihilation, which of the following is true?
A. The incident photon is scattered with reduced energy
B. Annihilation photons always have an energy of 0.511 MeV each
C. A pair of orbital electrons are ejected from the atom
D. Two positrons are emitted at 180 degrees
E. It cannot occur if the photon energy is above 1.02 MeV
Answer: B
Annihilation photons always have an energy of 0.511 MeV each, which is equal to the rest energy of the positron and electron.
2019-I9. Directly ionizing radiation includes all of the following EXCEPT:
A. Electrons
B. Positrons
C. Alpha particles
D. Neutrons
E. Betas
Answer: D
Neutrons are not charged particles and, therefore, cannot ionize atoms directly. They do, however, transfer some of their energy to protons or light nuclei, which then cause ionization. They are, therefore, indirectly ionizing.
2019-I10. Concerning fast neutron interactions with matter, which of the following is FALSE?
A. They do not interact with atomic electrons of biological media
B. They interact primarily with oxygen in water
C. They may cause the ejection of an alpha particle
D. They may activate the target nucleus.
E. They may transfer a large fraction of its energy in the process of elastic scattering.
Answer: B
Fast neutrons with kinetic energy between a few and several tens of MeV
are slowed down in biological media mainly by elastic collisions with hydrogen nuclei (protons) of the cellular water.
A fraction of energy lost by fast neutrons in elastic collision with oxygen nuclei is less than 10% of that which occurs with hydrogen nuclei. For the beams of neutrons used in radiation therapy, recoil protons from elastic collistions produce a large density of ionizations along their tracks.
Neutrons do not interact with atomic electrons but, instead, interact with atomic nuclei.
Alpha particles can be produced by neutron capture reactions with isotopes of both carbon and oxygen, but the probability is strongly dependent on the neutron energy and target material.
Example:
17O + n → 14C + α
Neutron absorption in a target nucleus is called activation. This is a process by which neutron radiation induces radiaoactivity in materials. It occurs when atomic nuclei capture free neutrons, becoming heavier and entering excited states. The excited nucleus often decays immediately by emitting gamma rays, beta particles, alpha particles, fission products, and/or neutrons (in nuclear fission).
Neutron activation is a potential health hazard in therapy with high energy photons because when photons with energy > 10 MeV are utilized, neutrons are generated in linacs via the interaction of photons with nuclei of high atomic number materials within the linac head and the beam collimator systems.
These photoneutrons can have an energy of 0.1 to 2 MeV, are highly penetrating, have a quality factor of 20, and can significantly add to a patient’s off-field dose.
2019-I11. Which of the following results from the recombination of the initial water radiolysis products?
A. Solvated electron
B. Solvated proton
C. Hydrogen ion
D. Water
E. Only A and B
Answer: D
The main initial products of resulting from irradiation of pure water
are the short-lived free radicals, hydrogen radical (H•) (10%), hydroxyl radical (•OH) (45%), and the solvated electrons (e-aq ) (45%). These react with DNA or with each other.
Therefore,
•OH + H• → H2O
The remaining recombination reactions of free radicals are:
e-aq + e-aq +2 H2O → H2 + 2 OH-
•OH + •OH → H2O2
H• + H• → H2
These reactions always compete with reactions that lead to direct damage of the biological molecules. The relative efficiency of the recombinations will depend on the separation of the short-lived free radicals after the passage of the charged particle, and therefore depend on LET. At low LET values, the spacing of the ionizations is large. As a result, •OH radicals are widely separated thereby decreasing the probability of recombination to form H2O2.
As LET increases, the spacing between ionizations decreases and the probability of production of an •OH from one ionization event as well as an •OH from another ionization event along a single track increases. The yield of hydrogen peroxide increases rapidly with LET of about 20 - 150 keV/μm, the range of LET where direct damage to DNA dominates over indirect damage fro the free radicals.
2019-I12. When a live human cell is irradiated by gamma-rays, which one of the following events may eventually cause most of the damage to DNA?
A. Absorption of radiation energies by the chemical bonds in the DNA molecules
B. Ionization and excitation on atoms within the DNA structure
C. Ionization and excitation on atoms within the histones that are bound to DNA
D. Ionization and excitation of the water molecules that surround DNA
E. Direct damage to the lipids that may later oxidize DNA
Answer: D
The indirect effect mediated by free-radical reactions involving water are most responsible to cause DNA damage upon low LET irradiation
2019-II1. The SF2 (surviving fraction at 2 Gy) for an irradiated population of cells is most closely correlated with the:
A. Level of y-H2AX 30 minutes after irradiation
B. Level of y-H2AX present 24 hours after irradiation
C. Acetylation of H2AX on lysine 4
D. Rate of DNA single-strand break repair
E. Rate of thymine glycol repair
Answer: B
The nucleosome contains an octamer of core histones: H3, H4, H2A, and H2B. Histone variants and their post-translational modifications regulate chromosomal functions; the post-translational modifications include acetylation, methylation, and phosphorylation. Histone H2A has nine subtypes, among them the H2AX variant, which is involved in the response to DNA damage.
Production of DNA double-strand breaks (DSBs) by ionizing radiation leads to the rapid phosphorylation of histone H2AX on serine 139 (-H2AX). The specificity of this reaction provides a reliable yardstick for DSBs and the means to spatially localize DSBs within the nuclei of cells (the -H2AX focus assay). The degree of H2AX phosphorylation measured at a specific time after induction of the DSBs represents a balance between the rate of phosphorylation following DNA damage and the dephosphorylation that occurs as DNA repair progresses. S
F2, the cell surviving fraction after 2 Gy, is a model-independent measure of radiation sensitivity. The numbers of phosphorylated gammnaH2AX foci shortly after the irradiation represent the initial level of DNA damage, but the number of phosphorylated H2AX foci at 24 hours after irradiation represent the residual level of unrepaired DNA double strand break at this time. It has been shown that the number of phosphorylated sites remaining 24 hours after irradiation directly correlates with intrinsic radiosensitivity.
In contrast, after a 30 minute incubation, H2AX has been phosphorylated, but there has been little time for repair. A correlation between cell survival and the repair of either DNA single-strand breaks or thymine glycols has not been observed.
2019-II2. Which of the following statements about ionizing radiation (IR) induced DNA damage is correct?
A. IR causes only DNA double-strand breaks
B. IR may produce thymine glycols, but much less frequently than DNA double strand breaks
C. IR can cause more clustered lesions at low dose rates than at high dose rates
D. IR cannot cause oxidization of nucleotide bases
E. IR is unlikely to produce pyrimidine dimers
Answer: E
In contrast with the other forms of damage listed, pyrimidine dimers are principally produced following absorption of photons in the ultraviolet wavelength range and are not produced by X-rays.
Pyrimidine dimers are cytotoxic, but more of these DNA lesions are required in order to achieve cell death compared to the DNA lesions produced by X-rays. It is estimated that the number of DNA lesions per cell from X-rays necessary to kill 63% of the cell population (thereby allowing 37% to survive) is 40 double-stranded DNA breaks (DSBs).
In comparison, 1,000,000 pyrimidine dimers from ultraviolet (UV) radiation are needed to kill 63% of the cell population. IR can produce not only DSBs, but also other forms of damage including single strand breaks, thymine glycols, and base damage. These other forms of DNA damage, however, are more readily repaired and are less likely to result in cell death.
2019-II3. A clustered lesion:
A. Results from the creation of multiple DNA double-strand breaks (DSBs) within a particular exon of a gene following exposure to high LET radiation
B. Involves the formation of several DNA lesions within a highly localized region of DNA
C. Occurs more frequently as the LET of the radiation decreases
D. Represents the repair of multiple lesions within a gene
E. Results from transcription-coupled DNA repair
Answer: B
A clustered lesion, which has been hypothesized to play an important role in cell lethality, involves the formation of several DNA damages within a highly localized region of DNA.
2019-II4. Which one of the following assays would be the most appropriate to use for quantitative measurement of DNA double-strand breaks (DSBs) in cells immediately following exposure to ionizing radiation?
A. Alkaline elution
B. Western blotting
C. Neutral comet assay
D. PCR
E. BrdU incoporation assay
Answer: C
The neutral comet assay is used to measure DNA double-strand breaks (DSBs). The comet assay is the electrophoresis of single-cells in order to detect DNA damage and its repair. Cells are exposed to ionizing radiation, embedded in agarose, and then subjected to an electrical gradient to move the DNA into the gel. The negatively charged DNA in the cell moves through the agarose toward the positive electric pole. If there are no breaks, the cell’s DNA moves all together in a small ball. Double-strand DNA breaks creates DNA fragments that are smaller than the unbroken DNA and migrate further into the agarose making what appears like a comet’s tail. Alkaline conditions cause the separation of the two strands of the DNA helix and allows the visualization of DNA fragments created by both double-strand and single-strand DNA breaks. In neutral pH conditions, the DNA helix is intact so single-strand breaks do not result in separate fragments and you can only see the fragments created by double-strand DNA breaks.
Alkaline elution is used to measure single-strand breaks and some base damages (Answer Choice A)
Western blotting is for detection of proteins (Answer Choice B). Polymerase chain reaction (PCR) is used to amplify DNA sequences (Answer Choice D). The BrdU incorporation
2019-II5. Which statement regarding radiation-induced nuclear foci is correct?
A. ATR is the main apical kinase that responds to radiation-induced double-strand breaks
B. ERCC1-containing foci indicate the presence of radiation-induced single-strand breaks
C. Gamma-H2AX foci can be detected within 15 minutes of radiation exposure
D. p53 forms ATR-dependent foci within minutes of radiation exposure
Answer: C
One important characteristic of the cellular response to DNA lesions is the spatiotemporal manner by which repair and other proteins are recruited to the site of damaged DNA.
Frequently, these protein accumulations can be visualized as subnuclear “foci” using immunofluorescence microscopy. Ionizing radiation-induced DNA double-strand breaks activate ATM kinase, which phosphorylates multiple damage response and repair proteins.
ERCC1 is involved in nucleotide excision repair, in addition to roles in homologous recombination and replication fork repair but does not form subnuclear foci.
Histone H2AX is phosphorylated by ATM within 15 minutes after irradiation and can be visualized using a phospho-specific antibody. These gamma-H2AX foci are regarded a marker for radiation-induced DNA double-strand breaks in cells. p53 itself does not form foci, though specific ATM-dependent phospho-forms of p53 might be detected as foci. ATM functions in response to double strand breaks.
By contrast, ATR is activated during every S-phase to regulate the firing of replication origins, the repair of damaged replication forks and to prevent the premature onset of mitosis. Although ATR is activated in response to many different types of DNA damage including double strand breaks (DSB), a single DNA structure that contains a single-stranded DNA may be responsible for its activation. Furthermore, p53 does not form ATR-dependent foci.
2019-II6. Which of the following has been shown to be a reliable surrogate marker for DNA double strand breaks (DSBs) in the cells?
A. Phosphorylated histone variant H2AX (or y-H2AX)
B. Degraded histone H2AX
C. Dephosphorylated H2AX
D. Cleavage of Caspase 3
E. DNA methylation
Answer: A
The level of phosphorylated H2AX has been shown correlate with the level of DNA double strand breaks.
2019-III1. Double-strand DNA breaks caused by ionizing radiation trigger the transcription of DNA damage response genes. Which of the following proteins is a transcriptional transactivator?
A. p21 (CDKN1A)
B. p53 (TP53)
C. ATM
D. CHK1 (CHEK1)
E. TRAIL (TNFSF10)
Answer: B
In response to various forms of DNA damage, including double-strand breaks, p53 is stabilized and binds to the promoters of numerous target genes, including p21, activating their transcription. This transcriptional transactivation by p53 is an important component of the cellular DNA damage response.
ATM and CHK1 are protein kinases that are activated in response to double-strand breaks (Answer Choices C and D).
TRAIL is a ligand that induces cell death through the extrinsic apoptosis pathway (Answer Choice E).
2019-III2. Which of the following molecular events occurs earlier than the other events following the creation of a double-strand DNA break?
A. Destabilization of the mitochondrial outer membrane
B. Inactivation of the CDC25 phosphatases
C. Phosphorylation of CHK1 (CHEK1)
D. Activation of p21 (CDKN1A) transcription
E. Phosphorylation of histone H2AX
Answer: E
Phosphorylation of histone H2AX to y-H2AX occurs within several minutes of a cell being irradiated.
This modification is triggered by ATM and serves to mark the chromosomal site of the DNA break for the subsequent recruitment of signaling proteins, such as CHK1 kinase.
Activated CHK1 phosphorylates and inactivates CDC25 proteins, thereby causing the arrest of the cell cycle. P21 transcription is induced several hours after DNA damage, following the stabilization of p53 (TP53).
2019-III3. Which of the following statements is FALSE?
A. DNA repair by homologous recombination occurs preferentially in the G1 phase of the cell cycle
B. Non-homologous end joining is an error-prone repair pathway that involves DNA-PKcs (PRKDC)-associated repair of DNA double-strand breaks
C. The DNA repair proteins MRE11, NBS1 (NBN) and RAD50, localize at nuclear foci corresponding to presumed sites of DNA damage following exposure to DNA-damaging agents
D. A defect in nucleotide excision repair is the basis for the hereditary disorder xeroderma pigmentosum and can lead to increased rates of skin cancer
E. Following the production of DNA double-strand breaks, ATM is converted from an inactive dimer to an active monomer form
Answer: A
Homologous recombination requires a second copy of the relevant DNA duplex.
Although homologous recombination can take place in G1 phase, using the homologous chromosome as the template for repair, it occurs much more frequently after replication when the template strand is the sister chromatid located in close proximity to the damaged strand.
The sister chromatid is created during S-phase and serves as a template from which to copy the intact DNA sequence to the site of the damaged strand of DNA. It has been estimated that homologous recombination occurs 1000-fold more frequently in S and G2 than in G1. In G1, the principal form of DNA double-strand break repair is non-homologous recombination
2019-III4. Which of the following proteins is most involved in homologous recombinational repair of radiation-induced DNA double-strand breaks?
A. RAD51
B. XPG (ERCC5)
C. DNA-PKcs (PRKDC)
D. CHK1 (CHEK1)
E. TFIIH
Answer: A
RAD51 is a recombinase and plays a critical role in homologous recombinational repair of DNA double-strand breaks.
XPG is an endonuclease that cleaves the DNA strand on the 3’ side of the damage site. It also stabilizes the nucleotide excision repair pre-incision complex that is essential for the 5’ incision by the XPF (ERCC4) endonuclease (Answer Choice B).
The catalytic unit of DNA protein kinase (DNA-PKcs) plays a central role in non-homologous end joining (NHEJ) of DNA double-strand breaks through its recruitment by the KU70 (XRCC6)/80 (XRCC5) heterodimer to sites of DNA double-strand breaks, forming the DNA-dependent protein kinase holo-enzyme complex (DNA-PK; Answer Choice C).
CHK1 is a serine/threonine protein kinase and a key mediator of the DNA damage-induced checkpoint pathway (Answer Choice D).
TFIIH is associated with nucleotide excision repair (Answer Choice E).
2019-III5. An agent that inhibits non-homologous end joining (NHEJ) repair of radiation-induced DNA double-strand breaks might be expected to do all of the following, EXCEPT:
A. Affect the immune response
B. Sensitize cells to low dose rate irradiation
C. Decrease normal tissue tolerance during fractionated radiotherapy
D. Increase cellular radioresistance
E. Inhibit sublethal damage recovery
Answer: D
Inhibition of non-homologous end joining (NHEJ) would be expected to decrease cellular radioresistance.
An effect on immune response would be anticipated because inhibition of NHEJ would affect V(D)J recombination, thereby affecting antigen recognition (Answer Choice A).
Cells and tissues would be sensitized to low dose-rate irradiation since the recovery that occurs at low dose-rates depends at least in part upon repair of double-strand breaks by NHEJ (Answer Choice B).
Normal tissue tolerance doses would likely decrease due to radiosensitization (Answer Choice C).
Sublethal damage recovery would be inhibited since this process depends at least in part on the repair of double-strand breaks (Answer Choice E).
2019-III6. All of the following proteins are involved in non-homologous end-joining of DNA double-strand breaks, EXCEPT:
A. XRCC4
B. RAD52
C. Artemis (DCLRE1C)
D. KU70 (XRCC6)/KU80 (XRCC5)
E. DNA ligase IV (LIG4)
Answer: B
RAD52 plays a central role in homologous recombinational repair (HR) of DNA double-strand breaks through recruitment of RAD51 to single-stranded DNA complexed with RPA. RAD52 does not appear to be involved in NHEJ.
XRCC4 is an adaptor protein that tightly complexes with DNA ligase IV, which directly mediates DNA-strand joining by NHEJ (Answer Choice A).
The KU70/KU80 heterodimer recruits DNA-PKcs (PRKDC) to the site of DNA double-strand breaks to form a multiprotein complex that keeps broken DNA ends in close proximity and provides a platform for the enzymes required for end processing and ligation (Answer Choice D).
DNA-PKcs phosphorylate the Artemis protein, thereby activating it for endonucleolytic activity. The Artemis:DNA-PKcs complex cleaves 5´ and 3´ nucleotide overhangs, which prepares double-strand breaks for ligation by XRCC4 and DNA ligase IV (Answer Choice C and E).
2019-III7. A mutation in which of the following genes is LEAST likely to cause an increase in sensitivity to ionizing radiation:
A. NBS1 (NBN)
B. BRCA1
C. ATM
D. MRE11
E. XPC
Answer: E
XPC is a gene whose product is involved in nucleotide excision repair (NER). Mutations in XPC result in the human genetic disease xeroderma pigmentosum, which is characterized by extreme sensitivity to ultraviolet light.
Mutations in all of the other genes result in human genetic diseases characterized by sensitivity to ionizing radiation, including Nijmegen breakage syndrome (NBS1), familial breast cancer (BRCA1), ataxia telangiectasia (ATM), and ataxia telangiectasia-like disorder (MRE11).
2019-III8. Which of the following statements concerning DNA repair is CORRECT?
A. Cells deficient in nucleotide excision repair tend to display hypersensitivity to ionizing radiation
B. A person with LIG4 syndrome is radiation sensitive
C. Mismatch repair involves the action of a DNA glycosylase and an AP endonuclease
D. People with Fanconi anemia exhibit normal sensitivity to DNA cross-linking agents
E. A mutation in p53 (TP53) produces an immune deficient phenotype in SCID mice
Answer: B
People diagnosed with LIG4 syndrome are radiation sensitive because these individuals are deficient in the DNA ligase IV enzyme (LIG4), which plays a central role in non-homologous end joining (NHEJ) of double-strand breaks.
Cells deficient in nucleotide excision repair exhibit normal sensitivity to ionizing radiation, since this repair process plays little or no role in the repair of damages induced by ionizing radiation, but are very sensitive to UV radiation (Answer Choice A).
Base excision repair (BER), not mismatch repair, involves the action of a DNA glycosylase and an AP endonuclease (Answer Choice C).
People with Fanconi anemia are highly sensitive to DNA cross-linking agents due to inhibition of the mono-ubiquitination of FANCD2, a downstream Fanconi anemia protein, following genotoxic stress (Answer Choice D).
NOTE: FANconi = DNA CROSS linking sensitivity (think of the wooden links of a fan crossed together)
The immune deficient phenotype in SCID mice is caused by a defect in XRCC7 (DNA-PKcs), which is critical for NHEJ as well as V(D)J rejoining. As a result, a defect in XRCC7 leads to a radiosensitive phenotype as well as the immune deficits seen in the SCID mouse. Defects in several genes are now known to cause SCID phenotypes; the mutation in the common human disease of the same name (severe combined immunodeficiency) differs from that in the well-known mouse strain.
2019-III9. Two of the main proteins involved in mismatch repair are:
A. MSH2/MLH1
B. DNA ligase IV (LIG4)/XRCC4
C. KU70 (XRCC6)/KU80 (XRCC5)
D. XPA/XPG (ERCC5)
E. DNA-PKcs (PRKDC)/Artemis
Answer: A
MSH2 and MLH1 play a central role in mismatch repair. XPA/XPG are involved in nucleotide excision repair (Answer Choice D).
DNA Ligase IV (XRCC4), Ku70, and DNA-PKcs all play roles in NHEJ (Answer Choices B, C, and E).
NOTE: XPA = Xeroderma pigmentosum (XP); both this and Cockayne Syndrome (CS) genes are involved in NER
2019-III10. Which of the following best describes the action of an exonuclease enzyme?
A. Seals breaks in a DNA strand
B. Adds a new nucleotide to the end of DNA during DNA synthesis.
C. Produces nicks within intact DNA strands
D. Generates new species of mRNA
E. Removes nucleotides from the ends of DNA strands
Answer: E
An exonuclease cleaves one nucleotide at a time beginning at the end of a DNA strand.
2019-III11. Which of the following statements is CORRECT? Base excision repair (BER):
A. May increase mutation rate when defective, but usually does not dramatically alter cellular radiosensitivity
B. Is the principal pathway responsible for the repair of UV-induced DNA damage
C. Involves the XP and CS genes
D. Acts primarily on bulky DNA lesions induced by polycyclic aromatic hydrocarbons
E. Is defective in patients with Li-Fraumeni Syndrome
Answer: A
Defects in base excision repair (BER) may increase mutation rate but generally do not alter cell survival after ionizing radiation with the exception of mutation of the XRCC1 gene, which would confer a slight increase in radiation sensitivity.
Defects in nucleotide excision repair (NER) increase sensitivity to UV radiation but not to ionizing radiation (Answer Choice B).
The xeroderma pigmentosum (XP) and Cockayne Syndrome (CS) genes are involved in NER (Answer Choice C).
BER acts to remove damaged bases from DNA, including those damaged by ionizing radiation, but NER acts on pyrimidine dimers, single-strand breaks, and bulky adducts (Answer Choice D).
The gene defective in most patients with Li-Fraumeni Syndrome is p53, although some patients with that condition have mutations in CHK2 (Answer Choice E).
2019-III12. Which statement regarding the roles of non-homologous end-joining (NHEJ) and homologous recombination (HR) in the repair of ionizing radiation-induced DNA double-strand breaks (DSBs) is TRUE?
A. HR removes DSBs from the genome at a faster rate than NHEJ
B. Defects in HR compromise DSB repair but do not affect the repair of damage at DNA replication forks
C. NHEJ requires homologies of 200-600 nucleotides between broken ends of DNA
D. Defects in NHEJ increase radiosensitivity more than defects in HR in mammalian cells.
Answer: D
Two principal recombinational DNA repair pathways have been identified, homologous recombination (HR) and non-homologous end-joining (NHEJ), each of which employs separate protein complexes.
DSB repair by HR requires an undamaged template molecule that contains a homologous DNA sequence, typically derived from the sister chromatid in the S and G2 phase cells.
In contrast, NHEJ of double-stranded DNA ends, which can occur in any cell-cycle phase, does not require an undamaged partner and does not rely on extensive homologies between the recombining ends (typically 2-6 bp of microhomology are used).
Defective HR can be causally linked to impaired DNA replication, genomic instability, human chromosomal instability syndromes, cancer development, and cellular hypersensitivity to DNA damaging agents. Cells with genetic defects in NHEJ (such as mutation of DNA-PK, XRCC4, or DNA ligase IV) display a more pronounced hypersensitivity to ionizing radiation than cells defective in HR (such as mutation of BRCA1, BRCA2, or RAD51).
2019-III13. Chemotherapeutic agents frequently produce DNA double-strand breaks (DSBs) by causing stalling and collapse of DNA replication forks. Which of the following pathways has a dominant role in the repair of replication-associated double-strand breaks?
A. Non-homologous end-joining (NHEJ)
B. Homologous recombination (HR)
C. Single-strand annealing (SSA)
D. Translesional DNA synthesis (TLS)
E. Nucleotide excision repair (NER)
Answer: B
Several DNA repair pathways, including translesional DNA synthesis (TLS), nucleotide excision repair (NER), and homologous recombination (HR) can be mobilized at stalled DNA replication forks depending on the type of fork-blocking lesion.
Chemotherapy-induced DNA lesions, such as interstrand crosslinks, interfere with the progress of the replicative DNA helicase or DNA polymerases, thereby leading to replication fork blockage or demise and producing DNA gaps or one-sided DNA double-strand breaks (DSBs).
Uncoupling of the replicative DNA helicase from the polymerases may occur generating excessive single-stranded DNA, which could in turn be the target of endonucleoytic processing, resulting in a one-sided DSB. In addition, single-stranded breaks induced by endogenous and exogenous sources may lead to the formation of one-sided DSBs due to runoff of the replication fork.
In the repair of one-sided DSBs, HR appears to be the only pathway leading to their productive resolution. This entails resection of the DSB to form a 3′-tailed end for Rad51 filament assembly and DNA strand invasion and ultimately reconstruction of the replication fork.
2019-III14. A human disorder thought to be due to a DNA repair deficiency is which of the following:
A. Lesch-Nyhan syndrome
B. Xeroderma pigmentosum
C. Tay-Sachs disease
D. Phenylketonuria
E. Down syndrome
Answer: B
Xeroderma pigmentosum is associated with NER
NOTE:
Lesch-Nyhan syndrome - deficiency of the enzyme hypoxanthine-guanine phosphoribosyltransferase (HGPRT)
Tay-Sachs disease - mutation of HEXA gene on chromosome 15, which codes for a subunit of the hexosaminidase enzyme known as hexosaminidase
Phenylketonuria - phenylalanine hydroxylase deficit
Down syndrome - trisomy 21
2019-III15. Which of the following statements is TRUE regarding BRCA1 and BRCA2:
A. BRCA1 and BRCA2 mutations account for only a few cases of familial hereditary breast and ovarian cancer
B. BRCA1-deficient cells are resistant to the DNA crosslinking agent mitomycin C
C. The prevalence of BRCA1 mutation is higher than that of BRCA2 mutations
D. BRCA1 and BRCA2 predominantly regulate homologous recombination as opposed to non-homologous end joining
E. The breast cancer risks for carriers of BRCA1 and BRCA2 mutations are similar but with later age of disease onset for the BRCA1 mutation
Answer: D
BRCA1 and BRCA2 predominantly regulate homologous recombination (HR) as opposed to non-homologous end joining (NHEJ)
2019-III16. Which of the following gene mutations would be expected to cause the greatest increase in sensitivity after exposure to a DNA damaging agent that induces double-strand breaks (DSBs)?
A. DNA-PKcs null mutation
B. P53 null mutation
C. Activiating K-Ras mutation
D. MLH1 nonsense mutation
E. XRCC1 null mutation
Answer: E
Ionizing radiation produces multiple types of DNA damage, including DNA double strand breaks (DSB), single strand breaks (SSBs), and base damage.
Many more instances of SSBs and base damage are induced compared to DSBs. Although DSBs are the most lethal form of damage, this is not due to an inability to detect or repair these lesions, but instead because DSBs have the strongest impact on cell viability through the generation of lethal chromosome aberrations.
NOTE:
XRCC1 is involved in the efficient repair of DNA single-strand breaks formed by exposure to ionizing radiation and alkylating agents
XRCC2/3 encodes a member of the RecA/Rad51-related protein family that participates in homologous recombination to maintain chromosome stability and repair DNA damage
XRCC4 is the key protein that enables interaction of LigIV to damaged DNA
2019-IV1. Which of the following statements concerning chromosome aberrations produced in cells after whole body X-irradiation is TRUE?
A. The formation of terminal deletions follows an exponential dose response
B. Translocations are an unstable type of chromosome aberration
C. The number of dicentric chromosomes detected in peripheral blood lymphocytes remains relatively constant with time
D. SKY (spectral karyotyping) is a useful method for detection of stable aberrations decades following irradiation
E. The minimum dose that can be estimated by scoring dicentric chromosomes is 2 Gy
Answer: D
Spectral karyotyping (SKY) uses fluorescence staining of chromosomes employing uniquely-colored probes specific for individual chromosomes, thus allowing them to be distinguished from each other on the basis of color. Stable translocations are revealed using SKY as a single chromosome that appears to be multi-colored. The formation of terminal deletions follows a linear dose response since these are single-hit aberrations (Answer Choice A). Translocations can be stable aberrations since they do not necessarily lead to cell death (Answer Choice B). The number of dicentric chromosomes detected in peripheral blood lymphocytes decreases with time after irradiation since these are unstable aberrations that ultimately cause the death of the lymphocyte progenitors and stem cells (Answer Choice C). The minimum dose that can be detected through scoring dicentric chromosomes is roughly 0.25 Gy (Answer Choice E).
2019-IV2. Which of the following types of chromosome aberrations is most responsible for the formation of micronuclei observed after irradiation?
A. Sister chromatid exchanges
B. Chromatid gaps
C. Inversions
D. Quadriradials
E. Acentric fragments
Answer: E
Micronuclei are created due to the presence of acentric fragments, which form in the progeny of irradiated cells that undergo mitosis in the presence of one or more asymmetrical chromosome aberrations.
Sister chromatid exchanges are reciprocal exchanges between chromatids of the same chromosome that are not readily induced by ionizing radiation (Answer Choice A).
Chromatid gaps appear as loss of genetic material from a single chromatid arm and may be caused by incomplete breaks (Answer Choice B).
Inversions result when two breaks are produced in a single chromosome and the resulting excised chromosomal fragment reinserts itself back into the chromosome, but with the opposite polarity (Answer Choice C).
A quadriradial is a chromatid-type aberration that may arise from illegitimate interchromosomal recombination, accompanied by crossing-over (Answer Choice D).
2019-IV3. Which of the following is the best measure for the presence of radiation-induced chromosome aberrations in interphase cells?
A. Reciprocal translocations
B. Ring chromosomes
C. Dicentric chromosomes
D. Micronuclei
E. Chromatid breaks
Answer: D
Individual chromosome aberrations can, in general, be detected readily only during mitosis. However, some chromosome aberrations lead to the formation of micronuclei, which develop when a pseudo nuclear membrane forms around acentric chromosome fragments or whole chromosomes that did not segregate properly into daughter cells during the previous mitosis.
Micronuclei are observed in peripheral lymphocytes and thus can be seen in interphase cells.
2019-IV4. Which one of the following statements concerning the induction of chromosome aberrations is INCORRECT?
A. Primary radiation-induced breaks can reconstitute without apparent morphological change to the chromosome, rejoin illegitimately with another break site to produce an intra- or inter-chromosomal aberration, or remain “open,” leading to a simple break
B. The induction and interaction of DNA double-strand breaks is the principal mechanism for the production of chromosome aberrations
C. Dicentrics, centric rings, and translocations are formed following X-irradiation of cells in the G0/G1 phase of the cell cycle, and their formation follows a linear-quadratic dose response
D. Fluorescence in situ hybridization (FISH) using multi-colored probes has allowed chromosome aberration complexity to be studied in detail
E. Chromatid type aberrations are observed when cells are irradiated during the G1 phase of the cell cycle
Answer: E
Chromatid type aberrations are produced in cells only when irradiation follows DNA synthesis in S phase.
NOTE: Interphase is the portion of the cell cycle that is not accompanied by observable changes under the microscope, and includes the G1, S and G2 phases. During interphase, the cell grows (G1), replicates its DNA (S) and prepares for mitosis (G2).
2019-IV5. The formation of dicentric chromosome aberrations follows a linear-quadratic dose response curve. This has been interpreted to mean that the production of dicentric chromosomes results from:
A. Two chromosome breaks, produced either by one or by two separate radiation tracks
B. Two chromosome breaks produced by two separate radiation tracks
C. Two chromosome breaks produced by a single radiation track
D. One chromosome break produced by two separate radiation tracks
E. One chromosome break produced by a single track of radiation
Answer: A
The formation of dicentric chromosomes is linear at low radiation doses but follows a quadratic function at higher doses. Two distinct mechanisms are thought to be responsible for these two components of the linear-quadratic dose response curve.
The linear portion of the dose response relationship is assumed to result from the simultaneous induction of two chromosome breaks by a single track. The quadratic portion is assumed to result from the two chromosome breaks being produced by two separate radiation tracks.
2019-IV6. Which of the following statements concerning chromosome aberrations is TRUE?
A. A ring chromosome is an example of a chromatid-type aberration
B. A dicentric is a stable chromosome aberration
C. Breakage of a single chromatid in G2 often leads to the formation of an anaphase bridge
D. Terminal deletions are induced as a linear function of dose
E. For low LET radiation, the yield of dicentric chromosomes is inversely proportional to the dose-rate
Answer: D
Terminal deletions are induced as a linear function of dose since they result from a single chromosomal break.
A ring chromosome is an example of a chromosome-type aberration, not a chromatid-type aberration (Answer Choice A).
A dicentric is an unstable aberration since it results in the formation of an acentric fragment and ultimately causes cell death (Answer Choice B).
Breaks in two chromatids, followed by illegitimate rejoining, produce an anaphase bridge (Answer Choice C).
The yield of dicentric chromosomes increases with increasing dose-rate for low LET radiation (Answer Choice E)
2019-IV7. Increased numbers of chromosome aberrations, especially quadriradials, are frequently found even in the absence of radiation in which of the following human syndromes?
A. Xeroderma pigmentosum
B. Fanconi anemia
C. Cockayne’s syndrome
D. Niemann-Pick disease
E. Li-Fraumeni syndrome
Answer: B
Blood cells from individuals with Fanconi anemia are often found to have high numbers of chromosome aberrations, especially quadriradials.
These complex aberrations increase dramatically with exposure to DNA cross-linking agents such as mitomycin c.
NOTE: “FAN”coni = DNA cross “X”-linking; “QUAD”riradials
2019-V1. Pathways that trigger apoptosis culminate in widespread intracellular proteolysis. Which of the following proteases is a downstream executioner that directly participates in the breakdown of numerous cellular proteins?
A. caspase-8 (CASP8)
B. caspase-9 (CASP9)
C. caspase-3 (CASP3)
D. caspase-10 (CASP10)
E. XIAP (BIRC4)
Answer: C
Apoptotic signals trigger a series of proteolytic events known as the caspase cascade. There are at least 14 human caspases, which fall into two categories:
The initiator caspases (caspases-_2, -8, -9 and -10), which activate the downstream caspases, and the executioner caspases (caspases-_3, -6 and -7), which cleave cellular substrates.
The actions of the executioner caspases produce the cellular effects that distinguish apoptosis from other forms of cell death.
XIAP is a protein that binds to and inhibits the action of caspases.
2019-V2. Which of the following statements regarding cell death following radiotherapy is TRUE?
A. The majority of solid epithelial tumors regress during treatment because of radiation-induced apoptosis
B. The intrinsic apoptotic pathway can be triggered either by radiation-induced DNA damage or by sphingomyelin-mediated damage to the outer plasma membrane
C. A novel drug that abolishes the G1 checkpoint would be expected to reduce the incidence of mitotic catastrophe in irradiated cells.
D. Cells that undergo replicative senescence following radiotherapy are characterized by increased membrane blebbing and DNA fragmentation
E. The presence of γ-H2AX histone foci in irradiated cells is indicative of sphingomyelin activation
Answer: B
The intrinsic apoptotic pathway can be triggered either by damage to DNA or by damage to the plasma membrane. Radiation acts directly on the plasma membrane, activating acid sphingomyelinase, which generates ceramide by enzymatic hydrolysis of sphingomyelin. Ceramide then acts as a second messenger in initiating an apoptotic response via the mitochondrial system.
Mitotic catastrophe, and not apoptosis, is the major mechanism of cell death in epithelial tumors. Inhibition of the G1 checkpoint in irradiated cells may increase the probability of mitotic catastrophe since cells are more likely to enter mitosis with damaged chromosomes.
Radiation-induced senescent cells cease dividing and can remain metabolically active for extended periods before dying, but do not show membrane blebbing and DNA fragmentation, which are characteristic of apoptosis.
γ-H2AX foci noted in the nuclei of irradiated cells are indicative of the presence of DNA double-strand breaks.
2019-V3. Radiation-induced cellular senescence is often the result of:
A. Cellular nutrient deprivation
B. Oxidative stress secondary to mitochondrial dysfunction
C. p16-mediated cell cycle arrest
D. Telomere shortening
E. Mitotic catastrophe
Answer: C
The term “senescence” refers to the loss of cellular replicative potential leading to a reduced capability to repopulate a tissue after exposure to genotoxic agents, including ionizing radiation. Senescence is most often the result of a permanent arrest in G1, associated with elevated expression of the cell cycle inhibitors p16INK4A (CDKN2A) and p21 (CDKN1A, WAF1/CIP1). Importantly, senescence is not a type of cell death per se because cells remain morphologically intact and metabolically active when senescent.
Depending on the level of tumor suppressor proteins and the oncogenic signal, senescence can be reversible in a small subset of cells though in most cells this process is irreversible.
A clinically relevant scenario for radiation-induced senescence is the loss of salivary gland function and xerostomia commonly seen in head and neck cancer patients undergoing radiotherapy. Another one is radiation-induced premature senescence in fibroblasts that triggers proinflammatory and profibrotic senescence associated secretory phenotype (SASP) and ultimately drives fibrosis in the lung.
Mitochondrial dysfunction is a hallmark of apoptotic cell death, not senescence (Answer Choice B).
Telomere shortening occurs in most normal somatic cells as part of each cell cycle (“end replication problem”) and triggers senescence once a critical low threshold is reached, but telomere shortening tends not to be the cause for radiation-induced senescence which is driven by DNA-damage and cell cycle arrest (Answer Choice D).
Nutrient deprivation can lead to autophagy, and ultimately autophagic death cell distinct from apoptosis (Answer Choice A).
2019-V4. The extrinsic pathway of apoptotic cell death requires:
A. Signals derived from changes in chromatin conformation
B. Activation of death receptors that translocate from the plasma membrane to the nucleus and degrade DNA
C. Engagement of death receptors located on the plasma membrane that lead to activation of the initiator caspase-8 (CASP8)
D. p53 (TP53) activation
E. The triggering of changes in mitochondrial membrane potential
Answer: C
There are two principal pathways that can lead to apoptotic death. One of these, the extrinsic pathway, involves extracellular signaling through death receptors located on the plasma membrane such as TRAILR-1 (TNFRSF10A), TRAILR-2 (TNFRSF10B) or FAS (CD95/APO-1).
These death receptors are activated in response to ligand binding of TRAIL (TNFSF10) or FAS ligand (FASLG/CD95-L) and signal through a series of adapter molecules such as the adapter molecule Fas- associated death domain (FADD) within the death-inducing signalling complex (DISC). Upon recruitment and oligomerization FADD then binds pro-caspases-8 and -10, causing their homodimerization and activation.
The activation of procaspase-8 is thought to occur via an induced proximity model leading to its conversion to the active enzyme, caspase-8. Ionizing radiation can elicit activation of the extrinsic pathway leading to apoptosis. The other pathway by which ionizing radiation can elicit an apoptotic response is the intrinsic pathway. This can be stimulated by DNA damage leading to signaling to mitochondria, changes in mitochondrial membrane potential, release of cytochrome c, and activation of procaspase-9.
In most cases, activated caspase-8 induces apoptosis through activation of pro-caspase-3 at the DISC independently of mitochondria. However, in some cells, especially when only a low amount of active caspase-8 is generated (and hence not sufficient amounts of pro-caspase-3), caspase-8 cleaves the ‘Bcl-2 homology (BH) 3-only protein’ Bid, generating an active fragment (tBid) that activates the (intrinsic) mitochondrial death pathway. In this manner, the extrinsic death signal may be amplified through formation and activation of the apoptosome which contributes to effector caspase activation.
In other words, the extrinsic pathway can feed into the intrinsic one and additionally change mitochondrial membrane potential.
2019-V5. One hallmark of the apoptotic process is the display of phosphatidylserine residues on the outer surface of the plasma membrane. This is an important event in terms of the tissue response to ionizing radiation because it:
A. Helps recruit death ligands expressed by neighboring cells to receptors on the cell surface
B. Stimulates an inflammatory response to remove dying cells from the tissue
C. Signals the recruitment of phagocytes that engulf the dying cells without causing an inflammatory response
D. Is required for DNA condensation and fragmentation
E. Leads to increased ceramide levels
Answer: C
Apoptosis helps maintain tissue homeostasis because cells that are undergoing an apoptotic response recruit phagocytes that clear the dying cells, also known as “apoptotic corpses”, from the tissue without stimulating an inflammatory response.
In fact, uptake of apoptotic cells by macrophages can actually lead to the release of anti-inflammatory mediators such as TGF-β and IL-10, and the attenuation of the RIG-I/IRF-3 pathway and the cGAS/STING pathway through proteolytically inactivating RIP kinase 1 or the degradation of cytoplasmic DNA. As a result, apoptosis can decrease the expression of interferons and other inflammatory factors.
Of note, the concept that apoptosis is entirely non-inflammatory isn’t always strictly true. An example is the induction of apoptosis in hepatocytes following FAS activation that causes a strong inflammatory response probably because they can’t get cleared fast enough by phagocytes.
The exposure of phosphatidylserines (phospholipids) on the exterior of the plasma membrane is the signal that initially recruits phagocytes. Ordinarily, phosphatidylserine is sequestered within the phospholipid bilayer and is not displayed on the cell’s surface. The process of necrosis, which involves rupture of the cell membrane and the leakage of cellular contents into the surrounding tissue, does elicit an inflammatory response.
While DNA condensation and fragmentation are important steps in the apoptotic process, they are not coordinated directly through the exposure of phosphatidylserine on the plasma membrane. A number of stimuli lead to increased ceramide levels, including TNF, FasL and ionizing radiation, but not phosphatidylserine.
2019-V6. Regarding the regulation of apoptosis, which of the following pairs of mammalian proteins and their apoptosis-related functions is FALSE?
A. p53 (TP53) — upregulation of PUMA
B. DIABLO — caspase activation
C. XIAP (BIRC4) — caspase inhibition
D. BAX — cytochrome c release
E. caspase-3 (CASP3) — initiator caspase
Answer: E
The characteristic changes associated with apoptosis are due to activation of a family of intracellular cysteine proteases, known as caspases.
Initiator caspases are the first to be activated, and include caspases-2, -8, -9 and -10. Initiator caspases cleave and activate the effector/executioner caspases, including caspases-3, -6, and -7, which then cleave, degrade or activate other cellular proteins.
Activation of caspases is regulated by members of the BCL2 family and by the inhibitors of apoptotic protein (IAP) family. BAX is one of a series of pro-apoptotic members of the BCL2 family. These pro-apoptotic BCL2 family members regulate the release of cytochrome c from mitochondria and elicit the subsequent activation of caspases.
Another important function of p53 is that it causes upregulation of pro-apoptotic PUMA.
X-linked IAP (XIAP) inhibits the activity of caspases directly.
DIABLO is a pro-apoptotic protein that prevents IAPs from inhibiting caspases.
BAX and p53 are required for some forms of DNA damage-induced apoptosis.
2019-V7. Which ONE of the following is a morphological or biochemical feature of apoptosis?
A. Random cleavage of DNA
B. Cellular swelling
C. Lack of dependence on ATP as an energy source
D. Chromatin condensation
E. Rupture of the plasma membrane
Answer: D
During the apoptotic process, endonucleases cut the DNA at precise sites corresponding to the linker region between nucleosomes. This leads to the formation of fragments that are multiples of 80 bp units. There is no cell swelling, such as occurs in necrosis, but rather cell shrinkage after the apoptotic process begins followed by condensation of chromatin at the periphery of the nucleus.
Apoptosis is an energy-dependent process requiring ATP. During the apoptotic process, the plasma membrane initially remains intact but later fragments and surrounds the apoptotic bodies.
2019-V8. The TUNEL assay used to identify apoptotic cells detects:
A. The action of BAX on the mitochondria
B. Membrane integrity
C. Mitochondrial release of cytochrome c
D. Binding of TNFα to its receptor
E. DNA fragmentation
Answer: E
The terminal deoxynucleotidyl transferase (TdT) mediated deoxyuridine triphosphate (dUTP) nick end-labeling (TUNEL) technique has been used to identify apoptotic cells.
It is based upon the binding of TdT to the exposed 3’-OH terminal ends of DNA fragments generated during apoptosis and catalyzes the addition of modified deoxynucleotides, conjugated with biotin or fluorescein, to the DNA termini.
2019-V9. Which of the following best describes radiation-induced bystander effects?
A. Damage to unirradiated normal tissue noted after irradiation of a tumor
B. Cell killing that results from irradiation of the cell’s cytoplasm in the absence of direct irradiation of the nucleus
C. Radiation-induced increase in cell membrane permeability that causes increased sensitivity to cytotoxic drugs
D. DNA and/or chromosomal damage that occurs in unirradiated cells that are nearby irradiated cells
E. Intercellular communication that modifies the shoulder region of the radiation survival curve
Answer: D
While damage to cellular DNA was long considered the major initiator of cellular responses to ionizing radiation, more recent evidence suggests the involvement of non-targeted pathways, including radiation-induced bystander effects. Bystander effects are defined as radiation-like effects observed in cells that are not themselves irradiated, but that are in communication with irradiated cells through their location near these cells or by stimuli transferred from the irradiated cells through the intracellular medium. Various endpoints have been measured as bystander effects, including enhanced cell killing, induction of apoptosis, presence of chromosome aberrations and micronuclei, presence of DNA double-strand breaks, increased oxidative stress, genetic effects (including induction of mutations, and neoplastic transformation) and altered gene expression
2019-V10. Mitotic death in irradiated cells results primarily from:
A. The mis-rejoining of DNA single strand breaks.
B. DNA ladder formation.
C. Stimulation of the extrinsic death pathway.
D. Mis-assortment of genetic material into daughter cells.
E. An alteration in cell membrane permeability.
Answer: D
Mitotic death in most irradiated cells results primarily from mis-assortment of genetic material into daughter cells as a result of the formation of asymmetrical chromosome aberrations. This aberrant mitosis triggers mitotic catastrophe, which is characterized by cells exhibiting multiple tubulin spindles and centrosomes as well as the formation of multinucleated giant cells that contain uncondensed chromosomes. Mitotic death can be of any molecular mechanism, including apoptosis or necrosis.
Single strand breaks are repaired rapidly and do not appear to play an important role in cell lethality (Answer Choice A).
DNA ladder formation is characteristic of apoptosis (Answer Choice B).
An alteration in cell permeability occurs in cells undergoing necrosis (Answer Choice E).
2019-V11. Which of the following concerning autophagy is INCORRECT?
A. Autophagy is a reversible process that can contribute both to tumor cell death and survival
B. Anti-malarial drugs, chloroquine and hydroxychloroquine, are the only U.S. Food and Drug Administration–approved inhibitors of autophagy
C. Autophagy contributes to cellular metabolism by degradation of damaged protein aggregates and organelles
D. Mitophagy referes to autophagy in mitotic cells
E. Autophagy is controlled by the Atg family of proteins
Answer: D
Autophagy can be nonselective or selective. Nonselective, bulk degradation of cytoplasm and organelles by autophagy provides material to support metabolism during periods of cellular stress. For example, autophagy provides internal nutrients, when external ones are unavailable.
Whether mechanisms exist to prevent bulk autophagy from consuming essential components, such as a cell’s final mitochondrion, remains unclear, and in some cases such consumption may lead to cell death. Selective autophagy of proteins and of organelles such as mitochondria (mitophagy), ribosomes (ribophagy), endoplasmic reticulum (reticulophagy), peroxisomes (pexophagy), and lipids (lipophagy) occurs in specific situations.
Autophagy (‘self-eating’) tends to refer to macroautophagy: the sequestration process of cytoplasmic material for degradation. (Microautophagy and chaperone-mediated autophagy are other types.) After initiation, an isolation membrane encloses a small portion of cytoplasmic material, including damaged organelles and unused proteins, to form a double-membraned structure called an “autophagosome” that subsequently fuses with lysosomes to become an “autolysosome”, in which the cytoplasmic material is degraded by lysosomal enzymes.
The whole process is tightly regulated through at least 30 Atg-autophagy related genes that orchestrate initiation, cargo recognition, packaging, vesicle nucleation expansion and fusion and breakdown. The initial steps center around the Atg1 complex that translocates to the ER, (thought to be the major membrane source for autophagy).
This leads to recruitment of the autophagy-specific form of the PI(3)K complex, which includes Vps34, Vps15, Atg6/Beclin-1 and Atg14, to the ER. To form an autophagosome, elongation and closure of the isolation membrane requires 2 protein conjugation systems, the Atg12–Atg5–Atg16 complex and the Atg8/LC3–phosphatidylethanolamine (PE) complex. Detection of autophagy relies on the redistribution of GFP-LC3 fusion proteins into vesicular structures (which can be autophagosomes or autolysosomes).
‘Autophagic cell death’ is the excessive version of autophagy, that occurs in the absence of chromatin condensation. In contrast to apoptotic cells, there is little or no association of autophagic cells with cells phagocytes. Although the expression ‘autophagic cell death’ is a linguistic invitation to believe that cell death is executed by autophagy, the term simply describes cell death with autophagy.
2019-VI1. Which of the following in vivo assays of radiation response does NOT depend on a functional endpoint?
A. LD50
B. Skin nodule formation
C. Myelopathy
D. Breathing rate
E. Cognitive impairment
Answer: B
A functional endpoint for radiation response is a measured endpoint that is downstream of clonogenic survival and may involve measurement of tissue/organ function, the incidence of toxicity, or whole animal survival.
Clonogenic endpoints directly measure the replicative capacity of cells (e.g., colony formation). Skin nodule formation is not a functional endpoint; it is a clonogenic assay measuring survival of individual epidermal cells regrowing in situ.
All of the other assays cited represent non-clonogenic, functional endpoints for assaying radiation damage.
2019-VI2. Using the linear-quadratic survival curve model, what would the cell surviving fraction be following a dose of 2 Gy delivered acutely (use α=0.3 Gy-1 and β=0.1 Gy-2)?
A. 0.01
B. 0.10
C. 0.37
D. 0.50
E. 0.90
Answer: C
Using the equation SF = e-(αD+βD^2)
The surviving fraction would be:
e-[(0.3)(2)+(0.1)(2)^2]
= e-[(0.6)+(0.4)]
= e-1
= 0.37
2019-VI3. For α=0.3 Gy-1 and β=0.1 Gy-2, what would be the approximate surviving fraction if a 2 Gy dose were delivered at a low dose rate over a 6 hour period instead of acutely (assume no repopulation takes place during the irradiation)?
A. 0.10
B. 0.20
C. 0.37
D. 0.55
E. 0.90
Answer: D
If the dose was delivered at a low dose rate, the surviving fraction would increase due to repair of sublethal damage during the course of irradiation.
If one assumes that there is full repair of sublethal damage during the 6 hr irradiation (which is probably an oversimplification), sublethal damage would not contribute to cell killing.
The β component of the LQ equation would therefore approach zero, leaving the α component to dominate.
The surviving fraction can therefore be estimated as e-(0.3)(2) = e-0.6 = 0.55
2019-VI4. Which clonogenic assay has been used to measure the radiation sensitivity of bone marrow stem cells in vivo?
A. Dicentric assay
B. BrdU (BrdUrd) assay
C. Endpoint dilution assay
D. In vivo/in vitro excision assay
E. Spleen colony assay
Answer: E
The spleen colony assay involves the ability of donated bone marrow stem cells, injected intravenously into lethally-irradiated recipient mice, to form discrete splenic colonies.
The higher the radiation dose received by the donated marrow, the fewer colonies (relative to the number of cells injected) will form in the recipients’ spleens. This technique allows a cell survival curve to be generated in vivo.
2019-VI5. The components typically required for the analysis of a standard, adherent cell clonogenic survival assay require all of the following, EXCEPT:
A. Calculation of a plating efficiency
B. Colony formation rates at a range of cell densities, for several radiation doses
C. A cell line capable of multiple cell divisions
D. Intact apoptosis pathways
E. Nonirradiated control
Answer: D
The clonogenic survival assay measures the ability of single cells to divide continuously after a given exposure, and typically measures colony formation 7-14 days after exposure to the agent. It requires a normalization in which the number of colonies formed is divided by the number of cells seeded (in the absence of any DNA damaging agent), which yields the plating efficiency.
Surviving fraction is then calculated for each dose of a given agent by dividing the number of colonies formed by the number of cells seeded and normalizing to the “0 Gy” plating efficiency. Multiple doses and cell densities typically are needed for the adequate analysis of cell survival. This is not a short-term growth delay assay, and thus a cell capable of multiple cell divisions is needed.
DNA damaging-agents induce cell death via a number of pathways, including apoptosis. However, apoptosis is not the sole cell death pathway.
2019-VII1. If a cell line exhibiting a strictly exponential radiation survival curve is exposed to a dose that produces an average of one lethal “hit” per cell, the surviving fraction after this dose would be approximately:
A. 0.01
B. 0.10
C. 0.37
D. 0.50
E. 0.90
Answer: C
Assuming that all cells in the cell population are identical and that cell killing is a random, probabilistic process that follows a Poisson distribution, one model that can calculate the radiation dose that produces an average of one lethal hit is the single-target single-hit model.
From the equation that describes this model,
S = e-D/D0
The dose, D, at which there would be an average of one hit per cell would be equal to D0, the constant of proportionality.
Therefore, S = e-1 ~ 0.37.
2019-VII2. The α/β ratio is equal to the:
A. Surviving fraction at which the amount of cell killing caused by the induction of irreparable damage equals the amount of cell killing caused by the accumulation of sublethal damage
B. Optimal fraction size to use in a fractionated radiotherapy regimen
C. Dose below which a further decrease in fraction size will not affect the surviving fraction for a particular total dose
D. Dq
E. Dose at which the aD component of cell kill is equal to the bD2 contribution to cell killing
Answer: E
The α/b ratio represents the dose at which the αD component of cell killing, assumed to result from single hit killing, is equal to the lethality produced by the βD2 component of cell killing that results from multi hit killing.
2019-VII3. Cells from individuals diagnosed with which of the following diseases/syndromes would be expected to have an X-ray survival curve with a relatively large D0?
A. Nijmegen breakage syndrome
B. LIG4 syndrome
C. ATR-Seckel syndrome
D. Xeroderma pigmentosum
E. Ataxia telangiectasia
Answer: D
D0 is a measurement of radiosensitivity made on the exponential part of the survival curve. It is defined as a dose that gives an average of one lethal hit per cell. A dose of D0 Gy reduces survival from 1 to 0.37. Hence, the smaller D0 is, the more radiosensitive the cells are.
Cells derived from an individual diagnosed with xeroderma pigmentosum are defective in nucleotide excision repair (NER). These cells are sensitive to UV radiation because this form of radiation produces damages such as pyrimidine dimers that are removed through the nucleotide excision repair pathway.
Because DNA double-strand breaks are important lesions responsible for lethality in cells exposed to X-rays and because DNA double-strand break repair is generally normal in cells derived from a person diagnosed with xeroderma pigmentosum, the D0 determined from a radiation survival curve for these cells would not be particularly small.
People with Nijmegen breakage syndrome, LIG4 syndrome, ATR-Seckel syndrome and ataxia telangiectasia, who possess mutations in either NBS1, LIG4, ATR or ATM, respectively, are all characterized by defects in strand break repair or repair-related signaling.
Therefore, at least a small increase in radiosensitivity (a decrease in the D0) would be expected in cells derived from people with these syndromes.
2019-VII4. Which of the following statements concerning cell survival curve analysis is TRUE?
A. The β parameter generally increases as the radiation dose rate decreases
B. The inverse of the Dq corresponds to the final slope of the survival curve
C. The extrapolation number, n, of a survival curve increases with increasing LET of the radiation
D. D0 is a measure of the incremental increase in cell survival when a given dose is fractionated
E. If n = 1, then D37 = D0
Answer: E
Parameters to define a radiation cell survival curve include: the initial slope (D1), a final slope (D0), and some quantity that is a measure of the width of the shoulder. This quantity can be the extrapolation number (n) or the quasi-threshold dose (Dq).
If n=1, the survival curve has no shoulder and D37 (dose resulting in a survival fraction of 0.37) equals the D0.
For the same radiation dose, radiation delivered at a lower dose rate may produce less cell killing than radiation delivered at a higher dose rate because sublethal damage repair occurs during the protracted exposure.
As the dose rate is reduced, the slope of the survival curve becomes shallower and the shoulder tends to disappear because α does not change significantly but β trends to zero in the linear quadratic model.
The inverse of the D0, not the Dq, is equal to the final slope of the survival curve.
For densely ionizing radiation (increasing LET), the shoulder of the survival curve tends to disappear. N, therefore, decreases, until it reaches a value of 1.0 for very high LET radiations.
The D0 would not necessarily be a good predictor for the effect of fractionation on survival; Dq or n would be better.
2019-VII5. Reducing the dose rate at which a continuous y-irradiation is delivered may affect its cell killing efficacy due to several different biological processes. For a total dose of 6 Gy, which pair of dose rate ranges and biological processes resulting in altered cell killing is INCORRECT?
A. 10 - 1 Gy/min : reoxygenation
B. 1 - 0.1 Gy/min : repair
C. 0.1 - 0.01 Gy/min : redistribution
D. 0.01 - 0.001 Gy/min : repopulation
Answer: A
Reoxygenation generally occurs over a period of hours to days. Little to no reoxygenation of hypoxic cells is therefore likely during irradiation performed at dose rates in the 1-10 Gy/min range since, for a total dose of 6 Gy, the irradiation times would only vary from 0.6-6 minutes.
If the total treatment time is long enough that significant repair of sublethal damage (half-time on the order of 0.5-1.0 hour) can occur during irradiation, repair does influence cell survival. The irradiation time would vary from 6-60 minutes for dose-rates in the range of 1-0.1 Gy/min and significant repair would occur (Answer Choice B).
Movement of the surviving cells through the cell cycle (causing redistribution of viable cells from resistant phases into sensitive phases) can influence the radiosensitivity of the cell population when irradiation times are increased to several hours (for the dose-rate range of 0.1-0.01 Gy/min, times of 1-10 hours would be needed to produce 6 Gy; Answer Choice C).
Repopulation can lead to an increase in the number of cells during irradiation and, hence, to an increase in the total number of surviving cells when a radiation dose is delivered over days (10-100 hours are required to produce a total dose of 6 Gy over a range of 0.01-0.001 Gy/min; Answer Choice D).
2019-VII6. The survival curve for a cell population irradiated with a form of high LET radiation is characterized by a D10 of 3 Gy. For a starting population of 108 cells, approximately how many cells will survive when a single dose of 18 Gy is given?
A. 100
B. 101
C. 102
D. 103
E. 104
Answer: C
For high LET radiation it can be assumed that the survival curve is exponential, or near exponential, and cell survival can be modeled using the single-target, single-hit equation (S = e), or the simplified form of the linear quadratic equation in which β is zero (S = e-αD).
Using either of these equations, 3 Gy reduces the surviving fraction to 10-1, and a dose of 18 Gy therefore would reduce survival to 10-6. Therefore, irradiating 108 cells with 18 Gy would result in the survival of:
(108 cells) x (10-6 surviving fraction)
= 102 cells.
2019-VII7. A total dose of 12 Gy of X-rays delivered in 3 Gy fractions reduces cell survival to 10-4. Assuming that cell killing can be modeled using an exponential survival curve, what dose would be required to reduce the surviving fraction to 10-6?
A. 9 Gy
B. 18 Gy
C. 24 Gy
D. 36 Gy
E. 72 Gy
Answer: B
An exponential survival curve can be modeled using the single-target, single-hit equation (S = e), or the simplified form of the linear quadratic equation in which β is zero (S = e-αD).
Since four 3 Gy fractions reduce the surviving fraction to 10-4, and assuming an equal effect per fraction, each 3 Gy fraction reduces the surviving fraction by 10-1.
Accordingly, two additional 3 Gy fractions (producing a total dose of 18 Gy) would yield a surviving fraction of 10-6.
2019-VII8. In an attempt to generate a radiation survival curve for a new cell line, four cell culture dishes were seeded with 102, 103, 104 and 105 cells, and X-irradiated with 0, 3, 6 and 9 Gy, respectively. At the end of a two-week incubation period, a total of 40 colonies was counted on each dish. Which one of the following statements is TRUE?
A. The D0 for this cell line is 3 Gy
B. The survival curve for this cell line is exponential
C. The n and Dq values for this survival curve are large
D. The cell surviving fraction after a dose of 3 Gy is 0.04
E. The alpha-beta ratio for this cell line is small
Answer: B
The survival curve for this cell line is exponential because each incremental dose of 3 Gy decreased the surviving fraction by an additional factor of 0.1.
Thus, this survival curve can be modeled using an exponential equation which can be expressed as either S= e-αD (linear-quadratic model) or S = e-D/D0 (target theory model).
The D0 is equal to the D10/2.3, or 1.3 Gy, not 3 Gy. The n and Dq values for this survival curve are equal to 1 and 0 Gy, respectively, and are therefore small, not large.
The surviving fraction after a single dose of 3 Gy can be calculated from the colony forming efficiency of the irradiated cells (40/1000), divided by the plating efficiency (PE) of the unirradiated cells (40/100), which is equal to 0.04/0.4 or 0.1.
Since this survival curve can be represented by S = e-αD, the β term of the linear-quadratic equation must approach zero, so the α/β would be very high, and in fact will be undefined if the β term is actually zero.
2019-VII9. What is the approximate eD10 (effective D10) for a particular cell line if the eD0 is 4 Gy?
A. 2 Gy
B. 4 Gy
C. 6 Gy
D. 9 Gy
E. 12 Gy
Answer: D
eD10 is the dose required to kill 90% of population
eD10 = 2.3 x D0
The eD10 is equal to the eD0 multiplied by 2.3, or:
4 Gy x 2.3 = 9.2 Gy
2019-VII10. When irradiating a cell population with a dose that causes an average of one lethal event per cell, this will likely result in a survival fraction of:
A. 0% of cell survival.
B. 10% of cell survival.
C. 37% of cell survival
D. 63% of cell survival
E. 100% of cell survival
Answer: C
When the irradiated cell population receives an average of 1 lethal hit, it results in 37% cell survival based on Poisson statistics.
2019-VII11. A 1 cm-diameter tumor that contains 107 clonogenic cells is irradiated with daily dose fractions of 1.8 Gy. The effective dose response curve has been determined and is exponential with a D10 of 8 Gy. What total dose will correspond to the TCD90 (90% probability of tumor control), assuming no cell proliferation between dose fractions?
A. 32 Gy
B. 40 Gy
C. 48 Gy
D. 56 Gy
E. 64 Gy
Answer: E
VII-11 through VII-13: Answers and Explanations are missing.
NOTE:
D10 = 8 Gy
107 x 10 -x = 10 -1 (90% tumor control)
x = 8, so 8 x D10 = 64 Gy
2019-VII12. Based on the information presented in the previous question (1 cm-diameter tumor that contains 107 clonogenic cells is irradiated with daily dose fractions of 1.8 Gy. The effective dose response curve has been determined and is exponential with a D10 of 8 Gy), what would be the TCD90 if a surgical excision removed 99% of the tumor clonogens prior to radiotherapy (assume that the surgery did not otherwise affect the growth fraction of the tumor).
A. 24 Gy
B. 32 Gy
C. 40 Gy
D. 48 Gy
E. 56 Gy
Answer: D
VII-11 through VII-13: Answers and Explanations are missing.
99% = 10-2; so an initial surgery removing 99% of the 107 tumor would lead to 105 cells at start of radiation
D10 = 8 Gy
105 x 10-x = 10-1 (90% tumor control)
x = 6, so 6 x D10 = 48 Gy
2019-VII13. For a tumor that requires 18 days to double its diameter, what is the approximate cell cycle time of its constituent cells (assume no cell loss and that all cells are actively dividing)?
A. 6 days
B. 9 days
C. 12 days
D. 15 days
E. 18 days
Answer: A
VII-11 through VII-13: Answers and Explanations are missing.
NOTE: To simplify - consider a cube. In order to double the diameter, you need 8 cubes (23).
Thus 3 doublings or 18/3 = 6 days
2019-VIII1. Concerning RBE, OER, and LET, which of the following statements is TRUE?
A. Maximum cell killing per dose delivered occurs at an LET corresponding to approximately 1000 keV/μm
B. RBE changes the most over the LET range of 0.1 to 10 keV/μm
C. The relationship between OER and LET is bell-shaped
D. RBE decreases with increasing LET above about 100 keV/μm
E. OER increases with LET
Answer: D
RBE decreases with increasing LET above approximately 100 keV/μm. This is thought to be due to the “overkill” effect in which many more ionizations (and damage) are produced in a cell traversed by a very high LET particle than are minimally necessary to kill it, thereby “wasting” some of the energy.
Maximum cell killing occurs at an LET of approximately 100 keV/μm, not 1000 keV/μm (Answer Choice A).
RBE shows the greatest changes for LET values between roughly 20 and 100 keV/μm (Answer Choice B).
OER decreases slowly with increasing LET for low LET values, but falls rapidly after LET exceeds about 60 keV/μm and, therefore, does not follow a bell-shaped curve (Answer Choices C and E).
2019-VIII2. Which of the following statements is correct? Compared with damage from low LET radiation, damage from high LET radiation:
A. Is reduced to a greater extent in the presence of sulfhydryl compounds
B. Shows more potentially lethal damage recovery
C. Exhibits a greater OER
D. Is less subject to split-dose recovery
E. Shows greater sparing when the irradiation is given at a low dose rate
Answer: D
There is little or no split-dose recovery following high LET radiation exposure because the single dose survival curves for high LET radiations have little or no shoulder. There is also little or no potentially lethal damage recovery, oxygen effect or radioprotection afforded by the presence of sulphydryl compounds. Delivery of a radiation dose at a low dose rate leads to less sparing for a high LET radiation compared with a low LET radiation.
2019-VIII3. Which of the following statements concerning RBE is TRUE? The RBE:
A. Is lower for neutrons than for protons over the therapeutic energy range
B. Is greater for high LET particles in hypoxic cells as compared to oxygenated cells of the same type
C. Is diminished for carbon ions when delivered over several fractions as compared to a single dose
D. Is greatest for heavy charged particles at the beginning of the particle track
E. Increases for MeV alpha-particles with increasing dose
Answer: B
Relative Biological Effectiveness (RBE) is defined as:
Doseof Reference Radiation (250 keV X-Rays)/Doseof Test Radiation to give the same biological effect
The reference radiation for calculation of RBE is low LET radiation, such as 250 keV X-rays or Co-60.
The dose of the reference radiation that will achieve the same level of cell killing as high LET particles in hypoxic cells will be greater because there is little to no oxygen effect for high LET radiation (Answer Choice B).
The RBE is greater for neutrons than it is for protons in the therapeutic energy range because the high energy protons used in radiotherapy are of a relatively low LET and therefore possess an RBE of approximately 1.1 (Answer Choice A).
The RBE for carbon ions, or any other type of high LET radiation, is greater for fractionated irradiation compared with an acute exposure due
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to the substantial sparing exhibited with reference X-rays with fractionation (Answer Choice C).
The RBE for charged particles is low at the beginning of the particle track and greatest near the end of the track, in the Bragg peak region (Answer Choice D).
RBE does show a fractionation dependence; it decreases with increasing fraction size. The RBE for 4 MeV alpha particles will decrease with increasing dose because there is more sublethal damage repair with low-LET X-rays at lower doses, and therefore more survival compared with high-LET radiation (Answer Choice E).
2019-VIII4. Which of the following pairs of radiation type and approximate LET value is CORRECT?
A. 150 MeV protons – 0.5 keV/μm
B. 1 GeV Fe ions – 20 keV/μm
C. 60Co y-rays – 15 keV/μm
D. 2.5 MeV a-particles – 5 keV/μm
E. 250 kV X-rays – 10 keV/μm
Answer: A
60Co y-rays 0.2 keV/μm
150 MeV protons 0.5 keV/μm.
250 kV X-rays 2 keV/μm
1 GeV Fe ions 143 keV/μm
2.5 MeV a-particles 166 keV/μm
2019-VIII5. Which of the following statements concerning LET is INCORRECT?
A. LET is proportional to charge density of a medium
B. LET is proportional to charge (squared) of the particle moving through a medium
C. LET is inversely proportional to speed (squared) of the particle
D. LET is inversely proportional to mass of the particle moving through a medium
E. LET is related to density of ionizations along the particle’s track
Answer: D
LET is a measure of local energy deposition along a track of medium. It is inversely proportional to the energy of a given charged particle. The local transfer of energy to medium is more probable at lower energies.
2019-VIII6. Which statement concerning the linear energy transfer (LET) is CORRECT?
A. LET is equal to the energy transferred by ionizing radiation to soft tissue per unit mass of soft tissue
B. LET is equal to the number of ion pairs formed per unit track length
C. Once a photon transfers all its energy to an electron, the LET is that of the electron
D. LET is the quotient of the average energy that a particle lost in causing ionization to the average distance it travels between two consecutive ionizations
E. The track average method and the energy average method for calculating LET give different numerical values for therapy protons in soft tissue
Answer: D
Photons, such as 250 KV X-rays, in passing through tissue produce no ionizations directly but only by setting in motion atomic electrons of tissue molecules. Electrons set in motion by incident photons have a broad energy distribution which is dissipated in tracts with LET ranging from about 0.4 to 40 keV/μm.
Radiation therapy high energy photons can generate neutrons with energy between 0.1 to 2 MeV through photon interactions with nuclei of high atomic number materials that constitute the linac head and collimator systems. These neutrons in passing through tissue also produce no ionization directly but by setting protons in motion by knock on collisions with hydrogen nuclei of the cellular water molecules.
Protons set in motion by photoneutrons dissipate energy over a range of LET up to about 90 keV/μm.
Answer choice E mainly pertains to neutrons, not protons, where the average method and the energy method for calculating LET give significantly different numbers.
2019-VIII7. How many ion clusters are formed by 55 keV/μm silicon ion along a 1 μm segment of the ion trajectory through the cell nucleus? Assume silicon ion irradiation with the beam parallel to a cellular monolayer and that ion clusters are uniformly spaced along the silicon ion track
A. 0.5 cluster every 1 μm or 1 cluster every 2 μm
B. 5.5 clusters every 1 μm
C. 500 clusters every 1 μm
D. 5,500 clusters every 1 μm
E. 55,000 clusters every 1 μm
Answer: C
On average, the formation of a three-ion cluster requires dissipation of 110 eV.
Therefore,
55 𝑘𝑒𝑉/𝜇 × 1000 𝑒𝑉/ 1 𝑘𝑒𝑉 × 1 𝑖𝑜𝑛 𝑐𝑙𝑢𝑠𝑡𝑒𝑟/ 110 𝑒𝑉
= 500 𝑖𝑜𝑛 𝑐𝑙𝑢𝑠𝑡𝑒𝑟𝑠/𝜇
or 1 cluster every 20 Ǻ (1 μm = 10,000 Ǻ). This spacing of ion clusters along the silicon ion track corresponds to a 20 Ǻ diameter of of the DNA helix.
2019-IX1. What is the approximate maximum diffusion distance of oxygen from a normally-oxygenated capillary through a typical respiring tissue?
A. 5 nm
B. 15 μm
C. 200 μm
D. 900 μm
E. 2.6 mm
Answer: C
In a typical respiring tissue, the approximate distance that oxygen can diffuse from a normally oxygenated capillary before cellular hypoxia is detectable is ranges from approximately 70-200 μm.
The oxygen diffusion distance will depend on the partial pressure of oxygen in the capillary and on the rate of oxygen consumption by the tissue, and therefore shows some variability.
Thomlinson and Gray measured 150 μm in their landmark experiments in 1955.
Olive et al. (IJROBP 1992) determined that the maximum oxygen diffusion distance using solid tumor cubes incubated with fluorescent probes and found it to range from 107 um to 192 μm, depending on the cell line.
Torres Filho et al. (Proc. Natl. AcadUSA 1994) measured in vivo oxygen concentration in a SCID mouse model and found hypoxia to occur at distances >200 μm.
2019-IX2. A dose of 10 Gy of X-rays reduces the tumor cell surviving fraction to 0.001 in an animal irradiated while breathing air, and to 0.1 in an animal irradiated under nitrogen. An estimate of the hypoxic fraction for this tumor in the air-breathing mice would be:
A. 0.0001
B. 0.01
C. 0.25
D. 10
E. 25
Answer: B
The fraction of cells in a tumor that are hypoxic can be estimated using the paired survival curve method.
This corresponds to the surviving fraction of cells irradiated in normally oxygenated tumors divided by the surviving fraction of cells from a tumor made fully hypoxic by asphyxiating the host with nitrogen immediately prior to irradiation, which is assumed to render all of the tumor cells radiobiologically hypoxic.
Thus, the estimate for the fraction of hypoxic cells would be 0.001/0.1 = 0.01.
2019-IX3. The Km for radiosensitization by oxygen (the oxygen concentration at which cellular radiosensitivity is halfway between the fully aerobic and fully hypoxic response) corresponds to an oxygen concentration of approximately:
A. 0.02%
B. 0.5%
C. 3%
D. 15%
E. 30%
Answer: B
The Km value occurs at an oxygen concentration of roughly 0.5-1% or 3-8 mm Hg.
2019-IX4. The most dramatic change in radiation sensitivity occurs over which of the following ranges of oxygen tension (in units of mm Hg or Torr)?
A. 0-20
B. 20-60
C. 60-100
D. 100-250
E. 250-760
Answer: A
The most dramatic change in radiation sensitivity occurs over an oxygen tension range of 0-30 mm Hg (Torr).
Cells irradiated under an oxygen partial pressure at the low end of this range are maximally radioresistant, whereas irradiation at 30 mm Hg oxygen results in near maximum radiosensitization.
2019-IX5. Which of the following statements concerning the oxygen effect is TRUE?
A. OER values obtained for high energy protons used in radiotherapy are similar to those measured for X-rays
B. During irradiation, an oxygen partial pressure of about 30 mm Hg is required to produce full radiosensitization.
C. The OER is defined as the ratio of the surviving fraction of cells irradiated with a particular X-ray dose under hypoxic conditions divided by the surviving fraction of cells irradiated with the same dose under aerated conditions
D. Tumors are thought to contain regions of both acute and chronic hypoxia; however, only chronically hypoxic cells can reoxygenate
E. The oxygen effect is principally a manifestation of the reaction of O2 with organic radicals (R•) to form ROO
Answer: A
Since the high energy protons used in radiotherapy have an LET similar to that of X-rays, their OER values are also similar.
An oxygen partial pressure greater than about 2-3% during irradiation will result in essentially full radiosensitization (Answer Choice B).
The OER is defined as the ratio of the radiation dose needed to cause a certain biological effect in hypoxic cells divided by the dose needed to produce the same effect in aerated cells (Answer Choice C).
Both acutely and chronically hypoxia cells can reoxygenate (Answer Choice D).
The increased cell killing resulting from irradiation in the presence of oxygen is thought to be the result of increased radical damage and damage fixation by oxygen. The initial number of ionizations produced by radiation in the aerated and hypoxic cells would be the same (Answer Choice E).
2019-IX6. For single, large radiation doses delivered at a high dose rate, the ratio of the OER for X-rays divided by the OER for 15 MeV neutrons is approximately:
A. 0.3
B. 1
C. 2
D. 4
E. 10
Answer: C
Since the X-ray OER is typically about 3 and the OER for 15 MeV neutrons is about 1.6, the ratio of the OERs is about 2.
NOTE: Alpha particles approach OER of 1
2019-IX7. Which of the following statements concerning the effect of oxygen is TRUE?
A. Oxygen acts as a radiosensitizer because it inhibits chemical repair of DNA
B. The OER and RBE both increase with increasing LET
C. Based on pO2 microelectrode measurements, few human tumors contain regions of hypoxia
D. At an oxygen partial pressure of about 20 mmHg
E. Exposure of cells to hypoxia may stimulate gene transcription
Answer: E
Exposure of cells to hypoxia, as in other stress situations, leads to changes in expression of a number of stress genes, many of which are responsive to the transcription factor, hypoxia-inducible factor-1α (HIF-1α) (HIF1A).
Under normoxic conditions, HIF-1α is hydroxylated on proline residues by oxygen-dependent prolyl hydroxylases. The hydroxylated prolines bind to the von Hippel-Lindau (VHL) protein, which is a component of the E3 ubiquitin-protein ligase complex that ubiquitinates HIF-1 and targets it for degradation.
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Oxygen acts as a radiosensitizer principally through its ability to “fix” radiation-induced DNA damage and does not inhibit DNA repair (Answer Choice A).
The OER decreases with increasing LET, whereas the RBE increases with LET until reaching a maximum at approximately 100 keV/μm, and then decreases (Answer Choice B).
Measurements with pO2 microelectrodes and bioreductive probes have demonstrated that hypoxic cells are often present in human tumors (Answer Choice C).
The Km of radiosensitivity for cells (i.e., the concentration at which there is 50% radiosensitivity compared to oxic conditions) is close to 0.5-1%, not 3%.
2019-IX8. Which of the following statements is FALSE when describing tumor hypoxia?
A. In rodent tumors, the hypoxic cell fraction is generally within the range of 5-50%
B. Hypoxia is rarely observed in common human solid tumors
C. Oxygen diffusion and delivery is limited in some parts of tumors
D. Hypoxia can enhance tumor progression by means of hypoxia-related changes in gene expression
E. Hypoxia induces gene amplification and mutation
Answer: B
Hypoxia in tumors has been detected using both imaging and direct electrode measurements. The other statements are true.
2019-IX9. Which chemical or compound CANNOT be used to mitgate hypoxia-related radioresistance?
A. Nicotinamide and carbogen
B. Perfluorocarbon
C. Amifostine
D. Misonidazole
E. Nimorazole
Answer: C
Amifostine is a drug whose active metabolite contains a sulfhydrl moiety and acts as a free radical scavenger. It has been studied as a radioprotectant in several clinical and preclinical settings. As a radioprotectant, it does not sensitize hypoxic cells.
Carbogen is a mixture of 95% oxygen and 5% carbon dioxide and has been used to mitigate chronic hypoxia. Nicotinamide, used concurrently with carbogen, is intended to mitigate acute intermittent hypoxia seen in tumor vessels by preventing intermittent vessel closure (Answer Choice A).
Perfluorocarbons such as perflubron have been shown to improve tumor oxygenation in preclinical cancer models but have not yet shown clinical utility
Both misonidazole and nimorazole are nitroimidazoles that have radiosensitizing properties in hypoxia cells (Answer Choice E).
2019-IX10. Oxygen enhancement ratio (OER) changes depend on the type of radiation. Which of the following combinations is FALSE?
A. OER 3.0 for x-rays
B. OER 1.6 for neutrons
C. OER 3.0 for protons
D. OER 0.5 for carbon ions
E. OER 1.0 for alpha-particles
Answer: D
OER for energized ions should be 1.0. By definition, OER cannot be smaller than 1.0.
2019-IX11. Tirapazamine (a hypoxic cytotoxin) has recently been developed. Which of the following statements is FALSE when describing the mechanisms and effects of tyrapazamine?
A. If it loses one electron in hypoxic conditions it becomes cytotoxic
B. When two electrons are extracted in aerobic conditions, it becomes less toxic
C. In normoxic conditions, it can also sensitize cells to radiation
D. Its uptake is greater for cells in hypoxic conditions than cells in aerobic conditions
E. The potency of some chemotherapy agents can be enhanced by the presence of this cytotoxin
Answer: D
There is no difference in the uptake of the chemical by aerobic and hypoxic cells, but there is an obvious difference in the action of cell kill due to the amount of oxygen available in the cells.
2019-IX12. Which of the following statements about tirapazamine is FALSE?
A. A trial examining the utility of the drug in the definitive treatment of locally advanced cervical cancer was conducted but failed to fully accrue due to lack of drug availability.
B. Addition of tirapazamine failed to improve 2-year overall survival in head and neck patients treated with cisplatin-based chemoradiation compared to chemoradiation alone without tirapazamine.
C. The use of tirapazamine was not associated with higher rates of esophagitis in limited stage small cell lung cancer patients treated with definitive chemoradiation compared to historical controls.
D. In GOG 219, tirapazamine increased gastrointestinal toxicity while having no effect on progression free survival.
Answer: C
In the SWOG 0222 trial, tirapazamine was associated with higher rates of esophagitis compared to historical estimates. Tirapazamine was developed as a hypoxic cytotoxin to potentially enhance tumor responses and showed promise in single-arm phase II trials in several disease sites. However, the phase III trials failed to show efficacy above the control arms.
These include GOG 219, which examined the efficacy of the drug when added to standard chemoradiation for locally advanced cervical cancer. The trial failed to accrue due to lack of drug availability. Of a planned accrual of 750 patients, only 379 were eligible and evaluable; no difference in overall or progression-free survival was seen, but there was an increase in grade 3+ leukopenia, GI toxicity, and hepatic/renal dysfunction.
TROG examined tirapazamine’s utility and safety in head and neck cancers in the HeadSTART TROG 02.02 Trial. There was no overall survival or failure-free survival benefit of tirapazamine when added to cisplatin-based chemoradiation in this trial. The authors reported more frequent muscle cramps, diarrhea, and skin rash in the experimental arm.
2019-IX13. Which of the following is TRUE regarding the use of the hypoxic radiosensitizer nimorazole in treating head and neck cancers with radiotherapy?
A. The DAHANCA 5-85 trial showed improved locoregional control but not overall survival when nimorazole.
B. Low plasma concentrations of osteopontin were associated with worse outcomes compared to higher concentrations in the DAHANCA 5 trial but was also associated with a higher benefit from the addition of nimorazole.
C. The addition of nimorazole improves locoregional control in p16-positive tumors but not p16-negative tumors.
D. The addition of nimorazole to radiation is associated with increased acute mucositis and late fibrosis compared to placebo.
Answer: A
The DAHANCA 5-85 Trial examined the addition of nimorazole to conventionally fractionated radiotherapy for pharyngeal and supraglottic larynx cancers. The trial demonstrated that nimorazole improved locoregional control and disease-specific survival compared to placebo but did not significantly improve overall survival.
Nimorazole was reasonably well-tolerated in the DAHANCA 5-85 Tria, although only 60% of patients completed the treatment, and it was associated with a higher rate of nausea/vomiting, flushing, dizziness, and skin rash; patients also had trouble swallowing the large capsules of the drug. There was no increase in mucositis or late complications with the addition of nimorazole (Answer Choice D).
Additional subset analyses of the trial showed that high osteopontin concentration was associated with worse disease-specific mortality, but also improved response to nimorazole in terms of locoregional control and disease-specific mortality (Answer Choice B).
The benefit of nimorazole also appears to be isolated to p16-negative tumors, whereas p16-positive tumors did not appear to benefit (Answer Choice C).
2019-X1. An exponentially-growing, asynchronous population of cells is maintained under normal physiological conditions. Which of the following experimental manipulations would potentiate cell killing following radiotherapy as measured by a clonogenic assay?
A. Cell synchronization in S-phase at the time of irradiation
B. Irradiation under hypoxic conditions
C. Irradiation with the dose split into two fractions with a 24-hour interval between fractions rather than given as an acute exposure to the same total dose
D. Incorporation of bromodeoxyuridine into the DNA prior to irradiation
E. Addition of cysteine to the cellular growth medium prior to irradiation
Answer: D
Bromodeoxyuridine incorporated into cellular DNA in place of thymidine acts as a radiation sensitizer, so cell killing would be enhanced, not reduced.
S-phase is the most radioresistant phase of the cell cycle, so cell killing would be decreased relative to that for an asynchronous population (Answer Choice A).
Oxygen is a radiation sensitizer, so cell killing would decrease in cells made hypoxic before irradiation (Answer Choice B).
Splitting the dose into two fractions separated by 24 hours would allow sublethal damage recovery and possibly enable cellular proliferation to take place between fractions. Cell killing would therefore be less than if the total dose had been delivered acutely (Answer Choice C).
Cysteine is a sulfhydryl-containing compound that scavenges radiation-induced free radicals; it therefore acts as a radioprotector and reduces cell killing (Answer Choice E).
2019-X2. For irradiation with X-rays, the increased cell survival observed when a given total dose is delivered at a low dose-rate (~1 Gy/hr) versus high dose-rate (~1 Gy/min) is due primarily to:
A. Repair of DNA double-strand breaks
B. Decreased production of DNA double-strand breaks
C. Induction of free radical scavengers
D. Activation of cell cycle checkpoints
E. Down-regulation of apoptosis
Answer: A
Therapeutic radiation at a low dose-rate of ~1 Gy/hr is associated with increased cell survival compred to higher dose-rates primarily due to cellular repair - and is generally assumed to be secondary to repair of DNA double-strand breaks produced as a result of radiation.
2019-X3. Relative to the surviving fraction of cells maintained in a non-cycling state for several hours after irradiation, decreased cell survival observed in cells forced to re-enter the cell cycle immediately following irradiation is evidence for:
A. Rejoining of chromosome breaks
B. Sublethal damage recovery
C. Cell cycle reassortment
D. Translesion of DNA synthesis
E. Expression of potentially lethal damage
Answer: E
Potentially lethal damage recovery is operationally defined as an increase in cell survival after delivery of a large, single radiation dose under environmental conditions not conducive to progression of cells through the cell cycle for several hours after irradiation. If non-cycling cells are forced to re-enter the cell cycle immediately after irradiation, rather than remaining quiescent, potentially lethal damage will be “expressed” and therefore the surviving fraction will be lower.
Sublethal damage recovery is operationally defined as an increase in cell survival noted when a total radiation dose is delivered as two fractions with a time interval between the irradiations, as opposed to a single exposure (Answer Choice B).
Repair of DNA damage and rejoining of chromosome breaks presumably underlie both the sublethal and potentially lethal damage recovery (Answer Choice A).
Cell cycle reassortment has a sensitizing effect on a population of cells receiving multi-fraction or protracted irradiation regimens. This is because surviving cells that were in a resistant phase of the cell cycle during the initial irradiation may progress through the cell cycle between fractions and reassert into a more sensitive phase of the cell cycle by the time of delivery of the next fraction. This process is irrelevant under the conditions described here, in which only a single radiation dose was administered (Answer Choice C).
Translesion DNA synthesis is an error-prone process during which certain DNA polymerases synthesize DNA using a damaged DNA strand as a template, resulting in error-prone DNA synthesis (Answer Choice D).
2019-X4. 5 Gy of X-rays is delivered at a high dose rate (1 Gy/min) rather than a low dose rate (1 Gy/hr). Which of the following statements about the effects of this change on cell survival is TRUE?
A. The surviving fraction would change the least for a cell line with a radiation survival curve characterized by a low α/β ratio
B. Treatment of cells during irradiation with an agent that inhibits DNA repair would have a greater impact on the surviving fraction of cells irradiated at the high dose rate
C. More cell killing would occur following treatment at the high dose rate
D. The difference in the surviving fractions between the two protocols results primarily from repopulation
E. The total number of ionizations produced is decreased with treatment at the high dose rate
Answer: C
When a dose of 5 Gy is delivered at a dose rate of 1 Gy/min, irradiation requires 5 minutes. When 5 Gy is delivered at 1 Gy/hr, irradiation requires 5 hours. Extensive repair of sublethal damage will occur during the low dose rate, but will not be able to occur during high dose rate irradiation. As a result, the component of cell killing will decrease and result in a strictly exponential and shallow survival curve. More cell killing would therefore occur when a dose of 5 Gy is delivered at a high dose rate rather than a low dose rate.
The surviving fraction would change the least for a cell line with a radiation survival curve characterized by a high, not low, / ratio (Answer Choice A).
Treatment with an agent that inhibits DNA repair would have little impact during the 5 minute period of irradiation that would occur at the high dose rate (Answer Choice B).
In contrast, such a treatment would markedly reduce cell survival for the 5 hour irradiation required at the low dose rate since, in the absence of the agent, substantial repair would take place during the course of the irradiation. The increase in the surviving fraction for this low dose rate irradiation is primarily a consequence of sublethal damage recovery and not repopulation, as the repopulation would only occur for overall treatment times on the order of days (Answer Choice D).
The number of ionizations produced is a reflection of the total dose delivered and does not vary with the dose rate (Answer Choice E).
2019-X5. Exponentially growing cells were maintained at 37oC in 95% air/ 5% CO2 and irradiated with either a single dose of 8 Gy of X-rays or two 4 Gy fractions separated by either 2 hours or 8 hours. The surviving fractions for the 3 treatments were 0.02, 0.15, and 0.08, respectively. The two processes that best account for these differences in survival are:
A. Reassortment and repopulation
B. Repair and reassortment
C. Reoxygenation and repair
D. Repopulation and reassortment
E. Repair and reoxygenation
Answer: B
Compared to the cell surviving fraction after the single 8 Gy dose, the increase in cell survival noted for the two 4 Gy doses delivered with a 2 hour interfraction interval was due to sublethal damage repair (SLDR). Although SLDR also occurred when the interfraction interval was 8 hours, cells surviving the first dose also reassorted from the radioresistant phases they were in at the time of the initial irradiation (e.g. late S) into more radiosensitive phases (e.g. G2 and M), thereby resulting in an overall surviving fraction for the 8 hour interval that was lower than that for the split dose protocol with a 2 hour interval between fractions. It is unlikely that much repopulation would take place during the total time of 8 hours needed to complete the irradiations. Reoxygenation would not be an issue for cells maintained in a well-aerated 95% air environment.
2019-X6. Which of the following pairs of radiobiological process and corresponding assay method is CORRECT?
A. Reoxygenation – HIF-1α phosphorylation by ATM
B. Potentially lethal damage recovery – tritiated thymidine uptake
C. Cell cycle “age response” – paired survival curve method
D. Sublethal damage recovery – split dose experiment
E. Repopulation – mitotic shakeoff procedure
Answer: D
Sublethal damage recovery is operationally defined and demonstrated using a split dose protocol.
Potentially lethal damage repair is detected by changing the post-irradiation environment and observing the effect on survival (Answer Choice B). Incorporation of tritiated thymidine into DNA would not specifically measure PLDR, but would reflect DNA synthesis and other forms of DNA repair.
Reoxygenation would best be assayed by performing repeat measurements during the course of radiotherapy by using an oxygen electrode or by teating with a hypoxia maker, such as pimonidazole, that is metabolized and incorporated exclusively into hypoxic cells (Answer Choice A).
Cell cycle age response is best demonstrated by performing cell synchronization followed by irradiation of cohorts of cells in particular cell cycle phases and then performing the clonogenic survival assay as a readout (Answer Choice C).
Repopulation can be assayed in vitro by counting the number of cells present as a function of time after irradiation. The mitotic shake off technique is used to collect synchronous populations of cells for use in experiments examining age response functions (Answer Choice E).
2019-X7. Which of the following is a phosphoinositol 3-kinase like kinase (PIKK) that serves as the central orchestrator of the signal transduction response to DSBs?
A. Ku70/80
B. ATM
C. Rad50
D. MSH2
E. p53 (TP53)
Answer: B
Ataxia Telangictasis Mutated (ATM) serves as the central orchestrator of the signal transduction response to DSBs.
Cells deficient in ATM activity display cell cycle checkpoint defects and sensitivity to ionizing radiation.
2019-X8. Which of the following is TRUE for potentially lethal radiation damage (PLD)?
A. It is irreversible and irreparable.
B. It is the damage that can be repaired efficiently if cells are allowed to progress through the cell cycle immediately following IR.
C. It is thought to be primarily complex or “dirty” double strand breaks.
D. It can be observed in a “split dose” experiment.
E. It cannot be detected in tumors in vivo.
Answer: C
PLD is believed to be complex double strand breaks (DSBs) that are repaired slowly as compared to simple DSBs. Therefore, cells that are left in stationary phase after irradiation display enhanced survival as they have time to repair complex DSBs before resuming progression through the cell cycle.
2019-X9. Which of the following is FALSE for the split dose experiment and sublethal damage (SLD)?
A. The survivors of the first dose are mainly S phase.
B. The fraction of cells surviving a split dose increases as the time interval between the two doses decreases due to the repair of SLD.
C. When cells are cycling during the split dose experiments, there is a dip (decrease) in cell survival caused by reassortment.
D. SLD can be repaired before they can interact to form lethal chromosomal damage.
E. SLD is demonstrated by low-LET radiation
Answer: B
The fraction of cells surviving a split dose increases with increasing time between the two doses because of the repair of SLD.
2019-XI1. Which of the following assays would NOT be useful for the purpose of quantifying the response of a tumor to irradiation?
A. Lung colony assay
B. Number of tumors per animal
C. Time to reach a certain size
D. Growth delay
E. Colony forming ability of cells explanted from the tumor
Answer: B
An increase in the number of tumors per animal would be a reflection of metastatic spread of the tumor, and would not necessarily reflect the radiation response of the primary tumor per se. All of the other assays can be used to quantify the response of tumors to irradiation.
2019-XI2. The TCD50 assay:
A. Measures radiation-induced tumor growth delay
B. Can be conducted using mouse tumors but not human tumor xenografts
C. Gives a measure of the number of cells required to produce a tumor in a mouse
D. Yields results independent of the immune competence of the host animal
E. Measures tumor cure, making it a relevant endpoint for extrapolation to the clinic
Answer: E
The TCD50 assay quantifies the dose required to cure 50% of a group of matched tumors and is therefore a highly relevant endpoint for extrapolation to the clinic.
The assay can be conducted using mouse tumors or human tumor xenografts (Answer Choice B), although suppression of the host immune system when using xenografts is crucial in order to minimize misleading results due to rejection of implanted cells (Answer Choice D).
The TD50 assay can be used to measure the number of cells required to cause a tumor in mice and has historically been used to determine tumor cell survival curves, to assess the number of clonogens in a tumor, and to study host factors that influence tumor development (Answer Choice C).
2019-XI3. The number of cells required to produce a tumor “take” in mice is indicated in the graph below for a glioblastoma cell line (bulk) and two sub-lines derived from it (CD133+ and CD133-). Which of the following statements would best explain the experimental findings?
A. CD133- cells comprise only a small fraction of the total tumor
B. CD133 is a putative marker for cancer stem cells
C. Unsorted bulk cells contain a large fraction of cancer stem cells
D. CD133- cells are more radiosensitive than CD133+ cells
E. CD133+ cells are more radiosensitive than CD133- cells
Answer: B
CD133 has been described as a putative marker for cancer stem cells in glioblastoma. Injection of 100 CD133+ cells is sufficient to initiate tumor formation in >30% of nude mice, supporting CD133 as a potential cancer stem cell marker.
The unsorted bulk cells contain cancer stem cells; given, however, that 100-1,000 more cells are required in order to form the same number of tumors as that seen when purified CD133+ cells are injected, it appears that <1% of the cells in the tumor are stem cells.
CD133- cells were derived from a glioblastoma and are therefore not normal, although they possess a very limited ability to form tumors de novo. In this experiment, no data are provided regarding the sensitivity of the cell lines (or sublines) to radiation.
2019-XI4. A local tumor recurrence after radiotherapy can be caused by:
A. Any surviving cancer cell
B. Any proliferating cancer cell
C. Only cancer cells with the ability to form colonies in vitro
D. Only cancer cells with unlimited proliferative potential
E. Only cancer cells that were well-oxygenated during irradiation
Answer: D
Results from tumor transplantation experiments indicate that only a small proportion of all cancer cells have an unlimited proliferative capacity and demonstrate the capacity of self-renewal.
In analogy to in vitro assays, tumor cells that demonstrate the ability to achieve a local recurrence following radiotherapy have been termed “clonogenic cells” and correspond to putative “cancer stem cells”. The existence of cancer stem cells, defined by the ability for self-renewal and generation of the heterogeneous lineage of cells within a tumor, has been hypothesized.
2019-XI5. Which assay or endpoint would provide the best estimate of the radiation response of putative cancer stem cells?
A. Time to first evidence of tumor shrinkage following irradiation
B. Tumor regrowth delay
C. Determining the fraction of proliferating tumor cells 2 weeks after irradiation
D. 50% tumor control dose
E. Quantifying the number of apoptotic tumor cells 6 hours after irradiation
Answer: D
It has been suggested that a small proportion (< 1%) of all cells in a tumor are cancer stem cells. If correct, this hypothesis suggests that all cancer stem cells must be inactivated in order to achieve permanent local tumor control.
In theory, one surviving cancer stem cell would be sufficient to cause a local recurrence following irradiation. Thus, the rate of permanent local tumor control is a direct measure of radiation response of cancer stem cells.
In contrast, tumor shrinkage and growth delay are dominated by the response of the bulk of cancer cells and not specific for the radiation response of cancer stem cells. Cancer cells with a limited proliferative capacity, as well as doomed cancer stem cells, might undergo a number of cell divisions before they permanently stop proliferating and ultimately die.
Determination of proliferating cells will therefore not provide information regarding the radiation response of cancer stem cells. Cancer cells can die following exposure to radiation in different ways, including interphase death (i.e. apoptosis) and mitotic catastrophe (apoptosis, autophagy, or necrosis). None of these modes of cell death is likely to be specific for cancer stem cells.
Given that many solid tumors exhibit resistance to undergoing apoptosis and the controversial data from studies comparing the rate of apoptosis with radiation response of tumors, it is unlikely that the rate of apoptosis after irradiation will be a proper parameter to determine the response of irradiated cancer stem cells.
2019-XI6. In some experiments, tumors treated with radiation and concurrent molecularly-targeted drugs against EGFR and VEGFR displayed longer regrowth delays, but not higher tumor control probabilities, compared to tumors that were treated with radiation only. Which of the following statements provides the most likely explanation for this?
A. The treatment is effective for the bulk of tumor cells, but not for cancer stem cells.
B. The drug did not reach most of the cells due to poor vascular perfusion in the tumor.
C. Experimental error accounts for this, because growth delay and tumor control assays usually yield similar results.
D. Tumor cells generally do not express receptors that are targeted by these drugs.
E. The radiosensitivity of tumor cells does not depend on vascular supply or physiology.
Answer: A
There are indeed some examples in the literature showing a discrepancy between growth delay and tumor control probability. In these experiments, various molecular targeting approaches in combination with radiation were investigated. Though difficult to prove, the assumption of a differential effect on cancer stem cells and non-cancer stem cells is the mostly likely explanation for these results. It is likely that the drug reached the tumor since there was an effect on tumor growth.
Cancer cells generally express EGFR and cell survival following irradiation is affected by vascular supply. The observed discrepancy between growth delay and local tumor control in some experimental settings suggests that the latter assay is the preferable endpoint to evaluate new therapeutic approaches with curative intent.
2019-XI7. Which of the following is NOT a feature of apoptosis?
A. Chromatin condensation
B. Cell shrinkage
C. DNA fragmentation
D. Rapid engulfment by neighboring cells
E. Inflamatory response
Answer: E
There is no inflammatory response in apoptosis. While an inflammatory response is more a feature of necrosis than apoptosis, there are situations in which apoptosis can stimulate an inflammatory response.
2019-XI8. Which of the following statements is TRUE regarding the appearance of giant multinucleated cells following radiation?
A. Giant multinucleated cells exhibit an exponential survival curve
B. Giant multinucleated cells are a characteristic seen just prior to mitotic catastrophe
C. The presence of giant multinucleated cells is associated with dose response
D. Giant multinucleated cells are radioresistant mitotic cells
E. Giant multinucleated cells demonstrate increased apoptosis
Answer: B
Radioresistant cells display mitotic catastrophe caused by aberrant mitosis, which is associated with the formation of giant multinucleated cells that contain uncondensed chromosomes.
2019-XI9. Which of the statements is TRUE regarding the activation of one type of apoptotic pathway?
A. Apoptosis is initiated by PARP
B. Fas ligand binding its receptor initiates apoptosis
C. Caspases involved in the execution of apoptosis are also involved in the execution of necrosis
D. Bcl2 is a pro-apoptotic protein.
E. Anti-apoptotic Bax dimerizes and translocates to the mitochondria.
Answer: B
Fas ligand binds its receptor and triggers the external death receptor pathway.
Cleavage of PARP-1 by caspases is considered to be a biochemical hallmark of apoptosis (Answer Choice A).
Different caspases play different roles in the initiation and execution of apoptosis and are not involved in necrosis. Necrosis is the unregulated digestion of cellular components as a result of external factors (Answer Choice C).
Bcl-2 inhibits apoptosis while Bax stimulates apoptosis (Answer Choices D and E).
2019-XII1. Which of the following statements concerning tumor hypoxia is TRUE?
A. Hypoxic regions in tumors may be detected using labeled bortezomib
B. As a tumor increases in size, the hypoxic fraction of cells decreases
C. Regions of chronic hypoxia may develop in tumors due to the intermittent opening and closing of blood vessels
D. In the absence of reoxygenation it is unlikely that all hypoxic cells would be eliminated from a tumor following a typical course of radiotherapy
E. Acutely hypoxic tumor cells usually exhibit slow reoxygenation while chronically hypoxic tumor cells reoxygenate rapidly
Answer: D
In the absence of reoxygenation it is unlikely that all hypoxic cells would be eliminated following a typical course of radiotherapy from a tumor possessing even a small percentage of hypoxic cells because hypoxic cells demonstrate approximately 3-fold greater radioresistance compared with aerated cells.
Hypoxic regions in tumors can be detected using a labeled nitroimidazole compound. Bortezomib (Velcade) is a proteasome inhibitor (Answer Choice A).
Although not without exceptions, as tumors increase in size the hypoxic fraction generally also increases. This is because the typically abnormal tumor vasculature is insufficient to maintain oxygen demand (Answer Choice B).
Regions of acute or transient hypoxia may develop in tumors due to intermittent blood flow via the intermittent closing down of blood vessels. Chronic hypoxia, on the other hand, is defined as diffusion-limited hypoxia due to the inability of oxygen to diffuse farther than 100 m from a blood vessel (Answer Choice C).
Acutely hypoxic cells that tend to exhibit rapid reoxygenation, whereas chronically hypoxic cells generally reoxygenate more slowly (Answer Choice E).
2019-XII2. Bevacizumab (avastin) is a monoclonal antibody that targets:
A. Basic fibroblast growth factor (bFGF; FGF2)
B. Hypoxia-inducible factor-1α (HIF-1α)
C. Von Hippel-Lindau (VHL) protein
D. Ras
E. Vascular endothelial growth factor (VEGF; VEGFA)
Answer:
Bevacizumab (avastin) binds to and neutralizes vascular endothelial growth factor (VEGF)-A ligand, thereby preventing its interaction with cell surface receptors, including the VEGF receptor (VEGFR).
The fibroblast growth factors (FGFs) are a family of pluripotent growth factors that stimulate proliferation of mesodermal or neuroectodermal cells and can play a role in angiogenesis (Answer Choice A). FGFs have yet to be successfully targeted pharmaceutically.
Hypoxia-inducible factor (HIF)-1α is a transcription factor that detects hypoxia and enhances angiogenesis (Answer Choice B).
The Von Hippel-Lindau (VHL) protein belongs to a complex that is involved in the ubiquitination and degradation of HIF (Answer Choice C).
The Ras proteins are a family of small GTPases involved in the activation of signaling cascades following activation of receptors. Ras is frequently mutated in human cancers but is difficult to target pharmacologically (Answer Choice D).
2019-XII3. Which of the following responses is LEAST likely to be observed?
A. Exposure to hypoxia increases the expression of angiogenesis-promoting genes
B. Anti-angiogenic therapy improves tumor oxygenation
C. A chronically hypoxic environment increases the metastatic potential of tumor cells
D. Hypoxia inhibits apoptosis in tumor cells
E. Exposure to hypoxia inhibits cell proliferation
Answer: D
An increased apoptotic index is often observed in hypoxic regions of tumors.
The gene for vascular endothelial growth factor (VEGF/VEGFA) is one of the major genes under the control of the hypoxia responsive promoter, HRE, which binds the transcription factor, hypoxia-inducible-factor (HIF)-1 (Answer Choice A).
Studies in animal models have indicated that treatment with anti-angiogenics can cause “normalization” of tumor blood vessels and result in a transient improvement in tumor oxygenation before vessels start to deteriorate (Answer Choice B).
Pre-clinical studies with animal models as well as clinical studies have linked increased hypoxia in tumors to increased tumor aggressiveness and metastatic potential (Answer Choice C).
Exposure to severe hypoxia halts progression of cells through the cell cycle and therefore inhibits proliferation (Answer Choice E).
2019-XII4. Which of the following statements concerning chronically hypoxic cells in tumors is TRUE? Chronically hypoxic cells:
A. Can be selectively targeted for killing with certain bioreductive drugs
B. Are resistant to hyperthermia
C. Are located within 10 um of capillaries
D. Exist in a high pH microenvironment
E. Are a consequence of intermittent blood flow
Answer: A
Chronically hypoxic regions in a tumor can be targeted for elimination by administering certain bioreductive drugs that preferentially kill hypoxic, but not aerobic, cells.
Chronically hypoxic cells tend to be sensitive to hyperthermia. This is because they exist in an acidic (low pH) microenvironment (Answer Choices B and D).
Acutely hypoxic cells are a consequence of intermittent blood flow in tumors (Answer Choice E).
It has been shown via model calculations of oxygen consumption rates in respiring tissues and through the use of hypoxia markers that chronically hypoxic cells rarely appear closer than about 70 um from capillaries (Answer Choice C).
2019-XII5. Which of the following statements concerning tumor angiogenesis is TRUE?
A. Even without angiogenesis, tumors can grow up to 2 cm in diameter
B. For most tumor types a high microvessel density has been negatively correlated with metastatic spread
C. Vascular endothelial growth factor (VEGF) is induced under hypoxic conditions
D. Angiostatin and endostatin are stimulators of angiogenesis
E. Basic fibroblast growth factor (bFGF) is a negative regulator of angiogenesis
Answer: C
Expression of vascular endothelial growth factor (VEGF) and downstream angiogenesis is induced under hypoxic conditions via hypoxia-inducible transcription factors that bind to the VEGF promoter to stimulate its transcription.
In the absence of angiogenesis, tumors would only be expected to reach a diameter of about 2 mm, not 2 cm (Answer Choice A).
Microvessel density, a measure of angiogenesis, has been correlated positively with metastatic spread for most tumor types (Answer Choice B).
Angiostatin and endostatin are inhibitors of angiogenesis while basic fibroblast growth factor (bFGF) is a positive regulator of angiogenesis (Answer Choice D and E).
2019-XII6. The regulation of hypoxia-inducible factor-1 alpha (HIF-1a; HIF1A) by oxygen concentration is best described by which of the following statements?
A. Under hypoxic conditions, HIF-1a transcription and translation are upregulated as well as translocation of HIF-1a from the cytosol to the nucleus
B. Under aerobic conditions, the HIF-1a heme moiety becomes oxygenated. This drives a conformational change in the protein that limits DNA binding and prevents upregulation of target genes
C. Under hypoxic conditions, HIF-1a is activated by bioreduction, thereby promoting the up-regulation of target genes
D. Under hypoxic conditions, the HIF-1a heme moiety becomes deoxygenated. This induces a conformational change in the protein that leads to enhancing DNA binding and subsequent upregulation of target genes
E. Under aerobic conditions, HIF-1a is hydroxylated by HIF prolyl hydroxylases that target the protein for ubiquitination and subsequent proteosomal degradation, thereby preventing the up-regulation of target genes
Answer: E
Hypoxia-inducible factor-1 (HIF-1) is a heterodimer that acts as a key regulator of several oxygen-responsive proteins, including erythropoietin and vascular endothelial growth factor (VEGF). HIF-1 was first identified as a DNA-binding protein that mediated the up-regulation of the erythropoietin gene under hypoxic stress. Subsequent studies have implicated HIF-1 in the regulation of a broad range of oxygen responsive genes including VEGF, VEGF receptors, angiopoietins, nitric oxide synthase, fibroblast growth factors and platelet-derived growth factor (PDGF). Under aerobic conditions, HIF-1 is hydroxylated by HIF prolyl hydroxylases. Hydroxylation at two prolyl residues targets HIF-1 to the von Hippel-Lindau (VHL) E3 ubiquitin ligase and results in HIF-1 ubiquitination and subsequent proteosomal degradation, thereby limiting upregulation of target genes. Because the hydroxylation catalyzed by prolyl hydroxylases requires molecular oxygen, HIF-1 escapes inactivation under hypoxic conditions.
2019-XII7. Which of the following statements best describes the “normalization hypothesis” proposed to explain the survival benefit associated with combining anti-angiogenics with traditional chemotherapy agents?
A. Anti-angiogenic therapy stimulates the formation of leaky blood vessels thereby enhancing access of chemotherapy agents to the tumor parenchyma
B. Anti-angiogenic therapy transiently reduces pericyte coverage of tumor blood vessels, which would otherwise form a significant mechanical and biochemical barrier to the delivery of chemotherapy to the tumor
C. Tumor cell-derived pro-angiogenic factors render endothelial cells resistant to chemotherapy-induced apoptosis. Anti-angiogenic therapy eliminates this protection and restores endothelial cell sensitivity to chemotherapeutic agents
D. Anti-angiogenic therapy reduces the secretion of anti-apoptotic factors by vascular endothelial cells that would otherwise render nearby cancer cells relatively resistant to chemotherapeutic agents
E. Anti-angiogenic therapy transiently restores the normal balance of pro- and anti-angiogenic factors in tumor tissue thereby reducing tumor vessel leakiness, dilation, and tortuosity as well as increasing pericyte coverage
Answer: E
When administered as a single agent, several anti-angiogenic drugs have not yielded a long-term survival benefit. In contrast, delivery of anti-angiogenic agents with chemotherapy has produced a significant survival benefit in colon cancer and previously untreated lung and breast cancers.
If the anti-angiogenic agent were destroying tumor vasculature in combination regimens, one would expect decreased tumor blood flow and compromised delivery of chemotherapy to the tumor. The survival benefits produced by the addition of an anti-angiogenic drug to traditional chemotherapeutic regimens therefore appears paradoxical. One possible explanation for this has been termed the “normalization hypothesis.”
Under the pressure of pro-angiogenic factors, tumor vasculature is structurally and functionally abnormal. Anti-angiogenic therapy (transiently) restores the balance of pro- and anti-angiogenic factors. Consequently, immature and leaky blood vessels are pruned, pericyte coverage increases, and the basement membrane becomes more homogenous and normalized.
As a result, the resultant vascular bed achieves greater organization by being less leaky, dilated, and tortuous. These morphological changes also result in functional changes, including decreased interstitial fluid pressure, increased tumor oxygenation, and improved penetration of drugs into the tumor parenchyma.
Due to improved drug delivery, chemotherapy is more efficacious. Sustained or high-dose anti-angiogenic therapy, however, may drive an imbalance favoring anti-angiogenic factors leading to inadequate tumor blood supply and compromise of the efficacy of systemic therapies.
2019-XII8. At a distance of 150 μm from the nearest tumor blood vessel, one might expect all of the following microenvironmental conditions, EXCEPT:
A. Increased hypoxia
B. Decreased pH
C. Decreased interstitial fluid pressure
D. Decreased glucose
Answer: C
Solid tumors develop regions of increased hypoxia (decreased pO2), decreased pH, decreased glucose, and increased (not decreased) interstitial fluid pressure. Oxygen can diffuse about 70 μm from the arterial end of a capillary
2019-XII9. Paclitaxel appears to be effective in radiosensitizing tumors in vivo for all the following mechanisms, EXCEPT:
A. Induction of apoptosis
B. Upregulation of HIF-1
C. Oxygenation of radioresistant hypoxic cells
D. Arrest of cells in the radiosensitive G2/M phase
E. Decrease of interstitial fluid pressure
Answer: B
Paclitaxel stabilizes microtubule polymers and protects them from disassembly. As a result, mitosis is consequently blocked and apoptosis is activated.
Paclitaxel has been shown to increase the radiation sensitivity of tumors by inducing apoptosis, increasing oxygenation of hypoxic cells in tumors, arresting cells in the radiosensitive G2/M phase of the cell cycle, and decreasing interstitial fluid pressure.
Some of these studies have been conducted in animal models while others have been performed in clinical trials of human breast cancer patients. No studies have demonstrated upregulation of HIF-1 following treatment with paclitaxel; indeed, one might expect HIF-1 to be degraded more rapidly following reoxygenation.
2019-XII10. All of the following statements as to why larger tumors are more difficult to control with radiotherapy than smaller tumors are true, EXCEPT:
A. Larger tumors generally contain a greater number radioresistant hypoxic cells than smaller tumors
B. In order to deliver a curative total dose to a large tumor, the volume of irradiated adjacent normal tissue may become so large as to exceed normal tissue tolerance
C. A larger primary tumor volume is associated with a higher risk of regional and distant metastatic spread
D. The fraction of rapidly proliferating cells tends to increase with the size of the tumor
Answer: D
Larger tumours tend to have large necrotic centers where hypoxic radioresistant cells largely reside. In almost all cancer types, the size of a primary tumour correlates with the risk of regional and distant metastatic spread. Hence, in the AJCC staging system, the T stage is often defined by size.
The fraction of proliferating cells therefore tends to decrease with increasing tumor volume.
2019-XII11. Which of the following statements regarding the tumor microenvironment is FALSE?
A. Blood vessel supply is heterogeneous and irregular
B. Blood flow through micro-vessels may be sluggish
C. There tends to be an increase in vessel density compared to normal tissue
D. There are a greater number of hypoxic regions within the microenvironment of a tumor compared to that seen in normal tissue
E. Nutritional support to the tumor microenvironment is adequate and homogeneous
Answer: E
Tumor masses exhibit abnormal blood vessel networks that fail to provide adequate and homogeneous nutritional support
2019-XII12. Which of the following statements regarding angiogenesis is FALSE?
A. For a multi-cellular organism to grow, it must have the capacity to recruit new blood vessels via angiogenesis
B. Angiogenesis is normally regulated by pro-angiogenic, but not anti-angiogenic, molecules
C. Angiogenesis is dysregulated in a neoplastic environment
D. Without a nearby blood vesse or effective angiogenesis, a tumor cannot grow beyond a critical size or metastasize to other organs
E. Viable cells are located within 70 um of blood vessels due to the diffusion limits of oxygen
Answer: B
Since oxygen is unable to diffuse more than approximately 70 μm from the arterial end of a capillary, tumours require the ability to develop new blood vessels in order to grow (Answer Choice A). This process is normally regulated by a balance of both pro-(including VEGF, FGF, PDGF) and anti-(thrombospondin, angiostatin, endostatin) angiogenic molecules.
2019-XII13. Which of the following statements regarding anti-angiogenic therapy strategies is FALSE?
A. Anti-angiogenic therapies interfere with activators of angiogenesis
B. Anti-angiogenic therapies target receptor tyrosine kinases and related signal transductions
C. Anti-angiogenic therapies seek to amplify endogenous suppressors of angiogenesis.
D. Anti-angiogenic therapies use colchicine as an anti-angiogenic agent.
E. Anti-angiogenic therapies ultimately target VEGFR-1 in order to achieve inhibition of angiogenesis.
Answer: C
Colchicine is an anti-inflammatory agent that binds tubulin. Colchicine itself induces vascular damage but only at doses that are limited by toxicity and therefore not used in the clinical setting for this purpose.
2019-XII14. Which of the following statements best describes the “abscopal effect?”
A. Localized irradiation of a tumor is associated with improved disease-related symptoms that are not associated with the treated tumor.
B. Localized irradiation of a tumor is associated with a decrease in size not only of the irradiated tumor but also of a tumor that is far from the irradiated area
C. Localized irradiation of a tumor is associated with a decrease in size not only of the irradiated tumor but also of a tumor that is far from the irradiated area when given with concurrent immunotherapy
D. Localized irradiation of a tumor is associated with a decrease in size not only of the irradiated tumor but also of a tumor that is far from the irradiated area when given with concurrent chemotherapy
Answer: B
The abscopal response is defined as the influence of ionizing radiation on cell killing outside of a field of radiation. For example, an abscopal response would be considered if an unirradiated tumor is shown to have decreased in size following the irradiation of a tumor located in a distant area.
2019-XII15. Which of the following statements concerning vasculogenesis is TRUE?
A. The process of vasculogenesis is specific to the developing embryo
B. Vasculogenesis refers to a subset of angiogenesis, in that it describes the formation of only venous vessels as tumors grow beyond 1-2 mm3
C. Tumors use vasculogenesis or angiogenesis in a mutually exclusive fashion
D. Vasculogenesis is critical for a tumor to achieve local tumor recurrence following radiotherapy
E. Vasculogenesis utilizes pre-existing blood vessels during the early stages of tumor development to facilitate tumor growth
Answer: D
Whereas angiogenesis is the sprouting of endothelial cells from nearby blood vessels, vasculogenesis is the formation of blood vessels from circulating endothelial progenitor cells (i.e. the bone marrow).
Vasculogenesis is of particular importance following treatment with anti-angiogenic agents as well as following irradiation. During early tumor development, both vasculogenesis and angiogenesis are likely utilized.
Because tumor irradiation abrogates local angiogenesis, the tumor must rely on the vasculogenesis pathway for re-growth following irradiation. Tumor irradiation produces a marked influx of CD11b+ myeloid cells into tumors and are criticial in order for a tumor to achieve formation of blood vessels after irradiation as well as for tumor recurrence.
2019-XII16. Antigen recognition by T cells is imperative for the development of cellular adaptive immunity. How does a T cell recognize an antigen?
A. T cells recognizes antigens via pattern recognition receptors
B. T cells recognizes antigenic determinants presented in the MHC cleft by the T cell receptor
C. T cells recognize antigens via the Fc receptor binding to membrane-bound IgD antibodies
D. T cells recognize antigens via PD-1 binding
Answer: B
The surface of each T cell has approximately 30,000 antigen-receptor polypeptide chains, T-cell receptor α (TCRα) and β (TCRβ), that are linked by a disulfide bond.
The TCR is very similar in structure to the Fab fragment of an immunoglobulin molecule and account for antigen recognition by most T cells. A minority of T cells has an alternative, but structurally similar, receptor made up of a different pair of polypeptide chains designated γ and δ.
Unlike B cells, which can recognize a protein antigen in its native state, T cells recognize an antigen via the TCR only after it has been processed into peptides and loaded onto major histocompatibility complex (MHC) molecules.
During the endogenous antigen presentation pathway (performed by the majority of cells), an intracellular antigen is degraded by the proteasome into peptides (typically 9-10 amino acids long) and then loaded onto MHC class I molecules in the Endoplasmic Reticulum (ER). In the exogenous pathway, an extracellular antigen is taken up by antigen presenting cells (APCs) and degraded into peptides (typically 11-19 amino acids long) within endosomes and then bound to MHC class II molecules.
In both pathways, full MHC-peptide complexes are transported to the cell surface for recognition by the TCR on CD8+ cytotoxic T cells (MHC I) or CD4+ T helper cells (MHC II). Cross-presentation is the display of peptides from extracellular antigens on MHC class I and is only performed by APCs..
Pattern recognition receptors (PPRs) such as Toll-like receptors (TLRs) are predominantly found on APCs and other innate immune cells and are used for the detection of danger signals such as pathogen-associated molecular patterns (PAMPs) or damage-associated molecular patterns (DAMPs). The engagement of PPRs initiates the maturation of APCs, especially dendritic cells, thereby allowing them to stimulate T cells by providing the first signal (signal 1: antigen) to the TCR and the second signal (signal 2: co-stimulation) to CD28, which then amplifies signal 1.
PD-1 is an immune checkpoint that inhibits proximal signaling of the TCR by sequestering Src Homology Region-Containing Protein Tyrosine Phosphatase-2 (SHP-2) and facilitating Csk-mediated inhibitory phosphorylation of Lck.
2019-XII17. Which of the following molecules is NOT an immune checkpoint receptor protein?
A. LAG3
B. PD-1
C. TIM3
D. OX40
E. CTLA-4
Answer: D
Lymphocyte Activating 3 (LAG3) is a cell surface immune checkpoint receptor protein that is expressed on activated T cells and negatively regulates cellular prol and is activated by Major Histocompatibility Class (MHC) II.
Programmed cell death protein (PD)-1 is a cell surface immune checkpoint receptor protein that is bound by its two ligands, PD-L1 and PD-L2 and functions to suppress T cell inflammatory activity and promotes self tolerance.
Cytotoxic T-lymphocyte-associated protein 4 (CTLA-4) is another immune checkpoint receptor protein that is activated by CD80 and CD86 and functions to downregulate the immune response.
More recently identified as a potential immune checkpoint target, T-cell immunoglobulin and mucin domain-3 (TIM-3) is activated by Galectin-9 (Gal-9), Phosphatidylserine, HMGB-1, and Cecam-1 and plays a role in T cell exhaustion. LAG3, PD-1, CTLA-4, and TIM3 all act as co-inhibitory receptors that limit or inhibit the activation of T cells even if the TCR is engaged.
In contrast, OX40 (OX40L) is a co-stimulatory receptor that does promotes T cell activation by driving T-cell proliferation, memory, cytotoxic effector function, and cytokine production. Other examples of co-stimulatory molecules are 4-1BB, CD40L, GITR, ICOS, and CD27.
Of these, CTLA-4 and PD-1 are probably the most studied to date. CTLA4 counteracts the activity of the T cell co-stimulatory receptor, CD28, both sharing identical ligands: namely, CD80 (B7.1) and CD86 (B7.2). Although CTLA-4 is active on CD8+ T cells, it seems that most of its effects are derived from down-modulation of helper T cell activity and enhancement of Treg immunosuppressive activity. On the other hand, PD-1 limits the activity of T cells in peripheral tissues at the time of an inflammatory response to infection and functions to limit autoimmunity by, for example, suppressing immune system activation within the tumor microenvironment.
PD-1 expression is induced following T cell activation. When engaged by one of its ligands, PD-1 inhibits kinases that are involved in T cell activation through the phosphatase SHP250, although additional signaling pathways are also likely induced. The general concept is that blocking CTLA-4 affects early T cell activation whereas blockade of PD-1 signaling is more relevant later, at the tissue site, thereby explaining why the CTLA-4 inhibitors are associated with more significant toxicity.
2019-XII18. Which radiation-induced immune effect would be counterproductive to effective anti-tumor immunity?
A. Radiation-induced release of danger signals
B. Radiation-induced increase in regulatory T cells
C. Radiation-induced increase in MHC class I expression
D. Release of pro-inflammatory cytokines
E. Radiation-induced epitope spreading
Answer: B
Danger signals, MHC class I expression, pro-inflammatory cytokines and epitope spreading would support the development of an immune response with the purpose of achieving anti-tumor immunity, while activation of regulatory T cells would not.
Regulatory (suppressor) T cells are a subset of CD4+ T cells that express the transcription factor forkhead box P3 (FOXP3). FOXP3 is a potent suppressor of immune responses to self and non-self and is essential to the maintenance of peripheral immunological tolerance.
Tregs suppress the activation, proliferation, and cytokine production of CD4+ and CD8+ T cells, and are additionally thought to suppress B cells and dendritic cells. They exert their suppressive activity through cell-to-cell contact and via the production of soluble suppressive/inhibitory messengers (i.e. TGF-beta, IL-10 and adenosine). Loss of function mutations in the Foxp3 gene underlie the lymphoproliferative disease of the Scurfy mouse and the homologous autoimmune lymphoproliferative disorder in man, termed Immune dysregulation Polyendocrinopathy Enteropathy−X (IPEX) linked syndrome. Of note, despite the immune suppressive function, the infiltration of tumors by Tregs doesn’t necessarily indicate worse prognosis as it is also an indicator for a T cell inflamed tumor phenotype, (i.e. a (positive) sign for immune reactivity). Recognition of danger signals such as pathogen-associated molecular patterns (PAMPs) or damage-associated molecular patterns (DAMPs) by pattern recognition receptors on APCs initiates cellular maturation and confers to these cells the ability to process and present antigens to T cells (signal 1) as well as achieve co-stimulation (signal 2), both of which are necessary to induce T cell activation. Antigen presentation withough signal 2 can cause T cell anergy. Antigenic peptides are presented within the MHC cleft on the cell surface with the assumption is that an increase in MHC class I expression would be supportive of better antigen presentation thereby making tumor cells more ‘visible’ to T cells. Tumor cell death in response to immunotherapy may lead to the release of secondary (nontargeted) tumor antigens that prime subsequent immune responses. Epitope spreading (or antigen cascade, antigen spread, determinant spread) describes a phenomenon where the immune response evolves and expands from focusing on a single antigenic epitope, into a multi-epitopic response be it naturally or following therapeutic intervention e.g. vaccination or radiotherapy. This process is dynamic and may continue to expand over time. Antigen spreading of the anti-tumor immune response from one antigen to another antigen has been linked to superior clinical outcome with the assumption that it counteracts tumor immune evasion.
2019-XII19. Tumor cells may escape the host’s immune response by a plethora of innate and adaptive mechanisms. Which of the following would NOT be considered such a mechanism? A. Loss of β2-microtubulin expression leading to decreased MHC class I expression B. Tumor cell intrinsic alterations in signaling pathways such as WNT/β-catenin, loss of PTEN, and IFNγ that inhibit T cell priming and infiltration C. Recruitment of myeloid suppressor cells D. Loss of antigen expression through immune selection E. Increased expression of immune inhibitory factors such as Indoleamine 2,3-Dioxygenase (IDO) and PD Ligand 1 (PD-L1)
Answer: A
Beta-2-microglobulin (β2-microglobulin, B2M; not β2-microtubulin) is a crucial component of major histocompatibility complex (MHC) class I molecules.
MHC class I molecules are heterodimers made of two, non-covalently linked polypeptide chains, α and B2M. The conformation of the MHC class I protein is highly dependent on the presence of B2M.
B2M is essential for proper MHC class I folding and transport to the cell surface. Its deficiency has long been recognized as a genetic mechanism of acquired resistance to immunotherapy.
2019-XII20. Which immune-mediated mechanism plays a role in cancer prevention?
A. Detection and elimination of tumor cells
B. Allergic responses
C. Prevention of chronic inflammation
D. Protection against viral infection and integration
E. A, C and D
Answer: E
2019-XII21. What does PD-1 stand for?
A. Programmed cell death 1 receptor
B. Presentation determinant 1
C. Pre-determinant molecule 1
D. Pattern determinant 1
E. Principal determinant 1
Answer: A
2019-XIII1. Which of the following CDK or cyclin is paired with the correct phase transition?
A. CDK1 (CDC2) – G2 into M
B. CDK4 – S into G2
C. cyclin A – G2 into M
D. cyclin B – S into G2
E. cyclin D – M into G1
Answer: A
CDK1 (and cyclin A/B) is associated with the G2 to M cell cycle phase transition.
The other CDKs and Cyclins are appropriately paired as follows:
- G1 phase → S phase: CDK4 and Cyclin D1; CDK2 and Cyclin E
- S phase → G2 phase: CDK2 and Cyclin A
- G2 phase → M phase: CDK1 and Cyclin B/A
2019-XIII2. Irradiation of an exponentially-growing population of cells in culture with a dose that kills 90% of cells tends to select surviving cells that are initially in which phase of the cell cycle?
A. G0
B. G1
C. S
D. G2
E. M
Answer: C
A dose that kills 90% of the cells in the population would leave a surviving cell population heavily enriched in the radioresistant cells in late S phase. Radiosensitivity across the cell cycle is ranked as follows from least to most sensitive: Late S, Early S, G1, G2≈M.
2019-XIII4. Which of the following statements concerning the cell cycle kinetics of tumors is TRUE?
A. Often, the cell loss factor (Φ) decreases several weeks after the start of radiotherapy
B. The growth fraction (GF) is the ratio of the number of viable cells to the sum of viable and non-viable cells
C. If the volume doubling time (TD) is 60 days and the potential doubling time (Tpot) is 3 days, then the cell loss factor is 5%
D. Tpot has proven useful in predicting tumor response to accelerated radiotherapy
E. Typically, the cell loss factor is not of major importance in determining a tumor’s volume doubling time
Answer: A
The cell loss factor (Φ) often appears to decrease several weeks after the start of radiotherapy, which has the net effect of slowing tumor regression.
The growth fraction is the ratio of the number of proliferating cells to the sum of proliferating and quiescent cells (Answer Choice B).
If the the observed tumor volume doubling time (TD) is 60 days and the potential doubling time (Tpot) calculated from the cell cycle time and the growth fraction is 3 days then the cell loss factor is 95% (Answer Choice C).
Although Tpot (as measured from a tumor biopsy derived from patients previously given bromodeoxyurdine) has not proven to be a robust predictor of long-term outcome after accelerated radiotherapy, it might still be useful for the pre-selection of patients most likely to benefit from accelerated treatment (Answer Choice D).
For carcinomas, the cell loss factor is usually the major determinant of the discrepancy between a tumor’s potential doubling time and its overall volume doubling time (Answer Choice E).
NOTE:
Cell loss factor (Φ) = 1 - ( Tpot/ TD)
2019-XIII5. Exponentially growing cells are pulse-labeled with tritiated thymidine and sampled as a function of time thereafter. The time required for the percent of labeled mitoses to reach 50% of its maximum value corresponds approximately to:
A. TS
B. TC
C. TG2
D. TG1 + TS/2
E. TG2 + TM/2
Answer: E
The length of time required for the first radioactively-labeled S phase cells to first enter mitosis, as measured using the percent-labeled mitosis technique, would correspond to the duration of the G2 phase (TG2).
The additional time required for the cells to completely fill the mitotic compartment (i.e., 100% labeled mitoses) would be equal to the length of M (TM).
The time to reach 50% of the maximum point, therefore, corresponds to TG2 plus TM/2.
2019-XVI3. Which of the following pairs of total body radiation effects and approximate threshold dose is CORRECT?
A. Gastrointestinal syndrome – 2 Gy
B. LD50 (no medical intervention) – 3.5 Gy
C. LD50 (best current medical treatment) – 15 Gy
D. Cerebrovascular syndrome – 5 Gy
E. Hematopoietic syndrome – 0.2 Gy
Answer: B
The human LD50 (dose to result in lethality in 50% of an irradiated population) in the absence of medical intervention is estimated at 3.5 Gy. The dose thresholds for the hematopoietic, gastrointestinal and cerebrovascular syndromes are roughly 2 Gy, 8 Gy and 20 Gy, respectively. The LD50 with the best current medical treatment is about 7 Gy.
2019-XIX7. Your clinic has a fancy new treatment planning system that uses radiobiologic tumor control probability (TCP) and normal tissue complication probability modeling (NTCP) to help select the best treatment plan. Your dosimetrist has come up with 4 plans to review. Based on the TCP/NTCP curves below, which of the following plans would you choose?
Answer: D
Tumor control probability (TCP)/Normal tissue complication probability modeling (NTCP) curves are a convenient way to assess the therapeutic ratio of a treatment as it relates to the prescribed dose. The probability of either tumor control or tissue complication is typically sigmoidal. The therapeutic ratio can be estimated by the widest vertical separation of the TCP and NTCP curves. The ideal treatment plan (i.e., a high therapeutic ratio) demonstrates a high likelihood (e.g., >90%) of tumor control while maintaining TCP as low as possible.
The example in A is the worst case scenario, in which normal tissue toxicity is nearly certain at doses that would provide meaningful tumor control. Such is sometimes the case in situations where reirradiation is being considered.
Examples B and C show suboptimal scenarios in which NTCP is still highly likely if trying to achieve curative doses. Of the depicted curves, example D demonstrates the highest therapeutic ratio.
2019-XX2. Assuming no difference in overall treatment time, which of the following statements is CORRECT concerning isoeffect curves?
A. Tissues with a greater repair capacity have steeper isoeffect curves.
B. Increased proliferation of the critical cell population during the course of radiotherapy will decrease the slope of the isoeffect curve.
C. Tissues with steep isoeffect curves have high 𝛼/𝛽 ratios.
D. Isoeffect curves for tumor control will be steeper if significant reoxygenation occurs between dose fractions.
Answer: A
An isoeffect curve describes the relationship between total dose for a given level of tissue effect and the different fractionation parameters (overall time, dose per fraction, number of fractions, etc). Isoeffect curves are often plotted with the log of the total dose on the y-axis and the log of the fraction size (from high to low) on the x-axis. Tissues with a greater repair capacity will show greater sparing with increasing fractionation (smaller fraction sizes) and therefore will have steeper isoeffect curves.
Increased proliferation will cause an increase in the slope of an isoeffect curve because it would take a higher total dose to kill the larger number of cells produced during the course of treatment (Answer Choice B).
Tissues with steep isoeffect curves have low, not high, 𝛼/𝛽 ratios (Answer Choice C).
Reoxygenation decreases the slope of the isoeffect curve because it decreases the number of radioresistant hypoxic cells and hence reduces the total dose required to control the tumor, everything else being equal (Answer Choice D).
2019-XXI1. Which of the following equations would be most appropriate to use when calculating the BED for treatment involving a permanent radioactive implant?
Answer: E
An equation that takes into account the complete decay of the brachytherapy source is most appropriate for calculation of the BED for a permanent radioactive implant.
Choice A is the correct equation to use for standard external beam therapy when the dose is delivered typically over a 1-2 minute period or for high dose-rate brachytherapy.
Choice B takes into account repopulation during the course of radiotherapy and should be used to compare fractionated protocols of different durations.
Choice C is used for treatment with closely-spaced, multiple fractions per day when incomplete repair may be an issue.
Choice D is used to calculate the BED for a brachytherapy treatment at a constant, low-dose rate
2019-XXIV8. Rapamycin, everolimus and temsirolimus may act as a radiosensitizers by inhibiting:
A. K-ras
B. mTOR
C. MAPK
D. p38
E. EGFR
Answer: B
Rapamycin, everolimus and temsirolimus are all inhibitors of the mTOR (mammalian target of rapamycin) protein.
mTOR functions downstream of PI(3)K to promote cell survival. Inhibition of mTOR blocks these pro-survival pathways.
2019-XXIV13. Which of the following is FALSE regarding superoxide dismutase (SOD) mimetics?
A. The chemical structure of a SOD mimetic contains a redox active metal ion that is oxidized in the presence of superoxide (O2●-)
B. Their mechanism of action involves converting superoxide (O2●-) into hydrogen peroxide (H2O2)
C. Their ability to protect against radiation- and chemotherapy-related oral mucositis in head and neck cancer is currently being evaluated in clinical trials
D. An SOD mimetic has yet to be identified that protects against radiation toxicity in normal tissues without also protecting tumor tissue
Answer: D
One SOD mimetic, GC4419, that is currently in clinical trials is able to selectively protect normal cells from radiation- and chemotherapy-related toxicity without protecting tumor cells. Superoxide dismutase (SOD) mimetics contain a redox active metal ion as part of their chemical structure (Answer Choice A). In the presence of superoxide (O2●-), the redox active metal ion is oxidized (loses an electron) and the superoxide is reduced to hydrogen peroxide (Answer Choice B). SOD mimetics have been evaluated in phase 1 and 2 clinical trials regarding their ability to protect against radiation- and chemotherapy- induced oral mucositis; a phase 3 clinical trial is currently underway (Answer Choice C).
2019-XXV1. Which of the following statements concerning the Arrhenius analysis of mammalian cell killing by heat is TRUE?
A. An Arrhenius curve plots the log of the slope (1/D0) of heat survival curves as a function of temperature
B. The break point in the Arrhenius plot is defined as the temperature at which the slope of the plot significantly increases.
C. The Arrhenius relationship has been used to define the temperature dependence of mechanisms of cell killing
D. The results of these analysis suggested the nuclear matrix may be a target of heat-induced cell killing
E. The break point in the Arrhenius plot is different between rodent and human cancer cells.
Answer: C
The Arrhenius plot demonstrates the temperature at which the mechanisms underlying cell killing changes, potentially reflecting different targets for cytotoxicity above the break point (43oC)
The Arrenhius analysis plots survival data of cell cultures exposed to increasing temperatures. The X-axis plots 1/D0, where D0 represents the time at a given temperature required to reduce the fraction of surviving cells to 37% of the initial population. The Y-axis plots 1/T, where T is the absolute temperature (Answer Choice A).
The slope of the Arrhenius plot provides the activation energy of the chemical process involved in the cell killing.
At some point (1/T), corresponding to approximately 43o, there is a significant and abrupt decrease in the slope (Answer Choice B).
The results of these analysis suggested that the target for heat cell killing may be a protein.
The break point in the Arrhenius plot occurs at a temperature of approximately 43oC, and is the same across mammalian cells (Answer Choice E).
2019-XXVI8. Which of the following is an example of a stochastic effect of exposure to high-dose radiation:
A. Mental retardation following exposure of the fetus in utero
B. Acute mucositis
C. Development of breast cancer 20 years following exposure to radiation as a teenager
D. Cardiac toxicity
E. Cataracts
Answer: C
Development of cancer after high dose radiation is an example of a stochastic effect.
A stochastic effect fulfills two criteria:
1) the probability of an outcome of interest increases with increasing dose, typically without a threshold dose; and
2) the severity of an outcome of interest is not altered by dose (all or none)
In contrast, deterministic effects, such as mental retardation, cardiac toxicity, cataracts and acute toxicity, have a threshold dose below which it does not occur and have increased severity with increased dose.
2019-XXVI12. Approximately 10,000-20,000 cases of lung cancer each year in the United States are attributed to alpha-particles produced by:
A. Nuclear weapon testing
B. Decrease in the ozone layer
C. Radon gas
D. Chemical contamination
Answer: C
Alpha-particles are emitted during the decay of radionuclides, such as radon, that occur in nature. Radon gas escapes from the soil and builds up inside homes where it is inhaled and can cause lung cancer.
2019-XXVIII2. Which of the following pairs of gestational stage and radiation-induced developmental defect is CORRECT?
A. preimplantation – congenital malformations
B. organogenesis – prenatal death
C. early fetal period – mental retardation
D. late fetal period – neonatal death
E. entire gestation period – malformations of the kidney
Answer: C
Irradiation during the early fetal period, corresponding to weeks 8-15 of gestation in humans, is associated with the greatest risk for mental retardation. The main risks during preimplantation, organogenesis, and the late fetal period are prenatal death, congenital malformations, growth retardation and carcinogenesis, respectively. There is an increased risk of carcinogenesis following irradiation throughout the gestation period
2019-XXVIII4. Once a pregnancy is declared, the maximum permissible dose to the fetus is:
A. 0.005 mSv per month
B. 0.05 mSv per month
C. 0.5 mSv per month
D. 5 mSv per month
E. 50 mSv per month
Answer: C
Once a pregnancy is declared, the maximum permissible dose to the fetus is 0.5 mSv per month subsequent to this declaration. Prior to declaration,
there are no special limits except for the general limits for radiation workers.
NOTE:
For radiation workers
- 50 mSv/ year (~4 mSv/month)
For general public
- 1 mSv/ year
2019-XXIX1. A woman begins working at a nuclear power plant on her 18th birthday. According to current NCRP guidelines, once she reaches her 20th birthday she will have been permitted a total work-related lifetime effective dose equivalent of:
A. 5 mSv
B. 50 mSv
C. 100 mSv
D. 200 mSv
E. 300 mSv
Answer: C
A radiation worker is permitted either 50 mSv per year for each year that the person was engaged in radiation work or else a lifetime dose equal to his/her age multiplied by 10 mSv, whichever is less. Based on the lifetime dose rule, this woman would have been permitted 200 mSv as of her 20th birthday. The 50 mSv per year rule dictates that her maximum allowable dose would be only 100 mSv.
2019-XXIX2. A woman begins working at a nuclear power plant on her 18th birthday. Suppose that on her 21st birthday, she declared that she was 3 months pregnant. What additional dose limit to the fetus has the NCRP recommended for the duration of her pregnancy?
A. She would not be allowed any additional radiation exposure once the pregnancy was declared
B. 1 mSv
C. 10 mSv
D. 50 mSv, assuming that she had no measurable exposure yet that year
E. 0.5 mSv per month
Answer: E
The NCRP recommendations state that a worker who has declared a pregnancy may receive a maximum dose of 0.5 mSv per month to the fetus.
2019-XXIX3. What are the NCRP maximum permissible annual dose limits for the eye and to localized skin areas for radiation workers?
A. 50 mSv to the eye and skin
B. 150 mSv to the eye and skin
C. 50 mSv to the eye and 150 mSv to the skin
D. 50 mSv to the eye and 500 mSv to the skin
E. 500 mSv to the eye and 150 mSv to the skin
Answer: D
A radiation worker is currently permitted 50 mSv to the eye and 500 mSv to the skin in any given year.
This is a recent change from the previous NCRP recommendations for which the annual dose equivalent limit for the lens the eye was 150 mSv.
These limits are based on risk estimates for the production of radiation-induced deterministic effects.
The ICRP recommends an annual equivalent absorbed dose limit for the lens of the eye to be 15 mSv for the public.
For chronic occupational exposures, the ICRP recommends an equivalent dose limit for the lens of the eye of 20 mSv in a year, averaged over defined periods of 5 years, with no single year exceeding 50 mSv.
2019-XXIX4. In the United States, the average annual effective dose equivalent from all sources of radiation is closest to:
A. 0.2 mSv
B. 1 mSv
C. 3 mSv
D. 6 mSv
E. 15 mSv
Answer: D
The average annual effective dose for a person residing in the US is approximately 6 mSv. This total includes an average radon contribution of 2 mSv; cosmic, terrestrial and internal radioactivity of 1 mSv. In addition, an average of 3 mSv from man-made sources, primarily from diagnostic and nuclear medicine procedures, is received. This increase from the often quoted 1980 figure of 0.5 mSv associated with medical procedures is the result primarily from the significant increase in the use of CT scans over the past 30 years. CT scans deliver relatively high radiation doses compared with other imagining modalities.
2019-XXXI4. Which of the following statements concerning computed tomography (CT) is CORRECT?
A. Tissues that strongly absorb X-rays appear black while others that absorb poorly appear white on CT images.
B. Iodine-based contrast agents are mainly used in the imaging of the digestive system via CT scanning.
C. Water has an X-ray attenuation of 0 Hounsfield units (HUs).
D. Organ-specific radiation doses from CT scans are negligibly low compared to those associated with conventional radiography.
E. CT devices and image reconstruction software are regulated by the U.S. Nuclear Regulatory Commission (NRC).
Answer: C
The Hounsfield unit (HU) scale relates X-ray attenuation in various tissue types (μ) to X-ray attenuation in water μ(water) through the equation:
HU = 1000 × [𝜇−𝜇(𝑤𝑎𝑡𝑒𝑟)] / 𝜇(𝑤𝑎𝑡𝑒𝑟)
Based on the above equation, the HU of water is 0.
Other choices are incorrect. Barium-based rather than iodine-based contrast agents are used in the imaging of the digestive system by CT. Organ doses from diagnostic CT procedures are typically estimated to be in the range of 1 cGy per scan and as much as 10 cGy from multiple CT scans.
For example, cumulative doses from 2-3 head CTs to the brain are 5-6 cGy. For comparison, X-ray doses from chest radiography are 0.01 cGy (0.1 mGy) and X-ray doses from mammography are 0.04 cGy (0.4 mGy). The Nuclear Regulatory Commission (NRC) is responsible for the regulation of radioactive materials used medically; this includes the diagnostic radionuclides used in positron emission tomography (PET) imaging.
On the other hand, CT scanners, image reconstruction software, as well as other “medical devices” such as linear accelerators and associated treatment planning software are regulated by the United States Food and Drug Administration (FDA).
