ASTRO 2023 (1,2,3,26-32) Flashcards

Question 1

Q

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

A

(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.

-PE increases with the atomic number of the absorber

-The predominant interaction of 10 keV photons in soft tissue is the photoelectric effect.

-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

-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

Question 2

Q

Which one of the following is a radiolysis product of water responsible for the molecular damage caused by the indirect action of ionizing radiation?
A. eaq
B. O2
C. OH
D. OH*
E. O2

Answer

A

-(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)

-O2 is produced primarily by photosensitizers and, rarely, by ionizing radiation.

-Neither OH- nor O2 are primary radiolysis products, although O2 can be produced secondarily by reaction of eaq with O2

Question 3

Q

The approximate minimum photon energy required to cause ionization of a water molecule 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.

Question 4

Q

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

A

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.

Question 5

Q

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

A

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.

Question 6

Q

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 gamma 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

A

-High linear energy transfer (LET), or densely ionizing, radiations include particles such as 290 MeV carbon ions, GeV cosmic 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).

Question 7

Q

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

A

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.

Question 8

Q

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

A

Annihilation photons always have an energy of 0.511 MeV each, which is equal to the rest energy of the positron and electron.

Question 9

Q

Directly ionizing radiation includes all of the following EXCEPT:
A. Electrons
B. Positrons
C. Alpha particles
D. Neutrons
E. Betas

Answer

A

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.

Question 10

Q

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

A

-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.
-(B)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.
Neutron beams used in radiation therapy can be made using a cyclotron by accelerating protons into a beryllium target.
-The fast neutrons then recoil protons in target tissue from elastic collisions and 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
-Neutron absorption in a target nucleus is called activation. This is a process by which neutron radiation induces radioactivity 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.

Question 11

Q

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

A

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 (eaq ) (45%). These react
with DNA or with each other.
OH + H = H2O

The remaining recombination reactions of free radicals are:
eaq + eaq +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 from the free radicals.
-Note that LET for photons is in the ~1 keV/µm range, while protons are approximately 10 keV/µm, carbon atoms are 10-100 keV/µm, and alpha particles or heavy charged particles are >100 keV/µm.

Question 12

Q

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

A

The indirect effect mediated by free-radical reactions involving water are
most responsible to cause DNA
damage upon low LET irradiation.

Question 13

Q

The SF2 (surviving fraction at 2 Gy) for
an irradiated population of cells
is most closely correlated with the:
A. Level of gamma-H2AX 30 minutes after irradiation
B. Level of gamma-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

A

-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 (gamma-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.
-SF2, the cell surviving fraction after 2 Gy, is a model-independent measure of radiation sensitivity. The numbers of phosphorylated gamma-H2AX
foci shortly after the irradiation represent the initial level of DNA damage, but the number of phosphorylated gamma-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.

Question 14

Q

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

A

In contrast with the other forms of damage listed, pyrimidine dimers are principally produced following absorption of photons in the ultraviolet (UV) 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 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.

Question 15

Q

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

A

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.

Question 16

Q

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

A

-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 create 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 assay
measures the amount of new DNA synthesis (Answer Choice E).

Question 17

Q

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 (H2AX) foci can be detected within 15 minutes of radiation exposure

D. p53 forms ATR-dependent foci within minutes of radiation exposure

E. The remaining gamma H2AX foci within the nucleus at 24 hours after irradiation is not related to the cellular sensitivity radiation.

Answer

A

-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.

An elevated level of remaining H2AX foci at an extended time (such as 24
hours) after the initial DNA damage is indicative of an impaired DNA repair and cell sensitivity to radiation.

Question 18

Q

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 gammaH2AX)
B. Degraded histone H2AX
C. Dephosphorylated H2AX
D. Cleavage of Caspase 3
E. DNA methylation

Question 19

Q

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

A

-Transcriptional transactivators increase the expression of a protein by binding to the promoter region of the target gene and turning on transcription. 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).

Question 20

Q

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

A

Phosphorylation of histone H2AX to  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

Question 21

Q

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

Question 22

Q

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 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).

Question 23

Q

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. Impact 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

A

-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).

Question 24

Q

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

A

-RAD52 plays a central role in homologous recombinational repair (HR) of DNA double-strand breaks through recruitment of RAD51 to singlestranded 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).

Question 25

Q

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

A

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).

Question 26

Q

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 crosslinking agents
E. A mutation in p53 (TP53) produces an immune deficient phenotype in SCID mice

Answer

A

-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).

-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.

Question 27

Q

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, Ku70, and DNA-PKcs all play roles in NHEJ (Answer Choices B, C,
and E)

Question 28

Q

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

A

-An exonuclease cleaves one nucleotide at a time beginning at the end of a DNA strand.
-Ligases seal breaks in the DNA strand (Answer choice A).
-DNA polymerases add a new nucleotide to the end of DNA during DNA synthesis (Answer choice B).
-Endonucleases produce nicks within intact DNA strands (Answer choice C).
-RNA polymerase generates new species of mRNA (Answer choice D).

Question 29

Q

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, as it is also involved in single-stranded DNA break repair.

-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).

Question 30

Q

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.
E. The cell cycle status does not affect the choice between HR and NHEJ
to repair a DSB.

Answer

A

Two principal recombinational DNA repair pathways have been identified, homologous recombination (HR) and non-homologous endjoining (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)

Question 31

Q

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

A

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 doublestrand 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 onesided 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.

Question 32

Q

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

A

Xeroderma pigmentosum is a genetic disorder due to mutation in genes involved in nucleotide excision repair (NER). This phenotype is characterized by extreme sensitivity to ultraviolet light.

Question 33

Q

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

A

BRCA1 and BRCA2 predominantly regulate homologous recombination (HR) as opposed to non-homologous end joining (NHEJ). Fewer than 10% of patients with breast cancer are found to have mutations in these breast cancer susceptibility genes. Due to the negative effect that BRCA1 mutation has on HR, BRCA1-deficient cells are more SENSITIVE, not resistant, to mitomycin C and other DNA crosslinking agents (Answer
choice A).
The prevalence of BRCA1 mutation is slightly lower than that of BRCA2 mutations in the general US population (1:500 for BRCA1 vs 1:222 for BRCA2). The breast cancer risks for carriers of BRCA1 and BRCA2 mutations are similar at about 70% by the age of 80,
but with earlier age of disease onset for the BRCA1 mutation (peaks in 4th decade) compared to BRCA2 carriers (peaks in 5th decade).

Question 34

Q

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. Activating K-Ras mutation
D. MLH1 nonsense mutation
E. XRCC1 null mutation

Answer

A

Among the 5 genes listed, all may affect radiation or DSB sensitivity at
different levels. However, DNA-PKcs is directly involved in DSB repair and
DNA-PKcs mutation is known to cause hypersensitivity to DSB damage and radiation.

Question 35

Q

Which statement is TRUE concerning the role of p53 (TP53) and p21
(CDKN1A) in the response of the cells to radiation?
A. p21 phosphorylates NBS1 (NBN), thereby stimulating homologous
recombinational repair of DNA double-strand breaks
B. p53-mediated G1 phase arrest results from the inactivation of p21
C. A decrease in the amount of p53 can trigger apoptosis or G1 arrest
D. p21 inhibits CDK-cyclin activity thereby decreasing phosphorylation
of RB1
E. DNA damage initiates a signal transduction pathway that results in a
marked increase in transcription of the p53 gene

Answer

A

p21 inhibits CDK-cyclin activity, which has the effect of decreasing the phosphorylation of RB1. ATM, and not p21, phosphorylates NBS1 thereby
stimulating homologous
recombinational repair. p53-mediated G1 arrest results from transactivation of p21 by p53. An increase in the amount of p53 can result in apoptosis or G1 arrest. DNA damage does initiate
a signal transduction pathway that
results in increased amounts of p53,
however this occurs by stabilization of the existing protein, rather than by increased transcription of the gene that encodes it.

Question 36

Q

Which statement is CORRECT concerning the ataxia telangiectasia mutated (ATM) gene and Rad3-related (ATR) genes and proteins?
A. Ionizing radiation induced phosphorylation of Chk1 requires either ATM or ATR.
B. ATM is recruited to double strand breaks by the Mre11-Rad50-Mbs1
complex
C. ATR activation and Chk1 phosphorylation occurs prior to ATM
activation
D. Cells derived from patients with AT typically display increased levels
of p53 (TP53) phosphorylation
E. Irradiation causes autophosphorylation of ATM which converts it from an active monomer to an inactive dimer

Answer

A

The ATM protein contains a highly conserved C-terminal kinase domain
resembling a phosphatidylinositol-3 kinase (PI(3)K); this kinase is an
important component of a number of DNA damage repair pathways. Both ATM and ATR are required for IR-induced Chk1 phosphorylation.
ATM is recruited to double strand breaks by the Mre11-Rad50-Nbs1 complex. ATR is recruited to single
stranded DNA at sites of stalled replication forks by ATR-interacting protein (ATRIP).
Cells derived from patients with AT typically display decreased levels of p53 phosphorylation. Irradiation causes autophosphorylation of ATM which
converts it from an inactive dimer into the active monomeric form, not vice versa.
ATM activation and Nbs1 recruitment to damaged DNA occurs prior to ATR
recruitment and Chk1 phosphorylation

Question 37

Q

Cancer Immunology

Which of the following organ systems may be affected by an immune adverse event in a lung cancer patient receiving radiotherapy plus immune check point blockade?

Question 38

Q

Question 39

Q

Which of the following is an example of adaptive immune resistance?
A. Process whereby a patient is tolerant to a tumor associated antigen (i.e. NY-ESO or PSCA) before starting immunotherapy but develops immunity to it once beginning treatment.
B. Process by which tumor cells change phenotype in response to an immune response (cytotoxicity or inflammation) in an attempt to avoid recognition (i.e. the induction of PD-1, PD-L1, and IDO following antigen recognition and the production of IFNγ).
C. Process by which tumors, following radiotherapy, undergo accelerated proliferation with an increased incidence of failure.
D. Process by which there is increased resistance to radiotherapy due to reactive response by the immune system.
E. Process by which pattern recognition receptors detect damage associated molecular patterns (DAMPs) induced by radiation.

Answer

A

B

The process by which tumor cells change phenotype in response to an immune response (cytotoxicity or inflammation) in an attempt to avoid recognition (i.e. the induction of PD-1, PD-L1, and IDO following antigen recognition and the production of IFNγ) is characteristic of an adaptive immune response

Question 40

Q

Which of the following may increase a patient’s susceptibility to experience an immune related adverse event?
A. History of autoimmune disease
B. Abnormal thyroid function
C. Previous use of checkpoint blockade therapy
D. Previous radiation therapy
E. All of the above
F. A, C and D

Question 41

Q

Immune related adverse events (irAEs) describe a range of immune mediated toxicities that can result from treatment with immune checkpoint inhibitors. Which statement regarding irAEs is NOT correct:
A. Skin, gut, endocrine, lung and musculoskeletal irAEs are relatively common, whereas cardiovascular, hematologic, renal, neurologic and ophthalmologic irAEs occur much less frequently.
B. irAEs typically have a delayed onset and prolonged duration compared to adverse events from chemotherapy.
C. irAEs are discrete toxicities caused by tissue-specific inflammation and activation of the immune system and can affect almost any organ system.
D. The overall incidence of irAEs following treatment with anti-CTLA4 monotherapy tends to be lower than those with anti-PD-1/PD-L agents
E. The incidence of irAEs with ipilimumab and pembrolizumab is dose dependent, with greater toxicity at higher dose levels;

Answer

A

D

Section 1: Regulation of T Cell Activation by CD28-CTLA4
Naive and memory T cells express high levels of CD28 but lack surface CTLA4. CTLA4 is initially sequestered within intracellular vesicles. Upon TCR activation, CTLA4 is transported to the cell surface. The strength of TCR signaling determines the level of CTLA4 expression. Stronger signals lead to higher CTLA4 surface expression.
Role of CTLA4: Acts as a negative regulator of T cell activation. Dampens T cell responses to maintain appropriate activation levels despite varying antigen stimulation. Prevents excessive T cell activation in response to strong TCR signals.

Section 2: Role of PD-1 Pathway in T Cell Regulation
PD-1 pathway primarily regulates T cell function in peripheral tissues, not during initial activation.
Activated T cells express PD-1.
Inflammatory signals, particularly IFN-γ produced by TH1 cells, induce the expression of PD-L1 on target cells. Binding of PD-1 on T cells to PD-L1 on target cells downregulates T cell activity.
This limits tissue damage and maintains immune homeostasis.
Chronic Antigen Exposure: Excessive PD-1 signaling in the context of chronic antigen exposure can lead to T cell exhaustion or anergy, impairing their function.

Question 42

Q

Which of the following agents does not target the PD1 / PDL-1 axis?
A. Pembrolizumab
B. Avelumab
C. Ipilimumab
D. Durvalumab
E. Nivolumab

Answer

A

C

Inhibitors of programmed cell death 1 (PD-1) receptor: pembrolizumab and nivolumab.

Inhibitors of programmed death ligand 1 (PD-L1): atezolizumab, durvalumab, and avelumab.

Ipilimumab is an example of a Cytotoxic T-Lymphocyte Antigen 4 (CTLA-4) inhibitor.

Question 43

Q

Which of the following statements is CORRECT when comparing the abscopal effect versus the bystander effect in the context of radiation responses?

Answer

A

The abscopal effect describes a situation whereby a patient being treated with radiation therapy to a site of metastatic disease experiences concurrent regression of a distant site of metastatic disease that is not being directly irradiated. The abscopal hypothesis was first described in 1953 to refer to the effects of ionizing radiation occurring “at a distance from the irradiated volume but within the same organism.”

In contrast, the bystander effect describes the induction of biologic effects in cells that are in close proximity to cells that are directly traversed by a charged particle

Question 44

Q

Antigen recognition by T cells is imperative for the development of cellular adaptive immunity. How does a T cell recognize an antigen?

Answer

A

B. T cells recognizes antigenic determinants presented in the MHC cleft by the T cell receptor

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.

Question 45

Q

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

A

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 (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.

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.

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.

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.

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. 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.

Question 46

Q

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

A

B

Regulatory (suppressor) T cells are a subset of CD4+ T cells that express the transcription factor foxhead 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).
Antigen presentation without signal 2 can cause T cell anergy.

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

Question 47

Q

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 B2-microtubulin expression leading to
decreased MHC class I expression
B. Tumor cell intrinsic alterations in signaling pathways such as WNT/- 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

MHC class I molecules are heterodimers made of two, noncovalently 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.

Question 48

Q

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

Question 49

Q

What does PD-1 stand for?

Answer

A

A. Programmed cell death 1 receptor

Question 50

Q

Molecular Imaging

The most commonly used biologically active molecule for positron emission tomography (PET) scanning is a fluoridinated analog of which of the following:
A. Phosphate
B. Glucose
C. Calcium
D. Albumin
E. Sphingomyelin

Question 51

Q

The following nucleoside has been radiolabeled in an effort to image DNA synthesis using positron emission tomography (PET):
A. Adenosine
B. Guanosine
C. Thymidine
D. Uridine
E. Cytidine

Question 52

Q

The Hounsfield unit scale is:

Answer

A

The Hounsfield unit scale is a standardized approach to interpreting reconstructed images obtained with a computerized tomography (CT) scanner. CT is a technique that relies on differential levels of X-ray attenuation by tissues within the body to produce digital images reflecting anatomy. Hounsfield units (HU) are numerical values that reflect these differences in density and composition, and thus X-ray attenuation, between various tissue types. Radiologists use software that automatically assigns HUs to every voxel of a CT scan to enable efficient scan interpretation. In oral radiology, approximate HUs can be derived using grayscale levels in cone beam CT (CBCT) images.

Question 53

Q

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

A

C

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).

B- barium
E- FDA

Question 54

Q

Which of the following statements concerning the prognostic significance of pre-therapy [18-F] fluorodeoxyglucose (FDG) positron emission tomography (PET) imaging in patients is CORRECT?
A. SUV score values are directly correlated with local tumor control.
B. SUV score values are inversely proportional to a patient’s body weight.
C. SUV score values can be used to distinguish between quiescent and proliferating tumors.
D. SUV score values are insensitive to extending the time between radioisotope injection and completion of the PET scan.
E. Typical doses of FDG for the clinically useful PET imaging are in a range of 15 Ci.

Answer

A

C

The standard uptake value (SUV) is a standard method of quantifying the radioactive uptake observed in a positron emission tomography (PET) scan image.

As a cancer detection method, (FDG) PET is based on the observation that in normoxic conditions tumor cells primarily use glycolysis for energy production instead of mitochondrial oxidative phosphorylation as normal cells do. This phenomenon is known as the Warburg effect.

SUV scores of >15 g/mL usually indicate a tumour that is highly dependent on glucose metabolism and is therefore more aggressive rather than indolent. It therefore follows that SUV scores inversely correlate with local tumor control

The SUV is calculated as follows: SUV [g/ml] = (Tissue activity (mCi/ml))/(injected dose (mCi)) × patient’s body weight (g). SUV score values are therefore proportional to a patient’s body weight

SUV values increase in value if the PET scan is delayed after FDG injection.
The typical dose of FDG administered to adult patient is approximately 10-18 mCi.

Question 55

Q

Which of the following statements is correct regarding PET-PSMA scans?
A. Physiological uptake is seen in the brain, heart and kidneys.
B. PSMA stands for Phosphatidyl Serine Membrane Antagonist.
C. PSMA scans are a good method of detecting micrometastases from prostate cancer
D. The tracer is based upon the peptido mimetic Glu-NH-CO-NH-Lys(Ahx)-HBED-CC combined with the radionuclide 68Ga.
E. The radionuclide gallium-68 (68Ga) decays through electron emission to the stable isotope zinc-68 (68Zn)

Answer

A

D

PSMA is expressed in certain non-prostate tissues, such as the kidneys, small intestine and salivary glands, consequently these tissues also demonstrate uptake on PET-PSMA scan. Conversely, On PET-FDG scans physiological uptake is seen in the brain, heart and urinary system and other metabolically active tissues.

PSMA stands for Prostate-specific membrane antigen. PSMA is a membrane- bound antigen that is highly specific for benign and malignant prostatic epithelium, although endothelial cells in multiple organs are also immunoreactive. PET PSMA is not reliable for lesions less than 5 mm in size.

The radionuclide gallium-68 (68Ga) decays through positron emission to the stable isotope zinc-68 (68Zn). The tracer is indeed based upon the peptidomimetic Glu-NH-CO-NH-Lys(Ahx)-HBED-CC combined with the radionuclide 68Ga.

Question 56

Q

Molecular Techniques used in Radiation and Cancer Biology

siRNAs and miRNAs:
A. bind to and inhibit the replication of specific genes
B. stimulate RNA synthesis
C. are typically 1 kb in size
D. stimulate protein synthesis
E. inhibit the translation of specific genes

Answer

A

E

The use of microRNAs (miRNAs) and small interfering RNAs (siRNAs): important tool for “gene silencing” or RNA interference (RNAi). miRNAs and siRNAs bind to and inhibit the transcription of specific genes and/or they can silence cytoplasmic mRNAs either by stimulating their cleavage or by inhibiting translation.

miRNAs are produced from transcripts that form stem-loop structures, whereas siRNAs are produced from long double-stranded RNA precursors.

In the initiation phase of RNAi, the ribonuclease-III enzyme Dicer cleaves double-stranded RNA molecules into 21–23-nt short interfering siRNA duplexes. In the effector phase of RNAi, the siRNA becomes unwound and assembles into RISC (RNA-induced silencing complex). The activated effector complex recognizes the target by siRNA–mRNA base pairing and cleaves the mRNA strand with its endoribonuclease activity.

Question 57

Q

Which of the following statements is TRUE concerning the use of PET imaging?
A. 18F-2-deoxy-2-fluoro-D-glucose (18F-FDG) has a radioactive half-life of approximately 10 days
B. A PET imaging camera detects positrons generated from the decay of radiopharmaceuticals
C. The uptake of 18F-FDG is typically lower in areas of inflammation
D. An important advantage to using 18F-FDG-PET/CT fusion images for radiotherapy treatment planning is that they provide both functional and anatomical information
E. Tumors tend to show a reduced uptake of 18F-FDG

Answer

A

D

The radioactive half-life of 18F is 110 minutes,
PET imaging cameras detect the 0.51 MeV photons produced by the annihilation resulting from the interaction of a positron and
electron.

Question 58

Q

Which of the following statements concerning gel electrophoresis is TRUE?
A. DNA molecules are negatively charged so they will migrate toward the positive electrode of the electrophoresis apparatus
B. SDS is a detergent used for the separation of DNA molecules of different size
C. The higher the concentration of agarose in a gel, the faster DNA molecules will migrate
D. Polyacrylamide gels are used to separate large DNA molecules whereas agarose gels are used for smaller-sized DNAs
E. The higher the molecular weight of the molecule, the faster it will migrate through a gel

Answer

A

In gel electrophoresis, DNA molecules are negatively charged and therefore migrate towards the positive electrode.

Sodium dodecyl sulfate (SDS), a detergent, is used to denature proteins, not DNA, so that the proteins can be separated by size on a gel.

The higher the concentration of agarose in the gel, the slower DNA molecules will migrate.

Polyacrylamide gels are generally used to separate small DNA molecules whereas agarose gels are used for large sized DNA.

The lower the molecular weight of the molecule, the more rapidly it will migrate through a gel.

Question 59

Q

Which of the following assays would best determine whether a particular radiation sensitivity syndrome is characterized by defective repair of DNA double-strand breaks?
A. quantitation of GAMMA-H2AX foci
B. western blot
C. alkaline comet assay
D. southern hybridization
E. northern hybridization

Question 60

Q

Which statement regarding next generation sequencing (NGS) is FALSE:
A. Unlike capillary sequencing, NGS requires the cloning and amplification of DNA sequence-containing phage libraries
B. NGS is a massively parallel process with a million or more simultaneous DNA sequence reads.
C. NGS is not hampered by homopolymer repeat sequences.
D. NGS generally performs short DNA reads of less than 100 bases.
E. In NGS, bases are read by sequential computer-mediated image analysis.

Answer

A

Capillary sequencing requires in vivo cloning and amplification whereas next generation sequencing (NGS) utilizes adaptor ligation of DNA fragments and binding to a matrix for DNA sequencing.

Question 61

Q

Pulsed-field gel electrophoresis can be used in order to:
A. determine a cell’s karyotype
B. detect DNA interstrand crosslinks
C. separate protein molecules on the basis of both molecular weight and charge
D. monitor the repair/rejoining of large pieces of DNA after the production of double-strand breaks
E. determine the rate of base versus nucleotide excision repair

Question 62

Q

Which one of the following is NOT a method for studying gene expression at the protein level?
A. immunohistochemistry
B. ELISA
C. northern blots
D. western blots
E. two-hybrid screening

Answer

A

C

Northern blotting is used to study RNA.

Question 63

Q

A scientist is planning an experiment in which he wants to determine whether his exponential cultures of human HCT116 colorectal cancer cells express wild-type p53 protein. Which of the following experimental
assays would NOT be an effective readout for this purpose?

A. Western blotting with the p53 protein-specific antibody

B. Performing flow cytometry in order to analyze the cycling characteristics of HCT116 cells 12 h after exposure to 6 Gy X-Rays

C. Northern blotting to measure WAF1/CIP1 mRNA expression in HCT116 cells 3 h after exposure to 6 Gy X-Rays.

D. Immunoblotting with a cyclin E-specific antibody in order to detect cyclin E activity 12 hours after exposure to 6 Gy X-Rays.

E. Immunoblotting with an antibody against p53, phosphorylated at the serine 15 residue, 1 hour after exposure to 6 Gy X-Rays.

Answer

A

Wild-type p53 protein is not detectable, because its mRNA is short-lived (T½ = 8 min) in unstressed cells.

The induction of DNA double strand breaks by X-ray irradiation initiates a p53-dependent signal transduction
cascade. One downstream target of this cascade includes the induction of WAF1/CIP1 mRNA, which encodes the p21 protein. Upregulation of WAF1/CIP1
protein inhibits the cyclin E/cyclin-dependent kinase 2
complex, an event that is able to stop cells from progressing through G1.

Phosphorylation of p53 at serine-15 in response to ionizing radiation correlates with both accumulation of total p53 as well as its transactivation of downstream genes

Question 64

Q

Clonogenic survival assays are commonly used to assess in vitro cellular response to ionizing radiation and cytotoxic agents. Which of the following statements is INCORRECT about the clonogenic survival assay?

A. A feeder layer consists of irradiated cells that no longer divide but still produce growth-stimulating factors for the colony forming cells

B. Trypsinizing donor culture cell is necessary to produce a single-cell suspension

C. Plating efficiency is the ratio of the number of formed colonies to the number of cells plated

D. The survival fraction is the number of colonies formed multiplied by the plating efficiency and divided by the number of cells plated

E. Survival curves are plotted with surviving fraction on the Y-axis and therapeutic dose (radiation or cytotoxic agent) on the X-axis

Answer

A

D

A clonogenic survival assay is an in vitro cell survival assay based on the ability of a single cell to grow into a colony (~50 cells). This approach is commonly used by radiation biologists to assess 248 reproductive death after ionizing radiation or treatment with other cytotoxic agents. Following treatment with radiation and/or cytotoxic agents, only a fraction of cells retain reproductive capacity and can form colonies.

A feeder layer consists of cells (fibroblasts) that have been irradiated (typically ~30 Gy) and can no longer divide but still produce growth factors for the colony forming cells.

Trypsin is a protease that breaks down cellular adhering proteins to facilitate release of cells from each other and detach cells from a tissue culture dish.

Survival curves are semi-log plots, with the survival fraction (logarithmic scale) on the Y-axis and therapeutic dose on the X-axis (linear scale).

Answer 56

A

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.

Answer 57

A

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 mother’s fetal monitor worn at the mother’s waist, under any protective apron worn by the woman.

Answer 58

A

D

NCRP recommendations: A radiation worker is currently permitted 50 mSv to the eye and 500 mSv to the skin in any given year & annual dose equivalent limit for the lens the eye was 150 mSv.

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.

Answer 59

A

B

A member of the general public is permitted 1 mSv per year for “chronic” radiation exposure over extended periods of time, or 5 mSv per year for an infrequent exposure.

Radiation workers, including radiation oncologists and nuclear power plant employees, may receive 50 mSv per year

A person under the age of 18 may be exposed to radiation up to 1 mSv per year if the potential exposure occurs as part of an educational or training program

A patient’s relative transporting a patient to and from radiotherapy treatment, presumably an infrequent event, would be considered to be a member of the general public and therefore would be allowed 5 mSv per year

Answer 60

A

C

The Maximum Permissible Dose (MPD) defines the recommended occupational exposure dose limits and does not include any dose received from medical procedures or natural background radiation. Radiation workers (including residents) are considered subject to occupational exposure limits. Medical students are considered subject to the education and training exposure limits.

In nearly all cases, the MPD is greater than the dose that would be obtained with strict adherence to the principles of ALARA, which stipulate that personnel should receive doses “as low as reasonably achievable”.

The MPD recommendations for radiation workers are typically 10-50 fold higher than for members of the general public.

The NCRP and ICRP guidelines treat age differently in establishing the MPD. The effective dose limit for occupational exposure per the NRCP guidelines is 10 mSv per year of age or 50 mSv per year and per the ICRP guidelines is 20mSv per year (averaged over 5 years) or 50mSv per year.

The MPD for younger workers is therefore greater under NCRP guidelines than under ICRP guidelines, but the MPD for older workers is greater under ICRP guidelines than under the NCRP guidelines.

Answer 61

A

B

The “committed dose equivalent” is the dose equivalent to a tissue or organ that will be received over a 50-year period from the ingestion of radioactive material(s)

Answer 62

A

The maximum dose with fluoroscopy is to the skin. Early transient erythema may occur with doses of 2 Gy, dry desquamation with single doses of 14 Gy, and moist desquamation with single doses of 18 Gy or more.

Answer 63

A

Dose and dose-rate effectiveness factor is used to take into consideration dose and dose-rate when considering exposures for cancer risk.
LNT is the linear non-threshold model.
HSR is the hypoxia sensitizer radio.
LDRR is a usual abbreviation for low dose rate radiation.

Answer 64

A

The most sensitive period during gestation is when radiation exposure may cause embryonic lethality. Based on animal studies, this period of time occurs immediately following conception but prior to implantation within the uterine wall.

Answer 65

A

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.

Preimplantation (0-1.5 weeks) Prenatal Death
Organogenesis (1.5-6 weeks) Congenital Malformations
Early Fetal Period
(6-8 weeks) NIL
8-15 weeks Mental Retardation (High Risk)
(Risk ~0.4 per Gy ~25 IQ points per Gy)
additional risk: Microcephaly

Late Fetal Period (16-25) weeks
Mental Retardation (Lower Risk) Risk ~0.1 per Gy
additional risk: Growth Retardation, Carcinogenesis

Answer 66

A

B
Spina bifida is a neural tube defect typically associated with folate deficiency, not ionizing radiation exposure.

Answer 67

A

The LD50 for oocytes has been shown to be approximately 0.5 Gy

Answer 68

A

C

Low dose radiation of the fetus during the third trimester, not the first trimester, causes an increased risk of childhood cancers, especially leukemias.

The Oxford Survey of Childhood Cancers suggested an association between childhood cancer and in utero exposure to diagnostic x-rays. This was a retrospective case-controlled study of 7,649 children who died of leukemia of childhood cancers. The relative cancer risk estimate assuming radiation to be the causative agent was estimated at 1.52 (Lancet 1:1185, 1970).

Answer 69

A

C

The genetically significant dose (GSD) is the annual average gonadal dose to a population adjusted for the relative child expectancy of that population.

Exposure to radon does not contribute significantly to the GSD because the decay products of radon are deposited almost entirely in the lung.

The GSD resulting from medical procedures performed annually in the United States is estimated to be 0.3 mSv, not 1 Sv.

Although the GSD can be utilized to estimate the number of children born each year with a radiation-induced mutation, the GSD itself is an estimate of the average gonadal dose to the population (including potential parents), not an estimate of the effects on offspring

The GSD is an annual population dose, not an individual lifetime dose

Answer 70

A

D

The dose required to double the incidence of mutations in humans has been estimated to be approximately 1-2 Sv.

Radiation does not induce characteristic mutations; it only increases the incidence of mutations that are known to occur spontaneously

A higher incidence of genetic abnormalities was not found in the children with at least one parent who previously received treatment with ionizing radiation prior to conception.
The best estimates are that no more than 1-6% of spontaneous mutations in humans are due to exposure to background radiation

The absolute mutation rate for humans has been estimated to be approximately 0.1-0.6% per Sv.

Answer 71

A

‘Studies of the Japanese A-bomb survivors by RERF have served as a “gold standard” for radiation epidemiology. One of the key findings is that there has NOT been found to be a statistically significant increase in mutations identified in the F1 generation (approximately 70,000 individuals), despite the original expectation that there might be based on animal experiments. The doubling-dose estimate for radiation-induced genetic mutations in humans is therefore based on mouse data coupled with estimates of human spontaneous mutation rates.

A majority of the survivor cohort received relatively low radiation exposure of less than 100 mSv.

A recently revised dosimetry model (DS02) provides improved estimates of individual exposures received by individuals who survived the Japanese A-bomb.

The Adult Health Study cohort members even today continue to undergo a thorough clinical exam every two years. By providing data and biological samples these participants remain an important resource for future analyses

Answer 72

A

B

The mutation rate decreased significantly when the dose rate was reduced. This was attributed to repair processes that take place during irradiation at low dose rates. Interestingly, this is different than what was observed in the fruit fly study, which demonstrated that dose rate had no effect on the mutagenesis rate.

The dose response curve for radiation-induced mutagenesis was found to be linear WITHOUT a threshold.

Males were found to be MORE susceptible to radiation induced mutation than females.

Mutation rates at the different loci studied DID vary widely

The estimated doubling dose for mutations was approximately 1 Gray

Answer 73

A

D

The dose that will lead to oligospermia and reduced fertility in the male is estimated to be 0.15 Gy.

There is neither a latent period nor temporary sterility in the female following exposure to radiation.

Radiation sterility does not affect hormone balance, libido, or physical capability in the male, but can induce permanent ovarian failure and menopausal symptoms in the female

The dose that will lead to permanent sterility in the female is 12 Gy in the prepubertal woman and 2 Gy in the premenopausal (mature) woman

Answer 74

A

D

Low dose-rate exposure usually results in fewer mutations than the same dose given at a high dose rate.
T to A transitions are usually found following exposure to ultraviolet (UV) light, but not to ionizing radiation

High LET radiation tends to cause large deletions, while low LET radiation tends to cause small deletions

The types of mutations observed following exposure to ionizing radiation can differ from the spectrum of mutations observed following exposure to UV radiation

The relative dose to double the rate of mutagenesis is estimated to be 1 Gy

Answer 75

A

B

Testis is a dose-rate sensitive organ with low dose rates causing more damage than the same dose at a high dose rate.

A dose-rate effect is described as occuring when a decreased response is observed following a decreased dose rate.

Exposure of the fetus in utero causes more damage if the exposure is at a low dose rate than if the same dose is given at a high dose-rate; this is likely due to the fact that more stages of gestation are affected at a low dose-rate.

FLASH technology is based on using an ultra-fast dose-rate to enhance tumor killing and somehow lowering normal tissue toxicity.

Answer 76

A

Cancer survivors constitute 3.5% of the US population, but second primary malignancies among this high-risk group now account for 16% of all cancers diagnosed. A high frequency of second primary tumors among patients diagnosed with soft tissue sarcoma patients has been reported, with a particularly high risk of developing a new soft tissue sarcoma

Radiotherapy to the breast or chest wall of young women is associated with long-term cardiotoxicity and an increased risk of second breast cancers.

Genetic factors, as well as the potential carcinogenic effects of treatment, can affect the probability of second cancers in survivors.

Patients with the BRCA2 mutation demonstrate an increased risk of subsequent ovarian cancer, as well as cancers in the irradiated and unirradiated breast.

Patients with Li-Fraumeni syndrome and other familial cancer syndromes would likewise be at increased risk of developing second malignancies unrelated to the carcinogenic effects of their initial treatments.

Children who receive cranial irradiation as part of their treatment for leukemia are at a significant increased risk for developing meningiomas.

Answer 77

A

D

Both benign nodules and malignant tumors of the thyroid can be induced by radiation. None of the other sites shows such an increased incidence of benign tumors following treatment with ionizing radiation.

Answer 78

A

C

Approximately 15% of the fatal cancers diagnosed among patients previously treated with total body irradiation are leukemias.

Answer 79

A

C

Among the population of children who were treated for tinea capitis (ringworm) using ionizing radiation, an excess incidence was not detected for head and neck cancers.

Brain cancers, thyroid cancers, adenomas, (non-CLL) leukemias, and late development of breast cancer were all observed at an excess incidence among children that were previously treated with X-Rays compared with children who only received topical medications.

Answer 80

A

E

The susceptibility to radiation-induced cancer decreases with increasing age at the time of irradiation.

Radium dial painters ingested significant quantities of radium-containing paint by repeatedly licking the paint brushes they used. These women subsequently developed an excess number of osteosarcomas due to the incorporation of radium into their growing bones and the continuous low dose-rate irradiation received by these tissues over the next decades.

At this time, cancers induced by radiation cannot be distinguished from cancers that occur naturally, although molecular markers for radiation exposure may eventually be identified

The current consensus among radiation protection organizations is that the most appropriate dose response curve for radiation carcinogenesis is one that increases linearly with increasing radiation dose and without a dose threshold (linear no-threshold or LNT model). This hypothesis, however, has been challenged by those who believe that exposure to low radiation doses may be less harmful than what is predicted by the LNT model, and possibly even beneficial (often referred to as hormesis).

The LNT model has also been criticized by those who believe that bystander effects may result in an increased risk at low doses over those predicted by the LNT model

Hematological malignancies have shorter latency periods compared to solid tumors

Answer 81

A

D

Statistically significant increases in non-cancer disease mortality with increasing radiation dose have been observed, particularly for diseases of the circulatory, digestive, and respiratory systems. Survivors who received less than 5 Gy demonstrate an increased risk of heart disease.

Among the Japanese A-bomb survivors, susceptibility to radiation induced breast cancer was found to dramatically decrease with increasing age at time of exposure, with women over 50 years of age showing little or no excess.

The latency period for the appearance of most radiation-induced solid tumors is far greater than 1-3 years, ranging from 10-60 years post exposure.

It is estimated that 8% of people exposed to 1 Sv would die from a radiation-induced cancer. Thus, in a population of 1,000 people, approximately 80 would develop and die from a fatal cancer

Answer 82

A

C

The risk estimates based on the Radiation Effects Research Foundation (RERF) analyses for solid tumors are well-fit using a linear model; a linear-quadratic model provides a much better fit to the leukemia dose response data. Although the BEIR VII Committee conducted an analysis of the data related to the DDREF, and used a value of 1.5 for its own risk estimations, the factor of 2.0 has historically been applied to adjust for lower doses and dose-rates.

RERF data clearly indicate that radiation risk is dependent on gender, as well as age at exposure and time since exposure

Population studies frequently have more limitations compared to more quantitative case-control studies, including smaller population sizes and uncertainties associated with dose estimations, confounding factors, and lack of relevant control populations

The BEIR VII estimates the lifetime additional cancer risk is about 1% following 100 mSv

Answer 83

A

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.

Answer 84

A

C

Administration of potassium iodide (KI) to the children that had been exposed to 131 Iodide (131I) as a result of the accident would have reduced the number of thyroid cancers by decreasing exposure to the radioactive iodide. While 137 Cesium (137Cs) was released as part of the accident, 131I was the cause of the thyroid cancers that occurred.

Most of the tumors demonstrated rearrangements of the RET and PTC genes

The majority of the tumors occurred predominantly in children within 7- 10 years following exposure.

Answer 85

A

D

Data from the atomic bomb survivors demonstrate that breast cancer, thyroid cancer, bladder cancer, and non-CLL leukemia were all significantly induced following exposure to radiation

Cancer of the cervix is tightly linked to HPV viral infection and is not considered to be a highly radiogenic tumor

Answer 86

A

Medical exposure contributes more than 50% of the average annual effective radiation dose, slightly higher than that contributed by environmental exposures, and has significantly increased since the 1980s. The average annual effective dose related to population and occupational exposure from nuclear reactors remains minimal, at 0.0005 mSv to 0.005mSv

Following medical exposures, natural background radiation is the next largest contributor to the average annual effective dose, contributing approximately 37% of all radiation exposure

The US population has a higher average annual effective dose from medical procedures compared to that of other developed countries. CT scans represent the greatest contributer followed by nuclear medicine procedures.

Answer 87

A

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.

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ASTRO 2023 (1,2,3,26-32) Flashcards

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