ASTRO 2023 (1,2,3,26-32) Flashcards
(101 cards)
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
- (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
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
-(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
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
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.
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
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.
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
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.
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
-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).
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
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.
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
Annihilation photons always have an energy of 0.511 MeV each, which is equal to the rest energy of the positron and electron.
Directly ionizing radiation includes all of the following EXCEPT:
A. Electrons
B. Positrons
C. Alpha particles
D. Neutrons
E. Betas
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.
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.
-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.
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
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.
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
The indirect effect mediated by free-radical reactions involving water are
most responsible to cause DNA
damage upon low LET irradiation.
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
-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.
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
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.
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
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.
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
-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).
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.
-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.
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
A
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)
-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).
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
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
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
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
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
-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).
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
-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).
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)
-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).