Author Questions

Question 1

MHC class 2 receptors present antigens to be recognized by the following immune cell type:

A) Myeloid-derived suppressor cells
B) Natural Killer cells
C) CD8+ “Cytotoxic” T cells
D) CD4+ “Helper” T cells
E) B cells

Answer 1

D) CD4+ “Helper” T cells

Question 2

Anti-PD1 and Anti-CTLA4 target the following type of tumor reactive immune cells:

A) Myeloid-derived suppressor cells
B) CD8+ “Cytotoxic” T cells
C) CD4+ “Helper” T cells
D) B cells
E) B and C

Answer 2

E) B and C
Both CD8+ “Cytotoxic” T cells and CD4+ “Helper” T cells

Question 3

A radiation oncology resident is designing a clinical trial combining radiotherapy and immune checkpoint inhibitor for melanoma. Which of the following approaches have shown promise in clinical trials:

A) SBRT, 8Gy x 3, immediately followed by anti- PD1 or anti-PDL1
B) Conventionally fractionated radiation with concurrent anti-PD1 or anti-PDL1
C) Conventionally fractionated radiation, immediately followed by anti-PD1
D) All of the above
E) A and C

Answer 3

E) A and C
Conventionally fractionated radiation with concurrent anti-PD1 or anti-PDL1 does not work.

Question 4

A scientist has developed a radioprotector compound that acts by scavenging free radicals in the cell. For which type of radiation would the relative radioprotective effect likely be the greatest?

A – Photons
B – Neutrons
C – Equal protection for photons and neutrons
D – Neither would be radioprotected by this compound

Answer 4

A - Photons
Photons are low LET, and their greatest (~70%) effects are indirect (via formation of free radicals). Thus, free radical scavengers would be more effective on those.

Radiation interaction with matter

Question 5

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 - e (aq)
B - 1O2
C - OH -
D. OH*
E - O2-

Answer 5

D 65-75% of the damage caused by indirect action is mediated by the hydroxyl radical, OH*. Little biological damage is caused by the hydrated electron (eaq; Answer Choice A). 1O2 is produced primarily by photosensitizers and, rarely, by ionizing radiation (Answer Choice B). Neither OH- nor O2- are primary radiolysis products, although O2- can be produced secondarily by reaction of eaq with O2 (Answer Choices C and E). Mitchell JB, et al. Radiation, Radicals, and Images. Ann N Y Acad Sci. 899:28-43, 2000. Pubmed

Huber, K, Woloshak, G, Rosenstein B. 2023 Astro Radiation and Cancer Biology Study Guide. 2023. American Society for Radiation Oncology. Available from: https://www.astro.org/ASTRO/media/ASTRO/AffiliatePages/arro/PDFs/RadBio_StudyGuide_23.pdf

Radiation interaction with matter

Question 6

Which of the following ionization processes represents the principal interaction with tissue for X-rays used in radiotherapy?

A - Pair production
B - Photoelectric effect
C - Compton process
D - Photodisintegration
E - Coherent scattering

Answer 6

C

For photons in the energy range used typically in radiotherapy, the Compton process is predominant. In the Compton process, a high-energy photon interacts with an atom to cause ejection of an outer shell electron (referred to as a recoil electron) and a scattered photon. The energy of the incident photon is distributed between the scattered photon and the kinetic energy of the recoil electron. The Compton interaction may occur when photon energies range from 150 keV to 3 MeV although it also occurs to some extent at lower energies of 100-150 keV. Pair production occurs when a photon of greater than 1.02 MeV interacts with a nucleus to form an electron-positron pair. This amount of energy is just sufficient to provide the rest mass of the electron and positron, 0.51 MeV each. Excess of energy above 1.02 MeV will be possessed by these two particles, which produce ionizations as they travel in the material. As the positron comes to rest, it interacts with an electron in an annihilation reaction and is replaced by two photons, each having an energy of 0.51 MeV and moving in opposite directions. Pair production becomes an important form of interaction above about 10 MeV. The photoelectric effect is predominant for photons that have energies less than approximately 100-150 keV, typical of X-rays used in diagnostic radiology. In the photoelectric process, a photon interacts with an inner orbital electron and is completely absorbed. The electron is ejected from the atom becoming a free photoelectron. The kinetic energy of the ejected electron is equal to the energy of the incident photon minus the binding energy of the electron that has been ejected. The vacancy left in the shell by the ejected electron is filled in by the transition of an electron from an outer shell and is accompanied by the emission of a characteristic X-ray, whose energy represents the difference in the energy levels of the shells involved in the electron transition. When the excess energy derived from the transition of the electron from the higher to the lower energy state is transferred to an orbital electron that is ejected, this is referred to as an Auger electron. Photodisintegration occurs at photon energies much higher than those used in either diagnostic radiology or radiation therapy. In this process, a high-energy photon interacts with the nucleus of an atom resulting in the emission of one or more nucleons. An electron is not ejected through coherent scattering and not ejected through coherent scattering and no energy is transferred in this type of interaction, only the direction of the incident photon is altered.

(ASTRO 2020 Study Guide: Radiation-Matter Interactions)

Woloshak, G, Huber, K, Barker, C, Rosenstein B, et al. 2020 Astro Radiation and Cancer Biology Study Guide. 2020. American Society for Radiation Oncology. Available from: https://www.astro.org/ASTRO/media/ASTRO/AffiliatePages/arro/PDFs/Radiobiology_StudyGuide21.pdf

Radiation interaction with matter

Question 7

The largest contributor to radiation exposure of the US population each year is:

A - Radon
B - Cosmic radiation
C - Computed Tomography
D - Industrial activity
E - Consumer products

Answer 7

A
Per the National Council on Radiation Protection and Measurements (NCRP), Report No. 160, the average annual radiation dose per person in the United States is approximately 6.2 millisieverts (mSv) or 620 millirem (mrem). The majority (37%) of this dose can be attributed to background radiation sources including Radon and Thoron. These gases are created when other naturally occurring elements undergo radioactive decay.
Cosmic radiation contributes 5% of the average annual dose (Answer Choice B). An additional 48% of the average dose to an individual in the United States is from medical procedures (not including dose received during therapeutic radiation). Of these, Computed Tomography (CT) scans comprise approximately 24% of radiation dose (Answer Choice C). Industrial activity contributes only a very small amount of the average annual dose to the average American (<0.1% or 0.003mSv).

Consumer products contributes approximately 2% of the average annual dose (Answer Choice E) Radiation Sources and Doses. United States Environmental Protection Agency. https://www.epa.gov/radiation/radiation-sources-and-doses

Huber, K, Woloshak, G, Rosenstein B, et al. 2023 Astro Radiation and Cancer Biology Study Guide. 2023. American Society for Radiation Oncology. Available from: https://www.astro.org/ASTRO/media/ASTRO/AffiliatePages/arro/PDFs/RadBio_StudyGuide_23.pdf

Radiation interaction with matter

Question 8

Question 4: 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 8

D

The main initial products of resulting from irradiation of pure water are the short-lived free radicals, hydrogen radical (H) (10%), hydroxyl radical (OH) (45%), and solvated electrons (e-aq ) (45%).

These react with DNA or with each other. Therefore, OH + H –> H2O

The remaining recombination reactions of free radicals are:
e-aq + e-aq +2 H2O –>H2 + 2 OH-
OH + OH –>H2O2
H + H –>H2

These reactions always compete with reactions that lead to direct damage of the biological molecules.

Modified from question from the ASTRO 2023 study guide
Original question from: Huber, K, Woloshak, G, Rosenstein B, et al. 2023 Astro Radiation and Cancer Biology Study Guide. 2023. American Society for Radiation Oncology. Available from: https://www.astro.org/ASTRO/media/ASTRO/AffiliatePages/arro/PDFs/RadBio_StudyGuide_23.pdf

Radiation interaction with matter

Question 9

5. Which of the following particles is most efficient at directly ionizing the deoxyribonucleic acid backbone?

A) Carbon ion
B) Proton
C) Alpha particle
D) Co60 gamma ray

Answer 9

C

High LET radiation types have an increased propensity for direct ionization of biomolecules, including the induction of clustered lesions. Low LET radiation types are more likely to impact biomolecules (including DNA) through indirect ionization via hydroxyl radicals or other radiolysis species.

Approximate LET values according to Hall (keV/uM)
- 14 MeV Carbon ion: 12
- 150 MeV Proton: 0.5
- 2.5 MeV alpha particle: 166
- Co60 gamma ray: 0.2

Radiation interaction with matter

Question 10

6. 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 10

B
The initial ionization process takes approximately 10-15 second (this can be either ionization of water (radiolysis) or direct ionization of a biomolecule like DNA. If radiolysis occurs, the primary radicals (H20+, e-aq) produced by the ejection of an electron from water typically have a lifetime of 10-10 second. Subsequent interaction of these primary radicals with water results in H3O+ and OH* , which has a lifetime of approximately 10-9 second. The DNA radicals formed through direct or indirect ionization of DNA (base radicals or sugar-phosphate ionization) have a lifetime of approximately 10-5 second.

Timescales by process
Biology (e.g., DNA damage) – microseconds (usec) to decades
Chemistry (e.g., radiolysis species and derivatives) - picoseconds to nanoseconds
Physics (e.g., initial ionization event) – femtoseconds

Modified from ASTRO study guide question
Original question from:
Huber, K, Woloshak, G, Rosenstein B, et al. 2023 Astro Radiation and Cancer Biology Study Guide. 2023. American Society for Radiation Oncology. Available from: https://www.astro.org/ASTRO/media/ASTRO/AffiliatePages/arro/PDFs/RadBio_StudyGuide_23.pdf

Radiation interaction with matter

Question 11

Radiobiology refers to which of the following:

A) The study of different physical sources of radiation
B) The study of how electromagnetic radiation interacts with biological
systems
C) The use of radiological imaging to define the anatomy of living
organisms
D) None of the above

Answer 11

B) The study of how electromagnetic radiation interacts with biological

Question 12

Which of the following statements is true regarding clinical responses to radiation

A) Effects are fully realized within a matter of weeks after radiation is
administered
B) The tumor’s genetic profile is the sole determinant of radiation
sensitivity
C) Both tumor cell and normal tissue responses contribute to the clinical
response to radiation, which can take years to fully manifest
D) None of the above

Answer 12

C) Both tumor cell and normal tissue responses contribute to the clinical

Question 13

Clinical scenarios when you might use your radiation and cancer
biology knowledge

A) Estimating normal tissue dose tolerances during a course of re-irradiation
B) Making clinical decisions about dose/fractionation
C) Determining whether radiation plans need to be modified when a patient’s germline genetic testing identifies an unexpected pathogenic mutation in TP53
D) Deciding how to manage timing of a recently approved targeted cancer therapy drug with a course of palliative radiotherapy
E) All of the above

Answer 13

E) All of the above

Question 14

Which of the following proteins inhibits the intrinsic pathway of apoptosis?

1) BAX
2) P53
3) PUMA
4) BCL-2

Answer 14

4)BCL-2

Question 15

What is the primary mechanism by which radiation therapy kills follicular lymphoma cells?

1)Senescence
2)Pre-mitotic apoptosis
3)Ferroptosis
4)Necroptosis

Answer 15

2) Pre-mitotic apoptosis

Question 16

What is the difference between the survival curves of high LET and low LET irradiations?

Answer 16

High LET (neutrons): more direct, closer together in hypoxic and oxygenated conditions.
Low LET (x-rays): more indirect, farther apart in hypoxic and oxygenated conditions.
The oxygen effect is absent for alpha particles.

Hypoxia and Tumor Angiogenesis

Question 17

A tumor’s hypoxic fraction is measured/estimated to be 15%. After irradiating with 2 Gy × 5 treatments, the hypoxic fraction estimated to be ~15%. Which of the 4 “R”s of radiobiology accounts for this observation?

A) Repopulation
B) Reoxygenation
C) Repair is measured and again
D) Redistribution

Answer 17

B) Reoxygenation

Hypoxia and Tumor Angiogenesis

Question 18

A scientist evaluates a series of tumors in the lab and determines that the more radioresistant tumors tended to have an increased density of microvessels when evaluated by immunofluorescence. What explains this discrepancy?

A) These vessels facilitated increased diffusion of oxygen to surrounding tumor
B) These vessels were aberrant and often nonfunctional due to occlusion or decreased transit
C) These vessels were products of vasculogenesis rather than angiogenesis
D) The tumors with increased microvascular density tended to be more

Answer 18

Vascular density may not reflect effective perfusion or oxygen transfer. Often tumors have increased vascular structures (see right) that are not functional or excessively leaky, leading to increased interstitial pressure that may subsequently occlude nearby vessels.

Hypoxia and Tumor Angiogenesis

Question 19

What happens if Cdc25C is phsphorylated?

Answer 19

Higher Cdc2 phorphorylation

Cdc25C dephosphorylates Cdc2, so if it is phosphorylater, its activity is decreased and Cdc2 remains phosphorylated.

Increased ATM kinase activity ->
Increased Chk2 kinase activity ->
Phospho Cdc25c ->
Decreased Cdc25c phosphatase activity ->
Increased Cdc2 phosphorylation (less G2/M progression)

Cell cycle

Question 20

Why adding ribociclib improves overall survival in patients with breast cancer?

Answer 20

Impaired G1/S cell cycle progression
ribociclib/palbociclib inhibit CDK 4/6

Cell cycle

Question 21

Time in G1 is highly variable amongst normal cells, time in G1 is fairly similar (and short) within cancer cells. What may genetic explanation may account for this observation?

A - ~40-50% of cancers have loss of TP53 signaling
B - Tumor suppressor loss (e.g., p16) and oncogene signaling (Myc) converge on G1/S regulation
C - Cdc25a/c is often overexpressed in cancers
D - Some cancers lose Rb tumor suppressor activity through direct mutation or other means of inactivation
E - Any of A, B, and D
F- All of the above could contribute

Answer 21

F - All can contribute

Cell cycle

Question 22

Homologous recombinational repair of DNA double strand breaks is most likely to occur:

a. In G0
b. In G1
c. In early S phase
d. In late S phase
e. Throughout the cell cycle

Answer 22

D Homologous recombinational repair requires the presence of a homologous DNA template, and is therefore most likely to occur following DNA replication in late S phase (when a sister chromatid is available as a template) or G2 phases of the cell cycle.

Woloshak, G, Huber, K, Barker, C, Rosenstein B, et al. 2020 Astro Radiation and Cancer Biology Study Guide. 2020. American Society for Radiation Oncology. Available from: https://www.astro.org/ASTRO/media/ASTRO/AffiliatePages/arro/PDFs/Radiobiology_StudyGuide21.pdf

Cell cycle

Question 23

Which of the following flow cytometry methods or the combination of methods are used to estimate cell cycle distribution in mammalian cells?

a. Analysis of annexin V stained cells
b. Analysis of cells treated with a high dose (2 mM or more) thymidine
c. Analysis of propidium iodide stained cells
d. Analysis of cells labeled with an phospho-serine-H3 antibody
e. Analysis of cells pulse-labeled with 3H-thymidine

Answer 23

C
The cell cycle can be analysed by flow cytometry using a fluorescence-activated cell sorter (FACS). Cells are treated with a DNA-binding dye, such as propidium iodide. The amount of cellular DNA is proportional to the amount of fluorescence detected by dye binding. For a proliferating cell population, a plot of cell number versus DNA content yields the percentage of cells in G1 (with 1 arbitrary unit of DNA or 2N), in G2 + M (with 2 arbitrary unit of DNA or 4N), and in S (cells with more than 1 but less 2 units of DNA. Other assays: Apoptosis assays - Annexin V conjugates provide a method for studying the externalization of phosphatidylserine, an indicator of intermediate stages of apoptosis, by flow cytometry. Nuclear fragmentation, mitochondrial membrane potential flux, and caspase-3 activation precede phosphatidylserine “in-out flipping” during apoptosis, whereas permeability to propidium iodide occur later. There are caspase activity assays that can measure the ability of active caspases (3/7, 8, 9, etc) to cleave fluorescence or luminescent caspase substrates Direct assessment of nuclear fragmentation is a classical way to assess for Proliferation assays – Pulse labeling with small amounts of radiolabeled (3H) thymidine or bromodeoxyuridine, BrdU will result in incorporation of these labeled nucleosides into replicating DNA. The amount of incorporation in a cell or cell population reflects how much DNA synthesis occurs during the exposure to the thymidine/deoxyuracil. This pulse labeling is DIFFERENT than using a “thymidine block” which is where cells are flooded with high doses of thymidine (usually NOT radiolabeled with 3H), which can result in cell cycle arrest in G1S due to deoxyribonucleotide feedback inhibition. This will lead to a build-up of cells at the G1/S checkpoint. Staining of cells for markers of S phase, like phospho-histone H3 or Ki67 reveals the proportion of a cell population that is in S phase at that time, giving an indication of the likely proportion of cells in S phase at the moment of staining.

Cell cycle

Question 24

Which of the following observations is most closely associated with G1/S arrest in wild type-p53 cells following X-irradiation?

a. Increase in bromouridine incorporation
b. Increase in the labeling index
c. Formation of the cyclin E/A-Cdk2 complex
d. Increased transcription of the p21Waf1 (CDKN1A) gene
e. Increased transcription of the p53 (TP53) gene

Answer 24

D
p21inhibits cyclin A-Cdk2 and cyclin E-Cdk2 activity, which in turn prevents G1 cells from entering into S phase. The cyclin E/A-Cdk2 complex is required for entry of G1 cells into S- phase. The induction of p21 in X-irradiated cells is dependent on functional p53. p53 is a transcription factor which is activated in response to a wide variety of genotoxic stresses, frequently via post-translational modification. In response to ionizing radiation-induced DNA damage the existing p53 protein is modified by phosphorylation at multiple sites. The modified p53 becomes more stable (that is, its half-time is significantly increased), which results in increased amounts of the p53 protein, and confers its activity as a transcription factor. The labeling index is the fraction of cells in S-phase, relative to the total number of cells in a proliferating cell population. Choices A and B imply the paradoxical increase in the number of the S-cell population following G1/S arrest.

Cell cycle

Question 25

Ionizing radiation induced chromatid-type aberrations are a consequence of failed or mis-repaired DNA double strand breaks produced during which phase of the cell cycle?

a. G1 and G2 phase
b. G1 and S phase
c. S and G2 phase
d. Only in M phase
e. G0 phase

Answer 25

C
Chromatid type of chromosome aberrations are caused by chromatid breaks in S or G2 phase; these result in non-symmetric impacts on the two post-S chromatids. The breaks induced in G1 phase often display as chromosome type of aberrations. These impact PRE-replication chromosomes, and thus these abnormalities should be symmetrically represented in the post-S chromosome.

Cell cycle

Question 26

Radiation-induced cellular senescence is often the result of:

A. Cellular nutrient deprivation
B. Oxidative stress secondary to mitochondrial dysfunction
C. p16-mediated cell cycle arrest
D. Telomere shortening
E. Mitotic catastrophe

Answer 26

C The term “senescence” refers to the loss of cellular replicative potential leading to a reduced capability to repopulate a tissue after exposure to genotoxic agents, including ionizing radiation. Senescence is most often the result of a permanent arrest in G1, classically associated with elevated expression of the cell cycle inhibitor p16INK4A (CDKN2A), as well as p21 (CDKN1A, WAF1/CIP1). Other features of senescence include flattened morphology of cells with dominant cytoplasm (in culture) and expression of beta-galactosidase enzyme. Importantly, senescence is not a type of cell death per se because cells remain morphologically intact and metabolically active when senescent. In cancer cells, depending on the level of tumor suppressor proteins and the oncogenic signal, senescence can be reversible in a small subset of cells though classically, and in most cells, this process is irreversible. A clinically relevant scenario for radiation-induced senescence is the loss of salivary gland function and xerostomia commonly seen in head and neck cancer patients undergoing radiotherapy. Another one is radiation induced premature senescence in fibroblasts that triggers proinflammatory and profibrotic senescence associated secretory phenotype (SASP) and ultimately drives fibrosis in the lung. In both of these scenarios, radiation induces senecence (not death) of the target cells; in the case of the lung fibrosis, the continued metabolism of the senescent cells influences the microenvironment and contributes to the pathophysiology of lung fibrosis.

Other options:
Mitochondrial dysfunction is a hallmark of apoptotic cell death, not senescence (Answer Choice B).
Telomere shortening occurs in most normal somatic cells as part of each cell cycle (“end replication problem”) and triggers senescence once a critical low threshold is reached, but telomere shortening tends not to be the cause for radiation-induced senescence which is driven by DNAdamage and cell cycle arrest (Answer Choice D).
Nutrient deprivation can lead to autophagy, and ultimately autophagic death cell distinct from apoptosis (Answer Choice A).
Mitotic catastrophe is a major mechanism by which radiation-induced cell death is precipitated, not senescence (answer choice E)

Ohtani N, Mann DJ, Hara E. Cellular senescence: its role in tumor suppression and aging. Cancer Sci. 100:792-7, 2009. Pubmed Munoz-Espin D and Serrano M. Cellular senescence: from physiology to pathology. Nat Rev Mol Cell Biol. 15(7):482-96, 2014. Pubmed Kuilman T, et al. The essence of senescence. Genes Dev. 24:2463–2479, 2010. Pubmed 48 Nguyena HQ, et al. Ionizing radiation-induced cellular senescence promotes tissue fibrosis after radiotherapy. A review. Crit Rev Oncol Hematol. 129:13-26, 2018. Pubmed

Cell cycle

Question 27

An unsynchronized culture of lymphoid-lineage cells is irradiated with a dose of radiation that kills ~90% of cells via immediate apoptosis. The surviving cells are immediately evaluated by cell cycle analysis, showing an enrichment of cells in which cell cycle phase?

A. G0
B. G1
C. S
D. G2
E. M

Answer 27

C
A dose that immediately kills 90% of the cells in the population would leave a surviving cell population heavily enriched in the radioresistant cells in late S phase. Radiosensitivity across the cell cycle is ranked as follows from least to most sensitive: Late S, Early S, G1, Early G2, late G2, M. G0 is most likely to be similar to G1 as it’s typically pre-S phase. Moreover, these cells are intrinsically blocked from progressing through the cell cycle so there may be more time for DNA damage repair prior to division and propagation of DNA damage.

Sinclair R. Steady-state suspension culture and metabolism of strain L mouse cells in simple defined medium. Radiat Res. 41(1):20-33. 1966. Pubmed

Cell cycle

Question 28

A set of data defining the survival of cells irradiated with graded doses of X-rays is well-fitted by the mathematical expression for a single-hit survival curve having an SF2 of 0.37. The best estimate for the “α” parameter that describes this survival response is:

A) 0.1 Gy’
B) 0.01 Gy’
C) 0.05 Gy-‘
D) 0.5 Gy 1
E) 2.0 Gy1

Answer 28

0.5
Single hit model: SF = e^-αD
0.37 = e^-1
aD=1
a=0.5

Survival curves

Question 29

According to classical target theory, DO is a measure of the:

a. Amount of sublethal damage a cell can accumulate before lethality occurs
b. total number of targets that must be inactivated to kill a cell
c. dose required to produce an average of one lethal lesion per irradiated cell
d. width of the shoulder region of the cell survival curve
e. total number of hits required per target to kill a cell

Answer 29

c. dose required to produce an average of one lethal lesion per irradiated cell

Survival curves

Question 30

The DO for most mammalian cells irradiated with X-rays in vitro under well-aerated conditions falls in the range of:

a. 0.1 - 0.2 Gy
b. 0.2 - 1 Gy
c. 1 - 2 Gy
d. 2 - 4 Gy
e. 4 - 8 Gy

Answer 30

C. You mainly have to memorize this one. 37% survival occurs with relatively low doses, most often in the range of 1-2Gy

Survival curves

Question 31

For a particular cell line characterized by a DO of 1 Gy and n equal to 1, what would be the approximate percentage of cells survived by a dose of 3 Gy?

a. 5
b. 10
c. 37
d. 50
e. 95

Answer 31

E.
First notice that n = 1 means it is a single hit model, with only a linear component.
DO = 1Gy, and for each additional 1 Gy there will be an additional 0.37 reduction in SF.
SF36y = (0.37)(0.37)(0.37) = 0.05 = 5%
Also, you can use the equation SF = e-AD. When DO = 1Gy, a=1, SO SF36y = e-1+3 = 05
100%-5%=95%

Survival curves

Question 32

Following an X-ray dose of 8 Gy, a clonogenic assay revealed that 20 colonies arose from an initial cell population of 2,000 cells. When 200 unirradiated cells were assayed for clonogenic survival, 40 colonies grew.
What is the percent survival following the 8 Gy dose?

Answer 32

SF = (20/2000)/(40/200)=0.1/0.2=0.05=5%

Survival curves

Question 33

What would be the estimated surviving fraction of V79 Chinese hamster cells irradiated with an X-ray dose of 5 Gy delivered acutely? (Assume a=0.2 Gy-‘ and ß=0.05 Gy-2)

Answer 33

B. SF = e^-(αD+ßD^2)= e-(2.25) = 0.1

Survival curves

Question 34

What is the approximate surviving fraction following 5 doses of 0.5 Gy of carbon ions, assuming that the surviving fraction following one dose is 0.4?

Answer 34

A. High LET radiation is thought to be predominantly driven by the single hit
model. So the cumulative effect will be: SF = 0.4^5 = .01

Survival curves

Question 35

Tumor-bearing mice are randomized into a control group and groups treated with localized irradiation of the tumor alone, an anticancer drug alone, or radiation in combination with the drug. Which of the following represents the most rigorous, reliable and informative approach to comparing the effectiveness of the different treatments?

a. Killing the mice at a predetermined time after treatment, removing and weighing the tumors, and calculating the ratio of the volumes of the treated and control tumors
b. Measuring three diameters of the tumors with calipers at a predetermined time after treatment, calculating the volume and computing the ratio of the volumes of the treated and control tumors
c. Measuring the tumors 3x per week until the treated tumors return to their pre-irradiation volume and calculating the mean time needed for each group to reach that volume
d. Measuring the tumors 3x per week until the control tumors reach 4 times the volume at the time of treatment, and comparing the mean volume of the tumors in each treatment group at that time
e.
Measuring the tumors 3x per week until each tumor reaches 4 times the volume at the time of treatment and calculating the mean time needed for the tumors in each group to reach that volume

Answer 35

c (&e)
Time to tumor regrowth is the best measure of RT effectiveness.

Survival curves

Question 36

For a group of tumors identical in size and homogeneous with respect to cellular radiosensitivity, what would be the general shape of the curve in a linear-linear graph defining the increase in tumor control probability with increasing radiation dose?

a. Step function from 0 - 100% at the dose that kills all of the cells
b. Linear increase from 0 - 100% over a narrow range of doses
c. Logarithmic increase from 0 - 100% over a wide range of doses
d. Sigmoidal increase from 0 - 100% over a narrow range of doses
e. Exponential increase from 0 - 100% over a narrow range of doses

Answer 36

d. Sigmoidal increase from 0 - 100% over a narrow range of doses

Survival curves

Question 37

Genetically engineered mouse models (GEMMs) are becoming more popular for preclinical tumor studies with and without radiation. Which of the following is the most correct regarding the advantages of GEMMs over xenograft-based studies?

a. Intact immune system
b. Tumors are orthotopic in location
c. Tumors contain mutations that are relevant for human tumors
d. Intact extracellular tumor microenvironment
e. Tumors develop more rapidly

Answer 37

a. Intact immune system

Survival curves

Question 38

For a population of patients with identical tumors, assuming that: 1) each tumor has 2x10^9 cells, 2) after treatment with 25 fractions of 2 Gy, the effective survival is 10^-9; 3) there is no regrowth of the tumor during the treatment, 4) the tumor may recur even if there is only one viable tumor cell survive. What is a patient’s probability to be recurrence-free from the original tumor?

a. 50.0% (or 1/2)
b. 13.5%, (or e-2)
c. 0.01 (or 10-2)
d. 0%, because the each of the patients will have 2 surviving cells e.
None of the above

Answer 38

b - e -x

Survival curves

Question 39

Which type of the following genetic alterations is not a feature of radiation-associated solid tumors compared with radiation naïve tumors

1) Increased single-nucleotide variants
2) Increased structural variants
3) Increased small deletions
4) Increased complex genomic rearrangements

Answer 39

1) Increased single-nucleotide variants

Question 40

43F with ER+/HER2- T1CNOMO IDC of right breast with FH notable for deceased daughter (2yr) from brain tumor of unknown molecular/pathologic type. Germline testing reveals NBS1 heterozygous inactivating mutation.
This patient desires breast conserving therapy but has
is concerned at her isolate and late toxicity incide second malignancis.
How do you advise her to proceed?

A - She has a high risk of both acute and late toxicity. She should proceed with mastectomy.
Second malignancy. She should be advised to prodiated with maste coding risk for toxicity and
C - She has a theoretical but unproven risk of radiation sensitivity including risk for toxicity and second malignancy. Atter counseling, she should be offered breast conserving therapy as an option (including RT) with a standard, non hypotractionated dose/fractionation
D - She has a theoretical but unproven risk of radiation sensitivity including risk for toxicity and setion inaliding A sing anselinatio ate hele breast east ach similar to sTART-s
E - She has a theoretical but unproven risk of radiation sensitivity including risk for toxicity and second malignancy. After counseling, she should be offered breast conserving therapy as an option (including RT), with consideration for plan/dose adjustment partial breast treatment options or immobilization that minimizes non-target irradiation (e.g., prone treatment)

Answer 40

C, D, and E.
She is heterozygous, so she is probably OK.
C - She has a theoretical but unproven risk of radiation sensitivity including risk for toxicity and second malignancy. After counseling, she should be offered breast conserving therapy as an option (including RT) with a standard, non hypotractionated dose/fractionation
D - She has a theoretical but unproven risk of radiation sensitivity including risk for toxicity and second malignancy. After counseling, she should be offered breast conserving therapy as an option (including RT), using a hypofractionated whole breast approach similar to START-B.
E - She has a theoretical but unproven risk of radiation sensitivity including risk for toxicity and second malignancy. After counseling, she should be offered breast conserving therapy as an option (including RT), with consideration for plan/dose adjustment partial breast treatment options or immobilization that minimizes non-target irradiation (e.g., prone treatment)

DNA repair

Question 41

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
Е - ХРС

Answer 41

E - XPC is a gene whose product is involved in nucleotide excision repair(NER). Mutations in XPC result in the human genetic disease xeroderma pigmentosum, which is characterized by extreme sensitivity to ultraviolet light. Mutations in all of the other genes result in human genetic diseases characterized by sensitivity to ionizing radiation, including Nijmegen breakage syndrome (NBS), familial breast cancer (BRCAI), ataxia telangiectasia (ATM), and ataxia telangiectasia-like disorder (MRE11).

DNA repair

Question 42

Which off 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 42

• B - A person with LIG4 syndrome is radiation sensitive

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-PKs), 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 see several genes are now known to cause SCID phenotes, the mutatoin in the common human disease of the same name (severe combined immunodeficiency) differs from that in the well-known mouse strain.

DNA repair

Question 43

Increased numbers of chromosome aberrations, especially quadriradials are frequently found in the absence of radiation in which of the following human syndromes?

• A - Xeroderma pigmentosum
• B - Fanconi anemia
• C - Cockayne’s syndrome
• D - Niemann-Pick disease
• E - Li- Fraumeni syndrome

Answer 43

B
Blood cells from individuals with Fanconi anemia are often found to have
high numbers of chromosome aberrations, especially quadriradials. These complex aberrations increase dramatically with exposure to DNA cross-linking agents such as mitomycin c.

Xeroderma and Cockayne are NER pathways and not associated with chromosomal abnormalities (or overt RT sensitivity)
Li Fraumeni is an autosomal DOMINANT syndrome involving
TP53 mutations. It does not result in chromosome breakage
Niemann-Pick disease is a lysosomal storage disease

DNA repair

Question 44

A patient with Cowden syndrome develops colon cancer. The colon cancer is sequenced and has an inactivating mutation in only ONE allele of PTEN, amongst other tumor-specific mutations. Through what mechanism did the inherited PTEN mutation likely contribute to tumor development?

A)Oncogene activation
B)Haploinsufficiency
C)Epigenetic dysregulation
D)Loss of heterozygosity

Answer 44

B
Haploinsufficiency is where the function of a SINGLE gene copy can be insufficient for the function of a gene (here a tumor suppressor gene). In some fraction of cells with INADEQUATE function of a tumor suppressor, a tumor may eventually develop

Cancer Hallmarks: Oncogenes, Tumor Suppressors, Epigenetics

Question 45

Which of the following genetic abnormalities in colorectal cancer results in oncogene addiction?

A)BRAF V600E
B)TP53 R175H
C)PTEN copy number loss
D)APC I1307K

Answer 45

A
Here if you identify the only oncogene you’re on the right track. Would be difficult to discriminate which oncogenes result in “addiction” or not, as this can be debateable, but you would not have “oncogene addiction in a tumor suppressor…

Cancer Hallmarks: Oncogenes, Tumor Suppressors, Epigenetics

Question 46

Which of the following combinations and corresponding descriptor is correct:

A) Tumor suppressor genes—activated in many human tumors
B) Exon— the non-coding region of a gene
C) Promoter—involved in regulating gene transcription
D) DNA repair gene— EGFR
E) Oncogene—activated through loss of heterozygosity

Answer 46

c - Promoter-Involved in regulating gene transcription is the correct pairing.

“The promoter region is the regulatory portion of a gene that plays a critical role in directing whether a gene is transcribed or not. Tumor suppressor genes are generally inactivated in many cancers, typically resulting in a loss of control over cell proliferation. Exons are the expressed, or coding, regions of genes, whereas introns are the non-coding sequences. The protein encoded by the EGFR (epidermal growth factor receptor) gene is a cell surface tyrosine kinase receptor that is activated by epidermal growth factor (EGF) ligand, among others, and is important for cell proliferation. Loss of heterozygosity is a common mechanism by which tumor suppressor genes are inactivated. Oncogenes are generally activated by mechanisms including deletion/point mutation, chromosome rearrangement, retroviral integration, or gene amplification” (Huber et al, 2023, ASTRO Radiation and Cancer Biology Study Guide).

Huber, K, Woloshak, G, Rosenstein B, et al. 2023 Astro Radiation and Cancer Biology Study Guide. 2023. American Society for Radiation Oncology. Available from: https://www.astro.org/ASTRO/media/ASTRO/AffiliatePages/arro/PDFs/RadBio_StudyGuide_23.pdf.

Cancer Hallmarks: Oncogenes, Tumor Suppressors, Epigenetics

Question 47

Which is an example of a tumor suppressor gene?

N-myc
Her-2
c-kit
p16
Ras

Answer 47

Answer d - p16 is the correct answer.
p16 blocks cyclin D kinase activity. Cyclin D kinase maintains Rb in a hypophosphorylated state to prevent transcription driven by E2Fs. p16 is increased in a feedback loop by an increase in E2Fs. Without p16, cyclin D is unregulated. Palbociclib is a CyclinD-CDK4/6 inihibitor, often used for patients with metastatic ER+ breast cancer.
N-myc (normal copy of the wild type gene in the genome) is amplified in 40% of advanced neuroblastoma.
Her-2/neu is amplified in some types of breast cancer. This can be targeted with monoclonal antibody directed at Her2+ such as trastuzumab (Herceptin).
c-kit is a point mutation in the gene encoding a tyrosine kinase receptor in gastrointestinal stromal tumors (GIST). Imatinib, a tyrosine kinase inhibitor, is used in the treatment of GIST.
Point mutations in Ras keep Ras in the active (GTP bound) state. Overexpressed ras then promotes tumorigenesis via multiple pathways

Cancer Hallmarks: Oncogenes, Tumor Suppressors, Epigenetics

Question 48

Temozolomide improves survival in patients with glioblastoma that receive radiation therapy, particularly if the tumor demonstrates:

A) Epigenetic silencing of O6-methylguanine-DNA methyltransferase (MGMT)
B) Epigenetic silencing of microRNA expression
C) Epigenetic silencing of PTEN
D) Expression of the mutant receptor EGFRvIII
E) Expression or amplification of Her2/neu

Answer 48

Answer a) Epigenetic silencing of O6-MGMT is the correct answer.
“Methylation of the promoter for MGMT (O6 -methylguanine-DNA methyltransferase) via an epigenetic mechanism (not via a gene mutation) decreases expression of this DNA repair gene. When tumor cells do express MGMT they are able to repair the alkylation of DNA caused by temozolomide. Therefore, patients with MGMT-expressing glioblastomas derive little benefit from concurrent temozolomide and radiation therapy. In contrast, when MGMT is silenced, temozolomide is able to achieve significant DNA damage via alkylation, which ultimately increases its radiosensitivity” (Huber et al, 2023, ASTRO Radiation and Cancer Biology Study Guide)
Huber, K, Woloshak, G, Rosenstein B, et al. 2023 Astro Radiation and Cancer Biology Study Guide. 2023. American Society for Radiation Oncology. Available from: https://www.astro.org/ASTRO/media/ASTRO/AffiliatePages/arro/PDFs/RadBio_StudyGuide_23.pdf.
Stupp R, Mason WP, van den Bent MJ, et al., Radiotherapy Plus Concomitant and Adjuvant Temozolomide for Glioblastoma. NEJM 352(10): 987-996, 2005. Pubmed
Hegi ME, Diserens A, Gorlia T, et al., MGMT Gene Silencing and Benefit from Temozolomide in Glioblastoma. NEJM 352(10): 997-1003,

Cancer Hallmarks: Oncogenes, Tumor Suppressors, Epigenetics

Question 49

Pathogenic mutations of the KRAS oncogene function by:

A) Recycling GTP
B) Phosphorylating B-Raf
C) Preventing GTP hydrolysis
D) Preventing KRas protein degradation
E) Localizing KRas to the plasma membrane

Answer 49

Answer - C
Preventing GTP hydrolysis

Common mutations in KRAS function by preventing GTPase activity. The KRas protein is active in the GTP-bound form and inactive in the GDP-bound form. Guanine nucleotide exchange factors (GEFs) exchange GDP for GTP, thereby activating KRas. GTPase activating proteins (GAPs) are proteins that, when bound to KRas, can induce GTP hydrolysis into GDP. Pathologic mutations of KRAS, including those at positions G12 and G13, prevent KRas from hydrolyzing GTP. Thus KRas remains in the active GTP-bound state, with resultant downstream signaling. Reference: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC4869631/

Woloshak, G, Huber, K, Barker, C, Rosenstein B, et al. 2020 Astro Radiation and Cancer Biology Study Guide. 2020. American Society for Radiation Oncology. Available from: https://www.astro.org/ASTRO/media/ASTRO/AffiliatePages/arro/PDFs/Radiobiology_StudyGuide21.pdf

Cancer Hallmarks: Oncogenes, Tumor Suppressors, Epigenetics

Question 50

Which of the following pathways have been implicated in the loss of clonogenic capacity in irradiated tumor cells harboring wild type p53?

A) Dedifferentiation
B) Sublethal damage repair
C) Senescence
D) Telomere inversion
E) Oncogene activatio

Answer 50

Answer: CSenescence is a possible response to genotoxic insult that can result in loss of replication capacity. This can be observed in normal tissues and in tumor cells with a wild type p53.
Tumor cells that harbor a mutant p53 may be at least partially immune to radiation induced senescence.

Refs:
Purvis JE, Karhohs KW, Mock C, Batchelor E, Loewer A, Lahav G. p53 dynamics control cell fate. Science. 2012 Jun 15;336(6087):1440-4.
Woloshak, G, Huber, K, Barker, C, Rosenstein B, et al. 2020 Astro Radiation and Cancer Biology Study Guide. 2020. American Society for Radiation Oncology. Available from: https://www.astro.org/ASTRO/media/ASTRO/AffiliatePages/arro/PDFs/Radiobiology_StudyGuide21.pdf

Cancer Hallmarks: Oncogenes, Tumor Suppressors, Epigenetics

Question 51

Which of the following statements is TRUE concerning the retinoblastoma protein (RB1)? RB1:

A. Is an important downstream effector controlling the G2 checkpoint
B. Once phosphorylated, releases E2F
C. Is encoded by an oncogene
D. Is phosphorylated by ATM
E. Activity is altered in approximately 10% of cancers

Answer 51

Answer: B
RB1 is the product of the RB1 tumor suppressor gene (not an oncogene). Once phosphorylated by CDK4/6, RB1 releases E2F, which then activates genes associated with the G1 checkpoint. RB1 is functionally inactivated in virtually all human cancers, either directly or indirectly, via p53 (TP53). p53-dependent induction of p21 (CDKN1A) regulates cyclin E/CDK2 and cyclin A/CDK2 complexes, both of which phosphorylate RB1. The RB1 and p53 signaling pathways are dysregulated in the majority of human cancers.
Refs:
Mittnacht S. The retinoblastoma protein – from bench to bedside. Eur J Cell Biol 84:97-107, 2005.
Massague J. G1 Cell-cycle control and cancer. Nature 432:298-306, 2004.
Classon M, Harlow E. The retinoblastoma tumour suppressor indevelopment and cancer. Nat Rev Cancer 2:910-917, 2002.
Huber, K, Woloshak, G, Rosenstein B, et al. 2023 Astro Radiation and Cancer Biology Study Guide. 2023. American Society for Radiation Oncology. Available from: https://www.astro.org/ASTRO/media/ASTRO/AffiliatePages/arro/PDFs/RadBio_StudyGuide_23.pdf.