Modifiers of cell survival: Repair Flashcards
An exponentially-growing, asynchronous population of cells is maintained under normal physiological conditions. Which of the following experimental manipulations would potentiate cell killing following radiotherapy as measured by a clonogenic assay?
A. Cell synchronization in S-phase at the time of irradiation
B. Irradiation under hypoxic conditions
C. Irradiation with the dose split into two fractions with a 24-hourinterval between fractions rather than given as an acute exposure to the same total dose
D. Incorporation of bromodeoxyuridine into the DNA prior to irradiation
E. Addition of cysteine to the cellular growth medium prior to irradiation
D
Bromodeoxyuridine, a synthetic nucleoside analogue, when incorporated into cellular DNA in place of thymidine acts as a radiation sensitizer, so cell killing would be enhanced.
S-phase is the most radioresistant phase of the cell cycle, so cell killing would be decreased relative to that for an asynchronous population (Answer Choice A).
Oxygen is a radiation sensitizer, so cell killing would decrease in cells made hypoxic before irradiation (Answer Choice B).
Splitting the dose into two fractions separated by 24 hours would allow sublethal damage recovery and possibly enable cellular proliferation to take place between fractions. Cell killing would therefore be less than if the total dose had been delivered acutely (Answer Choice C).
Cysteine is a sulfhydryl-containing compound that scavenges radiationinduced free radicals; it therefore acts as a radioprotector and reduces cell killing (Answer Choice E).
For irradiation with X-rays, the increased cell survival observed when a given total dose is delivered at a low dose-rate (~1 Gy/hr) versus high dose-rate (~1 Gy/min) is due primarily to:
A. Repair of DNA double-strand breaks
B. Decreased production of DNA double-strand breaks
C. Induction of free radical scavengers
D. Activation of cell cycle checkpoints
E. Down-regulation of apoptosis
A
Therapeutic radiation at a low dose-rate of ~1 Gy/hr is associated with increased cell survival compared to higher dose-rates primarily due to DNA repair - especially the repair of DNA double-strand breaks produced as a result of radiation.
Compared to the surviving fraction of cells maintained in a non-cycling state for several hours after irradiation, decreased cell survival observed in cells forced to re-enter the cell cycle immediately following irradiation is evidence for:
A. Rejoining of chromosome breaks
B. Sublethal damage recovery
C. Cell cycle reassortment
D. Translesion of DNA synthesis
E. Expression of potentially lethal damage
E
Potentially lethal damage recovery is operationally defined as an increase in cell survival under environmental conditions not conducive to progression of cells through the cell cycle for several hours after delivery of a large, single radiation dose. If non-cycling cells are forced to re-enter the cell cycle immediately after irradiation, rather than remaining
quiescent, potentially lethal damage will be “expressed” and therefore the surviving fraction will be lower.
Sublethal damage recovery is operationally defined as an increase in cell survival noted when a total radiation dose is delivered as two fractions with a time interval between the irradiations, i.e. splitting the dose, as opposed to a single exposure (Answer Choice B).
Repair of DNA damage and rejoining of chromosome breaks presumably underlie both the sublethal and potentially lethal damage recovery (Answer Choice A).
Cell cycle reassortment has a sensitizing effect on a population of cells receiving multi-fraction or protracted irradiation regimens. This is because surviving cells that were in a resistant phase of the cell cycle during the initial irradiation may progress through the cell cycle between fractions and reassert into a more sensitive phase of the cell cycle by the time of delivery of the next fraction. This process is irrelevant under the conditions described here, in which only a single radiation dose was administered (Answer Choice C).
Translesion DNA synthesis is an error-prone process during which certain DNA polymerases synthesize DNA using a damaged DNA strand as a template, resulting in error-prone DNA synthesis (Answer Choice D).
5 Gy of X-rays is delivered at a high dose rate (1 Gy/min) rather than a low dose rate (1 Gy/hr). Which of the following statements about the effects of this change on cell survival is TRUE?
A. The surviving fraction would change the least for a cell line with a radiation survival curve characterized by a low a/B ratio
B. Treatment of cells during irradiation with an agent that inhibits DNA repair would have a greater impact on the surviving fraction of cells irradiated at the high dose rate
C. More cell killing would occur following treatment at the high dose rate
D. The difference in the surviving fractions between the two protocols results primarily from repopulation
E. The total number of ionizations produced is decreased with treatment at the high dose rate
C
When a dose of 5 Gy is delivered at a dose rate of 1 Gy/min, irradiation requires 5 minutes. When 5 Gy is delivered at 1 Gy/hr, irradiation requires 5 hours. Extensive repair of sublethal damage will occur during the low dose rate, but will not be able to occur during high dose rate irradiation. Essentially, the chance for separate lesions to interact with one another forming more complex, lethal lesions increases at the higher dose rate. As a result, the B component of cell killing will increase. More cell killing would therefore occur when a dose of 5 Gy is delivered at a high dose rate rather than a low dose rate.
The surviving fraction would change the least for a cell line with a radiation survival curve characterized by a high, not low, a/b ratio (Answer Choice A).
Treatment with an agent that inhibits DNA repair would have little impact during the 5-minute period of irradiation that would occur at the high dose rate (Answer Choice B).
In contrast, such a treatment would markedly reduce cell survival for the 5-hour irradiation required at the low dose rate since, in the absence of the agent, substantial repair would take place during the course of the irradiation. The increase in the surviving fraction for this low dose rate irradiation is primarily a consequence of sublethal damage recovery and not repopulation, as the repopulation would only occur for overall treatment times on the order of days (Answer Choice D).
The total number of ionizations produced is a reflection of the total dose delivered and does not vary with the dose rate (Answer Choice E).
Exponentially growing cells were maintained at 37 C in 95% air/5% CO2 and irradiated with either a single dose of 8 Gy of X-rays or two 4 Gy fractions separated by either 2 hours or 8 hours. The surviving fractions for the three treatments were 0.02, 0.15, and 0.08, respectively. The two processes that best account for these differences in survival are:
A. Reassortment and repopulation
B. Repair and reassortment
C. Reoxygenation and repair
D. Repopulation and reassortment
E. Repair and reoxygenation
B
Compared to the cell surviving fraction after the single 8 Gy dose, the increase in cell survival noted for the two 4 Gy doses delivered with a 2- hour interfraction interval was due to sublethal damage repair (SLDR). Although SLDR also occurred when the interfraction interval was 8 hours, cells surviving the first dose start progressing through the cell cycle and reassort from the radioresistant phases they were in at the time of the initial irradiation (e.g. late S) into more radiosensitive phases (e.g. G2 and M), thereby resulting in an overall surviving fraction for the 8-hour interval that was lower than that for the split dose protocol with a 2-hour interval between fractions. It is unlikely that much repopulation would take place during the total time of 8 hours needed to complete the irradiations. Reoxygenation would not be an issue for cells maintained in a well-aerated 95% air environment
Which of the following pairs of radiobiological process and corresponding assay method is CORRECT?
A. Reoxygenation – HIF-1a (HIF1A) phosphorylation by ATM
B. Potentially lethal damage recovery – tritiated thymidine uptake
C. Cell cycle “age response” – paired survival curve method
D. Sublethal damage recovery – split dose experiment
E. Repopulation – mitotic shakeoff procedure
D
Sublethal damage recovery is operationally defined and demonstrated using a split dose protocol.
Potentially lethal damage repair is detected by changing the post- irradiation environment and observing the effect on survival (Answer Choice B). Incorporation of tritiated thymidine into DNA would not specifically measure PLDR, but would reflect DNA synthesis and other forms of DNA repair.
Reoxygenation would best be assayed by performing repeat measurements during the course of radiotherapy by using an oxygen electrode or by treating with a hypoxia maker, such as pimonidazole, that is metabolized and incorporated exclusively into hypoxic cells (AnswerChoice A).
Cell cycle age response is best demonstrated by performing cell synchronization followed by irradiation of cohorts of cells in particular cell cycle phases and then performing the clonogenic survival assay as a readout (Answer Choice C).
Repopulation can be assayed in vitro by counting the number of cells present as a function of time after irradiation. The mitotic shake off technique is used to collect synchronous populations of cells for use in experiments examining age response functions (Answer Choice E).
Which of the following is a phosphoinositol 3-kinase like kinase (PIKK) that serves as the central orchestrator of the signal transduction response to DSBs?
A. Ku70/80
B. ATM
C. Rad50
D. MSH2
E. p53 (TP53)
B
Ataxia Telangiectasis Mutated (ATM) serves as the central orchestrator of the signal transduction response to DSBs. Cells deficient in ATM activity display cell cycle checkpoint defects and sensitivity to ionizing radiation.
Which of the following is TRUE for potentially lethal radiation damage (PLD)?
A. It is irreversible and irreparable.
B. It is the damage that can be repaired efficiently if cells are allowed to progress through the cell cycle immediately following IR.
C. It is thought to be primarily complex or “dirty” double strand breaks.
D. It can be observed in a “split dose” experiment.
E. It cannot be detected in tumors in vivo.
C
PLD is believed to be complex double strand breaks (DSBs) that are repaired slowly as compared to simple DSBs. Therefore, cells that are left in stationary phase after irradiation display enhanced survival as they have time to repair complex DSBs before resuming progression through the cell cycle.
Which of the following is FALSE for the split dose experiment and sublethal damage (SLD)?
A. The survivors of the first dose are mainly S-phase.
B. The fraction of cells surviving a split dose decreases as the time interval between the two doses increases
C. When cells are cycling during the split dose experiments, there is a dip (decrease) in cell survival caused by reassortment.
D. SLD can be repaired before they can interact to form lethal chromosomal damage.
E. SLD is demonstrated by low-LET radiation
B
Sublethal damage (SLD) is the term used to describe the increase in cell survival that is observed if a given radiation dose is split into two fractions separated by a time interval, i.e. allowing time for repair. However, this benefit in survival can be offset when the time between doses extends so much that reassortment, reoxygenation, and repopulation kick in (answer C). Ultimately, the outcome of split-fraction experiments depends on the interplay between Repair of DNA damage, Reassortment through the cell cycle and Repopulation (requires a split time period greater than the cell cycle length).
S-phase, especially the latter part of S-phase is the most radioresistant cell cycle phase therefore most survivors of the first fraction are generally in S phase. SLD is seen to a far higher-degree in low-LET irradiated cells than in high-LET irradiation.
The fraction of cells surviving a split dose increases with increasing time between the two doses because of the repair of SLD, hence B is false.
The survival of mammalian cycling cells at 37 C increases dramatically if a X-ray dose is split into two fractions, given within 1 hour of one another and reaches its maximum if the time between each fraction is about 2h. What is the mechanism driving this phenomenon?
A. Enhanced survival in split-dose experiments such as this indicate recovery from potentially lethal damage.
B. The main mechanism driving this is the repair of sublethal damage in the G1 cell cycle phase.
C. The mechanism that enhances survival is sublethal damage repair that in cycling mammalian cells increases to a maximum between about 1-2h after the first exposure.
D. Potentially lethal damage repair tends to be complete by 1h and survival does not increase further by increasing time allowed beyond 1h.
E. The initial, dramatic benefit in survival is due to sublethal damage repair while potentially lethal damage repair drives the survival benefits at intervals of about 2h.
C
The effect of a given dose of radiation is less if it is split into two fractions, delivered a few hours apart which is due to the repair of sublethal damage. Most drastically this is observed within the first hour when most damage is repaired, starting to plateau after 2h for most proliferating mammalian cells.
The majority of surviving cells will come from S-phase as this is the most radioresistant cell cycle phase.
Potentially lethal damage repair is not assessed in split dose experiments but is more about post-exposure conditions and the time available for repair in arrested populations before cell cycle resumes. (single-dose experiment)
