Models of cell survival Flashcards
If a cell line exhibiting a strictly exponential radiation survival curve is exposed to a dose that produces an average of one lethal “hit” per cell, the surviving fraction after this dose would be approximately:
A. 0.01
B. 0.10
C. 0.37
D. 0.50
E. 0.90
C
Assuming that all cells in the cell population are identical and that cell killing is a random, probabilistic process that follows a Poisson distribution. One model that can calculate the radiation dose that produces an average of one lethal hit is the single-target single-hit model. From the equation that describes this model, S = e^(-D/D0) , the dose, D, at which there would be an average of one hit per cell would be equal to D0, the constant of proportionality. Therefore, S = e^-1 ~ 0.37.
The a/B ratio is equal to the:
A. Surviving fraction at which the amount of cell killing caused by the induction of irreparable damage equals the amount of cell killing caused by the accumulation of sublethal damage
B. Optimal fraction size to use in a fractionated radiotherapy regimen
C. Dose below which a further decrease in fraction size will not affect the surviving fraction for a particular total dose
D. Dq
E. Dose at which the aD component of cell kill is equal to the BD^2 contribution to cell killing
E
The a/B ratio represents the dose at which the aD component of cell killing, assumed to result from single hit killing, is equal to the lethality produced by the BD^2 component of cell killing that results from multi hit killing.
Cells from individuals diagnosed with which of the following diseases/syndromes would NOT be expected to have increased sensitivity to X-ray irradiation?
A. Nijmegen breakage syndrome
B. LIG4 syndrome
C. ATR-Seckel syndrome
D. Xeroderma pigmentosum
E. Ataxia telangiectasia
D
Cells derived from an individual diagnosed with xeroderma pigmentosum are defective in nucleotide excision repair (NER). These cells are sensitive to UV radiation because this form of radiation produces damage such as pyrimidine dimers that are removed through the nucleotide excision repair pathway. Because DNA double-strand breaks are important lesions responsible for lethality in cells exposed to X-rays and because DNA double-strand break repair is generally normal in cells derived from a person diagnosed with xeroderma pigmentosum, their cells are not expected to be especially sensitive to X rays. People with Nijmegen breakage syndrome, LIG4 syndrome, ATR-Seckel syndrome and ataxia telangiectasia, who possess mutations in either NBS1, LIG4, ATR or ATM, respectively, are all characterized by defects in strand break repair or repair-related signaling. Therefore, their cells would be expected to be sensitive to X rays.
Which of the following statements concerning cell survival curve analysis is TRUE?
A. The B parameter generally increases as the radiation dose rate decreases
B. The inverse of the Dq corresponds to the final slope of the survival curve
C. The extrapolation number, n, of a survival curve increases with increasing LET of the radiation
D. D0 is a measure of the incremental increase in cell survival when a given dose is fractionated
E. If n = 1, then D37 = D0
E
Parameters to define a radiation cell survival curve include: the initial slope (D1), a final slope (D0), and some quantity that is a measure of the width of the shoulder. This quantity can be the extrapolation number (n) or the quasi-threshold dose (Dq). If n=1, the survival curve has no shoulder and D37 (dose resulting in a survival fraction of 0.37) equals the D0.
D0 is a measurement of radiosensitivity based on the exponential part of the survival curve. In a target theory, it can be defined as a dose that gives an average of one lethal hit per cell in a population. Thus, an increase of an exposed dose of D0 would result in a 37% further reduction of survival. A smaller D0 value indicates a higher sensitivity to radiation.
For the same radiation dose, radiation delivered at a lower dose rate may produce less cell killing than radiation delivered at a higher dose rate because sublethal damage repair occurs during the protracted exposure. As the dose rate is reduced, the slope of the survival curve becomes shallower and the shoulder tends to disappear because alpha does not change significantly but B trends to zero in the linear quadratic model.
The inverse of the D0, not the Dq, is equal to the final slope of the survival curve.
For densely ionizing radiation (increasing LET), the shoulder of the survival curve tends to disappear. N, therefore, decreases, until it reaches a value of 1.0 for very high LET radiations. The D0 would not necessarily be a good predictor for the effect of fractionation on survival; Dq or n would be better
Reducing the dose rate at which a continuous gamma-irradiation is delivered may affect its cell killing efficacy due to several different biological processes. For a total dose of 6 Gy, which pair of dose rate ranges and biological processes resulting in altered cell killing is INCORRECT?
A. 10 - 1 Gy/min : reoxygenation
B. 1 - 0.1 Gy/min : repair
C. 0.1 - 0.01 Gy/min : redistribution
D. 0.01 - 0.001 Gy/min : repopulation
A
Reoxygenation generally occurs over a period of hours to days. Little to no reoxygenation of hypoxic cells is therefore likely during irradiation performed at dose rates in the 1-10 Gy/min range since, for a total dose of 6 Gy, the irradiation times would only vary from 0.6-6 minutes.
If the total treatment time is long enough that significant repair of sublethal damage (half-time on the order of 0.5-1.0 hour) can occur during irradiation, repair does influence cell survival. The irradiation time would vary from 6-60 minutes for dose-rates in the range of 1-0.1 Gy/min and significant repair would occur (Answer Choice B).
Movement of the surviving cells through the cell cycle (causing redistribution of viable cells from resistant phases into sensitive phases) can influence the radiosensitivity of the cell population when irradiation times are increased to several hours (for the dose-rate range of 0.1-0.01 Gy/min, times of 1-10 hours would be needed to produce 6 Gy; Answer Choice C).
Repopulation can lead to an increase in the number of cells during irradiation and, hence, to an increase in the total number of surviving cells when a radiation dose is delivered over days (10-100 hours are required to produce a total dose of 6 Gy over a range of 0.01-0.001 Gy/min (Answer Choice D).
The survival curve for a cell population irradiated with a form of high LET radiation is characterized by a D10 of 3 Gy. For a starting population of 10^8 cells, approximately how many cells will survive when a single dose of 18 Gy is given?
A. 10^0
B. 10^1
C. 10^2
D. 10^3
E. 10^4
C
For high LET radiation it can be assumed that the survival curve is exponential, or near exponential, and cell survival can be modeled using the single-target, single-hit equation (S = e), or the simplified form of the linear quadratic equation in which B is zero (S = e^-aD). Using either of these equations, 3 Gy reduces the surviving fraction to 10^-1, and a dose of 18 Gy therefore would reduce survival to 10^-6. Therefore, irradiating 10^8 cells with 18 Gy would result in the survival of: (10^8
cells) x (10^-6 surviving fraction) = 10^2 cells.
A total dose of 12 Gy of X-rays delivered in 3 Gy fractions reduces cell survival to 10^-4. Assuming that cell killing can be modeled using an exponential survival curve, what dose would be required to reduce the surviving fraction to 10^-6?
A. 9 Gy
B. 18 Gy
C. 24 Gy
D. 36 Gy
E. 72 Gy
B
An exponential survival curve can be modeled using the single-target, single-hit equation (S = e), or the simplified form of the linear quadratic equation in which B is zero (S = e^-aD). Since four 3 Gy fractions reduce the surviving fraction to 10^-4, and assuming an equal effect per fraction, each 3 Gy fraction reduces the surviving fraction by 10^-1. Accordingly, two additional 3 Gy fractions (producing a total dose of 18 Gy) would yield a surviving fraction of 10^-6.
In an attempt to generate a radiation survival curve for a new cell line, four cell culture dishes were seeded with 10^2, 10^3, 10^4 and 10^5 cells, and X-irradiated with 0, 3, 6 and 9 Gy, respectively. At the end of a two-week incubation period, a total of 40 colonies was counted on each dish. Which one of the following statements is TRUE?
A. The D0 for this cell line is 3 Gy
B. The survival curve for this cell line is exponential
C. The n and Dq values for this survival curve are large
D. The cell surviving fraction after a dose of 3 Gy is 0.04
E. The alpha-beta ratio for this cell line is small
B
The survival curve for this cell line is exponential because each incremental dose of 3 Gy decreased the surviving fraction by an additional factor of 0.1. Thus, this survival curve can be modeled using an exponential equation which can be expressed as either S= e^-aD (linear-quadratic model) or S = e^-D/D0 (target theory model). The D0 is equal to the D10/2.3, or 1.3 Gy, not 3 Gy. The n and Dq values for this survival curve are equal to 1 and 0 Gy, respectively, and are therefore small, not large. The surviving fraction after a single dose of 3 Gy can be calculated from the colony forming efficiency of the irradiated cells (40/1000), divided by the plating efficiency (PE) of the unirradiated cells (40/100), which is equal to 0.04/0.4 or 0.1. Since this survival curve can be represented by S = e^-aD, the B term of the linear-quadratic equation must approach zero, so the a/B would be very high, and in fact will be undefined if the B term is actually zero.
What is the approximate eD10 (effective D10) for a particular cell line if the eD0 is 4 Gy?
A. 2 Gy
B. 4 Gy
C. 6 Gy
D. 9 Gy
E. 12 Gy
D
eD10 is the dose required to kill 90% of population
eD10 = 2.3 x D0
The eD10 is equal to the eD0 multiplied by 2.3, or 4 Gy x 2.3 = 9.2 Gy
When irradiating a cell population with a dose that causes an average of one lethal event per cell, this will likely result in a survival fraction of:
A. 0% of cell survival.
B. 10% of cell survival.
C. 37% of cell survival
D. 63% of cell survival
E. 100% of cell survival
C
When the irradiated cell population receives an average of 1 lethal hit, it results in 37% cell survival based on Poisson statistics.
A 1 cm-diameter tumor that contains 10^7 clonogenic cells is irradiated with daily dose fractions of 1.8 Gy. The effective dose response curve has been determined and is exponential with a D10 of 8 Gy. What total dose will correspond to the TCD90 (90% probability of tumor control), assuming no cell proliferation between dose fractions?
A. 32 Gy
B. 40 Gy
C. 48 Gy
D. 56 Gy
E. 64 Gy
E
In order to achieve a 90% tumor control rate, the total dose delivered must reduce the number of surviving clonogenic cells to an average of 0.1. This is based on the equation P = e-(N)(SF), where P is the probability of tumor cure (90% or 0.9 in this case), N is the initial number of tumor clonogens (10^7) and SF is the surviving fraction resulting from the irradiation protocol. Thus, for 10^7 clonogenic cells, a total dose that reduces the surviving fraction to 10^-8 (i.e., an average of 0.1 clonogen surviving in each tumor or 1 cell surviving out of every 10 tumors irradiated) must be used to achieve a 90% control rate. Since the effective survival curve is exponential with a D10 of 8 Gy, it would be necessary to use a dose of 64 Gy.
A 1 cm-diameter tumor that contains 10^7 clonogenic cells is irradiated with daily dose fractions of 1.8 Gy. The effective dose response curve has been determined and is exponential with a D10 of 8 Gy. What would be the TCD90 if a surgical excision removed 99% of the tumor clonogens prior to radiotherapy (assume that the surgery did not otherwise affect the growth fraction of the tumor).
A. 24 Gy
B. 32 Gy
C. 40 Gy
D. 48 Gy
E. 56 Gy
D
A surgical excision of 99% of the tumor would reduce the initial number of clonogens to 10^5. Thus to achieve a 90% control rate, a dose of 48 Gy would be required, corresponding to a final clonogen surviving fraction
of 10^-6.
For a tumor that requires 18 days to double its diameter, what is the approximate cell cycle time of its constituent cells (assume no cell loss and that all cells are actively dividing)?
A. 6 days
B. 9 days
C. 12 days
D. 15 days
E. 18 days
A
A doubling of a tumor’s diameter reflects about an 8-fold increase in cell number, which would require 3 cell divisions to accomplish. Thus, if 18 days were required to complete 3 cell cycles, the cell cycle time must be 6 days.
