Physics 561
Radiation Biophysics

Final Examination Key
7-8 May 2003

1.(a) In the figure shown, C is the survival curve for a cell line irradiated under anoxic conditions. Two other conditions are tested: low but nonzero P[O2], and atomospheric oxgen. Which correspond to these two conditions? Why?
A corresponds to atmospheric oxygen and B to the low but nonzero oxygen pressure. Oxygen tends to fix (stabilize) the radicals that cause much of the biological effect of radiation, so any given dose has the effect of being more hazardous under oxic conditions than under anoxic conditions. Conditions of low but nonzero P[O2] give rise to intermediate levels of damage fixation by oxygen.
5 points for recognizing which one is A and which is B; 3 points for describing the oxygen enhancement effect; 2 more for explicitly mentioning fixation of free-radical damage or describing it.

(b) Given the above, would you expect that the low-but-nonzero P[O2] mentioned above be closer to 0.1 atm or 0.01 atm ? Why? (hint: atmospheric P[O2] = 0.19 atm, roughly.)
It's probably 0.01 atm. As is shown in fig. 9.3 and discussed in the text, the conditions that give rise to the oxygen enhancement remain almost fully potent until the oxygen pressure falls to quite low levels. Fig. 9.3 shows that the oxygen enhancement effect remains at 50% of its effectiveness at P[O2] = 0.1%, i.e. closer to 0.01 atm than to 0.1 atm.
5 points for seeing that it's closer to 0.01 atm; 5 points for referring to the results in fig. 9.3 or something similar.

2. The results of a ficticious lung cancer study are as follows. Three populations were identified: nonsmokers, light smokers, and heavy smokers. Radiation exposure was monitored for the three groups and it was found that the cancer incidence in the three groups showed stochastic sensitivity to dose, behaving according to linear-quadratic statistics. The linear-quadratic coefficients associated with Alpen's eqn. (12.2) were:
 
population In a1, Sv-1 a2, Sv-2
nonsmokers 2*10-4 1*10-4 0.5*10-4
light smokers 10*10-4 10*10-4 1.33*10-4
heavy smokers 20*10-4 20*10-4 2*10-4

(a) Do these data fit a relative risk model or an absolute risk model for radiation carcinogenesis? Explain.
These data fit a relative risk model, because the risk of cancer associated with varying background levels of risk (represented by the three different In values) also entail increasing probabilities of risk for the population exposed to radiation. That is, insofar as the coefficients a1 and a2 themselves are larger for the groups with higher background cancer risk levels, the risk of cancer to the radiation-exposed population is not merely shifted upward but in fact shows greater sensitivity to radiation as the background risk goes up.
5 points for recognizing that these data imply a relative-risk model. 2 points for recognizing that the increases in In as a function of amount of smoking constitute a change in the background level of risk. 2 points for explaining that the relative-risk model involves a higher sensitivity rather than a simple shift in baseline; 1 point for the remainder of the explanation.
(b) Describe the ways in which these results are consistent with the initiator-promotor model of carcinogenesis.
Radiation often acts by increasing the rate at which mutations arise, i.e. it acts as an initiator. Cigarette smoking contains chemicals that are known promotors of cancer, as well as containing mutagens that can act as initiators. Because it contains both mutagens and promotors, the background incidence of cancer is higher for the heavy smokers than for the nonsmokers, and the light smokers have an intermediate level of background incidence. But in addition, the initiative events that the radiation exposure causes have a higher probability of actually engendering cancer in the smokers, because the promotors found in the smoke make it more likely that the mutations will become expressed as cancer.
2 points for (one way or another) making it evident that the student understands what the initiator-promotor model means. 2 points for describing radiation as an initiator. 2 points for recognizing that smoking is a promotor, and 1 more for recognizing that it's also an initiator. 2 points for seeing that smoking's role as both an initiator and as a promotor gives rise to the increase in background incidence through the three groups. 1 point for seeing that an initiator's probability of actually giving rise to a tumor in the radiation-exposed groups is enhanced by the chemicals with promotor activity in the cigarette smoke.

3. Explain why a plot of relative biological effectiveness versus linear energy transfer for a constant dose shows RBE increasing as a function of increasing LET, up to a point but no further. Include in your explanation the reason that d(RBE)/d(LET) < 0 at very high values of LET.
High-LET radiation deposits most of its energy in a small area--typically, in the area occupied by a single cell. Therefore as LET goes up over a range from 0.1 to 20 keV/µm or so, the probability of doing enough damage to actually make a cell die clonogenically goes up. In particular, the forms of damage that the DNA surveillance and repair systems in the cell cannot recognize and repair become more common as LET increases over this range. As LET goes up even more, though, the damage becomes so localized that the likelihood is that the cell will be killed with only a fraction of the energy actually deposited. The remainder of the deposited energy, in effect, goes into killing a cell that is already dead. This has little biological significance, so d(RBE)/d(LET) may actually go down over this range of LET.
4 points for demonstrating that the student understands what LET is. 4 points for demonstrating that the student understands what RBE is. 4 points for describing the positive d(RBE)/d(LET) values in low to moderate values of LET in terms of increased likelihood of clonogenic death.4 points for tying this increased probability to the reduced effectiveness of DNA repair mechanisms. 4 points for recognizing that very high LET gives rise to conditions in which the cell is already dead and can't get any deader.

4. Patients afflicted with ataxia telangiectasia (AT) are at hgih risk for numerous health problems arising from their tissues' inability to repair certain kinds of DNA damage. Homozygotes for AT rarely survive to adulthood. Heterozygotes rarely show symptoms while young but are at substantially increased risk for contracting several forms of cancer as adults. Explain.
Patients with AT will show deficiencies in DNA repair. Homozygotes are completely lacking in an enzyme that is crucial to DNA repair, and as such are completely at the mercy of mutagenic events in their DNA. This means that the normal levels of error in DNA replication that can be found even in cells not exposed to mutagens or radiation are likely to give rise to problems in cell replication, and the individual will die from an accumulation of DNA damage. Heterozygotes have one correct  gene for the DNA repair enzyme carried on one copy of the relevant chromosome and a faulty gene for the repair enzyme carried on the other copy. The result is that in any given cell there is a reduced level of that enzyme that can be produced to fight off mutations. Thus it is more likely for such an individual to be unable to keep up with errors in replication than would be true for people with two functional genes for this enzyme in each cell. The heterozygotes are therefore at a higher risk for contracting cancer and other diseases associated with sensitivity to mutation.
4 points for showing that the student understands what "heterozygous" and "homozygous" mean. 4 points for tying increased risk for cellular damage arising from mutations to inadequate DNA repair. 3 points for recognizing that enzymes are involved in this and 3 for recognizing that some DNA codes for these enzymes. 6 for the remainder of the explanation.

5. Describe two different mechanisms by which fetal damage might arise if the uterus is irradiated during the early stages of pregnancy.
Briefly: mutations to somatic DNA in the fetus itself, and damage to the uterine environment in which the fetus is growing. More specifically, the irradiation might cause cells in the fetus to suffer mutations that cannot be repaired quickly enough to prevent their being passed on through the rapid cell divisions that the fetus is experiencing. These mutations could give rise to abnormalities in the fetus. The other route by which the fetus could suffer harm involves damage to the placenta. Since the placenta provides nutrients to the fetus as it grows, interference with the process by which those nutrients are provided could result in harm to the growing fetus.
Full credit for recognizing these two concepts. Up to 18 points will be awarded for an articulate exposition of two independent mechanisms on either the fetal-damage or placental-damage side.

6. Macromolecular crystallography consists of irradiating small crystals of highly hydrated proteins with X-rays and then measuring the diffraction patterns that emanate from the crystals. Protein crystals usually float, just barely. Suppose a cubical protein crystal of dimension 0.1mm is irradiated with 1011 X-ray photons/second, using 10KeV X-rays. Assume that 0.01% of the incident energy is actually absorbed in the protein crystal. Calculate how many Gy of dose is received by the crystal in a typical 1-hour experiment.
Dose = energy absorbed / kg = energy imparted * fraction absorbed / mass of absorber
energy imparted = number of photons * energy per photon
number of photons = (number per second) * number of seconds
The mass of the absorber is its density times its volume.
The volume of the absorber is its linear dimension cubed, since this is a cubic absorber.
Define:

  • N = number of photons per second
  • T = number of seconds of time elapsed
  • f = fraction absorbed
  • E = energy per photon
  • r = density of sample
  • L = linear dimension of sample

  • Putting all our thoughts together,
    Dose = NTEf/(rL3)
    The fact that the crystal just barely floats means that it has essentially the density of water, i.e. 1 g/mL. Since an mL is 10-3 (dm)3 = 10-6 m-3, this is 1 g/10-6m3 = 10-3kg/10-6m3 = 103 kg/m3.
    L = 0.1mm=10-4m
    N = 1011 Xph/sec
    T = 3600 sec
    f = 10-4
    E = 10 KeV/Xph = 104 eV/Xph = 104 * 1.609*10-19J/Xph = 1.609 * 10-15 J/Xph (an Xph is an X-ray photon, in case you're wondering).
    Thus Dose = 1011 Xph/sec * 3600 sec * 1.609 * 10-15 J /Xph * 10-4 / [103 kg/m3 * (10-4m)3]
    Dose = 3.6 * 1.609 * 10(11+3-15-4) / 103-4*3 J/kg = 5.79*10-5/10-9 J/kg = 5.79 * 104 Gy = 57900 Gy.
    Full credit for getting the right answer to within 5% even if it isn't very well explained.
    Minimum of 10 points for recognizing what the right approach is.
    3 points for recognizing that the fact that the crystals just barely float tells you what the density is; 1 more for knowing that the density of water is 1 g/mL.

    7. Define fixation of damage by oxygen and similar molecules.
    Fixation of damage refers to the stabilization of an chemical species owing to the proximity of a compound like O2 that can complex with an unstable species like a free radical. This allows the unstable molecule to remain present long enough to do actual damage to a biological entity.
    Partial credit for mentioning free radicals even if the rest of the answer isn't right.

    8. In an Elkind-Sutton experiment, does the slope of the linear portion of the log(survival) curve depend on the time delay between the inital dose and the final dose?
    No. See fig. 8.3, where all the slopes are the same.
    Since I didn't ask for an explanation, you get 5 points if you say no; 0 if you say yes; 1-2 points if you say yes but provide some sort of scientifically sensible explanation.

    9. Describe the continuous slowing down approximation.
    This approximation describes the fate of a fast-moving charged particle as it passes through a medium. It states that the energy  is deposited as it moves through the medium in a continuous stream rather than in a discrete fashion as it interacts with specific target atoms. The latter is actually closer to the truth, but it is convenient to approximate the herky-jerky path as if it were continuous.
    Since this definition is given in the text, it really ought to be right. Partial credit for badly worded versions of the correct answer.

    10.  A chromosome is damaged by radiation in such a way as to produce a product that has two centromeres. What is this product called?
    A dicentric. See fig. 13.2(c).

    11. If a human inhales radionuclides adsorbed onto particles of diameter greater than 10 micrometers, is he or she more likely to develop tumors in the throat or in the deep lung? Why?
    The throat, because the particles are large enough get stopped in the cilia lining the throat. They should get carried back out of the throat in mucus and swallowed. Thus the exposure of the throat to the carcinogenic effects of the radionuclides will be substantial, whereas the exposure of the deep lung will be negligible.
    2 points for getting the right answer; 3 more for a plausible explanation.

    12. A patient receives a 0.5 Gy whole-body dose of 10MeV neutrons as part of a therapeutic regimen. What is the equivalent dose in Sv?
    According to table 16.1 the radiation weighting factor for neutrons with energies between 2 and 20 MeV is 10. Therefore the equivalent dose in Sv is 10 times the actual dose in Gy, so the equivalent dose is 10*0.5 = 5 Sv.
    3 points for recognizing that the equivalent dose is the product of the weighting factor and the whole-body dose. 2 more for doing the arithmetic right.

    13. Contrast the roles of the p53 gene product and the bcl-2 gene product in the context of apoptosis.
    This one is not only in the book, it's in the index. The p53 gene product induces apoptosis in many cell lines; the bcl-2 gene product inhibits or prevents apoptosis.
    3 points for p53, 2 points for bcl-2. If you get them backward, you get 2 points.

    14. Why does the testis weight loss assay not follow simple exponential dose dependence, e.g. ln(W/W0) = -k*D?
    The testes contain at least two different populations of cells with differing sensitivities to radiation, so the weight loss follows a two-component exponential form WD= Wsexp(-ksD) + WIexp(-kID) (eqn. 10.3). If either Ws or WI were zero, i.e. if the initial population of cells contained only radiation-sensitive or radiation-insensitive cells, then the form given in the question would be right.
    3 points for recognizing that there are two different populations of cells with different radiosensitivities; 2 for describing the phenomenon correctly.

    15. (Extra credit)
    (a) Show that dose rate is equal to power absorbed per unit mass of absorber.
    Dose is defined energy absorbed per unit mass of absorber, i.e. D = dE/dm. The rate of anything means the time rate of change of that thing. Assuming that the mass of the absorber isn't changing, then
    dD/dt = d(dE/dm)/dt = d2E/dmdt = d/dm(dE/dt)
    But dE/dt is by definition power, so this is the power absorbed per unit mass.
    Full credit (5 points) for doing it in this elegant way; up to 5 points for a less mathematical but conceptually equivalent explanation.
    (b) Dose rate is given in units of dose per unit time, e.g. Gy s-1. Express this unit in basic MKS (meter-kilogram-second) units. You may treat the coulomb as a basic unit.
    Since a Gy is a J/kg and a J is a kg-m2/s2, one Gy s-1 = 1 (kg-m2/s2) / (kg-s) = 1 m2/s3.
    The coulomb is a red herring; I figure I'm allowed to put red herrings into extra-credit questions.
    3 points for understanding how to do this; 2 more for getting the final answer right.