(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:
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.