Transcript Illinois Institute of Technology

Illinois Institute of Technology
PHYSICS 561
Radiation Biophysics
Lecture 5:
Survival Curves and Modifiers of
Response
Andrew Howard
17 June 2014
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Looking forward
Be alert for changes in the posted
assignments: I may add a few things
Midterm will cover chapters 1-7 and the bit of
extra material on free radicals that we
discussed last week; therefore, it will not
include the material from today’s lecture
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Survival Curves
We discussed models for cell survival last time
We looked at various ln(S/S0) vs. dose models
and the logic behind them
Today we’ll focus on the graphical implications
and how we can look at the numbers
Then we’ll talk about cell cycles and other good
solid cell-biology topics.
(Warning: I’m more of a biochemist than a cell
biologist, so don’t expect high expertise in this
later section!)
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Errata in Chapter 8
Page 169, Paragraph 2, 1st sentence:
Until the later 1950’s it was not possible to use …
Two sentences later:
Bacteroides  Bacillus
Fig. 8.1, p. 173:
The label that says Dq is pointing at the wrong thing:
it should be pointing at the place where the dashed
line crosses the (Surviving faction = 1.0) value.
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What Fig. 8.1 should have said
100
n
Surviving Fraction
Dq = D0lnn = quasi-threshold
10
Dose, Gy
1
0
2
4
8
10
12
14
Slope = k = 1/D0
0.1
0.01
6
A
B
0.001
0.0001
0.00001
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Shoulder of the Survival Curve
Shoulder of the Survival Curve
ecognize
that
We recognize
thatdosewith MTSH
MTSH
onsedose-response
we have a
we have a region
n where
wherethe
the slope
is closetotozero.
zero.
e is close
We describe
describe
that that
region as a
n asshoulder:
a shoulder:
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Shoulder
region
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Slopes in the MTSH model
Remember that the MTSH model says
ln(S/S0) = ln(1-(1-exp(-D/D0))n)
Because S/S0 = 1-(1-exp(-D/D0))n
So what is the slope of the S/S0 vs. D curve?
… and … what is the slope of the ln(S/S0) vs. D curve?
In particular, what is the slope’s behavior at low dose?
Answer: calculate dS/S0/ dD and d(ln(S/S0))/dD and
investigate their behavior at or near D = 0.
Note: we’re looking here at the n>1 case.
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Slope investigation, part I
For S/S0 itself,
d(S/S0)/dD = d/dD(1-(1-exp(-D/D0))n)
The 1 out front doesn’t affect the derivative:
d(S/S0)/dD = -d/dD(1-exp(-D/D0))n)
We’ll do the rest of this calculation now
based on the general formulas
–
–
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dun/dx = nun-1du/dx
deu/dx = eudu/dx
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Arithmetic & Calculus
of Survival Models
MTSH says S/S0 = 1 - (1-e -D/D0)n
What I want to investigate is the slope at low
dose, I.e. for D << D0,
And at high dose, I.e. for D >> D0.
But are we interested in the slope of S/S0 vs. D
or ln(S/S0) vs. D?
Both!
Slope = derivative with respect to D. So
Slope = d/dD(1 - (1-e -D/D0)n) = -d/dD(1-e -D/D0)n
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MTSH slope, continued
Recalling that dun/dx = nun-1du/dx, for n>1,
Slope = -n(1-e -D/D0)n-1 d/dD (1-e -D/D0)
= -n(1-e -D/D0)n-1 [-d/dD(e -D/D0 )]
But we know deu/dx = eu du/dx,
so d/dD(e -D/D0 ) = e -D/D0 (-1/D0) = -1/D0 e -D/D0
Therefore Slope = -n(1-e -D/D0)n-1 (-)(-1/D0) e -D/D0
i.e. Slope = (-n/D0)(1-e -D/D0)n-1e -D/D0
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Slope at D << D0
This formula for the slope is valid for all
values of D,
including D << D0 and D >> D0
For small D, i.e. for D << D0,
Slope is -n/D0 (1-e -0/D0)n-1e -0/D0
= (-n/D0) (1-e0)n-1e-0; for n > 1 this is
= (-n/D0)(1-1)n-11 = 0. Shazam.
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Slope of ln(S/S0) vs D
The behavior of the slope of the ln(S/S0) vs D
curve is not much harder to determine.
Recall d lnu /dx = (1/u)du/dx. We apply this here:
d ln(S/S0) / dD = (1/(S/S0)) d(S/S0)/dD.
For very small D, S/S0 = 1,
so d ln(S/S0) / dD = (1/1) * d(S/S0)/dD =
d(S/S0)/dD .
But we’ve just shown that that derivative is zero,
so d ln(S/S0) / dD = 0.
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High-dose case
We’ve covered the low-dose case.
What happens at high dose, i.e. D >> D0?
What we’d like to show is that the slope of
lnS/S0 vs. D is -1/D0. Let’s see if we can do that.
Slope = d ln(S/S0) / dD = (1/(S/S0) d/dD(S/S0)
Thus slope =
(1 - (1-e -D/D0)n-1)d/dD(1 - (1-e -D/D0)n)
For D >> D0, D/D0 is large and
-D/D0 is a large negative number;
therefore e-D/D0 is close to zero.
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High-dose case, continued
Slope = limD∞
{(1 - (1-e -D/D0)n)-1(-n/D0)(1-0)n-1e-D/D0)}
That’s messy because the denominator and the
numerator both go to zero. There are ways to do
that using L’Hôpital’s rule, but there are simpler
ways that don’t involve limits.
The trick is to recognize that we can do a
binomial expansion
I’ve done that in the HTML notes
The result will be slope = -1/D0 and ln n is the Y
intercept of the extrapolated curve
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What constitutes a high dose here?
The only scaling of the dose that
occurs in the formula is the value of
D0, so we would expect that we are
in that high-dose regime provided
that D >> D0.
In practice the approximation that
the slope is -1/D0 is valid if D > 5 D0.
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Linear-Quadratic Model
This is simpler. ln(S/S0) = aD + bD2
Therefore slope = d/dD (ln(S/S0)) = d/dD(aD + bD2)
Thus slope = a + 2bD. That’s a pretty simple form.
At low dose, |a| >> 2|b|D, so slope = a.
At high dose (what does that mean?) |a| << 2|b|D,
so slope = 2bD.
What constitutes a high dose?
Well, it’s a dose at which |a| << 2|b|D, so D >> |a / (2b|
Thus if dose >> |a / 2b|, then slope = 2bD.
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Implications of this model
At low dose slope = a is independent of dose
but is nonzero; thus ln(S/S0) is roughly linear
with dose.
At high dose slope = 2bD,
i.e. it’s roughly quadratic.
How can we represent this easily?
We discussed this last time:
ln(S/S0) / D = a + bD,
so by plotting ln(S/S0) / D versus D, we can get
a simple linear relationship.
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LQ: Plot of ln(S/S0) / D versus D
ln(S/S0)/D
Y-intercept = a
Slope = b
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D
By observation, both a and b
are negative. a < 0 tells us
that radiation harms cells;
b < 0 is an observed fact.
Thus the intercept is below
the axis and the slope is
negative.
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LQ graphical analysis: one step further
Dz
ln(S/S0)/D
D
Y-intercept = a
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We can read -a/b directly
off an extrapolation of the
plot: at D = Dz,
ln(S/S0)/D = 0,
a + bDz = 0,
Slope = b
a = -bDz,
Dz = -a / b.
Note a < 0, b < 0,
so -a/b < 0.
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Where do linear and quadratic
responses become equal?
At what dose does the linear response
equal the quadratic response?
At that dose, aD = bD2, D = a/b
So the value we read off the X-intercept of
the previous curve is simply the opposite of
the dose value at which the two influences
are equal.
We mentioned this last time, but we’re
reminding you now
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How plausible is all this?
Model studies suggest reasons to think
that ln(S/S0) = aD + bD2 is a good
approach.
Much experimental data are consistent
with the model
Some of these LQ approaches allow for
time-dependence to be built in.
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LQ vs MTSH and thresholds
Response
How does the question of comparing the
LQ model to the MTSH model relate to
the question of threshold doses?
A typical real-world question is a doseresponse relationship for which the only
reliable experimental results are obtained
at high doses; at lower doses the
confounding variables render the
experiments uninformative.
Dose
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Dose-Response in Epidemiology
Human
health
effect
Experimental data
Extrapolations
#1
#2
#3
Threshold
Baseline
Limit of reliable measurement
#2 is the linear nonthreshold (LNT) model
Is #3 more realistic?
This has regulatory
consequences!
Dose
Tolerable
dose #1
Tolerable
dose #2
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Tolerable
dose #3
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Studying repair
We’ve been suggesting that LQ models
and even some MTSH models are
dependent on the idea that some DNA
damage can be repaired accurately.
Let’s look for approaches to studying
DNA damage that might provide a fuller
understanding of the effects of repair.
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The Elkind-Sutton Experiment
Provides a way of probing repair functions in cells
Procedure:
–
–
Irradiate and establish survival curve
(“conditioning dose”)
Take cells surviving at S=0.1 and subject them to
further irradiation at varying time intervals after
reaching S=0.1
If repair is taking place, then the appearance of a
curve similar to the original shoulder is indicative of
full recovery
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Results
Results
Survival
Curves;
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Physics
561, Summer 2006,
Lecture
6 6/21 Modifiers
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Interpretation
If slope and implied n value are
equivalent to the original curve, then
repair is complete
Smaller n values indicate insufficient
time has elapsed
n=1 implies repair has not begun
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Elkind-Sutton and LET
We might expect more complicated
results if we vary the LET for the two
dosing regimens
Low-LET first, high-LET second gives
two lines of different slope,
independent of the time interval
High-LET first, low-LET second gives
line followed by usual Elkind-Sutton
distribution
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Low-LETfollowed
followedbyby
High-LET
Low-LET
High-LET
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High-LET,
High-LET,then
thenLow-LET
Low-LET
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Physics 561, Summer 2006, Lecture 6 10/21
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The Cell Cycle
Cells have a definite cycle over which specific activities occur.
Particular activities are limited to specific parts of the cycle
Howard and Pelc (1953) characterized four specific phases:
–
–
–
–
M (mitosis, i.e cell division)
G1 (growth prior to DNA replication)
S (synthesis, i.e DNA replication)
G2 (preparation for mitosis)
Mitosis
(M)
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Presynthetic
(G1)
Synthesis
(S)
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Post-synthetic
(G2)
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What happens in S phase?
DNA is replicated; thus, we have twice as much DNA
at the end of S as at the beginning.
During mitosis the two duplexes of DNA can separate
One goes to one daughter cell, the other to the other
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How much time do these segments
take?
Depends on the overall mitotic rate and the
type of cell:
Cell Cycle Times in hours:
Segment
CHO
HeLa
M
1
1
G1
1
11
S
6
8
G2
3
4
-----------------------------------------TOTAL
11
24
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Pie charts of cycle percentages
The point is that different kinds of cells spend
differing amounts of time in the various phases
M
G1
M
G2
G2
G1
HeLa (human)
Chinese Hamster Ovary
S
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S
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Phase Sensitivity
Many cells are much more sensitive to radiation in
some parts of the cell cycle than they are in others.
Why?
–
–
–
Repair is more vigorous in some stages
Unrepaired damage has more opportunity to
manifest itself as clonal alteration close to mitosis
Access of repair enzymes to damaged DNA is
sometimes influenced by how organized the DNA is.
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What phases are sensitive?
In general, cells are radioresistant when they are
synthesizing DNA.
Cells that are synthesizing DNA are taking up label;
the time that they’re doing that is correlated with survival:
% of labeled cells
% of labeled cells
0.4
Surviving Fraction
0
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Time in Hours
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Surviving Fraction
90
24
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Survival Curves in Various Phases
See fig. 8.8:
– Late S is least radiosensitive
– Early S next least
– G-1 somewhat sensitive
– G-2 and M most radiosensitive
M and G2 curves are essentially straight lines
(log-linear dose-response), suggesting that
repair is unavailable or of little influence
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Figure 8.8, reimagined with LQ model
CHO V79 survival curves (fig. 8.8)
Dose, Gy
0
0
log10 of survival fraction
-0.4
2
4
6
8
10
12
Late S:
a=0.16, b=0.024
-0.8
-1.2
Early S:
a=0.26, b=0.022
-1.6
-2
-2.4
G-1:
a=0.4, b=0.02
M and G2:
a=0.75, b=0
-2.8
-3.2
-3.6
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Radiation-Induced
Cell Progression Delay
Note that various biochemical signals regulate
progression from one phase of the cycle to another.
To study this, you need synchronized cells . . .
Sample study (Leeper, 1973):
–
–
–
CHO cells exposed to 1.5 Gy in mid-G1 experienced a
delay of 0.5 h in cell division
1.5 Gy in late S or early G2 caused a delay of 2-3 h
Dose-dependent: (4h for 3Gy, 6-7h for 6 Gy)
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Shape of the culture matters!
How the cells grow influences how much the
cells’ progression is altered by radiation
Monolayers’ progression is altered less than cells
in a multicellular spheroid geometry
0.53 hours per Gray
0.22 hours per Gray
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Is that such a big deal?
Probably not:
The cells in the spheroidal mass divide
half as fast even in the absence of
radiation, possibly due to contact
inhibition.
Therefore it may simply be that the whole
mitotic clock has been slowed down,
including the clock as it’s been influenced
by radiation.
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Causes for these effects
Why are cells more radiosensitive in M and G2?
–
–
Reduced availability of repair enzymes
Repackaged DNA is hard to repair
How is cell progression influenced by radiation?
–
–
–
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Damage to protein kinases and cyclins involved
in cellular checkpoints
Premature degradation of p21, maybe…
Sample 1994 study:
Edgar et al, Genes Dev. 440: 52 (1994)
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Effectors of Radiation Sensitivity
Biological
–
Cells go through life cycles & are
much more sensitive to radiation
damage at some stages than at
others
Chemical
Physical
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Assignment related to amino acids
1. There are exactly twenty amino acids that
serve as the building blocks for proteins in
almost all organisms. The general formula for
19 of these 20 amino acids is +NH3-CHR-COO-,
where R is any of 19 different side-chain
groups. The simplest of these R groups is H, for
which the amino acid is called glycine; the most
complicated is a moiety known as indole, for
which the amino acid is called tryptophan.
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Upcoming Problem 1, cont’d
Much of the chemistry that these R groups participate
in in proteins is ionic in nature, involving charges or
partial charges; but usually these ionic interactions
involve pairs of electrons rather than unpaired
electrons. An exception is the amino acid tyrosine,
which can participate in free-radical (unpaired-electron)
interactions. Draw a structure of the tyrosyl free radical
and explain why it might have a reasonably long
lifetime, as compared to a hydroxyl radical or some
other short-lived free radical.
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Second upcoming problem
2. Most biochemical oxidation-reduction reactions involve
transfers of pairs of electrons and therefore do not involve free
radical mechanisms. A sizeable minority, however, do involve
free radicals. Which of the following biochemical oxidizing or
reducing agents are capable of participating in single-electron
(free-radical) reactions, and which are not? Explain briefly. You
may need to look up the structures of some of these compounds.
(a) ferric iron, Fe3+
(b) nicotinamide adenine dinucleotide, oxidized form (NAD)
(c) flavonamide mononucleotide (FMN)
(also spelled flavin amide mononucleotide; it’s an instance of
what are generally known as flavin prosthetic groups)
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Errata
Page 205, in the EXAMPLE:
4.0 µM l-1 should be 4.0 µM, or 4.0 µmol l-1
Page 206, last sentence:
If the lesions produced by high LET radiation are
predominantly of type II (irrepairable), then m-1 will be
disappearingly small and no oxygen sensitization will
be detectable.
Page 213, last paragraph: cysteine, not crysteine
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Cellular Life Cycles (review)
Phases
Mitotic
–
–
G1
(variable)
Radiation causes 1/2 h delay here
Synthetic
–
(short)
Sensitive
Presynthetic
–
M
S
(4 - 8 h)
G2
(usually short 1 - 2h)
DNA synthesis
Least sensitive
Postsynthetic
Radiation causes 3 - 4 h delay here
– End of G2 sensitive
__________
–
Overall Process
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What happens in G1?
Routine cellular metabolism
Both buildup of new cellular structures and gathering
energy to do so, i.e. both catabolism and anabolism:
Metabolism as a whole consists of:
Catabolism
Anabolism
Energy-producing
Energy-requiring
Breakdown of complex
Build-up of complex
molecules into simpler
molecules from simpler
ones, producing ATP
precursors, using ATP
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A mitotic cell
Other organelles
Mitochondrion
(energy metabolism)
Mitotic Spindle
Ribosomes
(protein synthesis)
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Timescales of damage (fig. 4.1)
Physics (~10-16 s)
– Primary interaction event
with biomolecule or H2O
– Excitations & ionizations
Fast Chemistry (10-15 - 10-7 s)
– Water radicals and other
quasi-stable species form
– Reactive species diffuse and
cause damage
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Slower chemistry
(10-7 to 10-3 s)
– Further damage via
diffusion
– Chemical restitution &
repair
Biochemistry (msec-min):
enzymatic repair of
damage
Biology (hrs-days):
biological repopulation
from surviving cells
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Physical & Chemical Effectors
Water
hn + H2O
free radicals
•OH, •H
•O2-, etc
ionized species
Pure water 1kg
1L
MW (H2O ) = 18
water 18g / mole
1000g
1 kg H2O =
= 55.5 moles
18g / mole
1kg
moles
r=
= 55.5
= 55.5 M
L
L
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How does water matter?
In a dry setting, the damage must be direct
In a wet environment, water-derived free
radicals and ions are the source of much of
the chemical damage
Experiments that can eliminate secondary
(radical-mediated) damage show much
reduced radiosensitivity
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An example from my field
Protein crystals must be irradiated wet because
they fall apart when they’re dry.
Protein crystals irradiated at 300K are
destroyed:
–
–
–
In days on a conventional X-ray source
In minutes on a 2nd-generation synchrotron
In seconds on a 3rd-generation synchrotron
They’re essentially immortal at 110K except on
a 3rd-generation synchrotron source, where
they live for 5-100 minutes
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Radiosensitivity of Enzymes
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11
Radiosensitivity and Hydration
10
D37, dose in megarad
Most enzymes are more
radiosensitive in the
presence of substantial
hydrations than when dried
Data can’t readily be
measured below ~4%
hydration
But those aren’t very
physical anyway (except
maybe with integral
membrane proteins)
9
8
D37, arylesterase
D37, Cholinesterase
7
6
2
8
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14
20
26
Water Content, %
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Why does this work this way?
Normal response
–
–
Arylesterase: less water means less damage
Dry conditions mean that radiosensitivity
depends entirely on direct, not indirect damage
Abnormal response: Cholinesterase
–
–
–
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harder to explain
Augustinsson suggests: at low [H2O], free
radicals are detoxified by nonfunctional
sulfhydryls in the protein
At higher water concentrations the sulfhydryls
can’t get to the damage in time
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Do we care?
Not much:
–
–
We can’t really alter the hydration of most cell
systems without destroying them
We definitely can’t influence the hydration states
of whole organisms without doing damage that is
probably much more significant than that of the
radiation
But these studies may help us understand
mechanisms of damage, and that could be
relevant
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Do we care, vol. II
DNA and RNA are heavily hydrated because they’re
charged and because they tend to associate directly
with solvent or with soluble proteins like histones
Proteins vary a lot in their hydration:
–
–
Soluble proteins are heavily hydrated
Integral membrane proteins have very little water
around them
Is there a difference in radiosensitivity?
–
–
Not much research that I could find
One instance: Takáts et al. (1993) J.Rad.Res. 34:141
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Temperature-sensitivity
Direct damage will be essentially
temperature-independent
Indirect damage should be temperature
dependent because it relies on diffusion of
radicals and ions from the site of their
production to the macromolecule
Ionizations and excitations may display
different temperature dependencies because
once the molecule is excited, its chemistry
may depend on thermal interactions
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Arrhenius plots
We can examine temperature dependence via the Arrhenius
plot, wherein we expect k = Qexp(-G‡/RT)
Thus ln k = ln[Qexp(-G‡/RT)] = lnQ - G‡/RT, so if we plot
ln(k) as a function of 1/T then the relationship should be
linear, and the slope will be -G‡/R, i.e it will be proportional
to the activation energy G‡.
If multiple processes are involved we may get a non-loglinear response.
Over temperature ranges typical of the internals of
homeothermic organisms, we’re not going to see much
effect!
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Physical meaning of G‡
Chemical reactions are
characterized by an activation
energy barrier, i.e., an energy
that must be put into the
system in order to get from
reactant to product or vice
versa.
G‡ is the height of that
activation barrier.
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Energy
G‡
Product
Reactant
Reaction coordinate
Survival Curves; Modifiers
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Radiation & Temperature
Kinetics
lnQ - DG/RT
3.5
0.0028
3
37oC = 310K
27oC = 300K
0.0030
ln(k)
2.5
0.0032
0.0033
2
1.5
0.0037
1
0.0038
0.5
0.0040
p199
T < 100K
100 < T < 170 K
170 < T < 420 K
hn
0
0.0025
0.0027
0.0029
0.0031
0.0033
0.0037
0.0039
0.0041
1/T, K-1
·+
R - H ¾¾® R - H + e - ® R· + H+ + e R - H + ·OH ® R· + H2O
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0.0035
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(direct )
(indirect )
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Radiation and Temperature
Effects in various temperature ranges:
Tmin
Tmax
Effects
0K
100K
temperature-insensitive;
no charge migration, target must be hit.
100K
170K
excitation localized at site;
exciton migration is crucial
170K
420K
disruption of disulfides, ionization
Does this matter much with biological systems (esp. in
homeotherms), where T ~ 310K almost all the time?
Probably not, other than for understanding mechanisms.
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Oxygen and Radiation
FACT: O2 is a radiation-sensitive molecule
e.g. O2 + e-  O2-• (superoxide) 
interactions with macromolecules
It’s tempting to think that all biology occurs in the
presence of 19-21% O2, but it doesn’t!
H2O free-radical chemistry in the presence of O2 is
different from H2O free radical chemistry in the
absence of O2.
(recall Fricke dosimetry story)
P(O2) in tissue varies widely
–
–
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Hemoglobin transports O2
Myoglobin stores O2
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Damage Fixation by Oxygen
R• + O2  RO2•
semi stable
10-7 - 10-3 sec?
_______________________________________
_
Mitigators of
O2 fixation
radical presence
R• + R´SH  R-H + R´-S•
R´-S• + R´-S•  R´-S-S-R´(dimerization)
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Minor syntactic point
Beware of the word fixation
It doesn’t mean correction:
it refers to stabilization
Stabilizing a radical doesn’t make it less
dangerous;
it makes it more dangerous!
So if we say that oxygen is involved in
fixation of damage, we mean that it makes
the damage worse, not better!
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Experiments on
Oxygen Sensitivity
More cells are killed in air than in N2:
Oxygen Enhancement Ratio
1
0
5
10
15
20
25
30
Survival in the
absence of oxygen
Surviving Fraction
0.1
Survival in the
presence of oxygen
0.01
0.001
Dose, Gy
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LQ: oxic vs. anoxic
LQ models: oxic vs anoxic
1
0
5
10
15
20
25
Survival Fraction
0.1
0.01
0.001
Dose, Gy
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Radiosensitivity as function of P(O2)
Shigella experiments show threshold in P(O2)
below which few cells are killed
Alpen, fig. 9.3
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Oxygen Enhancement Ratio
OER 
[dose
in N2 for surviving fraction S/So] /
_____________________________
dose in O2 for surviving fraction S/So
if S = So/10,
then OER = DN / DO
This definition is
somewhat
arbitrary,
but it works!
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Quantitative Oxygen Sensitivity
Paul Howard-Flanders and Tikvah Alper:
Define “S/SN” to be the ratio of the 10% survival
dose under experimental conditions to the 10%
survival dose in N2, i.e. in the absence of oxygen
“S/SN” = (m[O2] + K) / ([O2] + K)
– for any concentration of oxygen.
– m and K are separately determined for each
system
– m is dimensionless, K is a concentration
– m represents maximum relative sensitivity
so m  1.
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Extrema of “S/SN”
At [O2] = 0, S/SN = (m*0 + K) / (0 + K) = 1
For [O2] >> K,
S/SN = (m[O2] + K) / ([O2] + K)
= m[O2] / [O2] = m
Thus justifying description of m as
maximum relative sensitivity
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How to compute m and K
m is available from asymptotic behavior
–
–
It’s equal to S/SN for [O2] >> K.
In practice most systems are equally radiosensitive
from about 1% [O2] on up
To compute K, note that if [O2] = K, then
SK/SN = (mK + K) / (K+K) = (m+1)/2
Therefore if we know m from asymptotic behavior,
we can examine a curve like 9.3 to find the point where
S/SN = (m+1)/2 and read off K from the abscissa
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Howard-Flanders Coefficients
Coefficients m and K for various organisms:
Organism
m
K, µM
Shigella flexneri Y6R
2.9
4.0
Escherichia coli B/r
3.1
4.7
Saccharomyces cerevisiae
2.4
5.8
 Can be read off curves like fig. 9.3
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Curve-fits for Organismal Data
These are idealizations, of course!
Note that 11.3 µM = 1%
3
Shigella
E.coli
Yeast
S/SN
2.5
2
Oxygen Enhancement
1.5
1
0.5
P[O2], µM
0
0
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20
40
60
80
100
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120
p. 75 of 96
Why does this happen?
Alper’s model:
2 types of damage from primary radiation event
–
–
Type I: lesion requires oxygen for lethality
Type II: always lethal, independent of oxygen
Thus: type I can be chemically restituted
Restitution competes with oxygen fixation
Posit: n1 type I lesions, n2 type II lesions,
krep = rate of repair, kfix = rate of O2 fixation, then:
m-1 = n1/n2 and K = krep/kfix
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What happens with high-LET?
Essentially all damage is type 2
Nothing depends on restitution
Therefore n1/n2 << 1, m-1=0, m=1
Thus S/SN = ([O2] + K) / ([O2] + K) = 1,
independent of K.
The fact that high-LET produces this behavior is mildly
supportive of the appropriateness of Alper’s model,
although it’s far from conclusive
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Time-Dependence
Study the kinetics via short bursts of radiation
We look to see whether providing O2 at a certain
time-point before or after irradiation sensitizes cells
–
–
–
Oxygen sensitizes the cell if present before
irradiation
If available till ~3 msec after irradiation, it still matters
If oxygen is made available later than a few msec
post-irradiation, it doesn’t sensitize the cell
This suggests that the oxygen-dependent free
radicals have lifetimes shorter than 3 msec.
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Michael’s results with Serratia
Time constant ~ 0.5 msec for sensitization
Serratia time-dependence
Time, ms
-2.5
-4
-3
Dose: 280 Gy
of electrons
-2
-1
0
1
2
3
4
5
-3.5
log(Surviving Fraction)
-5
-4.5
-5.5
-6.5
-7.5
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What radicals are involved?
The short time-span shown in this experiment suggests
that free radicals are involved:
but which ones?
Michael suggests that superoxide (O2-•), hydroperoxyl
(H-O=O•), and eaq- have shorter lifetimes than is
consistent with 0.5 msec timing
Other researchers continue to plug for superoxide and
hydroperoxyl as candidates
OH• doesn’t interact with O2 that much except via CO2•-:
OH• + HCO2- -> CO2•- + H2O
CO2•- + O2 -> CO2 + O2-•
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Superoxide
We’re still not sure if superoxide is
a major actor in oxygen-mediated damage
But this is how superoxide is detoxified
by superoxide dismutase (SOD):
–
–
2 O2•- + 2H+  H2O2 + O2 (catalyzed by SOD)
Human Cu-Zn SOD
H2O2  H2O + (1/2)O2 (catalyzed by catalase)
Exogenous SOD doesn’t help alleviate damage
(but maybe that’s because it isn’t tied to catalase, so
peroxide can build up to toxic levels?)
Manipulating constitutive levels of SOD by genetic
means gives muddy results
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Thiol Mitigators
We expect that oxygen competes with thiols, such
that the more thiol mitigation is involved, the less
oxygen-dependent fixation of damage can occur:
1. [macro]-R• + R´SH  [macro]-R-H + R´S•
2R´S•  R´-S-S-R´
2. [small]• + R´SH  [small]-H + R´S•
2R´S•  R´-S-S-R´
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Glutathione as mediator
Glutathione can readily dimerize
under oxidizing conditions:
R-SH + HS-R  R-S-S-H
Reasonably prevalent in cells
(millimolar: [glut] >> [cys])
In principle cysteine could also
operate this way, but its cellular
concentration is too low
glutathione
Cysteine
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Are thiols really important?
Some argument about that
Revesz’s results support the competition model
Maybe only exogenous thiols really influence
radiosensitivity
Introducing cysteine or synthetic thiols does definitely
mitigate damage:
RSH + OH•  RS• + H2O
RS- + OH•  RS• + OH. . . And then these RS• radicals recombine as disulfides
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Dose reduction factors
We envision the effects of these exogenous thiols as
involving dose reduction
The quantitation is equivalent to reducing the available
dose to influence the biological system
Example, WR2721 or amifostine:
–
–
–
Works well in vitro
Limited utility in vivo
Fairly high toxicity
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Sensitization by Nitroaromatics


How do we make the cells in the rapidly growing tumor
as rad-sensitive as they would be if PO2 were higher?
Nitroaromatics react with radicals to “fix” (stabilize) the
damage
metronidazole
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misonidazole
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Nitroaromatic action
Like oxygen, nitroaromatics react with short-lived
radicals to produce longer-lived and therefore
more reactive radicals
Concentrations required to enhance
radiosensitivity are much higher than with O2
These compounds have other applications, but
they can be used therapeutically as potentiators
of radiation damage
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Oxygen vs. Mizonidazole (fig. 9.5)
Sensitizers
3
2.8
Enhancement ratio
2.6
2.4
2.2
2
1.8
1.6
oxygen
misonidazole
1.4
1.2
Log([Sensitizer]), M
-7
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-6
-5
-4
-3
1
-2
Response Modifiers; Tumors
-1
0
p. 88 of 96
5-Halogen-substituted pyrimidines
These are molecules that resemble thymine
The halogen at the 5- position looks like the
methyl group in thymine and can be
incorporated in place of thymine in DNA
Most common: 5-bromodeoxyuridine
Sensitization produced by ready reaction with
the aqueous electron
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The 5-BrdU radical
Semi-stable radical is actually resident in the DNA
and can influence chemistry in neighboring bases
Other mechanisms are probably acting too.
O
O
Br
HN
+ eaq
N
R
7/7/2015
-
O
Br
HN
•-

•
HN
+ Br-
N
N
R
R
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Cell Death
Clonogenic cell death:
inability to produce several generations’ worth of
progeny
Acute pathological cell death: necrosis
–
–
Cells typically swell, then lyse
Accompanied by inflammation
Apoptosis
–
–
–
Programmed cell death
Shrinkage, fragmentation, phagocytosis
p53 is activator of genes that regulate it
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Gilbert & Lajtha’s cell types
See next slide for explanation!
A
D
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B
E
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C
F
p. 92 of 96
Cell populations in Tissue
A. Simple transit population
Cells in, cells out
Spermatozoa, blood cells
B. Decaying population
(e.g. oocytes)
C. Closed, static population
(neurons?)
D. Dividing, transit population
–
–
E. Stem cell population
(many kinds)
F. Closed,
dividing population
–
–
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Some cell division,
so more leave than enter
Differentiating blood cells
No cells in or out—
just a lot of division
Tumors, eye-lens
epithelial cells
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Cell population kinetics
Cell types that divide are the most sensitive.
Cells are most sensitive during G2 and M,
so cells that spend a lot of time
in G2 and M are more sensitive
If a cell population is exposed to radiation,
the outcome depends on there being an
adequate number of (clonogenically)
surviving cells.
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Growth Fraction
Lajtha (1963): described “G0” phase in cell cycle:
cell is not engaged in proliferation but could later re-enter
proliferative stage
Growth fraction is defined as fraction of total cellular
population that is clonogenically competent and actually in
the process of DNA replication and cell division.
Measurement: uses 3H-thymidine uptake
Significance: cells in G0 have time to repair DNA damage
–
–
Works even if [repair enzymes] is low during G0
This is suspected but not proven
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The expanded cell cycle (Lajtha)
G0 is seen as an alternative to normal cycling
Cells may re-enter the cycle after a change in
environmental conditions or upon receiving a signal
M
G0
G2
G1
S
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