Transcript Radiation Biophysics: Introduction

Illinois Institute of
Technology
PHYSICS 561
RADIATION BIOPHYSICS:
Review of chapters 1-7
ANDREW HOWARD
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Tonight’s plan
• I want you to prepare calmly for Thursday’s
midterm, so I’ll clarify a couple of points that
may have been left unclear.
• This is not a complete review: it’s a review of
some concepts that I wanted to reemphasize,
and a bit of extra mathematical exposition
• If you don’t happen to see these notes before
Thursday, you won’t have missed anything
essential.
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Literature Reviews
• The first of the homework assignments
involving readings from the peerreviewed literature in the field is
nominally due in a few days
• PDFs of two of the papers are posted
• Don’t rely on every paper being
available that way: if you have access to
a library, use it!
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Radiation Measurement
Units
Quantity
Dose
Definition
Exposure
(e.m. only)
Q/m
ED/m
Energy
Imparted
ED
SI Unit
C/kg
Gray
Joule
Joule/kg
kg-m2/sec2
Definition
Old Unit
Roentgen
Rad
Erg
Definition
1 esu/cm3
100 erg/g
g-cm2/sec2
1 Gy=100 Rad
1 J = 107 erg
Conversion 1 R =
2.58 á10-4C/kg
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Converting ergs to Joules
• 1 J = 1 kg m2 sec-2
• 1 erg = 1 g cm2 sec-2
• 1 kg = 103 g, 1 m = 100 cm,
1 m2 = 104 cm2
• 1 kg m2 = 103 g * 104 cm2 = 107 g cm2
• 1 kg m2 sec-2 = 107 g cm2 sec-2
• 1 J = 107 erg.
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Charged Particle Equilibrium
• CPE exists at a point p
centered in a volume V if
each charged particle
carrying a certain energy
out of V is replaced by
another identical charged
particle carrying the same
energy into V.
• If CPE exists, then dose =
kerma.
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p
Volume = V
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Region of stability and
nuclear decay
N=Z
a
b-
N
• b- decays: down&right.
np+e-+n
_
+
• b decays: up&left: pn+e-+n
• a2+: 2 units down&left: AZQ
A-4Z-2R
• g: no effect here: Q*  Q
b+
Z
88
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Beta Decays
Negative Electron Decay
A
X A Y + b - + n + Q
Positive Electron Decay
A
X A Y + b + + n + Q
p  n 
Spontaneous annihilation
b + + e-  2g
0.5 1 1MeV + 0.5 1 1MeV  1.0 2 2MeV
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Decay of mixtures
• Suppose we have several nuclides
present in the same sample.
• The most common circumstance of this
kind involves an emitter that decays into
something else that decays further, but it
doesn’t have to be that way.
• Total activity is the sum of the activities of
the individual nuclides:
Atotal = A1 + A2 + A3 + …
= l1N1 + l2N2 + l3N3 + …
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Example of mass decrement
•
•
•
•
•
•
4He
is highly stable.
W = 2*1.007276+2*1.008650+2*0.0005486=4.032949 amu
Measured M = 4.00260 so d = W-M = 0.030349 amu
That corresponds to 28.27 MeV, since 1 amu ~ 931 MeV
– ~3% of the rest energy of a nucleon
– ~55.3 * the rest energy of an electron
That’s the energy that would be released if 2 protons, 2
neutrons, and 2 electrons were brought together to form a
helium atom
Mass decrement doesn’t, by itself, serve as a predictor of
stability. But it helps.
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Beta decay for 41Ar to 41K
• Product’s isotopic mass M(41K) =
40.9784 amu
• Starting isotopic mass is M(41Ar) =
40.98108 amu
• Difference in d is therefore 0.00268 amu
• This is spread between the b and the g
• b is 0.00129 amu and g is 0.00139 amu.
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Energy Transferred and
Absorbed
Energy in, out, absorbed, and leaving:
Ein  Etr + Eout
Etr = Eabs + Eleave
so transferred energy is greater than absorbed
energy
We define separate attenuation coefficients:
• Energy transfer attenuation coefficient
• Energy absorbed attenuation coefficient
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Compton Scattering
• The most important of the e-g processes
for hn > 100 KeV is Compton scatter,
especially if the matter is water or tissue
• See fig. 5.2(B) in the text to see why:
µab/r (Compton) predominates above
100KeV
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Attenuation Coefficients for
Molecules (and mixtures)
• Calculate mole fraction fmi for each atom type i in
a molecule or mixture, subject to Sifmi = 1
• Recognize that, in a molecule, fmi is proportional
to the product of the number of atoms of that type
in the molecule, ni, and to the atomic weight of
that atom, mi:
fmi = Qni mi (Q a constant to be determined)
• Thus Sifmi = Si Qni mi = 1 so Q = (Si ni mi)-1
• Then (s/r) for the compound will be
(s/r)Tot = Sifmi(s/r)i
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Calculating Mole Fractions and
Attenuation Coefficients
• Example 1: Water (in book):
– H2: n1 = 2, m1 = 1; O: n2 = 1, m2 = 16
– Q = (Si ni mi)-1= (2*1 + 1 * 16)-1 = 1/18
– Thus fH2 = 2/18, fO = 16/18,
– (s/r)Tot = Sifmi(s/r)I = (2/18)*(0.1129cm2g-1) +
(16/18)(0.0570 cm2g -1)= 0.0632
• Benzene (C6H6):
– C6: n1 = 6, m1 = 12; H6: n2 = 6, m2 = 1
– Q = (6*12+6*1) = 1/78, fC6 = 72/78, fH6 = 6/78
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Interaction of Charged
Particles with Matter
• Recall diagram 5.3, p.84.
• The crucial equation is for E(b), the energy
imparted to the light particle:
E(b) = z2r02m0c4M/(b2E)
where E is the kinetic energy of the moving
particle = (1/2)Mv2.
• Thus it increases with decreasing impact
parameter b
• Energy imparted is inversely proportional to
the kinetic energy E of the incoming heavy
particle!
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Dose
Remember Dose = energy deposited per unit mass.
What is the meaningful size scale for a mammalian
cell?
We’ll need to know this to estimate dose on a cell.
size scales 5mm
r  1 g/cm3 for water or soft tissue
mass of (5mm)3  r
=(5 * 10-4cm)3  r
=125 * 10-12cm3  1g/cm3
=125 * 10-12g = 1.25 * 10-13kg
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Interactions of Energetic Electrons
With Biological Tissue
biol response
• Direct
log - linear
dose - response
e-fast + DNA  DNAbroken+e-fast
e-fast + Protein  Proteinbroken+e-fast
dN
 cons tan t  D
N
Nundamaged  No e -kD
• Indirect Action
H2O* + e-fast
e-fast + H2O
H2O+• + e-H2O+e-fast
further
radical
chemistry
 water *

es  biom olecul
es + H 2O products

 +biom olecul
m olecules

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Indirect action of radiation
• Initial absorption of radiative energy gives rise to
secondary chemical events
• Specifically, in biological tissue
• R + H2O  H2O* (R = radiation)
H2O* + biological macromolecules 
damaged biological macromolecules
• The species “H2O*” may be a free radical or an ion,
but it’s certainly an activated species derived from
water.
• Effects are usually temperature-dependent, because
they depend on diffusion of the reactive species to
the biological macromolecule.
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What’s an immortalized cell
line?
• Certain transformed cell lines lose their
responsiveness to cell-cell communication and to the
apoptotic count
• These cells can replicate without limit
• Often this kind of transformation is associated with
cancer
• It’s always questionable whether experiments on
transformed cell lines are telling us anything useful
about the behavior of untransformed cells
• But we’re somewhat stuck with this kind of system
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Lea’s model for cellular
damage
• Four basic propositions (1955):
– Clonogenic killing is multi-step
– Absorption of energy in some critical volume is step 1
– Deposition of energy as ionization or excitation in the
critical volume will give rise to molecular damage
– This molecular damage will prevent normal DNA
replication and cell division
• Alpen argues that this predates Watson & Crick.
That’s not really true, but it probably began
independent of Watson & Crick
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Lea’s assumptions
• There exists a specific target for the action of
radiation
• There may be more than one target in the
cell, and the inactivation of n of these targets
will lead to loss of clonogenic survival
• Deposition of energy is discrete and random
in time & space
• Inactivation of multiple targets does not
involve any conditional probabilities,
i.e., P(2nd hit) is unrelated to P(1st hit)
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The cellular damage model
• Cell has volume V; target volume is v << V
• Mechanistically we view v as the volume
surrounding the DNA molecule such that
absorption of energy within v will cause DNA
damage.
Sensitive
volume v
Cell, volume V
5 µm
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Single-target, single-hit
model
• In this instance, each hit within the volume
v is sufficient to incapacitate the cell
• Define S(D) as the survival fraction upon
suffering the dose D. Define S0 = survival
fraction with no dose.
• Note that S0 may not actually be 1:
some cells may lack clonogenic capacity
even in the absence of insult
• Then: S/S0 = exp(-D/D0)
• D0 = dose required to reduce survival by
1/e.
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STSH model: graphical
behavior
0
• Slope of curve = -1/D0
• Y intercept = 0
(corresponds to S/S0 = 1)
-1
Slope = -1/D0
D0
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Multi-target, single-hit model
• Posits that n separate targets must be hit
• Probabilistic algebra given in Alpen
• Outcome: S/S0 = 1 - (1 - exp(-qD))n, or for D0=1/q,
S/S0 = 1 - (1 - exp(-D/D0))n
• This model looks at first glance to involve a very
different formula, but it doesn’t, really:
• For n = 1, this is S/S0 = 1 - (1 - exp(-D/D0))1
• But that’s just S/S0 = exp(-qD), i.e. ln(S/S0) = -qD
• That’s the same thing as STSH.
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MTSH algebra
ln(n)
• Physical meaning of exponent n:
• Based on the derivation, it’s the
number of hits required to inactivate
the cell.
• Graphical meaning for n>1: ln(n) =
extrapolation to D=0 of the linear
portion of the ln(S/S0) vs. D curve.
Dose, Gy
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MTSH Asymptotic behavior
• Midway through Tuesday’s lecture we
discussed the fact that the MTSH model
provides for a log-linear relationship
between dose and response for high
doses, namely D >> D0.
• Today I’ll show that.
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Reminder: Taylor series
• Remember that Taylor’s theorem says
that for a function f(x) that is continuous
and has continuous derivatives between
x0 and x, we may write
f(x) = Sk=0∞ f(k)(x)|x=x0 (x-x0)k / k!
• Where f(k)(x)|x=x0 means the kth derivative of
f with respect to x evaluated at x= x0.
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Applying this to (a+bx)n
• Say our function f(x) = (a+bx)n
where n is not necessarily an integer,
and a and b are constants.
• Then for x0 = 0, f(x)|x=x0  f(0)(x)|x=x0 = an
• df/dx = n(a+bx)n-1*b = nb(a+bx)n-1
so f(1)(x)|x=x0 = nban-1
• Similarly d2f/dx2 = nb(n-1)(a+bx)n-2*b =
n(n-1) b2(a+bx)n-2, so f(2)(x)|x=x0 = n(n-1)b2an-2
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General formula for f(k)(x)|x=x0
• f(k)(x)|x=x0 = n(n-1)…(n+1-k)bkan-k
• But in fact that product n(n-1)…(n+1-k)
can be more tidily written n!/(n-k)!
• Thus the Taylor-formula result is
f(x) = Sk=0∞ f(k)(x)|x=x0 (x-x0)k / k!
f(x) = Sk=0∞{ n!/(n-k)! } bkan-k xk / k!
f(x) = Sk=0∞ n!/((n-k)!k!)bkan-k xk
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Polynomials, continued
• That expression n!/((n-k)!k!) appears
routinely in combinatorics: it’s the
number of combinations of n objects
taken k at a time, or nCk. Thus
• f(x) = Sk=0∞ nCk bkan-kxk. This simple
form is known as a binomial expansion,
much-loved by 19th-century thinkers.
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Formulating MTSH result
• Remember that in MTSH,
S/S0 = 1 - (1 - exp(-D/D0))n;
for D >> D0, -D/D0is a large negative
number and exp(-D/D0) is very small.
Therefore an expansion like the one we
just did makes sense,
using x = exp(-D/D0): S/S0 = 1 - (1 - x)n
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Apply to MTSH equation
• S/S0 = 1 - (1 - x)n
• This looks like the form we’ve been using,
with a=1, b=-1, so
• S/S0 = 1 - (Sk=0∞ nCk bkan-kxk)
• S/S0 = 1 - (Sk=0∞ nCk (-1)kxk)
• The first few terms here are
• 1 - {1 - nx + [n(n-1)/2]x2 - [n(n-1)(n-2)/6]x3 +
[n(n-1)(n-2)(n-3)/24]x4 - … }
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Limit for small x, I.e. D >> D0
• If x is small, which corresponds to D >>
D0, we can ignore all but the first two
terms because the subsequent terms
are small compared to the first ones:
• S/S0 = 1 - {1 - nx} = nx = nexp(-D/D0)
• Thus lnS/S0 = ln(nexp(-D/D0)) =
ln(n) + ln(exp(-D/D0)) = lnn - D / D0.
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Defining the threshold dose
• We define the threshold dose Dq to be
the value of D for which the
extrapolated line goes through the X
axis (i.e. ln(S/S0) = 0). Thus:
• ln(S/S0) = 0 = ln(n) - Dq/D0; thus
• Dq = D0ln(n)
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Graphical significance
3
Survival for n=5, D0=2 Gy
ln(n ) = ln(5) = 1.61
1
5
0
15
10
20
25
ln(S/S0)
-1
-3
Dq = D 0 ln(n ) =
2ln(5) = 3.22 Gy
-5
-7
-9
Dose, Gy
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How does S/S0 @ Dq vary
with n?
• Obviously Dq = 0 for n = 1
• Dq > 0 for n > 1.
• At D=Dq, S/S0 = 1-(1-exp(-Dq/D0))n =
1-(1-exp(- D0lnn/D0))n = 1-(1-exp(-ln n))n
• Thus S/S0 = 1-(1-1/n)n
• Obviously this is 1 at n=1 and goes down
from there:
it’s asymptotic to 1-1/e = 0.63212.
• That’s not an accident: e = limn∞(1+1/n)n
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(S/S0 at Dq ) versus n
1
S /S
0
at D
q
Survival fraction
0.9
0.8
0.7
multiplicity, n
0.6
1
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11
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Extrapolating to D=0
3
Survival fo r n=5, D0=2 G y
ln(5)
1
0
5
10
15
20
25
ln(S/S0)
-1
-3
-5
-7
Note that the low-dose limit
doesn’t correspond to
physical reality because the
line is based on D>>D0, but
it’s good to look at it
-9
Dose, Gy
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Low-dose limit for MTSH
with n > 1
• At exactly D=0, S/S0 = 1 as we would expect
• Curve departs from linearity, though
• Slope of ln(S/S0) vs. D curve at low dose:
ln(S/S0) = ln(1 - (1 - exp(-D/D0))n)
• Remembering that d(ln(u))/dx = (1/u)du/dx,
d/dD [(ln(S/S0)] = (1-(1-exp(-D/D0))n)-1*
(0 - (1 - exp(-D/D0))n-1)*(-1/D0)*exp(-D/D0) =
(1-(1-exp(-D/D0))n)-1(- (1 - exp(-D/D0))n-1))*
(-1/D0) exp(-D/D0). For D = 0, this is
• d/dD[ln(S/S0)] = (1-(1-1)n)-1(-(1-1)n-1))(-1/D0)1
= 0.
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So what if the slope is zero?
• It’s been routinely claimed that the flat slope at low
dose is a deficiency in the MTSH model:
• It implies that at very low dose, the exposure has
no effect
• That’s politically unpalatable, and it flies in the
face of some logic.
• BUT it is consistent with the notion that there
might be a “threshold” dose below which not much
happens
• There are a number of circumstances where that
appears to be valid!
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Tobias: Repair-Misrepair
Model
• Posit: linear and quadratic mechanisms up
front
for repair, with explicit time-dependence
• Time-independent formulas arise at times that
are long compared with cell-cycle times
• In those cases
S = exp(-aD)(1+aD/)
where  = l/k is the ratio of the repair rates of
linear damage to quadratic damage.
• This gives roughly quadratic behavior in ln S.
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What are the units of a, b,
and a/b?
• In order for aD to be unitless,
a must be measured in terms of inverse
dose, e.g. a is in Gy-1
• In order for bD2 to be unitless,
b must be measured in terms of inverse
dose squared, e.g. b is in Gy-2.
• Therefore a/b must have dimensions of
dose, i.e units of Gray or rad.
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Modeled significance of a/b
• Suppose we expose a cell line to a dose
equal to a/b.
• Then ln(S/S0) = aD + bD2
= a(a/b) + b(a/b)2 = a2/b + a2/b
• Thus at dose D = a/b,
influence from linear term and
influence from quadratic term are equally
significant
• Thus it’s the crossover point:
– Linear damage predominates for D < a / b
– Quadratic damage predominates for D > a / b
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Homework help
• Recall that one of the problems we
assigned was chapter 5, problem 4:
• Establish that the dimensions given in
Eq. 5.4 for the “classical electron radius”
are correct and show that the value of r0
is 2.817*10-15m.
• That equation is r0 = ke2/(m0c2)
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Dimensions
• Dimensions aren’t units: they’re simply
expressions of the kind of quantity we are
looking at.
• In the equation r0 = ke2/(m0c2), the only tricky
one is k:
• The electron charge e has dimensions of
charge Q; m0 has dimensions of mass M; c
has dimensions of length per unit time, i.e. LT1.
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Dimensions of k
• Recall that the Coulomb-law constant
fits into the equation
F = kq1q2r -2
• So k = Fr2q1-1q2-1
• Therefore the dimensions of k are
dim(k) = dim(Fr2q1-1q2-1)
• But the dimensions of force, F,
are MLT-2, so dim(k)= MLT-2L2Q-2
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Dimensions of r0
• Therefore dim(r0) = dim(ke2/(m0c2))
= MLT-2L2Q-2 * Q2/(M(LT-1)2)
= ML3T-2Q-2Q2/(ML2T-2)
=L
• So we’ve convinced ourselves that the
dimensions of r0 are those of length.
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Getting a value for r0
• We repeat: r0 = ke2(m0c2)-1
• We note: k = 8.98*109 Nm2C-2 =
8.98*109 kg m3s-2C-2 because 1 N = 1 kg ms-2
• e = 1.602*10-19 C
• m0 = 9.11*10-31 kg,
• c = 3.000*108 ms-1 (roughly!)
• Therefore r0 =
8.99*109 kg m3s-2C-2 * (1.602*10-19 C)2 /
(9.11*10-31 kg * (3.000*108 ms-1)2)
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Finishing off the calculation
• r0 = 8.99*(1.602)2/(9.11*3.0002)*
109-19-19+31-8-8) kg m3s-2C-2 C2 kg-1 m-2 s2
• r0 = 0.282 * 1040-54 m = 2.82 * 10-15 m.
• Life is good.
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