Illinois Institute of Technology

04/19/23 Rad Bio: Review of chs. 1-7

p. 1 of 51

Illinois Institute of Technology

PHYSICS 561

RADIATION BIOPHYSICS:Review of chapters 1-7ANDREW HOWARD


p. 2 of 51

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.


p. 3 of 51

Literature Reviews

• The first of the homework assignments involving readings from the peer-reviewed 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!

p. 4 of 51Rad Bio: Review of chs. 1-7

04/19/23

Quantity Exposure

(e.m. only)

Dose Energy

Imparted

Definition ΔQ/Δm ΔED/Δm ED

S I Unit C/kg Gray Joule

Definition Joul /ekg k -g m2/sec2

Old Unit Roentgen Rad Erg

Definition 1 esu/cm3 100 erg/g g-cm2/sec2

Conversion 1 =R2.58 ·10-4C/kg

1 Gy=100 Rad 1 J = 107 erg

Radiation Measurement Units


p. 5 of 51

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.


p. 6 of 51

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.

p

Volume = V


p. 7 of 51

Region of stability and nuclear decay

• decays: down&right. np+e-+

• decays: up&left: pn+e-+• 2+: 2 units down&left: A

ZQ A-

4Z-2R

• : no effect here: Q* Q

N

Z

N = Z

88

-

+

_


p. 8 of 51

Beta DecaysNegative Electron Decay

Positive Electron Decay

Spontaneous annihilation

€

A X →A Y + β − + ν + Q

A X →A Y + β + + ν + Q

p → n( )

β + + e− → 2γ

0.511MeV + 0.511MeV =1.022MeV


p. 9 of 51

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 + …= 1N1 + 2N2 + 3N3 + …


p. 10 of 51

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 = 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.


p. 11 of 51

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 is therefore 0.00268 amu

• This is spread between the and the • is 0.00129 amu and is 0.00139 amu.


p. 12 of 51

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


p. 13 of 51

Compton Scattering

• The most important of the e- processesfor h > 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/ (Compton) predominates above 100KeV


p. 14 of 51

Attenuation Coefficients for Molecules (and mixtures)

• Calculate mole fraction fmi for each atom type i in a

molecule or mixture, subject to ifmi = 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 ifmi = i Qni mi = 1 so Q = (i ni mi)-1

• Then (/) for the compound will be(/)Tot = ifmi(/)i


p. 15 of 51

Calculating Mole Fractions and Attenuation Coefficients

• Example 1: Water (in book):– H2: n1 = 2, m1 = 1; O: n2 = 1, m2 = 16

– Q = (i ni mi)-1= (2*1 + 1 * 16)-1 = 1/18

– Thus fH2 = 2/18, fO = 16/18,

– (/)Tot = ifmi(/)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


p. 16 of 51

Interaction of Charged Particles with Matter

• Recall diagram 5.3, p.84.• The crucial equation is for ΔE(b), the energy

imparted to the light particle: ΔE(b) = z2r0

2m0c4M/(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!


p. 17 of 51

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 5m 1 g/cm3 for water or soft tissue

mass of (5m)3 =(5 * 10-4cm)3 =125 * 10-12cm3 1g/cm3

=125 * 10-12g = 1.25 * 10-13kg


p. 18 of 51

Interactions of Energetic Electrons With Biological Tissue

• Direct

e-fast + DNA DNAbroken+e-

fast

e-fast + Protein Proteinbroken+e-

fast

• Indirect Action

H2O* + e-fast

e-fast + H2O

H2O+• + e-H2O+e-

fast

log - lineardose - response

biol response

furtherradicalchemistry

€

water

molecules

⎛

⎝ ⎜

⎞

⎠ ⎟*

+biomolecules → biomolecules( )∗

+ H2O products

kDoundamaged eNNDttancons

N

dN −=•=


p. 19 of 51

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.


p. 20 of 51

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


p. 21 of 51

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


p. 22 of 51

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)


p. 23 of 51

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.

Cell, volume VNucleusSensitive

volume v

5 µm


p. 24 of 51

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.


p. 25 of 51

STSH model: graphical behavior

• Slope of curve = -1/D0

• Y intercept = 0(corresponds to S/S0 = 1)

ln(S

/S0)

Dose, Gy

0

Slope = -1/D0-1

D0


p. 26 of 51

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.


p. 27 of 51

MTSH algebra

• 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.

ln(S

/S0)

Dose, Gy

ln(n)


p. 28 of 51

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.


p. 29 of 51

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) = k=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.


p. 30 of 51

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


p. 31 of 51

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) = k=0∞ f(k)(x)|x=x0

(x-x0)k / k!

f(x) = k=0∞{ n!/(n-k)! } bkan-k xk / k!

f(x) = k=0∞ n!/((n-k)!k!)bkan-k xk


p. 32 of 51

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) = k=0∞

nCk bkan-kxk. This simple form is known as a binomial expansion, much-loved by 19th-century thinkers.


p. 33 of 51

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


p. 34 of 51

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 - (k=0∞

nCk bkan-kxk)

• S/S0 = 1 - (k=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.


p. 36 of 51

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)


p. 37 of 51

Graphical significanceSurvival for n=5, D0=2 Gy

-9

-7

-5

-3

-1

1

3

0 5 10 15 20 25

Dose, Gy

ln(S/S0)

ln(n ) = ln(5) = 1.61

Dq = D0 ln(n ) =2ln(5) = 3.22 Gy


p. 38 of 51

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


p. 39 of 51

(S/S0 at Dq ) versus n

S /S 0 at D q

0.6

0.7

0.8

0.9

1

1 6 11 16 21

multiplicity, n

Survival fraction


p. 40 of 51

Extrapolating to D=0Survival for n=5, D0=2 Gy

-9

-7

-5

-3

-1

1

3

0 5 10 15 20 25

Dose, Gy

ln(S/S0)

ln(5)

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


p. 41 of 51

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.


p. 42 of 51

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!


p. 43 of 51

Tobias: Repair-Misrepair Model

• Posit: linear and quadratic mechanisms up frontfor repair, with explicit time-dependence

• Time-independent formulas arise at times that are long compared with cell-cycle times

• In those casesS = exp(-D)(1+D/)where = /k is the ratio of the repair rates of linear damage to quadratic damage.

• This gives roughly quadratic behavior in ln S.


p. 44 of 51

What are the units of , , and /?

• In order for D to be unitless, must be measured in terms of inverse dose, e.g. is in Gy-1

• In order for D2 to be unitless, must be measured in terms of inverse dose squared, e.g. is in Gy-2.

• Therefore / must have dimensions of dose, i.e units of Gray or rad.


p. 45 of 51

Modeled significance of /

• Suppose we expose a cell line to a dose equal to /.

• Then ln(S/S0) = D + D2

= (/) + (/)2 = 2/ + 2/• Thus at dose D = /,

influence from linear term andinfluence from quadratic term are equally significant

• Thus it’s the crossover point:– Linear damage predominates for D < / – Quadratic damage predominates for D > /


p. 46 of 51

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)


p. 47 of 51

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. LT-1.


p. 48 of 51

Dimensions of k

• Recall that the Coulomb-law constant fits into the equationF = kq1q2r -2

• So k = Fr2q1-1q2

-1

• Therefore the dimensions of k aredim(k) = dim(Fr2q1

-1q2-1)

• But the dimensions of force, F,are MLT-2, so dim(k)= MLT-2L2Q-2


p. 49 of 51

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.


p. 50 of 51

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)


p. 51 of 51

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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