Illinois Institute of Technology PHYSICS 561 RADIATION BIOPHYSICS: Review of chapters 1-7 ANDREW HOWARD 02/26/20 Rad Bio: Review of chs. 1 p. 1 of 51 Tonights plan I want you to prepare calmly for Thursdays midterm, so Ill clarify a couple of points that may have been left unclear. This is not a complete review: its a review of some concepts that I wanted to reemphasize, and a bit of extra mathematical exposition

If you dont happen to see these notes before Thursday, you wont have missed anything essential. 02/26/20 Rad Bio: Review of chs. 1 p. 2 of 51 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 Dont rely on every paper being available that way: if you have access to a library, use it! 02/26/20 Rad Bio: Review of chs. 1

p. 3 of 51 Radiation Measurement Units Quantity Dose Definition Exposure (e.m. only) Q/m E /m Energy Imparted E

SI Unit C/kg G ray Joule Joule/kg kg-m 2/sec2 efinition Old Unit R oentgen R ad

Erg efinition 1 esu/cm 3 100 erg/g g-cm 2/sec2 1 Gy=100 Rad 1 J = 107 erg Conversion 1 R = 2.58 10-4C/kg 02/26/20

Rad Bio: Review of chs. 1 p. 4 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. 02/26/20 Rad Bio: Review of chs. 1 p. 5 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. 02/26/20 Rad Bio: Review of chs. 1 p Volume = V p. 6 of 51 Region of stability and nuclear decay

N=Z a - N - decays: down&right. np+e-+n _ -+n decays: up&left: pn+e a2+: 2 units down&left: AZQ A4 Z-2R g: no effect here: Q* Q + Z

88 02/26/20 Rad Bio: Review of chs. 1 p. 7 of 51 Beta Decays Negative Electron Decay A X A Y + + + Q Positive Electron Decay A X A Y + + + + Q ( p n)

Spontaneous annihilation + + e 2 0.511MeV + 0.511MeV = 1.022 MeV 02/26/20 Rad Bio: Review of chs. 1 p. 8 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 doesnt have to be that way. Total activity is the sum of the activities of the individual nuclides:

Atotal = A1 + A2 + A3 + = l1N1 + l2N2 + l3N3 + 02/26/20 Rad Bio: Review of chs. 1 p. 9 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 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 Thats the energy that would be released if 2 protons, 2 neutrons, and 2 electrons were brought together to form a helium atom Mass decrement doesnt, by itself, serve as a predictor of stability. But it helps.

02/26/20 Rad Bio: Review of chs. 1 p. 10 of 51 Beta decay for Ar to K 41 41 Products 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 and the g is 0.00129 amu and g is 0.00139 amu. 02/26/20

Rad Bio: Review of chs. 1 p. 11 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 02/26/20 Rad Bio: Review of chs. 1 p. 12 of 51

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: ab/r (Compton) predominates above 100KeV 02/26/20 Rad Bio: Review of chs. 1 p. 13 of 51 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 02/26/20 Rad Bio: Review of chs. 1 p. 14 of 51 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 02/26/20 Rad Bio: Review of chs. 1 p. 15 of 51 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! 02/26/20 Rad Bio: Review of chs. 1 p. 16 of 51 Dose Remember Dose = energy deposited per unit mass. What is the meaningful size scale for a mammalian cell? Well 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 02/26/20

Rad Bio: Review of chs. 1 p. 17 of 51 Interactions of Energetic Electrons With Biological Tissue biol response Direct log - linear e-fast + DNA DNAbroken+e-fastdose - response e-fast + Protein Proteinbroken+e-fast dN =cons tan t D N Nundamaged =No e kD Indirect Action H2O* + e-fast

e-fast + H2O HO+e +e further radical chemistry + 2 H2O fast water * +biomolecules (biomolecules) + H 2O products molecules 02/26/20

Rad Bio: Review of chs. 1 p. 18 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 its 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. 02/26/20

Rad Bio: Review of chs. 1 p. 19 of 51 Whats 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 Its always questionable whether experiments on transformed cell lines are telling us anything useful about the behavior of untransformed cells But were somewhat stuck with this kind of system 02/26/20 Rad Bio: Review of chs. 1

p. 20 of 51 Leas 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. Thats not really true, but it probably began independent of Watson & Crick 02/26/20

Rad Bio: Review of chs. 1 p. 21 of 51 Leas 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) 02/26/20 Rad Bio: Review of chs. 1 p. 22 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. Sensitive volume v cle Nu us Cell, volume V 5 m 02/26/20

Rad Bio: Review of chs. 1 p. 23 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. 02/26/20 Rad Bio: Review of chs. 1

p. 24 of 51 STSH model: graphical behavior 0 Slope of curve = -1/D0 Y intercept = 0 (corresponds to S/S0 = 1) -1 ln(S/S0) Slope = -1/D0 D0 02/26/20 Dose, Gy

Rad Bio: Review of chs. 1 p. 25 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 doesnt, really: For n = 1, this is S/S0 = 1 - (1 - exp(-D/D0))1 But thats just S/S0 = exp(-qD), i.e. ln(S/S0) = -qD Thats the same thing as STSH. 02/26/20 Rad Bio: Review of chs. 1

p. 26 of 51 MTSH algebra ln(S/S0) ln(n) Physical meaning of exponent n: Based on the derivation, its 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 02/26/20 Rad Bio: Review of chs. 1

p. 27 of 51 MTSH Asymptotic behavior Midway through Tuesdays 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 Ill show that. 02/26/20 Rad Bio: Review of chs. 1 p. 28 of 51 Reminder: Taylor series Remember that Taylors 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. 02/26/20 Rad Bio: Review of chs. 1 p. 29 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 02/26/20 Rad Bio: Review of chs. 1 p. 30 of 51 General formula for f (x)|x=x0 (k) 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

02/26/20 Rad Bio: Review of chs. 1 p. 31 of 51 Polynomials, continued That expression n!/((n-k)!k!) appears routinely in combinatorics: its 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. 02/26/20 Rad Bio: Review of chs. 1 p. 32 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 02/26/20 Rad Bio: Review of chs. 1 p. 33 of 51 Apply to MTSH equation S/S0 = 1 - (1 - x)n This looks like the form weve 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 - } 02/26/20 Rad Bio: Review of chs. 1 p. 34 of 51 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. 02/26/20

Rad Bio: Review of chs. 1 p. 35 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) 02/26/20 Rad Bio: Review of chs. 1 p. 36 of 51 Graphical significance 3

Survival for n=5, D0=2 Gy ln(n ) = ln(5) = 1.61 1 0 5 10 15 20 25 -1

Dq = D 0 ln(n ) = 2ln(5) = 3.22 Gy -3 ln(S/S0) -5 -7 -9 Dose, Gy 02/26/20 Rad Bio: Review of chs. 1 p. 37 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: its asymptotic to 1-1/e = 0.63212. Thats not an accident: e = limn(1+1/n)n 02/26/20 Rad Bio: Review of chs. 1 p. 38 of 51 (S/S0 at Dq ) versus n 1 S /S

0 at D q 0.9 0.8 Survival fraction 0.7 multiplicity, n 0.6 1 02/26/20

6 11 16 Rad Bio: Review of chs. 1 21 p. 39 of 51 Extrapolating to D=0 3 Survival for n=5, D0=2 Gy ln(5) 1

0 5 10 15 20 25 -1 -3 Note that the low-dose limit doesnt correspond to physical reality because the

line is based on D>>D0, but its good to look at it ln(S/S0) -5 -7 -9 Dose, Gy 02/26/20 Rad Bio: Review of chs. 1 p. 40 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. 02/26/20 Rad Bio: Review of chs. 1 p. 41 of 51 So what if the slope is zero? Its 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 Thats 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! 02/26/20 Rad Bio: Review of chs. 1 p. 42 of 51 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/e)e where e = l/k is the ratio of the repair rates of linear damage to quadratic damage. This gives roughly quadratic behavior in ln S. 02/26/20 Rad Bio: Review of chs. 1 p. 43 of 51 What are the units of a, , and a/? 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 D2 to be unitless,

must be measured in terms of inverse dose squared, e.g. is in Gy-2. Therefore a/ must have dimensions of dose, i.e units of Gray or rad. 02/26/20 Rad Bio: Review of chs. 1 p. 44 of 51 Modeled significance of a/ Suppose we expose a cell line to a dose equal to a/ . Then ln(S/S0) = aD + D2 = a(a/) + (a/)2 = a2/ + a2/ Thus at dose D = a/, influence from linear term and influence from quadratic term are equally significant Thus its the crossover point:

Linear damage predominates for D < a / Quadratic damage predominates for D > a / 02/26/20 Rad Bio: Review of chs. 1 p. 45 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) 02/26/20 Rad Bio: Review of chs. 1

p. 46 of 51 Dimensions Dimensions arent units: theyre 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. 02/26/20 Rad Bio: Review of chs. 1 p. 47 of 51 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 02/26/20 Rad Bio: Review of chs. 1 p. 48 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 weve convinced ourselves that the dimensions of r0 are those of length.

02/26/20 Rad Bio: Review of chs. 1 p. 49 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) 02/26/20 Rad Bio: Review of chs. 1

p. 50 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. 02/26/20 Rad Bio: Review of chs. 1 p. 51 of 51