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Illinois Institute of Technology

Illinois Institute of Technology. PHYSICS 561 RADIATION BIOPHYSICS: Review of chapters 1-7 ANDREW HOWARD. 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.

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Illinois Institute of Technology

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  1. Illinois Institute of Technology PHYSICS 561 RADIATION BIOPHYSICS:Review of chapters 1-7ANDREW HOWARD Rad Bio: Review of chs. 1-7

  2. 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. Rad Bio: Review of chs. 1-7

  3. 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! Rad Bio: Review of chs. 1-7

  4. Radiation Measurement Units Rad Bio: Review of chs. 1-7

  5. 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. Rad Bio: Review of chs. 1-7

  6. 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 Rad Bio: Review of chs. 1-7

  7. Region of stability and nuclear decay • 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 N = Z a b- b+ N _ Z 88 Rad Bio: Review of chs. 1-7

  8. Beta Decays Negative Electron Decay Positive Electron Decay Spontaneous annihilation Rad Bio: Review of chs. 1-7

  9. 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 + … Rad Bio: Review of chs. 1-7

  10. 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. Rad Bio: Review of chs. 1-7

  11. 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. Rad Bio: Review of chs. 1-7

  12. 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 Rad Bio: Review of chs. 1-7

  13. Compton Scattering • The most important of the e-g processesfor 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 Rad Bio: Review of chs. 1-7

  14. 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 Rad Bio: Review of chs. 1-7

  15. 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 Rad Bio: Review of chs. 1-7

  16. Interaction of Charged Particles with Matter • Recall diagram 5.3, p.84. • The crucial equation is for DE(b), the energy imparted to the light particle:DE(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! Rad Bio: Review of chs. 1-7

  17. 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 forwaterorsofttissue mass of (5mm)3 r =(5 * 10-4cm)3  r =125 * 10-12cm3  1g/cm3 =125 * 10-12g = 1.25 * 10-13kg Rad Bio: Review of chs. 1-7

  18. Interactions of Energetic Electrons With Biological Tissue biol response • 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 - linear dose - response further radical chemistry Rad Bio: Review of chs. 1-7

  19. 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. Rad Bio: Review of chs. 1-7

  20. 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 Rad Bio: Review of chs. 1-7

  21. 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 Rad Bio: Review of chs. 1-7

  22. 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) Rad Bio: Review of chs. 1-7

  23. 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 V Nucleus Sensitive volume v 5 µm Rad Bio: Review of chs. 1-7

  24. 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. Rad Bio: Review of chs. 1-7

  25. STSH model: graphical behavior • Slope of curve = -1/D0 • Y intercept = 0(corresponds to S/S0 = 1) 0 -1 Slope = -1/D0 ln(S/S0) Dose, Gy D0 Rad Bio: Review of chs. 1-7

  26. 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. Rad Bio: Review of chs. 1-7

  27. 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(n) ln(S/S0) Dose, Gy Rad Bio: Review of chs. 1-7

  28. 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. Rad Bio: Review of chs. 1-7

  29. 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 writef(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. Rad Bio: Review of chs. 1-7

  30. Applying this to (a+bx)n • Say our function f(x) = (a+bx)nwhere 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-1so 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 Rad Bio: Review of chs. 1-7

  31. 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 isf(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 Rad Bio: Review of chs. 1-7

  32. 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. Rad Bio: Review of chs. 1-7

  33. 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 Rad Bio: Review of chs. 1-7

  34. 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 - … } Rad Bio: Review of chs. 1-7

  35. 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. Rad Bio: Review of chs. 1-7

  36. 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) Rad Bio: Review of chs. 1-7

  37. Graphical significance Rad Bio: Review of chs. 1-7

  38. 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 Rad Bio: Review of chs. 1-7

  39. (S/S0 at Dq ) versus n Rad Bio: Review of chs. 1-7

  40. Extrapolating to D=0 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 Rad Bio: Review of chs. 1-7

  41. Low-dose limit for MTSH with n > 1 • At exactlyD=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. Rad Bio: Review of chs. 1-7

  42. 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! Rad Bio: Review of chs. 1-7

  43. 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. Rad Bio: Review of chs. 1-7

  44. 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. Rad Bio: Review of chs. 1-7

  45. Modeled significance of / • Suppose we expose a cell line to a dose equal to /. • Then ln(S/S0) = D + D2= (/) + (/b)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 >  /  Rad Bio: Review of chs. 1-7

  46. 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 r0is 2.817*10-15m. • That equation is r0 = ke2/(m0c2) Rad Bio: Review of chs. 1-7

  47. 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. Rad Bio: Review of chs. 1-7

  48. 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 Rad Bio: Review of chs. 1-7

  49. 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. Rad Bio: Review of chs. 1-7

  50. 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) Rad Bio: Review of chs. 1-7

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