1 / 51

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.

wynter-graham
Download Presentation

Illinois Institute of Technology

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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

More Related