Illinois Institute of Technology

Illinois Institute of Technology

Illinois Institute of Technology RADIATION BIOPHYSICS Fourth Lecture: Radiation and Molecular Biology; Survival Models ANDREW HOWARD Thursday 12 June 2014 01/26/20 Rad Bio: mol bio and survival p. 1 of 97 Survival Cells sometimes survive radiation exposure, and sometimes they dont; we need to develop mathematical models for survival but first, I want to say a few things about DNA and chromosomes

01/26/20 Rad Bio: mol bio and survival p. 2 of 97 What well discuss Discussion of homework and relativity DNA, chromosomes, and DNA repair What we mean by survival Models for survival Target theory STSH MTSH Molecular models

Linear-quadratic models of various sorts 01/26/20 Rad Bio: mol bio and survival p. 3 of 97 Bookkeeping Generally Im happy with your homework performance: most of you have turned in the first assignment Dont obsess over getting them in on time, but get caught up by the first midterm Midterm (Wed-Thu 6/18 6/19)

Get a proctor! Closed book, closed notes, but a help-sheet will be provided Covers chapters 1-7 My midterms are long but not hard: budget your time accordingly I will proctor Wed 11am-12:15pm and Thu 1-2:15pm Chuck Scott will proctor late Wednesday afternoon 01/26/20 Rad Bio: mol bio and survival p. 4 of 97 Typos in Alpen Alpen seems to have replaced D with D! at least twice

Several terms have too many factorial signs in them; Eqn. 7.3 should be P(,h,D) = (DCh)(h)(1-)(D-h)(H(h)) Eqn. 7.4 should be S(,D) = h=0h=D P(,h,D) Axis labels are faulty sometimes too: Pp. 136-137: the lowest number on the Y axis should be 0.01, not 0.001 01/26/20 Rad Bio: mol bio and survival p. 5 of 97 The APS problem

Remember we said we have 7 GeV electrons, and we asked you to calculate c-v. Keep the calculation in algebra until the very end! So: E = m0c2 = (1-v2/c2)-1/2m0c2 Therefore (1-v2/c2)-1/2 = E/m0c2 Squaring both sides, (1-v2/c2)-1 = (E/m0c2)2 Inverting, 1-v2/c2 = (m0c2/E)2, so1 - (m0c2/E)2 = v2/c2 Hence v = c(1 - (m0c2/E)2 )1/2 Therefore c - v = c - c(1 - (m0c2/E)2 )1/2, i.e. c - v = c[1 - (1 - (m0c2/E)2 )1/2] 01/26/20 Rad Bio: mol bio and survival p. 6 of 97 APS problem, continued

Use our preloaded knowledge that the rest energy of the electron, m0c2, is 0.511004 MeV, so m0c2/E = 0.511004 MeV/7000 MeV = 7.300*10-5 (unitless!) Therefore (m0c2/E)2 = 5.3291 * 10-9, so 1 - (m0c2/E)2 = 0.99999999467, (1 - (m0c2/E)2)1/2 = 0.99999999734, 1- (1 - (m0c2/E)2)1/2 = 2.6645418 * 10-9. Hence c-v = 2.99797*108 ms-1 * 2.6645418*10-9 i.e. c - v = 0.7988 ms-1, a moderate walking pace. 01/26/20 Rad Bio: mol bio and survival p. 7 of 97 APS acceleration

This part doesnt need to be viewed relativistically: Acceleration a = v2/r Weve just convinced ourselves that v is pretty close to c, so we can ignore that tiny difference. Circumference of the APS is 1100m, so radius = C/2 = 1100m/2 a = c2/r = (2.99797*108 ms-1)2 / (1100m /2) = 2 * 8.171 * 1013 ms-2 = 5.11 * 1014 ms-2 = 5.239 * 1013 * g The force exerted on the electron is F = ma = m0a = (1-v2/c2)-1/2m0a 01/26/20 Rad Bio: mol bio and survival p. 8 of 97 Relativistic kinetic energy Ive mentioned in passing that, according to special relativity, the formula for energy is E = m0c2 = (1-2)-1/2m0c2 = (1-v2/c2)-1/2m0c2

Where = (1-2)-1/2 = (1-v2/c2)-1/2m0c2 and = v/c Our intuition should tell us that this total energy consists of mass energy, m0c2, and something else that is related to movement, namely, kinetic energy. The kinetic energy, then, should be KE = E - m0c2 = m0c2 - m0c2 = (-1)m0c2 01/26/20 Rad Bio: mol bio and survival p. 9 of 97 So what? Our nonrelativistic formula for kinetic energy, which you all learned in the sixth week of freshman physics, is

KE = (1/2)mv2, or using the notation were using here, KE = (1/2)m0v2. Does our relativistic formula reduce to that in the low-speed limit, i.e. when v << c? Lets convince ourselves of that, both because its interesting and to give us some practice using Taylor (or Maclaurin) expansions 01/26/20 Rad Bio: mol bio and survival p. 10 of 97 Maclaurin expansion Remember that the Taylor expansion of a continuous function f(x) about a point x0 is f(x) = f(x0) + [(df/dx)|x0](x-x0) + (1/2)[d2f/dx2|x0](x-x0)2 + (1/6)[d3f/dx3|x0](x-x0)3 + higher-order terms,

Where z|y means z evaluated at x=y. Or, more precisely, f(x) = f(x0) + k=1(1/k!)[dkf/dxk|x0](x-x0)k In the specific case x0 = 0, its a Maclaurin expansion f(x) = f(0) + k=1(1/k!)[dkf/dxk|0]xk 01/26/20 Rad Bio: mol bio and survival p. 11 of 97 Using this for kinetic energy

To apply this to the kinetic energy problem, we define a variable u = v2/c2 = 2, so that KE = (-1)m0c2 = [(1-u)-1/2 - 1]m0c2 And we will expand that in u around u = 0: KE = KE(0) + d/du { [(1-u)-1/2 - 1]m0c2}|0}*u + higherorder terms. But KE(0) = 0, clearly, since if u=0, = 1. So KE = d/du { [(1-u)-1/2 - 1]m0c2|0}*u + higher-order terms. 01/26/20 Rad Bio: mol bio and survival p. 12 of 97 So does this work?

d/du[(1-u)-1/2] = (-1/2)(1-u)-3/2(-1) = (1/2)(1-u)-3/2 Evaluating that at u=0, d/du[(1-u)-1/2]|0 = (1/2)(1) = 1/2. Therefore KE = d/du { [(1-u)-1/2 - 1]m0c2}|0}*u + h.o.t. = (1/2)m0c2 * u = (1/2) m0c2 * (v2/c2) = (1/2) m0v2 which is what we were hoping for. You can do this directly as an expansion in v, but its a little bit harder. 01/26/20 Rad Bio: mol bio and survival p. 13 of 97 APS problem reconsidered Remember that we wound up with a number very

close to zero for (m0c2/E)2, and we were at the mercy of our calculators ability to keep a whole lot of zeros. Is there a way to do this that doesnt depend on 10digit calculators? Yes: For small x, (1-x)n = 1 - nx This is a binomial expansion, (W.S. Gilbert: About binomial theorem Im teeming with a lot of news) which is just a special case of a Taylor series. 01/26/20 Rad Bio: mol bio and survival p. 14 of 97 How do we use this? Well, remember that our equation is c - v = c[1 - (1 - (m0c2/E)2 )1/2] But were saying x = (m0c2/E)2 << 1, so

c - v = c[1 - (1 - x)1/2] = c[1 - (1- x/2)] = cx/2 Remember that x = 5.3291 * 10-9 Therefore c - v = 2.99797 * 108 ms-1 * 5.3291 * 10-9 / 2 Thus c - v = 0.7988 ms-1 Thats identical to what we got with all those 9s on our calculator (0.7988 ms-1) differences are in the next decimal place or two 01/26/20 Rad Bio: mol bio and survival p. 15 of 97 Force on the electron We said F = ma = m0a = (1-v2/c2)-1/2m0a

But (1-v2/c2)-1/2 = E / m0c2 = 7000 MeV/0.511004 MeV = 13698, so F = (1-v2/c2)-1/2m0a = 13698 * 9.11*10-31 kg * 5.11 * 1014 ms-2 = 637695 * 10-17 N = 6.38 * 10-12 N Comparable to the forces measured in some atomic force microscopy experiments Thats a small force, but its acting on a very small object! 01/26/20 Rad Bio: mol bio and survival p. 16 of 97 Some definitions from biochemistry A catalyst is a species that increases the rate of a reaction without ultimately being changed in the net reaction

An enzyme is a biological macromolecule capable of catalysis Most enzymes are proteins In the last two decades we have begun to study RNA molecules that function as enzymes (may do damage surveillance in RNA?) 01/26/20 Rad Bio: mol bio and survival p. 17 of 97 Catalytic vs. noncatalytic mechanisms Uncatalyzed reaction: A B (slow) Catalyzed reaction, with E as the catalyst: A + E A-E B-E B + E

All three of these reactions are likely to be much faster than the uncatalyzed reaction (~ 107 times faster) 01/26/20 Rad Bio: mol bio and survival Hermann Emil Fischer p. 18 of 97 DNA nucleotides W. H. Brown & J. A. McClarin, Introduction to Organic and Biochemistry, 3rd Ed., 1981 01/26/20 Rad Bio: mol bio and survival p. 19 of 97 Protein Backbones and Nucleic Acid Backbones

R1, R2 = one of 20 amino acid sidechains (H, CH3, CH2OH, ) Image courtesy McGrawHill Education. The circled P indicates a phosphate linkage, i.e.. 01/26/20 Rad Bio: mol bio and survival p. 20 of 97 Realities of DNA damage Covalent damage to one strand doesnt always result in failure to replicate correctly

but it increases the rate of copying errors Base-pairing can be destroyed by covalent damage A-T pairs (2 hydrogen bonds per base pair) are more fragile than C-G pairs (3 H-bonds / pair) 01/26/20 Rad Bio: mol bio and survival H.J. Muller: characterized DNA damage from X-radiation p. 21 of 97 Reactions of Radiation with DNA Single-strand breaks and double-strand breaks SSB High-reliability

enzymatic repair DSB caused by a single event Lower-reliability enzymatic repair 01/26/20 Rad Bio: mol bio and survival p. 22 of 97 DSBs from two events DSB caused by two neighboring events Moderate-reliability enzymatic repair If the breaks are not at the same position, they can be more readily repaired: 5-C-G-A-T-C-C-G-A-3 5-C-G-A- -C-C-G-A-3 3-G-C-T-A-G-G-C-T-5 3-G-C-T-A- -G-C-T-5

01/26/20 Rad Bio: mol bio and survival p. 23 of 97 Chemistry of DNA damage Damage to sugars and bases (not removed but covalently damaged) Loss of base (apurination or apyrimidation) Strand scission due to radical chemistry on a nucleotide Single-strand breaks on the backbone Double-strand breaks (see above) on the backbone

01/26/20 Rad Bio: mol bio and survival p. 24 of 97 Chromatin This refers to DNA in a cell. In between cell divisions, DNA is spread out in the nucleus. At a particular stage in the cell cycle, the DNA becomes highly coiled and organized in preparation for replication. Thomas Hunt Morgan, pioneer in the understanding of the role of chromosomes in heredity 01/26/20 Rad Bio: mol bio and survival

p. 25 of 97 How does chromatin become organized? At the lowest level of organization, ~200 base-pairs of DNA wrap themselves around a group of nitrogen-rich proteins called histones which have been organized into the nucleosome core particle That interaction is stabilized by charge-charge interactions between the negatively-charged phosphate groups in the DNA and positively charged amino acids in the histone 01/26/20 Rad Bio: mol bio and survival p. 26 of 97

Higher levels of organization Neighboring nucleosomes group together to form even higher levels of coiling through an interaction with another histone, H1.This forms a solenoidlike structure. 01/26/20 Rad Bio: mol bio and survival p. 27 of 97 Why does this matter? DNA tends to be more radiationsensitive when it is more organized

More tightly packedharder for the repair enzymes to get access to the lesions Closer to the time of replication So we need to be conscious of these levels of organization 01/26/20 Rad Bio: mol bio and survival p. 28 of 97 DNA repair All organisms have some DNA repair mechanisms. Eukaryotes have more and nimbler repair systems than prokaryotes DNA repair enzymes can recognize and repair

Single-strand breaks (SSBs) Double-strand breaks (DSBs) Chemically altered bases Chemically altered sugars Damage to DNA-related proteins (e.g. histones) 01/26/20 Rad Bio: mol bio and survival p. 29 of 97 DNA repair, continued Some mechanisms are more error-prone than others Certain kinds of damage are effectively

irreparable Repair in eukaryotes is much more effective than in prokaryotes 01/26/20 Rad Bio: mol bio and survival p. 30 of 97 Excision repair The least error-prone type of repair It relies on the complementary strands base to define which base (dA, dC, dG, or dT) to insert Characteristic of repair of certain kinds of damage, like pyrimidine dimers; but it can come into use whenever the damage involves a single base on one strand Missing Base Template Strand

01/26/20 Rad Bio: mol bio and survival p. 31 of 97 Other forms of repair Error-prone repair: recA and similar mechanisms Underlying notion: any base is better than no base. This is particularly true if the replaced base happens to be the wobble (3rd) base: changes at the 3rd base often are harmless, or nearly so Recombination repair: see figure 6.5.

Recombination repair: image from answers.com 01/26/20 Rad Bio: mol bio and survival p. 32 of 97 How do we define survival? Its harder at the cellular level than you might think. It takes a lot of radiation to destroy metabolism. It takes a lot less to compromise DNA replication badly enough to either: Prevent replication after 1-4 generations Produce large changes in morphology or function, again after 1-4 generations

Therefore: we concentrate on clonogenic survival as a definition for cell survival 01/26/20 Rad Bio: mol bio and survival p. 33 of 97 What kind of experiments do we envision here? A few cells placed on a growth medium Cells are exposed to a toxicant or to radiation Cells allowed to divide for several generations Compare number of progeny in the treated cell group to number in untreated group Damage is said to be significant if the treated group produces fewer progeny than the control group

01/26/20 Rad Bio: mol bio and survival p. 34 of 97 What do we mean by clonogenic survival? Clean definition: clonogenic survival is the ability to produce six generations of viable offspring This works well for prokaryotic cells and cultured eukaryarotic cells, particularly immortalized ones 01/26/20 Rad Bio: mol bio and survival p. 35 of 97

Clonogenic survival in differentiated eukaryotic cells The definition works less well for differentiated eukaryotic cells: A respectable eukaryotic cell has a chromosomal component called a telomere that regulates the number of cell divisions before the cell undergoes programmed cell death (apoptosis) If the cell youre studying is close to its natural cutoff point for cell divisions, its unfair to blame the treatment for its inability to produce 5 generations of progeny! 01/26/20 Rad Bio: mol bio and survival p. 36 of 97 Six generations

0 Roughly corresponds to 50 surviving progeny 1 2 01/26/20 3 4 5 6 Rad Bio: mol bio and survival p. 37 of 97 Contact inhibition

Many cells change behavior when they come into contact with neighbors Often the change involves inhibition of replication That complicates the definition of clonogenic survival: If the cells stop dividing because theyre getting too crowded, its unfair to blame that on the treatment! Changes 01/26/20 Rad Bio: mol bio and survival p. 38 of 97 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 01/26/20 Rad Bio: mol bio and survival p. 39 of 97 Mechanisms of Reproductive Cell Survival and Death Up until around 1970 there were two

highly disparate lines of research surrounding these issues: ln(Survival fraction) Modelers, who carried out mathematical studies of dose-response; Biologists, who sought understanding of the mechanisms of the cellular response Enzymatic Molecular-biological Dose Since 1970 there has been better communication between these two communities

01/26/20 Rad Bio: mol bio and survival p. 40 of 97 Sorting out multiple causes can be tricky. Ancient study of uranium mine workers: Status Smoking Non-smoking Miner 1 2 Non-miner 3 4 Result: cancer(1) > cancer(3) >> cancer(2) ~ cancer(4)

So the effect of mining is potentiated by smoking Wed like to know why! 01/26/20 Rad Bio: mol bio and survival p. 41 of 97 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 More details about his assumptions on next slide. Alpen argues that this predates Watson & Crick. Thats not really true, but it probably began independent of Watson & Crick 01/26/20 Rad Bio: mol bio and survival p. 42 of 97 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) 01/26/20 Rad Bio: mol bio and survival p. 43 of 97 The role of DSBs We will eventually want to emphasize unrepairable DNA damage as the true bad actor in all of this We saw at the end of last class that double strand breaks are harder to repair with high fidelity

So DSBs are likely to be the real issue here You can begin to see the utility of an interaction between the modelers and the biologists! 01/26/20 Rad Bio: mol bio and survival p. 44 of 97 Log-linear response ln(Survival fraction) 0 With cells that are distinctly deficient in DSB repair (e.g., bacterial cells): Log-linear dose-response to

radiation over several logs ln(N/N0) = -D/D0 N0 is the number of cells in the absence of treatment -12 0 01/26/20 Dose, Gy 40 Rad Bio: mol bio and survival p. 45 of 97 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 01/26/20 Rad Bio: mol bio and survival p. 46 of 97 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. 01/26/20 Rad Bio: mol bio and survival p. 47 of 97 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 01/26/20 Dose, Gy Rad Bio: mol bio and survival p. 48 of 97 When is STSH insufficient? Any situation in which more than one hit might be necessary, or in which the repair capacity of the cell

includes the ability to eliminate certain kinds of damage, may result in a plot of ln(S/S0) vs. D that isnt linear. We seek plausible mechanistic mathematical models that can account for this behavior In practice the Multitarget Single-hit (MTSH) and Linear-Quadratic (LQ) models have been used most 01/26/20 Rad Bio: mol bio and survival p. 49 of 97 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 01/26/20

Rad Bio: mol bio and survival p. 50 of 97 STSH as a special case 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; so STSH is a special case of MTSH with n=1. 01/26/20

Rad Bio: mol bio and survival p. 51 of 97 MTSH algebra ln(n) ln(S/S0) 0 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 01/26/20

Rad Bio: mol bio and survival p. 52 of 97 A semi-real case: n=5, D0= 2 Gy 3 Survival for n=5, D0=2 Gy 1 0 5 10 15 20 25 ln(S/ S0) -1

-3 -5 -7 -9 Dose, Gy 01/26/20 Rad Bio: mol bio and survival p. 53 of 97 Behavior of this function for D >> D0 For D >> D0, exp(-D/D0) << 1 so we can expand it: (1- exp(-D/D0))n = (1 - x)n ~ 1 - nx for x = exp(-D/D0) << 1

Therefore 1 - (1- exp(-D/D0))n = nx = nexp(-D/D0) Thus ln(S/S0) = ln(1 - (1- exp(-D/D0))n) = ln(nexp(-D/D0) = ln n - D / D0 So the behavior for high doses is log-linear with slope = -1/D0, just as in the STSH model, But with Y intercept = ln n rather than 0. 01/26/20 Rad Bio: mol bio and survival p. 54 of 97 Extrapolating to D=0

3 Survival for n=5, D0=2 Gy ln(5) 1 0 5 10 15 20 25 ln(S/ S0) -1 -3 -5

-7 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 -9 01/26/20 Dose, Gy Rad Bio: mol bio and survival p. 55 of 97 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 = 1-1 * 0 * (-1/D0) = 0. 01/26/20 Rad Bio: mol bio and survival p. 56 of 97 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! 01/26/20 Rad Bio: mol bio and survival p. 57 of 97 MTSH Quasi-Threshold Dose We note that the curve stays close to linear

until we get to fairly low doses. We describe Dq = dose at which the linear extrapolation hits ln(S/S0) = 0, i.e. S=S0: Since the line is ln(S/S0) = ln n - D/D0, 0 = ln n - Dq / D0, so Dq = D0ln n 01/26/20 Rad Bio: mol bio and survival p. 58 of 97 Quasi-Threshold Dose Graphically 3 Survival for n=5, D0=2 Gy 1 0 5 10 15

20 25 ln(S/ S0) -1 -3 Dq=D0 ln(n) = 2*ln(5) = 3.21Gy -5 -7 -9 01/26/20 Dose, Gy Rad Bio: mol bio and survival p. 59 of 97

Deficiencies in the MTSH model Zero slope at zero dose (is that really bad?) We can tweak this if we need to: S/S0 = exp(-q1D)(1 - (1-exp(-D/D0)n) ln(S/S0) has slope -q1 at D=0. High-dose behavior: Does it remain truly linear at D >> D0? Some suggestions that it doesnt: maybe D0 gets bigger, i.e. the slope gets steeper, at very high dose (saturating repair mechanisms?) Derivation may or may not match realities

01/26/20 Rad Bio: mol bio and survival p. 60 of 97 What do we do about this? Maybe we need to set aside MTSH! Late 1970s through today: other more explicitly repair-based models were concocted. Most wind up proposing linear-quadratic solutions, i.e. ln(S/S0) = D + D2 The logic behind this varies from derivation to derivation, but the final results are hauntingly similar 01/26/20

Rad Bio: mol bio and survival p. 61 of 97 Repair-based models Introduction Poisson statistics 2-term Taylor Expansions Linear-Quadratic Models Molecular Model Dual Radiation Action Model Repair-misrepair model Lethal-Potentially Lethal model Graphical Implications Limitations of Applicability 01/26/20 Rad Bio: mol bio and survival

p. 62 of 97 Poisson survival Alpens comment: A cell can be killed only once, and further action on remaining cells is constrained to that smaller number of cells. This is equivalent to saying dN = N * d(f(D)), I.e. dN/N = d(f(D)), or ln(N) = f(D) + lnN0, I.e. ln(N/N0) = f(D) But S = N/N0, so we have a basic formalism: ln(S) = f(D), where f(D) is some function of dose. Lets seek out the appropriate functional form. 01/26/20 Rad Bio: mol bio and survival p. 63 of 97

Linear-Quadratic Model: Generalized Form Back away temporarily from mechanistic approaches, and say that given Poisson statistics for lethality ln(S) = f(D), where f is some function For an arbitrary function f(D), we Taylor expand in D: ln(S) = a0 + a1D + a2D2 + . . . + anDn + . . . Where ai are the Taylor coefficients (including the factorials in the denominator) But we take a0 = 0 because at D = 0 the survival fraction is 1, i.e. ln(S) = 0 Thus the second-order expansion is ln(S) = a1D + a2D2 01/26/20

Rad Bio: mol bio and survival p. 64 of 97 Molecular Model This emphasizes double-stranded breaks in DNA as a source of lasting damage Distinguishes between single hits causing DSBs and pairs of hits causing DSBs: Ultimately, the pairs of hits give rise to the quadratic dependency on D in the formulas The derivation in Alpen is okay, but we wind up with a few parameters that arent independently determinable 01/26/20 Rad Bio: mol bio and survival p. 65 of 97

Dual Radiation-Action Formulation Emphasizes that a single interaction between a high-LET radiation event and a cell produces a DSB, whereas low-LET radiation requires pairs of events Gives rise to a linear-quadratic model: The one-event DSB (linear) coefficient predominates for high-LET radiation The two-event (quadratic) coefficient predominates for low-LET radiation 01/26/20 Rad Bio: mol bio and survival p. 66 of 97

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(-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. 01/26/20 Rad Bio: mol bio and survival p. 67 of 97

Lethal - Potentially Lethal Model Sets up a three-state system: Undamaged cells (A) Potentially-lethally-damaged cells (B) Lethally damaged cells (C) Eurepair returns state B to state A B automatically becomes C at long times Gives rise to explicit quadratic formulation ln(S) = D + D2 with and having explicit timedependence 01/26/20 Rad Bio: mol bio and survival A B

C p. 68 of 97 LQ Graphical Analysis At low dose the linear dependence predominates; at higher doses the quadratic dominates Crossover from mostly-linear to mostly-quadratic behavior 01/26/20 ln(S/S0) = D + D2 Can we assign physical significance to and , or perhaps to ? Here = -0.05 Gy-1 = -0.01 Gy-2 So / = 5 Gy

Rad Bio: mol bio and survival p. 69 of 97 How do we linearize the relationship? Dose, D ln(S/S0) / D Slope = 01/26/20 Pretty simple, actually: ln(S/S0) = D + D2 means ln(S/S0) / D = + D So: plot ln(S/S0) / D versus D. Y-intercept =

Slope= By convention < 0 (radiation kills!) is generally < 0 also Rad Bio: mol bio and survival p. 70 of 97 What are the units of , , and /? In order for D to be dimensionless, must be have dimensions of inverse dose, e.g. is in Gy-1 In order for D2 to be dimensionless, must have dimensions of inverse dose squared, e.g. is in Gy-2. Therefore / must be in dimensions of dose, e.g. units of Gray or rad. 01/26/20 Rad Bio: mol bio and survival

p. 71 of 97 Modeled significance of / Suppose we expose a cell line to a dose equal to /. Then ln(S/S0) = D + D2 = (/) + (/)2 = 2/ + 2/2/ Thus at dose D = /, influence from linear term and influence from quadratic term are equally significant Thus its the crossover point: Linear damage predominates for D < / Quadratic damage predominates for D > /

Obviously that only works if & have the same sign! 01/26/20 Rad Bio: mol bio and survival p. 72 of 97 Clonogenic survivability Even unirradiated cells dont provide 100% survival; Survival for irradiated cells has to be normalized against whats happening to the controls This sets an upper limit on the precision of the determinations You also need to set up a lot of plates (Poisson statistics) This limits ones ability to distinguish between two LQ models or between an LQ model and an MTSH model on the basis of the resulting muddy data 01/26/20

Rad Bio: mol bio and survival p. 73 of 97 Another limitation on accuracy and applicability: feeder cells Often the treated cells survive poorly if they arent provided with metabolites from neighboring cells So we irradiate a set of cells enough that they cannot divide but they can metabolize Plate out the cells you wish to study atop those This provides a feeder-cell layer that will supply the cells we wish to study This limits applicability because the feeders can be problematic Recent advances make this less of an issue than before 01/26/20

Rad Bio: mol bio and survival p. 74 of 97 Looking forward Be alert for changes in the posted assignments: I may add a few things Midterm will cover chapters 1-7 and the bit of extra material on free radicals that we discussed last week; therefore, it will not include the material from todays lecture 01/26/20 Survival Curves; Modifiers p. 75 of 97 Survival Curves

We discussed models for cell survival last time We looked at various ln(S/S0) vs. dose models and the logic behind them Today well focus on the graphical implications and how we can look at the numbers Then well talk about cell cycles and other good solid cell-biology topics. (Warning: Im more of a biochemist than a cell biologist, so dont expect high expertise in this later section!) 01/26/20 Survival Curves; Modifiers p. 76 of 97 Errata in Chapter 8

Page 169, Paragraph 2, 1st sentence: Until the later 1950s it was not possible to use Two sentences later: Bacteroides Bacillus Fig. 8.1, p. 173: The label that says Dq is pointing at the wrong thing: it should be pointing at the place where the dashed line crosses the (Surviving faction = 1.0) value. 01/26/20 Survival Curves; Modifiers p. 77 of 97 What Fig. 8.1 should have said n Surviving Fraction 01/26/20

Dq = D0lnn = quasi-threshold Dose, Gy Slope = k = 1/D0 A B Survival Curves; Modifiers p. 78 of 97 Shoulder of the Survival Curve Shoulder of the Survival Curve ecognize that We recognize thatdosewith MTSH MTSH onsedose-response we have a

we have a region n where wherethe the slope e is close is closetotozero. zero. We describe describe that that region as a n asshoulder: a shoulder: 01/26/20 Shoulder region Survival Curves; Modifiers p. 79 of 97 Slopes in the MTSH model

Remember that the MTSH model says ln(S/S0) = ln(1-(1-exp(-D/D0))n) Because S/S0 = 1-(1-exp(-D/D0))n So what is the slope of the S/S0 vs. D curve? and what is the slope of the ln(S/S0) vs. D curve? In particular, what is the slopes behavior at low dose? Answer: calculate dS/S0/ dD and d(ln(S/S0))/dD and investigate their behavior at or near D = 0. Note: were looking here at the n>1 case. 01/26/20 Survival Curves; Modifiers p. 80 of 97 Slope investigation, part I For S/S0 itself, d(S/S0)/dD = d/dD(1-(1-exp(-D/D0))n)

The 1 out front doesnt affect the derivative: d(S/S0)/dD = -d/dD(1-exp(-D/D0))n) Well do the rest of this calculation now based on the general formulas 01/26/20 dun/dx = nun-1du/dx deu/dx = eudu/dx Survival Curves; Modifiers p. 81 of 97 Arithmetic & Calculus of Survival Models

MTSH says S/S0 = 1 - (1-e -D/D0)n What I want to investigate is the slope at low dose, I.e. for D << D0, And at high dose, I.e. for D >> D0. But are we interested in the slope of S/S0 vs. D or ln(S/S0) vs. D? Both! Slope = derivative with respect to D. So Slope = d/dD(1 - (1-e -D/D0)n) = -d/dD(1-e -D/D0)n 01/26/20 Survival Curves; Modifiers p. 82 of 97 MTSH slope, continued

Recalling that dun/dx = nun-1du/dx, for n>1, Slope = -n(1-e -D/D0)n-1 d/dD (1-e -D/D0) = -n(1-e -D/D0)n-1 [-d/dD(e -D/D0 )] But we know deu/dx = eu du/dx, so d/dD(e -D/D0 ) = e -D/D0 (-1/D0) = -1/D0 e -D/D0 Therefore Slope = -n(1-e -D/D0)n-1 (-)(-1/D0) e -D/D0 i.e. Slope = (-n/D0)(1-e -D/D0)n-1e -D/D0 01/26/20 Survival Curves; Modifiers p. 83 of 97 Slope at D << D0 This formula for the slope is valid for all values of D, including D << D0 and D >> D0

For small D, i.e. for D << D0, Slope is -n/D0 (1-e -0/D0)n-1e -0/D0 = (-n/D0) (1-e0)n-1e-0; for n > 1 this is = (-n/D0)(1-1)n-11 = 0. Shazam. 01/26/20 Survival Curves; Modifiers p. 84 of 97 Slope of ln(S/S0) vs D The behavior of the slope of the ln(S/S0) vs D curve is not much harder to determine. Recall d lnu /dx = (1/u)du/dx. We apply this here: d ln(S/S0) / dD = (1/(S/S0)) d(S/S0)/dD.

For very small D, S/S0 = 1, so d ln(S/S0) / dD = (1/1) * d(S/S0)/dD = d(S/S0)/dD . But weve just shown that that derivative is zero, so d ln(S/S0) / dD = 0. 01/26/20 Survival Curves; Modifiers p. 85 of 97 High-dose case Weve covered the low-dose case. What happens at high dose, i.e. D >> D0? What wed like to show is that the slope of

lnS/S0 vs. D is -1/D0. Lets see if we can do that. Slope = d ln(S/S0) / dD = (1/(S/S0) d/dD(S/S0) Thus slope = (1 - (1-e -D/D0)n-1)d/dD(1 - (1-e -D/D0)n) For D >> D0, D/D0 is large and -D/D0 is a large negative number; therefore e-D/D0 is close to zero. 01/26/20 Survival Curves; Modifiers p. 86 of 97 High-dose case, continued Slope = limD {(1 - (1-e -D/D0)n)-1(-n/D0)(1-0)n-1e-D/D0)} Thats messy because the denominator and the numerator both go to zero. There are ways to do

that using LHpitals rule, but there are simpler ways that dont involve limits. The trick is to recognize that we can do a binomial expansion Ive done that in the HTML notes The result will be slope = -1/D0 and ln n is the Y intercept of the extrapolated curve 01/26/20 Survival Curves; Modifiers p. 87 of 97 What constitutes a high dose here? The only scaling of the dose that occurs in the formula is the value of D0, so we would expect that we are in that high-dose regime provided that D >> D0. In practice the approximation that

the slope is -1/D0 is valid if D > 5 D0. 01/26/20 Survival Curves; Modifiers p. 88 of 97 Linear-Quadratic Model This is simpler. ln(S/S0) = D + D2 Therefore slope = d/dD (ln(S/S0)) = d/dD(D + D2) Thus slope = + 2D. Thats a pretty simple form. At low dose, || >> 2||D, so slope = . At high dose (what does that mean?) || << 2||D, so slope = 2D. What constitutes a high dose? Well, its a dose at which || << 2||D, so D >> | /

(2| Thus if dose >> | / 2|, then slope = 2D. 01/26/20 Survival Curves; Modifiers p. 89 of 97 Implications of this model At low dose slope = is independent of dose but is nonzero; thus ln(S/S0) is roughly linear with dose. At high dose slope = 2D, i.e. its roughly quadratic. How can we represent this easily? We discussed this last time: ln(S/S0) / D = + D, so by plotting ln(S/S0) / D versus D, we can get a simple linear relationship.

01/26/20 Survival Curves; Modifiers p. 90 of 97 LQ: Plot of ln(S/S0) / D versus D ln(S/S0)/D Y-intercept = Slope = 01/26/20 D By observation, both and are negative.< 0 tells us that radiation harms cells; < 0 is an observed fact. Thus the intercept is below the axis and the slope is negative. Survival Curves; Modifiers

p. 91 of 97 LQ graphical analysis: one step further Dz ln(S/S0)/D Y-intercept = 01/26/20 D We can read -/ directly off an extrapolation of the plot: at D = Dz, ln(S/S0)/D = 0, Slope = + Dz = 0, = -Dz, Dz = - / . Note < 0, < 0, so -/ < 0. Survival Curves; Modifiers

p. 92 of 97 Where do linear and quadratic responses become equal? At what dose does the linear response equal the quadratic response? At that dose, D = D2, D = / So the value we read off the X-intercept of the previous curve is simply the opposite of the dose value at which the two influences are equal. We mentioned this last time, but were reminding you now 01/26/20 Survival Curves; Modifiers p. 93 of 97

How plausible is all this? Model studies suggest reasons to think that ln(S/S0) = D + D2 is a good approach. Much experimental data are consistent with the model Some of these LQ approaches allow for time-dependence to be built in. 01/26/20 Survival Curves; Modifiers p. 94 of 97 LQ vs MTSH and thresholds Response

01/26/20 How does the question of comparing the LQ model to the MTSH model relate to the question of threshold doses? A typical real-world question is a doseresponse relationship for which the only reliable experimental results are obtained at high doses; at lower doses the confounding variables render the experiments uninformative. Dose Survival Curves; Modifiers p. 95 of 97 Dose-Response in Epidemiology Human health effect Experimental data Extrapolations

#1 Limit of reliable measurement #2 #3 Threshold Baseline Tolerable dose #1 Tolerable dose #2 01/26/20 #2 is the linear nonthreshold (LNT) model Is #3 more realistic? This has regulatory

consequences! Tolerable dose #3 Survival Curves; Modifiers Dose p. 96 of 97 Studying repair Weve been suggesting that LQ models and even some MTSH models are dependent on the idea that some DNA damage can be repaired accurately. Lets look for approaches to studying DNA damage that might provide a fuller understanding of the effects of repair. 01/26/20

Survival Curves; Modifiers p. 97 of 97

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