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RADIATION PROTECTION IN RADIOTHERAPY

RADIATION PROTECTION IN RADIOTHERAPY

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RADIATION PROTECTION IN RADIOTHERAPY

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  1. RADIATION PROTECTION IN RADIOTHERAPY IAEA Training Material on Radiation Protection in Radiotherapy Part 3: Radiation Biology PRACTICAL EXERCISE

  2. Objectives of Part 3 • To understand the various effects of radiation on human tissues • To appreciate the difference between high and low dose; deterministic and stochastic effects • To gain a feel for the order of magnitude of dose and effects • To appreciate the risks involved in the use of ionizing radiation as a starting point for a system of radiation protection Part 3, Practical 2

  3. Part 3: Radiation Biology IAEA Training Material on Radiation Protection in Radiotherapy Practical 2: Calculations using the linear quadratic model

  4. Contents + Objective • Be familiar with the the mathematical framework of the linear quadratic model • for dose fractionation • for time corrections • Perform calculations using information given in the lecture Part 3, Practical 2

  5. What Minimum Equipment is Needed? • Paper, pocket calculator • Whiteboard • Handout and lecture notes Part 3, Practical 2

  6. Please calculate what additional total dose can be given to a prostate tumor when the dose per fraction is reduced from 2Gy to 1.8Gy. Assume the critical organ is the rectum with a/b = 3Gy , the dose given in 2Gy fractions is 70Gy and time effects play no role Q1 Q1 Part 3, Practical 2

  7. Q1 Assumptions • No time effects (i.e. time between fractions is large enough to allow full repair and the overall treatment time is short enough to prohibit significant repopulation during the treatment). The biologically effective dose (BED) of the treatment schedules can then be calculated as: • Rectum is the dose limiting structure and the maximum dose to the rectum is identical to tumour dose • BED = nd (1 + d/(a/b)) with n number of fractions, d dose per fraction and a/b the alphabeta ratio Part 3, Practical 2

  8. Q1 Answer • BED = nd (1 + d/(a/b)) with n number of fractions, d dose per fraction and a/b the alphabeta ratio • BED (rectum, 35fx of 2Gy) = 35 * 2 (1 + 2/3) = 117Gy (comment: this is often referred to as BED3 to specify the a/b ratio used) • BED (rectum, 1.8Gy/fx) = 117Gy = x * 1.8 (1 + 1.8/3) • x = 117/(1.8 (1 + 1.8/3)) = 40.5 fractions or 73Gy • One can give 3Gy more dose using 1.8Gy fractions instead of 2Gy fractions to expect the same rectal complication probability Part 3, Practical 2

  9. Would the change in fractionation still be a good approach if the a/b ratio for prostate cancer turned out to be confirmed as 1.5Gy? Q2 Q2 Part 3, Practical 2

  10. Q2 Answer • The difference in BED for the prostate tumour by going from 70Gy in 35fx to 73.8 in 41fx is for a/b=2Gy: • BED = nd (1 + d/(a/b)) • BED1.5 (tumor, 35fx of 2Gy) = 35 * 2 (1 + 2/1.5) = 163Gy • BED (tumor, 41fx of 1.8Gy) = 41 * 1.8 (1 + 1.8/1.5) = • x = 117/(1.8 (1 + 1.8/3)) = 40.5fraction or 73Gy = 163Gy • In other words despite the higher physical dose the tumour control probability would be even lower. Part 3, Practical 2

  11. Please calculate what additional physical dose must be given to a head and neck tumour with a Tp of 3 days to compensate for a treatment interruption of one week due to holiday closure? Assume that 2Gy fractions are used, that the treatment time before the holiday has been longer than the kick-off time, that alpha is 0.35Gy-1 and that a/b ratio for the tumour is 10Gy. Q3 Q3 Part 3, Practical 2

  12. Q3 Answer • Need the LQ time model. E = - ln S = n * d (α + βd) - γT • or BED = (1 + d/(α/β)) * nd - (ln2 (T - Tk)) / αTp Part 3, Practical 2

  13. Q3 Answer • If the BED shall be unchanged, the difference between the schedules is the time factor: ((ln2((T+ tbreak)-Tk))/αTp)- ((ln2(T-Tk))/αTp) • Assuming Tp = 3days, alpha = 0.35Gy-1 and T>Tk, the difference is ln2 tbreak / αTp = 4.6Gy Part 3, Practical 2

  14. Q3 Answer • The break of 1 week in the head and neck treatment is equal to a loss of 4.6Gy given in 2Gy fractions due to repopulation of the tumor. • In other words one loses between 0.5 and 1Gy per day treatment prolongation in fast growing tumors • This is clinically proven for head and neck and cervix tumors • If the additional dose is not given on the same day but in more fractions (increasing the time further) one needs to give about 3 more fractions to allow for the break Part 3, Practical 2

  15. Q3 Answer • Some comments: • It is often a good strategy to try and finish the treatment prior to a break, e.g. by giving two fractions per day. • The minimum time between two fractions must be sufficient for repair of normal tissues (i.e. 6hours or more in the case of spinal cord) • Due to the accelerated repopulation after treatment commences it may be better to commence treatment after a scheduled break rather than starting for a couple of days and then having a break. Once commenced, the treatment for fast growing tumours must be completed within the prescribed time. Part 3, Practical 2

  16. Assume in the previous example, 10 fractions are remaining after the break - what increase in dose per fraction is required to make up for the repopulation during the break? Q4 Q4 Part 3, Practical 2

  17. Q4 Answer • Using BED1 (break + 10 fractions with higher dose) = BED2 (no break + 10 x 2Gy fractions) one can derive: • with d=2Gy (fractions prior to change), n2= 10 fractions remaining, R = 7days the time for the break and beta = alpha/(a/b) = 0.35/10 = 0.035Gy-2 Part 3, Practical 2

  18. Q4 Answer • The result is: dnew = - 10/2 + sqrt((10/2+2)2+7/3 ((ln2/0.035)/10)) = = 2.32Gy • One can also make up for the increased repopulation by increasing the dose per fraction in 10 fractions by more than 15% • Note that this will increase the risk of significant late effects dramatically…. Part 3, Practical 2

  19. Questions? Let’s get started... Part 3, Practical 2