05 Biology
Incomplete-repair BED (two fractions)
θ = e^{−μ Δt}. BED = n d [1 + g d/(α/β)] with g = 1 + 2θ/(n−…) for two doses: 2d(1 + (d(1+θ))/(α/β)).
Listen
Listen · English
Simulation
Incomplete-repair BED (two fractions) — Change the numbers; the scene follows.
Where it works
Linear accelerator
Isocenter
In the treated volume — tumour and OARs at isocenter, after the dose has been delivered.
Open this machineFormula
Variables
Results
θ
Remaining fraction
0.0625
BED
BED of the pair
6.8333Gy
BED_∞
BED at infinite gap
6.6667Gy
Curve
Explanation
What it means
When two fractions (or two HDR pulses, or two daily IMRT beams many hours apart? usually not) are separated by a gap Δt comparable to the repair half-time, some sublethal damage remains and the β term is inflated by 1+θ, θ = e^{−μ Δt}. Repair T½ is ~0.5–1 h for early tissues and ~1.5–4 h for late. This is why twice-daily treatments need ≥6 h and why pulsed-dose-rate HDR is not equivalent to a single LDR dwell of the same total dose. This is a working relation in Radiobiology.
Where it is used
Clinically it sits on the Linear accelerator — Isocenter. In the treated volume — tumour and OARs at isocenter, after the dose has been delivered. Radiobiology sits between the prescription and the organ-at-risk: LQ, BED, EQD2, and why 2 Gy is not 2 Gy if the fraction size changed. Use it to compare regimens, not to invent one.
Linear accelerator · Open this machineHow to use it
Enter d (each), gap Δt (h), repair T½ (h) and α/β. Read θ, the extra β factor (1+θ), and BED for the pair (n=2). Δt = 0 → fully additive (one 2d fraction); Δt → ∞ → two independent fractions. Change one input and watch the curve and the simulation follow.
Symbols
- dDose each2 Gy
- ΔtGap6 h
- T½Repair half-time1.5 h
- α/βAlpha/beta3 Gy
Worked example
A typical case from the default values: d = 2 Gy (Dose each); Δt = 6 h (Gap); T½ = 1.5 h (Repair half-time); α/β = 3 Gy (Alpha/beta). Substituting into the relation gives θ = 0.0625; BED = 6.8333 Gy; BED_∞ = 6.6667 Gy. These are teaching numbers — align them with your machine.
Typical values give
- θ = 0.0625
- BED = 6.8333Gy
- BED_∞ = 6.6667Gy
Where it comes from
The displayed formula is the working relation. θ = e^{−μ Δt}. BED = n d [1 + g d/(α/β)] with g = 1 + 2θ/(n−…) for two doses: 2d(1 + (d(1+θ))/(α/β)). Usual reference: Thames / Dale / Lea–Catcheside. Derive it in the specialty lesson, then return here to pin the numbers.
Reference: Thames / Dale / Lea–Catcheside
Assumptions & limits
Mono-exponential repair, two equal fractions, no repopulation. Multi-fraction g is the Lea–Catcheside sum (see the g-factor calculator). Not a full PDR / HDR schedule.
Pitfalls
α/β is a model parameter, not a measured organ. BED from incomplete repair or a changed overall time is not the simple n·d·(1+d/(α/β)). Never EQD2 a stereotactic dose with an α/β you did not state. Mono-exponential repair, two equal fractions, no repopulation. Multi-fraction g is the Lea–Catcheside sum (see the g-factor calculator). Not a full PDR / HDR schedule.
Keep this
Always write α/β and the fraction size next to a BED or EQD2. Mono-exponential repair, two equal fractions, no repopulation. Multi-fraction g is the Lea–Catcheside sum (see the g-factor calculator). Not a full PDR / HDR schedule.
In this specialty