I have a question about the pdb4dna example, And it is about Figure b. I attempt to get the smiliar results from Figure b by injecting 10000 protons of 1 MeV (The root histogram I did not post because new user can only send 1 picture each time, and I will give relative description when it is needed).
The LET can be calculated from the Mean Energy Deposition divide the Track Distance in volume. Here is 214.1eV (as shown in root histogram, the mean energy deposition in the volume is 214.1eV) /30nm (The box size is 13.0 × 15.2 × 25.4 nm, the particle source comes from the outer sphere of the box, let’s assume it is 30nm), and the result is 7.13 KeV/um, I think it matches the x-axis.
But I don’t know how to get the SSB yeild. So 1Gy=1J/1kg, the SSB threshold is 8.22 eV as the author set, I assume it is 10.0 eV when it acctually happens. The energy deposition would be 2144/10000 × 1.266 × 10=2.7eV=4.310^-19J (2144 times happen SSB out of 10000 injections, and for each time it happened, it makes 1.266 SSB), the mass would be 1000 × 13.0 × 15.2 × 25.4 × 10^-27=510^-21kg. So the dose would be 4.310^-19/510^-21=86Gy. It is impossible for this very small size to absorb 86 Gy, so I know my perspective is incorrect. I would like to know if you could have the right soulution about the y-axis result from Figure b. I am really appreciate for your reply!!!
I am not familiar with that example but calculations of dose can be sensitive to the regions you choose to average which is a quirk of how dose is calculated. You should be averaging over gigabasepairs and not specifically supposing all of the energy goes into breaking DNA strands.
I mean a crude approximation would be LET * (SSB yield) * (DNA width) / (mass of all exposed DNA). That denominator would be the gigabasepair mass. I have a feeling the values will be much more reasonable.
I have a problem with the equation:LET * (SSB yield) * (DNA width) / (mass of all exposed DNA).
Is the LET the same vaule as the x-axis? And SSB yield, DNA width, and mass of all exposed DNA need to use the value of 1 Gbp, or just for mass uses 1 Gbp?
I have a problem with the equation:LET * (SSB yield) * (DNA width) / (mass of all exposed DNA).
Is the LET the same vaule as the x-axis? And SSB yield, DNA width, and mass of all exposed DNA need to use the value of 1 Gbp, or just for mass uses 1 Gbp?
I misread the units of the vertical axis but the idea is still that the numbers will be very sensitive to what you mean by volume. Estimate the range assuming water-equivalent sample. There are lookup tables for this. With this range you can get an LET. The chart then gives you the yield. Geant4 will let you have an infinitely narrow proton beam but a more realistic simulation/beam area would be 1mm or greater. This times the range is the “volume” irradiatied. With this volume of “water” you can get the mass irradiated and thus get the dose D.
Then you estimate the average number of gigabasepairs G in that volume. This will depend on the “sample” you want to investigate and is not something the simulation could give you. That will then give you the yield:
SSB/Gy/Gbp(LET of 1 MeV protons in water) * D * G = SSBs for 10000 protons of 1 MeV in sample
I assume that it is strange if those data come from the actual samples or experiments, but the authors didn’t give the corresponding explanations? I assume that those data are quite difficult to get just from this simulation in the article as well.
Thank you for your answer again, and I still have problems with this equation.
SSB/Gy/Gbp(LET of 1 MeV protons in water) * D * G = SSBs for 10000 protons of 1 MeV in sample
So assume that I get the LET from the table, and I need to calculate D, right? Does G mean the gain factor of radiation in this case? Do SSBs for 10000 protons of 1 MeV in sample mean the SSBs of 1 Gbp? If so, basically, there is no data extracted from the simulation?
Proton LET is rapidly changing near the end of its range but as an approximation you can just divide the range of protons in water (or run a simulation to determine this value) by its initial energy to get an average LET.
This table gives 20 microns->last column divided by density of water in g/cm3 which is 1. This gives an average LET of 50 keV/um. This range times the beam spot size (cross sectional area) gives you the volume irradiated. The dose D in that volume is just Energy of one proton * number of protons / mass of the irradiated volume (water). G is not a gain factor. It is the number of gigabasepairs you would expect in that volume.
There are about 3.1 Gbp in human DNA so that would be per cell. A human cell is typically 30 microns. A 1 MeV proton will cover about 2/3 of the cell so the protons will encounter on average (2/3) * 3 Gbp = 2 Gbp. Or if the proton energy is higher, calculate how many cells you would expect to cover. And if its also a petri dish, some estimate of the number of cells per volume since there will be a lot of “empty” space. This is more of an approximation tool.
Thank you! Your answer is very helpful, but I am still a little bit confused.
Assume that 1 proton irradiate 1 cell and I use the equation to give a rough estimation:
SSB/Gy/Gbp(LET of 1 MeV protons in water) * D * G = SSBs for 1 protons of 1 MeV in sample
The inital proton energy is 1 MeV, to get the mass, I use the 0.002 * 1 * cross sectional area per cell, which is ((310^-3 cm))^2=1810^-9g=1.810^-11 kg 1Mev=1.610^-13 J D=1.610^-13 / 1.810^-11 = 8.89*10^-3 Gy G is 2 gigabasepairs in the volume, and SSBs for 1 protons of 1 MeV in sample is around 0.25 per SSB for each 1 MeV proton (according to the simulation of 1000 protons, 2144 each and 1.266 SSB per time, we roughly estimate it as 0.25 for each).
It would be:
SSB/Gy/Gbp(LET of 1 MeV protons in water) * 8.89*10^-3 * 2 = 0.25
SSB/Gy/Gbp(LET of 1 MeV protons in water) would be140.6, Is it correct?
I think you have the idea. If you are in a single proton regime than yes, 0.25 sounds about right. Things would get trickier with actual beams since would be very difficult in the real world to make a beam of protons with that spot size.
Dear all,
May I know how can I calculate the LET for pdb4dna example, please? the output gives the energy deposited, SSB, DSB. LET is calculated as dE (energy deposited) / dx (per unit length), how can I know the unit length? The dimension is 12 x 15 x 25 nm3.
If I calculate LET for neuron example at 30 MeV of proton, can I use its value for energy 30 MeV of proton for Pdb4dna?
Also, how can I reach to low doses (1 or 2 Gy) for pdb4dna ? I work on energy ranges (30 - 300 MeV), The energy deposited that I have from output file is large due to the no of events or beam ON which in pdb4dna.in.
In the 1 MeV proton case, the beam path is basically straight in the volume of 12 x 15 x 25 nm3, so I use 20 nm as an estimation as an averge. If you are very familiar with Geant4, I guess you can also track the length of particle ID 0, this would be accurate. By the way, it’s just my opinion, please correct this if anything I provided is wrong.
Sorry, what do you mean using 20 nm as an average? Do you mean instead of the dimension 12 x 15 x 25 nm3, I use 20 nm3?
Also, what do you mean the length ID 0, please?
I work on energies (30 - 300 ) MeV for protons.
There is no fixed LET value for protons as it depends on the step length you choose to use and changes as it slows down (dE/dX proportional to mass * velocity_squared). One way to get the “unit length” would be to divide the incident energy by the range of the proton. This can either be done with a proton range lookup table as shown earlier in the thread or by having Geant output this value for you. I do not know what is output from the file but you could get that information any number of ways by tracking the protons position when it enters the volume to when it stops. This would be easiest to do in the tracking action.
The dose you want to simulate depends on the volume you irradiate. See earlier discussion above. The beam width * proton range.
I have a question about my LET calculation method. I assume that to calculate LET, the first thing I need to do is to get the volume comparison. If current volume equals to the volume of the previous step, then going through the following step. The second step is to get the trackID, if the trackID shows that it is the mother particle, then going through the final step, which is the sum of energy deposition divide the sum of step length, then I get the LET of this particle in the volume.
I have 2 questions. One is that the LET is based on the mother (injected) particle (so just sum of energy deposition of mother particle divide sum of step length of mother particle) instead of all particles (include primary, secondary, etc.) right? The second is that this calculation does not need to consider the physics process, like transportation,ionization, etc. Right?
LET refers to the particle in question. And it already implicitly accounts for all possible processes - ionization for charged particles, nuclear reactions for neutrons/protons, elastic collisions, etc. It is an average. Some components have functional forms such as the Bethe-Bloch formula.
Thank you, " Analysis of the track- and dose-averaged LET and LET spectra in proton therapy using the geant4 Monte Carlo code." This article gave calculation method of LET in Geant4. Problem solved.
