C0
|
106 cells/mL |
Maximum PCC density, estimated |
P0
|
105 cells /mL |
Maximum PSC density, estimated |
kc |
7.5 × 10−2 cells1/4mL−1/4day−1
|
Estimated |
μc |
20kc /P0
|
Estimated |
Kc |
0.1 |
We assume that the killing rate of cancer increases by a factor of 5, when R increases from 0.1 to 0.9. |
λc |
10−7 mL per cell per day |
In experiments in Seki et al. (2002), 107 CTL cells were found to kill half of the renal cancer cells in about 16 hours. Therefore the maximum rate of CTL in killing cancer cells can be approximated as ln2/16/107 · 24 ≈ 10−7 per cell per day. At the same time, this number is given by λc/(Kc +(1−R)) in a patient. Assume that R is approximately 0.1 in patients, we obtain λc = 10−7 per cell per day. |
kp |
0.2 per day |
Estimated |
μp |
20 kp
|
Estimated |
Kp |
C0/100 |
Estimated |
λp |
0.15 per day |
The half-life T
1/2 of PSC is 2–5 days. The relation of λp and T1/2 is λp = ln 2/T1/2. |
kr |
0.2 per day |
From the full model, we have kr = α + k1λM/kM. Taking the value that α = 0.2/day, k1 = 40 cells/ml/day (Sichert et al., 2007), kM = 228 cells/ml/day (Sichert et al., 2007; Day et al., 2009), and λM = 0.02/day (Day et al., 2009), we obtain kr ≈ 0.2. |
λr |
0.22 per day |
From the full model, we have λr = λM + α. |
γp |
0.02λr /Ps
|
Ps is the PSC density in a healthy person, which satisfies Ps = P0(1−λp/kp). |
γc |
= γp
|
Estimated |
kt |
3300 cells per mL per day |
Estimated |
Kt |
= Kc
|
Estimated |
λt
|
0.3 per day |
Day et al. (2009) |