HA Concentration | Molecular Weight Between Cross-Links (kg/mol) |
---|---|
1% | 4.047 |
2% | 2.210 |
3% | 1.028 |
HA Concentration | Mesh size, m |
---|---|
1% | 1.570*10-8 |
2% | 1.224*10-8 |
3% | 0.835* 10-8 |
Abbreviation | Full name | Value | Units | Source |
---|---|---|---|---|
Environment | Tumor site Environment | 965 | uL | - |
E.coli | E.coli BL21 (DE3) | 5 | uL | - |
Tumor | Tumor cells | 524 | uL | - |
ColE1_env | Colicin E1 in Environment | 0 | M | - |
Lactate_IN | Lactate inside E.coli BL21 (DE3) | 20 | mM | [23] |
LldR | LldR protein in E.coli BL21 (DE3) | 4.4 x 10-9 | M | Calculated |
P11 | P11 promoter | 4.4 x 10-9 | M | Calculated |
ColE1mRNA | Colicin E1 mRNA | 0 | M | - |
T7 | T7 polymerase | 4.98 x 10-6 | M | - |
ColE1 | Colicin E1 in E.coli BL21 (DE3) | 0 | M | - |
Abbreviation | Full name | Value | Units | Source |
---|---|---|---|---|
k_1 | LldR and lactate affinity value | 0.8 | uL | - |
k_2 | P11 promoter release rate from LldR | 1 | - | Assumption |
kf_0 | Colicin E1 mRNA degradation rate | 0.0078 | 1/s | [27] |
kf_1 | Colicin E1 translation rate | 0.0383 | 1/s | [28] |
kf_2 | Colicin E1 degradation rate | 0.069 | 1/s | [29] |
kf_3 | Colicin E1 escape rate from bacteria | 1 | 1/s | Assumption |
kf_4 | Colicin E1 mRNA transcription rate | 0.1147 | 1/(M/s) | [28] |
kf_5 | Rate of lactate decrease | 0.0238 | 1/hour | [19] |
k_ColE1_threshold | Efficient anti-tumor concentration of Colicin E1 | 2.19 x 10-8 | M | [19] |
Vtumor = (4/3) * π * r3 = (4/3) * π * 53 ≈ 524 μL
SAtumor = 4 * π * r2 = 4 * π * 52 ≈ 314 mm2
For now, we can find the volume of the hydrogel needed to be injected into the tumor site to cover it completely. The volume of the hydrogel would be:
Vhydrogel = -0.5533 + 1.4087 * SAtumor = -0.5533 + 1.4087 * 314 ≈ 442 μL
Now, we will find the total volume of the system, in which the required concentration should be obtained.
Vtotal = Vhydrogel + Vtumor = 442 + 524 = 966 μL
To identify the volume of bacteria needed, we refer to MatLab Simbiology.
To produce the required amount of Colicin E1 in the volume of 966 uL, we need to incorporate 5 uL of the transformed bacteria.
966 μL with 21.9 nM → 5 μL of bacteria
So, the % load of the hydrogel can be calculated now:
% load = Vbacteria / (Vbacteria + Vhydrogel) * 100% = 1.119%