This post relates to previous post about biotransfer factor in previous post. If you haven’t visited that post yet, I suggest you do so. I would like to provide an example on how to calculate the concentration of chemical A if not 100% is removed in the liver organ.

The keyproblem to this matter is whether BTF is dependent or independent for chemical removal in the liver or not. Depends or independent is determined by toxicology study. We can know only if the figure is confirmed as a linear function between concentration of chemical in the body with the dose as follows:

Concentration of chemical (in the blood, liver etc) = BTF x dose (mg A/day)

Suppose we would like to calculate the equilibrium concentration in blood, liver and kidneys under the following conditions. Dose is 0.4 mg/day, and it is found out that liver only manage to metabolize 0.08 mg A/day with maximum concentration A in the liver is 1.6 mg A/kg_liver. If BTF is independent, then there will be no problem, we can just straight away using the formula above.

We already know the BTF value from previous post. The example on how to calculate the BTF is posted on BioTransfer Factor post, please visit that post to grab the idea on how this will work, and to get the number of BTF for each organs, blood, kidneys and liver.

Since 10% of chemical dose is absorbed in the stomach and 90% is absorped in digestive tract, thus the dose here will equal to 0.36 mg A/day.

Concentration of A = BTF x dose

C_blood = 8.1 (mg A/kg_blood)/(mg A/day) x 0.36 mg A/day = 2.9 (mg A/kg_blood)

C_kidneys = 3.2 (mg A/kg_kidneys)/(mg A/day) x 0.36 mg A/day = 1.15 (mg A/kg_kidneys)

And the last is C_liver which is 1.6 mg A/kg_liver.

It must be noted thate the examples above only work if BTF is independent from chemical removal in liver, because if it is dependent, there will be a transformation of unmetabolized chemical to other organs to be processed, and this will of course increase the chemical concentration in liver, kidneys and blood as well.

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