Supplement - Quantification of drug loading and release kinetics of nanocarriersI Richard Schwarzla,∗, Fang Dub , Rainer Haagb , Roland R. Netza b Freie
a Freie Universit¨ at Berlin, Fachbereich Physik, Berlin,Germany Universit¨ at Berlin, Institut f¨ ur Chemie und Biochemie, Berlin, Germany
1. MATERIALS AND METHODS Solution for the three-state system. The simplified (utilizing the law of conservation of mass) set of rate equations that describes the release of dexamethasone in the absence of nanocarriers can be written as Φ˙ I (t) = −rIM ΦI + rMI (1 − ΦI − ΦO ) ,
(1)
Φ˙ O (t) = −rOM ΦO + rMO (1 − ΦI − ΦO ) .
(2)
A standard method for solving such a system of coupled ordinary differential equations uses the Laplace transform which is defined as Z ∞ ˜ f (s) := e−st f (t)dt.
(3)
0
Applying the Laplace transform yields 0 ˜ ˜ I (s) = −srMI ΦO (s) + sΦI + rMI , Φ s (rIM + rMI + s)
(4)
0 ˜ ˜ O (s) = −srMO ΦI (s) + sΦO + rMO . Φ s (rMO + rOM + s)
(5)
I Fully
documented templates are available in the elsarticle package on CTAN. Schwarzl Email address:
[email protected] (Richard Schwarzl)
∗ Richard
Preprint submitted to European Journal of Pharmaceutics and BiopharmaceuticsJuly 22, 2016
˜ O , which can be further simplified using the fact that at t = 0 We can solve for Φ the full amount of dexamethasone is inside the dialysis bag (Φ0I ≡ ΦI (0) = 1): ˜ O (s) = Φ
rMO rIM . s [rIM (rMO + rOM + s) + rMI (rOM + s) + s (rMO + rOM + s)]
(6)
If we now use the fact that rMI = rMO due to symmetry (as explained in the main article), eq. (6) reduces to ˜ O (s) = Φ
s3
+
s2
rIM rMI . (rIM + 2rMI + rOM ) + s (rIM rMI + rIM rOM + rMI rOM )
(7)
To be able to back transform the above equation, we need to find the roots of the denominator. The two nontrivial roots are s1 and s2 : q 1 2 + 4r 2 − 2r 2 s1 = r + r −rIM − 2rMI − rOM − rIM IM OM MI OM , 2 1 s2 = 2
q 2 2 2 −rIM − 2rMI − rOM + rIM + 4rMI − 2rIM rOM + rOM .
(8)
(9)
Using the roots of the denominator we can rewrite eq. (7) as ˜ O (s) = Φ
rIM rMI . s (s − s1 ) (s − s2 )
(10)
We can then back transform eq. (10) as ΦO (t, rIM , rMI , rOM ) = rIM rMI
s1 − s2 + s2 es1 t − s1 es2 t . s1 s2 (s1 − s2 )
(11)
Lastly we reduce the number of parameters by invoking the relation VI rIM = VO rOM ,
(12)
so only two fitting parameters remain, namely rIM and rMI . Solution for the four-state system. The experiment measuring the dexamethasone release including CMS nanocarriers inside the solution in the dialysis bag can be modeled by a four-state system. The following set of rate equations describes the dynamics of the system: Φ˙ N (t) = −rNI ΦN + rIN ΦI , 2
(13)
Φ˙ I (t) = − (rIM + rIN ) ΦI + rMI (1 − ΦI − ΦO − ΦN ) + rNI ΦN ,
(14)
Φ˙ O (t) = −rOM ΦO + rMO (1 − ΦI − ΦO − ΦN ) .
(15)
If we go to Laplace space we get the following relations: 0 ˜ ˜ N (s) = rIN ΦI (s) + ΦN , Φ rNI + s
(16)
0 ˜ ˜ ˜ ˜ I (s) = −srMI ΦO (s) − srMI ΦN (s) + srNI ΦN (s) + sΦI + rMI , Φ s (rIM + rIN + rMI + s)
(17)
0 ˜ ˜ ˜ O (s) = −srMO ΦI (s) − srMO ΦN (s) + rMO + sΦO . Φ s (rMO + rOM + s)
(18)
By solving this set of three equations we find a closed form for the Laplace transform of the time dependent fraction of dexamethasone outside the dialysis bag (including the assumptions rMO = rMI , Φ0M = 0 and Φ0O = 0), ˜ O (s) = f (s) , Φ g(s)
(19)
where the numerator is given by f (s) = rIM rMI sΦ0I + rNI
(20)
and the denominator divided by s is a third order polynomial, g(s)/s = s3 + αs2 + βs + γ,
(21)
α ≡ rIM + rIN + 2rMI + rNI + rOM ,
(22)
with the abbreviations:
β ≡ rIM (rMI + rNI + rOM ) + rIN (2rMI + rOM ) + 2rMI rNI + rOM (rMI + rNI ) , (23)
3
γ ≡ rIM rNI (rMI + rOM ) + rMI rOM (rIN + rNI ) .
(24)
If we define the following term, √ p ξ ≡ (−2α3 + 3 3 4α3 γ − α2 β 2 − 18αβγ + 4β 3 + 27γ 2 + 9αβ − 27γ)1/3 , (25) the roots s1 , s2 and s3 of the polynomial in eq. (21) can be written as √ 3 α 2(3β − α2 ) ξ s1 = − − + √ , 3 3ξ 332
(26)
√ √ 1 + i 3 3β − α2 (1 − i 3)ξ α √ √ − , s2 = − + 3 3 3 4ξ 632
(27)
√ √ 1 − i 3 3β − α2 α (1 + i 3)ξ √ √ s3 = − + − . 3 3 3 4ξ 632
(28)
Once we found all roots of the denominator we rewrite the Laplace transform ˜ O as Φ ˜ O (s) = Φ
rIM rMI Φ0I rIM rMI rNI + , (s − s1 )(s − s2 )(s − s3 ) s(s − s1 )(s − s2 )(s − s3 )
(29)
which yields the Laplace back transform " ΦO (t) = rIM rMI
es2 t rNI + s2 Φ0I es1 t rNI + s1 Φ0I + s1 (s1 − s2 )(s1 − s3 ) s2 (s2 − s1 )(s2 − s3 ) # es3 t rNI + s3 Φ0I rNI + − . (30) s3 (s3 − s1 )(s3 − s2 ) s1 s2 s3
The above equation states the full solution for the time dependent fraction of dexamethasone outside the dialysis bag.
2. Results Dexamethasone loading and release. Fig. 1 shows the release data for three individual runs (marked as blue, yellow and green). Circles mark the release data of the experiments in the absence of CMS nanocarriers. Triangles that point upwards mark the release data of the experiments including CMS nanocarriers 4
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released fraction ΦO
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Figure 1: Data of the three individual runs (blue, yellow, green) of the same release experiment. The data for the experiments in the absence of nanocarriers are marked by circles. The data for the experiments including CMS nanocarriers are marked as triangles pointing upwards (30 s loading) and triangles pointing downwards (3 min loading).
and a loading time of 30 s. Triangles that point downwards mark the release data of the experiments including CMS nanocarriers and a loading time of 3 min. The release data in the main article have been calculated as the arithmetic mean of the data sets of three individual runs.
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