Farmakokinetika Tugas Soal PPT Ekstravaskuler [PDF]

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Nama : Fitria Pujiastuti NIM



: 18334009



FARMAKOKINETIK KELAS K



1. Fenitoin diberikan secara oral pada pasien dengan dosis 300 mg kemudian ditentukan AUC iv = 130 µg/ml jam, kadar obat dalam darahnya sbb T(jam)



Cp(µg/ml) Ln



Cp’



Cp



Cp’-



Ln



Cp



CpCp’



1



0,65



7,07



6,42



1,86



2



2



6,76



4,76



1,56



5



3,55



5,91



2,36



0,86



10



4,05



4,72



0,67



15



3,6



1,28



20



3,2



1,16



30



2



0,69



40



1,2



0,18



50



0,8



-0,22



1.4 1.2 1 0.8



0.6 Series1 0.4 0.2 0 -0.2 -0.4



0



10



20



30



40



50



60



Hitunglah a.



t1/2



dari data T sebagai x dan ln Cp (fase eliminasi) sebagai y, diperoleh persamaan regresi y = 1,9975-0,045x (r=0.99) (diekstrapolasikan menjadi 7,4e-0,045t). T1/2 = 0,693/0,045 = 15 jam. b.



Ke



ke = 0,045 jam-1 c.



t1/2ab



dari data T sebagai x dan ln Cp-Cp’(Cp’ = 7,4e-0,045t) sebagai y, diperoleh persamaan regresi y = 2,09-0,25x. T1/2ab = 0,693/ka = 0,693/0,25) = 2,8 jam. d.ka ka = 0,25 jam-1. e.Z Z = (FxDosisxka) / Vd(ka-ke) = 0,9 x 300 x 0,25 / 157 (0,25-0,045) = 2,097 mg/L = 2,097 µg/ml. f.AUC ƩAUC = Ʃ [(Cn+Cn-1)/2 x (tn-tn-1)] = AUC2,1 = 1,325. AUC 5,2 = 8,325. AUC 10,5 = 19. AUC 15,10



= 19,125. AUC 20,15 = 17. AUC 30,20 = 26. AUC 40,30 = 16. AUC 50,40 = 10. ƩAUC = 116,775



µg jam/ml. g.F F = AUC oral / AUC iv = 116,775 / 130 = 0,9. h.Clp Clp = FxDosis / AUC oral = 0,9 x 300 / 116,775 = 2,3 L/jam. Vd = (FxDosisxka) / a(ka-ke) = 0,9 x 300 x 0,25 / 2,097 (0,25-0,045) = 157 L.



QUESTION 1 Table P2.1 gives the plasma drug concentrations (Cp) that were obtained following the oral administration of 1 g dose of a drug. Plot the data and, using the plot, determine the following.



a.



The elimination half life (t½) and the elimination rate constant (K).



b.



The absorption half life, (t½) abs.



c.



The absorption rate constant (Ka).



d.



The observed and calculated peak time (tmax).



e.



The observed and calculated peak plasma concentrations, (Cp) max.



f.



The y-axis intercept. Compare :



g.



The observed peak time (tmax) with the calculated peak time.



h.



The observed and the calculated peak plasma concentrations (Cp) max.



Diketahui : Dosis yang diberikan 1 gram.



t (jam) 0,25 0,5 1 1,5 2 3 4 5 6 7



Cp (mg %) 3 4,6 5,7 5,6 4,8 3,2 2 1,2 0,75 0,46



Ditanya : a.



The elimination half life (t½) and the elimination rate constant (K).



b.



The absorption half life, (t½) abs.



c.



The absorption rate constant (Ka).



d.



The observed and calculated peak time (tmax).



e.



The observed and calculated peak plasma concentrations, (Cp) max.



f.



The y-axis intercept. Compare :



g.



The observed peak time (tmax) with the calculated peak time.



h.



The observed and the calculated peak plasma concentrations (Cp) max.



Jawab :



t (jam) 0,25 0,5 1 1,5



Cp (mg %) 3 4,6 5,7 5,6



Log Cp



Cp’



Cp’ – Cp



Log (Cp’ – Cp)



0,47712 0,66275 0,75587 0,74818



12,22479 10,82611 8,490534 6,658823



9,224786 6,226112 2,790534 0



0,964956 0,794217 0,445687 0



Profil Obat Dalam Darah 100



2 3 4 5 6 7



4,8 3,2 2 10 1,2 0,75 0,46



0,68124 0,50514 0,30102 0,07918 -0,12493 -0,33724



5,222278 3,21207 1,97565 1,215164 0,747412 0,459711



0 0 0 0 0 0



0 0 0 0 0 0



1



0.1 0



1



2



3



4



5



t (jam) Cp (mg%)



Fase Eliminasi, Diperoleh : a = 1,14001 b = -0,211075 r2 = -0,99992 → orde pertama a.



t ½ = 0,693/k = 0,693/0,48547 = 1,42748 jam Ke = -b x 2,3 = -(-0,211075) x 2.3



Cp'



Cp'-Cp



6



7



8



= 0,211075 x 2,3 = 0,48547 jam –1



Persamaan Fase Eliminasi : -0,211075t + 1,14001 Perhitungan Cp Ekstrapolasi (Cp’) -



t=0,25 → -0,211075(0,25) + 1,14001= 1,08724, anti log = 12,22475



-



t=0,5 → -0,211075(0,5) + 1,14001= 1,03447 , anti log = 10,82613



-



t=1 → -0,211075(1) + 1,14001= 0,928935 , anti log = 8,49249



Cari kadar residual (Cr) dengan cara : Cr = Cp’ – Cp -



t=0,25 → 12,22475 – 3 = 9,22475



-



t=0,5 → 10,82613 – 4,6 = 6,22613



-



t=1 → 8,49249 – 5,7 = 2,79249



Fase absorpsi : t (jam) 0,25 0,5 1



Cp (mg%) 3 4,6 5,7



Diperoleh : a = 1,13906 b = -0,69258 r2 = 0,99998 b. t ½ abs = 0,693/K = 0,693/1,59293 = 0,43504 jam



c.



Ka = -b x 2,3 = -(-0,69258) x 2,3 = 1,59293 jam–1



Log Cp



Cp’



Cp’-Cp



0,47712 0,66275 0,75587



12,22475 10,82613 8,49249



9,22475 6,22613 2,79249



Log (Cp’Cp) 0,96495 0,79421 0,44599



d. T maks =



2,3 log



Ka K



Ka−K 2,3 log



1,59293 0,48547



= 1,59293− 0,48547 =



2,3 log 3,28121 1,10746 1,18687



= 1,10746 = 1,07170 jam e.



Cp maks = [Ae– k.tmax] – [Be–ka.tmax] = [13,8038 e– 0,48547.1,07170] – [13,8038 e–1,59293.1,07170] = [13,8038 e–0,52027] – [13,8038 e–1,70714] = [13,8038 . 0,59436] – [13,8038 . 0,18138] = [8,20442] – [2,50373] = 5,70069 mg%



f.



y-axis intesep, perpotongan digaris y. y = bx + a y=a Log Cp0 = a Log Cp0 = 1,140011 →13,80419 ~ 14 mg%



g.



Membandingkan hasil pengamatan dari kurva dengan perhitungan Tmax Dari hasil pengamatan dapat dilihat bahwa puncak konsentrasi tertinggi adalah pada jam ke 1 dengan konsentrasi 5,7 mg% (satu titik waktu dimana darah dikumpulkan). Sedangkan berdasrkan perhitungan Tmax nya adalah 1,07170 jam. Perhitungan Tmax lebih akurat.



h. Membandingkan hasil pengamatan dari kurva dengan perhitungan Cp max Dari hasil pengamatan dapat dilihat bahwa puncak konsentrasi tertinggi adalah 5,7 mg% pada jam ke 1. Sedangkan berdasrkan perhitungan Cp max nya adalah 5,70069 mg%. Perhitungan Cp max lebih akurat.



QUESTION 2 Table 2.2



Plot the data and, using plot, determine the following a. The elimination half (t 1/2) for each dose. b. The elimination rate constant (K) for each dose. c. The absorption half (t 1/2) for each dose. d. The absorption rate constant (Ka) for each dose. e. The observed and computed peak time (t max) for each dose. f. The observed and computed peak plasma concentrations, (Cpmax) for each dose. g. The y-axist intercept for each dose. h. The apperent volume of distribution (V).



i.



The fraction of drugs absorbed (F).



j.



The characteristic of a plot on rectilinier paper of peak time (t max) against the administered dose (than make a important observation)



k. The characteristic of a plot on rectilinier paper of peak plasma concentrations, (Cp max).



Diketahui : t (jam)



0.5 1 1.5 2 3 4 6 8 10 12 24



25mg tablet Plasma Concentration ( mg /% ) 0.12 2.2 5.38 6.8 6.91 6.32 4.25 3.6 2.72 2.3 0.67



50 mg tablet Plasma Concentration ( mg /% ) 0.26 3.62 6.65 10.74 12.54 11.2 8.54 6.48 4.85 4.05 1.7



Ditanya: l.



The elimination half (t 1/2) for each dose.



m. The elimination rate constant (K) for each dose. n. The absorption half (t 1/2) for each dose. o. The absorption rate constant (Ka) for each dose. p. The observed and computed peak time (t max) for each dose. q. The observed and computed peak plasma concentrations, (Cpmax) for each dose. r. The y-axist intercept for each dose.



s. The apperent volume of distribution (V). t. The fraction of drugs absorbed (F). u. The characteristic of a plot on rectilinier paper of peak time (tmax) against the administered dose (than make a important observation) v. The characteristic of a plot on rectilinier paper of peak plasma concentrations, (Cp max).



Jawab :



t (jam)



0.5 1 1.5 2 3 4 6 8 10 12 24



25mg tablet Plasma Concentration ( mg /% ) 0.12 2.2 5.38 6.8 6.91 6.32 4.25 3.6 2.72 2.3 0.67



50 mg tablet Plasma Concentration ( mg /% ) 0.26 3.62 6.65 10.74 12.54 11.2 8.54 6.48 4.85 4.05 1.7



100



10



cp 25 mg cp 50 mg 10



15



20



25



30



Diperoleh : Dosis 25 mg : a = 0,8999 b = -0,0448 r2 = -0,9983 → orde pertama y = -0,0448x + 0,8999 log Cp = -0,0448t + 0,8999



Dosis 50 mg : a = 1,0502 b = -0,0345 r2 = -0,9858 → orde pertama y = -0,0345x + 1,0502 log Cp = -0,0345t + 1,0502



a.



The elimination rate constant (Ke) for each dose Dosis 25 mg: Ke = -b x 2,303 = -(-0,0448) x 2,303 = 0,103174 jam-1 Dosis 50 mg: Ke = -b x 2,303



= -(-0,0345) x 2,303 = 0,07935 jam-1 b. The elimination half life (t1/2) for each dose Dosis 25 mg: t1/2 el =



0,693 Ke 0.693



= 0,103174 = 6,716782 jam



Dosis 50 mg: t 1/2 el =



0,693 Ke 0,693



= 0,07935 = 8,733459 jam



Persamaan Fase Eliminasi 25 mg : -0,0448t + 0,8999 Perhitungan Cp Ekstrapolasi (Cp’) -



t=0,5 → -0,0448(0,5) + 0,8999 = 0,8775, anti log = 7,542234



-



t=1 → -0,0448(1) + 0,8999 = 0,8551 , anti log = 7,163083



-



t=1,5 → -0,0448(1,5) + 0,8999 = 0,8327 , anti log = 6,802993



Cari kadar residual (Cr) dengan cara : Cr = Cp’ – Cp -



t=0,5 → 7,5422 – 0,12 = 7,422234



-



t=1 → 7,1631 – 2,20 = 4,963083



-



t=1,5 → 6,8030 – 5,38 = 1,422993 t (jam ) 0.5 1.0 1.5



Cp (mg%)



Log Cp



Cp’



Cp’-Cp



Log (Cp’Cp)



0.12 ± 0.45 2.20 ± 1.76 5.38 ± 4.26



-0,9208



7,542234



7,422234



0,870535



0,3424



7,163083



4,963083



0,695752



0,7308



6,802993



1,422993



0,153203



Persamaan Fase Eliminasi 50 mg : -0,0345t + 1,0502



Perhitungan Cp Ekstrapolasi (Cp’) -



t=0,5 → -0,0345(0,5) + 1,0502 = 1,0329, anti log = 10,7869



-



t=1 → -0,0345(1) + 1,0502 = 1,0157 , anti log = 10,3681



-



t=1,5 → -0,0345(1,5) + 1,0502 = 0,9985 , anti log = 9,9655



Cari kadar residual (Cr) dengan cara : Cr = Cp’ – Cp -



t=0,5 → 10,7869 – 0,26 = 10,5269



-



t=1 → 10,3681 – 3,62 = 6,7481



-



t=1,5 → 9,9655 – 6,65 = 3,3155



t (jam) 0.5 1.0 1.5



Cp (mg%)



Log Cp



Cp’



Cp’-Cp



0.26 ± 0.75 3.62 ± 3.05 6.65 ± 4.15



-0,5850



10,7869



10,5269



Log (Cp’Cp) 1,0223



0,5587



10,3681



6,7481



0,8292



0,8228



9,9655



3,3155



0,5205



Fase Absorbsi : Dosis 25 mg : y = -0,7173x + 1,2904 a = 1,2904 b = -0,7173 r2 = 0,9194 Dosis 50 mg : y = -0,5018x+ 1,2925 a = 1,2925 b = -0,5018 r2 = 0,9826



c. The absorption rate constant Kabs for each dose. Tablet 25 mg Ka = -b x 2,303 = - (-0,7173) x 2,303 = 1,64979 jam-1



Tablet 50 mg Ka = -b x 2,303 = - (-0,5018) x 2,303 = 1,15564 jam-1 d. The absorption half life (t1/2)abs for each dose. Tablet 25 mg t 1/2 abs =



0,693 Ka 0,693



= 1,64979 = 0,42005 jam Tablet 50 mg t 1/2 abs =



0,693 Ka 0,693



= 1,15564 = 0,599668 jam e. The observed and computed peak time (tmax) for each dose. 25 mg tablet: tmax =



(2,303 log e) Kabs−Ke (2,303 log 1,64979)



= 1,64979 −0,103174 = =



(2,303 .log 15,99036) 1,546616 2,303 ×1,203858 1,546616



= 1,792614 jam 25 mg tablet: tmax = = = =



(2,303 log e) Kabs−Ke (2,303 log 1,15564 ) 1,15564− 0,07935 (2,303 .log 15,99036) 1,546616 2,303 ×1,1632756 1,07629



= 2,48913 jam f. The observed and computed peak plasma concentrations (Cpmax) for each dose. Cpmax 25 mg tablet:



Intersep fase eliminasi = 0,8999 B = antilog a = antilog 0,8999 B = 7,9414 Intersep fase absorbsi = 1,2925 A = anti log a = anti log 1,2925 A = 19,6110 Cp max = B.e k.tmax - A.e k.tmax Cp max = 7,9414e-0,1032x0,2824 – 19,6110e-1,6521x0,2824 Cp max = 7,7135– 12,2409 Cp max = 12,2409 - 7,7135 = 4,5274 mg% Cpmax 50 mg tablet: Intersep fase eliminasi = 1,0502 B = antilog a = antilog 1,0502 B = 11,2253 Intersep fase absorbsi = 1,2931 A = anti log a = anti log 1,2931 A = 19,6381 Cp max = B.e-k.tmax – A.e-ka.tmax Cp max = 11,2253e-0,0795x0,070 – 19,6381e-1,1556x0,070 Cp max = 11,1630 – 18,1121 Cp max = 18,1121 - 11,1630 = 6,9490 mg% g. The y-axis intercept for each dose. y-intersep (25 mg): y = -0,0448x + 0,8999 y = antilog 0,8999 y = 7,9415 y-intersep (50 mg): y = -0,0345x + 1,0502 y = antilog 1,0502 y = 11,2254 h. The apparent volume of distribution (Vd). Vd (25 mg) = (F.Do.Ka)/a(Ka-k) = (1,3118 × 25 x 1,6520) / 1,2905 (1,6520-0,1032) = 27,1059 mg/ml Vd (50 mg) = (F.Do.Ka)/a(Ka-k) = (1,3118 x 50 x 1,1556) / 1,2931 (1,1556-0,0795) = 76,3043 = 54,4703 mg/ml



i. The fraction of drug absorbed (F). Intersep fase eliminasi= 0,8999 Intersep fase absorpsi = 1,2905 AUCa = (Intersep/k) – (intersep/ka) = (0,8999/0,1032) – (1,2905/1,6520) = 8,7120 – 0,7811 = 7,9309 𝜇𝑔.jam/ml AUCb = (Intersep/k) – (intersep/ka) = (1,0502/0,0795) – (1,2925/1,1556 ) = 13,2100 – 1,1185 = 12,0915 𝜇𝑔.jam/ml F = [AUC]a . Dosis B [AUC]b . Dosis A 7,9309 μg.jam/ml .50 mg



F = 12,0112 μg.jam/ml .25 mg F = 396,545/300,28 F = 1,3118 mg



j. The characteristic of a plot on rectilinear paper of peak time (tmax) against the administered dose (then make an important observation). Tmaks 25 mg = 4 jam Tmaks 50 mg = 4 jam



k. The characteristics of a plot on rectilinear paper of peak plasma concentrations (Cp)maks (Cp)maks 25 mg = 7 mg% (Cp)maks 50 mg = 12,25 mg%