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Pharmacokinetics Calculations

Pharmacokinetics Calculations. Prof. Dr. Basavaraj K. Nanjwade M. Pharm., Ph. D Department of Pharmaceutics KLE University’s College of Pharmacy BELGAUM- 590010, Karnataka, India Cell No: 0091 974243100 E-mail: bknanjwade@yahoo.co.in. Introduction :. Pharmacokinetic Parameters:

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Pharmacokinetics Calculations

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  1. Pharmacokinetics Calculations Prof. Dr. Basavaraj K. Nanjwade M. Pharm., Ph. D Department of Pharmaceutics KLE University’s College of Pharmacy BELGAUM- 590010, Karnataka, India Cell No: 0091 974243100 E-mail: bknanjwade@yahoo.co.in KLECOP, Nipani

  2. Introduction : • Pharmacokinetic Parameters: • Elimination rate constant • Biological Half life • Rate constant of absorption • Apparent volume of distributions • Area under the curve KLECOP, Nipani

  3. Example : 1 • The plasma concentration after the 250 mg intravenous bolus dose of an antibiotic is given below. Plot the data and describe the pharmacokinetic model. KLECOP, Nipani

  4. Solution : • Graph KLECOP, Nipani

  5. Elimination rate constant : Suppose we choose the following two points to determine the slope of the straight line : • x1= 0 hr, y1 = 10.0 mcg/ml, and x2 = 7.0hr, y2 = 2.0 mcg/ml. then ln2.0 – ln 10.0 ln y2 – ln y1 0.6931 – 2.3026 = = Slope = 7.0 hr 7.0 hr – 0 hr x2 – x1 - 1.6095 - 0.2299/ hr = = 7.0 hr Therefore Ke = - slope = - (-0.2299/hr) = 0.2299/hr KLECOP, Nipani

  6. 0.693 Ke = t1/2 , therefore 0.693 t½ = Ke 0.693 = 3.01 hr = 0.2299/hr • Biological half-life : KLECOP, Nipani

  7. Area under curve : Area from 0 to 7.0 hours – AUC0-7.0 by trapezoidal rule = 34.85 mcg.hr/ml AUC0-7.0 by counting squares = 34.85 mcg.hr/ml AUC0-7.0 by Cutting and Weighing = 34.85 mcg.hr/ml KLECOP, Nipani

  8. 2.0mcg/ml Ct AUC 7.0- ∞= = 8.7 mcg. hr/ml = Ke 0.2299/hr C0 10mcg / ml = = 43.55 mcg.hr /ml AUC0-∞= 0.2299/ hr Ke Total area under curve : • This is a two step method, first determine AUC0-7.0 , then determine AUC7.0-∞ Adding this value to AUC 0-7.0, we have AUC0-infi = AUC0-7.0 + AUC7.0-∞ = 34.85mcg.hr/ml + 8.7mcg.hr/ml = 43.50mcg.hr/ml KLECOP, Nipani

  9. Dose Vd = Co 250 mg = 10 mcg / ml 250 mg = 10 mg/L = 25 L • Volume of distribution : KLECOP, Nipani

  10. intravenous 250 mg injection 0.2299 / hr Description of model : • It shoes that a 250mg dose is administered intravenously. The apparent volume of distribution is 25 L and the rate constant of elimination (Ke) is 0.2299 / hr. since biological half-life is 3.01 hr. 25 LITRES KLECOP, Nipani

  11. Example 2 • The plasma concentration versus time data following the administration of a single 250 mg rapid intravenous bolus dose of a drug is represented by the biexponential equation; C = 1.5e-0.13t + 12.5 e–1.3t. Draw a schematic of the pharmacokinetic model, assuming concentration is in mcg / ml and time is in hours. KLECOP, Nipani

  12. Solution • From the biexponential equation, the following parameters of the two compartment pharmacokinetic model are deduced : b= 0.13/hr (because the smallest hybrid rate constant always b), and B = 1.5 mcg/ml (because B is y-intercept corresponding to b). therefore a must be equal to 1.3/hr, and A = 12.5mcg/ml. In order to draw a schematic of the pharmacokinetic model, the following parameters need to be calculated: rate constants K10, K12, K21,, and apparent volumes of distribution Vc, and Vt. KLECOP, Nipani

  13. 1.625 + 1.95 3.575 Ab + Ba = 0.2554/hr K 21 = = = 14.0 1.5 + 12.5 B + A 0.169/hr (1.3/hr) ( o.13/hr) ab = K10 = = =0.6617/hr 0.2554/hr 0.2554/hr K21 Rate constants K12 = a + b – K21 – K10 K12 = 1.3 /hr + 0.13 /hr – 0.2554 / hr – 0.6617 / hr K12 = 0.5433 / hr KLECOP, Nipani

  14. D Vc = = = 17.857 L 14 mcg ml (17.857 L) (0.5433 / hr+ 0.2554 / hr) 14.2624 L = 55.843 L = Vd = 0.2554 / hr 0.2554 (Vc) (K12) (17.857 L) (0.5433 / hr) Vt = = 37. 986 L = K21 0.2554 /hr Apprent volume of distribution 250 mg B + A Vc (K12 + K21) Vd = K21 Vt = Vd – Vc = 55.843 L – 17.857 L = 37.986 L KLECOP, Nipani

  15. Schematic representation : This schematic shows that the 250 mg dose can was given intravenously. The apparent volume of the central and tissue compartment are 17.857 L and 37.986 L, respectively. The first-order rate constant of transfer of the from the central compartment into the tissue compartment is 0.5433 /hr and the first-order rate constant of transfer of drug from the tissue compartment in to the central compartment is 0.2554 / hr. the first-order rate constant of elimination of drug from the central compartment is 0.6617 / hr. KLECOP, Nipani

  16. 0.5433/hr intravenous 37.986 L 250 mg injection 0.6617 /hr Schematic of the two compartment model 17.857 L 0.2554/hr KLECOP, Nipani

  17. Example -3 • The following data were obtained when a 500 mg dose of an antibiotic was given orally. Calculate the pharmacokinetic parameters, assuming 100% of the administered dose was absorbed. KLECOP, Nipani

  18. Graph KLECOP, Nipani

  19. Solution: Elimination rate constant: The rate constant of elimination is calculated from the terminal linear portion of plasma profile. To determine it, we need to calculate slope of the straight line having y-intercept = B. if natural log are used the rate constant of elimination (b) = negative slope of this straight line. Therefore b = - slope ln 5.734 - ln 4.343 0.2778 = 0.139/ hr = - = - 2 hr (18 - 20) hr KLECOP, Nipani

  20. Biological half life: The biological half life (t1/2) is determined using the equation t1/2 = 0.693/b 0.693 0.139/hr = 4.98 hr = KLECOP, Nipani

  21. The Y-intercept, B The Y-intercept of this straight line is B and is determined using the first order equation ln Ct = ln B – bt Which upon rearrangement gives ln B = ln Ct + bt = ln 4.343 + (0.139/hr)(20hr) =1.4686 + 2.78 = 4.2486 B = Inverse ln 4.2486 = 70.0 mcg/ml KLECOP, Nipani

  22. Feathering the curve : • To obtain the straight line which represents absorption phase, the technique of feathering or the method of residuals is used. for example, to feather the first plasma conc. point at 1 hr, the plasma conc. at 1 hr on the straight line having the y intercept = B is subtracted from the plasma conc. data provided in the data. • ln Ct = ln B - bt ln Ct = ln 70 – (0.139) (1) = 4.2485 – 0.139 = 4.1095 Ct = inverse ln 4.109 = 60.916 mcg/ml KLECOP, Nipani

  23. Graph KLECOP, Nipani

  24. The residual conc. at 1 hr is obtained by subtracting from this concentration at 1 hr provided in the data. therefore the residual concentration at 1hr is, 1 hr = 60.916 – 26.501 = 34.415mcg/ml KLECOP, Nipani

  25. Rate constant of absorption • The rate constant of absorption is obtained from the slope of the straight line which represent absorption as follows; ln 70 – ln 0.1 Ka = a = - slope = - 0 hr – 9.22 hr 6.5511 = - - 9.22 hr = - 0.71/ hr KLECOP, Nipani

  26. (F)(D)(a) B = (Vd)(a-b) (1)(500mg)(0.71/hr) 70 mcg/ml= (Vd)(0.71/hr – 0.139 /hr) 621.72 mg (Vd) = = 8.88 L 70mcg / ml Apparent volume of distribution • Since 100% of the administstered dose was absorbed, F = 1. substittuting the values of B= 70mcg/ml, D= 50mg, a = Ka = 0.71/hr, b Ke = 0.139/hr, KLECOP, Nipani

  27. Area under the curve 70 mcg/ml 70 mcg/ml B A = - - AUC = a b 0.139/hr 0.71/hr AUC = 503.597 mcg – hr/ml – 98.592 mcg – hr /ml = 405.005 mcg – hr/ml KLECOP, Nipani

  28. 0.71 /hr 500 mg 0.139 / hr Description of the model • Schematic shows that a 500 mg of the dose of the drug was administered by an extravascular route. The first-order rate constant of absorption is 0.71/hr and the first order rate constant elimination is 0.139/hr. the apparent volume of the central compartment is 8.88 L. 8.88 LITRS KLECOP, Nipani

  29. Example -3.1 • From the data given Calculate the time when administered drug dose reaches its maximum concentration in the plasma. KLECOP, Nipani

  30. ln Ka – ln Ke t max = Ka – Ke ln 0.71 /hr – ln 0.139 /hr t max = 0.71 /hr – 0.139 /hr – 0.3425 – (- 1.9733) 1.6308 t max = = = 2.856 hr 0.571 /hr 0..571 /hr • From the pharmacokinetic parameters found, the first-order rate constant of absorption, Ka = 0.71/hr and the first order rate constant elimination, Ke = 0.139/hr. KLECOP, Nipani

  31. Example -3.2 • From the data given Calculate the maximum concentration of the drug in plasma attained after the administration of the dose. KLECOP, Nipani

  32. B = 70 mcg /ml, Ka = 0.71 /hr, Ke = 0.139 /hr, and tmax = t’ = 2.856 hr B (e-bt’ – e-at’) C max = C max = (70 mcg/ml) (e-(0.139/hr))(2.856 hr) – e-(0.71/hr)(2.856hr)) C max = (70 mcg/ml) (0.623 – 0.1316) C max = (70 mcg/ml) (0.5407) = 37.85 mcg ml KLECOP, Nipani

  33. Example -4 • The following data were obtained when a 500 mg dose of an antibiotic was given orally calculate the pharmacokinetic parameters, assuming 100% of the administered dose was absorbed. KLECOP, Nipani

  34. graph KLECOP, Nipani

  35. Elimination rate constant: The rate constant of elimination (b) is calculated using the terminal two points of the plasma profile as follows; Therefore b = - slope ln 0.546 mcg/ml - ln 0.300 mcg/ml = - (24 - 28) hr 0.5988 = 0.15/ hr = - 4 hr KLECOP, Nipani

  36. The y-intercept, b, of this straight line is determined using the first-order rate equation : B = Ct e bt = (0.3 mcg/ml) e(0.15/hr)(28hr) B = (0.3 mcg/ml) (66.6863) B = 20 mcg/ml KLECOP, Nipani

  37. To obtained the straight line which represents absorption phase, the technique of feathering is used. The plasma profile is feathered with respect to the straight line having y-intercept = B. To feather the first concentration point, the concentration at 2 hr on the straight line having y- intercept = B is subtracted from the data concentration at 2 hr. C = Be-bt = (20 mcg/ml)e-(1.5/hr)(2hr) C = (20 mcg/ml) (0.7408) = 14.816 mcg/ml Therefore, residual concentration at 2 hr is : 14.816 mcg/ml – 3.915 mcg/ml = 10.901 mcg/ml KLECOP, Nipani

  38. Biological half life: The biological half life (t1/2) is determined using the equation t1/2 = 0.693/b 0.693 0.15/hr = 4.62hr = KLECOP, Nipani

  39. Rate constant of absorption • The rate constant of absorption is obtained from the slope of the straight line having the y-intercept = A. It is calculated as follows; ln 40 mcg/ml – ln 0.221 mcg/ml Ka = a = - slope = - 0 hr – 8 hr 5.198 = 8hr = 0.65/ hr KLECOP, Nipani

  40. Lag-Time • Since the value of the y-intercept A is not equal to the value of the y-intercept B, the dosage from exhibits lag-time. The lag-time (L) is determined using equation ln A – ln B L = a – b ln 40 mcg/ml – ln 0.20 mcg/ml = 0.65 – 0.15 0.693 = 0.5 / hr = 1.386 hr KLECOP, Nipani

  41. The equation for calculating the time of maximum concntration of drug in plasma in presence of lag-time, tmax (L), is ln A – ln B + ln a – ln b t max = a - b ln 40 – ln 20 + ln 0.65 – ln 0.15 t max = 0.65 /hr – 0.15 /hr 1.4663 t max = = 4.319 hr 0.5 /hr KLECOP, Nipani

  42. Be-bt” – Ae-at”) C max (L) = C max (L) = (20 mcg/ml) (e-(0.15/hr))(4.319 hr) – e-(0.65 /hr)(4.319hr)) C max (L) = (20 mcg/ml) (0.5253) – (40 mcg /ml) (0.0604) C max (L) = 10.463 mcg/ml – 2.415 mcg /ml C max (L) = 8.048 mcg /ml KLECOP, Nipani

  43. Pharmacokinetics and Pharmacodynamics Parameters KLECOP, Nipani

  44. Measurement of bioavailability • Pharmacokinetic methods ( indirect ) 1. Blood analysis 2. Urinary excretion data • Pharmacodynamic methods ( direct ) 1. Acute pharmacological response 2. Therapeutic response KLECOP, Nipani

  45. Blood analysis • Plasma level time studies or The plasma concentration – time curve or blood level curve. • A direct relationship exists concentration of drug at the site of action & concentration of drug in the plasma. • Serial blood samples are taken after drug administration & analyzed for drug concentration. • A typical blood level curve obtained after oral administration of drug. KLECOP, Nipani

  46. KLECOP, Nipani

  47. Parameters determined • Peak Plasma Concentration (Cmax) • Time of Peak concentration (tmax). • Area Under Curve (AUC) • Minimum Effective Concentration (MEC) / Minimum Inhibitory Concentration (MIC). • Maximum Safe Concentration (MSC) / Maximum Safe Dose (MSD). • Duration of action • Onset of action. • Intensity of action. Pharmacokinetic parameters Pharmacodynamics parameters KLECOP, Nipani

  48. Parameters determined • AUC or Extent of absorption can be measured by 3 methods… 1.Planimeter Instrument for mechanically measuring the area 2. Cut & weigh method AUC is cut & weighed on analytical balance. The weight obtained is converted to proper unit by dividing it by the wt of a unit area of same paper. 3. Trapezoidal method KLECOP, Nipani

  49. Parameters determined 3. Trapezoidal method AUC = ½ ( C1 + C2) (t2 – t1) + ½ (C2 + C3) (t3 – t2) +……. ½ (C n-1 + C n ) (tn – tn-1 ) C = Concentration t = time subscript= sample number AUC = Area Under Curve KLECOP, Nipani

  50. Parameters determined Relative bioavailability F rel = ( AUC) drug . (Dose) standard (AUC) standard .(Dose) drug Absolute bioavailability Fab = (AUC)drug . (Dose) IV (AUC)IV . (Dose) drug KLECOP, Nipani

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