calculations of bioavailability

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Calculations of bioavailability & bioequivalence from urinary excretion data & plasma concentration-time curve:

10 March 2013 dept of pharmaceuticus 1 Calculations of bioavailability & bioequivalence from urinary excretion data & plasma concentration-time curve saiesh phaldesai

Contents :

10 March 2013 dept of pharmaceuticus 2 Contents Calculation of bioavailability & bioequivalence from given data. Calculation of (C max ),(t max ), & biological half life from given data. Calculation of area under curve (AUC) from given data. Calculation of bioavailability from urinary excretion data. Calculation of urinary excretion rate constant & elimination rate constant.

Plasma concentration-time profile:

10 March 2013 dept of pharmaceuticus 3 Plasma concentration-time profile

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10 March 2013 dept of pharmaceuticus 4 Pharmacokinetics parameter :- C max T max AUC Pharmacodynamics parameter :- MEC MSC Onset of action Onset of time Duration of action Intensity of action Therapeutic range

Calculation of bioavailability & bioequivalence from given data. :

10 March 2013 dept of pharmaceuticus 5 Calculation of bioavailability & bioequivalence from given data. Following data is obtained from formulation in volunteers of average weight 50 kg. Drug product Dose (mg/kg) AUC (mg/l) I.V solution 1.2 450 Oral solution 4.0 822 Oral capsule 4.0 736 Oral SR tablet 8.0 1040

Formula::

10 March 2013 dept of pharmaceuticus 6 Formula: (AUC) oral D iv (AUC) iv D oral (AUC) test D std (AUC) std D test Fr = F =

Absolute bioavailability:

10 March 2013 dept of pharmaceuticus 7 Absolute bioavailability (AUC) oral D iv (AUC) iv D oral Capsule:( AUC) oral =736 µg hr/ml (AUC) iv =450 µg hr/ml D oral =4 mg/kg D iv =1.2 mg/kg 736 × 1.2 450 × 4 % F =0.49 × 100 =49% F = F = = 0.49

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10 March 2013 dept of pharmaceuticus 8 Oral SR tablet : ( AUC) oral =1040 µg hr/ml (AUC) iv =450 µg hr/ml D oral =8 mg/kg D iv =1.2 mg/kg 1040 × 1.2 450 × 8 % F =0.346 × 100 =34.6% F = = 0.346

Relative bioavailability:

10 March 2013 dept of pharmaceuticus 9 Relative bioavailability (AUC) test D std (AUC) std D test Capsule : (AUC) test =736 µg hr/ml D test = 4 mg/kg (AUC) std =822 µg hr/ml D std = 4 mg/kg 736 × 4 822 × 4 % Fr =0.895 × 100 =89.5 % F r = F r = =0.895

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10 March 2013 dept of pharmaceuticus 10 SR tablet : (AUC) test =1040 µg hr/m D test =8 mg/kg (AUC) std =822 µg hr/ml D test = 4 mg/kg Fr = 1040 × 4 =0.632 822 × 8 % Fr =0.632 × 100 =63.2% Result: absolute bioavailability of drug Capsule=40% SR tab =34.6% Relative bioavailability of drug Capsule=89.5% SR tablet =63.2% Two solid formulations are not bioequivalent.

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10 March 2013 dept of pharmaceuticus 11 The equation that best fits the pharmacokinetic of paracetamol after oral administration of 500mg dose is: C = 1.18 (e -0.24t –e -1.6t ) assuming one-compartment kinetic determine- Peak time, peak plasma concentration, elimination half life.

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10 March 2013 dept of pharmaceuticus 12 Solution :- C = 1.18 (e -0.24t –e -1.6t ) where A= 1.18, k E =0.24, K a =1.6, 2.303 × log ( K a / k E ) K a – k E 2.303 × log (1.6 /0.24) 1.6 -0.24 t max =1.4hrs Peak plasma concentration (C max ) = C = 1.18 e -0.24×1.4 - e -1.6×1.4 = 1.18 (0.714 – 0.106) =1.18 (0.6075) C max =0.717 µg /ml peak time (t max ) = =

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10 March 2013 dept of pharmaceuticus 13 Elimination half life = where k E =0.24, t 1/2 =0.693 /k E =0.693 /0.24 =2.88 hrs Result :-assuming one compartment kinetic the values was found to be- Peak time = 1.4 hrs Peak plasma concentration = 0.717 µg /ml Elimination half life =2.88 hrs

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10 March 2013 dept of pharmaceuticus 14 AUC can be measured by the following methods. Trapezoidal method. Integration method. Physical method. Planimeter. Counting the squares.

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10 March 2013 dept of pharmaceuticus 15 Trapezoidal method It involves dividing the AUC of the plasma concentration time curve into several trapezoids. The total area of the trapezoids will approximate the area under the curve. Average area at first time interval is obtained by adding the concentration at the beginning and at the end of the first time interval. i.e. area of a trapezoids is half the product of the sum of the heights times the width .

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10 March 2013 dept of pharmaceuticus 16 More complex method for assessment of AUC. This method involves fitting an equation to the observations made and then calculating the area by integrating the fitted equation. Physical method (cut and weigh method): This method involves, first preparing the calibrated (reference) plot by cutting the squares of the graph and their weights are recorded and plotted as weight Vs area. Then the sample curve to be estimated is cut and its weight is recorded. Once the weight is known then by interpolation method area of the sample graph can be found from the reference plot. Advantage: Irregularly shaped curve can be estimated. Integration method:

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10 March 2013 dept of pharmaceuticus 17 It is precision instrument for measurement of AUC. It measures the area by tracing the outlines of the curve. The disadvantages of this method is that the degree of error is high due to instrument and human errors . Counting the Squares: In this method, the total number of squares enclosed by the plasma conc-Vs time curve is counted. The area of each square is determined using the relationship: Area of the square =Height x Width. The Height and Width are measured in the units of concentration of Y-axis and time on X-axis. Planimeter:

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10 March 2013 dept of pharmaceuticus 18 For example: If 10 division on X-axis were used to represent 5 hrs. the each division on X-axis is equal to 0.5 hrs. Similarly, if 9 division on Y-axis were used to represent 45  g/ml, then each division on Y-axis is equal to 5  g/ml. Thus, the are of a square composed of each division on the two axis is =0.5 x 5 =2.5  g-hr/ml for one square Total AUC= number of squares x area of square = 30 x 2.5 = 75  g-hr/ml .

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10 March 2013 dept of pharmaceuticus 19 From the following blood data obtained after the oral administration of 50 mg of drug A, calculate its AUC? Time in hr Plasma drug concentration in mcg/ml 1 5.5 2 9.2 3 14.9 4 10.3 5 7.1 6 2.2

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10 March 2013 dept of pharmaceuticus 20 16 2 4 6 8 10 12 14 1 2 3 4 5 6 7 8 ---------------- ---------------------------- --------------------------------------------- ------------------------------- ---------------------- ------- Time in hour Plasma drug conc. In mcg/ml t 1 c 1 c 2 c 3 c 4 c 5 c 6 t 2 t 3 t 4 t 5 t 6

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10 March 2013 dept of pharmaceuticus 21 According to trapezoidal rule Formula=AUC =1/2 (C 1 +C 2 ) (t 2 -t 1 ) + 1/2s (C 2 +C 3 ) (t 3 -t 2 )…..+1/2 (C n-1 + C n ) (t n – t n-1 ) where C = drug concentration t = time AUC = ½ (5.5 + 9.2) (2-1) + ½ (9.2 +14.9) (3-2)+ ½(14.9 + 10.3 ) (4-3) + ½(10.3 +7.1) (5-4) + ½ (7.1 +2.2) (6-5) AUC = 45.35 mcg/ml hr.

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10 March 2013 dept of pharmaceuticus 22 The AUC (between 0 to 24) of Diclofenac sodium when given as oral tablet was found to be 183.24 mcg/ml hr.the last point of concentration was found to be 2.2 mcg/ml at 24 hr.Given the value of k e =0.125 hr-1, report the AUC o ∞ of the drug. Solution:- AUC o ∞ =AUC 0 t +C* /k e AUC o ∞ =183.24 +2.2/0.125 AUC o ∞ =200.84 mcg/ml hr. Result :- AUC o ∞ =200.84 mcg/ml hr of the drug.

Urinary excretion study:

10 March 2013 dept of pharmaceuticus 23 Urinary excretion study Adv. :- i) Avoid inconvenience & hazards of withdrawal of i.v. blood samples. ii) A less sensitive analytic method is required for determining urine drug conc. as compared to plasma conc. Disadv. :- i) Erroneous in calculations of the absorption rate because, After each collection the Remaining amount of drug in the residual urine is added to the next point of collection. ii) frequent collection of samples can be difficult & collection time may be extended to long periods of time.

Calculation of urinary excretion rate constant & elimination rate constant.:

10 March 2013 dept of pharmaceuticus 24 Calculation of urinary excretion rate constant & elimination rate constant. The amount of drug excreted in urine after i.v. dose of 400 mg of norfloxacin was as follows: After completing the table & using rate of excretion method, determine-elimination rate constant & half life of drug, urinary excretion rate constant. t(hours) X u (mg) dx u /dt (mg/hrs) t* 0 0 2 49.05 4 80.0

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10 March 2013 dept of pharmaceuticus 25 solution:- Time (hr) t dt X in (mg) X u dx u dx u /dt (mg/hrs) Mid pt logdx u /dt 0 _ _ _ 0 0 _ 2 2 49.05 49.05 24.75 1 (1.3935) ~1.4 4 4 80.0 80.0 13.81 3 1.1402

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10 March 2013 dept of pharmaceuticus 26 0.2 0.4 0.6 0.8 1.2 1.0 1.4 1.6 1.8 1 2 3 4 ------------------------ ------------------------------ ------- -------------------------------------- 1.59 (y- intercept) (Log KeXo) Log dXu/dt Mid point time for urine collection t* 5

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10 March 2013 dept of pharmaceuticus 27 Estimation of rate constant=k E from graph slope = -k E /2.303 eq (1) slope =1.1-1.3 =-0.2/2 =-0.1 -k E = 2.303 × slope = 2.303 × (-0.1) -k E =0.2303/hr Half life (t 1/2 ) =0.693/ eq (2) =0.693/0.2303 = 3 hrs 3 – 1 - k E

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10 March 2013 dept of pharmaceuticus 28 Urinary excretion rate constant = k e y intercept = log k e x 0 (3) where k e = urinary excretion rate x 0 = iv dose (400 mg) Putting the value in eq..(3) y intercept =1.59 1.59 = log k e × 400 log k e =1.59/400 =0.2013/2.6020 =0.77/hr Result :-The elimination rate constant & the excretion rate constant was found to be 0.2303/hr & 0.77/hr respectively.

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10 March 2013 dept of pharmaceuticus 29 The three pharmacokinetic parameter from urinary excretion data of a drug given as 50 mg oral formulation of two different companies, of which A is the innovator’s product, are as follows: parameters formulations A B (dx u /d t ) max (mg/hr) 6.0 8.0 (t u ) max (hour) 2.0 1.0 x u ∞(mg) 39.1 35.9

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10 March 2013 dept of pharmaceuticus 30 Solution :- Relative bioavailability = (x u ∞ ) test D std (x u ∞ ) std D test (x u ∞ ) test =35.9 (x u ∞ ) std =39.1 D std = 50mg D test = 50mg Relative bioavailability of formulation ‘B’ against ‘A’ f r A =39.1 × 50/35.9 × 50 =1.08 % f r A = 1.08 × 100 =108 % f r B = 35.9 × 50/39.1 × 50 =0.92 % f r B = 0.92 × 100 =92% Result :- The bioavailability of formulation ‘B’ against ‘A’ is 92 % & two formulations are not bioequivalent. f r =

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10 March 2013 dept of pharmaceuticus 31 Consider a drug that is eliminated by first order renal excretion & hepatic metabolism. the drug follows a one compartment model & is given in single oral dose (100mg).the drug is 90%systematically available. The total amount of unchanged drug recovered in the urine is 30mg,the elimination half life for drug is 3.3 hr & its apparent volume of distribution is 1000mL.from this given information find (a) Total body clearance (b) Renal clearance (c) Nonrenal clearance of the drug

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10 March 2013 dept of pharmaceuticus 32 Solution:- (a) Total body clearance:Cl T =kV D Cl T =0.693/3.3(1000) =210 mL/hr. (b)Renal clearance. first find k e : k e /k=D u ∞ /fd 0 =D u ∞ /M u ∞ + D u ∞ k e =(0.693/3.3)(60/30+60) =0.14 Then, renal clearance=Cl R =k e V D =(0.14)(1000) =140 mL /hr (c) Nonrenal clearance :Cl h =Cl T –Cl R =210 -140 =70 mL /hr. Result :-from the given information the following values was found to be Total body clearance =210 mL/hr Renal clearance =140 mL/hr Nonrenal clearance =70 mL/hr

References :

10 March 2013 dept of pharmaceuticus 33 References Biopharmaceutics and Pharmacokinetics- A Treatise, by D.M.Brahmankar Sunil B.Jaiswal, Vallabh Prakashan A text book of Biopharmaceutics and Pharmacokinetics, by Shobha Rani,R.Hiremath Applied Biopharmaceutics and pharmacokinetics by Shargel.L.2 nd edition

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10 March 2013 dept of pharmaceuticus 34 THANK YOU

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