Drug Clearance

Clearance lecture

1. DRUG CLEARANCE Dr. Rajib Bhattacharjee Asstt . Professor Dept of Pharmacy, NSU

2. The extent of ionization of acidic and basic drugs in renal tubules depends on ( i ) pK a of the compound and (ii) pH of the tubular fluid that is urine pH. Effect of urine pH on tubular reabsorption & renal drug secretion For a weakly acidic drug, Fraction of drug ionized = antilog (pH- pKa ) / 1 + antilog (pH- pKa ) For a weakly basic drug, Fraction of drug ionized = 1 / 1 + antilog (pH- pKa )

3. Acidification of urine : promotes reabsorption of weak acids and retards reabsorption of weak bases. Alkalization of urine : retards reabsorption of weak acids and promotes reabsorption of weak bases. Therefore, excretion of weak acids is increased in ………….. urine and excretion of weak bases is increased in ………….. urine. pH of urine % ionized of a weak acidic drug % ionized of a weak basic drug pK a 3.0 pK a 5.0 pK a 7.5 pK a 10.5 4.5 96.93 24.03 99.90 100.00 6.0 99.90 90.91 96.93 100.00 7.5 100.00 99.68 50.00 99.90 Ionization of acidic & basic drugs at different urine pH The effect of pH on urinary drug excretion is significant only when…………..

4. Effect of rate of urine flow on tubular reabsorption and renal drug secretion Rate of urine flow ( normal 1-2 ml / min ) will influence the amount of drug which is reabsorbed. Urine flow increases  time for reabsorption decreases  drug excretion promoted Large fluid intake, use of diuretics etc. increases urine flow and decreases drug reabsorption .

5. Clearance : Drug clearance is the fixed volume of body fluid which is cleared of drug per unit time. Here body fluid indicates apparent volume of distribution. Example - A drug has a clearance of 50 ml / min means 50 ml of the body fluid ( V D ) becomes free of drug per minute. (ii) Drug clearance is the ratio of drug elimination rate to plasma drug concentration. Plasma drug concentration Drug elimination rate Cl T = dD E / dt = C p Unit = mass / time  mass / volume = volume / time Clearance values are often expressed as per kilogram body weight basis since this pharmacokinetic parameter varies with body weight. Thus, unit of Cl T becomes ml / min / kg.

6.  Cl T = = C p dD E / dt C p k E C p V D k E V D = This equation shows that Cl T is the product of two constants k E and V D . In the above definitions, Cl T will remain constant as long as drug elimination follows first order kinetics. The volume of distribution and elimination rate constant are both affected by blood flow. So blood flow will influence the magnitude of drug clearance. Now drug elimination rate dD E / dt  D B  dD E / dt = k E x D B where k E = elimination rate constant. Again D B = C p x V D  dD E / dt = k E C p V D

7. Drug excretion rate Glomerular filtration rate + + = Active secretion rate Passive reabsorption rate  Cl R = = C p dD e / dt C p filtration secretion reabsorption rate rate rate – + From physiologic viewpoint, ultimate drug excretion is the combined effect of three processes – i ) filtration ii) secretion and iii) reabsorption

8. In real cases, drug excretion is often measured by comparing its clearance value with that of a standard reference compound such as inulin or creatinine . These compounds are only excreted by filtration neither secreted nor reabsorbed. So there clearance is equal to GFR that is 125 ml /min. The clearance ratio of a drug is its clearance value divided by clearance value of the standard ( inulin ). The value of the ratio gives a gross idea about the mechanism involved in renal excretion. Clearance ratio : Cl R (drug) / Cl R ( inulin ) Probable mechanism of renal excretion < 1 Drug is partially reabsorbed or reabsorption dominates over secretion. = 1 Drug is filtered only or secretion and reabsorption rates are similar. > 1 Drug is actively secreted or secretion dominates over reabsorption . For a substance which is only filtered in the kidneys but neither secreted nor reabsorbed Cl R = = C p dD e / dt C p drug filtration rate

9. Cl R = dD e / dt C p A B C p dD e \ dt  Determination of renal clearance - ( i ) Slope of the line gives the clearance value ( Cl R ). Drug A has higher clearance than drug B (ii) A B AUC D e Plot of cumulative drug excretion vs AUC slope = Cl R Drug A has higher clearance than drug B

10. (iii) Any type of clearance value can be obtained if the volume of distribution and the corresponding elimination, excretion and biotransformation rate constants are known. Cl T = k E V D Cl R = k e V D Cl b = k m V D Again Cl T = Cl R + Cl b  k E = k e + k m  Fraction of drug excreted unchanged by urine - f e = k e / k E  Cl R = ( k e / k E ) Cl T = f e Cl T Relationship among clearance, elimination half-life and volume of distribution – Cl T = k E VD and k E = 0.693 / t½ Cl T = 0.693V D / t½ and t½ = 0.693V D / Cl T This equation shows that if Cl T of a drug is changed due to renal problems, the elimination half-life of the drug will change accordingly  Determination of renal clearance -

11. Problems : 1. A new antibiotic secreted by the kidney ; V D is 35 L in the normal adult. The clearance of this drug is 650 ml/min. What is the usual t 1/2 of the drug? What would be the new t 1/2 of this drug in an adult with partial renal failure whose clearance of the antibiotic was only 75 ml/min? 2. An antibiotic is given by IV bolus injecton at a dose of 500 mg. The apparent volume of distribution was 21 liter and the elimination half-life was 6 hrs. Urine was collected for 48 hrs and 400 mg of the unchanged drug was recovered. What is the fraction of the dose excreted unchanged in the urine? Calculate k, ke , Cl T , Cl R and Cl h .