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Clearance

1. 1. Clearance CHAPTER 9 OBJECTIVES 1. Given patient information regarding organ function, the student will calculate (III) changes in clearance and other pharmacokinetic parameters inherent in compro- mised patients. 2. Determine the total clearance based on Dose and AUC. 3. Determine clearance of an organ based on dose, AUC, and fraction of drug elimi- nated by the organ 4. Determine change in clearance due to functional changes in an organ. 5. Determine change in clearance due to change in blood flow through an organ. 6. Prepare a professional consult (V) and justify (VI) modifications in drug therapy based on clearance of a drug. 9-1 Basic Pharmacokinetics REV. 99.4.25 Copyright © 1996-1999 Michael C. Makoid All Rights Reserved http://kiwi.creighton.edu/pkinbook/
2. 2. Clearance 9.1 Equations Rate of Elimination- Cl = -------------------------------------------------- (EQ 9-1) Serum Concentration Cl tot = f ⋅ D os --------------- - (EQ 9-2) AUC Cl r = Cl tot ⋅ ( Fraction of drug that is renally eliminated ) (EQ 9-3) Cl H = Cl tot ⋅ ( Fraction of drug that is hepatically eliminated ) (EQ 9-4) L min L Q r = 0.0191 ------------------- renal blood perfusion ⋅ 70kg ⋅ 60 --------- ≈ 80 -----blood - (EQ 9-5) min ⋅ kg hr hr L min L Q H = 0.0238 ------------------- hepatic blood perfusion ⋅ 70kg ⋅ 60 --------- ≈ 100 -----blood - (EQ 9-6) min ⋅ kg hr hr Er = ( Cl r ) ⁄ Q r (EQ 9-7) E H = ( Cl H ) ⁄ Q H (EQ 9-8) Q ⋅ Cl f u ⋅ Cl int = --------------- - (EQ 9-9) Q – Cl Q ⋅ Cl- --------------- Q – Cl Cl int = ---------------- (EQ 9-10) fu f u∗ ⋅ Cl∗ int F i = ------------------------- - (EQ 9-11) f u ⋅ Cl int ∗ FR = Q - ------ (EQ 9-12) Q Fi ⋅ FR F Cl = ----------------------------------------- (EQ 9-13) F R + Er ( Fi – FR ) Cl∗ tot Cl∗ H + Cl∗ r ∗∗ FCl tot = ------------- = k ⋅ V = ----------------------------- - ---------------- (EQ 9-14) k⋅V Cl tot Cl H + Cl r 0.80 ≤ FCl t o t ≤ 1.20 (EQ 9-15) 9-2 Basic Pharmacokinetics REV. 99.4.25 Copyright © 1996-1999 Michael C. Makoid All Rights Reserved http://kiwi.creighton.edu/pkinbook/
3. 3. Clearance 9.2 Definitions and Terms Clearance: The hypothetical volume of a fluid from which a substance is totally and irreversibly removed per unit time. 3 L ⁄T Dimensions: Examples of fluids: blood, serum, plasma, bile, gut contents, CSF. ( Cl ), ( Cl tot ) Systemic or Total Body Clearance: Removal process is elimination (excretion and metabolism). Fluid is usually plasma or serum (rarely blood). ( Cl r ) Renal Clearance: Removal process is urinary excretion of unchanged drug. Fluid is usually plasma or serum (rarely blood). ( Cl m ) Metabolic Clearance: Removal process is metabolism. Fluid is usu- ally blood (rarely plasma or serum). ( Cl H ) Hepatic Clearance: This is when the liver is the metabolic Cl m organ. ( Cl cr ) Creatinine Clearance: This is Cl r applied to endogenous creatinine. It is used to monitor renal function, and thus is a valuable parameter for calculating dosage regimens in elderly patients or those suffering from renal dysfunction. Cre- atinine t 1 ⁄ 2 = 231min Value for normal males: 117 ± 20 ml/min Value for normal females: 108 ± 20 ml/min Inulin Clearance: This is for inulin, and yields the glomerular Cl inulin Cl r filtration rate. Value for normal males: 124.5 ± 9.7 ml/min Value for normal females: 108.8 ± 13.5 ml/min 9-3 Basic Pharmacokinetics REV. 99.4.25 Copyright © 1996-1999 Michael C. Makoid All Rights Reserved http://kiwi.creighton.edu/pkinbook/
4. 4. Clearance 9.3 Measurement of Creatinine Clearance The mass of endogenous creatinine excreted into the urine collected over a given time interval ( ∆t ) is determined. The mean serum creatinine concentration ( Cs )cr over that interval is calculated from sample determinations; this should be the con- centration halfway through the interval. In practice, ∆t = 24hr and, as ( C s ) cr is rela- tively constant, the serum sample is taken at any convenient time. Let “a” be a volume of serum having a creatinine concentration of ( C s ) cr . The mass of creatinine in the serum will be a ⋅ ( Cs ) cr . If this creatinine is totally and irrevers- ibly removed from the serum to the urine in the time interval, ∆t , then a ⋅ ( C s ) cr = ∆Xu (EQ 9-16) ∆Xu a = -------------- - Thus, (EQ 9-17) ( C s )cr a ⁄ ∆t ; The volume of serum from which this creatinine is removed in unit time is this is the definition of clearance. Hence, ( ∆Xu ⁄ ∆t ) a T Cl cr = ----- = -------------------------- - (EQ 9-18) ∆t ( C s ) cr Siersbaek-Neilson et al. report a value of 11.1 µg ⁄ ml for ( Cs )cr in 149 males (aged 20-99). The value of ( ∆X u ⁄ ∆t ) T decreased with age from 16.53 µg ⁄ min per Kg body µg ⁄ min weight (age 20-29) to 6.53 per Kg body weight (age 90-99). For a 25 year ml old 70Kg male, equation 9-18 yields Cl cr = 104.2 --------- min 9-4 Basic Pharmacokinetics REV. 99.4.25 Copyright © 1996-1999 Michael C. Makoid All Rights Reserved http://kiwi.creighton.edu/pkinbook/
5. 5. Clearance 9.4 Model Correlations Although intrinsically model independent, clearance can also be related to com- partmental models. 9.4.1 RENAL CLEARANCE The plasma renal clearance of a drug may be measured analogously to creatinine clearance: ( ∆Xu ⁄ ∆t ) T Cl r = -------------------------- - (EQ 9-19) Cp The practical versions of and states: ( ∆X u ⁄ ∆t ) = k u X = k u VC p (EQ 9-20) T Comparing equation 9-19 and equation 9-20, Cl r = k u V (EQ 9-21) This relates clearance to model parameters. What is the slope of a plot of ( ∆Xu ⁄ ∆t ) T against Cp ? Note that if Cl r > 117 ± 20ml ⁄ min (males), it may indicate active secretion of the drug into the kidney tubules. If Cl r < 108 ± 20ml ⁄ min (females), it may indicate reabsorp- tion of the drug from the kidney tubules. 9.4.2 SYSTEMIC CLEARANCE AND METABOLIC CLEARANCE How could you measure Cl ? and Cl m By analogy, 0.693V Cl tot = KV = ----------------- (EQ 9-22) t1 ⁄ 2 and , so . K = ku + km Cl tot = Cl r + Cl m Consequently, fractional changes in clearance, 9-5 Basic Pharmacokinetics REV. 99.4.25 Copyright © 1996-1999 Michael C. Makoid All Rights Reserved http://kiwi.creighton.edu/pkinbook/
6. 6. Clearance ° ° ° ° K ⋅V Cl h + Cl r F Cl = ------------------------ = ---------------- - (EQ 9-23) K⋅V Cl h + Cl r tot where X° is new or altered variable. Hepatic function and renal function are not a priori connected, although some physiological functional changes might result in similar changes in clearance of both organs. We can see from equation 9-12 that changes in total body clearance can result in changes in either K, V, or both. The consequences of that will be discussed in the section on dosage regimens. 9.4.3 USE IN PHARMACOKINETIC EQUATIONS Systemic Clearance (Cl) can be used in many equations where the drug is removed by elimination (renal excretion and metabolism). If renal excretion is the only removal process, use Cl r l if metabolism, use Cl m . Some examples: Q- ( C p ) ss = ----- Intravenous infusion: Cl f( Xa) 0 fD fD Oral and Intravenous Bolus: Cl = --------------- = --------------- = ------------ ∞ ∞ AUC ∫ C p dt ∫ C p dt 0 0 This equation becomes a means of calculating Cl from plasma data. fD - Dosage Regimen: ( Cp ) ss = ------------ τ ⋅ Cl These examples are all model-independent expressions, which are very useful in calculating dosage regimens. The importance of clearance terms rests on their abil- ity to account for variations in both t 1 ⁄ 2 and / or V simultaneously, as both these parameters can change in disease states and with age. 9-6 Basic Pharmacokinetics REV. 99.4.25 Copyright © 1996-1999 Michael C. Makoid All Rights Reserved http://kiwi.creighton.edu/pkinbook/
7. 7. Clearance 9.5 Physiological Factors Affecting Clearance INTRINSIC CLEARANCE ( Cl int ) 9.5.1 Intuitively, it may be recognized that two factors will affect the clearance of a drug: 1. The rate at which blood is presented to the eliminating organ. 2. The intrinsic ability of the eliminating organ to clear the drug. Mathematically, a hyperbolic equation has been derived to illustrate the relative effect of these factors. (Note: this is one model of clearance. There are several oth- ers which also illustrate the effect of these factors.) Liver Drug Metabolism QH ⋅ f u ⋅ ( Cl H )int Cl H = ------------------------------------------ (EQ 9-24) Q H + f u ⋅ ( Cl H )int Where Q H is the rate of blood flow through the liver (assumed 23.8 ml/min/Kg body weight in normal adult), fu is the fraction unbound of the drug, and ( Cl H ) int is the intrinsic hepatic clearance of the drug. If there were no physiological limits to the rate of blood flow( Q H → ∞ ), hence equa- tion 9-24 becomes Cl H = ( Cl H )int (EQ 9-25) This equation provides a definition for intrinsic clearance, namely the clearance of a drug were there to be no physiological limits on the rate of blood flow through the clearing organ. 9-7 Basic Pharmacokinetics REV. 99.4.25 Copyright © 1996-1999 Michael C. Makoid All Rights Reserved http://kiwi.creighton.edu/pkinbook/
8. 8. Clearance By analogy for excretion of unchanged drug by the kidney: Kidney drug excretion Q r ⋅ f u ⋅ ( Cl r )int Cl r = --------------------------------------- (EQ 9-26) Q r + fu ⋅ ( Cl r )int Where Q r is the rate of blood flow through the kidney (assumed 19.1 ml/min per Kg body weight in normal adults), and ( Cl r ) int is the intrinsic renal clearance of the drug. Note that the value of Cl cr (assumed 1.75 ml.min per Kg body weight in normal adults) is about 9% of Q r . 9.5.2 EXTRACTION RATIO (E) This is defined as “the ratio of the clearance of a drug compared to the rate of blood flow through the clearing organ.” As such, it indicates what fraction of the drug in the blood is cleared (extracted) on each passage through the clearing organ. Note: when using clearance to calculate extraction ratio, blood flow must be used. Drug metabolism by the liver Cl H E H = -------- - (EQ 9-27) QH Where is the steady-state hepatic extraction ratio. EH By comparison with equation 9-24, f u ⋅ ( Cl H ) int E H = ------------------------------------------ (EQ 9-28) Q H + f u ⋅ ( Cl H ) int Thus, the range of values of E H is from zero, when ( Cl H )int = 0 , to one, when Q H = 0 or ( Cl H ) int » Q H . For example, propanolol has E H = 0.75 , yielding 17.9ml 71.4ml ( Cl H )int = ---------------------------------------------------- and in normal adult males. Cl H = ---------------------------------------------------- - - min ⋅ Kg body weight min ⋅ Kg body weight 9-8 Basic Pharmacokinetics REV. 99.4.25 Copyright © 1996-1999 Michael C. Makoid All Rights Reserved http://kiwi.creighton.edu/pkinbook/
9. 9. Clearance Kidney excretion of unchanged drug Cl r Er = ------- - (EQ 9-29) Qr f u ⋅ ( Cl r ) int Er = --------------------------------------- (EQ 9-30) Q r + f u ⋅ ( Cl r ) int where is the steady-state renal extraction ratio. Er ( Cl r ) int = 0 Thus the range of values is from zero, when , to one, when or Qr = 0 1.72ml ( Cl r ) int » Q r . For example, digoxin has , yielding E r = 0.09 Cl r = ---------------------------------------------------- - min ⋅ Kg body weight 1.89ml ( Cl r ) int = ---------------------------------------------------- and in normal adult males. In this case, note that - min ⋅ Kg body weight Cl r ≈ Cl c r 9-9 Basic Pharmacokinetics REV. 99.4.25 Copyright © 1996-1999 Michael C. Makoid All Rights Reserved http://kiwi.creighton.edu/pkinbook/
10. 10. Clearance 9.6 Hepatic Function and Clearance 9.6.1 ALTERATIONS IN HEPATIC BLOOD FLOW For a given drug, equation 9-24 predicts that alterations in the hepatic blood perfu- sion rate will cause a change in drug clearance, assuming the intrinsic hepatic clearance is unaltered. A general equation may be derived relating the ratio of hepatic clearances at two blood perfusion rates to the fractional change in perfu- sion rate and the extraction ratio of the drug. Cl∗ H FR ----------- = ---------------------------------------- - - (EQ 9-31) F R + EH ( 1 – F R ) Cl H where denotes normal hepatic clearance, Cl H Cl ∗ H denotes altered hepatic clearance Q H∗ is the new flow rate over the old flow rate, the fractional change in blood F R = ---------- QH perfusion rate, and is the hepatic extraction ratio under normal conditions. EH The equation predicts that, for any given decrease in blood perfusion rate, drugs having a large normal extraction ratio will experience a proportionally greater reduction in clearance than drugs having a small normal extraction ratio. Liver blood flow can be reduced by congestive heart failure, for example. The intrinsic hepatic clearance can be represented by the inherent activity of the enzymes responsible for drug metabolism. 9.6.2 ALTERATIONS IN HEPATIC INTRINSIC CLEARANCE For any given drug, equation 9-24 predicts that alterations in the intrinsic hepatic clearance will cause a change in drug clearance, assuming the blood flow rate is unchanged. A general equation may be derived relating the ratio of hepatic clear- ance at two intrinsic hepatic clearances to the fractional change in intrinsic hepatic and the extraction ratio of the drug. 9-10 Basic Pharmacokinetics REV. 99.4.25 Copyright © 1996-1999 Michael C. Makoid All Rights Reserved http://kiwi.creighton.edu/pkinbook/
11. 11. Clearance Cl∗ H Fi ----------- = ----------------------------------- - (EQ 9-32) 1 + EH ( F i – 1 ) Cl H f u∗ ⋅ ( Cl int )∗ where is the fractional change in fraction unbound times the frac- F i = ------------------------------- f u ⋅ ( Cl int ) tional change in intrinsic hepatic clearance. The equation predicts that, for any given decrease in intrinsic hepatic clearance, drugs having a small normal extraction ratio will experience a proportionately greater reduction in clearance than drugs having a large normal extraction ratio. The intrinsic hepatic clearance of a drug can be reduced by cirrhosis or increased by enzyme inducers, such as phenobarbitol. 9.6.3 TABULATED OR GRAPHICAL ALTERATIONS A table or graph of clearance changes when the hepatic blood flow (but not the intrinsic hepatic clearance) is altered shows that drugs having a low extraction ratio ( EH = 0.1 ) need little adjustment in dosage. Even if the hepatic blood flow were halved ( FR = 0.5 ) , the hepatic clearance is still 91% of its normal value. Con- versely, dosage adjustment is necessary for drugs having a high extraction ratio and predominantly eliminated by hepatic metabolism (e.g., propanolol). A table or graph of clearance changes when the intrinsic hepatic clearance (but not the hepatic blood flow) is altered shows that drugs having a high extraction ratio ( E H = 0.9 ) need little adjustment in dosage. Even if the intrinsic hepatic clearance were halved ( Fi = 0.5 ) , the hepatic clearance is still 91% of its normal value. Con- versely, dosage adjustment is necessary for drugs having a low extraction ratio and predominately eliminated by hepatic metabolism (e.g., phenylbutazone). 9-11 Basic Pharmacokinetics REV. 99.4.25 Copyright © 1996-1999 Michael C. Makoid All Rights Reserved http://kiwi.creighton.edu/pkinbook/
12. 12. Clearance 9.7 Renal Function and Clearance Approximately 25% of cardiac output goes to the kidneys or approximately 735 ml/min of plasma is presented to the kidneys of a 70 kg man (19.1 mL/min/kg x 70 kg). Approximately 125 ml/min (1.8 mL/min/kg) of that goes to the glomeruli for filtration (Glomerular Filtration Rate, GFR). Unbound drug is filtered into the proximal renal tubule at this point. The remaining plasma (as blood) is shunted around the tubule in the arterioles adjascent to the proximal tubule where drug may be actively secreted from the arteriol into the proximal tubule or actively reabsorbed in the opposite direction. As the blood flows down the vessels adjas- cent to the loop of Henle, the drug may be also passively reabsorbed into the blood vessel as the water in the urine is being reabsorbed and the urine is being concen- trated. This leads to some interesting possibilities: Cl r = f u ⋅ GFR . It is likely that the drug is filtered only, in this case, . It is also possible that 1. secretion and reabsortion balance and cancel each orther out but are still occurring. The actual clearance of the drug may be low as the drug may be bound to plasma protiens or red blood cells. 11. Cl r > f u ⋅ GFR . Net active secretion is infered in this case. These active mechanisms are non- specific and consequently, drugs actively secreted compete with each other. Secretion, if it occurs, occurs on the unbound drug and thus is also effected by changes in free fraction. In cases where secretion is very rapid and as a consequence, virtually all of the drug is removed by the single pass through the kidney (Er ~1), the disssociation of the drug from the protien or out of the red blood cells is not a hinderance. Some reabsorption may occur but it is less than secre- tion. 12. Cl r < f u ⋅ GFR . Net active reabsorbtion is infered in this case. Active reabsorption occurs for many exogenous compounds, including glucose and vitamins. For many compounds, reabsorp- tion is passive, occurring only as a consequence of the concentration gradient produced as water is removed from the urine as is proceeds down the renal tubule. Since the membrane is lipoidal in nature, polar compounds, ionized acids and ionized bases are less likely to be reabsorbed. Thus changing the pH of the urine would result in changing the reabsorption characteristics of weakly acidic or basic drugs. For low molecular weight drugs (<2,000 dalton) , filtration always occurs. Active secretion, active reabsorption and passive reabsorption may occur. It has been found that renal blood flow is little affected by changes in blood flow elsewhere. However, in chronic renal dysfunction there are two effects which exhibit a parallel decline. One is a decrease in glomerular filtration rate (GFR), as measured by Cl cr , and the other is the net secretion of drugs into the kidney tubules. Note that p-amino hippurate (PAH) clearance measures the sum of both effects. 9-12 Basic Pharmacokinetics REV. 99.4.25 Copyright © 1996-1999 Michael C. Makoid All Rights Reserved http://kiwi.creighton.edu/pkinbook/
13. 13. Clearance For any given drug, equation 9-26 predicts that alterations in both the renal blood perfusion rate (as manifest by the GFR) and the intrinsic renal clearance will cause a change in drug clearance. A general equation may be derived relating the ratio of renal clearances at two different blood flow rates and two different intrinsic renal clearances. Cl∗ r ---------- = F i - (EQ 9-33) Cl r where F i is the fractional change in drug unbound times the fractional change in intrinsic renal clearance. In this case, f u∗ F i = ------ ⋅ F - (EQ 9-34) fu R where is the fractional change in blood flow rate (or GFR). FR Thus, equation 9-33 shows that the renal clearance of a drug is reduced by a con- stant fraction, independent of the renal extraction ratio ( E r ) . This fractional decrease can be estimated by changes in creatinine clearance: Cl∗ cr FR = ------------ - (EQ 9-35) Cl cr Cl∗ cr where is the altered creatinine clearance Substituting, equation 9-33 through equation 9-35, we get: Cl∗ r Cl ∗ cr ---------- = ------------ - - (EQ 9-36) Cl r Cl cr This equation shows why, in cases of chronic renal dysfunction, a change in the measured creatinine clearance indicated a likely change in drug renal clearance. Hence, dosage adjustments are made on this basis, particularly for drugs predomi- nantly eliminated by renal filtration (e.g., gentamicin, digoxin). 9-13 Basic Pharmacokinetics REV. 99.4.25 Copyright © 1996-1999 Michael C. Makoid All Rights Reserved http://kiwi.creighton.edu/pkinbook/
14. 14. Clearance 9.8 General Equations for Changes in Clearance For each clearing organ, FI ⋅ FR Cl ° F Cl = ----- = --------------------------------------- - - (EQ 9-37) F R + E ( F I – FR ) Cl When a drug has a high extraction ratio, E ≈ 1 , then equation 9-26 becomes F Cl ≈ F R (EQ 9-38) and when a drug has a low extraction ratio, E ≈ 0 , then equation 9-26 becomes F Cl ≈ F I (EQ 9-39) Thus, the clearance of drugs with a high extraction ratio are more effected by phys- iological changes in flow of blood to the clearing organ, while drugs with a low extraction ratio are more effected by physiological changes in the function of the organ. 9.8.1 PLASMA/BLOOD RATIO Calculation of Extraction Ratio requires measurement in whole blood by defini- tion. Since most clinical measurements are done in plasma, knowledge of the plasma/blood ratio is necessary. Blood in made up of plasma and red blood cells (RBCs). Thus the amount of drug in the blood is made up of the amount of drug in the plasma and the amount of drug in the RBCs. C b ⋅ Vb = C p ⋅ Vp + C rbc ⋅ V rbc (EQ 9-40) C x ⋅ V x = AMOUNT x where and b = blood, p = plasma, rbc = red blood cell If we define the ratio of the concentration of the drug in the RBCs to the concentra- tion of the free drug in plasma as C rbc ρ = -------------- - (EQ 9-41) fu ⋅ C p and V rbc = H ⋅ V b (EQ 9-42) 9-14 Basic Pharmacokinetics REV. 99.4.25 Copyright © 1996-1999 Michael C. Makoid All Rights Reserved http://kiwi.creighton.edu/pkinbook/
15. 15. Clearance Vp = ( 1 – H) ⋅ Vb (EQ 9-43) Pluging equation 9-41 thru equation 9-43 into equation 9-40 results in: Cb ⋅ Vb = ( 1 – H ) ⋅ V b ⋅ Cp + f u ⋅ ρ ⋅ H ⋅ Vb ⋅ Cp (EQ 9-44) Rearranging and simplifying results in: Cb ----- = 1 + H ⋅ ( fu ⋅ ρ – 1 ) - (EQ 9-45) Cp H – 1 + ( Cb ⁄ C p ) ρ = ----------------------------------------- (EQ 9-46) fu ⋅ H Thus equation 9-46 determines the affinity of the drug for the RBCs. Drugs with a high affinity for the RBCs should result in a smaller volume of distribution. For drugs that are primarily filtered by the glomeruli, the renal extraction ratio is: Rate of filtration - GFR ⋅ f u ⋅ C p E rf = ----------------------------------------------- = ------------------------------- - (EQ 9-47) Qr ⋅ Cb Rate of presentation Putting equation 9-45 into equation 9-47 results in: GFR ⋅ f u E rf = ----------------------------------------------------------- (EQ 9-48) Q r ⋅ ( 1 + H ⋅ ( fu ⋅ ρ – 1 ) ) Thus: Cl r If the the ratio of ------- to calculated Erf from equation 9-48 is one, it is likely that the drug is - 1. Qr filtered only. Clr If the the ratio of ------- to calculated Erf from equation 9-48 is greater than one, active secretion - 13. Qr is infered in this case. Cl r If the the ratio of ------- to calculated Erf from equation 9-48 is less than one, active reabsorb- - 14. Qr tion is infered in this case. 9-15 Basic Pharmacokinetics REV. 99.4.25 Copyright © 1996-1999 Michael C. Makoid All Rights Reserved http://kiwi.creighton.edu/pkinbook/
16. 16. Clearance 9.8.2 HALF LIFE AND ELIMINATION RATE CONSTANT IN RELATIONSHIP TO CLEARANCE The elimination rate constant is related to the volume of distribution and the total body clearance by equation 9-22 above which when rewritten yields: ( Mass ) ⁄ ( Time ) ( Volume ) ⁄ ( Time ) Rate of Elimination Cl K = ----------------------------------------------- = ---------------------------------------- = ----- = ---------------------------------------------- - - (EQ 9-49) Amount in the body Mass V Volume Cl pw Cl b K = Cl = ------- = ------------u ----- - - - (EQ 9-50) V Vb Vpw u where b = blood and pwu = unbound plasma water. Clearance of drug from blood (Clb) is useful in considering drug extraction in the eliminating organs. Volume and clearance terms based on unbound drug concentration are particularly useful in therapeutics, because only the unbound drug is thought to cause the therapeutic action. 9.8.3 EFFECTS OF ALTERATIONS IN PROTEIN BINDING ON CLEARANCE Protein binding of drugs may be altered in disease states and by interferance bind- ing by other drugs on the protein. These changes in binding effect Fi , the frac- tional change in intrinsic clearance in both renal and hepatic clearances even though the actual intrinsic clearance, the indicator of organ function, may be unef- fected as shown in equation 9-11: f u∗ ⋅ Cl ∗int F i = ------------------------- - f u ⋅ Cl int where Cl∗int = Cl int . Thus, a change in protein binding will cause a proportional change in Fi of both clearances. There is more discussion in the chapter on protein binding. 9-16 Basic Pharmacokinetics REV. 99.4.25 Copyright © 1996-1999 Michael C. Makoid All Rights Reserved http://kiwi.creighton.edu/pkinbook/
17. 17. Clearance 9.9 Problems For each of the problems, do questions A through R now and S through W after completing chapter 10. In doing questions S through W, please try to obtain a ss plasma concentration of free drug within 120 % of normal and 80 % of Cp max free ss . Cp min free 9-17 Basic Pharmacokinetics REV. 99.4.25 Copyright © 1996-1999 Michael C. Makoid All Rights Reserved http://kiwi.creighton.edu/pkinbook/
18. 18. Clearance Acebutolol (Problem 9 - 1) Problem Submitted By: Maya Leicht AHFS 00:00.00 Problem Reviewed By: Vicki Long GPI: 0000000000 Piquette-Miller, M., et. al., “Effect of aging on the pharmacokinetics of acebutolol enantiomers”, Journal of Clinical Pharmacol- ogy, Vol. 32, (1992), p. 148 - 156. Kukes, VG; Gneushev ET; Mamedov TS; Gneusheva IA; “Acebutolol and diacetolol: thier bind- ing to plasma and erythrocytes and secretion with saliva.” Farmakol-Toksikol. 1991 Jan-Feb; 54(1) Acebutolol is a beta-adrenergic blocking agent which is often used in the treatment of hypertension.(Use Qr = 72 L/hr; PROBLEM TABLE 9 - 1. Acebutolol Patient Condition Normal Normal FRR=0.5 FIR=0.5 FRH=0.5 FIH=0.5 A Dose(mg) 200 400 B f 0.4 C fu 0.867 ρ D 1.93 E Vd (L) k (hr-1) F G T 1/2 (hr) 5.5 H %Clr 40 I %Clnr 60 J AUC (mg/L*hr) 3.97 K Cl tot (L/hr) L Cl h (L/hr) M Cl r (L/hr) L Eh O Er P (L/hr) Cl h int Q (L/hr) Cl r int R FCL 1 τ (hr) S 12 T N U µg ss  ------- Cp - max free  mL V ss  µg  Cp ------- - avg free  mL  W ss  µg  Cp ------- - min free  mL  Qh = 90 L/hr) 9-18 Basic Pharmacokinetics REV. 99.4.25 Copyright © 1996-1999 Michael C. Makoid All Rights Reserved http://kiwi.creighton.edu/pkinbook/
19. 19. Clearance TABLE 9 - 1. Answers for Acebutolol Patient Condition Normal Normal FRR=0.5 FIR=0.5 FRH=0.5 FIH=0.5 A Dose(mg) 200 400 BID 400 BID 400 BID 400 BID 200 QID FRR 1 1 0.5 1 1 1 FIR 1 1 0.5 0.5 1 1 FRH 1 1 1 1 0.5 1 FIH 1 1 1 1 1 0.5 B f 0.4 C fu 0.867 ρ D 1.93 E Vd (L) 160 k (hr-1) F 0.126 0.101 0.102 0.12 0.091 G T 1/2 (hr) 5.5 6.87 6.77 5.89 7.64 H %Clr 40 25 26.1 42.9 55.5 I %Clnr 60 75 73.9 57.1 44.6 J AUC (mg/L*hr) 3.97 9.93 9.78 8.51 5.5 20.2 16.1 16.4 18.8 14.5 K Cl tot (L/hr) L 12.1 12.1 12.1 10.7 6.5 Cl h (L/hr) M 8.1 4.03 4.27 8.1 8.1 Cl r (L/hr) L 0.126 Eh O 0.112 Er P 16 (L/hr) Cl h int Q 10.5 (L/hr) Cl r int R FCL 1 0.8 0.81 0.93 0.72 τ (hr) S 12 12 12 12 6 T N 2.18 1.75 1.77 2.04 0.785 U 1.11 1.24 1.23 1.15 1.03 µg ss  ------- Cp -  mL max free V 0.57 0.72 0.71 0.61 0.80 µg ss  -------  Cp -  mL  avg free W 0.25 0.37 0.36 0.28 0.6 ss  µg  Cp ------- - min free  mL  9-19 Basic Pharmacokinetics REV. 99.4.25 Copyright © 1996-1999 Michael C. Makoid All Rights Reserved http://kiwi.creighton.edu/pkinbook/
20. 20. Clearance Bisoprolol (Problem 9 - 2) Problem Submitted By: Maya Leicht AHFS 00:00.00 Problem Reviewed By: Vicki Long GPI: 0000000000 Kirch, W., et. al., “Pharmacokinetics of bisoprolol during repeated oral administration to healthy volunteers and patients with kid- ney or liver disease”, Clinical Pharmacokinetics, Vol. 13, (1987), p. 110 - 117. Bisoprolol (comes as 5 and 10 mg tablets) is a - selective adrenergic antagonist. It is used in the treatment of hyperten- sion and angina pectoris.(Use Qr = 72 L/hr; Qh = 90 L/hr) PROBLEM TABLE 9 - 2. Bisoprolol Patient Condition Normal Normal FRR=0.5 FIR=0.5 FRH=0.5 FIH=0.5 A Dose(mg) 10 10 TID B f 0.7 C fu 1 ρ D 1 E Vd (L) k (hr-1) F G T 1/2 (hr) 10 H %Clr 50 I %Clnr 50 J AUC (mg/L*hr) 0.661 K Cl tot (L/hr) L Cl h (L/hr) M Cl r (L/hr) L Eh O Er P (L/hr) Cl h int Q (L/hr) Cl r int R FCL 1 τ (hr) S T N U µg ss  ------- Cp - max free  mL V ss  µg  Cp ------- - avg free  mL  W µg ss  -------  Cp - min free  mL  9-20 Basic Pharmacokinetics REV. 99.4.25 Copyright © 1996-1999 Michael C. Makoid All Rights Reserved http://kiwi.creighton.edu/pkinbook/
21. 21. Clearance TABLE 9 - 2. Answers for Bisoprolol Patient Condition Normal Normal FRR=0.5 FIR=0.5 FRH=0.5 FIH=0.5 A Dose(mg) 10 10 10 10 10 10 FRR 1 1 0.5 1 1 1 FIR 1 1 0.5 0.5 1 1 FRH 1 1 1 1 0.5 1 FIH 1 1 1 1 1 0.5 B f 0.9 C fu 0.7 ρ D E Vd (L) 152.8 k (hr-1) F 0.0693 0.052 0.0526 0.067 0.0525 G T 1/2 (hr) 10 13.3 13.2 10.3 13.2 H %Clr 50 33.3 34 51.3 66 I %Clnr 50 66.7 66 48.7 34 J AUC (mg/L*hr) 0.661 0.85 0.87 0.69 0.87 10.6 7.94 8.0 10.3 8.0 K Cl tot (L/hr) L 5.3 5.3 5.3 5.0 2.7 Cl h (L/hr) M 5.3 2.65 2.8 5.3 5.3 Cl r (L/hr) L 0.055 Eh O 0.074 Er P 5.6 (L/hr) Cl h int Q 5.7 (L/hr) Cl r int R FCL 1 0.75 0.76 0.97 0.76 τ (hr) S 8 12 12 8 12 T N 0.8 0.9 0.9 0.78 0.91 U 0.108 0.099 0.098 0.11 0.098 ss  µg  Cp ------- - max free  mL V 0.083 0.073 0.073 0.085 0.073 µg ss  -------  Cp -  mL  avg free W 0.062 0.052 0.052 0.064 0.052 µg ss  -------  Cp -  mL  min free 9-21 Basic Pharmacokinetics REV. 99.4.25 Copyright © 1996-1999 Michael C. Makoid All Rights Reserved http://kiwi.creighton.edu/pkinbook/
22. 22. Clearance Cefonicid (Problem 9 - 3) Problem Submitted By: Maya Leicht AHFS 00:00.00 Problem Reviewed By: Vicki Long GPI: 0000000000 Fillastre, J., et. al., “Pharmacokinetics of cefonicid in uraemic patients”, Journal of Antimicrobial Chemotherapy, Vol. 18, (1986), p. 203 - 211. Cefonicid is a beta-lactamase resistant cephalosporin which is useful in treating many infections caused by Gram-posi- tive and Gram-negative organisms. Cefonicid is 80% renally excreted.(Use Qr = 72 L/hr; Qh = 90 L/hr) PROBLEM TABLE 9 - 3. Cefonicid Patient Condition Normal Normal FRR=0.5 FIR=0.5 FRH=0.5 FIH=0.5 A Dose(mg) 1000 1000 TID B f 1 C fu 0.06 ρ D E Vd (L) k (hr-1) F G T 1/2 (hr) 5.3 H %Clr 80 I %Clnr 20 J AUC (mg/L*hr) 654 K Cl tot (L/hr) L Cl h (L/hr) M Cl r (L/hr) L Eh O Er P (L/hr) Cl h int Q (L/hr) Cl r int R FCL 1 τ (hr) S T N U µg ss  ------- Cp - max free  mL V ss  µg  Cp ------- - avg free  mL  W µg ss  -------  Cp - min free  mL  9-22 Basic Pharmacokinetics REV. 99.4.25 Copyright © 1996-1999 Michael C. Makoid All Rights Reserved http://kiwi.creighton.edu/pkinbook/
23. 23. Clearance TABLE 9 - 3. Answers for Cefonicid Patient Condition Normal Normal FRR=0.5 FIR=0.5 FRH=0.5 FIH=0.5 A Dose(mg) 1000 1000 1000 1000 1000 1000 FRR 1 1 0.5 1 1 1 FIR 1 1 0.5 0.5 1 1 FRH 1 1 1 1 0.5 1 FIH 1 1 1 1 1 0.5 B f 1 C fu 0.06 ρ D E Vd (L) 11.7 k (hr-1) F 0.131 0.078 0.079 0.131 0.118 G T 1/2 (hr) 5.3 8.8 8.8 5.3 5.9 H %Clr 80 66.7 66.9 80 88.9 I %Clnr 20 33.3 33.1 20 11.1 J AUC (mg/L*hr) 654 1090 1083 654 727 1.53 0.917 0.93 1.5 1.4 K Cl tot (L/hr) L 0.3 0.31 0.31 0.3 0.15 Cl h (L/hr) M 1.22 0.612 0.62 1.22 1.22 Cl r (L/hr) L 0.0032 Eh O 0.017 Er P 5.11 (L/hr) Cl h int Q 20.7 (L/hr) Cl r int R FCL 1 0.6 0.6 1 0.9 τ (hr) S 8 12 12 8 8 T N 1.51 1.4 1.4 1.51 1.36 U 7.9 8.4 8.4 7.9 8.4 µg ss  ------- Cp -  mL max free V 4.9 5.5 5.4 4.9 5.4 µg ss  -------  Cp -  mL  avg free W 2.8 3.3 3.3 2.8 3.3 ss  µg  Cp ------- - min free  mL  9-23 Basic Pharmacokinetics REV. 99.4.25 Copyright © 1996-1999 Michael C. Makoid All Rights Reserved http://kiwi.creighton.edu/pkinbook/
24. 24. Clearance Cefpirome (Problem 9 - 4) Problem Submitted By: Maya Leicht AHFS 00:00.00 Problem Reviewed By: Vicki Long GPI: 0000000000 Lameire, N., et. al., “Single-dose pharmacokinetics of cefpirome in patients with renal impairment”, Clinical Pharmacology and Therapeutics, Vol. 52, (1992), p. 24 - 30. Cefpirome is a third-generation, broad-spectrum cephalosporin which is useful against many cephalosporin-resistant organisms. Cefpirome PROBLEM TABLE 9 - 4. Patient Condition Normal Normal FRR=0.5 FIR=0.5 FRH=0.5 FIH=0.5 A Dose(mg) 2000 IV 2000 TID B f 1 C fu ρ D E Vd (L) k (hr-1) F G T 1/2 (hr) 2.6 H %Clr 85 I %Clnr 15 J AUC (mg/L*hr) 342 K Cl tot (L/hr) L Cl h (L/hr) M Cl r (L/hr) L Eh O Er P (L/hr) Cl h int Q (L/hr) Cl r int R FCL 1 τ (hr) S T N U µg ss  ------- Cp - max free  mL V ss  µg  Cp ------- - avg free  mL  W µg ss  -------  Cp - min free  mL  9-24 Basic Pharmacokinetics REV. 99.4.25 Copyright © 1996-1999 Michael C. Makoid All Rights Reserved http://kiwi.creighton.edu/pkinbook/
25. 25. Clearance TABLE 9 - 4. Answers for Cefpirome Patient Condition Normal Normal FRR=0.5 FIR=0.5 FRH=0.5 FIH=0.5 A Dose(mg) B f C fu ρ D E Vd (L) k (hr-1) F G T 1/2 (hr) H %Clr I %Clnr J AUC (mg/L*hr) K Cl tot (L/hr) L Cl h (L/hr) M Cl r (L/hr) L Eh O Er P (L/hr) Cl h int Q (L/hr) Cl r int R FCL 1 τ (hr) S T N U µg ss  ------- Cp - max free  mL V ss  µg  Cp ------- - avg free  mL  W ss  µg  Cp ------- - min free  mL  9-25 Basic Pharmacokinetics REV. 99.4.25 Copyright © 1996-1999 Michael C. Makoid All Rights Reserved http://kiwi.creighton.edu/pkinbook/
26. 26. Clearance Cefprozil (Problem 9 - 5) Problem Submitted By: Maya Leicht AHFS 00:00.00 Problem Reviewed By: Vicki Long GPI: 0000000000 Shyu, W., et. al., “Pharmacokinetics of cefprozil in healthy subjects and patients with renal impairment”, Journal of Clinical Phar- macology, Vol. 31, (1991), p. 362 - 371. Cefprozil is a broad-spectrum oral cephalosporin. Cefprozil PROBLEM TABLE 9 - 5. Patient Condition Normal Normal FRR=0.5 FIR=0.5 FRH=0.5 FIH=0.5 A Dose(mg) 1000 B f 0.95 C fu 0.7 ρ D E Vd (L) k (hr-1) F G T 1/2 (hr) 1.2 H %Clr 75 I %Clnr 25 J AUC (mg/L*hr) 58.1 K Cl tot (L/hr) L Cl h (L/hr) M Cl r (L/hr) L Eh O Er P (L/hr) Cl h int Q (L/hr) Cl r int R FCL 1 τ (hr) S T N U ss  µg  Cp ------- - max free  mL V µg ss  -------  Cp - avg free  mL  W ss  µg  Cp ------- - min free  mL  9-26 Basic Pharmacokinetics REV. 99.4.25 Copyright © 1996-1999 Michael C. Makoid All Rights Reserved http://kiwi.creighton.edu/pkinbook/
27. 27. Clearance TABLE 9 - 5. Answers for Cefprozil Patient Condition Normal Normal FRR=0.5 FIR=0.5 FRH=0.5 FIH=0.5 A Dose(mg) B f C fu ρ D E Vd (L) k (hr-1) F G T 1/2 (hr) H %Clr I %Clnr J AUC (mg/L*hr) K Cl tot (L/hr) L Cl h (L/hr) M Cl r (L/hr) L Eh O Er P (L/hr) Cl h int Q (L/hr) Cl r int R FCL 1 τ (hr) S T N U µg ss  ------- Cp - max free  mL V ss  µg  Cp ------- - avg free  mL  W ss  µg  Cp ------- - min free  mL  9-27 Basic Pharmacokinetics REV. 99.4.25 Copyright © 1996-1999 Michael C. Makoid All Rights Reserved http://kiwi.creighton.edu/pkinbook/
28. 28. Clearance Chloramphenicol (Problem 9 - 6) Problem Submitted By: Maya Leicht AHFS 00:00.00 Problem Reviewed By: Vicki Long GPI: 0000000000 Ambrose, P., “Clinical pharmacokinetics of chloramphenicol and chloramphenicol succinate”, Clinical Pharmacokinetics, Vol. 9, (1984), p. 222 - 238. Chloramphenicol succinate is a prodrug which is converted in vivo to the active form, chloramphenicol. Chloramphenicol PROBLEM TABLE 9 - 6. Patient Condition Normal Normal FRR=0.5 FIR=0.5 FRH=0.5 FIH=0.5 A Dose(mg) 1000 B f 1 C fu 0.4 ρ D E Vd (L/kg) 2.8 k (hr-1) F G T 1/2 (hr) 0.6 H %Clr 30 I %Clnr 70 J AUC (mg/L*hr) K Cl tot (L/hr) L Cl h (L/hr) M Cl r (L/hr) L Eh O Er P (L/hr) Cl h int Q (L/hr) Cl r int R FCL 1 τ (hr) S T N U ss  µg  Cp ------- - max free  mL V µg ss  -------  Cp - avg free  mL  W ss  µg  Cp ------- - min free  mL  9-28 Basic Pharmacokinetics REV. 99.4.25 Copyright © 1996-1999 Michael C. Makoid All Rights Reserved http://kiwi.creighton.edu/pkinbook/
29. 29. Clearance TABLE 9 - 6. Answers for Chloramphenicol Patient Condition Normal Normal FRR=0.5 FIR=0.5 FRH=0.5 FIH=0.5 A Dose(mg) B f C fu ρ D E Vd (L) k (hr-1) F G T 1/2 (hr) H %Clr I %Clnr J AUC (mg/L*hr) K Cl tot (L/hr) L Cl h (L/hr) M Cl r (L/hr) L Eh O Er P (L/hr) Cl h int Q (L/hr) Cl r int R FCL 1 τ (hr) S T N U µg ss  ------- Cp - max free  mL V ss  µg  Cp ------- - avg free  mL  W ss  µg  Cp ------- - min free  mL  9-29 Basic Pharmacokinetics REV. 99.4.25 Copyright © 1996-1999 Michael C. Makoid All Rights Reserved http://kiwi.creighton.edu/pkinbook/
30. 30. Clearance Enalapril (Problem 9 - 7) Problem Submitted By: Maya Leicht AHFS 00:00.00 Problem Reviewed By: Vicki Long GPI: 0000000000 Ohnishi, A., et. al., “Kinetics and dynamics of enalapril in patients with liver cirrhosis”, Clinical Pharmacology and Therapeutics, Vol. 45, (1989), p. 657 - 665. Enalapril is an ACE inhibitor which is a prodrug that is metabolized in the liver to the active metabolite, enalaprilat. PROBLEM TABLE 9 - 7. Enalapril Patient Condition Normal Normal FRR=0.5 FIR=0.5 FRH=0.5 FIH=0.5 A Dose(mg) 10 mg po 10 BID1 B f 0.65 C fu 0.55 ρ D E Vd (L) k (hr-1) F G T 1/2 (hr) 0.63 H %Clr 27 I %Clnr 73 J AUC (mg/L*hr) 0.123 K Cl tot (L/hr) L Cl h (L/hr) M Cl r (L/hr) L Eh O Er P (L/hr) Cl h int Q (L/hr) Cl r int R FCL 1 τ (hr) S T N U ss  µg  Cp ------- - max free  mL V µg ss  -------  Cp - avg free  mL  W ss  µg  Cp ------- - min free  mL  9-30 Basic Pharmacokinetics REV. 99.4.25 Copyright © 1996-1999 Michael C. Makoid All Rights Reserved http://kiwi.creighton.edu/pkinbook/
31. 31. Clearance TABLE 9 - 7. Answers for Enalapril Patient Condition Normal Normal FRR=0.5 FIR=0.5 FRH=0.5 FIH=0.5 A Dose(mg) B f C fu ρ D E Vd (L) k (hr-1) F G T 1/2 (hr) H %Clr I %Clnr J AUC (mg/L*hr) K Cl tot (L/hr) L Cl h (L/hr) M Cl r (L/hr) L Eh O Er P (L/hr) Cl h int Q (L/hr) Cl r int R FCL 1 τ (hr) S T N U µg ss  ------- Cp - max free  mL V ss  µg  Cp ------- - avg free  mL  W ss  µg  Cp ------- - min free  mL  9-31 Basic Pharmacokinetics REV. 99.4.25 Copyright © 1996-1999 Michael C. Makoid All Rights Reserved http://kiwi.creighton.edu/pkinbook/