Drug Clearance PDF

Drug Clearance Winter, Part 1 Objectives To learn about clearance of drugs. To understand mathematical equations of clearance. To learn different mechanisms of clearance. To understand how to calculate Maintenance Dose To learn about Factors that alter Clearance(Cl) Definition...

Drug Clearance Winter, Part 1 Objectives To learn about clearance of drugs. To understand mathematical equations of clearance. To learn different mechanisms of clearance. To understand how to calculate Maintenance Dose To learn about Factors that alter Clearance(Cl) Definition clearance Intrinsic ability of body or its organs of elimination - > (usually kidneys and liver) to remove drug from the blood or plasma. = - Clearance not an indicator of how much drug is - being removed 2 impunit of clearance # Only represents theoretical G volume of blood or plasma which is completely cleared of drug in a givenO period. > time  Amount of drug removed depends on de & ↳ plasma conc. of drug and Ri - ↳ clearance. - - Steady state, = RA RZ = administration rate rate of drug administration (RA) ② and the rate of = drug elimination ①(RE) must be equal. Rate elimenation e m Drug clearance : ↳ measurement ofE drug elimination from body without reference to mechanism of the process. a e i  ↑ Body or organ tissues #compartment of fluid witha definite volume ( - apparent volume of distribution) in which the drug is dissolved. Expressed in (nits) &ml/min or litres/hour dependent for volum & piride Mathematically * clearance : rate of drug elimination divided by plasma drug conc. at that time point. Rate of elimination Cl = plasma drug conc.  T excretion rate & Clearance = ------------------- plasma conc. 3 (1) refure & rug in d Du / dt refure > - time to & urine Cl = ------------------- (2) C Multiplying both sides of the equation by C gives d Du * ClC = ------------------- (3) dt  Eqn 3 shows Wit Y rate of drug elimination is- - directly proportional to plasma drug conc., C. => - Clearance, however, is constant for any given I plasma drug conc. w - True as long as -rate of drug elimination is a first- order process. [ Clearance · Ke product of#first-order elimination rate constant and & apparent volume of distribution. Vd ↳ Cl = KeX Vo  From eqn 2 d Du / dt Cl = ------------------- (2) & C & Rate of drug elimination CL Kel XVd = & (d Du / dt) = CKe Vd This on substitution in eqn 2 yields ke vd =. & CKe Vd Cl = ------------------- = Ke Vd (4 ) C · ⑨ = The clearance calculated by the equation is termed total drug clearance or total body clearance. Objectives To understand what is Total body clearance To understand Renal Clearance It's part one is clearance (kidny) that is  Total body clearance ↳ Osum total of all clearance pathways in the body& clearance of drug through the kidney (renal => - clearance) - clearance of drug through the liver (hepatic = - clearance). Clearance : Y expressed on - per kilogram body weight Both - pharmacokinetic parameters are constant under normal conditions. & -  To obtain clearance for an individual patient, multiply clearance- - per kilogram times the- body weight of the patient. - Clearance For - Clearence (per kg) * Body wight patients  Renal Clearance mine (Amout) ↳ Renal excretion of drugs quantitatively described - - - by renal clearance value for the drug. Renal excretion : major route of elimination for - many drugs. on : Drugs eliminated by renal excretion Dependent G ⑳ - = water soluble,- * - have low mol. weight( less than or equal to 300), or - are - * slowly biotransformed by liver  What is 3 ⑭ a Mechanisms of Renal Clearance?! ~♦Glomerular filtration ~♦ Active tubular secretion ~ ♦ Tubular reabsorption  Foul ponizea mined Paros - sure - ⑮ Glomerular filtration ① Unidirectional processcone direction > occurs for most small molecules (MW less - than 500), including undissociated (non- ionized) and dissociated ( ionized) drugs. Petra 3 > - - => · Protein bound drugs behave as large mol. - - N or Donot get filtered at the glomerulus. - Major driving force for glomerular filtration: ↳Hydrostatic pressure within glomerular - ↑capillaries. From higher to tower is What the maine of filtration ? Fourse glomerular  Kidneys receive large blood supply (approx. 25% of the cardiac output) via renal artery with very = - little decrease in hydrostatic pressure. * - Glomerular filtration rate measured by using a drug ↑ that is eliminated by filtration only( i.e. it is neither - reabsorbed nor secreted) - Examples of such drugs : absorbedand ⑳ Inulin and Creatinine. > - not only filtar to D a not chanism &  Clearance of inulin will be equal to the glomerular ⑨ filtration rate which is# 125-130 ml/min. Value for glomerular filtration rate correlates fairly well with body surface area. Change > - if also change GF is Glomerular filtration of drugs is directly related to I the free or non-protein-bound drug concs. in the plasma. only free drug is filter (direct relationship As the free drug conc. in the plasma increases, the glomerular filtration for the drug will increase proportionately.  Glomerular Filtration Summary Clearance of inulin will(one * Uni-Direction proceses direction) be equal to the glomerular * filtration rate which Fours - Hydrostatic prusseris 125-130 ml/min. (High to Low Value for glomerular filtration rate correlates fairly * well with Occure For body > Small - surface area. - Less than 500 nw Molecule Glomerular↳ filtration of drugs is directly related to Ionized and non-lonized the free or non-protein-bound drug concs. in the plasma. · so Large protine drug find don't get filliter As the free drug conc. in the plasma increases, the ↳filtration Happen for free/unbond drug glomerular onlytheindrug will increase proportionately. free drug Con in. plasm * glumerular Filtration. Example Inulin #Creatinine (only Filter /Secreation N * Drug Reabsorption - · · * Filtrat rate of Inulin - 125-130 mL/min => Active tubule secreation Active renal secretion => Need Energy (enzyme-carrier) An active process from Low mu - - High) to Carrier mediated system which &requires energy - input,· Drug is transported against a conc. gradient. s The-carrier system is capacity limited and may be - saturated.- m Drugs with similar structures may compete for the - same carrier system.# - Two active renal secretion systems are -1. Systems for weak acids - ↓2. Systems for weak bases. -  · Probencid will compete > ~ with penicillin for the same carrier system (weak acids) - - Active tubular secretion rate is dependent on renal -- plasma flow. - Drugs commonly used to measure active tubular - S secretion : PAH · p-amino-hippuric acid(PAH) and - 3 iodopyracet(Diodrast). I iodopyracet # & These substances are both filtered by the glomeruli - and secreted by the tubular cells. - Active secretion is extremely rapid for these drugs, - and practically all the drugs carried to the kidney are  mi 650 mit 425- -- * eliminated in a single pass. The clearance for these drugs therefore reflects the & effective -a renal plasma flow. ERPF varies from 425-650 ml/min. - - For a drug that is excreted solely by glomerular filtration Elimination half life may change markedly in ↳ accordance with the binding affinity of the drug for - plasma proteins. Protein binding has very little effect on the - - elimination half-life of a drug excreted mostly by - active secretion.  Since drug-protein binding is reversible, the bound drug and free drug are excreted by active secretion during the first pass through the kidney. - · Some of the penicillins are extensively protein bound, but their elimination half lives are short due - to rapid elimination by active secretion.  Tubular reabsorption - - Occurs after the - drug is filtered through the - glomerulus - - ~ Can be active or passive. - If a drug is completely reabsorbed (e.g. glucose) & - then the value for the clearance of the drug is - - approx. zero. For drugs which are partially reabsorbed, clearance - values will be- less than the GFR of 125-130 ml/min. otumar Fate -  I Reabsorption of drugs which are acids or weak m j m bases is influenced by : ↳2 pH of the fluid in the renal tubule( urine pH) and the pka of the drug. & D the Both these factors together determine percentage of dissociated and => 3 undissociated drug. & Tigh reabsorbed in body  - Undissociated species is more lipid soluble( less water soluble) and has greater membrane permeability. - - - > Undissociated drug is easily reabsorbed from the renal tubule back in the body. Process of drug reabsorption can significantly - reduce the amount of drug excreted, depending on the pH of the urinary fluid and the pKa of the drug. - - - The pka of the drug is constant, - & Normal urinary pH may vary from 4.5 to 8.0 - depending on the diet, pathophysiology and drug - intake. & Vegetable - diets or diets rich in carbohydrates will result N in - higher urinary pH & pH. - - Diets rich in protein will result in lower urinary Drugs such as ascorbic acid and antacids such as - = sodium carbonate may alter urinary pH when -  Most imp. changes in urinary pH are caused by fluids administered intravenously. Intravenous fluids such as solutions of bicarbonate - or ammonium chloride are used in- acid-base -therapy. Excretion of these solutions may drastically change urinary pH and alter drug reabsorption. Objectives To understand Henderson-Hesselbalch eqn. For weak acids and weak bases  ionized Fekseid > -- PH = pKa + Log ionized - - The # percentage of ionized weak acid drug corresponding ↓ to a given pH can be obtained from the Henderson-Hesselbalch eqn.: - [ionized] & pH = pKa + log --------------- ↓ [ unionized] (5) ↳ wiRearrangement [ionized] of this equation gives Tweak Acid --------------- = 10 pH - pKa (6) PKa(2 [unionized] ↓ ionized in Urine pH  [ionized] Percent of drug ionized = -------------------- [ionized]+[unionized] 10 pH - pKa[ unionized] = --------------------------------------- [ unionized] + 10 pH - pKa[ unionized] 10 pH - pKa = --------------------- (7) 1+ 10 pH - pKa net out  # Application of equation 58 important The percentage of weak acid drug ionized in any pH - environment can be calculated by eqn 7. % [ 15 o ① The extent of dissociation is more greatly affected by changes in urinary pH with a# pka of 5 than with a⑤ > - - pka of 3. i - Weak acids with pka values of less than 2 are highly ionized = at all urinary pH values and are only slightly affected by pH variations. For a Oweak base drug, the Henderson-Hesselbalch eqn.: 5 [ unionized ] - Tweak Acid pH = pKa + log ------------------- (8) PKa(2 [ ionized ] ↓ ionized in Urine pH  D [ionized ] pKa = pH + log ------------------- [ unionized ] [ionized ] pKa - pH = log ------------------- [ unionized ] [ionized] Percent of drug ionized = -------------------- [ionized]+[unionized]  - [ unionized ] 10 pKa - pH = [ionized ] [ unionized ] 10 pKa - pH = -------------------------------- [ unionized ] 10 pKa - pH + [ unionized ] 10 pKa – pH -------------------- 1+ 10 pKa – pH  * weak acid 22 ↳ pka gs). a ionized in c) &↓ weak base # * 7 5-10 5. Percent of drug ionized ↳ pla. 10 pKa – pH O -------------------- x 100 1+ 10 pKa – pH o Greatest effect of urinary pH on reabsorption occurs with weak bases with pka of 7.5-10.5 From the Henderson-Hesselbalch relationship, a conc. ratio for the distribution of a weak acid or basic drug between urine and plasma may be derived 10 5. per effecti for compination #Metable - r Objectives To study the effect of ionisation. To understand Renal clearance. To learn how to determine Clearance. To learn how to calculate Maintenance Dose  The urine: plasma ratios for these drugs are as follows: For weak acids: Urine U 1+ 10 pHurine - pKa ------------- = --------- = -------------------------- ( 10) Plasma P 1+ 10 pHplasma - pKa For weak bases: U 1+ 10 pKa - pHurine ------------- = --------------------- ( 11 ) P 1+ 10 pKa - pHplasma  Alkaline I weakt ↓ ⑳ Y * reabsorbed 2- unionized Urine pH made alkaline Lipid Soluble S - drug bat the weak base reabsorbed : unionized > - in it's unabsorbed state so - - more lipid soluble unionized species formed. Acidification of the urine : weak base become more ionized ( form a wea - - se salt). -Acidification - reabsorbed Salt form [W Form ·ionized wate more water soluble - Eless likely to be reabsorbed = - excreted into the urine more quickly. - -  Weak Base ↳ Acidification : Excretion Lionized) ↳ Alkalinization : reabsorbed Lunionized) Weak acids acidification of urine : greater reabsorption = - alkalinization of urine : more rapid excretion Conized - mine Besides pH of the urine: Rate of urine flow : influence amount of filtered drug which is reabsorbed. ⑧ Normal flow of urine : approx.6 - 1-2 ml/min.  Non-polar and unionized drugs 3 normally reabsorbed in renal tubules sensitive to changes in rate of urine flow. - Drugs which increase urine flow (Excreationnee Decrease time for drug reabsorption - promote their excretion. & time Forced diuresis : diuretics useful adjunct for removing excessive ↑ drug in an intoxicated patient. increase renal drug excretion. & Renal clearance [ Rate of elimination by kidney ClR = ------------------------------------------ (12) C &Filtration rate + secretion rate- reabsorption rate ↳ ClR = ------------------------------------------------- (13) C (plasma drug conc) reabsortation - Filtration Rate Secreation rate LR = 8 Con (Plasma diy. )  Renal clearance Total amount of drug excreted over some time interval ________________________Mixed Plasma conc. measured at midpoint of time interval. ↓ should take pintto Determination of Clearance At steady state adm elim. RA = RE -Clearance : proportionality constant --- makes average steady-state plasma drug level equal to the rate of drug administration = (RA) RA = (Cl)(Css ave) form ( 14 ) salt Where RA : (S)(F)(Dose)/τ OS : fraction of administered dose that is - active drug.  OF : bioavailability factor -Css ave : average steady-state me drug conc,. Clearance : (S)(F) (Dose/τ) * Cl = -------------------------- (15) Cssave E i S- > Fraction #-> of RA Bioarilibity Facter state Issave- > steady  Maintenance Dose Maintenance dose which will produce desired average plasma conc. at steady state. * (Cl)(Cssave)(τ) Maintenance Dose = ---------------------- (16) (S)(F) (T) Issave) 57 -  (RA) & Administration rate - Mass/time = mglter Drug conc. Mass/volume. = mg/L - > Clearance Volume/time = Inr - Drug administration rate : mg/hr and conc. : - mg/L, then clearance ------L/hr. * & ↳ Administration rate : mg/day , conc : mg/L, then clearance ------ in 6 L/day. Objectives To study the Factors that alter Clearance(Cl) To learn about Renal and Hepatic Functions  si · Factors that=> alter Clearance(Cl) ga Body O weight 50 Body surface area Cardiac output 80 Drug-drug interactions o Extraction ratio Genetics Live Hepatic function 20 Binding Plasma protein binding less plassma protine more clearence Renal function Kidnig Factors that alter Clearance(Cl) 1. Body Surface Area (BSA) various charts. S Equation & Patient’s wt. in kg - 0.7 BSA in m2 = ( --------------------------) 1.73m2 (17) - 70 kg =>  · Patient’s Cl = ( Literature Cl per m2) (Patient’s BSA) (18) Patient’s Cl = ( Literature Cl per 70kg) (Patient’s BSA) ------------------------- (19) 1.73 m2 Patient’s Cl = ( Literature Cl per 70kg) (Patient’s weight in kg) (20) ---------------------- 70kg - - Patient’s Cl = ( Literature Cl per kg)- (Patient’s wt. in kg) (21)  ↑ Eqns 20 and 21 : clearance in proportion to weight, Eqns 18 and 19 : clearance in proportion to body surface area. Patient’s liver and kidney size vary in proportion to these measurements. If patient’s weight differs significantly from 70kg, use of weight or surface area likely to generate substantially different estimates of patient’s clearance. Renal and Hepatic Function Eliminated or cleared as unchanged drug kidney (renal clearance) metabolism in the liver (metabolic clearance). Two routes are assumed to be independent of one another and additive. Clt = Clm + Clr ( 23) Where Clt is total clearance  Liver in F - Clm : metabolic clearance or the fraction cleared by metabolism. Clr : renal clearance or the fraction cleared by the renal route. - 95 Kidneys and liver function Eindependently, - assumed that a change in one does ② - - not affect the other. -  Clt : estimated in presence of renal or hepatic failure or both. Metabolic function difficult to quantitate, Clt most commonly adjusted when there is decreased renal function. E Cl adjusted = ( Cl m) + [( Cl r) (Fraction of normal renal function remaining)] ( 24 ) Clm (0) (1 0)x(0 8) o Dadjusted = = +.. 2)v = 80 % 30 8. = 0. 8 Fru = 100 % - 1 0.  Clearance adjusted for renal function Used to estimate the maintenance dose for a patient with diminished renal function.(eqn 16) Adjusted clearance eqn only valid if drug’s metabolites are inactive and metabolic clearance unaffected by renal dysfunction When is * ⑧ decrease in function of an organ of elimination mostnumen significant? Organ primary route of drug elimination. -  As the major elimination pathway becomes increasingly compromised, - the “minor” pathway becomes more significant Y ge means the function become dec Why? -. · It assumes a greater proportion of the total clearance. se Drug : usually 67% eliminated by the renal route and 33% by the metabolic route - Will be ↳ s 100% metabolized in the event of complete renal failure: Crea l not work all Total at fulyer) renal = ① Total clearance : one-third of the normal value. -ie  ⑫ reninormal val Cl = em + dr bc2 & renal => 0. 33 + 0 failure function = 0. 33 33 % For adjusting Clt to calculate dosing rate, one can substitute fraction of the total clearance that is metabolic and renal for Cl m and Cl r  Using this technique the eqn below can be derived: - Dosing rate adjustment factor = I Used to adjust maintenance dose for a - patient with altered renal function. - - ( Fraction Eliminated metabolically)+ [ ( Fraction eliminated renally)( Fraction of normal renal function remaining)] Dosing rate adjustment factor - 99 % Drug 25% metabolized and - 75% renally cleared and - normally administered as 100 mg every 12 hours Calculate the dosing rate adjustment factor Patient has only 33% of normal renal function  ↓ Rather dosing rate adjustment factor would be 0.5. Dosing rate adjustment factor = - ( Fraction eliminated metabolically)+ [( Fraction eliminated renally)( Fraction of normal renal function remaining)] = ( 0.25)+ [( 0.75) (0.33)] = ( 0.25) + [( 0.25)] = 0.5 - - 3 so , now i give 0. 5 mg (half at regulal) for patient  What does Dosing rate adjustment factor of 0.5 suggest? ⑧ Drug should be administered at half the usual rate. Accomplished by Decreasing the dose and maintaining the same interval (e.g. 50 mg every 12 hours) or Maintaining the same dose and increasing the interval (e.g. 100 mg every 24 hours).  Depending on situation and therapeutic intent Either method (or a combination of dose and dosing interval adjustment) might be appropriate. Most pharmacokinetic adjustments for drug elimination are based on renal function we Why ? ↓ - Hepatic function more difficult to quantitate. ~g) - 3  Elevated liver enzymes reflect liver damage but not a good measure of function. How is Hepatic function often evaluated ? Prothrombin time, serum albumin conc., and serum bilirubin conc. Unfortunately, each of these lab. tests is affected by variables other than altered hepatic function. # 0 Serum albumin may be low Due to decreased protein intake or Increased renal or GI loss, as well as Decreased hepatic function. Liver function tests donot provide quantitative data, Pharmacokinetic adjustments must still take into consideration liver function Why? This route of elimination is imp. for a significant number of drugs. Objectives To study the effect of Cardiac output on clearance. To understand how Protein Bound Drugs are cleared.  Cardiac output: ↳ Affects drug metabolism.* re s e a r Hepatic or metabolic clearance for some - drugs decreased by 25% to 50% in patients - with congestive- - heart failure. & Metabolic clearances of ↳ theophylline and digoxin - ↓ 1/2 Reduced by approx. one – half in patients with congestive - - heart failure. 30  Metabolic clearance for both these drugs is - much lower than the hepatic blood or plasma flow(low extraction ratio), The decreased cardiac ouput and resultant hepatic congestion must, in some way, decrease the intrinsic metabolic capacity of the liver.  Protein Bound Drugs Not eliminated by glomerular filtration. Only the free drug is excreted by a linear process. Bound drugs usually excreted by active secretion, following capacity limited determind the kinetics. Determination of clearance which separates two components would result in a hybrid - e drug have When the clearance.=> Icomponent way that have different thisnot of elimination , have No simple way to overcome this problem. 2 example : one have => fitration and other secretion => have  Clearance values for protein-bound drugs Rate of unbound drug excretion ClR = ----------------------------------------------------- Conc. of unbound drug in the plasma  Certain limitations with the above eqn Rate of drug excretion is usually determined after collecting urine samples. Drug excreted in urine =sum of drug excreted by tubular secretion and by passive glomerular filtration. Not possible to distinguish the amount of bound drug actively secreted from the amount of drug which is excreted by glomerular filtration. · Plasma protein binding Very little effect on renal clearance of drugs like penicillin which are actively - - secreted. For these drugs Free drug fraction is filtered at the glomerular; Protein bound drug appears to be stripped from the binding sites and actively secreted into the renal tubules.  Clearance ratio Actual physiologic process for renal clearance of a drug is not generally obtained by direct measurement. Comparing clearance value for the drug to that of a standard reference drug Example: Inulin which is cleared through the kidney by glomerular filtration only, Physiologic clearance process for the first drug may be inferred.  Clearance Ratio Probable mechanism of renal excretion CL drug * I Re-absorption CLinuline & Cl drug --------- < 1 Drug is partially reabsorbed - - Cl inulin Cl drug & --------- = 1 Drug is filtered only Cl inulin -  Cl drug · --------- > 1 Drug is actively secreted > Cl inulin  1. If intravenous lidocaine is infused continuously ⑧ at a rate of 2 mg/min and if the conc of lidocaine at - steady state is 3 mg/L., calculate lidocaine - clearance in L/min and L/hr.. - O 2. If theophylline clearance is 2.8 L/hr, calculate the rate of intravenous administration for theophylline that will produce a steady-state plasma theophylline conc of 10mg/L. Also calculate the dose if theophylline were to be given every 12 hours. ⑤ 1. If intravenous lidocaine is infused continuously at a rate of 2 mg/min and if the conc of lidocaine at steady state is 3 mg/L., calculate lidocaine = 6 clearance in L/min and L/hr.. 6 2. If theophylline clearance is 2.8 L/hr, calculate - the rate of intravenous administration for theophylline that will produce a steady-state plasma theophylline conc of 10mg/L. Also calculate the dose if theophylline were to be given every 12 hours.  1. If intravenous lidocaine is infused continuously at a rate of 2 mg/min and if the conc of lidocaine at steady state is 3 mg/L., calculate lidocaine clearance in L/min and L/hr.. 2. If theophylline clearance is 2.8 L/hr, calculate the rate of intravenous administration for theophylline that will produce a steady-state plasma theophylline conc of 10mg/L. Also calculate the dose if theophylline were to be given every 12 hours.  # Calculat kn , when c = 101/day and Volumo of dustribution is 100 I Learning outcomes = 10 kn - = o V 100 # found relation between tyes V , CL At the end of this chapter the students will be able 693X du 0 693 tyz. 0 +. = y= - 9 C) K = to: Cl k = - Vo - Understand various mechanisms andcleared factors rout · 17 renal clearance and is other metabolised by % 33 % and 50 % # A Drug , affecting clearance. adjesment 60 Calcat rate function only Dosing %, kidny and is Calculate maintenance Dosing rate adjustment factor = dose for various drugs - Adjust doses in conditions of renal impairment ( Fraction Eliminated metabolically)+ [ ( Fraction eliminated renally) # M ( Fraction of normal renal function remaining)] = 0. 33 + (0 5)(0 6).. + 0. 17 = 0. 33 + 0. 3 + 0. 17 = 0. 8