please can someone help me out with problem 12.3? 278 MULTIPLE INTRAVENOUS BOLUS
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please can someone help me out with problem 12.3?
278 MULTIPLE INTRAVENOUS BOLUS INJECTIONS IN THE ONE-COMPART concentrations in the range 1000-100 g/L are therapeutic. The maior netic parameters of disolvprazole are CI = 12 L/h, Vd-35 L/70 k pharmak g, and S: l. 12.3 A synthetic analgesic has the following pharmacokinetic and pharmacodynanti parameters: Vd = 0.2 Lkg, Cl = 0.012 L/h/kg, and S = 1; the therapeutic rang is between 0.5 and 1.5 mg/L (a) Determine an appropriate dosing regimen for a 75-kg male who has just une gone major orthopedic surgery (b) Would you recommend a loading dose? If so, calculate one. (e) Afier 10 days of treatment, the patient develops toxicity, and therapy is wi drawn: (1) How long will it take to eliminate 99% of the drug in the body? (2) What is the Cp value 24 h after the last dose? 70-kg man is to receive quinidine, an antiarrthymic drug used in the 70 kg man is to receive quinidine, an antiarrthymic drug use kinetic parae 12.4 of atrial fibrillation and other cardiac arrhythmias ters in this man are estimated to be CI = 4 mL/min/kg and intravenous bolus formula to devise a dosing regimen fo (s-082 and F = 0.73) that will maintain the plasma c 3-15 mg/l. Erele 12n The pharmac Use 3.0LgU concentrations in the ra REFERENCE 1. Winter, M. E. (2010) Basie Clinical Pharmacokinetics, 5th e ed., Lippincott Wlliams Baltimore.Explanation / Answer
1. Understand the characteristics of the plasma concentration profile after multiple inter- mittent infusions 2. Understand the derivation of an expression for the steady-state plasma concentrations achieved by intermittent infusions 3. Use two steady-state plasma concentrations to determine a drug’s half-life, the steady- state peaks and troughs, and volume of distribution 4. Individualize doses for patients receiving multiple intermittent infusions13.1 INTRODUCTIONWhen drugs are administered as bolus injections, typically they are injected over a periodof 1 min or more to avoid very high initial concentrations. Even with this approach, theBasic Pharmacokinetics and Pharmacodynamics.
MULTIPLE INTERMITTENT INFUSIONSinitial plasma concentrations of some drugs may still be high enough to cause toxicity. Thiscan be especially problematic for drugs that display two-compartment pharmacokinetics,where the initial distribution volume is small. These drugs can be administered more safelyby extending the administration period and infusing the dose at a constant rate over a periodof anywhere from half an hour, up to 2 h or more. The administration is then repeated withthe same frequency that it would be for bolus injections. This type of drug administrationthus consists of multiple intermittent infusions or multiple short infusions. The aminogly-coside antibiotics are important examples of drugs administered in this way. These drugsare used to treat serious life-threatening gram-negative infections, and their use is com-plicated by their potential to cause serious renal impairment and ototoxicity (hearing andvestibular damage). As a result of both wide interindividual variability in their pharma-cokinetics and the serious consequences of either subtherapeutic or toxic concentrations,plasma concentrations of these drugs are monitored and doses individualized. A knowledgeand understanding of the pharmacokinetics and associated equations of multiple intermit-tent infusions is necessary to perform this process. The characteristics and equations for multiple intermittent infusions are very similarto those for multiple bolus injections. Figure 13.1 shows the typical plasma concentrationprofile associated with this type of administration, which in common with bolus injections,demonstrates fluctuation and an accumulation of drug in the buildup to steady state. Again incommon with multiple bolus injections, it takes 3–5 elimination half-lives to achieve steadystate. When the plasma concentrations during a dosing interval are studied, some importantdifferences between the two forms of drug administration become apparent (Figure 13.2).Although the trough concentration occurs at the end of the dosing interval, the peak plasmaconcentrations observed with multiple infusions do not occur at the beginning of the dosinginterval but when the infusion is stopped. The symbols used are the same as those used for bolus injections: t is the time that haselapsed since the dose was administered (start of the infusion), and is the dosing intervalor time between doses. Tau varies constantly from zero to , to zero to , and so on. Theequations for multiple infusions incorporate an additional time parameter, the duration ofthe infusion (tinf ). The dosing interval can be broken down into two parts: the period fromtime 0 to tinf when the infusion is running, during which plasma concentrations increase(A in Figure 13.2); and the period from the end of the infusion to the end of the dosing 2 Cp (mg/L) 1 0 0 8 16 24 32 40 48 tinf Time into therapy (h)FIGURE 13.1 Typical plasma concentration–time profile observed with multiple intermittentinfusions. The duration of the infusion is tinf and the interval between the start of consecutiveinfusions is .
STEADY-STATE EQUATIONS FOR MULTIPLE INTERMITTENT INFUSIONS 281Cp Cpmax,ss CpSS = Cp e k (ttinf) max, ss k (tinf) max,ss Cp = Cp emin,ss Mono exponential decaytinf t-tinf A B t: 0 8e.g., 0FIGURE 13.2 Multiple intermittent infusions: plasma concentrations during a steady-state dosinginterval. Time t is the time after the start of the infusion. The dosing interval can be separated intotwo periods: A and B. During period A, the infusion is running and plasma concentrations increase.Note that the peak occurs when the infusion stops (t = tinf ). During period B (from t = tinf to t = ),there is no ongoing drug input, and plasma concentrations decay monoexponentially.interval (period tinf to ) when the infusion is not running. During the latter period, plasmaconcentrations fall monoexponentially under the influence of first-order elimination (B inFigure 13.2). Note in Figure 13.2 that the peak occurs when t = tinf , and a trough is obtainedat the end of the dosing interval when t = . In a dosing interval, once the infusion stops, the plasma concentration falls monoexpo-nentially from the peak. As a result, the plasma concentration at any time during this period(B in Figure 13.2) can be expressed in terms of its decay from the peak and the time thathas elapsed from the peak (t tinf ). For example, at steady state: Cpss = Cpmax,ss ek(ttinf ) (13.1)where Cpss is the concentration at any time, t, during a steady state dosing interval andCpmax,ss is the steady-state peak. Cp at the trough during a steady-state dosing interval(Cpmin,ss) is expressed as: Cpmin,ss = Cpmax,ss ek(tinf ) (13.2)13.2 STEADY-STATE EQUATIONS FOR MULTIPLEINTERMITTENT INFUSIONSIn Chapter 12, we demonstrated that an equation for multiple dosing could be constructedas follows (Figure 12.6): Cpmultiple doses = Cpsingle doses × fraction of steady state anytime × final accumulation at steady state
282 MULTIPLE INTERMITTENT INFUSIONS Expressions for the fraction of steady state achieved at any time and the final accumula-tion at steady state were derived in Chapter 12 [equations (12.27) and (12.23), respectively].Thus,Cpmultiple doses = Cpsingle doses (1 enk) 1 1 (13.3) ek If the equation is to be limited to steady state, the fraction of steady state (1 enk)equals 1 and can be removed from the expression:Cpmultiple doses, ss = Cpsingle doses 1 1 (13.4) ek The equation for the plasma concentration at any time after a single infusion was pre-sented in Chapter 11 [equation (11.16)]:Cpsingle infusion = S F k0 (1 ekT ) ekt (13.5) Clwhere T is the time the infusion was terminated, and t is the time elapsed since termination.Next, we substitute the time symbols used for multiple infusions: T = tinf and t = t tinf .Substituting these symbols into equation (13.5) and then substituting into equation (13.4)yieldCpss = S F k0 (1 ektinf ) ek(ttinf ) (13.6) Cl (1 ek)where Cpss is the plasma concentration at any time during a steady-state dosing interval, k0is the infusion rate, S is the salt factor of the drug, F is the bioavailability, k is the eliminationrate constant, tinf is the duration of the infusion, t is the time elapsed since the infusionwas started, Cl is the clearance, and is the dosing interval or time between the start ofconsecutive infusions. A peak plasma concentration occurs at the end of the infusion when t = tinf :Cpmax,ss = S F k0 (1 ektinf ) (13.7) Cl (1 ek)A trough plasma concentration occurs at the end of the dosing interval (t = ), 0Cpmin,ss = S F k0 (1 ektinf ) ek(tinf ) (13.8) Cl (1 ek)Example 13.1 A drug (Cl = 8.67 L/h, Vd = 100 L, and S = 1) was administered asmultiple intravenous infusions. The dose (80 mg) was administered over a 1-h period every8 h. Calculate the estimated plasma concentrations on the third day of therapy, at (a) 1 h,(b) 2 h, (c) 4 h, and (d) 8 h after the start of the first infusion of the day.Solution The drug has a t12 = 0.693 × 1008.67 = 8 h. A dose of 80 mg is infused over1 h: k0 = 80 mg/h, S = 1, and F = 1. The problem is summarized in Figure 13.3. It will take
STEADY-STATE EQUATIONS FOR MULTIPLE INTERMITTENT INFUSIONS 283 Cpmax,ss Cpss = Cp e k (ttinf) max,ssCp ? ? ? ? Cpmin,ss = Cp ek ( tinf) max,ss 01 2 4 8t Infusion Infusion k0 = 80 mg/h starts stopsFIGURE 13.3 Example problem using multiple intermittent infusion equations. The drug is admin-istered at a rate of 80 mg/h over a 1-h period. Plasma concentrations at 1, 2, 4, and 8 h must becalculated after the first dose on the third day.about 24–40 h to get to steady state. By the third day, steady state will have been achievedand the profiles from the three infusions of the day will be exactly the same. (a) When t = 1 h, the infusion has just been terminated, Cp = Cpmax,ss. Substituting into equation (13.7) yields Cpmax,ss = 80(1 e1×0.6938) = 1.53 mgL 8.67(1 e8×0.6938) (b) When t = 2 h, it is 1 h after the infusion stops. Cpss,t=2 = 80(1 e1×0.6938) e(21)0.6938 8.67(1 e8×0.6938) or Cpss,t=2 = Cpmax,ss e(21)0.6938 = 1.53e(21)0.6938 = 1.40 mgL (c) When t = 4 h, it is halfway through the dosing interval and 3 h after the infusion has stopped: Cpss,t=4 = Cpmax,ss e(41)0.6938 = 1.53e(41)0.6938 = 1.18 mgL (d) When t = 8 h, it is the end of the dosing interval and 7 h from the peak: Cpss,t=8 = Cpmin,ss = Cpmax,ss e(81)0.6938 = 1.53e(81)0.6938 = 0.83 mgL
284 MULTIPLE INTERMITTENT INFUSIONSExample 13.2 A 0.5-h infusion is used to administer the dose (80 mg) of the drugdescribed in Example 13.1. Calculate the estimated plasma concentrations on the third dayof therapy, at (a) 0.5 h, (b) 1 h, (c) 2 h, and (d) 8 h after the start of the infusion.Solution The drug has a t12 = 0.693 × 1008.67 = 8 h. A dose of 80 mg is infused over0.5 h: k0 = 160 mg/h, S = 1, and F = 1. It will take about 24–40 h to get to steady state.By the third day, steady state will have been achieved. A diagram of the steady-state dosinginterval is shown in Figure 13.3 but in this example, tinf = 0.5 h. (a) When t = 0.5, the infusion has just been terminated, Cp = Cpmax,ss. Substituting into equation (13.7), we haveCpmax,ss = 160(1 e0.5×0.6938) = 1.58 mgL 8.67(1 e8×0.6938)(b) When t = 1, it is 0.5 h after the infusion stops. Cpss,t=1 = Cpmax,ss e(10.5)0.6938 = 1.58e(10.5)0.6938 = 1.51 mgL(c) When t = 2, it is 1.5 h after the infusion stops. Cpss,t=2 = Cpmax,ss e(20.5)0.6938 = 1.58e(20.5)0.6938 = 1.38 mgL(d) When t = 8, it is the end of the dosing interval and 7.5 h from the peak. Cpss,t=8 = Cpmin,ss = Cpmax,ss e(80.5)0.6938 = 1.58e(80.5)0.6938 = 0.82 mgL13.3 MONOEXPONENTIAL DECAY DURING A DOSING INTERVAL:DETERMINATION OF PEAKS, TROUGHS, AND ELIMINATION HALF-LIFE13.3.1 Determination of Half-LifeA drug’s elimination half-life can be determined from two plasma concentrations mea-sured during the period when the infusion is not running (period tinf to ), when plasmaconcentrations are falling monoexponentially as a result of first-order elimination. Duringthis period,Cpn = Cpmax,n ek(ttinf) (13.9)where Cpmax,n and Cpn are the peak and plasma concentration at time t, respectively, duringthe nth dosing interval. Taking the logarithmic of equation (13.9) yields ln Cpn = ln Cpmax,n k (t tinf ) (13.10)Thus, k can be determined: (13.11) k = ln(Cp1Cp2) t2 t1
MONOEXPONENTIAL DECAY DURING A DOSING INTERVAL 285and the half-life t12 = 0.693 (13.12) kExample 13.3 A patient is being treated with gentamicin. A 140-mg dose is adminis-tered as a short infusion over a 1-h period every 8 h. Plasma concentrations of the drugwere determined during the first dosing interval and are given in Table E13.3. Assume one-compartmental pharmacokinetics and calculate the drug’s half-life. TABLE E13.3 Gentamicin Plasma Concentrations Determined at Two Time Points After the Start of an Infusion Time After Start of Infusion (h) Gentamicin Concentration (mg/L) 1.5 6.1 6 2.2Solution The problem is summarized in Figure E13.3. The plasma samples were takenduring a period when the infusion was not running. Thus, the only process affecting plasmaconcentrations is first-order elimination, and equation (13.11) can be used to determine therate constant and half-life. k = ln(6.12.2) = 0.227 h1 6 1.5 t12 = 0.693 = 3.05 h 0.227 Cpmax,1 Cpn = Cpmax,nek (tt inf)Cp 2.2 ? 6.1 ? Cpmin,n = Cpmax,nek ( t inf) 0 1 1.5 6 8t Infusion Infusion starts stops140 mg/hFIGURE E13.3 Calculation of the half-life. If two plasma concentrations are known in the periodwhen the infusion is not running, the half-life can be calculated. Once the half-life is known, the peaksand troughs associated with this dosing interval may also be calculated.
286 MULTIPLE INTERMITTENT INFUSIONSAb Abmax,n= amount from most recent infusion + amount remaining from Abmin,(n-1) Abmin,(n-1) 0 tinf Infusion t stops Infusion startsFIGURE 13.4 Determination of the volume of distribution. The derivation of the equation for Vdis based on partitioning the amount of drug in the body at the time of a peak into two components:(1) the amount of drug that came from the most recent infusion and (2) the amount of drug remainingfrom the previous trough.13.3.2 Determination of Peaks and TroughsIf a drug’s half-life or elimination rate constant is known, and if at least one plasma con-centration during the dosing interval is known, the values of the peak and trough associatedwith the dosing interval can be calculated.Example 13.4 Continuing with Example 13.3, determine the trough and peak of the dos-ing interval described above.Solution The drug’s elimination rate constant is 0.227 h1. Once the infusion is stopped,plasma concentrations decay monoexponentially (Figure 13.4): Cpn = Cpmax,n ek(ttinf ) (13.13) Cpn = 6.1 mg/L when t = 1.5 h and t tinf = 0.5 h. Substitute in equation (13.13) todetermine the peak: 6.1 = Cpmax,n e0.227×0.5 Cpmax,n = 6.83 mgLThe trough occurs when t = and t tinf = 7 h: Cpmin,n = Cpmax,n e0.227×7 = 6.83e0.27×7 = 1.39 mgL13.4 DETERMINATION OF THE VOLUME OF DISTRIBUTIONThe volume of distribution can be calculated if a peak plasma concentration and its previoustrough are known.
DETERMINATION OF THE VOLUME OF DISTRIBUTION 287Theory The equation for the volume of distribution is derived by considering the amountof drug in the body at the time of a peak (Abmax,n) (Figure 13.4). The amount of drug inthe body at the time of a peak can be partitioned into two parts: drug from the most recentinfusion and drug remaining from previous infusions. At the peak: Abmax,n = drug from the most recent infusion (13.14) + the amount remaining from previous infusions The amount of drug from the most recent infusion is determined by multiplying Vd bythe equation for the plasma concentration from a single infusion [see equation (13.5)], withT = tinf and t = 0): Abrecent = Vd S F k0 (1 ektinf ) (13.15) ClAs Cl = k Vd, Abrecent = Vd S F k0 (1 ektinf ) (13.16) k Vd The amount remaining from previous infusion(s) is equal to the amount at the time ofthe previous trough (Abmin,(n1)) minus the amount eliminated during the recent infusion: Abprevious = Abprevious trough elimination during infusion (13.17) = Vd Cpmin,(n1) ektinf Substituting the expression of the amount from the most recent infusion (13.15) and theamount from previous infusion(s) [equation (13.17)] into equation (13.14) gives usAbmax,n = Vd S F k0 (1 ektinf ) + Vd Cpmin,(n1) ektinf (13.18) k VdSubstituting Cpmax,n Vd for Abmax,n results inVd Cpmax,n = Vd S F k0 (1 ektinf ) + Vd Cpmin,(n1) ektinf k Vd Cpmax,n = S F k0 (1 ektinf ) + Cpmin,(n1) ektinf (13.19) k Vd Vd = S F k0 (1 ektinf ) k(Cpmax,n Cpmin,(n1) ektinf )Practice From equation (13.19), it can be seen that the volume of distribution can becalculated if a peak concentration (Cpmax,n) and a previous trough (Cpmin,(n1)) are known.If the therapy is at steady state, any peak and trough may be used because the troughs andpeaks are exactly the same at steady state.Example 13.5 The suitability of a gentamicin regimen (110 mg every 8 h infusedover a 1-h period) is being assessed on the third day of treatment. A steady-state trough
288 MULTIPLE INTERMITTENT INFUSIONS 3.9 mg/L Cpmax,ss Cp ? 7.5 mg/L 3.9 mg/L 0 1 1.5 8t Infusion Infusionk0 = 110 mg/h starts stopsFIGURE 13.5 Example calculation of the volume of distribution. This parameter is determinedfrom a peak plasma concentration and its previous trough. Initially, the half-life must be estimated inorder to calculate the peak plasma concentration.concentration is found to be 3.9 mg/L. The plasma concentration 1.5 h after the start of themost recent infusion is found to be 7.5 mg/L. (a) Determine the elimination half-life. (b) Determine the peak plasma concentration after the most recent infusion. (c) Determine the volume of distribution.Solution The infusion is at steady state, so all the peaks and troughs will be the same.The profile associated with this problem is shown in Figure 13.5. The elimination rate constant may be calculated as: k = ln(7.53.9) = 0.101 h1 8 1.5(a) The elimination half-life is determined: t12 = 0.693 = 6.86 h k(b) The steady-state peak can be determined: 7.5 = Cpmax,n e0.101×0.5 Cpmax,n = 7.89 mgL(c) The volume of distribution may be calculated from equation (13.19):Vd = 110(1 e0.101×1) = 23.9 L 0.101(7.89 3.9 × e0.101×1)
SIMULATION 28913.5 INDIVIDUALIZATION OF DOSING REGIMENSThe methods used to design the dosing regimens discussed in Chapter 12 can be applied tomultiple intermittent infusions.Example 13.6 Continuing with Example 13.5, the trough concentration found in thepatient was considered to be too high. Design another regimen that will provide a gen-tamicin steady-state peak and trough of 8 and 0.5 mg/L, respectively. The pharmacokineticparameters in this patient were found to be k = 0.101 h1, Vd = 23.9 L, and Cl = k Vd =2.41 L/h.Solution Here the focus of the regimen is on the achievement of specific peaks andtroughs. Recall the equation introduced in Chapter 12 to calculate a dosing interval to pro-vide specific peaks and troughs with multiple intravenous injections: = 1 ln Cpmin,ss (13.20) k Cpmax,ss For multiple infusions, the time between the peak and the trough is tinf (Figure 13.2)and not as it is for multiple bolus injections. Thus, equation (13.20) will have to bemodified appropriately: tinf = 1 ln Cpmin,ss (13.21) k Cpmax,ss Using equation (13.21) to calculate the necessary dosing interval for peaks and troughsof 8 and 0.5 mg/L, respectively, yields tinf = 1 ln 0.5 = 27.5 h 0.101 8 = 28.5 h, which would be rounded to 24 h The appropriate dose could be calculated using the equation for either the peak (8 mg/L)or the trough (0.5 mg/L). Using the peak equation (13.7), we have Cpmax,ss = 8 = k0 (1 e1×0.101) 2.41(1 e24×0.101) k0 = 183 mg To obtain peaks and troughs of 8 and 0.5 mg/L, 183 mg of gentamicin should be infusedover a 1-h period every 24 h.13.6 SIMULATION: http://web.uri.edu/pharmacy/research/rosenbaum/simsOpen model 10, Multiple Intermittent Infusions, which can be found at http://web.uri.edu/pharmacy/research/rosenbaum/sims/Model10. Default settings for the model are dose = 80 mg, = 8 h, tinf = 1 h, Cl = 8.67 L/h, andVd = 20 L. 1. Review the objectives and the “Model Summary” page. 2. Go to the “Cp–Time Profile” page. Perform a simulation and note that the peaks occur when the infusion stops (t = tinf ) and a trough occurs just before the next dose
290 MULTIPLE INTERMITTENT INFUSION. Also note that accumulation occurs with successive doses but stops at steady state. 3. Go to the “Influence of Infusion Duration” page. The overall rate of drug administra- tion or the dose administered will remain constant (80 mg every 8 h), but the duration of the infusion will be altered. Since the individual dose will remain constant, the infu- sion rate will change in inverse proportion to the infusion duration. This adjustment will be made automatically by the software. Observe how the profile changes with infusion of duration 0.5, 2, and 4 h. (The corresponding infusion rates will be 160, 40, and 20 mg/h). As the duration of the infusion increases, the peak gets lower and the trough gets higher; that is, fluctuation decreases. 4. Go to the “Determine a Dosing Regimen” page. Use the drugs pharmacokinetic parameters (Cl = 8.67 L/h and Vd = 20 L) and a 1-h infusion to calculate a dose and a dosing interval that will provide a peak and trough of 10 and 1 mg/L, respec- tively. Check your answers using the model.PROBLEMS13.1 A drug (Cl = 1.73 L/h, Vd = 30 L, and S = 1) was administered as multiple short intravenous infusions. A dose of 40 mg was administered over a 1-h period every 12 h. (a) How long will it take to get to steady state? (b) Calculate the peak plasma concentration during a steady-state dosing interval. (c) Calculate the plasma concentration 6 h into a steady-state dosing interval. (d) Calculate the trough plasma concentration during a steady-state dosing interval.13.2 If the dose of the drug discussed above was infused over a 0.5-h period, calculate the steady-state peak and trough concentrations.13.3 If the dose of the drug above was infused over a 2-h period, calculate the steady-state peak and trough concentrations.13.4 A 38-year-old male patient (75 kg) is given gentamicin (120 mg every 8 h infused over a 1-h period). Plasma samples taken after the second dose were analyzed for gentamicin and the results are shown in Table P13.4. TABLE P13.4 Gentamicin Plasma Concentrations Determined at Two Time Points After the Start of an Infusion Time After Start of Infusion (h) Gentamicin Cp (mg/L) 2 3.84 7 1.10 (a) What is gentamicin’s half-life in this patient? (b) What is the peak plasma concentration of this dosing interval? (c) What is the trough plasma concentration of this dosing interval?13.5 Tobramycin (120 mg) is being infused over a 1-h period every 8 h. Three days into therapy, at steady state, tobramycin concentrations at 1 and 8 h after the start of an
PROBLEMS 291 infusion were found to be 7.12 and 0.75 mg/L, respectively. Calculate the drug’s half- life and its volume of distribution. Note that because steady state has been achieved, the trough concentration of the previous dosing interval can also be assumed to be 0.75 mg/L.13.6 Gentamicin (140 mg) is infused over a 1-h period every 8 h. A steady-state peak and trough of 10 and 0.25 mg/L, respectively, are required. At steady state, gentamicin concentrations at 2 and 8 h after the start of an infusion were found to be 5.5 and 0.75 mg/L, respectively. Is this regimen satisfactory? If not, recommend a regimen that would achieve the desired peaks and troughs of gentamicin concentrations.
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