Multicompartment Models Objectives Compartment Modeling of Drugs Models

Multi-compartment Models Objectives Compartment Modeling of Drugs Models describing drug concentration-time profiles

L/70 kg 50, 000 20, 000 10, 000 5, 000 1, 000 500 100 50 10 5 General Principles of Distribution Quinacrine Chloroquine Nortriptyline Digoxin Propranolol Quinidine Quinolones (1 - 2 L/kg), Tetracycline Phenobarbital Phenytoin To date, we have assumed that Theophylline (0. 45 L/kg) distribution of drug in the Aminoglycodises (0. 25 L/kg) ASA body was instantaneous. Warfarin What if that is NOT true?

1000 mg 10 L In this patient the drug “instantly” distributes throughout the body and the concentration is homogeneous. If we draw a “model” of the compartments, drug is loaded into the compartment … it rapidly distributes (instantaneously) The body acts as a single tub or compartment. and then begins to be eliminated.

1000 mg 1 L 10 L In this patient the drug distributes to In this patient the drug “instantly” distributes to the some fluids (tissues) quickly, but from there, drug distributes slowly entire volume and the concentration is homogeneous. to tissues that are poorly perfused. It takes some time before all tissues establish a concentration in The body acts as a equilibrium with other tissues. single tub or compartment. The body acts as if it contains multiple compartments.

1000 mg 1 L 10 L When the body appears to have a second compartment, distribution into each compartment occurs at different rates If we draw a “model” of the compartments, drug is loaded into the first compartment (central compartment) … it rapidly distributes (instantaneously) throughout the first compartment and then begins to distribute into the second compartment.

1000 mg 1 L 10 L While drug is distributing to the second compartment, it will still be eliminated (cleared) from the first compartment This means that the drug concentration in the second compartment will be rising as the concentration in the first is falling.

1000 mg 1 L 10 L While drug is distributing to the second compartment, it will still be eliminated (cleared) from the first compartment This means that the drug concentration in the second compartment will be rising as the concentration in the first is falling. But just as drug diffused into the second compartment it can diffuse back to the first.

2 Compartment Model: IV Dosing The rate constants describing drug movement are labeled based on their movement from one compartment to another. k 12 describes movement from compartment 1 to compartment 2. k 21 describes movement from compartment 2 to compartment 1. k 10 describes movement from compartment 1 out of the body.

2 Compartment Model: IV Dosing When the body appears to have a second compartment, distribution into the second compartment and elimination contribute an initial rapid decline in concentration. However, after some time, distribution into the second compartment is complete. The two compartments are in equilibrium with each other.

Although equilibrium has been established between the two compartments, concentrations in each compartment may be different. Equilibrium Established Here concentrations in the second compartment are in red. Equilibrium is established after ~4 -6 hours. After 4 -6 hrs concentrations in both compartments decline with the same half-life.

1000 mg When the body appears to have a second compartment, distribution into each compartment occurs at different rates. 1 L 10 L Distribution to tissues is possibly related to tissue blood flow. Distribution to tissues (2) with lower perfusion occurs more slowly. Distribution to tissues (1) with greater blood flow occurs more quickly … almost instantaneously.

k 12=0. 3 hr-1; k 21=0. 4 hr-1 k 12/k 21= 0. 75 k 12=0. 3 hr-1; k 21=0. 217 hr-1 k 12/k 21= 1. 38 However, differences in profiles is not solely related to the size of k 12. Here, both profiles have the same terminal elimination ( = 0. 2 hr-1) AND the same k 12 (0. 3 hr-1). Profiles have different k 21 values. Differences in profiles are created by different relative values of k 12 : k 21. Movement in and out

2 Compartment Model: IV Dosing Patients with different body compositions, Different proportions of highly perfused vs. lean muscle vs. poorly perfused adipose tissue Will have different rates and degrees of distribution to their tissues. Different relative values of k 12: k 21.

2 Compartment Model: IV Dosing Questions to Consider What is the difference between the profiles on the left? Assume that it is the same drug given to 3 different patients Is the terminal half-life the same? Is the initial volume the same? Is the final volume the same? Is the rate of distribution the same? Which profile has the largest k 12: k 21 ratio? Which has the smallest?

2 Compartment Model: IV Dosing Questions to Consider What happens if you change Input Rate? (bolus infusion) Can the profile change from a 2 -Compartment following IV bolus to a 1 -Compartment Model following IV Infusion? How would you explain this?

Analysis of 2 Compartment Model: IV Dosing Using semi-log paper, graph the data from these two patients. Each patient is given 1000 mg of a drug by IV bolus. Patient 1 Time Plasma Conc (hr) (mg/L) Patient 2 time Plasma Conc (hr) (mg/L) 0. 25 0. 50 1 2 3 4 6 9 12 0. 25 0. 5 1 2 3 4 6 9 12 94 89 79 63 50 40 25 12. 5 6. 25 60 41 26 18 14 11 7. 0 3. 5 1. 8

Plotting / Graphing Plasma Concentrations Do we need 1 cycle or 2 -cycle semi log paper? Using semi-log paper, graph the data from these two patients, each given 1000 mg of a drug. Patient 1 Time Plasma Conc (hr) (mg/L) 0. 25 94 0. 50 89 1 79 2 63 3 50 4 40 6 25 9 12. 5 12 6. 25

Graph Patient 1 Data 100 60 30 10 6 4 2 1 0 2 4 6 8 10 12 14 Using semi-log paper, graph the data from these two patients, each given a 1000 mg dose. Patient 1 Time Plasma Conc (hr) (mg/L) 0. 25 94 0. 50 89 1 79 2 63 3 50 4 40 6 25 9 12. 5 12 6. 25

Graph Patient 1 Data 100 60 30 10 6 4 2 1 0 2 4 6 8 10 12 14 What is the Half-life? … the Initial [ ]? What model best describes data? Patient 1 Time Plasma Conc (hr) (mg/L) 0. 25 94 0. 50 89 1 79 2 63 3 50 4 40 6 25 9 12. 5 12 6. 25

Graph Patient 1 Data 100 60 30 10 6 4 2 1 0 2 4 6 8 10 12 14 What is the Half-life? … the Initial [ ]? What model best describes data? Patient 1 Time Plasma Conc (hr) (mg/L) 0. 25 94 0. 50 89 1 79 2 63 3 50 4 40 6 25 9 12. 5 12 6. 25

Graph Patient 1 Data 100 60 30 10 6 4 2 1 0 2 4 6 8 10 12 14 What is the Half-life? 3 hr … the Initial [ ]? … 100 mg/L What model best describes data? Patient 1 Time Plasma Conc (hr) (mg/L) 0. 25 94 0. 50 89 1 79 2 63 3 50 4 40 6 25 9 12. 5 12 6. 25

Graph Patient 1 Data 100 IV Dose (bolus) 1 60 V 1 30 k 10 10 6 4 2 1 0 2 4 6 8 10 12 14 In Patient 1, distribution of drug to the body is instantaneous (rapid). Drug enters the blood and rapidly distributes to tissues. At 0. 25 hr drug is homogeneously distributed.

Graph Patient 1 Data 100 IV Dose (bolus) 1 60 V 1 30 K 10 Dose e-Kt Ct = -----V 6 4 2 1 0 2 4 6 8 10 12 14 Dose = 1000 mg V=10 L K= 0. 231 hr-1 -0. 231(t) Ct = 100 e

Graph Patient 2 Data 100 60 30 10 6 4 2 1 0 2 4 6 8 10 12 14 Using semi-log paper, graph the data from Patient 2, also given a 1000 mg dose. Patient 2 Time Plasma Conc (hr) (mg/L) 0. 25 60 0. 5 41 1 26 2 18 3 14 4 11 6 7. 0 9 3. 5 12 1. 8

Graph Patient 2 Data 100 60 What is different about this patient? 30 10 6 4 2 1 0 2 4 6 8 10 12 14 Patient 2 Time Plasma Conc (hr) (mg/L) 0. 25 60 0. 5 41 1 26 2 18 3 14 4 11 6 7. 0 9 3. 5 12 1. 8

Graph Patient 2 Data 100 60 What is the Half-life? … the Initial [ ]? 30 10 6 4 2 1 What is different about this patient? 0 2 4 6 8 10 12 14 Patient 2 Time Plasma Conc (hr) (mg/L) 0. 25 60 0. 5 41 1 26 2 18 3 14 4 11 6 7. 0 9 3. 5 12 1. 8

Graph Patient 2 Data 100 60 30 10 6 4 2 1 0 2 4 6 8 10 12 14 What is different about this patient? What is the Half-life? 3 hr … the Initial [ ]? . . 100 mg/L What model describes a patient like Patient 2 ? Time Plasma Conc (hr) (mg/L) 0. 25 60 0. 5 41 1 26 2 18 3 14 4 11 6 7. 0 9 3. 5 12 1. 8

Graph Patient 2 Data 100 60 30 In Patient 2, distribution of drug to the body occurs in 2 steps. Initial instantaneous distribution followed by a slower phase. This is possibly related to organ and tissue blood flow. 10 6 4 2 1 0 2 4 6 8 10 12 14 Drug begins to distribute, immediately but distribution is not complete until ~2 hours after the dose. Only then is drug is in each tissue in equilibrium with other tissues.

Graph Patient 2 Data 100 60 30 -αt Ct = A e 10 -βt + Be 6 4 The appearance of the second exponential in the [ ] – time profile is an indication that we have a 2 -compartment model. 2 1 0 2 4 6 8 10 12 14

2 Compartment Model: IV Dosing Rapid distribution [α phase] Drug distributes from highly perfused tissues to tissues in 2 nd compartment (fat, lean tissue). C= Ae -αt + Be -βt

2 Compartment Model: IV Dosing Slow Elimination [β phase] Distribution is complete Pseudo-equilibrium has been achieved, body behaving as 1 C. C= Ae -αt + Be -βt

2 Compartment Model: IV Dosing Slow Elimination [β phase] Distribution is complete Pseudo-equilibrium has been achieved, body behaving as 1 C. Concentration in 2 nd Compartment (in red) Equilibrium established between compartments at the end of the distribution phase. C= A e -αt + Be -βt

Analysis of 2 Compartment Model: IV Dosing Analysis of a 2 C model But what are the values A and B? Ct = A e -αt + Be -βt

Analysis of 2 Compartment Model: IV Dosing Analysis of a 2 C model Terminal Phase Slope B is the Zero time intercept of the back-extrapolated β phase B Ct = A e -αt + Be -βt

Analysis of 2 Compartment Model: IV Dosing Analysis of a 2 C model Terminal Phase Slope = A is the Zero time intercept of the Residual … … the difference between the back extrapolated concentrations on the β phase and the observed concentrations … in the phase. B Ct = A e -αt + Be -βt

Analysis of 2 Compartment Model: IV Dosing Analysis of a 2 C model Step 1 Method of residuals Back extrapolate β phase Ct = A e -αt + Be -βt

Analysis of 2 Compartment Model: IV Dosing Analysis of a 2 C model Step 2 Determine back-extrapolated concentrations at same time as observed concentrations Ct = A e -αt + Be -βt

Analysis of 2 Compartment Model: IV Dosing Analysis of a 2 C model Step 3 Calculate the difference between the observed concentration and the back-extrapolated concentrations at each time point and plot these concentrations ( ). Ct = A e -αt + Be -βt

Analysis of 2 Compartment Model: IV Dosing Analysis of a 2 C model Step 4 Calculate the slope of the line of the residual. This is done in the same way as you would calculate K, it is the slope of the log of concentration vs. time. This slope is proportional to alpha. Ct = A e -αt + Be -βt

Analysis of 2 Compartment Model: IV Dosing Analysis of a 2 C model A Step 5 B Zero time intercepts of back-extrapolated β phase and of residual (α), are equal to B and A respectively. α and β are “MACRO” constants in that they are made up of the “MICRO” constants k 12, k 21 and k 10. Ct = A e -αt + Be -βt

Analysis of 2 Compartment Model: IV Dosing Analysis of a 2 C model A Step 5 B α and β are “MACRO” constants in that they are made up of the “MICRO” constants (k 12, k 21 and k 10) and reflect a variety of simultaneous processes. [α + β = k 12 + k 21 + k 10] (Equations can be shown in either form) Dose * (k 21 -α) e -αt + ------------Dose * (k 21 -β) e -βt C = ----------V * ( β – α) V * (α - β)

Analysis of 2 Compartment Model: IV Dosing Analysis of a 2 C model A Step 6 B Begin to solve for micro constants. k 21 = (Aβ + Bα) / ( A + B) Ct = A e -αt + Be -βt

Analysis of 2 Compartment Model: IV Dosing Analysis of a 2 C model A Step 7 B Begin to solve for micro constants. k 10 = α β / k 21 Ct = A e -αt + Be -βt

Analysis of 2 Compartment Model: IV Dosing Analysis of a 2 C model A Step 8 B Begin to solve for micro constants. k 12 = α + β - k 21 – k 10 Ct = A e -αt + Be -βt

Analysis of 2 Compartment Model: IV Dosing Analysis of a 2 C model A Step 9 B Begin to solve for other parameters V 1 can be obtained from the initial concentration and dose V 1 = dose / C 0 Ct = A e -αt + Be -βt

Analysis of 2 Compartment Model: IV Dosing Analysis of a 2 C model A Step 10 B Begin to solve for other parameters Vss = V 1(1 + k 12/k 21) Ct = A e -αt + Be -βt

Analysis of 2 Compartment Model: IV Dosing Analysis of a 2 C model A Step 11 B Begin to solve for other parameters Calculate AUC by trap. rule or AUC = A/ α + B/β Ct = A e -αt + Be -βt

Analysis of 2 Compartment Model: IV Dosing Analysis of a 2 C model A Step 11 B Begin to solve for other parameters Calculate AUC by trap. rule or AUC = A/ α + B/β Ct = A e -αt + Be -βt

Graph Patient 2 Data 100 -αt Ct = A e -βt + Be 60 In this [ ] – time profile the βphase intercept (B) is 27. 8 mg/L and the α-phase intercept (A) is 72. 2 mg/L. -αt -βt Ct = 72. 2 e + 28. 8 e 30 10 6 What would the initial concentration be at time zero? 4 2 1 0 2 4 6 8 10 12 14

Graph Patient 2 Data 100 -αt Ct = A e -βt + Be 60 30 10 6 4 2 1 0 2 4 6 8 In this [ ] – time profile the βphase intercept (B) is 27. 8 mg/L and the α-phase intercept (A) is 72. 2 mg/L. -αt -βt Ct = 72. 2 e + 28. 8 e What would the initial concentration be at time zero? C 0=A+B and V 1 can be obtained from the initial [ ]and dose. V 1 = Dose / C 0 = 1000 mg/100 mg/L = 10 L 10 12 14

Graph Patient 2 Data 100 -αt Ct = A e -βt + Be 60 In this [ ] – time profile the βphase intercept (B) is 27. 8 mg/L and the α-phase intercept (A) is 72. 2 mg/L. -αt -βt Ct = 72. 2 e + 28. 8 e 30 10 6 AUC following an IV dose in a 2 C model can be calculated by standard methods 4 2 1 (trapezoidal rule + kinetic method from the last concentration: [ ]/β) 0 2 4 6 8 10 12 14 or based on the formula AUC 0 -∞ = (A/ α) + (B/ β)

Graph Patient 2 Data 100 60 -αt Ct = A e -βt + Be -αt -βt Ct = 72. 2 e + 28. 8 e 30 A 2 -C model is the result of a drug which has an initial small volume of distribution (V 1) which grows following the dose to achieve its final <larger> volume (Vss). 10 6 4 2 1 0 2 4 6 8 The micro-constants k 12 and k 21 represent the transfer constants between the central (C-1) and peripheral (C-2) compartments. 10 12 14

Graph Patient 2 Data 100 60 -βt + Be -αt -βt Ct = 72. 2 e + 28. 8 e 30 The final <larger> volume (Vss) is a function of the rate into the peripheral compartment (k 12) and the rate out of the peripheral compartment (k 21). 10 6 4 Vss = V 1(1 + k 12/k 21) 2 1 -αt Ct = A e 0 2 4 6 8 10 12 14

Analysis of 2 Compartment Model: IV Dosing Analysis of a 2 C model A B Zero time intercepts of back-extrapolated β phase and of residual (α), are equal to A and B respectively. Terminal Phase Slope = Residual Slope = α AUC 0 -∞ = (A/ α) + (B/ β) -αt Ct = A e -βt + Be

Analysis of 2 Compartment Model: IV Dosing Analysis of a 2 C model A B Terminal Phase Slope = What is the initial time zero concentration? Residual Slope = α Zero time intercepts of back-extrapolated β phase and of residual (α), are equal to A and B respectively. C= A e -αt + Be -βt

Analysis of 2 Compartment Model: IV Dosing Analysis of a 2 C model A B Terminal Phase Slope = What is the initial time zero concentration? Residual Slope = α Zero time intercepts of back-extrapolated β phase and of residual (α), are equal to A and B respectively. [ ]0 = A + B C= A e -αt + Be -βt

Analysis of 2 Compartment Model: IV Dosing What does the value of do to the concentration – time profile? k 12, k 21 and k 10 are micro constants α and β are more complex constants and reflect a variety of simultaneous processes. α + β = k 12 + k 21 + k 10 α = k 21*k 10/β and β 2 – (k 12+k 21+k 10)*β + k 21/k 10 = 0 -βt Dose * (k -α) -αt Dose * (k -β) 21 21 C = ----------- e + ------------ e V * ( β – α) V * (α - β)

Analysis of 2 Compartment Model: IV Dosing α = 2. 0 β = 0. 2 k 21 = 0. 25 K 12 = 0. 35 Here there is a 4 fold difference in α, but what else is different? Relative size of & as well as relative size of k 21&k 12. α = 0. 5 β = 0. 2 k 21 = 0. 25 K 12 = 0. 05 Recall … Vss = V 1(1 + k 12/k 21) The concentration changes rapidly. What does this imply? -βt Dose * (k -α) -αt Dose * (k -β) 21 21 C = ----------- e + ------------ e V * ( β – α) V * (α - β)

Analysis of 2 Compartment Model: IV Dosing α = 2. 0 β = 0. 2 k 21 = 0. 25 K 12 = 0. 35 V 1 = 10 V 2 = 14 α = 1. 0 β = 0. 2 k 21 = 0. 25 K 12 = 0. 15 V 1 = 10 V 2 = 6 α = 0. 5 β = 0. 2 k 21 = 0. 25 K 12 = 0. 05 V 1 = 10 V 2 = 2 -βt Dose * (k -α) -αt Dose * (k -β) 21 21 C = ----------- e + ------------ e V * ( β – α) V * (α - β)

Analysis of 2 Compartment Model: IV Dosing α = 2. 0 β = 0. 2 k 21 = 0. 25 K 12 = 0. 35 V 1 = 10 V 2 = 14 As increases, the rate of change of concentration during the distribution phase increases and the size of V 2 increases. Vss = V 1 + V 2 α = 0. 5 β = 0. 2 k 21 = 0. 25 K 12 = 0. 05 V 1 = 10 V 2 = 2 -βt Dose * (k -α) -αt Dose * (k -β) 21 21 C = ----------- e + ------------ e V * ( β – α) V * (α - β)

Summary Compartment Models: Describing [ ] - Time 1 compartment with IV bolus input [one exponential – K] Dose e-Kt Ct = -----V

Summary Compartment Models: Describing [ ] - Time 1 compartment with first order absorption [2 exponentials ka and K) ka*F*Dose e -kt - e -kat C = ---------V*(ka - k)

Summary Compartment Models: Describing [ ] - Time 2 compartment with IV bolus input [2 exponentials and ) Dose * (k 21 -α) e -αt + ------------Dose * (k 21 -β) e -βt C = ----------V * ( β – α) V * (α - β)

Summary Compartment Models: Describing [ ] - Time 2 compartment with first order absorption [3 exponentials , and ka) ka. FDose * (k 21 -ka) -kat C = ----------- e V * (α-ka)*(β-ka) ka. FDose *(k 21 -α) -αt + ------------ e V * (ka-α)*(β-α) ka. FDose *(k 21 -β) -βt + ------------ e V * (ka-β)*(α-β)

Summary Compartment Models: Describing [ ] - Time 3 compartment with IV bolus input [3 exponentials , and ) C= A*e -αt + B*e -βt +C*e - t