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

L/70 kg 50,000 20,000 10,000 5,000 1, General Principles of Distribution Quinacrine Chloroquine Nortriptyline Digoxin Propranolol Quinidine Quinolones (1- 2 L/kg), Tetracycline Phenobarbital Phenytoin Theophylline (0.45 L/kg) Aminoglycodises (0.25 L/kg) ASA Warfarin To date, we have assumed that distribution of drug in the body was instantaneous. What if that is NOT true?

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

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

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. 10 L 1000 mg 1L

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. 10 L 1000 mg 1L

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. 10 L 1000 mg 1L

The rate constants describing drug movement are labeled based on their movement from one compartment to another. k12 describes movement from compartment 1 to compartment 2. k21 describes movement from compartment 2 to compartment 1. k10 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. 2 Compartment Model:IV Dosing

Although equilibrium has been established between the two compartments, concentrations in each compartment may be different. 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. Equilibrium Established

When the body appears to have a second compartment, distribution into each compartment occurs at different rates. 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 greater blood flow occurs more quickly … almost instantaneously. 10 L 1000 mg 1L

However, differences in profiles is not solely related to the size of k12. Here, both profiles have the same terminal elimination (  = 0.2 hr-1) AND the same k12 (0.3 hr-1). Profiles have different k21 values. Differences in profiles are created by different relative values of k12 : k21. Movement in and out k12=0.3 hr-1; k21=0.217 hr-1 k12=0.3 hr-1; k21=0.4 hr-1 k12/k21= 0.75 k12/k21= 1.38

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 k12:k21. 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 k12:k21 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? 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 Patient 2 TimePlasma Conctime Plasma Conc (hr) (mg/L)(hr) (mg/L) Analysis of 2 Compartment Model:IV Dosing

Patient 1 TimePlasma Conc (hr) (mg/L) 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. Plotting / Graphing Plasma Concentrations

Graph Patient 1 Data Using semi-log paper, graph the data from these two patients, each given a 1000 mg dose. Patient 1 TimePlasma Conc (hr) (mg/L)

Graph Patient 1 Data What is the Half-life? … the Initial [ ]? What model best describes data? Patient 1 TimePlasma Conc (hr) (mg/L)

Graph Patient 1 Data Patient 1 TimePlasma Conc (hr) (mg/L) What is the Half-life? … the Initial [ ]? What model best describes data?

Graph Patient 1 Data Patient 1 TimePlasma Conc (hr) (mg/L) What is the Half-life? 3hr … the Initial [ ]? …100 mg/L What model best describes data?

Graph Patient 1 Data 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. IV Dose (bolus) V1 k

Graph Patient 1 Data IV Dose (bolus) V1 K 1 Dose V C t = e -Kt Dose = 1000 mg V=10 L K= hr -1 C t = 100 e (t)

Graph Patient 2 Data Patient 2 TimePlasma Conc (hr) (mg/L) Using semi-log paper, graph the data from Patient 2, also given a 1000 mg dose.

Graph Patient 2 Data What is different about this patient? Patient 2 TimePlasma Conc (hr) (mg/L)

Graph Patient 2 Data What is different about this patient? What is the Half-life? … the Initial [ ]? Patient 2 TimePlasma Conc (hr) (mg/L)

Graph Patient 2 Data What is different about this patient? What is the Half-life? 3hr … the Initial [ ]?..100 mg/L What model describes a patient like Patient 2? TimePlasma Conc (hr) (mg/L)

Graph Patient 2 Data 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. 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 C t = A e + B e -αt-αt-βt-βt The appearance of the second exponential in the [ ] – time profile is an indication that we have a 2-compartment model.

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 = A e + B e -αt-αt-βt-βt

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

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

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

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

Analysis of a 2C model 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 Terminal Phase Slope =  C t = A e + B e -αt-αt-βt-βt Analysis of 2 Compartment Model:IV Dosing

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

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

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

Analysis of a 2C 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. C t = A e + B e -αt-αt-βt-βt Analysis of 2 Compartment Model:IV Dosing

Analysis of a 2C model Step 5 Zero time intercepts of back-extrapolated β phase and of residual (α), are equal to B and A respectively. B A C t = A e + B e -αt-αt-βt-βt α and β are “MACRO” constants in that they are made up of the “MICRO” constants k12, k21 and k10. Analysis of 2 Compartment Model:IV Dosing

Analysis of a 2C model Step 5 B A Dose * (k 21 -α) Dose * (k21-β) V * ( β – α) V * (α - β) C = e e -αt-αt -βt-βt α and β are “MACRO” constants in that they are made up of the “MICRO” constants (k12, k21 and k10) and reflect a variety of simultaneous processes. [α + β = k12 + k21 + k10] (Equations can be shown in either form) Analysis of 2 Compartment Model:IV Dosing

Analysis of a 2C model Step 6 Begin to solve for micro constants. k21 = (Aβ + Bα) / ( A + B) B A C t = A e + B e -αt-αt-βt-βt Analysis of 2 Compartment Model:IV Dosing

Analysis of a 2C model Step 7 Begin to solve for micro constants. k10 = α β / k21 B A C t = A e + B e -αt-αt-βt-βt Analysis of 2 Compartment Model:IV Dosing

Analysis of a 2C model Step 8 Begin to solve for micro constants. k12 = α + β - k21 – k10 B A C t = A e + B e -αt-αt-βt-βt Analysis of 2 Compartment Model:IV Dosing

Analysis of a 2C model Step 9 Begin to solve for other parameters V1 can be obtained from the initial concentration and dose V1 = dose / C 0 B A C t = A e + B e -αt-αt-βt-βt Analysis of 2 Compartment Model:IV Dosing

Analysis of a 2C model Step 10 Begin to solve for other parameters Vss = V1(1 + k12/k21) B A C t = A e + B e -αt-αt-βt-βt Analysis of 2 Compartment Model:IV Dosing

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

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

Graph Patient 2 Data C t = A e + B e -αt-αt-βt-βt In this [ ] – time profile the β-phase intercept (B) is 27.8 mg/L and the α-phase intercept (A) is 72.2 mg/L. C t = 72.2 e e -αt-αt -βt-βt What would the initial concentration be at time zero?

Graph Patient 2 Data C t = A e + B e -αt-αt-βt-βt In this [ ] – time profile the β-phase intercept (B) is 27.8 mg/L and the α-phase intercept (A) is 72.2 mg/L. C t = 72.2 e e -αt-αt -βt-βt What would the initial concentration be at time zero? C 0 =A+B and V1 can be obtained from the initial [ ]and dose. V1 = Dose / C 0 = 1000 mg/100 mg/L = 10 L

Graph Patient 2 Data C t = A e + B e -αt-αt-βt-βt In this [ ] – time profile the β-phase intercept (B) is 27.8 mg/L and the α-phase intercept (A) is 72.2 mg/L. C t = 72.2 e e -αt-αt -βt-βt AUC following an IV dose in a 2C model can be calculated by standard methods (trapezoidal rule + kinetic method from the last concentration: [ ]/β) or based on the formula AUC 0-∞ = (A/ α) + (B/ β)

Graph Patient 2 Data C t = A e + B e -αt-αt-βt-βt C t = 72.2 e e -αt-αt -βt-βt A 2-C model is the result of a drug which has an initial small volume of distribution (V1) which grows following the dose to achieve its final volume (Vss). The micro-constants k12 and k21 represent the transfer constants between the central (C-1) and peripheral (C-2) compartments.

Graph Patient 2 Data C t = A e + B e -αt-αt-βt-βt C t = 72.2 e e -αt-αt -βt-βt The final volume (Vss) is a function of the rate into the peripheral compartment (k12) and the rate out of the peripheral compartment (k21). Vss = V1(1 + k12/k21)

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

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

Analysis of a 2C model What is the initial time zero concentration? 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 + B e -αt-αt-βt-βt B A Terminal Phase Slope =  Residual Slope = α Analysis of 2 Compartment Model:IV Dosing

Dose * (k 21 -α) Dose * (k 21 -β) V * ( β – α) V * (α - β) C = e e -αt-αt -βt-βt What does the value of  do to the concentration – time profile? k12, k21 and k10 are micro constants α and β are more complex constants and reflect a variety of simultaneous processes. α + β = k12 + k21 + k10 α = k21*k10/β and β 2 – (k12+k21+k10)*β + k21/k10 = 0 Analysis of 2 Compartment Model:IV Dosing

Dose * (k 21 -α) Dose * (k 21 -β) V * ( β – α) V * (α - β) C = e e -αt-αt -βt-βt α = 0.5 β = 0.2 k21 = 0.25 K12 = 0.05 α = 2.0 β = 0.2 k21 = 0.25 K12 = 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 k21&k12. Recall … Vss = V1(1 + k12/k21) The concentration changes rapidly. What does this imply? Analysis of 2 Compartment Model:IV Dosing

Dose * (k 21 -α) Dose * (k 21 -β) V * ( β – α) V * (α - β) C = e e -αt-αt -βt-βt α = 1.0 β = 0.2 k21 = 0.25 K12 = 0.15 V1 = 10 V2 = 6 α = 0.5 β = 0.2 k21 = 0.25 K12 = 0.05 V1 = 10 V2 = 2 α = 2.0 β = 0.2 k21 = 0.25 K12 = 0.35 V1 = 10 V2 = 14 Analysis of 2 Compartment Model:IV Dosing

Dose * (k 21 -α) Dose * (k 21 -β) V * ( β – α) V * (α - β) C = e e -αt-αt -βt-βt As  increases, the rate of change of concentration during the distribution phase increases and the size of V2 increases. Vss = V1 + V2 α = 2.0 β = 0.2 k21 = 0.25 K12 = 0.35 V1 = 10 V2 = 14 α = 0.5 β = 0.2 k21 = 0.25 K12 = 0.05 V1 = 10 V2 = 2

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

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

Summary Compartment Models:Describing [ ] - Time 2 compartment with IV bolus input [2 exponentials  and  ) Dose * (k 21 -α) Dose * (k21-β) V * ( β – α) V * (α - β) C = e e -αt-αt -βt-βt

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

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