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Handling Concentration – Time Data Determining Elimination Rate (K) AUC Calculations Back Calculation

Handling Concentration – Time Data Determining Elimination Rate (K) AUC Calculations Back Calculation. K. We have estimated AUC … now we need to calculate k. Dose = 1000 mg Time Conc AUC (hr) (mg/L) mg*hr/L 0 100.0 1 89.1 94.55

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Handling Concentration – Time Data Determining Elimination Rate (K) AUC Calculations Back Calculation

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  1. Handling • Concentration – Time • Data • Determining Elimination Rate (K) • AUC Calculations • Back Calculation

  2. K We have estimated AUC … now we need to calculate k Dose = 1000 mg Time Conc AUC (hr) (mg/L) mg*hr/L 0 100.0 1 89.1 94.55 2 79.4 84.25 4 60.0 139.40 12 25.0 340.00 18 12.5 112.50 24 6.25 56.25 8. Estimate the AUC from the last measured time point to infinity using the pharmacokinetic method: [ ]last / K. How do we calculate k? We first calculated K from Cl & V, but the purpose of estimating AUC was to determine Cl !

  3. K Determining Elimination Rate Constant Drugs are cleared from the body. Clearance can occur by several pathways, urinary, biliary, excretion into the air, biotransformation in the liver… Ca Cv

  4. K Determining Elimination Rate Constant Drugs are cleared from the body. Clearance can occur by several pathways, urinary, biliary, excretion into the air, biotransformation in the liver… Elimination can often be characterized by an apparent first order process. Rate of Elimination is proportional to the amount of drug in the body at that time.

  5. K Determining Elimination Rate Constant The proportionality constant relating the rate and amount is the first order elimination rate constant (K) with units time-1 (min-1, hr-1). Ca Cv

  6. K Determining Elimination Rate Constant The proportionality constant relating the rate and amount is the first order elimination rate constant (K) with units time-1 (min-1, hr-1). The first order rate constant characterizing overall elimination is often given the symbol K and it is often the sum of two or more rate constants which characterize individual elimination processes … K = ke + km + kl … Ca Cv ke

  7. K Determining Elimination Rate Constant Drug elimination from the body can therefore be described by dX dt where X is the amount of drug in the body at any time t after bolus iv administration. The negative sign indicates that drug is being lost from the body. = - KX Ca Cv

  8. K Determining Elimination Rate Constant Drug elimination from the body can therefore be described by dX dt where X is the amount of drug in the body at any time t after bolus iv administration. To describe the time course of the amount of drug in the body after bolus injection we must integrate this expression to yield: X = X0e-Kt where ‘e’ is the base of the natural log = - KX

  9. K Determining Elimination Rate Constant X = X0e-Kt X0 represented the amount at time 0 and X would represent the amount at any time t, thereafter… Xt = X0e-Kt This expression can be used to determine the amount in the body at any time following a bolus dose where the body resembles a homogeneous single compartment (container or tub).

  10. K Determining Elimination Rate Constant Xt = X0e-Kt X0 represented the amount at time 0 and Xt would represent the amount at any time t, thereafter. Taking the natural log of both sides ln(Xt) = ln(X0) – Kt or alternatively, since 2.303 log(a) = ln(a) then: log(Xt) = log (X0) – Kt/2.303

  11. K Determining Elimination Rate Constant Xt = X0e-Kt ln(Xt) = ln(X0) – Kt log(Xt) = log (X0) – Kt/2.303 but we cannot directly measure the amount of drug in the body at any time. Concentration in plasma is more directly measured. Volume of distribution relates the amount in the body to concentration. Therefore: sinceX = VCthenln(Ct) = ln(C0) –Kt and / orlog(Ct) = log(C0) –Kt / 2.303

  12. K Determining Elimination Rate Constant ln(Ct) = ln(C0) – Kt If the concentration is reduced to half of the initial concentration in time t then: ln(0.5*C0) = ln(C0) – Kt½ ln(0.5) / K = t½ 0.69315 / K = t½ Half-life is determined directly from K which can be determined from a change in concentration T½ = 0.69315 / K or T½ = 0.693 / K

  13. K Determining Elimination Rate Constant So why did we use logarithms? If a patient with a volume of 10 L is administered a bolus dose of 1000 mg, a plot of concentration vs. time would produce this graph. C = C0e-Kt Note: The initial concentration is 100 mg/L. However, if we convert each concentration to a common logarithm, the same concentration-time plot would now look like linear. 100 mg/L

  14. K Determining Elimination Rate Constant So why did we use logarithms? First order processes appear log-linear. A log-linear relationship is <generally> easier to interpret. Conversion can be done easily by using semi-log paper where only the y-axis is in a log scale. In Excel you can also easily change the scale to a log scale. 100 mg/L

  15. K Determining Elimination Rate Constant So why did we use logarithms? The slope of a straight – line is easier to evaluate. log(C) = log(C0) –Kt / 2.303 The slope of a log- concentration-time profile is: Slope = -K / 2.303 This will also us to determine the elimination rate constant (K). Slope = -K / 2.303

  16. K Determining Elimination Rate Constant Example from last day …? K ? Dose = 1000 mg Time Conc AUC (hr) (mg/L) mg*hr/L 0 100.0 1 89.1 94.55 2 79.4 84.25 4 60.0 139.40 12 25.0 340.00 18 12.5 112.50 24 6.25 56.25 8. Estimate the AUC from the last measured time point to infinity using the pharmacokinetic method: [ ]last / k. Estimation of K. K is the slope of the line calculated from a graph or by equation K = -2.303[log(C2) – log(C1)] / (t2 - t1)

  17. Determining Elimination Rate Constant K Estimating K using graph paper ! 100 Dose = 1000 mg Time Conc (hr) (mg/L) 0 100.0 1 89.1 2 79.4 4 60.0 12 25.0 18 12.5 24 6.25 Upper cycle 10 Lower cycle 2 cycle semi-log paper 1

  18. Determining Elimination Rate Constant K Estimating K using graph paper ! 100 89.1 Dose = 1000 mg Time Conc (hr) (mg/L) 0 100.0 1 89.1 2 79.4 4 60.0 12 25.0 18 12.5 24 6.25 60 6.25 2 cycle semi-log paper 0 4 8 12 16 20 24 28

  19. Determining Elimination Rate Constant K Estimating K using graph paper ! Dose = 1000 mg Time Conc (hr) (mg/L) 0 100.0 1 89.1 2 79.4 4 60.0 12 25.0 18 12.5 24 6.25 Plot the points What is the half-life? 0 4 8 12 16 20 24 28

  20. Determining Elimination Rate Constant K Estimating K using graph paper ! Dose = 1000 mg Time Conc (hr) (mg/L) 0 100.0 1 89.1 2 79.4 4 60.0 12 25.0 18 12.5 24 6.25 What is the half-life? C0 = 100 mg/L 1 half-life later = [ ? ] = T½ 0 4 8 12 16 20 24 28

  21. Determining Elimination Rate Constant K 50 mg/L Estimating K using graph paper ! Dose = 1000 mg Time Conc (hr) (mg/L) 0 100.0 1 89.1 2 79.4 4 60.0 12 25.0 18 12.5 24 6.25 What is the half-life? C0 = 100 mg/L 1 half-life later = [ ? ] = T½ = 50 mg/L Time? … 0 4 8 12 16 20 24 28

  22. Determining Elimination Rate Constant K 50 mg/L Estimating K using graph paper ! Dose = 1000 mg Time Conc (hr) (mg/L) 0 100.0 1 89.1 2 79.4 4 60.0 12 25.0 18 12.5 24 6.25 What is the half-life? C0 = 100 mg/L 1 half-life later = [ ? ] = T½ = 50 mg/L Time? … 6 hours. = T½ 0 4 8 12 16 20 24 28

  23. Determining Elimination Rate Constant K 50 mg/L Estimating K using graph paper ! Dose = 1000 mg Time Conc (hr) (mg/L) 0 100.0 1 89.1 2 79.4 4 60.0 12 25.0 18 12.5 24 6.25 Check… If the half-life is 6 hr, what will the [ ] be at 12 hours? 0 4 8 12 16 20 24 28

  24. Determining Elimination Rate Constant K 50 mg/L Estimating K using graph paper ! 25 mg/L Dose = 1000 mg Time Conc (hr) (mg/L) 0 100.0 1 89.1 2 79.4 4 60.0 12 25.0 18 12.5 24 6.25 12.5 mg/L Check… If the half-life is 6 hr, what will the [ ] be at 12 hours? … 25 mg/L … 12.5 mg/L 0 4 8 12 16 20 24 28

  25. Determining Elimination Rate Constant K 50 mg/L Estimating K using graph paper ! 25 mg/L Dose = 1000 mg Time Conc (hr) (mg/L) 0 100.0 1 89.1 2 79.4 4 60.0 12 25.0 18 12.5 24 6.25 12.5 mg/L If the half-life is 6 hr, what is K? K = 0 4 8 12 16 20 24 28

  26. Determining Elimination Rate Constant K 50 mg/L Estimating K using graph paper ! 25 mg/L Dose = 1000 mg Time Conc (hr) (mg/L) 0 100.0 1 89.1 2 79.4 4 60.0 12 25.0 18 12.5 24 6.25 12.5 mg/L If the half-life is 6 hr, what is K? K = 0.693 / T½ 0 4 8 12 16 20 24 28

  27. Determining Elimination Rate Constant K 50 mg/L Estimating K using graph paper ! 25 mg/L Dose = 1000 mg Time Conc (hr) (mg/L) 0 100.0 1 89.1 2 79.4 4 60.0 12 25.0 18 12.5 24 6.25 12.5 mg/L If the half-life is 6 hr, what is K? K = 0.693 / T½ = 0.693 / 6 hr = 0.1155 hr-1 0 4 8 12 16 20 24 28

  28. K Determining Elimination Rate Constant Estimate K by equation … Dose = 1000 mg Time Conc AUC (hr) (mg/L) mg*hr/L 0 100.0 1 89.1 94.55 2 79.4 84.25 4 60.0 139.40 12 25.0 340.00 18 12.5 112.50 24 6.25 56.25 Estimation of K. K is the slope of the line (t=4 to 24 hr) K = -2.303[log(C2) – log(C1)] / (t2 - t1) =

  29. K Determining Elimination Rate Constant Estimate K by equation … Dose = 1000 mg Time Conc AUC (hr) (mg/L) mg*hr/L 0 100.0 1 89.1 94.55 2 79.4 84.25 4 60.0 139.40 12 25.0 340.00 18 12.5 112.50 24 6.25 56.25 Estimation of K. K is the slope of the line (t=4 to 24 hr) K = -2.303[log(C2) – log(C1)] / (t2 - t1) = -2.303[log(6.25) – log(60)] / (24 – 4) = -2.303

  30. K Determining Elimination Rate Constant Estimate K by equation … Dose = 1000 mg Time Conc AUC (hr) (mg/L) mg*hr/L 0 100.0 1 89.1 94.55 2 79.4 84.25 4 60.0 139.40 12 25.0 340.00 18 12.5 112.50 24 6.25 56.25 Estimation of K. K is the slope of the line (t=4 to 24 hr) K = -2.303[log(C2) – log(C1)] / (t2 - t1) = -2.303[log(6.25) – log(60)] / (24 – 4) = -2.303[0.796 - 1.778] / (20) = -2.303

  31. K Determining Elimination Rate Constant Estimate K by equation … Dose = 1000 mg Time Conc AUC (hr) (mg/L) mg*hr/L 0 100.0 1 89.1 94.55 2 79.4 84.25 4 60.0 139.40 12 25.0 340.00 18 12.5 112.50 24 6.25 56.25 Estimation of K. K is the slope of the line (t=4 to 24 hr) K = -2.303[log(C2) – log(C1)] / (t2 - t1) = -2.303[log(6.25) – log(60)] / (24 – 4) = -2.303[0.796 - 1.778] / (20) = -2.303[ - 0.9823]/20 = 0.1131 hr-1 T½ = 0.693/0.1131 = ~6.12 hr.

  32. K Determining Elimination Rate Constant Methods of estimating K … Dose = 1000 mg Time Conc AUC (hr) (mg/L) mg*hr/L 0 100.0 1 89.1 94.55 2 79.4 84.25 4 60.0 139.40 12 25.0 340.00 18 12.5 112.50 24 6.25 56.25 • Methods of Estimating K. • Visual inspection of concentration –time data • Plotting the log [ ] vs. time and determining the half-life  K • Determining K by equation from the log of [ ] of any 2 points. • all methods should produce the same estimate when points line on the line.

  33. K Determining Elimination Rate Constant Now we have an estimate of k and can determine the area by the kinetic method from the last point to infinity. Dose = 1000 mg Time Conc AUC (hr) (mg/L) mg*hr/L 0 100.0 1 89.1 94.55 2 79.4 84.25 4 60.0 139.40 12 25.0 340.00 18 12.5 112.50 24 6.25 56.25 8. Estimate the AUC from the last measured time point to infinity using the pharmacokinetic method: [ ]last / k. K = 0.1131 hr-1 AUC LP -  = [ ]last / k. =

  34. K Determining Elimination Rate Constant Example from last day …K? Dose = 1000 mg Time Conc AUC (hr) (mg/L) mg*hr/L 0 100.0 1 89.1 94.55 2 79.4 84.25 4 60.0 139.40 12 25.0 340.00 18 12.5 112.50 24 6.25 56.25 8. Estimate the AUC from the last measured time point to infinity using the pharmacokinetic method: [ ]last / k. K = 0.1131 hr-1 AUC LP -  = [ ]last / k. = 6.25 / 0.1131 =6.25 / 0.1155 = 55.25 mg*hr/L =54.11 mg*hr/L

  35. Determining Volume & Clearance AUC Estimation of AUC0- Dose = 1000 mg Time Conc AUC (hr) (mg/L) mg*hr/L 0 100.0 1 89.1 94.55 2 79.4 84.25 4 60.0 139.40 12 25.0 340.00 18 12.5 112.50 24 6.25 56.25 24- 55.25 ______ Total AUC0- (mg*hr/L) Sum all of the individual areas to obtain the total AUC0-

  36. Determining Volume & Clearance Cl Estimation of AUC0- Dose = 1000 mg Time Conc AUC (hr) (mg/L) mg*hr/L 0 100.0 1 89.1 94.55 2 79.4 84.25 4 60.0 139.40 12 25.0 340.00 18 12.5 112.50 24 6.25 56.25 24- 55.25 ______ Total AUC0- (mg*hr/L)881.24 Sum all of the individual areas to obtain the total AUC0- With K and AUC0- calculated, determine Clearance

  37. Determining Volume & Clearance Cl Estimation of AUC0- Dose = 1000 mg Time Conc AUC (hr) (mg/L) mg*hr/L 0 100.0 1 89.1 94.55 2 79.4 84.25 4 60.0 139.40 12 25.0 340.00 18 12.5 112.50 24 6.25 56.25 24- 55.25 ______ Total AUC0- (mg*hr/L)881.24 With K and AUC0- calculated, determine Clearance Clearance =

  38. Determining Volume & Clearance Cl Estimation of AUC0- Dose = 1000 mg Time Conc AUC (hr) (mg/L) mg*hr/L 0 100.0 1 89.1 94.55 2 79.4 84.25 4 60.0 139.40 12 25.0 340.00 18 12.5 112.50 24 6.25 56.25 24- 55.25 ______ Total AUC0- (mg*hr/L)881.24 With K and AUC0- calculated, determine Clearance Clearance = Dose / AUC0- =

  39. Cl Determining Volume & Clearance Estimation of AUC0- Dose = 1000 mg Time Conc AUC (hr) (mg/L) mg*hr/L 0 100.0 1 89.1 94.55 2 79.4 84.25 4 60.0 139.40 12 25.0 340.00 18 12.5 112.50 24 6.25 56.25 24- 55.25 ______ Total AUC0- (mg*hr/L)881.24 With K and AUC0- calculated, determine Clearance Clearance = Dose / AUC0- = 1000 mg / 881.24 mg*hr/L = 1.13 L/hr

  40. Determining Volume & Clearance V Estimation of AUC0- Dose = 1000 mg Time Conc AUC (hr) (mg/L) mg*hr/L 0 100.0 1 89.1 94.55 2 79.4 84.25 4 60.0 139.40 12 25.0 340.00 18 12.5 112.50 24 6.25 56.25 24- 55.25 ______ Total AUC0- (mg*hr/L)881.24 Pharmacokinetic Summary: Volume (L) = 10 L

  41. Determining Volume & Clearance V Estimation of AUC0- Dose = 1000 mg Time Conc AUC (hr) (mg/L) mg*hr/L 0 100.0 1 89.1 94.55 2 79.4 84.25 4 60.0 139.40 12 25.0 340.00 18 12.5 112.50 24 6.25 56.25 24- 55.25 ______ Total AUC0- (mg*hr/L)881.24 Pharmacokinetic Summary: Volume (L) = 10 L Elim. Rate (K) = 0.1131 hr  T½ = 0.693/K = 6.12 hr

  42. Determining Volume & Clearance V Estimation of AUC0- Dose = 1000 mg Time Conc AUC (hr) (mg/L) mg*hr/L 0 100.0 1 89.1 94.55 2 79.4 84.25 4 60.0 139.40 12 25.0 340.00 18 12.5 112.50 24 6.25 56.25 24- 55.25 ______ Total AUC0- (mg*hr/L)881.24 Pharmacokinetic Summary: Volume (L) = 10 L Elim. Rate (K) = 0.1131 hr  T½ = 0.693/K = 6.12 hr AUC0- (mg*hr/L) = 881.24 mg*hr/L Clearance (L/hr) = 1.13 L/hr

  43. Dealing with Concentration –time Data (ii) Calculating AUC Time Conc Conc (hr) (mg/L) (mg/L) Calculate the AUC by trapezoidal rule 1 100 100 For these two patients. The second. 2 50 --- is missing the 2 hr sample 3 25 25 OPENING PROBLEM

  44. CONSIDER THIS PROBLEM AUC Calculate the AUC by trapezoidal rule for these two patients. The second is missing the 2 hr sample. Patient Patient 1 2 Time Conc Conc (hr) (mg/L) (mg/L) 1 100 100 2 50 --- 3 25 25 Equations Conc = Dose / V V = Dose/Conc Cl = Q x ER ER = Cl / Q AUC = -------- (t2-t1) Cl = Dose / AUC K = Cl / V AUC from 1-2hr: =((C1 + C2)/2)(t2 – t1) = (C1+C2) 2

  45. CONSIDER THIS PROBLEM AUC Calculate the AUC by trapezoidal rule for these two patients. The second is missing the 2 hr sample. Patient Patient 1 2 Time Conc Conc (hr) (mg/L) (mg/L) 1 100 100 2 50 --- 3 25 25 Equations Conc = Dose / V V = Dose/Conc Cl = Q x ER ER = Cl / Q AUC = -------- (t2-t1) Cl = Dose / AUC K = Cl / V AUC from 1-2hr: =((C1 + C2)/2)(t2 – t1) = ((100+50)/2)(2-1) = 75 AUC from 2-3hr: =((C1 + C2)/2)(t2 – t1) = (C1+C2) 2

  46. CONSIDER THIS PROBLEM AUC Calculate the AUC by trapezoidal rule for these two patients. The second is missing the 2 hr sample. Patient Patient 1 2 Time Conc Conc (hr) (mg/L) (mg/L) 1 100 100 2 50 --- 3 25 25 Equations Conc = Dose / V V = Dose/Conc Cl = Q x ER ER = Cl / Q AUC = -------- (t2-t1) Cl = Dose / AUC K = Cl / V AUC from 1-2hr: =((C1 + C2)/2)(t2 – t1) = ((100+50)/2)(2-1) = 75 AUC from 2-3hr: =((C1 + C2)/2)(t2 – t1) = ((50+25)/2)(3-2) = 37.5 Total AUC 1-3 hr: = (C1+C2) 2

  47. CONSIDER THIS PROBLEM AUC Calculate the AUC by trapezoidal rule for these two patients. The second is missing the 2 hr sample. Patient Patient 1 2 Time Conc Conc (hr) (mg/L) (mg/L) 1 100 100 2 50 --- 3 25 25 Equations Conc = Dose / V V = Dose/Conc Cl = Q x ER ER = Cl / Q AUC = -------- (t2-t1) Cl = Dose / AUC K = Cl / V AUC from 1-2hr: =((C1 + C2)/2)(t2 – t1) = ((100+50)/2)(2-1) = 75 AUC from 2-3hr: =((C1 + C2)/2)(t2 – t1) = ((50+25)/2)(3-2) = 37.5 Total AUC 1-3 hr: = 75 + 37.5 = 112.5 mg*hr/L (C1+C2) 2

  48. CONSIDER THIS PROBLEM AUC Calculate the AUC by trapezoidal rule for these two patients. The second is missing the 2 hr sample. Patient Patient 1 2 Time Conc Conc (hr) (mg/L) (mg/L) 1 100 100 2 50 --- 3 25 25 Equations Conc = Dose / V V = Dose/Conc Cl = Q x ER ER = Cl / Q AUC = -------- (t2-t1) Cl = Dose / AUC K = Cl / V Patient 2 AUC from 1-3hr: =((C1 + C2)/2)(t2 – t1) = ((100+25)/2)(3-1) = (125/2)(2) = (C1+C2) 2

  49. CONSIDER THIS PROBLEM AUC Calculate the AUC by trapezoidal rule for these two patients. The second is missing the 2 hr sample. Patient Patient 1 2 Time Conc Conc (hr) (mg/L) (mg/L) 1 100 100 2 50 --- 3 25 25 Equations Conc = Dose / V V = Dose/Conc Cl = Q x ER ER = Cl / Q AUC = -------- (t2-t1) Cl = Dose / AUC K = Cl / V Patient 2 AUC from 1-3hr: =((C1 + C2)/2)(t2 – t1) = ((100+25)/2)(3-1) = (125/2)(2) Pt 2; AUC 1-3 hr: = 125.0 mg*hr/L Pt 1; AUC 1-3 hr: = 112.5 mg*hr/L (C1+C2) 2

  50. CONSIDER THIS PROBLEM AUC Calculate the AUC by trapezoidal rule for these two patients. The second is missing the 2 hr sample. Patient Patient 1 2 Time Conc Conc (hr) (mg/L) (mg/L) 1 100 100 2 50 --- 3 25 25 Equations Conc = Dose / V V = Dose/Conc Cl = Q x ER ER = Cl / Q AUC = -------- (t2-t1) Cl = Dose / AUC K = Cl / V Patient 2 AUC from 1-3hr: =((C1 + C2)/2)(t2 – t1) = ((100+25)/2)(3-1) = (125/2)(2) Pt 1; AUC 1-3 hr: = 112.5 mg*hr/L Pt 2; AUC 1-3 hr: = 125.0 mg*hr/L Why the difference? (C1+C2) 2

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