Determining Elimination Rate (K)

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

We have estimated AUC … now we need to calculate k K
Dose = 1000 mg Time Conc AUC (hr) (mg/L) mg*hr/L 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 !

Determining Elimination Rate Constant
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

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.

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

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

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

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

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).

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

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: since X = VC then ln(Ct) = ln(C0) –Kt and / or log(Ct) = log(C0) –Kt / 2.303

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 / K = t½ Half-life is determined directly from K which can be determined from a change in concentration T½ = / K or T½ = / K

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

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

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 log- concentration-time profile is: Slope = -K / 2.303 This will also us to determine the elimination rate constant (K). Slope = -K / 2.303

K Determining Elimination Rate Constant Example from last day …? K ? Dose = 1000 mg Time Conc AUC (hr) (mg/L) mg*hr/L 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 = [log(C2) – log(C1)] / (t2 - t1)

K Estimating K using graph paper ! 100 Dose = 1000 mg Time Conc (hr) (mg/L) Upper cycle 10 Lower cycle 2 cycle semi-log paper 1

K Estimating K using graph paper ! 100 89.1 Dose = 1000 mg Time Conc (hr) (mg/L) 60 6.25 2 cycle semi-log paper

K Estimating K using graph paper ! Dose = 1000 mg Time Conc (hr) (mg/L) Plot the points What is the half-life?

K Estimating K using graph paper ! Dose = 1000 mg Time Conc (hr) (mg/L) What is the half-life? C0 = 100 mg/L 1 half-life later = [ ? ] = T½

K 50 mg/L Estimating K using graph paper ! Dose = 1000 mg Time Conc (hr) (mg/L) What is the half-life? C0 = 100 mg/L 1 half-life later = [ ? ] = T½ = 50 mg/L Time? …

K 50 mg/L Estimating K using graph paper ! Dose = 1000 mg Time Conc (hr) (mg/L) What is the half-life? C0 = 100 mg/L 1 half-life later = [ ? ] = T½ = 50 mg/L Time? … 6 hours. = T½

K 50 mg/L Estimating K using graph paper ! Dose = 1000 mg Time Conc (hr) (mg/L) Check… If the half-life is 6 hr, what will the [ ] be at 12 hours?

K 50 mg/L Estimating K using graph paper ! 25 mg/L Dose = 1000 mg Time Conc (hr) (mg/L) 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

K 50 mg/L Estimating K using graph paper ! 25 mg/L Dose = 1000 mg Time Conc (hr) (mg/L) 12.5 mg/L If the half-life is 6 hr, what is K? K =

K 50 mg/L Estimating K using graph paper ! 25 mg/L Dose = 1000 mg Time Conc (hr) (mg/L) 12.5 mg/L If the half-life is 6 hr, what is K? K = / T½

K 50 mg/L Estimating K using graph paper ! 25 mg/L Dose = 1000 mg Time Conc (hr) (mg/L) 12.5 mg/L If the half-life is 6 hr, what is K? K = / T½ = / 6 hr = hr-1

K Determining Elimination Rate Constant Estimate K by equation … Dose = 1000 mg Time Conc AUC (hr) (mg/L) mg*hr/L Estimation of K. K is the slope of the line (t=4 to 24 hr) K = [log(C2) – log(C1)] / (t2 - t1) =

K Determining Elimination Rate Constant Estimate K by equation … Dose = 1000 mg Time Conc AUC (hr) (mg/L) mg*hr/L Estimation of K. K is the slope of the line (t=4 to 24 hr) K = [log(C2) – log(C1)] / (t2 - t1) = [log(6.25) – log(60)] / (24 – 4) =

K Determining Elimination Rate Constant Estimate K by equation … Dose = 1000 mg Time Conc AUC (hr) (mg/L) mg*hr/L Estimation of K. K is the slope of the line (t=4 to 24 hr) K = [log(C2) – log(C1)] / (t2 - t1) = [log(6.25) – log(60)] / (24 – 4) = [ ] / (20) =

K Determining Elimination Rate Constant Estimate K by equation … Dose = 1000 mg Time Conc AUC (hr) (mg/L) mg*hr/L Estimation of K. K is the slope of the line (t=4 to 24 hr) K = [log(C2) – log(C1)] / (t2 - t1) = [log(6.25) – log(60)] / (24 – 4) = [ ] / (20) = [ ]/20 = hr-1 T½ = 0.693/ = ~6.12 hr.

K Determining Elimination Rate Constant Methods of estimating K … Dose = 1000 mg Time Conc AUC (hr) (mg/L) mg*hr/L 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.

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 8. Estimate the AUC from the last measured time point to infinity using the pharmacokinetic method: [ ]last / k. K = hr-1 AUC LP -  = [ ]last / k. =

K Determining Elimination Rate Constant Example from last day …K? Dose = 1000 mg Time Conc AUC (hr) (mg/L) mg*hr/L 8. Estimate the AUC from the last measured time point to infinity using the pharmacokinetic method: [ ]last / k. K = hr-1 AUC LP -  = [ ]last / k. = 6.25 / =6.25 / = mg*hr/L =54.11 mg*hr/L

Determining Volume & Clearance
AUC Estimation of AUC0- Dose = 1000 mg Time Conc AUC (hr) (mg/L) mg*hr/L 24- ______ Total AUC0- (mg*hr/L) Sum all of the individual areas to obtain the total AUC0-

Estimation of AUC0- Dose = 1000 mg Time Conc AUC (hr) (mg/L) mg*hr/L 24- ______ Total AUC0- (mg*hr/L) Sum all of the individual areas to obtain the total AUC0- With K and AUC0- calculated, determine Clearance

Estimation of AUC0- Dose = 1000 mg Time Conc AUC (hr) (mg/L) mg*hr/L 24- ______ Total AUC0- (mg*hr/L) With K and AUC0- calculated, determine Clearance Clearance =

Estimation of AUC0- Dose = 1000 mg Time Conc AUC (hr) (mg/L) mg*hr/L 24- ______ Total AUC0- (mg*hr/L) With K and AUC0- calculated, determine Clearance Clearance = Dose / AUC0- =

Estimation of AUC0- Dose = 1000 mg Time Conc AUC (hr) (mg/L) mg*hr/L 24- ______ Total AUC0- (mg*hr/L) With K and AUC0- calculated, determine Clearance Clearance = Dose / AUC0- = 1000 mg / mg*hr/L = 1.13 L/hr

Estimation of AUC0- Dose = 1000 mg Time Conc AUC (hr) (mg/L) mg*hr/L 24- ______ Total AUC0- (mg*hr/L) Pharmacokinetic Summary: Volume (L) = 10 L

Estimation of AUC0- Dose = 1000 mg Time Conc AUC (hr) (mg/L) mg*hr/L 24- ______ Total AUC0- (mg*hr/L) Pharmacokinetic Summary: Volume (L) = 10 L Elim. Rate (K) = hr  T½ = 0.693/K = 6.12 hr

Estimation of AUC0- Dose = 1000 mg Time Conc AUC (hr) (mg/L) mg*hr/L 24- ______ Total AUC0- (mg*hr/L) Pharmacokinetic Summary: Volume (L) = 10 L Elim. Rate (K) = hr  T½ = 0.693/K = 6.12 hr AUC0- (mg*hr/L) = mg*hr/L Clearance (L/hr) = 1.13 L/hr

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

CONSIDER THIS PROBLEM Equations AUC
Calculate the AUC by trapezoidal rule for these two patients. The second is missing the 2 hr sample. Patient Patient Time Conc Conc (hr) (mg/L) (mg/L) 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

Calculate the AUC by trapezoidal rule for these two patients. The second is missing the 2 hr sample. Patient Patient Time Conc Conc (hr) (mg/L) (mg/L) 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

Calculate the AUC by trapezoidal rule for these two patients. The second is missing the 2 hr sample. Patient Patient Time Conc Conc (hr) (mg/L) (mg/L) 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

Calculate the AUC by trapezoidal rule for these two patients. The second is missing the 2 hr sample. Patient Patient Time Conc Conc (hr) (mg/L) (mg/L) 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: = = mg*hr/L (C1+C2) 2

Calculate the AUC by trapezoidal rule for these two patients. The second is missing the 2 hr sample. Patient Patient Time Conc Conc (hr) (mg/L) (mg/L) 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

Calculate the AUC by trapezoidal rule for these two patients. The second is missing the 2 hr sample. Patient Patient Time Conc Conc (hr) (mg/L) (mg/L) 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: = mg*hr/L Pt 1; AUC 1-3 hr: = mg*hr/L (C1+C2) 2

Calculate the AUC by trapezoidal rule for these two patients. The second is missing the 2 hr sample. Patient Patient Time Conc Conc (hr) (mg/L) (mg/L) 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: = mg*hr/L Pt 2; AUC 1-3 hr: = mg*hr/L Why the difference? (C1+C2) 2

PROBLEM – AUC AUC Trapezoidal rule assumes a linear decline in [ ] with time. Examine Patient 1 Data: Given Actual Time Conc Calc (hr) (mg/L) Conc

PROBLEM – AUC AUC Trapezoidal rule assumes a linear decline in [ ] with time. Examine Patient 1 Data: Given Actual Arith Time Conc Calc Calc (hr) (mg/L) Conc Conc

PROBLEM – AUC AUC Trapezoidal rule assumes a linear decline in [ ] with time. Examine Patient 1 Data: Given Actual Arith Time Conc Calc Calc (hr) (mg/L) Conc Conc 50 mg/L 25 mg/L

PROBLEM – AUC AUC Trapezoidal rule assumes a linear decline in [ ] with time. Examine Patient 1 Data: Given Actual Arith Time Conc Calc Calc (hr) (mg/L) Conc Conc Line in red shows the actual concentration that would be present given the initial and half-life. 50 mg/L 25 mg/L

PROBLEM – AUC AUC Trapezoidal rule assumes a linear decline in [ ] with time. Examine Patient 1 Data: Given Actual Arith Time Conc Calc Calc (hr) (mg/L) Conc Conc 50 mg/L 25 mg/L Calculated concentration given by red line in previous slide

Calculated concentration using
PROBLEM – AUC AUC Trapezoidal rule assumes a linear decline in [ ] with time. Examine Patient 1 Data: Given Actual Arith Time Conc Calc Calc (hr) (mg/L) Conc Conc 50 mg/L 25 mg/L Calculated concentration using Ct = Co e(-Kt)

Arithmetically calculated concentration
PROBLEM – AUC AUC Trapezoidal rule assumes a linear decline in [ ] with time. Examine Patient 1 Data: Given Actual Arith Time Conc Calc Calc (hr) (mg/L) Conc Conc 50 mg/L Notice 37.5 vs. 35.4 Arithmetically calculated concentration

recall AUC Calc from 2-3 hr
PROBLEM – AUC AUC Trapezoidal rule assumes a linear decline in [ ] with time. Examine Patient 1 Data: Given Actual Arith Time Conc Calc Calc (hr) (mg/L) Conc Conc 50 mg/L Notice 37.5 vs. 35.4 recall AUC Calc from 2-3 hr AUC = ((C1 + C2)/2)(t2 – t1) = ((50+25)/2)(3-2) = 37.5

PROBLEM – AUC AUC Trapezoidal rule assumes a linear decline in [ ] with time. Examine Patient 1 Data: Given Actual Arith Time Conc Calc Calc (hr) (mg/L) Conc Conc 50 mg/L Notice 37.5 vs. 35.4 Over-estimation of conc. using arithmetic formula Trap rule produces additional (green) area

PROBLEM – AUC AUC Trapezoidal rule assumes a linear decline in [ ] with time. Examine Patient 2 Data: Given Actual Arith Time Conc Calc Calc (hr) (mg/L) Conc Conc 100 mg/L 25 mg/L

PROBLEM – AUC AUC Trapezoidal rule assumes a linear decline in [ ] with time. Examine Patient 2 Data: Given Actual Arith Time Conc Calc Calc (hr) (mg/L) Conc Conc Again, RED Line shows the actual concentration that would be present given the initial and half-life. 100 mg/L 25 mg/L

PROBLEM – AUC AUC Trapezoidal rule assumes a linear decline in [ ] with time. Examine Patient 2 Data: Given Actual Arith Time Conc Calc Calc (hr) (mg/L) Conc Conc Conc calc Using Ct=Coe(-Kt) This set of conc is identical to Pt 1. 100 mg/L 25 mg/L

Arithmetically calculated concentrations
PROBLEM – AUC AUC Trapezoidal rule assumes a linear decline in [ ] with time. Examine Patient 2 Data: Given Actual Arith Time Conc Calc Calc (hr) (mg/L) Conc Conc Notice 62.5 vs. 50.0 25 mg/L Arithmetically calculated concentrations

PROBLEM – AUC AUC Trapezoidal rule assumes a linear decline in [ ] with time. Examine Patient 2 Data: Given Actual Arith Time Conc Calc Calc (hr) (mg/L) Conc Conc Notice 62.5 vs. 50.0 25 mg/L Over-estimation of conc. using arithmetic formula (Trap rule) produces additional (blue) area

PROBLEM – AUC AUC Trapezoidal rule assumes a linear decline in [ ] with time. Patient 1 & 2 Arithmetic Data: Actual Pt 1 Pt 2 Time Conc Arith Arith (hr) (mg/L) Conc Conc Over-estimation of conc. using arithmetic formula (Trap rule) produces additional area

PROBLEM – AUC AUC Previous graphs were Log-linear. NOTICE Y-AXIS SCALE
Patient 1 & 2 Arithmetic Data: Actual Pt 1 Pt 2 Time Conc Arith Arith (hr) (mg/L) Conc Conc Actual Conc. Over-estimation of conc. using arithmetic formula (Trap rule) produces additional area

Patient 1 & 2 Arithmetic Data: Actual Pt 1 Pt 2 Time Conc Arith Arith (hr) (mg/L) Conc Conc Actual Conc. Over-estimation of conc. using arithmetic formula (Trap rule) produces additional area pt 1.

Patient 1 & 2 Arithmetic Data: Actual Pt 1 Pt 2 Time Conc Arith Arith (hr) (mg/L) Conc Conc Actual Conc. Over-estimation of conc. using arithmetic formula (Trap rule) produces additional area pt 1 & 2.

PROBLEM – AUC AUC Trapezoidal rule assumes a linear decline in [ ] with time. Patient Patient Time Conc Conc (hr) (mg/L) (mg/L) Trapezoidal rule assumes a linear decline in [ ] with time and over-estimates AUC. Patient 1; AUC 1-3 hr: = mg*hr/L Patient 2; AUC 1-3 hr: = mg*hr/L

PROBLEM – AUC AUC Trapezoidal rule assumes a linear decline in [ ] with time. Patient Patient Time Conc Conc (hr) (mg/L) (mg/L) AUC mg*hr/L So … if concentrations are declining in log-linear fashion, can we not estimate AUC by a method which more closely approximates the change in concentration? … PCK method ?

PROBLEM – AUC AUC Patient Patient 1 2 Time Conc Conc
Time Conc Conc (hr) (mg/L) (mg/L) Calculation of AUC by the pharmacokinetic method: [ ]t / K What is K? … what is the half-life? T½ =

Time Conc Conc (hr) (mg/L) (mg/L) Calculation of AUC by the pharmacokinetic method: [ ]t / K What is K? … what is the half-life? T½ = 1 hr  K = 0.693/ T½  = 0.693 How will be calculate AUC..?

Time Conc Conc (hr) (mg/L) (mg/L) Calculation of AUC by the pharmacokinetic method: [ ]t / K What is K? … what is the half-life? T½ = 1 hr  K = 0.693/ T½  = 0.693 How will be calculate AUC..? AUC1 = 100 / 0.693 = mg*hr/L AUC3 =

Time Conc Conc (hr) (mg/L) (mg/L) AUC1 = 100 / 0.693 = mg*hr/L AUC3 = 25 / 0.693 = mg*hr/L AUC13 = – mg*hr/L = mg*hr/L … another different answer!

Time Conc Conc (hr) (mg/L) (mg/L) Summary: Kinetic method AUC13 = mg*hr/L (Patients 1 & 2) Trap. Rule; Patient 1; AUC 1-3 hr: = mg*hr/L Trap. Rule; Patient 2; AUC 1-3 hr: = mg*hr/L

Summary: PROBLEM – AUC Equations AUC
Calculate the AUC by trapezoidal rule for these two patients. The second is missing the 2 hr sample. Patient Patient Time Conc Conc (hr) (mg/L) (mg/L) Equations Conc = Dose / V V = Dose/Conc Cl = Q x ER ER = Cl / Q AUC = (t2-t1) Cl = Dose / AUC K = Cl / V T½ = / K Summary: AUC AUC Trap Rule PCK mg*hr/mL mg*hr/mL Pt 1; AUC 1-3 hr: Pt 2; AUC 1-3 hr: How many ways could we estimate AUC? (C1+C2) 2

PROBLEM – AUC Equations AUC
Calculate the AUC by trapezoidal rule for these two patients. The second is missing the 2 hr sample. Patient Patient Time Conc Conc (hr) (mg/L) (mg/L) Equations Conc = Dose / V V = Dose/Conc Cl = Q x ER ER = Cl / Q AUC = (t2-t1) Cl = Dose / AUC K = Cl / V T½ = / K Methods of Estimating AUC Trapezoidal Rule Pharmacokinetic Method Trapezoidal Rule using log [ ] Trapezoidal Rule using exponentials (C1+C2) 2

Calculate the AUC by trapezoidal rule for these two patients. The second is missing the 2 hr sample. Patient Patient Time Conc Conc (hr) (mg/L) (mg/L) Equations Conc = Dose / V V = Dose/Conc Cl = Q x ER ER = Cl / Q AUC = (t2-t1) Cl = Dose / AUC K = Cl / V T½ = / K Methods of Estimating AUC Trapezoidal Rule using log [ ]. AUC = [10^(log(C1) – log(C2)/2] x (t2- t1) (C1+C2) 2

Calculate the AUC by trapezoidal rule for these two patients. The second is missing the 2 hr sample. Patient Patient Time Conc Conc (hr) (mg/L) (mg/L) Equations Conc = Dose / V V = Dose/Conc Cl = Q x ER ER = Cl / Q AUC = (t2-t1) Cl = Dose / AUC K = Cl / V T½ = / K Methods of Estimating AUC Trapezoidal Rule using log [ ]. AUC = [10^(log(C1) + log(C2)/2)] x (t2- t1) = [10^(log(100)+log(25)/2)] x (3-1) = [10^ (1.699)] x (2) = 100 mg/hr*/L or = [(C1 x C2) ] x (t2- t1) = [(25 x 100 ] x (3-1) = 100 mg*hr/L Geometric (C1+C2) 2

Calculate the AUC by trapezoidal rule for these two patients. The second is missing the 2 hr sample. Patient Patient Time Conc Conc (hr) (mg/L) (mg/L) Equations Conc = Dose / V V = Dose/Conc Cl = Q x ER ER = Cl / Q AUC = (t2-t1) Cl = Dose / AUC K = Cl / V T½ = / K Methods of Estimating AUC 4. Trapezoidal Rule using log Exponential AUC = [(t2 –t1) / ln(C1) – ln(C2)] x (C1- C2) (C1+C2) 2

Calculate the AUC by trapezoidal rule for these two patients. The second is missing the 2 hr sample. Patient Patient Time Conc Conc (hr) (mg/L) (mg/L) Equations Conc = Dose / V V = Dose/Conc Cl = Q x ER ER = Cl / Q AUC = (t2-t1) Cl = Dose / AUC K = Cl / V T½ = / K Methods of Estimating AUC 4. Trapezoidal Rule using log Exponential AUC = [(t2 –t1) / ln(C1) – ln(C2)] x (C1- C2) = [(3-1) / (ln(100)-ln(25))] x (100-25) = [2/( )] (75) = [2/1.386](75) = 1.44(75) = mg*hr/L (C1+C2) 2

Summary: Arithmetic AUC
Calculate the AUC by trapezoidal rule for these two patients. The second is missing the 2 hr sample. Patient Patient Time Conc Conc (hr) (mg/L) (mg/L) Summary: Arithmetic AUC AUC AUC AUC Trap Rule PCK Geometric Exponential mg*hr/mL mg*hr/mL mg*hr/mL mg*hr/mL Pt 1; AUC 1-3 hr: Pt 2; AUC 1-3 hr: 4 methods … so which one should we use?

AUC Summary: So which one should we use? AUC AUC AUC AUC AUC
Trap Rule PCK Geometric Exponential mg*hr/mL mg*hr/mL mg*hr/mL mg*hr/mL Pt 1; AUC 1-3 hr: Pt 2; AUC 1-3 hr: So which one should we use? In log-linear regions the PCK method is accurate, simple and quick, but arithmetic trapezoidal rule is still a reasonable estimate.. In “other regions”, where true knowledge of the rate of change in concentration is not known, arithmetic trapezoidal rule is a simple, quick & a reasonable estimate of AUC and may be the best.

Use Arithmetic Trapezoidal Rule
AUC AUC Summary: So which one should we use? If the conc.-time profile is log linear you can use the kinetic method… [ ]/k. … but if it is not log-linear, if you are unsure, use the arithmetic trapezoidal rule. It is a simple, quick and a reasonable estimate of AUC. Use Arithmetic Trapezoidal Rule

Back Extrapolation Dealing with [ ] –time Data (3)
How do you calculate Volume if you do not have an initial concentration? (a time-zero concentration)

Dealing with [ ] –time Data
Back Extrap What happens if you do not have a time zero [ ]? How do you calculate V? Dose = 1000 mg Time Conc (hr) (mg/L) 1 2 Volume (L) = Dose / [ ]t=0 What is the concentration at time zero … or what would it have been?

Back Extrap What happens if you do not have a time zero [ ]? How do you calculate V? Dose = 1000 mg Time Conc (hr) (mg/L) 1 2 Volume (L) = Dose / [ ]t=0 Plot the data to observe the rate of change in [ ]. Is it linear ? … log-linear? If so extrapolate or back-extrapolate to t=0.

Back Extrap What happens if you do not have a time zero [ ]? How do you calculate V? Dose = 1000 mg Time Conc (hr) (mg/L) 1 2 Volume (L) = Dose / [ ]t=0 Extrapolate by one of two methods: Graphical, using semi-log paper … using slope or equation Or using Excel “Intercept” function.

What happens if you do not have a time zero [ ]?
Dealing with [ ] –time Data Back Extrap What happens if you do not have a time zero [ ]? How do you calculate V? Dose = 1000 mg Time Conc (hr) (mg/L) 1 2 Extrapolate by Equation: Ct = C0 e-kt Equation determines concentration at any time following a given initial concentration C12 = C4 e-K(8) where K = (T½ = 6 hr) C12 = 25 mg/L Negative sign (-K) indicates loss of concentration

What happens if you do not have a time zero [ ]?
Dealing with [ ] –time Data Back Extrap What happens if you do not have a time zero [ ]? How do you calculate V? Dose = 1000 mg Time Conc (hr) (mg/L) 1 2 Extrapolate by Equation: Ct = C0 e+kt A Positive sign (+K) would indicates INCREASING conc. C0 = C4 e+K(4) where K = (T½ = 6 hr) C0 = 100 mg/L An example is shown in the Excel tutorial slides 40 & 41.

at the end of the slideshow.
What happens if you do not have a time zero [ ]? 60 mg/L Dose = 1000 mg Time Conc (hr) (mg/L) 1 2 Graphically …. Time zero Intercept should be exactly (very close) to 100 mg/L Excel® example shown at the end of the slideshow.

Back Extrap What is the volume of distribution following a 1000 mg dose, if the following conc. were observed? Dose = 1000 mg Time Conc (hr) (mg/L) 1 2 Step by Step: What do we need to calculate first? Volume, AUC, Clearance, half-life or K?

Back Extrap What is the volume of distribution following a 1000 mg dose, if the following conc. were observed? Dose = 1000 mg Time Conc (hr) (mg/L) 1 2 Step by Step: 2. K or T½, by either visual inspection of data or equation. T½ by visual inspection is 6 hr  K = 0.693/6= hr-1

Back Extrap What is the volume of distribution following a 1000 mg dose, if the following conc. were observed? Dose = 1000 mg Time Conc (hr) (mg/L) 1 2 Step by Step: 3. Back – extrapolate using K to determine C0. Ct = C0 e+kt C0 = C4 e+K(4) where K = hr-1 & C4 = 60 mg/L  C0 = 100 mg/L

Back Extrap What is the volume of distribution following a 1000 mg dose, if the following conc. were observed? Dose = 1000 mg Time Conc (hr) (mg/L) 1 2 Step by Step: Determine volume using the Dose (1000 mg) and the back extrapolated concentration. (100 mg/L) Volume = Dose / Conc = 1000 mg / 100 mg/L = 10 L.

Back Extrap What is the volume of distribution following a 1000 mg dose, if the following conc. were observed? Dose = 1000 mg Time Conc (hr) (mg/L) 1 2 Step by Step: You could now calculate AUC and then clearance. Remember, AUC MUST include the C0 concentration. Do not start calculating AUC from 4 hours. !!

Brief Tutorial on the use of Spreadsheets (Excel®)
Excel Tutorial Slides Using Slope or Intercept DEMANDS that you Convert Raw Concentrations to log concentration and back again. The log of a concentration can be obtained using the Excel function ‘LOG(##)’. The value in parenthesis (##) may be either an actual number or a cell reference. Using a Cell Reference allows the formula to be copied more easily. Use of Spreadsheets (Excel®) Not covered in class

Excel Tutorial Slides Using Slope or Intercept DEMANDS that you Convert Raw Concentrations to log concentration and back again. The log of a concentration can be obtained using the Excel function ‘LOG(##)’. The value in parenthesis (##) may be either an actual number or a cell reference. Using a Cell Reference allows the formula to be copied more easily.

Excel Tutorial Slides Converting Raw Concentrations to log concentration and back again. If you have the log of a number and wish to convert it back to the ‘raw’ concentration, this can be done by computing the value of 10x where x is the log value you wish to convert. To do this in Excel the format is 10^x Where ‘^’ is the Excel operator for power.

Excel Tutorial Slides (iii) Back Extrapolation Using the Excel ‘INTERCEPT’ function Selecting at least 2 points in the terminal phase phase to determine ‘SLOPE’. You can also determine the intercept using the and the same pairs of conc. & time values. In the worksheet on the left the Initial intercept value of 100 was obtained using the equation in Excel: =10^INTERCEPT(C$9:C$10,A$9:A$10) for the last 2 points.

Excel Tutorial Slides (iii) Back Extrapolation (b) Using the Excel ‘SLOPE’ function. In Excel when the slope is calculated on log-conc. & time data, and the line is straight we can estimate the concentration anywhere on the line as it is in the form of y = mx = b. A concentration at any time (t1)can be used and the concentration at another time (t2) can be determined. LOG [ ]t2 = LOG [ ]t1 + SLOPE * (t2 – t1) The log of concentration at t2 (LOG [ ]t2) can be convert to a raw concentration.

Excel Tutorial Slides (iii) Back Extrapolation (b) Using the Excel ‘SLOPE’ function. For example, if the concentration at time zero was to be calculated from the given data, t2 would = 0. t1 could be any other given time. We will use 18 hours. The concentration at 18 hours is 12.5 mg/L (as a log:1.097). LOG [ ]t2= LOG [ ]t1 + SLOPE * (t2 – t1) = ( * (0-18) = ( ) = 2.00 and converting to raw concentration [ ]t2=0 = 10^2.00 = Deviation of the concentration from the line of best fit may result in small deviations from the expected value of 100 if other concentrations and times are used. This method can be used to calculate a concentration at any time on the extrapolated line.

Determining Elimination Rate (K)

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Determining Elimination Rate (K)

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