handling concentration – time data (i)determining elimination rate (k) (ii) auc calculations (iii)...

Handling Concentration – Time

Data(i) Determining Elimination Rate (K)

(ii) AUC Calculations(iii) Back Calculation

We have estimated AUC … now we need to calculate k

Dose = 1000 mgTime 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 !

KKKK

DeterminingDetermining 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…

CaCv

KKKK

DeterminingDetermining Elimination Rate ConstantDrugs 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.

KKKK

DeterminingDetermining Elimination Rate ConstantThe proportionality constant relating the rate and amount

is the first order elimination rate constant (K) with units time-1

(min-1, hr-1).

CaCv

KKKK

DeterminingDetermining Elimination Rate ConstantThe 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 eliminationis 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 …

CaCv

ke

KKKK

DeterminingDetermining Elimination Rate ConstantDrug elimination from the body

can therefore be described bydX 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.Ca

Cv

= - KX

KKKK

DeterminingDetermining Elimination Rate ConstantDrug elimination from the body

can therefore be described bydX 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

KKKK

DeterminingDetermining Elimination Rate ConstantX = X0e-Kt

X0 represented the amount at time 0and 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 ahomogeneous single compartment

(container or tub).

KKKK

DeterminingDetermining Elimination Rate ConstantXt = X0e-Kt

X0 represented the amount at time 0and 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

KKKK

DeterminingDetermining Elimination Rate ConstantXt = 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

KKKK

DeterminingDetermining 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

KKKK

DeterminingDetermining Elimination Rate ConstantSo 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

KKKK


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 easilychange the scale to a log scale.

100 mg/L

KKKK


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

KKKK

DeterminingDetermining Elimination Rate ConstantExample from last day …? K ?Dose = 1000 mgTime 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)

KKKK


Dose = 1000 mgTime 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

Estimating K using graph paper !

2 cycle semi-log paperLower cycle

Upper cycle

1

10

100

KKKK




2 cycle semi-log paper

60

89.1

100

6.25

0 4 8 12 16 20 24 28

KKKK




Plot the points

What is the half-life?

0 4 8 12 16 20 24 28

KKKK




What is the half-life?C0 = 100 mg/L1 half-life later = [ ? ]

= T½

0 4 8 12 16 20 24 28

KKKK





= T½= 50 mg/L

Time? …0 4 8 12 16 20 24 28

50 mg/L

KKKK





= T½= 50 mg/L

Time? … 6 hours. = T½0 4 8 12 16 20 24 28

50 mg/L

KKKK




Check…If the half-life is 6 hr, what will the [ ] be at 12 hours?

0 4 8 12 16 20 24 28

50 mg/L

KKKK




Check…If the half-life is 6 hr, what will the [ ] be at 12 hours? … 25 mg/L … 12.5 mg/L0 4 8 12 16 20 24 28

50 mg/L

25 mg/L

12.5 mg/L

KKKK




If the half-life is 6 hr, what is K? K =

0 4 8 12 16 20 24 28

50 mg/L

25 mg/L

12.5 mg/L

KKKK




If the half-life is 6 hr, what is K? K = 0.693 / T½

0 4 8 12 16 20 24 28

50 mg/L

25 mg/L

12.5 mg/L

KKKK




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

50 mg/L

25 mg/L

12.5 mg/L

KKKK

DeterminingDetermining Elimination Rate ConstantEstimate K by equation …Dose = 1000 mgTime 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)

=

KKKK



= -2.303[log(6.25) – log(60)] / (24 – 4)= -2.303

KKKK



= -2.303[log(6.25) – log(60)] / (24 – 4)= -2.303[0.796 - 1.778] / (20)= -2.303

KKKK



= -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.

KKKK

DeterminingDetermining Elimination Rate ConstantMethods of estimating K …Dose = 1000 mgTime 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. 1. Visual inspection of concentration –time data2. Plotting the log [ ] vs. time and determining the half-life K3. 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.

KKKK


Dose = 1000 mgTime 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


K = 0.1131 hr-1

AUC LP - = [ ]last / k. =

Now we have an estimate of k and can determine thearea by the kinetic method from the last point to infinity.

KKKK

DeterminingDetermining Elimination Rate ConstantExample from last day …K?Dose = 1000 mgTime 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


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

KKKK

Determining Volume & Clearance

Estimation of AUC0-Dose = 1000 mgTime 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-

AUCAUCAUCAUC



______

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

ClClClCl



______


With K and AUC0- calculated, determine Clearance Clearance =

ClClClCl



______


With K and AUC0- calculated, determine Clearance Clearance = Dose / AUC0-

=

ClClClCl



______


With K and AUC0- calculated, determine Clearance Clearance = Dose / AUC0-

= 1000 mg / 881.24 mg*hr/L= 1.13 L/hr

ClClClCl



______

Total AUC0- (mg*hr/L) 881.24Pharmacokinetic Summary:Volume (L) = 10 L

VVVV



______

Total AUC0- (mg*hr/L) 881.24Pharmacokinetic Summary:Volume (L) = 10 LElim. Rate (K) = 0.1131 hr T½ = 0.693/K = 6.12 hr

VVVV



______

Total AUC0- (mg*hr/L) 881.24Pharmacokinetic Summary:Volume (L) = 10 LElim. Rate (K) = 0.1131 hr T½ = 0.693/K = 6.12 hrAUC0- (mg*hr/L) = 881.24 mg*hr/LClearance (L/hr) = 1.13 L/hr

VVVV

Dealing with Concentration –time Data

Time Conc Conc (hr) (mg/L) (mg/L)

Calculate the AUC by trapezoidal rule 1 100 100For these two patients. The second. 2 50 ---is missing the 2 hr sample 3 25 25

OPENING PROBLEM

(ii) Calculating AUC

Calculate the AUC by trapezoidal rule for these two patients. The second is missing the 2 hr sample.

CONSIDER THIS PROBLEM

Patient Patient 1 2

Time Conc Conc (hr) (mg/L) (mg/L) 1 100 100 2 50 --- 3 25 25

AUC from 1-2hr: =((C1 + C2)/2)(t2 – t1)=

Equations

Conc = Dose / V

V = Dose/Conc

Cl = Q x ER

ER = Cl / Q

AUC = -------- (t2-t1)

Cl = Dose / AUCK = Cl / V

(C1+C2) 2

AUCAUCAUCAUC

Calculate the AUC by trapezoidal rule for these two patients. The second is missing the 2 hr sample. Patient Patient

1 2Time Conc Conc (hr) (mg/L) (mg/L) 1 100 100 2 50 --- 3 25 25

AUC from 1-2hr: =((C1 + C2)/2)(t2 – t1)= ((100+50)/2)(2-1)= 75

AUC from 2-3hr: =((C1 + C2)/2)(t2 – t1)=

Equations

Conc = Dose / V

V = Dose/Conc

Cl = Q x ER

ER = Cl / Q

AUC = -------- (t2-t1)


(C1+C2) 2

CONSIDER THIS PROBLEM AUCAUCAUCAUC



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: =

Equations

Conc = Dose / V

V = Dose/Conc

Cl = Q x ER

ER = Cl / Q

AUC = -------- (t2-t1)


(C1+C2) 2




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

Equations

Conc = Dose / V

V = Dose/Conc

Cl = Q x ER

ER = Cl / Q

AUC = -------- (t2-t1)


(C1+C2) 2




Patient 2AUC from 1-3hr: =((C1 + C2)/2)(t2 – t1)

= ((100+25)/2)(3-1)= (125/2)(2)=

Equations

Conc = Dose / V

V = Dose/Conc

Cl = Q x ER

ER = Cl / Q

AUC = -------- (t2-t1)


(C1+C2) 2





= ((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

Equations

Conc = Dose / V

V = Dose/Conc

Cl = Q x ER

ER = Cl / Q

AUC = -------- (t2-t1)


(C1+C2) 2





= ((100+25)/2)(3-1)= (125/2)(2)

Pt 1; AUC 1-3 hr: = 112.5 mg*hr/LPt 2; AUC 1-3 hr: = 125.0 mg*hr/L

Why the difference?

Equations

Conc = Dose / V

V = Dose/Conc

Cl = Q x ER

ER = Cl / Q

AUC = -------- (t2-t1)


(C1+C2) 2


Examine Patient 1 Data:Given Actual

Time Conc Calc(hr) (mg/L) Conc1.00 100 1001.25 84.11.50 70.71.75 59.52.00 50 50.02.25 42.12.50 35.42.75 29.73.00 25 25.0

Trapezoidal rule assumes a linear decline in [ ] with time.

PROBLEM – AUC AUCAUCAUCAUC

Examine Patient 1 Data:Given Actual Arith

Time Conc Calc Calc(hr) (mg/L) Conc Conc1.00 100 100 1001.25 84.1 87.51.50 70.7 75.01.75 59.5 62.52.00 50 50.0 50.02.25 42.1 43.752.50 35.4 37.502.75 29.7 31.253.00 25 25.0 25.00







50 mg/L

25 mg/L





50 mg/L

25 mg/L

Line in red showsthe actual

concentration that would be

present given the initial concentration and half-life.





50 mg/L

25 mg/L

Calculated concentration given by red line in previous slide





50 mg/L

25 mg/L

Calculated concentration using Ct = Co e(-Kt)





50 mg/L

Notice 37.5 vs. 35.4

Arithmetically calculated concentration





50 mg/L


recall AUC Calc from 2-3 hrAUC = ((C1 + C2)/2)(t2 – t1)

= ((50+25)/2)(3-2) = 37.5





50 mg/L


Over-estimation of conc. using arithmetic formula Trap rule

produces additional (green) area


Time Conc Calc Calc(hr) (mg/L) Conc Conc1.00 100 100 1001.25 84.1 90.61.50 70.7 81.31.75 59.5 71.92.00 -- 50.0 62.52.25 42.1 53.12.50 35.4 43.82.75 29.7 34.43.00 25 25.0 25.00



25 mg/L

100 mg/L





25 mg/L

Again,RED Line

showsthe actual

concentration that would be

present given the initial concentration and half-life.

100 mg/L





25 mg/L

Conc calc

UsingCt=Coe(-Kt)

This set of

conc is identical to Pt 1.

100 mg/L





25 mg/L


Arithmetically calculated concentrations





25 mg/L


Over-estimation of conc. using arithmetic formula (Trap rule)produces additional (blue) area

Patient 1 & 2 Arithmetic Data:Actual Pt 1 Pt 2

Time Conc Arith Arith(hr) (mg/L) Conc Conc1.00 100 100 1001.25 84.1 87.5 90.61.50 70.7 75.0 81.31.75 59.5 62.5 71.92.00 50.0 50.0 62.52.25 42.1 43.75 53.12.50 35.4 37.5 43.82.75 29.7 31.25 34.43.00 25 25.0 25.00



Over-estimation of conc. using arithmetic formula (Trap rule)

produces additional area



Previous graphs were Log-linear. NOTICE Y-AXIS SCALE



produces additional area

Actual Conc.





Over-estimation of conc. using arithmetic formula (Trap rule)produces additional area pt 1.

Actual Conc.






produces additional area pt 1 & 2.

Actual Conc.



Patient Patient 1 2


Trapezoidal rule assumes a linear decline in [ ] with time and over-estimates AUC. Patient 1; AUC 1-3 hr:

= 112.5 mg*hr/LPatient 2; AUC 1-3 hr:

= 125.0 mg*hr/L



Patient Patient 1 2


AUC 112.5 125.0 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 ?

Patient Patient 1 2


Calculation of AUC by the pharmacokinetic method: [ ]t / KWhat is K? … what is the half-life?

T½ =


Patient Patient 1 2



T½ = 1 hr K = 0.693/ T½ = 0.693How will be calculate AUC..?


Patient Patient 1 2



T½ = 1 hr K = 0.693/ T½ = 0.693How will be calculate AUC..?

AUC1 = 100 / 0.693 = 144.3 mg*hr/L

AUC3 =


Patient Patient 1 2


AUC1 = 100 / 0.693 = 144.3 mg*hr/L

AUC3 = 25 / 0.693 = 36.08 mg*hr/L

AUC13 = 144.3 – 36.08 mg*hr/L = 108.22 mg*hr/L

… another different answer!


Patient Patient 1 2


Summary:Kinetic method AUC13 = 108.22 mg*hr/L (Patients 1 & 2)

Trap. Rule; Patient 1; AUC 1-3 hr: = 112.5 mg*hr/LTrap. Rule; Patient 2; AUC 1-3 hr: = 125.0 mg*hr/L




Summary: AUC AUCTrap Rule PCKmg*hr/mL mg*hr/mL

Pt 1; AUC 1-3 hr: 112.5 108.2Pt 2; AUC 1-3 hr: 125.0 108.2

How many ways could we estimate AUC?

Equations

Conc = Dose / V

V = Dose/Conc

Cl = Q x ER

ER = Cl / Q

AUC = -------- (t2-t1)


T½ = 0.693 / K

(C1+C2) 2




Methods of Estimating AUC1. Trapezoidal Rule2. Pharmacokinetic Method3. Trapezoidal Rule using log [ ]4. Trapezoidal Rule using exponentials

Equations

Conc = Dose / V

V = Dose/Conc

Cl = Q x ER

ER = Cl / Q

AUC = -------- (t2-t1)


T½ = 0.693 / K

(C1+C2) 2




Methods of Estimating AUC3. Trapezoidal Rule using log [ ]. AUC = [10^(log(C1) – log(C2)/2] x (t2- t1)

Equations

Conc = Dose / V

V = Dose/Conc

Cl = Q x ER

ER = Cl / Q

AUC = -------- (t2-t1)


T½ = 0.693 / K

(C1+C2) 2




Methods of Estimating AUC3. 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

Equations

Conc = Dose / V

V = Dose/Conc

Cl = Q x ER

ER = Cl / Q

AUC = -------- (t2-t1)


T½ = 0.693 / K

(C1+C2) 2




Methods of Estimating AUC4. Trapezoidal Rule using log Exponential

AUC = [(t2 –t1) / ln(C1) – ln(C2)] x (C1- C2)

Equations

Conc = Dose / V

V = Dose/Conc

Cl = Q x ER

ER = Cl / Q

AUC = -------- (t2-t1)


T½ = 0.693 / K

(C1+C2) 2




Methods of Estimating AUC4. 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/(4.605-3.219)] (75)= [2/1.386](75)= 1.44(75)= 108.2 mg*hr/L

Equations

Conc = Dose / V

V = Dose/Conc

Cl = Q x ER

ER = Cl / Q

AUC = -------- (t2-t1)


T½ = 0.693 / K

(C1+C2) 2




Summary: Arithmetic AUC AUC AUC AUCTrap Rule PCK Geometric Exponential mg*hr/mL mg*hr/mL mg*hr/mL mg*hr/mL

Pt 1; AUC 1-3 hr: 112.5 108.2 106.1 108.2Pt 2; AUC 1-3 hr: 125.0 108.2 100.0 108.2

4 methods … so which one should we use?

AUCAUCAUCAUC

AUC Summary: AUC AUC AUC AUCTrap Rule PCK Geometric Exponential mg*hr/mL mg*hr/mL mg*hr/mL mg*hr/mL

Pt 1; AUC 1-3 hr: 112.5 108.2 106.1 108.2Pt 2; AUC 1-3 hr: 125.0 108.2 100.0 108.2

So which one should we use?In log-linear regions the PCK methodis 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 AUCand may be the best.

AUCAUCAUCAUC

AUC Summary:So which one should we use?

1. If the conc.-time profile is log linear you can use the kinetic method… [ ]/k.

2. … 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

AUCAUCAUCAUC

Dealing with [ ] –time Data (3)

How do you calculate Volume if you do not

have an initial concentration?(a time-zero concentration)

Back Extrapolation

Dealing with [ ] –time DataWhat happens if you do not have a time zero [ ]?

Dose = 1000 mgTime Conc (hr) (mg/L) 0 1 2 4 60.0 12 25.0 18 12.5 24 6.25

Volume (L) = Dose / [ ]t=0

What is the concentration at time zero … or what would it have been?

How do you calculate V?

Back ExtrapBack ExtrapBack ExtrapBack Extrap




Plot the data to observe the rate of change in [ ].Is it linear ? … log-linear?

If so extrapolate or back-extrapolate to t=0.






Extrapolate by one of two methods:Graphical, using semi-log paper … using slope or equation

Or using Excel “Intercept” function.





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 = 0.1155 (T½ = 6 hr) C12 = 25 mg/LNegative sign (-K) indicates loss of concentration





Extrapolate by Equation:Ct = C0 e+kt A Positive sign (+K) would indicates INCREASING conc.

C0 = C4 e+K(4) where K = 0.1155 (T½ = 6 hr) C0 = 100 mg/L

An example is shown in the Excel tutorial slides 40 & 41.



What happens if you do not have a time zero [ ]?Dose = 1000 mgTime Conc(hr) (mg/L) 0 1 2 4 60.0 12 25.0 18 12.5 24 6.25

Graphically ….

Time zero Intercept should be exactly

(very close)to 100 mg/L

Excel® example shown at the end of the slideshow.0 4 8 12 16 20 24 28

60 mg/L

Dealing with [ ] –time DataWhat is the volume of distribution following a

1000 mg dose, if the following conc. were observed?Dose = 1000 mgTime Conc (hr) (mg/L) 0 1 2 4 60.0 12 25.0 18 12.5 24 6.25

Step by Step:1. What do we need to calculate first?

Volume, AUC, Clearance, half-life or K?




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=0.1155 hr-1




Step by Step:3. Back – extrapolate using K to determine C0.

Ct = C0 e+kt C0 = C4 e+K(4) where K = 0.1155 hr-1 & C4 = 60 mg/L C0 = 100 mg/L




Step by Step:4. 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.




Step by Step:5. 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®)

Using Slope or Intercept DEMANDS that you

ConvertRaw 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 31 - 34

Use of Spreadsheets (Excel®)Not covered in class


Using Slope or Intercept DEMANDS that you

ConvertRaw 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.



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.


(iii) Back Extrapolation(a) Using the Excel

(b) ‘INTERCEPT’ function

Selecting at least 2 points in the terminal phase

phase to determine ‘SLOPE’.

You can also determine the intercept using the ‘INTERCEPT’ function

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.

Brief Tutorial on the use of Spreadsheets (Excel®)Excel Tutorial Slides 36 - 41


(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.



(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)= 1.097 + (-0.050172 * (0-18)

= 1.097 + (0.90309)= 2.00

and converting to raw concentration[ ]t2=0 = 10^2.00

= 100.00Deviation 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.


handling concentration – time data (i)determining elimination rate (k) (ii) auc calculations (iii)...

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