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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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slide1

Handling

  • Concentration – Time
  • Data
  • Determining Elimination Rate (K)
  • AUC Calculations
  • Back Calculation
slide2

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 !

slide3

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

slide4

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.

slide5

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

slide6

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

slide7

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

slide8

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

slide9

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).

slide10

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

slide11

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

slide12

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

slide13

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

slide14

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

slide15

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

slide16

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)

slide17

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

slide18

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

slide19

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

slide20

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

slide21

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

slide22

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

slide23

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

slide24

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

slide25

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

slide26

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

slide27

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

slide28

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)

=

slide29

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

slide30

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

slide31

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.

slide32

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.
slide33

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.

=

slide34

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

slide35

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-

slide36

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

slide37

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 =

slide38

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-

=

slide39

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

slide40

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

slide41

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

slide42

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

slide43

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

slide44

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

slide45

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

slide46

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

slide47

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

slide48

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

slide49

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

slide50

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

slide51

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

1.00 100 100

1.25 84.1

1.50 70.7

1.75 59.5

2.00 50 50.0

2.25 42.1

2.50 35.4

2.75 29.7

3.00 25 25.0

slide52

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

1.00 100 100 100

1.25 84.1 87.5

1.50 70.7 75.0

1.75 59.5 62.5

2.00 50 50.0 50.0

2.25 42.1 43.75

2.50 35.4 37.50

2.75 29.7 31.25

3.00 25 25.0 25.00

slide53

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

1.00 100 100 100

1.25 84.1 87.5

1.50 70.7 75.0

1.75 59.5 62.5

2.00 50 50.0 50.0

2.25 42.1 43.75

2.50 35.4 37.50

2.75 29.7 31.25

3.00 25 25.0 25.00

50 mg/L

25 mg/L

slide54

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

1.00 100 100 100

1.25 84.1 87.5

1.50 70.7 75.0

1.75 59.5 62.5

2.00 50 50.0 50.0

2.25 42.1 43.75

2.50 35.4 37.50

2.75 29.7 31.25

3.00 25 25.0 25.00

Line in

red shows

the actual

concentration

that would be

present

given the initial

concentration

and half-life.

50 mg/L

25 mg/L

slide55

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

1.00 100 100 100

1.25 84.1 87.5

1.50 70.7 75.0

1.75 59.5 62.5

2.00 50 50.0 50.0

2.25 42.1 43.75

2.50 35.4 37.50

2.75 29.7 31.25

3.00 25 25.0 25.00

50 mg/L

25 mg/L

Calculated concentration given by

red line in previous slide

slide56

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

1.00 100 100 100

1.25 84.1 87.5

1.50 70.7 75.0

1.75 59.5 62.5

2.00 50 50.0 50.0

2.25 42.1 43.75

2.50 35.4 37.50

2.75 29.7 31.25

3.00 25 25.0 25.00

50 mg/L

25 mg/L

Calculated concentration using

Ct = Co e(-Kt)

slide57

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

1.00 100 100 100

1.25 84.1 87.5

1.50 70.7 75.0

1.75 59.5 62.5

2.00 50 50.0 50.0

2.25 42.1 43.75

2.50 35.4 37.50

2.75 29.7 31.25

3.00 25 25.0 25.00

50 mg/L

Notice 37.5 vs. 35.4

Arithmetically calculated concentration

slide58

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

1.00 100 100 100

1.25 84.1 87.5

1.50 70.7 75.0

1.75 59.5 62.5

2.00 50 50.0 50.0

2.25 42.1 43.75

2.50 35.4 37.50

2.75 29.7 31.25

3.00 25 25.0 25.00

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

slide59

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

1.00 100 100 100

1.25 84.1 87.5

1.50 70.7 75.0

1.75 59.5 62.5

2.00 50 50.0 50.0

2.25 42.1 43.75

2.50 35.4 37.50

2.75 29.7 31.25

3.00 25 25.0 25.00

50 mg/L

Notice 37.5 vs. 35.4

Over-estimation of conc. using arithmetic formula Trap rule

produces additional (green) area

slide60

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

1.00 100 100 100

1.25 84.1 90.6

1.50 70.7 81.3

1.75 59.5 71.9

2.00 -- 50.0 62.5

2.25 42.1 53.1

2.50 35.4 43.8

2.75 29.7 34.4

3.00 25 25.0 25.00

100 mg/L

25 mg/L

slide61

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

1.00 100 100 100

1.25 84.1 90.6

1.50 70.7 81.3

1.75 59.5 71.9

2.00 -- 50.0 62.5

2.25 42.1 53.1

2.50 35.4 43.8

2.75 29.7 34.4

3.00 25 25.0 25.00

Again,

RED Line

shows

the actual

concentration

that would be

present

given the initial

concentration

and half-life.

100 mg/L

25 mg/L

slide62

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

1.00 100 100 100

1.25 84.1 90.6

1.50 70.7 81.3

1.75 59.5 71.9

2.00 -- 50.0 62.5

2.25 42.1 53.1

2.50 35.4 43.8

2.75 29.7 34.4

3.00 25 25.0 25.00

Conc

calc

Using

Ct=Coe(-Kt)

This

set of

conc is

identical

to Pt 1.

100 mg/L

25 mg/L

slide63

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

1.00 100 100 100

1.25 84.1 90.6

1.50 70.7 81.3

1.75 59.5 71.9

2.00 -- 50.0 62.5

2.25 42.1 53.1

2.50 35.4 43.8

2.75 29.7 34.4

3.00 25 25.0 25.00

Notice 62.5 vs. 50.0

25 mg/L

Arithmetically calculated concentrations

slide64

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

1.00 100 100 100

1.25 84.1 90.6

1.50 70.7 81.3

1.75 59.5 71.9

2.00 -- 50.0 62.5

2.25 42.1 53.1

2.50 35.4 43.8

2.75 29.7 34.4

3.00 25 25.0 25.00

Notice 62.5 vs. 50.0

25 mg/L

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

produces additional (blue) area

slide65

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

1.00 100 100 100

1.25 84.1 87.5 90.6

1.50 70.7 75.0 81.3

1.75 59.5 62.5 71.9

2.00 50.0 50.0 62.5

2.25 42.1 43.75 53.1

2.50 35.4 37.5 43.8

2.75 29.7 31.25 34.4

3.00 25 25.0 25.00

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

produces additional area

slide66

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

1.00 100 100 100

1.25 84.1 87.5 90.6

1.50 70.7 75.0 81.3

1.75 59.5 62.5 71.9

2.00 50.0 50.0 62.5

2.25 42.1 43.75 53.1

2.50 35.4 37.5 43.8

2.75 29.7 31.25 34.4

3.00 25 25.0 25.00

Actual Conc.

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

produces additional area

slide67

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

1.00 100 100 100

1.25 84.1 87.5 90.6

1.50 70.7 75.0 81.3

1.75 59.5 62.5 71.9

2.00 50.0 50.0 62.5

2.25 42.1 43.75 53.1

2.50 35.4 37.5 43.8

2.75 29.7 31.25 34.4

3.00 25 25.0 25.00

Actual Conc.

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

produces additional area pt 1.

slide68

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

1.00 100 100 100

1.25 84.1 87.5 90.6

1.50 70.7 75.0 81.3

1.75 59.5 62.5 71.9

2.00 50.0 50.0 62.5

2.25 42.1 43.75 53.1

2.50 35.4 37.5 43.8

2.75 29.7 31.25 34.4

3.00 25 25.0 25.00

Actual Conc.

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

produces additional area pt 1 & 2.

slide69

PROBLEM – AUC

AUC

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

Patient Patient

1 2

Time Conc Conc

(hr) (mg/L) (mg/L)

1 100 100

2 50 ---

3 25 25

Trapezoidal rule assumes a linear decline in [ ] with time and over-estimates AUC.

Patient 1; AUC 1-3 hr:

= 112.5 mg*hr/L

Patient 2; AUC 1-3 hr:

= 125.0 mg*hr/L

slide70

PROBLEM – AUC

AUC

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

Patient Patient

1 2

Time Conc Conc

(hr) (mg/L) (mg/L)

1 100 100

2 50 ---

3 25 25

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 ?

slide71

PROBLEM – AUC

AUC

Patient Patient

1 2

Time Conc Conc

(hr) (mg/L) (mg/L)

1 100 100

2 50 ---

3 25 25

Calculation of AUC by the pharmacokinetic method: [ ]t / K

What is K? … what is the half-life?

T½ =

slide72

PROBLEM – AUC

AUC

Patient Patient

1 2

Time Conc Conc

(hr) (mg/L) (mg/L)

1 100 100

2 50 ---

3 25 25

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..?

slide73

PROBLEM – AUC

AUC

Patient Patient

1 2

Time Conc Conc

(hr) (mg/L) (mg/L)

1 100 100

2 50 ---

3 25 25

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

= 144.3 mg*hr/L

AUC3 =

slide74

PROBLEM – AUC

AUC

Patient Patient

1 2

Time Conc Conc

(hr) (mg/L) (mg/L)

1 100 100

2 50 ---

3 25 25

AUC1 = 100 / 0.693

= 144.3 mg*hr/L

AUC3 = 25 / 0.693

= 36.08 mg*hr/L

AUC13 = 144.3 – 36.08 mg*hr/L

= 108.22 mg*hr/L

… another different answer!

slide75

PROBLEM – AUC

AUC

Patient Patient

1 2

Time Conc Conc

(hr) (mg/L) (mg/L)

1 100 100

2 50 ---

3 25 25

Summary:

Kinetic method AUC13 = 108.22 mg*hr/L (Patients 1 & 2)

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

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

slide76

PROBLEM – AUC

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

T½ = 0.693 / K

Summary:

AUC AUC

Trap Rule PCK

mg*hr/mL mg*hr/mL

Pt 1; AUC 1-3 hr: 112.5 108.2

Pt 2; AUC 1-3 hr: 125.0 108.2

How many ways could we estimate AUC?

(C1+C2)

2

slide77

PROBLEM – AUC

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

T½ = 0.693 / K

  • Methods of Estimating AUC
  • Trapezoidal Rule
  • Pharmacokinetic Method
  • Trapezoidal Rule using log [ ]
  • Trapezoidal Rule using exponentials

(C1+C2)

2

slide78

PROBLEM – AUC

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

T½ = 0.693 / K

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

(C1+C2)

2

slide79

PROBLEM – AUC

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

T½ = 0.693 / 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

slide80

PROBLEM – AUC

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

T½ = 0.693 / K

Methods of Estimating AUC

4. Trapezoidal Rule using log Exponential

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

(C1+C2)

2

slide81

PROBLEM – AUC

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

T½ = 0.693 / 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/(4.605-3.219)] (75)

= [2/1.386](75)

= 1.44(75)

= 108.2 mg*hr/L

(C1+C2)

2

slide82

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

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: 112.5 108.2 106.1 108.2

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

4 methods …so which one should we use?

slide83

AUC

AUC Summary:

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: 112.5 108.2 106.1 108.2

Pt 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 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.

slide84

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

slide85

Dealing with [ ] –time Data (3)

Back Extrapolation

How do you calculate Volume

if you do not

have an initial concentration?

(a time-zero concentration)

slide86

Back Extrap

Dealing with [ ] –time Data

What happens if you do not have a time zero [ ]?

How do you calculate V?

Dose = 1000 mg

Time 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?

slide87

Back Extrap

Dealing with [ ] –time Data

What happens if you do not have a time zero [ ]?

How do you calculate V?

Dose = 1000 mg

Time Conc

(hr) (mg/L)

0

1

2

4 60.0

12 25.0

18 12.5

24 6.25

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.

slide88

Back Extrap

Dealing with [ ] –time Data

What happens if you do not have a time zero [ ]?

How do you calculate V?

Dose = 1000 mg

Time Conc

(hr) (mg/L)

0

1

2

4 60.0

12 25.0

18 12.5

24 6.25

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.

slide89

Back Extrap

Dealing with [ ] –time Data

What happens if you do not have a time zero [ ]?

How do you calculate V?

Dose = 1000 mg

Time Conc

(hr) (mg/L)

0

1

2

4 60.0

12 25.0

18 12.5

24 6.25

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/L

Negative sign (-K) indicates loss of concentration

slide90

Back Extrap

Dealing with [ ] –time Data

What happens if you do not have a time zero [ ]?

How do you calculate V?

Dose = 1000 mg

Time Conc

(hr) (mg/L)

0

1

2

4 60.0

12 25.0

18 12.5

24 6.25

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.

slide91

What happens if you do not have a time zero [ ]?

60 mg/L

Dose = 1000 mg

Time 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

slide92

Back Extrap

Dealing with [ ] –time Data

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)

0

1

2

4 60.0

12 25.0

18 12.5

24 6.25

  • Step by Step:
  • What do we need to calculate first?
  • Volume, AUC, Clearance, half-life or K?
slide93

Back Extrap

Dealing with [ ] –time Data

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)

0

1

2

4 60.0

12 25.0

18 12.5

24 6.25

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

slide94

Back Extrap

Dealing with [ ] –time Data

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)

0

1

2

4 60.0

12 25.0

18 12.5

24 6.25

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

slide95

Back Extrap

Dealing with [ ] –time Data

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)

0

1

2

4 60.0

12 25.0

18 12.5

24 6.25

  • 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.
slide96

Back Extrap

Dealing with [ ] –time Data

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)

0

1

2

4 60.0

12 25.0

18 12.5

24 6.25

  • 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. !!
slide97

Brief Tutorial on the use of Spreadsheets (Excel®)

Excel Tutorial Slides 31 - 34

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

slide98

Brief Tutorial on the use of Spreadsheets (Excel®)

Excel Tutorial Slides 31 - 34

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.

slide99

Brief Tutorial on the use of Spreadsheets (Excel®)

Excel Tutorial Slides 31 - 34

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.

slide100

Brief Tutorial on the use of Spreadsheets (Excel®)

Excel Tutorial Slides 36 - 41

  • (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
  • ‘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.
slide101

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.

slide102

Brief Tutorial on the use of Spreadsheets (Excel®)

Excel Tutorial Slides 36 - 41

(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

= 100.00

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.