Transcript Lecture 5 (Concentration-time data: determining elimination rate)

Handling
Concentration – Time
Data
(i) Determining Elimination Rate (K)
(ii) AUC Calculations
(iii) Back Calculation
K
We have estimated AUC
… now we need to calculate k
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
AUC
mg*hr/L
94.55
84.25
139.40
340.00
112.50
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 !
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
K
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.
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
K
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
ke
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 …
Determining Elimination Rate Constant
Drug elimination from the body
can therefore be described by
dX = - KX
dt
where X is the amount of drug
in the body at any time t after
bolus iv administration.
Ca
Cv
The negative sign indicates that
drug is being lost from the body.
K
Determining Elimination Rate Constant
Drug elimination from the body
can therefore be described by
dX = - KX
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
K
Determining Elimination Rate Constant
K
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).
Determining Elimination Rate Constant
K
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
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
K
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
Determining Elimination Rate Constant
K
So why did we use logarithms?
100 mg/L
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.
Determining Elimination Rate Constant
K
So why did we use logarithms?
First order processes
appear log-linear.
A log-linear relationship is
<generally> easier to interpret.
100 mg/L
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.
Determining Elimination Rate Constant
K
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
Slope = -K / 2.303
This will also us to determine
the elimination rate constant (K).
Determining Elimination Rate Constant
Example from last day …? K ?
Dose = 1000 mg
Time
(hr)
0
1
2
4
12
18
24
8.
Conc
(mg/L)
100.0
89.1
79.4
60.0
25.0
12.5
6.25
AUC
mg*hr/L
94.55
84.25
139.40
340.00
112.50
56.25
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)
K
Determining Elimination Rate Constant
Estimating K using
graph paper !
100
Upper cycle
10
Lower cycle
1
Dose = 1000 mg
Time
(hr)
0
1
2
4
12
18
24
Conc
(mg/L)
100.0
89.1
79.4
60.0
25.0
12.5
6.25
2 cycle semi-log paper
K
Determining Elimination Rate Constant
Estimating K using
graph paper !
100
89.1
Dose = 1000 mg
Time
(hr)
0
1
2
4
12
18
24
60
6.25
Conc
(mg/L)
100.0
89.1
79.4
60.0
25.0
12.5
6.25
2 cycle semi-log paper
0
4
8
12
16 20 24
28
K
Determining Elimination Rate Constant
K
Estimating K using
graph paper !
Dose = 1000 mg
Time
(hr)
0
1
2
4
12
18
24
Conc
(mg/L)
100.0
89.1
79.4
60.0
25.0
12.5
6.25
Plot the points
What is the half-life?
0
4
8
12
16 20 24
28
Determining Elimination Rate Constant
Estimating K using
graph paper !
Dose = 1000 mg
Time
(hr)
0
1
2
4
12
18
24
Conc
(mg/L)
100.0
89.1
79.4
60.0
25.0
12.5
6.25
What is the half-life?
C0 = 100 mg/L
1 half-life later = [ ? ]
= T½
0
4
8
12
16 20 24
28
K
Determining Elimination Rate Constant
50 mg/L
Estimating K using
graph paper !
Dose = 1000 mg
Time
(hr)
0
1
2
4
12
18
24
Conc
(mg/L)
100.0
89.1
79.4
60.0
25.0
12.5
6.25
What is the half-life?
0
4
8
12
16 20 24
28
C0 = 100 mg/L
1 half-life later = [ ? ]
= T½
= 50 mg/L
Time? …
K
Determining Elimination Rate Constant
50 mg/L
Estimating K using
graph paper !
Dose = 1000 mg
Time
(hr)
0
1
2
4
12
18
24
Conc
(mg/L)
100.0
89.1
79.4
60.0
25.0
12.5
6.25
What is the half-life?
0
4
8
12
16 20 24
28
C0 = 100 mg/L
1 half-life later = [ ? ]
= T½
= 50 mg/L
Time? … 6 hours. = T½
K
Determining Elimination Rate Constant
50 mg/L
Estimating K using
graph paper !
Dose = 1000 mg
Time
(hr)
0
1
2
4
12
18
24
Conc
(mg/L)
100.0
89.1
79.4
60.0
25.0
12.5
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
K
Determining Elimination Rate Constant
50 mg/L
Estimating K using
graph paper !
25 mg/L
12.5 mg/L
Dose = 1000 mg
Time
(hr)
0
1
2
4
12
18
24
Conc
(mg/L)
100.0
89.1
79.4
60.0
25.0
12.5
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
… 25 mg/L … 12.5 mg/L
K
Determining Elimination Rate Constant
50 mg/L
Estimating K using
graph paper !
25 mg/L
12.5 mg/L
Dose = 1000 mg
Time
(hr)
0
1
2
4
12
18
24
Conc
(mg/L)
100.0
89.1
79.4
60.0
25.0
12.5
6.25
If the half-life is 6 hr,
what is K?
K =
0
4
8
12
16 20 24
28
K
Determining Elimination Rate Constant
50 mg/L
Estimating K using
graph paper !
25 mg/L
12.5 mg/L
Dose = 1000 mg
Time
(hr)
0
1
2
4
12
18
24
Conc
(mg/L)
100.0
89.1
79.4
60.0
25.0
12.5
6.25
If the half-life is 6 hr,
what is K?
K = 0.693 / T½
0
4
8
12
16 20 24
28
K
Determining Elimination Rate Constant
50 mg/L
Estimating K using
graph paper !
25 mg/L
12.5 mg/L
Dose = 1000 mg
Time
(hr)
0
1
2
4
12
18
24
Conc
(mg/L)
100.0
89.1
79.4
60.0
25.0
12.5
6.25
If the half-life is 6 hr,
what is K?
K = 0.693 / T½
0
4
8
12
16 20 24
28
= 0.693 / 6 hr
= 0.1155 hr-1
K
Determining Elimination Rate Constant
K
Estimate K by equation …
Dose = 1000 mg
Time
(hr)
0
1
2
4
12
18
24
Estimation of K.
K
Conc
(mg/L)
100.0
89.1
79.4
60.0
25.0
12.5
6.25
AUC
mg*hr/L
94.55
84.25
139.40
340.00
112.50
56.25
K is the slope of the line (t=4 to 24 hr)
= -2.303[log(C2) – log(C1)] / (t2 - t1)
=
Determining Elimination Rate Constant
K
Estimate K by equation …
Dose = 1000 mg
Time
(hr)
0
1
2
4
12
18
24
Estimation of K.
K
Conc
(mg/L)
100.0
89.1
79.4
60.0
25.0
12.5
6.25
AUC
mg*hr/L
94.55
84.25
139.40
340.00
112.50
56.25
K is the slope of the line (t=4 to 24 hr)
= -2.303[log(C2) – log(C1)] / (t2 - t1)
= -2.303[log(6.25) – log(60)] / (24 – 4)
= -2.303
Determining Elimination Rate Constant
K
Estimate K by equation …
Dose = 1000 mg
Time
(hr)
0
1
2
4
12
18
24
Estimation of K.
K
Conc
(mg/L)
100.0
89.1
79.4
60.0
25.0
12.5
6.25
AUC
mg*hr/L
94.55
84.25
139.40
340.00
112.50
56.25
K is the slope of the line (t=4 to 24 hr)
= -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
Determining Elimination Rate Constant
K
Estimate K by equation …
Dose = 1000 mg
Time
(hr)
0
1
2
4
12
18
24
Estimation of K.
K
Conc
(mg/L)
100.0
89.1
79.4
60.0
25.0
12.5
6.25
AUC
mg*hr/L
94.55
84.25
139.40
340.00
112.50
56.25
K is the slope of the line (t=4 to 24 hr)
= -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.
Determining Elimination Rate Constant
K
Methods of estimating K …
Dose = 1000 mg
Time
(hr)
0
1
2
4
12
18
24
Conc
(mg/L)
100.0
89.1
79.4
60.0
25.0
12.5
6.25
AUC
mg*hr/L
94.55
84.25
139.40
340.00
112.50
56.25
Methods of Estimating K.
1. Visual inspection of concentration –time data
2. Plotting the log [ ] vs. time and determining the half-life  K
3. 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.
Determining Elimination Rate Constant
K
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
(hr)
0
1
2
4
12
18
24
8.
Conc
(mg/L)
100.0
89.1
79.4
60.0
25.0
12.5
6.25
AUC
mg*hr/L
94.55
84.25
139.40
340.00
112.50
56.25
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.
=
Determining Elimination Rate Constant
Example from last day …K?
Dose = 1000 mg
Time
(hr)
0
1
2
4
12
18
24
8.
Conc
(mg/L)
100.0
89.1
79.4
60.0
25.0
12.5
6.25
AUC
mg*hr/L
94.55
84.25
139.40
340.00
112.50
56.25
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
= 55.25 mg*hr/L
=6.25 / 0.1155
=54.11 mg*hr/L
K
AUC
Determining Volume & Clearance
Estimation of AUC0-
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
24-
AUC
mg*hr/L
94.55
84.25
139.40
340.00
112.50
56.25
55.25
______
Total AUC0- (mg*hr/L)
Sum all of the individual areas to obtain the total AUC0-
Cl
Determining Volume & Clearance
Estimation of AUC0-
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
24-
AUC
mg*hr/L
94.55
84.25
139.40
340.00
112.50
56.25
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
Cl
Determining Volume & Clearance
Estimation of AUC0-
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
24-
AUC
mg*hr/L
94.55
84.25
139.40
340.00
112.50
56.25
55.25
______
Total AUC0- (mg*hr/L)
881.24
With K and AUC0- calculated, determine Clearance
Clearance =
Cl
Determining Volume & Clearance
Estimation of AUC0-
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
24-
AUC
mg*hr/L
94.55
84.25
139.40
340.00
112.50
56.25
55.25
______
Total AUC0- (mg*hr/L)
881.24
With K and AUC0- calculated, determine Clearance
Clearance = Dose / AUC0-
=
Cl
Determining Volume & Clearance
Estimation of AUC0-
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
24-
AUC
mg*hr/L
94.55
84.25
139.40
340.00
112.50
56.25
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
V
Determining Volume & Clearance
Estimation of AUC0-
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
24-
AUC
mg*hr/L
94.55
84.25
139.40
340.00
112.50
56.25
55.25
______
Pharmacokinetic Summary:
Volume (L)
=
10 L
Total AUC0- (mg*hr/L)
881.24
V
Determining Volume & Clearance
Estimation of AUC0-
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
24-
AUC
mg*hr/L
94.55
84.25
139.40
340.00
112.50
56.25
55.25
______
Pharmacokinetic Summary:
Volume (L)
=
10 L
Elim. Rate (K)
= 0.1131 hr 
Total AUC0- (mg*hr/L)
881.24
T½ = 0.693/K = 6.12 hr
V
Determining Volume & Clearance
Estimation of AUC0-
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
24-
AUC
mg*hr/L
94.55
84.25
139.40
340.00
112.50
56.25
55.25
______
Total AUC0- (mg*hr/L)
Pharmacokinetic Summary:
Volume (L)
=
10 L
Elim. Rate (K)
= 0.1131 hr 
AUC0- (mg*hr/L) = 881.24 mg*hr/L
Clearance (L/hr) = 1.13 L/hr
881.24
T½ = 0.693/K = 6.12 hr
Dealing with
Concentration –time Data
(ii) Calculating AUC
OPENING PROBLEM
Calculate the AUC by trapezoidal rule
For these two patients. The second.
is missing the 2 hr sample
Time Conc Conc
(hr) (mg/L) (mg/L)
1
100
100
2
50
--3
25
25
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
Equations
(hr) (mg/L)
(mg/L)
1
100
100
Conc = Dose / V
2
50
--V = Dose/Conc
3
25
25
Cl = Q x ER
ER = Cl / Q
(C1+C2)
AUC = -------- (t2-t1)
2
Cl = Dose / AUC
K = Cl / V
AUC from 1-2hr:
=((C1 + C2)/2)(t2 – t1)
=
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
Equations
(hr) (mg/L)
(mg/L)
1
100
100
Conc = Dose / V
2
50
--V = Dose/Conc
3
25
25
Cl = Q x ER
AUC from 1-2hr:
ER = Cl / Q
(C1+C2)
AUC = -------- (t2-t1)
2
Cl = Dose / AUC
K = Cl / V
AUC from 2-3hr:
=((C1 + C2)/2)(t2 – t1)
= ((100+50)/2)(2-1)
= 75
=((C1 + C2)/2)(t2 – t1)
=
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
Equations
(hr) (mg/L)
(mg/L)
1
100
100
Conc = Dose / V
2
50
--V = Dose/Conc
3
25
25
Cl = Q x ER
AUC from 1-2hr:
ER = Cl / Q
(C1+C2)
AUC = -------- (t2-t1)
2
AUC from 2-3hr:
Cl = Dose / AUC
K = Cl / V
Total AUC 1-3 hr:
=((C1 + C2)/2)(t2 – t1)
= ((100+50)/2)(2-1)
= 75
=((C1 + C2)/2)(t2 – t1)
= ((50+25)/2)(3-2)
= 37.5
=
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
Equations
(hr) (mg/L)
(mg/L)
1
100
100
Conc = Dose / V
2
50
--V = Dose/Conc
3
25
25
Cl = Q x ER
AUC from 1-2hr:
ER = Cl / Q
(C1+C2)
AUC = -------- (t2-t1)
2
AUC from 2-3hr:
Cl = Dose / AUC
K = Cl / V
Total AUC 1-3 hr:
=((C1 + C2)/2)(t2 – t1)
= ((100+50)/2)(2-1)
= 75
=((C1 + C2)/2)(t2 – t1)
= ((50+25)/2)(3-2)
= 37.5
= 75 + 37.5
= 112.5 mg*hr/L
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
Equations
(hr) (mg/L)
(mg/L)
1
100
100
Conc = Dose / V
2
50
--V = Dose/Conc
3
25
25
Cl = Q x ER
ER = Cl / Q
(C1+C2)
AUC = -------- (t2-t1)
2
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)
=
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
Equations
(hr) (mg/L)
(mg/L)
1
100
100
Conc = Dose / V
2
50
--V = Dose/Conc
3
25
25
Cl = Q x ER
Patient 2
AUC from 1-3hr:
ER = Cl / Q
(C1+C2)
AUC = -------- (t2-t1)
2
Cl = Dose / AUC
K = Cl / V
Pt 2; AUC 1-3 hr:
Pt 1; AUC 1-3 hr:
=((C1 + C2)/2)(t2 – t1)
= ((100+25)/2)(3-1)
= (125/2)(2)
= 125.0 mg*hr/L
= 112.5 mg*hr/L
AUC
CONSIDER THIS PROBLEM
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
Equations
(hr) (mg/L)
(mg/L)
1
100
100
Conc = Dose / V
2
50
--V = Dose/Conc
3
25
25
Cl = Q x ER
Patient 2
AUC from 1-3hr:
ER = Cl / Q
(C1+C2)
AUC = -------- (t2-t1)
2
Cl = Dose / AUC
K = Cl / V
Pt 1; AUC 1-3 hr:
Pt 2; AUC 1-3 hr:
=((C1 + C2)/2)(t2 – t1)
= ((100+25)/2)(3-1)
= (125/2)(2)
= 112.5 mg*hr/L
= 125.0 mg*hr/L
Why the difference?
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
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
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
50 mg/L
1.75
59.5
62.5
2.00
50
50.0
50.0
25 mg/L
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
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) ConcLine in Conc
1.00
100
100red shows
100
the actual87.5
1.25
84.1
concentration
1.50
70.7
75.0
that would62.5
be
50 mg/L
1.75
59.5
2.00
50
50.0present 50.0
given
25 mg/L
2.25
42.1 the initial
43.75
concentration
2.50
35.4
37.50
and half-life.
2.75
29.7
31.25
3.00
25
25.0
25.00
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
50 mg/L
1.75
59.5
62.5
2.00
50
50.0
50.0
25 mg/L
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
Calculated concentration given by
red line in previous slide
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
50 mg/L
1.75
59.5
62.5
2.00
50
50.0
50.0
25 mg/L
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
Calculated concentration using
Ct = Co e(-Kt)
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
50 mg/L
1.75
59.5
62.5
2.00
50
50.0
50.0
Notice 37.5 vs. 35.4
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
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
1.00
100
100
100
1.25
84.1
87.5
1.50
70.7
75.0
50 mg/L
1.75
59.5
62.5
2.00
50
50.0
50.0
Notice 37.5 vs. 35.4
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
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
1.00
100
100
100
1.25
84.1
87.5
1.50
70.7
75.0
50 mg/L
1.75
59.5
62.5
2.00
50
50.0
50.0
Notice 37.5 vs. 35.4
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
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
1.00
100
100
100
100 mg/L
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
25 mg/L
2.75
29.7
34.4
3.00
25
25.0
25.00
PROBLEM – AUC
AUC
Trapezoidal rule assumes a linear decline in [ ] with time.
Examine Patient 2 Data:
Given
Actual
Arith
Again,
Time Conc
Calc
Calc
RED Line
(hr) (mg/L) Conc
Conc
shows
1.00
100
100
100
the actual
100 mg/L
1.25
84.1
90.6
concentration
1.50
70.7
81.3
that would be
1.75
59.5
71.9
present
2.00
-50.0
62.5
given the initial
2.25
42.1
53.1
concentration
2.50
35.4
43.8
and half-life.
25 mg/L
2.75
29.7
34.4
3.00
25
25.0
25.00
PROBLEM – AUC
AUC
Trapezoidal rule assumes a linear decline in [ ] with time.
Examine Patient 2 Data:
Given
Actual
Arith
Time Conc
Calc
Calc
Conc
(hr) (mg/L) Conc
Conc
calc
1.00
100
100
100
Using
100 mg/L
1.25
84.1
90.6
Ct=Coe(-Kt)
1.50
70.7
81.3
1.75
59.5
71.9
This
2.00
-50.0
set62.5
of
2.25
42.1 conc
53.1
is
2.50
35.4 identical
43.8
25 mg/L
2.75
29.7 to Pt
34.4
1.
3.00
25
25.0
25.00
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
Notice 62.5 vs. 50.0
2.00
-50.0
62.5
2.25
42.1
53.1
2.50
35.4
43.8
25 mg/L
2.75
29.7
34.4
3.00
25
25.0
25.00
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
1.00
100
100
100
1.25
84.1
90.6
1.50
70.7
81.3
1.75
59.5
71.9
Notice 62.5 vs. 50.0
2.00
-50.0
62.5
2.25
42.1
53.1
2.50
35.4
43.8
25 mg/L
2.75
29.7
34.4
3.00
25
25.0
25.00
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
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
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
Actual Conc.
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
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
Actual Conc.
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 pt 1.
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
Actual Conc.
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 pt 1 & 2.
PROBLEM – AUC
AUC
Trapezoidal rule assumes a linear decline in [ ] with time.
Patient
1
Time Conc
(hr) (mg/L)
1
100
2
50
3
25
Patient
2
Conc
(mg/L)
100
--25
Trapezoidal rule assumes a linear
decline in [ ] with time and overestimates AUC.
Patient 1; AUC 1-3 hr:
= 112.5 mg*hr/L
Patient 2; AUC 1-3 hr:
= 125.0 mg*hr/L
PROBLEM – AUC
AUC
Trapezoidal rule assumes a linear decline in [ ] with time.
Time
(hr)
1
2
3
Patient
1
Conc
(mg/L)
100
50
25
Patient
2
Conc
(mg/L)
100
--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 ?
PROBLEM – AUC
AUC
Patient
1
Time Conc
(hr) (mg/L)
1
100
2
50
3
25
Patient
2
Conc
(mg/L)
100
--25
Calculation of AUC by the pharmacokinetic method: [ ]t / K
What is K? … what is the half-life?
T½ =
PROBLEM – AUC
AUC
Patient
1
Time Conc
(hr) (mg/L)
1
100
2
50
3
25
Patient
2
Conc
(mg/L)
100
--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..?
PROBLEM – AUC
AUC
Patient
1
Time Conc
(hr) (mg/L)
1
100
2
50
3
25
Patient
2
Conc
(mg/L)
100
--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 =
PROBLEM – AUC
AUC
Patient
1
Time Conc
(hr) (mg/L)
1
100
2
50
3
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!
Patient
2
Conc
(mg/L)
100
--25
PROBLEM – AUC
AUC
Patient
1
Time Conc
(hr) (mg/L)
1
100
2
50
3
25
Summary:
Kinetic method AUC13 = 108.22 mg*hr/L
Patient
2
Conc
(mg/L)
100
--25
(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
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
Equations
Time Conc
Conc
(hr) (mg/L)
(mg/L)
Conc = Dose / V
1
100
100
V = Dose/Conc
2
50
--3
25
25
Cl = Q x ER
ER = Cl / Q
Summary:
1+C2) (t -t )
AUC = (C
-------2 1
2
Cl = Dose / AUC
K = Cl / V
T½ = 0.693 / K
Pt 1; AUC 1-3 hr:
Pt 2; AUC 1-3 hr:
AUC
Trap Rule
AUC
PCK
mg*hr/mL
mg*hr/mL
112.5
125.0
108.2
108.2
How many ways could we estimate AUC?
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
Equations
Time Conc
Conc
(hr) (mg/L)
(mg/L)
Conc = Dose / V
1
100
100
V = Dose/Conc
2
50
--3
25
25
Cl = Q x ER
ER = Cl / Q
1+C2) (t -t )
AUC = (C
-------2 1
2
Cl = Dose / AUC
K = Cl / V
T½ = 0.693 / K
Methods of Estimating AUC
1.
2.
3.
4.
Trapezoidal Rule
Pharmacokinetic Method
Trapezoidal Rule using log [ ]
Trapezoidal Rule using exponentials
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
Equations
Time Conc
Conc
(hr) (mg/L)
(mg/L)
Conc = Dose / V
1
100
100
V = Dose/Conc
2
50
--3
25
25
Cl = Q x ER
Methods of Estimating AUC
3. Trapezoidal Rule using log [ ].
ER = Cl / Q
AUC = [10^(log(C1) – log(C2)/2] x (t2- t1)
1+C2) (t -t )
AUC = (C
-------2 1
2
Cl = Dose / AUC
K = Cl / V
T½ = 0.693 / K
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
Equations
Time Conc
Conc
(hr) (mg/L)
(mg/L)
Conc = Dose / V
1
100
100
V = Dose/Conc
2
50
--3
25
25
Cl = Q x ER
Methods of Estimating AUC
3. Trapezoidal Rule using log [ ].
ER = Cl / Q
AUC = [10^(log(C1) + log(C2)/2)] x (t2- t1)
1+C2) (t -t )
= [10^(log(100)+log(25)/2)] x (3-1)
AUC = (C
-------2 1
2
= [10^ (1.699)] x (2)
= 100 mg/hr*/L
Cl = Dose / AUC
or
= [(C1 x C2) ] x (t2- t1)
K = Cl / V
= [(25 x 100 ] x (3-1)
T½ = 0.693 / K
= 100 mg*hr/L
Geometric
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
Equations
Time Conc
Conc
(hr) (mg/L)
(mg/L)
Conc = Dose / V
1
100
100
V = Dose/Conc
2
50
--3
25
25
Cl = Q x ER
Methods of Estimating AUC
ER = Cl / Q
4. Trapezoidal Rule using log Exponential
1+C2) (t -t ) AUC = [(t2 –t1) / ln(C1) – ln(C2)] x (C1- C2)
AUC = (C
-------2 1
2
Cl = Dose / AUC
K = Cl / V
T½ = 0.693 / K
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
Equations
Time Conc
Conc
(hr) (mg/L)
(mg/L)
Conc = Dose / V
1
100
100
V = Dose/Conc
2
50
--3
25
25
Cl = Q x ER
Methods of Estimating AUC
ER = Cl / Q
4. Trapezoidal Rule using log Exponential
1+C2) (t -t ) AUC = [(t2 –t1) / ln(C1) – ln(C2)] x (C1- C2)
AUC = (C
-------2 1
2
= [(3-1) / (ln(100)-ln(25))] x (100-25)
= [2/(4.605-3.219)] (75)
Cl = Dose / AUC
= [2/1.386](75)
K = Cl / V
= 1.44(75)
T½ = 0.693 / K
= 108.2 mg*hr/L
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
Pt 1; AUC 1-3 hr:
Pt 2; AUC 1-3 hr:
mg*hr/mL
mg*hr/mL
112.5
125.0
108.2
108.2
mg*hr/mL
mg*hr/mL
106.1
100.0
108.2
108.2
4 methods … so which one should we use?
AUC
AUC Summary:
AUC
Trap Rule
Pt 1; AUC 1-3 hr:
Pt 2; AUC 1-3 hr:
AUC
PCK
mg*hr/mL
mg*hr/mL
112.5
125.0
108.2
108.2
AUC
AUC
Geometric Exponential
mg*hr/mL
mg*hr/mL
106.1
100.0
108.2
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.
AUC
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
Dealing with [ ] –time Data (3)
Back Extrapolation
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)
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?
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)
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.
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)
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.
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)
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
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)
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.
What happens if you do not have a time zero [ ]?
Dose = 1000 mg
Time
Conc
(hr)
(mg/L)
0
1
2
4
60.0
12
25.0
18
12.5
24
6.25
60 mg/L
Graphically ….
Time zero Intercept
should be exactly
(very close)
to 100 mg/L
0
4
8
12
16 20 24
28
Excel® example shown
at the end of the slideshow.
Dealing with [ ] –time Data
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)
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?
Dealing with [ ] –time Data
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)
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
Dealing with [ ] –time Data
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)
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
Dealing with [ ] –time Data
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)
0
1
2
4
60.0
12
25.0
18
12.5
24
6.25
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.
Dealing with [ ] –time Data
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)
0
1
2
4
60.0
12
25.0
18
12.5
24
6.25
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®)
Excel Tutorial Slides 31 - 34
Using Slope or Intercept
DEMANDS that you
Convert
Raw Concentrations
to log concentration
Use of Spreadsheets (Excel®)
and back again.
Not covered in class
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.
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.
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.
Brief Tutorial on the use of Spreadsheets (Excel®)
Excel Tutorial Slides 36 - 41
(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.
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.