Handling Concentration – Time Data (i)Determining Elimination Rate (K) (ii) AUC Calculations (iii)...
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Transcript of 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
DeterminingDetermining Elimination Rate ConstantSo 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 easilychange the scale to a log scale.
100 mg/L
KKKK
DeterminingDetermining Elimination Rate ConstantSo 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
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
DeterminingDetermining Elimination Rate Constant
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
DeterminingDetermining Elimination Rate Constant
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 paper
60
89.1
100
6.25
0 4 8 12 16 20 24 28
KKKK
DeterminingDetermining Elimination Rate Constant
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 !
Plot the points
What is the half-life?
0 4 8 12 16 20 24 28
KKKK
DeterminingDetermining Elimination Rate Constant
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 !
What is the half-life?C0 = 100 mg/L1 half-life later = [ ? ]
= T½
0 4 8 12 16 20 24 28
KKKK
DeterminingDetermining Elimination Rate Constant
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 !
What is the half-life?C0 = 100 mg/L1 half-life later = [ ? ]
= T½= 50 mg/L
Time? …0 4 8 12 16 20 24 28
50 mg/L
KKKK
DeterminingDetermining Elimination Rate Constant
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 !
What is the half-life?C0 = 100 mg/L1 half-life later = [ ? ]
= T½= 50 mg/L
Time? … 6 hours. = T½0 4 8 12 16 20 24 28
50 mg/L
KKKK
DeterminingDetermining Elimination Rate Constant
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 !
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
DeterminingDetermining Elimination Rate Constant
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 !
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
DeterminingDetermining Elimination Rate Constant
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 !
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
DeterminingDetermining Elimination Rate Constant
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 !
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
DeterminingDetermining Elimination Rate Constant
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 !
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
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)
= -2.303[log(6.25) – log(60)] / (24 – 4)= -2.303
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)
= -2.303[log(6.25) – log(60)] / (24 – 4)= -2.303[0.796 - 1.778] / (20)= -2.303
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)
= -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
DeterminingDetermining Elimination Rate Constant
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.
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
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
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
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) 881.24
Sum all of the individual areas to obtain the total AUC0-
With K and AUC0- calculated, determine Clearance
ClClClCl
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) 881.24
With K and AUC0- calculated, determine Clearance Clearance =
ClClClCl
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) 881.24
With K and AUC0- calculated, determine Clearance Clearance = Dose / AUC0-
=
ClClClCl
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) 881.24
With K and AUC0- calculated, determine Clearance Clearance = Dose / AUC0-
= 1000 mg / 881.24 mg*hr/L= 1.13 L/hr
ClClClCl
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) 881.24Pharmacokinetic Summary:Volume (L) = 10 L
VVVV
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) 881.24Pharmacokinetic Summary:Volume (L) = 10 LElim. Rate (K) = 0.1131 hr T½ = 0.693/K = 6.12 hr
VVVV
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) 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)
Cl = Dose / AUCK = Cl / V
(C1+C2) 2
CONSIDER THIS PROBLEM 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)= ((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)
Cl = Dose / AUCK = Cl / V
(C1+C2) 2
CONSIDER THIS PROBLEM 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)= ((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)
Cl = Dose / AUCK = Cl / V
(C1+C2) 2
CONSIDER THIS PROBLEM 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
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)
Cl = Dose / AUCK = Cl / V
(C1+C2) 2
CONSIDER THIS PROBLEM 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
Patient 2AUC 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
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
CONSIDER THIS PROBLEM 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
Patient 2AUC 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/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)
Cl = Dose / AUCK = Cl / V
(C1+C2) 2
CONSIDER THIS PROBLEM AUCAUCAUCAUC
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
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
Trapezoidal rule assumes a linear decline in [ ] with time.
PROBLEM – AUC AUCAUCAUCAUC
50 mg/L
25 mg/L
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
Trapezoidal rule assumes a linear decline in [ ] with time.
PROBLEM – AUC AUCAUCAUCAUC
50 mg/L
25 mg/L
Line in red showsthe actual
concentration that would be
present given the initial concentration and half-life.
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
Trapezoidal rule assumes a linear decline in [ ] with time.
PROBLEM – AUC AUCAUCAUCAUC
50 mg/L
25 mg/L
Calculated concentration given by red line in previous slide
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
Trapezoidal rule assumes a linear decline in [ ] with time.
PROBLEM – AUC AUCAUCAUCAUC
50 mg/L
25 mg/L
Calculated concentration using Ct = Co e(-Kt)
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
Trapezoidal rule assumes a linear decline in [ ] with time.
PROBLEM – AUC AUCAUCAUCAUC
50 mg/L
Notice 37.5 vs. 35.4
Arithmetically calculated concentration
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
Trapezoidal rule assumes a linear decline in [ ] with time.
PROBLEM – AUC AUCAUCAUCAUC
50 mg/L
Notice 37.5 vs. 35.4
recall AUC Calc from 2-3 hrAUC = ((C1 + C2)/2)(t2 – t1)
= ((50+25)/2)(3-2) = 37.5
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
Trapezoidal rule assumes a linear decline in [ ] with time.
PROBLEM – AUC AUCAUCAUCAUC
50 mg/L
Notice 37.5 vs. 35.4
Over-estimation of conc. using arithmetic formula Trap rule
produces additional (green) area
Examine Patient 2 Data:Given Actual Arith
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
Trapezoidal rule assumes a linear decline in [ ] with time.
PROBLEM – AUC AUCAUCAUCAUC
25 mg/L
100 mg/L
Examine Patient 2 Data:Given Actual Arith
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
Trapezoidal rule assumes a linear decline in [ ] with time.
PROBLEM – AUC AUCAUCAUCAUC
25 mg/L
Again,RED Line
showsthe actual
concentration that would be
present given the initial concentration and half-life.
100 mg/L
Examine Patient 2 Data:Given Actual Arith
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
Trapezoidal rule assumes a linear decline in [ ] with time.
PROBLEM – AUC AUCAUCAUCAUC
25 mg/L
Conc calc
UsingCt=Coe(-Kt)
This set of
conc is identical to Pt 1.
100 mg/L
Examine Patient 2 Data:Given Actual Arith
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
Trapezoidal rule assumes a linear decline in [ ] with time.
PROBLEM – AUC AUCAUCAUCAUC
25 mg/L
Notice 62.5 vs. 50.0
Arithmetically calculated concentrations
Examine Patient 2 Data:Given Actual Arith
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
Trapezoidal rule assumes a linear decline in [ ] with time.
PROBLEM – AUC AUCAUCAUCAUC
25 mg/L
Notice 62.5 vs. 50.0
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
Trapezoidal rule assumes a linear decline in [ ] with time.
PROBLEM – AUC AUCAUCAUCAUC
Over-estimation of conc. using arithmetic formula (Trap rule)
produces additional area
Patient 1 & 2 Arithmetic Data:Actual Pt 1 Pt 2
Time Conc Arith Arith(hr) (mg/L) Conc 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
Previous graphs were Log-linear. NOTICE Y-AXIS SCALE
PROBLEM – AUC AUCAUCAUCAUC
Over-estimation of conc. using arithmetic formula (Trap rule)
produces additional area
Actual Conc.
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
Previous graphs were Log-linear. NOTICE Y-AXIS SCALE
PROBLEM – AUC AUCAUCAUCAUC
Over-estimation of conc. using arithmetic formula (Trap rule)produces additional area pt 1.
Actual Conc.
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
Previous graphs were Log-linear. NOTICE Y-AXIS SCALE
PROBLEM – AUC AUCAUCAUCAUC
Over-estimation of conc. using arithmetic formula (Trap rule)
produces additional area pt 1 & 2.
Actual Conc.
Trapezoidal rule assumes a linear decline in [ ] with time.
PROBLEM – AUC AUCAUCAUCAUC
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/LPatient 2; AUC 1-3 hr:
= 125.0 mg*hr/L
Trapezoidal rule assumes a linear decline in [ ] with time.
PROBLEM – AUC AUCAUCAUCAUC
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 ?
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 / KWhat is K? … what is the half-life?
T½ =
PROBLEM – AUC AUCAUCAUCAUC
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 / KWhat is K? … what is the half-life?
T½ = 1 hr K = 0.693/ T½ = 0.693How will be calculate AUC..?
PROBLEM – AUC AUCAUCAUCAUC
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 / KWhat is K? … what is the half-life?
T½ = 1 hr K = 0.693/ T½ = 0.693How will be calculate AUC..?
AUC1 = 100 / 0.693 = 144.3 mg*hr/L
AUC3 =
PROBLEM – AUC AUCAUCAUCAUC
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
AUC13 = 144.3 – 36.08 mg*hr/L = 108.22 mg*hr/L
… another different answer!
PROBLEM – AUC AUCAUCAUCAUC
Patient Patient 1 2
Time Conc Conc (hr) (mg/L) (mg/L) 1 100 100 2 50 --- 3 25 25
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
PROBLEM – AUC 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
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)
Cl = Dose / AUCK = Cl / V
T½ = 0.693 / K
(C1+C2) 2
PROBLEM – AUC 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
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)
Cl = Dose / AUCK = Cl / V
T½ = 0.693 / K
(C1+C2) 2
PROBLEM – AUC 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
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)
Cl = Dose / AUCK = Cl / V
T½ = 0.693 / K
(C1+C2) 2
PROBLEM – AUC 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
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)
Cl = Dose / AUCK = Cl / V
T½ = 0.693 / K
(C1+C2) 2
PROBLEM – AUC 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
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)
Cl = Dose / AUCK = Cl / V
T½ = 0.693 / K
(C1+C2) 2
PROBLEM – AUC 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
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)
Cl = Dose / AUCK = Cl / V
T½ = 0.693 / K
(C1+C2) 2
PROBLEM – AUC 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
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
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
Plot the data to observe the rate of change in [ ].Is it linear ? … log-linear?
If so extrapolate or back-extrapolate to t=0.
How do you calculate V?
Back ExtrapBack ExtrapBack ExtrapBack Extrap
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
Extrapolate by one of two methods:Graphical, using semi-log paper … using slope or equation
Or using Excel “Intercept” function.
How do you calculate V?
Back ExtrapBack ExtrapBack ExtrapBack Extrap
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
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
How do you calculate V?
Back ExtrapBack ExtrapBack ExtrapBack Extrap
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
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.
How do you calculate V?
Back ExtrapBack ExtrapBack ExtrapBack Extrap
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?
Back ExtrapBack ExtrapBack ExtrapBack Extrap
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: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
Back ExtrapBack ExtrapBack ExtrapBack Extrap
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: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
Back ExtrapBack ExtrapBack ExtrapBack Extrap
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: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.
Back ExtrapBack ExtrapBack ExtrapBack Extrap
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:5. You could now calculate AUC and then clearance.
Remember, AUC MUST include the C0 concentration. Do not start calculating AUC from 4 hours. !!
Back ExtrapBack ExtrapBack ExtrapBack Extrap
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
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
Brief Tutorial on the use of Spreadsheets (Excel®)
Converting Raw Concentrations to log concentration
and back again. If you have the log of a number
and wish to convert it back to the ‘raw’ concentration, this can be done by computing
the value of 10x where x is the log value
you wish to convert.
To do this in Excel the format is 10^x
Where ‘^’ is the Excel operator for power.
Excel Tutorial Slides 31 - 34
(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
Brief Tutorial on the use of Spreadsheets (Excel®)
(iii) Back Extrapolation(b) Using the Excel ‘SLOPE’ function.
In Excel when the slope is calculated on log-conc. & time data,
and the line is straight we can estimate the concentration
anywhere on the line as it is in the form of y = mx = b.
A concentration at any time (t1)can be used and the
concentration at another time (t2) can be determined.
LOG [ ]t2 = LOG [ ]t1 + SLOPE * (t2 – t1)
The log of concentration at t2 (LOG [ ]t2) can be convert to a raw concentration.
Excel Tutorial Slides 36 - 41
Brief Tutorial on the use of Spreadsheets (Excel®)
(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.
Excel Tutorial Slides 36 - 41