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Pharmacokinetics lecture 5 Contents . Single i.v. bolus into a one compartment system Area Under the Curve. Single i.v. bolus dose into one compartment. D. Dose = 400 mg V = 100 Litres K = 0.3 h -1 Initial conc (C 0 ) = 4 mg/L. V. K. D = 400 mg V = 100 L K = 0.3 h -1.

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pharmacokinetics lecture 5 contents
Pharmacokinetics lecture 5Contents ...
  • Single i.v. bolus into a one
  • compartment system
  • Area Under the Curve
single i v bolus dose into one compartment
Single i.v. bolus dose into one compartment

D

Dose = 400 mg

V = 100 Litres

K = 0.3 h-1

Initial conc (C0) = 4 mg/L

V

K

slide3

D = 400 mg V = 100 L K = 0.3 h-1

Mass = 4 mg/L x 100 L = 400 mg

Rate of elim = 400 mg x 0.3 h-1

= 120 mg/h

4

3

2

1

0

Mass = 2 mg/L x 100 L = 200 mg

Rate of elim = 200 mg x 0.3 h-1

= 60 mg/h

Conc (mg/L)

Mass = 1 mg/L x 100 L = 100 mg

Rate of elim = 100 mg x 0.3 h-1

= 30 mg/h

Exponential curve

0 2 4 6 8 10

Time (h)

half life and k
Half-life and K

4

3

2

1

0

C0 = 4 mg/L

Inefficientlyeliminated

Efficiently eliminated

Conc (mg/L)

Low K / Long t ½

t ½ =

2.3 h

t ½ =

5.4 h

High K / Short t ½

0 2 4 6 8 10

Time (h)

predicting concentration at a given time point
Predicting concentration at a given time point

C0

4

3

2

1

0

Ct = C0.e-Kt

Conc (mg/L)

Ct

t

0 2 4 6 8 10

Time (h)

predicting concentration at a given time point6
Predicting concentration at a given time point

D = 10 mg V = 50 L K = 0.05 h-1

What conc. after 12 hours?

C0 = D/V = 10mg/50L = 0.2mg/L = 200µg/L

Ct = C0 .e-Kt

= 200µg/L . e-0.05h-1 x 12h

= 200µg/L . e-0.6

= 200µg/L . 0.55

= 110 µg/L

predicting the time to reach a given concentration
Predicting the time to reach a given concentration

C0 = 5 mg/L K = 0.02h-1

How long until C = 1 mg/L?

Ct = C0 .e-Kt

Ct / C0 = e-Kt

1/5 = e-Kt

0.2 = e-Kt (Take natural logs of both sides)

Ln(0.2) = -Kt

-1.609 = -Kt (Drop minus from both sides)

1.609 = 0.02h-1 x t

t = 1.609 / 0.02h-1

= 80.5 hours

area under the curve auc
Area Under the Curve (AUC)

3

2

1

0

  • Area under the concentration
  • versus time curve
  • ‘Area Under the Curve’
  • AUC

Conc (mg/L)

0 2 4 6 8 10

Time (h)

units of auc
Units of AUC

3

2

1

0

Area is height x width

Units are mg/L x h

mg.L-1.h

Usually rearranged as:

mg.h .L-1

In general: Mass.Time.Volume-1

Conc (mg/L)

0 2 4 6 8 10

Time (h)

slide10

AUC between time limits

3

2

1

0

‘AUC’ assumed to mean AUC 0-¥ (The complete area under the whole curve)

AUC2-4h

Conc (mg/L)

0 2 4 6 8 10

Time (h)

relationship between auc dose and clearance
Relationship between AUC, Dose and Clearance

Total amount of drug entering the body. (More drug = greater AUC)

AUC = F.D

Cl

Efficiency of removal. (Greater clearance = lower AUC)

calculation of auc following single i v bolus
Calculation of AUC following single i.v. bolus

AUC = F.D

Cl

e.g. D = 5 mg & Cl = 4 L/h

AUC = 1.0 x 5mg

4 L/h

= 1.25 mg.h.L-1

Conc (mg/L)

0 2 4 6 8 10

Time (h)

example calculation
Example calculation

1mg of drug is administered (i.v) at time zero.

Vol Dis = 50 L.

Elim rate constant = 0.015h-1

What will blood concentration be 2 days later?

Tip: Calculate C0 first, then Ct.

answer
Answer

C0 = D/V

= 1mg / 50L

= 0.02 mg/L

= 20 µg/L

Ct = C0.e-Kt

= 20µg/L x e-0.015h-1 x 2days

= 20µg/L x e-0.015h-1 x 48h

= 20µg/L x e-0.72

= 20µg/L x 0.487

= 9.7µg/L

Mixed units

£2 calculator is

no good.

Need a good one!

what you should be able to do
What you should beable to do
  • For a single i.v. bolus injection into a one compartment system ...
    • Describe why the graph of concentration versus time adopts its distinctive shape
    • Calculate remaining concentration from initial concentration, elimination rate constant and time elapsed (etc)
  • Apply appropriate units to an AUC
  • Calculate AUC following single i.v. injection