1. ADME lecture 3 Contents ... Elimination rate constant Half life
Clearance
Single i.v. bolus injection into one compartment

2. Elimination rate constant
Part 1
Elimination rate constant

3. Elimination rate constant
It is assumed that for a given drug in a given patient, a fixed proportion of the dose is eliminated every hour (or day etc).
If this distinctive rate was (say) 10% per hour, then if the patient’s body contained 50mg of the drug, elimination would occur at a rate of 5mg/h and if 100mg was present, the elimination rate would be 10mg/h and so on.

4. Units for elimination rate constant (K)
This distinctive proportion removed is called the elimination rate constant and is usually represented as K (or sometimes (Kel).
If the rate of removal was 10% per hour (as on the previous slide), K would be written as 0.1h-1. The reason for using ‘h-1’ may not be immediately obvious, but it means ‘per hour’. So as 0.1 is the same as 10%, ‘0.1h-1’ is simply shorthand for ‘10% per hour’.

5. Units for elimination rate constant (K)
For drugs that are eliminated very slowly, it may be more convenient to quote the proportion eliminated per day. In such a case units of ‘day-1’ would be used. Thus, elimination at a rate of 5% per day would be written as 0.05 day-1.
Other units such as Min-1 or Sec-1 are also acceptable, however therapeutically useful drugs are not generally eliminated so quickly as to require such units.

6. Interconverting units of K
If it is necessary to convert K from one set of units to another, the obvious approach is in fact the correct one.
Example: Re-express K = 0.48day-1 in units of h-1
A rate of 48% per day is equivalent to 2% per hour, so …
0.48day-1 = 0.02h-1

7. K varies between drugs and between patients
Some drugs are eliminated much more quickly than others. Examples of average elimination rate constants are …
Phenobarbitone 0.007h-1 (Very slow)
Theophylline h-1
Propranolol h-1 (Quick)

8. K varies between drugs and between patients
Elimination rate also varies between patients. For example, with gentamicin, K might be 0.3h-1 in a patient with good renal function but only 0.015h-1 in a patient with severely compromised kidneys.

9. Putting K into an equation
Putting what we have already said in more formal mathematical terms:
If the rate of elimination was 2mg/h and the body load was 40mg, then …
K = 2mg/h mg
= 0.05h-1
K = Rate of elimination Dose present in body

10. Part 2
Half-life

11. Half-life
The time required for a 50% reduction in plasma concentrations of drug.
Half-life is independent of how high or low the initial concentration may be.

12. Half-life and K
There will be an inverse relationship between K and half life.
K = t½ = t½ K

13. Half-life and K If t½ = 5.4 hours ... K = 0.693 t½ = 0.693 5.4 h

14. Part 3
Clearance

15. Extraction ratio (E) Cin Cout Liver E = Cin - Cout Cin
(10 mg/L)
(4 mg/L)
E =
Cin - Cout
Cin
E = = = 0.6 or 60%

16. Hepatic blood flow (QH) and Clearance (Cl)
Cin
Cout
Liver
(10 mg/L)
(0 mg/L)
E = 1.0
Assume 2 litres of blood flow through the liver per minute (QH = 2 L/min).
2 litres of blood are completely cleared of drug every minute. (Clearance = 2 L/min)

17. Clearance with incomplete extraction
Cin
Cout
(10 mg/L)
Liver
(5 mg/L)
E = 0.5
Assume 2 litres of blood flow through the liver per minute (QH = 2 L/min).
In effect, 1 litre of blood is completely cleared of drug every minute. (Clearance = 1 L/min)

18. Generally ... Clearance = QH x E Cin Cout Liver QH
Clearance is volume of blood effectively cleared of drug in unit time.
Clearance = QH x E

19. Clearance, Volume and K Vol dis = 20L Clearance = 4L/h
We have already defined K as the proportional rate of removal of drug. If K = 0.2h-1, we know that drug is eliminated at a rate equivalent to 20% of the total body load per hour.
In this diagram, the total volume throughout which the drug is distributed is 20L and 4 out of the 20L is shown as being cleared of drug every hour. As 20% of the volume is cleared every hour, we are also removing 20% of the drug every hour and K must be 0.2h-1.
K therefore equals Clearance / Vol dist.
K = Cl/V or Cl = K x V
Vol dis = 20L
Clearance = 4L/h

20. Use of Cl = K.V
Q1) Calculate Cl if elim rate constant = 0.015h-1 and vol dist = 80L.
Q2) Calculate K if Clearance = mL/h and vol dist = 20L.

21. Use of Cl = K.V (1)
Calculate Cl if elim rate constant = 0.015h-1 and vol dist = 80L.
Cl = K.V = 0.015h-1 x 80L = 12 L/h

22. Use of Cl = K.V (2)
Calculate K if Clearance = 200mL/h and vol dist = 20L.
Cl = K.V K = Cl/V = 200mL/h L = 0.2L/h L = h-1
Always check that units match. mL & L don’t!

23. Single i.v. bolus injection into one compartment

24. Single i.v. bolus dose into one compartment
Dose = 400 mg
V = 100 Litres
K = 0.3 h-1
Initial conc (C0) = 4 mg/L V K

25. D = 400 mg V = 100 L K = 0.3 h-1 Conc (mg/L) Exponential curve
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
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
Time (h)

26. Half-life and K Inefficiently eliminated Efficiently eliminated4 3 2 1
Inefficiently eliminated
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
Time (h)

27. Terms with which you should be familiar ...
Elimination rate constant
Half-life
Extraction ratio
Clearance
Exponential curve

28. What you should be able to do
Calculate elimination rate constant from rate of elimination and body load of drug (etc)
Apply appropriate units to an elim. rate constant
Calculate elim. rate constant from half life (etc)
Describe the information conveyed by a drug’s clearance
Calculate clearance from the volume of distribution and the elimination rate constant (etc)