ADME lecture 3 Contents ...

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ADME lecture 3Contents ...

• Elimination rate constant
• Half life
• Clearance
• Single i.v. bolus injection into one
  compartment

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Part 1
Elimination rate constant
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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 patients 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.
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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.
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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.
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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
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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 0.09h-1 Propranolol
0.18h-1 (Quick)
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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.
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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 40mg 0.05h-1
K Rate of elimination Dose present in
body
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Part 2
Half-life
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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.
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Half-life and K
There will be an inverse relationship between K
and half life. K 0.693 t½
0.693 t½
K
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Half-life and K
If t½ 5.4 hours ... K 0.693 t½
0.693 5.4 h 0.13 h-1
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Part 3
Clearance
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Extraction ratio (E)
Cin
Cout
Liver
(10 mg/L)
(4 mg/L)
Cin - Cout Cin
E
E 10 - 4 6 0.6 or 60 10
10
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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)
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Clearance with incomplete extraction
Cin
Cout
Liver
(10 mg/L)
(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)
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Generally ...
Cin
Cout
Liver
QH
Clearance is volume of blood effectively cleared
of drug in unit time.
Clearance QH x E
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Clearance, Volume and K
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
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Use of Cl K.V
Q1) Calculate Cl if elim rate constant
0.015h-1 and vol dist 80L. Q2) Calculate K
if Clearance 200mL/h and vol dist
20L.
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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
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Use of Cl K.V (2)
Calculate K if Clearance 200mL/h and vol dist
20L. Cl K.VK Cl/V 200mL/h 20L
0.2L/h 20L 0.0125h-1
Always check that units match. mL L dont!
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Part 4
Single i.v. bolus injection into one compartment
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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
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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
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Time (h)
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Half-life and K
4 3 2 1 0
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
0 2 4 6 8
10
Time (h)
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Terms with which you should be familiar ...
Elimination rate constant Half-life Extraction
ratio Clearance Exponential curve
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What you should beable 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 drugs
  clearance
• Calculate clearance from the volume of
  distribution and the elimination rate constant
  (etc)