The document discusses how urine pH affects the tubular reabsorption and renal excretion of weak acids and bases. It states that acidification of urine promotes reabsorption of weak acids and retards reabsorption of weak bases, while alkalization of urine has the opposite effects. The extent of ionization of drugs depends on the drug's pKa value and urine pH. It also discusses how urine flow rate and drug clearance influence renal drug excretion. Clearance is defined as the volume of fluid cleared of drug per unit of time and can be used to determine the mechanism of renal drug excretion.

1. DRUG CLEARANCE
Dr. Rajib Bhattacharjee
Asstt. Professor
Dept of Pharmacy, NSU

2. EFFECT OF URINE PH ON
TUBULAR REABSORPTION & RENAL DRUG SECRETION
The extent of ionization of acidic and basic drugs in renal tubules
depends on
(i) pKa of the compound and
(ii) pH of the tubular fluid that is urine pH.
For a weakly acidic drug,
Fraction of drug ionized = antilog (pH- pKa) / 1 + antilog (pH- pKa)
For a weakly basic drug,
Fraction of drug ionized = 1 / 1 + antilog (pH- pKa)

3. Ionization of acidic & basic drugs at different urine pH
pH
of urine
4.5 96.93 24.03 99.90 100.00
6.0 99.90 90.91 96.93 100.00
7.5 100.00 99.68 50.00 99.90
• Acidification of urine :
% ionized of a
weak acidic drug
% ionized of a
weak basic drug
pKa
3.0
pKa
5.0
pKa
7.5
pKa
10.5
promotes reabsorption of weak acids and retards reabsorption of weak bases.
• Alkalization of urine :
retards reabsorption of weak acids and promotes reabsorption of weak bases.
Therefore,
excretion of weak acids is increased in ………….. urine and excretion of weak bases is
increased in ………….. urine.
The effect of pH on urinary drug excretion is significant only when…………..

4. Effect of rate of urine flow on
tubular reabsorption and renal drug secretion
Rate of urine flow ( normal 1-2 ml / min ) will influence
the amount of drug which is reabsorbed.
Urine flow increases  time for reabsorption decreases  drug excretion promoted
Large fluid intake, use of diuretics etc.
increases urine flow and decreases drug reabsorption.

5. Clearance :
Drug clearance is the fixed volume of body fluid
which is cleared of drug per unit time.
Here body fluid indicates apparent volume of distribution.
Example - A drug has a clearance of 50 ml / min
means 50 ml of the body fluid ( VD ) becomes free of drug per minute.
(ii) Drug clearance is the ratio of
drug elimination rate to plasma drug concentration.
Drug elimination rate
Plasma drug concentration
ClT =
dDE / dt
=
Cp
Unit = mass / time  mass / volume = volume / time
Clearance values are often expressed as per kilogram body weight basis
since this pharmacokinetic parameter varies with body weight.
Thus, unit of ClT becomes ml / min / kg.

6. Now drug elimination rate dDE / dt  DB
 dDE / dt = kE x DB where kE = elimination rate constant.
Again DB = Cp x VD  dDE / dt = kECpVD
 ClT = =
Cp
dDE / dt
Cp
kECpVD
= kEVD
This equation shows that ClT is the product of two constants kE and VD .
In the above definitions, ClT will remain constant
as long as drug elimination follows first order kinetics.
The volume of distribution and elimination rate constant
are both affected by blood flow.
So blood flow will influence the magnitude of drug clearance.

7. From physiologic viewpoint, ultimate drug excretion is
the combined effect of three processes –
i) filtration
ii) secretion and
iii) reabsorption
Drug
excretion rate
Glomerular
Active
= filtration rate + secretion rate
+
Passive
reabsorption rate
ClR = =
Cp
dDe / dt
Cp
filtration secretion reabsorption
rate + rate – rate

8. In real cases, drug excretion is often measured
by comparing its clearance value with that of a standard reference compound
such as inulin or creatinine.
These compounds are
- only excreted by filtration
- neither secreted nor reabsorbed.
So there clearance is equal to GFR that is 125 ml /min.
The clearance ratio of a drug is
its clearance value divided by clearance value of the standard ( inulin ).
The value of the ratio gives a gross idea about the mechanism involved in renal excretion.
Clearance ratio :
ClR(drug) / ClR(inulin)
Probable mechanism of renal excretion
< 1
Drug is partially reabsorbed or reabsorption
dominates over secretion.
= 1
Drug is filtered only or
secretion and reabsorption rates are similar.
> 1
Drug is actively secreted or
secretion dominates over reabsorption.
For a substance which is
only filtered in the kidneys but neither secreted nor reabsorbed
ClR = =
Cp
dDe / dt
Cp
drug filtration rate

9. ClR =
dDe / dt
Cp
A
B
Cp
dDedt
 Determination of renal clearance -
(i)
Slope of the line gives
the clearance value (ClR).
Drug A has higher clearance than drug B
(ii)
A
B
AUC
De
Plot of cumulative drug excretion vs AUC
slope = ClR
Drug A has higher clearance than drug B

10.  Determination of renal clearance -
(iii) Any type of clearance value can be obtained
if the volume of distribution and the corresponding elimination,
excretion and biotransformation rate constants are known.
ClT = kEVD ClR = keVD Clb = kmVD
Again ClT = ClR + Clb  kE = ke + km
 Fraction of drug excreted unchanged by urine -
fe = ke / kE  ClR = ( ke / kE )ClT = feClT
Relationship among clearance, elimination half-life and volume of distribution –
ClT = kEVD and kE = 0.693 / t½
ClT = 0.693VD / t½ and t½ = 0.693VD / ClT
This equation shows that if ClT of a drug is changed due to renal problems,
the elimination half-life of the drug will change accordingly

11. Problems :
1. A new antibiotic secreted by the kidney ; VD is 35 L in the normal adult.
The clearance of this drug is 650 ml/min.
•What is the usual t1/2 of the drug?
•What would be the new t1/2 of this drug in an adult with partial renal failure
whose clearance of the antibiotic was only 75 ml/min?
2. An antibiotic is given by IV bolus injecton at a dose of 500 mg. The
apparent volume of distribution was 21 liter and the elimination half-life was
6 hrs. Urine was collected for 48 hrs and 400 mg of the unchanged drug was
recovered.
• What is the fraction of the dose excreted unchanged in the urine?
• Calculate k, ke, ClT, ClR and Clh.