BS2560 Pharmacology and Toxicology - D R Davies
Data Handling Question Answers
(a) Determine the biological half-life, elimination rate constant , volume of
distribution and plasma clearance of the drug.
Plot data on semi-log graph paper and determine t1/2 graphically
– it should be 2.4 – 2.6 hr. Assuming the former value:
– Elimination Rate constant = 0.693/2.4 = 0.29 hr-1
– In norder to calculate the volume of distribution you must first determine Co
(plasma concn at zero time) and the dose. From the graph extrapolate the linear
portion of the curve (5 – 8 hr). The intercept is at 49 g/ml or 42 mg/l
– The drug was given orally but only 67% was bioavailable. Therefore the effective
dose was 3 * 67/100 = 2.01g or 2010 mg
Volume of distribution = 2010/49 = 41.02 litres
Plasma clearance = Vd * K = 41.02 * 0.29 = 11.89 litre hr -1
(b) What conclusions can be drawn about the body distribution of this drug?
Total body water is about 60% of body weight
i.e. 60/100*80 = 48 litres
Since the calculated volume of distribution is only slightly less than the total body
water we can conclude that the drug is extensively distributed, and is able to traverse
the blood vessels and diffuse through cell membranes
(c) If the renal excretion constant of the drug is 0.075 hr –1 determine the renal
clearance.
Renal clearance Q = ke*Vd = 0.075 * 41.02 = 3.075 litre hr-1
or 51.25 ml min-1
(d) What processes are likely to be involved in the renal excretion of this drug?
The drug is not protein–bound and will thus undergo glomerular filtration. Since the
renal clearance rate is less than the glomerular filtration rate (130 ml.min –1) we can
assume that the drug undergoes some passive reabsorbtion.
(e) Increasing the urinary pH by the administration of sodium bicarbonate doubled
the renal clearance. Explain this effect.
Since the drug is a weak acid, increasing the pH of the urine will increase its
ionsiation, preventing reabsorbtion. This explains why the renal clearance was
doubled.
(f) If the minimum effective plasma concentration is 3.5 g/ml, comment on the
efficiency of this preparation.
If the minimum effective concentration is 3,5 g/ml the drug becomes effective after
1 – 1.5 hr and the effect is maintained for at least 8 hr indicating a clearly efficient
drug.