Pharmacokinetic Impact on Health - Assignment Example

Topic: Pharmacokinetics Name: Registration No: Institution: Date of Submission: Question 1: Plasma Concentrations following IV Bolus Administration of Compound A i. Elimination Rate Constant This is the rate at which the drug is being eliminated from the body of the animal in the case study (rat) it is obtained by dividing the initial plasma concentration by the time taken for elimination as shown below. Elimination Rate Constant = 60.9/10.1 = 6.03 ng/h ii. Half-life (Finkel, Clark and Cubeddu 2009) states that half-life is the time taken for the drug to be eliminated to half its original volume. In this case, it is the time taken for the drug to be eliminated to a volume of 30.45. According to the above graph, the time taken is 4.1 hours. iii. Volume of Distribution This is the ratio of the total amount of drug in the body to the concentration of blood in plasma. The total amount of drug in the body =1000ng/kg, while the concentration of blood in plasma =60.9 ng/mL. Hence, volume distribution = 1000/60.9. =16.42 ng/mL iv. Area under the plasma concentration versus time curve This can be obtained by use of the trapezoidal rule.∆/2[(f(x1) + 2f (x2) + 2f(x3)….+ 2f(xn-1) + f(xn)] ∆ = (60-0)/2 =30; Area =30/2[(0 +2(10) +2(20) +2(30) +2(40) +2(50) +60)] =15*[0+20+40+60+80+100+60] =5400 sq units. v. Clearance This is a measure of the rate at which waste substances are cleared from blood. It is obtained by the formula CL= Vd*(0.693/t 1/2) where Vd = volume distribution = 16.42 ng/mL and t ½ = half-life = 4.1 hours. Substituting into the above equation, we get CL =16.42*0.693/4.1 = 3.747 ng/h. i. Elimination Rate Constant This is the rate of removal of blood from the body of the dog and it is obtained by dividing the initial plasma concentration by the time taken for elimination as shown below. Elimination Rate Constant = 127-10.7/8.0-0.5 = 15.51 ng/h ii. Half-life Half life is the amount of time taken for half the volume of the drug to be eliminated from the body of the dog. In this case, it is the time taken for the drug to be eliminated to a volume of 63.5 ng/mL. According to the above graph, the time taken is 3.5 hours. iii. Volume of Distribution This is the ratio of the total amount of drug in the body to the concentration of blood in plasma. The total amount of drug in the body =1000ng/kg, while the concentration of blood in plasma =127 ng/mL. Hence, volume distribution = 1000/127. =7.87 ng/mL iv. Area under the plasma concentration versus time curve This can be obtained by use of the trapezoidal rule.∆/2[(f(x1) + 2f (x2) + 2f(x3)….+ 2f(xn-1) + f(xn)] ∆ = (127-0)/2 =63.5; Area =63.5/2[(0 +2(20) +2(40) +2(60) +2(80) +2(100) +120)] =31.75*[0+40+80+120+160+200+120] =22860 sq units. v. Clearance This is a measure of the rate at which waste substances are cleared from blood. It is obtained by the formula CL= Vd*(0.693/t 1/2) where Vd = volume distribution = 7.87 ng/mL and t ½ = half-life = 3.5 hours. Substituting into the above equation, we get CL =7.87*0.693/3.5 = 1.55 ng/h. i. Elimination Rate Constant Elimination Rate Constant is the rate at which the body removes the drug and is computed dividing the initial plasma concentration by the time taken for elimination (Craig and Stitzel 2004). Elimination Rate Constant = 396/6.5 = 60.92 ng/h ii. Half-life This is the duration that the drug takes before it is eliminated to half its original volume and it is obtained by looking at the point where the drug is at half its original concentration. According to the above graph, the time taken is 3.5 hours. iii. Volume of Distribution This is the ratio of the total amount of drug in the body to the concentration of blood in plasma. The total amount of drug in the body =1000ng/kg, while the concentration of blood in plasma =396 ng/mL. Hence, volume distribution = 1000/396. = 2.525 ng/mL iv. Area under the plasma concentration versus time curve This can be obtained by use of the trapezoidal rule.∆/2[(f(x1) + 2f (x2) + 2f(x3)….+ 2f(xn-1) + f(xn)] ∆ = (400-0)/2 =200; Area =200/2[(0 +2(50) +2(100) +2(150) +2(200) +2(250) +2(300) + 2(350) + 400)] =100*[0+100+200+300+400+500+600+700] =280000 sq units. v. Clearance This is a measure of the rate at which waste substances are cleared from blood. It is obtained by the formula CL= Vd*(0.693/t 1/2) where Vd = volume distribution = 2.525 ng/mL and t ½ = half-life = 3.5 hours. Substituting into the above equation, we get CL =2.525*0.693/3.5 = 0.5 ng/h. Question 2. Plasma Concentration following a single Oral administration of Compound A I. Maximum plasma concentration is Cmax 120 ng/mL and the corresponding time tmax is 6.4 hours II. Elimination rate constant is the rate at which the drug is eliminated from the body of the rat. It is obtained by (120-20)/(10-6.4) = 27.78 ng/mL The corresponding half life is 1.8 hours III. Absorption rate constant is the rate at which the body of the rat absorbs the drug. It is obtained by (120-2)/(6.4-1) = 18.52 ng/mL Half life is the time taken for the drug to be absorbed to half its highest absorprion concentration of 60 ng/mL and is 2 hours in this case IV. Area by trapezoidal rule ∆/2[(f(x1) + 2f (x2) + 2f(x3)…. + 2f (xn-1) + f (xn)] where ∆= (10-1)/2 = 4.5 Area = 4.5/2[(40 + 2(60) + 2(80) + 2(100) +2(120) +2(100) + 2(80) + 2(60) + 40)] = 2.25*1520 = 3420 sq units. Comment on values of Cmax and AUCoral The Cmax value is 120 ng/mL which shows that the maximum amount of AUCoral that can be absobed by the body of the rate is 120ng/mL I. Maximum plasma concentration is Cmax 700 ng/mL and the corresponding time tmax is 5.4 hours II. Elimination rate constant is the rate at which the drug is eliminated from the body of the dog. It is obtained by (700-100)/(10-5.4) = 130.43 ng/mL The corresponding half life is 2.1 hours III. Absorption rate constant is the rate at which the body of the dog absorbs the drug. It is obtained by (700-300)/(5.4-1) = 136.36 ng/mL Half life is the time taken for the drug to be absorbed to half its highest absorprion concentration of 350 ng/mL and is 0.18 hours in this case IV. Area by trapezoidal rule ∆/2[(f(x1) + 2f (x2) + 2f(x3)…. + 2f (xn-1) + f (xn)] where ∆= (10-1)/2 = 4.5 Area = 4.5/2[(300 + 2(400) + 2(500) + 2(600) +2(700) +2(600) + 2(500) + 2(400) + 2(300) + 2(200) + 100)] = 2.25*8800 = 19800 sq units. Comment on values of Cmax and AUCoral The maximum value of Cmax is 700 ng/mL which implies that the maximum amount of AUCoral that can be absorbed by the body of the dog is 700 ng/mL I. Maximum plasma concentration is Cmax 2000 ng/mL and the corresponding time tmax is 5.0 hours II. Elimination rate constant is the rate at which the drug is eliminated from the body of the monkey. It is obtained by (2000-10)/(10-6) = 497.5 ng/mL The corresponding half life is 2 hours. III. Absorption rate constant is the rate at which the body of the monkey absorbs the drug. It is obtained by (2000-610)/(6-1) = 278 ng/mL Half life is the time taken for the drug to be absorbed to half its highest absorprion concentration of 1000 ng/mL and is 2.5 hours in this case IV. Area by trapezoidal rule ∆/2[(f(x1) + 2f (x2) + 2f(x3)…. + 2f (xn-1) + f (xn)] where ∆= (10-1)/2 = 4.5 Area = 4.5/2[(500 + 2(1000) + 2(1500) + 2(2000) +2(1500) +2(1000) + 500)] = 2.25*15000 = 33750 sq units. Comment on values of Cmax and AUCoral The Cmax value is 1980 ng/mL which implies that the body of the monkey is able to aborb 1980 ng/mL of AUCoral drug in a period of 5 hours. Question Three Bioavailability calculation for drug A by use of urinary excretions   Urinary Excretion (% dose)       Compound A (Unchanged) Metabolites(Total) Drug utilised Route of Administration IV Oral IV Oral IV Oral Species             Rat 17.1 5.6 28.2 20.6 11.1 15 Dog 13.4 6.4 34.7 38.8 21.3 32.4 Monkey 14.8 10.3 - 35.8 0 25.5 Man - 30.1 - 12.8 0   Bioavailability = Drug absorbed/Drug administered For rat Oral drug bioavailability = (15/20.6)*100% = 72.81% IV Bioavailability = (11.1/28.2)*100% = 39.36% For Dog Oral drug bioavailability = (32.4/38.8)*100% = 83.5% IV Bioavailability = (21.3/34.7)*100% = 61.38% For monkey Oral drug bioavailability = (25.5/35.8)*100% = 71.23% IV Bioavailability = 0% Comparison with those obtained from plasma data Oral drug utilisation is highest in dog and lowest in rats. IV bioavailability is highest in monkeys and lowest in rats. It is observed that all the values in the above calculations are generally higher than those obtained in the plasma data. 1. The type of transformation that drug A has undergone to produce the metabolites is chemoselective chemistry. This is where the drug reacts with carboxyl groups to give high yields. For instance, when there is a reaction with an electrophilic compound, aldehyde and ketone are formed. The intermediate compounds which may be involved in the pathways include ketone and aldehyde groups when there is a reaction with –OH or –CO2H groups. The enzymes which may be involved in the drug reactions include carbonic anhydrase, flavine adenine dinucleotide (FAD) and nicotinamide adenine dinucleotide (NAD). Their cofactors include, zinc and organic molecules 2. The tentative conclusion that can be drawn concerning phase I functionalization metabolism of Compound A in the four species is that during bolus administration of a drug, metabolism of the drug takes places faster than when the drug is taken in dose. 3. The significance of this knowledge of metabolism in interpretation of pharmacokinetic data is that it enables understanding the changes through which a drug undergoes when it is administered to a patient (Troy 2005). In addition, it enables understanding of the time taken by a drug to reach its maximum absorption by the body and the amount of drug that a body absorbs as a fraction of the amount ingested. 4. The metabolic data are of high significance in interpretation of animal toxicity in a number of ways. It shows the maximum amount of a drug that the body of an animal can absorb so that any amount that is higher than the maximum amount results into toxicity. In addition, it enables understanding the changes through which a drug undergoes so that it is possible to determine if there is any drug chemical product that can be toxic to the animal. References Finkel, R., Clark, M. A., and Cubeddu, L. X. 2009. Pharmacology. Philadelphia: Lippincott Williams & Wilkins. Troy, D. B. 2005. Remington: The science and practice of pharmacy. Philadelphia, PA: Lippincott, Williams & Wilkins. Craig, C. R., & Stitzel, R. E. 2004. Modern pharmacology with clinical applications. Philadelphia: Lippincott Williams & Wilkins. Read More