# Analysis Of Drug Plasma Concentration And Dosing Of Phenytoin To Control Seizures

• December 28, 2023/

## Distribution phase and elimination phase of drug concentration

A patient weighing 73 kg received a 250 mg dose of drug via an IV injection. Blood samples were taken from the patient at regular intervals and the drug plasma concentration was determined for each sample (see the table below).

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 Time (hr) Cp(mg/L) 0.25 11.6 Save Time On Research and Writing Hire a Pro to Write You a 100% Plagiarism-Free Paper. 0.5 8.4 0.75 7.2 1.00 6.1 1.50 4.2 2.00 3.2 3.00 1.9 4.00 1.0 5.00 0.3

Plot the data and determine the elimination rate constant for this drug.

Rate =  = 5.571 mg/L/hr

• Determine both the half-life (in both minutes and hours) and the volume of distribution (in both L and L/kg) for this drug.

Half of Cp =   =

Half – life = time take for formation of 5.8 mg/L

From the graph (light blue line), HALF – LIFE = 1.1 hrs

• The minimum therapeutically active concentration of the drug is 10 μg/mL. What is the duration of action of this drug?

Conversation of 10 10 μg/mL  to mg/L

= 10 mg/L

From the graph(black line) = 0.4 hours

• If the dose of the drug was doubled, what will be the increase in its duration of action?

If the initial dosage was  250 mg  the duration taken was 5 hours and elimination from the blood was 11.6 mg/L

Y = 12.826

Doubling dosage will double elimination too

0.3 *2 = 12.826

= 0.04678

= 0.04678

0.698x = 3.0623

X = 4.39 hours

• How long would it take for: (i) 50.0%, (ii) 75.0 %, and (ii) 99.9% of this drug to be eliminated from the body?

Duration 50% of the drug to be eliminated = half- life = 1.1 hrs

Duration 75% of the drug to be eliminated (green line) = 75% *11.6 mg/L = 8.7 mg/L = 0.5 hours

Duration 75% of the drug to be eliminated (purple line) = 99.9% *11.6 mg/L = 11.5884 mg/L = 0.25 hoursPlot the data and estimate the clearance of the drug using the area under the curve method

Using excel to solve for area under the cover

 Time (hr) Cp(mg/L) Area under the cove 0.25 11.6 2.5 0.5 8.4 1.95 0.75 7.2 1.6625 1 6.1 2.575 1.5 4.2 1.85 2 3.2 2.55 3 1.9 1.45 4 1 0.65 5 0.3 Total 15.1875

15.1875 unit-square

• If 45% of the drug is excreted unchanged in the urine, determine both the renal and metabolic clearances.

Rate of elimination = 5.571 mg/L/hr

45% = 2.50695 mg/L/hr

2.50695 mg = 1 L/hr

1 mg = 0.34 L/hr

Renal and metabolic clearance = 0.34 L/hr

Q2)

An 80 kg patient is to be given a 1.5 mg/kg body weight dose of gentamicin intravenously over 1 hour every 8 hours. Assume the half-life of gentamicin and volume of distribution in the patient are 2.6 hours and 0.25 L/kg respectively.

• List the main reasons why a patient would be prescribed IV administration of gentamicin.
1. It helps highly in killing bacteria and
2. Prevents bacteria growth.
• How long will it take for gentamicin to reach steady state plasma concentration?

Half – life =

• hours = =

ke =

ke = 0.267

Concentration = C =  =

= 6.0 mg/L

t =

t =

= 19.5 hours

• Calculate the peak and trough plasma concentrations of gentamicin after: (i) 1 dose, (ii) 2 doses, and (iii) 5 doses.

Peak

Cmax =

1 dose

Cmax =

= 0.204

1. ii) 2 doses

Cmax =

= 0.408

iii) 5 doses

Cmax =

= 1.02

• Calculate the steady state peak and trough plasma concentrations of gentamicin. What is the purpose of giving a loading dose to a patient? In the light your peak and trough plasma concentration calculations, do you think it is necessary for a loading dose of gentamicin to be used in treating this patient?

Cmax= C*max *

Cmax= 6.0 *

= 0.0329 mg/L

No, since the concentration is low as compared to the initial concentration

• If the dose of gentamycin given in a 24 hour period remains the same, but dosing frequency is decreased to 12 hour intervals, determine the steady state peak and trough plasma concentrations. Comment on your results and whether changing the dosing interval has affected steady state peak and trough plasma concentrations.

Cmax= C*max *

Cmax= 6.0 *

= 0.00989 mg/L

Changing of the dosing interval has not affected the steady peak and through plasma concentration, since value achieved is lower

• If the dose of gentamycin was halved, what is the change observed in the steady state peak and trough plasma concentrations (providing the elimination rate constant, volume of distribution, and dosing interval remain the same as in part (a) of the question)?

## Calculation of dosing for controlling seizures using Michaelis-Menten kinetics

Cmax= C*max *

Half of the dose = 6/2 = 3.0 mg/L

Cmax= 3.0 *

= 0.01644 mg/L

The steady state peak and through plasma concentration decreases by 50%

(g) If the volume of distribution decreases by 50%, what is the change observed in the steady state peak and trough plasma concentrations (providing the dose elimination rate constant, and dosing interval remain the same as in part (a) of the question)?

Concentration = C =  =

= 12.0 mg/L

Cmax= C*max *

Cmax= 12.0 *

= 0.0658 mg/L

The steady state peak and through plasma concentration increase by 50%

Q3)

A 70.0 kg patient is given a 2.6 mg/kg dose of a drug by intravenous injection. Blood samples were taken from the patient at regular intervals and the drug plasma concentration was determined for each sample (see the table below).

 Time (min) Cp(mg/L) 2 93.08 7 69.11 10 63.82 15 54.79 20 48.73 30 38.63 45 27.85 60 22.92 75 19.12 90 13.62 105 11.43 120 9.04

The pharmacokinetics of the drug displays two-compartment model kinetics. Plot the data for the drug and determine the rate constants for the both distribution phase and the elimination phase.

Distribution phase (upper part)  =   = 4.33 mg/L/min

Elimination phase (lower part)  =   = 0.3125 mg/L/min

• Give a brief description of the two-compartment model. In your answer include definitions for the terms “distribution phase” and “elimination phase” and indicate on your graph where these regions occur.

Distribution phase is the initial steep phase of the cur the dissemination into tissues principally decides the early quick decrease in plasma concentration. With time, distribution equilibrium of medication in tissue with that in plasma is built up in an ever increasing number of tissues, lastly changes in plasma concentration reflect a relative change in the concentration of medication in all tissues and in this manner in the measure of medication in the body. Amid this proportionality stage, the body demonstrations dynamically as a solitary compartment.. Amid this stage, the decay of the plasma concentration is related exclusively with elimination of medication from the body, at the end of the day it compares to the energy of medication end. Along these lines, this stage is frequently called the elimination stage.

(c) Explain why the distribution phase rate constant is greater than the elimination phase rate constant.

It is the steep phase of the curve, since distribution into tissues primarily determines the early rapid decline in plasma concentration

Q4)

Phenytoin is being taken by a patient to control seizures. A daily dose of 300 mg at bedtime results in a steady state plasma concentration of 8 mg/L. When the dose is increased to 400 mg, the steady state plasma concentration changes to 22 mg/L. Assuming Michaelis-Menten kinetics, determine the daily dose that would give a steady state plasma concentration changes of 15 mg/L. Include a plot of the data and the Michaelis-Menten parameters that you used to calculate your answer

 R, mg Css, mg/L R/Css 300 8 38 400 22 18

## Effect of the surface charge on particle dispersion

Equation

Y = -5x + 490

Y = -5*15 + 490

= 415 mg

Q5)

• A small drop of oleic acid is placed on the surface of a beaker of water. Determine the spreading coefficient and the contact angle of the oleic acid with the surface of the water. Does the oleic acid will spread spontaneously over the surface of the water? Give a brief explanation to support your answer. Note: ϒ(water) = 72.9 mN m-1 ; ϒ(oleic acid) = 32.5 mN m-1 ; ϒ(oleic acid-water) = 15.6 mN m-1 at 20 °C.

S = Wa – Wc = (gamma)s – (gamma)l – (gamma)ls

72.9 – 32.5 -15.6 = 24.8 mNm-1

A positive number means that the material will spread

• You have 120.0 g of a 30.0 % w/w paraffin aqueous mixture in a sealed bottle and the bottle is shaken to from an emulsion. Determine the amount of work that would be required to form an emulsion where the disperse phase consisted of paraffin droplets with an average particle diameter of 25 μm. ϒ(water-liquid paraffin) = 52 mN m-1 at 20 °C.

Energy = 6 γ V/d

V = 120/0.8 = 150 ml

Diameter = 0.0025 cm

Work done = 6*52 *150/0.0025

= 18720000

• A small amount of polysorbate 80 is then added to the mixture (in part (b) of the question). The interfacial tension is measured and is now 22.4 mN m-1 . Determine the work now required to produce an emulsion with paraffin droplets with an average particle diameter of 25 μm. Explain why the presence of polysorbate 80 changes the amount of work required to produce an emulsion with an average particle diameter of 20 μm.

Energy = 6 γ V/d

V = 120/0.8 = 150 ml

Diameter = 0.0025 cm

Work done = 6*24 *150/0.0025

= 8,640,000

(d) Draw the structure of polysorbate 80 and explain why it is more soluble in water at lower temperatures?

Q6)

• You have been asked to prepare a suspension of a drug that is available in tablet form. To prepare your suspension you first crush the tablets in a pestle and mortar. What excipients would you use to prepare a pharmaceutical suspension of this drug that is suitable for oral use? Provide reasons for your choice of excipients. Include in your answer a description of the general requirements for an acceptable pharmaceutical suspension.

Requirements

Tablet in fine powder

Gelatin Benzoic acid

Amaranth solution

Chloroform water and raspberry syrup BP

Acacia gum powder

Compound tragacanth powder

Tamarind gum

For preparation of tablet suspensions, polysaccharide powder of about 10 g of tablet were triturated together with a volume of about 20 ml of Raspberry syrup to form a smooth paste. Benzoic acid solution (2 ml) and 1ml of amaranth solution were added gradually with constant stirring and then mixed with 50 ml of chloroform water double strength. The mixture is then transferred into a 100 ml amber bottle, made up to volume with distilled water and then shaken vigorously for about 2 min (thus making 0.5%w/v of the gum in the preparation).

• List all of the physical parameters that affect the rate at which solid spherical particle settles out in a suspension.
• a)Thevdiameter (d) of the particles

V α d2

The velocity (v) of sedimentation is directly proportional to the square of diameter of particle.

1. b) Density difference between dispersed phase and dispersion media

(ρ s – ρo) V α (ρ s – ρo)

1. c) Viscosity of dispersion medium (η )

V α 1/ ηo The velocity of the sedimentation is inversely proportional to viscosity of dispersion medium. Therefore increasing the viscosity of medium will decrease the settling, therefore the particles will achieve a good dispersion system.

• Unfortunately, the tablet powder in your suspension is very hard to re-suspend even with considerable shaking. However, after consulting several references you decide to add a small amount of sodium citrate to the suspension. The sediment that forms is now much less dense and also much easier to re-suspend. From your knowledge of charged surfaces in solution, explain how the addition of sodium citrate to the suspension improves the ability of the tablet powder to be re-suspended.

The ormation of  net charge on the particle surface will affects the ions  in the surrounding interfacial region, this will resulit to increase in conter ionswhich are close o the surface.  Therefore their eill be a formstion an electrical layer which will exists around each every particle. Hence the ion that be beyond this boundary will remain in bulk dispersant.

(d) Often, a surfactant is added to a suspension mixture in order to make it easier to re-suspend the solid particles. With the aid of diagrams, describe the mechanisms to explain how this occurs.

Ionic surfactants act by neutralizing the charge on the particles. The surfacant will be absorbed onto more than one particle because of their long structure hence, formation of a loose flocculated structure.