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Chapter: Pharmaceutical Drugs and Dosage: Biopharmaceutical considerations

Pharmaceutical Drugs and Dosage: Biopharmaceutical considerations - Review questions and answers


Review questions

4.1 The characteristics of an active transport process include all the fol-lowing, except:

A.      Active transport moves drug molecules against a concentration gradient.

B.      It follows Fick’s law of diffusion.

C.      It is a carrier-mediated transport system.

D.      It requires energy.

E.       Active transport of drug molecules may be saturated at high drug concentrations.

4.2 The passage of drug molecules from a region of high drug concentra-tion to the region of low drug concentration is known as:

A.      Active transport

B.      Simple diffusion or passive transport

C.      Pinocytosis

D.      Bioavailability

E.       Biopharmaceutics

4.3 Which of the following is true about Fick’s first law of diffusion?

A.      It refers to a nonsteady-state flow.

B.      The amount of material flowing through a unit cross-section of a barrier in unit time is known as the concentration gradient.

C.      Flux of material is proportional to the concentration gradient.

D.      Diffusion occurs in the direction of increasing concentration.

E.       All of the above.

4.4 Which equation describes the rate of drug dissolution from a tablet?

A.      Fick’s law

B.      Henderson–Hasselbalch equation

C.      Michaelis–Menten equation

D.      Noyes–Whitney equation

E.       All of the above

4.5 The diffusion coefficient of a permeant depends on:

A.      Diffusion medium

B.      Diffusion length

C.      Temperature

D.      All of the above

4.6 The rate of drug dissolution from a tablet dosage form will increase with:

A.      The particle size of the drug

B.      The surface area of drug particles

C.      The disintegration time

D.      The amount of excipients to dilute the drug

4.7 The permeability coefficient of a weak electrolyte through a biological membrane will increase if:

A.      The particle size of the drug increases.

B.      The surface area of drug particles increases.

C.      The partition coefficient increases.

D.      The drug dissolution rate increases.

4.8 Indicate which statement is true and which is false.

A.      Fick’s first law of diffusion states that the amount of material flowing through a unit cross-section of a barrier in unit time is proportional to the concentration gradient.

B.      The diffusion rate of molecules with a larger particle size is less than that of those with a smaller particle size.

C.      Under the sink condition, the drug concentration in the receptor compartment is lower than that in the donor compartment.

4.9 Define Fick’s first law of diffusion. Describe how Fick’s first law is expressed in the Noyes–Whitney equation for dissolution. Calculate the diffusion coefficient of the new diet drug Lipidease® across a diffu-sion cell, given the following information: mass rate of diffusion = 5 × 10–4 g/s, cross-section of barrier = 1.0 cm2, and concentration gradi-ent = –175 g/cm4.

4.10 Calculate the rate of dissolution (dM/dt) of drug particles with a sur-face area of 2.5 × 103 cm3 and a saturated solubility of 0.35 mg/mL at room temperature. The diffusion coefficient is 1.75 × 10–7 cm2/s, and the thickness of the diffusion layer is 1.25 μm. The drug concentra-tion in the bulk solution is 2.1 × 10–4 mg/mL.

4.11 The diffusion coefficient of tetracycline in a hydroxyethyl methacrylate—methyl methacrylate copolymer film is D = 8.0 (±4.7) × 10–9 cm2/s and the partition coefficient, k, for tetracycline between the membrane and the reservoir is 6.8 (±5.9) × 10–3. The membrane thickness, h, of the trilaminar device is 1.40 × 10–2 cm, and the con-centration of tetracycline in the concentration, C0, is 0.02 g/cm3 of the core material. Calculate the release rate, Q/t, in units of mg/cm2 of tetracycline per day.

4.12 Drug A weighs 0.5 g and has a total surface area of 0.3 m2. In an experiment, it was found that 0.15 g of A (C) dissolved in 1000 mL of water in the first 2 min. Sink conditions were present. The saturation solubility was found to be 1.2 × 10–3 g/cm3. Calculate the dissolution rate constant in cm/min. Assume that saturation solubility, Csat, is much greater than the value C.

 

Answers:

4.1 B.

4.2 B. In passive transport, a drug travels from high concentration to a low concentration, whereas active transport moves drug molecules against a concentration gradient and requires energy.

4.3 C. Fick’s first law of diffusion states that the amount of material flow through a unit cross section of a barrier in unit time, which is known as the flux, is proportional to the concentration gradient. Fick’s first law of diffusion describes the diffusion process under steady-state conditions when the concentration gradient does not change with time.

4.4 D. The Noyes–Whitney equation describes the rate of drug dissolution from a tablet. Fick’s first law of diffusion is similar to the Noyes–Whitney equation in that both equations describe drug movement due to a concentration gradient. The Michaelis–Menten equation involves enzyme kinetics, whereas Henderson–Hasselbalch equations are used for determination of pH of the buffer and the extent of ionization of a drug molecule.

4.5 D. Diffusion coefficient is not a constant. It is affected by changes in the concentration, temperature, pressure, solvent properties, and chemical nature of the diffusant.

4.6 B. According to the Noyes–Whitney equation, the rate of drug dissolution from a solid dosage form will increase with increase in surface area, which will increase with decrease in particle size or molecular weight of a drug.

4.7 C. The permeability of a weak electrolyte through a biological mem- brane depends on the degree of its ionization; the more lipophilic drug will permeate more, which is possible when its partition coefficient increases.

4.8 A. True

B . True

C. True

4.9 Fick’s first law of diffusion states that the amount of material (M) flowing through a unit cross section (S) of a barrier in unit time (t) is proportional to the concentration gradient (dC/dx).

J= 1/S dM/dt = − D dC/dx = D(C1 − C2) / h

Because K = C1/Cd = C2/Cr, we can rewrite this equation as

dM/ dt = DSK(Cd − Cr) / h  = DSKCd / h

The rate at which a solid dissolves in a solvent can be determined using the

Noyes–Whitney equation:

dM/ dt = kS(Cs − C)

Under sink conditions, when the drug concentration (C) is much less than the solubility of the drug (Cs), we can ignore C (C → 0). A sim-plified Noyes– Whitney equation can be used to measure dissolution rates:

dM/ dt = KSCs = DSCs/h

or

dC/dt = kSCs / V = DSCs / Vh

dM/ dt = DS(Cd − Cr) / h

D = 5× (10−4/175) ×1

D = 2 . 86 ×10−6 cm2 /s

4.10 dM / d t = S × D × (C1 C2)/h, dM/ dt = (2 .5 ×103 ) × (1 . 75 ×107 ) × (0 .35 2.1 ×104 )/(1 .25 × 10 4) = 1.225mg/s

4.11 Q/t = kDC0/h = (6 .8× 103 )(8 ×109 cm2 /s)

 (0 .02 g/cm3 )/(1.40 ×102 cm) = 7 .77 ×1011 gcm 2 s1

To obtain the results in micrograms per day, one must multiply the result by 106 μg/g and 86,400 s/24 h day.

Q/t = (7 .77 ×10 11) gcm2 s 1(106 µg/g)(86, 400 s/day)

= 6 .71 µg/cm2day

4.12 Cs >> C, dM/dt = kS(Cs C) = kSCs


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