Solution Manual For Fundamentals and Design of Drug Delivery Systems, A Textbook with Lab Activities Edition 1 By Wujie Zhang Chapters 1-16
Answers for selected questions Chapter 1 Q2: 1. Interact with nucleic acids through ionic interactions (forming lipid-nucleic acid complex plus protection); 2. Promote cellular uptake (e.g., cell membrane fusion/destabilization and endocytosis).
Chapter 2 Q2: Because of the acidic or alkaline end groups, they are pH-responsive. For example, at low pH conditions, -COOH groups are protonated (no charge) and are more hydrophobic (less soluble); while at high pH conditions, -COOH groups are deprotonated, becoming -COO- groups, which is more hydrophilic (more soluble).
Chapter 3 Q3: a. For IV injection, Cp0 = Cmax = 7.5 mg/L (y-axis intercept of the Cp vs time curve). b. Vd = dose/ Cp0 = 200 mg / 7.5 mg/L = 26.7 L (remains constant for the same drug) when Cp0 = 10 mg/L dose = Vd × Cp0 = 26.7 L × 10 mg/L = 267 mg Q5: carbonyl group – H-bond; positively-charged N – ionic interactions; methyl groups – VDW. Q6: a. E = F – R + S S = 0 (100 % inhibition) R = 71 mg/min E = Cu × Vu = 1 mg/mL × 1 mL/min = 1 mg/min ClR = E/Cp = 1 mg/min / 0.6 mg/mL = 1.67 mL/min b. F = E + R = 72 mg/min 1
GFR = F/Cp = 72 mg/min / 0.6 mg/mL = 120 mL/min c. E = 15 mg/mL × 1 mL/min = 15 mg/min S = E – F + R = (15 -72 +71) = 14 mg/min d. ClR = E/Cp = 15 mg/mL / 0.6 mg/mL = 25 mL/min * Change of S doesn’t affect F and R. Q8:
Q9:
Q10: drug - nicotinamide
Q12: a. Ct = C0 ‒ kt Ct @5days = 100 mg/mL ‒ 0.1 mg/(mL×hr) × 24 hrs/day × 5days = (100 – 0.1× 24×5) mg/mL = 88 mg/mL b. Ct = 0.9C0 = C0 – kt = (100 – 0.1×t) mg/mL = 90 mg/mL t = 100 hrs or use the equation: t90 = 0.1C0/k = 0.1× 100 / 0.1 = 100 hrs Chapter 4 Q1: Gap junctions and desmosomes. 2
Q2: Small and hydrophobic but not highly hydrophobic.
Q4: Structures of spironolactone and glycopyrronium bromide are shown below.
Spironolactone is hydrophobic and has high passive absorption rates independent of pH (I); while glycopyrronium bromide is hydrophilic and tends to be un-absorbable. Q5: Since trimethoprim is a weak base, use the weak base version of the Henderson-Hasselbalch equation: [ ]
Stomach:
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, predominately in the ionized form (BH+).
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Small intestine:
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, un-ionized form is much more than the ionized form.
Hence, trimethoprim is dominantly absorbed (passive) in small intestine. Q6: Length and large surface area. Q7: Hepatic portal vein. Q8: Method I: 3
F = 90 mg/200 mg = 0.45 F = (1 – Eg) × (1 – Eh) 0.45 = (1 – 0.3) × (1 – Eh) (1 – Eh) = 0.45/0.7 Eh = 1 - 0.45 × 0.7 = 0.357 Method II: Use the amounts of drugs at different stages: the amount of drug absorbed and entered the first-pass metabolism was: 200 mg × (1 – Fh) = 200 × 0.7 mg =140 mg
and 140 mg × (1 – Eh) = 90 mg (1 – Eh) = 90 mg/140 mg = 0.643 Eh = 1 – 0.643 = 0.357
Q13: ATP binding: substrate release ATP hydrolysis: conformational change (outward-facing to inward-facing).
Chapter 5 Q5:
Q6:
Q7:
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Chapter 6 Q4: deacetylation Q5: Collagen: At physiological conditions (37 ºC), collagen molecules (helical fibers) aggregate and form hydrogel. Gelatin: Gelatin dissolves in hot water and spontaneously forms gels on cooling. Below the solgel transition temperature (20 – 30 ºC), gelatin molecules go from random coil to helical structure, forming a hydrogel. Q7: above CMC (critical micelle concentration)
Q8: Microcapsules are core-shell structured. Q11: W/O: PGPR (Polyglycerol polyricinoleate; HLB value: 3-4) O/W: Tween 80 (HLB value: 15) Q14: hydrophobic interaction (effect) Q15: a. 18:1 b. 16:0-14:0 c. 16:0-22:6 Q20:
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