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BIOD 102 Portage Learning - Lab 5 Exam: Pharmacokinetics 2026/2027 | Verified Q&A | 100% Correct | Pass Guaranteed - A+ Graded

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Master your BIOD 102 Lab 5 Exam on Pharmacokinetics with this comprehensive Portage Learning resource. This A+ Graded resource for the BIOD 102 Portage Learning - Lab 5 Exam (Pharmacokinetics) (2026/2027) contains verified questions and answers that are 100% correct for complete exam preparation. Featuring comprehensive coverage of the four fundamental processes of pharmacokinetics – Absorption, Distribution, Metabolism, and Excretion (ADME) , including routes of drug administration, bioavailability, volume of distribution, half-life calculation, first-pass metabolism, hepatic clearance, renal excretion, enzyme induction/inhibition, and factors affecting drug absorption and elimination, it provides the laboratory-based knowledge and pharmacological understanding needed to mirror Portage Learning's official lab exam format and rigor. Each answer is carefully verified to ensure accuracy, helping you master key concepts such as drug concentration-time curves, steady-state principles, loading and maintenance doses, therapeutic drug monitoring, and pharmacokinetic parameter calculations. With fully verified Q&A and our Pass Guarantee, this is the definitive tool to ace your BIOD 102 Lab 5 Exam on Pharmacokinetics on the first attempt and excel in your biology lab course. Get instant access now and start studying today.

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Institución
BIOD 102
Grado
BIOD 102

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BIOD 102 Portage Learning - Lab 5 Exam
(Pharmacokinetics) (2026/2027)

Section 1: Pharmacokinetic Parameter Calculations


Q1: A 750 mg dose of Drug A is administered as an intravenous bolus. The concentration
immediately after injection (C0) is 15 mg/L. What is the Volume of Distribution (Vd)?

A. 0.02 L

B. 50 L [CORRECT]

C. 100 L

D. 11.25 L

Correct Answer: B

Rationale: Vd is calculated as Dose / C0. Calculation: 750 mg / 15 mg/L = 50 L.



Q2: A researcher plots plasma concentration data on a semi-logarithmic graph. The slope of the
terminal straight line is calculated to be -0.231 hours^-1. What is the elimination rate constant
(ke)?

A. -0.231 hours^-1

B. 0.231 hours^-1 [CORRECT]

C. 0.693 hours^-1

D. 0.333 hours^-1

Correct Answer: B

Rationale: The elimination rate constant (ke) is the absolute value (positive value) of the slope
found on a semi-log plot of concentration versus time.




.

,Q3: Using the elimination rate constant (ke) from Question 2 (0.231 hours^-1), calculate the
half-life (t1/2) of the drug.

A. 0.693 hours

B. 1.5 hours

C. 3.0 hours [CORRECT]

D. 3.33 hours

Correct Answer: C

Rationale: Half-life is calculated as t1/2 = 0.693 / ke. Calculation: 0..231 hours^-1 = 3.0
hours.



Q4: A patient has a Volume of Distribution (Vd) of 40 L and an elimination rate constant (ke) of
0.1 hours^-1. Calculate the Total Body Clearance (CL) in L/hour.

A. 4.0 L/hour [CORRECT]

B. 0.25 L/hour

C. 400 L/hour

D. 40.1 L/hour

Correct Answer: A

Rationale: Clearance is the product of the elimination rate constant and volume of
distribution: CL = ke × Vd. Calculation: 0.1 hours^-1 × 40 L = 4.0 L/hour.



Q5: A drug has a half-life of 8 hours. Approximately how long will it take to reach 93.75% of
steady state (approximately 4 half-lives) if administered via continuous infusion?

A. 8 hours

B. 16 hours

C. 24 hours

D. 32 hours [CORRECT]

Correct Answer: D

Rationale: It takes 4 to 5 half-lives to reach steady state. Calculation: 4 half-lives × 8 hours/half-
life = 32 hours.
.

, Q6: The plasma concentration of a drug drops from 50 mg/L to 12.5 mg/L in 12 hours. How
many half-lives have passed, and what is the half-life?

A. 2 half-lives; 6 hours [CORRECT]

B. 3 half-lives; 4 hours

C. 4 half-lives; 3 hours

D. 1 half-life; 12 hours

Correct Answer: A

Rationale: 50 to 25 mg/L is one half-life; 25 to 12.5 mg/L is a second half-life. Total drop
represents 2 half-lives. 12 hours / 2 = 6 hours per half-life.



Q7: Calculate the Loading Dose required to achieve a target concentration (Ctarget) of 6 mg/L
immediately. The patient has a Volume of Distribution (Vd) of 35 L and the drug is administered
intravenously (F = 1).

A. 5.83 mg

B. 210 mg [CORRECT]

C. 41.67 mg

D. 35 mg

Correct Answer: B

Rationale: Loading Dose = (Ctarget × Vd) / F. Calculation: (6 mg/L × 35 L) / 1 = 210 mg.



Q8: You need to calculate an oral Maintenance Dose. The desired average steady-state
concentration is 10 mg/L. The drug has a Clearance (CL) of 5 L/hour and a bioavailability (F) of
0.8. The dosing interval (τ) is 12 hours.

A. 600 mg

B. 480 mg

C. 750 mg [CORRECT]

D. 62.5 mg

Correct Answer: C
.

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Institución
BIOD 102
Grado
BIOD 102

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Subido en
5 de abril de 2026
Número de páginas
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Escrito en
2025/2026
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