GW BGZ2026 Practical - Beaker glass pharmacokinetics

What is the main goal of the BGZ2026 pharmacokinetics practical?

The goal is to understand how clearance (CL) and volume of distribution (VD) determine the time course of drug concentration in the body.

• The experiment simulates pharmacokinetics using a colored compound in water.

• It helps link theoretical PK parameters to a physical model system.

What pharmacokinetic processes are simulated in the beaker experiment?

• Dose → adding colored compound

• Volume of distribution (VD) → unknown water volume in beaker

• Clearance (CL) → removal and replacement of liquid

• Concentration changes → measured over time via absorbance

What is clearance (CL) in pharmacokinetics?

Clearance is the volume of plasma (or system) completely cleared of a substance per unit time.

• In this experiment:

  • Removal of a fixed volume each minute mimics clearance.

• Higher clearance → faster decrease in concentration

What is volume of distribution (VD)?

Volume of distribution describes how a drug distributes in the body relative to blood concentration.

• In this experiment:

  • The beaker’s water volume represents VD.

• Large VD → more dilution → lower concentration

What is the relationship between absorbance and concentration?

Beer–Lambert Law:

• A = ε × c × l

  • A = absorbance

  • ε = extinction coefficient

  • c = concentration

  • l = path length

• Used to convert spectrophotometry readings into concentration

Calculate the concentration in mM from the absorbance (A) of a solution of a colored compound when A = 0.5, using the molecular extinction coefficient ( = 43000 M-1 cm-1 ). Convert this concentration from mM to mg/L. The molecular weight of the colored compound is 854 g/mol.

Given:

• A = 0.5

• ε = 43000 M⁻¹ cm⁻¹

• l = 1 cm (standard cuvette assumed)

• MW = 854 g/mol

Step 1: Beer–Lambert Law A= ε ⋅ c ⋅ l

c = A/ε ⋅ l

c = 0.5/43000 ⋅ 1

c = 1.16 × 10^−5 mol/L

Step 2: Convert to mM

1 mM = 10^−3 mol/L

c = 1.16 × 10^−5 × 1000

c = 0.0116 mM

Step 3: Convert to mg/L

mg/L = c (mol/L) × MW (g/mol) × 1000

= 1.16 ×10^−5 × 854 × 1000 = 9.91 mg/L

Final answer (a):

• 0.0116 mM

• 9.91 mg/L

Calculate the concentration of the colored compound solution of c = 1 mg/ml in mM

Given:

• c = 1 mg/mL = 1000 mg/L

• MW = 854 g/mol = 854 mg/mmol

Step 1: Use conversion

c (mM) = 1000 mg/L / 854 mg/mmol

c = 1.17 mmol/L

Final answer (b):

• 1.17 mM

After i.v. application of a drug (D = 1 ml of a solution with a concentration of c = 1 mg/ml) to a patient, a plasma concentration at t = 0 min of c0 = 8.54 mg/L is measured. Calculate the volume of distribution (VD).

Given:

• Dose = 1 mL × 1 mg/mL = 1 mg

• C₀ = 8.54 mg/L

Formula: VD = Dose/C0

Step 1:

VD = 1 mg / 8.54 mg/L

VD = 0.117 L

Final answer (c):

• VD = 0.117 L (≈ 117 mL)

Assume an elimination rate constant of ke = 0.01 min-1 . Calculate the clearance (CL) and elimination half time (t1/2) based on the VD calculated at 3(c).

Given:

• ke = 0.01 min⁻¹

• VD = 0.117 L

Step 1: Half-life

t1/2 = 0.693/ke

t1/2 = 0.693/0.01

t1/2 ​= 69.3 min

Half-life result:

• t½ = 69.3 min

Step 2: Clearance

CL = ke ⋅ VD

CL = 0.01 × 0.117

CL = 0.00117 L/min

Clearance result:

• CL = 0.00117 L/min

What happens when dose increases?

• Dose ↑ → initial concentration (C₀) ↑ proportionally

• The entire concentration–time curve shifts upwards

• Shape of the curve does NOT change

Key relationships:

• C₀ = Dose / VD

• If VD and CL stay constant:

  • doubling dose → doubles concentration at all time points

• Dose affects only the height of the curve

• It does not change slope, clearance, or half-life

→ More drug = higher concentrations, same elimination speed

What happens when clearance increases?

• Clearance = how much volume is “cleaned” per time

If clearance increases:

• Faster decline in concentration

• Steeper curve downward

• Shorter half-life

• Lower drug exposure over time

Key relationships:

• CL ↑ → ke ↑ → faster elimination

• t½ = 0.693 / ke → decreases

• ln(concentration) slope becomes steeper

• Clearance controls the speed of elimination

• Higher clearance = drug disappears faster

→ Fast clearance = shorter action duration

What happens when VD increases?

Volume of distribution reflects how widely drug spreads in the system

If VD increases:

• Drug becomes more diluted

• Lower initial concentration (C₀ decreases)

• Drug is “hidden” in a larger space

Key relationships:

• C₀ = Dose / VD

• CL = ke × VD

• t½ = 0.693 × VD / CL

Important consequence:

If clearance stays constant:

• Higher VD → longer half-life

• Drug stays in the body longer

• VD controls how diluted and distributed the drug is

• Larger VD → lower starting concentration but often longer persistence

→ Big body space = drug spreads more and lasts longer

What does a linear ln(concentration) vs time plot indicate?

It indicates first-order elimination kinetics.

• Straight line → exponential decay

• Slope = −ke

How do clearance, volume of distribution, and dose affect concentration-time curves?

• Clearance ↑ → faster decline

• VD ↑ → lower initial concentration

• Dose ↑ → higher initial concentration

What is the main principle learned from all three experiments?

Pharmacokinetic profiles are determined by:

• Dose (amount given)

• Clearance (removal rate)

• Volume of distribution (distribution space)

Together they determine:

• concentration-time curve shape

• drug exposure

• half-life

Why is spectrophotometry used in this experiment?

Because it allows quantitative measurement of drug concentration.

• Based on light absorption

• Converts absorbance → concentration

• Enables construction of PK curves