⬡ Pharmacokinetics Practical

One-Compartment
IV Bolus Model

From standard textbooks · RGUHS Practical Notes

10
Solved Problems
60+
Formulae
8
Viva Q&As
12
PK Categories
A01 Definition & Assumptions
  • Pharmacokinetics: the study of the time course of drug ADME — quantifying the relationship between dose and drug concentration in the body over time. S&Y Ch.1
  • One-compartment open model: the body is treated as a single, homogeneous, well-stirred compartment. Changes in plasma concentration are proportional to changes in all tissues. S&Y Ch.4
  • IV bolus: drug injected as a rapid single dose directly into systemic circulation. Absorption is complete and instantaneous — only distribution and elimination require modelling. S&Y Ch.4

Assumptions

  • Drug distributes instantaneously and uniformly throughout the body (well-stirred)
  • Elimination begins immediately upon drug entry into the body
  • Elimination follows first-order kinetics — rate is proportional to drug concentration at that time
  • Volume of distribution (Vd) is constant; plasma is the sampling compartment
  • First-order elimination: k remains constant regardless of dose → straight line on semilog plot. S&Y Ch.4
  • Zero-order elimination: rate constant regardless of concentration (saturable, e.g. phenytoin). Clearance is NOT constant for zero-order processes. S&Y Ch.4
A02 Key Pharmacokinetic Parameters
  • k — Elimination rate constant: k = 0.693/t½ = ln(Cp1/Cp2)/(t2−t1). From semilog slope: k = −slope × 2.303. Units: h⁻¹ S&Y Ch.4
  • t½ — Half-life: t½ = 0.693/k = (0.693 × Vd)/Cl. After 4 t½: ~94% eliminated; after 5 t½: ~97%. Time to steady state = 4–5 t½. DiPiro L3
  • Vd — Volume of Distribution: Vd = Dose/Cp⁰. NOT a real anatomical space — reflects partitioning between plasma and tissues. S&Y Ch.4
  • Cl — Clearance: Cl = k × Vd = Dose/AUC₀→∞. Total Cl = ClR + ClH + ClOther. S&Y Ch.4
  • AUC — Area Under the Curve: AUC₀→∞ = Cp⁰/k = Dose/Cl. Units: mg·h/L. S&Y Ch.4

Vd Reference Table

Vd RangeBody Compartment% Body WeightDrug Examples
~0.04–0.06 L/kgPlasma only~4–6%Heparin, warfarin (highly protein-bound)
~0.1–0.3 L/kgExtracellular fluid~10–30%Aminoglycosides (gentamicin ~0.25 L/kg)
~0.5–0.7 L/kgTotal body water~50–70%Theophylline, ethanol
>1 L/kgExtensive tissue binding>100%Digoxin (7 L/kg), chloroquine, TCAs
A03 Core Equations — One-Compartment Model
IV Bolus — One-Compartment S&Y Ch.4
ParameterEquationNotes
Plasma conc. (exponential)Cp = Cp⁰ × e^(−kt)Monoexponential decay
Plasma conc. (logarithmic)log Cp = log Cp⁰ − (k/2.303)×tStraight line on semilog; slope = −k/2.303
Initial concentrationCp⁰ = Dose / VdExtrapolated from y-intercept
Elimination rate constantk = −slope × 2.303From semilog plot
Half-lifet½ = 0.693 / k= (0.693 × Vd) / Cl
Volume of distributionVd = Dose / Cp⁰Units: L or L/kg
Total clearanceCl = k × Vd= Dose / AUC₀→∞
AUC (total)AUC₀→∞ = Cp⁰ / k= Dose/Cl; units: mg·h/L
AUC (trapezoidal)AUC₀→t = Σ[(Cn+Cn+1)/2 × Δt]Sum of trapezoids
AUC (terminal)AUCt→∞ = Cp(last) / kAdded to trapezoid AUC
IV Infusion — One-Compartment S&Y Ch.6
ParameterEquationNotes
Cp during infusionCp = (R₀/Cl)(1 − e^(−kt))R₀ = infusion rate (mg/h)
Steady-state conc.Css = R₀ / Cl = R₀ / (k × Vd)Reached after 4–5 t½
Post-infusion declineCp = Css × e^(−k×(t−T))T = total infusion duration
Loading doseDL = Css × VdIV bolus at infusion start
Oral Absorption — One-Compartment S&Y Ch.8
ParameterEquationNotes
Plasma conc. (oral)Cp = (F×D×ka)/[Vd(ka−k)] × (e^(−kt) − e^(−kat))F = bioavailability; ka = absorption rate constant
TmaxTmax = ln(ka/k) / (ka − k)Time of peak plasma concentration
BioavailabilityF = AUC(oral) / AUC(IV)Dose-normalise if doses differ
Flip-flop kineticsTerminal slope = −ka/2.303When ka << k; depot formulations
A04 Graphical Analysis — Semilog Plot
  • Semilog plot (log Cp vs t) is the key tool for one-compartment analysis. S&Y Ch.4
  • One-compartment IV bolus → single straight line (monoexponential decay)
  • Slope = −k/2.303; Y-intercept = log Cp⁰
  • R² ≥ 0.99 indicates good fit to one-compartment model. DiPiro
  • Two-compartment drugs: biexponential curve — initial steep α (distribution) then shallower β (elimination). Method of residuals separates phases. S&Y Ch.5 + DiPiro L6
  • Slope calculation: pick two widely separated points on the best-fit line. Slope = (log C2 − log C1)/(t2 − t1). k = −slope × 2.303. DiPiro L3
A05 Clinical Significance & TDM Applications
  • TDM indications: narrow therapeutic index drugs — aminoglycosides, vancomycin, digoxin, phenytoin, lithium, theophylline, cyclosporine. Also for unpredictable PK or suspected toxicity. DiPiro L1
  • Aminoglycosides (gentamicin, tobramycin, amikacin): one-compartment model; peak (Cmax) and trough (Cmin) monitored; once-daily dosing exploits concentration-dependent killing + PAE. DiPiro L12

Effect of disease on PK DiPiro L11

  • Renal failure: ↓ ClR → ↑ t½ → accumulation; adjust dose/interval for aminoglycosides
  • Hepatic failure: ↓ ClH → ↑ t½ for hepatically metabolised drugs (e.g. propranolol)
  • Oedema/ascites: ↑ Vd for water-soluble drugs → may need higher loading dose
A06 Worked Example — Gentamicin 80 mg IV Bolus

Given — plasma concentrations after 80 mg IV bolus

t=1h → 6.0 mg/L t=2h → 4.5 mg/L t=4h → 2.5 mg/L t=6h → 1.4 mg/L
2Slope = (log 1.4 − log 6.0) / (6 − 1) = (0.146 − 0.778) / 5 = −0.126 h⁻¹
3k = 0.126 × 2.303 = 0.290 h⁻¹
4t½ = 0.693 / 0.290 = 2.39 h
5Back-extrapolate → Cp⁰ = ~8.0 mg/L
6Vd = 80 / 8.0 = 10 L (0.14 L/kg in 70 kg patient)
7Cl = 0.290 × 10 = 2.90 L/h (48.3 mL/min)
8AUC = 8.0 / 0.290 = 27.6 mg·h/L [verify: 80/2.90 = 27.6 ✓]
∴ k=0.290 h⁻¹ · t½=2.39 h · Vd=10 L · Cl=2.90 L/h · AUC=27.6 mg·h/L
B01 Aim

To determine the pharmacokinetic parameters — elimination rate constant (k), half-life (t½), apparent volume of distribution (Vd), total clearance (Cl), and area under the curve (AUC) — of a drug following one-compartment model kinetics after intravenous bolus administration.

B02 Requirements
  • Drug / Subject: As specified (e.g. gentamicin 80 mg IV bolus in a 70 kg patient)
  • Samples: Serial venous blood samples (e.g. at 0.5, 1, 2, 4, 6, 8, 12 h post-dose)
  • Analysis: Validated drug assay (HPLC, immunoassay, or fluorescence polarisation)
  • Materials: Semilogarithmic graph paper, ruler, calculator or PK software (PK Solutions, Excel)
B03 Method — 10 Steps
1

Administer the IV bolus dose at t = 0. Record exact dose, time, and patient weight.

2

Collect blood samples at pre-specified time intervals. Label each sample with exact collection time.

3

Centrifuge at 3000 rpm for 10 min; separate plasma. Store at −20°C until assay.

4

Measure plasma drug concentrations (Cp) by validated assay. Record in observation table.

5

Calculate ln Cp and log Cp for each time point. Compute AUC segments using trapezoidal rule.

6

Plot log Cp (y-axis, logarithmic scale) vs time (x-axis, linear) on semilog paper. Single straight line confirms one-compartment model.

7

Draw best-fit line. Read Cp⁰ from y-intercept (t = 0 extrapolation).

8

Calculate slope from two points on best-fit line: Slope = (log C2 − log C1)/(t2 − t1). k = −slope × 2.303.

9

Calculate all PK parameters using equations in the results table.

10

State clinical interpretation: Vd distribution compartment, t½ relevance to dosing interval, Cl vs normal values.

B04 Observation Table
Time (h) Cp (mg/L) ln Cp log Cp Δt (h) AUC segment (mg·h/L)
0.50.5
10.5
21
42
62
82
124

AUC₀→t (total trapezoidal) = Σ all AUC segments = ________ mg·h/L

B05 Calculations & Results
ParameterFormulaCalculated valueUnits
Cp⁰ (y-intercept)Dose / Vd_______mg/L
k (elimination rate constant)−slope × 2.303_______h⁻¹
t½ (half-life)0.693 / k_______h
Vd (volume of distribution)Dose / Cp⁰_______L or L/kg
Cl (total clearance)k × Vd_______L/h
AUC₀→t (trapezoidal)Σ [(Cp1+Cp2)/2 × Δt]_______mg·h/L
AUCt→∞ (terminal)Cp(last) / k_______mg·h/L
AUC₀→∞ (total)AUC₀→t + AUCt→∞_______mg·h/L
B06 Graphical Record — Semilog Plot

Paste or draw your semilog plot here

Mark Cp⁰, slope calculation points, and best-fit line.
State whether data confirm one-compartment or two-compartment model.

Graph: log Cp vs Time (h) — Semilog Plot

B07 Discussion Points — Viva Questions
Why is the one-compartment model the simplest PK model? What assumptions make it clinically useful even though the body is not truly a single compartment?
S&Y Ch.4
What does a biexponential curve on a semilog plot indicate? How would you use the method of residuals to identify a two-compartment drug?
S&Y Ch.5 + DiPiro L6
A drug has Vd = 0.05 L/kg. What does this tell you about its distribution? Give an example and explain why Vd is small.
S&Y Ch.4
Why does t½ depend on both Vd and Cl? Give an example where t½ is long because of large Vd, not slow clearance.
DiPiro L3
If a patient develops acute renal failure, which PK parameter changes first, and how does this affect aminoglycoside dosing?
DiPiro L12
What is the difference between clearance and elimination rate constant? Why is clearance preferred in clinical PK?
S&Y Ch.4 + DiPiro L3
Explain why AUC is directly proportional to dose for a first-order drug but not for phenytoin.
S&Y Ch.10
What is flip-flop kinetics? How would you distinguish it from a slow-elimination drug without IV data?
S&Y Ch.8

Simple Problems — P01 to P05

P01 · Simple Calculate k and t½ from two plasma concentrations

Given

IV bolus t = 2 h → Cp = 12 mg/L t = 8 h → Cp = 1.5 mg/L

Find: k and t½

1k = ln(Cp1/Cp2) / (t2 − t1) = ln(12/1.5) / (8 − 2) = ln(8) / 6 = 2.079 / 6
2k = 0.347 h⁻¹
3t½ = 0.693 / k = 0.693 / 0.347 = 2.0 h
∴ k = 0.347 h⁻¹ · t½ = 2.0 h  S&Y Ch.4
P02 · Simple Calculate Vd and interpret distribution compartment

Given

Dose = 600 mg IV bolus Cp⁰ = 15 mg/L Weight = 80 kg

Find: Vd; state body compartment

1Vd = Dose / Cp⁰ = 600 / 15 = 40 L
2Vd per kg = 40 / 80 = 0.5 L/kg
30.5 L/kg → total body water distribution (e.g. theophylline-like)
∴ Vd = 40 L (0.5 L/kg) — total body water distribution  S&Y Ch.4
P03 · Simple Calculate clearance and verify with AUC

Given

k = 0.15 h⁻¹ Vd = 25 L Dose = 500 mg IV bolus Cp⁰ = 20 mg/L

Find: Cl and AUC₀→∞ (two methods)

1Cl = k × Vd = 0.15 × 25 = 3.75 L/h
2AUC Method 1: AUC = Cp⁰ / k = 20 / 0.15 = 133.3 mg·h/L
3AUC Method 2: AUC = Dose / Cl = 500 / 3.75 = 133.3 mg·h/L ✓
∴ Cl = 3.75 L/h · AUC₀→∞ = 133.3 mg·h/L  S&Y Ch.4
P04 · Simple Calculate Css during constant IV infusion

Given

R₀ = 60 mg/h Vd = 30 L k = 0.2 h⁻¹

Find: Css and time to reach Css

1Cl = k × Vd = 0.2 × 30 = 6 L/h
2Css = R₀ / Cl = 60 / 6 = 10 mg/L
3t½ = 0.693 / 0.2 = 3.47 h → Css reached in ~5 × t½ ≈ 17.3 h
4At 1 t½: Cp = Css × (1 − e⁻¹) = 10 × 0.632 = 6.32 mg/L (63.2% of Css)
∴ Css = 10 mg/L · reached after ~17 h (5 × t½)  S&Y Ch.6
P05 · Simple Loading dose design for target Css

Given

Target Css = 8 mg/L Vd = 50 L Cl = 5 L/h

Find: Loading dose (DL) and maintenance infusion rate (R₀)

1DL = Css × Vd = 8 × 50 = 400 mg (single IV bolus at start)
2R₀ = Css × Cl = 8 × 5 = 40 mg/h (maintenance infusion)
3With DL: Css achieved immediately. Without DL: Css approached gradually over 5 t½
4k = Cl/Vd = 5/50 = 0.1 h⁻¹ · t½ = 6.93 h → without DL, ~35 h to Css
∴ Loading dose = 400 mg · Maintenance rate = 40 mg/h  S&Y Ch.6

Moderate Problems — P06 to P10

P06 · Moderate Full PK parameter determination from plasma concentration dataset

Given — 100 mg IV bolus

t (h)1246810
Cp (mg/L)8.67.04.63.02.01.3

Find: k, t½, Cp⁰, Vd, Cl, AUC₀→∞

1Semilog plot: plot log Cp vs t; confirm linear relationship (R² ≥ 0.99)
2Slope (using t=1, Cp=8.6 and t=10, Cp=1.3): slope = (log 1.3 − log 8.6) / (10 − 1) = (0.114 − 0.934) / 9 = −0.0911 h⁻¹
3k = −slope × 2.303 = 0.0911 × 2.303 = 0.210 h⁻¹
4t½ = 0.693 / 0.210 = 3.30 h
5Back-extrapolate Cp⁰: log Cp⁰ = log 8.6 + 0.0911 × 1 = 0.934 + 0.091 = 1.025 → Cp⁰ = 10.6 mg/L
6Vd = 100 / 10.6 = 9.4 L (0.13 L/kg in 70 kg → plasma/highly bound)
7Cl = k × Vd = 0.210 × 9.4 = 1.97 L/h
8AUC = Cp⁰/k = 10.6/0.210 = 50.5 mg·h/L [verify: 100/1.97 = 50.8 ≈ 50.5 ✓]
∴ k=0.210 h⁻¹ · t½=3.30 h · Vd=9.4 L · Cl=1.97 L/h · AUC=50.5 mg·h/L  S&Y Ch.4
P07 · Moderate Oral one-compartment — calculate Tmax and Cmax

Given

500 mg oral tablet F = 0.80 ka = 1.5 h⁻¹ k = 0.3 h⁻¹ Vd = 40 L

Find: Tmax and Cmax

1Tmax = ln(ka/k) / (ka − k) = ln(1.5/0.3) / (1.5 − 0.3) = ln(5) / 1.2
2Tmax = 1.6094 / 1.2 = 1.34 h
3Coefficient A = (F×D×ka) / [Vd×(ka−k)] = (0.80×500×1.5) / (40×1.2) = 600/48 = 12.5 mg/L
4Cmax = A × (e^(−k×Tmax) − e^(−ka×Tmax)) = 12.5 × (e^(−0.402) − e^(−2.010))
5= 12.5 × (0.669 − 0.134) = 12.5 × 0.535 = 6.69 mg/L
∴ Tmax = 1.34 h · Cmax = 6.69 mg/L  S&Y Ch.8
P08 · Moderate Absolute bioavailability from oral vs IV AUC

Given

50 mg IV bolus → AUC = 40 mg·h/L 200 mg oral tablet → AUC = 100 mg·h/L

Find: Absolute bioavailability (F)

1F = (AUCoral / AUCiv) × (Div / Doral)  [dose-normalised formula]
2F = (100 / 40) × (50 / 200) = 2.5 × 0.25 = 0.625
3Cross-check: AUC per mg (IV) = 40/50 = 0.80 mg·h/L per mg
4Expected oral AUC if F=1: 200 × 0.80 = 160 mg·h/L. Observed = 100. F = 100/160 = 0.625 ✓
∴ Absolute bioavailability F = 0.625 (62.5%)  S&Y Ch.7
P09 · Moderate Vancomycin: loading dose + maintenance + trough prediction

Given

70 kg patient Target Css(avg) = 20 mg/L Vd = 0.7 L/kg Cl = 4.2 L/h Interval: q12h IV

Find: Total Vd, DL, maintenance dose, expected trough

1Total Vd = 0.7 × 70 = 49 L
2k = Cl / Vd = 4.2 / 49 = 0.0857 h⁻¹ · t½ = 0.693/0.0857 = 8.09 h
3Maintenance dose: Dm = Css(avg) × Cl × τ = 20 × 4.2 × 12 = 1008 mg ≈ 1000 mg q12h
4Loading dose: DL = Css(avg) × Vd = 20 × 49 = 980 mg ≈ 1000 mg
5Trough: Css(min) = (Dm/Vd) / (e^(k×τ) − 1) = (1000/49) / (e^1.028 − 1) = 20.4 / 1.796 = 11.4 mg/L
∴ DL = 980 mg · Dm = 1000 mg q12h · Expected trough ≈ 11.4 mg/L  DiPiro L12
P10 · Moderate Aminoglycoside dose adjustment in renal failure

Given

Normal: k = 0.35 h⁻¹, Vd = 16 L, 80 mg q8h ClCr drops 100 → 25 mL/min Gentamicin 90% renally excreted

Find: Adjusted k, t½; new dosing regimen to maintain same peak

1Normal Cl = k × Vd = 0.35 × 16 = 5.6 L/h
2Fraction of normal renal function remaining = 25/100 = 0.25
3Adjusted Cl = 5.6 × [(1 − 0.90) + 0.90 × 0.25] = 5.6 × 0.325 = 1.82 L/h
4New k = 1.82 / 16 = 0.114 h⁻¹
5New t½ = 0.693 / 0.114 = 6.08 h  [was 2.0 h — 3× longer due to renal failure]
6Option A — keep dose 80 mg, extend interval to ≈5 × t½ ≈ 30 h → q36h
7Option B — reduce dose proportionally: 80 × 0.325 = 26 mg q8h
8ALWAYS monitor levels: target peak 5–10 mg/L (once-daily) or 4–8 mg/L (traditional)
∴ New k=0.114 h⁻¹ · t½=6.1 h · Regimen: 80 mg q36h OR 26 mg q8h · Monitor levels  DiPiro L12
Parameter / Formula Equation Notes & Source
1. First-Order Elimination Kinetics
Plasma conc. at time t Cp(t) = Cp⁰ × e^(−kt) Monoexponential decay; one-compartment IV bolus S&Y Ch.4
Plasma conc. (semilog form) log Cp = log Cp⁰ − (k/2.303)×t Straight line on semilog paper; slope = −k/2.303 S&Y Ch.4
k from semilog slope k = −slope × 2.303 Slope is negative; k is always positive; units: h⁻¹ S&Y Ch.4
k from two time points k = ln(Cp1/Cp2) / (t2 − t1) t2 > t1; equivalent to slope×2.303 method DiPiro L3
Half-life t½ = 0.693 / k Independent of dose for first-order kinetics S&Y Ch.4
Half-life (extended form) t½ = (0.693 × Vd) / Cl Explains why t½ changes with disease altering Vd or Cl DiPiro L3
% drug remaining at time t % = e^(−kt) × 100 After 1 t½: 50%; 2 t½: 25%; 4 t½: 6.25% DiPiro L3
% drug eliminated % = (1 − e^(−kt)) × 100 4 t½: 93.75%; 5 t½: 96.9% DiPiro L3
Amount after n half-lives A(n) = A₀ × (0.5)^n n = number of half-lives elapsed S&Y Ch.2
2. Volume of Distribution (Vd)
Vd (IV bolus) Vd = Dose / Cp⁰ Cp⁰ = initial conc. extrapolated to t=0 S&Y Ch.4
Vd (from AUC) Vd = Dose / (k × AUC₀→∞) Model-independent; does not require Cp⁰ S&Y Ch.4
Vd (protein binding effect) Vd = Vp + Vt×(fu/fut) fu = plasma unbound fraction; fut = tissue unbound; Vp≈3 L S&Y Ch.9
Vd (steady state, 2-cpt) Vdss = Vc + Vt Vc = central; Vt = peripheral; Vdss unaffected by kinetics S&Y Ch.5
3. Clearance (Cl)
Total clearance Cl = k × Vd = Dose / AUC₀→∞ Model-independent second form S&Y Ch.4
Cl relationship Cl = ClR + ClH + ClOther Sum of all routes DiPiro L3
Renal clearance ClR = Cl × fe fe = fraction excreted unchanged in urine DiPiro L11
Hepatic clearance ClH = Q × ER Q = hepatic blood flow (~90 L/h); ER = extraction ratio DiPiro L9
High ER drugs (flow-limited) ClH ≈ Q (hepatic blood flow) ER > 0.7; sensitive to changes in Q (e.g. propranolol) DiPiro L9
Low ER drugs (capacity-limited) ClH = fu × CLint ER < 0.3; sensitive to protein binding (e.g. warfarin) DiPiro L9
Creatinine clearance (C-G) ClCr = [(140−age)×wt] / (72×SCr) Multiply by 0.85 for females; wt in kg; SCr in mg/dL DiPiro L11
4. Area Under the Curve (AUC)
AUC₀→∞ (IV bolus) AUC = Cp⁰ / k = Dose / Cl Total drug exposure from t=0 to infinity S&Y Ch.4
AUC (linear trapezoidal) AUC₀→t = Σ [(Cn+Cn+1)/2 × Δt] Linear trapezoidal; use for ascending or flat sections DiPiro L3
AUC (log-linear trapezoidal) AUC = (Cn−Cn+1)/ln(Cn/Cn+1) × Δt Better for declining log-linear segments S&Y Ch.4
AUC (terminal extrapolation) AUCt→∞ = Cp(last) / k Added to trapezoid AUC for total AUC₀→∞ DiPiro L3
AUC (oral dose) AUC = F×Dose / Cl F = bioavailability; accounts for incomplete absorption S&Y Ch.8
AUMC (1st moment) AUMC₀→∞ = Cp⁰ / k² For non-compartmental analysis; used to calc MRT S&Y Ch.9
Mean Residence Time MRT = AUMC / AUC = 1/k 1/k for one-compartment IV bolus; 1/k + τ/2 for infusion S&Y Ch.9
5. IV Bolus — One-Compartment
Cp at time t Cp = Cp⁰ × e^(−kt) Most fundamental equation S&Y Ch.4
Initial concentration Cp⁰ = Dose / Vd Concentration if distribution were truly instantaneous S&Y Ch.4
Cp (log form) log Cp = log Cp⁰ − (k/2.303)t For semilog plot; y-intercept = log Cp⁰ S&Y Ch.4
6. IV Infusion — One-Compartment
Cp during infusion (t < T) Cp = (R₀/Cl)(1 − e^(−kt)) R₀ = infusion rate (mg/h); approaches Css asymptotically S&Y Ch.6
Steady-state concentration Css = R₀ / Cl = R₀ / (k×Vd) Reached after 4–5 t½ regardless of R₀ S&Y Ch.6
% of Css at time t % = (1 − e^(−kt)) × 100 At 1 t½: 50%; 3.32 t½: 90%; 4 t½: 93.75% DiPiro L4
Post-infusion decline (t > T) Cp = Css × e^(−k(t−T)) T = total infusion duration S&Y Ch.6
Loading dose DL = Css × Vd Give as IV bolus before starting infusion S&Y Ch.6
Required infusion rate R₀ = Css(target) × Cl Design equation for target-driven dosing DiPiro L4
7. Oral Absorption — One-Compartment
Cp after oral dose Cp = (F×D×ka)/[Vd(ka−k)]×(e^−kt−e^−kat) Biexponential; ka > k for normal absorption S&Y Ch.8
Time of peak (Tmax) Tmax = ln(ka/k) / (ka − k) At Tmax, absorption rate = elimination rate S&Y Ch.8
Peak concentration (Cmax) Cmax = Cp at t = Tmax Substitute Tmax into Cp(t) equation S&Y Ch.8
Flip-flop kinetics Terminal slope = −ka/2.303 When ka << k; absorption is rate-limiting; depot formulations S&Y Ch.8
Lag time (tlag) Modified Tmax = tlag + ln(ka/k)/(ka−k) Add tlag to Tmax formula if absorption not immediate S&Y Ch.8
8. Bioavailability & Bioequivalence
Absolute bioavailability F = (AUCoral×DIV) / (AUCIV×Doral) Dose-normalise if different doses used S&Y Ch.7
Relative bioavailability Frel = AUCtest / AUCreference For generic vs innovator (must use same doses) S&Y Ch.7
Hepatic first-pass F F = 1 − ER ER = hepatic extraction ratio; high-ER drugs have low F DiPiro L9
Overall oral F F = fa × fg × fh fa = fraction absorbed; fg = gut wall; fh = hepatic fraction S&Y Ch.7
Bioequivalence criterion 90% CI of Cmax, AUC ratio: 80–125% FDA/EMA standard for generic approval S&Y Ch.7
9. Multiple Dosing & Steady State
Css(max) at steady state Css,max = (F×D/Vd)/(1−e^(−kτ)) τ = dosing interval; instantaneous absorption assumed S&Y Ch.9
Css(min) at steady state Css,min = Css,max × e^(−kτ) Just before the next dose S&Y Ch.9
Average Css Css(avg) = F×D / (Cl×τ) = AUCss/τ; fundamental target for dosing design DiPiro L4
Accumulation factor (R) R = 1 / (1 − e^(−kτ)) Ratio of Css to single-dose Cp at same time S&Y Ch.9
Time to steady state 4–5 × t½ Independent of dose, route, or frequency DiPiro L4
Dosing interval design τ = ln(Cmax/Cmin) / k Design equation: choose target Cmax and Cmin DiPiro L4
Loading dose (multiple dosing) DL = Dm / (1 − e^(−kτ)) Dm = maintenance dose; gives Css from first dose S&Y Ch.9
10. Non-linear (Michaelis-Menten) Pharmacokinetics
Rate of drug elimination v = Vmax×C / (Km + C) Vmax = max rate (mg/h); Km = Michaelis constant (mg/L) S&Y Ch.10
High conc. (C >> Km) v ≈ Vmax (zero-order) Saturated enzymes; rate is constant regardless of Cp S&Y Ch.10
Low conc. (C << Km) v ≈ (Vmax/Km)×C (first-order) Approximates linear kinetics; clearance = Vmax/Km S&Y Ch.10
Phenytoin Css Css = Km×DR / (Vmax−DR) Css rises disproportionately with dose rate S&Y Ch.10
Half-life (non-linear) Not constant; increases with Cp Cannot use t½ for dosing interval design S&Y Ch.10
AUC (non-linear) Not proportional to dose Small dose increase → large AUC increase (phenytoin toxicity risk) S&Y Ch.10
11. Protein Binding
Fraction unbound in plasma fu = Cu / Ctotal Cu = free conc.; warfarin fu = 0.01 (99% bound) S&Y Ch.9
Fraction bound % bound = (1 − fu) × 100 Only unbound drug crosses membranes and exerts effect S&Y Ch.9
Effect of binding on Vd Vd ↑ as fu ↑ (or fut ↓) More unbound drug leaves plasma for tissues S&Y Ch.9
Effect on Cl (capacity-limited) Cl = fu × CLint Low-ER drugs: Cl proportional to fu; binding affects Cl DiPiro L9
Effect on Cl (flow-limited) Cl ≈ hepatic blood flow Q High-ER drugs: protein binding does NOT significantly alter Cl DiPiro L9
12. Pharmacodynamics (PK/PD)
Emax model E = Emax × C / (EC50 + C) Emax = max effect; EC50 = conc. at 50% effect DiPiro L7
Sigmoid Emax (Hill equation) E = Emax × C^n / (EC50^n + C^n) n = Hill coefficient; n>1: steep; n<1: shallow S-curve DiPiro L7
Log-linear model E = m × log C + b Valid for 20–80% of effect range; m = slope DiPiro L7
Therapeutic index TI = TD50 / ED50 Ratio of toxic to effective median doses; narrow TI needs TDM S&Y
Conc.-effect at Css E = Emax × Css / (EC50 + Css) At steady state; basis for target concentration approach DiPiro L7
PK/PD target (bactericidal) AUC/MIC or Cmax/MIC Aminoglycosides: Cmax/MIC > 8–10; fluoroquinolones: AUC/MIC > 125 DiPiro
PK/PD target (time-dependent) T > MIC (% of dosing interval) β-lactams: T>MIC > 40–70% for bactericidal activity DiPiro
No formulae match your search.

From standard textbooks