Laplace%20transformation

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Laplace transformation
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Laplace transformation

• The Laplace transform is a mathematical technique
  used for solving linear differential equations
  (apparent zero order and first order) and hence
  is applicable to the solution of many equations
  used for pharmacokinetic analysis.

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Laplace transformation procedure

1. Write the differential equation
2. Take the Laplace transform of each differential
   equation using a few transforms (using table in
   the next slide)
3. Use some algebra to solve for the Laplace of the
   system component of interest
4. Finally the 'anti'-Laplace for the component is
   determined from tables

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Important Laplace transformation (used in step 2)
Expression Transform
dX/dt
K (constant)
X (variable)
KX (K is constant)
where s is the laplace operator, is the
laplace integral , and X0 is the amount at time
zero
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Anti-laplce table (used in step 4)
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Anti-laplce table continued (used in step 4)
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Laplace transformation example

• The differential equation that describes the
  change in blood concentration of drug X is
• Derive the equation that describe the amount of
  drug X??

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Laplace transformation example

1. Write the differential equation
2. Take the Laplace transform of each differential
   equation

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Laplace transformation example

1. Use some algebra to solve for the Laplace of the
   system component of interest
2. Finally the 'anti'-Laplace for the component is
   determined from tables

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Laplace transformation example

• The derived equation represent the equation for
  IV bolus one compartment

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Continuous intravenous infusion(one-compartment
model)

• Dr Mohammad Issa

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Theory of intravenous infusion

• The drug is administered at a selected or
  calculated constant rate (K0) (i.e. dX/dt), the
  units of this input rate will be those of mass
  per unit time (e.g.mg/hr).
• The constant rate can be calculated from the
  concentration of drug solution and the flow rate
  of this solution, For example, the concentration
  of drug solution is 1 (w/v) and this solution is
  being infused at the constant rate of 10mL/hr
  (solution flow rate). So 10mL of solution will
  contain 0.1 g (100 mg) drug.
• The infusion rate (K0) equals to the solution
  flow rate multiplied by the concentration
• In this example, the infusion rate will be
  10mL/hr multiplied by 100 mg/10 mL, or 100mg/hr.
  The elimination of drug from the body follows a
  first

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IV infusion
During infusion
Post infusion
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IV infusion during infusion

• where K0 is the infusion rate, K is the
  elimination rate constant, and Vd is the volume
  of distribution

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Steady state
steady state concentration (Css)
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Steady state

• At steady state the input rate (infusion rate) is
  equal to the elimination rate.
• This characteristic of steady state is valid for
  all drugs regardless to the pharmacokinetic
  behavior or the route of administration.

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Fraction achieved of steady state concentration
(Fss)

• since , previous equation
  can
• be represented as

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Fraction achieved of steady state concentration
(Fss)
or
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Time needed to achieve steady state

• time needed to get to a certain fraction of
  steady state depends on the half life of the drug
  (not the infusion rate)

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Example

• What is the minimum number of half lives needed
  to achieve at least 95 of steady state?
• At least 5 half lives (not 4) are needed to get
  to 95 of steady state

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Example

• A drug with an elimination half life of 10 hrs.
  Assuming that it follows a one compartment
  pharmacokinetics, fill the following table

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Example
Time Fss
10
30
50
70
90
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Example
Time Number of elapsed half-lives Fss
10 1 0.5
30 3 0.875
50 5 0.969
70 7 0.992
90 9 0.998
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IV infusion Loading IV bolus

• During constant rate IV administration, the drug
  accumulates until steady state is achieved after
  five to seven half-lives
• This can constitute a problem when immediate drug
  effect is required and immediate achievement of
  therapeutic drug concentrations is necessary such
  as in emergency situations
• In this ease, administration of a loading dose
  will be necessary. The loading dose is an IV
  holus dose administered at the time of starting
  the IV infusion to achieve faster approach to
  steady state. So administration of an IV loading
  dose and starting the constant rate IV infusion
  simultaneously can rapidly produce therapeutic
  drug concentration. The loading dose is chosen to
  produce Plasma concentration similar or close to
  the desired plasma concentration that will be
  achieved by the IV infusion at steady state

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IV infusion Loading IV bolus
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IV infusion Loading IV bolus

• To achieve a target steady state conc (Css) the
  following equations can be used
• For the infusion rate
• For the loading dose

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IV infusion Loading IV bolus

• The conc. resulting from both the bolus and the
  infusion can be described as
• Ctotal Cinfusion Cbolus

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IV infusion Loading IV bolus Example

• Derive the equation that describe plasma
  concentration of a drug with one compartment PK
  resulting from the administration of an IV
  infusion (K0 CssCl ) and a loading bolus (LD
  CssVd) that was given at the start of the
  infusion

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IV infusion Loading IV bolus Example

• Ctotal Cinfusion Cbolus
• Cinfusion
• Cbolus

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IV infusion Loading IV bolus Example

• Ctotal Cinfusion Cbolus

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Scenarios with different LD
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Changing Infusion Rates
Increasing the infusion rate results in a new
steady state conc. 5-7 half-lives are needed to
get to the new steady state conc
Concentration
Half-lives
5-7 half-lives are needed to get to steady state
Decreasing the infusion rate results in a new
steady state conc. 5-7 half-lives are needed to
get to the new steady state conc
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Changing Infusion Rates

• The rate of infusion of a drug is sometimes
  changed during therapy because of excessive
  toxicity or an inadequate therapeutic response.
  If the object of the change is to produce a new
  plateau, then the time to go from one plateau to
  anotherwhether higher or lowerdepends solely on
  the half-life of the drug.

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Post infusion phase
During infusion
Post infusion
C (Concentration at the end of the infusion)
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Post infusion phase data

• Half-life and elimination rate constant
  calculation
• Volume of distribution estimation

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Elimination rate constant calculation using post
infusion data

• K can be estimated using post infusion data by
• Plotting log(Conc) vs. time
• From the slope estimate K

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Volume of distribution calculation using post
infusion data

• If you reached steady state conc (C CSS)
• where k is estimated as described in the previous
  slide

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Volume of distribution calculation using post
infusion data

• If you did not reached steady state (C
  CSS(1-e-kT))

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Example 1

• Following a two-hour infusion of 100 mg/hr plasma
  was collected and analysed for drug
  concentration. Calculate kel and V.

Time relative to infusion cessation (hr) 1 3 7 10 16 22
Cp (mg/L) 12 9 8 5 3.9 1.7
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Time is the time after stopping the infusion
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Example 1

• From the slope, K is estimated to be
• From the intercept, C is estimated to be

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Example 1

• Since we did not get to steady state

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Example 2

• Estimate the volume of distribution (22 L),
  elimination rate constant (0.28 hr-1), half-life
  (2.5 hr), and clearance (6.2 L/hr) from the data
  in the following table obtained on infusing a
  drug at the rate of 50 mg/hr for 16 hours.

Time (hr) 0 2 4 6 10 12 15 16 18 20 24
Conc (mg/L) 0 3.48 5.47 6.6 7.6 7.8 8 8 4.6 2.62 0.85
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Example 2
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Example 2

• Calculating clearance
• It appears from the data that the infusion
  has reached steady state
• (CP(t15) CP(t16) CSS)

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Example 2

• Calculating elimination rate constant and half
  life
• From the post infusion data, K and t1/2 can be
  estimated. The concentration in the post infusion
  phase is described according to
• where t1 is the time after stopping the infusion.
  Plotting log(Cp) vs. t1 results in the following

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Example 2
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Example 2

• K-slope2.3030.28 hr-1
• Half life 0.693/K0.693/0.28 2.475 hr
• Calculating volume of distribution

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Example 3

• A drug that displays one compartment
  characteristics was administered as an IV bolus
  of 250 mg followed immediately by a constant
  infusion of 10 mg/hr for the duration of a study.
  Estimate the values of the volume of distribution
  (25 L), elimination rate constant (0.1hr-1),
  half-life (7), and clearance (2.5 L/hr) from the
  data in the following table

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Example 3
Time(hr) 0 5 20 45 50
Conc(mg/L) 10 7.6 4.8 4.0 4.0
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Example 3

• The equation that describes drug concentration
  is
• a- Calculating volume of distribution
• At time zero,

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Example 3

• b- Calculating elimination rate constant and half
  life
• Since the last two concentrations (at time 45 and
  50 hrs) are equal, it is assumed that a steady
  state situation has been achieved.
• Half life 0.693/K0.693/0.1 6.93 hr

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Example 3

• c- Calculating clearance

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Example 4

• For prolonged surgical procedures,
  succinylcholine is given by IV infusion for
  sustained muscle relaxation. A typical initial
  dose is 20 mg followed by continuous infusion of
  4 mg/min. the infusion must be individualized
  because of variation in the kinetics of
  metabolism of suucinylcholine. Estimate the
  elimination half-lives of succinylcholine in
  patients requiring 0.4 mg/min and 4 mg/min,
  respectively, to maintain 20 mg in the body. (35
  and 3.5 min)

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Example 4

• For the patient requiring 0.4 mg/min
• For the patient requiring 4 mg/min

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Example 5

• A drug is administered as a short term infusion.
  The average pharmacokinetic parameters for this
• drug are
• K 0.40 hr-1
• Vd 28 L
• This drug follows a one-compartment body model.

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Example 5

• 1) A 300 mg dose of this drug is given as a
  short-term infusion over 30 minutes. What is the
  infusion rate? What will be the plasma
  concentration at the end of the infusion?
• 2) How long will it take for the plasma
  concentration to fall to 5.0 mg/L?
• 3) If another infusion is started 5.5 hours after
  the first infusion was stopped, what will the
  plasma concentration be just before the second
  infusion?

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Example 5

• 1) The infusion rate (K0) Dose/duration 300
  mg/0.5 hr 600 mg/hr.
• Plasma concentration at the end of the infusion
• Infusion phase

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Example 5

• 2) Post infusion phase
• The concentration will fall to 5.0 mg/L 1.66 hr
  after the infusion was stopped.

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Example 5

• 3) Post infusion phase (conc 5.5 hrs after
  stopping the infusion)

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