PK all 2011

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PHARMACOKINETICSPHARMACOKINETICS

John Orton

Royal Oldham Hospital

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What have the following got incommon?

V1V2

k12

k21

kel

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What have the following got incommon?

V1V2

k12

k21

kel’

!

They

re allmo

dels

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Models

Created to predict from limitedinformation what will happen in the realworld

Dependant of data Complexity can be at the cost of

accuracy

Often aims to predict in individualcircumstances from data obtained froma population

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Why bother?

To adjust the dose of drug so that theresponse of the body is as much andfor as long as required

But...Problems measuring drug at biophaseDrug concentration may not correlate with

response

May not be possible to measure responseaccurately

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Qualitative ApproachQualitative ApproachYou can get quite far without much maths

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In (or uptake)

Intravenous Inhalational

main routes for anaesthesia

Intramuscular Oral

Bioavailability

○ Effective dose / Administered dose Transdermal

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Around (or distribution)(Drug factors)

Lipid solubility Water solubility Molecular size pKa

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Around (or distribution)(Tissue factors)

Lipid content○ Brain, CSF

Blood vessel content Highly perfused

○ Lung, kidney, spleen, thyroid, adrenal,brain

Poorly perfused○ Bone, fat

Variable○ Muscle, skin, gut

Tissue binding Tissue barriers

Blood brain barrier Placenta Glomerulus “Leaky Capillaries”

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Around (or distribution)(Blood factors)

Plasma proteins Variable: depends on drug Effect is to slow down rate of diffusion Types

Albumin Acidic drugs Lipo- and GlycoproteinsBasic Drugs α 1 acid glycoprotein

Red blood cells Cardiac output Sequestration

Lungs and fentanyl StomachpH trap Muscle

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Distribution: summary

Depends on: Molecular size

Ionisation Protein binding Tissue perfusion

eg Relaxants Low Protein binding

(<60%) so perfusion

affects distribution alot.

Highly ionised so tendto stay in ECF

BUT dTc undergoestissue binding, soVd is unexpectedlyhigher than ECF

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“Diffusible Fraction”

Consists of: Drug unbound to plasma protein

which is also

Unionised

The more protein bound and ionised adrug is, the more it tends to stay in the

plasma.

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Out (Metabolism, excretion andelimination) Liver

Lipid soluble drugs Kidney

Polar drugs Lung

Vapours and gases

Metabolism

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Out (Liver)

First pass effectLipid soluble drugsOrally administered

HER = (Ca - Cv)/CaHER = 1

○ clearance is flow dependant○ liver has spare capacity

HER approx 0.3○ clearance is flow independant○ liver is working at maximum rate

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A brief diversionA brief diversionNow for some maths

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BASIC MATHS FORBASIC MATHS FOR

PHARMACOKINETIPHARMACOKINETI

CSCS

Its easy – honest!

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10 + 10

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+1 0 1 0

is 1 0 x 2

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1 0 x 1 0

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10 x 10

is102

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10x10 x 10x10 is

102 x 102 is 104 10(2+2)

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103 x 106

is 10 +3 6

109

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t3 x t6

is t +3 6

t9

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tn x tm

is

t +n m

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/10000 100

is

100

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104 / 10 2

is

102

10000 / 100

is

100

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10 -42

is

102

10000 / 100

is

100

104 / 102

is

102

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t -42

is

t2

t4 / t2

is

t2

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Logarithms

Multiply by adding the powers Divide by subtracting the powers

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Logarithms

Any positive number can be expressed( )as a number usually 10 raised to

a power Eg

100 is 102

64 is 10 .1806

2 is 10 .0301

So how do you calculate 2 x 64 using

logarithms?

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Logarithms

Any positive number can be expressed( )as a number usually 10 raised to

a power Eg

100 is 102

64 is 10 .1806

2 is 10 .0301

( )+ ( ) = .log 64 log 2 1 806+0.301 = 2.107 Antilog(2.107)=128 (ie 128 = 102.107 )

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Wierdness

What is 10-1? 103/104 (103-4 ) [10 x 10 x 10] / [10 x 10 x 10 x 10] [10 x 10 x 10] / [10 x 10 x 10 x 10] 1/10 t-1 = 1/t

If t-n = y, 0<y<1 (y is between 0 and 1)

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Wierdness

What is 100.5 ? =√10 t0.5 = √t

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Wierdness

What is 100? =1 t0 = 1

10 x 10 / 10 x 10 = 1 = 102/102=10(2-2)

=100!

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Wierdness

What is 100? =1 t0 = 1

10 x 10 / 10 x 10 = 1 = 102/102=10(2-2)

=100!

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Summary

100 is 102

64 is 101.806

1 is 100

0.1 is 10-1

0.001 is 10-2

Notice you cannot express a negative as

a logarithm 100.5 is √10

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eeWhat is it and where does it come from?

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e

=2.718

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As frequencyapproaches infinity thisbecomes e!(2.718)

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y = 10t

Simplify y = 10t

Use a logarithmic transform log y = t

Strictly speaking log10y = t

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y = et

Simplify using logs to base e instead of base 10, ie loge instead of log10

y = et becomes loge(y) = t

Also written as ln instead of loge ie ln(y)=t

Natural logarithms

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y = ek.t

Logarithmic transform: y = ek.t becomes Ln(y) = k.t

Logarithmic transform:

Ct = C0.e-k.t becomes Ln(Ct) = (Ln C0) - k.t

Ct = C0.e-k.t

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y = mx + c

General eqation for a straight line graph m = slope of line c = intercept of y axis

y

x

m

c

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Logarithmic transform

Lets you convert a complicated equation: Ct = C0e-tk

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Lets you convert a complicated equation: Ct = C0e-tk

Into a linear equation Ln(Ct) = Ln(C0)-t.k

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Into a linear equation Ln(Ct) = Ln(C0) - t . k

= – .y c x m

=y – . +m x c

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pK a

Ka = [H+].[Drug-][H.Drug]

Since

log 1/[H+] = -log[H+] = pHandlog 1/Ka = -logKa = pKa

1 1 [Drug-][H+] Ka [H.Drug].=i.e

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pK a

pH = pKa+log [Drug- / H.Drug] or

pH = pKa+log [Ionised / Unionised]

or pH = pKa+log [Proton acceptor / Proton Donor ]

If Ionised = Unionised

pH = pKa (since log(1)=0)

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pKa

So what? If pKa is near

physiological pHthen patients

condition willinfluence degreeof ionisation andtherefore clinical

effect cf local anaesthetics

% unionised at differentpH

Lignocaine pKa 7.9pH 7.0 11.2pH 7.4 24pH 7.8 44

Bupivacaine pKa 8.1pH 7.0 7pH 7.4 16pH 7.8 33

k h l l l l

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Pk at the molecular levelDiffusion

Law of mass actionFlow along concentration

gradients

Major assumption in Pk models○ no active processes influencing

distribution

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Rt α Ct

Or

Rt α Ct1

That is, elimination is proportional toconcentration raised to the first

power, hence ‘first order kinetics’

So…

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But if….

Rt α Ct0 and Rt = k.Ct

0

Anything raised to the power of zero is 1

So Ct0

=1, therefore Rt α 1 Rt = k.1

So, Rt = k

In other words elimination is constant. Hence ‘zero order kinetics’.

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Clearance, Vd, k, T1/2 and

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Clearance

Rate of elimination (R) (mg.sec-1) Concentration (C) (mg.ml-1)

Law of mass action dictates:Rt α Ct

Rate of elimination at time t is proportional to concentration at time t

(assuming system isn’t saturated)

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Clearance

Rt α Ct

Rt = Ct.k, k=Rt/Ct

k is a constant – what sort of constant? Rt/Ct units are [mg].[sec]-1/ [mg].[ml] -1

Rearranging: [mg].[sec]-1.[mg]-1.[ml]

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Rt α Ct

Rt = Ct.k

k is a constant – what sort of constant? Rt/Ct units are [mg].[sec]-1/ [mg].[ml] -1

Rearranging: [mg].[sec]-1.[mg]-1.[ml]

Cancelling out: ml.sec-1

It’s Clearance!

Clearance

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Volume of Distribution (Vd)

Theoretical; rarely relates to aphysiological volume

After an IV bolus:

Dose (D)Concentration at time zero (C0)VVdd==

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How to find C0CC00

C o n

c e

n t r a t i o

n

C o n c

e n

t ra t i o

n

TimeTime

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Or...

CC00

L o

g

C o

n c

e n t r

a t i o

n

L o

g

C o

n c e

n t r a t

i o n

TimeTime

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Vd and Cl

Rate of drug elimination is constantlychanging

But…

Percentage (or fraction) eliminated in unittime is constant (k)actually it’s

the natural log (ln) of the fraction ln{ C 0 / C t }

where t=1 Units are min-1

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k = Cl / Vd

Another way of looking at k The fraction of Vd cleared of drug in unit

time

i.e. Cl / Vd

k = Clearance expressed as a fraction of Volume of Distribution

k is the Rate Elimination Constant

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k, T1/2 and With k, the denominator is unit time

(minute, second, hour, whatever), thevariable is the fraction

With T1/2 , the denominator is a fraction(which in this case is 1/2), the variableis time

But T1/2 is not 1/k (or not exactly)

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k, T1/2 and

Ct =C0e-k.t (basic decay function)

Ct/C0 =e-k.t

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k, T1/2 and

Ct/C0 =e-k.t

When Ct/C0 =0.5(C has dropped to half original conc)

e-k.t =0.5

Ln{e-k.t } = Ln{0.5} = -k.t

Ln{0.5 } = -0.693 k.t = 0.693, k=0.693 / t Since t=T1/2 , k = 0.693 / T1/2

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k, T1/2 and

τ is the time constant

It is the time taken for to reach zero if rate of fall were constant or the time taken for Ct to fall to 1/e

(instead to 1/2)

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k, T1/2 and

Ct/C0 =e-k.t

When t=τ , Ct/C0 =1/e e-k.t

=1/e

Ln{e-k.t } = -k.t

Ln{1/e } = -1 k.t = 1 or k=1 / t

Since t=τ , k = 1/τ

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k, T1/2 and

Since k = 1/τ And since k = 0.693 / T1/2

0.693 / T1/2 = 1/τ

T1/2 = τ x 0.693

= τ x ln(2)

V k Cl T

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Vd k elCl T1/2

Vd volume of distribution

○ If <5L corresponds to vascular space○ If >15L corresponds more to ECF (or an organ is

concentrating the drug)

kel proportion of drug eliminated in unit time (or more

precisely the logarithm: ln(Ct / Co)

Cl Clearance, ml/second

T1/2 Time taken for drug concentration to change by 1/2 T1/2 = 0.693 / kel

Cl = Vd.ke = Total Dose / AUC

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Different Vd

Vd

VdssSteady state

Vdβ

elimination phase

Infusion Kinetics

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Infusion KineticsHow to calculate rate of infusion to achieve a given blood

concentration

Assume that at steady statein = outi.e. rate of infusion = rate of elimination

Rate of elimination = Clearance x Concentration Concentration you know Clearance you either look up or calculate from

Cl = Vd.k For an infusion you will require Vdss (not ordinary Vd from a

bolus) Calculate this by doing an actual study, infusing at a constant

rate to steady state and measuring blood levels (see nextslide)

Vd

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Vdss

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Exponential Change(decay in this example)

Ct=C0e-kt

Universal process

Where something changes according to its own magnitude

l d l

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Compartmental Modelsand some basic Pk terms

One compartmentRapid bolusRapid mixingOpen (ie drug leaves as 1st order process)

D

Central CompartmentVd

k elCl

Estimation of Pk parameters

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Estimation of Pk parameters

in a one compartment model

0

5

10

15

20

25

0 1 2 3 4 5 6 7 8 9 1 0

1 1

1 2

1 3

1 4

1 5

1 6

1 7

1 8

1 9

2 0

Time

0.001

0.01

0.1

1

10

100

Drug Concentration

Drug Concentration

(Log plot)

Co

Co

Gradient = k el

Ct = Coe-k el

.t

Drug Concentration Ct Drug concentration (Log Axis)

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Two Compartment Model

0.01

2.01

4.01

6.01

8.01

10.01

12.01

14.01

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

Co = Ae-α t

+ Be-β t

V1

V2

k12

k21

k10

Drug Concentration (Log plot:: looks flattish because of log transform)t

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TwoCompartment

ModelLogarithmic Plot

• NB α and β (calledhybrid rate

constants) are notk 12 , k 21 , k 10 in the

model

•They are hybrids of k 12 , k 21 and k 10

0.001

0.01

0.1

1

10

100

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

g ( g p g )

A (Distribution)

B (Elimination)

C- t -β

o = Aeα

+ Bet

A

B

Gradient = βBe-

β t Gradient = α

Ae-

α t

V1V2

k12

k21

kel

Documents

PK all 2011