### Present Remotely

Send the link below via email or IM

CopyPresent to your audience

Start remote presentation- Invited audience members
**will follow you**as you navigate and present - People invited to a presentation
**do not need a Prezi account** - This link expires
**10 minutes**after you close the presentation - A maximum of
**30 users**can follow your presentation - Learn more about this feature in our knowledge base article

Do you really want to delete this prezi?

Neither you, nor the coeditors you shared it with will be able to recover it again.

### Make your likes visible on Facebook?

Connect your Facebook account to Prezi and let your likes appear on your timeline.

You can change this under Settings & Account at any time.

# Pharmacokinetics

No description

by

Tweet## Harendra Patel

on 27 July 2011#### Transcript of Pharmacokinetics

ADME Pharmacokinetics (PK) can be defined as:

'the mathematical analysis of the time courses of the processes of adsorption, distribution, metabolism and excretion of a drug administered to a subject'

PK describes the relationship between drug levels in various regions ('compartments') of the body and time. Absorption Rate of absorption Rate of absorption = P x A x (C1 - C2) where:

P = partition coefficient

A = effective surface area

(C1 - C2) = concentration gradient Is the process of getting the drug from the delivery site to the blood stream. % absorption refers to the amount absorbed compared to the amount compared to the amount administered. There are several routes of absorption from the gastric lumen and factors mitigating against absorption from the gastric lumen. For oral absorption to be effective, the drug MUST be UN-IONISED and IN SOLUTION to be absorbed. Factors which can affect this are:

- the pKa of the drug

- the pH of the gastric site

- the Log P of the drug Distribution Is a description of where in the body the drug goes to after administration and absorption. It is dependent on the drugs physico-chemical characterisitcs (generally logP and pKa) and biological factors. pH of body compartments The pH of different aqueous phase tissue compartments varies greatly. It is therefore important to take this into account. If a drug is ionisable, it may affect the amount partioning into lipid phase and into the next aqueous compartment. Relative size of aqueous compartments Plasma has a pH of 7.4.

In normal acidosis or alkalosis it may range from 7.2 to 7.5.

In severe cases it may range from 6.8 to 7.8. The total amount of body water varies in patients, with the average being 66% in adults (exc. adipose tissue). Total body water can be divided into intracellular and extracellular water. Intracellular : Extracellular water ratio

in adults is 1 : 0.5

in infants is 1 : 1.5 Extracellular water volume in adults is approx 15L.

plasma: extravascular fluid is approx 1 : 4

blood: extravascular fluid is approx 1 : 2 Blood volume in adults is approx 77mL/kg in men and 65mL/kg in women. It is approx 80mL/kg in children. REMEMBER - children are not small adults! Assuming the pH of each aqueous compartment is the same, the conc of the drug in each compartment will also be the same. The total amount of the drug in each compartment will be dependent on the relative size of the compartment. This effect must be considered when looking at the effect on drug action. Relative blood flow to tissues Equilibrium between tissues and plasma will be reached more quickly if blood flow to the tissues is increased, eg kidneys always show a faster rate of equilibration than muscle. Need to consider this effect on drug action. Proportion of body fat Adult women tend to have more body fat than adult men. Lipid soluble drugs will be preferentially absorbed by body fat.

- rate of deposition is dependent on blood flow to adipose tissue

- total deposition is dependent on the quantity of adipose tissue

This effect must be taken into account when looking at drug action as drug doses may need to be adjusted based on body fat. Plasma protein binding Plasma protein content is approx 70g/L consisting of globulins and albumin Globulins

- sparely soluble in water

- soluble in dilute salt solutions

- ppt in conc salt solutions

- approx 25g/L

- important bilogically but not for PK Albumin

- soluble in water

- soluble in dilute salt solutions

- binds to anionic, cationic and non-ionic drugs

- approx 45g/L

High affinity : low capacity - "specific" binding sites

and

low affinity : high capacity - "non-specific" binding sites

Binding is generally a mixture of ionic and hydrophobic interactions. Prescence of the drug may alter conformation of albumin and increase binding. The binding is generally reversible.

Only unbound drug can be distributed to other sites. One drug may also be displaced by another - theraputic interactions.

Disease has a major effect on protein binding and drug activity. Children may show different drug disposition to adults. Important factors in distribution:

- Log P of the drug (see earlier)

- pKa of the drug (see earlier)

- pH of aqueous compartment

- relative size of aqueous compartment (blood, CSF, vitreous humour etc

- relative blood flow to tissue

- proportion of body fat

- plasma protein binding Metabolism Is how the body chemically changes foreign compounds so that they can be more easily removed from the body. See Prof Searcey's lectures Excretion Is how the body removes drugs or metabolites. AKA Elimination Refers to the removal of intact (unchanged) drug through several routes:

Renal - from the plasma to the urine

Faecal - including entero-hepatic recycling

Lungs - eg ethanol

Sweat/milk etc - minor amounts

Metabolites need to be removed from the body and may use one or more of these routes. The kidney function is to maintain blood volume and composition, remove excess electrolytes and water from plasma and remove waste products from plasma. It does this via:

- Glomerular filtration

- Passive tubular reabsorption of water and solutes (into plasma)

- Active tubular secretion of solutes (into urine) Is essentially a passive process. Glomerular filtration rate:

- depends on hydrostatic pressure within the gloerulus

- depends on osmotic pressure of plasma proteins

- approx 700mL/min plasma pass over glomerular surface

- approx 125 to 130mL/min appear as filtrate

- hence GFR = 125 to 130 mL/min

Glomerular filtration of drugs is dependent on:

- the drug molecular weight (size)

- free concentration in plasma Is a active transport process - so requires ATP. One transporter for anions and one for cations. There may be competition between drugs for the same transporter - eg Probenecid and Penicillins. Is the active transport process of solutes - eg glucose and a passive process for water. Approx 99% of the volume of glomerular filtrate will be reabsorbed - final urine production rate is about 1 to 2 mL/min.

It is a passive process for most drugs:

- attempt to balance conc in plasma and urine

- only non-ionised drugs can pass back into the plasma

- dependent on pKa

- dependent on pH of plasma and urine

- dependent on urine flow Simply put, is the removal of non-abosrbed, orally administered material. Passive diffusion across enterocytes back into the gut. Conjugation with glucuronides:

- occurs in liver

- drug complex is ejected via the bile into the gut

- may be excreted via faeces

- complex may be hydrolysed by gastic bacteria and the drug reabsorbed - enterohepatic recycling Major route for volatile anaesthetics is dependent on relative conc in plasma and alveolar air and dependent on blood/gas partition coefficient.

Minor route of excretion for most drugs, even for ethanol. Not deliberate routes of excretion, more a case of the drug just happening to be there. If present in breast milk can be significant for mothers who are breast feeding. Glomerular filtration Tubular secretion Tubular reabsorption Faecal excretion Lung excretion Other routes of excretion IV Drugs PK of an IV bolus injection One compartment model - simplest possible representation of PK

1) drug goes in

2) drug comes out

No complications ---> Drug is injected IV as a bolus solution.

It is then removed passively through the kidneys into the urine for excretion. No metabolism, no protein binding, no sequestration by other tissues. However some major assumptions are made:

- at time = 0, drug is completely and evenly distributed throughout the plasma

- you know the dose of drug (mg)

- you can measure the concentration of drug in plasma (mg/mL) at different times (t)

- you can draw a graph of concentration against time and calculate various important PK parameters. So lets look at a example.

Drug X

- normal healthy adult volunteer weighing 70kg

- given an IV bolus dose of 300mg

- smallest possible volume for the injection

- plasma volume = 3000L Plasma conc vs Time graph looks exponential. To confirm this we need to plot the graph as ln of plasma conc vs time. We can see the graph looks linear. We can then easily derive an equation for the line using y = mx + c.

So for Drug X

y = -0.2 x + 4.6052 so substituting 'real descriptors' for x and y

In (plasma concentration) = -0.2 time + 4.6052 We can also plot the graph as log10 of plasma conc vs time. Intercept = the initial plasama conc Select any large conc (eg 80 micro g/mL). Draw a horizontal line from y = 80 micro g/mL to the curve. Draw a verticle line from the intercept to the x axis and read off the time (t=80) 4.58hrs.

Calculate half the original large conc (eg 40 micro g/mL). Draw a horizontal line from y = 40 micro g/mL to the curve. Draw a verticle line from the intercept to the x axis and read off the time (t=40) 1.12hrs.

The t1/2 of the drug in the plasma is therefore:

t=80 - t=40 = 4.58 - 1.12 = 3.46hrs Use

t1/2 = 0.693 / kelim

t1/2 is dependent ONLY on kelim and NOT the original plasma conc.

(to see how to derive this equation refer to your lecture notes) To resurrect the original exponential equation governing the elimination of drug from the plasma

Use:

Ct = C0 . exp(-t . kelim)

This is a typical first order exponential decay

(to see how to derive this equation refer to your lecture notes) To resurrect the original differential equation governing the elimination of drug from the plasma

Use:

dCt / dt = -kelim . Ct

This is a typical first order differential equation

(to see how to derive this equation refer to your lecture notes) We generally assume that in PK we are dealing with first order reactions. The rate of decrease of conc is directly related to the amount left. So far we have:

- dosed a volunteer with a known amount of drug

- measured plasma levels of the drug over time

- constructed a concentration-time profile (graph)

- calculated the half life of the drug in the plasma

- calculated the elimination constant of the drug from the plasma

- reconstructed the original exponential equation governing the elimination of drug from the plasma

- reconstructed the original differential equation governing the elimination of the drug from the plasma We need to consider some other PK concepts:

- Volume of distribution

- AUC (exposure)

- Clearance This is a theoretical construct that helps in comparing different drugs and in manipulating PK equations. It is defined as:

'the total volume of fluid that the drug would occupy if the total amount of drug in the body was in solution at the same concentration it is in the plasma'

Vd = Total amount in body = Dose

Plasma conc Plasma conc

Units = L or L/kg (varies if the dose is expressed as mg or mg/kg) Think about it like this:

- you dose a patient with an IV bolus of a drug

D = dose, Y= volume of plasma, Cp = conc measured in plasma

If the drug is only in the plasma, then the maximum volume it can occupy is the volume if the plasma.

Total amount of drug in the body = D

Cp = D / Y

Vd = D / Cp = D / (D/Y) = Y

so Vd = Y If the drug diffuses into the adipose tissue, then there is less in the plasma. For example, at equilibrium 50% of the total drug is in the plasma and 50% is in the adipose tissue.

Total amount of drug in the body = D

Cp = 0.5 D / Y

Vd = D / Cp = D / (0.5 D / Y) = Y / 0.5 = 2Y

so Vd = 2Y LOW values of Vd indicate that the drug remains predominantly associated with the vascular system (eg plasma)

INTERMEDIATE values of Vd indicate that the drug is distributed to other tissues

VERY HIGH values of Vd indicate that the drug is tightly bound to very specific tissues

Plasma volume = approx 0.05L/kg To calculate Vd from a conc-time graph

You can use two methods:

- Using the initial plasma conc: Vd = Dose / Initial plasma conc

For Drug X,

Vd = 300mg / 100 micro g/mL = 3L

Vd = 300mg / 100 micro g/mL / 70kg = 0.043L/kg

- Or use AUC (exposure) Ultimately, both patients will clear the drug, but patient B has a greater exposure to the drug as it is a slower process (smaller kelim). Exposure can be measured using AUC - Area Under the conc time Curve. It may be calculated from the graph or by integration of Ct using:

AUC = C0 = Dose

kelim Kelim . Vd

(to see how to derive this equation refer to your lecture notes)

so for Drug X:

AUC = Co = 100 micro g/mL

kelim 0.2 per hour

= 500 micro g . h/mL = 0.5mg . h/mL

NOTE!!! Units of AUC Can be defined as:

a means of comparing how fast two drugs are 'cleared' from the body or how fast two patients 'clear' the same dose of drug.

For patients A and B we know that the same dose was administered, they have different kelim and different AUC (exposure).

Clearance is a way of relating the AUC to the total dose and is defined mathematically as:

CL = Dose

AUC Clearnace can also be defined as:

the volume of plasma from which the drug is completely removed per unit of time.

This then brings in the concepts of kelim and Vd.

- a higher kelim will result in more drug being removed from the plasma per unit time

- a higher Vd will effectively mean that more drug is removed from the plasma

CL = Vd . kelim

This means that there are two formulas for CL. For Drug X, CL = 600mL/h

NOTE!!! Units of CL sometimes include a kg term if dose is expressed per kg So, to summarise, from the plasma conc data we now know:

kelim = 0.2 per hour

t 1/2 = 3.46 hours

Vd = 3L = 0.043 L/kg

AUC = 500 micro g.h/mL = 0.5mg.h/mL

CL = 600mL/hours Log P P = partition coefficient concentration in oil concentration in water solubility in oil solubility in water = Log P = zero ---> equally soluble in both

Log P = -ve ---> more soluble in water

Log P = +ve ---> more soluble in However, Log P calculations ONLY consider un-ionised material. It is more correct to discuss Log Papp and specify pH Papp = conc un-ionised in oil (conc un-ionised in water + conc ionised in water) logPapp = log P - log (1 + 10 (pH - pKa)) log P = logPapp + log (1 + 10 (pH - pKa)) pKa of the pKb of the pKa of the drug The graph looks linear, so we can easily derive an equation of the line using y = mx + c.

So for Drug X

y = -0.0869 x + 2 so substituting 'real descriptors' for x and y

log10 (plasma concentration) = -0.0869 time + 2 The same information is contained in both graphs.

- the ln plot is best for equation manipulation

- the log10 plot is best for simple parameter extraction What is the significance of the intercept and the gradient? Gradient = elimination constant (kelim) For Drug X,

on the ln plot: y = -0.2 x + 4.6052, so we need to anti ln 4.6052, exp(4.6052) = 100

on the log10 plot: y = -0.0869 x + 2, so we need to anti-log10 2, 10 squared = 100

CHECK UNITS !!! - should be a concentration, ie weight per volume eg micro g/mL and expressed as ln or log For Drug X,

on the ln plot: y = -0.2 x + 4.6052, so kelim = 0.2, no conversion is needed.

on the log10 plot: y = -0.0869 x + 2, so we need to apply a conversion factor (ie from log10 to ln) so multiply by 2.303

kelim = -0.0869 x 2.303 = 0.2

CHECK UNITS !!! - should be per time eg per min So now, from the graph we know:

- the rate of removal of the drug from the plasma is exponential

- the initial conc of the drug in plasma

- the elimination rate constant, kelim

We can use this to calculate:

- the conc remaining in the plasma at any time, Ct

- the half life of the drug in the plasma, t1/2 So which plot should we use? log10 of plasma conc vs time or ln of plasma conc vs time To calculate t1/2 from the 'normal' graph To calculate t1/2 from an equation Assume two patients were identically dosed, but plasma conc time graphs were different due to different kelim (Apparent) Volume of distribution, Vd AUC Clearance (CL) Repeated IV bolus dosing In most cases, repeat doses of drugs are required. We need to maintain plasma concentrations between specified limits for maximum therapeutic effect. Generally, the next dose of drug is given while there is still some drug left in the plasma from the previous dose.

Two factors need to be considered

- Repeat dose

- Dose interval

Both are assumed to be constant We are still using the same volunteer and Drug X from the previous lecture.

Plasma volume = 3000mL

Dose = 300mg IV bolus

kelim = 0.2 per hour

t 1/2 = 3.46 hours

Vd = 3L = 0.043 L/kg

AUC = 500 micro g.h/mL = 0.5mg.h/mL

CL = 600mL/hours

Now we are repeating the dose after x hours. We make some major assumptions here. Linear pharmacokinetics.

- repeated doses are additive

- increased dose results in increased plasma levels

- increased plasma levels have NO EFFECT on kelim

- drug is removed from the plasma according to first order kinetics Going back to the plasma conc time graph and its equation, we need to construct a table of plasma levels after x time following a single dose to work out how much of the drug is left using:

Ct = C0 . exp (-t . kelim)

e.g. when t = 2 hours

100 x exp (-2 x 0.2) = 67.03 micro g/mL

For Drug X, with a dose of 300 mg IV bolus, a kelim of 0.2 per hour and a C0 of 100micro g/mL: With repeated doses, we usually have some drug left in the plasma from the earlier doses - this needs to be accounted for. To do this, we need to construct a table of plasma levels immediately following each dose of a repeated series. We assume the dose and dosing interval is fixed and constant. We can then plot a graph of the plasma conc against time using a seperate curve for each individual dose, assuming that each dose is treated seperately is shown in black. The 'total' curve for the overall value is shown in pink. As you can see, the plasma conc reaches a 'plateau' - this occurs when the rate of drug input equals the rate of drug excretion ---> 'steady state'

The graph shows a 'saw tooth' profile of min and max plasma concs. We need a way of making a genearl description of this so that it can be applied for all cases. We need to calculate p - the proportion remaining after a single dose at a specific timepoint.

p = Ct = Co . exp (-t . kelim) = exp (-t . kelim)

C0 C0

so for Drug X:

p = exp (-2 x 0.2) = exp (-0.4) = 0.6703 To predict the maximum plasma concentration, Cmax, that will ever be reached on repeated dosing (same dose, same dose interval) we use:

Cmax = Co

(1 - p)

Cmax is also known as the peak concentration To predict the plasma conc immediately before dosing, (once steady state conditions have been reached) we simply subtract Co from Cmax. This conc is also known as trough conc or the minimum steady state conc.

Ctrough = Cmax - Co

Ctrough = p . Cmax

Peak and trough conc at steady state are important in determining whether the drug is within the therapeeutic window.

so for Drug X:

Co = 100 micro g/mL and p = 0.6703

Cmax = 100 = 303 micro g/mL

(1 - 0.6703)

Ctrough = 100 x 0.6703 = 203 micro g/mL

(1 - 0.6703)

Ctrough = 0.6703 x 303 = 203 micro g/mL

Ctrough = 303 - 100 = 203 micro g/mL We still need to consider the effect of the dose interval. Using our example, but with dose intervals of 2, 4 and 6 hours we get the following graphs. Note the difference in the x and y axis scale. As you can see from the graphs, steady state is reached in fewer doses when the dosing interval is greater and Cmax increases as the dosing interval decreases.

We need to formalise this using maths So now we need to decide on a dosing schedule for a particular drug. How do we go about doing this?

- Need to define standard PK parameters

- Need to define the ideal therapeutic window

---> target plasma level

---> minimum and maximum plasma levels

- Need to balance rate of drug administration with drug elimination

- Need to define dose and dose interval

Going back to our original example for Drug X:

Healthy male volunteer

Plasma volume = 3000mL

Dose = 300mg

kelim = 0.2 per hour

t 1/2 = 3.46 hours

AUC = 500 micro g.h/mL = 0.5 mg.h/mL

C0 = 100 micro g/mL

Vd = 3L = 0.043L/kg

CL = 600mL/hour

New information is now available from clinical studies.

- ideal plasma conc is 175 micro g/mL to 250 micro g/mL

- patients may be taking this drug for a long time

We need to work on steady state conditions. We can then calculate the dosing interval using:

tint = -1 . t 1/2 . ln (Ctrough / Cmax)

0.693

(to see how to derive this equation refer to your lecture notes)

so for Drug X:

tint = -1 x 3.46 x ln (187.5 / 237.5)

0.693

tint = 1.18 hours = 1 hour and 11 minutes

So for Drug X:

Repeated dose = 150mg IV bolus

tint = 1.18 hours = 1 hour 11 minutes

Is this suitable? We need to draw a plasma conc vs time graph As you can see from the graph, it takes too long for the drug to reach the theraputic window. To combat this we use:

A loading dose - give a high initial dose to get high plasma levels

And then a maintenance dose - using the predefined repeated dose We need to decide on the target initial plasma conc (C0 load) and then work out the dose required to give this plasma conc. We also need to remember we have two factors to consider:

- the decrease in plasma levels from the loading dose

- the increase in plasma levels from the maintenance doses

Loading dose kinetics:

Ct = C0 exp(-t . kelim)

Maintenance dose kinetics:

Cn = C0 (1-pn+1)

(1-p)

We cant mathematically combine these as the loading dose kinetics are time-limited and the maintenance dose kinetics are 'infinite' so we must use trial and error.

Simple logic would suggest that the loading dose gives a plasma concentration equal to the maximum desired plasma concentration For Drug X: C0 load = Cmax = 237.5 micro g/mL

Define the dose giving this plasma concentration - assume linear pharmacokinetics as before

An IV bolus dose of 300 mg gave a C0 of 100 micro g/mL, therefore an IV bolus dose of 712.5 mg will give a C0 of 237.5 micro g/mL

Using the Vd equation: Dose = Vd x target plasma conc

IV bolus dose = 3000 mL x 237.5 g/mL = 712.5 mg Using C0,load = Cmax :

- The target plasma concentration was achieved after the loading dose ---> GOOD

- Plasma concentrations greater than maximum permitted levels were observed after approx 3 maintenance doses ---> NOT SO GOOD

- Plasma concentrations will eventually revert to Cmax after the loading dose has been totally eliminated ---> GOOD We can try reducing C0 load, = midpoint of theraputic range = 212 micro g/mL

Dose = 637.5mg Using C0,load = midpoint of therapeutic range :

- Plasma concentrations go below the minimum desired level for significant time periods ---> BAD

- The target plasma concentration was achieved for most of the time ---> GOOD

- Plasma concentrations never go over the maximum permitted level ---> GOOD You need to use your clinical judgement. How sick is the patient?

What is more important? The slightly higher plasma level or the slightly lower levels?

How wide is your margin of error? What effect does this have on the total plasma levels? What effect is their on the plasma levels of Drug X? - the highest possible dose will give a C0 equal to the difference between the maximum and minimum theraputic plasma levels.

- highest possible C0 = Cmax therapeutic level - Cmin therapeutic level

so for Drug X:

Highest possible C0 = 250 - 175 = 75 micro g/mL Dose fluctuations should not really go near the limits of the ideal plasma concentrations. This means that we should use a narrower window, e.g. 187.5 to 237.5 micro g/mL, therefore we will use a lower C0, e.g. 50 micro g/mL.

We can then calculate the dose which gives this C0 - assuming a linear relationship between dose and C0. Two methods are possible depending on what information you have.

1) Use the original IV bolus dose data - A dose of 300mg gave a C0 of 100 micro g/mL, therefore a IV bolus dose of 150mg will give the desired C0 of 50 micro g/mL

2) Use the Vd equation

Vd = Dose / Initial plasma conc - Use the target C0 value and rearrange the equation to give:

Dose = Vd x Target plasma conc

3000mL x 50 micro g/mL = 150 000 micro g = 150mg (IV bolus) IV infusion Continuous IV infusions are used when the t 1/2 life of a drug is very short as repeated doses are troublesome. They are generally given at a constant rate and plasma levels will rise until a plateau is reached. Two factors need to be considered:

- input, ie infusion, ie plasma levels increase

- output, ie kelim, ie plasma levels decrease Theoretically, at any time after starting the infusion, the plasma concentration can be calculated by comparing input and output parameters:

Ct = amount going in at time t - amount going out at time t

We know the amount going in BUT the amount going out is dependent on the plasma conc at that particular time, so straight forward addition is not possible. We need to use rate equations (differential equations) A = the infusion rate. Units in mg / time

CA,rate = rate of increase in plasma conc due to the infusion. Units in micro g/mL/time

The plasma conc at any time t after the start of a continuous IV infusion can be calculated by using:

Ct = CA,rate [1-exp(-t . kelim)]

kelim

To calculate the maximum plasma conc use:

Cmax = CA,rate . t 1/2

0.693

Remember kelim = 0.693 / t 1/2

(to see how to derive this equations refer to your lecture notes) Back to our example using Drug X using a healthy male volunteer:

Plasma volume = 3000mL

Dose = 300 mg IV bolus

Co = 100 micro g/mL

kelim = 0.2 per hour

t 1/2 = 3.46 hours

AUC = 500 micro g.h/mL = 0.5 mg.h/mL

Vd= 3 L = 0.043 L/kg

CL = 600mL/hour

Continuous IV infusion rates of 60mg/hour, 120mg/hour and 180mg/hour Cmax is dependent on infusion rate as expected. Steady state is reached when t = 5 t 1/2 (approx) irrespective of infusion rate. This is the same as for repeat bolus doses. Decide on the ideal Cmax and apply the Cmax equation:

Cmax = CA, rate . t 1/2 ==> CA,rate = 0.693 x Cmax

0.693 t 1/2 so for Drug X:

- ideal plasma conc is 175 micro g/mL to 250 micro g/mL but we use a narrower range: 187.5 micro g/mL to 237.5 micro g/mL.

- select the midpoint (212.5 micro g/mL)

CA,rate = 0.693 x 212.5

3.46

= 42.6 micro g/mL/hour We then need to calculate the infusion rate (mg/hour) that will give this rate of increase in the plasma concentration. We assume:

- a linear relationship between a single dose and the resulting plasma concentration

- this applies with a rate

The original IV bolus dose of 300mg gave a C0 of 100 micro g/mL - assuma a dose rate of 300mg/hour will give a rate of increase of the plasma conc of 100 micro g/mL/hour.

Therefore, a dosing rate of 127.8mg/hour will give a rate of increase of the plasma conc of 42.6 micro g/mL/hour. We can also use a modification of the Vd equation.

Vd = Dose becomes Vd = Dosing rate

C0 CA,rate

Dosing rate = Vd . CA,rate

3000 mL x 42.6 micro g/mL/hour = 127,800 = 127.8mg/hour CA,rate = 42.6 micro g/mL/hour and dosing rate = 127.8mg/hour are both awkward numbers and are therefore prone to error. To combat this we can try rounding up the CA,rate up to 45 micro g/mL/hour and recalculate the dosing rate to 135mg/hour. Recalculate the Cmax to 225 micro g/mL and then plot a plasma conc graph. As you can see from the graph, it takes too long for the drug to reach the therapeutic window. We need to use a IV bolus loading dose. We must decide on the target initial plasma conc, C0,load, and work out the dose required to give this plasma concentration. Remember that we have two factors to consider, the decrease in plasma levels from the loading dose and the increase in plasma levels from the infusion.

Loading dose kinetics:

Ct = C0 exp(-t . kelim)

Infusion kinetics:

Ct = CA,rate [1-exp(-t . kelim)]

kelim Cant mathematically combine these as the loading dose kinetics are time limited and the infusion kinetics are infinite, so we must use trail and error.

Try using a loading dose which will give the maximum therapeutic level, in this case 237.5 micro g/mL, dose = 712.5 mg Success!

- plasma levels never go above the maximum therpeutic level

- plasma levels never go below the minimum therapeutic level Pharmacokinetics Oral dosing Patients generally prefer the oral dosing to other routes. For oral dosing, three factors need to be considered:

- formulation - ie the rate of drug release from the tablet

- absorption (input) - ie plasma levels increase

- output - ie kelim, ie plasma levels decrease

Mathematically this is more complex (unfortunatley) We make assumptions:

- that the drug is 100% in solution and non-ionised at the moment of administration and remains so

- there is 100% absorption

- there is instantaneous movement from GI tract to plasma Back to our example, Drug X. We give a oral dose of 300mg and assume a 100% absorption. We then plot a plasma conc vs time graph. There are two parts to this graph. Firstly, the absorption and elimination phase where plasma levels increase - kabs and kelim. The second is just elimination where plasma levels decease - essentially just kelim. The second part can be used to define kelim. Theoretically, at any time after oral administration, the plasma conc can be calculated by comparing input and output parameters.

Ct = amount going in at time t - amount going out at time t

Complications:

- the amount going in (being absorbed) is dependent on the amount remaining unabsorbed at that particular time

- the amount going out is dependent on the plasma concentration at the particular time

This means straight forward addition is not possible. We need to use rate equations (differential equations) Absorption phase Plot ln of plasma conc vs time. We can then use this to derive the equation of the line for the STRAIGHT bit from a few points after the peak to the last time point. By analogy to the situation with an IV bolus dose the gradient is equal to kelim

Gradient of the line = 0.2 per hour (as expected) 1) Calculate the difference between kelim and the combined kelim and kabs at the beginning of the curve.

2) Back calculate the predicted Ct values before Cmax along the tangent line drawn earlier

3) Subtract values of the actual Ct from the calculated Ct (use real values, not ln values)

4) Calculate ln of the difference

In this case, kabs = 1 per hour Is useful to define the time to maximum plasma conc. This is calculated using:

tmax = ln (kabs/kelim)

(kabs - kelim)

(to see how to derive this equations refer to your lecture notes)

For Drug X:

tmax = ln (1.0/0.2) = 2.0 hours

(1.0-0.2) Uses the terminal linear portion of the ln Ct vs time plot. For Drug X, the intercept = 4.828, but what does this mean?

We are looking at the elimination phase only so assume that absorption is complete. We use this value in the equation for Ct.

Gradient = -kelim

Intercept = ln [ D0 . kabs ]

[ Vd (kabs - kelim ) ]

Do = amount of drug administered orally at time 0

(to see how to derive this equations refer to your lecture notes)

The intercept can be used to derive Vd if kabs and kelim are known but assumes 100% absorption

For Drug X:

Intercept = 4.828

4.828 = ln [ 300 x 1.0 ] = ln (375/Vd)

[ Vd x (1.0 - 0.2) ]

exp (4.828) = 375/Vd

Vd = 375/exp(4.828 = 375/125 = 3L Calculate tmax and substitute into the Ct equation to calculate Cmax at tmax using:

Cmax = D0 . kabs . [exp(-tmax . kelim) - exp (-tmax . kabs) ]

Vd (kabs - kelim) For Drug X:

Cmax = 300 x 1.0 x [exp(-2.0 x 0.2) - exp(-2.0 x 1.0)

3 x (1.0 - 0.2)

Cmax = 66.875 micro g/mL To calculate AUC use:

AUC = D0

Vd . kelim

This assumes 100% absorption

(to see how to derive this equations refer to your lecture notes)

For Drug X:

AUC = 300 / 3 x 0.2 = 500 micro g.h/mL = 0.5 mg.h/mL Remember, there are two equations for CL:

CL = Dose / AUC & CL = Vd . kelim

For Drug X:

CL = 300 / 0.5 = 600 mL/h & CL = 3 x 0.2 = 600mL/h So in summary we have found:

kelim = 0.2 per hour

kabs = 1.0 per hour

Vd = 3L

tmax = 2 hours

Cmax = 66.875 micro g/mL

AUC = 500 micro g.h/mL = 0.5 mg.h/mL

CL = 600 mL/h This is very simple to cope with. We define the fraction absorbed as f. This can range from 0.0 to 1.0. We apply f to the equations pertaining to oral absorption. Plasma concentration equation becomes:

Ct = f . D0 . kabs . [exp(-t . kelim) - exp(-t . kabs)]

Vd (kabs - kelim) Applies also to Cmax:

Cmax = f. D0 . kabs . [exp(-tmax . kelim) - exp(-tmax . kabs)]

Vd (kabs - kelim) Intercept on the lnCt vs time graph becomes:

Intercept = ln [ f . D0 . kabs ]

Vd (kabs - kelim) AUC equation equation becomes:

AUC = f . Do / Vd . kelim We can derive kelim, kabs and tmax as before. For all other parameters, we need to derive f first by running a comparison to an IV bolus dose. We need to compare the AUC for oral and IV bolus doses:

AUC oral = f . D0 / Vd . kelim

AUC IV bolus = C0 / kelim

Combining gives = D0 / Vd . kelim We then need to calculate the ratio of AUC values for oral and IV bolus doses

f = AUC (oral) / AUC (IV bolus)

Ideally f should be as close as possible to 1.0 kelim kabs tmax Vd Cmax AUC CL But what about incomplete absorption? Renal metabolism Up until now, we have assumed that our drug is exclusively renally excreted, ie:

kelim = kelim,renal & CL = CLrenal

However, many drugs are partially metabolised and partially renally excreted, ie:

kelim = kelim,renal + kelim,metab

CL = CLrenal + CLmetab Assuming first order kinetics, the rate of appearance of drug in the urine is dependent on the concentration of drug in the plasma. This means we cant use urine concentration. We need to use total amount of drug in urine - Durine.t and so also need to refer to total amount of drug in the body - Dbody.t

Using:

ln [dDurine.t] = ln [kelim,renal . Dose] -t . kelim

dT

(to see how to derive this equations refer to your lecture notes)

We can graw a graph of ln (dDurine,t / dt) against time which has:

- a gradient of -kelim & a intercept of ln [kelim,renal . Dose] Therefore, we can estimate kelim,renal if Dose and kelim are known, but we need urine data. This is collected intermittently as it cant be collected constantly. We use the midpoint of collection period for calculations.

For Drug X:

- IV bolus of 300mg

- Measure plasma levels and urine colections

- calculate kelim and kelim,renal We first take plasma and urine samples at various timepoints We can then calculate the rate of appearance of drug in urine using:

rate = amount of drug in collection interval / time of interval

Then we can plot ln rate against mid point of collection interval The equation of the line: y = -0.2 x + 4.1

We know that the gradient = kelim, so kelim = 0.2 per hour as expected.

We know that the intercept = ln [kelim.renal . Dose] so,

ln [kelim.renal . Dose] = 4.1

kelim.renal . Dose = exp(4.1) = 60.34

kelim.renal = 60.34/300 = 0.2 per hour

As kelim.renal = kelim, we can say that Drug X is exclusively excreted renally. Now lets look at Drug Q.

Drug Q is believed to be partially metabolised as well as being partially excreted renally. Drug Q was administered as an IV bolus of 90mg. Plasma and urine samples were taken at various time points. We then need to calculate the rate of appearance of drug in urine as before and then plot a ln graph as before. The equation of the line from the plasma conc graph was: y = -0.3 x +3.40, so kelim = 0.3 per hour.

The equation of the line from the urinary rate graph was: y = -0.3 x +2.6, so kelim = 0.3 per hour, as expected.

ln [kelim.renal . Dose] = 2.6

kelim.renal . Dose = exp(2.6) = 13.46

kelim.reanl = 13.46/90 = 0.15 per hour So for Drug Q:

kelim = 0.3 per hour

kelim.renal = 0.15 per hour

The difference between the two is the non-renal excretion assumed to be metabolism, ie:

kelim = kelim.renal + kelim.metab

kelim.metab = kelim - kelim.renal

= 0.3 - 0.15 = 0.15 per hour. Removal via kidneys occurs in three ways:

- glomerular filtration

- active tubular secretion

- passive tubular reabsorption

Total removal rate = Glomerular filtration rate + Secretion rate - Reabsorption rate We need a method for quantifying renal parameters.

Glomerular filtration rate (GFR) - measures the clearance of creatinine or inulin. Essentially eliminated by glomerular filtration only, so CLcreatinine and CLinsulin = GFR

Effective renal plasma flow (ERPF) - measures the clearance of p-amino-hipppuric acid (PAH). Eliminated by both glomerular filtration and active secretion. Essentially all PAH is removed from the plasma in one pass so removal is dependent on plasma, so CLpah = ERPF After viewing this Prezi, you will be experts in the field of pharmacokinetics Its Dr Barker here! Welcome to the PK Prezi. Hi, Here are a few simple instructions on how to use and navigate around the Prezi. (to see how these equations were derived, refer to your lecture notes) Renal excretion AR = Cmax/C0

AR = Accumulation Ratio Steady state is reached when the elapsed time is approx 5 t1/2 irrespective of the dosing interval.

AR is dependent on dosing interval irrespective of dose. Ctint = effect of the dose interval assuming the dose is identical

Ctint = C0 exp (-tint . kelim) ie AR is related to the ratio tint / t1/2

AR rises very sharply if tint <t/2

AR is negligible if tint/t1/2 >5 But how do we choose a loading dose? Elimination phase The equations are then solved using complicated maths to give: We can then extract the following parameters graphically or by calculation:

1) kelim, 2) kabs, 3) Tmax, 4) Vd, 5) Cmax, 6) AUC, 7) CL

In that order. We can estimate the method of renal excretion of the drug by comparing drug clearance values to that of creatinine:

Cldrug / CLcreatinine = 1 ---> elimination is by glomerular filtration only

CLdrug / CLcreatinine > 1 ---> elimination is by glomerular filtration and active tubular secretion (may also have some tubular reabsorption)

CLdrug / CLcreatinine < 1 ---) elimination is by glomerular filtration with some reabsorption (may also have some active tubular secretion) Liver disease affects metabolism so affect kelim,metab. Affects drugs which have significant metabolism (high excretion ratio)

CL = Vd . kelim and CLmetab = Vd . kelim

We assume that Vd stays constant in liver disease. It is difficult to predict the exact effect of liver disease on drug metabolism as the liver funstion tests do not adeduately predict metabolis function.

Generally:

- choose drugs which are predominantly renally excreted

- choose drugs with a wide therapeutic window

- use with caution and monitor the patients response Kidney disease affects renal elimination and drugs which have significant renal excretion. We need to assess the effect on kelim,renal.

CL = Vd . kelim and CLrenal = Vd . kelim,renal

We assume Vd stays constant in renal disease. We need to adjust the dose given but we first need to asses the extent of kidney damage by looking at creatinine (inulin) clearance and PAH clearance Creatinine is slightly different for men and women and dependent on muscle mass. It can be predicted for average normal people of all ages using the Cockcroft and Gault equation: So how do we adjust the dose in kidney disease? We assume that renal clearance of the drug is proportional to creatinine clearance and that this ratio is the same in both normal individuals and patients with impaired kidney function.

e.g. if CLcreatinine = 50% of normal, then CLdrug = 50% of normal

CLdrug,renal,patient = CLdrug,renal,normal x CLcreatinine,patient / CLcreatinine,normal So, if the drug is 100% renally excreted we can use:

Dose,patient = Dose,normal x CLcreatinine,patient / CLcreatinine,normal

Remember that this assusmes 100% renal excretion

(to see how this equation was derived, refer to your lecture notes) If the drug is only PARTIALLY renally cleared, we only need to adjust for that portion. The dose of drug in a normal individual can be considered as:

R = the proportion of the drug normally renally eliminated (can take values of 0.0 to 1.0)

The equation then becomes: So for Drug X, we know that it is entirely renally excreted. We need to calculate the IV bolus dose required for patient H, if the normal dose is 300mg as an IV bolus dose, and patient H has a creatinine clearance 30% that of a normal matched patient After viewing this Prezi, you will be able to use graphs and equations to calculate pharmacokinetic parameters. This Prezi will follow a path around the presentation. Use the scroll wheel on your mouse to zoom in and out of the Prezi. Alternatively, use the magnifying glass on the right hand side of the screen. This can be used to view images and smaller text. If you get lost in the Prezi or want to view the whole Prezi, click on the Home button on the right hand side of the screen to zoom out and view the whole Prezi. Feel free to move around the Prezi to focus on the areas that you are interested in or are having difficulty with. Lets get started! Whilst following the path, click and hold the left mouse button to move around anywhere in the Prezi. When you are ready to move on, click or press the right arrow to continue on the path. Click & Hold Zoom Zoom in to view in more detail Good luck for the exam! You should now be able to use graphs and equations to calculate pharmacokinetic parameters. There are also refresher lectures available on Blackboard if you are having difficulty with any of the maths used in this module. If you have any questions regarding pharmacokinetics, please feel free to email me.

Full transcript'the mathematical analysis of the time courses of the processes of adsorption, distribution, metabolism and excretion of a drug administered to a subject'

PK describes the relationship between drug levels in various regions ('compartments') of the body and time. Absorption Rate of absorption Rate of absorption = P x A x (C1 - C2) where:

P = partition coefficient

A = effective surface area

(C1 - C2) = concentration gradient Is the process of getting the drug from the delivery site to the blood stream. % absorption refers to the amount absorbed compared to the amount compared to the amount administered. There are several routes of absorption from the gastric lumen and factors mitigating against absorption from the gastric lumen. For oral absorption to be effective, the drug MUST be UN-IONISED and IN SOLUTION to be absorbed. Factors which can affect this are:

- the pKa of the drug

- the pH of the gastric site

- the Log P of the drug Distribution Is a description of where in the body the drug goes to after administration and absorption. It is dependent on the drugs physico-chemical characterisitcs (generally logP and pKa) and biological factors. pH of body compartments The pH of different aqueous phase tissue compartments varies greatly. It is therefore important to take this into account. If a drug is ionisable, it may affect the amount partioning into lipid phase and into the next aqueous compartment. Relative size of aqueous compartments Plasma has a pH of 7.4.

In normal acidosis or alkalosis it may range from 7.2 to 7.5.

In severe cases it may range from 6.8 to 7.8. The total amount of body water varies in patients, with the average being 66% in adults (exc. adipose tissue). Total body water can be divided into intracellular and extracellular water. Intracellular : Extracellular water ratio

in adults is 1 : 0.5

in infants is 1 : 1.5 Extracellular water volume in adults is approx 15L.

plasma: extravascular fluid is approx 1 : 4

blood: extravascular fluid is approx 1 : 2 Blood volume in adults is approx 77mL/kg in men and 65mL/kg in women. It is approx 80mL/kg in children. REMEMBER - children are not small adults! Assuming the pH of each aqueous compartment is the same, the conc of the drug in each compartment will also be the same. The total amount of the drug in each compartment will be dependent on the relative size of the compartment. This effect must be considered when looking at the effect on drug action. Relative blood flow to tissues Equilibrium between tissues and plasma will be reached more quickly if blood flow to the tissues is increased, eg kidneys always show a faster rate of equilibration than muscle. Need to consider this effect on drug action. Proportion of body fat Adult women tend to have more body fat than adult men. Lipid soluble drugs will be preferentially absorbed by body fat.

- rate of deposition is dependent on blood flow to adipose tissue

- total deposition is dependent on the quantity of adipose tissue

This effect must be taken into account when looking at drug action as drug doses may need to be adjusted based on body fat. Plasma protein binding Plasma protein content is approx 70g/L consisting of globulins and albumin Globulins

- sparely soluble in water

- soluble in dilute salt solutions

- ppt in conc salt solutions

- approx 25g/L

- important bilogically but not for PK Albumin

- soluble in water

- soluble in dilute salt solutions

- binds to anionic, cationic and non-ionic drugs

- approx 45g/L

High affinity : low capacity - "specific" binding sites

and

low affinity : high capacity - "non-specific" binding sites

Binding is generally a mixture of ionic and hydrophobic interactions. Prescence of the drug may alter conformation of albumin and increase binding. The binding is generally reversible.

Only unbound drug can be distributed to other sites. One drug may also be displaced by another - theraputic interactions.

Disease has a major effect on protein binding and drug activity. Children may show different drug disposition to adults. Important factors in distribution:

- Log P of the drug (see earlier)

- pKa of the drug (see earlier)

- pH of aqueous compartment

- relative size of aqueous compartment (blood, CSF, vitreous humour etc

- relative blood flow to tissue

- proportion of body fat

- plasma protein binding Metabolism Is how the body chemically changes foreign compounds so that they can be more easily removed from the body. See Prof Searcey's lectures Excretion Is how the body removes drugs or metabolites. AKA Elimination Refers to the removal of intact (unchanged) drug through several routes:

Renal - from the plasma to the urine

Faecal - including entero-hepatic recycling

Lungs - eg ethanol

Sweat/milk etc - minor amounts

Metabolites need to be removed from the body and may use one or more of these routes. The kidney function is to maintain blood volume and composition, remove excess electrolytes and water from plasma and remove waste products from plasma. It does this via:

- Glomerular filtration

- Passive tubular reabsorption of water and solutes (into plasma)

- Active tubular secretion of solutes (into urine) Is essentially a passive process. Glomerular filtration rate:

- depends on hydrostatic pressure within the gloerulus

- depends on osmotic pressure of plasma proteins

- approx 700mL/min plasma pass over glomerular surface

- approx 125 to 130mL/min appear as filtrate

- hence GFR = 125 to 130 mL/min

Glomerular filtration of drugs is dependent on:

- the drug molecular weight (size)

- free concentration in plasma Is a active transport process - so requires ATP. One transporter for anions and one for cations. There may be competition between drugs for the same transporter - eg Probenecid and Penicillins. Is the active transport process of solutes - eg glucose and a passive process for water. Approx 99% of the volume of glomerular filtrate will be reabsorbed - final urine production rate is about 1 to 2 mL/min.

It is a passive process for most drugs:

- attempt to balance conc in plasma and urine

- only non-ionised drugs can pass back into the plasma

- dependent on pKa

- dependent on pH of plasma and urine

- dependent on urine flow Simply put, is the removal of non-abosrbed, orally administered material. Passive diffusion across enterocytes back into the gut. Conjugation with glucuronides:

- occurs in liver

- drug complex is ejected via the bile into the gut

- may be excreted via faeces

- complex may be hydrolysed by gastic bacteria and the drug reabsorbed - enterohepatic recycling Major route for volatile anaesthetics is dependent on relative conc in plasma and alveolar air and dependent on blood/gas partition coefficient.

Minor route of excretion for most drugs, even for ethanol. Not deliberate routes of excretion, more a case of the drug just happening to be there. If present in breast milk can be significant for mothers who are breast feeding. Glomerular filtration Tubular secretion Tubular reabsorption Faecal excretion Lung excretion Other routes of excretion IV Drugs PK of an IV bolus injection One compartment model - simplest possible representation of PK

1) drug goes in

2) drug comes out

No complications ---> Drug is injected IV as a bolus solution.

It is then removed passively through the kidneys into the urine for excretion. No metabolism, no protein binding, no sequestration by other tissues. However some major assumptions are made:

- at time = 0, drug is completely and evenly distributed throughout the plasma

- you know the dose of drug (mg)

- you can measure the concentration of drug in plasma (mg/mL) at different times (t)

- you can draw a graph of concentration against time and calculate various important PK parameters. So lets look at a example.

Drug X

- normal healthy adult volunteer weighing 70kg

- given an IV bolus dose of 300mg

- smallest possible volume for the injection

- plasma volume = 3000L Plasma conc vs Time graph looks exponential. To confirm this we need to plot the graph as ln of plasma conc vs time. We can see the graph looks linear. We can then easily derive an equation for the line using y = mx + c.

So for Drug X

y = -0.2 x + 4.6052 so substituting 'real descriptors' for x and y

In (plasma concentration) = -0.2 time + 4.6052 We can also plot the graph as log10 of plasma conc vs time. Intercept = the initial plasama conc Select any large conc (eg 80 micro g/mL). Draw a horizontal line from y = 80 micro g/mL to the curve. Draw a verticle line from the intercept to the x axis and read off the time (t=80) 4.58hrs.

Calculate half the original large conc (eg 40 micro g/mL). Draw a horizontal line from y = 40 micro g/mL to the curve. Draw a verticle line from the intercept to the x axis and read off the time (t=40) 1.12hrs.

The t1/2 of the drug in the plasma is therefore:

t=80 - t=40 = 4.58 - 1.12 = 3.46hrs Use

t1/2 = 0.693 / kelim

t1/2 is dependent ONLY on kelim and NOT the original plasma conc.

(to see how to derive this equation refer to your lecture notes) To resurrect the original exponential equation governing the elimination of drug from the plasma

Use:

Ct = C0 . exp(-t . kelim)

This is a typical first order exponential decay

(to see how to derive this equation refer to your lecture notes) To resurrect the original differential equation governing the elimination of drug from the plasma

Use:

dCt / dt = -kelim . Ct

This is a typical first order differential equation

(to see how to derive this equation refer to your lecture notes) We generally assume that in PK we are dealing with first order reactions. The rate of decrease of conc is directly related to the amount left. So far we have:

- dosed a volunteer with a known amount of drug

- measured plasma levels of the drug over time

- constructed a concentration-time profile (graph)

- calculated the half life of the drug in the plasma

- calculated the elimination constant of the drug from the plasma

- reconstructed the original exponential equation governing the elimination of drug from the plasma

- reconstructed the original differential equation governing the elimination of the drug from the plasma We need to consider some other PK concepts:

- Volume of distribution

- AUC (exposure)

- Clearance This is a theoretical construct that helps in comparing different drugs and in manipulating PK equations. It is defined as:

'the total volume of fluid that the drug would occupy if the total amount of drug in the body was in solution at the same concentration it is in the plasma'

Vd = Total amount in body = Dose

Plasma conc Plasma conc

Units = L or L/kg (varies if the dose is expressed as mg or mg/kg) Think about it like this:

- you dose a patient with an IV bolus of a drug

D = dose, Y= volume of plasma, Cp = conc measured in plasma

If the drug is only in the plasma, then the maximum volume it can occupy is the volume if the plasma.

Total amount of drug in the body = D

Cp = D / Y

Vd = D / Cp = D / (D/Y) = Y

so Vd = Y If the drug diffuses into the adipose tissue, then there is less in the plasma. For example, at equilibrium 50% of the total drug is in the plasma and 50% is in the adipose tissue.

Total amount of drug in the body = D

Cp = 0.5 D / Y

Vd = D / Cp = D / (0.5 D / Y) = Y / 0.5 = 2Y

so Vd = 2Y LOW values of Vd indicate that the drug remains predominantly associated with the vascular system (eg plasma)

INTERMEDIATE values of Vd indicate that the drug is distributed to other tissues

VERY HIGH values of Vd indicate that the drug is tightly bound to very specific tissues

Plasma volume = approx 0.05L/kg To calculate Vd from a conc-time graph

You can use two methods:

- Using the initial plasma conc: Vd = Dose / Initial plasma conc

For Drug X,

Vd = 300mg / 100 micro g/mL = 3L

Vd = 300mg / 100 micro g/mL / 70kg = 0.043L/kg

- Or use AUC (exposure) Ultimately, both patients will clear the drug, but patient B has a greater exposure to the drug as it is a slower process (smaller kelim). Exposure can be measured using AUC - Area Under the conc time Curve. It may be calculated from the graph or by integration of Ct using:

AUC = C0 = Dose

kelim Kelim . Vd

(to see how to derive this equation refer to your lecture notes)

so for Drug X:

AUC = Co = 100 micro g/mL

kelim 0.2 per hour

= 500 micro g . h/mL = 0.5mg . h/mL

NOTE!!! Units of AUC Can be defined as:

a means of comparing how fast two drugs are 'cleared' from the body or how fast two patients 'clear' the same dose of drug.

For patients A and B we know that the same dose was administered, they have different kelim and different AUC (exposure).

Clearance is a way of relating the AUC to the total dose and is defined mathematically as:

CL = Dose

AUC Clearnace can also be defined as:

the volume of plasma from which the drug is completely removed per unit of time.

This then brings in the concepts of kelim and Vd.

- a higher kelim will result in more drug being removed from the plasma per unit time

- a higher Vd will effectively mean that more drug is removed from the plasma

CL = Vd . kelim

This means that there are two formulas for CL. For Drug X, CL = 600mL/h

NOTE!!! Units of CL sometimes include a kg term if dose is expressed per kg So, to summarise, from the plasma conc data we now know:

kelim = 0.2 per hour

t 1/2 = 3.46 hours

Vd = 3L = 0.043 L/kg

AUC = 500 micro g.h/mL = 0.5mg.h/mL

CL = 600mL/hours Log P P = partition coefficient concentration in oil concentration in water solubility in oil solubility in water = Log P = zero ---> equally soluble in both

Log P = -ve ---> more soluble in water

Log P = +ve ---> more soluble in However, Log P calculations ONLY consider un-ionised material. It is more correct to discuss Log Papp and specify pH Papp = conc un-ionised in oil (conc un-ionised in water + conc ionised in water) logPapp = log P - log (1 + 10 (pH - pKa)) log P = logPapp + log (1 + 10 (pH - pKa)) pKa of the pKb of the pKa of the drug The graph looks linear, so we can easily derive an equation of the line using y = mx + c.

So for Drug X

y = -0.0869 x + 2 so substituting 'real descriptors' for x and y

log10 (plasma concentration) = -0.0869 time + 2 The same information is contained in both graphs.

- the ln plot is best for equation manipulation

- the log10 plot is best for simple parameter extraction What is the significance of the intercept and the gradient? Gradient = elimination constant (kelim) For Drug X,

on the ln plot: y = -0.2 x + 4.6052, so we need to anti ln 4.6052, exp(4.6052) = 100

on the log10 plot: y = -0.0869 x + 2, so we need to anti-log10 2, 10 squared = 100

CHECK UNITS !!! - should be a concentration, ie weight per volume eg micro g/mL and expressed as ln or log For Drug X,

on the ln plot: y = -0.2 x + 4.6052, so kelim = 0.2, no conversion is needed.

on the log10 plot: y = -0.0869 x + 2, so we need to apply a conversion factor (ie from log10 to ln) so multiply by 2.303

kelim = -0.0869 x 2.303 = 0.2

CHECK UNITS !!! - should be per time eg per min So now, from the graph we know:

- the rate of removal of the drug from the plasma is exponential

- the initial conc of the drug in plasma

- the elimination rate constant, kelim

We can use this to calculate:

- the conc remaining in the plasma at any time, Ct

- the half life of the drug in the plasma, t1/2 So which plot should we use? log10 of plasma conc vs time or ln of plasma conc vs time To calculate t1/2 from the 'normal' graph To calculate t1/2 from an equation Assume two patients were identically dosed, but plasma conc time graphs were different due to different kelim (Apparent) Volume of distribution, Vd AUC Clearance (CL) Repeated IV bolus dosing In most cases, repeat doses of drugs are required. We need to maintain plasma concentrations between specified limits for maximum therapeutic effect. Generally, the next dose of drug is given while there is still some drug left in the plasma from the previous dose.

Two factors need to be considered

- Repeat dose

- Dose interval

Both are assumed to be constant We are still using the same volunteer and Drug X from the previous lecture.

Plasma volume = 3000mL

Dose = 300mg IV bolus

kelim = 0.2 per hour

t 1/2 = 3.46 hours

Vd = 3L = 0.043 L/kg

AUC = 500 micro g.h/mL = 0.5mg.h/mL

CL = 600mL/hours

Now we are repeating the dose after x hours. We make some major assumptions here. Linear pharmacokinetics.

- repeated doses are additive

- increased dose results in increased plasma levels

- increased plasma levels have NO EFFECT on kelim

- drug is removed from the plasma according to first order kinetics Going back to the plasma conc time graph and its equation, we need to construct a table of plasma levels after x time following a single dose to work out how much of the drug is left using:

Ct = C0 . exp (-t . kelim)

e.g. when t = 2 hours

100 x exp (-2 x 0.2) = 67.03 micro g/mL

For Drug X, with a dose of 300 mg IV bolus, a kelim of 0.2 per hour and a C0 of 100micro g/mL: With repeated doses, we usually have some drug left in the plasma from the earlier doses - this needs to be accounted for. To do this, we need to construct a table of plasma levels immediately following each dose of a repeated series. We assume the dose and dosing interval is fixed and constant. We can then plot a graph of the plasma conc against time using a seperate curve for each individual dose, assuming that each dose is treated seperately is shown in black. The 'total' curve for the overall value is shown in pink. As you can see, the plasma conc reaches a 'plateau' - this occurs when the rate of drug input equals the rate of drug excretion ---> 'steady state'

The graph shows a 'saw tooth' profile of min and max plasma concs. We need a way of making a genearl description of this so that it can be applied for all cases. We need to calculate p - the proportion remaining after a single dose at a specific timepoint.

p = Ct = Co . exp (-t . kelim) = exp (-t . kelim)

C0 C0

so for Drug X:

p = exp (-2 x 0.2) = exp (-0.4) = 0.6703 To predict the maximum plasma concentration, Cmax, that will ever be reached on repeated dosing (same dose, same dose interval) we use:

Cmax = Co

(1 - p)

Cmax is also known as the peak concentration To predict the plasma conc immediately before dosing, (once steady state conditions have been reached) we simply subtract Co from Cmax. This conc is also known as trough conc or the minimum steady state conc.

Ctrough = Cmax - Co

Ctrough = p . Cmax

Peak and trough conc at steady state are important in determining whether the drug is within the therapeeutic window.

so for Drug X:

Co = 100 micro g/mL and p = 0.6703

Cmax = 100 = 303 micro g/mL

(1 - 0.6703)

Ctrough = 100 x 0.6703 = 203 micro g/mL

(1 - 0.6703)

Ctrough = 0.6703 x 303 = 203 micro g/mL

Ctrough = 303 - 100 = 203 micro g/mL We still need to consider the effect of the dose interval. Using our example, but with dose intervals of 2, 4 and 6 hours we get the following graphs. Note the difference in the x and y axis scale. As you can see from the graphs, steady state is reached in fewer doses when the dosing interval is greater and Cmax increases as the dosing interval decreases.

We need to formalise this using maths So now we need to decide on a dosing schedule for a particular drug. How do we go about doing this?

- Need to define standard PK parameters

- Need to define the ideal therapeutic window

---> target plasma level

---> minimum and maximum plasma levels

- Need to balance rate of drug administration with drug elimination

- Need to define dose and dose interval

Going back to our original example for Drug X:

Healthy male volunteer

Plasma volume = 3000mL

Dose = 300mg

kelim = 0.2 per hour

t 1/2 = 3.46 hours

AUC = 500 micro g.h/mL = 0.5 mg.h/mL

C0 = 100 micro g/mL

Vd = 3L = 0.043L/kg

CL = 600mL/hour

New information is now available from clinical studies.

- ideal plasma conc is 175 micro g/mL to 250 micro g/mL

- patients may be taking this drug for a long time

We need to work on steady state conditions. We can then calculate the dosing interval using:

tint = -1 . t 1/2 . ln (Ctrough / Cmax)

0.693

(to see how to derive this equation refer to your lecture notes)

so for Drug X:

tint = -1 x 3.46 x ln (187.5 / 237.5)

0.693

tint = 1.18 hours = 1 hour and 11 minutes

So for Drug X:

Repeated dose = 150mg IV bolus

tint = 1.18 hours = 1 hour 11 minutes

Is this suitable? We need to draw a plasma conc vs time graph As you can see from the graph, it takes too long for the drug to reach the theraputic window. To combat this we use:

A loading dose - give a high initial dose to get high plasma levels

And then a maintenance dose - using the predefined repeated dose We need to decide on the target initial plasma conc (C0 load) and then work out the dose required to give this plasma conc. We also need to remember we have two factors to consider:

- the decrease in plasma levels from the loading dose

- the increase in plasma levels from the maintenance doses

Loading dose kinetics:

Ct = C0 exp(-t . kelim)

Maintenance dose kinetics:

Cn = C0 (1-pn+1)

(1-p)

We cant mathematically combine these as the loading dose kinetics are time-limited and the maintenance dose kinetics are 'infinite' so we must use trial and error.

Simple logic would suggest that the loading dose gives a plasma concentration equal to the maximum desired plasma concentration For Drug X: C0 load = Cmax = 237.5 micro g/mL

Define the dose giving this plasma concentration - assume linear pharmacokinetics as before

An IV bolus dose of 300 mg gave a C0 of 100 micro g/mL, therefore an IV bolus dose of 712.5 mg will give a C0 of 237.5 micro g/mL

Using the Vd equation: Dose = Vd x target plasma conc

IV bolus dose = 3000 mL x 237.5 g/mL = 712.5 mg Using C0,load = Cmax :

- The target plasma concentration was achieved after the loading dose ---> GOOD

- Plasma concentrations greater than maximum permitted levels were observed after approx 3 maintenance doses ---> NOT SO GOOD

- Plasma concentrations will eventually revert to Cmax after the loading dose has been totally eliminated ---> GOOD We can try reducing C0 load, = midpoint of theraputic range = 212 micro g/mL

Dose = 637.5mg Using C0,load = midpoint of therapeutic range :

- Plasma concentrations go below the minimum desired level for significant time periods ---> BAD

- The target plasma concentration was achieved for most of the time ---> GOOD

- Plasma concentrations never go over the maximum permitted level ---> GOOD You need to use your clinical judgement. How sick is the patient?

What is more important? The slightly higher plasma level or the slightly lower levels?

How wide is your margin of error? What effect does this have on the total plasma levels? What effect is their on the plasma levels of Drug X? - the highest possible dose will give a C0 equal to the difference between the maximum and minimum theraputic plasma levels.

- highest possible C0 = Cmax therapeutic level - Cmin therapeutic level

so for Drug X:

Highest possible C0 = 250 - 175 = 75 micro g/mL Dose fluctuations should not really go near the limits of the ideal plasma concentrations. This means that we should use a narrower window, e.g. 187.5 to 237.5 micro g/mL, therefore we will use a lower C0, e.g. 50 micro g/mL.

We can then calculate the dose which gives this C0 - assuming a linear relationship between dose and C0. Two methods are possible depending on what information you have.

1) Use the original IV bolus dose data - A dose of 300mg gave a C0 of 100 micro g/mL, therefore a IV bolus dose of 150mg will give the desired C0 of 50 micro g/mL

2) Use the Vd equation

Vd = Dose / Initial plasma conc - Use the target C0 value and rearrange the equation to give:

Dose = Vd x Target plasma conc

3000mL x 50 micro g/mL = 150 000 micro g = 150mg (IV bolus) IV infusion Continuous IV infusions are used when the t 1/2 life of a drug is very short as repeated doses are troublesome. They are generally given at a constant rate and plasma levels will rise until a plateau is reached. Two factors need to be considered:

- input, ie infusion, ie plasma levels increase

- output, ie kelim, ie plasma levels decrease Theoretically, at any time after starting the infusion, the plasma concentration can be calculated by comparing input and output parameters:

Ct = amount going in at time t - amount going out at time t

We know the amount going in BUT the amount going out is dependent on the plasma conc at that particular time, so straight forward addition is not possible. We need to use rate equations (differential equations) A = the infusion rate. Units in mg / time

CA,rate = rate of increase in plasma conc due to the infusion. Units in micro g/mL/time

The plasma conc at any time t after the start of a continuous IV infusion can be calculated by using:

Ct = CA,rate [1-exp(-t . kelim)]

kelim

To calculate the maximum plasma conc use:

Cmax = CA,rate . t 1/2

0.693

Remember kelim = 0.693 / t 1/2

(to see how to derive this equations refer to your lecture notes) Back to our example using Drug X using a healthy male volunteer:

Plasma volume = 3000mL

Dose = 300 mg IV bolus

Co = 100 micro g/mL

kelim = 0.2 per hour

t 1/2 = 3.46 hours

AUC = 500 micro g.h/mL = 0.5 mg.h/mL

Vd= 3 L = 0.043 L/kg

CL = 600mL/hour

Continuous IV infusion rates of 60mg/hour, 120mg/hour and 180mg/hour Cmax is dependent on infusion rate as expected. Steady state is reached when t = 5 t 1/2 (approx) irrespective of infusion rate. This is the same as for repeat bolus doses. Decide on the ideal Cmax and apply the Cmax equation:

Cmax = CA, rate . t 1/2 ==> CA,rate = 0.693 x Cmax

0.693 t 1/2 so for Drug X:

- ideal plasma conc is 175 micro g/mL to 250 micro g/mL but we use a narrower range: 187.5 micro g/mL to 237.5 micro g/mL.

- select the midpoint (212.5 micro g/mL)

CA,rate = 0.693 x 212.5

3.46

= 42.6 micro g/mL/hour We then need to calculate the infusion rate (mg/hour) that will give this rate of increase in the plasma concentration. We assume:

- a linear relationship between a single dose and the resulting plasma concentration

- this applies with a rate

The original IV bolus dose of 300mg gave a C0 of 100 micro g/mL - assuma a dose rate of 300mg/hour will give a rate of increase of the plasma conc of 100 micro g/mL/hour.

Therefore, a dosing rate of 127.8mg/hour will give a rate of increase of the plasma conc of 42.6 micro g/mL/hour. We can also use a modification of the Vd equation.

Vd = Dose becomes Vd = Dosing rate

C0 CA,rate

Dosing rate = Vd . CA,rate

3000 mL x 42.6 micro g/mL/hour = 127,800 = 127.8mg/hour CA,rate = 42.6 micro g/mL/hour and dosing rate = 127.8mg/hour are both awkward numbers and are therefore prone to error. To combat this we can try rounding up the CA,rate up to 45 micro g/mL/hour and recalculate the dosing rate to 135mg/hour. Recalculate the Cmax to 225 micro g/mL and then plot a plasma conc graph. As you can see from the graph, it takes too long for the drug to reach the therapeutic window. We need to use a IV bolus loading dose. We must decide on the target initial plasma conc, C0,load, and work out the dose required to give this plasma concentration. Remember that we have two factors to consider, the decrease in plasma levels from the loading dose and the increase in plasma levels from the infusion.

Loading dose kinetics:

Ct = C0 exp(-t . kelim)

Infusion kinetics:

Ct = CA,rate [1-exp(-t . kelim)]

kelim Cant mathematically combine these as the loading dose kinetics are time limited and the infusion kinetics are infinite, so we must use trail and error.

Try using a loading dose which will give the maximum therapeutic level, in this case 237.5 micro g/mL, dose = 712.5 mg Success!

- plasma levels never go above the maximum therpeutic level

- plasma levels never go below the minimum therapeutic level Pharmacokinetics Oral dosing Patients generally prefer the oral dosing to other routes. For oral dosing, three factors need to be considered:

- formulation - ie the rate of drug release from the tablet

- absorption (input) - ie plasma levels increase

- output - ie kelim, ie plasma levels decrease

Mathematically this is more complex (unfortunatley) We make assumptions:

- that the drug is 100% in solution and non-ionised at the moment of administration and remains so

- there is 100% absorption

- there is instantaneous movement from GI tract to plasma Back to our example, Drug X. We give a oral dose of 300mg and assume a 100% absorption. We then plot a plasma conc vs time graph. There are two parts to this graph. Firstly, the absorption and elimination phase where plasma levels increase - kabs and kelim. The second is just elimination where plasma levels decease - essentially just kelim. The second part can be used to define kelim. Theoretically, at any time after oral administration, the plasma conc can be calculated by comparing input and output parameters.

Ct = amount going in at time t - amount going out at time t

Complications:

- the amount going in (being absorbed) is dependent on the amount remaining unabsorbed at that particular time

- the amount going out is dependent on the plasma concentration at the particular time

This means straight forward addition is not possible. We need to use rate equations (differential equations) Absorption phase Plot ln of plasma conc vs time. We can then use this to derive the equation of the line for the STRAIGHT bit from a few points after the peak to the last time point. By analogy to the situation with an IV bolus dose the gradient is equal to kelim

Gradient of the line = 0.2 per hour (as expected) 1) Calculate the difference between kelim and the combined kelim and kabs at the beginning of the curve.

2) Back calculate the predicted Ct values before Cmax along the tangent line drawn earlier

3) Subtract values of the actual Ct from the calculated Ct (use real values, not ln values)

4) Calculate ln of the difference

In this case, kabs = 1 per hour Is useful to define the time to maximum plasma conc. This is calculated using:

tmax = ln (kabs/kelim)

(kabs - kelim)

(to see how to derive this equations refer to your lecture notes)

For Drug X:

tmax = ln (1.0/0.2) = 2.0 hours

(1.0-0.2) Uses the terminal linear portion of the ln Ct vs time plot. For Drug X, the intercept = 4.828, but what does this mean?

We are looking at the elimination phase only so assume that absorption is complete. We use this value in the equation for Ct.

Gradient = -kelim

Intercept = ln [ D0 . kabs ]

[ Vd (kabs - kelim ) ]

Do = amount of drug administered orally at time 0

(to see how to derive this equations refer to your lecture notes)

The intercept can be used to derive Vd if kabs and kelim are known but assumes 100% absorption

For Drug X:

Intercept = 4.828

4.828 = ln [ 300 x 1.0 ] = ln (375/Vd)

[ Vd x (1.0 - 0.2) ]

exp (4.828) = 375/Vd

Vd = 375/exp(4.828 = 375/125 = 3L Calculate tmax and substitute into the Ct equation to calculate Cmax at tmax using:

Cmax = D0 . kabs . [exp(-tmax . kelim) - exp (-tmax . kabs) ]

Vd (kabs - kelim) For Drug X:

Cmax = 300 x 1.0 x [exp(-2.0 x 0.2) - exp(-2.0 x 1.0)

3 x (1.0 - 0.2)

Cmax = 66.875 micro g/mL To calculate AUC use:

AUC = D0

Vd . kelim

This assumes 100% absorption

(to see how to derive this equations refer to your lecture notes)

For Drug X:

AUC = 300 / 3 x 0.2 = 500 micro g.h/mL = 0.5 mg.h/mL Remember, there are two equations for CL:

CL = Dose / AUC & CL = Vd . kelim

For Drug X:

CL = 300 / 0.5 = 600 mL/h & CL = 3 x 0.2 = 600mL/h So in summary we have found:

kelim = 0.2 per hour

kabs = 1.0 per hour

Vd = 3L

tmax = 2 hours

Cmax = 66.875 micro g/mL

AUC = 500 micro g.h/mL = 0.5 mg.h/mL

CL = 600 mL/h This is very simple to cope with. We define the fraction absorbed as f. This can range from 0.0 to 1.0. We apply f to the equations pertaining to oral absorption. Plasma concentration equation becomes:

Ct = f . D0 . kabs . [exp(-t . kelim) - exp(-t . kabs)]

Vd (kabs - kelim) Applies also to Cmax:

Cmax = f. D0 . kabs . [exp(-tmax . kelim) - exp(-tmax . kabs)]

Vd (kabs - kelim) Intercept on the lnCt vs time graph becomes:

Intercept = ln [ f . D0 . kabs ]

Vd (kabs - kelim) AUC equation equation becomes:

AUC = f . Do / Vd . kelim We can derive kelim, kabs and tmax as before. For all other parameters, we need to derive f first by running a comparison to an IV bolus dose. We need to compare the AUC for oral and IV bolus doses:

AUC oral = f . D0 / Vd . kelim

AUC IV bolus = C0 / kelim

Combining gives = D0 / Vd . kelim We then need to calculate the ratio of AUC values for oral and IV bolus doses

f = AUC (oral) / AUC (IV bolus)

Ideally f should be as close as possible to 1.0 kelim kabs tmax Vd Cmax AUC CL But what about incomplete absorption? Renal metabolism Up until now, we have assumed that our drug is exclusively renally excreted, ie:

kelim = kelim,renal & CL = CLrenal

However, many drugs are partially metabolised and partially renally excreted, ie:

kelim = kelim,renal + kelim,metab

CL = CLrenal + CLmetab Assuming first order kinetics, the rate of appearance of drug in the urine is dependent on the concentration of drug in the plasma. This means we cant use urine concentration. We need to use total amount of drug in urine - Durine.t and so also need to refer to total amount of drug in the body - Dbody.t

Using:

ln [dDurine.t] = ln [kelim,renal . Dose] -t . kelim

dT

(to see how to derive this equations refer to your lecture notes)

We can graw a graph of ln (dDurine,t / dt) against time which has:

- a gradient of -kelim & a intercept of ln [kelim,renal . Dose] Therefore, we can estimate kelim,renal if Dose and kelim are known, but we need urine data. This is collected intermittently as it cant be collected constantly. We use the midpoint of collection period for calculations.

For Drug X:

- IV bolus of 300mg

- Measure plasma levels and urine colections

- calculate kelim and kelim,renal We first take plasma and urine samples at various timepoints We can then calculate the rate of appearance of drug in urine using:

rate = amount of drug in collection interval / time of interval

Then we can plot ln rate against mid point of collection interval The equation of the line: y = -0.2 x + 4.1

We know that the gradient = kelim, so kelim = 0.2 per hour as expected.

We know that the intercept = ln [kelim.renal . Dose] so,

ln [kelim.renal . Dose] = 4.1

kelim.renal . Dose = exp(4.1) = 60.34

kelim.renal = 60.34/300 = 0.2 per hour

As kelim.renal = kelim, we can say that Drug X is exclusively excreted renally. Now lets look at Drug Q.

Drug Q is believed to be partially metabolised as well as being partially excreted renally. Drug Q was administered as an IV bolus of 90mg. Plasma and urine samples were taken at various time points. We then need to calculate the rate of appearance of drug in urine as before and then plot a ln graph as before. The equation of the line from the plasma conc graph was: y = -0.3 x +3.40, so kelim = 0.3 per hour.

The equation of the line from the urinary rate graph was: y = -0.3 x +2.6, so kelim = 0.3 per hour, as expected.

ln [kelim.renal . Dose] = 2.6

kelim.renal . Dose = exp(2.6) = 13.46

kelim.reanl = 13.46/90 = 0.15 per hour So for Drug Q:

kelim = 0.3 per hour

kelim.renal = 0.15 per hour

The difference between the two is the non-renal excretion assumed to be metabolism, ie:

kelim = kelim.renal + kelim.metab

kelim.metab = kelim - kelim.renal

= 0.3 - 0.15 = 0.15 per hour. Removal via kidneys occurs in three ways:

- glomerular filtration

- active tubular secretion

- passive tubular reabsorption

Total removal rate = Glomerular filtration rate + Secretion rate - Reabsorption rate We need a method for quantifying renal parameters.

Glomerular filtration rate (GFR) - measures the clearance of creatinine or inulin. Essentially eliminated by glomerular filtration only, so CLcreatinine and CLinsulin = GFR

Effective renal plasma flow (ERPF) - measures the clearance of p-amino-hipppuric acid (PAH). Eliminated by both glomerular filtration and active secretion. Essentially all PAH is removed from the plasma in one pass so removal is dependent on plasma, so CLpah = ERPF After viewing this Prezi, you will be experts in the field of pharmacokinetics Its Dr Barker here! Welcome to the PK Prezi. Hi, Here are a few simple instructions on how to use and navigate around the Prezi. (to see how these equations were derived, refer to your lecture notes) Renal excretion AR = Cmax/C0

AR = Accumulation Ratio Steady state is reached when the elapsed time is approx 5 t1/2 irrespective of the dosing interval.

AR is dependent on dosing interval irrespective of dose. Ctint = effect of the dose interval assuming the dose is identical

Ctint = C0 exp (-tint . kelim) ie AR is related to the ratio tint / t1/2

AR rises very sharply if tint <t/2

AR is negligible if tint/t1/2 >5 But how do we choose a loading dose? Elimination phase The equations are then solved using complicated maths to give: We can then extract the following parameters graphically or by calculation:

1) kelim, 2) kabs, 3) Tmax, 4) Vd, 5) Cmax, 6) AUC, 7) CL

In that order. We can estimate the method of renal excretion of the drug by comparing drug clearance values to that of creatinine:

Cldrug / CLcreatinine = 1 ---> elimination is by glomerular filtration only

CLdrug / CLcreatinine > 1 ---> elimination is by glomerular filtration and active tubular secretion (may also have some tubular reabsorption)

CLdrug / CLcreatinine < 1 ---) elimination is by glomerular filtration with some reabsorption (may also have some active tubular secretion) Liver disease affects metabolism so affect kelim,metab. Affects drugs which have significant metabolism (high excretion ratio)

CL = Vd . kelim and CLmetab = Vd . kelim

We assume that Vd stays constant in liver disease. It is difficult to predict the exact effect of liver disease on drug metabolism as the liver funstion tests do not adeduately predict metabolis function.

Generally:

- choose drugs which are predominantly renally excreted

- choose drugs with a wide therapeutic window

- use with caution and monitor the patients response Kidney disease affects renal elimination and drugs which have significant renal excretion. We need to assess the effect on kelim,renal.

CL = Vd . kelim and CLrenal = Vd . kelim,renal

We assume Vd stays constant in renal disease. We need to adjust the dose given but we first need to asses the extent of kidney damage by looking at creatinine (inulin) clearance and PAH clearance Creatinine is slightly different for men and women and dependent on muscle mass. It can be predicted for average normal people of all ages using the Cockcroft and Gault equation: So how do we adjust the dose in kidney disease? We assume that renal clearance of the drug is proportional to creatinine clearance and that this ratio is the same in both normal individuals and patients with impaired kidney function.

e.g. if CLcreatinine = 50% of normal, then CLdrug = 50% of normal

CLdrug,renal,patient = CLdrug,renal,normal x CLcreatinine,patient / CLcreatinine,normal So, if the drug is 100% renally excreted we can use:

Dose,patient = Dose,normal x CLcreatinine,patient / CLcreatinine,normal

Remember that this assusmes 100% renal excretion

(to see how this equation was derived, refer to your lecture notes) If the drug is only PARTIALLY renally cleared, we only need to adjust for that portion. The dose of drug in a normal individual can be considered as:

R = the proportion of the drug normally renally eliminated (can take values of 0.0 to 1.0)

The equation then becomes: So for Drug X, we know that it is entirely renally excreted. We need to calculate the IV bolus dose required for patient H, if the normal dose is 300mg as an IV bolus dose, and patient H has a creatinine clearance 30% that of a normal matched patient After viewing this Prezi, you will be able to use graphs and equations to calculate pharmacokinetic parameters. This Prezi will follow a path around the presentation. Use the scroll wheel on your mouse to zoom in and out of the Prezi. Alternatively, use the magnifying glass on the right hand side of the screen. This can be used to view images and smaller text. If you get lost in the Prezi or want to view the whole Prezi, click on the Home button on the right hand side of the screen to zoom out and view the whole Prezi. Feel free to move around the Prezi to focus on the areas that you are interested in or are having difficulty with. Lets get started! Whilst following the path, click and hold the left mouse button to move around anywhere in the Prezi. When you are ready to move on, click or press the right arrow to continue on the path. Click & Hold Zoom Zoom in to view in more detail Good luck for the exam! You should now be able to use graphs and equations to calculate pharmacokinetic parameters. There are also refresher lectures available on Blackboard if you are having difficulty with any of the maths used in this module. If you have any questions regarding pharmacokinetics, please feel free to email me.