Pharmacokinetics 2

Question 1

how are drugs eliminated from the body?

Answer 1

Elimination of drugs from the body is by metabolism and excretion
- Elimination is associated with a rate, KE

Question 2

what order are the majority of drugs eliminated by?

Answer 2

first order

Question 3

what does the Cp against t plots look like for a 1st order elimination,using the two compartment model?

Answer 3

1) Cp against t plots look : double exponential - this would be hard to distinguish from a first order one compartment model.
2) but if ln(Cp) against t is plotted is made up of two segments rather than one that are both straight.
- α phase: distribution andelimination-this is the initial fast phase
- β phase - elimination only - this is the second slower phase on the graph

Question 4

describe the Phases of a two compartment curve

Answer 4

1) During the α phase, drug is distributing from the central compartment to the peripheral whilst also being eliminated from the central
2) In the β phase, the central and peripheral compartments are at equilibrium and we only see elimination from the central compartment
3) If distribution is very rapid, the observed profiles resemble one compartment kinetics

Question 5

where does most drug excretion occur in the body?

Answer 5

1) Kidneys, to urine

2) Biliary, to faeces

Question 6

explain the concept of drug clearance and state where it occurs in the body

Answer 6

1) A drug is considered to be cleared when it is removed irreversibly from the systemic circulation by metabolism or excretion
- So reversibly bound drug does not count
2) Clearance occurs as blood flows through an organ of elimination: Liver (metabolism) and kidneys (excretion) are the main ones

Question 7

The clearance of a drug is the volume of plasma from which the drug is completely removed per unit time. give the formula for total clearance

Answer 7

1) CL total = CL renal + CL non-renal (most non-renal is hepatic)
2) Note that clearance is a volume per unit time, and hence it has units such as: mL / minute, L / hour

Question 8

1) what formula would you use to estimate the clearance of a drug?
2) why is the clearance of a drug a constant?

Answer 8

1) CLs = KᴇVd
- Kᴇ is the first order elimination rate constant (units of time-1)
- Vd is the volume of distribution (units L)
- ‘s’ subscript denotes ‘systemic’
2) Clearance of a drug is thus a constant, because KE and Vd are both constants

Question 9

A drug has a clearance of 5 L/h. If Cp is 5 mg/L, how much drug is cleared in one hour?

Answer 9

1) Five litres of plasma are cleared of drug every hour. As each litre contains 5 mg of drug, the amount of drug cleared in one hour must be 25mg, because:
5 L/h x 5 mg/L x 1 h = 25 mg.

Question 10

explain what is meant by the term Elimination half life?

Answer 10

Denoted t½ , this is the time taken for Cp to fall to Cp/2 (i.e. half its starting value)
- amount of time it takes for the drug concentration to fall by half its current value

Question 11

how do we calculate Elimination half life?

Answer 11

t½ = 0.693/Kᴇ (REMEMBER)

Question 12

how do we determine the half life or Kᴇ of a drug when there is no previous data?

Answer 12

1) Get some ln(Cp) measurements at various times
2) plot a graph of ln(Cp) against t
3) put a straight line through it and work out the slope, which is equal to -Kᴇ
4) t½ = 0.693/Kᴇ can then be used to work out the half life

Question 13

explain how would you work out the elimination half life when you are given two data points on a graph?

Answer 13

1) you need to work out the slope of a line
- slope of a line = y1 - y2/ x1 - x2 (remember)
2) apply the above formula to the data so :
- lnC1 - lnC2/ t1 - t2 = -Kᴇ
3) now that you have the Kᴇ (use the possitive of the -Kᴇ value) you can use the elimination half life equation t½ = 0.693/Kᴇ

Question 14

Patient has digoxin level Cp=4.5 μg/L. Assuming no more drug is administered, how long before Cp falls to 1.5 μg/L?
- Digoxin’s t1/2 is 60 hours.

Answer 14

We have two (t, Cp) data points: (0, 4.5 μg/L) and(x, 1.5 μg/L) , where x is the time at which the level hits 1.5 μg/L. First, calculate KE from t1/2

1) t½ = 0.693/Kᴇ gives us = 0.01155h-1
2) -Kᴇ = -0.01155h-1 = ln4.5 - ln1.5/ o - x
3) so X = ln4.5 - ln1.5/ 0.01155 when solved gives 95.118h which is 3.96 days

Question 15

define salt factor

Answer 15

1) The “salt factor”, designated “S”, is the fraction of the administered dose that is the active drug
2) Some drugs (mainly for reasons of solubility) are formulated as salts or hydrates. Accordingly, only a fraction of the administered dose will actually be the drug that has the therapeutic effect

Question 16

Promazine is the CNS-active component. It is formulated as a hydrochloride, C17H20N2S1.HCl which has a MWt=320.88. The MWt of promazine is 284.42, calculate the salt factor

Answer 16

Fraction that is promazine hydrochloride that is actually promazine is 284.42 ÷ 320.88 = 0.89
- i.e. S = 0.89

Question 17

the single dose calculation can take into account the salt factor. the original equation for Cp was Cp = mass of drug in the body ÷ Vd. state and explain the new equation that takes into account the salt factor.

Answer 17

1) Cp= S x F x D ÷ Vd (REMEMBER)

• Cp=concentration in the plasma
• S = Salt factor (can be ignored if not dealing with a salt)
• F = Bioavailability (can be ignored for IV i.e. when F=1)
• D = Dose administered
• Vd= Volume of distribution

Question 18

explain what is meant by steady state ?

Answer 18

1) “Steady state” is a state where the amount of drug administered in a given time period equals the amount of drug eliminated over the same period
2) The steady state plasma concentration (Cpss ) is similar to the peak and trough levels in any dosing interval

Question 19

what does the time taken to reach steady state depend on ?

Answer 19

1) The time taken to reach steady state depends upon t½ for the drug
2) In fact, time to steady state is roughly equal to five times the half life.
- Time to steady state is also independent of dosing interval. The higher the dose, the higher the CpSS, and the bigger the difference between peak and trough levels, but the time taken to achieve steady state is not affected

Question 20

state the formula used to work out Cp at steady state (Cpˢˢ)

Answer 20

1) Cpˢˢ = S x F x D ÷ CL x τ (REMEMBER)

• Cpˢˢ = Cp at steady state
• S = Salt factor (drops out of equation if not dealing with a salt i.e. when S=1)
• F = Bioavailability (drops out of equation for IV i.e. when F=1)
• D = Dose
• CL = Clearance
• τ (tau) = Dosing interval

Question 21

drug with has a t½ of 7hrs. how long will it take to reach steady state?

Answer 21

1) The time taken to achieve steady state is ca. 5 x t½

2) Steady state is achieved after ca. 35 hrs of infusion

Question 22

let us assume that Cpˢˢ =50mg/L is a good level, but we want to achieve it faster. explain how we could achieve this faster ?
- Vd=20L

Answer 22

1) Use a loading dose!
2) loading dose = Cp x Vd
3) So loading dose = 50 mg/L x 20 L = 1000 mg
4) Then start continuous infusion

Question 23

outline the possible problems with multiple dosing

Answer 23

Oral administration, problems:

1) doses not often taken at even intervals (e.g. sleep!)
2) dose are frequently missed
3) two doses sometimes taken at once (maybe to make up for a missed dose)
4) changes in rate of absorption