• Shuffle
    Toggle On
    Toggle Off
  • Alphabetize
    Toggle On
    Toggle Off
  • Front First
    Toggle On
    Toggle Off
  • Both Sides
    Toggle On
    Toggle Off
  • Read
    Toggle On
    Toggle Off
Reading...
Front

Card Range To Study

through

image

Play button

image

Play button

image

Progress

1/17

Click to flip

Use LEFT and RIGHT arrow keys to navigate between flashcards;

Use UP and DOWN arrow keys to flip the card;

H to show hint;

A reads text to speech;

17 Cards in this Set

  • Front
  • Back
drug excretion

renal excretion
most important for the majority of drugs
- glomerular filtration
- active secretion (non-linear, carrier-mediated, for low drug concentrations can be linearized in the same way as the enzymatic rxn
- tubular reabsorption
remaining excretion processes:
are linear or can be linearized (biliary excretion) with respect to Ca

* except fecal excretion
clearance of excretion

here, the clearance due to excretion, Clx, is the summation of the clearances for:
-glomerular filtration + CLf
-active secretion + Cls
-tubular reabsorption - Clr
other excretion processes + Clo

+Clf, + Cls, -Clr = Clk
clearance of excretion

equation (10)
- dnx/dt = Clx * ca
drug elimination I

the total eliminated drug amount is:
the sum of metabolized and excreted drug amounts

-dn/dt = dnm/dt - dnx/dt

after substitution from 8 and 10
-dn/dt = (Clm + Clx) * ca
drug elimination II

the metabolic and excretion clearances:
can be summed up

-dn/dt = Cl * ca

cannot be integrated because of different variables: the eliminated drug amount n and the free drug concentration, ca. the same variable on both sides of 11 is needed
dose and free drug concentration I

the relation between n and ca can be found from:
the definition of the partition coefficient (eq 5-1)

in each aqueous phase, the drug exists as free and bound to proteins (binding to enzymes is neglible)
ci = ca +[DP]i
dose and free drug concentration II

for linear binding, the concentration of the drug/protein complex is:
eq 6-2

[DP] = K * ca * po

then the total drug concentration (free and bound) in the ith aqueous drug phase is:
ci= ca + [DP]i= ca * (1+Ki * Pi)
dose and free drug concentration III
see slide
final equation
see slide
clearance

total clearance is:
the sum of clearances for all elimination processes (eq 11)
clearance of a process is:
the rate constant of the process multiplied by the volume of the phase where the process takes place
clearance equations
see slide
elimination rate constant I

from eq 14
ke = Cl/Vdf
elimination rate constant is:
the sum of individual rate constants weighted by the factor

volume of elimination phase/ distribution volume
elimination rate constant II

clearance can be summed up:
regardless of the volume of the phase where the elimination is taking place
elimination rate constant II

elimination rate constants can be summed up:
only if the elimination processes take place in the same phase

otherwise, the individual rate constants have to be scaled (weighed) by the ratio of the phase volume and the distribution volume