Absorption & Distribution, cont.

Effects of Ionization on Distribution, cont.

So what does pH partition theory tell us in a "practical" sense? For weak acids, only HA will cross to significant extent, for weak bases only A. Look at acid example. Can estimate the probability that a weak acid will cross at given pH using the Henderson-Hasselbalch equation.

Consider the stomach @ pH ~ 2 , pH= pK_a+log [A^-]/[HA],

or log [A^-]/[HA] = pH - pK_a, then for pH = 2

pKa log [A-]/[HA] [A-]/[HA] % unionized 1 1 10 9.1 2 0 1 50 3 -1 10-1 91 4 -2 10-2 99.1 6 -4 10-4 99.991 8 -6 10-6 99.9991

Similarly can calculate for bases - if you don't remember how to do logs on a calculator ­ see me.

Transport across membranes can be complicated by ionization when pH's are different on either side. One place where this will generally occur is in GI tract absorption.

Example: absorption of benzoic acid:

I) Stomach - assume pH=2, pK_a = 4 for benzoic acid.

So benzoic acid should readily cross membrane (largely nonpolar) but for plasma, pH = 7.4,

and log A^-/HA = 7.4 - 4 = 3.4

therefore: A^-/HA = 10^3.4 & have only 0.04% HA, and will shift toward absorption:

 

K_overall = 10^5.4 = 2.5 X 10⁵! This is dependent on the difference in pH only!

II) Intestinal tract: assume pH = 7, pK_a = 4 and log [A^-]/[HA] = 7- 4 = 3, so 10^-3 and have 0.10% HA. Thus K = (10^-3)(10^3.4) = 10^0.4 = 2.5 and transfer is much less.

Now so far this discussion has been based on equilibrium - generally do not get equilibrium; first, blood is constantly refreshed keeping plasma concentration low and second, the GI tract is also active, things are flowing through it. Thus transfer can be much greater or less due to Kinetic factors.

Summarizing, we looked at equilibrium distributions of acids and bases at different pH's and we concluded that the equilibrium distribution across a lipid membrane will depend only on:

• pH (pH difference across membrane)
• Whether the compd is an acid (HA) or a base (HA⁺) which show, respectively, reciprocal distributions

And the equilibria favor the compartment with the greatest degree of ionization.

Finally, strict acid-base equilibrium and passive diffusion will allow any acid or base to cross a membrane and reach expected equilibrium concentrations, so if absorption is less than equililibrium, other factors must be coming into play!

Distribution of Absorbed Compounds

Proteins and Distribution

Look at serum proteins as a reservoir for toxicants and resultant equilibrium. Simply add another equilibrium in coming to the overall distribution. However, proteins can be rather complex in their interactions, so today I want to look at how to analyse & intrepret data relevant to binding phenomena.

For the simplest situation Protein (P) has only one binding site for the adduct (A), and the dissociation can be written as:

PA P + A

To summarize then - plasma protein binding may affect

• distribution by shifting equilibria
• prolong the half life by providing a stable compartment
• explain some instances of toxic dose thresholds by providing a
  saturable compartment.

Tissue Localization

Can again account for some localization due to simple equilibrium phenomena as we have previously discussed. Thus highly lipid soluble compounds are frequently sequestered in fat depots, often for some time.

Example - thiopental (anesthetic):

Rapidly sequested in body fat, accounts for short duration of action

Example - Variety of halogenated hydrocarbons are found in body fat deposits of both people and animals - number of incidents now where animals destroyed due to such sequestrations ( Mich 1973, hexapoly Br-biphenyls; fire retardant).

Can also have specific binding

• to proteins: especially in liver and kidney proteins. Find with various kinds of organic acids, Azo-dyes, and cortico steroids; Heavy metals
• to bone: may bind heavy metals , tetracyclines. Bone storage generally removes substances from toxic interactions, but can be released.

Toxicokinetics

lst order models: dX/dt = -k_eX, X(0) = X_o, where X_o = initial amount. {Compartmentation models overhead, 3-2}

Describes simple first order elimination assuming body is simple homogeneous unit with toxin distributed uniformally at all times.

k_e = apparent first-order elimination constant. Apparent emphasizes fact that underlying processes may vary and are only approximately 1st order. (Example: biliary secretion may be active and exhibit 0 - order saturation kinetics under suitable conditions)

FYI Single Compartment Models (See the chapter by D.B. Tuey, "Toxicokinetics" in Introduction to Biochemical Toxicology (Ernest Hodgeson and Frank E. Guthrie, eds.), Elsevier, New York (1980) pp 40-66 for more detail.) {One compartment model overhead, Figure 3.3} For rapid introduction of amount D with no prior concentration get{Overhead Figure a}: X(O) = D = dose and Xt = De-ket Where Xt = amount left in body @ time = t For a dynamic equilibria between all body compartments, monitoring [plasma] get: X=VC, where C= [ ] in plasma V= constant, which happens to have units of volume so called the "apparent volume of distribution"! Then V=D/Co and substitute Ct = (D/V)e-ket taking logs gives a linear equation: log C (t) = log (D/V) - (0.434)ket can use to estimate ke and t1/2 for toxin elimination from serial blood or plasma samples. Alternatively, one can monitor excreta instead of plasma concentration: dXe/dt = keX, where Xe(0) = 0 substituting and integrating over time then gives: Xe(t) = D(1 - e-ket) Of course for most situations we get non-instantaneous absorption {Overhead Figure b}. For first-order absorption and elimination processses: dX/dt = -keXe + kaXa where ka is the apparent first-order rate constant and Xa = [ ] at the absorption site. Solving for X X(t) = ((fDka)/(ka-ke)) ( e-ket - e-kat) where f = the fraction of the dose (D) absorbed. If Xo << fD, and ka>ke, then at t > 0, e-kat << e-ket, and X fDe-ket A third common situation is constant input {Overhead Figure c}(that is, continuous, chronic exposure). For first-order absorption and elimination processses: dX/dt = -keX + D where x(0) = 0 and D = the dose rate. then: X(t) = (D/ke) (1 - e-ket) for t > 5 t1/2 e-ket 0 and we get a steady state CSS = D/Vke And elimination = intake. This is a convenient way of determining V (apparent volume of distribution), since at steady state (SS0 V is not dependent on any particular model for number of compartments etc.

Last modified 18 February 2010

© RA Paselk 2001