jag009 ★★★ NJ, 20130913 15:48 Posting: # 11492 Views: 3,327 

Hi all, Can someone explain to me the following (or a kick in the butt if this is too simple?) from Csizmadia & Endrenyi? I don't have the Gibaldi PK book on hand (they referenced it from his book). Basis of the Proposed Methods. If a drug exhibits firstorder absorption with oneexponential disposition and the absorption (ka) and elimination (k) rate constants are equal, then the plasma concentration (C) depends on time according to: C=kAtexp(kt), where A=FD/V For the above model, k=1/Tmax , where Tmax is the time when the plasma concentration reaches maximum.How did he arrive to k=1/Tmax ?Thanks John ___________________________________________________________________________ Csizmadia F, Endrenyi L. ModelIndependent Estimation of Lag Times with FirstOrder Absorption and Disposition. J Pharm Sci Vol. 87:5. May 1998 
Helmut ★★★ Vienna, Austria, 20130913 16:16 @ jag009 Posting: # 11494 Views: 3,178 

Hi John, after administration of an IR formulation (if we have no IV data in the same subject) we assume that k_{a} > k_{el} (absorption is faster than elimination). In a controlled release formulation we try to slow down the absorption process. We reach flipflop PK if k_{a} = k_{el}. Note that in the common formula we have FDk_{a} in the numerator and V(k_{a}–k_{el}) in the denominator. FDk_{a} divided by zero? Oops! Therefore, we use the given modification. Note, that in Phoenix/WinNonlin for the same reason we have Model #5 (and #6 with a lagtime). » How did he arrive to k=1/Tmax ?Calculus followed by curve sketching and a little algebra. Start with the first derivative of the concentrationtime curve. Its root is at t_{max} (concentrations increasing before t_{max} = positive slope, decreasing after t_{max} = negative slope). Stepbystep: The flipflop models is (1) C = kFD/Vtℯ^{–kt} The first derivative is(2) dC/dt = kFD/Vℯ^{–kt} – k^{2}tFD/Vℯ^{–kt} We know that at t_{max} dC/dt = 0. Therefore, we can write(3) kFD/Vℯ^{–ktmax} = k^{2}t_{max}FD/Vℯ^{–ktmax} Solving for t_{max} (shortening some stuff and dividing by k), we get(4) k = k^{2}t_{max} Example: k 0.25, F 1, D 100, V 10 k = 1/4 = 0.25, Q.E.D. BTW, like in the usual onecompartment model the inflection point – where the tangent chances the “side” – is at the minimum of slopes (or the root of the 2^{nd} derivative) = 2×t_{max}. If you go for a semilogplot you will notice that we have no linear phase. Therefore, in PK modeling we have a problem. If we aim for a onecompartment model and get similar estimates of k_{a} and k_{el} (generally with large variances and overlapping CIs) it’s time to try a flipflop model. » I don't have the Gibaldi PK book on hand. An overrated book, IMHO. Gibaldi/Perrier work most of the time with decadic logarithms and therefore have to introduce this stupid t_{½} = 2.303/k… Not quite elegant didactics. — Cheers, Helmut Schütz The quality of responses received is directly proportional to the quality of the question asked. 🚮 Science Quotes 
jag009 ★★★ NJ, 20130913 19:51 @ Helmut Posting: # 11495 Views: 2,666 

Thanks Helmut 