BEAZ ★ India, 20080711 13:35 Posting: # 2015 Views: 16,545 

hai, can any one clarify that, is there any relation between Tmax and t1/2. for some products the Tmax found to be very close to t1/2. how it is possible? regards Edit: Category changed. [Helmut] — Alhas 
Helmut ★★★ Vienna, Austria, 20080711 17:08 @ BEAZ Posting: # 2016 Views: 14,563 

Dear Alhas! » can any one clarify that, … Hopefully! » … is there any relation between Tmax and t1/2. The regulatory definition of BE contains the phrase "… rate and extent of absorption …". Unfortunately the rateconstant of absorption (different names: k_{a}, k_{in}, k_{01}, …) often is difficult to estimate even with compartmental modeling. In NCA for a onecompartment model one can start with the WagnerNelson method. t_{max} is a rather insensitive metric of the rate of absorption (as shown by László Endrényi about 15 years ago). That’s the reason it’s only supportive (based on clinical grounds) in some regulations and partial AUCs are preferred by the FDA instead. For an IR formulation the ‘slow’ phase k_{el}, k_{10}, …) is assumed to be the elimination of drug. Let’s have a look at the simple onecompartment open model:
D × F_{a} k_{01} We can calculate C_{max} and t_{max} in the following way:
1 and (by substituting t in (1) by (2) after rearrangement)
D × F_{a} Unfortunately it is not possible to differentiate the functions of higher compartment models (more than two exponential terms). In other words, there’s no analytical solution – curve fitting programs approximate the solution by numeric integration of the differential equations by means of variants of the GaussNewton algorithm. So it’s nothing for M$Excel (unless you trust in the ‘Solver’AddIn – which you shouldn’t). » for some products the Tmax found to be very close to t1/2. how it is possible? Chance? I prepared an example: D = 100 V_{d} = 5 F_{a} = 1 k_{01} = 2 k_{10} = 0.025 – 1.999 If the elimination is slow (k_{10} = 0.050) we get t_{max} 1.89 and t_{1/2} 13.9, if the elimination approaches the absorption (k_{10} = 1.999), we get t_{max} 0.500 and t_{1/2} 0.347… Only for k_{10} = 1 we get t_{max} = t_{1/2} = 0.693 – which is ln(2)… Therefore only for drugs with relatively fast elimination (as compared to absorption) t_{max} ~ t_{1/2}; t_{max} equals t_{1/2} for k_{10} = k_{01}/2. For drugs with a biphasic elimination similar t_{1/2} and t_{max} are almost impossible. A further remark on the insensitivity of t_{max} to changes in the absorption rate: essentially the same example as above, but the elimination kept constant with k_{10} = 0.15 and the absorption varies between 0.50 and 1.999. Setting a k_{01} of 1.0 as the reference value (t_{max} 2.23), a 10% change (k_{01} 1.1) results in a 6% decrease in t_{max} (to 2.10). If we increase k_{01} further by 50% to 1.5, t_{max} decreases only by 24% to 1.71. That’s why we use t_{max} as a PK metric in BE only if unavoidable. — Cheers, Helmut Schütz The quality of responses received is directly proportional to the quality of the question asked. 🚮 Science Quotes 
BEAZ ★ India, 20080715 07:54 @ Helmut Posting: # 2033 Views: 14,232 

Dear Helmut, Thank you for the information. still i am not clear. Tmax is the time reach the maximum concentration, logically after reaching a Tmax only the drug shall startes its elimination phase, if the T1/2 comes first then the conc of the drug in systemic circulation is reduced to its half and after this how we get a Tmax kindly clarify Thanks and Regards — Alhas 
Ohlbe ★★★ France, 20080715 10:33 @ BEAZ Posting: # 2036 Views: 14,200 

Dear Alhas, The body does not press a button at Tmax to stop absorbing the drug and to start eliminating it. Elimination starts at the second the drug enters the body. Tmax is the time when you reach a balance between absorption and elimination: before Tmax the drug enters the body faster than it is eliminated, and concentrations increase. After Tmax the drug is eliminated faster than it is absorbed, and concentrations decrease. Regards Ohlbe 
Helmut ★★★ Vienna, Austria, 20080716 01:18 @ Ohlbe Posting: # 2038 Views: 14,481 

Dear Ohlbe & Alhas! » Elimination starts at the second the drug enters the body. Exactly. There are many processes involved, all of them are stochastic ones. In PK we retreat into the concept of ‘compartments’, because we cannot better model our body consisting of 10^{13}–10^{14} cells. Speaking of a onecompartment model, we treat blood and all highly perfused organs (liver, brain, kidneys; partly muscles) as a (wellstired) beaker with a small hole in the bottom. We assess these processes statistically; if we talk about the half life of two hours of a drug with a molecular mass of 200 g/mol after a bolus dose of 10 mg on the average 1.506×10^{19} molecules will have left the systemic circulation within the first two hours. Or modifying your example (∆ input of zero duration), about 2.9×10^{15} molecules leave the system in the first second!^{1} – or 962 ng if you prefer to weigh a bunch of molecules… If we split the equation we can see that after an oral dose the elimination process itself is exactly the same as after a fast bolus (the concentration at t=0 is D/V_{d}, which I have set throughout the examples in this thread to 100/5). An nice property of the function is the intersection of the elimination with the profile at t_{max}/C_{max}. It’s clear why the starting point for the estimation of the halflife in the TTTmethod (see this post) was chosen. Also one of Kamal Midha’s Mantras (‘Once absorption is over, formulation differences no longer apply.’)^{2} is obviously justified (valid for ≥2×t_{max}). » Tmax is the time when you reach a balance between absorption and elimination: before Tmax the drug enters the body faster than it is eliminated, and concentrations increase. After Tmax the drug is eliminated faster than it is absorbed, and concentrations decrease. Almost… In a onecompartment system both absorption and elimination are treated as constant. Therefore, there’s no faster/slower absorption/elimination throughout the profile. In the logplot the parallel course after about 2×t_{max} is even more evident (in more mathematical terms after the inflection point of the first derivative; where the root of the first derivative = t_{max}). @Alhas: I strongly suggest to study one of the basic textbooks on pharmacokinetics – otherwise it will be very difficult for you to design and evaluate BA/BEstudies. I’m not always in the mood to answer ‘neophytes’. If you want a quick solution I recommend David Bourne’s A First Course in Pharmacokinetics and Biopharmaceutics.
— Cheers, Helmut Schütz The quality of responses received is directly proportional to the quality of the question asked. 🚮 Science Quotes 
BEAZ ★ India, 20080724 14:09 @ Helmut Posting: # 2078 Views: 14,024 

Dear Ohlbe & Helmut! thanks a lot. @Helmut: » I strongly suggest to study one of the basic textbooks on pharmacokinetics – otherwise it will be very difficult for you to design and evaluate BA/BEstudies. » I’m not always in the mood to answer ‘neophytes’. » If you want a quick solution I recommend David Bourne’s A First Course in Pharmacokinetics and Biopharmaceutics. — Alhas 