Bioequivalence and Bioavailability Forum

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 rate-constant of absorption (different names: k_a, k_in, k₀₁, …) often is difficult to estimate even with compartmental modeling. In NCA for a one-compartment model one can start with the Wagner-Nelson 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, k₁₀, …) is assumed to be the elimination of drug.
Let’s have a look at the simple one-compartment open model:$$C(t)=\frac{D\times F_{\textrm{a}}}{V_{\textrm{d}}}\times \frac{k_{01}}{k_{01}-k_{01}}\times\left (\exp (-k_{10}\times t)-\exp(-k_{01}\times t) \right ) \tag{1}$$We can calculate C_max and t_max in the following way:

1. The slope of the function is positive before t_max (increasing values), zero at t_max, and negative (decreasing) afterwards.
   
2. Differentiate the function.
   
3. Find the root (i.e., the intersection of the derivative with the abscissa). This is t_max.

Formulas:$$t_{\textrm{max}}=\frac{1}{k_{01}-k_{10}}\times \log_{e}(k_{01}/k_{10}) \tag{2}$$and by substituting \(t\) in \((1)\) with \((2)\) after some rearrangement (not shown)$$C_{\textrm{max}}=\frac{D\times F_{\textrm{a}}}{V_{\textrm{d}}}\times(k_{01}/k_{10})^{k_{10}/(k_{10}-k_{01})}\tag{3}$$Unfortunately it is not possible to differentiate 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 Gauss-Newton algorithm. So it’s nothing for M$Excel (unless you trust in the ‘Solver’-Add-In – 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
k₀₁ = 2
k₁₀ = 0.025 – 1.999

If the elimination is slow (k₁₀ = 0.050) we get t_max 1.89 and t_½ 13.9, if the elimination approaches the absorption (k₁₀ = 1.999), we get t_max 0.500 and t_½ 0.347…

Only for k₁₀ = 1 we get t_max = t_½ = 0.693 – which is log_e(2)…

Therefore only for drugs with relatively fast elimination (as compared to absorption) t_max ~ t_½;
\(t_\textrm{max}=t_{1/2}\; \leftrightarrow\; k_{10}=k_{01}/2\: \small{\square}\)



For drugs with a biphasic elimination similar t_½ 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 k₁₀ = 0.15 and the absorption varies between 0.50 and 1.999.

Setting a k₀₁ of 1.0 as the reference value (t_max 2.23), a 10% change (k₀₁ 1.1) results in a 6% decrease in t_max (to 2.10).
If we increase k₀₁ 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.