T max calculation[1]
- 1. Determination of 𝒕 𝒎𝒂𝒙 Prepared by: Imran Ahammad chowdhury
- 2. The rate of change of the amount of drug in the body after oral administration is a fraction of both the absorption rate 𝑑AA 𝑑𝑡 and elimination rate 𝑑AE 𝑑𝑡 . That is- 𝑑𝐴 𝑑𝑡 = 𝑑𝐴 𝐴 𝑑𝑡 − 𝑑𝐴 𝐸 𝑑𝑡 or, 𝑑𝐶𝑝 𝑑𝑡 = 1 𝑉𝐷 𝑑𝐴 𝐴 𝑑𝑡 − 𝑑𝐴 𝐸 𝑑𝑡 …………..1 Here, 𝑉𝐷 = 𝐴 𝐶 𝑝 ∴ 𝐴 = 𝑉𝐷 𝐶 𝑝 A=Amount of drug in the body
- 3. Now How much drug is absorbed from the GIT, that means rate of absorption from the GIT is given by the following expression: 𝑑𝐷 𝐺𝐼 𝑑𝑡 = −𝐾 𝑎 𝐷 𝐺𝐼 = 𝑑𝐴 𝐴 𝑑𝑡 ……. 2 So, from 1. 𝑑𝐶𝑝 𝑑𝑡 = 1 𝑉 𝐷 −𝐾 𝑎. 𝐷 𝐺𝐼 −𝐾𝐷 𝐵 … … . . 3 Similarly, rate of elimination 𝑑𝐴 𝐸 𝑑𝑡 = −k𝐷 𝐵
- 4. From equation 2 in case of multiple oral dosage regimen, we can write, 𝑑𝐷 𝐺𝐼 𝑑𝑡 =-𝐾 𝑎 𝑛.τ. 𝐷 𝐺𝐼 𝑑𝐷 𝐺𝐼 𝑑𝑡 =-𝐾 𝑎 𝑛.τ.dt 𝑑𝐷 𝐺𝐼 𝑑𝑡 = 𝐾𝑎 𝑛.τ.dt 𝐷 𝐺𝐼 = −𝐷 𝑜 𝑒−𝐾 𝑎 𝑛.τ.𝑡 𝐷 𝐺𝐼 = −𝐹𝐷 𝑜 𝑒−𝐾 𝑎 𝑛.τ.𝑡 if fraction of dose is absorbed. Where, n=number of dose τ =dosing interval
- 5. So, putting this value in equation 3 we get- 𝑑𝐶𝑝 𝑑𝑡 = 1 𝑉 𝐷 𝐹𝐾 𝑎. 𝐷 𝑜 𝑒−𝑛.𝐾 𝑎.τ.𝑡 − 𝐾𝐷 𝐵 … … … .4 Now, integration this equation we get, the general oral absorption equation for calculating the drug concentration (𝐶 𝑝) in the plasma, at any time t, 𝐶 𝑝= 𝐹𝐾 𝑎 𝐷 𝑜 𝑉 𝐷(𝑘−𝑘 𝑎) 1−𝑒−𝑛.𝐾 𝑎.τ.𝑡 1−𝑒−𝐾 𝑎.τ 𝑒−𝑘 𝑎 𝑡 − 1−𝑒−𝑛𝑘τ 1−𝑒−𝑘τ 𝑒−𝑘𝑡 ….5
- 6. For a single oral dose, the equation can be written as follows- 𝐶 𝑝= 𝐹𝐾 𝑎 𝐷 𝑜 𝑉 𝐷(𝑘−𝑘 𝑎) 𝑒−𝑘 𝑎 𝑡 − 𝑒−𝑘𝑡 …. ……6 The time needed to reach maximum concentration is independent of dose & is dependent on the rate constants for absorption (𝑘 𝑎) & elimination (k). At the maximum concentration, the rate of drug absorbed is equal to the rate of drug eliminated. Therefore, the rate of concentration change is equal to zero.
- 7. The rate of concentration change can be obtained by differentiating the equation 6. 𝑑𝐶 𝑝 𝑑𝑡 = 𝐹𝑘 𝑎 𝐷 𝑜 𝑉 𝐷(𝑘−𝑘 𝑎) −𝑘 𝑎. 𝑒−𝑘 𝑎 𝑡 + 𝑘. 𝑒−𝑘𝑡 The rate of concentration change can be obtained by differentiating the equation 6. 𝑑𝐶 𝑝 𝑑𝑡 = 𝐹𝑘 𝑎 𝐷 𝑜 𝑉 𝐷(𝑘−𝑘 𝑎) −𝑘 𝑎. 𝑒−𝑘 𝑎 𝑡 + 𝑘. 𝑒−𝑘𝑡 But at 𝐶∞ 𝑚𝑎𝑥 𝑑𝐶 𝑝 𝑑𝑡 = 0 0= 𝐹𝑘 𝑎 𝐷 𝑜 𝑉 𝐷(𝑘−𝑘 𝑎) −𝑘 𝑎. 𝑒−𝑘 𝑎 𝑡 + 𝑘. 𝑒−𝑘𝑡 𝐹𝑘 𝑎 𝐷 𝑜 𝑉𝐷(𝑘 − 𝑘 𝑎) ≠ 0
- 8. −𝑘 𝑎. 𝑒−𝑘 𝑎 𝑡 + 𝑘. 𝑒−𝑘𝑡 = 0 or, 𝑘 𝑎. 𝑒−𝑘 𝑎 𝑡 = 𝑘. 𝑒−𝑘𝑡 or, ln 𝑘 − 𝑘𝑡 = ln 𝑘 𝑎 − 𝑘 𝑎t or, 𝑘 𝑎t − 𝑘𝑡 = ln 𝑘 𝑎 − ln 𝑘 or, t(𝑘 𝑎 − 𝑘)= ln 𝑘 𝑎 − ln 𝑘 or, 𝑡 𝑚𝑎𝑥 = ln 𝑘 𝑎−𝑘 𝑘 𝑎−𝑘 = 2.303 log 𝑘 𝑎 𝑘 𝑘 𝑎−𝑘
- 9. Thank you