1 Exercise 15 Drug binding and investigation of the Michaelis Menten equation; pitfall of methods of linearisation.

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Presentation transcript:

1 Exercise 15 Drug binding and investigation of the Michaelis Menten equation; pitfall of methods of linearisation

2 The Michaelis Menten equation

3 Equation relating bound vs Free drug concentration where Bound is the bound drug concentration, Bmax the maximal binding capacity of the transport protein(s) Kd the steady state dissociation constant. –Kd is the free concentration of the drug for which the bound concentration is equal to Bmax/2. –The inverse of Kd is Ka, the steady-state constant of affinity. The unit of ka is the inverse of concentration. For example a Ka equal to 5X109 M means that to obtain half the saturation of Bmax, one mole of ligand should be diluted in a volume of 5X109 L.

4 Free fraction vs free concentration In blood, a drug (ligand) coexists under two forms: bound [B] and unbound. The unbound drug is also called free [Free]. The proportion (fraction) of drug that is free depends on the specific drug affinity for plasma proteins. – This proportion is noted fu (from 0 to 1 or 0 to 100%) with u meaning unbound

5 Free vs total concentrations In PK, the most commonly used analytical methods measure the total (plasma) drug concentrations i.e. [Total=B+Free]. As generally only the unbound part of a drug is available at the target site of action including pathogens, the assessment of blood and plasma protein binding is critical to evaluate the in vivo free concentrations of drug i.e. [Free].

6 Computation of the Free concentration from total concentration

7 Determination of fu

8 Determination of Bmax & Kd

9 Determination of fu

10 Determination of fu A more advanced protocol consists in repeating this kind of experiment but with very different values of drug concentrations in order to exhibit a possible saturation of the tested proteins.

11 Data to analyse

12 Plot of data Linear portion

13 Equation to fit Where NS is the proportionality costant for the Non Saturable binding

14 Estimation of Bmax, Kd and NS using Non-Linear regression MODEL remark ****************************************************** remark Developer:PL Toutain remark Model Date: remark Model Version:1.0 remark ****************************************************** remark remark - define model-specific commands COMMANDS NSECO 3 NFUNCTIONS 1 NPARAMETERS 3 PNAMES 'Bmax', 'Kd', 'NS' SNAMES 'fu5', 'fu100','fu1000' END remark - define temporary variables TEMPORARY Free=x END remark - define algebraic functions FUNCTION 1 F= Free+(Bmax*Free)/(Kd+Free)+NS*Free END SECO remark: free fraction for free=5 or 100 ng/ml fu5=5/(5+(5*Bmax)/(kd+5)+NS*5)*100 fu100=100/(100+(100*Bmax)/(kd+100)+NS*100)*100 fu1000=1000/(1000+(100*Bmax)/(kd+1000)+NS*1000)*100 END remark - define any secondary parameters remark - end of model EOM

15 The command block COMMANDS NSECO 3 NFUNCTIONS 1 NPARAMETERS 3 PNAMES 'Bmax', 'Kd', 'NS' SNAMES 'fu5', 'fu100','fu1000' END

16 The temporary block TEMPORARY Free=x END

17 The function block FUNCTION 1 F= Free+(Bmax*Free)/(Kd+Free)+NS*Free END

18 The SECONDARY block SECO remark: free fraction for free=5 or 100 ng/ml fu5=5/(5+(5*Bmax)/(kd+5)+NS*5)*100 fu100=100/(100+(100*Bmax)/(kd+100)+NS*100) *100 fu1000=1000/(1000+(100*Bmax)/(kd+1000)+NS *1000)*100 END

19 Results

20 Linearization of the MM equation Before nonlinear regression programs were available, scientists transformed data into a linear form, and then analyzed the data by linear regression. There are numerous methods to linearize binding data; the two most popular are the methods of Lineweaver-Burke and the Scatchard/Rosenthal plot.

21 Linearization of the MM equation The Lineweaver Burke transformation

22 The Lineweaver Burke transformation

23 Linear regression analysis of binding data

24 Estimate parameters after transformation

25 Pitfall of the linear regression analysis Now repeat your estimation of these two parameters (Bmax, Ka) by linear and non- linear regression but after altering a single value (replace the first value B=61.7 by B=50).