Tham khảo tài liệu 'thermodynamics interaction studies solids, liquids and gases part 14', kỹ thuật - công nghệ, cơ khí - chế tạo máy phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | 770 Thermodynamics - Interaction Studies - Solids Liquids and Gases Assuming that all n binding sites in the target molecule are identical and independent it is possible to establish b n i 13 where k is the constant for binding to a single site. According to this equation this system follows the hyperbolic function characteristic for the one-site binding model. To define the model n and k can be evaluated from a Scatchard plot. The affinity constant k is an average over all binding sites it is in fact constant if all sites are truly identical and independent. A stepwise binding constant Kst can be defined which would vary statistically depending on the number of target sites previously occupied. It means that for a target with n sites will be much easier for the first ligand added to find a binding site than it will be for each succesive ligand added. The first ligand would have n sites to choose while the nth one would have just one site to bind. The stepwise binding constant can be defined as number of free target sites n-b 1 Kst -----7--77- 7 k 7 - k 14 number of bound sites b It is interesting to notice that a deviation from linearity in the Scatchard plot and to a lesser extent in the Benesi-Hildebrand gives information on the nature of binding sites. A curved plot denotes that the binding sites are not identical and independent. Allosteric interactions Another common situation in biological systems is the cooperative effect in that case several identical but dependent binding sites are found in the target molecule. It is important to define the effect of the binding of succesive ligands to the target to describe the system. An useful model for that issue is the Hill plot Hill 1910 . In this case the number of ligands bound per target molecule will be take into account that the situation in this system for equation 2 is m l and n l p _ n PLn PLJ P if equation 5 is solved for PLn and substitute into equation 15 then p _ nKa L n Ka L n l This expression can
