Question

Think of a toy model of a dimeric receptor binding with a divalent ligand. The K_Ds for binding between a two-armed compound LL1 (this is the ligand) and individual receptors R1 and R2 are:

R1 … 7.9 nM             R2 … 50.1 nM

Now think of a protein R1-R2, which is a protein comprised of both R1 and R2. At a temperature 300K, if there is no loss of energy/entropy to induce conformational changes of the protein and ligand during the binding, what is the expected K_D between LL1 and R1-R2 and corresponding deltaG?

Answer 1

equilibria among monomeric proteins and a bivalent ligand

equilibrium between a monovalent ligand (l), a monomeric protein (p), and a complicated of protein and ligand (p • l). this equilibrium is characterised via the dissociation regular kd in which...

Kd=[L][P]/[P⋅L]

p and p • l typically have units of concentration (i.e., molar) in solution or units of density (i.e., moles area−1) while sure to a surface. The fee for the dissociation steady in answer may be significantly extraordinary from the dissociation consistent while both the receptor or ligand is tethered to a surface. this difference calls for that kd be determined for the particular equilibrium under have a look at.

to maximize the conversion of proteins (p) to dimeric complexes (p • l2 • p), one must design ligands with low values of kd and, if feasible, with fantastic cooperativity (α >1). a simple protocol for screening bivalent ligands may be advanced based on the homodimerization model provided on this paper. as an instance, one could display bivalent ligands with extraordinary linkers (e.g., linkers of different length) by using measuring θppl2pmax. Assuming that kd is equivalent for all ligands, the biggest price of θppl2pmaxobserved inside the experimental curves might imply the bivalent ligand that binds with the most favorable fee of α.