Question

A researcher is analyzing the kinetics of binding between insulin-like growth factor (IGF) and IGF receptor (IGFr) on the cell membrane. In this analysis, he incubated cells in the media supplemented with IGF, and the IGF concentration was 500 µM. At equilibrium, he found that 62.5% of IGFr associated with IGF. According to the literature, the association constant is 0.75 s1M-1 .Calculate percentages of cellular receptors bound with ligands after 10 min from equilibrium state, while all free IGF was washed away at equilibrium state?

Answer 1

SOL)

Given data

[IGF] = 500

[IGF-IGFr] = 62.5%

                     =62.5/100

                      =0.625

Association constant (Ka) = 0.75 /s/M

we know Dissciation constant (Kd) is the reciprocal of Association constant

Hence, Kd = 1/Ka

                   =1/0.75

                   = 1.33 sM

--> According to law of mass action, at equilibrium, Ka = Kd

Ka [IGF][IGFr] = Kd [IGF-IGFr]

where [IGF] and [IGFr} are the free molar concentrations of ligand and receptor respectively.

Percentage of bound receptors after 10 minutes after equilibrium = 500 / 0.625*1.33                                                                                            =500 / 0.83                                                                                             = 61.50 %