Question

The calcium ion binds to a certain protein to form a 1:1 complex. When 96μM protein and 60μM Ca2+ are mixed together, the concentration of free ligand (calcium) was found to be 31μM.

a. Calculate the dissociation constant of the Ca2+−proteinCa2+−protein complex.

b. What concentration of Ca2+Ca2+ would you need to dissolve in the solution to ensure that 95% of the protein is bound to calcium?

c. What percentage of the calcium ions are bound to the protein in the conditions of part B?

Answer 1

* To Binding ratio = 1:1 Concentration of protein: 96 UM concentrati of Ca2+ = 60 MM co- € Ginen gren is cant + Protein = cat portein The concentration of free digand (Ca)= 311M i Co2t + Protein ca 24 - bootein intially cone 60 96 (cin-avo r o Afte mixing 60-20=31 96-29=67 0+29=29 & Keg for eq 0 be is Association Content, keq = [cat protein] 0 . Rubofituting all values in egin D, adulation content Kea = 29 = 29 31867 2077 :. dinnoc ation constant of Co2 portein - reverse of anno ciation constant ca + portein , will be the will be the ! I dissociations 2071 - 0162 29

② The initial protein cone = 96km only 95% of protein should be bound to calleren : 95% of 964M protein = 96895 - 91.24M 100 unreacted protein = 96-9112 = 4.8 MM equilibrium eq å may be written as a 1929 + postein - cat portein initial conc. x 96 at 95% -91.2 96-91.2 91.2 D x-91.2 ated for i keq = [cat protein ato 4.B 91-2 U votos 29 2077 91.2 41862-91.2) P l ata 200e X=1452 MM cone of cq 24 = 1452 MM) . . Conc of Cart at equilibrium z a-912=1452-91.22 1360.84 % of Ca21 = on 1360.8 xioo 1452 - 93.7%)