Answer :
To answer this item, we must take note that the ligand that binds the tightest is the one with the lowest dissociation constant, Kd. Kd's for both A and B are already given so, we only need to solve Kds for C and D.
Kd of C
0.3 = (1x10⁻⁶)/(1x10⁻⁶ + Kd) ; Kd = 2.3x10⁻⁶
Kd of D
0.8 = (1x10⁻⁹)/(1x10⁻⁹ + Kd) ; Kd = 2.5x10⁻10
Since Ligand D has the least value of dissociation constant then, it can be concluded that it binds the tightest.
Kd of C
0.3 = (1x10⁻⁶)/(1x10⁻⁶ + Kd) ; Kd = 2.3x10⁻⁶
Kd of D
0.8 = (1x10⁻⁹)/(1x10⁻⁹ + Kd) ; Kd = 2.5x10⁻10
Since Ligand D has the least value of dissociation constant then, it can be concluded that it binds the tightest.