Need help, please. ### Protein-Ligand Binding AffinityA protein has a binding affinity for its ligand with a K\_A of \(2 \times 10^5 \, \text{1/M}\) at pH 5.0 and room temperature.**a.** What fraction of the protein is bound at a ligand concentration of 1.25 μM?**b.** At what concentration is 80% of the ligand bound?**c.** When the pH is raised to 6.5, the \(K_D\) increased to 20 μM. Is the binding tighter or weaker at this pH compared to a pH of 5.0? Explain your answer.

Kd is the dissociation constant for a protein-ligand complex that measures the affinity with which a protein binds its ligand. Fractional Saturation (θ) is the fraction of protein molecules saturated with ligands. (since it is a fraction, its value can only be between 0 and 1) θ = ...eq 1Also, Kd is the ligand concentration where the 50% of the total protein or the total ligand binding sites are occupied by ligand. Mathematically it is expressed as : Kd = [L] when θ=0.5.Recall that Kd = 1/Ka. So,Kd = 1/2x105 M-1 = 0.5 x 10-5 M or 5 uM {1 M = 10-6 uM}It is given that [L] = 1.25 uM.We substitute these values in the equation in eq 1:Fractional saturation = Fractional saturation = Fractional saturation = 0.2Fractional saturation of the protein is 0.2 when the ligand concentration is 1.25 uM. As per the question: θ = 80% or 0.8Kd = 5 uMRearranging equation 1 we get: [L] = θ Kd / (1 - θ)[L] = 0.8 x 5 uM / (1 - 0.8)[L] = 0.8 x 5 uM / 0.2[L] = 20 uM20 uM of ligand is sufficient to 80% saturate the protein. Lower Kd means stronger affinity for ligand. As per the text, Kd increases from 5 uM to 20 uM as pH is raised from 5 to 6.5. Since Kd increases as pH increases, protein's affinity for the ligand decreases, that is, the binding gets weaker. Fractional saturation of the protein is 0.2 when the ligand concentration is 1.25 uM. 20 uM of ligand is sufficient to 80% saturate the protein. Binding gets weaker as pH increases from 5.0 to 6.5, as indicated by an increase in Kd.