Looking at avidity between a ligand-receptor, you're looking at an enhancement of the $K_d$ compared to a lone ligand.

Is it more appropriate to compare the $K_d$ using the concentration of the "free" ligand or the concentration of the free complex?

Using the ligand is odd at the lower regimes since the ligand efficency is fairly low but using the nanoparticle changes how you have to calculate the effective concentration. I've seen papers that do both and I'm curious which is most appropriate/correct

  • $\begingroup$ What, precisely, do you mean by avidity? Your question seems to be asking about cooperative binding (multiple ligand-macromolecule events at discrete sites) rather than avidity (sum of interactions of a single ligand-macromolecule across a contiguous binding interface). $\endgroup$
    – KG3
    Jul 26, 2018 at 17:49

1 Answer 1


In the case where there is not cooperativity between binding events, the Scatchard Equation is the generalized solution to multiple ligand binding to a macromolecule. The Scatchard plot uses the free ligand concentration.

There are multiple models used to address cooperative binding. The Hill Equation is the most frequently encountered in biochemistry literature: $$ \theta = \frac{[L]^n}{K_d+[L]^n} $$ Where $[L]$ is the concentration of free ligand, $\theta$ is the fraction of bound ligand-receptor complex, $K_d$ is the apparent dissociation constant, and $n$ is the Hill coefficient, describing cooperativity.


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