Question

Usually a protein-binding curve is a hyperbolic function, with theta on the y-axis and [total ligand] on the x-axis. We can only assume that [Free L]=[L total] when the ligand is in excess of the protein. For example the [protein] would be 0.001 nM and you start adding ligand in .05nm increments. But what would the binding curve above look like if the [receptor]=1 nM: the ligand concentration is no longer in excess of the protein concentration? Would you still be able to measure the Kd? Please explain using the equation for theta and draw what the curve would look like! Thank you!!

1.0 r 0.8 0.6 0.4 0.2 0.2 Kd 0.1 nM [receptor] 0.001 nM 0.4 0.6 0.8 [Insulin]total nM 1.0

Answer 1

The curve would follow the equation format of Hill equation which suggests

$\theta =\frac{[L]^{n}}{Kd+[L]^{n}}$

Where theta represents ligand-receptor fraction, L free ligand concentration, Kd is dissociation constant and n is Hill coefficient.

Hill coefficient tells about the nature of binding.

In the situation where receptor concentration is >>> ligand concentration

The free ligand concentration would be zero because all ligand will bind to receptors, so [L]ⁿ = zero and then θ= zero too. Therefore, we the plot will remain at 0 point and it will not move until the concentration of ligand is not reach to 1nM. That will reach after 20 times we add ligand in .05nm increments [1nM /0.05 nM = 20].