Question

Three membrane receptor proteins bind tightly to a hormone. The free hormone concentrations and their corresponding fractions of hormone binding, Y, are given in the table. [Hormone), nM Protein 1 (Y) Protein 2 (Y) Protein 3 (Y)| 0.20 0.048 0.29 0.17 0.50 0.11 0.50 0.33 1.0 0.20 0.67 0.50 0.50 0.89 0.80 0.71 0.95 0.91 0.83 0.97 0.93 0.99 0.98 0.95 50 Select the K, for hormone binding by protein 2. 0.50 nM 0.99 nM O 0.20 nM 50 nM Select the protein that binds most tightly to the hormone. O protein 3 O protein 1 O protein 2

Answer 1

1) when single subunit protein like myoglobin binds with single substrate (hormone) then binding graph is hyperbola type.

And mathematical expression for binding is

---- equation no.1

O=L/Kd+L)

Where $\Theta$ is fraction of hormone binding

L is ligand= here hormone

Kd = dissociation constant

So when $\Theta$ is 50% or 1/2 value then, equation 1 become

1/2= L/(Kd+L) => Kd+L =2 L => Kd = L

So when $\Theta$ is equal to 50% then Kd = L

Now, for protein 2, $\Theta$ become 50% or 0.5 when hormone concentration is 0.5 nM so answer is 0.5 nM which is option 1

2) similarly hormone conc. Is 4 nM for protein 1 at $\Theta$ 0.5 value while for protein 3 its value is 1 nM and for protein 2 we already calculated above which is 0.5 nM

Now on comparing ligand concentration which is hormone here for all three different protein at same $\Theta$ = 0.5 value, we found that lowest for protein 2 which is only 0.5 nM followed by protein 3 which is 1 nM and highest value is 4 nM for protein 1

So, at very low concentration of hormone concentration, protein 2 acheived$\Theta$ 0.5 or 50% site occupied so it's affinity is highest while protein 1 showed least affinity for its ligand which is hormone.

So more affinity means more strongly binds so answer wil be protein 2 which is option no. 3

Hope it's clear..thanks