The dissociation constant for the binding of the ligand to a GPCR is 10^-10 M. I

Question

The dissociation constant for the binding of the ligand to a GPCR is 10^-10 M. It was found that ligand binding to only 10% of the around 1000 GPCRs on the cell is required for the maximal activation of the G protein. Calculate what is the ligand concentration required to bring about the maximal activation. Now you mutate the G-protein a subunit to increase its affinity to GTP. What will be the effect on ligand binding to the receptor and why? If these mutant G-proteins are now treated with a non-hydrolyzable analog of GTP, what is going to happen to the ligand binding and why?

Explanation / Answer

Answer (a). Kd for ligand binding was 10-10

Kd= [GPCR][ligand] / [GPCRLigand]

For maximal activity 10% of 1000 i.e., 100 should bind

Therefore, 10-10= (1000-100)[ligand]/ 1000

or (1000-100)[ligand] =10-10 x 1000

900 x [ligand] = 10-7

[Ligand]=1/(900x107)

[ligand] = 1x 10-10 M

Ligand concentration to carry out maximum binding is 0.1 nM

(b). Mutation of G protein alpha subunit to enhance binding to GTP would increase the action of secondary messengers and need for further ligand binding would be delayed. G alpha protein has the ability to slowly hydrolyze GTP to GDP, due to which alpha-GDP would attach to beta and gamma proteins to return to inactive state and wait for ligand binding.

In case if these mutant G alpha proteins were treated with a non-hydrolyzable analog of GTP, the binding of G alpha protein to these analogs would be permanent and this would lead to negative effect on the ligand binding.

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