SBA — Single Best Answer · Review
Question 494
Endocrinology → General
Classification
Question
SSBAQuestion Header
A peptide hormone binds to a seven-transmembrane cell-surface receptor and activates adenyl cyclase. Which intracellular event follows most directly from this receptor activation?
Question Stem
Select the single best answer.
Options
A
Direct binding of the hormone-receptor complex to nuclear DNA
B
Displacement of HSP-90 from an intracellular steroid receptor
C
Generation of cAMP with subsequent activation of protein kinases
D
Direct conversion of cholesterol to pregnenolone
E
Inhibition of all intracellular phosphorylation
Explanation
Seven-transmembrane G-protein-coupled receptors can activate adenyl cyclase, increasing intracellular cAMP. cAMP acts as a second messenger and activates protein kinases, leading to phosphorylation of intracellular proteins. Phosphodiesterase limits excessive accumulation of cAMP.
Option Validity
A) Direct nuclear DNA interaction is characteristic of intracellular steroid-hormone receptor signalling rather than this membrane-receptor pathway.
B) HSP-90 displacement is associated with ligand binding to steroid receptors.
D) Cholesterol-to-pregnenolone conversion is a steroidogenic step and is not the immediate consequence of adenyl cyclase activation described here.
E) cAMP activates protein kinases and promotes phosphorylation rather than inhibiting it.
Further Reading
Saxena, MRCOG-1 Basic Sciences, Endocrinology: Mechanism of Action of Hormones, pp. 301–302.