Answer: 1. Cyclic AMP(cAMP) concentration increases;
2. cAMP then binds to the two regulatory subunits;
3. A change in the conformation of enzyme results. The autoinhibitory sequences move away from the active site of the catalytic subunits;
4. The catalytic sites are now free to catalyse the various reactions of PKA.
(Note: The question did not list the steps to be placed in the correct order)
Explanation:
The inactive form of Protein Kinase A (PKA) contains two identical sub-units (C) and two identical regulatory sub-units (R). Two auto-inhibitory sequences of the regulatory subunits occupy the catalytic sites of the catalytic subunits. After the reaction of adenylyl cyclase, this inactive form of PKA is activated in the following steps:
1.<em> </em><em>Cyclic AMP</em><em>(</em><em>cAMP</em><em>) concentration increases;</em>
<em>2. </em><em>cAMP</em><em> then binds to the two regulatory subunits;</em>
<em>3. A change in the conformation of enzyme results. The autoinhibitory sequences move away from the active site of the catalytic subunits;</em>
<em>4. The catalytic sites are now free to catalyse the reactions various of </em><em>PKA</em><em> which includes the regulation of glycogen, sugar, and lipid metabolism. </em>