Question

Activated protein kinase C (PKC) can lead to the modification of the membrane lipids in the vicinity of the active PKC. You have discovered the enzyme activated by PKC that mediates the lipid modification. You call the enzyme Rafty and demonstrate that activated PKC directly phosphorylates Rafty, activating it to modify the plasma membrane lipids in the vicinity of the cell where PKC is active. Explain how each of the following will affect the lipid modification:

A. a Ca2+ channel in the endoplasmic reticulum with an increased affinity for IP3

B. a mutation in the gene that encodes Rafty such that the enzyme can no longer be phosphorylated by PKC

Answer 1

a) IP3 (Inositol Triphosphate), is produced when PIP2 (Phosphatidyl inositiol- 4,5-bisphosphate) breaks into IP3 and DAG. IP3 causes an increase in Calcium ion concentration. PKC is activated by calcium ions and DAG (DIacylglycerol). Increase in calcium ions because of IP3 will activate more PKCs and hence more of Rafty proteins will be recruited and lipid modification will increase

b) Mutated Rafty protein produced as a result of mutated Rafty gene will be unable to be activated by PKC. Hence control of lipid modification and the cell signaling pathways thereof will be impaired. Since calcium ions, IP3 and ligands binding to IP3 receptors are the major players, these control will be nullified and lipid modification will remain unregulated by these factors.