You are interested in cell-size regulation and discover that signaling through an enzyme-coupled receptor is important for the growth (enlargement) of mouse liver cells

Question

Activation of the receptor activates adenylyl cyclase, which ultimately leads to the activation of PKA, which then phosphorylates a transcription factor called TFS on threonine 42. This phosphorylation is necessary for the binding of TFS to its specific sites on DNA, where it then activates the transcription of Sze2, a gene that encodes a protein important for liver cell growth. You find that liver cells lacking the receptor are 15% smaller than normal cells, whereas cells that express a constitutively activated version of PKA are 15% larger than normal liver cells. Given these results, predict whether you would expect the cell's size to be bigger or smaller than normal cells if cells were treated in the following fashion.
A. You change threonine 42 on TFS to an alanine residue.
B. You create a version of the receptor that is constitutively active.
C. You add a drug that inhibits adenylyl cyclase.
D. You add a drug that increases the activity of cyclic AMP phosphodiesterase.
E. You mutate the cAMP-binding sites in the regulatory subunits of PKA, so that the complex binds cAMP more tightly.

Answer 1

A. Smaller. This mutation will make a TFS that cannot be phosphorylated by PKA.
B. Bigger.
C. Smaller.
D. Smaller. cAMP phosphodiesterase is involved in converting cAMP to AMP and will down-regulate this signaling pathway.
E. Bigger. Higher affinity of the PKA complex for cAMP will increase its activity, and thus cells will be bigger.