The tumor suppressor pRB functions in the cell cycle by regulating progression from G1 to S-phase. In a cell line derived from a colon carcinoma it is found that the activity of pRB is defective. Which of the following proteins is most likely to be hyperactive in these tumor cells as a consequence of the defective function of pRB?
A) anaphase-promoting complex
B) E2F
C) p21
D) p53
E) securin

Question

Quizplus · Accepted Answer

E2F is a transcription factor that is normally inhibited by pRB. When pRB is defective, E2F becomes hyperactive, leading to uncontrolled progression from G1 to S-phase.  In the context of cell cycle regulation, the transcription factor E2F activates the expression of cyclin A, cyclin E, and CDK2. The activity of E2F is itself controlled via interaction with pRB. When pRB binds E2F it can no longer function as a transcription factor as it is sequestered in the cytosol. Interaction of pRB and E2F correlates to the state of phosphorylation of pRB and the affinity between the 2 proteins is highest when pRB is hypophosphorylated. Phosphorylation of pRB is maximal at the start of S-phase and lowest after mitosis and entry into G1. When pRB is phosphorylated by G1 cyclinCDK complexes (cyclin D-CDK4/6), it releases E2F allowing E2F to transcriptionally activate its target genes.

The Tumor Suppressor PRB Functions in the Cell Cycle by Regulating