Question

Given the situation where normal cells grown under standard conditions (without ligand) are 14 µm in diameter while normal cells exposed to TRK ligand are 10.5 µm in diameter, which of the following conditions do you predict will more likely lead to smaller cells?

1) addition of TRK ligand and a drug that stimulates the GTPase activity of Ras
2) addition of TRK ligand and a drug that inhibits the activity of the phosphatase that acts on SZE
3) addition of TRK ligand and a drug that stimulates the degradation of Pig1
4) addition of TRK ligand and a drug that inhibits Pig1 binding to DNA

Answer 1

The condition that would most likely lead to smaller cells is the addition of TRK ligand and a drug that inhibits the activity of the phosphatase that acts on SZE, as this would prolong the growth-reducing signal transduction activated by the TRK ligand.

Step-by-step explanation:

Given the situation where normal cells grown under standard conditions are 14 µm in diameter while normal cells exposed to TRK ligand are 10.5 µm in diameter, we can evaluate which condition would likely lead to the smallest cells. The second option, addition of TRK ligand and a drug that inhibits the activity of the phosphatase that acts on SZE, is predicted to lead to smaller cells because inhibiting a phosphatase likely prolongs the signal transduction activated by the TRK ligand, thereby enhancing its effect of reducing cell size.

Introduction of the TRK ligand activates receptor tyrosine kinases (RTKs) which then initiate a signaling pathway involving a G-protein called RAS, leading to the activation of the MAP kinase pathway, which stimulates cell division. Since adding TRK ligand already leads to a reduction in cell size, further inhibiting the deactivation of this pathway (which could be the role of the phosphatase on SZE) would likely result in even smaller cells due to increased signaling activity.