Question

Which of the following conditions would lead to signal-independent modification of the membrane lipids by Rafty?

1) the expression of a constitutively active phospholipase C
2) a mutation in the GPCR that binds the signal more tightly
3) a Ca₂⁺ channel in the endoplasmic reticulum with an increased affinity for IP3
4) a mutation in the gene that encodes Rafty such that the enzyme can no longer be phosphorylated by PKC

Answer 1

The conditions that could lead to signal-independent modification of membrane lipids by Rafty include the expression of a constitutively active phospholipase C, a mutation in the GPCR that binds the signal more tightly, and a mutation in the gene that encodes Rafty.

Step-by-step explanation:

The conditions that would lead to signal-independent modification of the membrane lipids by Rafty are:

1. The expression of a constitutively active phospholipase C: Phospholipase C generates second messengers such as inositol triphosphate (IP3) and diacylglycerol (DAG) that can modify membrane lipids independently of external signals.
2. A mutation in the GPCR that binds the signal more tightly: This would increase the activation of phospholipase C, leading to signal-independent modification of membrane lipids.
3. A mutation in the gene that encodes Rafty: If the enzyme Rafty can no longer be phosphorylated by PKC, it can modify membrane lipids independently of PKC activation.