Consider the following sequence of micro-operations. MBR $\leftarrow$ PC MAR $\leftarrow$ X PC $\leftarrow$ Y Memory $\leftarrow$ MBR Which one of the following is a possible operation performed by this sequence?

Let us analyze the micro-operations step-by-step:

1. $\text{MBR} \leftarrow \text{PC}$: The current content of the Program Counter (PC), which is the return address, is moved to the Memory Buffer Register (MBR).

2. $\text{MAR} \leftarrow \text{X}$: A memory location $\text{X}$ (such as a stack pointer or reserved vector address) is loaded into the Memory Address Register (MAR).

3. $\text{PC} \leftarrow \text{Y}$: The starting address $\text{Y}$ of the Interrupt Service Routine (ISR) is loaded into the PC.

4. $\text{Memory} \leftarrow \text{MBR}$: The return address stored in MBR is saved to memory at address $\text{X}$.

This sequence of saving the return address and branching to the ISR address is the standard procedure for the Initiation of interrupt service.

Hence, option (D) is correct.