TPS631010: Designing with TPS631010YBGR

Hello Team,
I am using the buck-boost converter TPS631010YBGR in one of our projects.
The input voltage is 3V from a LiMnO2 primary battery.
The required output voltage is 1.8V at 100mA.
The 1.8V is to power an MCU.
The MCU will be in sleep mode most of the time and the power consumption will be less than 400uA.
To measure the power consumed by the MCU alone we are planning to give a 1.8V output of the TPS631010YBGR keeping the regulator disabled and no voltage at the input of the TPS631010YBGR.
The TPS631010 doesn't have an active discharge circuit.
So this procedure may not damage the IC.
Can you please comment on this?

Also, the IC has a feature of Reverse Current Operation.
Can anyone please explain what is Reverse Current Operation?
Does this mean if there is no voltage at the input and if a voltage is present at the output of the regulator, the regulator will reverse boost/Buck the voltage?

Also what will happen to the Reverse Current Operation if the chip is disabled?



Looking for your reply