A tree $T$ on $\omega$ is *recursively pointed* if it is recursive in each of its branches. We can consider a variant of Sacks forcing where the conditions are recursively pointed perfect trees ordered by inclusion. Given a $V$-generic filter $G$ for this forcing, we can define the real $x\_G = \bigcap\_{T\in G} [T]$. 
For every real $x \in V$ we have $x <\_\text{T} x\_G$. Given a real $x \in V[G]$ with $x <\_\text{T} x\_G$, must we have $x \in V$?


Probably the answer to this question is well-known, but I'm afraid I don't see how the proof of the minimality property for Sacks forcing (as given by Jech) might adapt to recursively pointed trees.



Does forcing with recursively pointed perfect trees add a Turing degree that is minimal over $V$?
If you two haven't worked this out already, it seems to me that $x\_G$ is not minimal as a Turing degree above $V$, but is a minimal $V$ degree. For the first, since $x\_G$ computes $0'$, it follows that $x\_G$ is the join of two mutually 1-generic reals, neither of which computes $x\_G$ and at least one of which is not in $V$. However, if $x\_G$ computes $x$ and $x$ is not in $V$, then there is an $a$ in $V$ such that $x\oplus a$ computes $x\_G$. This is by the usual minimality argument: $a$ is the degree of the splitting tree for the functional used to compute $x$ from $x\_G$.


