TITLE: Martin's cone theorem and recursion theory
QUESTION [9 upvotes]: Martin's remarkable cone theorem in the theory of determinacy says the following: 

Suppose $A\subseteq \omega^\omega$ is Turing invariant and determined. If $\forall x\exists y(x\le_T y\& y\in A)$ then $A$ contains a cone.

Let me explain what this means: $A$ is Turing invariant iff $\forall x\in A\forall y(x\equiv_T y\Rightarrow y\in A)$. Here, $\le_T$ is the relation of Turing reducibility and $\equiv_T$ is the corresponding equivalence relation.
"Determined" is in the usual sense of infinite games on integers. 
A cone is a set of the form $$C_y=\{x\mid y\le_T x\}.$$ Clearly, cones are Turing invariant. We say that $y$ is the base of the cone $C_y$. 
If $\forall x\exists y(x\le_T y\& y\in A)$ we say that $A$ is cofinal.
When Martin proved his theorem, he thought that it would be a quick way of showing determinacy fails (in ZF), by considering explicit sets coming from recursion theory. Instead, he found several results in recursion theory as a consequence.
Here are some examples: 

For every $x$ we have $x<_T x'$, where $x'$ is the Turing jump of $x$. This means that the set of jumps is cofinal. By Borel determinacy, it follows that there is a $y$ such that if $y\le_T x$, then $x\equiv_T z'$ for some $z$. Well known recursion theoretic results show that in fact we can take $y=0'$.
Again by Borel determinacy, there is a real $x$ such that any $y$ with $x\le_T y$ is a minimal cover above some $z$. Again, recursion theoretic arguments show that we can take $x=0^{(\omega)}$.

I do not know many examples coming from recursion theory, but maybe somebody here does. 

Are there (natural) examples of sets $A$ defined recursive theoretically that we know need to contain a cone, but for which we do not know of a (natural) base? 

"Need to contain a cone" could be taken to mean, say, that they are Turing invariant and cofinal and Borel.
Naively, a negative answer would mean we have very strong abstract basis results. But I would be interested in natural candidates for a positive answer as well.

REPLY [8 votes]: I would like to add another example.
Given a sentence $\phi$ from partial order language, then for any Turing degree $x$, either $D(\leq x)\models \phi$ or $D(\leq x)\models \neg\phi$. By the BD, there is a Turing degree $x_{\phi}$ so that either for all $y\geq_T x_{\phi}$, $D(\leq y)\models \phi$ or  for all $y\geq_T x$, $D(\leq y)\models \neg \phi$.
Let $z$ be a Turing degree above all the $x_{\phi}$'s, then for every $y\geq_T z$, $D(\leq y)$ is elementary equivalent to $D(\leq z)$. 
I don't know a natural base for this.