Suppose $(M, \mathcal X) \models ACA_0$. Recall that a subset $A \subseteq M$ is $inductive$ over $M$ if $M$ satisfies all instances of induction in the expanded language with a predicate for $A$. Suppose $A$ is inductive and let us denote by $(M, \mathcal X[A])$ the model whose second order part consists of all sets definable from parameters in $\mathcal X$ and $A$. What I want to know is whether or not this is necessarily again a model of $ACA_0$.

If it helps, my intended application concerns countable models ($M$ and $\mathcal X$ are both countable) and I know that the $A$ I have in mind is in fact inductive in the larger language i.e. $M$ satisfies all instances of induction in the language with predicates for $A$ and all elements of $\mathcal X$ (note that every set in a model of $ACA_0$ is inductive).

I suppose more generally I'm interested in knowing, given $A \notin \mathcal X$ which is inductive over $M$, what do I need to require of a $\mathcal Y$ so that $A \in \mathcal Y$, $\mathcal X \subseteq \mathcal Y$ and $(M, \mathcal Y) \models ACA_0$?