Comparison between MA and $P^{NP}$

Question

I am aware that $MA \subseteq ZPP^{NP} \subseteq \Sigma_2^{P}\cap \Pi_{2}^{P}$ and also $NP^{BPP} \subseteq MA \subseteq AM \subseteq \Pi_{2}^{P}$.

My question is: In this whole thing, where does $P^{NP}$ lie? (Obviously, $P^{NP} \subseteq ZPP^{NP}$) But, what about the relationship between $MA$ and $P^{NP}$? Which is contained in which one?

Intuitively, It is known that $MA_{EXP} \not\subset P/poly$, but for $EXP^{NP}$, no polynomial size circuit lower bound is known. Does that mean $P^{NP}$ is contained in $MA$?

Answer 1

Apparently, no relationship is known between $MA$ and $P^{NP}$. Even $BPP$ ($\subseteq MA$) vs $P^{NP}$ problem is still open. It is conjectured though that $MA = NP$ (Under standard hardness assumptions). So, it is likely that $MA \subseteq P^{NP}$.