Let $P,Q \subseteq \Sigma^*$ be languages such that $P$ and $Q$ are both in NP. Assume that $P$ is NP-complete under Karp reductions (there is a polynomial time many-one reduction from 3-SAT to $P$) and there is a Turing-reduction from $P$ to $Q$, i.e. there is a polynomial-time algorithm which decides $P$ when given oracle-access to $Q$. Intuitively this means that deciding $Q$ is as hard as deciding any problem in NP, while at the same time $Q$ is in NP, so Q cannot be ``harder'' than NP. Does this imply that $Q$ is NP-complete under Karp reductions, i.e. does this imply the existence of a polynomial-time many-one reduction from 3-SAT to Q? Any pointers are much appreciated.