Let $X$ be a (sufficiently nice) scheme. Then the inclusion $\varphi$ of quasicoherent sheaves on $X$ into all $\mathcal O_X$-modules admits a right adjoint $Q$, the coherator. As a right adjoint, this functor is left exact. But is it right exact?
In Thomason-Trobaugh, paragraph B.17, it is casually mentioned that $Q$ is right exact. This is puzzling, coming as it does right after Proposition B.16 where something strictly weaker was just shown. Namely, it's shown that $R^q Q(\varphi F) = 0$ for $q > 0$ and $F$ quasicoherent, whereas to say that $Q$ is right exact is to say that $R^q Q (G) = 0$ for all $q> 0$ and all $\mathcal O_X$-modules $G$.
Moreover, Thomason-Trobaugh also show (B.15) that $Q$ preserves filtered colimits, so if $Q$ were right exact, it would preserve all colimits, and so have a further right adjoint by the adjoint functor theorem. Since nobody talks about such a functor, I very much suspect it doesn't exist.