if and only if $\mathfrak{Quad}^4_{g, 2a+3}(C)\neq \emptyset$. Noting that $\rho(g, 4, 2a+3)=2a-9$, whereas $h^0(C, L^{\otimes 2})=2a+6$, one computes that the virtual dimension of $\mathfrak{Quad}_{g, 2a+3}^4(C)$ as a determinantal variety, is equal to $-1$. Since it is not difficult to provide examples of *embedded* curves $C\subset \PP^4$ of genus $g(C)=2a+1$ and $\mbox{deg}(C)=2a+3$, which lie on a single quadric such that $\OO_C(1)\in \mathfrak{Quad}_{g, 2a+3}^4(C)$ is an isolated point, one infers that only two scenarios are possible:

1.  $\mathfrak{Syz}_{g, 2a+3}^4$ is a divisor inside $\cM_g$, that is, $K_{1, 1}(C, L)=0$ for a general curve $[C]\in \cM_g$ and for *every* $L\in W^4_{2a+3}(C)$, or

2.  $\mathfrak{Syz}_{g, 2a+3}^4=\cM_g$.

Conjecture $(MRC)_{2a+1, 2a+3}^4$ predicts that the second possibility does not appear. In any event, the case of $\PP^4$ ought to be one of the more manageable situations for testing MRC in arbitrary genus. We can confirm this expectation for $a=5$. *Proof of Theorem \[gen11\].* Assume by contradiction that for a general curve $[C]\in \cM_{11}$ there exists a linear series $L\in W^4_{13}(C)$ such that $C\stackrel{|L|}\hookrightarrow \PP^4$ lies on a quadric $Q\subset \PP^4$. We claim that $Q$ must be smooth, because otherwise, $\mbox{rank}(Q)\leq 4$, and then $L$ is expressible as the sum of two pencils. This contradicts the fact that 