Curves and trisecant lines We know that rational normal curves and elliptic normal curves have no trisecant lines. For the "next" case, this is still true. That is, a nondegenerate curve of degree $d\geq 5$ and genus $2$ in $\mathbb{P}^{d-2}$ also has no trisecant lines, because if there were one, by projecting from the trisecant line we would get a degree $d-3$ curve in $\mathbb{P}^{d-4}$ of genus $\geq 2$.
I am wondering if this phenomenon keeps holding for the next cases. Precisely: Are there nondegenerate curves $C\subseteq \mathbb{P}^{d-g}$ of degree $d$ and genus $g$ having a trisecant line for $g\geq 3$?
The expected dimension of trisecant lines to a given curve $C\subseteq \mathbb{P}^r$ is $4-r$ (Yes, it is independent of the degree and the genus of the curve). Thus, as the dimension of the projective space grows, it is less and less probable to find any curve with a trisecant line. However, as the genus grows the dimension of the parameter space (Hilbert scheme) grows too and my feeling is that from some genus onwards we can start getting curves that have a trisecant line.
Any help will be appreciated.
 A: Edit.  Actually, every nondegenerate curve in $\mathbb{P}^3$ except for twisted cubics and elliptic quartics has a $1$-parameter family of trisecant lines: Curve of 3-secant lines
I am just expanding my comment into an answer.  Let $\Sigma \cong \mathbb{P}^1\times \mathbb{P}^1$ be a smooth quadric surface in $\mathbb{P}^3$.  For every integer $g\geq 2$, let $C$ be a smooth curve in $\Sigma$ of bidegree $(g+1,2)$.  Then the genus of the curve $C$ is $(g+1-1)(2-1) = g$.  Also, the total degree of $C$ equals $d=g+3$.  The lines of ruling of $\Sigma$ that have intersection number $g+1$ with $C$ give a $1$-parameter family of $(g+1)$-secant lines to $C$.  In particular, a degree $5$ curve of genus $2$ can have a $1$-parameter family of trisecant lines.
A: Take any curve $C$ of genus $g\geq 5$. Let $\eta $ be an element of $\mathrm{Pic^o}(C)$ which is linearly equivalent to $A-B$, with $A,B\in \mathrm{Sym}^3(C)$, but not to $A'-B'$ with  $A,B\in \mathrm{Sym}^2(C)$ (such elements exist for dimension reasons). Then the linear system $|K_C+\alpha |$ is very ample, and embeds $C$ as a curve of degree $d=2g-2$ in $\mathbb{P}^{g-2}$. Since $h^0(K+\alpha -A)=h^0(K-B)=h^0(K+\alpha )-2$, the three points of $A$ are collinear in this embedding, hence span a trisecant.
