If C in P3 is an elliptic quintic, then h0(IC(2))=0. This follows from the fact that the system: a + b = 5, (a-1)(b-1) = 1 has no integral solutions. It follows that h1(IC(2))=0 and, by Castelnuovo-Mumford's lemma: h1(IC(k)) = 0, k > 1 (n.b. of course h1(IC(1)) = 1) and the homogeneous ideal I(C) is generated by its degree 3 elements.
The minimal free resolution looks like:
0 -> +O(-ct) -> +O(-bj) -> 5.O(-3) -> IC -> 0
Now, just using general facts, we can determine the whole minimal free resolution. We have c+ := max{ct} = c(C) + 4 = 5 and 5 = c+ > b+ > 3, so b+ = 4. Since we also have c_ > b_ > 3 (where c_ := min{ct} etc...), we conclude that ct = 5 for all t, bj = 4 for all j. Now card{t / ct = 5} = h1(IC(1)) = 1. Counting the ranks, we conclude that the minimal free resolution is of the form:
0 -> O(-5) -> 5.O(-4) -> 5.O(-3) -> IC -> 0
An elliptic quintic has no 4-secant line (such a 4-secant line would be contained in any cubic containing C, but the intersection of all these cubics is C; stated differently a curve of degree d with a (d-1)-secant line is rational), but through any point of C, there passes a 3-secant line to C (otherwise projecting from that point would yield a smooth quartic curve of genus 1, but a smooth plane quartic has genus 3).
The general plane section, C.H, of C consists of 5 points in general position. In particular h0(ICH(2)) = 1. This means that the conic containing C.H doesn't come from a quadric surface containing C. In other words the conics containing C.H, do not glue, as H varies, to build a quadric surface containing C. Think about it...