Moduli of pointed Curves Every curve of genus $g\leq 2$ has non trivial automorphisms. So the fiber of the forgetful morphism $\pi:\bar M_{g,1}\rightarrow\bar M_{g}$ over $[C]$ is not isomorphic to $C$ but to $C/Aut(C)$ (I am considering the coarse moduli space instead of the stack).
The general fiber of $\pi:\bar M_{1,2}\rightarrow\bar M_{1,1}$ is isomorphic to $\mathbb{P}^{1}$ and the same is true for $\pi:\bar M_{2,1}\rightarrow\bar M_{2}$. Now $\bar M_{1,2}$ is a rational surface and $\bar M_{2,1}$ is rational $4$-fold.
The question is the following:
Does anyone know an explicit description of $\bar M_{1,2}$ and $\bar M_{2,1} ?$
My feeling is that $\bar M_{1,2}$ should be a blow-up of a ruled surface.
 A: The moduli space $\overline{M}_{1,2}$ can not be the blow up of $\mathbb{P}^{2}$ in $2$ points because its rational Picard group has rank $2$, indeed it is generated by the divisor parametrizing genus $0$ irreducible nodal curves with $2$ marked points and the divisor parametrizing reducible curves whose components are a smooth genus $1$ curve and a smooth genus $0$ curve with $2$ marked points.
The moduli space of genus $1$ stable curves with $2$ marked points is a rational surface with four singular points. Two singular points lie in $M_{1,2}$, and are:
a singularity of type $\frac{1}{4}(2,3)$ representing an elliptic curve of Weierstrass representation $C_{4}$ with marked points $[0:1:0]$ and $[0:0:1]$;
a singularity of type $\frac{1}{3}(2,4)$ representing an elliptic curve of Weierstrass representation $C_{6}$ with marked points $[0:1:0]$ and $[0:1:1]$.
The remaining two singular points lie on the boundary divisor $\Delta_{0,2}$, and are:
a singularity of type $\frac{1}{6}(2,4)$ representing a reducible curve whose irreducible components are an elliptic curve of type $C_{6}$ and a smooth rational curve connected by a node;
a singularity of type $\frac{1}{4}(2,6)$ representing a reducible curve whose irreducible components are an elliptic curve of type $C_{4}$ and a smooth rational curve connected by a node.
From this one can prove that:
The moduli space of genus $1$ stable curves with $2$ marked points is isomorphic to a weighted blow up of the weighted projective plane $\mathbb{P}(1,2,3)$ in its smooth point $[1:0:0]$. In particular $\overline{M}_{1,2}$ is a toric variety.
A: Since an elliptic curve is determined by its double branched cover to $P^1$, the coarse space $\overline{M}_{1,1}$ is isomorphic to $\overline{M}_{0,4}/S_3$ where the quotient by $S_3$ is because only one of the branched points is marked. Thus $\overline{M}_{1,2} $ is isomorphic to $ \overline{M}_{0,5}/S_3$, at least on the level of coarse spaces. 
$\overline{M}_{0,5}$ is famously the blowup of $CP^2$ at four points and the map from $\overline{M}_{0,5}$ to $\overline{M}_{0,4}$ is the family given by the linear system of conics through the 4 points. If I take those 4 points in $CP^2$ to be (1:0:0), (0:1:0), (0:0:1), (1:1:1), then I can perform the quotient pretty explicitly --- it is induced by the permutation of the homogeneous coordinates $(x,y,z)$. This quotient is proj of the ring of symmetric functions which is generated by $u=x+y+z$, $v=xy+xz+yz$, and $w=xyz$. This is the weighted projective space $CP(1,2,3)$ and the orbits of the 4 points in $CP^2$ become two points $(1:0:0)$ and $(3:2:1)$. 
So I think (if all my above logic is correct) that $\overline{M}_{1,2}$ is isomorphic (as coarse spaces) to the blowup of the weighted projective space $CP(1,2,3)$ at the two points $(1:0:0)$ and $(3:2:1)$. Note that $CP(1,2,3)$ has only two singular points at (0:1:0) and (0:0:1) (of type $A_1$ and $A_2$) which are away from the blowup points.
The same kind of argument should work on $\overline{M}_{2,1}$ since by similar logic the coarse space of is isomorphic to $\overline{M}_{0,7}/S_6$ and $\overline{M}_{0,7}$ has a description as a blowup of $CP^4$ (with the ruling given by a pencil of rational normal curves). 
