Criterion for transverse boundary intersection of one-parameter family in $\overline M_{g,n}(X,\beta)$ Suppose we have a projective, nonsingular, convex variety $X$, $\beta \in H_2(X,\mathbb{Z})$ and a family 
$$
\begin{array}{ccc}
\mathcal{C}& \to & X \cr 
\downarrow& &  &  \cr
B &  & 
\end{array}.
$$
together with sections $\sigma_1, \ldots, \sigma_n: B \to \mathcal{C}$ making the diagram above a family of stable maps from $n$-pointed genus $g$ curves over $B$, inducing a map $B \to \overline M_{g,n}(X,\beta)$.
Assume $B$ is a smooth curve and all fibres $\mathcal{C}_b$ are smooth except for one fibre over a point $0 \in B$ which is a curve with one node. 
Question: Is $\mathcal{C}$ being smooth a sufficient criterion for $B \to \overline M_{g,n}(X,\beta)$ being transverse to the boundary? If not, how could I try showing transversality (or computing the multiplicity of the intersection)?
If this helps: I only need the answer for $g=0$ and $X=\mathbb{P}^1 \times \mathbb{P}^1$.
 A: The following argument should answer the original question. However it still seems that the right way to do it is via the Artin stack of $n$-pointed, at-worst-nodal curves as proposed by Jason Starr. 
My argument why $\mathcal{C}$ being smooth suffices for obtaining transversality  goes as follows:


*

*Over the locus $\Delta_1 \subset \overline M_{0,n}(X, \beta)$ of $q=(f:C \to X; p_1, \ldots, p_n)$ with $C$ having exactly one node, the universal family $\mathcal{U} \to \overline M_{0,n}(X, \beta)$ has a section corresponding to this node (where the fibre over $q$ is identified with $C$). This can be shown using the inductive boundary structure via gluing maps.

*Locally around $0 \in B$ the family $\mathcal{C} \to B$ is the pullback of this universal family (if $0$ has no automorphisms). 

*If $B$ was tangent to the boundary at $0$ then every tangent vector $v$ of $0$ in $B$ would be in the tangent space of $\Delta_1$. 

*Via pushforward by the section of the universal family over $\Delta_1$ we obtain a tangent vector in $\mathcal{C}$ based at the node $N$ of $\mathcal{C}_0$ which maps to $v$ under $\pi$.

*This is a contradiction: the derivative of $\pi$ at $N$ must vanish because otherwise, by the implicit function theorem, $\mathcal{C}_0 = \pi^{-1}(0)$ would be smooth around $N$.
