Let $\Gamma$ be a connected graph. By (Kleitman-West, 1991), if every vertex of $\Gamma$ has degree $\geq 3$, then $\Gamma$ has a spanning tree with $\geq n/4+2$ leaves, where $n$ is the number of vertices of $\Gamma$.

It is relatively forward (though not completely trivial) to deduce that, if every vertex of $\Gamma$ has degree $\geq 2$, then $\Gamma$ has a spanning tree with $\geq n/4+2$ leaves, where $n$ is the number of vertices of $\Gamma$ of degree $\geq 3$.

Question: can the assumption on the degree of all vertices be dropped altogether? That is, is it true that every connected graph $\Gamma$ with $n$ vertices of degree $\geq 3$ has a spanning tree with $\geq n/4+2$ leaves? If not, can you give a counterexample?

Note 1: The one case in doubt is that where there is exactly one vertex of degree $1$. All other cases follow from (Bankevich-Karpov, 2011), which gives the lower bound $\geq m/4+3/2$, where $m$ is the number of vertices of $\Gamma$ of degree not $2$. Alternatively, one may reduce the general problem to the case where exactly one vertex has degree $1$ as follows: given two vertices $v_1$, $v_2$ of degree $1$, we may identify them (not changing the number of vertices of degree $\geq 3$ thereby) and apply the bound we are proving, recursively (since the number of vertices of degree $1$ has decreased); if the spanning tree contains the new vertex $v$ as a leaf, it is valid as a spanning tree of the original graph; if it contains $v$ as an internal vertex, we separate $v$ again into $v_1$ and $v_2$ (thus increasing the number of leaves by $2$), and find that we have two trees, covering all vertices of $\Gamma$; there is some edge of $\Gamma$ connecting them, and we may add it at a cost of at most $2$ leaves.

Note 2: It obviously follows from Bankevich-Karpov that, when there is exactly one vertex of degree $1$, the bound $\geq n/4+7/4$ holds. It then follows from (Karpov, 2012) that a counterexample to $\geq n/4 + 2$ would need to have no vertices of degree $>3$.