Spanning trees: the last darn $1/4$

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?

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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 $>3$.