We will use the terminology of graph theory. Here the vertices of our main graph $G$ are the bacteria and there is an edge between two precisely when they are friends. The degree $d(v)$ of a vertex $v$ of $G$ is the number of neighbours of $v$. The minimum degree $\delta(G)$ of $G$ is the smallest amongst all $d(v)$ for vertices $v$ of $G$. The chromatic number $\chi(G)$ of $G$ is the number of colours needed in order to colour the vertices such that neighbouring vertices get distinct colours.
It suffices to establish the following:

Claim. Let $k$ be a positive integer and let $G$ be a graph on $n > k$ vertices with $\delta(G) \ge 1$ and $\chi(G) = k$. Suppose that merging one pair or two pairs of vertices results in a graph $G'$ with $\chi(G') \le k - 1$. Then $\delta(G) \ge k$.

We establish this in a series of claims.

Claim 1. $\delta(G) \ge k - 1$.

Proof. Suppose for contradiction that we have a vertex $v$ of degree $r \le k - 2$ and denote its neighbours by $v_1, ..., v_p$. (Note that, by assumption, $v$ has at least one neighbour.)

Suppose we merge $v$ with $v_i$. We denote the new vertex by $v_0$, and we colour the obtained graph in $k - 1$ colours. Note that at most $r \le k - 2$ colours can appear in the set $S_1 = \{v_0, v_1, ..., v_{i-1}, v_{i+1}, ..., v_p\}$. Therefore we can get a $(k - 1)$-colouring of $G$ by assigning the colour of $v_0$ to $v_i$ and an unused colour (from the $k - 1$ available) to $v$, thus contradicting the assumption that $\chi(G) = k$. $\square$

So from now on we may assume that there is a vertex $v$ of $G$ with $\deg(v) = k - 1$, as otherwise the proof is complete. We denote its neighbours by $v_1, ..., v_{k-1}$.

Claim 2. The set of neighbours of $v$ induces a complete graph.

Proof of Claim 2. Suppose $v_i v_j \notin E(G)$. Merge $v$ with $v_i$, giving a next vertex $w$, and then merge $w$ with $v_j$, denoting the newest vertex by $v_0$. Then colour the resulting graph in $k-1$ colours. Note that at most $k-2$ colours can appear in the set $S_2 = \{v_0, v_1, \dots, v_{k-1}\} \setminus \{v_i, v_j\}$. So we can get a $(k-1)$-colouring of $G$ by assigning the colour of $v_0$ to $v_i$ and $v_j$ and an unused colour (from the $k-1$ available) to $v$, thus contradicting the assumption that $\chi(G) = k$. $\square$

Claim 3. For every edge $uw$, both $u$ and $w$ belong in the set $\{v, v_1, ..., v_{k-1}\}$.

Proof. Otherwise merge $u$ and $w$ and call the new vertex $z$. If $u, w \notin \{v, v_1, ..., v_{k-1}\}$ then by Claim 2 the resulting graph contains a complete graph on $\{v, v_1, ..., v_{k-1}\}$ and so its chromatic number is at least $k$, a contradiction. If one of $u, w$ belongs in the set $\{v, v_1, ..., v_{k-1}\}$, say $u = v_i$, then the resulting graph contains a complete graph on $\{v, v_1, ..., v_{k-1}, z\} \setminus \{v_i\}$. This is again a contradiction. $\square$

From Claim 3 we see that $G$ consists of a complete set on $k$ vertices together with $n-k > 0$ isolated vertices. This is a contradiction as $\delta(G) \ge 1$.