recursively extracting a MIS from an undirected simple graph $G$ does produce a minimal coloring for $G$ ?

I searched extensively the internet and found a paper [1] which answer partially to this question.

In this paper is shown a counterexample which is the graph depicted at the end of this post. In this graph if we extract the independent set $\{4,5,6\}$ we get necessarily a 4-colors coloring, while $\chi(G)=3$ for example by using the coloring $\{1,5,6\},\{2,4\},\{3\}$.
However, it can be noted, that in this particular instance every minimum coloring can be produced from MIS extraction by selecting the proper MIS at every step.

So the question is, is there always at least one proper ordering of MIS extraction which results in a minimum coloring for $G$ ?

The answer should be no, because the following related simpler statement is false for a generic graph either, but I have no authoritative quotation for this.

*"there exist a minimum coloring for $G$ in which one color class is a MIS for $G$ ?"*

However it's easy to see that a coloring in which a color class is a MIS requires at most $\chi(G)+1$ colors:

Extract a MIS $S$ from $G$ and color it with the same color. We have $\chi(G)-1\le\chi(G\setminus S)\le\chi(G)$, so a minimum coloring for $G\setminus S$ plus
the color used for $S$ is a feasible coloring for $G$ which uses at most $\chi(G)+1$ colors.

1---------2 |\ /| | \ / | | \ / | | 3 | | /|\ | | / | \ | | / | \ | |/ | \| 4 5 6

[1] S.I. Butenko, C.W. Commander, and P.M. Pardalos. On the complexity of the broadcast scheduling problem, University of Florida Technical Report, 2004