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Is this true that nonplanar graphs which chromatic numbers is less than 5 is almost(!) bipartite?

I think the nonplanarity is because of minor of $K_{3,3}$ because they can't have $K_5$ (am I right?) so it must be some kind of bipartite with some additional edges. It means it have a giant max cut?

Is there some results concerning this?

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    You certainly can have a $K_5.$ think of one with edges subdivided. – 2017-01-02
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    By Brook's theorem (https://en.wikipedia.org/wiki/Brooks'_theorem) almost every graph with degree $\leq 5$ has chromatic number $\leq 5$. So, definitely no. – 2017-01-02

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I don't know what "almost" bipartite means here, but it seems that you can make examples that are quite far from being bipartite.

For example let you graph has node set $\{1,2,3,4\}\times\{1,2,\ldots n\}$ for a sufficiently large $n$, and edges between any two nodes with different first components. This has chromatic number $4$ and is non-planar (since it has $K_5$ as a minor) for $n\ge 2$

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    Why it has $K_5$ as a minor? As I know for finding a minor you must partition your graph to stable sets and join to sets iff there is an edge between those edges. – 2017-01-02
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    @Bellmondo: $K_5$ is the minor resulting for the partitioning into $\{(1,1)\}, \{(2,1)\}, \{(3,1)\}, \{(4,1)\}$, and then everything else. – 2017-01-02
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    Everything else is not a stable set! – 2017-01-02
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    @Bellmondo: It doesn't have to be (or if it does, your concept of a graph minor is quite nonstandard). – 2017-01-02
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    But if it does not $C_n$ is nonPlanar too! Consider a large enough n and partition it to five random parts. With high probability these part have edge between hence it would be a $K_5$ minor – 2017-01-02
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    @Bellmondo: Very confused now. Exactly what does a "stable set" mean to you here? The sets in the partition have to be _connected_, but "everything else" _is_ connected. – 2017-01-02
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    Oops! That's it. I confused the definition of minor! Thanks & apology. – 2017-01-02