Let $G$ be a graph in which any two odd cycles have a common vertex. It is easy to see that $\chi(G)\leq 5$ (choose minimal odd cycle $C$, use two colors for $G\setminus C$ and three colors for $C$). And this is sharp as $K\_5$ shows.
May we improve this bound to $\chi(G)\leq 4$ assuming additionally that $G$ does not contain $K\_5$? Or to $\chi(G)\leq 3$ assuming that $G$ does not contain $K\_4$? I can not prove even $\chi(G)\leq 4$ for $G$ without $K\_3$.



Graphs in which any two odd cycles have a common vertex
The claim on graphs without $K\_5$ is a particular case of the (still open in general) [Erdos--Lovasz Tihany conjecture](http://www.math.ucsd.edu/~erdosproblems/erdos/newproblems/DecomposeToIncreaseChromaticNumber.html). (Tihany is not a surname, but the name of a peninsula on Balaton lake in Hungary.)


This particular case has been proved in [W.G. Brown and H.A. Jung, On odd circuits in chromatic graphs, Acta Math. Acad. Sci. Hungarica, V. 20(1), pp. 129-134](http://link.springer.com/article/10.1007%2FBF01894573). In fact, this paper contains more general results.


As I already mentioned in the comments, this bound cannot be lowered to 3 even if the graph contains no $K\_3$. An example is provided by the Mycielski graph of $C\_5$. To see that it does not contain two disjoint odd cycles, notice that each odd cycle either contains the apex and an edge of $C\_5$, or contains at least one of any two neighbors in $C\_5$. 


