How to determine Gauge Boson charge? Given the field strength:
$$F^a_{\mu\nu} = \partial_\mu A^a_\nu - \partial_\nu A^a_\mu + g \; f^{abc} A^b_\mu A^c_\nu$$
We'd get interactions between three and four bosons, former one will look like:
$$g \; f^{abc} (\partial_\mu A^a_\nu - \partial_\nu A^a_\mu) A^{b\mu} A^{c\nu} $$
And the latter one will have $g^2$ and four fields.
For fermions, on other hand the charge enters through this term:
$$i g \; A^a_\mu T^a \psi$$
In this case we say that fermions have $g$ as a charge.
How does one determine charges of gauge bosons in Gauge Theories?
Can charge of bosons be related to the charge of fermions?

Update (using answer below):
Apparently it has something to do with global gauge transformation of the fields. Fermions transform as:
$$\psi \rightarrow (1 + i \alpha^a T^a) \psi$$
Gauge bosons:
$$A_\mu^a T^a \rightarrow A_\mu^a T^a + i [\alpha^a T^a, A^b_\mu T^b]$$
In Electroweak theory we are interested following transformation for fermions:
$$exp\left(i \alpha \left( \tau^3 + Y \; \mathbb{1} \right) \right) \psi$$
The infinitesimal transformation of $Z^\pm$ 
(given $\tau^\pm = \frac{1}{\sqrt{2}} \left( T^1 \pm i T^2 \right)$ and
$[\tau^3, \tau^\pm ] = \pm \tau^\pm$) is:
$$Z^\pm \tau^\pm \rightarrow Z^\pm \tau^\pm \pm i \alpha \tau^\pm Z^\pm$$
Thus all in all gauge field should transform as:
$$e^{\pm i \alpha} Z^\pm$$
Is this why we say that $Z^\pm$ boson has charge $\pm 1$? Does it generalize for Yang-Mills in some way?
 A: The concept of charge is usually defined in the case of an Abelian (=commutative) gauge group. If the group is non-commutative, one can find a maximal commutative subgroup, and define the charge with respect to this subgroup. 
But I think a better perspective is the following. When you say that a field has charge $n$ under the group $U(1)$, you mean that a gauge transformation $e^{i \theta}$ acts on the field by multiplication by $e^{i n \theta}$. In other words, a charge $n$ field transforms in the representation of $U(1)$ labeled by $n \in \mathbb{Z}$. The charge just tells you in which representation of the gauge group the field transforms. 
Now the non-abelian generalization is clear : you just have to say in which representation your field transforms. In your example, I think you use $SU(N)$ as gauge group. Then the quarks transform in the fundamental representation, of dimension $N$ (this is why there are three quarks for $SU(3)$, one for each color), the anti-quarks transform in the antifundamental representation, also of dimension $N$. Finally, the gluons transform in the adjoint representation, of dimension $N^2-1$ (for $N=3$, you have 8 gluons). 
How then to see that on the Lagrangian? Well, just by looking for the right representation! The fundamental acts by multiplication on the left by a matrix of the gauge algebra, while the adjoint is given by the commutator. 
