TITLE: Are all classes Stiefel-Whitney classes?
QUESTION [10 upvotes]: When I thought of this question, I was sure it must have been asked before on this site, but I could't find anything. Maybe my search skills are lacking, or maybe the question is obvious and it's my math skills that are lacking. Anyway, here it goes.
For a $CW$-complex $X$ let $sw^*X$ be the subring of $H^*(X,\mathbb{F}_2)$ generated by all classes which are Stiefel-Whitney classes of some vector bundle over $X$. It is not hard to see that $sw$ is a proper subfunctor of mod 2 homology. For example (and this might be overkill) if you take the right dimensional sphere $S^n$, then by Bott periodicity, $KO(S^n)=0$, so $sw^*S^n=0$.
Now let $SW^*X$ be the subring generated by all classes which are either Stiefel-Whitney classes of some vector bundle over $X$, or suspensions or desuspensions of such classes. 
$\textbf{Edit}$: Perhaps it wasn't clear from context, but I want $SW^*$ to be a functor, so I force it to be closed under pullbacks. For that reason I am puzzled by Nicholas Kuhn's suggested answer below. Also, we know in retrospect that $H\mathbb{F}_2^*X$ is a summand in $MO^*X$, and that thing is sort of tautologically built out of characteristic classes...
Is $SW^*X=H^*(X,\mathbb{F}_2)$?
I suppose the question is equivalent to something like: does the identity map of $K(\mathbb{F}_2,n)$ factor, stably, through some $BO(m)$?

REPLY [9 votes]: A 1968 paper in Topology by Anderson and Hodgkin shows that 
$KO^*(K(\mathbb F_2, n)) = 0$ if $n \geq 2$. This implies that if $n \geq 2$, then no nonzero classes in $H^*(K(\mathbb F_2,n);\mathbb F_2)$ are SW classes.  (And of course, $BO(1) = K(\mathbb F_2,1)$.)