Beilinson's version of Hodge conjecture has the following form. For any quasi-projective smooth complex variety $X$ the following map is surjective: $$H^i_{\mathcal{M}}(X, \mathbb{Q}(j))\rightarrow \text{Hom}_{\text{MHS}}(\mathbb{Q}(0), H^i(X,\mathbb{Q}(j)))=\Gamma_H(H^i(X,\mathbb{Q}(j)))$$ On the left side we have the Motivic cohomology groups and on the right hand we have the hom in the category of mixed Hodge structures between $\mathbb{Q}(0)$ and cohomology of $X$ with various Tate twsits $\mathbb{Q}(n)$. This conjecture is known to be false for quasi-projective varieties (the first counter-example is due to Uwe Jannsen). For projective varieties for $i=2j$ one recovers the Hodge conjecture. Let's assume $X$ has this property that we know its cohomology ring is generated by the first cohomology group then does Beilinson-Hodge conjecture hold for $X$?
As far as I can see with this assumption one can show that $\Gamma_H(H^{2j}(X,\mathbb{Q}(j)))$ is generated by $\Gamma_H(H^{2}(X,\mathbb{Q}(1)))$. (Here the fact that weight $k$ part of the $k$-th cohomology becomes the smallest weight in the mixed Hodge structure plays a role in the proof) and so this implies the classical Hodge conjecture. I was not able to figure out whether the assumption mentioned on the cohomology ring, necessarily imply the conjecture for $i\neq 2j$ or not.
