Let $G$ be a semisimple and simply connected linear algebraic group over $\mathbb{C}$.
Let $H$ be a connected, Zariski closed and semisimple linear algebraic $\mathbb{Q}$-subgroup of $G$.


Is $H$ a simply connected linear algebraic group?


Here "simply connected" means every central isogeny to $G$ is an isomorphism. 



Are Zariski connected and closed semisimple subgroups of semisimple and simply connected algebraic groups again simply connected?
As Venkataramana says, the answer is no. In fact, every linear algebraic $\mathbb{Q}$-group $H$, whether simply connected or not, is (isomorphic to) a Zariski closed $\mathbb{Q}$-subgroup of some $SL\_n$ (which is semisimple and simply connected). (By definition of being an algebraic group, $H$ is a Zariski closed $\mathbb{Q}$-subgroup of $GL\_k$, for some $k$. Furthermore, it is well known that $GL\_k$ can be embedded as a Zariski-closed $\mathbb{Q}$-subgroup of $SL\_{k+1}$.)


