Galois group and traslations by rational numbers. Is a well known result that, for every $n \in \mathbb{N}$, there exist an irreducible polynomial $p \in \mathbb{Q}[x]$ such that the Galois Group of its splitting field is $S_n$.
Now my question: 

Given a polynomial $g(x) \in \mathbb{Q}[x]$ of degree $n$ is true
  that there exist a rational number $q$ such that the Galois Group of
  $g(x)+q$ is $S_n$?

Edit: A more complicated question:

Let's fix $g(x) \in \mathbb{Q}[x]$ of degree $n$ and define $GAL(n)$
  as the set of groups that can be realized as Galois Group of a
  polynomial of degree $n$.
Consider the map from $\mathbb{Q}$ to $GAL(n)$ that sends $q \in\mathbb{Q}$ to $Gal (g(x)+q)$.
Is this map surjective?


No.  Take $g(x)=x^4$, and consider the polynomial $x^4+q$, where $q$ is a rational number.
If $q$ is negative, then the roots of $x^4+q$ in $\mathbb{C}$ are $\sqrt[4]{-q}, i\sqrt[4]{-q}, -\sqrt[4]{-q}$ and $-i\sqrt[4]{-q}$, where $\sqrt[4]{-q}$ is the positive real fourth root of $-q$.  Then the decomposition field of $x^4+q$ is
$$ \mathbb{Q}( \sqrt[4]{-q}, i\sqrt[4]{-q}) =  \mathbb{Q}( \sqrt[4]{-q}, i)$$
which has degree at most $8$ over $\mathbb{Q}$.  Thus the Galois group has order at most $8$, and cannot be $S_4$.
Now, if $q$ is positive, then take $\zeta = e^{i\pi/4}$, a primitive eight root of unity.  Then the roots of $x^4+q$ are $\zeta \sqrt[4]{q}, \zeta^3 \sqrt[4]{q}, \zeta^5 \sqrt[4]{q}$ and $\zeta^7 \sqrt[4]{q}$.  The field of decomposition is then
$$ \mathbb{Q}(\zeta \sqrt[4]{q}, i),$$
which again is of degree at most $8$.
