Addendum
Since several people pointed out the same thing, I may as well add some value to the answer by including the standard proof of my remark about universal functions. A universal two-place function, for some system, is a function $g(i,k)$ such that for every one-place function $f(k)$ in the system there is some natural number $e$ with $\lambda k .f(k) = \lambda k. g(e,k)$. Suppose that $g$ is such a function; define a function $h$ as $h(k) = g(k,k) + 1$. Then $h$ has some index $e$. Thus $g(e,e) = h(e)$ by the definition of $g$ and $e$, and $h(e) = g(e,e) + 1$ by construction of $h$. This is impossible, so any system of total functions that allows me to form the functions like $h$ cannot have a universal function $g$.
In particular, for any class of functions $A$ and any function $g(j,i)$ in $\operatorname{PA}(A)$, the function $h(n) = g(n,n)+1$ is in $\operatorname{PR}(A)$. So this class will not have a universal function provided that every function in $A$ is total. Of course if you let $A$ be the set of all partial computable functions then $\operatorname{PR}(A)$ is also the set of all partial computable functions and so it does contain a universal function (which is not total).
From this point of view, the limitation is not on whether particular computable functions can be included; the limitation is on the internal structure of any particular effective class of functions.

