Consider the inclusion $k\subset A$ of the field $k$ in the domain $A$ and the fraction field $K=Frac(A)$ of $A$.     
Obviously if a family $(a_i)_{i\in I}$ of elements $a_i \in A$  is algebraically independant over  $k$ it will remain algebraically independant in $K$.    
Consider  however  a family $(\alpha _i) _{i \in I}$ of elements $\alpha _i \in K$ algebraically independent over $k$.     
To my puzzlement, I can't construct from it an algebraically independent family $(a_i)_{i\in I}$ of elements $a_i \in A$. Although my real question is whether it is possible to actually construct  such a family in a natural way, I'll ask something  more precise:    
   
**Precise question** Given the $k$- algebraically independent set $(\alpha _i) _{i \in I}$ in $K$, does there exist in $A$ some  $k$- algebraically independent set $(a_i)_{i\in I}$  ( with the  same index set $I$) ?

The answer is "yes" if $A$ is finitely generated over $k$., thanks to E.Noether's normalization theorem. Interestingly the proof of that theorem is not purely field-theoretic,  since it makes use of Krull dimension.   

**NB**  I'm not sure (despite the title of the question!) that I know what the transcendence degree of $A$ is:  the "correct" definition might follow from the answers to this question!