I think I understand what you're asking, but it depends on the quantization type, and your question lacks the details necessary to a definite answer.
However, in general, if you have an analog signal $x(t)$ that takes on values in a certain range, and a $m$-bit quantizer, your quantized (and sampled!) signal $x[n]$ can take on $M = 2^m$ values to represent the analog values within that range. This is done through a mapping operation. Which mapping is used depends on the quantizer.
For example, for an analog signal $x(t)$ in the range $[-1, 1]$, with $m = 2$ bits, the discrete signal $x[n]$ can take on $M = 2^2 = 4$ discrete values that could map to the analog values in the following ways:
\begin{align}
&\text{4-bit representation} &\qquad &\text{1st mapping} &\qquad &\text{2nd mapping}\\
&\tt{11} &\qquad &0.5 &\qquad &0.75\\
&\tt{10} &\qquad &0 &\qquad &0.25\\
&\tt{01} &\qquad -&0.5 &\qquad -&0.25\\
&\tt{00} &\qquad -&1 &\qquad -&0.75\\
\end{align}
Let's look at the result $x[n_0]$ of these mappings on a couple analog values $x(t_0)$ and $x(t_1)$:
Analog value $x(t_0) = 1$:
- 1st mapping: $x(t_0)$ would be represented by $\tt{11}$ which maps to $0.5$ (round down).
- 2nd mapping: $x(t_0)$ would be represented by $\tt{11}$ which maps to $0.75$ (round down).
Analog value $x(t_1) = 0$:
- 1st mapping: $x(t_1)$ would be represented by $\tt{10}$ which maps to $0$ (round down).
- 2nd mapping: $x(t_1)$ would be represented by $\tt{10}$ which maps to $0.25$ (round up) or by $\tt{01}$ which maps to $-0.25$ (round down).