How to rationalise the protonation state of amino acids
Let's assume you drop the amino acid in water at $\ce{pH} = 7$.
You'll have $$\ce{ H3N^+{}-CH(R)-COO^-}.$$
Then you can go to $\ce{pH}= 12$ and the amino acid loose the proton.

You don't get rid of the second proton because the $\mathrm{p}K_a$ is defined as $$\mathrm{p}k_{a}=-\lg\frac{[\ce{NH2}]\cdot[\ce{H+}]}{[\ce{NH3+}]}.$$
This means $\ce{NH3+ -&gt; NH2}$.

You could define another $\ce{p}K_a$ for $\ce{NH2 -&gt; NH^{-}}$.
The $\mathrm{p}K_a$ for this would be much higher than 12 - at least around 35.
You may look at this site: How to use a $\mathrm{p}K_a$ table
You can compare the amine $\ce{NH3}$ (without the rest of the amino acid) with your $\ce{NH2}$ group of the amino acid.