Calculating charge of amino acids (am I doing it right?)
I second Lighthart's "Looks good to me"


  Let's break out the Henderson-Hasselbalch equation to be sure.


The Henderson-Hasselbalch equation is usually used to determine the pH of a buffer, but we can work backwards to get the ratio of an acid to its conjugate base at any pH.

$$\ce{pH}=\ce{p}K_a +\log{\left(\dfrac{[\ce{A-}]}{[\ce{HA}]}\right)}$$
$$\log{\left(\dfrac{[\ce{A-}]}{[\ce{HA}]}\right)}=\ce{pH}-\ce{p}K_a$$
$$\dfrac{[\ce{A-}]}{[\ce{HA}]}=e^{\ce{pH}-\ce{p}K_a}$$

Let's take the three possible cases:


  $\ce{pH}&gt;\ce{p}K_a$


If the pH is greater than the pKa, then 

$$\ce{pH}-\ce{p}K_a&gt;0$$

$$e^{\ce{pH}-\ce{p}K_a}&gt;1$$
$$\dfrac{[\ce{A-}]}{[\ce{HA}]}&gt;1$$
$$[\ce{A-}]&gt;[\ce{HA}]$$

For OH and SH functional groups in proteins, when $\ce{pH}&gt;\ce{p}K_a$, then the group is deprotonated and has a negative charge.

For amino groups, $\ce{HA}$ is $\ce{RNH3+}$, so for amino groups at $\ce{pH}&gt;\ce{p}K_a$, they are not protonated and neutral.


  $\ce{pH}&lt;\ce{p}K_a$


If the pH is less than the pKa, then 

$$\ce{pH}-\ce{p}K_a&lt;0$$

$$e^{\ce{pH}-\ce{p}K_a}&lt;1$$
$$\dfrac{[\ce{A-}]}{[\ce{HA}]}&lt;1$$
$$[\ce{A-}]&lt;[\ce{HA}]$$

For OH and SH functional groups in proteins, when $\ce{pH}&lt;\ce{p}K_a$, then the group is protonated and neutral.

For amino groups at $\ce{pH}&lt;\ce{p}K_a$, they are protonated and positively charged.


  $\ce{pH}=\ce{p}K_a$


In this case you will find that:
$$[\ce{A-}]=[\ce{HA}]$$

Half of the groups are protonated and the other half are not.