# Calculating charge of amino acids (am I doing it right?)

Please consider this picture of a peptide chain: I've done some examining on the different amino acids and concluded that the chain is as follows:

Lys-Leu-Gly-Ser-Citrullin (variant of arginine).

I would now like to calculate the charge of this peptide at neutral pH.

I'm using this table to orient myself: http://www.sigmaaldrich.com/content/dam/sigma-aldrich/life-science/biochemicals/migrationbiochemicals1/common-amino-acids-table.jpg

So here's my train of thought:

As we're at neutral pH the +1 and -1 of the mandatory amino and acid groups will cancel out for all amino acids.

Lysine has a pKx of barely 11 which is some higher than 7 so it is basic in our solution. It means that it will accept a proton so Lysine has a charge of +1 in our respective solution.

Leucine, Glycine and Serine have no charge on their x group so their charge is 0 and does not contribute anything to the total charge.

Citrullin, I read, is uncharged at neutral pH, opposed to Arginine which is, so it does not contribute either.

So my total answer is +1. Is this correct? I am in no way 100% sure of my method but it would be great to get a confirmation that I am doing it correctly or to be corrected on my possibly faulty method.

Thank you!

• Looks good to me. Feb 22 '15 at 6:16

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:

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

If the pH is greater than the pKa, then

$$\ce{pH}-\ce{p}K_a>0$$

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

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

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

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

If the pH is less than the pKa, then

$$\ce{pH}-\ce{p}K_a<0$$

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

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

For amino groups at $\ce{pH}<\ce{p}K_a$, 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.