7
$\begingroup$

Amino acids with non‐ionizable side chains are zwitterions when they are at physiological pH, pH 7.4. This is what my book says. But I do not understand why. The Pka for a carboxyl group is around 3 and the pKa for an amino group is around 9. So the isoelectric point should be around ( 9+3) /2 = 6 for an amino acid with no ionizable side chain. At pH 7.4 it would be slightly negative, would it not?

Also what exactly is a basic amino acid? Is that an amino acid that has a PI value of more than 7?

Thank you.

$\endgroup$
3
$\begingroup$

You are right, most of the amino acids have pI around 6, so they are slightly negative at pH 7.

The isoelectric point depends on the amino and carboxylic-acid groups and the groups of the side chain. E.g. lysine has an amino group on the side chain, so it has a pI of 9.74.

$\endgroup$
2
$\begingroup$

If an amino acid has two amino groups and only one carboxyl group (like arginine and lysine), then they are positively charged (++ vs. -) at neutral pH. On the other hand, if amino acids have two carboxyl groups and only one amino group (like glutamic acid and aspartic acid), they are negatively charged at pH7 (-- vs. +). You are not correct that the groups are truly neutral. The carboxyl group is negatively charged in a solution much higher than the pKa of the carboxyl group (which is pH 2.48 for arginine). Same is true for the amino groups (pKb is at pH 11.5 for arginine). So, at pH 9.5, the ratio of the positively charged amino group to a neutral amino group is 100:1 (at the pKb it would be 50:50). At a pH of 4.48, the carboxyl group of arginine would be 1:100 negatively charged, versus neutral. So, at pH7, all of the carboxyl group is negatively charged (COO-) and all of the amino groups (2 of them) are fully positively charged (NH3+). Net charge is plus 1, in the case of arginine at pH7.

$\endgroup$
1
$\begingroup$

At pH 7.4 a carboxyl group is mostly present as COO-, while an amino group is present as NH3+. So when a molecule has both groups, such as an amino acid, the molecule as a whole is neutral. As it is a Zwitterion (equal + and - charges) the hydrogen can also flip flop between amino and carboxyl group. Making the molecule 'truly' neutral as both groups are not charged.

A basic amino acid has a basic group such as an amino group (Arginine) on its side chain.

$\endgroup$
  • 1
    $\begingroup$ The truly neutral (no positive charge on the amino group and no negative charge on the carboxyl group) does not really exist, since pH7 is far away from the pK of the individual groups (amino and carboxyl). Probability is much less than 1:100. $\endgroup$ – Stefan Gruenwald Feb 20 '16 at 17:29
0
$\begingroup$

Technically yes, most amino acids will be slightly negatively charged at physiological pH. But we need to understand what that means. Let's take alanine as an example. Any individual alanine molecule can either be uncharged or have an integer net charge- it's either protonated or it's not. You will never find a "partially charged" alanine molecule. However, if you have a bunch of alanine molecules in solution physiological pH, they won't all behave in exactly the same way at exactly the same time. One molecule may have a protonated N-terminus while another has a deprotonated N-terminus. At physiological pH, the C-terminus is pretty much completely deprotonated, so we just have to pay attention to the N-terminus. We can use the Henderson-Hasselbach equation to figure out what percentage of alanine molecules will have a protonated (positively charged) N-terminus and what percentage is deprotonated. pH = pKa + log(deprotonated/protonated). If we substitute in our known numbers, we get 7.4 = 9.5 + log(deprotonated/protonated)

Therefore -2.1 = log(deprot/prot) and 10^-2.1 = deprot/prot = 0.008. That means that out of 1000 alanine molecules, about 8 of them are deprotonated and therefore negatively charged. The other 992 are zwitterions. So for all intents and purposes alanine is neutral at physiological pH.

$\endgroup$
  • $\begingroup$ Please add some references to your answer. $\endgroup$ – another 'Homo sapien' Dec 22 '16 at 8:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.