pI of amino acids

[Hemichordate]

I know that for the majority of amino acids, to find the pI you just take the average of the pKa of both the carboxyl and amino group. But for amino acids with two or more ionizable groups, which pKa's do you use to find the pI?

---

Members don't see this ad.

 

[G1SG2]

I know that for the majority of amino acids, to find the pI you just take the average of the pKa of both the carboxyl and amino group. But for amino acids with two or more ionizable groups, which pKa's do you use to find the pI?

I think it's the one before and after the points where you have the zwitterion. I could be wrong. Maybe someone else can chime in.

 

[RogueUnicorn]

i believe that for acidic AAs (aspartate & glutamate), you average the two LOWEST pKa, for basic (his, lys, arg), the two HIGHEST pka.

 

[austinap]

I know that for the majority of amino acids, to find the pI you just take the average of the pKa of both the carboxyl and amino group. But for amino acids with two or more ionizable groups, which pKa's do you use to find the pI?

This is beyond the scope of the MCAT, just so you know.

If you are interested, the pI will be the average of pKa1 and pKa2 for negatively charged (acidic) residues, or the average of pKa2 and pKa3 for positively charged (basic) residues.

 

[SuperSaiyan3]

This is beyond the scope of the MCAT, just so you know.

If you are interested, the pI will be the average of pKa1 and pKa2 for negatively charged (acidic) residues, or the average of pKa2 and pKa3 for positively charged (basic) residues.

that doesn't even make sense, because a negatively charged amino acid is when an amino acid is in a HIGH pH. vice versa.

care to explain?

 

[G1SG2]

that doesn't even make sense, because a negatively charged amino acid is when an amino acid is in a HIGH pH. vice versa.

care to explain?

Here's an explanation by Will Hunting in a similar thread:

http://forums.studentdoctor.net/showthread.php?t=648448

 

[austinap]

that doesn't even make sense, because a negatively charged amino acid is when an amino acid is in a HIGH pH. vice versa.

care to explain?

Let's try Histidine, pKa1 ~ 2, pKa (side chain) ~ 6, pKa3 (amine) ~ 9,

pH vs. charge (carboxylic acid, side chain, amine) [total]
0 --> (0, +1, +1) [+2]
4 --> (-1, +1, +1) [+1]
7.5 --> (-1, 0, +1) [0]
10 --> (-1, 0, 0 ) [-1]

So somewhere in the pH range between the side chain and the amino group (in this case), the amino acid has a net zero charge. This happens in the pH range between the carboxylate and the side chain for negatively charged residues.

 

[Will Hunting]

that doesn't even make sense, because a negatively charged amino acid is when an amino acid is in a HIGH pH. vice versa.

care to explain?

The key is relative to the pI. An acidic amino acid at low pH has a +1 charge. Above the pI it is negative and at pI it is neutral. Isoelectric focusing takes advantage of this by having a gradient of pH.

Austinap knows what he is talking about but his wording wasn't good. The pH matters only relative to the pI.

If the pH is greater than the pI then the ENVIRONMENT is BASIC relative to the amino acid or protein so it has a NEGATIVE charge. On the contrary, if the pH is less than the pI, then the ENVIRONMENT is acidic relative to the amino acid or protein and it has a positive charge. This is what Austin meant.

Otherwise, what he said isn't always true because you can have an alkaline pH with a basic amino acid like arginine, and still have a positive charge. Why? Well, we know the carboxyl is gone so it has a -1 charge. However, the amino has a pka of 9 and the side change a pka of of 13. So, relative to the pka 2 and pka 3 the environment is acidic. A pH of 8.5 is alkaline. However, relative to the carboxyl terminal, the environment is basic so it has a negative charge. Relative to the pI, the environment is acidic so it has a positive charge.

However, at high pH, then it is almost always invariably negative because the environment is basic relative to the amino acid and this is what austinap was hinting at. However, it seems supersayin what I wrote answers your qualms.

Good discussion guys.

 

[boaz]

Here's what I wrote for a similar question:

Say you have 3 acidic protons (2 -COOH's and one -NH3) in 1 molecule with the following data:

pK1=1
pK2=3
pK3=9

At a pH of:

14: [OH-] is so high and [H+] is so low that we can rest assured that all sites are deprotonated and the predominant species is A2-.

10: Much the same as 14 but also some HA-. Keep in mind that the sites with pKa =1 and 3 are still completely deprotonated because the pH is still way too high for that.

9: pH=pK3 so [A2-]=[HA-]. As above, the other two sites are still completely deprotonated. This means that at this pH there's an overall -1 charge on the average molecule.

Since we need the average charge to be zero for isoelectric pt, we still have a way to go. Let's skip for a moment to pH=4.

4: At this [H+]we can rest assured that site 3 is "completely" protonated. There are essentially no molecules having a deprotonated site 3 at this pH. As for the other two sites: for site 1, pK1=1, we can still assume that it is completely deprotonated. Even site 2 can be assume to be predominantly in the deprotonated form because pH is one unit higher than pK2, which means (from Henderson-Hasselbach) that the base form is 10 times more predominant that the acidic form. So at this point we have a +charge on site 3 and two - charges on sites 1 and 2, giving an overall charge of -1. NOT MUCH CHANGED between pH 8 and pH 4. The -NH3 remains positively charged below pH 8 no matter what.

We need to get a -1 charge to balance the +1 charge on site 3. What this means is that we need an average of 0 and -2. If pH=0, all sites are protonated, leaving us with an overall charge of +1. So we need to find the average between pH 0 and 4: (0+4)/2=2. This means that at pH=2, the average charge between sites 1 and 2 is -1. Since -1 balances the +1 the isoelectric point is 2.

Remember, that after we move from pH 8 on downwards we do not need to worry about the status of the -NH3 group. It remains protonated (and more so!) all along.


Hope that helps.