pI

From what I recall the isoelectric point of an amino acid is the point at which its overall charge is zero. Let's just break this down:

At pH = 1.88 everything but the peptide carboxylic acid is protonated. As for the peptide carboxylic acid, 1/2 of them are protonated (see Henderson-Hasselbach and set pH = pKa = 1.88). The total charge is:

Peptide CA: -0.5
Side Chain CA: 0.0
Amide: +1.0
Total Charge: 0.5

Now at pH 3.65, the peptide CA is more or less completely dissociated, while the side chain CA is only half dissociated. The amide is still protonated. The overall charge is then:

Peptide CA: -1.0
Side Chain CA: -0.5
Amide: +1.0

Total charge: -0.5

But wait, we went from 0.5 to -0.5...so we must have crossed zero somewhere. Notice that the Amide did not play a role; so we can consider just the two carboxylic acids. The peptide CA went from -0.5 to -1.0 (change: -0.5) while the side chain CA went from 0.0 to -0.5 (change: -0.5). Since the electric changes of both are equal, the mean pKa provides the pI.

sweetsaja said:

I have a question about calcualting the isolectric point of amino acids.
Functional groups are peptide carboxylic acid-1.88, side chain carboxylic acid-3.65, and amid 9.60. The correct answer is 2.76. Why do yo take the avg of the two lowest pkas to get the answer?

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If it's an acid you take the two lowest pkas. If it's a base you take pka2 and pka3.

Here's why. The PI is where the amino acid is neutral or the protein. In order to do this, we hand the pH at which the concentration of the zwitterion is maximized. So, we have to use the pka where it goes from -1 to 0 and the pka where it goes from 0 to 1. For an acid at low pH everything is protonated so you have +1 charge. So at pka 1 for the carboxyl terminyl, you have 50-50 protonated/deprontonated. Once you go above the pka1, then it becomes primarily deprotonated until you reach pretty much 100%. so Now, the original H3A+ is now H2A. It is neutral. The next H+ to be lost is from the side change carboxyl which is pka2. Once the pH is raised past pka2 then this is primarily lost until it is basically all deprotonated to give you HA-. This is why you take the average of (pka1 +Pka2/2)

For a basic there are 3 H+, the amino 9, the carboxyl termino 2, and the side change 10. At low pH there is a charge of +2. So the first one pka1 gives us +1. Losing the amino pka2 gives us 0. Then losing the basic side change give you -1. Do you see why? The key is to figure out if it's basic or not. The pka's are average over the +1 to 0 to -1. This range is different for acids and bases.

Naijaba said:

From what I recall the isoelectric point of an amino acid is the point at which its overall charge is zero. Let's just break this down:

At pH = 1.88 everything but the peptide carboxylic acid is protonated. As for the peptide carboxylic acid, 1/2 of them are protonated (see Henderson-Hasselbach and set pH = pKa = 1.88). The total charge is:

Peptide CA: -0.5
Side Chain CA: 0.0
Amide: +1.0
Total Charge: 0.5

Now at pH 3.65, the peptide CA is more or less completely dissociated, while the side chain CA is only half dissociated. The amide is still protonated. The overall charge is then:

Peptide CA: -1.0
Side Chain CA: -0.5
Amide: +1.0

Total charge: -0.5

But wait, we went from 0.5 to -0.5...so we must have crossed zero somewhere. Notice that the Amide did not play a role; so we can consider just the two carboxylic acids. The peptide CA went from -0.5 to -1.0 (change: -0.5) while the side chain CA went from 0.0 to -0.5 (change: -0.5). Since the electric changes of both are equal, the mean pKa provides the pI.

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i love this explanation haha...really good stuff, thanks!