# Isoelectric point problem

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#### Rob Biomed

##### Full Member
10+ Year Member
I have a question about TBR's CBT 4, question #137.

The passage provides the chemical structure of ADH, which is shown with an overall charge of +2. ADH has 3 dissociable hydrogens, and the pKa's are given:

pKa of N terminus is 8.0
pKa of phenolic hydroxyl is 10.0
pKa of the guanidino group is 12.5

I thought that the molecule would attain a neutral charge once two hydrogens were removed (via base titration), which in this case would be the two hydrogens with the lowest pKa.

I thought that the pI is then:

pI = (8.0 + 10.0) / 2 = 9.0

The solution says that instead you have to take the average pKa of the second and third hydrogens to determine the pI: (10 + 12.5) / 2 = 11.25

I'm not too confident when it comes to isoelectric point problems, but the solution that TBR provided isn't making a whole lot of sense to me.

Can someone help explain this?
Cheers.

I have a question about TBR's CBT 4, question #137.

The passage provides the chemical structure of ADH, which is shown with an overall charge of +2. ADH has 3 dissociable hydrogens, and the pKa's are given:

pKa of N terminus is 8.0
pKa of phenolic hydroxyl is 10.0
pKa of the guanidino group is 12.5

I thought that the molecule would attain a neutral charge once two hydrogens were removed (via base titration), which in this case would be the two hydrogens with the lowest pKa.

I thought that the pI is then:

pI = (8.0 + 10.0) / 2 = 9.0

The solution says that instead you have to take the average pKa of the second and third hydrogens to determine the pI: (10 + 12.5) / 2 = 11.25

I'm not too confident when it comes to isoelectric point problems, but the solution that TBR provided isn't making a whole lot of sense to me.

Can someone help explain this?
Cheers.
Some amino acids contain acidic or basic side chains. Like Lysine, Arginine, Histidine are considered basic amino acids with basic side chains and Glutamic acid and Aspartic acid which exist as salts called glutamate and aspartate are considered acidic as they have 2 COOH groups. Now the question says ADH has a charge of +2. So this means the amino acid in ADH is protonated and in an acidic environment. Amino acids in basic environments are deprotonated and develop -ve charge.

pI for amino acid with acidic side chain, pI = (pKa(OH) + pKa(side chain)) / 2
pI for amino acid with basic side chain, pI = (pKa(N) + pKa(side chain)) / 2
pI for amino acid with no side chain, essentially non-polar, pI = (pKa(OH) + pKa(N))/2

So for this problem, pI = (10+12.5)/2 = 11.25

Let me know if my reasoning is wrong as I am in the same boat as you. It is easier to reason things out once you know the ans, but difficult otherwise......

If you have the pKas and the overall charge, you can shortcut to the answer really quickly. These titrations start at the most acidic pH possible (~1) and base is added. The overall charge of the protein can only decrease by -1 at a time.

+2 -> pKa1 -> +1 -> pKa2 -> +0 -> pKa3 -> -1

You start at +2 and three hydrogens are taken off, decreasing the charge by 1 each time. The +0 zwitterion is found between pKa2 and pKa3, so those two are what you average. The pKas are arranged from most acidic to most basic. In this case, pKa1 = 8, pKa2 = 10, pKa3 = 12.5, which gives you 11.25.

So as you add base and pass the pH of 8, the N terminus goes from -NH3+ to -NH2 (-1 charge).
As you pass the pH of 10, tyrosine gets deprotonated from OH to O- (-1 charge).
As you ass the pH of 11.25, Arginine gets deprotonated from =NH2+ to =NH (-1 charge).

If you have the pKas and the overall charge, you can shortcut to the answer really quickly. These titrations start at the most acidic pH possible (~1) and base is added. The overall charge of the protein can only decrease by -1 at a time.

+2 -> pKa1 -> +1 -> pKa2 -> +0 -> pKa3 -> -1

You start at +2 and three hydrogens are taken off, decreasing the charge by 1 each time. The +0 zwitterion is found between pKa2 and pKa3, so those two are what you average. The pKas are arranged from most acidic to most basic. In this case, pKa1 = 8, pKa2 = 10, pKa3 = 12.5, which gives you 11.25.

So as you add base and pass the pH of 8, the N terminus goes from -NH3+ to -NH2 (-1 charge).
As you pass the pH of 10, tyrosine gets deprotonated from OH to O- (-1 charge).
As you ass the pH of 11.25, Arginine gets deprotonated from =NH2+ to =NH (-1 charge).

Thank you, that helps a lot.