How to calculate the pI of a peptide?

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To calculate the pI of the peptide ATLDAK, one must identify the relevant pKa values of its amino acids. The peptide has two positive charges at low pH, requiring two equivalents of base for deprotonation to achieve a neutral charge. The first deprotonation occurs at the C-terminal carboxyl group, followed by the side chain carboxyl of aspartic acid. The pI can be estimated as the average of the pKa values of these two groups, resulting in approximately 6.885. This estimate provides a useful starting point, though various factors may influence the actual pI.
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How to calculate the pI of a peptide??

Hello everyone, I am hoping someone can shed some light on this question. I know the general idea behind it but i can't seem to put the nail in the coffin.

The sequence is ATLDAK and it asks to : Calculate its approximate pI

This is what i have so far:

...nT-A...T...L... D... A... K-... cT
pK...9.87......3.9...10.54...2.16

I know that there are 2 positive charges so this results in requiring to be deprotinated with two equivalents of base. I also know that the pI is given by the two amino acids that give the zwiterionic charge, or simply zero charge. However, how to identify those amino acids...?

Help please
 
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Use the pKa and what you know about titration. In this case, you have correctly noted that the peptide will have a +2 charge at low pH. Adding 2 equivalents of base will therefore give a neutral species. Examining the pKa’s, we see that the first species to deprotonate will be the C-terminal carboxyl group, and the second species to deprotonate will be the side chain carboxyl on the aspartic acid residue. This will give us a neutral species between roughly the pKa of the aspartate residue and the N-terminal amine group. We can then estimate the pI as the average of these two pKa’s:
$$pI\approx\frac{3.9+9.87}{2}=6.885$$
Keep in mind that this is just a handy estimate for pI. Lots of different factors will affect it, but this quick estimate will usually get you in the ballpark.
 
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