What the the $\mathrm{pI}$ of a tripeptide of lysine?

$$ \begin{array}{l|ccc} \hline \text{Group} & \ce{Lys{-}α{-}NH3+} & \ce{Lys{-}ε{-}NH3+} & \ce{Lys{-}α{-}COOH} \\ \hline \mathrm{p}K_\mathrm{a} & 8.0 & 10.5 & 2.2 \\ \hline \end{array} $$

I have written out the tripeptide, and I know when the hydrogens would come off of it depending on the $\mathrm{pH}.$ I know it has to do with the $\ce{α{-}NH3+}$ and the $\ce{ε{-}NH3+}$:

$$ \begin{align} \mathrm{pI} &= \frac{\mathrm{p}K_\mathrm{a}(\ce{Lys{-}α{-}NH3+}) + \mathrm{p}K_\mathrm{a}(\ce{Lys{-}ε{-}NH3+})}{2} \\ &= \frac{8.0 + 10.5}{2} \\ &= 9.25, \end{align} $$ but it was incorrect. Can anyone please help and explain the process to me?

Swiss-Prot — Compute pI/Mw calculator gives $\mathrm{pI} = 10.30$ for KKK sequence in single letter code.

  • $\begingroup$ The tripeptide has 1+3 NH2 and 1 COOH. Calculate the average partial charge for each of them as function of [H+] and of respective acidity constant. And then calculate [H+] resp. pH when their charges are balanced with zero net charge. $\endgroup$
    – Poutnik
    Jun 30 at 7:12
  • 2
    $\begingroup$ Another thing is acidity constants for alpha NH2 and COOH will be different, compared to a single lysine, as the peptidic bond has different induction effect than amino or carboxy group. But the task may ignore this. $\endgroup$
    – Poutnik
    Jun 30 at 9:15

I think there is not enough data to provide an exact answer theoretically. You can figure out an estimate value by virtually titrating the tripeptide and figuring out after what step an isolectric point likely situates:

$$ \def\VEQ{{\scriptsize-\ce{H+}}{\Large\downharpoonleft\!\!\upharpoonright}\scriptsize{+\ce{H+}}} \begin{array}{rcc} \mathrm{pH} < 2.2 &\quad &\ce{H3\overset{+}{N}-\overset{+}{Lys}-\overset{+}{Lys}-\overset{+}{Lys}-COOH} \\ & & \VEQ \\ 2.2 < \mathrm{pH} < 8.0 &\quad &\ce{H3\overset{+}{N}-\overset{+}{Lys}-\overset{+}{Lys}-\overset{+}{Lys}-COO-} \\ & & \VEQ \\ 8.0 < \mathrm{pH} < 10.5 &\quad &\ce{H2N-\overset{+}{Lys}-\overset{+}{Lys}-\overset{+}{Lys}-COO-}\\ & & \VEQ \\ 10.5 < \mathrm{pH} = \mathrm{pI} &\quad &\ce{H2N-\overset{+}{[Lys-Lys-Lys]}-COO-}\\ \end{array} $$

As you can see, the formal net zero charge doesn't occur between any given $\mathrm{p}K_\mathrm{a}$s. Rather, at the highest $\mathrm{p}K_\mathrm{a}(\ce{Lys{-}ε{-}NH3+}) = 10.5$ the tripeptide is still going to have $ 0.5 + 0.5 + 0.5 -1 = +0.5$ formal charge considering $50\,\%$ deprotonation of residue amine groups at $\mathrm{pH} = \mathrm{p}K_\mathrm{a}$ and $-1$ charge at the $\ce{-COO^-}$ group.

This implies that $\mathrm{pI}$ of trilysine must be slightly higher, about $10.8:$

Titration curve for trilysine. Drawn in Inkscape 1.1

For polylysines $\mathrm{pI}$ should increase with the chain length. ThermoFisher's Peptide Analyzing Tool suggests theoretical $\mathrm{pI}$ values of $11.0,$ $11.2$ and $11.3$ for tetra-, penta- and heptalysine, respectively. This tool, however, refuses to calculate isoelectric points for tripeptides (Input sequence is too short). This can be an indication of insufficient quality of applied algorithm resulting in apparently wrong data suggested by Swiss-Prot in your question.

Note that we didn't corrected $\mathrm{p}K_\mathrm{a}$ values for the tripeptide. I wouldn't assume they change drastically, but it's not the identical values either.


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