I have a question regarding how to find the isoelectric point (pI) of a peptide after amidation of the alpha-carboxyl group. The peptide I am working with is KQMP. I understand how to find the pI...

I have a question regarding how to find the isoelectric point (pI) of a peptide after amidation of the alpha-carboxyl group. The peptide I am working with is KQMP. I understand how to find the pI of the peptide before amidation given pk alpha-carboxyl = 2.0, pk alph-amino = 9.0, and pk epsilon-amino = 10.5. (pI=19.5/2)

Upon amidation, I understand the amide group is not ionizable. Therefore, the alpha-amino group is deprotonated at pH=9, giving an equilibrium charge between +1 and +2, the epsilon-amino is deprotonated at pH=10.5 giving an equilibrium of charge between 0 and +1. There is no pKa remaining and no groups left to deprotonate. The net charge at pH=10.5 would be +1/2, so the pI would lie somewhere between a pH of 10.5 and 14. Is that all that can be determined, or is there a way to determine the pI exactly?

1 Answer

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jchumzrz | (Level 1) Adjunct Educator

Posted on

Before answering anything I have a brief disclaimer: I will be using pKa = 9.16 for the n-terminus of lysine and pKa = 10.67 for the epsilon amino in lysine. I am using these values because they are the ones given in my references. That being said, pKa values are hard to measure exactly and so your values of 9 and 10.5 are not necessarily wrong.

You are correct in your understanding of the KQMP (Lysine-Glutamine-Methionine-Proline) peptide sequence, as well as amidation of the c-terminus. However, a more quantitative and exact answer can be obtained for the pI of this molecule by using at a powerful equation derived from simple acid-base chemistry. I have created and attached a molecular diagram of this peptide's charge when the pH is significantly below the pKa of the n-terminus (you already described this in your question but I am posting it so that everyone reading this is on the same page in terms of the chemistry). As can be seen in the diagram, the two circled groups (n-terminus and epsilon amino) are the only ones of interest in determining the pI because they are the only groups in this molecule that can be charged (at any pH). Because the pKa of each group is known, the fraction of a specific group that is protonated can be calculated with the following formula:

`([HA])/([HA]+[A-])=(1)/(1+10^(pH-pKa))`

If you want the derivation of this equation, please see my reference links. For this problem all that is important is that the equation describes the percentage of an acidic group that is protonated given the pH and the pKa of that group (where [HA] is the concentration of protonated acid and [A-] is the concentration of conjugate base). In the case of the peptide KQMP, pronation of each of the two groups of interest results in a positive charge (see my diagram). So for these two groups, the equation above actually gives us an "average charge" of the group in question because protonation correlates to positive charge (remember that when considering things like pI and pKa, overall trends in all of the molecules are being generalized through a representation of one molecule). These observations allow for development of the following expression to represent the effective charge of the peptide (with substitution of the respective pKa values):

KQMP charge =

` (1)/(1+10^(pH-9.16))+(1)/(1+10^(pH-10.67)) `

 

Because we are interested in the pI, which is defined as the pH at which the peptide effectively has no charge in solution, we want to solve for the pH at which the above equation is equal to zero. The easiest way to do this is to look at the limit of the equation as the pH increases. In the extreme case, as the pH increases to infinity the denominators of each term approaches infinity and the fractions go to zero. However, because pH is naturally capped at pH = 14, this is the best answer that can be obtained for the pI. In other words, at pH = 14, the net charge of KQMP is as close to zero as it can get (which, from the equation above evaluated at pH = 14 is about +0.0005). So your guess of 10.5 < pI < 14 was good, but a mathematically rigorous analysis shows that pI = 14 is a better answer.

I highly suggest plotting the equation above across the entire pH spectrum (graphs always express more than words can). If you want to get into the semantics of the definition of pI, then pH = 12 seems to be where the charge curve takes sharp changes and begins to "resemble" zero. But the validity of this reasoning depends on how accurate a value of pI is needed (i.e. can +0.05 be considered zero charge or must it be +0.0005). This is precisely why pI values are often hard to obtain exactly, because the average of many molecules and state of just one are essentially being described at the same time. The main takeaway is that pI = 14 for this peptide is the most mathematically stringent answer, which was obtained by using the equation that relates charge, pH, and pKa for a given molecular group.

 

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