R group effects on Isoelectric point

[manutdfan]

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my ek organic book says that acidic R groups lower the pI of amino acids whereas basic R groups increase the pI.... can anyone explain why this occurs??

 

[phEight]

my ek organic book says that acidic R groups lower the pI of amino acids whereas basic R groups increase the pI.... can anyone explain why this occurs??

Well, the definition of isoelectric point is the pH at which an amino acid carries no net charge. It's equation is pI = (pKi + pKj) / 2 where pKi and pKj are disassociation constants of the ionizations. This may be getting into far more detail than necessary for the MCAT, but I think it will help you understand your question. Let me explain further.

So, let's say for an acidic R group such as the one found in aspartic acid or glutamic acid, the isoelectric point would be the average of the ionization of the COOH group of the amino acid (the COOH that is in all amino acids) and the COOH group in the R group (see image). Looking up the values for these pK values (for aspartic acid), I find that pK1 is 1.99 and pKr is 3.90, where pK1 is COOH on amino acid backbone, pKr is COOH on the R group. pI = 2.95

Let's do the same for an amino acid containing a basic R group. Lysine. In this, we use pK2 and pKr. pK2 being the value of the NH3 group on the amino acid backbone, and pKr being the NH3 on the R group of lysine. pK2 = 9.06, pKr = 10.54 ==> pI = 9.8

I think just showing you the equation would've been sufficient enough to explain the question, or maybe even just the definition of what the isoelectric point is... but maybe it helped more going through it step by step. I had to reference by good ol biochem for this. Bases obviously have a higher pH, therefore the pH at which they will carry no electric charge will also be higher. Hope it helps. 🙂

 

[manutdfan]

in acidic AA's why do you use the pka values of the COOH and the acidic R group, instead of the COOH and the NH3+

and for basic AA's why use the pka values for the NH3+ and the basic R group?

thats what i dont understand.

 

[boaz]

Say you have 3 acidic protons (2 -COOH's and one -NH3) in 1 molecule with the following data:

pK1=1
pK2=3
pK3=9

At a pH of:

14: [OH-] is so high and [H+] is so low that we can rest assured that all sites are deprotonated and the predominant species is A2-.

10: Much the same as 14 but also some HA-. Keep in mind that the sites with pKa =1 and 3 are still completely deprotonated because the pH is still way too high for that.

9: pH=pK3 so [A2-]=[HA-]. As above, the other two sites are still completely deprotonated. This means that at this pH there's an overall -1 charge on the average molecule.

Since we need the average charge to be zero for isoelectric pt, we still have a way to go. Let's skip for a moment to pH=4.

4: At this [H+]we can rest assured that site 3 is "completely" protonated. There are essentially no molecules having a deprotonated site 3 at this pH. As for the other two sites: for site 1, pK1=1, we can still assume that it is completely deprotonated. Even site 2 can be assume to be predominantly in the deprotonated form because pH is one unit higher than pK2, which means (from Henderson-Hasselbach) that the base form is 10 times more predominant that the acidic form. So at this point we have a +charge on site 3 and two - charges on sites 1 and 2, giving an overall charge of -1. NOT MUCH CHANGED between pH 8 and pH 4. The -NH3 remains positively charged below pH 8 no matter what.

We need to get a -1 charge to balance the +1 charge on site 3. What this means is that we need an average of 0 and -2. If pH=0, all sites are protonated, leaving us with an overall charge of +1. So we need to find the average between pH 0 and 4: (0+4)/2=2. This means that at pH=2, the average charge between sites 1 and 2 is -1. Since -1 balances the +1 the isoelectric point is 2.

Remember, that after we move from pH 8 on downwards we do not need to worry about the status of the -NH3 group. It remains protonated (and more so!) all along.


Hope that helps.