The isoelectric point (pl) of a peptide is the pH at which the peptide does not

Question

The isoelectric point (pl) of a peptide is the pH at which the peptide does not migrate in an electric field. Since the peptide is zwitterionic, there are the same number of positive charges as negative charges on the peptide population. The pl can be estimated fairly accurately (within 0.1 or 0.2 pH units) from the pK values of all the proton dissociable groups in the peptide. Using pK values from the table at the right, estimate the pl value of the following hexapeptide: Amino Acid 12.5 3.7 8.2 4.3 6.0 10.5 10.5 8.0 Arg CyS Trp-His-Glu-Tyr-Gly-Asp STRATEGY His Step 1: Determine the total positive charge on the peptide (n) when all acidic Step 2: List the pK values of all acidic and basic groups in order from lowest (pKi) Step 3: Calculate the pl as the average of the values for pK(n), the proton and basic groups are fully protonated to highest (pK6) dissociation forming the neutral species from the +1 species, and pKn+), yr peptide- NH3 *peptide- 3.4 the proton dissociation forming the -1 species from the neutral species. COOH Step 1: Determine the total positive charge on the peptide (n) when all acidic and*Amino and carboxv basic groups are fully protonated Notice there is only one basic amino acid in the peptide, His. So the total positive charge is 2 terminal values differ from the amino and carboxy values of a single amino acid 8.0 H3N-CH CNHCH C NH-CH CNH CH CNH CH C-NH CH C-OH 3.4 CH2 CH2 H2C CH CH2 CH HN OH 3.7 NH OH NH 6.0 4.3 OH 10.5

Explanation / Answer

pI = (3.7 + 4.3)/ 2 = 4.0

How to calculate:

pI is the pH of a peptide when it is neutral that is, net charge on it is zero. After following step 1 above, we see that the charge on the peptide when all groups are protonated is +2. Now, in step 2, all our pK values have been set in increasing order.

Now, remember that if the pH is above the pKa value for a group, that group is deprotonated.

a) Let's consider pK1= 3.4. It belongs to the terminal COOH. If the pH is above 3.4 (and below pK2=3.7), the COOH group will be deprotonated to COO-. So, the net charge on the peptide between pH= 3.4 and 3.7 will be +2 - 1 = +1.

b) Let's consider the next value pK2= 3.7. It belongs to Asp COOH. If the pH is above 3.7 and below 4.3, this COOH will also be deprotonated in addition to the COOH mentioned above. Net charge on peptide is now 1-1= 0. This is our neutral species being formed from +1 species and so 3.7 will represent pKn.

c) Let's consider the next value pK2= 4.3. It belongs to Glu-COOH. Above pH 4.3, this COOH will be deprotonated to give COO-. Next charge= 0-1= -1. This is the proton dissociation forming -1 species from neutral species and so 4.3 will represent pKn+1.

So, pKn = 3.7

pKn+1 = 4.3

pI= (pKn + pKn+1) / 2 = 4.0

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