[Pchar]

Hi, hopefully I'm posting this in the right subforum, this is from a biophysics class, but the problem deals with the chemistry.

"Consider a pentapeptide consisting of asparatic acid residues, forming a closed ring. Indicate the ring's charge at the following pH values: 1, 2, 3, 4, 7, 9, 10, 11, 12"

pKa for aspartic acid, -COOH: 2.09, -NH3+: 9.82, -R group: 3.86

Because there are no carboxyl and  amino groups to deprotonate on the ring (I assume), it it would sense to me that at pHs 1,2,3 the charge would be zero, while at 4, 7, 9, 10, 11, 12, the charge should be -5 as the R groups lose their proton. When I submitted these as answers (this is an online class, so they were marked right away), it came back that the proper charges were:

 0 at 1 and 2
-5 at 3
-10 at 4, 7, 9, 11, 12

Which is consistent with each amino acid in the chain having its carboxyl group, but not its amino group attached, which doesn't make sense given what I've learned about peptides so far.

Thanks in advance, if anyone can shed any light on this for me.

[Babcock_Hall]

Just looking at it casually, I agree with you.  However (and not as important), I don't think that it is entirely realistic to say that the side chain carboxylate groups will fully deprotonate at 4.

[aHerraez]

I also agree. Only the sidechain (R) groups are ionizable in a cyclic peptide. So there is no basis for -10