September 29, 2008 at 8:28 pm #10180
How do you calculate the the net charge of glutamate amino acid at a pH of 99
September 29, 2008 at 9:49 pm #86177mithParticipant
just assign charges.
September 29, 2008 at 11:01 pm #86180
For the alpha amino group since pka>pH the charge would be +1. For the alpha carboxyl group since the pKa<pH the charge would be -1. I’m confused what to do with the pKA(s) of the side chain groups??
September 30, 2008 at 1:10 am #86182
The side chain of Alanine is aliphatic–essentially a hydrocarbon. The -CH3 protons don’t dissociate. The side chain of Alanine is electrically neutral at all pH’s
September 30, 2008 at 1:44 am #86184
Does this mean that based on the information I gave, the net charge of alanine at pH=8 is (+1) from alpha amino group plus (-1) from alpha carboxyl group which is equal to 0. (+1)+(-1)=0. Is this correct?
September 30, 2008 at 2:05 am #86186
Yes. Free Alanine will have a net charge of zero between pH 2 to pH 9, roughly.
September 30, 2008 at 3:32 am #86189
Something is not right because none of the answers are 0. So the net charge of alanine based on the information I gave cannot be zero. I’m missing something.
What am I missing? Don’t you use the Henderson Hasselbach equation somehow to calculate the net charge?
September 30, 2008 at 12:07 pm #86192
I’m a little puzzled, I guess. You haven’t been given the specific pKa’s for alanine and yet you are supposed to come up with some sort of fractional charge, as in % ionized or something like that? The pKa’s for alanine are 2.3 for the acid group and 9.7 for the amino group. The formal pI of alanine is 6.0. The pI is calculated from the two pKa’s as ( pKa1 + pKa2 )/2. If there were three titratable groups this would involve all three pKa’s and the denomimator would be "3" rather than "2" and so on.
The only ionic species present when the pH = pI is the zwitterion where both the amino group and the carboxylate are ionzed–the amino group is +1 and the carboxylate is -1. As you move above the pI, the proton is lost from the amino group and the alanine has a net -1 charge. But not all molecules will instantly lose the proton. Only when you get above the basic pKa (9.2) will all of the alanines carry a net charge of -1. If you need to calculate the ratio of (alanine)-0 to (alanine)-1 then you could use the Henderson-Hasselbach equation, but it could get complicated as you have two, coupled acid-base equilibria going on. You could assume that the pKa at 2.3 is irrelevant and only use the equilibrium at pH 9.7 (or 9, if that’s what your prof intends). It didn’t seem to me like the question was asking for such a detailed analysis of the titration curve. At pH’s between the two pKa’s, it is true, at least to a first approximation, that the net charge of alanine is roughly zero. It is only strictly zero at the pI. As you move away from the pI and get closer to one of the pKa’s that becomes less and less true. Within, say, 2 pH units of the pKa you could use the Henderson-Hasselbach equation to estimate the ratio of charged to uncharged species. The net charge per molecule, though, can only be 0,-1,+1.
September 30, 2008 at 12:29 pm #86193
If you assume the data of pKa = 9, and that the acidic equilibrium is not important, at pH 8 roughly 10% of the alanine carries a net charge of -1 while 90 % remain zwitterionic (net charge = 0). If you use the actual pKa for alanine, then even less alanine carries a net charge–maybe 1-2% of the alanines are net -1.
September 30, 2008 at 3:19 pm #86196
So, how would you find the net charge of a peptide such as MARSA at ph of 99?
September 30, 2008 at 5:16 pm #86197
You’ve got to try it first. Because it is a peptide, you only need to consider the terminal alpha-amino and terminal alpha carboxylate. Then you need to consider all the side chains and decide whether or not they will contribute to any charged species. This means putting together the amino acid name with it’s one-character symbol and then associating the right chemistry with the side chain.
October 1, 2008 at 2:34 am #86205
I’ve been trying to solve the last question for a couple of hours and I got a net charge of -4 which is wrong. Can someone please help me with this problem?
October 1, 2008 at 8:41 am #86208
How did you arrive at -4?
October 2, 2008 at 2:51 am #86220
K=Lysine has a charged R group
A=ALanine nonpolar so not charged
P=Proline (nonpolar) so not charged
R=Arginine R group positive)
using this formula log[COO-]\[COOH]=pH-pK or log[NH2]\[NH3+]=pH-PK
the alpha amino group has a charge of +1 based on the above formula
Lysine has a charge of +1 based on the above formula
Arginine has a charge of +1 based on the above formula
the alpha carboxyl group has a charge of -1 based on the above formula
so (+1)+(1)+1+(-1)=+2 so net charge of the peptide is +2 not -4
Am i doing this right??? Is my work right??
October 2, 2008 at 9:42 am #86227
Looks good to me. You only need to consider the terminal NH2 and COOH groups, as you’ve done. All the others are forming amide bonds with the next amino acid in the polypeptide and don’t substantially contribute to the pH equilibria. The only other titratable groups are on the side chains and you’ve correctly identified K and R as the only relevant residues for this peptide. There are trhee positive charges (the free amino terminus and the 2 basic amino acid side chains) and one negative charge (the C-terminal carboxylate) for a net charge of +2. I don’t know that you need to use the other formulae you listed but if they help you in some way, fine. These look like part of the Henderson-Hasselbach equation or part of the expression for the Ka or Kb, but I’m not sure what you mean by them. You’re getting the correct answer, at least in my opinion. For just the net charge on a peptide, in general you just do the arithematic, saving Henderson-Hasselbach for questions about "% ionized" or "fraction of sample in some or the other ionization state" etc.
Sorry, you do have the complete Henderson-Hasselbach equation there. I didn’t look at the formulae carefully enough. As I said, though, I doubt you really have to use them for this kind of problem, but it doesn’t hurt and it certainly is a useful equation to understand.
October 4, 2008 at 12:52 pm #86249
I should correct a mistake I made. The way I said to calculate pI’s is partially wrong. For amino acids with non-ionizable side chains, it is correct that the pI is the average of the pKa’s of the alpha-amino and alpha-carboxyl groups. It is NOT true, though, that the pI of an amino acid with a titratable side chain is the average of the three pKa’s. The pI is the average pKa of the two "like" groups. So, for glutamic acid, for example, the pI is the average of the pKa’s of the alpha-carboxyl and the gamma-carboxyl groups, not the average of all three pK’s. Sorry if I misled anyone.
May 28, 2013 at 8:12 pm #113872debbie4christParticipant
please, i need help with this question. Estimate the net charge of a polypeptide chain at physiological pH 7.4 and at pH 5.0 Ala-Arg-Val-His-Asp-Gln
May 29, 2013 at 1:56 pm #113879JackBeanParticipant
list the prononizable groups, find their pKa and define their charge at desired pH. Then just sum the charges and you get your result.
July 8, 2013 at 6:58 am #114056Richard02Participant
I have difficulty in evaluating the bands sspecially lower than 500 bp.
July 8, 2013 at 12:13 pm #114058JackBeanParticipant
What? This is about amino acid’s charges, not electrophoresis.
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