Predicting Peptide Charge – MCAT Biochemistry | MedSchoolCoach

Predicting Peptide Charge

MCAT Biochemistry - Chapter 2 - Section 1.4 - Peptides and Proteins - Predicting Peptide Charge
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Sample MCAT Question - Predicting Peptide Charge

What is the charge of this peptide chain at pH = 8: TNMFDKR?

a) +2

b) +1

c) 0

d) -1

B is correct. +1.
To determine the peptide charge, look at the two termini and the side chains of the individual amino acids. There is one free amino group at the N-terminus and one free carboxyl group at the C-terminus. If the pH is greater than the pKa of a group, then the group is deprotonated. If the pH is lower than the pKa of a group, then the group is protonated. At pH = 8, the amino group is protonated and has a +1 charge. At pH = 8, the carboxyl group is deprotonated and has a -1 charge. These two charges balance each other out. Next, we look at the acidic and basic side chains. This peptide has one acidic amino acid, D, and two basic amino acids, K and R. The aspartic acid (D) has a pKa of around 3, so it is deprotonated at pH = 8 and has a -1 charge. Both arginine (R) and lysine (K) have high pKa values (12 and 10, respectively) and are protonated at pH = 8. They both have +1 charges. Therefore, the charge of the peptide comes out to +1.

Predicting Peptide Charge

On the MCAT, you’re not always going to be working with individual amino acids. Sometimes, you’re going to get questions asking about the charge of peptides, which are simply polymeric structures of 2 or more amino acids linked together via peptide bonds

 

Let’s take a look at the example in Figure 1. What is the net charge of this peptide at pH 8? Notice first that we have only one N-terminus and one C-terminus, and all other amines and carboxylates of the backbone are currently participating in peptide bonds. This means, we only need to consider the net charge of these two termini and the charges on any acidic groups.

N-ANDLIFPE-C

Figure 1: A simple peptide in single-letter notation

In an earlier post, we discussed how to determine the charge state of certain amino acid functional groups, given the pKa of those groups and the pH of the surrounding solution. Recall from that post that the carboxylic acid terminus on an amino acid has a pKa of approximately 2, while the terminal amine has a pKa of approximately 9. So, if we are in an environment of pH of 8, we can expect the carboxyl group to be deprotonated and the amino group to still be almost exclusively protonated. Consequently, we can expect a -1 charge on the carboxylate group and a +1 on the amine, for a total of zero for the two termini. This also suggests that we are likely to only care about the contribution from our N and C termini in extreme environments, in cyclized peptides or in post-translationally modified ones.

 

So now that we’ve dealt with the N and C termini, we need to consider the charges on our residues. Logically, any polar and nonpolar amino acids will not be carrying labile protons and, therefore, will not be charged within our solution. In our example, these amino acids are: Alanine (A), asparagine (N), leucine (L), isoleucine (I), phenylalanine (F), and proline (P). Because the side chains of these amino acids will never be charged, we can ignore them when looking for the net charge of the peptide. We only need to take note of the basic and acidic amino acids which, in this case, are aspartic acid (D) and glutamic acid (E). Both have acidic side chains that will be deprotonated in a solution of pH 8, and therefore contribute a -1 charge each, for a net charge of -2. Since the termini had a net charge of 0, the overall net charge of the given peptide is also -2.

N-WFPRDIKM-C

Figure 2: Another simple peptide in single-letter notation

Let’s take a look at the peptide in Figure 2. If the peptide is also placed in a solution with a pH of 8, we can still assume that the N and C termini will combine for a net charge of zero. We just have to pay attention to the acidic and basic amino acids. Tryptophan (W), phenylalanine (F), isoleucine (I), methionine (M) and proline (P) all have no charge. Arginine and lysine can be expected to contribute one +1 charge each, while aspartate will contribute one -1 charge. The overall net charge of this peptide is then 2(+1) – 1 = +1.

 

N-CEHQYESK-C

Figure 3: A final peptide in single-letter notation

Let’s look at one final example at pH 8. As previously, the N and C termini will cancel each other out for a net contribution of 0. Cysteine (C), Glutamine (Q), Tyrosine (Y) and Serine (S) will all be charge neutrally charged. This leaves only two glutamates (E) for a total of -2, and a lysine for a charge of +1 at a pH of 8. Notice that we have left the discussion of histidine for last. Among the basic amino acids, histidine has by far the lowest pKa, at about 6. This suggests that at a pH of 8, it should be in it’s neutrally charged, basic form. Therefore, histidine, despite being a basic amino acid, contributes no charge to the net charge of this peptide. We can now, with just some simple addition, arrive at the conclusion that the net charge of the peptide from Figure 3 must be -1.

Peptide Charge and Gel Electrophoresis

The MCAT may also be interested in how the charge on a peptide may be useful in experimental design. One such instance is gel electrophoresis, in which peptides are separated by charge within a gel. One end of the gel is in contact with a positive electrode, the other with a negative one. Between the two electrodes, a potential gradient exists, and negatively charged molecules will move towards the positive pole in proportion to the magnitude of their net charge, and positively charged molecules will do the opposite. This movement occurs until each peptide or amino acid reaches a pH where they are neutrally charged (the isoelectric point). Therefore, we could expect to see the peptide from Figure 2 moving towards the negative electrode, while those from Figures 1 and 3 will move towards the positive one. And since the negative charge on the peptide from Figure 1 is far greater than that from Figure 3, we expect the one from Figure 1 to migrate further within the gel.

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