Determining Net Charge of a Peptide - Online Tutor, Practice Problems & Exam Prep

The net charge of a peptide or protein is determined by the ionizable groups of its amino acid residues, influenced by the pKa values compared to the solution's pH. Ionizable groups include the alpha amino and carboxyl groups, as well as specific R groups. At physiological pH (7.4), the charge of each group is assessed to estimate the overall net charge. For example, a peptide with two positive charges and two negative charges results in a net charge of zero. Accurate pKa values for amino acid residues are crucial for these calculations.

Determining Net Charge of a Peptide

In this video, we're going to talk about how to determine the net charge of a peptide or a protein. The net charge of a protein is actually dictated by the net charges of all of its ionizable groups. Recall that the way we determine the ionization of an ionizable group is by comparing the pKa of the ionizable group to the pH of the solution. We'll be comparing whether the amino acid residues in the chain have free or ionizable alpha amino or alpha carboxyl groups, respectively. The internal amino acid residues lack alpha amino or alpha carboxyl ionizable groups. The last thing I want to leave you with before we get to our example is the fact this idea that the net charge of a known polypeptide can only be estimated unless you determine it experimentally. The reason is that there is a unique microenvironment for every single peptide. The microenvironment is the immediate vicinity surrounding an atom or a molecule and can shift the pKa values of amino acid residues by several units. When we compare the pKas to the pH, if the pKas are shifted by several units, that will affect the net charge.

All I really want you to know is that I want you to double check and make sure that you're using the correct set of pKa values. You want to make sure that you're using the pKa values for amino acid residues and not for free amino acids. If your professor uses the same set of pKas for free amino acids and amino acid residues, that's totally fine. You're good to go, and you don't have to worry about anything. But if your professor expects you to use an entirely different set of pKas for amino acid residues than for free amino acids, then you need to zone in a little bit more and make sure that you're using the right set of pKas.

Let's go to our example. In this example, it says to estimate the net charge of the peptide at physiological pH. Here in this blue chart, we have the pKa values specifically for amino acid residues. Normally, the carboxyl group has a pKa of about 2, but for amino acid residues, notice the pKa is 3.5, which is shifted 1.5 units. This shift could affect the charges when we try to determine them. Make sure you're using the correct set of pKa values. We'll be using this table here for all our pKas for amino acid residues.

We're going to estimate the net charge of this peptide. We need to consider its ionizable groups. We know arginine here has an amino group that we need to consider that's ionizable, and arginine is one of the positively charged basic amino acids. We also need to consider histidine, aspartic acid, and glutamine. Glutamine, being at the end, will have an ionizable carboxyl group.

We'll be comparing the pKas of each of these ionizable groups to the given pH of 7.4. Starting with the amino group on the far left, the N terminus has a pKa of 8. Since the pKa of 8 is greater than the pH of 7.4, the conjugate acid form predominates, so it's going to be NH₃⁺. Arginine's R group has a pKa of 12.5, also greater than the pH of 7.4, so it will remain positively charged.

Histidine has a pKa of 6, which is smaller than the pH of 7.4, so the conjugate base form will predominate and will be neutral. Aspartic acid has a pKa of 3.9, smaller than the pH of 7.4, so the conjugate base form of aspartic acid will be negatively charged, as COO^-.

The total of the charges are, initially, two positive charges. Then we have zero charge from histidine, a negative charge from aspartic acid, and another negative charge from the carboxyl group of glutamine. This results in a net charge of zero. So, the estimation of the net charge for this peptide at pH 7.4 is zero. This multiple-step process doesn't require drawing all the R groups; just determine the net charge of each of its ionizable groups. This example confirms the correct answer is b. We will practice determining the net charges of peptides in our practice video. See you there.

What is the net charge if you drop the peptide above into bleach (pH 12)?

• A. -1
• B. 0
• C. +1
• D. +2

Answer the following questions (A, B & C) relating to the 4 tripeptides.

i) Tyr-Lys-Met                    ii) Asp-Trp-Tyr                    iii) Asp-His-Glu                  iv) Leu-Val-Phe

A. Which tripeptide is most negatively charged at pH = 7? _________

B. Which tripeptide contains the largest number of nonpolar R groups? __________

C. Which tripeptide contains sulfur? ____________

So this practice problem wants us to answer the following questions a, b, and c, which are below, which relate to the 4 tripeptides, which are labeled with Roman numerals 1, 2, 3, and 4. And so question a asks, which tripeptide is most negatively charged at pH 7? And so we're going to need to consider all of the ionizable groups and recall that there are 7 amino acids with an ionizable r group, and the mnemonic to help us memorize those 7 amino acids is just "yucky crazy dragons eat knights riding horses." So if it's not one of these 7 amino acids, then we're not going to be interested in its r group because the r group is not going to contribute to the net charge of the peptide. And so, let's take a look by starting with the amino groups. So the amino groups, on all of these far ends over here are going to have a potential ionization. So let's take a look and consider that for all of these peptides here. We'll do it all at once in one go. And so let's go up and take a look at our pKa chart to see what the pKa is of the amino group. So notice that the pKa of the amino group is 8. And so, a pKa of 8 is actually greater than the pKa, I'm sorry, a pKa of 8 is greater than the pH of 7. And so that means that the conjugate acid form is going to predominate and the amino group is going to contribute a positive charge because the conjugate acid form has a positive charge. So we can put a positive charge for the amino group for all of the peptides. So next, let's check the carboxyl group. And the carboxyl group for all of the peptides, we can also do all at once because we know that this end is going to have a carboxyl group that could be potentially ionized. So if we go up and check the pKa of the carboxyl group, the C terminus, notice that it's 3.5. And a pKa of 3.5 is less than the pH of 7. When the pKa is less than the pH, that means the conjugate base is going to predominate and the conjugate base is going to have a negative charge. So this is going to have a negative charge for all of these, carboxyl groups. Alright. So now, what we can do is focus on the IR groups that are ionized. And so, tyrosine over here is going to have an R group that could potentially ionized. And methionine does not fall into this group, so it doesn't have an ionizable R group. So there's only 2 charges we need to consider here. Let's start with tyrosine. And so when we look at tyrosine, you can see that its pKa is 10.5 and a pKa of 10.5 is greater than the pH of 7, which means it's going to be protonated, it's going to be in its conjugate acid form. And the conjugate acid form of tyrosine is just going to have a neutral net charge of 0. It's going to be 0. And so, let's check lysine, and you can see lysine's pKa is also 10.5. And so that means that the conjugate acid form is going to predominate, but the conjugate acid form of lysine has a positive charge. So it's going to be a positive charge. So that means that the total net charge on this peptide over here is going to be a sum of all of these charges. So these, these two charges here cancel each other out, this one and this one. And so the only thing that's left is this positive charge. So this one's going to have a plus one charge, which we'll do in a different color. Let's do it in green, just to say this one's got a plus one charge. So now we can check aspartic acid over here or the next peptide number 2, and we know aspartic acid could potentially have a negative charge. Tryptophan is nonionizable it's not one of these 7, so we don't care about its R group. Tyrosine, again, could potentially contribute a negative charge. And so we already did tyrosine, we know it's going to contribute a net charge of 0 we just checked it. But let's go up and check aspartic acid. And aspartic acid is right here, and you can see it has a pKa of 3.9. And so a pKa of 3.9 is less than the pH of 7. And when the pKa is less than the pH, that means that the conjugate base is going to predominate and the conjugate base of aspartic acid is going to have a negative charge. So we can put a negative charge in here. So the total net charge on this peptide here is going to be, you can see that this, charge and this charge cancel each other out. So the only charge that's available is this one, and that's going to make it a negative one charge. So this peptide has a negative one charge. So now we've got these 2 peptides. Let's check this one over here. And so the good thing is we already did aspartic acid over here and so we know it's going to have a negative charge. Histidine could potentially have a positive charge, so we'll put it in blank here. And glutamic acid could potentially have a negative charge. And glutamic acid could potentially have a negative charge. And so let's go up and check histidine first. So as you can see, histidine up here has a pKa of 6 and the pKa of 6 is less than the pH of 7. And so that means that the conjugate base form is going to predominate and the conjugate base of histidine has a neutral net charge of just 0. And so glutamic acid, we can check that, and glutamic acid has a pKa of 4.1. And so a pka of 4.1 is less than the pH of 7. And when the pka is less than the pH, that means the conjugate base predominates and the conjugate base of glutamic acid is going to have a negative charge on it. And so, when we total the net charges on this guy, what we'll see is that these two charges cancel each other out. So the only 2 charges that are available are this one and this one. So this one's going to have a negative two charge. So it's going to be negative 2. We'll put that in green. And then this last one over here, notice that leucine, valine and phenylalanine do not have ionizable R groups. They're not part of these 7. And so that means that their total net charge is just going to be 0. It's just going to be 0 because this positive charge here cancels out with this negative charge over here. So then, that makes it pretty clear. For a, it says, which tripeptide is most negatively charged at pH 7? That makes it pretty clear that 3 here is the one that's most negatively charged, so we can put peptide 3 here for our answer. Now, b says, which tripeptide contains the largest number of nonpolar R groups? And so recall that our mnemonic to memorize the nonpolar R groups is just GAV LIMP. So GAV LIMP. These are our nonpolar R groups. So if we take a look and count them up, what we'll see is that this first one over here, what we'll do is we'll highlight them in blue if they are nonpolar. Methionine here is the m, so you can see that this one does have one nonpolar R group, but the other 2 are not considered in this nonpolar group. So there's only one over here. Over here in this peptide number 2, there are none that fall into the nonpolar group, so this one has 0. This one over here also has 0 that fall into the nonpolar group, and then this one over here actually has leucine and valine that both fall in here. So leucine and valine are both right here. So this one has 2 peptides that fall into the nonpolar group. So that means that b here is going to be peptide number 4. So peptide number 4 has the largest number of nonpolar R groups. And then c asks, which tripeptide contains sulfur? And we know there are only 2 amino acids that contain a sulfur atom, and those are going to be methionine with the methyl thiol group and cysteine with the SH group, the sulfhydryl group. So there are none of these amino acids have cysteine in them, but there is one that has a methionine and that's this one over here. And so that means that peptide 1 contains sulfur, so we can put peptide 1 over here. And so this concludes this practice problem and I'll see you guys in our next one.

Estimate the net charge for a His-His-His-His peptide at pH 6 (His pK  _R = 6).

• A. -1
• B. 0
• C. +1
• D. +2
• E. +4

Estimate the net charge for the following peptide at pH 7: ATLDAK.

• A. -1
• B. 0
• C. +1
• D. +2

A. Draw the predominant structure of the following peptide at pH 9: Asn-Arg-Cys. What is its net charge?

(Asn pK_a1 = 8.8, Arg pK_R = 12.48, Cys pK_a2 = 1.96, Cys pK_R = 8.18).

B. What is the net charge of the same peptide if the pH is lowered to pH = 2? Draw the newly ionized peptide.

C. Within what pH range would the net charge on the peptide above be approximately +1?

pH: ______________

Alright. So here's a 3 part practice problem broken up into parts a, b, and c down below. Part a asks us to draw the predominant structure of the following peptides, specifically at pH 9. The peptide is asparagine, arginine, and cysteine. It also asks, what is its net charge? It gives us the pKas for all of the relevant ionizable groups. First, we're going to start off by drawing the peptide. The first step in drawing a peptide is to draw the peptide backbone and identify alpha carbons. We know that we're going to draw a nitrogen carbon carbon set for each of the amino acid residues. Since we have 3 amino acid residues, we draw 3 nitrogen carbon carbon sets. So let's go ahead and do that first. Here's a nitrogen carbon carbon set, here's another nitrogen carbon carbon set, and then here's another nitrogen carbon carbon set. Perfect. Now to identify the alpha carbons, we know that it's going to be the middle carbon in each NC C set. So it'll be this carbon here, this carbon here, and this carbon over here. Alright. Great. Next, we're going to draw in all of the carbonyl groups and consider amino acid chirality. The carbonyl groups will be on the carbons that are not labeled with alpha carbons, so there'll be one right here, one over here, and one over here. Considering amino acid chirality, we remember that life predominantly uses L amino acids. To ensure that we're drawing L amino acids, we note that when the R group is going down, it should be on a dash, and when the R group is going up, it should be on a wedge. Notice that this alpha carbon is pointing in an upwards fashion, so its R group is going up, and it should be on a wedge. This alpha carbon is pointing downward, so it should be down and on a dash. This one's pointing up, so it should be up and on a wedge. In the next step, we need to draw in all of the hydrogens on our nitrogen atoms. We have one here and one over here. This nitrogen over here is the free amino group that is potentially ionizable. Before we actually draw in the hydrogens, we first need to compare its pKa to the pH. So, we'll hold off on drawing the hydrogens on this nitrogen for now. Now, we need to draw in the R groups. Recall that asparagine's one-letter amino acid code is N, arginine's one-letter amino acid code is R, and cysteine's one-letter amino acid code is C. The mnemonic that helps us memorize the seven amino acids with ionizable R groups is "Yucky crazy dragons eat knights riding horses." Because we want to make sure that we're drawing our R groups in the appropriate ionization state, it's important to know which amino acids have ionizable R groups. Notice that asparagine does not fall into this mnemonic, so it does not have an ionizable R group. But arginine and cysteine both fall into this mnemonic, so they do have ionizable R groups. Thus, we're going to need to hold off on drawing the ionization of those R groups until we compare the pKas with the pH. First, let's draw asparagine's R group, which we know is literally just alanine with an amide group. So it's literally just CH₂, the carbon double-bonded to an oxygen with an NH₂ coming off. Again, it does not have an ionizable R group, so we can just leave it as is. This asparagine pKa that we're given, the only ionizable group that asparagine has in this residue, is its amino group. So we can label this pKa up here as the amino group pKa. Now, we can move on to arginine's R group. Arginine features resemble a sword; we count how many pointy ends it has in its one-letter code, and there are 3 pointy ends, which remind us that it has a 3 carbon start. So CH₂CH₂. That also reminds us there's a nitrogen, triangular nitrogen structure at the bottom. The way that works is there's a nitrogen, there's a carbon, there's a nitrogen, and then the carbon's double-bonded to a nitrogen. We know that this nitrogen has 1 hydrogen, this one has 2, and then this one is the one that's potentially ionizable. So we need to hold off and compare its pKr to the pH before we can determine how many hydrogens it has here. So we'll hold off on that. Last but not least, we have cysteine. Cysteines spends time in the cysteine chapel, so he's always like, "Shh." It's literally just alanine, so CH₂, with a SH group. But again, cysteine is ionizable, so we're not sure if this sulfur here has a hydrogen on it or not, so we need to compare its pKr here to the pH to determine that. And then, of course, this last group over here is the C-terminal end, which we know is going to have an oxygen, and we need to compare its pKa to the pH to determine if it is ionized or not. Next, we will determine the net charge. We need to compare the pH of the solution to the pKas of all the ionizable groups. So down here, what we have is a reminder for you on how that works. We will go ahead and compare the first group here, the amino group pKa. A pKa of 8.8 is smaller than a pH of 9, which means that the pH is greater. When the pH is greater than the pKa, this means that the concentration of the conjugate base, abbreviated as CB, is greater than the concentration of the conjugate acid. Thus, the conjugate base form of the amino group is just an NH₂, and it is not ionized. Now we move on to the R group of arginine. Arginine's pKr is 12.48, which is greater than the pH of 9. The pH is smaller, and if the pH is smaller than the pKa, the conjugate acid predominates, and the conjugate acid form of lysine's R group will be protonated, so it will have extra hydrogens and be charged positively. Next, we look at cysteine's R group. Cysteine's R group has a pKr of 8.18, which is smaller than the pH. The pH is greater. So, if the pH is greater, the conjugate base predominates. The conjugate base is going to be ionized and will have a negative charge on it. Last but not least, the pKa of the C-terminal carboxyl group is 1.96, smaller than the pH, so the pH is greater. If the pH is greater, the conjugate base predominates, and the conjugate base of the carboxyl group will be charged negatively. When determining the net charge of our peptide, we need to sum up all of the charges on these groups. We have two negative charges and one positive charge, so that comes out to a total net charge of negative one. This is essentially the answer for part a, a net charge of negative one. Now moving on to part b, it asks, what is the net charge of the same peptide if the pH is lowered to a pH of 2? Then it wants us to draw this newly ionized peptide. To save some time here, we can take the same peptide from above and bring it down below. This peptide will be exactly the same, the only thing that will be different is its ionizable groups. So we need to determine its ionizable groups all over again so we can address them. Essentially, we will do the same exact thing that we did earlier, just comparing the pKas that are given to this new pH. We start with the amino group, and the pKa of 8.8 for the amino group is greater than the pH, so the pH is smaller. When the pH is smaller than the pKa, the conjugate acid predominates. The conjugate acid of the amino group this time will be positively charged, an NH₃⁺. Next, we have Arginine's pKr, which is 12

Here’s what students ask on this topic:

How do you determine the net charge of a peptide at a given pH?

To determine the net charge of a peptide at a given pH, you need to consider the ionizable groups of its amino acid residues. Compare the pKa values of these groups to the pH of the solution. If the pKa is greater than the pH, the group will be in its conjugate acid form (protonated). If the pKa is less than the pH, the group will be in its conjugate base form (deprotonated). Sum the charges of all ionizable groups to estimate the net charge. For example, at physiological pH (7.4), a peptide with two positive charges and two negative charges will have a net charge of zero.

What are the key ionizable groups in amino acids that affect the net charge of a peptide?

The key ionizable groups in amino acids that affect the net charge of a peptide include the alpha amino group (N-terminus), the alpha carboxyl group (C-terminus), and specific R groups of certain amino acids. For example, the R groups of arginine, lysine, and histidine are positively charged at physiological pH, while the R groups of aspartic acid and glutamic acid are negatively charged. The ionization state of these groups depends on their pKa values relative to the pH of the solution.

Why is it important to use the correct set of pKa values when determining the net charge of a peptide?

Using the correct set of pKa values is crucial because the pKa values of amino acid residues can differ from those of free amino acids due to the unique microenvironment within the peptide. This microenvironment can shift the pKa values by several units, affecting the ionization state and, consequently, the net charge of the peptide. Incorrect pKa values can lead to inaccurate estimations of the net charge. Always ensure you are using the pKa values specific to amino acid residues in peptides.

How does the pH of a solution influence the ionization state of amino acid residues in a peptide?

The pH of a solution influences the ionization state of amino acid residues in a peptide by determining whether the ionizable groups are protonated or deprotonated. If the pH is lower than the pKa of a group, the group will be protonated (conjugate acid form). If the pH is higher than the pKa, the group will be deprotonated (conjugate base form). This affects the overall charge of the peptide. For example, at pH 7.4, the carboxyl group (pKa ~3.5) will be deprotonated and negatively charged, while the amino group (pKa ~8) will be protonated and positively charged.

Can the net charge of a peptide be determined experimentally, and if so, how?

Yes, the net charge of a peptide can be determined experimentally using techniques such as isoelectric focusing or electrophoresis. Isoelectric focusing separates peptides based on their isoelectric points (pI), where the net charge is zero. Electrophoresis separates peptides based on their charge-to-mass ratio. These methods allow for the experimental determination of the net charge by observing the migration patterns of peptides in an electric field. However, these techniques require specialized equipment and conditions.