pI for glutamate vs valine

[H4LfAd1ME]

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A mutant form of hemoglobin contains a valine residue at the sixth position while the wild-type form contains a glutamate residue. In the mutant form, the pI is expected to be:

A. equal to the pI for the wild-type form
B. slightly greater than the pI for the wild-type form
C. slightly less than the pI for the wild-type form
D. representative of a nonzwitterionic species

fyi:
valine has two ionizable groups = COO (-) and NH3 (+)
glutamate has three ionizable groups = COO (-), NH3 (+), and the R group has a COO (-)

Since the pI is the pH at which the molecule is neutral, it would be when the COO group gets ionized and becomes negative to cancel out the positive charge from the NH3 in the valine. The same for the glutamate--the R group will remain unionized and stay neutral, so the potential extra negative charge in the R group is not applicable here(?). So, I thought the answer was A, but the book says it is B. Can someone care to elaborate for me?

 

[FeinMS]

Let's assume pKa values of carboxyl group, amino group, and side chain carboxyl group to be 2, 9, 3 respectively. I can't recall exactly, but I think those values are similar to the real values.

It's true that neutral molecule is achieved by deprotonated carboxyl (-), protonated amino (+) group in the case of valine, and deprotonated alpha-carboxyl (-), protonated R-carboxyl (0), protonated amino (+) group in case of glutamate. But pI of valine is simply average value of pKa of carboxyl and amino group, while pI of glutamate is average value of pKa of alpha carboxyl and R carboxyl.

Since R-carboxyl group has lower pKa than amino group, pI of glutamate is lower.

 

[pm1]

A mutant form of hemoglobin contains a valine residue at the sixth position while the wild-type form contains a glutamate residue. In the mutant form, the pI is expected to be:

A. equal to the pI for the wild-type form
B. slightly greater than the pI for the wild-type form
C. slightly less than the pI for the wild-type form
D. representative of a nonzwitterionic species

fyi:
valine has two ionizable groups = COO (-) and NH3 (+)
glutamate has three ionizable groups = COO (-), NH3 (+), and the R group has a COO (-)

Since the pI is the pH at which the molecule is neutral, it would be when the COO group gets ionized and becomes negative to cancel out the positive charge from the NH3 in the valine. The same for the glutamate--the R group will remain unionized and stay neutral, so the potential extra negative charge in the R group is not applicable here(?). So, I thought the answer was A, but the book says it is B. Can someone care to elaborate for me?

The way I look at it is a bit more conceptual, I will try to explain it - let me know if it's clear.

The pI of a molecule is the pH where the molecule stop moving, or as you said becomes neutral - this way not being attracted to the cathode or anode anymore. In order for this to happen we need to neutralize the extra charges. Glutamate, being a negatively charged amino acid, will be attracted to where there is more H+ (a lower pH) / positive side. Only when this COO(-) is bound to H+ it will stop moving. pI at a lower pH

While valine being non polar and having only the two regular ionizable groups should be closer to a neutral pH as its pI.

Consequently, the pI of glutamate should be lower than of the valine.

 

[sshah92]

I think of it conceptually too. When you get stuck by questions like these, it helps to think about it from the conceptual POV instead of too directly because it is easy to get mired in the details. This is a technique you could use on the PS section where it's easy to lose track on time, but it carries over to BS as well. 🙂

Firstly, think of valine's structure -- not including charge from the N and C terminals. It's side group is neutral regardless of pH.

Then, think of glutamate (or glutamic acid, not to be confused with glutamine). Apart from the charges on its terminals, it's usually either protonated or deprotonated. Even that is too complicated though. Just think "acid" when you think glutamate, and you already know acids have low pH's.

Now, we're going from a wild-type peptide chain that has glutamate to one that has valine. Glutamate naturally brings down the pI of the peptide chain because it is acidic -- biochemically speaking, it becomes deprotonated at a pretty low pH, but you don't even need to think of it too technically. So if you're replacing a side chain that brings down the pI to one that brings up the pI, your answer would be (B).

Hope that helps!

 

[H4LfAd1ME]

I think of it conceptually too. When you get stuck by questions like these, it helps to think about it from the conceptual POV instead of too directly because it is easy to get mired in the details. This is a technique you could use on the PS section where it's easy to lose track on time, but it carries over to BS as well. 🙂

Firstly, think of valine's structure -- not including charge from the N and C terminals. It's side group is neutral regardless of pH.

Then, think of glutamate (or glutamic acid, not to be confused with glutamine). Apart from the charges on its terminals, it's usually either protonated or deprotonated. Even that is too complicated though. Just think "acid" when you think glutamate, and you already know acids have low pH's.

Now, we're going from a wild-type peptide chain that has glutamate to one that has valine. Glutamate naturally brings down the pI of the peptide chain because it is acidic -- biochemically speaking, it becomes deprotonated at a pretty low pH, but you don't even need to think of it too technically. So if you're replacing a side chain that brings down the pI to one that brings up the pI, your answer would be (B).

Hope that helps!

Thanks everyone! I'm still confused as to why glutamate has the lower pI than valine though. I think I'm missing a key point here. I know pI is when the molecule is neutral, but I'm unable to make the connection as to why the extra negative charge on the glutamate makes the pI lower than on valine.

 

[H4LfAd1ME]

I think of it conceptually too. When you get stuck by questions like these, it helps to think about it from the conceptual POV instead of too directly because it is easy to get mired in the details. This is a technique you could use on the PS section where it's easy to lose track on time, but it carries over to BS as well. 🙂

Firstly, think of valine's structure -- not including charge from the N and C terminals. It's side group is neutral regardless of pH.

Then, think of glutamate (or glutamic acid, not to be confused with glutamine). Apart from the charges on its terminals, it's usually either protonated or deprotonated. Even that is too complicated though. Just think "acid" when you think glutamate, and you already know acids have low pH's.

Now, we're going from a wild-type peptide chain that has glutamate to one that has valine. Glutamate naturally brings down the pI of the peptide chain because it is acidic -- biochemically speaking, it becomes deprotonated at a pretty low pH, but you don't even need to think of it too technically. So if you're replacing a side chain that brings down the pI to one that brings up the pI, your answer would be (B).

Hope that helps!

Does that mean the R group COO in glutamate get deprotonated BEFORE the COO group in the carboxylic acid in glutamate?

 

[sshah92]

Only the very ends of a polypeptide have COOH and NH2+ capable of being deprotonated (at pH of 4 and 8, respectively). The R group in glutamate has a pH of around 3, so yes, it is deprotonated first.

None of the above information is relevant to answering the question though. The MCAT isn't asking you to calculate the pI in this question. The question is only asking you about a comparison of pIs before and after some side chain substitition. (The pI is simply the pH at which the peptide is electrically neutral as you increase the pH from 0 to 14.) If the glutamate side chain has a pH of 3, and the valine side chain doesn't protonate or deprotonate at ANY pH, then replacing glutamic acid with valine will bring up the pI.

Think about it this way. Let's say you have this peptide chain: Val-Ser-Glu. Go ahead and calculate the pI of this. Then replace the Glu with Val so you have Val-Ser-Val. Is the pI higher? Lower? (It should be higher now.) If you can't calculate that out, you might want to go back to your mcat biochem notes haha

 

[pm1]

Thanks everyone! I'm still confused as to why glutamate has the lower pI than valine though. I think I'm missing a key point here. I know pI is when the molecule is neutral, but I'm unable to make the connection as to why the extra negative charge on the glutamate makes the pI lower than on valine.

hmmm.. (I will try in a different way. It took me a while to get this down too) you see how the side chain of glutamate is negatively charged right?

Consequently it will be attracted to the positive side of the gel (I think it is a gel). Do you see how the positive side is the side where there is more H+ ? Consequently the one with lower pH? So the glu will "run" until it finds a place where it likes to stay, or where it is attracted to.. a lower pH.

While val doesn't have that negative side chain, so it will not be attracted to a positive side more than the GLU. Thus, it will stop (become neutral) in a higher pH, where there is not as much H+.

does that make sense?

 

[H4LfAd1ME]

Ah I think i get it now. Since glutamate is more negative it needs more H+ to become neutral compared to valine. Hence it will be in an environment that has a lower pH. Valine on the other hand does not need to be in an environment as acidic to be neutral since it has one less COO group. Yes?

 

[pm1]

Ah I think i get it now. Since glutamate is more negative it needs more H+ to become neutral compared to valine. Hence it will be in an environment that has a lower pH. Valine on the other hand does not need to be in an environment as acidic to be neutral since it has one less COO group. Yes?

yep!! 👍

 

[H4LfAd1ME]

Thanks everyone! 😀