Glycine pH question

[vsl5]

I was doing the prep101 chem questions and came across this:

Glycine passes through a very low pH membrane channel in which form?

A)H2N-CH2-COOH
B)H3N+ -CH2-COOH
C) H2N-CH2-COO-
D) H3N+ -CH2-COO-

I chose B, which I am fairly confident is correct but te answer says C because glycine will pass through as its conjugate base because the membrane itself is very acidic and that only the H bonded to COOH is acidic.

I think this answer is pretty dumb as glycine is amphiprotic and the H on NH3+ is also acidic. I just need someone else to confirm this.

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[Rabolisk]

I agree with you, but a possible explanation for the book's answer is that acid dissolves base best. At low pH, the negatively charged form of glycine is most soluble, sort of like extracting a base with an acid solution.

 

[vsl5]

yeah but the question implies the channel is at a lower pH than glycine, meaning the channel membrane acts as an acid and glycine acts as the base. If glycine acts at the base, both functional groups will accept protons and will be -NH3+ and -COOH.

 

[Rabolisk]

Yeah, I think you're right. And I'm pretty sure I've seen a problem like it where the answer was the most protonated form.

 

[MiCasaMD]

Glycine has two pKa values, pKa1= 2.3 and pKa2=9.6 (The numbers will vary a little depending on your source). Therefore, at a very low pH only the carboxyl group will be deprotonated (COO-). The amine group will remain protonated since its pKa is 9.6. Only at pKa = pI = 5.97, glycine will be amphoteric.

 

[vsl5]

Glycine has two pKa values, pKa1= 2.3 and pKa2=9.6 (The numbers will vary a little depending on your source). Therefore, at a very low pH only the carboxyl group will be deprotonated (COO-). The amine group will remain protonated since its pKa is 9.6. Only at pKa = pI = 5.97, glycine will be amphoteric.

Sorry but that is wrong. You are right that the amine group will remain protonated, but then you say the carboxyl group will be deprotonated? That logic is contradictory, as you would expect BOTH to be in its acidic form (protonated). The only way your answer makes sense is if "very low pH" was interpreted to be a pH between 2.3 and 9.6. But it should be interpreted as a very low pH, eg 1.0. I can prove this via Henderson Hasselbach equation

Take -COOH to be HA and -COO- to be A-

HA + H2O ---> A- + H3O+

pH = pKa + log ([A-]/[[HA])

if pH = 1 and pKa for COOH functional group is 2.3

10^(1-2.3) = [A-] / [HA] = 0.05

So its obvious that most of the species will be protonated (~95%). Same logic applies for the amine group (HA=-NH3+), except since its pKa is so much higher, it will be 100% protonated

 

[Rabolisk]

Even if one were to take "low" pH as above 2.3, the answer would be D, not C. My opinion is that the answer should be B.

 

[MiCasaMD]

Sorry but that is wrong. You are right that the amine group will remain protonated, but then you say the carboxyl group will be deprotonated? That logic is contradictory, as you would expect BOTH to be in its acidic form (protonated). The only way your answer makes sense is if "very low pH" was interpreted to be a pH between 2.3 and 9.6. But it should be interpreted as a very low pH, eg 1.0. I can prove this via Henderson Hasselbach equation

Take -COOH to be HA and -COO- to be A-

HA + H2O ---> A- + H3O+

pH = pKa + log ([A-]/[[HA])

if pH = 1 and pKa for COOH functional group is 2.3

10^(1-2.3) = [A-] / [HA] = 0.05

So its obvious that most of the species will be protonated (~95%). Same logic applies for the amine group (HA=-NH3+), except since its pKa is so much higher, it will be 100% protonated

Sorry, You are correct. I should wake up before I start posting. 🙂 At pH = 2.3, glycine will exist as +NH3-CH2-COOH and at pH = 9.6, it will exist as NH2-CH2-COO-. Therefore, it does make sense that 'B' would be the correct answer. Then again what's considered "low"? My sincere apologies for any confusion.

 

[pmahesh107]

I think this was actually a question on the last AAMC test I took - I chose B and it was right so I'm pretty sure it's a typo on whatever you're looking it.

 

[Don Draper]

I think this was actually a question on the last AAMC test I took - I chose B and it was right so I'm pretty sure it's a typo on whatever you're looking it.

agreed. Low pH would likely mean below the COOH pKa and therefore both would be protonated.

Also remember that pH = pKa means that there are equal amounts of acid/base (HH equation). Therefore saying that the pH is 2.3 or whatever would mean that there would be a 50/50 mix of that species (not protonated or unprotonated).

Easiest way to remember, if the pH is lower (acid) than pKa, then it will be protonated. If the pH is higher (basic) than pKa, then it will be deprotonated. MCAT doesn't get much deeper than that, just have a basic understanding of the HH equation.

 

[vsl5]

Yeah sorry I got these questions off prep101.com, and I know they steal them from the old AAMCs (which I also have, but I do not have the answer keys), but prep101 listed C as an answer. I've been suspicious of a few of their answers for awhile now so I guess its time to stop using them.

 

[web2linc]

I took Biochem last semester, the correct answer is B. You don't need to know the H-H equations to solve this problem on the MCAT, just know that at a very low pH, the molecule will be in it's fully protonated form. You might also need to know the basic structure of amino acids. I don't know if you have to remember the structure of the side chains though. Glycine is the most simplest one to remember because it is only a hydrogen. Other amino acids have more than one pKa. pI is it's isoelectric point, not sure if you need to know that on the MCAT.