Section Bank: C/P Q51 (SPOILER ALERT!)

[Dr. Stalker]

Ello, question 51 in the chemical/physics section bank kind of tripped me up. Hoping one of you kind souls could clarify why the correct answer is, well, correct! Below is the question (AND THE ANSWER). This was a free-standing question (so no passage/background necessary).

At pH 7, which of the following peptides will bind to an anion-exchange column and require the lowest concentration of NaCl for elution?

A) AVDEKMSTRGHKNPG
B) YPGRSMHEWDIKAQP
C) HIPAGEATEKALRGD
D) EAPDTSEGDLIPEVS

The correct answer is C (HIPAGEATEKALRGD). Why? Thanks

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[Formation]

You want to look at the sequences given and add up the net charges. At pH of 7, all the acidic residues have a -1 charge and all the basic residues have a +1 charge.

If I remember correctly, it's C and D that both have negative net charges. However, the question also asks you to chose the one that requires the least amount of NaCl to elute it, which means you want to chose the peptide with the smallest negative charge. C.

 

[aldol16]

To add more detail to the excellent response above, you want to remember that an anion exchange column will pull down anions. So you want the peptide that's negative.

Now you also want the one that requires the least salt to elute out. So what happens when you elute something? The salt competes with the stuff that's bound to the stationary phase, i.e. your anionic peptide. The more negative it is, the higher the salt concentration you'll need to outcompete the peptide. So here, you want the one that is the least negative (but negative nonetheless) because that will require a lower salt concentration to elute out than one with more negative charge.

 

[Mr. otcoD]

To add more detail to the excellent response above, you want to remember that an anion exchange column will pull down anions. So you want the peptide that's negative.

Now you also want the one that requires the least salt to elute out. So what happens when you elute something? The salt competes with the stuff that's bound to the stationary phase, i.e. your anionic peptide. The more negative it is, the higher the salt concentration you'll need to outcompete the peptide. So here, you want the one that is the least negative (but negative nonetheless) because that will require a lower salt concentration to elute out than one with more negative charge.

So for an anion exchange column, you want the peptide that is negative. This will be different for ion-exchange chromotagraphy correct?

 

[aldol16]

So for an anion exchange column, you want the peptide that is negative. This will be different for ion-exchange chromotagraphy correct?

Anion exchange chromatography is a type of ion-exchange chromatography. Anion exchange columns are packed with resins that contain positive charge which pulls down negatively charged peptides. Conversely, cation exchange columns are packed with resins that contain negative charge which pulls down positively charged peptides. I like to remember the names as what they pull down - anion exchange takes out anions and vice versa.

 

[jeep1010]

@aldol16 is the salt present so that the anionic peptide will flow through (elute), while the NaCl is what will be bound to the stationary phase so the anionic peptide won't be?

 

[aldol16]

@aldol16 is the salt present so that the anionic peptide will flow through (elute), while the NaCl is what will be bound to the stationary phase so the anionic peptide won't be?

Yes, basically. But I think you might be combining two steps here. Whenever you're running a column, you always have some sort of solvent gradient. So in this case, you'll load your peptide solution onto the column and elute with 0% NaCl. This gets rid of the stuff that doesn't stick to the column. Then you use an NaCl gradient where you ramp to, say, 10% NaCl over 5 minutes and collect the fractions. This will elute out the stuff that's not that negatively charged. Then maybe you ramp to 50% NaCl over 20 minutes. This will elute out stuff that has a larger negative charge. And so on and so forth.

 

[upth3PUNX]

How is C a charge of -1? I calculated the following charge of zero, assuming K,H,R had +1 charge and D, E had -1 charge.
HIPAGEATEKALRGD
+ - -+ + - 3 + and 3- cancel to zero charge.

Is there an amino acid I'm not considering?

 

[aldol16]

How is C a charge of -1? I calculated the following charge of zero, assuming K,H,R had +1 charge and D, E had -1 charge.

At pH 7, most of the His is deprotonated.

 

[upth3PUNX]

@aldol16, good to know. Thank you.

 

[kitkat23456789]

How is C a charge of -1? I calculated the following charge of zero, assuming K,H,R had +1 charge and D, E had -1 charge.
HIPAGEATEKALRGD
+ - -+ + - 3 + and 3- cancel to zero charge.

Is there an amino acid I'm not considering?

i am having the same problem!!... why is that ?

 

[aldol16]

i am having the same problem!!... why is that ?

Read the responses above carefully.