Peptide Charge

[anonymousername]

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All of my usual resources are letting me down b/c they all seem to explain in the same way...this seems like a fairly easy concept but i can't figure out how to determine the charge of a peptide at a certain pH.

Lets start out with an easy one. At pH = 6 find the charge on these di/tripeptides

lys-phe
glu-glu
phe-glu
lys-phe-glu

I have the answers but if you could explain the thought processes and how you got to the charge, i might get a better idea.

I'm usually pretty good at orgo, i don't know whats going on. Can someone please help?

 

[just one]

First you need to know pka for each amino acid and the terminal carboxyl and amine group. If the pH is lower than the pKA, then the protonated form exits (think of since the solution has excess hydrogen ions present, then the amino acid becomes protonated). If the pH is higher than the pka, the de-protonated form exists ( since there is a deficiency of hydrogen ions in solution, the amino acid donates Hydrogen ion). Individually determine each amino acid for charge and then do the terminal carboxyl and amine. Add all the numbers together and you get the answer, I do not have the pKa for each amino acid in front of me so I can't give you an answer.

Does that make sense?

 

[PiBond]

If pH>pKa then your functional group is deprotonated. Pka of C-terminus is 3-5ish, N-terminus is 9-11ish. The R groups are variable so you have to know which are acidic and basic.

For lysine and phenylalanine:
At a pH of 6 you'll have your carboxyl group deprotonated and you'll have your amino group protonated. Lysine is basic so it'll be protonated and phenylalanine is nonpolar so it's R group won't be affected. Thus, you'll have an overall +1 charge on the dipeptide. It got this from -1 on COO-, +1 on NH3+ and +1 from lysine.

Does that get the ball rolling?

 

[anonymousername]

yeah! I got it. Finally.

thanks a lot guys, took a while to see what you guys were saying but that really helped.

 

[anonymousername]

so just to reiterate how it's done...(in a really dumbed down way)

asn - lys - ser @ pH = 1

asn n-terminus is protonated because it's pKa is 8.8 but it's R-group isn't because it's an amide and not basic. asn gives a +1 charge.

lys is in between the other 2 so I only need to look at it's R-group. It's protonated. lys gives +1

ser is the c-term and it's amine is protinated. +1

overall charge is +3 at pH =1? am I thinking about it correctly?

 

[PiBond]

so just to reiterate how it's done...(in a really dumbed down way)

asn - lys - ser @ pH = 1

asn n-terminus is protonated because it's pKa is 8.8 but it's R-group isn't because it's an amide and not basic. asn gives a +1 charge.

lys is in between the other 2 so I only need to look at it's R-group. It's protonated. lys gives +1

ser is the c-term and it's amine is protinated. +1

overall charge is +3 at pH =1? am I thinking about it correctly?

Since asparagine is our N-terminus then serine must be our C-terminus. The amino group on serine is tied up in the peptide bond with lysine and therefore won't be protonated. The C-terminus will be neutral (COOH) since we are at such a low pH. Thus we should get an overall charge of +2 I believe

 

[anonymousername]

ahhh....yes. I'm not visualizing this correctly.

thanks!