a) Identify the 3 ionizable groups within the tripeptide. b) There are four diff
ID: 946832 • Letter: A
Question
a) Identify the 3 ionizable groups within the tripeptide.
b) There are four different structures of the tripeptide across a pH range from 1 to 12. What is the overall charge of each structure? (+3, +2, +1, 0, -1, -2, -3)
More explicitly....
- Charge of Structure 1 (at pH 1) = _____ (+3, +2, +1, 0, -1, -2, -3)
- Charge of Structure 2 = _____ (+3, +2, +1, 0, -1, -2, -3)
- Charge of Structure 3 = _____ (+3, +2, +1, 0, -1, -2, -3)
- Charge of Structure 4 (at pH 12) = _____ (+3, +2, +1, 0, -1, -2, -3)
c) Calculate the pI. Show calculations please.
d) Calculate the fraction of this tripeptide that has the C-terminal group deprotonated at pH 3.0. (percent of deprotonated C-terminal species relative to all protonated and deprotonated C-terminal species). Calculations appreciated.
HIOH F 3 2 H2C 2 H3C CH3 SH 3Explanation / Answer
a) The three ionizable groups are the groups marked A, E and F
b) At pH = 1 all the groups are protonated , giving structure 1 with total charge +2.
As the pH increases, the most acidic -COOH (F) loses its proton (COO-)to give structure 2 with total charge +1
As the pH further increases , next acidic proton from tryptophan(E) loses proton to give structure 3, the zwitter ion with no net charge.
As the pH further rises , the remaining acidic protn A is lost to give structure 4, with -1 charge.
c) calculation of pI.
The zwitter ion is formed between structures 3 and 4 to the extent of 100%. Thus the pI is the average of these two pka values = [9.39+9.62]/ 2= 9.5
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