1. You discover a small protein containing only 10 amino acids, called interleuk

Question

1. You discover a small protein containing only 10 amino acids, called interleukin-87 (IL-87), that is involved in modulating the immune response. You are interested in purifying this small peptide so that you can study its function further. The following analyses (A-E) were performed on the purified peptide to determine its sequence and structure. ture.er Peptid A. Upon reaction with fluro-2.4-dinitropbenol (FDNB or Sanger reagent), you obtain DNP-Arg and DNP-Asp, leading you to believe your protein may not be pure after all. B. You then perform isoelectric focusing on IL-87 and observe only one band, as expected, at pH 6.25. IL-87 is then treated with dithiothreitol (DTT) and the protein is again subject to isoelectric focusing. Much to your relief, you obtain two bands, one at pH of 10 and one at pH of 2.85. Explain what you know so far about the protein (after steps A and B) and why you are relieved. C. Each peptide is then analyzed for its amino acid composition. One peptide, peptide A, contains a 1:1:1:1:1 molar ratio of Y, I, K, R, C. The other pasptide, peptide B, has a molar ratio of 1:1:1:1:1 for C, D, I, N, E D. Peptide A is digested with chymotrypsin and only one product results which contains (R, C, Y, I, K). However digestion with trypsin results in one 3-mer containing (C, I, K) and also free tyrosine and free arginine. In the first round of Edman degradation on the 3-mer, PTH-lle is obtained. What is the sequence of peptide A E. Peptide B is subject to Edman degradation. After the first two amino acids have been removed, the remaining fragment now has a neutral isoelectric point. The remaining 3-mer is reacted with FDNB and DNP-Aso is isolated. The disulfide bridge between peptide A and B is thought to occur between the C-terminal amino acid of peptide B and some residue in peptide A What is the sequence of peptide B? Show how the isoelectric points given in B prove this sequence and structure are correct.

Explanation / Answer

The sequence of Peptide A -

N- terminus to C terminus the sequence is = R-I-C-K-Y

Now lets follow step by step-

The amino acids were - Cys , Ile, Tyr, Arg, Lys

Step 1 : FDNB gave DNP - Arg so the N-terminal amino acid has to be Arg (R). FDNB gives N-terminal animo acid of any sequence.

Step 2 : Enzyme chymotrysin gave no new fragment so the amino acid tyrosine must be at the C-terminal end of the entire sequence. Chymotrypsin cuts at the C-terminal side of the aromatic aminoacids in a peptide bond. Here there is no bond to cut as Y is at the free end.

so the sequence so far is = R _ _ _ Y

Step 3 : Tyrpsin cleaves at C-terminal of Lysine and Argenine. Here we obtained free Y and R and a 3-mer is liberated, so the C-terminus of the 3-mer needs to have a K

so the sequence becomes

R _ _ K Y

Step 4 : Edman gives N-terminal amino acid from any sequence which is PTH-Ile here so the final sequence is

RICKY from N-terminal end to C-terminal end.

Sequence of Peptide B -

The amino acids were - Cys C , Glu E, Asp D, Asn N, Ile I

N- terminus to C terminus the sequence is = D-E-N-I-C

Step 1 : FDNB gave DNP - Asp so the N-terminal amino acid has to be Asp (D). FDNB gives N-terminal animo acid of any sequence

Step 2 : Here a 3-mer was obtained after 2 steps of Edman reaction was reacted with FDNB to yield - DNP-Asn

so the sequence is = D _ N _ _

Step 3 : since disuphide bridge is at the C terminus of peptide B the sequnece now becomes

D _ N _ C

Step 4 : the 3-mer had neutral isoeletric point so the only remaining uncharged amino acid is Ile.

and finally after filling is Glutamic acid in the remaining position we get the sequence.

D E N I C from N-terminal end to C-terminal end.

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