Q1: You discover a small protein containing only 10 amino acids, called interleu

Question

Q1: 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.

A. Upon reaction with fluro-2,4-dinitrophenol (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 peptide, 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-Ile 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-Asn 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.

Q2. A. You are interested in studying resilin, a protein thought to be responsible for the ability of fleas to jump many times higher than their own length. This protein is purified and then analyzed as follows. Fractions are collected from a gel filtration column that is a run with a buffer containing relatively low salt concentration. You collect 8mg of protein from these pooled fractions and discover that the protein contains 0.124mg of isoleucine and 0.209mg of glutamate after complete hydrolysis of the protein. Isoleucine has a molecular weight of 131g/mol and glutamate of 147g/mol.

A standard curve is produced from this same gel filtration column using proteins of known weight and measuring their relative elution volumes. The molecular weight of resilin is determined to be 34kDa by measurement of its relative elution volume and use of the standard curve.

How many isoleucine and glutamate residues are present in 1 molecule of resilin? How do you know?

B. Your labmate repeats this experiment but does not use the buffer you gave her. She incubates the protein instead in a buffer with higher salt concentration before running the experiment. Now the purified resilin elutes into two separate fractions, with two different relative elution volumes. This confuses you initially.

You then perform N-terminal sequencing analysis by reacting the protein in each of these two separate fractions with FDNB and discover that the protein in the fraction with a lower elution volume, protein A, volume results in isolation of DNP-A. N-terminal analysis of the protein in the fraction with the higher elution volume, protein B, results in isolation of DNP-B.

What might you now conclude about resilin? Identify the DNP derivatives.

C. The first fraction, the one that eluted in less volume, contained ‘protein A’ and was found to be 10.48% isoleucine by weight after complete acid hydrolysis and contain no glutamate. The second fraction, the one that eluted in a greater volume, contained ‘protein B’ that was found to be 7.35% glutamate by weight after complete acid hydrolysis but contained no isoleucine.

Explain what new information this gives you about resilin. In other words what is the minimum molecular weight of proteins A and B? How might these organize into a native resilin molecule?

D. To determine the actual size of the subunits that might be present in resilin, you subject the native protein (34kDa) to SDS-PAGE. The following standard curve was obtained using proteins of known molecular weight.

You observe two different bands on the SDS-PAGE gel. The relative migration of one of the bands is 1.229 and the relative migration of the other band is 1.044. The band with the higher relative migration is three times as dark as the one with the lower relative migration. Note that the results are identical when DTT is present in the buffer.

Explain what you now know about the subunit composition of resilin in terms of proteins A and B and how the SDS-PAGE data can be interpreted to lead you to this result. Also explain which subunits contain glutamate, isoleucine, alanine and methionine.

Q3. The amino acid sequence of several leucine zipper proteins are indicated in the figure below. This family of proteins are homologous at the protein sequence level. The leucine zipper motif is comprised of a single a-helix. One portion of this a-helix, termed the leucine zipper, homodimerizes with another molecule of an identical protein when acting as a transcription factor. The other major domain is termed the DNA-binding domain, and as its name suggests, is the part of the a-helix that interacts with negatively charged DNA. Proteins with leucine zippers are often important transcription factors, or proteins that are involved in stimulating transcription of genetic material into mRNA. Highlighted residues are conserved among different proteins and species but many other positions have similar chemical character.

A. A portion of the DNA binding domain of the transcription factor Jun is shown below: Arg-Lys-Arg-Met-Arg-Asn-Arg-Ile-Ala-Ala-Ser-Lys-Cys-Arg-Lys-Arg-Lys

Is the isoelectric point of this domain acidic, basic, or neutral and how do you know? Why is this the expected isoelectric point for a protein domain that ‘binds DNA’? Recall that nucleic acid is housed in the nucleus, a compartment that is around pH 7.

Explanation / Answer

1) Sanger's reagent acts on the N terminal aminoacid of the polypeptide. In step A, treatment with Sanger's reagent gave two products. This may indicate that the protein may have contamination. But, isoelectric focussing on IL-87 at pH 6.25 produces only one band. This shows that there is no contamination, but the protein contains two peptides. Peptide A contains the amino acids Y,I,K,R,C. Peptide B contains C,D,I,N,E.

Trypsin digestion of protein A produces tripeptide C,I,K and free lysine and arginine. Trypsin cleaves carboxyl side of lysine and arginine. Edmans reagent produced isoleucine. So, the sequence is isoleucine, cysteine, lysine, Argenine, Tyrosine ( NH2 - I-C-K-R-Y - COOH).

Cysteine can form disulphide bridges. Edmans test gave asparagine at amino terminal. So, the sequence of peptide B is NH2 - D-E-N-I-C- COOH

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