Sequencing of a protein (polypeptide) involves the following steps: a) purification, (b) determinati — Physical Chemistry — Kinetics Chemistry Question
Protein sequencing
Sequencing of a protein (polypeptide) involves the following steps: a) purification, (b) determination of N-terminal amino acid, (c) cleavage of the polypeptide chain by chemical or enzymatic methods, (d) isolation of the peptide fragments and (e) determination of their sequence by an automated sequencing machine (sequenator). It is also possible to sequence the mixture of peptide fragments without resolving it. The final sequence could be determined by constructing overlapping sequences after analyzing the information on the positional data on amino acids in different fragments.
A small protein, made up of 40 amino acid residues was sequenced as follows:
- Edman degradation involves treatment with phenyl isothiocyanate, subsequent hydrolysis and spectrophotometric identification of the modified amino acid. This procedure identified aspartic acid (Asp) as the N-terminal residue.
- The protein was cleaved with CNBr (cyanogen bromide) which cleaves the peptide bond between methionine and any other amino acid on its C-terminal side. The resulting peptide fragments were not separated. This mixture of peptides was analyzed on the protein sequenator. Therefore, the sequenator would detect as many amino acids in the given position as the number of fragments. The results are shown in Table 1(a).
- The protein was digested with a proteolytic enzyme trypsin. This enzyme cleaves the peptide bond between a basic amino acid (Arg or Lys) and the next C-terminal residue. The resulting mixture of peptides was also analyzed as above. The results are shown in Table 1(b).
[VISUAL]
Table 1. Data from protein sequenator.
a) CNBr:
| Position | Amino Acids Detected |
|---|---|
| 1 | Arg, Asp, Glu, Gly |
| 2 | Gln, Pro, Thr, Tyr |
| 3 | Asn, Pro, Ser, Tyr |
| 4 | Arg, His, Ilu, Val |
| 5 | Asn, Ilu, Leu, Phe |
| 6 | Arg, His, Trp, Val |
| 7 | Ala, Gly, Phe, Thr |
| 8 | Ala, Lys, Met, Tyr |
b) Trypsin:
| Position | Amino Acids Detected |
|---|---|
| 1 | Asp, Gly, Gly, Phe, Tyr |
| 2 | Cys, His, Pro, Pro, Tyr |
| 3 | His, Met, Thr, Tyr |
| 4 | Ala, Asn, Glu, Val |
| 5 | Ilu, Leu, Thr, Trp |
| 6 | Arg, Phe, Ser, Ser |
| 7 | Cys, Lys, Ilu |
| 8 | Glu, Leu |
Deduce the amino acid sequence common to the first fragment (N-terminal) obtained by CNBr and trypsin treatments.
Model Answer
Asp – Pro – Tyr – Val – Ile
Deduce the sequence of the first fragment generated by CNBr treatment.
Model Answer
Asp – Pro – Tyr – Val – Ile – Arg – Gly – Tyr
Deduce the entire sequence in the original polypeptide. Indicate the CNBr-labile and trypsin-labile sites in this sequence.
Model Answer
The full sequence is:
Asp1 – Pro2 – Tyr3 – Val4 – Ile5 – Arg6 – Gly7 – Tyr8 – Met9 – Glu10 – Thr11 – Ser12 – Ile13 – Leu14 – Val15 – Ala16 – Met17 – Gly18 – Gln19 – Asn20 – Arg21 – Phe22 – His23 – Thr24 – Ala25 – Leu26 – Ser27 – Cys28 – Glu29 – Met30 – Arg31 – Tyr32 – Pro33 – His34 – Asn35 – Trp36 – Phe37 – Lys38 – Gly39 – Cys40
CNBr-labile sites (cleave after Met residues):
- Bond between Met9 and Glu10
- Bond between Met17 and Gly18
- Bond between Met30 and Arg31
Trypsin-labile sites (cleave after basic residues Arg and Lys):
- Bond between Arg6 and Gly7
- Bond between Arg21 and Phe22
- Bond between Arg31 and Tyr32
- Bond between Lys38 and Gly39
What percentage of the total residues are basic amino acids?
Model Answer
There are 6 basic amino acid residues in the polypeptide (Arg6, Arg21, Arg31, His23, His34, Lys38). Thus, 6 / 40 = 15%.
If the polypeptide were exist as an α helix, what will be the length of this helical structure?
Model Answer
An α-helix has 3.6 amino acid residues per turn of 5.4 Å. Therefore, the length is: (40 / 3.6) * 5.4 = 59.4 Å.
What will be the size of the DNA segment (exon) coding for this polypeptide of 40 amino acids? Give the size in base pairs as well as in daltons. (consider average molecular weight of a nucleotide in DNA = 330).
Model Answer
Since each amino acid is coded for by a triplet codon of nucleotides, the size of the double-stranded DNA segment is:
40 * 3 = 120 base pairs.
The molecular weight (in daltons) of this double-stranded DNA segment is calculated considering two nucleotides per base pair:
330 * 2 * 120 = 79,200 Da.
Assuming that the DNA corresponding to the exon contains equal numbers of Adenine and Cytosine, calculate the number of H-bonds which will hold this double helix.
Model Answer
In a 120 base pair DNA segment with equal amounts of Adenine (A) and Cytosine (C), there will also be equal amounts of Thymine (T) and Guanine (G). Hence, there are exactly:
- 60 A-T base pairs (each held by 2 H-bonds)
- 60 G-C base pairs (each held by 3 H-bonds)
Total number of H-bonds = (60 * 2) + (60 * 3) = 120 + 180 = 300.