关键词 > BIOSE-40
BIOS E-40- HA #2
发布时间:2023-12-14
Hello, dear friend, you can consult us at any time if you have any questions, add WeChat: daixieit
BIOS E-40- HA #2
DUE DATE: October 17 2023
PART 1: Undergraduate + Graduate, (20 points)
A tryptic peptide was analyzed with a triple-quadrupole mass spectrometer in ms/ms mode. The following figure shows the mass spectrum after fragmentation (Panel A) and an enlarged detail of the spectrum (Panel B) .
The following table shows the m/z ratio of the fragments F1 to F7 determined on the spectrum shown in Panel A.
![](/Uploads/20231214/657ab3d96f279.png)
a) Complete the table and identify the peaks that belong to the same series. The table showing the contributions of residues to the m/z ratio of a peptide can be found in your lecture notes. Briefly explain your answer (4 pts) .
b) Determine the partial sequence of the fragmented peptide. The trypsin digestion was complete and the fragmented peptide did not contain any missed cleavage site. (2 pts)
c) A careful analysis of the region of the spectrum indicates the presence of two satellites peak flanking F3 (see Panel B above). Determine the nature of the F3 fragment (y-ion or b-ion) based on the difference in m/z ratio between these satellite peaks and F3. Briefly explain. (4 pts)
You may need to following information to answer the question:
Atomic mass: Oxygen: 16; Nitrogen: 14; Hydrogen: 1; Carbon: 12
d) Based on your previous answer, write the sequence of tryptic peptide from its amino- end to its carboxy-end (2 pts)
e) Peak X (Panel A) is produced by a multi-protonated form of the ion responsible for the formation of the peak F2 (contain only 1 proton) . Calculate how many protons are associated with peak X knowing that the m/z ratio for this peak is: 160.08. Show your calculation (4 pts).
f) The fragmentation spectrum of a mutant form of the protein where the N-terminal residue of the F2 ion is replaced by alanine (A) shows a shift of peaks F2 and X.
Calculate the m/z shift of the peak X due to the mutation. Show your calculation (4 pts).
PART 2: Graduate only (30 points)
In proteins disulfide bonds are formed by the oxidative linkage of sulfhydryl groups (-SH) carried by the side chain of cysteine residues .
Mass spectrometer can detect a change in mass due to reduction and alkylation of disulfide bonds or free sulfhydryl groups. In the rest of this question we will use the following terminology:
Mnat = mass of the native protein
Mr = mass of the reduced protein
Mr+a = mass of the reduced and alkylated protein
Ma = mass of the alkylated protein without reduction
a) Based on the atomic mass of H (1), C (12), O (16), S(32) calculate the change in protein mass due to the reduction and alkylation of one cysteine residue engaged in a disulfide bond and the change in mass due to the alkylation of a free cysteine residue. (8 pts)
b) Determine the equation linking Mr+a , Mnat , NSS (total number of cysteine residues engaged in disulfide bonds) and NSH (total number of free cysteine). (6 pts)
c) Determine the equation linking Ma , Mnat and NSH . (6 pts)
d) By combining the two previous equations calculate the number of disulfide bonds contained in a protein that has the following characteristics: Mnat = 2000 Da, Mr+a = 2586 Da and Ma = 2232 Da. Show your calculation. (6 pts)
e) In which region(s) of the cell could you find a protein with a disulfide bond? Briefly explain (4 pts)