Interpretation
.../z of y-ion 每 water 每 1 (mass 每 19) is the amino acid, and we can see the band at 175.10 fits the mass of Arginine when substituted into the formula. So, the m/z of the y1-ion is 175.10. The bn-1 ion m/z is thus (1274.67 每 175.10 + 1) 1100.57. 4. There are 4 possibilities of the b2-ion, but after some substitution, we can see that the 2nd amino acid from the N-terminal is Serine (S), as: m/z of yn-1 ion = m/z of yn-2 ion + mass of the 2nd residue from the N-terminal. 1074.55 + 87.04 = 1161.59 And that band can be found in the spectrum. So, the 1st residue of the peptide should be Leucine or Isoleucine (L/I) and the 2nd is Serine (S). 5. By using the formula: m/z of yn-i ion = m/z of yn-i-1 ion + mass of the i-1 residue from the N-terminal then substituting the monoisotopic masses of each residue from Table 1 and finding inspecting the spectrum to make sure the yn-i ions and the corresponding bi ion exists in the spectrum, we can come up with following: i m/z of bi ion m/z of yn-i ion Residue i from N-terminal 1 / 1161.59 L/I 2 201.12 1074.55 S 3 348.18 927.49 F 4 477.23 798.44 E 5 574.28 701.39 P 6 673.35 602.32 V 7 744.39 531.29 A 8 930.46 345.21 W 9 1029.53 246.14 V 10 1143.59 175.10 A 11 / / R n = 11 6. This step is not needed 7. The mass of the peptide = The total mass of the 11 residues = 1255.67Da It is 19Da less than the measured mass. This is because of the additional water molecule and the extra proton included in the measured mass. The additional water and the proton is there because the y-ion from the 1st MS result was used in the 2nd MS experiment to obtain the resu...