[burningdesire]

Identify all y ions in the spectrum and describe what ions they are. What sequence information can you deduce from that?

Please help me. I have no idea how to find the y ions from the spectrum. How should I start?

[MOTOBALL]

Consider the protonated tetrapeptide,

[NH2R₁(C=O)NHR₂(C=O)NHR₃(C=O)NHR₄(C=O)OH + H⁺]⁺

where R1, R2, R3 and R4 correspond to the residual masses of the aminoacids (AA), i.e., AA mol. weight minus NH2 minus COOH.

Fragmentation of the [Tetrapeptide + H⁺]⁺ can give a series of b-ions and a series of y-ions, as below.

b-ions are formed from the N-terminus

b₁      NH2R₁C=O+
b₂      NH2R₁C=ONHR₂C=O+
b₃      NH2R₁C=ONHR₂C=ONHR₃C=O+

The m/z difference between b₂ and b₁ is then NHR₂C=O+, etc..

y-ions are formed from the C-terminus

y₁      ⁺NH3R₄COOH
y₂      ⁺NH3R₃CONHR₄COOH
y₃      ⁺NH3R₂CONHR₃CONHR₄COOH

The m/z difference between y₂ and y₁ is then R₃C=ON⁺, etc..

NOTES
1. If ALL sequence ions are present, then the polypeptide can be sequenced completely (cannot distinguish Leu/Ile)
2. Sequence ions may also lose small neutrals (e.g.,H2O, NH3, CO) to give an accompanying ion at 18, 17 or 28 lower m/z than the sequence ion

INTERPRETATION

1. Construct, as a reference, a table (Table 1) of the NHRC=O+ values for all 20 common AA's.
2. Assume that m/z 875 is [M+H]⁺ (NB. it may be a "b" or "y" type ion from a larger peptide !!!)
3. MW 874.5 is consistent with an ≈ 8/9-residue polypeptide.
4. Start at either low or high m/z end of spectrum; construct a table (Table 2) of the m/z difference between adjacent m/z values.
5. Compare the values from Table 2 with those from Table 1.
6. On a copy of the spectrum, pencil in the m/z differences between adjacent signals, and from step 5 annotate on the spectrum the possible AA that is indicated.
7. Assign m/z values as "b" or "y" type ions; e.g., m/z 86 can be either y₁ (⁺NH3R_≈8/9COOH) or b₁ (NH2R₁C=O⁺).

Good Luck.

[MOTOBALL]

Correction,

The m/z difference between y2 and y1 is then R3C=ON+, etc..

should read,

The m/z difference between y2 and y1 is then R3C=ONH+, etc..

[burningdesire]

b₁      NH2R₁C=O+
b₂      NH2R₁C=ONHR₂C=O+
b₃      NH2R₁C=ONHR₂C=ONHR₃C=O+

Are b ions supposed to have C≡O⁺ ends instead of C=O⁺ ends?

Thanks for the help.

[burningdesire]

[NH2R₁(C=O)NHR₂(C=O)NHR₃(C=O)NHR₄(C=O)OH + H⁺]⁺

where R1, R2, R3 and R4 correspond to the residual masses of the aminoacids (AA), i.e., AA mol. weight minus NH2 minus COOH.

INTERPRETATION

1. Construct, as a reference, a table (Table 1) of the NHRC=O+ values for all 20 common AA's.

Are the following right?
NHRC=O+ values = AA molar mass - H - OH = AA molar mass - 18
and residual masses of AA = AA molar mass - NH2 - COOH = AA molar mass - 33
hence NHRC=O+ values = residual masses of AA + 15?

I was given this table


When comparing the values, should I use AA mass, residual masses of AA or NHRC=O+ values?

My table 1 looks like this:


According to the answers, the first amino acid at the C-terminus is lysine - which corresponds to the AA mass (146.11 vs 146.10) but not the residual mass..

Please clarify what's the difference between (1) AA mass, (2) residual mass and (3) NHRC=O+ values and which should I use each of them..


Thanks again.

[MOTOBALL]

Pease ignore my initial, confusing explanation.

Take the peptide,

                             NH2-Gly-Ala-Val-Leu-COOH

This will give, using the residue masses from your table,

b1   NH2-Gly-CO+                                      m/z 58
b2   NH2-Gly-Val-CO+                                m/z 157
b2   NH2-Gly-Val-Ala-CO+                          m/z 228

and,

y1   +NH3-Leu-COOH                                   m/z 132
y2   +NH3-Ala-Leu-COOH                             m/z 203
y3   +NH3-Val-Ala-Leu-COOH                       m/z 302

If the spectrum is presented as an unknown, i.e.  m/z 58   132     157    203     228     302
the interpretation could be done as follows, and will answer your questions about which masses to use.

                        Difference (mass residue)        Assignment
   m/z

   58                            --------                      b1 (NH2Gly-CO+)

   132                           74                            there is no mass residue of 74; therefore cannot be a b2 ion
                                                                   therefore must be y1 (+NH3Leu-COOH)

   157                           25                            there is no mass residue of 25 ;but diff. = +99 (Val) from m/z 58
                                                                   therefore is b2 (NH2Gly-Val-CO+)

   203                          46                             there is no mass residue of 46; but diff. = +71 from m/z 132
                                                                   therefore must be y2 (+NH3Ala-Leu-COOH)

   228                           25                            there is no mass residue of 25 ;but diff. = +71 (Ala) from m/z 157
                                                                   therefore is b3 (NH2Gly-Val-Ala-CO+)

   302                          74                             there is no mass residue of 74; but diff. = +99 from m/z 203
                                                                   therefore must be y3 (+NH3Val-Ala-Leu-COOH)

b1, b2, b3 give NH2Gly-Val-Ala

y1, y2, y3 give Val-Ala-Leu-COOH

Together, the two ion series give the full sequence.  Use the mass residue values given in your table to interpret your spectrum.