8 Chapter 5
43. The amount of Edman reagent must exactly match the amount of N–termini in the
first reaction. If there is too little Edman reagent, some of the N–termini will not
44. In the first cycle, the first and second amino acids from the N-terminal end would
be reacted and released as PTH derivatives. You would get a double signal and
not know which one was the true N-terminus.
45. Val—Leu—Gly—Met—Ser—Arg—Asn—Thr—Trp—Met—
Ile—Lys—Gly—Tyr—Met—Gln—Phe
47. It is possible that your protein is not pure and needs additional purification steps
to arrive at a single polypeptide. It is also possible that the protein has subunits,
so multiple polypeptide chains could be yielding the contradictory results.
48. There are two fragments that have C-termini that are not lysine or arginine, which
is what trypsin is specific for. Normally there would be only one fragment ending
49. It would tell you a relative concentration of the various amino acids. This is
important because it would help you plan your sequencing experiment better. For
example, if you had a protein whose composition showed no aromatic amino
acids, it would be a waste of time to use a chymotrypsin digestion.
50. Cyanogen bromide would be useless, because there is no methionine. Trypsin
51. Chymotrypsin would be a good choice. There are more than four residues of
52. It would work best if the aromatic residues were spread out in the protein. In that
53. Electrospray Ionization (ESI-MS) and Matrix-Assisted Laser Desorption
Ionization– Time of Flight (MALDI-TOF MS).
54. MALDI-TOF MS is very sensitive and very accurate. Attomole (10–18) quantities of
a molecule can be detected.