Answers for the Problems from “Chapter 19 Extra Stuff”
1. Feline gastrin…
End-group analysis shows that N-terminus is Glu
– so fragment 1 has to be the first fragment.
Also shows that the C terminus is Phe – so fragment 3 has to be the last fragment.
We have no information in this problem as to whether the middle two are
2-4 or 4-2 (which order), so the two possible sequences are
1-2-4-3 OR 1-4-2-3
2. Assigning pK a s to groups…
We have learned that amino/ammonium groups have pK a s roughly around
10. Also, carboxyl groups have pK a s roughly around 4. So, the lower two pK a s should correspond to the two carboxyl groups, and the higher pK a
will correspond to the N-terminal NH
3
. (I’d draw out the dipeptide here, but I can’t. There are three ionizable groups: the amino group in the N terminus, the carboxyl group on the C terminus, and the carboxyl group which is part of the side chain of aspartic acid.)
The highest pK a
(8.60) thus gets assigned to the N terminus. To distinguish between the other two, the carboxyl group closer to the electronwithdrawing NH
3
+ is the one that is more acidic, with the lower pK a
. The one that is closer/more acidic is the one that is the C terminal group, and not in the side chain… therefore, 2.81 = pK a
of C-terminal carboxyl group, and 4.45 = pK a
of side chain carboxyl group.
3. Here, look at the chemistry of the side chains on the amino acids to figure out what type of interactions they might have with other parts of a protein, or other molecules they might bind to, or a s ubstrate if the molecule is an enzyme… a. Leu/Phe/Ile are all hydrophobic. They are likely to interact with other amino acids that are also hydrophobic. b. Gln is polar, but not charged. It is likely to hydrogen bond with someone else who is also polar, and it can also interact with another charged amino acid. c. Arg is positively charged. It will form a salt bridge with a negatively charged amino acid d. Lys is also positively charged. It will also form a salt bridge with a negatively charged amino acid.
4. Carboxypeptidase releases Arg (note the typo… oops) so this is the Cterminal amino acid. One round of Edman degradation released Phe, so this is the N-terminal amino acid. If you then figure that Arg has to be the C end, and
Phe has to be the N end, you have to line the sequence up so that the purple chunk comes first, the green part comes second, and the blue Arg is third.

5. Here is one that is digested by two different enzymes, and you have to line it up to figure out where your fragments overlap.
Gly-Ile-Val-Glu-Gln-Cys-Cys-Ala-Ser-Val-Cys-Ser-Leu-Tyr
Leu-Tyr-Gln
The last chunk (Leu-Glu-Asn-Tyr-Cys-Asn) goes at the C terminus, and in fact, from the thermolysin cleavage, there is an overlapping Gln. (there really isn’t enough space here to show this properly!)
Final sequence would be…
Gly-Ile-Val-Glu-Gln-Cys-Cys-Ala-Ser-Val-Cys-Ser-Leu-Tyr-Gln-Leu-Glu-
Asn-Tyr-Cys-Asn which contains 21 amino acids, just like the problem reads.