1. Chapter 19 “extra stuff”

2. Amino Acids as Acids and Bases… Revisited Think of an acid dissociation as a reaction that has an associated equilibrium: HAH + + A - There is an equilibrium constant for this reaction, called K a. K a = [products]/[reactants] = [H + ][A - ]/[HA]

3. Concept of pKa… A stronger acid has a larger K a value (remember: [products]/[reactants]) but K a values are usually still small to work with for weak acids. More convenient number: pK a = -log K a This typically gives you a non-exponent number. Comparing acids based on pK a, the smaller the number, the stronger the acid.

4. Typical pK a Values… A standard ammonium ion has a pK a of 10-11. A standard carboxylic acid has a pK a of 4-5. However, things change when you add other functional groups onto the molecule, as happens with an amino acid. Take a look at a few cases.

5. How AA Functional Groups Change Standard pK a Values Comparing acetic acid to glycine: the presence of the amino group in glycine reduces the pK a of the carboxylic acid function from 4.75 (in acetic acid) to 2.3 (in glycine). Inductive effect!!!

6. More, pK a s of AA Functional Groups On an amino acid with two carboxyl groups, the carboxyl group farther away from the positively charged amino group is less affected by the amino group. Therefore, the more distant carboxyls have pK a s closer to what one would expect for a usual carboxyl group.

7. Changing Gears… Peptide Sequence Analysis After a peptide has been isolated from whatever source, we need to determine its sequence… how? Analyze the sequence through Edman Degradation (see handout) React the N terminus with phenylisothiocyanate Treat this with mild acid, and the first amino acid is cleaved off, rearranging to form a phenylthiohydantoin. This can be compared to a standard, and the amino acid identified. Each amino acid is removed from the N terminus and identified this way… and the process is automated on a machine.

8. What If You Only Need to Know the Ends? Identify N terminal with either Sanger reagent or dansyl chloride (rest of peptide is lost, however, so no further analysis can be performed) Identify C terminal with carboxypeptidase enzyme. Most of the time, it will cleave successive C terminal amino acids. If you’re careful enough to catch the first reaction product, you can determine the original C terminal amino acid.

9. Additional Protein Sequence Analysis Using methods similar to the “partial acid hydrolysis” from the worksheet last week, enzymes can also cut the peptide at certain points: Trypsin cuts a peptide after positively charged amino acids (Arg or Lys) Chymotrypsin cuts a peptide after aromatic amino acids (Phe, Tyr, Typ)

10. Practice Problem 1 Feline gastrin has the empirical composition Ala 2, Asp, Gly 2, Glu 5, Leu, Met, Phe, Pro, Trp 2, Tyr. The peptide was digested with chymotrypsin and four peptide fragments were isolated, found to have the following sequences: 1. Glu-Gly-Pro-Trp2. Gly-Trp 3. Met-Asp-Phe4. Leu-Glu-Glu-Glu-Glu-Ala-Ala-Tyr End-group analysis revealed that the N-terminus is Glu and the C- terminus is Phe. What two sequences are compatible with this evidence?

11. Practice Problem 2 The dipeptide Gly-Asp has three known pK a values: 2.81, 4.45, 8.60. Associate each pK a with the appropriate functional group in the structure of this peptide.

12. Practice Problem 3 Describe or show (what type of interactions) how the side chains of the following amino acids may participate in ligand binding: a. Leu/Phe/Ile b. Gln c. Arg d. Lys

13. Practice Problem 4 The beta chain of insulin was digested with trypsin to yield the following two chains: Phe-Val-Asn-Gln-His-Leu-Cys-Gly-Ser- His-Leu-Val-Glu-Ala-Leu-Tyr-Leu-Val-Cys-Gly-Glu-Arg Arg Gly-Phe-Phe-Tyr-Thr-Pro-Lys Treatment of the beta chain with carboxypeptidase released Ala, and one round of Edman degradation released Phe. What is the primary structure of the beta chain of insulin?

14. Practice Problem 5 A peptide containing 21 amino acids was hydrolyzed by thermolysin (cleaves before Leu, Ile, and Val). The products of this treatment were Gly, Ile, Val-Cys-Ser, Leu-Tyr-Gln, Val-Glu- Gln-Cys-Cys-Ala-Ser, and Leu-Glu-Asn-Tyr-Cys-Asn. When the same peptide was hydrolyzed by chymotrypsin (cleaves after Phe, Trp, and Tyr), the products were Cys-Asn, Gln-Leu-Glu- Asn-Tyr, and Gly-Ile-Val-Glu-Gln-Cys-Cys-Ala-Ser-Val-Cys-Ser- Leu-Tyr. Give the amino acid sequence of this molecule.