Chapter 19
“extra stuff”
Amino Acids as Acids and Bases…
Revisited
Think of an acid dissociation as a reaction that has an
associated equilibrium:
HA
H+ + A-
There is an equilibrium constant for this reaction,
called Ka.
Ka = [products]/[reactants] = [H+][A-]/[HA]
Concept of pKa…
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A stronger acid has a larger Ka value (remember:
[products]/[reactants]) but Ka values are usually still
small to work with for weak acids.
More convenient number: pKa = -log Ka
This typically gives you a non-exponent number.
Comparing acids based on pKa, the smaller the
number, the stronger the acid.
Typical pKa Values…
A standard ammonium ion has a pKa of 10-11.
A standard carboxylic acid has a pKa 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.
How AA Functional Groups Change
Standard pKa Values

Comparing acetic acid to glycine: the presence of
the amino group in glycine reduces the pKa of the
carboxylic acid function from 4.75 (in acetic acid) to
2.3 (in glycine).
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Inductive effect!!!
More, pKas 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 pKas closer
to what one would expect for a usual carboxyl group.
Changing Gears… Peptide Sequence
Analysis
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After a peptide has been isolated from whatever
source, we need to determine its sequence… how?
Analyze the sequence through Edman Degradation
(see handout)
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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.
What If You Only Need to Know the
Ends?
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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.
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:
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Trypsin cuts a peptide after positively charged amino acids
(Arg or Lys)
Chymotrypsin cuts a peptide after aromatic amino acids
(Phe, Tyr, Typ)
Practice Problem 1
Feline gastrin has the empirical composition Ala2, Asp,
Gly2, Glu5, Leu, Met, Phe, Pro, Trp2, Tyr. The
peptide was digested with chymotrypsin and four
peptide fragments were isolated, found to have the
following sequences:
Glu-Gly-Pro-Trp 2. Gly-Trp
3. Met-Asp-Phe
4. Leu-Glu-Glu-Glu-Glu-Ala-Ala-Tyr
End-group analysis revealed that the N-terminus is Glu and the Cterminus is Phe. What two sequences are compatible with
this evidence?
1.
Practice Problem 2
The dipeptide Gly-Asp has three known pKa values:
2.81, 4.45, 8.60. Associate each pKa with the
appropriate functional group in the structure of this
peptide.
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
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-SerHis-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?
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-GluGln-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-GluAsn-Tyr, and Gly-Ile-Val-Glu-Gln-Cys-Cys-Ala-Ser-Val-Cys-SerLeu-Tyr. Give the amino acid sequence of this molecule.