PROTEINS - SELF STUDY QUESTIONS
1. This is the titration curve for which amino
acid?
2. A Ramachandran plot shows
A. The amino acid residues that have the greatest degrees of rotational freedom.
B. The sterically allowed rotational angles between the side chain groups in a peptide
and the peptide backbone.
C. The sterically limited rotational angles (domains) where phi and psi are allowed in
the protein backbone.
D. The angles that are allowed about the bonds connecting the amide nitrogen in a
peptide bond.
3. Consider a peptide with the sequence Glu-Val-His-Ser-Arg. What will be the net charge
at pH 4? At pH 9?
4. These data relate the concentration of a ligand [L] and the extent of saturation of its
binding site ( ) for two different proteins, A and B. Plot  vs. [L] and determine whether
the proteins exhibit hyperbolic or sigmodial binding behavior.
[L] mM
 for protein A
 for protein B
0.10
2.2
0.3
0.35
7.2
1.0
0.80
15.0
3.0
1.80
29.0
9.0
3.00
40.0
25.0
4.50
50.0
50.0
5.75
56.0
76.0
8.00
64.0
90.0
13.00
74.0
97.0
5. The following question deals with the properties of amino acid sidechains buried in the
hydrophobic interior of a protein. (A) Would the pKa of a buried lysine be higher or
lower than the pKa of a surface Lys? (B) Would the strength of a buried hydrogen bond
be stronger or weaker than a hydrogen bond on the surface? (C) Would the strength of the
electrostatic interaction between a buried ion pair be stronger or weaker than an ion pair
on the surface?
6. The isoelectric point (pI) of an amino acid is the pH at which the net electrical charge is
zero. The two structures shown for Ala
each have a net charge of zero. (A) Why
is the predominant form of Ala at its
isoelectric point the zwitterionic and not
the uncharged form? (B) Calculate the
ratio of the concentration of the
zwitterionic form/uncharged form at the pI.
7. Explain the following observation. At pH 7.0, polylysine, which is a polypeptide in
which all the residues are lysine, adopts a random coil conformation, i.e., it has no
secondary structure, but at pH 12 it is present as a  -helix.
8. The binding of a ligand (L) to a protein (P) is often a simple equilibrium, P + L  PL,
which is characterized by the dissociation constant Kd = [L][P]/[PL], where [L] is the
concentration of unbound or free ligand, [P] is the concentration of free protein and [PL]
is the concentration of the protein-ligand complex. If PT = [P] + [PL] = total protein
concentration and  = [PL]/PT, then  = [L]/{Kd + [L]}. However, a plot of  vs. [L] is a
hyperbole (see 4B) and it is difficult to evaluate Kd from such a plot. Derive a linear
transformation of  = [L]/{Kd + [L]}, plot the data from
4A and determine Kd.
9. A useful method for visualizing the properties of a
protein helix is to display the sequence on a helical wheel
diagram. This diagram is based on a five-turn helix (18
residues). Adjacent residues in the sequence are 1000
apart on the wheel and a residue occurs every 20o. The
sequence is written onto the wheel in the order shown.
Display the following sequence on a helical wheel and
explain what information this representation provides.
Ala-Phe-Asp-Lys-Met-Ile-Glu-Asn-Leu-Gln-Arg-LeuTrp-Ser-Glu-Phe-Leu-Gln.
10. Myoglobin and the subunits of hemoglobin are about the same size and the structure of a
subunit of hemoglobin is very similar to the structure of myoglobin. Given these facts,
explain why hemoglobin has a higher ratio of nonpolar/polar amino acids than does
myoglobin.
11. This table contains the torsion angles ( ,  ) for several residues in a protein sequence.
Consult a Ramachandran diagram and determine the conformation of residues 3-11, of
residues 43-51?
Residue
 (deg)
 (deg)
Number
Residue
 (deg)
 (deg)
Number
1
-60
147
42
-142
150
2
-49
-32
43
-154
121
3
-67
-34
44
-121
136
4
-58
-49
45
-110
174
5
-66
-32
46
-128
162
6
-82
-36
47
-135
156
7
-69
-44
48
-121
138
8
-61
-44
49
-131
157
9
-72
-29
50
-115
130
10
-66
-65
51
-126
146
11
-67
-23
52
-67
-10
12. The binding of oxygen to hemoglobin involves an allosteric transition from a weakbinding form (T-State) to strong-binding form (R-State). One can evaluate the affinity of
the R and T States for O2 using a Hill plot in which the log{[ ]/[1- ]} is plotted vs. log
{pO2}, where  = Fractional saturation and pO2 = partial pressure of O2. Use these data to
construct a Hill plot and determine the p50 (the pO2 at which  = 0.5) for the T and R
States.
pO2

pO2

pO2

0.1
0.00315
2.88
0.24
12.88
0.969
0.35
0.0099
4.7
0.50
29.51
0.99
0.79
0.031
5.75
0.76
67.60
0.997
1.75
0.091
7.94
0.909
ANSWERS
1. Lys. (a) It requires 3 equivalents of OH- to titrate the amino acid; (b) the amino acid is
basic because there are 2 groups with pKas > 7.0; (c) the titration is finished at pH < 12,
which eliminates Arg.
2. C
3. At pH 4 charge is + 1.5 -  -carboxyl group completely ionized (-1.0); side chain
carboxyl 50% ionized (-0.5); side chain of His fully protonated (+1.0); side chain of Arg
fully protonated (+1.0);  -amino group fully protonated (+1.0). At pH 9.0 charge is – 0.5
-  -carboxyl group completely ionized (-1.0); side chain carboxyl completely ionized (1.0); side chain of His fully titrated (0); side chain of Arg fully protonated (+1.0);  amino group 50% protonated (+0.5).
4. See plot at the right. Protein A exhibits
hyperbolic binding behavior; protein B
exhibits sigmodial binding behavior.
5. (A) The pKa would be lower because deprotonation would remove a positive charge from
the hydrophobic environment, a favorable process. (B) Stronger because the dielectric
constant in the interior is lower than in water. The hydrogen bond has a partial
electrostatic character and a lower dielectric constant leads to stronger electrostatic
interactions. (C) Stronger because the dielectric constant in the interior is lower than in
water, and a lower dielectric constant leads to stronger electrostatic interactions.
6. (A) The carboxyl group (pK1 = 2.3) a stronger acid than the amino group (pK2 = 9.7),
therefore ionization of the carboxyl group occurs before ionization of the amino group.
(B) If we represent the fully protonated form as +, the zwitterion as  and the uncharged
form as O, then Henderson-Hasselbalch equation for the ionization of the COOH group
to form the zwitterion is pH = pK1 + log{[ ]/[+]} and for the ionization of the amino
group to form the uncharged species it is pH = pK2 + log{[O]/[+]}. At the pI these two
equations must be equal, pK1 + log{[ ]/[+]} = pK2 + log{[O]/[+]}; log [+] cancels from
both sides which leaves log{[ ]/[O]} = pK2-pK1 and [ ]/[O] = 10(pK2-pK1) = 2.2 x 107.
7. At pH 7.0, the amino sidechains are charged which causes electrostatic repulsion between
the side chains whenever the polylysine tries to adopt a secondary structure. At pH 12,
the sidechain amino groups are deprotonated, which removes the electrostatic repulsion
between sidechains, thereby permitting the polylysine to adopt a  -helical structure.
8. A linear transformation of this equation is 1/ =
Kd/[L] + 1, and a plot of 1/ vs. 1/[L] should be a
straight line with slope = Kd. For this example, Kd
= 4.5mM.
9. The helical wheel representation clearly shows that this is
an amphipathic helix with one nonpolar face (amino acids
in CAPS) and one polar face (amino acids in lower case).
This property of the helix is not obvious from a simple
inspection of the sequence.
10. The contacts between the subunits of hemoglobin consist of hydrophobic interactions (=
nonpolar amino acids). The analogous regions in myoglobin are in contact with water and
are therefore occupied by polar amino acids.
11. Residues 3-11 clearly form a  -helix, whereas residues 43-51 form a  -sheet.
12. The Hill plot is shown to the right. By
drawing an asymptote with a slope of 1.0
to the points at high values of  and then
extrapolating back to  = 0.5, one obtains
p50 for the R-State. Using the same
procedure with the points at low  gives
p50 for the T-State.