The MOLECULES
of LIFE
Physical and Chemical Principles
Selected Solutions for Students
Prepared by James Fraser and Samuel Leachman
Chapter 4
Protein Structure
Problems and Solutions
True/False and Multiple Choice
2. The secondary structure of a protein refers to the
extent and order of its α helices and β sheets.
that of a normal strain of bacteria. This extra thickness
is entirely due to longer hydrophobic tails and not
due to changes in the charged head groups of the
phospholipids that comprise the membrane. If you
isolated a single transmembrane helix from a protein
from this strain, how long would you expect it to be?
True/False
4. Match the following proteins with their quaternary
structure:
a. hemoglobin
b. RNA polymerase
c. myoglobin
i. one subunit
ii. many structurally
similar subunits
iii. many subunits of
varied structure
Answer: a–ii, b–iii, c–i
6. The only genetically encoded amino acid without a
stereoisomer is:
a.
b.
c.
d.
e.
alanine
tryptophan
glycine
proline
lysine
Answer:
70 Å membrane/1.5 Å per residue of alpha helix =
at least 47 residues.
16. Using the hydrophobicity scale in Figure 4.74, calculate
the hydrophobicity index for the 19 contiguous residue
window defined by the following sequence:
Pro-Gly-Ala-Val-Val-Ile-Trp-Phe-Val-Val-Met-Ser-Ala-IleIle-Phe-Tyr-Ala-Thr
Could this segment be part of a transmembrane helix?
Answer:
Pro-Gly-Ala-Val-Val-Ile-Trp-Phe-Val-Val-Met-Ser-Ala-IleIle-Phe-Tyr-Ala-Thr.
8. The active site in open twisted α/β domains is in a
crevice outside the carboxyl ends of the β strands.
True/False
0.4 + 4.5 + 2.1 – 2.1 – 2.1 – 4.5 – 8.8 – 7.1 – 2.1 – 2.1 –
2.9 + 2.1 + 2.1 – 4.5 – 4.5 – 7.1 – 2.9 + 2.1 + 0.8
= –36.6; –36.6/19 = –1.9
This segment could be part of a transmembrane helix
as the hydrophobicity index is negative and there are no
charged residues.
Fill in the Blank
10. The organization of the protein subunits in multimeric
proteins is known as the ___________ structure.
Answer: quaternary
12. _______ residues form cis peptide bonds in proteins
with significant frequency.
Answer: Proline
18. Draw an alanine-alanine dipeptide. Indicate the peptide
bond. What factors restrict the rotation about the
4Q18 bond?
peptide
Answer:
O
H2N
Quantitative/Essay
14. A group of scientists isolate a novel strain of bacteria.
Although the bacteria have normal nucleic acids and
proteins, the strain has an abnormal membrane. In
this strain, the lipid bilayer is twice as thick (70 Å) as
CH
CH3
C
peptide bond
N
CH
O
C
OH
CH3
Rotations are strongly hindered because the peptide
bond has partial double bond character. Evidence for
this double bond character includes the observation
The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer © Garland Science
2
Chapter 4: Protein Structure
from high resolution crystal structures that the peptide
bond length is shorter than normal single N–C bond
lengths.
20. How are the large loop elements of scorpion toxin
stabilized without participating in the hydrophobic
core?
Answer:
In addition to the hydrophobic core, this particular
protein fold is held together by four disulfide bonds. In
the loop regions, the disulfide bonds provide covalent
linkages between different parts of the protein
backbone. Since the energy required to break these
covalent bonds is very large, these loop elements can
be stabilized without significant contributions from the
hydrophobic effect.
22. The ridges and grooves of an α helix form an angle of
~25° or ~45° to the helical axis. What are the spacings
between the residues that line these two types of
grooves? How do these geometric considerations
constrain the packing of pairs of helices at relative
angles of 50° and 20°?
Answer:
~25° grooves are separated by four residues; ~45°
grooves are separated by three residues. In order to
pack the two helices using two ~25° grooves, one
helix must be turned 180° and placed on top of the
other helix. In the interface between the two helices
the directions of the ridges and grooves must thus be
inclined by an angle of about 50° (25° + 25°) in order
for the ridges of one helix to fit into the grooves of the
other and vice versa.
24. In contrast to globular proteins and regions exposed
to the cytoplasm, the integral membrane portions of
membrane proteins contain very few residues not
ordered into secondary structure elements such as
β sheets and α helices. Why is satisfying backbone
hydrogen bonds by forming secondary structure
elements more important for protein folding in the
membrane than in solution?
Answer:
In solution, residues in loop segments can form
backbone–water hydrogen bonds and thus do not pay
a large energetic penalty for failing to satisfy these
interactions with other backbone atoms. In contrast, in
the membrane the lipid tails do not provide hydrogen
bond donor or acceptors. Additionally, there is no
energetic gain from burying a hydrophobic core in the
membrane since the hydrophobic environment negates
the importance of the hydrophobic effect. Secondary
structure formation is an efficient mechanism
for completely satisfying the hydrogen bonding
requirements of the peptide backbone. These hydrogen
bonds provide the major energetic force driving folding
in the membrane.
26. What is the conformational change that occurs when
bacteriorhodopsin absorbs light?
Answer:
The conformational change occurs on the covalently
bound cofactor retinal. When retinal absorbs a photon,
the bond between carbon atom 13 and carbon atom
14 is isomerized from a trans conformation to a cis
conformation.
To form an interface between ridges formed using a
~25° and a ~45° groove, one helix must be rotated 180°.
The helices are then on the same side of the helical axis
and make an angle of 20° (45° – 25°) when the ridges
and grooves fit into each other.
The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer © Garland Science