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The MOLECULES
of LIFE
Physical and Chemical Principles
Solutions Manual
Prepared by James Fraser and Samuel Leachman
Chapter 2
Nucleic Acid Structure
Problems and Solutions
True/False and Multiple Choice
Fill in the Blank
1. The tertiary structure of functional RNA molecules is
easily predicted.
7. The ______________ structure of a nucleic acid is the
sequence of nucleotides in the DNA or RNA molecule.
Answer: primary
True/False
2. Which of the following is not a stabilizing force for the
structure and stability of double-stranded nucleic acids?
a. b.
c.
d.
base stacking
hydrogen bonding
disulfide bonds
electrostatic forces
3. H-bond acceptor, H-bond donor, H-bond acceptor,
methyl group is the pattern of potential interactions at
the edge of which Watson-Crick base pair?
a.
b.
c.
d.
e.
A-T
G-C
A-A
C-G
T-A
4. Genomic DNA can become deformed from its normal
B-form by DNA binding proteins, such as the histone
proteins and the TATA-box binding protein.
True/False
5. Which nonstandard base pair is most likely to form a
wobble base pair?
a.
b.
c.
d.
e.
U-U
A-G
A-A
G-U
G-G
6. Classify the following RNA structural elements as
secondary or tertiary structure:
a.
b.
c.
d.
e.
f.
g.
coaxial helices
pseudoknot
hairpin
junction
adenosine platform
ribose zipper
bulge
Answer: Secondary: c, d, g; tertiary a, b, e, f
8. The modified RNA base in which two methyl groups are
added to guanine is __________.
Answer: N,N-dimethylguanine. The corresponding
nucleoside is N,N-dimethylguanosine.
9. B-form DNA has a C2ʹ ________ sugar pucker.
Answer: endo
10. Hoogsteen base pairs, where the hydrogen-bonding
interactions utilize the Watson-Crick base-pairing edge
on one base and the major groove edge in the other
base, can be utilized to form an RNA _______ helix.
Answer: triple
11. Metal ions, such as K+, Na+, and Mg2+, typically interact
with the __________ group on the backbone of nucleic
acids.
Answer: phosphate
Quantitative/Essay
12. What physical factors force RNA to adopt only the C3ʹ
endo sugar configuration, but allow DNA to adopt either
the C2ʹ endo or the C3ʹ endo sugar configurations?
Answer:
The C2ʹ endo configuration results in close contact
(1.9 Å) between one of the oxygen atoms of the
3ʹ phosphate group and the hydrogen atom at the
2ʹ position of the ribose ring in DNA. Compared with
DNA, RNA has the 2ʹ hydrogen replaced by hydroxyl
group (OH) and the two oxygen atoms (2ʹ ribose and
3ʹ phosphate) repel each other strongly. Rather than
distorting the structure away from the Watson-Crick
model, it is energetically favorable for RNA to switch to
the C3ʹ endo conformation.
2
Chapter 2: Nucleic Acid Structure
2Q14
Problem
question
2) (na_57_v1)
13. WhatSet
type(quantitative
of RNA interaction
is shown
below?
Answer:
H
H
minor groove
G
N
N
O
H
N
N
N
N
H
H
Label the major and minor grooves of the Watson-Crick
base pair and the potential hydrogen bonds between all
bases.
Answer:
This shows an A-minor motif in which the minor groove
edges of an adenine base interact with the minor
groove
2Q13 of a G-C base pair.
major groove
U
O
H
O
N
H
major groove
Interactions on minor groove: H-bond acceptor, H-bond
donor, H-bond acceptor.
Interactions on major groove: hydrogen atom, H-bond
acceptor, H-bond acceptor, H-bond acceptor.
The G-U base pair is not a standard Watson-Crick base
pair.
15. List at least three physical features that distinguish
A- and Z-form DNA.
Answer:
The main differences between A- and Z-form DNA are
as follows:
For A-form DNA, the repeating unit is one base;
however, in Z-form DNA, the repeating unit is two bases
(most often alternating pyrimidine and purines).
minor groove
A-form DNA is a right-handed double helix. Z-form DNA
is a left-handed double helix.
In A-form DNA, the sugar pucker of the bases is
C3ʹ endo, whereas Z-form DNA alternates between
C2ʹ endo and C3ʹ endo.
14. Label the atoms, the bases, the hydrogen bonds
across bases, the major and minor grooves, and the
interactions along the grooves for the following base
pair (oxygen and nitrogen atoms are not identified
explicitly):
Problem
set (quantitative question 2) (na_56_v1)
The minor groove of A-form DNA is wide and shallow,
whereas the Z-form minor grove is narrow. The major
groove of A-form DNA is deep and narrow, whereas the
Z-form major groove is relatively shallow.
In A-form DNA the base pairs are tilted to the helical
axis, whereas the Z-form base pairs are nearly
perpendicular to the helical axis.
16. In water, why is base stacking relatively more important
than hydrogen bonding to forming a DNA double helix?
Answer:
Water can make H bonds with the DNA base H-bond
donor and acceptors, so the energetic gain in
interactions when those H bonds are satisfied by other
nucleotides is not great. However, the many small polar
interactions along the nucleotide rings that are formed
upon base stacking are energetically more favorable
than the interactions with water.
Is the base pair a standard Watson-Crick base pair?
17. How are the interactions between a nucleic acid and
Mg2+ mediated by water?
PROBLEMS and solutions
Answer:
Water molecules can form a hydration shell around
Mg2+, facilitating indirect interactions between it and
the phosphate and functional groups in nucleic acids.
18. Why is the R (purine) in GNRA tetraloops required
rather than a pyrimidine?
Answer:
The R position, adjacent to the A in sequence and
space, is functionally significant because it provides
a large surface area (the base has the two rings of a
purine, as opposed to one ring for a pyrimidine) against
which the A can base stack.
19. Why do most transcription factors interact with the
major groove of B-form DNA rather than the minor
groove?
3
22. The structure of the large ribosomal subunit from
Haloarcula marismortui (PDB code: 1FFK) has been
solved by x-ray crystallography. The 23S RNA contains
2922 nucleotides (758 A, 889 G, 739 C, and 536 U).
a. Assuming a random distribution of nucleotides,
how many four-mer sequences with the sequence G-N
(any base)-R (purine)-A are possible?
b. There are actually 21 GNRA tetraloops in
the structure. What percentage of possible GNRA
sequences actually formed tetraloops in the structure?
Answer:
a. Probability of G = total G/total nucleotides =
889/2922
Probability of N = 1.
Probability of R = (total A + total G)/total nucleotides =
(758 + 889)/2922.
Answer:
Probability of A = total A/total nucleotides = 758/2922.
The major groove has more distinguishing interactions
than the minor groove. Each base pair has a unique
pattern of molecular interactions along the major,
but not the minor, groove. Since transcription factors
need to bind specifically to sequences of DNA, the
best potential for distinguishing different sequences is
to interact with the major groove. Further, the minor
groove is narrow compared with the major groove.
The depth of the major groove is compatible with the
interaction with protein structural elements such as an
α helix.
Each four-mer has a GRNA probability = (probability of
G) × (probability of N) × (probability of R) × (probability
of A) = 0.0445.
20. Which of the following DNA sequences is most likely to
adopt the Z-form? Why?
a.
b.
GCGCGCGCATATGCGCGCGCC
AGAGAGCTCTCTCTCTAAAAT
Answer:
Sequence a is more likely to adopt Z-form because
it alternates purine and pyrimidines in a GC-rich
sequence. This alternating pattern of 2ʹ endo and 3ʹ
endo puckers yields the zig-zag pattern, where the
smallest repeating unit is two base pairs, characteristic
of Z-form DNA.
21. Consider a relaxed, closed-circular DNA plasmid that
has 1040 base pairs with writhe = 0. An intercalator is
added, such that there is one intercalator per 104 base
pairs. The effect of the intercalator is to cause the twist
between the base pairs that flank it to be reduced
to zero. Will the resulting intercalator-bound DNA be
positively or negatively supercoiled?
Answer:
Positively supercoiled. The relaxed plasmid contains
100 turns (1040 bp × 1 turn/10.4 bp) and therefore
has linking number L = 100. Because it is not initially
supercoiled, it has writhe W = 0. By the relation
W = L – T, initial twist T is 100, with each base pair
contributing ~35°. The cumulative effect of binding
10 intercalators is a reduction of the twist by a full turn
(~35° × 10 ≈ 360°) to T = 99. As L is constant, positive
supercoiling with W = 1 results.
Total four-mers = 2922 – 3 = 2919.
Probable GRNA four-mers = probability × total fourmers = 0.0445 × 2919 = 130.
b. If only 21 GNRA tetraloops formed out of a
possible 130, then16% of the sequences form a stable
tertiary GNRA tetraloop motif.
23. A bacterial DNA polymerase moves at approximately
1000 base pairs per second when replicating DNA. The
polymerase holoenzyme is approximately
110 Å long. By what multiple of its length does the
polymerase move forward along the axis of the DNA
double helix in 3 seconds? Assume the DNA stays fixed
and ignore the rotational component of the motion of
the polymerase along the DNA.
Answer:
Each bp of DNA ~3.4 Å rise per bp.
In 3 s, the polymerase moves 3000 bp.
Length = 3000 bp × 3.4 Å•bp–1 = 10,200 Å.
Length/polymerase length = 10,200 Å/110 Å = 92.7
times the total length of the polymerase in 3 sec.