Essential Cell Biology 5th edition

Answers A:15
transcriptionally inactive heterochromatin, whereas most
of the chromatin in (B) is decondensed and therefore
potentially transcriptionally active. The nucleus in (A) is
from a reticulocyte, a red blood cell precursor, which is
largely devoted to making a single protein, hemoglobin.
The nucleus in (B) is from a lymphocyte, which is active in
transcribing many different genes.
ANSWER 5–15 Helix (A) is right-handed. Helix (C) is
left-handed. Helix (B) has one right-handed strand and
one left-handed strand. There are several ways to tell the
handedness of a helix. For a vertically oriented helix, like
the ones in Figure Q5–15, if the strands in front point up
to the right, the helix is right-handed; if they point up to
the left, the helix is left-handed. Once you are comfortable
identifying the handedness of a helix, you will be amused
to note that nearly 50% of the “DNA” helices shown
in advertisements are left-handed, as are a surprisingly
high number of the ones shown in books. Amazingly,
a version of Helix (B) was used in advertisements for a
prominent international conference, celebrating the 30-year
anniversary of the discovery of the DNA helix.
ANSWER 5–16 The packing ratio within a nucleosome core
is 4.5 [(147 bp × 0.34 nm/bp)/(11 nm) = 4.5]. If there is an
additional 54 bp of linker DNA, then the packing ratio for
“beads-on-a-string” DNA is 2.3 [(201 bp × 0.34 nm/bp)/
(11 nm + {54 bp × 0.34 nm/bp}) = 2.3]. This first level of
packing represents only 0.023% (2.3/10,000) of the total
condensation that occurs at mitosis.
Chapter 6
ANSWER 6–1
A. The distance between replication forks 4 and 5 is about
280 nm, corresponding to 824 nucleotides (= 280/0.34).
These two replication forks would collide in about 8
seconds. Forks 7 and 8 move away from each other and
would therefore never collide.
B. The total length of DNA shown in the electron
micrograph is about 1.5 μm, corresponding to 4400
nucleotides. This is only about 0.002%
[= (4400/1.8 × 10 8 ) × 100%] of the total DNA in a fly cell.
ANSWER 6–2 Although the process may seem wasteful,
it is not possible to proofread during the initial stages of
primer synthesis. To start a new primer on a piece of singlestranded
DNA, one nucleotide needs to be put in place
and then linked to a second, and then to a third, and so on.
Even if these first nucleotides were perfectly matched to
the template strand, they would bind with very low affinity,
and it would consequently be difficult for a hypothetical
primase with proofreading activity to distinguish the correct
from incorrect bases; the enzyme would therefore stall.
The task of the primase is to “just polymerize nucleotides
that bind reasonably well to the template without worrying
too much about accuracy.” Later, these sequences are
removed and replaced by DNA polymerase, which uses
newly synthesized, adjacent DNA—which has already been
proofread—as its primer.
ANSWER 6–3
A. Without DNA polymerase, no replication can take place
at all. RNA primers will be laid down at the origin of
replication.
B. DNA ligase links the DNA fragments that are produced
on the lagging strand. In the absence of ligase, the
newly replicated DNA strands will remain as fragments,
but no nucleotides will be missing.
C. Without the sliding clamp, the DNA polymerase will
frequently fall off the DNA template. In principle, it can
rebind and continue, but the continual falling off and
rebinding will be so time-consuming that the cell will be
unable to divide.
D. In the absence of RNA-excision enzymes, the RNA
fragments will remain covalently attached to the newly
replicated DNA fragments. No ligation will take place,
because the DNA ligase will not link DNA to RNA.
The lagging strand will therefore consist of fragments
composed of both RNA and DNA.
E. Without DNA helicase, the DNA polymerase will
stall because it cannot separate the strands of the
template DNA ahead of it. Little or no new DNA will be
synthesized.
F. In the absence of primase, RNA primers cannot be
made on either the leading or the lagging strand. DNA
replication therefore cannot begin.
ANSWER 6–4 DNA damage by deamination and
depurination reactions occurs spontaneously. This type
of damage is not the result of replication errors and is
therefore equally likely to occur on either strand. If DNA
repair enzymes recognized such damage only on newly
synthesized DNA strands, half of the defects would go
uncorrected. The statement is therefore incorrect.
ANSWER 6–5 If the old strand were “repaired” using the
new strand that contains a replication error as the template,
then the error would become a permanent mutation in
the genome. The old information would be erased in the
process. Therefore, if repair enzymes did not distinguish
between the two strands, there would be only a 50% chance
that any given replication error would be corrected.
ANSWER 6–6 You cannot transform an individual from
one species into another species simply by introducing
random changes into the DNA. It is exceedingly unlikely
that the 5000 mutations that would accumulate every day in
the absence of the DNA repair enzyme would be in the very
positions where human and chimpanzee DNA sequences
are different. It is very likely that, at such a high mutation
frequency, many essential genes would be inactivated,
leading to cell death. Furthermore, your body is made up
of about 10 13 cells. For you to turn into an ape, not just
one but many of these cells would need to be changed.
And even then, many of these changes would have to occur
during development to effect changes in your body plan
(making your arms longer than your legs, for example).
ANSWER 6–7
A. False. Identical DNA polymerase molecules catalyze
DNA synthesis on the leading and lagging strands
of a bacterial replication fork. The replication fork
is asymmetrical because the lagging strand is made
in pieces while the leading strand is synthesized
continuously.
B. False. Okazaki fragments initially contain both RNA
primers and DNA, but only the RNA primers are
removed by RNA nucleases.