Essential Cell Biology 5th edition

A:16 Answers
C. True. With proofreading, DNA polymerase has an error
rate of one mistake in 10 7 nucleotides polymerized;
99% of its errors are corrected by DNA mismatch repair
enzymes, bringing the final error rate to one in 10 9 .
D. True. Mutations would accumulate rapidly, inactivating
many genes.
E. True. If a damaged nucleotide also occurred naturally
in DNA, the repair enzyme would have no way of
identifying the damage. It would therefore have only a
50% chance of fixing the right strand.
F. True. Usually, multiple mutations of specific types need
to accumulate in a somatic cell lineage to produce a
cancer. A mutation in a gene that codes for a DNA repair
enzyme can make a cell more liable to accumulate these
mutations, thereby accelerating the onset of cancer.
ANSWER 6–8 With a single origin of replication, which
launches two DNA polymerases in opposite directions on
the DNA, each moving at 100 nucleotides per second, the
number of nucleotides replicated in 24 hours will be
1.73 × 10 7 (= 2 × 100 × 24 × 60 × 60). To replicate all the
6 × 10 9 nucleotides of DNA in the cell in this time,
therefore, will require at least 348 (= 6 × 10 9 /1.73 × 10 7 )
origins of replication. The estimated 10,000 origins of
replication in the human genome are therefore more than
sufficient to satisfy this minimum requirement.
ANSWER 6–9
A. Dideoxycytidine triphosphate (ddCTP) is identical to
dCTP, except it lacks the 3ʹ-hydroxyl group on the
sugar ring. ddCTP is recognized by DNA polymerase
as dCTP and becomes incorporated into DNA; because
it lacks the crucial 3ʹ-hydroxyl group, however, its
addition to a growing DNA strand creates a dead end
to which no further nucleotides can be added. Thus,
if ddCTP is added in large excess, new DNA strands
will be synthesized until the first G (the nucleotide
complementary to C) is encountered on the template
strand. ddCTP will then be incorporated instead of
C, and no further extension of this strand will occur.
This strategy is exploited by a drug, 3ʹ-azido-3ʹdeoxythymidine
(AZT), that is now commonly used in
HIV-infected patients to treat AIDS. AZT is converted
in cells to the triphosphate form and is incorporated
into the growing viral DNA. Because the drug lacks a
3ʹ-hydroxyl group, it blocks further DNA synthesis and
replication of the virus. AZT inhibits viral replication
preferentially because reverse transcriptase has a higher
affinity for the drug than for thymidine triphosphate;
human cellular DNA polymerases do not show this
preference and therefore still function in the presence of
the drug.
B. If ddCTP is added at about 10% of the concentration
of the available dCTP, there is a 1 in 10 chance of its
being incorporated whenever a G is encountered
on the template strand. Thus a population of DNA
fragments will be synthesized, and from their lengths
one can deduce where the G nucleotides are located
on the template strand. This strategy forms the basis of
methods used to determine the sequence of nucleotides
in a stretch of DNA (discussed in Chapter 10).
C. Dideoxycytidine monophosphate (ddCMP) lacks the
5ʹ-triphosphate group as well as the 3ʹ-hydroxyl group
of the sugar ring. It therefore cannot provide the energy
that drives the polymerization reaction of nucleotides
into DNA and therefore will not be incorporated into the
replicating DNA. Addition of this compound should thus
not affect DNA replication.
ANSWER 6–10
1. beginning of
synthesis of
Okazaki fragment
See Figure A6−10.
2. midpoint of synthesis of Okazaki fragment
Figure A6−10
ANSWER 6–11 The two strands of the bacterial
chromosome contain 6 × 10 6 nucleotides in total. During the
polymerization of nucleoside triphosphates into DNA,
two phosphoanhydride bonds are broken for each
nucleotide added: the nucleoside triphosphate is hydrolyzed
to produce the nucleoside monophosphate added to the
growing DNA strand, and the released pyrophosphate is
ECB5 EA6.10/A6.10
hydrolyzed to phosphate. Therefore, 1.2 × 10 7 high-energy
bonds are hydrolyzed during each round of bacterial DNA
replication. This requires 4 × 10 5 (= 1.2 × 10 7 /30) glucose
molecules, which weigh 1.2 × 10 –16 g [= (4 × 10 5 molecules)
× (180 g/mole)/(6 × 10 23 molecules/mole)], which is 0.01%
of the total weight of the cell.
ANSWER 6–12 The statement is correct. If the DNA
in somatic cells is not sufficiently stable (that is, if it
accumulates mutations too rapidly), the organism dies (of
cancer, for example), and because this may often happen
before the organism can reproduce, the species will die out.
If the DNA in reproductive cells is not sufficiently stable,
many mutations will accumulate and be passed on to future
generations, so that the species will not be maintained.
ANSWER 6–13 As shown in Figure A6−13, thymine
and uracil lack amino groups and therefore cannot be
deaminated. Deamination of adenine and guanine produces
purine rings that are not found in conventional nucleic
acids. In contrast, deamination of cytosine produces uracil.
Therefore, if uracil were a naturally occurring base in DNA