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DNA Metabolism
chapter
25
S-283
1. Conclusions from the Meselson-Stahl Experiment The Meselson-Stahl experiment (see Fig. 25–2)
proved that DNA undergoes semiconservative replication in E. coli. In the “dispersive” model of
DNA replication, the parent DNA strands are cleaved into pieces of random size, then joined with
pieces of newly replicated DNA to yield daughter duplexes. Explain how the results of Meselson and
Stahl’s experiment ruled out such a model.
2. Heavy Isotope Analysis of DNA Replication A culture of E. coli growing in a medium containing
15
NH
4
Cl is switched to a medium containing
14
NH
4
Cl for three generations (an eightfold increase in
population). What is the molar ratio of hybrid DNA (
15
N–
14
N) to light DNA (
14
N–
14
N) at this point?
3. Replication of the E. coli Chromosome The E. coli chromosome contains 4,639,221 bp.
(a) How many turns of the double helix must be unwound during replication of the E. coli chromosome?
(b) From the data in this chapter, how long would it take to replicate the E. coli chromosome at 37 C
if two replication forks proceeded from the origin? Assume replication occurs at a rate of 1,000 bp/s.
Under some conditions E. coli cells can divide every 20 min. How might this be possible?
(c) In the replication of the E. coli chromosome, about how many Okazaki fragments would be
formed? What factors guarantee that the numerous Okazaki fragments are assembled in the
correct order in the new DNA?
Answer
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S-284 Chapter 25 DNA Metabolism
4. Base Composition of DNAs Made from Single-Stranded Templates Predict the base composi-
tion of the total DNA synthesized by DNA polymerase on templates provided by an equimolar mixture
of the two complementary strands of bacteriophage fX174 DNA (a circular DNA molecule). The base
composition of one strand is A, 24.7%; G, 24.1%; C, 18.5%; and T, 32.7%. What assumption is neces-
sary to answer this problem?
5. DNA Replication Kornberg and his colleagues incubated soluble extracts of E. coli with a mixture of
dATP, dTTP, dGTP, and dCTP, all labeled with
32
P in the a-phosphate group. After a time, the incuba-
tion mixture was treated with trichloroacetic acid, which precipitates the DNA but not the nucleotide
precursors. The precipitate was collected, and the extent of precursor incorporation into DNA was
determined from the amount of radioactivity present in the precipitate.
(a) If any one of the four nucleotide precursors were omitted from the incubation mixture, would ra-
dioactivity be found in the precipitate? Explain.
(b) Would
32
P be incorporated into the DNA if only dTTP were labeled? Explain.
(c) Would radioactivity be found in the precipitate if
32
P labeled the bor gphosphate rather than
the aphosphate of the deoxyribonucleotides? Explain.
Answer
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6. The Chemistry of DNA Replication All DNA polymerases synthesize new DNA strands in the
5n3direction. In some respects, replication of the antiparallel strands of duplex DNA would be sim-
pler if there were also a second type of polymerase, one that synthesized DNA in the 3n5direction.
The two types of polymerase could, in principle, coordinate DNA synthesis without the complicated
mechanics required for lagging strand replication. However, no such 3n5-synthesizing enzyme has
been found. Suggest two possible mechanisms for 3n5DNA synthesis. Pyrophosphate should be one
product of both proposed reactions. Could one or both mechanisms be supported in a cell? Why or
why not? (Hint: You may suggest the use of DNA precursors not actually present in extant cells.)
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S-286 Chapter 25 DNA Metabolism
7. Activities of DNA Polymerases You are characterizing a new DNA polymerase. When the enzyme
is incubated with [
32
P]-labeled DNA and no dNTPs, you observe the release of [
32
P]dNMPs. This
release is prevented by adding unlabeled dNTPs. Explain the reactions that most likely underlie these
observations. What would you expect to observe if you added pyrophosphate instead of dNTPs?
8. Leading and Lagging Strands Prepare a table that lists the names and compares the functions of
the precursors, enzymes, and other proteins needed to make the leading strand versus the lagging
strand during DNA replication in E. coli.
9. Function of DNA Ligase Some E. coli mutants contain defective DNA ligase. When these mutants
are exposed to
3
H-labeled thymine and the DNA produced is sedimented on an alkaline sucrose den-
sity gradient, two radioactive bands appear. One corresponds to a high molecular weight fraction, the
other to a low molecular weight fraction. Explain.
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10. Fidelity of Replication of DNA What factors promote the fidelity of replication during the
synthesis of the leading strand of DNA? Would you expect the lagging strand to be made with the
same fidelity? Give reasons for your answers.
11. Importance of DNA Topoisomerases in DNA Replication DNA unwinding, such as that occur-
ring in replication, affects the superhelical density of DNA. In the absence of topoisomerases, the
DNA would become overwound ahead of a replication fork as the DNA is unwound behind it. A bacterial
replication fork will stall when the superhelical density (j) of the DNA ahead of the fork reaches
0.14 (see Chapter 24).
Bidirectional replication is initiated at the origin of a 6,000 bp plasmid in vitro, in the absence of
topoisomerases. The plasmid initially has a jof 0.06. How many base pairs will be unwound and
replicated by each replication fork before the forks stall? Assume that each fork travels at the same
rate and that each includes all components necessary for elongation except topoisomerase.
12. The Ames Test In a nutrient medium that lacks histidine, a thin layer of agar containing ~10
9
Salmo-
nella typhimurium histidine auxotrophs (mutant cells that require histidine to survive) produces
~13 colonies over a two-day incubation period at 37 C (see Fig. 25–20). How do these colonies arise
in the absence of histidine? The experiment is repeated in the presence of 0.4 mg of 2-aminoanthracene.
The number of colonies produced over two days exceeds 10,000. What does this indicate about
2-aminoanthracene? What can you surmise about its carcinogenicity?
13. DNA Repair Mechanisms Vertebrate and plant cells often methylate cytosine in DNA to form
5-methylcytosine (see Fig. 8–5a). In these same cells, a specialized repair system recognizes G–T
mismatches and repairs them to GqC base pairs. How might this repair system be advantageous to
the cell? (Explain in terms of the presence of 5-methylcytosine in the DNA.)
14. DNA Repair in People with Xeroderma Pigmentosum The condition known as xeroderma pig-
mentosum (XP) arises from mutations in at least seven different human genes (see Box 25–1). The
deficiencies are generally in genes encoding enzymes involved in some part of the pathway for human
nucleotide-excision repair. The various types of XP are denoted A through G (XPA, XPB, etc.), with a
few additional variants lumped under the label XP-V.
Cultures of fibroblasts from healthy individuals and from patients with XPG are irradiated with ul-
traviolet light. The DNA is isolated and denatured, and the resulting single-stranded DNA is character-
ized by analytical ultracentrifugation.
(a) Samples from the normal fibroblasts show a significant reduction in the average molecular weight
of the single-stranded DNA after irradiation, but samples from the XPG fibroblasts show no such
reduction. Why might this be?
(b) If you assume that a nucleotide-excision repair system is operative in fibroblasts, which step
might be defective in the cells from the patients with XPG? Explain.
Answer
15. Holliday Intermediates How does the formation of Holliday intermediates in homologous genetic
recombination differ from their formation in site-specific recombination?
16. Cleavage of Holliday Intermediates A Holliday intermediate is formed between two homologous
chromosomes, at a point between genes Aand B, as shown below. The chromosomes have different
alleles of the two genes (Aand a, Band b). Where would the Holliday intermediate have to be cleaved
(points X and/or Y) to generate a chromosome that would convey (a) an Ab genotype or (b) an ab
genotype?
X
Ab
aB
X
YY
S-290 Chapter 25 DNA Metabolism
17. A Connection between Replication and Site-Specific Recombination Most wild strains of
Saccharomyces cerevisiae have multiple copies of the circular plasmid 2(named for its contour
length of about 2 m), which has ⬃6,300 bp of DNA. For its replication the plasmid uses the host repli-
cation system, under the same strict control as the host cell chromosomes, replicating only once per
cell cycle. Replication of the plasmid is bidirectional, with both replication forks initiating at a single,
well-defined origin. However, one replication cycle of a 2plasmid can result in more than two copies
of the plasmid, allowing amplification of the plasmid copy number (number of plasmid copies per cell)
whenever plasmid segregation at cell division leaves one daughter cell with fewer than the normal com-
plement of plasmid copies. Amplification requires a site-specific recombination system encoded by the
plasmid, which serves to invert one part of the plasmid relative to the other. Explain how a site-specific
inversion event could result in amplification of the plasmid copy number. (Hint: Consider the situation
when replication forks have duplicated one recombination site but not the other.)
Data Analysis Problem
18. Mutagenesis in Escherichia coli Many mutagenic compounds act by alkylating the bases in DNA.
The alkylating agent R7000 (7-methoxy-2-nitronaphtho[2,1-b]furan) is an extremely potent mutagen.
NO2
CH3
O
O
R7000
(a) Why are some mutants produced even when no R7000 is present?
Quillardet and colleagues also measured the survival rate of bacteria treated with different
concentrations of R7000.
(b) Explain how treatment with R7000 is lethal to cells.
(c) Explain the differences in the mutagenesis curves and in the survival curves for the two types of
bacteria, uvr
and uvr
, as shown in the graphs.
The researchers then went on to measure the amount of R7000 covalently attached to the DNA
in uvr
and uvr
E. coli. They incubated bacteria with [
3
H]R7000 for 10 or 70 minutes, extracted the
DNA, and measured its
3
H content in counts per minute (cpm) per g of DNA.
Chapter 25 DNA Metabolism S-291
R7000 (g/mL)
Survival (%)
uvr
uvr
0.6 0.8 10.2 0.4
100
10
10
R7000 (g/mL)
uvr
uvr
0.6 0.8 10.20 0.4
1,000
100
1
10
Rifampicin-resistant mutants produced
3
H in DNA (cpm/g)
Time (min) uvr
ⴙ
uvr
ⴚ
10 76 159
70 69 228
In vivo, R7000 is activated by the enzyme nitroreductase, and this more reactive form covalently at-
taches to DNA—primarily, but not exclusively, to GmC base pairs.
In a 1996 study, Quillardet, Touati, and Hofnung explored the mechanisms by which R7000 causes
mutations in E. coli. They compared the genotoxic activity of R7000 in two strains of E. coli: the wild-
type (uvr
) and mutants lacking uvrA activity (uvr
; see Table 25–6). They first measured rates of
mutagenesis. Rifampicin is an inhibitor of RNA polymerase (see Chapter 26). In its presence, cells will
not grow unless certain mutations occur in the gene encoding RNA polymerase; the appearance of
rifampicin-resistant colonies thus provides a useful measure of mutagenesis rates.
The effects of different concentrations of R7000 were determined, with the results shown in the
graph following.
(d) Explain why the amount of
3
H drops over time in the uvr
strain and rises over time in the uvr
strain.
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Number of Lac
ⴙ
cells (average ⴞSD)
CC101 CC102 CC103 CC104 CC105 CC106
(AUT(GmC(GmC (GmC(AUT(AUT
R7000 to to to to to to
(g/mL) CmG) AUT) CmG) TUA) TUA) GmC)
02 2 10 93 34 26 10.5 1
17 6 21 98 3 23 15 13 11 1
54 3 15 7 22 2 68 25 67 14 1 1
Number of Lac
ⴙ
cells (average ⴞSD)
CC101 CC102 CC103 CC104 CC105 CC106
(AUT (GmC (GmC (GmC (AUT (AUT
R7000 to to to to to to
(g/mL) CmG) AUT) CmG) TUA) TUA) GmC)
06 3 11 92 15 32 11 1
0.075 24 19 34 38 4 82 23 40 14 4 2
0.15 24 4 26 29 5 180 71 130 50 3 2
Quillardet and colleagues then examined the particular DNA sequence changes caused by
R7000 in the uvr
and uvr
bacteria. For this, they used six different strains of E. coli, each with a
different point mutation in the lacZ gene, which encodes -galactosidase (this enzyme catalyzes the
same reaction as lactase; see Fig. 14–11). Cells with any of these mutations have a nonfunctional
-galactosidase and are unable to metabolize lactose (i.e., a Lac
phenotype). Each type of point
mutation required a specific reverse mutation to restore lacZ gene function and Lac
phenotype. By
plating cells on a medium containing lactose as the sole carbon source, it was possible to select for
these reverse-mutated, Lac
cells. And by counting the number of Lac
cells following mutagenesis
of a particular strain the researchers could measure the frequency of each type of mutation.
First, they looked at the mutation spectrum in uvr
cells. The following table shows the results
for the six strains, CC101 through CC106 (with the point mutation required to produce Lac
cells in-
dicated in parentheses).
(e) Which types of mutation show significant increases above the background rate due to treatment
with R7000? Provide a plausible explanation for why some have higher frequencies than others.
(f) Can all of the mutations you listed in (e) be explained as resulting from covalent attachment of
R7000 to a GmC base pair? Explain your reasoning.
(g) Figure 25–27b shows how methylation of guanine residues can lead to a GmC to APT mutation. Using a
similar pathway, show how a G–R7000 adduct could lead to the GmC to APT or TPA mutations shown
above. Which base pairs with the GOR7000 adduct?
The results for the uvr
bacteria are shown in the table below.
(h) Do these results show that all mutation types are repaired with equal fidelity? Provide a plausible expla-
nation for your answer.
Answer
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