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1. Conclusion Conclusions s from the Meselson-St Meselson-Stahl ahl Experiment Experiment The Meselson-Stahl experiment (see Fig. 25–2) proved that DNA undergoes semiconservative semiconservative replication in E. 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. Answer If random, dispersive replication takes place, the density of the first-generation DNA in the Meselson-Stahl experiment would be the same as was actually observed: a single band midway between heavy and light DNA. In the second generation, however, all the DNA would again have the same density and would appear as a single band, midway between the band observed in the first generation and that of light DNA; this was not observed. Two bands were obtained in the experiment, ruling out the dispersive model. 2. Heavy Isotope Isotope Analysis Analysis of DNA DNA Replicatio Replication n A culture of E. E. coli growing in a medium containing 15 NH4Cl is switched to a medium containing 14NH4Cl for three generations (an eightfold increase in population). What is the molar ratio of hybrid DNA (15N–14N) to light DNA (14N–14N) at this point? Answer This experiment is an extension of the Meselson-Stahl experiment, which demonstrates that replication in E. in E. coli is semiconservative. After three generations the molar ratio of 15N–14N DNA to 14N–14N DNA is 2/6 0.33. 3. Replic Replicati ation on of of the the E.
coli
The E. coli chromosome contains 4,639,221 bp. Chromosome The E.
the E. coli chromosome? (a) How many turns of the double helix must be unwound during replication of the E. the E. coli chromosome at 37 C (b) From the data in this chapter, how long would it take to replicate the E. if two replication forks proceeded from the origin? Assume replication occurs at a rate of 1,000 bp/s. Under some conditions E. conditions E. coli cells can divide every 20 min. How might this be possible? the E. coli chromosome, about how many Okazaki fragments would be (c) In the replication of the E. formed? What factors guarantee that the numerous Okazaki fragments are assembled in the correct order in the new DNA?
Answer (a) During DNA replication, the complementary strands must unwind completely to allow the synthesis of a new strand on each template. Given the 10.5 bp/turn in B-DNA, and approximating the E. the E. coli chromosome as 4.64 106 bp, the number of helical turns
number of base pairs number of base pairs per helical turn
4.64 106 bp 4.42 10.5 bp/turn
105 turns
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(b) Chromosomal DNA replication in E. coli starts at a fixed origin and proceeds bidirectionally. Each replication fork travels (4.64 106 bp)/2 2.32 106 bp during replication. If we assume a replication rate of 1,000 bp/s, the time required for the completion of DNA synthesis in each replication fork is (2.32
106 bp)/[(1000 bp/s)(60 s/min)]
40 min
This is about twice the time required for cell division. One possible explanation is that replication of an E. coli chromosome starts from two origins, each proceeding bidirectionally to yield four replication forks. In this mode, it would take 20 min to complete the replication of the chromosome. However, we know there is only one replication origin in the E. coli chromosome. An alternative explanation is that a new round of replication begins before the previous one is completed: for cells dividing every 20 min, a replicative cycle is initiated every 20 min, and each daughter cell receives a chromosome that is half-replicated. This latter mode has been experimentally verified.
(c) The Okazaki fragments in E. coli are 1,000 to 2,000 nucleotides long, so (4.64 106 nucleotides)/(2000 nucleotides) to (4.64 106 nucleotides)/(1000 nucleotides) 2,000 to 5,000 Okazaki fragments are formed. The fragments are firmly bound to the template strand by base pairing, and each fragment is quickly joined to the lagging strand by the successive action of DNA polymerase I and DNA ligase, thus preserving the correct order of the fragments. A mixed pool of Okazaki fragments, detached from their template, does not form during normal replication. 4. Base Composition of DNAs Made from Single-Stranded Templates Predict the base composition 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 necessary to answer this problem? Answer The sequence of a strand of duplex DNA is complementary to that of the other strand, as determined by Watson-Crick base pairing (A with T, and G with C). The DNA strand made from the given template strand has A, 32.7%; G, 18.5%; C, 24.1%; T, 24.7%. The DNA strand made from this complementary template strand has A, 24.7%; G, 24.1%; C, 18.5%; T, 32.7%. Thus, the composition of the total DNA synthesized is calculated as, for A, (32.7% 24.6%)/2 28.7%; similarly, G 21.3%; C 21.3%; T 28.7%. We are assuming that both template strands are completely replicated. 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 32P in the a-phosphate group. After a time, the incubation 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 radioactivity be found in the precipitate? Explain. (b) Would 32P be incorporated into the DNA if only dTTP were labeled? Explain. (c) Would radioactivity be found in the precipitate if 32P labeled the b or g phosphate rather than the a phosphate of the deoxyribonucleotides? Explain. Answer (a) No. Incorporation of 32P into DNA results from the synthesis of new DNA, which requires the presence of all four nucleotide precursors. (b) Yes. Although all four nucleotide precursors must be present for DNA synthesis, only one of them has to be radioactive in order for radioactivity to appear in the new DNA.
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(c) No. No radioactivity would be incorporated if the 32P label were not at the a phosphate because DNA polymerase, which catalyzes this reaction, cleaves off pyrophosphate—that is, the b and g phosphate groups. 6. The Chemistry of DNA Replication All DNA polymerases synthesize new DNA strands in the 5n3 direction. In some respects, replication of the antiparallel strands of duplex DNA would be simpler if there were also a second type of polymerase, one that synthesized DNA in the 3n5 direction. 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 3n5 DNA 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.) Answer Mechanism 1: 3-OH of an incoming dNTP attacks the phosphate of the triphosphate at the 5 end of the growing DNA strand, displacing pyrophosphate. This mechanism uses normal dNTPs, and the growing end of the DNA always has a triphosphate on the 5 end. Z
5 end
P
P
CH2 O
P
Z P
P
P
H
CH2 O H
H
H
OH
PPi
H
O
O
H
Y
O Y
CH2 O
CH2 O
P
H
O
H
H
O
H
P
H H
O
O
H
O
H
O . 3 end
. .
O
H
H
O X
CH2
O
H
P
H
O X
CH2
H
P
O
O
H
H
O
5 end
P
H
P
H
P
H
H H
O
O
H
H
O
H
P O .
.
3 end .
O
O
H
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Mechanism 2: This uses a new type of precursor, nucleotide 3 -triphosphates. The growing end of the DNA strand has a 5-OH, which attacks the phosphate of an incoming deoxynucleotide 3-triphosphate, displacing pyrophosphate. Note that this mechanism would require the evolution of new metabolic pathways to supply the needed deoxynucleotide 3-triphosphates. Z
5 end
CH2 O
HO Z HO
CH2 O H
H
H PPi
H
H
OH
O
H 5 end
P P
H
P
O
O
H O
O
H
. end .
O
H
H
O X
CH2 H
O
H
O
O
H
P
H
P
X
O .
3
Y
O
CH2
O
O
H
H
O
H
CH2 O H
H
O
H
O
Y
H
P
H
P
CH2 O
HO
H
H
H
O
O
H
H
O
H
P
H
O
O .
.
3 end .
7. 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. Answer In DNA replication, the leading strand is produced by continuous replication of the DNA template strand in the 5 → 3 direction. The lagging strand is synthesized in the form of short Okazaki fragments, which are then spliced together. The following table lists the participants in DNA replication in E. coli.
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Leading strand
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Lagging strand
Component
Function
dNTPs (dATP, dGTP, dGTP, dTTP)
Source of new nucleotides for new DNA strand; ATP energy source
X
X
Template
Parent strand provides information for identities of incoming nucleotides; stabilizes association of nucleotides in growing strand by base-pairing reactions.
X
X
DNA primer
Provides a free 3 OOH, the point of attachment for incoming nucleotides
X
DNA helicase
Unwinds double-stranded DNA just ahead of the replication fork; requires ATP
X
X
DNA gyrase
Topoisomerase that favors unwinding of the DNA at the replication fork by twisting the DNA; requires ATP
X
X
Single-stranded DNA–binding proteins
Prevents base pairing of unwound DNA strands; stabilizes single-stranded DNA
X
X
DNA polymerase III
Elongates the new DNA strand by adding nucleotides; requires the cofactors Mg2 and Zn2
X
X
Pyrophosphatase
Hydrolyzes the PPi released by polymerase activity; helps “pull” the reaction in the forward direction.
X
X
RNA primer
Same as DNA primer; used to start each Okazaki fragment
X
Primase
Synthesizes RNA primer
X
NTPs (ATP, GTP, CTP, UTP)
Used in synthesis of RNA primer
X
DNA polymerase I
Exonuclease that removes RNA primer by replacing NMPs with dNMPs in Okazaki fragments; requires the cofactors Mg2 and Zn2
DNA ligase
Joins the Okazaki fragments in the lagging strand by catalyzing the formation of a phosphodiester bond; requires NAD as energy source
X
X
8. Function of DNA Ligase Some E. coli mutants contain defective DNA ligase. When these mutants are exposed to 3H-labeled thymine and the DNA produced is sedimented on an alkaline sucrose density gradient, two radioactive bands appear. One corresponds to a high molecular weight fraction, the other to a low molecular weight fraction. Explain. Answer During replication of the DNA duplex, the leading strand is replicated continuously and the lagging strand is replicated in short fragments (Okazaki fragments), which are then spliced together by DNA ligase. Mutants with defective DNA ligase produce a DNA “duplex” in which one of the strands remains fragmented. Consequently, when this duplex is denatured by the alkaline conditions of the sucrose gradient, sedimentation results in one fraction containing the intact single strand (the high molecular weight band) and one fraction containing the unspliced fragments (the low molecular weight band).
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9. 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. Answer Fidelity of replication is ensured by Watson-Crick base pairing between the template and leading strand, and proofreading and removal of wrongly inserted nucleotides by the 3-exonuclease activity of DNA polymerase III. The same fidelity would be expected in the lagging strand—perhaps. The factors ensuring fidelity of replication are operative in the leading and the lagging strands, but the greater number of distinct chemical operations involved in making the lagging strand might provide a greater opportunity for errors to arise. 10. Importance of DNA Topoisomerases in DNA Replication DNA unwinding, such as that occurring 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 j of 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.
Answer About 1,200 bp are unwound, or about 600 in each direction. The DNA is initially negatively supercoiled, with an Lk of about 537 and Lk0 of about 571 (see Chapter 24). Unwinding 571 537 34 turns of DNA relaxes the DNA, and further unwinding adds positive supercoils. Unwinding another 79 turns (or 113 in all) brings the superhelical density in the remaining wound DNA to about 0.14—the stalling point. These 113 turns are equivalent to 113 turns 10.5 bp/turn 1,190 bp, or about 1,200 bp. 11. The Ames Test In a nutrient medium that lacks histidine, a thin layer of agar containing ~109 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. 25–21). 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 2aminoanthracene? What can you surmise about its carcinogenicity? Answer Occasionally, some of the histidine-requiring mutants spontaneously undergo backmutation and regain their capacity to synthesize histidine, and thus can grow in a medium lacking histidine. The observation that only 13 of about 109 bacteria produce colonies indicates that the rate of back-mutation is quite low. The addition of 2-aminoanthracene increases the rate of back-mutations more than 800-fold, indicating that 2-aminoanthracene is mutagenic. Because about 90% of 300 known carcinogens are mutagenic, these observations suggest that 2-aminoanthracene is likely to be carcinogenic. 12. 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.) Answer This is very similar to Problem 6 of Chapter 8. Spontaneous deamination of 5-methylcytosine produces thymine, and thus a G–T mismatched pair. Such G–T pairs are among the most common mismatches in the DNA of eukaryotes. The specialized repair system restores the GqC pair.
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13. DNA Repair in People with Xeroderma Pigmentosum The condition known as xeroderma pigmentosum (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 ultraviolet light. The DNA is isolated and denatured, and the resulting single-stranded DNA is characterized 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 (a) Ultraviolet irradiation of skin fibroblast DNA results in formation of pyrimidine dimers. In normal fibroblasts, the damaged DNA is repaired by excision of the pyrimidine dimer. One step in this process is cleavage of the damaged strand by a special excinuclease. Thus, the denatured single-stranded DNA isolated from normal cells after irradiation contains the many fragments caused by the cleavage, and the average molecular weight is lowered. These fragments of single-stranded DNA are absent from the XPG samples, as indicated by the unchanged average molecular weight. (b) The absence of fragments in the single-stranded DNA from XPG cells after irradiation suggests that the special repair excinuclease is defective or missing. 14. Holliday Intermediates How does the formation of Holliday intermediates in homologous genetic recombination differ from their formation in site-specific recombination? Answer During homologous genetic recombination, a Holliday intermediate may be formed almost anywhere within the two paired, homologous chromosomes. Once formed, the branch point of the intermediate can move extensively by branch migration. In site-specific recombination, the Holliday intermediate is formed between two specific sites, and branch migration is generally restricted by heterologous sequences on either side of the recombination sites. 15. 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 replication 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 2 plasmid 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 complement 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.)
Answer Once replication has proceeded from the origin to a point where one recombination site has been replicated but the other has not, site-specific recombination not only inverts the DNA between the recombination sites but also changes the direction of one replication fork relative to the other. The forks will chase each other around the DNA circle, generating many tandem copies of the plasmid. The multimeric circle can be resolved to monomers by additional site-specific recombination events.
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1
Origin initiation of replication
1
Origin recombination at sites 1 and 2 (inversion)
2
2
Plasmid with recombination sites 1 and 2
Origin
Partially replicated plasmid
Origin
continued unidirectional replication
Plasmid with reoriented replication forks
Multimeric plasmid
Data Analysis Problem 16. 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. CH3 O
NO2 O R7000
In vivo, R7000 is activated by the enzyme nitroreductase, and this more reactive form covalently attaches 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 wildtype (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.
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uvr
DNA Metabolism
uvr
0.2
0.4
0.6
0.8
1
R7000 (g/mL)
(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. 100 ) % ( l a v 10 i v r u S
1
uvr
uvr
0
0.2
0.4
0.6
0.8
1
R7000 (g/mL)
(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 [ 3H]R7000 for 10 or 70 minutes, extracted the DNA, and measured its 3H content in counts per minute (cpm) per g of DNA.
3
H in DNA (cpm/ g)
Time (min)
uvr
uvr
10 70
76 69
159 228
(d) Explain why the amount of 3H drops over time in the uvr strain and rises over time in the uvr strain. 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. 14–10). 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.
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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 indicated in parentheses).
Number of Lac cells (average CC101 (A
T
U
R7000
to
(g/mL)
CmG)
0
6
3
SD)
CC102
CC103
CC104
CC105
CC106
(GmC
(GmC
(GmC
(A
(A
to
to
to
A
T)
U
CmG)
T
5
11
9
2
1
T
U
A)
T
U
to
A)
U
3
T
U
to
2
1
GmC) 1
1
0.075
24
19
34
3
8
4
82
23
40
14
4
2
0.15
24
4
26
2
9
5
180
71
130
50
3
2
(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 25–28b shows how methylation of guanine residues can lead to a GmC to A T mutation. Using a similar pathway, show how a G–R7000 adduct could lead to the GmC to A T or T A mutations shown above. Which base pairs with the G R7000 adduct? P
P
P
O
The results for the uvr bacteria are shown in the table below.
Number of Lac cells (average CC101 (A
T
U
SD)
CC102
CC103
CC104
CC105
CC106
(GmC
(GmC
(GmC
(A
(A
to
to
(g/mL)
CmG)
0
2
2
10
9
3
3
4
2
6
1
0.5
1
1
7
6
21
9
8
3
23
15
13
1
1
1
5
4
3
15
7
22
2
68
25
67
14
1
1
T)
U
to
CmG)
T
to
T
U
R7000
A
to
T
U
A)
U
T
to
A)
U
GmC)
(h) Do these results show that all mutation types are repaired with equal fidelity? Provide a plausible explanation for your answer. Answer (a) Even in the absence of an added mutagen, background mutations occur due to radiation, cellular chemical reactions, and so forth. (b) If the DNA is sufficiently damaged, a substantial fraction of gene products are nonfunctional and the cell is nonviable. (c) Cells with reduced DNA repair capability are more sensitive to mutagens. Because they less readily repair lesions caused by R7000, uvr bacteria have an increased mutation rate and increased chance of lethal effects. (d) In the uvr strain, the excision-repair system removes DNA bases with attached [3H]R7000, decreasing the 3H in these cells over time. In the uvr strain, the DNA is not 3 3 repaired and H level increases as [ H]R7000 continues to react with the DNA.
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(e) All mutations listed in the table except APT to GmC show significant increases over background. Each type of mutation results from a different type of interaction between R7000 and DNA. Because different types of interactions are not equally likely (due to differences in reactivity, steric constraints, etc.), the resulting mutations occur with different frequencies. (f)
No. Only those that start with a GmC base pair are explained by this model. Thus APT to CmG and APT to TP A must be due to R7000 attaching to an A or a T.
(g) R7000-G pairs with A. First, R7000 adds to GmC to give R7000-GmC. (Compare this with what happens with the CH3-G in Fig. 25–28b, p. 1000.) If this is not repaired, one strand is replicated as R7000-GP A, which is repaired to TP A. The other strand is wildtype. If the replication produces R7000-GPT, a similar pathway leads to an APT base pair. (h) No. Compare data in the two tables, and keep in mind that different mutations occur at different frequencies. APT to CmG: moderate in both strains; but better repair in the uvr strain GmC to APT: moderate in both; no real difference GmC to CmG: higher in uvr; certainly less repair! GmC to TP A: high in both; no real difference APT to TP A: high in both; no real difference APT to GmC: low in both; no real difference Certain adducts may be more readily recognized by the repair apparatus than others and thus are repaired more rapidly and result in fewer mutations. Reference Quillardet, P., Touati, E., & Hof nung, M. (1996) Influence of the uvr-dependent nucleotide excision repair on DNA adducts formation and mutagenic spectrum of a potent genotoxic agent: 7-methoxy-2-nitronaphtho[2,1-b]furan (R7000). Mutat. Res. 358, 113–122.