LEHNINGER PRINCIPLES OF BIOCHEMISTRY - VOL. 3. INFORMATION PATHWAYS - 2017

PART III. INFORMATION PATHWAYS

25. DNA METABOLISM

Questions and Problems

1. Conclusions from the Meselson-Stahl experiment.

The Meselson-Stahl experiment (Fig. 25-2) demonstrates that DNA in E. coli Cells undergoes semi-conservative Replication. In the "dispersive" model of replication, parental DNA strands are cleaved into segments of random length and then joined with newly replicated DNA fragments, resulting in The formation of duplexes. Explain why the results of the Meselson-Stahl experiment rule out this model.

2. Analysis of METABOLISM/36.html">DNA replication using heavy isotopes.

An E. coli culture growing on a medium with ¹⁵NH₄Cl is transferred to a medium containing ¹⁴NH₄Cl, where the cells divide three times (i.e., The Cell population increases eightfold). What is the molar ratio of hybrid DNA (¹⁵N-¹⁴N) to light DNA (¹⁴N-¹⁴N) in the resulting culture?

3. Replication of the E. coli chromosome.

The E. coli chromosome contains 4,639,221 bp.

a) How many double-helix turns are unwound during the replication of the E. coli chromosome?

b) Based on the information presented in this chapter, determine how long it will take to replicate the E. coli chromosome at 37 °C if two replication forks start moving from THE ORIGIN OF replication? Assume a replication rate of 1,000 bp/s. Under these conditions, E. coli cells divide every 20 min. Is this possible?

c) Approximately how many Okazaki fragments are formed during the replication of the E. coli chromosome? What factors ensure the accurate assembly of numerous Okazaki fragments into new DNA?

4. Base composition of DNA synthesized on single-stranded templates.

Estimate the base COMPOSITION OF THE total DNA synthesized by DNA polymerase on templates consisting of an equimolar mixture of two complementary strands of bacteriophage φX 174 (circular DNA molecules). Base composition of one strand: A — 24.7%; G — 24.1%; C — 18.5%; T — 32.7%. What assumption must be made to answer this question?

5. DNA replication.

Kornberg and coworkers incubated soluble E. coli extracts with a mixture of dATP, dTTP, dGTP, and dCTP, labeled with ³²P at their α-phosphate groups. After a certain time, the incubation mixture was treated with trichloroacetic acid, which precipitates DNA but does not precipitate precursor NUCLEOTIDES. The precipitate was collected, and The amount of radioactive label in the precipitate was used to determine the amount of precursors incorporated into the DNA.

a) If one of the four precursors is removed from the incubation mixture, will the radioactive label be detected in the precipitate? Explain your answer.

b) Can ³²P be incorporated into DNA if only dTTP carries the label? Explain your answer.

c) Will radioactivity be detected in the precipitate if the ³²P label is incorporated into the β- or γ-phosphate groups rather than the α-phosphate groups? Explain.

6. Chemistry of DNA replication.

All DNA polymerases synthesize a new DNA strand in the 5' —> 3' direction. In a certain sense, replication of the antiparallel strands of a DNA duplex would be simpler if there were a second type of polymerase capable of synthesizing DNA in the 3' —> 5' direction. In principle, Two Types of polymerases could coordinate DNA Synthesis without the complex mechanisms required for lagging strand replication. However, no such 3' —> 5' synthesizing Enzymes have been found. Propose two possible mechanisms for DNA synthesis in the 3' —> 5' direction. One of the products in both cases would be pyrophosphate. Are these mechanisms feasible in the cell? Why or why not? (Hint: You may assume the existence of DNA precursors not found in modern cells.)

7. Leading and lagging strands.

Construct a table comparing the replication of the leading and lagging DNA strands in E. coli. List the names and Functions of the precursors, enzymes, and other required Proteins.

8. Function of DNA ligase.

Some E. coli mutants contain a defective DNA ligase. When such mutants are treated with ³H-labeled thymine and the resulting DNA is sedimented in an alkaline sucrose density gradient, two radioactive bands appear. One corresponds to a high-molecular-weight fraction and the other to a low-molecular-weight fraction. Explain these data.

9. Fidelity of DNA Replication.

What factors increase replication fidelity during leading-strand DNA synthesis? Would you expect the lagging strand to be synthesized with the same fidelity? Explain your answer.

10. Important Role of DNA Topoisomerases in DNA Replication.

Unwinding of DNA, such as during replication, affects the supercoiling density of the DNA. In the absence of topoisomerases, the DNA ahead of a Replication fork would

become overwound because the DNA behind the fork is unwound. A bacterial replication fork halts when the supercoiling density (σ) of the DNA ahead of the fork reaches a value of +0.14 (see Chapter 24).

Bidirectional Replication of a 6,000 bp plasmid in vitro is initiated at an origin of replication without the participation of topoisomerases. Prior to replication, σ = -0.06. How many Base Pairs will each replication fork unwind and replicate before halting? Assume that all forks move at the same rate and have everything necessary for elongation except topoisomerase.

11. The Ames Test.

A thin layer of an Agar growth medium lacking Histidine was inoculated with approximately 10⁹ histidine-auxotrophic Salmonella typhimurium Bacteria (mutant cells that require histidine for survival), and after two days of incubation at 37 °C, about 13 colonies appeared (Fig. 25-21). How could colonies have appeared on the medium without histidine? The experiment was repeated in the presence of 0.4 μg of 2-aminoanthracene. After two days, more than 10,000 colonies formed. What Conclusion can be drawn about The properties of 2-aminoanthracene? What can be said about the carcinogenicity of this substance?

12. Mechanisms of DNA Repair.

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 mismatched pairs and replaces them with G≡C pairs. What benefit does such a repair system confer on the cell? (Explain, taking into account The Role of 5-methylcytosine in DNA.)

■ 13. DNA Repair in Humans with Xeroderma Pigmentosum.

Xeroderma pigmentosum (XP) results from Mutations in at least seven different human genes. The defects typically affect genes encoding Enzymes of the nucleotide Excision Repair pathway. There are several types of the disease, designated by letters from A to G (XPA, XPB, etc.), as well as a separate type, XPV. Fibroblast cultures from healthy individuals and XP patients were irradiated with UV light. The DNA was isolated and denatured, and the resulting single-stranded DNA molecules were analyzed by ultracentrifugation.

a) In normal fibroblast samples after irradiation, the average Molecular Weight of single-stranded DNA molecules is significantly reduced, whereas in fibroblast samples from XPG patients, no such reduction is observed. Why?

b) Assuming that the nucleotide excision repair system operates in fibroblasts, which step is defective in the fibroblasts of XPG patients? Explain.

14. Holliday Structures.

How does the formation of Holliday structures in homologous genetic recombination differ from their formation in Site-Specific Recombination?

15. Relationship between Replication and Site-Specific Recombination.

Most wild-type strains of Saccharomyces cerevisiae carry multiple copies of the 2μ circular plasmid (named for its circumference of about 2 μm), consisting of ~6,300 bp. The plasmid utilizes the host replication machinery, replicating under the same strict control as the host chromosome, exactly once per Cell Cycle. Replication of the plasmid proceeds bidirectionally, and both replication forks originate at the same well-characterized origin of replication. However, a single round of 2μ plasmid replication can result in The production of more than two plasmid copies, which helps increase plasmid copy number (the number of copies of the plasmid per cell) so that upon plasmid segregation during Cell Division, no daughter cell receives fewer plasmid copies than normal. This Amplification requires a site-specific recombination system encoded within the plasmid, which allows one part of the plasmid to be inverted relative to another part. Explain how site-specific inversion can lead to an increase in plasmid copy number. (Hint: Consider the situation where replication forks duplicate only one of the two recombination sites.)

Analysis of Experimental Data

16. Mutagenesis in Escherichia coli Cells.

Many mutagenic compounds act by alkylating DNA bases. The alkylating agent R7000 (7-methoxy-2-nitronaphtho[2,1-b]furan) is a very potent mutagen.

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In vivo, R7000 is activated by the enzyme nitroreductase, and the resulting more active form forms a covalent bond with DNA, primarily with G≡C pairs, though not exclusively. In 1996, Quillardet, Touati, and Hofnung investigated The Mechanism of the mutagenic action of R7000 on E. coli cells. They compared the genotoxic activity of R7000 in two E. coli strains: a wild-type strain (uvr⁺) and a mutant strain lacking uvrA activity (uvr; see Table 25-6). First, the researchers determined the degree of mutagenicity. The antibiotic rifampicin inhibits RNA polymerase (see Chapter 26); in its presence, cells can grow only if specific mutations have occurred in the RNA polymerase Gene. Thus, the appearance of rifampicin-resistant colonies can serve as a measure of mutagenicity.

The Effect of R7000 at various concentrations was analyzed; the experimental results are presented below.

(a) Why do mutations occur even in the absence of R7000?

Quiillardet and co-workers also determined the survival rates of bacteria treated with various concentrations of R7000.

(b) Explain why Treatment with R7000 kills bacterial cells.

(c) Explain the differences in R7000 mutagenesis and survival rates between the two bacterial strains, uvr⁺ and uvr^- (see data).

Next, the researchers measured the amount of R7000 that formed covalent bonds with DNA in the uvr⁺ and uvr^- strains. They incubated the bacteria with [³H]R7000 for 10 or 70 min, extracted the DNA, and determined the ³H level in cpm per microgram of DNA.

(d) Explain why the ³H content decreases over time in the uvr⁺ strain and increases in the uvr^- strain.

Quiillardet and colleagues also investigated the DNA sequence changes caused by R7000 exposure in the uvr⁺ and uvr^- strains. To do this, they used six different E. coli strains, each containing a specific point mutation in the lacZ gene, which encodes β-galactosidase (this enzyme catalyzes the same reaction as lactase; see Fig. 14-10 in Vol. 2). Cells carrying any of these mutations have a nonfunctional β-galactosidase and cannot metabolize lactose (Lac^- phenotype). Restoring lacZ gene function and the Lac⁺ phenotype in each specific mutant required a specific reverse mutation. By plating cells on a medium containing lactose as the sole carbon source, it is possible to selectively isolate such back-mutated Lac⁺ cells. Counting the number of Lac⁺ cells following mutagenesis of each strain allows the mutation frequency for each type to be determined.

First, the researchers analyzed the mutation spectrum in uvr^- cells. The table presents the results for six strains, CC101 through CC106 (the point mutations required to generate the Lac⁺ phenotype in each case are shown in parentheses).

(e) Which mutations show a substantial increase in the number of Lac⁺ cells upon R7000 treatment? Explain why the frequency of this phenotype is higher in some strains than in others.

(f) Can all of the listed mutations be explained by the covalent binding of R7000 to a GC base pair? Explain your answer.

(g) Figure 25-28b illustrates how methylation of guanine residues can lead to GC-to-AT transitions. Using a similar pathway, show how the G-R7000 adduct could lead to GC-to-AT or GC-to-TA transitions in the example above. Which base pairs with the G-R7000 adduct?

The results for the uvr⁺ strain are presented below.

(h) Do these results indicate that all Selection/21.html">Types of mutations are repaired with equal fidelity? Explain your answer.