To determine the reproducibility of mutation frequency measurements, you do the

The different cells in your body rarely have genomes with the identical nucleotide sequence.

In E. coli, where the replication fork travels at 500 nucleotide pairs per second, the DNA ahead of the fork in the absence of topoisomerasewould have to rotate at nearly 3000 revolutions per minute.

In a replication bubble, the same parental DNA strand serves as the template strand for leading-strand synthesis in one replication fork and as the template for lagging-strand synthesis in the other fork.

When bidirectional replication forks from adjacent origins meet, a leading strand always runs into a lagging strand.

DNA repair mechanisms all depend on the existence of two copies of the genetic information, one in each of the two homologous chromosomes.

To determine the reproducibility of mutation frequency measurements, you do the following experiment. You inoculate each of 10 cultures with a single E. coli bacterium, allow the cultures to grow until each contains 106 cells, and then measure the number of cells in each culture that carry a mutation in your gene of interest. You were so surprised by the initial results that you repeated the experiment to confirm them. Both sets of results display the same extreme variability, as shown in Table Q51. Assuming that the rate of mutation is constant, why do you suppose there is so much variation in the frequencies of mutant cells in different cultures?

DNA repair enzymes preferentially repair mismatched bases on the newly synthesized DNA strand, using the old DNA strand as a template. If mismatches were instead repaired without regard for which strand served as template, would mismatch repair reduce replication errors? Would such a mismatch repair system result in fewer mutations, more mutations, or the same number of mutations as there would have been without any repair at all? Explain your answers.

Discuss the following statement: Primase is a sloppy enzyme that makes many mistakes. Eventually, the RNA primers it makes are replaced with DNA made by a polymerase with higher fidelity. This is wasteful. It would be more energy-efficient if a DNA polymerase made an accurate copy in the first place.

If DNA polymerase requires a perfectly paired primer in order to add the next nucleotide, how is it that any mismatched nucleotides escape this requirement and become substrates for mismatch repair enzymes?

The laboratory you joined is studying the life cycle of an animal virus that uses circular, double-strand DNA as its genome. Your project is to define the location of the origin(s) of replication and to determine whether replication proceeds in one or both directions away from an origin (unidirectional or bidirectional replication). To accomplish your goal, you broke open cells infected with the virus, isolated replicating viral genomes, cleaved them with a restriction nuclease that cuts the genome at only one site to produce a linear molecule from the circle, and examined the resulting molecules in the electron microscope. Some of the molecules you observed are illustrated schematically in Figure Q51. (Note that it is impossible to distinguish the orientation of one DNA molecule relative to another in the electron microscope.) You must present your conclusions to the rest of the lab tomorrow. How will you answer the two questions your advisor posed for you? Is there a single, unique origin of replication or several origins? Is replication unidirectional or bidirectional?

You are investigating DNA synthesis in tissue-culture cells, using 3H-thymidine to radioactively label the replication forks. By breaking open the cells in a way that allows some of the DNA strands to be stretched out, very long DNA strands can be isolated intact and examined. You overlay the DNA with a photographic emulsion, and expose it for 3 to 6 months, a procedure known as autoradiography. Because the emulsion is sensitive to radioactive emissions, the 3H-labeled DNA shows up as tracks of silver grains. Because the stretching collapses replication Table Q51 Frequencies of mutant cells in multiple cultures (Problem 56) Experiment Culture (mutant cells/106 cells) 1 2 3 4 5 6 7 8 9 10 1 4 0 257 1 2 32 0 0 2 1 2 128 0 1 4 0 0 66 5 0 2 original molecule bubbles H-forms Problems p5.13/5.09/Q5.1 Figure Q51 Parental and replicating forms of an animal virus (Problem 510). CHAPTER 5 END-OF-CHAPTER PROBLEMS 297 bubbles, the daughter duplexes lie side by side and cannot be distinguished from each other. You pretreat the cells to synchronize them at the beginning of S phase. In the first experiment, you release the synchronizing block and add 3H-thymidine immediately. After 30 minutes, you wash the cells and change the medium so that the total concentration of thymidine is the same as it was, but only one-third of it is radioactive. After an additional 15 minutes, you prepare DNA for autoradiography. The results of this experiment are shown in Figure Q52A. In the second experiment, you release the synchronizing block and then wait 30 minutes before adding 3H-thymidine. After 30 minutes in the presence of 3H-thymidine, you once again change the medium to reduce the concentration of radioactive thymidine and incubate the cells for an additional 15 minutes. The results of the second experiment are shown in Figure Q52B. A. Explain why, in both experiments, some regions of the tracks are dense with silver grains (dark), whereas others are less dense (light). B. In the first experiment, each track has a central dark section with light sections at each end. In the second experiment, the dark section of each track has a light section at only one end. Explain the reason for this difference. C. Estimate the rate of fork movement (m/min) in these experiments. Do the estimates from the two experiments agree? Can you use this information to gauge how long it would take to replicate the entire genome?

If you compare the frequency of the sixteen possible dinucleotide sequences in the E. coli and human genomes, there are no striking differences except for one dinucleotide, 5-CG-3. The frequency of CG dinucleotides in the human genome is significantly lower than in E. coli and significantly lower than expected by chance. Why do you suppose that CG dinucleotides are underrepresented in the human genome?

With age, somatic cells are thought to accumulate genomic scars as a result of the inaccurate repair of double-strand breaks by nonhomologous end joining (NHEJ). Estimates based on the frequency of breaks in primary human fibroblasts suggest that by age 70, each human somatic cell may carry some 2000 NHEJ-induced mutations due to inaccurate repair. If these mutations were distributed randomly around the genome, how many protein-coding genes would you expect to be affected? Would you expect cell function to be compromised? Why or why not? (Assume that 2% of the genome1.5% protein-coding and 0.5% regulatoryis crucial information.)

Draw the structure of the double Holliday junction that would result from strand invasion by both ends of the broken duplex into the intact homologous duplex shown in Figure Q53. Label the left end of each strand in the Holliday junction 5 or 3 so that the relationship to the parental and recombinant duplexes is clear. Indicate how DNA synthesis would be used to fill in any single-strand gaps in your double Holliday junction.

In addition to correcting DNA mismatches, the mismatch repair system functions to prevent homologous recombination from taking place between similar but not identical sequences. Why would recombination between similar, but nonidentical sequences pose a problem for human cells?

Cre recombinase is a site-specific enzyme that catalyzes recombination between two LoxP DNA sites. Cre recombinase pairs two LoxP sites in the same orientation, breaks both duplexes at the same point in each LoxP site, and joins the ends with new partners so that each LoxP site is regenerated, as shown schematically in Figure Q54A. Based on this mechanism, predict the arrangement of sequences that will be generated by Cre-mediated site-specific recombination for each of the two DNAs shown in Figure Q54B