What are the four possible fates of a replication fork that encounters a template strand with a break or some other type of unrepaired DNA lesion?

A branched, circular DNA substrate is constructed to mimic one possible structure of a stalled replication fork, as shown below. An enzyme is added that promotes regression of the fork structure.

Draw a Holliday intermediate and label the ends of each DNA strand so that the strand polarity is evident.

The RecBCD enzyme acts as a nuclease and a helicase in preparing DNA ends for RecA binding and strand invasion. RecBCD has several functions built into its three subunits. Indicate the subunit (RecB, RecC, or RecD) responsible for each of the following functions.

Describe the three steps that initiate almost all processes related to homologous genetic recombination (the recombinational DNA repair of double-strand breaks).

Replication forks of a bacterial species are found to stall at double-strand cross-links, yielding a stalled fork with the structure shown below. The pathway for repair of these stalled forks involves the formation of a Holliday intermediate. Draw one step that will convert the fork into a structure with a Holliday intermediate. Place an arrowhead on all 3′ ends (one end is so represented below). Note that the Holliday intermediate is formed without cleaving any covalent bonds in the DNA.

In E. coli cells with mutations that eliminate the RecBCD enzyme, about 20% of the cells have linearized chromosomes when grown under normal aerobic conditions. Under similar growth conditions, fewer than 3% of the chromosomes are linearized in wild-type cells. Suggest, in two or three sentences, why this difference is observed.

Eukaryotes have two RecA-class recombinases, called Dmc1 and Rad51. How are these enzymes utilized by cells?

During meiosis in yeast, if the diploid cell has alleles a and A of a particular gene, it normally forms two spores with A and two spores with a. Rarely, meiosis yields one spore with A and three with a, or three with A and one with a. How could this happen?

Unlike recombination, the repair of double-strand breaks by nonhomologous end joining creates mutations. Explain why.

At the yeast mating-type locus, the mating-type switch is initiated by introducing a double-strand break at the MAT locus. What would happen if the DSB were introduced at the HMLα locus instead?

In the study by Keeney, Giroux, and Kleckner (see this chapter’s How We Know section), Spo11 was identified as the protein that introduces double-strand breaks to initiate meiotic recombination. To identify candidate proteins, the DNA was first extracted to remove most noncovalently bound proteins, then filtered to isolate remaining protein-DNA complexes. The samples were then extensively treated with nucleases before the samples were loaded onto a polyacrylamide gel. Why was the nuclease treatment necessary?

A Holliday intermediate is formed between two chromosomes at a point between two genes, A and B, as shown below. The two chromosomes have different alleles of the two genes (A and a; B and b). Where would the Holliday intermediate have to be cleaved (points X and/or Y) to generate a chromosome with (a) an Ab genotype or (b) an ab genotype?

Ionizing radiation causes double-strand breaks in DNA, but the bacterium Deinococcus radiodurans is highly resistant to these DSB-generating effects. DSBs also occur (slowly) during prolonged cell desiccation, and desiccation is thought to be the selective pressure in the evolution of the extraordinary capacity of D. radiodurans for DNA repair. After heavy irradiation, the bacterium produces several novel proteins at high levels. One of these, called DdrA (DNA damage repair protein A), binds tightly to the 3′ ends of broken DNA strands and prevents their degradation by nucleases. A mutation that eliminates DdrA function has little effect on survival after irradiation, but a large effect on survival after desiccation. Suggest an explanation for the role of DdrA during desiccation. In your answer, consider the requirements of DNA repair versus DNA degradation.