Chapter 17 - DNA Replication, Repair, and Recombination
Shown in the figure below are three nonstandard nitrogenous bases that are formed by the deamination of naturally occurring bases in DNA. Indicate which base in DNA must be deaminated to form each of these bases.
Adenine/Hypoxanthine Guanine/Xanthine Thymine/none Cytosine/Uracil
Which of the following statements regarding Nucleotide Excision Repair (NER) and Base Excision Repair (BER) is true?
Both NER and BER involve the removal of one or more damaged bases by a nuclease. This statement is true. In both NER and BER a nuclease will target damaged or distorted regions of DNA.
What would the consequence of deleting telomerase from a human cell line?
The ends of each chromosome would shrink after each cell division.
What are the consequences of having pyrimidine dimers in DNA?
These dimers distort the DNA structure and result in errors during DNA replication.
Why is it important that none of the bases shown in the figure occurs naturally in DNA?
They cause base excision repair errors.
Which two domains/subdomains of the RecB subunit display typical helicase motor domain activity?
1A and 2A Subdomains 1A and 2A show motor function typically found in helicases, as they use the energy from ATP to help drive the RecBCD protein unidirectionally along duplexed DNA. As a consequence of their motor function, these subdomains help pull the duplexed DNA into the RecC pin causing the DNA to split into single strands. Subdomain 1B contains an insertion of residues that form an extended arm that runs adjacent to and contacts the duplex DNA. Subdomain 2B is the major interface between RecB and RecC. The third carboxy-terminal nuclease domain, responsible for cleaving both the 3'-tailed strand (darker purple) and the 5'-tailed strand (bright purple), is connected to the remainder of the RecB subunit by a long 70-residue linker.
If you begin with a single DNA template segment, how much theoretical amplification could be achieved after 15 cycles of PCR?
32,768
The diagram below shows a replication bubble with synthesis of the leading and lagging strands on both sides of the bubble. The parental DNA is shown in dark blue, the newly synthesized DNA is light blue, and the RNA primers associated with each strand are red. The origin of replication is indicated by the black dots on the parental strands.
As soon as the replication bubble opens and the replication machinery is assembled at the two replication forks, the two primers for the leading strands (primers a and h) are produced. The production of the first primers on the lagging strands (those closest to the origin of replication, b, and g) is delayed slightly because the replication forks must open up further to expose the template DNA for the lagging strands. After completion of the first segments of the lagging strands, additional template DNA must be exposed before the second primers (c and f) can be produced. And after completion of the second segments, additional template DNA must be exposed before the third primers (d and e) can be produced. In summary, because of the way the replication bubble expands, the lagging strand primers near the origin of replication were produced before the primers near the replication forks.
Meiotic recombination is responsible for generating diversity in gamete formation, and the molecular mechanism begins with a double-strand DNA break. How is this break generated?
Created by the Spo11 enzyme.
RecB and RecC are structurally similar; both contain three domains (1A and 1B, 2A and 2B and 3). However, the function of these domains differs between the two proteins. Domains 1 and 2 in both RecB and RecC are structurally homologous to SF1 helicases, but only RecB still shows helicase function. Domains 1 and 2 of RecC evolved a new function, which is to form a channel for directing the 3' DNA strand down to the RecB nuclease. Domain 3 of RecB is a nuclease that digests both 3' and 5' ends. This domain in RecC has the novel function of splaying the DNA duplex into single strands. Given this information, identify the specific domains of RecC. A labeled image of RecB is provided for comparison.
Domain 3 (1), subdomain 2B (2), subdomain 1B (3), subdomain 2A (4), subdomain 1A (5) A comparison of the RecB and RecC models reveals that one subunit is inverted relative to the other. In RecC, domain 3 splits the duplex DNA into 3' and 5' strands; this means it is located at the top of the image near the DNA molecule (1). The homologous domain in RecB is the active nuclease (domain 3), which is found at the opposite end where the split DNA strands will be digested. Although domains 1 and 2 of RecB and RecC are homologous, only RecB still exhibits helicase function. Domains 1 and 2 of RecC evolved a new function, which is to form a channel for directing the 3' DNA strand down to the RecB nuclease (not shown, but located at the bottom of the image).
The helicase motors of RecD and RecB pull the double-stranded DNA (dsDNA) into the pin of RecC, thus separating the duplex into 5'- and 3'-tailed single strands. The 3' tail follows a channel through the complex, emerging at the nuclease active site. The 5' tail is threaded through a different channel, also emerging near the nuclease active site. Before encountering a chi site, the RecB nuclease digests the 3' tail at a faster rate than the digestion of the 5' tail is digested, possibly because the 3' tail is more favorably located near the nuclease active site. Which of the following is true about nuclease digestion by RecB after the trimer encounters a chi site in the DNA?
Nuclease digestion of the 3' tail ceases while the digestion rate for the 5' tail increases. There are two channels through which the 3' tail can exit. The 3'-ending strand can pass along the 3' channel through the protein, exiting between subdomain 1A of RecC and the nuclease domain of RecB (purple arrow). A second exit (yellow arrow) would bring the 3'-ending strand past the active site of the nuclease (green), where digestion would occur. Once the RecBCD trimer encounters a chi site, digestion of the 3' tail ceases. The presence of a flexible loop (a short alpha helix flanked by long, flexible linkers; light blue) appears to block the channel past the nuclease active site. With the 3' tail no longer being digested, the 5' tail is free from competition for access to the nuclease active site and thus its rate of digestion increases.
Why are there only three bases shown, when DNA contains four bases?
Only three bases contain amino group.
The repair of double-strand breaks in DNA is essential for maintaining the integrity of the genetic material and for cell viability. The homologous recombination pathway serves as a primary means of double-strand break repair in eubacteria. An initial step in this pathway involves the RecBCD trimeric protein recognizing blunt-ended DNA at the site of a break, followed by complex processing of DNA inward from the break, finally resulting in a 3' overhang of single-stranded DNA (ssDNA) coated by RecA protein. This serves as the substrate for homologous recombination via the RecA pathway. Which subunits of the RecBCD trimer show helicase structure and function?
RecB and RecD Subunits RecB and RecD are functional helicases, as both use ATP to move unidirectionally along a DNA duplex and to help separate the DNA into single stranded form. RecB also has a functional nuclease domain that enables the RecBCD protein to cleave single-stranded DNA and to help load the RecA protein onto the DNA strand containing the chi site after RecBCD completes its main function. RecC also has structural similarities to helicases suggesting it evolved from a helicase subunit. However, RecC no longer exhibits helicase function.
Bacterial cells use at least three different pathways for carrying out genetic recombination. One of these three pathways utilizes an enzyme complex called RecBCD, which binds to breaks in DNA and exhibits both helicase and single-strand nuclease activities. Briefly describe a model showing how the RecBCD enzyme complex might set the stage for genetic recombination.
RecBCD binds to double-strand(ed) breaks in chromosomal DNA, and its helicase activity then unwinds the DNA double helix, thereby creating single-strand(ed) loops. The nuclease activity of RecBCD cleaves one of the looped DNA strands, creating a free, single-strand(ed) DNA end. RecA then catalyzes a "strand invasion" reaction in which this free, single-strand(ed) DNA end invades and displaces one of the two strands of an homologous DNA double helix
In contrast to the leading strand, the lagging strand is synthesized as a series of segments called Okazaki fragments. The diagram below illustrates a lagging strand with the replication fork off-screen to the right. Fragment A is the most recently synthesized Okazaki fragment. Fragment B will be synthesized next in the space between primers A and B.
Synthesis of the lagging strand is accomplished through the repetition of the following steps. Step 1: A new fragment begins with DNA polymerase III binding to the 3' end of the most recently produced RNA primer, primer B in this case, which is closest to the replication fork. DNA pol III then adds DNA nucleotides in the 5' to 3' direction until it encounters the previous RNA primer, primer A. Step 2: DNA pol III falls off and is replaced by DNA pol I. Starting at the 5' end of primer A, DNA pol I removes each RNA nucleotide and replaces it with the corresponding DNA nucleotide. (DNA pol I adds the nucleotides to the 3' end of fragment B.) When it encounters the 5' end of fragment A, DNA pol I falls off, leaving a gap in the sugar-phosphate backbone between fragments A and B. Step 3: DNA ligase closes the gap between fragments A and B. These steps will be repeated as the replication fork opens up. Try to visualize primer C being produced to the right (closest to the replication fork). Fragment C would be synthesized and joined to fragment B following the steps described here.
The figure below shows 5-methylcytosine, a pyrimidine that arises naturally in DNA when cellular enzymes methylate cytosine. Why is the presence of this base likely to increase the probability of a mutation?
The base 5-methylcytosine yields thymine upon deamination, and because it is a naturally occurring base, there is no mechanism to detect and excise these newly generated thymine bases. deamination of 5-methylcytosine will therefore go unrepaired.
Helicases are "motorized" proteins that help unwind DNA by separating duplexed DNA into single-stranded form. They are powered by the energy derived from ATP hydrolysis. Helicases have a polarity that is defined as the direction they travel along single-stranded DNA (i.e., either in a 5'→→ 3' or in a 3'→→ 5' direction). What are the polarities of the RecB and RecD subunits?
The polarity of RecB is 3'→→ 5', wheras the polarity of RecD is 5'→→ 3'. The RecC pin splits the DNA duplex into 5'- and 3'-tailed strands. The RecB and RecD helicase subunits bind to different single-stranded tails. The 3'-ending strand is fed to the RecB helicase, and the 5'-ending strand is fed to the RecD helicase.
What is the role of primase is the process of bacterial DNA replication?
To synthesize short RNA primers, providing the DNA polymerase with a free 3'-hydroxyl group at the site of dNTP addition based on the template sequence.
Thymine dimers can be repaired by Photoreactivation Repair or Nucleotide Excision Repair.
True Both Photoreactivation Repair and Nucleotide Excision Repair will target UV-induced pyrimidine dimers in DNA.
Which of the following does NOT correctly match the repair enzyme with the DNA repair mechanism that it is associated with?
bypass polymerase - nonhomologous end-joining
As the two parental (template) DNA strands separate at a replication fork, each of the strands is separately copied by a DNA polymerase III (orange), producing two new daughter strands (light blue), each complementary to its respective parental strand. Because the two parental strands are antiparallel, the two new strands (the leading and lagging strands) cannot be synthesized in the same way.
leading strand made continuously only one primer needed daughter strand elongates toward replication fork lagging strand made in segments multiple primers needed daughter strand elongates away from replication fork both strands synthesized 5' to 3 Because DNA polymerase III can only add nucleotides to the 3' end of a new DNA strand and because the two parental DNA strands are antiparallel, synthesis of the leading strand differs from synthesis of the lagging strand. The leading strand is made continuously from a single RNA primer located at the origin of replication. DNA pol III adds nucleotides to the 3' end of the leading strand so that it elongates toward the replication fork. In contrast, the lagging strand is made in segments, each with its own RNA primer. DNA pol III adds nucleotides to the 3' end of the lagging strand so that it elongates away from the replication fork. In the image below, you can see that on one side of the origin of replication, a new strand is synthesized as the leading strand, and on the other side of the origin of replication, that same new strand is synthesized as the lagging strand. The leading and lagging strands built on the same template strand will eventually be joined, forming a continuous daughter strand.