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Provide at least two observations suggesting that Pol I is not the chromosomal replicase.
Possible answers: Pol I is slow in DNA synthesis compared with the rate of replication in E. coli. Pol I can be mutated and the cells still survive. Pol I is not highly processive.
Explain how the enzymes that join together Okazaki fragments ensure that all the RNA is removed from the ends of fragments before they are sealed.
Ligase will not seal a nick in which the 5′-terminal nucleotide is a ribonucleotide. Sealing is delayed until all the RNA has been removed.
Eukaryotic origins are tightly regulated so as not to fire more than once during S phase. Explain the key points of regulation of this process in eukaryotic cells.
The preRC forms only in G1, not in other phases of the cell cycle. Cyclin kinases produced only in S phase are needed to assemble the remaining proteins to produce active replication forks. Origins do not fire a second time, because new preRC complexes cannot form until the cell completes its cycle and returns to G1.
Explain the role of the τ subunit of Pol III. (a) How many τ subunits must be present in Pol III to coordinate leading-and lagging-strand DNA synthesis at a replication fork? (b) How many τ subunits must be present to allow Pol III to extend an oligonucleotide primer processively on a single-stranded template in vitro? Explain your answer in both cases.
The τ subunits link together the leading-and lagging-strand core polymerases, one τ linked to each core, and both connected to DnaB.(a) Two. (b) Zero. The core polymerase, in conjunction with a sliding clamp, is capable of processive synthesis of a new DNA strand on a single-stranded DNA template without any other subunits being present. This is analogous to leading-strand synthesis without lagging-strand synthesis.
The AAA+ ATPases are the wrenches and crowbars of DNA metabolism.Briefly indicate what is accomplished when ATP is hydrolyzed by each of the following AAA+ ATPases: DnaA, DnaC, and the γ and τ subunits of Pol III.
DnaA: ATP hydrolysis inactivates the DnaA for replication initiation. DnaC: ATP hydrolysis helps release DnaB helicase as it is loaded onto the DNA. Pol III γ and τ subunits: ATP hydrolysis allows the β subunit (sliding clamp) to close around the DNA as it is loaded.
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?
(1) The fork may bypass the lesion. If the DNA backbone is intact, the fork may either (2) stall or (3) leave the lesion behind in a single-strand gap. (4) If the fork encounters a break in a template strand, one arm of the fork is lost.
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.
(a) 3′ → 5′ helicase motor
(c) 5′ → 3′ helicase
(d) Having a “pin” structure that helps separate DNA strands
(e) Binding to chi sites
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.
During normal growth, forks collapse at sites where there is a break in the template strand. The absence of RecBCD in the mutants will curtail the repair of such double-strand breaks, leading to the increase in linearized chromosomes.
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 witha. Rarely, meiosis yields one spore with A and three with a, or three withA and one with a. How could this happen?
If gene conversion occurs across the region where the sequence difference between A and a is located, this will create a heteroduplex intermediate that has an A–containing strand paired with an a-containing complement. The mismatch at this gene locus will be resolved one way or the other by mismatch repair, resulting in the loss or gain of information.
Unlike recombination, the repair of double-strand breaks by nonhomologous end joining creates mutations. Explain why.
NHEJ involves degradation by nucleases and processing of the DNA ends, leading to some loss of base pairs.
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?
The information at HMLα would be subject to change. Whatever information was present in MAT would be transferred to HMLα.
In the study by Keeney, Giroux, and Kleckner (see How We Know),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?
The polyacrylamide gel separates proteins on the basis of molecular weight. For a protein covalently linked to DNA, the molecular weight is that of the combined protein and DNA. If the size of the linked DNA varies, the protein-DNA complex will have a highly variable molecular weight and will appear as a smear in the gel; detecting a single protein band would be impossible. Nuclease digestion eliminated this variability and allowed Spo11 to be detected as a discrete band.
Holliday intermediates are associated with both homologous genetic recombination and some site-specific recombination systems. How does their formation differ in the two types of reactions?
In tyrosine-class site-specific recombination systems, the Holliday intermediate is generated in a precise set of cleavage and strand-transfer steps, all occurring at a single, unique DNA sequence. In homologous recombination, Holliday intermediates can appear at any sequence and are generated by the strand invasion and branch migration promoted by RecA recombinases.
Compare and contrast the DNA sequence requirements of homologous genetic recombination, site-specific recombination, and transposition.
Homologous recombination occurs only where two DNA molecules have identical or very similar sequences over a significant region; any sequence is permitted. Site-specific recombination occurs only at particular DNA sequences that are recognized, bound, and recombined by the recombinases. Transposition, with a few exceptions, can occur at almost any sequence.
Draw the products of Cre-lox mediated site-specific recombination reactions for the recombination target sites and orientations indicated by arrows in the illustrations below.
The human genome has more than a million copies of the SINE element Alu. These transposons are found in the DNA between genes, and often in the introns of genes, but very rarely in gene exons. Explain why this is so.
Exons are the coding regions of genes. Insertion of a transposon of any type into an exon would almost certainly disrupt the activity of the protein encoded by the gene.
There may be a palindromic sequence that allows formation of a hairpin structure during transcription, creating a ρ-independent terminator that is not perfectly efficient, or an imperfect rut sequence to guide the loading of the ρ protein. In both cases, the sequences would have to be imperfect so that some transcripts were elongated through gene B.
The −10 and −35 sequences in bacterial promoters are separated by about two turns of the DNA double helix. How would transcription be affected if a deletion were introduced in the promoter region that moved the −35 sequence to the −29 position?
The deletion would move the −35 sequence closer to the −10 sequence by half a helical turn of the DNA, putting the two elements on opposite faces of the DNA duplex. This would dramatically reduce binding of sigma factor to the promoter, thereby decreasing transcription efficiency.
The gene encoding the E. coli enzyme β-galactosidase begins with the sequence ATGACCATGATTACG. What is the sequence of the mRNA transcript specified by this part of the gene?
5′-AUGACCAUGAUUACG. A sequence reported for any gene is, by convention, that of the coding strand, and sequences are always written in the 5′→3′ direction.
The sequence of the consensus −10 region is TATAAT. If two genes,tesA and tesB, have identical promoter sequences except in the −10 region, where the tesA sequence is TAATAT and the tesB sequence is TGTCGA, which gene do you expect to be more efficiently transcribed,and why?
The tesB −10 sequence deviates more from the consensus sequence, and its higher G≡C content will be more difficult to melt. Assuming these genes use similar transcription initiation modes, more tesA mRNA is expected relative to tesB.
If a Pol II promoter were replaced with a promoter specific for Pol III in a human cell, what do you expect would happen to the number of transcripts produced?
The number of transcripts is expected to increase, because Pol III typically generates many more transcripts than Pol II for the genes it transcribes. tRNAs, which are synthesized by Pol III, are required in much greater quantities in the cell than are most mRNAs, which are made by Pol II.
Name the three major steps, and indicate their relative rates, in the transcription of a typical bacterial gene.
Initiation (including abortive initiation), the slowest step; elongation, the fastest step but punctuated by pauses; termination, requiring stalling and subsequent dissociation of the polymerase from the DNA.
People who ingest Amanita phalloides (the source of α-amanitin) initially experience gastrointestinal distress caused by other toxins also produced by this mushroom. α-Amanitin shuts down the action of RNA polymerase II, but death does not occur until about 48 hours after ingestion and usually involves liver dysfunction. Suggest a reason for the delay in lethality.
Many RNAs are present in the body, synthesized before ingestion of the α-amanitin. The mRNAs support cell and tissue function until they are degraded, and typically last for about two days.
A drug company has discovered a natural product, cupramycin, that efficiently intercalates into DNA. How might this compound affect transcription?
Intercalation means that the planar portion of a small molecule inserts into the double-helical DNA between successive base pairs, deforming the DNA and preventing movement of RNA polymerase along the template. Actinomycin D and acridine act in a similar way.
How might an investigator search for Pol II promoters in the DNA sequence of an entire organism? Is it possible to find all such promoters computationally?
This is difficult to do experimentally. The TATA box and the Inr sequence are often, but not always, found upstream from genes transcribed by Pol II. A significant minority of eukaryotic promoters lack a well-defined TATA box—the so-called “TATA-less” promoters—and this can confound identification by computer algorithms alone.
Gene A encodes protein A. A genetic engineer excises a promoter sequence for gene A from the DNA and reinserts it at the other end of gene A, oriented so that an RNA polymerase binding at the promoter will transcribe across gene A. Will the mRNA synthesized by the RNA polymerase still possess a sequence that produces a functional protein A? Why or why not?
No. The two strands of a DNA molecule are antiparallel and complementary (not identical). When the promoter is inverted, it will direct RNA synthesis that uses what was originally the coding strand as the template strand. The mRNA sequence, derived from a different DNA strand and synthesized in the opposite direction, would be very different from the mRNA produced by the original gene and might not even contain an open reading frame.
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