Genetics Ch. 11
What is the major difference between T1 and T2 ter sequences? A) Tus binds to T1 but not T2. B) Tus binds to T2 but not T1. C) T1 stops only counterclockwise-moving forks while T2 stops only clockwise-moving forks. D) T2 stops only counterclockwise-moving forks while T1 stops only clockwise-moving forks.
C
C14. Single-strand binding proteins keep the two parental strands of DNA separated from each other until DNA polymerase has an opportunity to replicate the strands. Suggest how single-strand binding proteins keep the strands separated and yet do not impede the ability of DNA polymerase to replicate the strands.
Primase and DNA polymerase are able to knock the single-strand binding proteins off the template DNA.
DNA polymerase moves toward the replication fork on both the lagging and leading strands. A) True B) False
A
DNA polymerase recognizes correct base pairs and incorporates new nucleotides which are correctly base paired to the template. A) True B) False
A
DNA polymerase uses energy contained in the linkage between phosphate groups to drive DNA synthesis. A) True B) False
A
DNA replication requires an origin, where the two strands can be separated. A) True B) False
A
Telomerase activity in humans is generally restricted to the germ line, cancerous cells, and a few specific adult cell types. How would this be expected to impact humans? A) Cancer cells can divide indefinitely. B) Because there is a difference in number of divisions, sperm would be expected to have shorter telomeres than oocytes. C) Rapidly dividing populations of cells have telomeres equal in size to slowly dividing populations of cells. D) All of these. E) None of these.
A
The fidelity of DNA replication in E. coli is about one mistake per 100 million incorporated nucleotides. A) True B) False
A
Which enzyme produces the first piece of nucleic acid synthesized in E. coli DNA replication? A) DNA primase B) DNA polymerase (polIII) C) DNA polymerase (polI) D) DNA ligase
A
Which of the three models of DNA replication proposed that the original double helix remained entirely intact throughout the process? A) Conservative model. B) Semiconservative model. C) Dispersive model. D) Destructive model. E) None of these.
A
What aspect of DNA polymerase function necessitates use of telomerase in eukaryotes? A) 3'-5' exonuclease function. B) 5'-3' processivity. C) Induced fit impact on incorporation of only correct bases. D) All of these. E) None of these.
B
What enzyme performs decatenation? A) Polymerase. B) Topoisomerase. C) Telomerase. D) Decatenase.
B
What is the major difference between the lagging and leading strands? A) On the leading strand, DNA synthesis occurs from 5' to 3', while DNA synthesis occurs from 3' to 5' on the lagging strand. B) DNA polymerase is able to continuously add new nucleotides on the leading strand while it must keep 'starting over' on the lagging strand. C) The lagging strand requires only a single primer while the leading strand requires many. D) Helicase opens the leading strand at a faster rate than the lagging strand.
B
Which of the following is a description of the proofreading function of DNA polymerase? A) Endonuclease cleavage. B) Exonuclease cleavage. C) Methylation. D) An induced-fit phenomenon. E) All of these.
B
Base pairing rules are critical for the ability of each DNA strand to act as a template. A) True B) False
A
What is the size range of Okazaki fragments?
1000-2000 nucleotides.
DNA replication proceeds only in one direction. A) True B) False
B
C24. What is a processive enzyme? Explain why this is an important feature of DNA polymerase.
A processive enzyme is one that remains clamped to one of its substrates. In the case of DNA polymerase, it remains clamped to the template strand as it makes a new daughter strand. This is important to ensure a fast rate of DNA synthesis.
C4. The compound known as nitrous acid is a reactive chemical that replaces amino groups (-NH2) with keto groups (=O). When nitrous acid reacts with the bases in DNA, it can change cytosine to uracil and change adenine to hypoxanthine. A DNA double helix has the following sequence: TTGGATGCTGG AACCTACGACC A. What would be the sequence of this double helix immediately after reaction with nitrous acid? Let the letter H represent hypoxanthine and U represent uracil. B. Let's suppose this DNA was reacted with nitrous acid. The nitrous acid was then removed, and the DNA was replicated for two generations. What would be the sequences of the DNA products after the DNA had replicated twice? Your answer should contain the sequences of four double helices. Note: During DNA replication, uracil hydrogen bonds with adenine, and hypoxanthine hydrogen bonds with cytosine.
A. TTGGHTGUTGG HHUUTHUGHUU B. TTGGHTGUTGG HHUUTHUGHUU ↓ TTGGHTGUTGG CCAAACACCAA AACCCACAACC HHUUTHUGHUU ↓ TTGGHTGUTGG TTGGGTGTTGG CCAAACACCAA CCAAACACCAA AACCCACAACC AACCCACAACC GGTTTGTGGTT HHUUTHUGHUU
C16. In the following drawing, the top strand is the template DNA, and the bottom strand shows the lagging strand prior to the action of DNA polymerase I. The lagging strand contains three Okazaki fragments. The RNA primers have not yet been removed. The top strand is the template DNA 3ʹ———————————————————————————5ʹ 5ʹ*************———***************———*************————3ʹ RNA primer ↑ RNA primer ↑ RNA primer |—————————||—————————||—————————| Left Okazaki Middle Okazaki Right Okazaki fragment fragment fragment A. Which Okazaki fragment was made first, the one on the left or the one on the right? B. Which RNA primer would be the first one to be removed by DNA polymerase I, the primer on the left or the primer on the right? For this primer to be removed by DNA polymerase I and for the gap to be filled in, is it necessary for the Okazaki fragment in the middle to have already been synthesized? Explain why. C. Let's consider how DNA ligase connects the left Okazaki fragment with the middle Okazaki fragment. After DNA polymerase I removes the middle RNA primer and fills in the gap with DNA, where does DNA ligase function? See the arrows on either side of the middle RNA primer. Is ligase needed at the left arrow, at the right arrow, or both? D. When connecting two Okazaki fragments, DNA ligase needs to use NAD+ or ATP as a source of energy to catalyze this reaction. Explain why DNA ligase needs another source of energy to connect two nucleotides, but DNA polymerase needs nothing more than the incoming nucleotide and the existing DNA strand. Note: You may want to refer to Figure 11.12 to answer this question.
A. The right Okazaki fragment was made first. It is farthest away from the replication fork. The fork (not seen in this diagram) would be to the left of the three Okazaki fragments, and moving from right to left. B. The RNA primer in the right Okazaki fragment would be removed first. DNA polymerase would begin by elongating the DNA strand of the middle Okazaki fragment and removing the right RNA primer with its 5' to 3' exonuclease activity. DNA polymerase I would use the 3' end of the DNA of the middle Okazaki fragment as a primer to synthesize DNA in the region where the right RNA primer is removed. If the middle fragment was not present, DNA polymerase could not fill in this DNA (because it needs a primer). C. You need DNA ligase only at the right arrow. DNA polymerase I begins at the end of the left Okazaki fragment and synthesizes DNA to fill in the region as it removes the middle RNA primer. At the left arrow, DNA polymerase I is simply extending the length of the left Okazaki fragment. No ligase is needed here. When DNA polymerase I has extended the left Okazaki fragment through the entire region where the RNA primer has been removed, it hits the DNA of the middle Okazaki fragment. This occurs at the right arrow. At this point, the DNA of the middle Okazaki fragment has a 5' end that is a monophosphate. DNA ligase is needed to connect this monophosphate with the 3' end of the region where the middle RNA primer has been removed. D. As mentioned in the answer to part C, the 5' end of the DNA in the middle Okazaki fragment is a monophosphate. It is a monophosphate because it was previously connected to the RNA primer by a phosphoester bond. At the location of the right arrow, there was only one phosphate connecting this deoxyribonucleotide to the last ribonucleotide in the RNA primer. For DNA polymerase to function, the energy to connect two nucleotides comes from the hydrolysis of the incoming triphosphate. In this location shown at the right arrow, however, the nucleotide is already present at the 5' end of the DNA, and it is a monophosphate. DNA ligase needs energy to connect this nucleotide with the left Okazaki fragment. It obtains energy from the hydrolysis of ATP or NAD+.
DnaA protein is unable to bind hemimethylated DNA because its binding site includes a GATC methylation site. A) True B) False
B
Meselson and Stahl's experiment was able to demonstrate in one round of division that DNA replication is a semiconservative process. A) True B) False
B
What DNA sequence is required for termination of replication in E. coli? A) oriC B) ter C) Tus D) Fork
B
C10. As shown in Figure 11.5, five DnaA boxes are found within the origin of replication in E. coli. Take a look at these five sequences carefully. A. Are the sequences of the five DnaA boxes very similar to each other? (Hint: Remember that DNA is double-stranded; think about these sequences in the forward and reverse direction.) B. What is the most common sequence for the DnaA box? In other words, what is the most common base in the first position, second position, and so on until the ninth position? The most common sequence is called the consensus sequence. C. The E. coli chromosome is about 4.6 million bp long. Based on random chance, is it likely that the consensus sequence for a DnaA box occurs elsewhere in the E. coli chromosome? If so, why aren't there multiple origins of replication in E. coli?
A. When looking at Figure 11.5, the first, second, and fourth DnaA boxes are running in the same direction, and the third and fifth are running in the opposite direction. Once you realize that, you can see the sequences are very similar to each other. B. According to the direction of the first DnaA box, the consensus sequence is TGTGGATAA ACACCTATT C. This sequence is nine nucleotides long. Because there are four kinds of nucleotides (i.e., A, T, G, and C), the chance of this sequence occurring by random chance is 4−9, which equals once every 262,144 nucleotides. Because the E. coli chromosome is more than 10 times longer than this, it is fairly likely that this consensus sequence occurs elsewhere. The reason why there are not multiple origins, however, is because the origin has five copies of the consensus sequence very close together. The chance of having five copies of this consensus sequence occurring close together (as a matter of random chance) is very small.
In Meselson and Stahl's experiment, what result would have supported the conservative model of DNA replication?
After each round of replication, two types of DNA would be obtained - one of high density and one of low density. Intermediate bands would not be seen.
What is a processive enzyme?
An enzyme that does not dissociate from the product (growing DNA strand in the case of DNA polymerase) after catalysis.
After DNA replication, the newly synthesized strand contains only newly synthesized histones. A) True B) False
B
Which of the following is not a mechanism by which E. coli can coordinate DNA replication with cell division? A) DnaA protein does not efficiently bind to hemimethylated binding sites in the oriC. B) DNA polymerase does not efficiently bind to mismatched base pairs. C) DnaA protein is rapidly degraded after the initiation of DNA replication. D) DnaA proteins must bind to a minimum number of binding sites in the oriC. E) None of these; all are such mechanisms.
B
Which statement about bacterial DNA replication is correct? A) DNA replication begins at several places along the chromosome. B) DNA replication begins at the origin and travels in both directions. C) DNA replication begins at the origin and travels around the chromosome back to the origin. D) DNA replication begins at a GC rich region of the chromosome. E) All of these are correct.
B
C2. With regard to DNA replication, define the term bidirectional replication.
Bidirectional replication refers to DNA replication in both directions starting from one origin
What explanation for Okazaki fragments is the most accurate? A) DNA polymerase is only able to synthesize 1000-2000 bonds before it falls off the strand. B) DNA polymerase must stop to check its work every 1000-2000 nucleotides. C) DNA polymerase requires a free 3' OH group to attach new nucleotides; primers are produced on the lagging strand every 1000-2000 nucleotides. D) All of these are accurate statements. E) None of these are accurate statements.
C
Why is an AT rich region part of the E. coli oriC? A) DnaA proteins bind to AT rich sites. B) DnaC proteins bind to AT rich sites. C) AT base pairs are more easily separated than GC base pairs. D) AT base pairs contain more hydrogen bonds than GC base pairs.
C
What is a key difference between DNA polIII and DNA ligase? A) Only DNA polIII synthesizes phosphoester bonds. B) Only DNA ligase synthesizes phosphoester bonds. C) DNA polIII can synthesize DNA from 3'-5'. D) DNA ligase can use energy from ATP rather than nucleotides.
D
Which is a correct description of the replisome? A) The replisome is a complex of several proteins. B) The replisome includes DNA polymerase, DNA helicase, and primase. C) The replisome can be found at the replication fork during DNA synthesis. D) All of these are correct descriptions. E) None of the above are correct.
D
Which of the following is a key difference between eukaryotic and prokaryotic DNA replication? A) Eukaryotic origins of replication contain a high % of AT base pairs and require protein binding for correct function. B) In eukaryotes, DNA replication proceeds bidirectionally from the origin of replication. C) Eukaryotes contain several different DNA polymerases. D) Telomerase is required to replicate the ends of eukaryotic chromosomes. E) None of these: They could all also describe prokaryotic DNA replication.
D
C8. A DNA strand has the following sequence: 5ʹ-GATCCCGATCCGCATACATTTACCAGATCACCACC-3ʹ In which direction would DNA polymerase slide along this strand (from left to right or from right to left)? If this strand was used as a template by DNA polymerase, what would be the sequence of the newly made strand? Indicate the 5ʹ and 3ʹ ends of the newly made strand.
DNA polymerase would slide from right to left. The new strand would be 3'-CTAGGGCTAGGCGTATGTAAATGGTCTAGTGGTGG-5'
Some restriction enzymes which recognize 5'-GATC-3' sites are only able to cut the DNA if those sites are fully methylated. Other restriction enzymes which recognize this site are able to cut whether it is methylated or not. This is useful for molecular biologists who wish to control enzyme digestion when engineering new plasmids (see Chapter 18). In order to take advantage of this, they grow the plasmids in Dam- bacteria, which lack Dam methylase. However, these strains of bacteria tend to grow very poorly in culture. Why?
Dam methylation of the 5'-GATC-3' sites in the oriC is required for efficient binding of the DnaA boxes by DnaA protein.
What is the primary function of the DnaA protein in bacterial DNA replication?
DnaA binds to DnaA boxes within the oriC. This binding is thought to stimulate strand separation of an adjacent AT rich sequences.
C28. If a eukaryotic chromosome has 25 origins of replication, how many replication forks does it have at the beginning of DNA replication?
Fifty, because two replication forks emanate from each origin of replication. DNA replication is bidirectional.
What proteins are found within the replisome?
Helicase and primase (together known as the primosome) and two molecules of DNA polymerase.
Why are two ter sequences required in E. coli?
They probably aren't. DNA synthesis will stop when the two replication forks meet. The ter sequences simply control where the termination is most likely to occur by forcing one replication fork to wait there.
Why are lesion-replicating DNA polymerases useful for the eukaryotic cell?
In their absence, the incidence of abnormal chromosomes would be expected to increase due to gaps in DNA replication caused by abnormal DNA structures. The ability to replicate DNA in spite of these abnormal DNA structures allows the cell time to correct the abnormality.
Why is the AT/GC rule critical for DNA replication?
This allows both strands of the double helix to act as templates for synthesis of the new DNA.
Why was the use of nitrogen isotopes important for the success of Meselson and Stahl's experiment?
It allowed for distinction between the three proposed models of DNA replication, as the original DNA and newly synthesized DNA would have different densities.
C6. The chromosome of E. coli contains 4.6 million bp. How long will it take to replicate its DNA? Assuming DNA polymerase III is the primary enzyme involved and this enzyme can actively proofread during DNA synthesis, how many base pair mistakes will be made in one round of DNA replication in a bacterial population containing 1000 bacteria?
Let's assume there are 4,600,000 bp of DNA and that DNA replication is bidirectional at a rate of 750 nucleotides per second. If there were just a single replication fork, 4,600,000/750 = 6133 seconds, or 102.2 minutes. Because replication is bidirectional, 102.2/2 = 51.1 minutes. Actually, this is an average value based on a variety of growth conditions. Under optimal growth conditions, replication can occur substantially faster. With regard to errors, if we assume an error rate of one mistake per 100,000,000 nucleotides, 4,600,000 × 1000 bacteria = 4,600,000,000 nucleotides of replicated DNA. 4,600,000,000/100,000,000 = 46 mistakes. When you think about it, this is pretty amazing. In this population, DNA polymerase would cause only 46 single mistakes in a total of 1000 bacteria, each containing 4.6 million bp of DNA.
How could Kornburg's "in vitro DNA replication" system be used to identify all the proteins involved in DNA replication?
Many answers are possible. One might be to isolate individual proteins from the cell extract, and then incubate template DNA with various proteins to identify those that allow DNA replication. For example, primase would be expected to incorporate some radioactivity into the new strand, but this would not be as effective as the full extract. Ultimately, efficient and accurate replication could be accomplished by adding all members of the replisome to a test tube with template DNA. Another possibility would be to compare the protein profiles of mutants known to be missing a particular function to that of wild type strains in order to identify the protein responsible for that activity.
What is meant by bidirectional replication?
The DNA is separated in both directions from the origin, and replicated in both directions.
Why must the lagging strand loop out of the replisome?
The DNA polymerase molecule is physically linked to the helicase but performs DNA synthesis in the direction away from the helicase. For each Okazaki fragment, the DNA polymerase must "hop" back to the next primer.
Describe exonuclease function and give two specific examples.
The ability to digest away a nucleotide at the end of a DNA strand. This may occur from either the 5' or the 3' end of the strand. The proofreading function of DNA polymerase is a 3'-5' exonuclease activity; removal of the primers is a 5'-3' exonuclease activity.
C22. Discuss the similarities and differences in the synthesis of DNA in the lagging and leading strands. What is the advantage of a primosome and a replisome as opposed to having all replication enzymes functioning independently of each other?
The leading strand is primed once, at the origin, and then DNA polymerase III synthesizes DNA continuously in the direction of the replication fork. In the lagging strand, many short pieces of DNA (Okazaki fragments) are made. This requires many RNA primers. The primers are removed by DNA polymerase I, which then fills in the gaps with DNA. DNA ligase then covalently connects the Okazaki fragments. Having the enzymes within a complex such as a primosome or replisome provides coordination among the different steps in the replication process and thereby allows it to proceed faster and more efficiently.
C26. What enzymatic features of DNA polymerase prevent it from replicating one of the DNA strands at the ends of linear chromosomes? Compared with DNA polymerase, how is telomerase different in its ability to synthesize a DNA strand? What does telo merase use as its template for the synthesis of a DNA strand? How does the use of this template result in a telomere sequence that is tandemly repetitive?
The reason is the inability to synthesize DNA in the 3' to 5' direction and the need for a primer prevent replication at the 3' end of the DNA strands. Telomerase is different than DNA polymerase in that it uses a short RNA sequence, which is part of its structure, as a template for DNA synthesis. Because it uses this sequence many times in row, it produces a tandemly repeated sequence in the telomere at the 3' ends of linear chromosomes.
