Chapter 6 - DNA Replication and Repair

Question 1

True or False:

In semiconservative replication, the two parental strands separate and each serve as a template for synthesis of a new strand.

Answer 1

True

Question 2

When compared to each other, the two replication forks that form at an origin of replication move in which direction?

• in the 5’-to-3’ direction
• toward the origin
• in the 3’-to-5’ direction
• toward the template strand
• in opposite directions

Answer 2

In opposite directions

(The two forks move away from the origin in opposite directions. As DNA replication continues, these forks move farther and farther apart.)

Question 3

When DNA replication proceeds along a template, which of the following best describes the directionality of synthesis?

• in both the 3’-to-5’ and the 5’-to-3’ directions
• in the 3’-to-5’ direction
• from telomere to telomere
• in the 5’-to-3’ direction
• from the centromere to the telomeres

Answer 3

In the 5’-to-3’ direction

(DNA synthesis proceeds in the 5’-to-3’ direction, meaning that nucleotides are added to the 3’ end of a growing DNA strand.)

Question 4

What kinds of bonds link the two strands of a double helix to each other? Choose the most specific answer.

• hydrogen
• covalent
• hydrophobic
• ionic

Answer 4

Hydrogen

Hydrogen bonds form between the nucleotide bases in each complementary DNA strand.

Question 5

Nucleotides in each strand of DNA are held together by what type of bonds?

• phosphodiester
• peptide
• hydrogen
• ionic

Answer 5

Phosphodiester

(Phosphodiester bonds, a type of covalent bond, link the sugar and phosphate groups that form the backbone of DNA molecules.)

Question 6

At a replication fork, how is the leading strand synthesized?

• in the incorrect 5’-to-3’ direction
• continuously
• in the correct 3’-to-5’ direction
• without the use of a template
• discontinuously

Answer 6

Continuously

(At a replication fork, the leading strand is synthesized continuously. A sliding clamp protein allows the DNA polymerase to move along the leading-strand template without falling off.)

Question 7

At a replication fork, how is the lagging strand synthesized?

• discontinuously
• continuously
• in the incorrect 3’-to-5’ direction
• in the correct 3’-to-5’ direction
• without the use of a template

Answer 7

Discontinuously

(At a replication fork, the lagging strand is synthesized discontinuously. The Okazaki fragments are then joined together to form a continuous new DNA strand.)

Question 8

What is the role of the sliding clamp during replication?

• It hydrolyzes ATP to push the DNA polymerase along the DNA template.
• It unwinds the double helix at the replication fork to allow DNA polymerase to progress along the DNA.
• It keeps the RNA primer attached to the lagging strand.
• It loads DNA helicase onto the replication fork.
• It keeps DNA polymerase attached to the template while the polymerase synthesizes a new strand of DNA.

Answer 8

It keeps DNA polymerase attached to the template while the polymerase synthesizes a new strand of DNA.

(In the absence of the sliding clamp, most DNA polymerase molecules will synthesize only a short segment of DNA before falling off the template strand.)

Question 9

The production of a continuous new strand of DNA using the many separate Okazaki fragments (in other words, the joining of the already made fragments) found on the lagging strand requires all of the following except which one?

• nuclease
• DNA primase
• ATP
• DNA ligase
• repair polymerase

Answer 9

DNA primase

(Primase is needed to place the RNA primer required to initiate DNA synthesis. The joining of Okazaki fragments occurs at a later time.)

Question 10

Using your knowledge of telomerase enzyme and chromosomal size, choose the statement that correctly describes the information in the figure.

• Telomerase activity would complicate the end-replication problem, as this enzyme is involved in telomere shortening.
• Telomerase activity would complicate the end-replication problem, as this enzyme is involved in telomere lengthening.
• Telomerase activity would solve the end-replication problem, as this enzyme is involved in telomere shortening.
• Telomerase activity would solve the end-replication problem, as this enzyme is involved in telomere lengthening.

Answer 10

Telomerase activity would solve the end-replication problem, as this enzyme is involved in telomere lengthening.

(Telomerase activity would solve the end-replication problem because this enzyme is involved in telomere lengthening. Telomerase prevents linear eukaryotic chromosomes from shortening with each cell division.)

Question 11

What does depurination refer to?

• the accumulation of mutations and subsequent loss of purity of a nucleotide sequence
• the loss of A or G bases from DNA
• the breaking of the DNA backbone
• the loss of thymine due to damage from UV radiation
• the loss of G or C bases from DNA

Answer 11

the loss of A or G bases from DNA

Depurination removes a purine base, leaving a gap in the DNA. Both adenine and guanine are purine bases.

Question 12

How does ultraviolet radiation in sunlight typically damage DNA?

• It breaks hydrogen bonds between the two strands of DNA.
• It removes bases from nucleotides in DNA.
• It causes two adjacent pyrimidine bases to become covalently linked.
• It breaks the sugar–phosphate backbone of DNA.
• It converts cytosine into uracil.

Answer 12

It causes two adjacent pyrimidine bases to become covalently linked.

(Ultraviolet radiation in sunlight typically damages DNA by causing two adjacent pyrimidine bases to become covalently linked. These covalently linked thymine bases are referred to as thymine dimers, or T-T dimers.)

Question 13

Which of the following does not cause a mutation?

• evolution
• replication errors
• UV radiation
• failure of DNA repair systems
• metabolic activity

Answer 13

Evolution

(Evolution by means of natural selection is dependent upon variation in the genetic makeup of populations, which results from mutation. However, mutations arise independently through a number of different mechanisms.)

Question 14

Which of the following best defines a mutation?

• permanent change in a DNA sequence
• mistake created by faulty mismatch repair
• by-product of natural selection
• harmful change in a DNA sequence
• change in DNA sequence that causes a change in an amino acid in a protein

Answer 14

A permanent change in a DNA sequence

(A mutation is a permanent change in a DNA sequence. Mutations can arise during DNA replication, and if the event is not repaired, then the genetic change will be inherited by the daughter cells.)

Question 15

What potential outcomes are possible after replication in a DNA molecule with a depurination modification that is left unrepaired?

Choose one or more:

• The DNA molecule contains the normal sequence.
• The DNA molecule is converted into RNA.
• The DNA molecule is missing one nucleotide pair.
• The DNA molecule contains an extra three nucleotide pairs.

Answer 15

The DNA molecule contains the normal sequence.
AND
The DNA molecule is missing one nucleotide pair.

(One potential outcome is a normal double-stranded molecule without mutation, arising from the parental strand that was not chemically modified. The other potential outcome is a mutated sequence that has a nucleotide base pair deleted, arising from the strand with the depurinated modification.)

Question 16

What type of enzyme removes damaged DNA from the rest of the DNA molecule?

• ligase
• nuclease
• polymerase
• primase
• helicase

Answer 16

Nuclease

A nuclease, as it implies, digests nucleic acids, including DNA.

Question 17

What type of enzyme fills in the gap after damaged DNA has been removed?

• polymerase
• primase
• nuclease
• ligase
• helicase

Answer 17

Polymerase

(Specifically, a “repair polymerase” fills in this gap. DNA polymerase is involved in the reaction of 5’-to-3’ addition of new DNA nucleotides to a polymer.)

Question 18

What type of enzyme seals the newly added (repaired) DNA to the rest of the DNA molecule?

• polymerase
• ligase
• helicase
• primase
• DNase

Answer 18

Ligase

Ligases join DNA fragments by catalyzing the formation of phosphodiester bonds.

Question 19

In addition to its role in DNA repair, homologous recombination is also responsible for generating genetic diversity during what process?

• DNA replication
• independent assortment of chromosomes
• mitosis
• fertilization
• meiosis

Answer 19

Meiosis

During meiosis, homologous recombination is responsible for generating genetic diversity.

Question 20

What generally is the fate of mutations to the genome that have harmful consequences to an organism?

• They are always fixed by DNA repair.
• They tend to persist and spread.
• They do not occur.
• They are usually eliminated from the population by natural selection.
• They give rise to new species.

Answer 20

They are usually eliminated from the population by natural selection.

(Mutations to the genome that have harmful consequences to an organism are usually eliminated from the population by natural selection.)

Question 21

Synthetic biologists are trying to create cells from raw material. One step in the process is encapsulating genetic material into a compartment. Researchers of the origin of life think that the earliest cells on Earth may have used RNA as their genetic material instead of DNA. As biologists consider which genetic material to use in creating their synthetic cells, which of the following characteristics of a single-stranded RNA genome should they keep in mind?

• A newly synthesized RNA strand is not identical to the template strand.
• RNA is highly chemically stable relative to DNA.
• RNA cannot serve as a template.
• No enzymes can produce RNA.

Answer 21

A newly synthesized RNA strand is not identical to the template strand.

(The newly synthesized strand will be complementary to the template strand, not identical to it.)

Question 22

If DNA replication were conservative (although we know it is not), what would Meselson and Stahl have seen following the first round of replication in E. coli that had been switched from a heavy (15N-containing) nutrient medium to a light (14N-containing) nutrient medium?

• Half the DNA would be heavy, and the other half would be light.
• Half the DNA would be intermediate in density, and one-quarter would be heavy while one-quarter would be light.
• Half the DNA would be heavy, and the other half would be intermediate in density.
• All the DNA would be light.
• All the DNA would be heavy.

Answer 22

Half the DNA would be heavy, and the other half would be light.

(If DNA replication were conservative, Meselson and Stahl would have observed half the DNA to be heavy and the other half to be light following the first round of replication in E. coli that had been switched from a heavy (15N-containing) nutrient medium to a light (14N-containing) one. In the conservative model, the heavy, parent molecule would remain intact after being used as a template to produce the light, newly synthesized DNA.)

Question 23

To determine if DNA replication is semiconservative, dispersive, or conservative (see image below), Matt Meselson and Frank Stahl performed a classic experiment in which they grew E. coli for many generations in a heavy isotope of nitrogen (15N) and then transferred the bacteria to media containing 14N for a single round of DNA replication. Density of the DNA was determined via centrifugation and the result of intermediate density DNA (50% 14N, 50% 15N within a double helix) was consistent with both semiconservative and dispersive DNA replication. Another round of replication can discriminate between semiconservative and dispersive DNA replication. What kind of DNA is expected if these E. coli are grown for a second generation in14N-containing media?

• 25% intermediate density, 75% light density
• 50% light density, 50% heavy density
• 50% intermediate density, 50% light density
• same as result after one round of replication; all intermediate-density DNA

Answer 23

50% intermediate density, 50% light density

(When the intermediate strands (with one strand 15N and one strand 14N) split and each serves as a template for a new strand, one daughter DNA helix will be intermediate again, and the other daughter strand will be light density, with both strands containing 14N.)

Question 24

The DNA helicase animation shows the bacteriophage T7 helicase unwinding DNA. Which of the following are critical components of the helicase mechanism of action necessary to unwind DNA?

• dissociation of the helicase subunits
• ATP binding and hydrolysis
• conformational changes of subunits
• oscillating loops pulling the single-stranded DNA through a central hole
• binding of four helicase subunits to the double-stranded DNA

Answer 24

ATP binding and hydrolysis.
Conformational changes of subunits.
Oscillating loops pulling the single-stranded DNA through a central hole.

(Six helicase subunits go through cycles of ATP binding and hydrolysis to change conformation. The conformational changes lead to a pulling of the single-stranded DNA through the central hole.)

Question 25

AMP-PNP is a non-hydrolyzable analog of ATP that can bind to proteins in a similar manner as ATP but is no longer hydrolyzed. Predict what would happen to helicase activity if AMP-PNP were added to a DNA replication reaction.

• Helicase would function normally, since AMP-PNP is not the same as ATP and would not bind to helicase.
• Helicase would function normally, since AMP-PNP is an analog of ATP and would function in the place of ATP.
• Helicase would function faster than normal, since the AMP-PNP is not hydrolyzed. ATP binding induces conformational changes, but hydrolysis slows DNA unwinding by helicase.
• Helicase would no longer function, since the AMP-PNP is not hydrolyzed. ATP binding and hydrolysis induce the conformational changes that facilitate DNA unwinding by helicase.

Answer 25

Helicase would no longer function, since the AMP-PNP is not hydrolyzed. ATP binding and hydrolysis induce the conformational changes that facilitate DNA unwinding by helicase.

(ATP binding and hydrolysis are necessary for helicase function. AMP-PNP is not hydrolyzed like ATP, so it will block helicase conformational changes and function.)

Question 26

You have joined a lab studying DNA replication in E. coli. The graduate student you are working with has identified a mutation in primase that makes primase very inefficient. Your project is to characterize the cells with this mutation. Predict the defects you would most likely see in the mutant E. coli cells.

• inefficient Okazaki fragment joining
• a delay in DNA polymerase beginning synthesis
• rapid lagging-strand synthesis but slow leading-strand synthesis
• a longer total time to replicate DNA
• a delay in the unwinding of DNA

Answer 26

A delay in DNA polymerase beginning synthesis.
A longer total time to replicate DNA.

(Primase synthesizes the RNA primer for DNA polymerase to use in starting synthesis. If primase is slow to synthesize the RNA primer, it will take longer for DNA replication to finish and there will be a delay in the start of DNA polymerase synthesizing DNA.)

Question 27

What is the function of helicase?

Answer 27

It unwinds the two strands of DNA.

Question 28

What is the function of the single-stranded binding proteins?

Answer 28

They bind to the single-stranded DNA to prevent the 2 strands from re-forming before DNA synthesis can occur.

(This is particularly important for the lagging strand, which is looped out before another primer is added and the next Okazaki fragment is synthesized.)

Question 29

What is the function of primase?

Answer 29

It adds an RNA primer to each strand. This provides the starting point for DNA polymerase.

(Primase works repeatedly on the lagging strand as the DNA continues to be unwound.)

Question 30

What is the function of DNA polymerase?

Answer 30

It synthesizes the new DNA strand.

Question 31

What is the function of the sliding clamp?

Answer 31

It binds and holds the DNA polymerase on the DNA strand, allowing it to continue for a longer distance along the DNA before releasing.

Question 32

The leading strand is synthesized _______.

Answer 32

Continuously

Question 33

The lagging strand is synthesized _______.

Answer 33

Discontinuously in a series of Okazaki fragments.

Question 34

The RNA primers are removed from the Okazaki fragments and replaced with _______.

Answer 34

DNA nucleotides

Question 35

What is the function of ligase?

Answer 35

It closes the final gap between Okazaki fragments so that the two final DNA copies are finished.

Question 36

Researchers sought to grow blood vessels in a lab for implantation into patients with clogged coronary arteries. To avoid immune rejection, they wanted to grow vessels by starting with patients’ own vascular smooth muscle cells (SMCs). However, most patients requiring coronary artery replacement are elderly, and their vascular SMCs do not divide often enough to grow a new vessel. To get around this problem, the researchers introduced into the cells the gene coding for a particular protein in a format through which the gene would be transcribed and the protein produced. What gene was introduced into the elderly patients’ SMCs to allow continued cell divisions?

• primase
• DNA ligase
• telomerase
• DNA polymerase

Answer 36

Telomerase.

(Linear chromosomes end in telomeres, which are repetitive DNA sequences that cap chromosome ends. In cells that don’t express telomerase, telomeres shorten with each cell division. Once telomeres reach a critically short length, cell division ceases. The presence of telomerase allows for telomere lengthening after cell division.)

Question 37

The loss of purine bases from a strand of DNA is typically caused by which of the following?

• double-strand DNA break
• deamination
• a spontaneous chemical reaction
• ultraviolet radiation
• replication error

Answer 37

A spontaneous chemical reaction.

This spontaneous damage to DNA is a result of the random thermal motions that all molecules in a cell undergo.

Question 38

In the absence of repair, what would the replication of a double helix containing a mismatch yield?

• two DNA molecules containing the mismatch
• one DNA molecule with the normal sequence and one DNA molecule with a mutated sequence
• one DNA molecule with the normal sequence and one DNA molecule with a mismatch
• two DNA molecules containing different mismatches at the site of the original error
• two DNA molecules with a mutated sequence

Answer 38

One DNA molecule with the normal sequence and one DNA molecule with a mutated sequence.

(In the absence of repair, replication of a double helix containing a mismatch would yield one DNA molecule with the normal sequence and one DNA molecule with a mutated sequence. The mutation in one of the daughter DNA molecules will be permanent and hence replicated each time that DNA molecule is copied.)

Question 39

The mismatch repair system recognizes mismatched base pairs, removes a portion of the DNA strand containing the error, and then resynthesizes the missing DNA using the correct sequence as a template. But what if the mismatch repair system instead removed a piece of the DNA strand that contained the correct sequence? What would replication of this improperly repaired sequence produce?

• two DNA molecules bearing the same mutation
• two DNA molecules with a gap where the correct sequence was excised
• two DNA molecules that are missing one nucleotide pair
• one DNA molecule with a mutation and one DNA molecule with a mismatch
• two DNA molecules with different mutations

Answer 39

Two DNA molecules bearing the same mutation

(Removal of the correct nucleotide would produce a DNA molecule with an altered sequence because the mutated strand would serve as the template to “correct” the error instead of being the strand that was repaired. Further replication would produce daughter DNAs with this same mutation.)

Question 40

Choose all of the following that correctly describe a characteristic of mismatch repair.

• DNA polymerase and ligase fill in the gap.
• Exonuclease removes the newly synthesized DNA in the mismatched area.
• Regions of improper base-pairing between parent and daughter strand are detected and repaired.
• Helicase unwinds the DNA in the mismatched area.

Answer 40

• DNA polymerase and ligase fill in the gap.
• Exonuclease removes the newly synthesized DNA in the mismatched area.
• Regions of improper base-pairing between parent and daughter strand are detected and repaired.
• Helicase unwinds the DNA in the mismatched area.

(After the improper base-pairing is detected, helicase, exonuclease, DNA polymerase, and ligase work together to remove the mismatched area and replace it with the correct nucleotides.)

Question 41

Everyone is exposed regularly to ionizing radiation found in the soil, water, and air and from cosmic rays. In fact, 80% of the ionizing radiation people are exposed to comes from naturally occurring sources. Ionizing radiation can cause double-strand breaks in the DNA. Often, the DNA breaks have missing nucleotides at the broken ends. What type of repair would likely be used, and what would be the result of repairing this type of damage?

• Nonhomologous end joining would be used and would lead to properly repaired DNA.
• Nonhomologous end joining would be used to join the DNA, but errors would still remain.
• Direct repair would be used and would lead to properly repaired DNA.
• Direct repair would be used to join the DNA, but errors would still remain.

Answer 41

Nonhomologous end joining would be used to join the DNA, but errors would still remain.

(Nonhomologous end joining is one mechanism for repairing double-strand DNA breaks; this type of repair is prone to errors.)

Question 42

Ionizing radiation can also cause damage to the nitrogenous bases of DNA. Repair of the damaged bases takes several steps. Which of the following repair mechanisms can be used to repair the nitrogenous bases of DNA damaged by ionizing radiation?

• direct repair
• excision repair
• mismatch repair
• nonhomologous end joining

Answer 42

Excision repair.

Excision repair is usually used to replace the nitrogenous bases damaged by ionizing radiation.

Question 43

Which of the following occurs when a cell repairs a double-strand DNA break by the process of nonhomologous end joining?

• The original DNA sequence at the site of repair is reconstituted with 100% accuracy.
• The DNA sequence at the site of repair matches that of a homologous chromosome.
• The DNA sequence at the site of repair contains a short segment of telomere DNA.
• The DNA sequence at the site of repair is altered by a short addition.
• The DNA sequence at the site of repair is altered by a short deletion.

Answer 43

The DNA sequence at the site of repair is altered by a short deletion.

(During nonhomologous end joining, enzymes “clean” the broken DNA ends before joining them back together, a process that can result in a loss of nucleotides at the site of repair.)

Question 44

Determine whether the following statement is true or false and why:

Homologous recombination occurs only between DNA molecules that are identical in nucleotide sequence without any variation.

• false, because crossing-over does not happen
• false, because only similar sequence is needed
• true, for all eukaryotic diploid cells
• true, but only for cells that have more than 10 chromosomes

Answer 44

False, because only similar sequence is needed.

(Homologous recombination occurs between DNA molecules that are similar in nucleotide sequence. Homologous recombination requires extensive regions of matching sequence, but the DNA molecules need not be entirely identical.)

Question 45

Determine whether the following statement is true or false:

Homologous recombination occurs only in eukaryotes.

Answer 45

False.

(The proteins that carry out homologous recombination have been conserved in virtually all cells on Earth. Both bacterial cells and eukaryotic cells can undergo homologous recombination.)

Question 46

When does homologous recombination most likely occur in order to flawlessly repair double-stranded DNA breaks?

• when the damage is overlooked by the fast-acting mismatch repair system
• after the cell’s DNA has been replicated
• before DNA is replicated, to avoid propagating the error
• whenever a double-strand break is detected
• after the broken ends have been “cleaned” by a nuclease

Answer 46

After the cell’s DNA has been replicated.

(Homologous recombination most often occurs shortly after DNA replication, when the duplicated chromosomes are still physically close to one another.)

Question 47

What problem with replication of linear chromosomes does telomerase address?

• The lagging strand continues farther during replication, so chromosomes would lengthen in each replication cycle without telomerase.
• The leading strand stops short of the 5’ end during replication, so chromosomes would shorten in each replication cycle without telomerase.
• The leading strand continues farther during replication, so chromosomes would lengthen in each replication cycle without telomerase.
• The lagging strand stops short of the 3’ end during replication, so chromosomes would shorten in each replication cycle without telomerase.

Answer 47

The lagging strand stops short of the 3’ end during replication, so chromosomes would shorten in each replication cycle without telomerase.

(The end-replication problem exists because lagging-strand synthesis does not reach the 3’ end of the chromosome. Telomerase ensures chromosome length is maintained.)

Question 48

Telomerase was first described in the ciliate Tetrahymena thermophila by Elizabeth Blackburn and her student Carol Greider. They, along with Jack Szostak, subsequently won a Nobel Prize for this discovery. As the animation shows, the template RNA sequence in Tetrahymena is 3’-ACCCCAAC-5’. The telomerase protein and RNA template together extend the 3’ end of the chromosome by adding 5’-GGGTTG-3’ repeats to the chromosome. The complementary strand is then synthesized by DNA polymerase α.

Blackburn’s lab altered the sequence of the telomerase RNA. Predict what might happen to telomeres in Tetrahymena cells if the Tetrahymena template RNA were mutated to 3’-ACCCCGAC-5’.

• Telomeres would become shorter every generation compared to normal cells.
• There would be no changes in telomeres compared to normal cells.
• Telomere sequence would be altered to 5’-GGGCTG-3’.
• Telomere sequence would be altered to 5’-GTCGGG-3’.

Answer 48

Telomere sequence would be altered to 5’-GGGCTG-3’.

(If the telomerase RNA sequence changes in the template section of the RNA, the DNA sequence will also change to remain complementary to the template RNA sequence. A cytosine (C) will be added to the DNA to base-pair with the G in the template RNA.)

Question 49

Predict what might happen to telomeres in Tetrahymena cells if the Tetrahymena template RNA were mutated from 3’-ACCCCAAC-5’ to 3’-AGCCCAAC-5’.

• Telomeres would become shorter in every generation compared to normal cells.
• There would be no changes in telomeres compared to normal cells.
• Telomere sequence would be altered to 5’-GGGTTC-3’.
• Telomere sequence would be altered to 5’-CTTGGG-3’.

Answer 49

Telomeres would become shorter in every generation compared to normal cells.

(If the telomere RNA sequence in this region changes, the telomere RNA will no longer base-pair with the 3’ end of the chromosome, and telomere repeats will not be added. The chromosome telomeres will shorten each generation.)

Question 50

One approach to determining the function of a protein is to inhibit protein function by mutating the gene coding for the protein and observing the resulting phenotype. Cleves et al. sought to uncover the role of fibroblast growth factor 1a (FGF1a) in the coral Acropora millepora by mutating the FGF1a gene via CRISPR technology (Cleves et al., 2018 CRISPR/Cas9-mediated genome editing in a reef-building coral PNAS 115 (20) 5235-5240). The CRISPR process uses the Cas9 endonuclease to cut DNA at a site targeted by a guide RNA (see image). What happens next to induce gene mutations?

• mismatch repair
• telomerase upregulation
• homologous recombination
• nonhomologous end joining

Answer 50

Nonhomologous end joining.

(During nonhomologous end joining of double-strand DNA breaks, the broken ends are processed, often resulting in nucleotide losses.)