Question 1

What do the enzymes topoisomerase I and topoisomerase II have in common?

Accepted Answer

Unlike topoisomerase I, which can only relieve the tension in DNA through introducing a nick (single-strand break) in one of the DNA strands, topoisomerase II can actively introduce or relieve tension by creating double-strand breaks (that remain tightly associated with the enzyme) using energy from ATP hydrolysis.

Question 2

You have found a strain of Escherichia coli that has an unusually short doubling time of only 15 minutes, despite the fact that its complete DNA replication should take almost 35 minutes. You also find that there is only one replication origin on its chromosome from which two forks originate, just like the normal process described for E. coli. However, you discover that the origin of replication in this strain has a significantly shorter "refractory period," resulting in the reactivation of the origin before the previous round of replication is over. Based on this model, if you examine the chromosomes of this strain (under conditions of fast growth), how many replication forks would you expect to observe per chromosome on average?

Accepted Answer

Even though each round of replication takes 35 minutes, the "firing" interval can be adjusted independently, and even go out of control. In this strain, possibly due to mutations in the methylation pathway, the origins can become activated every 15 minutes, resulting in an average of six forks (three bubbles) per chromosome, the newest of which have started about 30 minutes after the oldest ones.

Question 3

The Dam methylase is responsible for methylating the adenine base in GATC sequences in Escherichia coli. Imagine two

Accepted Answer

An inactive or hyperactive Dam methylase can interfere with strand-directed mismatch repair, as both make it harder to distinguish between the newly replicated and old DNA strands. Both situations are expected to increase the overall mutation rate, giving rise to a mutator phenotype.

Question 4

This protein is present at every replication fork and prevents DNA polymerase from dissociating, but does not impede the rapid movement of the enzyme. Which of the following is true regarding this protein?

Accepted Answer

The sliding clamp is normally loaded (more frequently on the lagging strand) on the primer-template duplex by the clamp loader, a process that requires ATP hydrolysis by the loader. As soon as the polymerase runs into a double-strand region downstream, the clamp releases the polymerase but is not itself immediately disassembled from DNA.

Question 5

At the replication fork, the template for the lagging strand is thought to loop around. This looping would allow the lagging-strand polymerase to move along with the rest of the replication fork instead of in the opposite direction. The single-strand part of the loop is bound by the single-strand DNA-binding (SSB) proteins. As each Okazaki fragment is synthesized toward completion, how does the size of the loop change? What about the size of the SSB-bound part of the loop?

Accepted Answer

The double-strand region of the loop represents the growing Okazaki fragment and gradually increases in size. Additionally, as each Okazaki fragment is growing, the fork is also traveling, exposing more single-stranded DNA in the loop. The loop size resets to zero upon initiating the synthesis of the next Okazaki fragment.

Question 6

The telomerase enzyme in human cells …

Accepted Answer

Telomerase uses its RNA component as a template to polymerize DNA sequences at the chromosome ends to solve the "end-replication" problem.

Question 7

If this protein complex does not function normally, the ends of the eukaryotic chromosomes would activate the cell's DNA damage response, causing chromosomal fusions and other genomic anomalies. What is this protein complex called?

Accepted Answer

With the help of the t-loop structure, shelterin "hides" the telomere ends from the cell's double-strand break detector and repair systems.

Question 8

Fill in the gap in the following paragraph using what you know about the activities of the proteins involved in DNA replication.
"Mitochondrial DNA replication requires a set of proteins similar to those used for the replication of the nuclear genome. However, mitochondria lack a dedicated DNA ... and use the mitochondrial RNA polymerase instead."

Accepted Answer

Like the RNA polymerases involved in transcription, primases do not require priming and can synthesize RNA molecules that can be used as primers for DNA synthesis by the replicative DNA polymerases. RNA polymerase is used to create RNA primers for mitochondrial DNA replication.

Question 9

In Escherichia coli, replication of DNA can occur throughout the cell cycle while the cell is also actively transcribing its genes. This means collisions between replication forks and RNA polymerases are inevitable. Depending on the orientation of the genes, collisions can be rear-end (when both machines are traveling in the same direction) or head-on (when they are traveling in opposite directions). In the following paragraph, match each of the letters (A to D) to one appropriate number below. Do not use a number more than once. Your answer would be a four-digit number composed of digits 1 to 5 only, e.g. 1253.
"Typically, in a rear-end collision, the (A) of RNA polymerase collides with the (B) in the replication fork. In contrast, in a head-on collision, the (C) of RNA polymerase hits the (D) in the fork."
1. front edge (of RNA polymerase)
2. rear edge (of RNA polymerase)
3. DNA helicase
4. leading-strand DNA polymerase
5. lagging-strand DNA polymerase

Accepted Answer

The replication fork in E. coli can progress almost 20 times faster than the RNA polymerase, meaning that rear-end collisions are typically between the rear edge of the RNA polymerase and the front edge of the leading-strand DNA polymerase that is traveling (faster) in the same direction on the same template strand. In a head-on collision, however, the front edge of RNA polymerase hits the DNA helicase that is traveling on the same (lagging) strand in the opposite direction. Note that the polymerases translocate with respect to their template in the 3?-to-5? direction, while the helicase in the fork translocates in the 5?-to-3? direction.

Question 10

On average, errors occur in DNA synthesis only once in every ten billion nucleotides incorporated. Which of the following does NOT contribute to this high fidelity of DNA synthesis?

Accepted Answer

Base-pairing is the basis of all DNA replication and repair. The other three mechanisms each increase accuracy by about 100-fold.

Question 11

In the following schematic drawing, two DNA molecules are shown before and after the action of a protein that is also involved in the process of DNA replication. What is this protein called?

Accepted Answer

This untangling of intertwined DNA is mediated by the enzyme topoisomerase II. This type of reaction helps solve the winding problem during replication.

Question 12

During DNA replication, the single-strand DNA-binding (SSB) proteins …

Accepted Answer

The SSB proteins bind cooperatively to the regions of exposed single-stranded DNA (which are routinely found in the lagging strand) and prevent the formation of hairpin helices without blocking the base-pairing potential.

Question 13

The mutation rate in bacteria is about 3 nucleotide changes per 10 billion nucleotides per cell generation. Under laboratory conditions, bacteria such as Escherichia coli can divide and double in number about every 40 minutes. If a single Escherichia coli cell is allowed to exponentially divide for 10 hours in this manner, how many mutations would you expect to observe on average in the genome (4.5 million nucleotide pairs) of each of the resulting bacteria compared to the original cell? Assume all mutations are neutral; that is, they do not affect the cell-division time.

Accepted Answer

About 15 cell generations pass in 10 hours of growth. Thus, the expected average number of mutations is equal to:
(15 generations) × (3 × 10-¹⁰ mutations/generation/nucleotide) × (4.5 × 10⁶ nucleotides/genome)
= ~2 × 10-² mutations per genome.
This is equivalent to about 98% of the bacterial cells having zero mutations and the remaining 2% having only one mutation each. Please note that the number of mutations in different cells in a given culture will vary widely around the average calculated above.

Question 14

Which of the following features is common between the replication origins in Escherichia coli and Saccharomyces cerevisiae?

Accepted Answer

The yeast origins of replication are similar to the bacterial origins in that they are A-T-rich and have particular sequences that attract replication initiator proteins. In higher eukaryotes, in contrast, the determinants of the replication origins are probably less sequence-specific. E. coli and S. cerevisiae both have relatively short origins compared to higher organisms.

Question 15

DNA ligases are used in both DNA replication and repair to seal breaks in the DNA. But DNA damage can result in single- or double-strand breaks that are not normal ligase substrates. These need to be processed first before a ligase can act on them. One of the enzymes that is recruited to some of such breaks is called PNK. It has two separate activities on the DNA, both of which can help provide a canonical ligase substrate. Which of the following activities would you expect PNK to have in this context?

Accepted Answer

The DNA ligase reaction normally requires a free 3′-OH group and a 5′-phosphate group on the two DNA ends to proceed. If a break has created 3′-phosphate and 5′-OH ends instead, for example, an enzyme such as PNK can dephosphorylate the 3′ end in the upstream molecule, and phosphorylate the 5′ end in the downstream molecule, to create the canonical ends.

Question 16

What is the main source of the free energy for the mechanical work performed by DNA helicases during DNA replication in our cells?

Accepted Answer

The helicase at the replication fork binds and hydrolyses ATP to move on the DNA in a stepwise fashion.

Question 17

The nuclear DNA polymerases in human cells …

Accepted Answer

Eukaryotic DNA polymerases are not as fast as their bacterial counterparts, presumably due to the difficulty in replicating through nucleosomes. All known DNA polymerases polymerize DNA from 5′ to 3′ and require a free 3′ end provided by a primer. The self-correcting polymerases have an additional 3′-to-5′ exonuclease activity that takes place at a distinct site in the enzyme.

Question 18

Which of the following is correct regarding the mutation rate of genomic DNA in different organisms?

Accepted Answer

The mutation rate per nucleotide pair per replication cycle in humans is similar to that of Escherichia coli, and both are extremely low, less than one in a billion. Nevertheless, current mutation rates are thought to limit the number of essential genes in an organism to about 30,000. The stability of the germ-line genome depends on mutation rates, as does somatic-cell stability. Cancers typically arise from somatic mutations.

Question 19

During DNA replication in the cell, DNA primase makes short primers that are then extended by the replicative DNA polymerases. These primers …

Accepted Answer

The DNA primase enzyme syntheses RNA primers (with free 3′-OH groups) that are extended and later replaced by DNA that is more accurately synthesized. This happens more frequently in the lagging strand.

Question 20

Which of the following schematic drawings better depicts the end of mammalian chromosomal DNA?

Accepted Answer

The telomeres have 3′ overhangs and are folded in structures called t-loops, which help protect the chromosome ends.

Quiz 5: DNA Replication, Repair, and Recombination