9 Replication

1. Which mechanism ensures that histone modifications are preserved during DNA replication, allowing for epigenetic inheritance?

Telomerase activity
Base Excision Repair
Mismatch repair directed by the template strand
The addition of new histone proteins and passing of modifications after the fork passes

2. What is the typical time duration for the S phase in a eukaryotic cell cycle?

8−12 hours
24 hours
1 hour
Less than 30 minutes

3. The 3′→5′ exonuclease proofreading step improves the fidelity of DNA replication by a factor of:

1,000
100
10^9
10^5

4. DNA topoisomerases are critical during replication because they:

Unwind the DNA double helix at the replication fork.
Identify and remove uracil bases from the DNA.
Join Okazaki fragments together by creating phosphodiester bonds.
Prevent DNA from tangling as it is unwound.

5. Why is it difficult to replicate the very ends of eukaryotic chromosomes?

Because they lack an origin of replication.
Because the DNA is linear and the lagging strand cannot be primed at the extreme 3′ end.
Because they are circular and cause the fork to stall.
Because telomerase inhibits the standard DNA polymerase.

6. Which enzyme is primarily responsible for preventing the ‘tangling’ of DNA by relieving the tension caused by unwinding at the replication fork?

DNA ligase
DNA topoisomerase
Single-strand DNA-binding protein
DNA helicase

7. Which energy source is utilized by DNA polymerases to incorporate deoxyribonucleoside monophosphates (dNMPs) into the growing DNA strand?

ATP hydrolysis to ADP
The proton motive force
dNTP hydrolysis to dNMP and pyrophosphate
GTP hydrolysis

8. Which process is described as a ‘cut and paste’ mechanism that uses the correct strand of DNA as a template to replace damaged regions?

Excision repair
Epigenetic inheritance
Semiconservative replication
Tautomeric shifting

9. True or False: In prokaryotic organisms, replication proceeds from a single origin of replication using two replication forks.

False
True

10. What distinguishes the error correction of 3′→5′ exonucleolytic proofreading from mismatch repair?

Proofreading occurs after the entire DNA strand is finished.
Mismatch repair uses a glycosylase, while proofreading uses a nuclease.
Proofreading removes an incorrectly incorporated base before the DNA strand is extended.
Proofreading is only found in prokaryotes.

11. A cell is observed to have a DNA content of 4N but is not currently undergoing mitosis or cytokinesis. In which phase of the cell cycle is this cell most likely residing?

S phase
G2 phase
G1 phase
G0 phase

12. DNA synthesis occurs through the incorporation of dNMPs into the growing strand. What provides the energy for this endergonic process?

The release and hydrolysis of pyrophosphate from dNTPs.
The formation of hydrogen bonds between complementary bases.
The activity of DNA ligase in the replisome.
The hydrolysis of ATP by the DNA polymerase.

13. Which phase of the cell cycle is typically the longest and can vary significantly in length, potentially leading to a resting state?

G1 phase
G2 phase
M phase
S phase

14. During the synthesis of Okazaki fragments, which eukaryotic enzyme is responsible for the initial elongation of the RNA primer for approximately 20 to 30 nucleotides before displacement?

DNA polymerase ϵ
DNA polymerase α
DNA polymerase δ
DNA helicase

15. Why is DNA replication described as ‘discontinuous’?

Because DNA polymerase ϵ falls off the template every 100 nucleotides.
Because the cell cycle stops at several checkpoints during the S phase.
Because the lagging strand must be synthesized in short fragments due to the 5′→3′ constraint.
Because the replication forks stop whenever they collide with histones.

16. During eukaryotic DNA replication, which specific polymerase is primarily responsible for the continuous synthesis of the leading strand?

DNA polymerase δ
DNA polymerase γ
DNA polymerase α
DNA polymerase ϵ

17. The carcinogen benzo-pyrene, found in cigarettes, causes bulky lesions in DNA. Which repair pathway is primarily responsible for addressing this damage?

Base Excision Repair (BER)
Direct Reversal
Mismatch Repair
Nucleotide Excision Repair (NER)

18. If the error rate of 5′→3′ polymerization is 1 in 10^5 and the combined error rate of replication is 1 in 10^9, what is the specific contribution of strand-directed mismatch repair to this fidelity?

1 error per 10^9 nucleotides
1 error per 10^7 nucleotides
1 error per 10^5 nucleotides
1 error per 10^2 nucleotides

19. Which enzyme is specifically responsible for removing the RNA primer during the maturation of Okazaki fragments?

DNA polymerase α
RNase H
DNA primase
DNA ligase

20. Which enzyme is responsible for ‘straightening out’ the DNA template to facilitate the activity of DNA polymerase after the helix is unwound?

Single-strand DNA-binding (SSB) protein
DNA Topoisomerase
DNA Primase
DNA Helicase

21. During lagging strand synthesis, what is the role of RNAse H?

To remove the RNA primers from the Okazaki fragments.
To unwind the DNA ahead of the polymerase.
To stabilize the replication bubble.
To synthesize a short 10-base RNA sequence.

22. In the process of Base Excision Repair (BER), which enzyme is responsible for the specific removal of the sugar-phosphate backbone after the nitrogenous base has been excised?

DNA ligase
Excision nuclease
AP endonuclease
Uracil DNA glycosylase

23. A ‘tautomeric shift’ in a nitrogenous base (e.g., Cytosine to C∗) can lead to a replication error because:

It results in the immediate activation of Nucleotide Excision Repair.
The rare form of the base can form hydrogen bonds with an incorrect partner.
It causes the DNA backbone to break spontaneously.
It prevents the DNA helicase from unwinding that section of DNA.

24. If a bacterial replication fork moves at a rate of 1000 nucleotides per second, and the error rate is approximately 1 in 10^9 nucleotides, how many errors would be expected on average after 1,000,000 seconds of continuous replication?

1000
1
10
0.1

25. How do eukaryotic origins of replication differ from prokaryotic origins?

Eukaryotes require a large number of origins to replicate the complete genome in a reasonable time.
Prokaryotic replication forks never collide, whereas eukaryotic ones do.
Prokaryotes have many origins, while eukaryotes have only one.
Eukaryotic origins are activated all at once at the start of S phase.

26. Which of the following describes the unique mechanism by which Nucleotide Excision Repair (NER) differs from Base Excision Repair (BER)?

NER is primarily used for damage caused by X-rays.
NER removes a multi-nucleotide patch of DNA using an excision nuclease complex.
NER does not require DNA polymerase to fill the resulting gap.
NER uses Uracil DNA glycosylase to identify pyrimidine dimers.

27. The term ‘Replisome’ refers to:

A complex of 12 nucleotides removed during DNA repair.
The protein shell that protects DNA during the resting G0 phase.
The specific site on a chromosome where replication begins.
The complete assembly of enzymes and proteins that cooperate to replicate DNA.

28. The DNA content of a eukaryotic cell is measured to be 4N. In which phase of the cell cycle could this cell potentially be located?

G2 phase
S phase
G1 phase
G0 phase

29. Which of the following describes the activity of DNA primase?

It requires a pre-existing primer with a 3′ OH.
It has high proofreading activity to ensure primer accuracy.
It synthesizes DNA in the 3′→5′ direction.
It catalyzes the synthesis of short RNA molecules about 10 bases long.

30. Eukaryotic replication origins are activated in clusters. What is the approximate spacing between individual origins within one of these clusters?

10 to 20 base pairs
100 to 200 base pairs
30,000 to 250,000 base pairs
1,000 to 5,000 base pairs

31. In humans, why are multiple origins of replication necessary compared to bacteria?

Eukaryotic genomes are much larger and the replication rate is slower.
Eukaryotic DNA polymerases are significantly more error-prone.
Eukaryotic DNA is circular and requires multiple start points to avoid knots.
Eukaryotes do not have a G1 phase to prepare for replication.

32. Which of the following describes the correct order of enzyme action during the repair of a deaminated cytosine via Base Excision Repair (BER)?

RNAse H → DNA polymerase δ → DNA ligase
DNA helicase → Excision nuclease → DNA ligase
Uracil DNA glycosylase → AP endonuclease → DNA polymerase → DNA ligase
Photolyase → DNA helicase → DNA polymerase

33. A tautomeric shift in a base (e.g., C to C∗) during replication typically leads to:

The immediate stoppage of the DNA helicase.
The incorporation of an incorrect base followed by exonucleolytic proofreading.
The initiation of Nucleotide Excision Repair.
Permanent mutation that cannot be repaired.

34. The tandem repeat sequence found at the ends of human chromosomes to facilitate end-replication is:

GGCCTG
TATAAA
GGGTTA
AAACCC

35. How long does the DNA synthesis (S) phase typically last in a human cell?

8 to 12 hours
24 hours
20 to 30 minutes
1 hour

36. What is the primary factor that allows the DNA replication process to achieve a combined error rate of 1 in 10^9?

Strict reliance on DNA primase for accuracy
High speed of bacterial DNA polymerase
The combination of 5′→3′ polymerization, 3′→5′ proofreading, and mismatch repair
The use of RNA primers instead of DNA primers

37. What is the primary role of DNA ligase in the replication process?

To remove RNA primers from the lagging strand template.
To add new nucleotides to the 3′ end of the leading strand.
To seal the nick between adjacent Okazaki fragments.
To facilitate the binding of SSB proteins.

38. What is the primary role of Single-strand DNA-binding (SSB) proteins during the replication process?

Ligating the sugar-phosphate backbone between Okazaki fragments.
Straightening the DNA template and preventing re-annealing to facilitate polymerase activity.
Synthesizing RNA primers on the lagging strand.
Removing torsional stress caused by the unwinding of the helix.

39. Which of the following is a major cause of ‘bulky lesions’ in DNA that are typically repaired by Nucleotide Excision Repair?

Exposure to low-dose X-rays
Normal metabolic oxygen damage
Spontaneous deamination of cytosine
Benzo-pyrene from cigarette smoke

40. Which enzyme complex is responsible for removing a ‘patch’ of DNA (at least 12 nucleotides in length) to repair bulky lesions like pyrimidine dimers?

RNAse H
Excision nuclease
AP endonuclease
Uracil DNA glycosylase

41. True or False: In eukaryotes, the chromatin structure is dismantled during replication and only reforms once the cell enters the G2 phase.

True
False

42. DNA polymerase requires a 3′ OH group to initiate synthesis. Which enzyme provides this requirement de novo by joining two nucleoside triphosphates together?

DNA helicase
DNA primase
DNA polymerase ϵ
DNA ligase

43. During the termination of replication on a linear chromosome, which problem is addressed by the enzyme telomerase?

The inability to replicate the very ends of the lagging strand.
The conversion of the DNA back into a circular form.
The collision of replication forks coming from opposite directions.
The removal of the very first RNA primer on the leading strand.

44. True or False: DNA polymerase requires a free 5′ OH group on the primer to begin incorporation of dNMPs.

False
True

45. Which of the following is a key difference between the leading strand and the lagging strand?

The leading strand is synthesized continuously, while the lagging strand is synthesized discontinuously.
The leading strand is synthesized 3′→5′, while the lagging strand is 5′→3′
The leading strand synthesis is slower than the lagging strand synthesis.
The leading strand requires multiple RNA primers, while the lagging strand needs only one.

46. During proofreading, if a DNA polymerase incorporates an incorrect base, which enzymatic activity allows it to remove the mismatch before continuing?

Endonuclease activity
5′→3′ exonuclease activity
3′→5′ exonuclease activity
Helicase activity

47. What is the typical rate of nucleotide addition in bacteria?

10^9 nucleotides per second
Up to 1,000 nucleotides per second
100 to 200 nucleotides per second
50 nucleotides per minute

48. In eukaryotes, the telomere sequence consists of tandem repeats of which short DNA sequence?

GGGTTA
AATTGG
GGCCTT
CCCAAT

49. What is the error rate attributed specifically to the strand-directed mismatch repair mechanism?

1 error per 10^5 nucleotides
1 error per 10^12 nucleotides
1 error per 10^2 nucleotides
1 error per 10^9 nucleotides

50. In the eukaryotic cell cycle, the length of which phase is noted as being particularly variable between different cell types?

S phase
G2 phase
G1 phase
M phase

51. During the elongation of the lagging strand, what follows the synthesis of the RNA primer by DNA primase?

DNA helicase rewinds the DNA to protect the single-stranded region
DNA polymerase α extends the primer for 20−30 nucleotides
Immediate joining by DNA ligase
DNA polymerase ϵ takes over to finish the Okazaki fragment

52. The ‘Replication Bubble’ is formed by:

The binding of histones to newly synthesized DNA.
The accumulation of Okazaki fragments on the leading strand.
Two replication forks proceeding in opposite directions from an origin.
A single replication fork moving in one direction.

53. Which specific DNA damage scenario is correctly matched with its primary repair pathway?

Ultraviolet light-induced pyrimidine dimers → Base Excision Repair
Benzo-pyrene from cigarette smoke → Base Excision Repair
Cytosine deamination to Uracil → Base Excision Repair
X-ray induced damage → Nucleotide Excision Repair

54. What happens to the DNA content of a cell during the M (Mitosis) phase?

It remains constant at 2N.
It remains constant at 4N.
It increases from 2N to 4N.
It decreases from 4N to 2N.

55. Which enzyme acts as a subunit of DNA polymerase α and initiates the lagging strand fragments?

DNA Primase
DNA Ligase
DNA Helicase
Exonuclease

56. What happens to the chromatin structure immediately following the passage of the replication fork?

The daughter strands are degraded if histone proteins are not immediately available.
The DNA remains permanently as 'naked' DNA to allow for easier transcription.
Telomerase converts the new DNA into histone-free telomeres.
It is re-formed by adding new histone proteins, allowing for epigenetic inheritance.