What is one major difference between bacterial and eukaryotic DNA replication?

Answer: Bacteria have a single origin per circular chromosome; eukaryotes have many origins per linear chromosome to finish replication on time. Because eukaryotic chromosomes are large and linear, replication is initiated from many origins simultaneously to complete S phase in a reasonable time. Bacterial circular chromosomes typically use a single origin (oriC).

Helicase functions during replication by

Answer: Unwinding the DNA double helix at the replication fork Helicase breaks H-bonds between base pairs, separating strands so each can serve as a template.

DNA ligase functions to

Answer: Seal the gap between Okazaki fragments by forming phosphodiester bonds Ligase joins the 3'-OH and 5'-phosphate ends of adjacent DNA fragments, completing the lagging strand.

AP Biology · Topic 6.2

Replication Practice

Part of Gene Expression and Regulation.(IST-1.J)

Practice questions

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Sample questions

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Sample 1difficulty 2/5
$What is one major difference between bacterial and eukaryotic DNA replication?$
- A
  $Bacteria replicate DNA only in mitochondria.$
- B
  $Bacteria use RNA primers; eukaryotes do not.$
- C
  $Eukaryotes lack DNA polymerases entirely.$
- D
  $Bacteria have a single origin per circular chromosome; eukaryotes have many origins per linear chromosome to finish replication on time.$
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Why
$Because eukaryotic chromosomes are large and linear, replication is initiated from many origins simultaneously to complete S phase in a reasonable time. Bacterial circular chromosomes typically use a single origin (oriC).$
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Sample 2difficulty 2/5
$Helicase functions during replication by$
- A
  $Unwinding the DNA double helix at the replication fork$
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- B
  $Synthesizing short RNA primers to initiate strand synthesis$
- C
  $Adding new nucleotides to the 3' end of the growing strand$
- D
  $Joining Okazaki fragments through phosphodiester bonds$
Why
$Helicase breaks H-bonds between base pairs, separating strands so each can serve as a template.$
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Sample 3difficulty 2/5
$DNA ligase functions to$
- A
  $Unwind the DNA double helix ahead of the advancing replication fork$
- B
  $Seal the gap between Okazaki fragments by forming phosphodiester bonds$
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- C
  $Cut DNA strands at specific sequences to allow recombination events$
- D
  $Add short RNA primers to initiate synthesis of new DNA strands$
Why
$Ligase joins the 3'-OH and 5'-phosphate ends of adjacent DNA fragments, completing the lagging strand.$
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Sample 4difficulty 2/5
$DNA polymerase adds nucleotides to the$
- A
  $Either end at random$
- B
  $Middle of the strand$
- C
  $3' end of the new strand$
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- D
  $5' end of the new strand$
Why
$Synthesis is always 5'\to3'; nucleotides are added to the free 3'-OH of the growing strand.$
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Sample 5difficulty 2/5
$A replication fork is shown with a leading and lagging strand. Helicase unwinds DNA at the fork.$
$Why is the lagging strand synthesized in short Okazaki fragments?$
- A
  $The lagging strand uses a different polymerase that synthesizes 3' to 5'.$
- B
  $DNA polymerase only adds nucleotides 5' to 3', so the strand opposite to fork movement must be made discontinuously.$
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- C
  $RNA primers are not required on the lagging strand.$
- D
  $Helicase repeatedly stops on the lagging strand template.$
Why
$DNA polymerase synthesizes only in the 5' to 3' direction. On the lagging-strand template, this requires repeated short syntheses (Okazaki fragments) that are later joined by ligase.$
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