A mutation completely eliminates primase activity in a bacterium. What is the most direct consequence for DNA replication?
AReplication stalls at the origin because helicase cannot unwind DNA without primase assistance
BDNA polymerase III cannot initiate synthesis on either strand, because it requires a free 3'-OH end that only primase can provide
COnly the lagging strand fails to replicate; the leading strand continues unaffected
DOkazaki fragments form normally but cannot be joined because ligase has no substrate
DNA polymerase III can only add nucleotides to an existing 3'-OH group — it cannot start a new chain from scratch. Primase synthesizes the short RNA primers that provide that 3'-OH end. Without primase, no primer can be laid down, so Pol III cannot begin synthesis on either strand (both strands need at least one primer at the origin). Option 2 is the classic misconception: students assume only the lagging strand is affected because it uses multiple primers, but the leading strand also requires an initial primer.
Question 2 Multiple Choice
Why is the lagging strand synthesized as discontinuous Okazaki fragments rather than as one continuous strand?
AThe lagging strand template has more secondary structure, forcing repeated re-initiation
BDNA polymerase can only synthesize in the 5'→3' direction, and the lagging strand template runs antiparallel to the direction of fork movement, so synthesis must proceed in short bursts away from the fork
CHelicase opens the DNA in short segments rather than continuously, limiting how far polymerase can extend
DDNA ligase can only join fragments below a certain length, so the cell produces short fragments to accommodate it
This is the central asymmetry in DNA replication. Both strands must be copied, but DNA polymerase can only add nucleotides 5'→3'. On the leading strand, this direction coincides with fork movement, so synthesis is continuous. On the lagging strand, it does not — synthesis would have to run backward relative to fork movement. Instead, as helicase exposes new template, primase lays down a fresh primer and Pol III synthesizes a new Okazaki fragment (5'→3') away from the fork. The fragments are later joined by Pol I (primer removal) and ligase (nick sealing).
Question 3 True / False
DNA polymerase III synthesizes both the leading and lagging strands continuously in the 5'→3' direction.
TTrue
FFalse
Answer: False
DNA polymerase III synthesizes continuously only on the leading strand. On the lagging strand, synthesis is discontinuous: Pol III repeatedly reinitiates at new RNA primers to produce short Okazaki fragments, each synthesized 5'→3' but in the direction opposite to overall fork movement. The fragments are later processed by Pol I (which removes primers and fills gaps) and sealed by DNA ligase. Pol III works on both strands, but its synthesis on the lagging strand is inherently fragmented.
Question 4 True / False
RNA primers used in DNA replication must ultimately be removed and replaced with DNA, because leaving RNA residues in the daughter strand would compromise genomic stability.
TTrue
FFalse
Answer: True
RNA primers serve only as starting points — they are temporary scaffolding. They must be replaced with DNA because RNA-DNA hybrid regions are less stable and because the sugar in ribonucleotides (2'-OH) makes RNA more susceptible to hydrolysis than DNA. In bacteria, DNA polymerase I uses its 5'→3' exonuclease activity to remove primers while simultaneously filling in with DNA. DNA ligase then seals the final nick. Leaving RNA in place would create structurally weak points and replication errors in subsequent cycles.
Question 5 Short Answer
Why does DNA replication require RNA primers, and what fundamental constraint on DNA polymerase does this requirement reveal?
Think about your answer, then reveal below.
Model answer: RNA primers are required because DNA polymerase cannot initiate a new polynucleotide chain from scratch — it can only extend an existing chain by adding nucleotides to a free 3'-OH group. Primase, an RNA polymerase, does not have this constraint and can begin a chain de novo, providing the 3'-OH that DNA polymerase needs. This constraint reveals that DNA polymerase is a highly specialized enzyme optimized for accurate extension rather than initiation, and the cell evolved a separate enzyme (primase) to handle the initiation problem.
The primase solution is elegant but creates a downstream problem: RNA must be removed. This is why the replication machinery includes DNA polymerase I (for primer removal and gap filling) and DNA ligase (for nick sealing). The entire Okazaki fragment lifecycle on the lagging strand — primer synthesis, Pol III extension, Pol I primer removal and gap filling, ligase sealing — exists because DNA polymerase cannot start chains. One enzymatic constraint ripples into a cascade of additional enzymatic requirements.