Questions: Primer Synthesis, Helicase, and Polymerase Function

Question 1

If primase is chemically inhibited in a bacterial cell after replication has already initiated on the leading strand, what is the most immediate consequence?

• A. All DNA synthesis halts immediately because DNA polymerase III loses processivity without primase
• B. Leading strand synthesis continues, but no new Okazaki fragments can be initiated on the lagging strand
• C. Helicase stops unwinding DNA because it is functionally coupled to primase activity
• D. DNA polymerase I compensates by synthesizing new RNA primers to replace primase

Answer: (B)

Primase synthesizes a new RNA primer for each Okazaki fragment on the lagging strand, so inhibiting primase prevents new fragments from initiating. However, the leading strand only needed one primer to get started — DNA polymerase III can continue extending that existing 3'-OH continuously. This asymmetry is a direct consequence of the mechanistic difference between leading and lagging strand synthesis. Option D is a misconception: DNA polymerase I removes and replaces RNA primers with DNA but cannot synthesize new RNA primers.

Question 2

What specific chemical limitation of DNA polymerase III necessitates the existence of primase?

• A. DNA polymerase III cannot read a single-stranded DNA template; it requires the helicase to remain bound
• B. DNA polymerase III can only add nucleotides to an existing 3'-OH group and cannot start a new chain from scratch
• C. DNA polymerase III synthesizes DNA in the 3' to 5' direction, which requires a pre-formed 5' end
• D. DNA polymerase III lacks proofreading activity and would introduce too many errors on a new chain

Answer: (B)

DNA polymerase requires a free 3'-hydroxyl group on an existing strand before it can add the next nucleotide. It cannot catalyze the formation of the first phosphodiester bond between two free nucleotides. Primase (an RNA polymerase) does not share this limitation — it can start a new chain de novo. The short RNA primer it synthesizes provides the 3'-OH that DNA polymerase needs to take over. This is the fundamental reason the cell needs a completely separate enzyme just to initiate synthesis. Option D is wrong: DNA polymerase III actually does have a 3'→5' proofreading exonuclease, but that is unrelated to the initiation problem.

Question 3

RNA primers synthesized by primase are eventually replaced with DNA, so the final replicated chromosome contains no RNA.

• T. True
• F. False

Answer: True

True. After the replication fork passes, DNA polymerase I uses its 5'→3' exonuclease activity to remove RNA primer nucleotides one at a time and replace them with DNA, using the adjacent Okazaki fragment as a primer itself. DNA ligase then seals the remaining nick. The final double-stranded DNA product is composed entirely of deoxyribonucleotides — the RNA primers are transient scaffolding that must be removed.

Question 4

The sliding clamp (β-clamp) allows DNA polymerase III to begin synthesizing a new strand without a primer by anchoring the polymerase directly to single-stranded DNA.

• T. True
• F. False

Answer: False

False. The sliding clamp increases DNA polymerase's *processivity* — it encircles the double-stranded DNA and tethers the polymerase so it stays attached, adding thousands of nucleotides without falling off. But it does nothing to solve the initiation problem. A primer providing a free 3'-OH is still absolutely required before the sliding clamp (or DNA polymerase) can function. The clamp is loaded onto the DNA at the primer-template junction by a separate clamp loader complex.

Question 5

Why is it a fundamental limitation that DNA polymerase cannot begin synthesis de novo, and how does the cell solve this problem?

Model answer: DNA polymerase can only extend an existing chain because its active site requires a properly base-paired 3'-OH to position the incoming nucleotide for catalysis. Without a pre-existing strand end, there is no geometric template for the first bond. The cell solves this by using primase — an RNA polymerase that does not share this requirement — to synthesize a short RNA primer complementary to the template strand. This primer provides the 3'-OH that DNA polymerase needs. The cost is that RNA must later be removed and replaced, adding complexity to lagging strand synthesis.

This question targets the core mechanistic insight: the initiation constraint is not a quirk but a direct consequence of the chemistry of polymerization. Understanding why the constraint exists — not just that it exists — explains the entire primase-primer system and makes the lagging strand architecture legible. Students who know 'DNA pol needs a primer' without knowing why will be lost when asked about the end-replication problem or replication fidelity.