The plasmid replicates unidirectionally. Molecules (c) and (d) are inverted relative to (a) and (b). Molecule (a) identifies the position of the origin relative to one end. The observation that (b), (c), and (d) have one forked end of similar size and the other forked end differing in size reveals that a single replication fork moves first through the short arm of (a) and then proceeds around the circular plasmid. The order of replication time is (a), (b), (d), (c).

The primer is a preexisting strand of RNA or DNA to which new nucleotides are added at the 3′ end. The template is a longer strand, paired with the primer, that determines the identity of each new nucleotide added via base pairing.

(a) No. In the absence of any one dNTP, the polymerase would stop incorporating the other three dNTPs as soon as it encountered a template residue that should pair with the missing dNTP, and incorporation of ³²P would be undetectable. (b) No. DNA synthesis releases the β and γ phosphates of dNTPs as pyrophosphate.

Possible answers: Pol I is slow in DNA synthesis compared with the rate of replication in E. coli. Pol I can be mutated and the cells still survive. Pol I is not highly processive.

The DNA polymerase contains a 3′→5′ exonuclease that degrades DNA to produce [³²P]dNMPs. The activity is not a 5′→3′ exonuclease, because addition of dNTPs inhibits [³²P]dNMP production: the polymerase extends radioactive 3′ termini by adding nonradioactive dNTPs, protecting the radioactive portion of DNA from the 3′→5′ exonuclease. This would not protect the 5′ terminus of radioactive DNA from a 5′→3′ exonuclease. Adding pyrophosphate would result in production of [³²P]dNTPs through reversal of the polymerase reaction.

Ligase will not seal a nick in which the 5′-terminal nucleotide is a ribonucleotide. Sealing is delayed until all the RNA has been removed.

The DnaA protein, assisted by the protein HU, creates a bubble of unwound DNA at the origin and also targets DnaB protein to that location. The DnaC protein loads the DnaB helicase onto the DNA at the origin.

(a) Either any combination of three A sites is sufficient for origin function, or three particular A sites are required. Construct four plasmids, each with a different mutant A site. Transfer the mutant plasmids into the host organism, and plate each transformed product on a medium containing the appropriate antibiotic. Use an unmutated plasmid and a plasmid without A sites as controls. If a particular A site is essential, the mutant plasmid will not form a colony. (b) Either the B sites are not essential, or one B site is needed but either one suffices. Construct a plasmid containing mutations in both B sites. If a particular B site is essential, a colony will not appear after transformation. Use an unmutated plasmid and a plasmid without B sites as controls.

The 3′→5′ exonuclease is a proofreading function, eliminating mismatched nucleotides incorrectly inserted into the growing DNA strand. DNA polymerases IV and V function in translesion DNA synthesis during DNA repair, a situation where the accuracy of DNA polymerization becomes less important (as described in Chapter 12).

A number of proteins and complexes, including the origin recognition complex (ORC), Cdc6, Cdt1, and the replicative helicase Mcm2–7, are loaded onto origins of replication in G₁ phase.

The preRC forms only in G₁, not in other phases of the cell cycle. Cyclin kinases produced only in S phase are needed to assemble the remaining proteins to produce active replication forks. Origins do not fire a second time, because new preRC complexes cannot form until the cell completes its cycle and returns to G₁.

The τ subunits link together the leading- and lagging-strand core polymerases, one τ linked to each core, and both connected to DnaB. (a) Two. (b) Zero. The core polymerase, in conjunction with a β sliding clamp, is capable of processive synthesis of a new DNA strand on a single-stranded DNA template without any other subunits being present. This is analogous to leading-strand synthesis without lagging-strand synthesis. (c) Three. Two of the three τ subunits, and their accompanying polymerase core subunits, act on the lagging strand and may permit more efficient DNA synthesis on the lagging strand.

DnaA: ATP hydrolysis inactivates the DnaA for replication initiation. DnaC: ATP hydrolysis helps release DnaB helicase as it is loaded onto the DNA. Pol III γ or τ subunits: ATP hydrolysis allows the β subunit (sliding clamp) to close around the DNA as it is loaded.

The two replication forks would never meet, and part of the chromosome near the terminus would remain unreplicated.

The RNA template for telomere synthesis is telomerase RNA (RT), which is bound tightly to the telomerase reverse transcriptase (TERT) to form the active telomerase holoenzyme. A very short segment of the TR, equivalent to an organism’s telomere repeat sequence (typically 6 or 7 nucleotides), is used over and over again as a template in telomere DNA synthesis.

(a) As the DNA strands are linked together, the singly bonded phosphoryl groups are converted to phosphodiester bonds (doubly bonded phosphoryl groups) that are no longer susceptible to alkaline phosphatase. (b) The substrate for the reaction is a DNA strand break in one strand of double-stranded DNA. Ligation of single strands is not observed. (c) The reaction halts only because the enzyme runs out of substrate. Addition of poly(dA) creates more of the correct substrate and the reaction can continue. (d) The DNA ligase of E. coli uses NAD+ as cofactor rather than ATP.

S-12