DNA Replication Questions and Answers

Answer: D

Correct answer: Ligase

DNA ligase seals a nick by forming the missing phosphodiester bond between an adjacent 3’-OH and 5’-phosphate. In replication that means joining the Okazaki fragments into a continuous lagging strand. Polymerase extends an existing chain but cannot close a nick; primase makes the RNA primers; an endonuclease cuts rather than joins.

Answer: A

Correct answer: DNA was responsible for heredity

Hershey and Chase labelled phage protein with ³⁵S and phage DNA with ³²P, then let the phages infect bacteria and sheared off what remained outside. The ³²P entered the cells while the ³⁵S stayed outside, so it was the DNA, not the protein coat, that carried the genetic instructions. The A:T ratio result belongs to Chargaff.

Answer: D

Correct answer: semi-conservative and semi-discontinuous

Two separate descriptions. Semi-conservative: each daughter duplex keeps one parental strand and one new one, as Meselson and Stahl showed. Semi-discontinuous: because synthesis is 5’→3’ only, one strand is made continuously (the leading strand) while the other is made in short Okazaki fragments (the lagging strand). Only one of the two is discontinuous, which is why it is semi-discontinuous.

Answer: A

Correct answer: DNA in 5’-3’ direction

Every known DNA polymerase adds nucleotides only to a free 3’-OH, so the new strand grows 5’→3’. The reason is chemical: the energy for the bond comes from the incoming nucleotide’s own triphosphate, so the growing end must be the one that attacks it. This one constraint is what forces the lagging strand to be made in Okazaki fragments.

Answer: B

Correct answer: semi-conservative replication

Meselson and Stahl grew E. coli in ¹⁵N, shifted it to ¹⁴N, and separated the DNA by density. After one generation all the DNA was of intermediate density - which no conservative mechanism could produce - and after two it was half intermediate and half light, which the dispersive model could not produce. Only semi-conservative replication fits both results.

Answer: D

Correct answer: synthesize an RNA primer to begin the elongation process

DNA polymerase can extend a chain but cannot start one, since it needs a free 3’-OH to attack. Primase supplies that by laying down a short RNA primer - once on the leading strand, and once for every Okazaki fragment. The primers are later removed and replaced with DNA.

Answer: A

Correct answer: slowly, so they appear nearer the top of a gel photograph

The gel is a mesh, and a larger fragment threads through it more slowly. Since the samples are loaded at the top and run towards the anode below, the large fragments are left behind near the wells at the top, while small ones run furthest down. That is why a size ladder reads large-to-small from top to bottom.

Answer: A

Correct answer: Polymerase I

Pol I alone has a 5’→3’ exonuclease, and that is what lets it chew away the RNA primer ahead of itself while its polymerase activity fills the gap behind - the process called nick translation. Pol III, which does the bulk of synthesis, has no such activity.

Answer: D

Correct answer: Tus protein

Replication in E. coli ends at the ter sites, where the Tus protein binds and acts as a one-way barrier: it blocks the approaching replication fork by arresting the DnaB helicase, but only from one direction. The two forks travelling round the circular chromosome are therefore trapped in the terminus region and meet there.

Answer: A

Correct answer: It can synthesize DNA in the 5’ to 3’ direction

DNA polymerase adds each nucleotide to a free 3’-OH, so the strand can only grow 5’→3’. It makes DNA, not mRNA - that is the job of RNA polymerase, which also works 5’→3’.