BIO 218 Chapter 10 Quiz

Ribonucleases H1 and FEN-1 are responsible for removing primers in eukaryotes. What enzyme removes the primers in prokaryotes? a. RNA pol I b. DNA pol α c. DNA Pol I d. Replication protein A e. DNA pol III f. RNA pol III

c. DNA Pol I

In the eukaryotic replication fork, discontinuous synthesis of the lagging strand is carried out by which DNA polymerase? a. DNA polymerase α (alpha) b. DNA polymerase β (beta) c. DNA polymerase δ (delta) d. DNA polymerase ε (epsilon) e. DNA polymerase γ (gamma)

c. DNA polymerase δ (delta)

Eukaryotic organisms need to have DNA Polymerase α at their replication fork because DNA polymerase δ and DNA polymerase ε don't have 5' to 3' exonuclease activity. True False

False It is true that DNA polymerase δ and ε do not have 5' to 3' exonuclease activity, however, the function of DNA polymerase α is to extend the primer during replication, not to remove it.

The leading strand in DNA replication requires more DNA pol I than the lagging strand. True False

False The lagging strand needs more DNA polymerase I because it needed more primers to sytnthesize the Okazaki fragments. Therefore it needs more DNA polymerase I to remove those RNA primers and fill in the gaps.

Proofreading by DNA Polymerase requires _______________ a. 3' to 5' exonuclease activity b. 3' to 5' endonuclease activity c. 5' to 3' exonuclease activity d. 5' to 3' polymerase activity e. 5' to 3' endonuclease activity f. 3' to 5' polymerase activity

a. 3' to 5' exonuclease activity Exonuclease activity is the excision of a nucleotide base for either degradation of the DNA strand or to replace the excised base with a proper base (proofreading). Therefore, exonuclease activity is, for all intents and puroposes, proofreading. Therefore in order to have proofreading activity, an enzyme must have the 3' - 5' exonuclease activity. Remember, too, that DNA is synthesized in the 5' - 3' direction, therefore the opposite direction is reserved for proofreading.

Which of the following statements is NOT true about DNA primase? a. It is an RNA polymerase. b. DNA primase does not require a DNA template. c. It generates 10-60 bp primers in prokaryotes, but only 10 bp primers in eukaryotes. d. It can synthesize the segments that serve as primer "de novo". e. More DNA primase is required on the lagging strand.

b. DNA primase does not require a DNA template. DNA primase does not require a previously existing primer to synthesize its RNA primer, so no requirement for a 3'OH end. However, it DOES still require template the know what ribonucleotides to add.

You grow a bacterial culture in a media containing N15 and transfer it to a media containing N14. After two rounds of replication, you perform a CsCl density gradient centrifugation of the DNA. How many bands will you observe and what will be their intensity? a. Two, with the lower one more intense than the upper band b. Two, equally intense bands c. Three, with the middle one more intense than the upper and lower d. Three, equally intense bands e. Two, with the upper one more intense than the lower band f. One, very intense band

b. Two, equally intense bands In semi-conservative replication initially, the DNA was N15-N15. After 1st round of replication, both strands were N14-N15, making only a single band as they have the same density. In the next round, we will get N14N14, and N14-N15 making two bands that are equally intense for having 1:1 concentration.

In an experiment, you generate a DNA pol III mutant that does NOT have its β subunit (sliding clamp). When all other requisites for replication are added, what will be the effect of this mutation on replication? Consider the leading strand only. a. DNA replication will be faster. b. DNA replication will stop. c. DNA replication will be slower. d. No difference in DNA replication. e. DNA replication will be error-prone.

c. DNA replication will be slower. As a critical component of the DNA polymerase III holoenzyme, the β clamp binds DNA polymerase and prevents polymerase from dissociating from the template DNA strand. Because one of the rate-limiting steps in the DNA synthesis reaction is the association of the polymerase with the DNA template, the presence of the β sliding clamp dramatically increases the number of nucleotides that the polymerase can add to the growing strand per association event. The presence of the β clamp can increase the rate of DNA synthesis up to 1,000-fold compared with DNA pol III without the β clamp.

Which of the following statements is TRUE about telomerase: a. It polymerizes DNA without a template. b. It extends telomeres using its RNA polymerase activity. c. It has a "built in" RNA template to direct DNA Synthesis. d. It removes and replaces DNA at the ends of chromosomes. e. It creates single stranded breaks in the DNA to aid in replication.

c. It has a "built in" RNA template to direct DNA Synthesis. Telomerase is an enzyme that extends the telomeres of chromosomes. It is an RNA-dependent DNA polymerase, meaning an enzyme that can make DNA using RNA as a template.

During replication, nucleosome duplication occurs via a ____________________method where nucleosomes have ___________________ histones transported to the nucleus by ________________. a. Semi-conservative; recycled; CAF-1 b. Semi-conservative; recycled; Nap-1 c. Conservative; new; Nap-1 d. Dispersive; new and old; Nap-1 e. Dispersive; new and old; CAF-1 f. Conservative; new; CAF-1

d. Dispersive; new and old; Nap-1 During replication, when the replication fork reaches near a nucleosome the nucleosome will split into two, half-nucleosomes. Each half will remain on a strand of DNA as the replication fork moves passed the complexes. Once the DNA is duplicated nucleosomes reassemble from old, half-nucleasomes and newly synthesized histones. Therefore nucleosome duplication is dispersive with each of the "newly" formed nucleasomes containing a portion of the original nucleosome. New histones are transported by Nap-1 (nucleosome assembly protein-1) to the nucleus where CAF-1 (chromatin assembly factor-1) delivers them to the site of replication.

In E. coli, formation of the replication bubble begins when the _______ protein binds to the four _______ bp repeats of the oriC. After additional molecules of that protein form a complex, strand separation starts at the ____-rich sites where DnaB, also known as _______, and DnaC assist in forming the replication forks. a. DnaA; 13 bp; AT; DNA helicase b. DnaC; 9 bp; GC; DNA telomerase c. DnaD; 13 bp; GC; DNA helicase d. DnaA; 9 bp; AT; DNA helicase e. DnaA; 9 bp; AT; DNA Ligase

d. DnaA; 9 bp; AT; DNA helicase To complete this sentence one must understand bacterial replication. The first step in prokaryotic replication occurs when DnaA binds to the 9 bp repeats after the OriC (the single origin of replication on a bacterial genome). DnaB (or DNA helicase) then starts unwinding the DNA at A-T rich sites, due to their double hydrogen bonds versus G-C triple bonds. This is the 13 bp consensus sequence upstream of the 9 bp repeats.

Which of the following eukaryotic proteins is similar in function to the ß-subunit seen in the prokaryotic replicase? a. Rp-A b. DNA Polymerase α (alpha) c. Rf-C d. PCNA e. Nap-1

d. PCNA Proliferating Cell Nuclear Antigen (PCNA) is an accessory protein to DNA polymerase delta δ that helps with processivity. Processivity is the measure at which an enzyme can catalyze its function without having to release its substrate. In the case of replication PCNA allows DNA polymerase δ to continuously add more nucleotides before it needs to be refreshed by a new holoenzyme. In this respect it is like the β subunit seen in bacterial replicase.

Mapping by Schnos & Inman using "A:T denaturation bubbles" as physical markers was used to demonstrate that replication of the lambda phage genome occurs: a. in a 3' to 5' direction b. in a 5' to 3' direction c. dispersively d. bidirectionally e. semi-conservatively f. unidirectionally

d. bidirectionally Schnos and Inman used these rich A-T sequences to identify replication bubbles. These bubbles were then explored and found that replication of the DNA occurs in a bidirectional manner.

DNA ______________________ travels slightly ahead of the replication fork and alleviates coiling (overwinding) of the double helix that is caused by the action of DNA __________________ as it separates the DNA strands. a. Topoisomerase II; Helicase b. Helicase; Telomerase II c. Topoisomerase II; Topoisomerase I d. Topoisomerase I; Topoisomerase II e. Topoisomerase I; Helicase f. Helicase; Topoisomerase I

e. Topoisomerase I; Helicase Topoisomerase I is the topoisomerase that travels downstream of the DNA helicase, which is unwinding the DNA. As the DNA is unwound it creates dynamic stress and tension in the downstream portions. Topoisomerase cuts ONE time. The broken phosphodiester bond releases energy that is stored in topoisomerase 1. As the tension becomes relieved topoisomerase I will then re-connect the broken bond utilizing the same energy that was released from severing it in the first place. This obviates the need for ATP, therefore topoisomerase I is ATP-indendependent.

A replication bubble initiated at a replication origin in a cellular chromosome would have ___________. a. two replication forks, one leading strand and multiple lagging strands b. two replication forks, two leading strands and two lagging strands c. one replication fork, one leading strand and multiple lagging strands d. two replication forks, one leading strand and one lagging strand e. one replication fork, one leading strand and one lagging strand

e. one replication fork, one leading strand and one lagging strand Because of bi-directional replication there are two replication forks, two leading strands and two lagging strands with in a single replication bubble.