In contrast to prokaryotes, eukaryotic cells need multiple licensing factors to initiate replication. The diagram below depicts the process of eukaryotic replication initiation. What are the names of the factors depicted in the diagram? Factor A: In mammalian cells this protein is phosphorylated during S phase so it is rapidly degraded and cannot initiate 2nd round of replication Factor B: Forms complex with another protein that protects it from degradation until needed to initiate replication. Factor C: Complex of proteins that forms ring around DNA and have helicase activity.

In contrast to prokaryotes, eukaryotic cells need multiple licensing factors to initiate replication. The diagram below depicts the process of eukaryotic replication initiation. What are the names of the factors depicted in the diagram? Factor A: In mammalian cells this protein is phosphorylated during S phase so it is rapidly degraded and cannot initiate 2nd round of replication Factor B: Forms complex with another protein that protects it from degradation until needed to initiate replication. Factor C: Complex of proteins that forms ring around DNA and have helicase activity.

Biochemistry

6th Edition

ISBN:9781305577206

Author:Reginald H. Garrett, Charles M. Grisham

Publisher:Reginald H. Garrett, Charles M. Grisham

Chapter28: Dna Metabolism: Replication, Recombination, And Repair

Section: Chapter Questions

Problem 22P

See similar textbooks

Similar questions

a) Under normal conditions E. coli produces three DNA polymerases. State their functional similarities and differences. b) List the other proteins and enzymes involved in DNA replication in E.coli and give their functions.
Discuss DNA replication of prokaryotes and please mention all of the enzymes and components listed below. DNA Primase – DNA directed “RNA Pol” which inserts nucleoside triphosphates (NTPs) Primers are oligonucleotides; priming process is the formation of primers DNA Helicase – separates the DNA in advance of the replication fork (in E. coli DNA Helicase II); binds at AT-rich region of DNA; ATP then binds the helicase Single-stranded DNA-binding Protein (SSBP) – no enzymatic activity; does not consume ATP Topoisomerase – alter the supercoiling of double-stranded DNA DNA Ligase – nicking of strands done for replication to continue Okazaki fragments DNA Polymerase – removes primer via 5’ to 3’ exonuclease activity; comes again for 5’ to 3’ polymerization activity (closes the gap between Okazaki fragments) Prokaryotes DNA Pol I – auxiliary enzyme to DNA Pol III; repairs damage; capable of excising pyrimidine dimers; polymerization via single active site that can bind all 4 dNTPs;…
In eukaryotes, the Replication factor C (RFC) is a clamp loader. In the absence of RFC, what would be the consequences to the outcome of DNA replication? Explain.
Some antibiotic drugs fight infection by interfering with DNA replication, transcription, or translation in bacteria. Indicate whether each of the following antibiotic drug effects is on replication, transcription, or translation. HINT Each answer (replication, transcription, and translation) is used only once for the following: a. Rifampin binds to bacterial RNA polymerase. b. Streptomycin binds bacterial ribosomes, disabling them. c. Quinolone blocks an enzyme that prevents bacterial DNA from unwinding.
Initiation of replication occurs at an "origin of replication" site that typically includes an AT-rich region. Initiation benefits from these AT-rich regions because: the AT-rich region recruits topoisomerase to begin the process of DNA replication. GC-rich regions are impossible to replicate because of their strong H-bonds. the AT-rich region recruits DNA polymerase to begin the process of DNA replication. adenine-thymine pairs are held together by two H-bonds, making them easier to separate during unwinding. None of the answers is correct.
Human Fbh1 helicase is important in the process of DNA replication. When a mutation occurs during the production of Fbh1, the result is a mutant Fbh1 that binds at the replication fork and prevents any helicase protein from attaching to the strand. Based on this information and the image shown, what would happen during DNA replication if this mutant helicase were present? A - Topoisomerase would unwind the DNA and an RNA primer would attach to the DNA molecule and initiate replication. The process would then stop at the blue triangle because helicase is needed to separate the strands of DNA. B - Topoisomerase would unwind the DNA, but then the process would stop at the blue triangle because helicase, the RNA primer, would not be able to attach to the DNA molecule and initiate replication. C - The process would begin at the blue triangle when topoisomerase unwinds the DNA and an RNA primer attaches to the DNA molecule and initiates replication. DNA polymerase would begin the synthesis…
Imagine a bacterial cell with a mutation that renders helicase completely nonfunctional (note that this would be a lethal mutation). What, precisely, would go wrong with replication in this cell? Please describe what goes wrong, as well as all downstream effects.
At eukaryotic origins of replication, helicase cannot be activated until the polymerase is also positioned on the DNA. Explain what would happen if the helicase became active in the absence of DNA polymerase. RNA synthesis is much less accurate than DNA synthesis. Why does this not harm the cell? What is the role of folate in the synthesis of nitrogenous bases of purines?
DNA polymerase occasionally incorporates the wrong nucleotide during DNA replication. If left unrepaired, the base-pair mismatch that results will lead to mutation in the next replication. As part of a template strand, the incorporated wrong base will direct the incorporation of a base complementary to itself, so the bases on both strands of the DNA at that position will now be different from what they were before the mismatch event. The MER-minus strain of yeast does not have a functional mismatch excision repair system, but it has normal base excision repair and nucleotide excision repair systems. Which of the following statements is correct about differences in the mutation spectrum between MER-minus and wildtype yeast? More than one answer is correct. Options: More point mutations will arise in MER-minus yeast. Fewer point mutations will arise in MER-minus yeast as compared with wildtype. Of the total point mutations that…
Under the restrictive conditions a conditional mutation expresses its mutant phenotype and under the permissive conditions expresses the normal phenotype. First type of conditional mutation is a temperature-sensitive mutation, that expresses the mutant phenotype only at certain temperatures. Strains of E.coli have been isolated that contain temperature-sensitive mutations in genes encoding different components of the replication machinery. In the each of these strains, the protein produced by the mutated gene is nonfunctional under the restrictive conditions. You grow these strains under the permissive conditions and then abruptly switch them to the restrictive conditions. After one round of replication under the restrictive conditions, you isolate DNA from each strain and analyze it. What characteristics would you expect to see in the DNA isolated from a strain with a temperature-sensitive mutation in the gene that encodes each of the following proteins? 1. DNA ligase 2. DNA…
What is/are the attributes that make nucleotide excision repair (NER) and base excision repair (BER) similar and/or different from each other? Select the correct response: The NER pathway is the only one that can remove DNA lesions in the strand regardless of their size which is followed by attaching the correct strand, then sealed by a DNA ligase. They both use the enzyme DNA glycosylases that recognizes the damaged DNA segments and proceed with repairing the faulty base in the strand. They differ NER only repairs purine bases while BER repairs pyrimidine bases. They both remove the damaged parts of the DNA where the BER pathway corrects only the identified damaged bases which are usually non-bulky lesions. The NER pathway, on the other hand, repairs the damage by removal of bulky DNA adducts which is a short-single stranded DNA segment. They both utilize the enzyme photolyase to reverse the damages created by the faulty section of the DNA. They both remove the damaged parts of the…
If deoxyribonucleotides that lack the 3’-OH groups are added during the replication process, what do you expect will occur? Describe what happens when a nonsense mutation is introduced into the gene encoding transposase within a transposon. A pure culture of an unknown bacterium was streaked onto plates of a variety of media. You notice that the colony morphology is strikingly different on plates of minimal media with glucose compared to that seen on trypticase soy agar plates. How can you explain these differences in colony morphology?