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Name: 3′-->5′ Exonuclease activity allows DNA polymerase III (Pol III) to ba
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Description: 3′-->5′ Exonuclease activity allows DNA polymerase III (Pol III) to back-up and fix a mismatched base pair that was just incorporated into a growing strand of new DNA.TrueOr FalseIn one of the four ways to regulate gene expression, positive control w

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3′-->5′ Exonuclease activity allows DNA polymerase III (Pol III) to back-up and fix a mismatched base pair that was just incorporated into a growing strand of new DNA. True Or False

Biology: The Dynamic Science (MindTap Course List)

4th Edition

ISBN:9781305389892

Author:Peter J. Russell, Paul E. Hertz, Beverly McMillan

Publisher:Peter J. Russell, Paul E. Hertz, Beverly McMillan

Chapter16: Regulation Of Gene Expression

Section: Chapter Questions

Problem 11TYK: Discuss Concepts In a mutant strain of E. coli, the CAP protein is unable to combine with its target...

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Consider the Rho-dependent terminator sequence 5’CCCAGCCCGCCUAAUGAGCGGCCUUUUUUUU-3’. What affect would a point mutation at any one of the bolded and underlined nucleotides disrupt termination of transcription? Group of answer choices 1.Mutation in one of these nucleotides would disrupt base pairing, but not affect the formation of the hairpin and termination proceeds. 2.Mutation in one of these nucleotides would have no affect on base pairing, so the termination hairpin is formed and termination proceeds. 3.Mutation in one of these nucleotides would not disrupt base pairing, but would prevent the formation of the hairpin and disrupt termination. 4.Mutation in one of these nucleotides would disrupt base pairing, preventing the formation of the hairpin and disrupting termination.
5 5 S 6 5 5 5 6 U 6 U 6 5:14 PM | 0.2KB/s HHHHH R R U RUUR ARU AP AP R U U R R AP R R R AP MOLECULAR...GENETICS. Describe gene regulation at transcription level. Explain the role of antsense RNA in control mechanism. Describe translational control mechanisms. Describe common DNA damages. Distinguish excision and mismatch repair. Describe the role of recA protein in recombination repair Elaborate on SOS repair mechanism. Define thymine dimer. How are they formed and repaired? Describe the molecular basis of mutation. 11 Leu+ Met+ Arg+ Write a detailed note on spontaneous mutation. Explain about mutant detection methods. Define reverse mutation. Describe the mechanism underlying Intragenic and intergenic suppressor mutations Describe the transposition mechanisms. 13 Vo LTE UNIT IV Time (Min) Describe the process of generalised transformation occurring in bacterial chromosome and plasmid. Elaborate on molecular mechanism and significance of transformation 22 Describe the process of…
Consider the Rho-dependent terminator sequence 5’CCCAGCCCGCCUAAUGAGCGGCCUUUUUUUU-3’. What affect would a point mutation at any one of the bolded and underlined nucleotides disrupt termination of transcription? Group of answer choices Mutation in one of these nucleotides would disrupt base pairing, preventing the formation of the hairpin and disrupting termination. Mutation in one of these nucleotides would have no affect on base pairing, so the termination hairpin is formed and termination proceeds. Mutation in one of these nucleotides would not disrupt base pairing, but would prevent the formation of the hairpin and disrupt termination. Mutation in one of these nucleotides would disrupt base pairing, but not affect the formation of the hairpin and termination proceeds.
Different sensitivities to the mushroom toxin a-amanitin distinguish the three RNA polymerases from one another. Which of the following properties listed below also distinguish RNA Polymerase II from Pol I and Pol III? Options: Only RNA Pol II possesses a large subunit RNA Polymerase I and RNA Polymerase III do not require TBP for optimal transcription efficiency only RNA Polymerase II requires an ATP-dependent helicase to melt the DNA around the transcription start site Only RNA Polymerase II resembles the prokaryotic RNA Polymerase RNA Pol II has an extended N terminal region that becomes phosphorylated during intiation
Trp operon of E. coli is an inducible sytem since it turns on in the presence of tryptophan. In most bacteria, protein synthesis is initiated with a modified methionine residue (N-formylmethionine), whereas unmodified methionines initiate protein synthesis in eukaryotes. Both DNA replication and transcription follow a 5’ to 3’ direction of polarity. Write T if the statement is true and write F if the statement is false
Negative supercoiling of DNA favors the transcription of genes because it facilitates unwinding. However, not all promoter sites are stimulated by negative supercoiling. The promoter site for topoisomerase II itself is a noteworthy exception. Negative supercoiling decreases the rate of transcription of this gene. Propose a possible mechanism for this effect and suggest a reason why it may occur.
e B 1_30*_SP23 - General Biology I (for majors)/11364 of € 2 A us page X F1 What conditions would we find on the gene of a prokaryote if there is low amounts of tryptophan within the cytoplasm? Select one: a. lactose would attach to the repressor removing it from the promoter promoting the transcription of lactase b. no repressor is on the promoter creating constant transcription of lactase c. tryptophan would attach to the repressor removing it from the promoter promoting the transcription of lactase O d. repressors would bind to the promoter stopping the transcription of lactase e. lactose would attach to the repressor removing it from the promoter promoting the transcription of tryptophan producing enzymes no repressor is on the promoter creating constant transcription of tryptophan producing enzymes tryptophan would attach to the repressor binding it to the promoter and stopping the transcription of tryptophan producing enzymes O f. O g. 2 Oh. tryptophan would attach to the…
In addition to Tc1, the C. elegans genome contains otherfamilies of DNA transposons such as Tc2, Tc3, Tc4, andTc5. Like Tc1, their transposition is repressed in thegerm line but not in somatic cells. Predict the behaviorof these elements in the mutant strains where Tc1 is nolonger repressed due to mutations in the RNAi pathway.Justify your answer.
How do you think that transcription randomizes positions of nucleosomes and repression restores the ordering after transcription? How might you test to see if there was an exchange of histone subunits during transcription or if the nucleosome is truly transferred as a single unit? Would you expect the DNA band representing the distance from the restriction enzyme site to the hypersensitive site to be a single band or a smear? Defend your answer.
(c) By binding one L-tryptophan molecule/monomer, the trp repressor binds to DNA to suppress syn- thesis of L-tryptophan in E. coli. Below is the amino acid sequence of the helix – (reverse) turn – helix region of the trp repressor that binds to DNA compared to the sequence of the corresponding DNA binding motif of the Prl protein, a different type of repressor protein. A diagram of the trp repressor dimer is also shown. reverse turn trp helix 4 70 Trp -Gly-Glu-Met-Ser-Gln-Arg-Glu-Leu-Lys-Asn-Glu-Leu-Gly-Ala-Gly- Ile- Prl -Ser-Glu-Glu-Ala-Lys-Glu-Glu-Leu-Ala-Lys-Lys-Cys-Gly-Ile-Thr- Val- Pri heilix trp helix 5 80 90 Trp Ala-Thr-Ile-Thr-Arg-Gly-Ser sgn-Ser-Leu-Lys-Ala-Ala- Prl Ser-Gln-Val-Ser-Asn-Trp-Phe-Gly-Asn-Lys-Arg-Ile-Arg- Prl helix
Editing during DNA replication is provided by DNA polymerase enzymatic activity of O 5'-> 3' polymerase 3'-> 5' polymerase O 3'-> 5' exonuclease O an extra enzyme that binds to DNA polymerase O 5'-> 3'exonuclease
True/False: In general, weak promoters that support active transcription have -10 and -35 boxes that are close to the consensus sequences and a spacer between the two boxes that is 17 +/- 1 bp.