DNA Polymerase Holoenzyme Il is represented below. It consists of several domains with distinct functions. The structure capable of 5' to 3' DNA polymerization is represented by the letter and the structure capable of 3' to 5' exonuclease activity is represented by the letter B. D DNA Pol III holoenzyme O D; A O A; C

DNA Polymerase Holoenzyme Il is represented below. It consists of several domains with distinct functions. The structure capable of 5' to 3' DNA polymerization is represented by the letter and the structure capable of 3' to 5' exonuclease activity is represented by the letter B. D DNA Pol III holoenzyme O D; A O A; C

Biochemistry

6th Edition

ISBN:9781305577206

Author:Reginald H. Garrett, Charles M. Grisham

Publisher:Reginald H. Garrett, Charles M. Grisham

Chapter29: Transcription And The Regulation Of Gene Expression

Section: Chapter Questions

Problem 17P

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DNA Polymerase Holoenzyme Ill is represented below. It consists of several domains with distinct functions. The structure that provides high processivity (processivity factor) is represented by the letter and the structure that provides DNA proofreading function is represented by the letter A В. -C -D DNA Pol III holoenzyme В; D А; А А; D D; A A; C
A certain section of the coding (sense) strand of some DNA looks like this: 5'- ATGGGCCACTCATCTTAG-3' It's known that a very small gene is contained in this section. Classify each of the possible mutations of this DNA shown in the table below. I Don't Know mutant DNA 5'- ATG GGCCACAGTTCTTAG-3' 5'- ATG GG CTCATCTTAG - 3' 5'- ATG GGCCACGCATCTTAG-3' Submit type of mutation (check all that apply) ооооо O point O silent O noisy ооооо insertion deletion insertion O deletion Opoint Osilent noisy insertion O deletion ооооо Opoint silent O noisy X S Ⓒ2023 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Center Accessibility
DNA is made of two strands that are antiparallel. If one strand runs from 3’ to 5’ direction the other one will go from 5’ to 3’ direction. During replication or transcription, whatever the process is, it will always follow the 5’ to 3’ direction using the 3’ to 5’ directed strand as the template strand. Therefore, if following is the DNA sequence5’-CCG ATC GCA CAA-3’a) Using this sequence as template after transcription no protein can be translated. Why? I. Presence of start codonII. Absence of start codonIII. Due to mutationb) If you want to start the translation, what change you need in the second codon (from 5’ to 3’ direction)?I. Substitution of C with GII. No changeIII. Deletion of CIV. Both I & III
DNA is made of two strands that are antiparallel. If one strand runs from 3’ to 5’ direction the other one will go from 5’ to 3’ direction. During replication or transcription, whatever the process is, it will always follow the 5’ to 3’ direction using the 3’ to 5’ directed strand as the template strand. Therefore, if following is the DNA sequence 5’-CCG ATC GCA CAA-3’ Using this sequence as template after transcription no protein can be translated. Why? Presence of start codon Absence of start codon Due to mutation If you want to start the translation, what change you need in the second codon (from 5’ to 3’ direction)? Substitution of C with G No change4 Deletion of Both I & III
The 5'-3' exonuclease activity involves all the following EXCEPT: o Both ribonucleotides and deoxyribonucleotides can be removed. O DNA repair can also be undertaken by this activity. o Groups of altered nucleotides can also be removed. o Removal of one nucleotide at a time in the properly base paired DNA. o Activity is possessed by both DNA polymerase I and III
Given the following template DNA strand, what is the correct complementary DNA sequence? 3' CGC AGT GGA CAT TTC 5' O 5' GCG TCA CCT GTA AAG 3' O 5' GAA ATG TCC ACT GCG 3' O 5' GCG UCA CCU GUA AAG 3' O 5' GAA AUG UCC ACU GCG 3'
vvnicn the following statements are correct about the repair of a DNA duplex containing the sequence below that is grown INE coli (select all that apply)? Strand A Strand B GATCTAGCCGGCATCCGAT CTAGATCGGACGTAGGCTA Methyl ✔A. MutH cleaves Strand A O B. DNA repair will result in the bold A in strand B being replaced with a C O C. DNA repair will result in the bold G in strand A being replaced with a T ✔ D. Defect will not be properly repaired in dam(-) E coli O E. The mammalian repair system would also correct the mismatch shown based on the methylation status of the DNA
The beginning of the hexose kinase gene's sequence can be found below, the +1 nucleotide is underlined and bolded. It also contains an origin of replication (ORI) which is found at position 30. 1 20 ORI 40 60 5'.TTCGAGCTCTCGTCGTCGAGATACGCGATGATATTACTGGTAATATGGGGATGCACTATC...3' 3'.AAGCTCGAGAGCAGCAGCTCTATGCGCTACTATAATGACCATTATACCCCTACGTGATAG...5' promoter 2a. Assume that replication has been initiated at that ORI. Provide the sequence of the primer that is complementary to the DNA in each of the following positions. d Site A - binding to the top strand of the DNA at position 20 – 30 5' 3' Site B - binding to the top strand of the DNA at position 31 -41 5' 3' 2b. Replication is occurring normally in these cells; would you expect to find a primer in both positions? Why or why not?
The anti-viral drug Acyclovir is a nucleotide analog that is lacking the 3’ OH group which is required to form a 3’→5’ phosphodiester bond. This drug is ineffective against DNA polymerases with proofreading abilities, which is why human DNA polymerases are not targeted. Acyclovir can be used to treatsevere cases of Epstein-Barr viral (EBV) infection, but has little to no effect under non-severe infections. Based on this information, EBV will use ________ DNA polymerase during severe infections and __________ DNA polymerase during non-severe infections. Human; Human EBV; Human EBV; EBV Human; EBV
this is the worst figure i have ever seen. Evidently A and B are different but the same color? From what I understand A would be a sliding clamp, and B is Primase? Is F rna primer (I think personally) or representing the SSBP? this is the worst diagram ever. the words that are for labeling are as follows: DNA polymerase III, sliding clamp, helicase, single-stranded binding protein (SSBPs), topoisomerase, RNA primer, newly synthesized DNA, primase, and DNA ligase.
This is part of the Escherichia coli DNA sequence that contains an inverted repeat. (Note: top strand is the coding strand). 5'-AACGCATGAGAAAGCCCCCCGGAAGATCACCTTCCGGGGGCTTTATATAATTAGC-3' 3'-TTGCGTACTCTTTCGGGGGGCCTTCTAGTGGAAGGCCCCCGAAATATATTAATCG-5' (i) Draw the structure of hairpin loop that will be formed during the end of transcription. (ii) Describe the function of the hairpin loop during transcription.
The beginning of the hexose kinase gene's sequence can be found below, the +1 nucleotide is underlined and bolded. It also contains an origin of replication (ORI) which is found at position 30. 1 20 ORI 40 60 5...ТTCGAGCтСТСGТСGTCGAGATACGCGATтGATATTACTGGTAATАTGGGGATGCACTATC...3' 3'..AAGCTCGAGAGCAGCAGCTCТАTGCGOТАСТАТААТGACCATTATACСССТАСGTGATA...5" promoter 2a. Assume that replication has been initiated at that ORI. Provide the sequence of the primer that is complementary to the DNA in each of the following positions. Site A - binding to the top strand of the DNA at position 20 – 30 5' 3' Site B - binding to the top strand of the DNA at position 31 – 41 5' 3' 2b. Replication is occurring normally in these cells; would you expect to find a primer in both positions? Why or why not?