Biochemistry
6th Edition
ISBN: 9781305577206
Author: Reginald H. Garrett, Charles M. Grisham
Publisher: Cengage Learning
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Question
Chapter 28, Problem 22P
Interpretation Introduction
Interpretation:
To explain the interaction of RFC with PCNA to recognize primer-template junction during
Concept introduction:
DNA or Deoxyribonucleic acid is a molecule made of two chains which coil around one another. These form a double helix which carries instructions genetical in nature like related to reproduction, growth, development, functioning of the living organisms.
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Students have asked these similar questions
From standpoint of replication and transcription, explain how RNA polymerase is allowed to incorporate the first nucleotide whereas DNA polymerase needs a primer. Explain how this difference impacts the process of replication and transcription.
Matching Type
Choose the directionality of the given process. (4 points)
What is the directionality of the given process? *
4 points
3'-5'
5'-3'
Exonuclease activity
Complementary strand of the
continuous strand
Addition of nucleotides going
to the replication fork
Addition of nucleotides away
from the replication fork
Below is a depiction of a replication bubble.
5' AGCTCCGATCGCGTAACTTT
3'
TCGAGGCTAGCGCATTGAAA
CTAAAGCTTCGGGCATTATCG 3'
GATTTCGAAGCCCGTAATAGC
TATCGACS
Consider the following primer which binds to the DNA replication bubble on the diagram above:
5'-GCUAUCG-3'
Identify the DNA sequence to which this primer would bind and the orientation.
If the replication fork moves to the right, will the primer be used to create the leading strand of replication
or the lagging strand? Explain your answer
b. If the replication fork moves to the left, will the primer be used to create the leading strand of replication or
a.
the lagging strand? Explain your answer.
What would the next five nucleotides added to the primer by DNA polymerase?
С.
Chapter 28 Solutions
Biochemistry
Ch. 28 - Semiconservative or Conservative DNA Replication...Ch. 28 - The Enzymatic Activities of DNA Polymerase I (a)...Ch. 28 - Multiple Replication Forks in E. coli I Assuming...Ch. 28 - Multiple Replication Forks in E. coli II On the...Ch. 28 - Molecules of DNA Polymerase III per Cell vs....Ch. 28 - Number of Okazaki Fragments in E. coli and Human...Ch. 28 - The Roles of Helicases and Gyrases How do DNA...Ch. 28 - Human Genome Replication Rate Assume DNA...Ch. 28 - Heteroduplex DNA Formation in Recombination From...Ch. 28 - Homologous Recombination, Heteroduplex DNA, and...
Ch. 28 - Prob. 11PCh. 28 - Prob. 12PCh. 28 - Chemical Mutagenesis of DNA Bases Show the...Ch. 28 - Prob. 14PCh. 28 - Recombination in Immunoglobulin Genes If...Ch. 28 - Helicase Unwinding of the E. coli Chromosome...Ch. 28 - Prob. 17PCh. 28 - Functional Consequences of Y-Family DNA Polymerase...Ch. 28 - Figure 28.11 depicts the eukaryotic cell cycle....Ch. 28 - Figure 28.41 gives some examples of recombination...Ch. 28 - Prob. 21PCh. 28 - Prob. 22P
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Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, biochemistry and related others by exploring similar questions and additional content below.Similar questions
- 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.arrow_forward"DNA polymerase I serves a secondary function in vivo,now believed to be critical to the maintenance of fidelityof DNA synthesis" Explain this statement ?arrow_forwardPRDM9 recruits SPO11, a topoisomerase type II like protein for inducing DNA double strand breaks. Illustrate how SPO11 releases SPO11-linked oligonucleotides. rarrow_forward
- Determine the complementary strand of DNA that forms on this template DNA fragment during replication: 5′GGTTTCTTCAAGAGA3′arrow_forwardUsing the numbered sites on the DNA double helix strands below, where would the DNA Primer ACTTGCGA bind to for DNA Amplification?arrow_forwardMistakes made by DNA polymerase are corrected either by proofreading mechanisms during DNA replication or by DNA repair systems that operate after replication is complete. The overall rate of errors in DNA replication is about 1 × 10−10, that is, one error in 10 million base pairs. RNA polymerase also has some proofreading capability, but the overall error rate for transcription is significantly higher (1 × 10−4) or one error in each 10,000 nucleotides). Why can organisms tolerate higher error rates for transcription than for DNA replication?arrow_forward
- Explain how DNA polymerase and topoisomerase 2 contribute to replication in E.coli and what is the role of the role of the metal ions in the polymerase activity. B)How does the use of an RNA primer rather than a DNA primer affect the fidelity of DNA replication in E.coli?arrow_forwardPRDM9 recruits SPO11, a topoisomerase type II like protein, to induce DNA double strand breaks. Illustrate Illustrate and explain and explain how SPO11 releases SPO11-linked oligonucleotides.arrow_forwardExplain why two metal ions play important role on the palm domain of DNA polymerase.arrow_forward
- In the following sequence, a cytosine was deaminated and is now a uracil (underlined). 5’-GGTAUTAAGC-3’ a. Which repair pathway(s) could restore this uracil to cytosine? b. If the uracil is not removed before a DNA replication fork passes through, what will be the sequences of the two resulting double helices? Provide the sequences of both strands of both helices. Label the old and new strands and underline the mutation(s). c. Could the mismatch repair pathway fix the mutations you’ve indicated in part b? d. If the cell undergoes mitosis, and the replicated DNAs are distributed into the two daughter cells. Will 0, 1, or 2 daughter cells have a mutation in this sequence?arrow_forwardDescribe how the mismatch excision-repair pathway corrects errors introduced during replication. The following steps can be used for description of this process: Step 1: Damage recognition; Step 2: Unwinding and excision; and Step 3: Gap repair by D N A polymerase and ligase.arrow_forwardProvide the complementary strand and the RNA transcription product for the following DNA template segment:5'-AGGGGCCGTTATCGTT-3'arrow_forward
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