Genetics: From Genes to Genomes
6th Edition
ISBN: 9781259700903
Author: Leland Hartwell Dr., Michael L. Goldberg Professor Dr., Janice Fischer, Leroy Hood Dr.
Publisher: McGraw-Hill Education
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Chapter 18, Problem 25P
Nonhomologous end-joining (NHEJ) of a double-strand break almost always results in perfect resealing of the DNA lesion, without the loss or gain of
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DNA polymerases are capable of editing and error correction, meaning it is able to edit and correct single base error so that the gene is not affected. However, RNA polymerase has a limited capacity for error correction. Given that a single base error in either replication or transcription can lead to error in protein synthesis, suggest a brief explanation for this difference in the capability of error correction between DNA polymerase and RNA polymerase.
The following illustrates a jagged double-strand DNA break resulting from Cas9 cleavage that
occurred in the first step of genome editing using CRISPR-Cas9 technology:
5'-GCGCCGTCC
3'-CGCGGC
CTGTCAGGCGACACT-3'
AGGGACAGTCCGCTGTGA-5'
Which of the double-stranded DNA sequences listed below (A-D) is expected to result from repair of
the break above via non-homologous end joining? Note: this question is not asking what kinds of
mutations result from NHEJ repair of Cas9 cleavage in general, but specifically what is expected to
result from repair of the jagged cut illustrated above? In the answer choices below, sequences that
are the same in all four options are shown in bold to help you spot the differences.
A. 5'-GCGCCGCTGTCAGGCGACACT-3'
3'-CGCGGCGACAGTCCGCTGTGA-5'
B. 5'-GCGCCGTCTGTCAGGCGACACT-3
3'-CGCGGCAGACAGTCCGCTGTGA-5'
C. 5'-GCGCCGTCCCTGTCAGGCGACACT-3'
3'-CGCGGCAGGGACAGTCCGCTGTGA-5'
D. 5'-GCGCCGAGACTGTCAGGCGACACT-3'
3'-CGCGGCTCTGACAGTCCGCTGTGA-5'
Restriction sites are palindromic; that is, they read the same in the5' to 3' direction on each strand of DNA. What is the advantage ofhaving restriction sites organized this way?
Chapter 18 Solutions
Genetics: From Genes to Genomes
Ch. 18 - Match each of the terms in the left column to the...Ch. 18 - Mice are usually gray, but a mouse geneticist has...Ch. 18 - Sometimes, genes transferred into the mouse genome...Ch. 18 - In mice, a group of so-called Hox genes encode...Ch. 18 - The fly eyes shown in Fig. 18.7 are malformed...Ch. 18 - This problem concerns a technique called enhancer...Ch. 18 - Fish and other organisms that live in the Arctic...Ch. 18 - a. Describe two ways you could potentially make a...Ch. 18 - Figure 18.6 shows a picture of Glofish ,...Ch. 18 - Some people are concerned about the possible...
Ch. 18 - The goal of the Knockout Mouse Project is to...Ch. 18 - Prob. 12PCh. 18 - Prob. 13PCh. 18 - a. Which genome manipulation technique would you...Ch. 18 - a. Diagram a knockin construct that could have...Ch. 18 - Prob. 16PCh. 18 - Prob. 17PCh. 18 - The transcription factor Pax6 is required...Ch. 18 - Mouse models for human genetic diseases are...Ch. 18 - One way to determine where inside a cell a protein...Ch. 18 - In Problem 5 in Chapter 17, you saw that a SNP...Ch. 18 - Scientists now routinely use CRISPR/Cas9 to make...Ch. 18 - Geneticists are currently considering using...Ch. 18 - a. Figures 18.9 and 18.12 demonstrated methods to...Ch. 18 - Nonhomologous end-joining NHEJ of a double-strand...Ch. 18 - One problem that researchers sometimes encounter...Ch. 18 - Researchers at the University of California at San...Ch. 18 - Prob. 28PCh. 18 - F. Port and S. Bullock at the University of...Ch. 18 - On Fig 18.14, locate the PAM site and identify the...Ch. 18 - Prob. 31PCh. 18 - Prob. 32PCh. 18 - Recall that Leber congenital amaurosis LCA, a form...Ch. 18 - One potential strategy for gene therapy to correct...Ch. 18 - Recently, scientists have used a mouse model for...
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- The 3' exonuclease activity of E. coli DNA polymerase I was found to show no discrimination between correctly and incorrectly base- paired nucleotides at the 3' terminus; properly and improperly base-paired nucleotides are cleaved at equal rates there. How can this observation be reconciled with the fact that the 3' exonuclease activity increases the accuracy with which template DNA is copied?arrow_forwardThe 3′-exonuclease activity of E. coli DNA polymerase I was found to show no discrimination between correctly and incorrectly base-paired nucleotides at the 3′-terminus; properly and improperly base-paired nucleotides are cleaved at equal rates there. How can this observation be reconciled with the fact that the 3′-exonuclease activity increases the accuracy with which template DNA is copied?arrow_forwardA short genetic sequence, which may be recognized by primase, is repeated many times throughout the E. coli chromosome. Researchers have hypothesized that primase may recognize this sequence as a site to begin the synthesis of an RNA primer for DNA replication. The E. coli chromosome is roughly 4.6 million bp in length. How many copies of the primase recognition sequence would be necessary to replicate the entire E. coli chromosome?arrow_forward
- With reference to the image below, discuss the process and principle involved for screening/selection of hosts (last stage of cloning) containing the intended recombinant plasmid. LacZ' = Gene for alpha-peptide of β-galactosidase.arrow_forwardTo test patients for COVID19, lab workers will first convert all the RNA molecules extracted from a nasal swab to a double-stranded DNA copy (dsDNA). If the virus is present, its genomic sequence should be in some of the new dsDNA molecules. Part 1) A region of COVID genomic DNA sequence is shown below. Following convention, only the top strand is shown. Copy/paste the sequence into the text box and create the second strand. Be sure to label its ends. (You may need to reduce the font size so that it doesn't wrap around) AAGATCACATTGGCACCCGCAATCCTGCTAACAATGCTGCAATCGTGCTACAACTTCCTC Part 2) To test for the presence of COVID DNA sequence, lab workers use single-stranded DNA oligonucleotides as probes (short pieces of DNA that do not have a partner strand). If the two strands of DNA that you drew were separated from each other, where would the shorter DNA strand shown below be able to form continuous base pairs? Highlight that region in your dsDNA model. TGTAGCACGATTGCAGCATTG Note: If you…arrow_forwardHundreds of DNA double strand breaks are created by SPO11, a topoisomerase type II like protcin. (i) Draw and explain how SPO11 can generate double strand breaks.arrow_forward
- How can the genetic defect phenylketonuria (PKU), which is caused by mutations in the phenylalanine hydroxylase (PAH) gene be corrected by using the CRISPR/Cas9 system and homology directed DNA repair.arrow_forwardA pyrimidine dimer which is a bulky lesion has mutated an E. coli cell's DNA. Describe both the photoreactivation enzyme repair (PRE) and Nucleotide excision repair describe how the cell uses Uvr A, B, C, D gene products to effect repair.arrow_forwardIn a clinical context, a scientist is working with a viral DNA which is about 24000bps long. There are two known variants of the virus that share almost the same DNA but for a final fragment; with reference to Figure Q2b, the regions A and B are conserved in both variants, while the region C differs and is either 320bps (variant 1) or 380bps (variant 2). The scientist wants to set up a procedure to identify the variant they are dealing with. Viral dsDNA (i) (ii) (iii) Stable region (A) Variable region (C) Figure Q2b Known sequence (B) 5-GACCTCAATGTCCAGCGGTACGCTCATAAA-3' 3'-CTGGAGTTACAGGTCGCCATGCGAGTATTT-5' The scientists want to design a primer to amplify the variable region and to do so, they sequence a small fragment (sequence B) the conserved region close to the variable region C. Why is the scientist targeting a region outside of the fragment of interest? [3] The sequence of the fragment B is reported in Figure Q2b. Suggest a primer that can efficiently target this region and…arrow_forward
- For the following short sequence of double stranded DNA and the given primers, there will be one major duplex DNA product after many cycles (imagine 10 cycles) of PCR. Provide the sequence of this one major duplex product and label the 5’ and 3’ ends of each strand. Sequence to be amplified: 5’- GGTATTGGCTACTTACTGGCATCG- 3’ 3’- CCATAACCGATGAATGACCGTAGC- 5’ Primers: 5’-TGGC-3’ and 5’-TGCC-3’arrow_forwardDNA polymerase I (Pol I) of E. coli consists of three functional parts (domains): an N-terminal domain with 5´ to 3´ exonuclease activities required for removal of the RNA primer, a central domain responsible for 3´ to 5´ exonuclease proofreading, and a C-terminal domain with polymerase activity. Pol I is thought to simultaneously remove RNA primers and fill in the gaps that result. A group of proteins known as RNaseH also have 5´ to 3´ exonuclease activity and can thus remove RNA primers. However, they lack the other two functions observed for Pol I. Predict the ability of the following mutants to replicate DNA: (1) a strain with a mutant gene encoding Pol I such that it no longer has polymerase activity (but retains both types of nuclease activities); (2) a strain without RNaseH proteins; (3) a strain with a mutant gene encoding Pol I such that it no longer has 5´ to 3´ exonuclease activities (but retains 3´ to 5´ nuclease and polymerase activities); (4) a strain with…arrow_forwardExtreme UV exposure leads to the SOS response in bacteria. By what mechanism does the SOS response function? Answer choices induction of photolyase and the addition of white light to remove the thymine dimer destruction of lexA, which leads to expression of an alternate, error-prone DNA polymerase homologous recombination repair non-homologous end joining exinuclease removal of a segment of DNA including a thymine dimer, followed by the replacement of DNA using the complementary strand of DNAarrow_forward
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