I. The DNA template, 3'–CGTTACCCGAGCOCGTACGATTAGG–5', was exposed to a mutagen, resulting in the following three templates. 2. A. 3'–CGTTACCCGAGCCGTAACGATTAGG-5' nserti on leuding to frame shift B. 3-CGTTACCCGATCCGTACGATTAGG–5' Point - Transversin C. 3'-CGTTACCCGAGCCGTTCGATTAGG–5' Transition I. Transcribe and translate each template and write down the primary sequence of the resulting polypeptide/protein. Describe how each of the mutations will affect the final protein product from the corresponding template. N met -gla - Ser- Ala -cys -Stup-d Predict whether gene expression (from initiation of transcription to final protein product) would be faster in a prokaryotic or eukaryotic cell. Explain your answer.

I. The DNA template, 3'–CGTTACCCGAGCOCGTACGATTAGG–5', was exposed to a mutagen, resulting in the following three templates. 2. A. 3'–CGTTACCCGAGCCGTAACGATTAGG-5' nserti on leuding to frame shift B. 3-CGTTACCCGATCCGTACGATTAGG–5' Point - Transversin C. 3'-CGTTACCCGAGCCGTTCGATTAGG–5' Transition I. Transcribe and translate each template and write down the primary sequence of the resulting polypeptide/protein. Describe how each of the mutations will affect the final protein product from the corresponding template. N met -gla - Ser- Ala -cys -Stup-d Predict whether gene expression (from initiation of transcription to final protein product) would be faster in a prokaryotic or eukaryotic cell. Explain your answer.

Human Anatomy & Physiology (11th Edition)

11th Edition

ISBN:9780134580999

Author:Elaine N. Marieb, Katja N. Hoehn

Publisher:Elaine N. Marieb, Katja N. Hoehn

Chapter1: The Human Body: An Orientation

Section: Chapter Questions

Problem 1RQ: The correct sequence of levels forming the structural hierarchy is A. (a) organ, organ system,...

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Explain the reason. why is the answer option D and not Option C
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1. (a) cleaves both strands of the target DNA CRISPR repeat sequences gRNA Cas9 nuclease tracrRNA (b) encodes Guide RNAs CRISPR repeat sequences gRNA Cas9 nuclease tracrRNA
1a.The complete genotype of the host yeast strain is MATa ade2 his3 leu2 met15 ura3. Name one advantage of using a yeast strain with auxotrophies for several genes (i.e. his3, leu2 and ura3). b.)After successful CRIPSR editing, the yeast can later be “cured” of the recombinant CRISPR plasmid – that is, the plasmid is lost but the CRISPR edit is stably inherited. Write the new genotype of the yeast based on ADE6 gene.
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DNA polymerase I (Pol I) of E. coli consists of three functional parts (domains): an N-terminal domain with 59 to 39 exonuclease activity required for removal of the RNA primer, a central domain responsible for 39 to 59 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 (figure 13.14). A group of proteins known as RNaseH also have 59 to 39 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 59 to 39 exonuclease activity (but retains 39 to 59 nuclease and polymerase activities); (4) a strain with the…
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Primers designing for epitope tagging: Design forward and reverse primers to amplify the following gene with 6×HIS-tag on the N-terminus of the protein. To be cleaved and inserted into the plasmid, add restriction sites for EcoRI and HindIII at 5' and 3'. ATGCTCTCCGCCCTCGCCCGGCCTGTCAGCGCTGCTCTCCGCCGCAGCTTCAGCACCTCAGCC CAGAACAATGCTAAAGTAGCTGTGCTAGGGGCCTCTGGAGGCATCGGGCAGCCACTTTCAC TTCTCCTGAAGAACAGCCCCTTGGTGAGCCGCCTGACCCTCTATGATATCGCGCACACACCC GGAGTGGCCGCAGATCTGAGCCACATCGAGACCAAAGCCGCTGTGAAAGGCTACCTCGGAC CTGAACAGCTGCCTGACTGCCTGAAAGGTTGTGATGTGTAA
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