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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25. The restriction enzymes Kpnl and Acc651 recognize and cleave the same 6-bp sequence. You have a plasmid and a linear DNA strand that both contain a Kpnl and Acc651 sequence in the same orientation as shown below. You digest both DNA pieces with both enzymes and then attempt to ligate the sticky ends, followed by treatment with DNA ligase. What will happen? 5' GGTACC3' 5' G G T ACC 3' 3' CCATGG 5' y CCATGGS Kpnl Acс651 A) You will produce sticky ends but the two types of ends will not ligate. Instead, you may produce a small amount of religated plasmid where the digested plasmid sequence re-inserts. B) You will produce a recombinant plasmid in which the linear DNA strand is ligated in between the two sites, suitable for cloning. C) You will produce blunt ends that will not ligate because the two restriction enzymes will both operate on both of the sites. D) All the DNA will be completely digested as if you had applied a general DNAse enzyme.
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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…
(a-32P) GMP INCORPORATED (pmol) N 111 (a) (b) (c) 20 5 (min) 26 10 The image above shows an experiment to investigate the effect of TFIIS on transcription by RNA polymerase II. Why were NTPs added to the reaction last?
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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Can you help with 1a please HELPFUL INFORMATION: When performing classical Sanger or "dideoxy" sequencing, you set up 4 parallel reactions per template to be sequenced from a specific primer, with each of the four reactions containing a different dideoxynucleotide, and then the four reactions were run in a separate, adjacent lanes on a gel. 1a. Why couldn't you combine all 4 dideoxynucleotides with the primer and the template and do the whole reaction in one tube, and then run the set of fragments produced by the reaction mixture on a single lane in an acrylamide gel?
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