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- To determine the reproducibility of mutation fre-quency measurements, you do the following experiment.You inoculate each of 10 cultures with a single E. coli bac-terium, allow the cultures to grow until each contains 106cells, and then measure the number of cells in each culturethat carry a mutation in your gene of interest. You were sosurprised by the initial results that you repeated the experi-ment to confirm them. Both sets of results display the sameextreme variability, as shown in Table Q5–1. Assuming thatthe rate of mutation is constant, why do you suppose thereis so much variation in the frequencies of mutant cells indifferent cultures?A cloning vector map is shown below. EcoRI Bam Ban Hind P-galactosidase Amp Bam Bam EcoRI Ori C Which restriction site is best for inserting a DNA fragment for selection of chimeric plasmid containing colonies? 1) They're all equally good. 2) Hindll 3) EcoRI 4) BamHIIn a typical microbiology laboratory, reasons for no bands from a gel of a polymerase chain reaction may bedue to errors relating to omission of ingredients in the reaction mix and absence of the target sequence inthe template DNA. Based on (i) primer problem and (ii) purity/potential contamination of the target sequence, explain the reason for non-appereance on bands.
- What is the actual biological purpose of the CRISPR/Cas9 system? 9.2. Explain the fundamental difference between CRISPR/Cas9 and in vivo adenovirus-based gene therapy Explain why there is a part of the LacZ gene in pET32a(+) and also how the natural biochemical process wherein it normally functions is used during cloning by molecular biology.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.Primer designing: A single-stranded DNA sequence (963 nucleotides) that codes for a hypothetical protein are shown below (lower case shaded blue). 1. Design a pair of forward and reverse primers (~18 nucleotides long each) with EcoRI and BamHI added at 5' and 3' ends, respectively, for the amplification and cloning of this a plasmid with the same restriction sites. gene into GTATCGATAAGCTTGATATCGAATTCatggctaaaggcggagct cccgggttca aagtcgcaat acttggcgct gccggtggcattggccagccccttgcgatgttgatgaagatgaatcctctggtttctgttctacatctatatgatgtagtcaatgcccctggtgtcaccgctgatatta gccacatggacacgggtgctgtggtgcgtggattcttggggcagcagcagctggaggctgcgcttactggcatggatcttattatagtccctgcaggtgttcctcg aaaaccaggaatgacgagggatgatctgttcaaaataaacgcaggaattgtcaagactctgtgtgaagggattgcaaagtgttgtccaagagccattgtcaacctg atcagtaatcctgtgaactccaccgtgcccatcgcagctgaagttttcaagaaggctggaacttatgatccaaagcgacttctgggagttacaatgctcgacgtagt cagagccaatacctttgtggcagaagtattgggtcttgatcctcgggatgttgatgttccagttgttggcggtcatgetggtgtaaccatttgccccttctatctcagg…
- Clone number in this case is number 196 which is shown in the images. State whether a BamHI site has been re-created at the forward- and the reverse-end junctions of the human DNA with the plasmid vector band sizes are shown in one of the images. (0.5, 1, 2, 3, 4, 5, 6, 8, 10kb)Section Name MAPPING PRACTICE #4 Plasmid pBR 607 is a 2.6 Kb plasmid containing Ampicillin and Tetracycline resistance markers, an origin of replication, and unique restriction sites for the restriction enzymes EcoRI, BamHI, and Pstl. Given the restriction map for pBR 607 for the enzymes EcoRI, BamHI, and Pstl, show on the gel diagram, where the approximate positions of the restriction fragments generated from the restriction digests would be located after carrying out electrophoresis. BamHI 0.2 Kb pBR 607 ECORI 1.94 Kb 0.46 Kb Pstl Size EcoRI EcoRI EcoRI + Standards BamHI + Pstl BamHI Pstl 4.0 Kb 2.2 Kb 2.0 Kb 0.5 KbMap of pUC18 is shown on the here. Describe how to select recombinant clones if a foreign DNA is inserted in to the polylinker site of pUC18 and then introduced into E. coli cells.. Hindll Sphl Sbfl Pstl BspMI Acci Hincli Sall Xbal BamHI Aval Smal Xmal Acc651 Kpnl Banll Eco53kl Sacl Apol ECORI lacz MCS LacR binding site Plac PUC18 Amp 2686 bps PMB1 ori
- Nonhomologous end-joining (NHEJ) of a doublestrand break almost always results in perfect resealingof the DNA lesion, without the loss or gain of nucleotide pairs. Yet CRISPR/Cas9, which produces doublestrand breaks, is a highly efficient method of makingsmall deletions or insertions at the targeted site. Howcan you resolve this apparent contradiction?NEED ASAP Using the list provided (note you may not need all the steps listed) state the steps in cDNA production 1. Confirm size of bands by comparing to molecular size marker. 2. Ligate oligonucleotides onto blunt ended product. 3. Cut the cDNA with a unique restriction enzyme. 4. Ligate the cDNA and vector followed by transformation of E. coli. 5. Plate cells onto LB-ampicillin plates containing X-gal. 6. Choose colonies that are white in colour. 7. Cut plasmid with unique restriction enzymes. 8. The size of the clone is small and it will ensure that the entire gene will be found in one fragment. 9. Employ replica plating technique and select cells that are ampicillin-sensitive and tetracycline resistant. 10. Blot petri plates with a nitrocellulose filter, lyse the cells and denature the DNA with a dilute sodium hydroxide solution. 11. Blot electrophoresis gel with nitrocellulose filter and denature the DNA with a dilute sodium hydroxide solution. 12. Flood the filters with…Utilizing Hind III and EcoR V Restriction Enzyme with Pet41 and the following gene of interest... a tgaaacaaca aaaacggctt tacgcccgat tgctgacgct gttatttgcg 61 ctcatcttct tgctgcctca ttctgcagca gcggcggcaa atcttaatgg gacgctgatg 121 cagtattttg aatggtacat gcccaatgac ggccaacatt ggaagcgttt gcaaaacgac 181 tcggcatatt tggctgaaca cggtattact gccgtctgga ttcccccggc atataaggga 241 acgagccaag cggatgtggg ctacggtgct tacgaccttt atgatttagg ggagtttcat 301 caaaaaggga cggttcggac aaagtacggc acaaaaggag agctgcaatc tgcgatcaaa 361 agtcttcatt cccgcgacat taacgtttac ggggatgtgg tcatcaacca caaaggcggc 421 gctgatgcga ccgaagatgt aaccgcggtt gaagtcgatc ccgctgaccg caaccgcgta 481 atttcaggag aacacctaat taaagcctgg acacattttc attttccggg gcgcggcagc 541 acatacagcg attttaaatg gcattggtac cattttgacg gaaccgattg ggacgagtcc 601 cgaaagctga accgcatcta taagtttcaa ggaaaggctt gggattggga agtttccaat 661 gaaaacggca actatgatta tttgatgtat gccgacatcg attatgacca tcctgatgtc 721 gcagcagaaa ttaagagatg gggcacttgg tatgccaatg aactgcaatt ggacggtttc 781…