n your experiment there will be 4 different experimental samples labeled A-D (please see Table 3 on page 12 of the Lab 4 handout). In sample D, if you saw some bacterial colonies that are blue and some colonies that are white, what would this tell you? Group of answer choices the sgRNA was completely non functional the CRISPR/Cas9 donor DNA repair is not 100% efficient there is no functional lacZ on this bacterial plate arabinose was not active on this bacterial plate
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- In your experiment there will be 4 different experimental samples labeled A-D (please see Table 3 on page 12 of the Lab 4 handout). If you saw no bacterial colonies growing in sample A (pLZDonor, no arabinose), what would this tell you? Group of answer choices the bacteria has been successfully edited to include the donor template DNA the sgRNA caused a double-strand DNA break which caused the bacteria to die the DNA transformation was not successful there is functional lacZ in this sampleMap 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 oriAnalyzing Cloned Sequences A base change (A to T) is the mutational event that created the mutant sickle cell anemia allele of beta globin. This mutation destroys an MstII restriction site normally present in the beta globin gene. This difference between the normal allele and the mutant allele can be detected with Southern blotting. Using a labeled beta globin gene as a probe, what differences would you expect to see for a Southern blot of the normal beta globin gene and the mutant sickle cell gene?
- 5' UTR 5' and 3' UTR This region is specific sequence surrounding the start codon. OK Os OT OP LEU2 KSTP 2μm plasmid DNA Gene of interest 2μm plasmid DNA GAPDI GAPDP SC Cu/Zn-SOD cDNA Amp oriE gene What is selectable gene in this vector? 3' UTRCoding With the given coding strand perform the following 1. supply the correct non- coding strand 2. Identify the location of following restriction enzyme by enderlining it in the coding strands 3. Supply the correct non-coding strands for the two restriction enzymes EcoRi - 5' GAATTC 3'BamH1 - 5' GGATTC 3' 5' ATGCATGGTACGTAGAGTTCCATGAATTCGCCCCTATAGGGTAGCCGAGGATTCTATGCCCGAATGTC 3'Table 1 shows a list of restriction endonucleases with their recognition sequence and the sites of cleavage indicated by arrows. Table 1 Enzyme name Recognition sequence and position of cut 5'GIAATTC3 5'G!GATCC3' 5'GIGTACC3 5'GCIGGCCGC3' 5'IGATC3' 5'GGTACIC3' 5'ALGATCT3 EcoRI ВатHI Аcс651 Notl Sau3A Kpnl BglII (i) Which restriction enzyme(s) produce blunt ends? (ii) Are there any pair of neoschizomers in the list? Explain. (iii) Are there any pair of isocaudomers in the list? Explain.
- Alternative splicing Template strand S F yIGUide1,mq00:c-00: Replisome Transforming principle Origin of replication (or)eleb al msxS ain Coding strand Transcription factors Leading strand Single nucleotide polymorphism Okazaki fragment Telomerase M Nucleoside RNA Polymerase I RNA Polymerase II RNA Polymerase IIIon & of qu 9ven UoY Insertion mutagenesis Spliceosome Transcription Unit SNP Reverse transcriptase 1 Seminal work by Oswald Avery and colleagues demonstrated that DNA is what Frederick Griffiths called this etniog OS dotsM bioW 1-2kb of newly synthesized DNA strands are called this ainiog PS Assembly of the replisome is an orderly process that begins at these precise sites Snoiteeu 4 Transcribes ribosomal RNA genes in eukaryotes 5. A large nucleoprotein complex that coordinates activity at the replication fork Single base pair differences between homologous genomic regions isolated from different members of a population Complex of proteins and snRNAs catalyzing the removal of…BM4_DNA AND PROTEIN S X /1FAIPQLSDP_g5B-629FSHNpGnTMIEppLS4A71zBd4vcUBqNUILubXONw/formResponse 4. What is the nitrogen base pair of Adenine in transcription? O Cytosine O Uracil O guanine O thymine 5. The central dogma of Molecular Biology states that There are four nitrogen bases in DNA, two purines (adenine and guanine) and two pyrimidines (cytosine and thymine). Which process is not included in the central dogma? duplication transcription translation O translocation LeadpleShown below is an E. coli's DNA sequence coding for XXR protein. The nucleotides are numbered 1 to 330. Transcription starts at the Transcription Start Site (TSS) that is the base located at position 57. -55 5' AATAAСTTGAGATTTTGATTGACАТССТТСТСАCAGAGCCTATAATACCТАТТТС 3' 3'TТАТTGAACTСТААААСТААСТGTAGCAACAGTGTCТСGGATATTATGGATAAAG 5' 56- 5' ТACGTATAGAСАСТСAGAGGAAAGACAGAGAGAGAGTTAGCATTGTACTАТСТСТ 3' 3' АTGCATATСТсTGAGTCTCCTTтстстстстстсТСААТСGTAACATGATAGAGA 5' 65- -105 --110 -140- 5' СТTTTAGATATATCTCТАТСТСТСТСАСТССАТСТТТСТCGTGTTAACACAAСА 3' 3' GAAAATCTATАTAGAGATAGAGAGAAGTGAGGTAGAAAGAGCACAAТТСTСТTGT 5' 111- -120- -130- -150- -160----165 166- --175- --185- -195 -205- -215-----220 ------- 5' GTCACAGACTCACAGATCTTTGTCGGTGATCGGAGATGGAGTTCCGGGAGAAGCT 3' 3' CAGTGTCTGAGTGTCТAGAAACAGCCACTАGССТСТАССТСААGGCCCTСТТCGA 5' 221- -230- 240 -250 -260 -270-----275 5' TTATAAGTTCAAGTTGCAATAGGTGTTTGCCTTTGTTTTATCTCTCCTCACCGTA 3' 3'ААТАТТСААGTTCAACGTTATCCАCAAACGGAAACAAAАТAGAGAGGAGTGGCAT 5' 276- -285- -295- -305 -315…
- Sal1 Xho1 Insert pUC cloning vector MCS Insert Digested with Sal1 on the left and Xho1 on the right Sal1 Ori Vector Digested with Sal1 Xho1: СТСGAG Sal1: GTCGAC RE cut site Part B) After 8 h of sleep, you realize your mistake above (Question 3 Part A), and you successfully transformed your plasmid and obtained a subset of white colonies and blue colonies on the plate. Question: You conclude that all the blue colonies contain a plasmid O With the insert in the LacZ screening marker With the insert in the Antibiotic selectable marker O Without an insert in the LacZ screening marker O None of the above + Lacz alpha AmpR1. You are investigating a protein that has the amino acid sequence N ... Ala – Thr - Asn – Trp – Lys - Arg - Gly – Phe – Thr ... C within its primary structure. You found that several of the mutations affecting this protein produced shortened protein molecules that terminated within this region. In one of the mutants, the Asn became the terminal (last) amino acid. (a) What DNA single-base changes(s) would cause the protein to terminate at the Asn residue? (b) What other potential sites do you see in the DNA sequence encoding this protein where mutation of a single base pair would cause premature termination of translation? >#4 BamI --- 5’ CCTAG ↓G 3’ 5’ ACGCCTAGGACGTATTATCCTAGGTAT CCGCCGCCGT CATCA 3’ 3’ TGCGGATCCTGCATAATAGGATCCATAGGCGGCGGCAGTAGT 5’ Restriction enzyme: Recognition sequence: Number of pieces of DNA: Type of cut: