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- 27. You have a DNA sequence of 1400 bp. To characterize it, you decide to make a restriction map. Cutting with the first enzyme, you show by electrophoresis that the fragments produced are 400 and 1000 bp, respectively. After using the second enzyme, you recover 600-bp and 800-bp fragments. After treatment with both enzymes, fragments were 200, 400, and 800 bp. Draw the correct restriction map.1. a. Draw the primers for this longer strand (consider optimal length): TTGCTTAAATTTAAATTTATGCCGTAAGCGCGCGC b. Explain which PCR step is most dependent on the genetic code for proper temperature programming and why.1) Restriction enzymes come in a concentration of U/ml, and it is recommended that 1U be used for each ug of DNA to be digested. If an enzyme you want to use comes in at 22,000 U/ml, and you want to digest 5 ug of DNA. How much volume will you have to use for the reaction and how will you be able to measure it with the pipettes we have in the lab? 2) An enzyme for ligation (Cip) comes at a concentration of 16000 units/ml. How many units will there be in 10 ul?
- What information and materials are needed to amplify region of DNA using PCR?Why are antibiotic resistance markers such as ampR important components of bacterial plasmid cloning vectors? a. The plasmid must have resistance to accept DNA inserts. b. They allow the detection of plasmids that contain an inserted DNA fragment. c. They ensure the presence of the ori site. d. They ensure that the plasmid can be cut by a restriction enzyme. e. They allow identification of bacteria that have taken up a plasmid.When Griffith injected mice with a combination of live rough-strain and heat-killed smooth-strain pneumococci, he discovered that (a) the mice were unharmed (b) the dead mice contained living rough-strain bacteria (c) the dead mice contained living smooth-strain bacteria (d) DNA had been transferred from the smooth-strain bacteria to the mice (e) DNA had been transferred from the rough-strain bacteria to the smooth-strain bacteria
- 1. Briefly explain why the total size of the pMBBS plasmid in the Restriction Enzymes practical is 3000bp (base pairs) 2. Briefly explain why the cut sites on the pMBBS plasmid in each Restriction Enzymes (EcoRI, BamHI, and XhoI) just 1 each 3. Briefly explain what led to the 5 fragments formed which linked to the 500bp, 1000bp, 1500bp, 2000bp, 2500bp sizes1. How does PFGE separate larger fragments more efficiently than standard electrophoresis? 2. Why is SYBR green less toxic than EtBr? 3. What are the similarities and differences between Manual and Automated Sanger Sequencing? 4. What is the relationship between DNA fragment length and the distance it will run in a gel? (Restriction Enzyme Digestion)#16) The restriction enzymes Xhol and SalI cut their specific sequences as shown below: XhoI 5' C | TCGAG 3' SalI | 5' GTCGAC 3' 3' GAGC | TC 5' 3' G | AGCTG 5' Can the sticky ends created by XhoI and SalI sites be ligated? If yes, can the resulting sequences be cleaved by either XhoI or SalI?
- 2) Considering the technologies that pave the way for the identification of molecules on the basis of hybridization in the classical sense, within the scope of Recombinant DNA Technology; (20P) a) Which of these is used to identify which macromolecule,b) What do you understand in the context of hybridization and between which molecules there are technique-specificinteractions take placec) “Nylon membrane” or “nitrosdulose” commonly used in these technologieswhy "membranes" are needed,d) Explain how to design an application example for each technique you mentioned.For each of the restriction enzymes listed below:(i) Approximately how many restriction fragmentswould result from digestion of the human genome(3 × 109bases) with the enzyme? (ii) Estimate theaverage size of the pieces of the human genomeproduced by digestion with the enzyme. (iii) Statewhether the fragments of human DNA produced bydigestion with the given restriction enzyme wouldhave sticky ends with a 5′ overhang, sticky endswith a 3′ overhang, or blunt ends. (iv) If the enzymeproduces sticky ends, would all the overhangs on allthe ends produced on all fragments of the humangenome with that enzyme be identical, or not? (Therecognition sequence on one strand for each enzymeis given in parentheses, with the 5′ end written atthe left. N means any of the four nucleotides; R is any purine—that is, A or G; and Y is any pyrimidine—that is, C or T. ^ marks the site of cleavage.)a. Sau3A (^GATC)b. BamHI (G^GATCC)c. HpaII (C^CGG)d. SphI (GCATG^C)e. NaeI (GCC^GGC)f. BanI (G^GYRCC)g. BstYI…Please asap Original DNA template: 3'-ACGGTCAATTTGCTG-5 a) Transcribe the sequence. b) Translate the sequence. c) What type of mutation is present in the strand 3 '- ACGGTCAATATTGCTG - 5 d) Provide the entire mutated sequence of amino acids. e) Explain the effect that this mutation will have.