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- You have another circular plasmid. Complete and effective digestion of this plasmid with a restriction enzyme yields three bands: 4kb, 2kb, and 1 kb. In comparing the band intensity on an ethidium bromide-stained gel, you notice that the 4 kb and the 2 kb bands have the exact same brightness. The 1 kb band is exactly one fourth as bright as each of these. (Assume there is uniform staining with ethidium bromide throughout the gel.) How many times did the enzyme cut the plasmid? What is the size of the plasmid? Justify your answers to a and b above using a clearly labeled diagram showing the relative location of the cut-sites on the plasmid.Restriction enzymes recognize palindromic DNA sequences (i.e. sequences that read the same on both strands). An example of such a sequence is GAATTC – work out the reverse complement of this sequence if you need to convince yourself that it’s a palindrome. Write the sequence of a different six nucleotide palindromic DNA sequence that uses all four nucleotides – you only have to give the sequence of one strand, but you’re welcome to write out both.5’-GATCAGCTGACTGGATCCGTCCTCAACGTCAGGATCCAGCTTCAAG-3’ 1. How many cuts do you expect this enzyme to make on the above DNA and how many fragments do you expect to see on your gel? Assume that they are all different sizes.
- The total (50ul) of your restriction digest contains 500ng of DNA how much DNA (ng) are you loading into the gel?Restriction digestion and Gel electrophoresis: A single strand of a double-stranded DNA sequence is shown below. Draw a complementary DNA strand and show the restriction digestion pattern of the double-stranded DNA with BamHI and Pst1. Show the separation pattern of the undigested and the digested DNA on your agarose gel. Label the gel appropriately. 5’ – CGAGCATTTGGATCCTGTGCAATCTGCAGTGCGAT – 3’After restriction enzymes cut, they contain unpaired bases. Type II restriction enzymes leave ends that may be 5' overhanging, 3' overhanging, or blunt. In all cases each end is left with a 3' OH and a 5' phosphate. All blunt ends, and any complementary overhanging ends may be re-ligated with T4 DNA ligase, as long as at least one 5'- phosphate is present. In the tables below G^AATTC means that the end after cutting with enzyme will be: -----G 3' -----CTTAA 5' GTGCA^C means that the end will be: -----GTGCA 3' -----C 5' Which RE’s from table below have a 5’ overhang? Which ones have a 3’ Overhang? AccI GT^CGAC BamHI G^GATCC ClaI AT^CGAT NsiI ATGCA^T PstI CTGCA^G BglII A^GATCT TaqI T^CGA
- 5'-GCGGTACGTTACGGCTTTACTGACCTGCAGGC-3' A. Convert this to a double strand DNA molecule by writing the complementary sequence. Be sure to correctly label the ends of the double stranded molecule B. Pick one of the two strands and write the sequence of an RNA molecule that could be transcribed from it. Again, you must correctly label the ends of the molecule.Given the DNA sequence of the restriction enzyme: gi|6329444|dbj|AB034757.1| Hynobius retardatus mRNA for larval beta-globin, complete cds GCAGAATCTGACTCAAGAAATCCCTCCTCACCCAACACCACCAGCAGCCATGGTTCACTGGACAGCAGAGGAGAAGGCAGCCATCAGCTCTGTGTGGAAGCAGGTGAACGTGGAGAGCGATGGACAGGAGGCCCTGGCCAGGTTGCTGATCGTCTACCCCTGGACCCAGAGATACTTCAGCTCTTTTGGGGACCTGTCGAGCCCAGCTGCCATTTGTGCCAACGCCAAGGTCCGTGCCCATGGCAAGAAGGTCCTGTCCGCCCTGGGAGCCGGCGCCAACCACCTGGATGACATCAAAGGCAACTTTGCTGATCTGAGCAAGCTTCACGCAGACACACTCCATGTGGACCCCAATAACTTCCTGCTCCTGGCAAACTGCCTGGTGATCGTCTTGGCCCGCAAGCTGGGAGCCGCCTTCAACCCTCAAGTCCATGCGGCCTGGGAGAAGTTCCTGGCCGTCTCCACCGCGGCTCTGTCCAGAAACTACCACTAGAGACTGGTCTTTGGGTTTAATTCTGTGAACGTCCCTGAGACAAATGATCTTTCAATGTGTAAACCTGTCATTACATCAATAAAGAGACATCTAACAAAAAAAAAAAAAAAAAAAAAAAAAA Identify two blunt-end cutters Identify two sticky-end cutters. For each, Provide the sequence of the Restriction enzyme, Highlight using a specific color where the DNA sequence where the restriction enzyme will cut the DNA Indicate the…Cynt Classifying mutations A certain section of the coding (sense) strand of some DNA looks like this: $-ATGTATATCTCCAGTTAG-3" It's known that a very small gene is contained in this section. Classify each of the possible mutations of this DNA shown in the table below. mutant DNA 5- ATGTATCATCTCCAGTTAG-3' S-ATGTATATCTCCAGTTAG-3 5- ATGTATATATCCAGTTAG-3' type of mutation (check all that apply) insertion deletion point silent noisy insertion O deletion point silent noisy insertion O deletion point silent Onoisy X G
- The structure of a typical pUC19/human DNA recombinant clone. Ensure that you clearly indicate the restriction enzyme sites at the ends of the human DNA insert. Hint: think about the compatibility of the ends generated by partial digestion of human DNA and complete digestion of the vector – will the original sites in the vector be regenerated or not, or it is impossible to predict?The table shows where different restriction endonucleases (restriction enzymes) cleave DNA. The abbreviation R represents the purines (adenine and guanine). The pyrimidines (cytosine, thymine, and uracil) are abbreviated as Y. The abbreviation W represents adenine or thymine. Enzyme Target sequence |Cleavage 5' GAATTC 3' |3' СТТААG 5' 5'G AATTC 3' EcoRI |3' СТТАА G 5' 5' GATATC 3' |3' СТАТАG 5' 5' GAT ATC 3' |3' СТА ТАG 5' EcoRV 5' GGCC 3' 3' CCGG 5' 5' GG CC 3' |3' СС GG 5' HaellI 5' AAGCTT 3' 3' TTCGAA 5' 5'A 3' TTCGA AGCTT 3' HindIII A 5' 5' RGGWCCY 3' 3' YCCWGGR 5' 5' RG 3' YCCWG GWCCY 3' PpuMI GR 5' Which restriction endonucleases would cleave a DNA molecule at the given sequences? The complementary DNA substrate strand is omitted for clarity. 5' ATCGAACTAGGCC 3' 5' AAAGCTTGTGATATC 3' EcoRI ЕcoRI EcoRV EcoRV HaellI HindIII HindIII HaellIIn the following gel showing stained bands of the Alu insertion sequence, what is the genotype of individual 2? 941 bp 641 bp->>> 1 2 3 4 5 6 Homozygous for the 641 bp sequence that does not contain in the Alu insertion Heterozygous, containing one 941 bp sequence and one 641 bp sequence O Homozygous for the 941 bp sequence containing the Alu insertion