Two sets of sequenced DNA fragments for question #13 (A) In this example, a DNA sequence of 150 bases is cut with two different restriction enzymes, producing the following fragments, each of which has been sequenced. Fragments from the first restriction enzyme: GGTCGGCTATGTAACGAGTTGCC тсттоттсстассттотсAлСССGGGAТGAATOTTTACTO САCCCGGACCGTсссттCAT GTCCCAGAGCстаттасалаллат GCCCACCTT TTATTGAGTTGATGCTCGACGTAGCCAGACTTAA Fragments from the second restriction enzyme: ACCGGGGATGAATGTTTACTGGTCGCAGAG ссТАттасаAGAAатсатсааста сттетсA TGATGCTCGACGT сотссаттсAт AGCCAGACTTAACACGCGGAC тоталсаAOттоссосСсАCсттттATталот тсттеттсстаG Try to piece these fragmentary sequences together and determine the entire sequence of 150 bases, before you turn the page. Figure 4-8a Biology Today, 3le (0 2004 Garland Science)

Two sets of sequenced DNA fragments for question #13 (A) In this example, a DNA sequence of 150 bases is cut with two different restriction enzymes, producing the following fragments, each of which has been sequenced. Fragments from the first restriction enzyme: GGTCGGCTATGTAACGAGTTGCC тсттоттсстассттотсAлСССGGGAТGAATOTTTACTO САCCCGGACCGTсссттCAT GTCCCAGAGCстаттасалаллат GCCCACCTT TTATTGAGTTGATGCTCGACGTAGCCAGACTTAA Fragments from the second restriction enzyme: ACCGGGGATGAATGTTTACTGGTCGCAGAG ссТАттасаAGAAатсатсааста сттетсA TGATGCTCGACGT сотссаттсAт AGCCAGACTTAACACGCGGAC тоталсаAOттоссосСсАCсттттATталот тсттеттсстаG Try to piece these fragmentary sequences together and determine the entire sequence of 150 bases, before you turn the page. Figure 4-8a Biology Today, 3le (0 2004 Garland Science)

Biology: The Dynamic Science (MindTap Course List)

4th Edition

ISBN:9781305389892

Author:Peter J. Russell, Paul E. Hertz, Beverly McMillan

Publisher:Peter J. Russell, Paul E. Hertz, Beverly McMillan

Chapter19: Genomes And Proteomes

Section: Chapter Questions

Problem 1ITD: Below is a sequence of 540 bases from a genome. What information would you use to find the...

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Question 8. You isolate the DNA from the bacterial cells and apply the Sanger dideoxy sequencing method. You then separate the products of the reactions by gel electrophoresis and obtain the following pattern. What is the sequence of the template and the DNA on the gel? ddATP ddTTP ddCTP ddGTP Template V [ Choose 3'-CTAGTCAAGG-5' 5'-CTAGTCAAGG-3' Sequence 3'-GATCAGTTCC-5' 5'-GATCAGTTCC-3' 5'-GGAACTGATC-3'
Restriction sites of Lambda (A) DNA - In base pairs (bp) The sites at which each of the 3 different enzymes will cut the same strand of lambda DNA are shown in the maps (see figure 3 B-D), each vertical line on the map is where the respective enzymes will cut. A DNA A (bp) 48502 10 000 20 000 30 000 40 000 9162 17 198 B Sal I 7059 14 885 28 338 35 603 42 900 (bp) Hae III 11 826 21 935 29 341 38 016 (bp) 11648 29,624 Eco R1 (bp) 10 592 16 246 28 915 41 864 Figure 3: Restrictrion site map showing the following A) inear DNA that is not cut as reference B) DNA CLt with Sal L C) DNA cut with Hae , D) DNA cut with Eco RI 1. Calculate the size of the resulting fragments as they will occur after digestion and write the sizes on the maps below. Note that linear DNA has a total size of 48 502 bp (see figure 3A). Page 3 of 7 9162 17 198 Sal i (bp) 7059 14 885 28 338 35 603 42 900 Hae I (bp) 11 826 21 935 29 341 38 016 11648 29,624 Eco R1 (bp) 10 592 16 246 28 915 41 864
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.
Question 4. What is the probability that a restriction enzyme will cut DNA if the recognition sequence for the enzyme is 5'-GGATCC-3'? a) Assuming that the four bases are equally likely to be found in a DNA molecule. O (1/4)6 O (1/4)4 O (1/6)4 O 1 in 6 O (1/6)6
Figure 9-22 shows the first steps in the process of making a DNA microarray, or DNA chip, using photolithography. Describe the remaining steps needed to obtain the desired sequences (a different fournucleotide sequence on each of the four spots) shown in the first panel of the figure. After each step, give the resulting nucleotide sequence attached at each spot.
I. You wish to perform an electrophoretic resolution of your restriction enzyme–digested DNA. The sizes of the expected fragments range from 100 to 500 bp. You discover two agarose gels polymerizing on the bench. One is 0.5% agarose; the other is 2% agarose. Which one might you use to resolve your fragments? I. What does it mean to “denature” a protein and why is this important for SDS PAGE? II. Describe how you denatured the proteins you used for SDS-PAGE. III. Describe the roles of the primary Antibody and the secondary Antibody during an ELISA test. IV. After the addition of the secondary Antibody, how did you determine which wells contain the Antigen bound to primary Antibody bound to secondary Antibody?
Question 4. What is the probability that a restriction enzyme will cut DNA if the recognition sequence for the enzyme is 5'-GGATCC-3' ? b) Assuming that the % GC is = 60%. O (2/10)4 (3/10)² O (3/10)4 (2/10)² O (1/6)4 O (1/4)6 O (6/10)4 (4/10)2
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
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…
The restriction endonuclease BamHI recognizes the sequence GGATCC and cleaves between the Gs (in both strands). The enzyme Bgl II recognizes AGATCT and cuts between AG (in both strands). Justify your answers. 1. Draw the double stranded sequences of the DNA pieces shown above and indicate with an arrow the cutting points for each of the enzymes. Use different two colors for the two different sequences (one color for BamHI and another one for BglII). Do they generate blunt ends or sticky ends? 2. Draw the two sequences arising from each the cut (there will be four sequences).
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 HaellI
The short DNA shown below is to be sequenced. Using your knowledge of how the Sanger method works, in the gel diagram, draw in the bands that will appear when DNA polymerase is added to the reaction along with the four different nucleotide mixtures indicated. Note that some of these mixtures are not what would normally be used in a sequencing reaction. Dideoxynucleatides (ddNTPs) are added in relatively small amounts. The asterisk represents a radioactive label. *5' - 3'-ОН 3' – -- ACGACGCAGGACATTAGAC-5' Nucleotide mixtures: A. DATP, DTTP, dCTP, DGTP, ddTTP (given) B. DATP, ATTP, dCTP, AGTP, ddATE C. dTTP, dGTP. ACTP, ddCTP, ddATP D. DATP, dCTP, dTTP, ddGTP A в с D | || ||