23:15 O O 0•4G+ ll 8%Restriction enzyme tutorial 2021_1_-3 - SavedA DNA(bp)A10,00020,00030,00040,00048,502Bamtl550522,34627,972 34,49941,732(bp)ECOBI21,226 26,104 31,74739,16844,972(bp)DHindu23,130 27,47925,15736,895 37,584 44,14137,459(bp)Figure 3: Restrictrion site map showing the following: A) linear DNA that is not cut as reference B) DNA cut with BamHI, C) DNA cut withECORI, D) DNA cut with Hindlll1. Calculate the size of the resulting fragments as they will occur after digestion and writethe sizes on the maps below. Note that linear DNA has a total size of 48 502 bp (seefigure 3A).ВBamHI550522,34627,97234,49941,732(bp)CECORI21,226 26,104 31,74739,16844,972(bp)DHindll23,13027,47936,895 37,58437,45944,141(bp)25,1572. How many fragments would you expect to see for each of the maps indicated inquestion 1 above?Мар В:Мар С:Map D:33. Use the sequence indicated in figure 1 (also refer to figure 2 and 3) and indicate howmany times does the sequence GAATTC occur in the A DNA sequence? What aboutB IUAII|||| Section A: Linear DNACommon restriction enzymes include: EcoRI, Hindll and BamHl and their sequences are asfollows, with the cut site indicated by the arrow (figure 1). Please note that A DNA refers tolinear DNA in this tutorial.HindIII 5'..A AGCT.3'3'....TTCGA A...5'EcoRI5'..G AATTC...3BamHI 5'....G GATCC...3'3'.....CTTAA G..53'...CCTAG G...5'Figure 1: Restriction sites of restriction enzymes.When DNA is cut with restriction enzymes, the fragments can be seen on an agarose gel (seefigure 2).Base Pairs2122025.00010.0008,0006.0005,00065574361164172334,0003,0002,50074212.0005804E6431,5001.0004878750-564S00E 3530125250A cut with EcORIA cut with HindilA cut with BamHIACDFigure 2: DNA fragments on an agarose gel showing the following: A) a molecular ladder, B) DNA cut with EcoB, C) DNA cut with Hipd,D) DNA cut with BamThe above figure shows the size of each of the fragments/bands produced when A DNA is cutwith each of these restriction enzymes. The sizes were determined by comparison to amolecular ladder (figure 2A) which has bands of known sizes when it is separated byelectrophoresis at the same time as the digested A DNA.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 DNAare shown in the maps (see figure 3 B-D), each vertical line on the map is where the respectiveenzymes will cut.||

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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).

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

Chapter18: Dna Technologies: Making And Using Genetically Altered Organisms, And Other Applications

Section: Chapter Questions

Problem 9TYK

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25. The restriction enzymes Kpnl and Acc651 recognize and cleave the same 6-bp sequence. You have a plasmid and a linear DNA strand that both contain a Kpnl and Acc651 sequence in the same orientation as shown below. You digest both DNA pieces with both enzymes and then attempt to ligate the sticky ends, followed by treatment with DNA ligase. What will happen? 5' GGTACC3' 5' G G T ACC 3' 3' CCATGG 5' y CCATGGS Kpnl Acс651 A) You will produce sticky ends but the two types of ends will not ligate. Instead, you may produce a small amount of religated plasmid where the digested plasmid sequence re-inserts. B) You will produce a recombinant plasmid in which the linear DNA strand is ligated in between the two sites, suitable for cloning. C) You will produce blunt ends that will not ligate because the two restriction enzymes will both operate on both of the sites. D) All the DNA will be completely digested as if you had applied a general DNAse enzyme.
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
Kha Vu Danels Include: 8DX : Safehy Jor bromie tnto lab repart! Name Section date sheet MAPPING PRACTICE #1 Below is a restriction map for the plasmid PGEN 101 (total length = 20 Kb). Using this map as a guide, give the number of restriction fraqments along with their associated lengths that would result from digesting PGEN 101 with the restriction enzymes EcoRI, BamHI and a combination of ECORI and BamHI. BamHI 3.2 Kb 1.7 Kb EcoRI BamHI PGEN 101 8.7 Kb 5.5 Kb .9 Kb EcoRI ECORI DIGESTION PERFORMED SIZES OF FRAGMENTS OBTAINED 10.4 kb , 0.9kb, 8.7 Kb EcoRI 3.2 Kb, 16. 8kb BamHI EcoRI + BamHI
1) The DNA fragment shown in Figure 1 is cleaved by the restriction enzyme EcoRI as indicated. The number in parenthesis shows the position of the cleavage site. The total length of the DNA fragment is 4000 bp. Small parts of the DNA sequence is known as shown. 5' ACCCTAGGTGTGACCGCGATCCGGCAGCATAAT 3' EcoRI (400) EcoRI (1300) EcoRI (3800) 3' CGCGAAATGCTTTAAGCGCTCTACGGGAGGG5' 3'AGCGTTAGAGTAGCCGGTAAAGGGTACGCGCCTTAA 5' Figure 1: DNA fragment with a total length of 4000 bp. The figure below depicts a gel on which marker DNA of known size has been run. Sketch the location of the bands that will appear, if the DNA fragment shown in Figure 1 is cleaved by EcoRI and afterwards run on the gel along with the marker DNA. 6000 5000 4000 3000 2000 1000 800 600 500 400 200
25. The restriction enzymes Kpnl and Acc651 recognize and cleave the same 6-bp sequence. You have a plasmid and a linear DNA strand that both contain a Kpnl and Acc651 sequence in the same orientation as shown below. You digest both DNA pieces with both enzymes and then attempt to ligate the sticky ends, followed by treatment with DNA ligase. What will happen? 5' GGTACC3' 5' GGTACC3' 3 CCATGGS 3 CCATGGS Kpnl Aсс651 A) You will produce sticky ends but the two types of ends will not ligate. Instead, you may produce a small amount of religated plasmid where the digested plasmid sequence re-inserts. B) You will produce a recombinant plasmid in which the linear DNA strand is ligated in between the two sites, suitable for cloning. C) You will produce blunt ends that will not ligate because the two restriction enzymes will both operate on both of the sites. D) All the DNA will be completely digested as if you had applied a general DNAse enzyme.
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
#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?
B 6000 5100 Number of Fragment(s) 4000 100 pX 6000 bp 3000 A A 1500 2000 A B 10. Based on the restriction map of the above plasmid, determine the number of DNA fragment(s) and the size(s) of the fragment(s). Digestion with Enzyme A Enzyme B Enzyme C Enzyme A + B Enzymes A + C Enzyme B+ C Enzyme A + B + C Fragment size(s) in base pairs
Wa... SLSU SVM Status... X ↓ O 650 O 550 O 500 200 O 850 Smal ↓ O 1550 Applicant Dashbo... 300 Msel 350 Probe Smal ↓ Caption: Restriction enzyme sites are indicated by the arrows. Smal binds to CCCGGG sequences. Msel binds to TTAA sequences. Enzyme 'X' refers to an unnamed restriction endonuclease that is not affected by DNA methylation. The distance between enzyme sites is indicated as number of bases. Assume only CpG methylation occurs. A human genomic segment (depicted above) was subjected to various experimental conditions (enzyme digestion, DNA methylation) AS INDICATED BELOW. 200 If the DNA was not methylated and digested with enzymes X, and Smal, what is the largest DNA fragment size that would show up on a Southern blot using the radiolabeled probe? Smal ↓ 500 MacBook Pro X ↓
4. Animation 12.4: This table shows results from a genetic test conducted on DNA samples from two patients. The patients are a newly married couple seeking genetic counseling about their risks of passing on a genetic disease to any children they may have in the future. The samples included the DNA region including and immediately surrounding a single gene. Restriction enzyme cleavage analyses were performed on the samples. Female Patient 1) 3400 bp fragment Male Patient 1) 1600 bp fragment 2) 1800 bp fragment 3) 3400 bp fragment DNA fragments on gel What do the genetic test results indicate about this case? O Both the woman and the man are homozygous for the mutation. O Both the woman and the man are heterozygous. O The woman is homozygous for the mutation and the man is homozygous for the normal gene. O The woman is homozygous for the mutation and the man is heterozygous.
1. You have the plasmid pUC18/19, which is a circular plasmid that consists of 2686 bp. What would the number of and length of the fragments be if you cut the plasmid with the following restriction enzymes or combination of enzymes? Give a schematic representation of the digestions. a. PscI & GsuI b. ScaI, PdmI & BsaXI c. ScaI, SspI & EheI 2. You have determined that the amplicon you want to clone has enzyme restriction sites for HindIII and EcoRI. After investigation you have seen that the pUC18/19 plasmid also have these enzyme restriction sites in its multiple cloning site (MCS on map above). After enzyme digestion your amplicon is 854 bp long. a. What length will the recombinant plasmid be after you have inserted your amplicon? Show your calculation. b. In the amplicon insert you have an enzyme restriction site for NdeI at 500 bp. If you digest the recombinant plasmid with this enzyme what length will the fragments be?
Map 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