Concept explainers
(a)
Interpretation:
The reason for the separation of DNA in the agarose gel electrophoresis and to describe the reason for the difference in the DNA in each band needs to be explained.
Concept introduction:
Gel electrophoresis is a method where particles are differentiated according to their size and charge. Agarose is a substance used in electrophoresis in order to make a gel where the biological macromolecule will run. This process is known as agarose gel electrophoresis.
(b)
Interpretation:
The types of DNA represented by the various bands.
Concept introduction:
Gel electrophoresis is a method where particles are differentiated according to their size and charge. Agarose is a substance used in electrophoresis in order to make a gel where the biological macromolecule will run. This process is known as agarose gel electrophoresis.
(c)
Interpretation:
The reason for the cause of slower moving DNA in the given gel electrophoresis needs to be stated.
Concept introduction:
Gel electrophoresis is a method where particles are differentiated according to their size and charge. Agarose is a substance used in electrophoresis in order to make a gel where the biological macromolecule will run. This process is known as agarose gel electrophoresis.
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SAPLINGPLUS F/BIOCHEM+ICLICKER REEF-CODE
- Please answer this asap. Thanks, You have discovered a new plasmid RK21 in a unique bacterial community. As a first step towardunderstanding this plasmid, you digest the plasmid with three restriction enzymes: SspI, XhoI andSmaI. You run the digested plasmid DNA on an agarose gel, along with an uncut sample of theRK21 plasmid DNA as a control.Unfortunately you forget to load a DNA ladder, and obtain the following results. Assumecomplete digestion of all samples or all the digests worked completelyarrow_forwardYou 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.arrow_forwardPrimer designing: A single-stranded DNA sequence (963 nucleotides) that codes for a hypothetical protein are shown below (lower case shaded blue). 1. Design a pair of forward and reverse primers (~18 nucleotides long each) with EcoRI and BamHI added at 5' and 3' ends, respectively, for the amplification and cloning of this a plasmid with the same restriction sites. gene into GTATCGATAAGCTTGATATCGAATTCatggctaaaggcggagct cccgggttca aagtcgcaat acttggcgct gccggtggcattggccagccccttgcgatgttgatgaagatgaatcctctggtttctgttctacatctatatgatgtagtcaatgcccctggtgtcaccgctgatatta gccacatggacacgggtgctgtggtgcgtggattcttggggcagcagcagctggaggctgcgcttactggcatggatcttattatagtccctgcaggtgttcctcg aaaaccaggaatgacgagggatgatctgttcaaaataaacgcaggaattgtcaagactctgtgtgaagggattgcaaagtgttgtccaagagccattgtcaacctg atcagtaatcctgtgaactccaccgtgcccatcgcagctgaagttttcaagaaggctggaacttatgatccaaagcgacttctgggagttacaatgctcgacgtagt cagagccaatacctttgtggcagaagtattgggtcttgatcctcgggatgttgatgttccagttgttggcggtcatgetggtgtaaccatttgccccttctatctcagg…arrow_forward
- PCHEM4321. An agarose gel electrophoresis pattern of the plasmid PSPM4321 digestion (restriction) is shown below. Draw a restriction map of a plasmid with the appropriate restriction sites based on the data given below. Hindlll Hindll BamHI +BamHI Figure 1: 1% agarose gel electrophoresis of pCHEM4321 40 24 16 12 12 8 4 4 + |arrow_forwardplease help me with thi question. What advantages do CRISPR‑Cas systems have over restriction enzymes and engineered nucleases for editing DNA? The options are attached. Multiple answers can be chosenarrow_forwardB. Restriction Mapping. Single and double digestion of plasmid pMCS326 were performed using the restriction enzymes Alulll and EcoRV. DNA fragments are shown in an electrophoretogram below. Construct a restriction map of plasmid pMCS326 for enzymes Alulll and EcoRV. 20 kb 11 kb 8 kb 6 kb kb 3 Alulll + Alull EcoRV ECORV | || Restriction Map:arrow_forward
- 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’arrow_forwardPreparing plasmid DNA (double stranded, circular) for Sanger sequencing involves annealing a complementary, single-stranded oligonucleotide DNA primer to one strand of the plasmid template. This is routinely accomplished by heating the plasmid DNA and primer to 90°C and then slowly bringing the temperature down to 25°C. Why does this protocol work? What enzyme is used and what other components are required in the sequencing reaction? How does the Sanger method determine the sequence?arrow_forwardKpnl 4. Plasmid Z has a size of 7 kb, and the map shows Kpnl (K) and Pstl (P) cut sites relative to each other. This plasmid was digested with three different restriction enzymes 2000 bp 3500 bp Kpnl (K), Pstl (P) and Bgll (B) either alone or in combination and the samples run on an agarose gel as shown below. Where does Bgll (B) cut this plasmid ? Does the plasmid have one recognition site or two for Bg|l? Describe the Bgll cut site in this plasmid relative to the Kpnl cut Plasmid Z -7 kb Pstl site. How many bases to the left or right of the Kpnl cut site would you observe the Bgll cut site. Explain briefly. 1500 bp Pstl Ladder Kpni Psti K/P Bgl K/B KPB 7000 bp 7000 bp 5600 bp 5500 bp 4900 bp 3500 bp 2000 bp 1500 bp 1500 bp 1500 bp 1400 bp 600 bp %3Darrow_forward
- 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 864arrow_forwardHow many sites? A researcher has isolated a restriction endonuclease that cleaves at only one particular 10-base-pair site. Would this enzyme be useful in protecting cells from viral infections, given that a typical viral genome is 50,000 base pairs long? Explainarrow_forwardIn the DNA extraction. What is the role of alcohol in the DNA extraction process?arrow_forward
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