Study Guide for Campbell Biology
11th Edition
ISBN: 9780134443775
Author: Lisa A. Urry, Michael L. Cain, Steven A. Wasserman, Peter V. Minorsky, Jane B. Reece, Martha R. Taylor, Michael A. Pollock
Publisher: PEARSON
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Textbook Question
Chapter 20, Problem 3IQ
Which of the following DNA sequences would most likely function as a restriction site for a restriction enzyme? Why?
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Restriction sites are palindromic; that is, they read the same in the5' to 3' direction on each strand of DNA. What is the advantage ofhaving restriction sites organized this way?
Table 21.3 describes the cleavage sites of five different restrictionenzymes. After these restriction enzymes have cleaved the DNA, four of them produce sticky ends that can hydrogen bond with complementary sticky ends, as shown in Figure 21.1. The efficiency of sticky ends binding together depends on the number of hydrogen bonds; more hydrogen bonds makes the ends “stickier” and more likely to stay attached. Rank these four restriction enzymes from Table 21.3 (from best to worst)with regard to the efficiency of their sticky ends binding to each other.
When circular DNA is sequenced, the nucleotide base pairs are numbered starting from a fixed position on the DNA, all the way around, usually in a clockwise manner. a DNA molecule that is 3133 base pairs long is digested with RsaI restriction enzyme recognition sites at base numbers 366, 1534, and 2207. What are the sizes of the DNA fragments that will be produced after the DNA is digested with RsaI?
Chapter 20 Solutions
Study Guide for Campbell Biology
Ch. 20 - In what ways would third-generation sequencing be...Ch. 20 - The following schematic diagram depicts an...Ch. 20 - Which of the following DNA sequences would most...Ch. 20 - a. When PCR is used to prepare a DNA fragment for...Ch. 20 - a. What are some of the benefits of determining...Ch. 20 - Prob. 6IQCh. 20 - What are some of the practical and ethical...Ch. 20 - Prob. 8IQCh. 20 - Prob. 1SYKCh. 20 - Fill in the table on the previous page on the...
Ch. 20 - Prob. 3SYKCh. 20 - Prob. 1TYKCh. 20 - Prob. 2TYKCh. 20 - Gel electrophoresis is a means of separating...Ch. 20 - Prob. 4TYKCh. 20 - Prob. 5TYKCh. 20 - The following segment of DNA has restriction sites...Ch. 20 - Prob. 7TYKCh. 20 - Prob. 8TYKCh. 20 - Prob. 9TYKCh. 20 - Prob. 10TYKCh. 20 - Prob. 11TYKCh. 20 - Which enzyme is used in the polymerase chain...Ch. 20 - Prob. 13TYKCh. 20 - STRs (short tandem repeats) are a valuable tool...Ch. 20 - Prob. 15TYKCh. 20 - Which of the following has the greatest potential...Ch. 20 - Prob. 17TYKCh. 20 - Petroleum-lysing bacteria are being engineered for...Ch. 20 - Prob. 19TYKCh. 20 - Prob. 20TYK
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- A molecule of double-stranded DNA that is 5 million base pairs long has a base composition that is 62% G + C. How many times, on average, are restriction sites for the following restriction enzymes likely to be present in this DNA molecule? a. HindIII (recognition sequence is AAGCTT)arrow_forwardA molecule of double-stranded DNA that is 5 million base pairs long has a base composition that is 62% G + C. How many times, on average, are restriction sites for the following restriction enzymes likely to be present in this DNA molecule? a. BamHI (recognition sequence is GGATCC)arrow_forwardIn making recombinant DNA molecules that combine restriction fragments from different organisms, researchers usually prefer restriction enzymes like BamHI or HindIII that generate fragments with “sticky ends” (ends with overhangs) rather than enzymes like HpaI or SmaI (Table 12.1) that generate fragments with “blunt ends” (ends without overhangs). Can you think of a reason for this preference?arrow_forward
- A molecule of double-stranded DNA that is 5 million base pairs long has a base composition that is 62% G + C. How many times, on average, are restriction sites for the following restriction enzymes likely to be present in this DNA molecule? a. HpaII (recognition sequence is CCGG)arrow_forwardA 10 kb DNA fragment digested with the restriction endonuclease EcoRI yields fragments of 4 kb and 6 kb. When the 10 kb fragment is digested with BamHI, three fragments of 1, 3.5 and 5.5 kb are generated. Digestion with both enzymes yields four fragments of 0.5, 1, 3 and 5.5 kb. Draw the restriction map for the 10 kb fragment based on the data. Label the cut sites for the two enzymes, and the lengths between the cut sites.arrow_forwardThe partial sequence of one strand of a double-stranded DNA molecule is 5'-GACGAAGTGCTGCAGAAAGTCCGCGTTATAGGCATGAATTCCTGAGG -3' EcoRI is a restriction enzyme that cleaves after G in the sequence 5'-GAATTC-3'. PstI is a restriction enzyme that cleaves after A in the sequence 5'-CTGCAG-3'. Write the sequence of both strands of the DNA fragment created when this DNA is cleaved with both EcoRI and PstI. The first strand of your duplex DNA fragment should be derived from the given strand sequence. 5'- -3' 3'- -5'arrow_forward
- 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^CGAarrow_forwardRestriction endonuclease and ligase are two types of enzymes used in the process of genetic engineering, i.e., the manipulation of genes. The restriction endonuclease differs from ligase in that it breaks the DNA at ends, while ligase causes the breaks in DNA from interior joins the fragments of DNA, while ligase breaks the DNA into fragments breaks the DNA at specific points, while the ligase joins the fragments of DNA breaks the DNA apart at each nucleotide, while ligase use the pieces to translatearrow_forwardIf restriction endonucleases are produced by bacteria within a host, why don’t these enzymes chew up the genomic DNA of their host? What is the role of DNA methyltransferase in this? Indicate the answerarrow_forward
- Assume that a plasmid is 4700 base pairs in length and has restriction sites for a given restriction enzyme at the following locations: 800, 1400, 2900, and 3600. List the fragments by size that are ! expected when the plasmid is fully digested the restriction enzyme.arrow_forwardA plasmid DNA and a linear DNA (both of the same size) have one site for a restriction endonuclease. When cut and separated on agarose gel electrophoresis, plasmid shows one DNA band while linear DNA shows two fragments. Explain.arrow_forwardIf the GAATTC palindrome repeats are randomly found along the DNA strand, then what can you say about the sizes of the fragments that will be produced when the DNA is digested with a restriction enzyme that recognizes that sequence? How does the total length of the fragments relate to the size of the original DNA fragment?arrow_forward
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