Genetic Analysis: An Integrated Approach (2nd Edition)
2nd Edition
ISBN: 9780321948908
Author: Mark F. Sanders, John L. Bowman
Publisher: PEARSON
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Chapter 10, Problem 18P
Following restriction digestion, DNA fragments produced by digestion with certain enzymes have “sticky ends,” while fragments produced by digestion using other enzymes have “blunt ends.” Distinguish the meaning of these two terms.
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Restriction endonucleases are bacterial enzymes that cleave duplex (double-stranded) DNA at specific nucleotide sequences. The mode of replication of the animal virus SV40 has been investigated by using restriction endonucleases that cleave SV40 DNA into a number of unique segments. Like most viruses, SV40 DNA is circular.
The map positions of the 11 fragments produced by a pair of restriction endonucleases are shown on the next page. Immediately following a 5 or 10 minute pulse of radioactively labeled thymidine, labeled SV40 molecules that have completed replication during the pulse are isolated. These newly replicated DNA molecules are digested by the restriction endonucleases and the resulting fragments are analyzed for the relative amounts of pulse label they contain. The results are in the table below. Assume that at the time the label was added there was a random population of replicating SV40 DNA molecules in all possible stages of synthesis.
From the information given below,…
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.
The enzymes mentioned below are used as tools during cloning, DNA sequencing and/or gene therapy. Explain what they are used for. Also mention the actual biological function of the respective enzymes.
1) RNaseH
Chapter 10 Solutions
Genetic Analysis: An Integrated Approach (2nd Edition)
Ch. 10 - Define the following terms as described in this...Ch. 10 - 2. Using sickle cell disease as an example,...Ch. 10 -
3. Compare and contrast the contributions of...Ch. 10 - Why do differences in protein electrophoretic...Ch. 10 - Prob. 5PCh. 10 - Prob. 6PCh. 10 - Prob. 7PCh. 10 - 8. Wildtype βglobin protein is composed of amino...Ch. 10 - Prob. 9PCh. 10 - Prob. 10P
Ch. 10 - 11. How is an autoradiograph produced from a...Ch. 10 - Prob. 12PCh. 10 - Prob. 13PCh. 10 - Prob. 14PCh. 10 - The family represented in the pedigree and...Ch. 10 - Suppose the mating couple (I-1 and I-2) shown in...Ch. 10 - What are restriction endonucleases, and why are...Ch. 10 - 18. Following restriction digestion, DNA fragments...Ch. 10 - 19. The doublestranded DNA sequence below is part...Ch. 10 - 20. Restriction enzymes recognize specific...Ch. 10 - Prob. 21PCh. 10 - Prob. 22PCh. 10 - Prob. 23PCh. 10 - Prob. 24PCh. 10 - 25. A second strain of dwarf plants has a...Ch. 10 - During gel electrophoresis of linear DNA...Ch. 10 - Prob. 27PCh. 10 - 28. In molecular biology, restriction...Ch. 10 - A complete plant gene containing four introns and...Ch. 10 - Prob. 30PCh. 10 - The map below illustrates three alleles in a...Ch. 10 - Prob. 32PCh. 10 - 33. Northern blot analysis is performed on mRNA...
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- Consider the following image of a stained agarose gel representing the separation of a plece of DNA that has been digested by a series of restriction enzymes. Assume no digest fragment is of identical size to another digest fragment. nucleotide pairs (kb) 8 S 45 3.5 AA size Hindill + markers EcoRI Hindill EcoRI - 1st attempt Based on all data presented in the stained agarose gel, which of the following statements is correct? Choose one: OA. The original piece of DNA was circular and was cut by EcoRI twice and Hindill only once. OB. The original piece of DNA was linear and was cut by EcoRI twice and Hindill only once. OC. The original piece of DNA was circular and was cut by EcoRI once and not at all by Hind!!!.arrow_forwardWhen 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?arrow_forwardRestriction 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?arrow_forward
- A new restriction enzyme is discovered. It recognizes a 6 base pair pallindromic sequence. The first three bases of the target sequence are are given below with the cut identified by ^. Answer the following questions: Target sequence: 5' G^TA _ _ _ 3'3' _ _ _ _ _ _ 5' 1) Complete the target sequence with all missing bases. Also include the position of the cut in the lower sequence(^) 2) What kind of clevage does the enzyme use?arrow_forwardNucleosomes can be assembled onto defined DNA segments. When a particular 225-bp segment of human DNA was used to assemble nucleosomes and then incubated with micrococcal nuclease, which digests DNA that is not located within the nucleosome, uniform fragments 147 bp in length were generated. Subsequent digestion of these fragments with a restriction enzyme that cuts once within the original 225-bp sequence produced two well-defined bands at 37 bp and 110 bp. Why do you suppose two well-defined fragments were generated by restriction digestion, rather than a range of fragments of different sizes? How would you interpret this result?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_forward
- A 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_forwardA) For this DNA fragment (from 5' to 3') "TGAATTCCCGGGTTCCGGGAATTCGCGCGAATTCCCGGTATA", what is its complementary strand B) What are the products when the DNA with the above sequence is incubated with the restriction enzyme EcoRI C) What are the products when the DNA with the above sequence is incubated with the restriction enzyme Mspl D) Draw the first two (2) base pairings of the DNA molecule from the 5' end and label all key elements of the molecule including the bonds involvedarrow_forwardA DNA strand was sequenced using the Sanger method (https://www.youtube.com/watch?v=KTstRrDTmWI). The reaction tube contained the DNA strand, fluorescently labelled dideoxynucleotide triphosphates (ddATP – yellow, ddGTP – green, ddCTP – blue, ddTTP - red), deoxynucleotide triphosphates, DNA polymerase, or its Klenow fragment. Synthesis of DNA is allowed to proceed, and the results are shown on the right: 15 14 13 12 11 10 (a) What is the sequence of the copy and the template strands? (b) If the template strand were in the 5'-3' direction, what will be the sequence of the DNA copy? Nucleotide Lengtharrow_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_forwardA 12 kb linear DNA fragment is subject to single or double RE digest and agarose gelelectrophoresis, to yield the gel profile shown below. The first lane contains the size marker(M).a) Explain how the name of the enzyme EcoRI is derived.b) How many sites are there for EcoRI and PvuII respectively on this DNA fragment?c) Use the sizes of the DNA bands on the gel to compile a restriction enzyme map of the DNAfragment. Indicate the positions of the restriction enzymes sites for EcoRI and PvuII on themap.arrow_forwardDescribe the possible outcome of a PCR experiment in which (a) there is a single-stranded break in the target DNA sequence, which is present in only one copy in the starting sample, and (b) there is a doublestranded break in the target DNA sequence, which is present in only one copy in the starting sample.arrow_forward
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