500 9. The restriction site sequence for the restriction enzyme Sau3AI and BamHI include four identical bases, making their sticky ends identical as shown in the figure. Let's say you have foreign DNA whose flanking regions were cut with Sau3AI. Also, you want to use a plasmid containing a BamHI restriction site. BamHI Lenhere (1) Would it be possible to ligate the foreign DNA into the BamHI site of the plasmid? Explain. (2) Would it be possible to cut the ligated sites with Sau3AI? What about BamHI? What problems will you expect if you use BamHI? Sau3Al T-A-G-G G-A-T-C C-T-A-G →! C-C-T-A-G C-T-A-G G-A-T-C-C G-A-T-C 10. An ampicillin-resistant, tetracycline-resistant plasmid, pBR322, is cut with Pstl, which cleaves within the ampi gene. The cut plasmid is ligated with Pstl digested Drosophila DNA to prepare a genomic library, and t

24. Sickle-cell anemia arises from a mutation in the gene for the B chain of human hemoglobin. The change from GAG to GTG in the mutant that is located within the cleavage site for the restriction enzyme Mstll, which recognizes the target sequence CCTGAGG. These findings form the basis of a diagnostic test for the sickle-cell gene. Which answer best explains how this test should work? A) In combination with DNA polymerase, application of Mstll will produce long DNA strands with the normal gene but not the mutant gene. B) In combination with DNA polymerase, application of Mstll will produce long DNA strands with the mutant gene but not the normal gene. C) Treatment with Mstll produces DNA fragments from the normal gene but cannot cleave the normal gene. D) Treatment with Mstll produces DNA fragments from the normal gene but cannot cleave the mutant gene. E) Restriction enzymes are non-specific relative to the binding requirements for such a test. 25. The restriction enzymes Kpnl and…

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.

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,…