Genetics: From Genes to Genomes
Genetics: From Genes to Genomes
6th Edition
ISBN: 9781259700903
Author: Leland Hartwell Dr., Michael L. Goldberg Professor Dr., Janice Fischer, Leroy Hood Dr.
Publisher: McGraw-Hill Education
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Chapter 9, Problem 15P

A recombinant DNA molecule is constructed using a plasmid vector called pMBG36 that is 4271 bp long. The pMBG36 plasmid contains a so-called polylinker that has a single site for each of several restriction enzymes, including BamHI (5′ G^GATCC 3′) and EcoRI (5′ G^AATTC 3′). The sequence of the polylinker region of pMBG36 is shown below; the dots indicate the large majority of the vector that is not shown. You now cut

pMBG36 with EcoRI and insert into it a fragment of the DNA previously shown in Problem 4 that is also cut with EcoRI.

Chapter 9, Problem 15P, A recombinant DNA molecule is constructed using a plasmid vector called pMBG36 that is 4271 bp long.

a. Write out as much of the DNA sequence of the resultant recombinant DNA molecule as is possible. Two answers are possible; you need to show only one.
b. Why are there two possible answers to part (a)?
c. How many recognition sites for BamHI will be found in the recombinant DNA molecule shown in your answer to part (a)?
d. If you cut this recombinant DNA molecule with BamHI and run the digest on a gel, how many bands would you see and how large would they be?
e. How many recognition sites for EcoRI will be found in the recombinant DNA molecule shown in your answer to part (a)?
f. If you cut this recombinant DNA molecule with EcoRI and run the digest on a gel, how many bands would you see and how large would they be?
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A circular plasmid of 10,000 base pairs (bp) is digested with two restriction enzymes, A and B, to produce a 3000 bp and a 2000 bp bands when visualized on an agarose gel. When digested with one enzyme at a time, only one band is visible at 5000 bp. If the first site for enzyme A (A1) is present at the 100h base, the order in which the remaining sites (A2, B1 and B2) are present is - (A) 3100, 5100, 8100 115. (B) 8100, 3100, 5100 (C) 5100, 3100, 8100 (D) 8100, 5100,. 3100
A plasmid vector pBS281 is cleaved by the enzymeBamHI (5′ G^GATCC 3′), which recognizes only onesite in the DNA molecule. Human DNA is digestedwith the enzyme MboI (5′ ^GATC 3′), which recognizes many sites in human DNA. These two digestedDNAs are now ligated together. Consider only thosemolecules in which the pBS281 DNA has been joinedwith a fragment of human DNA. Answer the following questions concerning the junction between thetwo different kinds of DNA. a. What proportion of the junctions between pBS281and all possible human DNA fragments can becleaved with MboI?b. What proportion of the junctions between pBS281and all possible human DNA fragments can becleaved with BamHI?c. What proportion of the junctions between pBS281and all possible human DNA fragments can becleaved with XorII (5′ C^GATCG 3′)?d. What proportion of the junctions between pBS281and all possible human DNA fragments can becleaved with BstYI (5′ R^GATCY 3′)? (R and Ystand for purine and pyrimidine, respectively.)e.…
A closed circular plasmid B-DNA (10.5 bp/turn) consists of 231 base pairs and has    Wr= -1 (ccDNAa). Then a topoisomerace acts upon ccDNAa leading to strain relaxation caused by supercoiling, thus, a topoisomere ccDNAb is formed. Finally, EtBr (Ethidium bromide) intercalator is added leading to ccDNAc which has 11bp/turn. a) Calculate superhelical densities σa, σb, σc of the three plasmids ccDNAa, ccDNAb and ccDNAc, respectively. b) Which of the three topoisomeres will move faster in agarose gel electrophoresis and why?
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