Ch.9 Lehringer solutions - 1. Cloning When joining two or...

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Unformatted text preview: 1. Cloning When joining two or more DNA fragments, a researcher can adjust the sequence at the junction in a variety of subtle ways, as seen in the following exercises. (a) (b) (c) (d) (e) (f) (g) (h) 0) Draw the structure of each end of a linear DNA fragment produced by an EcoRI restriction digest (include those sequences remaining from the ECORI recognition sequence). Draw the structure resulting from the reaction of this end sequence with DNA polymerase I and the four deoxynucleoside triphosphates (see Fig. 8436). Draw the sequence produced at the junction that arises if two ends with the structure derived in (b) are ligated (see Fig. 25-16). Draw the structure produced if the structure derived in (a) is treated with a nuclease that degrades only single-stranded DNA. Draw the sequence of the junction produced if an end with structure 00) is ligated to an end with structure ((1). Draw the structure of the end of a linear DNA fragment that was produced by a Pvull restriction digest (include those sequences remaining from the Pvull recognition sequence). Draw the sequence of the junction produced if an end with structure (b) is ligated to an end with structure (1‘). Suppose you can synthesize a short duplex DNA fragment with any sequence you desire. With this synthetic fragment and the procedures described in (a) through (g), design a protocol that would remove an EcoRI restriction site from a DNA molecule and incorporate a new BamHl restriction site at approximately the same location (see Fig. 9—3). Design four different short synthetic double—stranded DNA fragments that would permit ligation of structure (a) with a DNA fragment produced by a Pstl restriction digest. In one of these fragments, design the sequence so that the final junction contains the recognition sequences for both ECORI and Pstl. In the second and third fragments, design the sequence so that the junction contains only the EcoRI and only the Pstl recognition sequence, respectively. Design the sequence of the fourth fragment so that neither the ECORI nor the Pstl sequence appears in the junction. Answer Type II restriction enzymes cleave double—stranded DNA within recognition sequences to create either blunt-ended or sticky-ended fragments. Blunt—ended DNA frag— ments can be joined by the action of T4 DNA ligase. Sticky—ended DNA fragments can be joined by either E, colt or T4 DNA ligases, provided that the sticky ends are complementary. Sticky-ended fragments without complementary ends can be joined only after the ends are made blunt, either by exonucleases or by E. coli DNA polymerase I. (a) The recognition sequence for EcoRI is (5’)GAATTC(3’), with the cleavage site between G and A (see Table 9—2). Thus, digestion of a DNA molecule with one EcoRI site 8-72 Chapter 9 DNA-Based Information Technologies (5’)-—GAATTC——(3’) —CTTAAG— \ would yield two fragments: (5’)—G(3’) and (5')AATTC—(3’) #CTTAA G-— (b) DNA polymerase I catalyzes the synthesis of DNA in the 5’——> 3’ direction in the presence of the four deoxyribonucleoside triphosphates. Therefore, both fragments generated in (a) will be made blunt ended: (5’)—GAATT(3’) and —CTTAA (5’)AATTC——~(8’) TTAAG—— (c) The two fragments generated in (b) can be ligated by T4 DNA ligase to form (5 ’ )———GAATTAATTC—(3 ’) —CTTAATTAAG——— (d) The fragments in (a) have sticky ends, with a protruding single—stranded region. Treat- ment of these DNA fragments with a single—strand—specific nuclease will yield DNA frag- ments with blunt ends: (5’)—G(3’) and (5’)C——(3’) “c G__ (e) The leftahand DNA fragment in (b) can be joined to the right-hand fragment in (d) to yield (5’)——GAATTC——(3’) ——CTTAAG— The same recombinant DNA molecule is produced by joining the right-hand fragment in (b) to the left—hand fragment in (d). (f) The recognition sequence for PvuII is (5’)CAGCTG(3’), with the cleavage site between G and C (see Table 9—2). Thus, a DNA molecule with a Pvull site will yield two frag- ments when digested with PvulI: (5’)—-—CAG (3’) and (5’)CTG—(3’) ~GTC GAG-— (g) The left-hand DNA fragment in (b) can be joined to the right—hand fragment in (f) to yield (5’)—~GAATTCTG——(8’) ~CTTAAGAC— The same recombinant DNA is produced by joining the right—hand fragment in (b) to the left-hand fragment in (f): (5 ' )--CAGAATTC-— (3 ’) ~GTCTTAAG— (11) There are two ways to convert an EcoRI restriction site to a BamHI restriction site. Method 1: Digest DNA with ECORI, and then create blunt ends by using either DNA polymerase I to fill in the single—stranded region as in (b) or a single-strand-specific nu- clease to remove the single—stranded region as in (d). Ligate a synthetic linker that con— tains the BamHI recognition sequence (5’)GGATCC(3’) (see Table 9—2), (5’)GCGGATCCCG(3’) CGCCTAGGGC 0) Chapter 9 DNA-Based Information Technologies 8-73 between the two blunt-ended DNA fragments to yield, if the EcoRl-digested DNA is treated as in (b), (5')—GAATTGCGGATCCCGAATTC—-(3’) —-CTTAACGCCTAGGGCTTAAG— or, if the EcoRI-digested DNA is treated as in (d), (5’)—GGCGATCCCG(3’) —CCGCCTAGGGC Notice that the EcoRI site is not regenerated after ligation of the linker. Method 2: This method uses a “conversion adaptor” to introduce a BamHI site into the DNA molecule. A synthetic oligonucleotide with the sequence (5’)AATTGGATCC(3’) is partially self-complementary, and it spontaneously forms the structure (5’)AATTGGATCC CCTAGGTTAA The sticky ends of this adaptor are complementary to the sticky ends generated by EcoRl digestion, so the adaptor can be ligated between the two EcoRI fragments to form (5 ')—GAATTGGATCCAATT—— (8 ’) ———CTTAACCTAGGTTAA— Because ligation between DNA molecules with compatible sticky ends is more efficient than ligation between DNA molecules with blunt ends, Method 2 is preferred over Method 1. Joining of the DNA fragments in (a) to a fragment generated by Pstl digestion requires a conversion adaptor. This adaptor should contain a single-stranded region complementary to the sticky end of an EcoRI-generated DNA fragment, and a single—stranded region complementary to the sticky end generated by Pstl digestion. The four adaptor se- quences that fulfill this requirement are shown below, in order of discussion in the prob— lem (N = any nucleotide): (5’)AATTCNNNNCTGCA GNNNNG (5’)AATTCNNNNGTGCA GNNNNC (5’)AATTGNNNNCTGCA CNNNNG (5’)AATTGNNNNGTGCA CNNNNC For the first adaptor: Ligation of the adaptor to the EcoRI—digested DNA molecule would yield (5’)——GAATTCNNNNCTGCA(3’) ——CTTAAGNNNNG This product can now be ligated to a DNA fragment produced by a Pstl digest, which has the terminal sequence (5’) G—(3’) ACGTC— to yield (5’)——GAATTCNNNNCTGCAG——(3’) ——CTTAAGNNNNGACGTC— 8-74 Chapter 9 DNA-Based Information Technologies Notice that the EcoRI and P321 sites are retained. In a similar fashion, each of the other three adaptors can be ligated to the EcoRI— digested DNA molecule, and the ligated molecule joined to a DNA fragment produced by a Pstl digest. The final products are as follows. For the second adaptor.- (5’)——GAATTCNNNNGTGCAG—-(3’) "OTTAAGNNNNCACGTCw The EcoRI site is retained, but not the Pstl site. For the third adaptor: ' (5’)——GAATTGNNNNCTGCAG——(3’) ——CTTAACNNNNGACGTC—- The Pstl site is retained, but not the EcoRI site. For the fourth adaptor: (5’)——GAATTGNNNNGTGCAG——(3’) —CTTAACNNNNCACGTC——— Neither the EcoRI nor the Pstl site is retained. 2. Selecting for Recombinant Plasmids When cloning a foreign DNA fragment into a plasmid, it is often useful to insert the fragment at a site that interrupts a selectable marker (such as the tetracy- cline-resistance gene of pBR322). The loss of function of the interrupted gene can be used to identify clones containing recombinant plasmids with foreign DNA. With a bacteriophage )\ vector it is not nec- essary to do this, yet one can easily distinguish vectors that incorporate large foreign DNA fragments from those that do not. How are these recombinant vectors identified? Answer Bacteriophage )\ DNA can be packaged into infectious phage particles only if it is be- tween 40,000 and 53,000 bp long. The two essential pieces of the bacteriophage )\ vector have about 30,000 bp in all, so the vector is not packaged into phage particles unless the additional, foreign DNA is of sufficient length: 10,000 to 23,000 bp. . DNA Cloning The plasmid cloning vector pBR322 (see Fig. 9—4) is cleaved with the restriction endonuclease Pstl. An isolated DNA fragment from a eukaryotic genome (also produced by Pstl cleavage) is added to the prepared vector and ligated. The mixture of ligated DNAs is then used to transform bacteria, and plasmid-containing bacteria are selected by growth in the presence of tetracycline. (a) In addition to the desired recombinant plasmid, what other types of plasmids might be found among the transformed bacteria that are tetracycline resistant? How can the types be distin— guished? (b) The cloned DNA fragment is 1,000 bp long and has an EcoRI site 250 bp from one end. Three different recombinant plasmids are cleaved with EcoRI and analyzed by gel electrophoresis, giv- ing the patterns shown. What does each pattern say about the cloned DNA? Note that in pBR322, the Pstl and EcoRI restriction sites are about 750 bp apart. The entire plasmid with no cloned insert is 4,361 bp. Size markers in lane 4 have the number of nucleotides noted. Chapter 9 DNA-Based Information Technologies ; Nucleotide 2 3 4 length 5,000 3,000 Electrophoresis 1,500 1,000 750 500 250 Answer (a) (b) Ligation of the linear pBR322 to regenerate circular pBR322 is a unimolecular process and thus occurs more efficiently than the ligation of a foreign DNA fragment to the linear pBR322, which is a bimolecular process (assuming equimolar amounts of linear pBR322 and foreign DNA in the reaction mixture). The tetracycline-resistant bacteria would include recombinant plasmids and plasmids in which the original pBR822 was regenerated without insertion of a foreign DNA fragment. (These would also retain resistance to ampicillin.) In addition, two or more molecules of pBRBZZ might be ligated together with or without insertion of foreign DNA. The clones giving rise to the patterns in lanes 1 and 2 each have one DNA fragment inserted, but in different orientations (see the diagrams, which are not drawn to scale; keep in mind that the products on the gel are from EcoRI cleavage). The clone produc- ing the pattern in lane 8 has two DNA fragments, ligated such that the EcoRI—site proximal ends are joined. PstI EcoRI ECORI EcoRI PstI PstI Clone in lane 1 EcoRI PstI PstI Clone in lane 2 Clone in lane 3 3-75 8-76 Chapter 9 DNA~Based Information Technologies 4. Identifying the Gene for a Protein with a Known Amino Acid Sequence Using Figure 27—7 to translate the genetic code, design a DNA probe that would allow you to identify the gene for a protein with the following amino~terminal amino acid sequence. The probe should be 18 to 20 nucleotides long, a size that provides adequate specificity if there is sufficient homology between the probe and the gene. llaN+ Ala Pro Met Thr Trp Tyr Cys Met Asp Trp lle Ala Gly Gly Pro Trp Phe Arg Lys Asn—Thr~Lys— Answer Most amino acids are encoded by two or more codons (see Fig. 27—7). To minimize the ambiguity in codon assignment for a given peptide sequence, we must select a region of the peptide that contains amino acids specified by the smallest number of codons. Focus on the amino acids with the fewest codons: Met and Trp (see Fig. 27~7 and Table 27—3). The best possibility for a probe is a span of DNA from the codon for the first Trp residue to the first two nucleotides of the codon for lie. The sequence of the probe would be (5’)UGG UA(U/C) UG(U/C) AUG GA(U/C) UGG AU The synthesis would be designed to incorporate either U or C where indicated, producing a mixture of eight ZO—nucleotide probes. . Designing a Diagnostic Test for a Genetic Disease Huntington’s disease (HD) is an inherited neurodegenerative disorder, characterized by the gradual, irreversible impairment of psychological, motor, and cognitive functions. Symptoms typically appear in middle age, but onset can occur at almost any age. The course of the disease can last 15 to 20 years. The molecular basis of the disease is becoming better understood. The genetic mutation underlying HD has been traced to a gene encoding a protein (Mr 350,000) of unknown function. In individuals who will not develop HD, a region of the gene that encodes the amino terminus of the protein has a sequence of GAG codons (for glutamine) that is repeated 6 to 39 times in succession. In individuals with adult—onset HD, this codon is typically repeated 40 to 55 times. In individuals with childhood-onset HD, this codon is repeated more than 70 times. The length of this simple trinucleotide repeat indicates whether an individual will develop HD, and at approximately what age the first symptoms will occur. A small portion of the amino-terminal coding sequence of the 3,143-codon HD gene is given below. The nucleotide sequence of the DNA is shown, with the amino acid sequence corresponding to the gene below it, and the CAG repeat shaded. Using Figure 27—7 to translate the genetic code, outline a PCR—based test for HD that could be carried out using a blood sample. Assume the PCR primer must be 25 nucleotides long. By convention, unless otherwise specified, a DNA sequence encoding a protein is displayed with the coding strand (the sequence identical to the mRNA transcribed from the gene) on top such that it is read 5’ to 3’, left to right. 307 ATGGCGACCCTGGAAAAGCTGATGAAGGCCTTCGAGTCCCTCAAGTCCTTC lMATLEKLMKAFESLKSF 358 CAGCAGTTCCAGCAGCAGCAGCAGCAGCAGCAGCAGCAGCAGCAGCAGCAG 18QQFQQQQQQQQQQQQVQWQ 409 CAGCAGCAGCAGCAGCAGCAGCAACAGCCGCCACCGCCGCCGCCGCCGCCG 35QQQQQQQQQPPPPPPPP 460 CCGCCTCCTCAGCTTCCTCAGCCGCCGCCG 52PPPQLPQPPP Source: The Huntington’s Disease Collaborative Research Group. (1993) A novel gene containing a trinucleotide repeat that is expanded and unstable on Huntington’s disease chromosomes. Cell 72, 971~983. Chapter 9 DNA—Based Information Technologies 3-77 Answer Yoru‘ test would require DNA primers, a heat—stable DNA polymerase, deoxynucleo— side triphosphates, and a PGR machine (thermal cycler). The primers would be designed to amplify a DNA segment encompassing the CAG repeat. The DNA strand shown is the coding strand, oriented 5’—-> 3’ left to right. The primer targeted to DNA to the left of the repeat would be identical to any 25-nuc1eotide sequence shown in the region to the left of the GAG repeat. Such a primer will direct synthesis of DNA across the repeat from left to right. The primer on the right side must be complementary and antipamllel to a 25—nucleotide sequence to the right of the GAG repeat. Such a primer will direct 5’—> 8’ synthesis of DNA across the repeat from right to left. Choosing unique sequences relatively close to the GAG repeat will make the amplified region smaller and the test more sensitive to small changes in size. Using the primers, DNA including the GAG repeat would be amplified by PGR, and its size would be determined by comparison to size markers after electrophoresis. The length of the DNA' would reflect the length of the GAG repeat, providing a simple test for the disease. Such a test could be carried out on a blood sample and completed in less than a day. 6. Using PGR to Detect Circular DNA Molecules In a species of ciliated protist, a segment of genomic DNA is sometimes deleted. The deletion is a genetically programmed reaction associated with cellular mating. A researcher proposes that the DNA is deleted in a type of recombination called site- specific recombination, with the DNA on either end of the segment joined together and the deleted DNA ending up as a circular DNA reaction product. proposed reaction w > Suggest how the researcher might use the polymerase chain reaction (PGR) to detect the presence of the circular form of the deleted DNA in an extract of the protist. Answer Design PGR primers complementary to DNA in the deleted segment, but which would direct DNA synthesis away from each other. No PGR product will be generated unless the ends of the deleted segment are joined to create a circle. 7. RFLP Analysis for Paternity Testing DNA fingerprinting and RFLP analysis are often used to test for paternity. A child inherits chromosomes from the mother and the father, so DNA from a child dis~ plays restriction fragments derived from each parent. In the gel shown here, which child, if any, can be excluded as being the biological offspring of the putative father? Explain your reasoning. Lane M is the sample from the mother, F from the putative father, and 01, OZ, and G8 from the children. 8-78 Chapter 9 DNA-Based Information Technologies 3W3? Answer None of the children can be excluded. Each child has one band that could be derived from the father. 8. Mapping a Chromosome Segment A group of overlapping clones, designated A through F, is isolated from one region of a chromosome. Each of the clones is separately cleaved by a restriction enzyme and the pieces resolved by agarose gel electrophoresis, with the results shown in the figure. There are nine different restriction fragments in this chromosomal region, With a subset appearing in each clone. Using this information, deduce the order of the restriction fragments in the chromosome. Overlapping clones “a D ‘E i A B C L, .m, em 1 Nine m 2 restriction fragments m m 6 5' W, 8 :l 9 W Answer Solving a restriction fragment map is a logic puzzle. The agarose gel shows which fragments are part of each clone, but deducing their order on the chromosome takes some work. Clone A: fragments 1, 3, 5, 7, and 9 Clone B: fragments 2, 3, 4, 6, 7, and 8 Clone C: fragments l, 3, 4, 5,'and 7 Clone D: fragments 2, 3, , 5, and 7 Clone E: fragments l, 5, and 9 Clone F: fragments 2, 4, 6, and 7 Chapter 9 DNA—Based Information Technologies 8—79 Begin with clone E, which has the fewest fragments. Fragments 1, 5, and 9 must be adjacent, but may be in any order. However, clone C includes fragments 1 and 5, but not 9, so we can conclude that fragments l and 5 are adjacent; 9 could be adjacent to either 1 or 5, but is not between them (i.e., the order is 9-1-5 or 1-5-9). Clones C and D are identical except that C also includes fragment 1, and D also includes fragment 2; from this we can deduce that frag— ments 1 and 2 must be at opposite ends of the overlapping region, which includes fragments 8, 4, 5, and 7 (in an as yet undetermined order). Because we have already concluded that fragments 1 and 5 are adjacent, we can propose the follong sequence: 1-5—(3,4,7)—2. To find the order of fragments 3, 4, and 7, look for fragments that contain a subset of these with a flanking fragment. Clone A includes fragments 5, 3, and 7, but not 4 (thus, 5—(3,7)-4); clone F includes fragments 4, 7, and 2 (thus, (4,7)~2). Combining these possibili— ties allows us to deduce the order as 1—5-3~7~4—2. We concluded earlier that 9 is adjacent to either 1 or 5. If it were adjacent to 5, it would be a part of clones C and D; because it is not in C or D, it must be adjacent to 1. We can now propose the following sequence: 9-1—5-3—74-2. Because clone F includes these last three fragments and fragment 6, we can append 6 after 2: 9—1—5-3—7—4—2-6. Finally, clone B is the only one that includes fragment 8, so it must occur at either end of our deduced sequence. Given the other fragments in clone B, fragment 8 must be adjacent to fragment 6. Thus, we have the order 9—1—5~3—7~4-2—6—8. The fragments were numbered based on their migration distance in the gel, which corre- lates inversely with the size of the fragment. Fragment 1 is the longest; fragment 9 the short— est. The relative sizes and the positions of the fragments on the chromosome are shown be— low. Molecular weight markers in the gel would allow a better estimation of the sizes of the various fragments. 9. Cloning in Plants The strategy outlined in Figure 9—28 employs Agrobactertum cells that contain two separate plasmids, Suggest why the sequences on the two plasmids are not combined on one plasmid. Answer Simply for convenience; the 200,000 bp Ti plasmid, even when the T DNA is re- moved, is too large to isolate in quantity and manipulate in vitro. It is also too large to reintro« duce into a cell by standard transformation techniques. Single—plasmid systems in which the T DNA of a Ti plasmid has been replaced by foreign DNA (by low~efficiency recombination in vivo) have been used successfully, but this approach is very laborious. The m'r genes can facil- itate transfer of any DNA between the T DNA repeats, even if they are on a separate plasmid. The second plasmid in the two—plasmid system, because it requires only the T DNA repeats and a few sequences necessary for plasmid selection and propagation, is relatively small, easily isolated, and easily manipulated (foreign DNA is easily added and/or altered). It can be propa- gated in E. col/i or Agrobdctem'um and is readily reintroduced into either bacterium. 5-80 Chapter 9 DNA-Based Information Technologies 10. DNA Fingerprinting and RFLP Analysis DNA is extracted from the blood cells of two humans, indi— viduals l and 2. In separate experiments, the DNA from each individual is cleaved by restriction endonu- cleases A, B, and C, and the fragments separated by electrophoresis. A hypothetical map of a 10,000 bp segment of a human chromosome is shown (1 kbp = 1,000 bp). Individual 2 has point mutations that eliminate restriction recognition sites B* and 0*. You probe the gel with a radioactive oligonucleotide complementary to the indicated sequence and expose a piece of X-ray film to the gel. indicate where you would expect to see bands on the film. The gel lanes are marked in the accompanying diagram. M121212 A B C Answer Cleaving DNA with restriction enzyme A produces identical fragments in both indi- viduals: 65 kbp and 3.5 kbp fragments. The probe hybridizes to both 6.5 kbp fragments, re- sulting in two identical bands in column A. Restriction enzyme B produces different cleavage products: DNA from individual 1 is cleaved into 8, 2, and 4 kbp fragments; that from individual 2 (who has an altered B recognition sequence) into 3 and 6 kbp fragments. However, the probe binds to the 3 kbp fragments from both individuals and therefore produces the same pattern of bands on the gel (in column B). Restriction enzyme 0 cleaves DNA from individual 1 into 2.5 and 4.5 kbp fragments, and the probe labels the 2.5 kbp piece. DNA from individual 2, however, is cleaved to produce a single 7 kbp fragment, which hybridizes with the probe. Thus, only in column C does a difference in DNA sequence between individuals 1 and 2 be— come apparent. This exercise points out the importance of the choice of restriction enzymes, as well as the choice of probes, when performing DNA fingerprinting and RFLP analysis. Chapter 9 DNA-Based Information Technologies 5-81 bPO‘lCDQCOCDO OJ M121212 A B C 11. Use of Photofithography to Make a DNA Microarray Figure 9—21 shows the first steps in the process of making a DNA microarray using photolithography. Describe the remaining steps needed to obtain the desired sequences (a different four-nucleotide sequence on each of the four spots) shown in the first panel of the figure. After each step, give the resulting nucleotide sequence attached at each spot. Answer Cover spot 4, add solution containing activated T, irracliate, and wash. The resulting sequences are now 1. A-T 2. G-T . 3. AT 4. GO Cover spots 2 and 4, add solution containing activated G, irradiate, and wash. 1. A~T~G 2. G-T 3. A~T—G 4. GO Cover spot 3, add solution containing activated C, irradiate, and wash. 1. A~T-G—C 2. G—T—C 3. A—T—G 4. G-C—C Cover spots 1, 3, and 4, acid solution containing activated C, irradiate, and wash. 1. A-T—G—C 2. G~T~C—C 3. A-T—G 4. G-C—C Cover spots 1 and 2, add solution containing activated G, irradiate, and wash. 1. A—T—G-C 2. G—T—C-C 3. A-T—G—C 4. G—C-C~C 12. Cloning in Mammals The retroviral vectors described in Figure 9—32 make possible the efficient integration of foreign DNA into a mammalian genome. Explain how these vectors, which lack genes for replication and viral packaging (gag, pol, 67w), are assembled into infectious viral particles. Suggest Why it is important that these vectors lack the replication and packaging genes. Answer The retroviral vectors must be introduced into a cell infected with a helper virus that can provide the necessary replication and packaging functions but cannot itself be packaged. The vectors packaged into infectious viral particles are then used to introduce the recombi- nant DNA into a mammalian cell. Once this DNA is integrated into the target cells chromo- some, the absence of replication and packaging functions makes the integration very stable by preventing deletion or replication of the integrated DNA. ...
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This note was uploaded on 02/13/2008 for the course BIOBM 3310 taught by Professor Feigenson,gw during the Fall '07 term at Cornell University (Engineering School).

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Ch.9 Lehringer solutions - 1. Cloning When joining two or...

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