ch9 - .l 1; DNA-Based Information Technologies 951%“???...

Info iconThis preview shows pages 1–13. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 6
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 8
Background image of page 9

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 10
Background image of page 11

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 12
Background image of page 13
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: .l 1; DNA-Based Information Technologies 951%“??? it t i: i i. 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 foilowing exercises. {3) Draw the structure of each end of a iinear DNA fragment produced by an EcoRI restriction digest (include those sequences remaining from the EcoRI recognition sequence). (b) Draw the structure resulting from the reaction of this end sequence with DNA polymerase I and the four deoxynucieoside triphosphates (see Fig. 8—38}. (c) Draw the sequence produced at the junction that arises if two ends with the structure derived in (b) are ligated (see Fig. 25—17). ((1) Draw the structure produced if the structure derived in (a) is treated with a nuclease that degrades only single—stranded DNA. (e) Draw the sequence of the junction produced if an end with structure (b) is iigated to an end with structure ((1). (1’) Draw the structure of the end of a linear DNA fragment that was produced by a Pauli restriction digest (include those sequences remaining from the Pauli recognition sequence). (g) Draw the sequence of the iunction produced if an end with structure (b) is iigated to an end with structure (1’). (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 BomHl restriction site at approximately the same iocation. (See Fig. 9—2.) (i) Design four different short synthetic double-stranded DNA fragments that would permit ligation of structure (a) with a DNA fragment produced by a Perl 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 EcoRl nor the Pstl sequence appears in the junction. Answer Type II restriction enzymes cleave double-stranded DNA witlun recognition sequences to create either blunt—ended or sticky-ended fragments. Biunt—ended DNA frag- ments can be ioined by the action of T4 DNA ligase. Sticky-ended DNA fragments can be joined by either 15'. cold or T4 DNA ligases, provided that the sticky ends are complementary. Sticky—ended fragments without complementary ends can he joined oniy after the ends are made blunt, either by exonucieases or by E. colt DNA polymerase I. (a) The recognition sequence for EcoRl is (5')GAATTC{3'), with the cleavage site between G and A (see Table 9—2). Thus, digestion of a DNA moiecuie with one EcoRi site 5-100 Chapter 9 DNA-Based Information Technologies 0)) (c) (d) (e) (f) (8) (h) (5’)---GAATTC—-~(3’) (3’)-~-CTTAAG---(5') wouid yield two fragments: (5')~~G (3’) (5’)AATTC---(8') (3')--~CTTAA(5') (3') G—--(5') 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 biunt ended: (5')—~-GAATT(3') (5’)AATTC—~~(3') (3')---CTTAA(5') (3’)TTAAG---(5') The two fragments generated in (b) can be ligated by T4 DNA ligase to form ‘ (5')---GAATTAATTC---(3') (3’)~-CTTAATTAAG~~~(5’) The fragments in (a) have sticky ends, with a protruding singie—stranded region. Treat- ment of these DNA fragments with a singlewstrend-specific nuclease will yield DNA frag ments with blunt ends: ‘ (53-43(3) (5') C-"(3’D (3’)”"CC5'D (3’)G“*(5') The left-hand DNA fragment in (b) can be joined to the right-hand fragment in (d) to yield and end and {5’)---GAATTC~M(3’) (3')«-«CTTAAG-—-(5’) The same recombinant DNA molecule is produced by joining the righthand fragment in (b) to the ieftahand fragment in (d). The recognition sequence for Pouii is (5')CAGCTG(3'), with the cleavage site between G and C (see Table 9—2). Thus, a DNA molecule with a Pqu site will yield two frag- ments when digested with Poul}: (5')~«~CAG(3’) (5’)C'1‘Gw(3’) {8')-—-GTC{5') (3’)GAC--—(5') The left-hand DNA fragment in (b) can be joined to the right—hand fragment in (f) to yieid - (5')W~GAATTCTG-—-(3’) (3')---CTTAAGAC---(5') The same recombinant DNA is produced by joining the right~hend fragment in (b) to the ieft—hand fragment in (f): and (5’)--—CAGAATTCM(3') (8’)WGTCTTAAG-~(5’) There are two ways to convert an EcoRI restriction site to a Bamlii restriction site. Method 1: Digest DNA with EcoRI, and. then create blunt ends by using either DNA polymerase I to fiil in the singleetranded region as in (b) or e singleetrand—specific nu— clease to remove the singleetranded region as in (d). Ligate a synthetic linker that con tame the BamHi recognition sequence (5’}GGATCC{3') (see Table 9-2), (5’)GCGGATCCCG (3’) (3’)CGCGTAGGGC{5') (i) ChapterQ DNA-Basedlnformation Technologies 5—101 between the two brunt-ended DNA fragments to yield, if the EcoRI—digested DNA is treated as in (b), (5’)---GAATTGCGGATCCCGAATTC---(3') (3’}—-—CTTAACGCCTAGGGCTTAA «(5') or, if the EcoRi—digested DNA is treated as in (d), (5’)——-GGCGGATCCCG(3‘} (3')---coooomoooccs') Notice that the EcoRI site is not regenerated after ligation of the linker. Method 2: This method uses a “conversion adaptor” to introduce a BdeE site into the DNA molecule. A synthetic oligonucieotide with the sequence C5’)AATTGGATCC(8') is partiain selfmcomplementary, and it spontaneously forms the structure (5’)AATTGGATCG(3’) (3’) CCTAGGTTAA(5’) The sticky ends of this adaptor are complementary to the sticky ends generated by EcoRI digestion, so the adaptor can be ligated between the two EcoRl fragments to form (5’)-~—GAATTGGATCCAATTW(3’) (a')n-orTAAcc'rAeorTAA-uce) Because ligation between DNA molecules with compatibie sticky ends is more emcient 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 and of an EcoRI—generated DNA fragment, and a single~stranded region comptementary 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« Eem (N a any nucleotide): (somrrcnnnncrocscs') (3’)GNNNNG(5’) (5’)AATTCNNNNGTGCA(3') (3')GNNNNC{5’) (5’)AATTGNNNNOTGCA(3') (3’)CNNNNG(5') (5’)AATTGNNNNGTGCA{8’) (snowman) For theftrst adaptor: Ligation of the adaptor to the Boom-digested DNA molecule would yield (5’)——-GAATTCNNNNCTGCA(3') (3')—~~CTTAAGNNNNG(5’) This product can now be ligated to a DNA fragment produced by a P331 digest, which has the terminal sequence (roe-"(3') (3’)ACGTC--~(5') to yield (5’)---GAATTCNNNNCTGCAG---(3’) (3’)---CTTAAGNNNNGACGTC---(5’) 8-102 Chapterg DNA~Basedlnformation Technologies Notice that the EcoRl and Pstl 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 PstI digest. The final products are as follows. For the second adaptor.- (5o“can'rrcnwnorecaomrsr) (3’)~«-CTTAAGNNNNCACGTC~~(5’) The EcoRi site is retained, but not the Pstl site. For the third adaptor: {EU-"GAATTGNNNNCTGCA «"(3‘) (3’)--—CTTAACNNNNGACGTC—~- (5’) The Psti site is retained, but not the EcoRi site. For thefourth adaptor: ' (5’)—wwGAAT’l‘GNNNNGTGCAG—--(3’) (3')~—-CTTAACNNNNCACGTCw--(5’) 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 ioss of function of the interrupted gene can be used to identify clones containing recombinant plasmids with foreign DNA. With a bacteriophage it 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 A rector have about 30,000 bp in all, so the vector is not packaged into phage particles unless the additional, foreign DNA is of sufficient iength: 10,000 to 23,000 bp. 3. DNA Cloning The plasmid cioning vector pBR322 (see Fig. 9—3) is cleaved with the restriction endonuclease Pstl. An isolated DNA fragment from a eukaryotic genome (also produced by Psti cieavage) is added to the prepared vector and ligated. The mixture of ligated DNAs is then used to transform bacteria, and plasmid-contaming 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 3 among the transformed bacteria that are tetracycline resistant? How can the types be distin« " gnished? (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 get electrophoresis, givm 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 hp. Size markers in lane 4 have the number of nucleotides noted. Chapter 9 DNA-Based Information Technologies 8-103 Nucieotide length 5,000 3,000 Electrophoresis 1,500 1,000 750 500 250 Answer (a) 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 bimoiecuiar process (assuming equimolar amounts of linear pBR322 and foreign DNA in the reaction mixture). The tetracycline-resistant bacteria would include recombinant plasmids and piasmids in which the original pBR322 was regenerated Without insertion of a foreign DNA fragment. (These would aiso retain resistance to ampicillin.) in adoition, two or more molecules of pBR322 nfight be ligated together with or Without insertion of foreign DNA. (1)) The clones giving rise to the patterns in lanes 1 anti 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, iigated such that the EcoRl—site proximal ends are joined. EcoRI Pstl Clone in lane 2 Clone in lane 3 $404 ChapterQ DNAwBased information Technologies 4. Identifying the Gene for a Protein with 2 Known Amino Acid Sequence Using Figure 27%? to translate the genetic code, design a DNA probe that would aliow you to identify the gene for a protein with the toliowing amino-terminal amino acid sequence. The probe shouid be 18 to 20 nucleotides long, a size that provides adequate specificity if there is sufficient homoiogy between the probe and the gene. H3N+—AlawPromMet—Thr—Trp—'l‘yrm~Cys-Met—Asp~’lrpmllem Aia—G1y-Gly~Pro—TrpwPhemAJg—Lys~Asn—ThrmLys— 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 smaliest number of codons. Focus on the amino acids with the fewest codons: Met and Trp (see Fig. 27—7 and Tabie 27MB). The best possibility for a probe is a span of DNA from the codon for the first ’i‘rp residue to the first two nucleotides of the codon for lie. The sequence of the probe would be (5’)UGG UACU/C) UGCU/C) AUG GA(U/C) UGG AU The synthesis would be designed to incorporate either [i or C where indicated, producing a mixture of eight 20wnucleotide probes. 5. Designing 3 Diagnostic Test for a Genetic Disease Huntington’s disease (ED) is an inherited neurodegenerative disorder, characterized by the graduai, irreversible impairment of psychological, motor, and cognitive functions. Symptoms typicaliy appear in middle age, but onset can occur at almost any age. The course of the disease can last 15 to 20 years. The molecuiar basis of the disease is becoming better understood. The genetic mutation underiying Hi) has been traced to a gene encoding a protein {II/Ir 350,000) of unknown function. In individuais who wiil not dcveiop HI), 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 individuais with adult-onset HI), this codon is typically repeated 40 to 55 times. in individuals with childhoodpnset HI), this codon is repeated more than 70 times. The length of this simple trinucieotide repeat indicates whether an individual will develop HD, and at approximately what age the first symptoms wili occur. A small portion of the amino-terminal coding sequence of the 3,143-codon H1) 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"? to translate the genetic code, outline a FOR-based test for HD that could be carried out using a biood sample. Assume the PCR primer must be 25 nucleotides iong. 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 ATGGCGACCCTGGAAAAGCTGATGAAGGCCTTCGAG‘I‘CCCTCAAGTCCTTC lMATLEKLMKAFESLKSF 358 CAGCAorchAooioon ' Adelaide” ' " H 18 a Q F sigg__' a AGCCGCCACCGCCGCCGCCGCCGCCG 409 CAGCAGCA‘ CA "it 'A "A" ._ . “eggs P P P P P P P 460 CCGCCTCC‘I‘CAGC’ITCCTCAGCCGCCGCCG 52PPPQLP‘QPPP Source: The Huntington‘s Disease Collaborative Research Group. (1998) A novel gene containing a trinucleotide repeat that is expanded and unstable on Huntington’s disease chromosomes. Cell 72, 971—983. ChapterQ DNA-Based information Technologies 8-105 Answer Your test would require DNA primers, a heat~stable DNA polymerase, deoxynucleo~ side triphosphates, and a PCR 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’—> 8' left to right. The primer targeted to DNA to the left of the repeat would be identical to any 25-nucleotide sequence shown in the region to the left of the CAG 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—mic1eotide sequence to the right of the CAG repeat. Such a primer will direct 5’—> 3’ synthesis of DNA across the repeat from right to left. Choosing unique sequences relatively close to the CAG repeat Mil make the amplified region smaller and the test more sensitive to small changes in size. Using the primers, DNA including the CAG repeat would be amplified by PCR, and its size would be determined by comparison to size markers after electrophoresis. The length of the DNA would reflect the length of the CAG 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 PCR 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 ioined 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 (PCR) to detect the presence of the circuiar 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 PCR product will be generated uniess the ends of the deleted segment are ioined to create a circle. 7. Glowing Plants When grown in ordinary garden soil and watered normally, a plant engineered to express green fluorescent protein (see Fig. 9—15a) will glow in the dark, whereas a plant engineered to express firer luciferase (see Fig. 9—29) will not. Explain these observations. Answer The plant expressing firefly luciferase must take up luciferin, the substrate of ill- ciferase, "before it can “glow” (albeit weakly). The plant expressing green fluorescent protein glows without requiring any other compound. 8. 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 disw 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 median)? from the putative father, and Cl, 02, and 03 from the children. 3-106 Chapterg DNA-Based information 'fechnologies A"? 3?} t1??? 1'77; Electrophoresis Answer None of the children can be excluded. Each chiid has one band that could be derived from the father. 9. 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 below. 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 1 Nine 5; 2 restriction fragments 3 .2 ii i 1 3‘ ' 5 E ‘ 5 a '7 8 9 Answer Solving a restriction fragment map is a logic puzzle. The agarose gei 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, anti 8 Clone C: fragments 1, 3, 4, 5, and 7 Clone D: fragments 2, 3, 4, 5, and 7 Clone E: fragments i, 5, and 9 Clone F: fragments 2, 4, 6, and 7’ Cbapterg DNA-Based information Technologies 3-107 Begin with clone E, which has the fewest fragments. Fragments i, 5, and 9 must be adjacent, but may be in any order. However, clone 0 includes fragments 1 and 5, but not 9, so we can conclude that fragments 1 and 5 are adjacent; 9 could be adiacent to either 1 or 5, but is not between them (i.e., the order is 9-1-5 or i-5-9). Clones C and D are identicai 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 3, 4, 5, and 7 (in an as yet undetermined order). Because we have already concluded that fragments l and 5 are adjacent, we can propose the following sequence: l—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 flanldng 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)12). Combining these possibili- ties allows us to deduce the order as 1-5—8-7-4—2. We conciuded 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 I); because it is not in C or D, it must be adiacent to 1. We can now propose the following sequence: 9—1-5—3-7-4-2. 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. rl‘hus, 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 l is the longest; fragment 9 the short— est. The relative sizes and the positions of the fragments on the chromosome are shown be- iow. Molecular weight markers in the gel would allow a better estimation of the sizes of the various fragments. 915374268 H—_+WW~+-_—i—s———+———-—i——+—i :———~«~WA_————s l—-—-—-—-——F——-—i 10. Cloning in Hams The strategy outlined in Figure 9—28 employs Agrobdctcrrlum 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 iarge to isoiate in quantity and manipulate in vitro. It is also too iarge to reintro- duce into a cell by standard transformation techniques. Single-piasnud systems in which the T DNA of a T1 plasmid has been replaced by foreign DNA (by low—efficiency recombination in vivo) have been used successfully, but this approach is very laborious. The air genes can facil» itate transfer of any DNA between the ’1‘ 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 seiection 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. colt or Agrobdctertum and is readiiy reintroduced into either bacterium. 3—108 Chapter9 DNA-Based information Technoiogies 11. DNA Fingerprinting and RFLP Analysis DNA is extracted from the blood cells of two humans, indi— viduals 1 and 2. In separate experiments, the DNA from each individual is cieaved by restriction endemi- 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 m 1,000 hp). lndividuai 2 has point mutations that eiiminate restriction recognition sites 3* 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 ianes of the gel are marked in the accompanying diagram. M 1A2 1132 102 Answer Cleaving DNA with restriction enzyme A produces identical fragments in both indi- viduals: 6.5 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 cieavage products: DNA from individual I is cleaved into 3, 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 coiurnn B). Restriction enzyme C cleaves DNA from individual i into 2.5 and 4.5 kbp fragments, and the probe iabels 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, oniy in column C does a difference in DNA sequence between individuais 1 and 2 be- come apparent. This exercise points out the importance of the choice of restriction enzymes, as weli as the choice of probes, when performing DNA fmgerprinting and RFLP analysis. Chapter 9 DNA-Based Information Technoiogies 8-109 A B C M121212 ea exam-ammo N) 12. Use of Photofithography to Make a DNA Microarray Figure 9W2} shows the first steps in the process of making a DNA microarray, or DNA chip, using photoiithography. Describe the remaining steps needed to obtain the desired sequences (a different four-nucieotide sequence on each of the four spots) shown in the first panel of the figure. After each step, give the resulting nucleotide se- quence attached at each spot. Answer Cover spot 4, add solution containing activated T, irradiate, and wash. The resulting sequences are now 1. A—T 2. (HP 3. AT 4. G~C Cover spots 2 and 4, add solution containing activated G, irradiate, and wash. 1. A~T~G 2. GT 3. AM'IEG 4. G—C Cover spot 8, acid solution containing activated C, irradiate, and wash. 1. A~TnG~C 2. GwT~C 3. A~T—G 4. G-C—C (lever spots 1, 3, and 4, add solution containing activated 0, irradiate, and wash. 1. A—T—G-C .2. G—T—C—C 3. A-T—G 4. (3-0-0 Cover spots 1 and 2, acid 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 13. Cloning in Mammals The retroviral vectors described in Figure 9—32 make possible the efficient integration of foreign DNA into a mammalian genome. Expiain how these vectors, which lack genes for replication and viral packaging (gag, poi, ens), 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 nifectioos viral particles are then used to introduce the recombi- nant DNA into a mammalian celi. Once this DNA is integrated into the target cell’s chromom some, the absence of replication and packaging functions makes the integration very stable by preventing deietion or replication of the integrated DNA. 8-110 Chapter 9 DNA-Based information Technologies Bata Anaiysis Problem 14. Hincfl: The First; Restriction Endonuciease Discovery of the first restriction endonuclease to be of practical use was reported in two papers published in 1970. In the first paper, Smith and Wilcox de- scribed the isolation of an enzyme that cleaved double-stranded DNA. They initialiy demonstrated the enzymes nuclease activity by measuring the decrease in viscosity of DNA sampies treated with the enzyme. (3) Why does treatment with a nuclease decrease the viscosity of a soiution of DNA? The authors determined whether the enzyme was an endo— or an exonuclease by treating 32P—labeied DNA with the enzyme, then adding trichloroacetic acid (TCA). Under the conditions used in their experi- ment, single nucleotides would be TGA—soiubie and oiigonucleotides would precipitate. (b) No TCA-soluble 32P-labeied material formed on treatment of 32l9-labe1ed DNA with the nuciease. Based on this finding, is the enzyme an endo- or exonuclease? Explain your reasorung. When a polynucleotide is cleaved, the phosphate usually is not removed but remains attached to the 5' or 8' end of the resulting DNA fragment. Smith and Wilcox determined the iocatlon of the phosphate on the fragment formed by the nuclease in the following steps: 1. ri‘reat unlabeled DNA with the nuclease. 2. Treat a sample (A) of the product with yw32P~labeied ATP and polynucieotide ldnase (which can attach the y—phosphate of ATP to a 5’ OH but not to a 5’ phosphate or to a 8’ OH or 3’ phosphate). Measure the amount of 32]? incorporated into the DNA. 3. Treat another sample (B) of the product of step I with aikaline phosphatase {winch removes phosphate groups from tree 5’ and 3’ ends), followed by polynucleotide kinase and y—3ZP~labeled ATP. Measure the amount of d‘3P incorporated into the DNA. (c) Smith and Wilcox found that sample A had 136 counts/min of 3213; sample 3 had 3,740 counts/min. Did the nuciease cleavage leave the phosphate on the 5’ or the 3’ end of the DNA fragments? Explain your reasoning. (d) Treatment of bacteriophage T7 DNA with the nuclease gave approximately 40 specific fragments of various lengths. How is this result consistent with the enzymes recognizing a specific sequence in the DNA as opposed to making random double-strand breaks? At this point, there were two possibilities for the site-specific cleavage: the cleavage occurred either (1) at the site of recognition or (2) near the site of recognition but not within the sequence recognized. To address this issue, Kelly and Smith determined the sequence of the 5’ ends of the DNA fragments generated by the nuclease, in the foliowing steps: 1. Treat phage T7 DNA with the enzyme. 2. Treat the resulting fragments with alkaline phosphatase to remove the 5’ phosphates. 3. Treat the dephosphorylated fragments with polynucleotide kinase and y-32Palabeled ATP to iabel the 5’ ends. 4. Treat the labeled molecuies with DNases to break them into a nuxture of mono, di—, and trinucleotides. 5. Determine the sequence of the iaheled mono, di—, and trinucieotides by comparing them with oligonucleotides of known sequence on thinalayer chromatography. The labeled products were identified as follows: mononucleotides: A and G; dinucleotides: 5'«pApA—3’ and 5’—pGpA-3’; trinucleotides: 5’—pApApr3’ and 5’~pGpApC—3’. (e) Which model of cleavage is consistent with these results? Explain your reasoning. Kelly and Smith went on to determine the sequence of the 3’ ends of the fragments. ri‘hey found a mixture of 5’-prC-3’ and 5'«prT—3’. They did not determine the sequence of any trinucieotides at the 3’ end. (1‘) Based on these data, what is the recognition sequence for the nuclease and where in the sequence is the DNA backbone cleaved? Use ’l‘abie 9% as a model for your answer. Chapter 9 DNA-Based information Technologies 3-111 Answer (a) DNA solutions are highly viscous because the very long moiecules are tangled in solu- tion. Shorter moiecules tend to tangle iess and form a less viscous solution, so decreased viscosity corresponds to shortening of the poiymers—as caused by nuclease activity. (1)) An endonuclease. An exonuciease removes single nucleotides from the 5’ or 3’ end and would produce TCAwsolnble 32P—labeied nucleotides. An endonuciease cuts DNA into oligonucleotide fragments and produces little or no TCA—solnbie 3ZP—iabeled material. (c) The 5' end. If the phosphate were left on the 3’ end, the kinase would incorporate signif~ icant 32F as it added phosphate to the 5' end; treatment with the phosphatase would have no effect on this. In this case, samples A and B would incorporate significant amounts of 321?. When the phosphate is left on the 5’ end, the kinase does not incorpd rate any 32F: it catmot add a phosphate if one is aiready present. Treatment with the phosphatase removes 5' phosphate, and the kinase then incorporates significant amounts of 3213’. Sample A Wili have little or no 3213, and B wiil show substantial 33F incorporationwas was observed. ((1) Random breaks would produce a distribution of fragments of random size. The produc- tion of specific fragments indicates that the enzyme is site-specific. (e) Cleavage at the site of recognition. This produces a specific sequence at the 5’ end of the fragments. If cleavage occurred near but not within the recognition site, the se« quence at the 5’ end of the fragments wouid be random. (f) The results are consistent with two recognition sequences, as shown below, cieaved where shown by the arrows: t (sou-err Attic-"(3') (sou—can TTG-——(5’) T which gives the (5')pApApC and (83’?pr fragments; and i (sou-ore onsets) (samplers ore-"(5') T which gives the (5')pGpApC and (39(3pr fragments. References Kelly, '1‘.J. & Smith, RD. (1976) A restriction enzyme from Hoemopht‘tus influenzae: Ii. Base sequence of the recognition site. J. Mot. Biol. 51, 393M499. Smith, H1). 8: Wilcox, KW. (£970) A restriction enzyme from Haemopm'tus influences: 1. Purification and general properties. J. Mol. Biol. 51, 379—391. ...
View Full Document

This note was uploaded on 08/02/2011 for the course BCHM 461 taught by Professor Gerratana during the Spring '11 term at University of Maryland Baltimore.

Page1 / 13

ch9 - .l 1; DNA-Based Information Technologies 951%“???...

This preview shows document pages 1 - 13. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online