DNA-Based Information Technologies

2608T_ch09sm_S99-S111 2/21/08 11:45AM Page S-99 ntt Os9:Desktop Folder:TEMPWORK:FEBRUARY:21-02-08:WHQY028/soln: chapter DNA-Based Information Technologies 9 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) Draw the structure of each end of a linear DNA fragment produced by an restriction digest (include those sequences remaining from the recognition sequence). (b) Draw the structure resulting from the reaction of this end sequence with DNA polymerase I and the four deoxynucleoside triphosphates (see Fig. 8 33). (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). (d) 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 ligated to an end with structure (d). (f) Draw the structure of the end of a linear DNA fragment that was produced by a PvuII restriction digest (include those sequences remaining from the PvuII recognition sequence). (g) Draw the sequence of the junction produced if an end with structure (b) is ligated to an end with structure (f). (h) 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 restriction site from a DNA molecule and incorporate a new BamHI restriction site at approximately the same location. (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 restriction digest. In one of these fragments, design the sequence so that the final junction contains the recognition sequences for both and. In the second and third fragments, design the sequence so that the junction contains only the and only the recognition sequence, respectively. Design the sequence of the fourth fragment so that neither the nor the 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 fragments can be joined by the action of T4 DNA ligase. Sticky-ended DNA fragments can be joined by either E. coli 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 is (5 )GAATTC(3 ), with the cleavage site between G and A (see Table 9 2). Thus, digestion of a DNA molecule with one site S-99

2608T_ch09sm_S99-S111 2/21/08 11:45AM Page S-100 ntt Os9:Desktop Folder:TEMPWORK:FEBRUARY:21-02-08:WHQY028/soln: S-100 Chapter 9 DNA-Based Information Technologies would yield two fragments: (5 ),GAATTC,(3 ) (3 ),CTTAAG,(5 ) (b) (c) (d) (e) (f) (g) (5 ),G(3 ) and (5 )AATTC,(3 ) (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 blunt ended: (5 ),GAATT(3 ) and (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 single-stranded region. Treatment of these DNA fragments with a single-strand-specific nuclease will yield DNA fragments with blunt ends: (5 ),G(3 ) and (5 )C,(3 ) (3 ),C(5 ) (3 )G,(5 ) The left-hand DNA fragment in (b) can be joined to the right-hand fragment in (d) to yield (5 ),GAATTC,(3 ) (3 ),CTTAAG,(5 ) The same recombinant DNA molecule is produced by joining the right-hand fragment in (b) to the left-hand fragment in (d). 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 PvuII site will yield two fragments when digested with PvuII: (5 ),CAG(3 ) and (5 )CTG,(3 ) (3 ),GTC(5 ) (3 )GAC,(5 ) The left-hand DNA fragment in (b) can be joined to the right-hand fragment in (f) to yield (5 ),GAATTCTG,(3 ) (3 ),CTTAAGAC,(5 ) The same recombinant DNA is produced by joining the right-hand fragment in (b) to the left-hand fragment in (f): (h) (5 ),CAGAATTC,(3 ) (3 ),GTCTTAAG,(5 ) There are two ways to convert an restriction site to a BamHI restriction site. Method 1: Digest DNA with, 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 nuclease to remove the single-stranded region as in (d). Ligate a synthetic linker that contains the BamHI recognition sequence (5 )GGATCC(3 ) (see Table 9 2), (5 )GCGGATCCCG(3 ) (3 )CGCCTAGGGC(5 )

2608T_ch09sm_S99-S111 2/21/08 11:45AM Page S-101 ntt Os9:Desktop Folder:TEMPWORK:FEBRUARY:21-02-08:WHQY028/soln: Chapter 9 DNA-Based Information Technologies S-101 between the two blunt-ended DNA fragments to yield, if the -digested DNA is treated as in (b), (5 ),GAATTGCGGATCCCGAATTC,(3 ) (3 ),CTTAACGCCTAGGGCTTAAG,(5 ) or, if the -digested DNA is treated as in (d), (5 ),GGCGGATCCCG(3 ) (3 ),CCGCCTAGGGC(5 ) Notice that the 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(3 ) (3 ) CCTAGGTTAA(5 ) The sticky ends of this adaptor are complementary to the sticky ends generated by digestion, so the adaptor can be ligated between the two fragments to form (i) (5 ),GAATTGGATCCAATT,(3 ) (3 ),CTTAACCTAGGTTAA,(5 ) 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 digestion requires a conversion adaptor. This adaptor should contain a single-stranded region complementary to the sticky end of an -generated DNA fragment, and a single-stranded region complementary to the sticky end generated by digestion. The four adaptor sequences that fulfill this requirement are shown below, in order of discussion in the problem (N any nucleotide): (5 )AATTCNNNNCTGCA(3 ) (3 )GNNNNG(5 ) (5 )AATTCNNNNGTGCA(3 ) (3 )GNNNNC(5 ) (5 )AATTGNNNNCTGCA(3 ) (3 )CNNNNG(5 ) (5 )AATTGNNNNGTGCA(3 ) (3 )CNNNNC(5 ) For the first adaptor: Ligation of the adaptor to the -digested DNA molecule would yield (5 ),GAATTCNNNNCTGCA(3 ) (3 ),CTTAAGNNNNG(5 ) This product can now be ligated to a DNA fragment produced by a digest, which has the terminal sequence to yield (5 )G,(3 ) (3 )ACGTC,(5 ) (5 ),GAATTCNNNNCTGCAG,(3 ) (3 ),CTTAAGNNNNGACGTC,(5 )

2608T_ch09sm_S99-S111 2/21/08 11:45AM Page S-102 ntt Os9:Desktop Folder:TEMPWORK:FEBRUARY:21-02-08:WHQY028/soln: S-102 Chapter 9 DNA-Based Information Technologies Notice that the and sites are retained. In a similar fashion, each of the other three adaptors can be ligated to the digested DNA molecule, and the ligated molecule joined to a DNA fragment produced by a digest. The final products are as follows. For the second adaptor: (5 ),GAATTCNNNNGTGCAG,(3 ) (3 ),CTTAAGNNNNCACGTC,(5 ) The site is retained, but not the site. For the third adaptor: (5 ),GAATTGNNNNCTGCAG,(3 ) (3 ),CTTAACNNNNGACGTC,(5 ) The site is retained, but not the site. For the fourth adaptor: (5 ),GAATTGNNNNGTGCAG,(3 ) (3 ),CTTAACNNNNCACGTC,(5 ) Neither the nor the 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 tetracycline-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 l vector it is not necessary 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 l DNA can be packaged into infectious phage particles only if it is between 40,000 and 53,000 bp long. The two essential pieces of the bacteriophage l 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. 3. DNA Cloning The plasmid cloning vector pbr322 (see Fig. 9 3) is cleaved with the restriction endonuclease. An isolated DNA fragment from a eukaryotic genome (also produced by 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 distinguished? (b) The cloned DNA fragment is 1,000 bp long and has an site 250 bp from one end. Three different recombinant plasmids are cleaved with and analyzed by gel electrophoresis, giving the patterns shown. What does each pattern say about the cloned DNA? Note that in pbr322, the and 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.

2608T_ch09sm_S99-S111 2/21/08 11:45AM Page S-103 ntt Os9:Desktop Folder:TEMPWORK:FEBRUARY:21-02-08:WHQY028/soln: Chapter 9 DNA-Based Information Technologies S-103 1 2 3 4 Nucleotide length 5,000 Electrophoresis 3,000 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 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 pbr322 was regenerated without insertion of a foreign DNA fragment. (These would also retain resistance to ampicillin.) In addition, two or more molecules of pbr322 might be ligated together with or without insertion of foreign DNA. (b) 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 cleavage). The clone producing the pattern in lane 3 has two DNA fragments, ligated such that the -site proximal ends are joined. pbr322 Clone in lane 1 Clone in lane 2 Clone in lane 3

2608T_ch09sm_S99-S111 2/21/08 11:45AM Page S-104 ntt Os9:Desktop Folder:TEMPWORK:FEBRUARY:21-02-08:WHQY028/soln: S-104 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. H 3 N + Ala Pro Met Thr Trp Tyr Cys Met Asp Trp Ile 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 Ile. 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 20-nucleotide probes. 5. 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 (M r 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 CAG 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 1 M A T L E K L M K A F E S L K S F 358 CAGCAGTTCCAGCAGCAGCAGCAGCAGCAGCAGCAGCAGCAGCAGCAGCAG 18 Q Q F Q Q Q Q Q Q Q Q Q Q Q Q Q Q 409 CAGCAGCAGCAGCAGCAGCAGCAACAGCCGCCACCGCCGCCGCCGCCGCCG 35 Q Q Q Q Q Q Q Q Q P P P P P P P P 460 CCGCCTCCTCAGCTTCCTCAGCCGCCGCCG 52 P P P Q L P Q P P P 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.

2608T_ch09sm_S99-S111 2/21/08 11:45AM Page S-105 ntt Os9:Desktop Folder:TEMPWORK:FEBRUARY:21-02-08:WHQY028/soln: Chapter 9 DNA-Based Information Technologies S-105 Answer Your test would require DNA primers, a heat-stable DNA polymerase, deoxynucleoside 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 3 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 antiparallel to a 25-nucleotide 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 will 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 sitespecific 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 Suggest how the researcher might use the polymerase chain reaction (PCR) to detect the presence of the circular form of the deleted DNA in an extract of the protist. Answer Design PCR primers complementary to DNA in the deleted segment, but which would direct DNA synthesis away from each other. No PCR product will be generated unless the ends of the deleted segment are joined 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 firefly luciferase (see Fig. 9 29) will not. Explain these observations. Answer The plant expressing firefly luciferase must take up luciferin, the substrate of luciferase, 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 displays 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 C1, C2, and C3 from the children.

2608T_ch09sm_S99-S111 2/21/08 11:45AM Page S-106 ntt Os9:Desktop Folder:TEMPWORK:FEBRUARY:21-02-08:WHQY028/soln: S-106 Chapter 9 DNA-Based Information Technologies M F C1 C2 C3 Electrophoresis Answer None of the children can be excluded. Each child 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 A B C D E F Electrophoresis 1 2 3 4 5 6 Nine restriction fragments 7 8 9 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 1, 3, 4, 5, and 7 Clone D: fragments 2, 3, 4, 5, and 7 Clone E: fragments 1, 5, and 9 Clone F: fragments 2, 4, 6, and 7

2608T_ch09sm_S99-S111 2/21/08 11:45AM Page S-107 ntt Os9:Desktop Folder:TEMPWORK:FEBRUARY:21-02-08:WHQY028/soln: Chapter 9 DNA-Based Information Technologies S-107 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 1 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 fragments 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 1 and 5 are adjacent, we can propose the following 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 possibilities 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: 9-1-5-3-7-4-2. 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. 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 correlates inversely with the size of the fragment. Fragment 1 is the longest; fragment 9 the shortest. The relative sizes and the positions of the fragments on the chromosome are shown below. Molecular weight markers in the gel would allow a better estimation of the sizes of the various fragments. 9 1 5 3 7 4 2 6 8 E A C D B F 10. Cloning in Plants The strategy outlined in Figure 9 28 employs Agrobacterium 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 removed, is too large to isolate in quantity and manipulate in vitro. It is also too large to reintroduce 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 vir genes can facilitate 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 propagated in E. coli or Agrobacterium and is readily reintroduced into either bacterium.

2608T_ch09sm_S99-S111 2/21/08 11:45AM Page S-108 ntt Os9:Desktop Folder:TEMPWORK:FEBRUARY:21-02-08:WHQY028/soln: S-108 Chapter 9 DNA-Based Information Technologies 11. DNA Fingerprinting and RFLP Analysis DNA is extracted from the blood cells of two humans, individuals 1 and 2. In separate experiments, the DNA from each individual is cleaved by restriction endonucleases 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 C*. 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 lanes of the gel are marked in the accompanying diagram. A B C probe B C * B * A C kbp 0 1 2 3 4 5 6 7 8 9 10 A B C M 1 2 1 2 1 2 10 9 8 7 6 5 4 3 2 1 M 1 2 1 2 1 2 A B C Answer Cleaving DNA with restriction enzyme A produces identical fragments in both individuals: 6.5 kbp and 3.5 kbp fragments. The probe hybridizes to both 6.5 kbp fragments, resulting in two identical bands in column A. Restriction enzyme B produces different cleavage products: DNA from individual 1 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 column B). Restriction enzyme C 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 become 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.

2608T_ch09sm_S99-S111 2/21/08 11:45AM Page S-109 ntt Os9:Desktop Folder:TEMPWORK:FEBRUARY:21-02-08:WHQY028/soln: Chapter 9 DNA-Based Information Technologies S-109 A B C M 1 2 1 2 1 2 10 9 8 7 6 5 4 3 2 1 M 1 2 1 2 1 2 A B C 12. Use of Photolithography to Make a DNA Microarray Figure 9 21 shows the first steps in the process of making a DNA microarray, or DNA chip, 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, irradiate, and wash. The resulting sequences are now 1. A-T 2. G-T 3. A-T 4. G-C 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. G-C 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, add 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 13. 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, env), 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 recombinant DNA into a mammalian cell. Once this DNA is integrated into the target cell s chromosome, the absence of replication and packaging functions makes the integration very stable by preventing deletion or replication of the integrated DNA.

2608T_ch09sm_S99-S111 02/22/2008 2:50 am Page S-110 pinnacle OS9:Desktop Folder: S-110 Chapter 9 DNA-Based Information Technologies Data Analysis Problem 14. HincII: The First Restriction Endonuclease 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 described the isolation of an enzyme that cleaved double-stranded DNA. They initially demonstrated the enzyme s nuclease activity by measuring the decrease in viscosity of DNA samples treated with the enzyme. (a) Why does treatment with a nuclease decrease the viscosity of a solution of DNA? The authors determined whether the enzyme was an endo- or an exonuclease by treating 32 P-labeled DNA with the enzyme, then adding trichloroacetic acid (TCA). Under the conditions used in their experiment, single nucleotides would be TCA-soluble and oligonucleotides would precipitate. (b) No TCA-soluble 32 P-labeled material formed on treatment of 32 P-labeled DNA with the nuclease. Based on this finding, is the enzyme an endo- or exonuclease? Explain your reasoning. When a polynucleotide is cleaved, the phosphate usually is not removed but remains attached to the 5 or 3 end of the resulting DNA fragment. Smith and Wilcox determined the location of the phosphate on the fragment formed by the nuclease in the following steps: 1. Treat unlabeled DNA with the nuclease. 2. Treat a sample (A) of the product with - 32 P-labeled ATP and polynucleotide kinase (which can attach the -phosphate of ATP to a 5 OH but not to a 5 phosphate or to a 3 OH or 3 phosphate). Measure the amount of 32 P incorporated into the DNA. 3. Treat another sample (B) of the product of step 1 with alkaline phosphatase (which removes phosphate groups from free 5 and 3 ends), followed by polynucleotide kinase and - 32 P-labeled ATP. Measure the amount of 32 P incorporated into the DNA. (c) Smith and Wilcox found that sample A had 136 counts/min of 32 P; sample B had 3,740 counts/min. Did the nuclease 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 enzyme s 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 following 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 - 32 P-labeled ATP to label the 5 ends. 4. Treat the labeled molecules with DNases to break them into a mixture of mono-, di-, and trinucleotides. 5. Determine the sequence of the labeled mono-, di-, and trinucleotides by comparing them with oligonucleotides of known sequence on thin-layer chromatography. The labeled products were identified as follows: mononucleotides: A and G; dinucleotides: 5 -papa-3 and 5 -pgpa-3 ; trinucleotides: 5 -papapc-3 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. They found a mixture of 5 -ptpc-3 and 5 -ptpt-3. They did not determine the sequence of any trinucleotides at the 3 end. (f) Based on these data, what is the recognition sequence for the nuclease and where in the sequence is the DNA backbone cleaved? Use Table 9 2 as a model for your answer.

2608T_ch09sm_S99-S111 2/21/08 11:45AM Page S-111 ntt Os9:Desktop Folder:TEMPWORK:FEBRUARY:21-02-08:WHQY028/soln: Chapter 9 DNA-Based Information Technologies S-111 Answer (a) DNA solutions are highly viscous because the very long molecules are tangled in solution. Shorter molecules tend to tangle less and form a less viscous solution, so decreased viscosity corresponds to shortening of the polymers as caused by nuclease activity. (b) An endonuclease. An exonuclease removes single nucleotides from the 5 or 3 end and would produce TCA-soluble 32 P-labeled nucleotides. An endonuclease cuts DNA into oligonucleotide fragments and produces little or no TCA-soluble 32 P-labeled material. (c) The 5 end. If the phosphate were left on the 3 end, the kinase would incorporate significant 32 P 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 32 P. When the phosphate is left on the 5 end, the kinase does not incorporate any 32 P: it cannot add a phosphate if one is already present. Treatment with the phosphatase removes 5 phosphate, and the kinase then incorporates significant amounts of 32 P. Sample A will have little or no 32 P, and B will show substantial 32 P incorporation as was observed. (d) Random breaks would produce a distribution of fragments of random size. The production 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 sequence at the 5 end of the fragments would be random. (f) The results are consistent with two recognition sequences, as shown below, cleaved where shown by the arrows: g (5 ),GTT AAC,(3 ) (3 ),CAA TTG,(5 ) h which gives the (5 )papapc and (3 )TpTp fragments; and g (5 ),GTC GAC,(3 ) (3 ),CAG CTG,(5 ) h which gives the (5 )pgpapc and (3 )CpTp fragments. References Kelly, T.J. & Smith, H.O. (1970) A restriction enzyme from Haemophilus influenzae: II. Base sequence of the recognition site. J. Mol. Biol. 51, 393 409. Smith, H.O. & Wilcox, K.W. (1970) A restriction enzyme from Haemophilus influenzae: I. Purification and general properties. J. Mol. Biol. 51, 379 391.