Chapter 9—Deconstructing the Genome: DNA at High Resolution
Fill in the Blank
1. Sickle cell anemia is caused by a mutation in the _______________________ gene.
Ans:

-globin
Difficulty: 1
2. AvaI (CPyCGPuG, Py stands for pyrimidine, Pu for purine) will cut DNA approximately every _________ bp.
Ans: 1024
Difficulty: 3
3. A stored collection of DNA fragments in vectors is called a __________________.
Ans: library
Difficulty: 1
4. A short single-stranded piece of DNA that is labeled and used in hybridization experiments is called a ____________________.
Ans: probe
Difficulty: 1
5. _________________ techniques are based on the natural propensity of complementary single-stranded molecule to form stable double helices.
Ans: hybridization
Difficulty: 1
6. Vectors that contain plasmid-derived selectable markers and two segments of phage
 are called ________________________.
Ans: cosmids
Difficulty: 1
7. A __________________________ enzyme recognizes a specific sequence of bases anywhere within the genome and then severs two covalent bonds in the sugar-phosphate backbone at particular positions within or near that sequence.
Ans: restriction
Difficulty: 1
8. Minute circles of dsDNA called ___________________ can gain access to and replicate, independently of the endogenous chromosome, within many kinds of bacterial cells.
Ans: plasmid
Difficulty: 1
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9. A specialized vector that allows production of transcript and polypeptide is called an
_________________________ vector.
Ans: expression
Difficulty: 1
10. ___________________________ are used as chain terminators in the Sanger DNA sequencing process.
Ans: dideoxynucleotides
Difficulty: 2
Multiple Choice
11. The technique of primer walking begins with a vector-based primer. This is used to obtain DNA sequence for one end of a long insert. The partial sequence of the insert is used to:
A) find restriction enzymes that can be used to cut the insert for shotgun sequencing.
B) pinpoint the location of the insert within the vector.
C) make primers so the rest of the insert can be amplified by PCR for cloning.
D) synthesize a probe for Southern blotting to identify the insert in genomic DNA.
E) design a new primer that will sequence into the next unknown section of the insert.
Ans: E
Difficulty: 2
12. What is the reason for partial digestion of genomic DNA with restriction enzymes?
A) Generating blunt ends, when only enzymes that leave sticky ends are available.
B) Generating sticky ends, when only enzymes that leave blunt ends are available.
C) Cutting only one strand of a restriction site, leaving a nick instead of a break.
D) Producing fragments that are slightly shorter than a complete digest would produce.
E) Producing fragments that are slightly longer than a complete digest would produce.
Ans: E
Difficulty: 1
13. Which of the following is an example of a recombinant DNA molecule?
A) A PCR-amplified fragment.
B) A single-stranded RNA hybridized to a single-stranded DNA.
C) A genomic fragment of human DNA ligated to a bacterial plasmid vector.
D) A  chromosome in a bacterial cell.
E) A bacterial plasmid cut with a restriction enzyme.
Ans: C
Difficulty: 1
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14. The diagram shows the products from enzymatic extension sequencing. What is the sequence of the template strand (5' to 3' left to right)?
A) GACCGCT
B) TCGCCAG
C) AGCGGTC
D) CTGGCGA
Ans: C
Difficulty: 3
15. Figure 1 shows the products from enzymatic extension sequencing. If a primer is designed to hybridize to the template strand for PCR amplification, what will its last 4 bases be (5' to 3', left to right)?
A) GCGA
B) CTGG
C) CCGA
D) CGCT
E) TCGC
Ans: D
Difficulty: 4
16. One of the reasons that PCR is useful in diagnosis of HIV is because it:
A) detects antibodies against the virus.
B) interferes with viral replication, causing the virus to become latent.
C) automatically determines the sequence, as well as the presence of the viral genome.
D) detects an infection at an earlier stage than the current standard test.
E) detects both viral DNA and viral proteins.
Ans: D
Difficulty: 1
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17. PCR cannot be successfully performed without:
A) at least 100 starting DNA molecules.
B) at least some sequence information about the region to be amplified.
C) a cDNA version of the region to be amplified.
D) a section of at least 100 kb to amplify.
E) an undamaged, non-degraded DNA sample.
Ans: B
Difficulty: 2
18. What information CANNOT be determined from the DNA sequence of a cDNA clone?
A) exon sequences
B) sequence of the promoter
C) similarity to previously identified sequences
D) amino acid sequence of the encoded polypeptides
E) sequence of the spliced mRNA transcript
Ans: B
Difficulty: 2
19. In Sanger (enzymatic extension) sequencing, what causes DNA synthesis to terminate at a specific base?
A) Chemicals that cleave DNA after particular bases.
B) Fluorescent chemicals.
C) Nucleotide triphosphates that lack a base.
D) Nucleotide triphosphates that lack a hydroxyl.
E) Nucleosides that lack a 5' phosphate.
Ans: D
Difficulty: 2
20. What technique is used to separate DNA molecules that are the size of whole chromosomes?
A) PCR
B) Southern blotting
C) hybridization
D) transformation
E) pulsed field gel electrophoresis
Ans: E
Difficulty: 1
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21. In gel electrophoresis of DNA, fragments move at different rates because they have different:
A) charge densities.
B) sizes.
C) base sequences.
D) amino acid compositions.
E) electrical strengths.
Ans: B
Difficulty: 1
22. The recognition sites for the restriction enzymes BamHI, XbaI and BglII are
G^GATCC, T^CTAGA and A^GATCT, respectively. Which enzymes leave compatible ends that can be ligated together?
A) All of these enzymes leave ends that are compatible with ends generated by the others.
B) None of the enzymes produce compatible ends.
C) Only BamHI and BglII fragments are compatible.
D) Only BamHI and XbaI fragments are compatible.
E) Only BglII and XbaI fragments are compatible.
Ans: C
Difficulty: 3
23. A genomic library is made for an organism with a haploid genome of 5 × 10
7 bp. If the fragments are approximately 20 kb each, how many clones would be screened to ensure a 95% probability of a particular locus being represented at least once?
A) 1.25

10
6
B) 5

10
6
C) 5 
10 5
D) 2.5 
10 5
E) 1.25 
10 4
Ans: E
Difficulty: 4
24. What charge does DNA have, and what gives it this charge?
A) positive, from the phosphates
B) positive, from the bases
C) negative, from the phosphates
D) negative, from the bases
E) negative, from the ribose
Ans: C
Difficulty: 2
Table 9.1
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restriction enzyme recognition site
1. BglII
2. EcoRV
3. PvuI
A^GATCT
GAT^ATC
CGAT^CG
4. SacII
5. SmaI
CCGC^GG
CCC^GGG
25. From table 9.1, which combination of enzymes leaves ends that can, theoretically, be ligated together?
A) 1 and 2
B) 1 and 3
C) 2 and 3
D) 2 and 5
E) 4 and 5
Ans: D
Difficulty: 2
26. From table 9.1, which restriction enzyme leaves an end that could be a substrate for a polymerase?
A) 1
B) 2
C) 3
D) 4
E) 5
Ans: A
Difficulty: 3
27. T4 phage DNA polymerase has 2 activities. One is polymerization. The other is 3' to 5' exonuclease activity against 3' overhangs, which are sticky ends that have a singlestranded region at the 3' end of one of the strands. The enzyme converts sticky ends to blunt. DNA fragments cut with the enzymes in the above table are incubated with T4 polymerase and all 4 nucleotide triphosphates. Which statement is TRUE about how the fragments will be affected?
A) The exonuclease activity will blunt the BglII ends.
B) The exonuclease activity will blunt both the BglII ends and the PvuI ends.
C) The exonuclease activity will blunt both the PvuI and SacII ends.
D) The polymerase activity will blunt both the BglII and PvuI ends.
E) The polymerase activity will blunt both the PvuI and SacII ends.
Ans: C
Difficulty: 4
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28. An EcoRV (GAT^ATC) fragment that contains a promoter, is ligated to a SmaI
(CCC^GGG) fragment that contains most of a protein-coding gene. What is the probability that the newly introduced ATG will be in frame with the protein encoded in the SmaI fragment?
A) 1 in 2
B) 1 in 3
C) 1 in 4
D) 1 in 6
E) 1 in 9
Ans: B
Difficulty: 2
Table 9.2 restriction enzyme recognition site
AvaI C^PyCGPuG (Py stands for pyrimidine, Pu for purine)
SalI
TaqI
G^TCGAC
T^CGA
XhoI C^TCGAG
29. Use table 9.2 for enzyme recognition sites. A plasmid is cut with SalI, and combined with genomic DNA digested with XhoI. In those plasmids with a genomic fragment, the
SalI-XhoI junction can be cut with:
A) SalI only.
B) XhoI only.
C) both original enzymes.
D) either SalI or TaqI.
E) TaqI only.
Ans: E
Difficulty: 2
30. What proportion of AvaI sites will also be XhoI sites?
A) 1 in 2
B) 1 in 3
C) 1 in 4
D) 1 in 5
E) 1 in 6
Ans: C
Difficulty: 2
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31. What percentage of AvaI sites will be compatible with SalI sites?
A) 15%
B) 25%
C) 30%
D) 50%
E) 75%
Ans: B
Difficulty: 2
32. An AvaI fragment is ligated to a SalI fragment. The junction can be cut with:
A) XhoI or SalI, but not AvaI.
B) XhoI or AvaI, but not SalI.
C) SalI or AvaI, but not XhoI.
D) all three enzymes: SalI, AvaI or XhoI.
E) none of the three enzymes: SalI, AvaI or XhoI.
Ans: E
Difficulty: 3
33. Approximately how many fragments would you predict if the haploid human genome was cut with AvaI?
A) 3 
10 5
B) 3 
10 6
C) 3 
10
7
D) 7.5 
10 5
E) 7.5 
10
6
Ans: B
Difficulty: 4
34. In a human genomic DNA library made by complete SalI (G^TCGAC) digestion, approximately how many clones constitute a genomic equivalent?
A) 12,000,000
B) 700,000
C) 150,000
D) 46,000
E) 15,000
Ans: B
Difficulty: 3
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35. PCR is used to:
A) generate DNA fragments for genomic libraries.
B) generate DNA fragments for cDNA libraries.
C) determine the sequence of a DNA fragment by the Maxam-Gilbert procedure.
D) amplify minute quantities of a DNA fragment.
E) replicate recombinant plasmids.
Ans: D
Difficulty: 1
36. Restriction enzymes were named because they restrict:
A) the capacity for viral growth.
B) the growth of the bacteria that produces them.
C) bacterial translation.
D) bacterial transformation.
E) bacterial transcription.
Ans: A
Difficulty: 2
37. Which of the following would not have its gene represented in a cDNA library from red blood cell precursors?
A) ß-globin
B) a glycolysis enzyme
C) a ribosomal protein
D) an rRNA
E) a DNA polymerase subunit
Ans: D
Difficulty: 3
38. DNA molecules are separated by pulsed field gel electrophoresis when they are:
A) going to be sequenced by the Maxam-Gilbert procedure.
B) going to be sequenced by the Sanger (enzymatic) procedure.
C) too long to be separated by standard agarose gels.
D) too short to be separated by standard agarose gels.
E) are single-stranded.
Ans: C
Difficulty: 2
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39. If DNA had 5 different bases, a restriction enzyme with a 4-base recognition sequence would cut DNA approximately every:
A) 125 bp.
B) 256 bp.
C) 425 bp.
D) 625 bp.
E) 1056 bp.
Ans: D
Difficulty: 3
40. If DNA had 5 different bases, approximately how many fragments would result from cutting a 1 Mb piece of DNA with a 4-base cutter?
A) 125
B) 320
C) 625
D) 1050
E) 1600
Ans: E
Difficulty: 3
41. Two DNA fragments are separated by electrophoresis. How is the fragment that traveled farther different from the other fragment?
A) It is longer.
B) It is shorter.
C) It is more negatively charged.
D) It has blunt ends.
E) It has sticky ends.
Ans: B
Difficulty: 2
42. Which of the following would be found in a cDNA library, but not a genomic library?
A) centromeres
B) replication origins
C) promoters
D) exons
E) introns
Ans: D
Difficulty: 2
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43. A genomic library is a:
A) collection of DNA fragments from an organism, inserted into vectors.
B) collection of restriction enzymes that make regularly spaced cuts in a genome.
C) series of DNA fragments generated by chemical cleavage, used to determine sequence.
D) set of genes that can be inserted into vectors to confer antibiotic resistance.
E) collection of mRNA molecules cloned into vectors.
Ans: A
Difficulty: 1
44. The lacZ gene is sometimes included in a cloning vector. What does it encode?
A) A restriction enzyme that cuts the vector while it is in transformed cells.
B) A blue pigment that identifies any cell that has been transformed.
C) An enzyme that converts RNA to DNA.
D) An enzyme that confers antibiotic resistance.
E) A detectable enzyme that is not produced if the gene is interrupted by an insert.
Ans: E
Difficulty: 3
45. A plasmid has tet r
and lacZ genes for tetracycline resistance and ß-galactosidase, respectively. There is a single EcoRI site within lacZ, and a single Hind III site within tet r
. The EcoRI-HindIII cut plasmid is combined with a DNA fragment with compatible ends. Cells that contain the plasmid with the insert will be:
A) tetracycline resistant and blue if grown on X-gal.
B) tetracycline sensitive and blue if grown on X-gal.
C) tetracycline resistant and white if grown on X-gal.
D) tetracycline sensitive and white if grown on X-gal.
Ans: D
Difficulty: 3
46. Genomic DNA is isolated from bacteria. After a long digestion with a 6-base cutting enzyme that recognizes the sequence GAATTC, the DNA is not cut. The genome is 4
Mb, so there should be approximately 1,000 sites in the DNA. What is the most likely explanation?
A) There are no G's in the genome.
B) There are no A's in the genome.
C) The bacterium methylated the DNA, because it produces the 6-base cutting enzyme.
D) All the recognition sites are in genes, and the enzyme will not cut in genes.
E) All the recognition sites are in promoters and regulatory sequences, and the enzyme cuts only in genes.
Ans: C
Difficulty: 3
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47. A 6-base cutting enzyme should cut every 4kb, on average. Approximately how many sites are in the human genome (for simplicity, use 4 million kb as the size of the human genome)?
A) 100
B) 1,000
C) 10,000
D) 100,000
E) 1,000,000
Ans: E
Difficulty: 2
48. The basic problem in isolating/cloning a human gene is that the genome contains
__________ of base pairs, and a gene is usually several __________ base pairs in length.
A) trillions, million
B) trillions, thousand
C) billions, million
D) billions, thousand
E) millions, hundred
Ans: D
Difficulty: 2
49. What is the function of the amp r gene in a vector?
A) It is a restriction enzyme recognition site.
B) It is a selectable marker.
C) It is the origin of replication.
D) It makes the enzyme ß-galactosidase if it is intact, indicating vectors without inserts.
E) It is required for packing into viral shells.
Ans: B
Difficulty: 2
50. A circular DNA of 1 kb length is cut with a restriction enzyme whose precise recognition sequence is not known. The digest yields 4 fragments of approximately equal size. What is the most likely conclusion from these data?
A) The enzyme is a 4-base cutter.
B) The enzyme is a 6-base cutter.
C) The enzyme is an 8-base cutter.
D) The enzyme leaves blunt ends.
E) This was a partial digest.
Ans: A
Difficulty: 3
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51. Which choice is NOT a characteristic of a plasmid used as a cloning vector?
A) Confers a detectable property on a host cell.
B) Can be purified away from the host cell's genome.
C) Replicates independently.
D) Replicates itself and the inserted DNA.
E) Methylated to protect from digestion by host cell restriction enzymes.
Ans: E
Difficulty: 2
52. Where do restriction enzymes come from, and what is their normal physiological function?
A) Phage
 produces them, and they replicate viral DNA.
B) Phage
 produces them and they confer antibiotic resistance.
C) They are yeast DNA replication enzymes.
D) Bacteria produce them to protect against viral invasion.
E) They are bacterial DNA replication enzymes.
Ans: D
Difficulty: 2
53. How do bacteria protect their own DNA from the restriction enzymes they produce?
A) Their DNA never contains the recognition site.
B) Ligase connects their DNA as soon as it is cut.
C) Digestion is blocked by -CH3 groups added to recognition sequences in their DNA.
D) Phosphate groups added to the recognition sequences in their DNA block digestion.
E) Restriction enzymes are blocked from entry into the nucleus.
Ans: C
Difficulty: 3
54. If the restriction enyzme EcoRI cuts a linear 19 kb piece of DNA into fragments of 1, 3,
6 and 9 kb, what size fragment (in kb), would NOT be possible from a partial digest?
A) 12
B) 10
C) 7
D) 5
E) 4
Ans: D
Difficulty: 2
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55. A restriction enzyme with a recognition site of 10 bp cuts an average of every:
A) 4
10
bp.
B) 10 4 bp.
C) 2
10
bp.
D) 10
2
bp.
E) 2 4 bp.
Ans: A
Difficulty: 2
56. A restriction enzyme with a recognition site of CCNNGG (where N is any base) cuts an average of every:
A) 16 bp.
B) 250 bp.
C) 400 bp.
D) 4 kb.
E) 65 kb.
Ans: B
Difficulty: 2
57. Reverse transcriptase is an enzyme made by:
A) red blood cells.
B) phage

C) retroviruses.
D) yeast.
E) E. coli.
Ans: C
Difficulty: 1
58. Which vector is appropriate for cloning an insert of 1 Mb?
A) phage
 vector
B) YAC
C) cDNA
D) cosmid
E) amp r
plasmid
Ans: B
Difficulty: 1
Page 171

59. A 16 kb linear DNA is cut with EcoRI and BamHI. The EcoRI digest yields fragments of 2, 6 and 8 kb. The BamHI digest yields fragments of 5 and 11 kb. An EcoRI-BamHI double digest yields 2, 3, 5 and 6 kb fragments. The map of this DNA has:
A) 3 EcoRI sites and 2 BamHI sites.
B) 3 BamHI sites and 2 EcoRI sites.
C) 2 EcoRI sites and 2 Bam HI sites.
D) 2 EcoRI sites and 1 BamHI site.
E) 1 EcoRI site and 2 Bam HI sites.
Ans: D
Difficulty: 1
60. A 16 kb linear DNA is cut with EcoRI and BamHI. The EcoRI digest yields fragments of 2, 6 and 8 kb. The BamHI digest yields fragments of 5 and 11 kb. An EcoRI-BamHI double digest yields 2, 3, 5 and 6 kb fragments. If the 3 kb fragment from the double digest is used as the probe in a Southern blot of the EcoRI digest, to which fragment(s) will it hybridize?
A) 2 kb
B) 6 kb
C) 8 kb
D) 2 and 6 kb
E) 6 and 8 kb
Ans: C
Difficulty: 4
61. A 10 kb linear DNA is cut with EcoRI and HindIII. The EcoRI digest yields fragments of 2 and 8 kb. The HindIII digest yields fragments of 3 and 7 kb. What are the possible fragments (in kb) from an EcoRI-HindIII double digest?
A) 1, 3, 6 or 1, 4, 5
B) 1, 4, 5 or 1, 2, 7
C) 1, 4, 5, or 2, 3, 5
D) 1, 2, 7 or 1, 3, 6
E) 1, 2, 7 or 2, 3, 5
Ans: E
Difficulty: 3
Table 9.3 restriction enzyme recognition site (5' to 3')
BamHI
BstBI
BglII
G^GATCC
TT^CGAA
A^GATCT
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62. Digestion with BstBI (TT^CGAA) leaves:
A) blunt ends.
B) sticky ends with the single-stranded sequence 5' CGAA 3'.
C) sticky ends with the single-stranded sequence 5' CG 3'.
D) sticky ends with the single-stranded sequence 5' AAGC 3'.
E) sticky ends with the single-stranded sequence 5' GC 3'.
Ans: C
Difficulty: 4
63. In 3 separate reactions, DNA is cut with the 3 enzymes as in table (9.3). Do the fragments have complementary ends that can be ligated together?
A) Fragments from the BamHI and BstBI digests can ligate to each other.
B) Fragments from the BamHI and BglII digests can ligate to each other.
C) Fragments from the BstBI and BglII digests can ligate to each other.
D) All the ends are complementary, so all the fragments can ligate to each other.
E) None of the fragments have complementary ends, so none will ligate to each other.
Ans: B
Difficulty: 3
64. Approximately how many bp between BstBI recognition sites (TT^CGAA)?
A) 65,000
B) 4,000
C) 1,000
D) 250
E) 100
Ans: B
Difficulty: 2
65. DNA with the following sequence was cut with BstBI (TT^CGAA) and cloned into a vector. What is the sequence of the peptide that is encoded within the BstBI-BstBI fragment?
5' ACCTTTCGAACTGATGGAAGAGTGTCACTGACTGGTTCGAAAAT 3'
A) lys-leu-gly-gln-ser
B) thr-phe-arg-thr-asp
C) met-his-cys-glu-lys
D) met-lys-leu-gly-gln
E) met-glu-glu-cys-his
Ans: E
Difficulty: 3
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66. Approximately how many times is BamHI (G^GATCC) expected to cut a 5kb piece of
DNA?
A) 1
B) 4
C) 10
D) 65
E) 250
Ans: A
Difficulty: 2
67. The partial amino acid sequence of a protein is: met-cys-tyr-trp-gln. A mixture of oligonucleotides containing all possible codons for each amino acid is generated. How many different oligonucleotides in the mixture?
A) 8
B) 16
C) 32
D) 64
E) 128
Ans: A
Difficulty: 4
68. When bacterial transformants containing a library are transferred to nitrocellulose filters and screened, one of the steps is treatment with an alkaline solution. What does this do?
A) Attaches the library DNA to the nitrocellulose.
B) Promotes hybridization between the probe and complementary inserts.
C) Labels the DNA so it can be visualized.
D) Stimulates DNA replication, to increase the plasmid copy number.
E) Lyses the cells and denatures the released DNA.
Ans: E
Difficulty: 2
69. In references to the “ 
-globin locus” or “ß-globin locus”, the term “locus” me
A) a duplicated, ancestral gene.
B) a chromosome region with a cluster of related globin protein genes.
C) an 
-, or a ß-globin protein with a single amino acid change.
D) the location, in this case red blood cells, where a mutant phenotype is manifested.
E) a geographical region where a genetic disease like sickle cell anemia is common.
Ans: B
Difficulty: 1
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70. How are the genes in the ß-globin locus different from each other?
A) They use different template strands for transcription.
B) They are expressed at different times in development.
C) They have different evolutionary origins, from different ancestral genes.
D) They are dispersed on different chromosomes.
E) Some are expressed in all cells, some are expressed only in red blood cells.
Ans: B
Difficulty: 1
71. Pseudogenes are evidence for:
A) gene duplications.
B) the existence of integrated viruses, or proviruses.
C) the importance of TATA boxes in eukaryotic promoters.
D) the importance of the order of the globin genes in dictating their expression.
E) diseases that result from deletions in the

-globin genes.
Ans: A
Difficulty: 2
72. Which statements are true?
1. In thalassemias, globin proteins are produced, but contain a point mutation.
2. In sickle cell anemia, globin proteins are produced, but contain a point mutation.
3. Sickle cell anemia is caused by a mutation in the ß-globin gene.
4. Sickle cell anemia is caused by a mutation in the

-globin gene.
A) 1, 2
B) 1, 3
C) 2, 3
D) 1, 4
E) 2, 4
Ans: C
Difficulty: 1
73. In cosmids, what is the function of the cos sites?
A) origins of replication
B) restriction enzyme recognition sites
C) allow packaging into viral shells
D) selectable markers
E) encode enzymes that allow detection of insert-containing recombinant molecules
Ans: C
Difficulty: 2
Page 175

74. If a single DNA molecule is amplified by PCR, how many DNA molecules will there be after 6 cycles?
A) 6
B) 12
C) 24
D) 64
E) 86
Ans: D
Difficulty: 3
75. Which mutation in the ß-globin gene could, in theory, be detected with a genomic clone but not with a cDNA clone?
A) A point mutation that changes a single amino acid.
B) A mutation that prevents removal of one intron.
C) A promoter mutation that prevents transcription.
D) A 10 bp deletion in an exon.
E) A frameshift mutation.
Ans: C
Difficulty: 2
Figure 2 above shows a series of restriction digests of a circular plasmid. lane enzyme(s)
1 BamHI
2 EcoRI
3 HindIII
4 SalI
5 BamHI and EcoRI
6 BamHI and HindIII
7 BamHI and SalI
8 EcoRI and SalI
Page 176

76. Referring to figure 2, if the plasmid is cut with EcoRI and HindIII, which size fragment is NOT a possibility?
A) 0.5
B) 1
C) 7
D) 7.5
E) 9
Ans: E
Difficulty: 3
77. Which technique is required to generate a restriction map?
A) electrophoresis
B) partial digestion with restriction enzymes
C) hybridization to a synthetic oligonucleotide probe
D) selection of recombinant bacteria by antibiotic resistance
E) transformation
Ans: A
Difficulty: 1
Matching
Match the technique with one or more applications a. b.
DNA fingerprinting analyzing trace amounts of degraded, ancient DNA c. d. making a cDNA library identifying a gene in genomic DNA, if sequence is unknown, but a probe is available
78. Southern blotting
Ans: a, d
Difficulty: 2
79. reverse transcription of mRNA to DNA
Ans: c
Difficulty: 2
80. PCR
Ans: a, b
Difficulty: 2 a. b. c. d.
Match the technique with one or more applications identifying sequences complementary to a probe separation of large DNA molecules introducing recombinant DNA into hostcells using synthetic oligonucleotide primers to direct the PCR amplification of a DNA fragment
Page 177

81. pulsed-field electrophoresis
Ans: b
Difficulty: 1
82. hybridization
Ans: a, d
Difficulty: 1
83. transformation
Ans: c
Difficulty: 1
True or False
84. The genetically determined molecular composition of hemoglobin changes several times during development including changing from embroyic, to fetal, to adult hemoglogin.
Ans: True
Difficulty: 1
85. Hybridization is the covalent binding together of two complementary DNA strands.
Ans: False
Difficulty: 2
86. All diploid human cells contain two nearly identical sets of 3 billion base pairs of DNA, that when unwound, would extend 200 km.
Ans: True
Difficulty: 1
87. Size of the recognition site and base composition of the genome effect fragment length produced by restriction enzymes.
Ans: True
Difficulty: 1
88. The average length of a human chromosome is 130,000,000 bp.
Ans: True
Difficulty: 1
89. Probability of a six-base restriction enzyme site existing in a stretch of DNA is (1/6)
4
.
Ans: False
Difficulty: 2
90. The formation of a recombinant DNA molecule occurs through the cutting and splicing together of a vector and a fragment of DNA from another origin.
Ans: True
Difficulty: 1
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91. Anitbiotic resistance genes are often used as selectable markers in vectors.
Ans: True
Difficulty: 1
92. The genomic equivalent of a library is determined by multiplying the number of clones on a plate by the size of the average insert.
Ans: False
Difficulty: 2
93. PCR makes millions of perfect copies of the DNA being amplified.
Ans: False
Difficulty: 2
Short Answer
94. Refer to Figure 2. If the 1 kb EcoRI site is used as a probe in a Southern blot of the gel, to what will it in hybridize in lane 7?
Ans: the 9 kb fragment
Difficulty: 3
95. From the information in Figure 2, what size fragments would result from a triple digest with BamHI, EcoRI and SalI?
Ans: 1, 3, 5, 6.5
Difficulty: 4
96. What are the largest capacity vectors that are currently available?
Ans: yeast artificial chromosomes (YAC)
Difficulty: 1
97. List two reasons that partial digests are performed on genomic DNA when producing fragments for a library.
Ans: 1. To obtain fragments of intermediate length (e.g. between a 6-base, and 8-base cutter) 2. So genes that contain recognition sites within their coding regions might be obtained intact and uninterrupted.
Difficulty: 4
98. List two advantages of cDNA libraries over genomic libraries.
Ans: 1. Only expressed coding sequences are represented 2. Introns, non-coding and repetitive sequences are usually not included 3. They are tissue-specific, with highly expressed sequences represented more often 4. Obtained clones can be ligated to promoters for expression in bacteria.
Difficulty: 2
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99. List the components of a Sanger (enzymatic extension) sequencing reaction.
Ans: template DNA, primer, nucleotide triphosphates, fluorescently labeled dideoxynucleotide triphosphate, DNA polymerase.
Difficulty: 3
100. What enzyme is used to generate DNA from RNA transcripts, for cDNA libraries?
Ans: reverse transcriptase, or RNA dependent DNA polymerase.
Difficulty: 1
Experimental Design and Interpretation of Data
101. The restriction enzyme EcoRI cuts a linear 19 kb piece of DNA into fragments of 1, 3, 6 and 9 kb. How many EcoRI sites are there in the 19 kb piece of DNA?
Ans: 3
Difficulty: 1
102. A genomic library for an organism with a haploid genome of 5 × 10
7
bp has fragments of approximately 20 kb. How many clones represent a single genomic equivalent?
Ans: 2500
Difficulty: 3
103. The partial amino acid sequence of a protein is met-cys-tyr-trp-gln. Write the sequence of the DNA probe that would hybridize to the template strand of the gene for this protein (if more than one base is possible at a given position, use “/” to indicate the mixture of bases).
Ans: 5' ATG TGT/C TAT/C TGG CAA/G 3'
Difficulty: 3
104. You wish to produce an abundance of protein for a gene you have just recently cloned from genomic DNA. How would you do this?
Ans: Isolate mRNA, reverse transcribe and create a cDNA library using an expression vector. Probe the library with your genomic clone and use the expression vector to produce your protein in large quantities for purification and further study.
Keep in mind that the expression vector needs to be compatible with the organism in which you need to produce your protein. In many cases, prokaryotes can not properly produce eukaryotic proteins which may require production in a more expensive but more appropriate system.
Difficulty: 4
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