The correct answer is c B. Answer b is incorrect. Type II enzymes recognize and cut a specific site, not at random sites.

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1 1. A recombinant DNA molecules is one that is a. produced through the process of crossing over that occurs in meiosis b. constructed from DNA from different sources c. constructed from novel combinations of DNA from the same source d. produced through mitotic cell division A. Answer a is incorrect. Crossing over produces novel combinations of DNA from a single individual. The offspring are called recombinants. constructed from DNA from different sources B. Answer b is correct. Recombinant DNA is constructed in a laboratory and manipulated for a specific purpose. C. Answer c is incorrect. New genetic combinations from the same individual occur as a product of crossing over during meiosis. Recombinant DNA is an artificial construct that uses DNA from different sources. D. Answer d is incorrect. Mitosis is the asexual form of cell division that produces genetically identical daughter cells. There is no recombination of DNA during mitosis. 2. Type II restriction endonucleases are useful because a. they degrade DNA from the 5 end b. they cleave the DNA at random locations c. they cleave the DNA at specific sequences d. they only cleave modified DNA A. Answer a is incorrect. The ability to cleave DNA from an end is a property of an exonuclease. Endonucleases work by breaking bonds within the DNA molecule. B. Answer b is incorrect. Type II enzymes recognize and cut a specific site, not at random sites. they cleave the DNA at specific sequences C. Answer c is correct. A type II restriction endonuclease works by cleaving the DNA at specific sequences. This property is used in many experimental techniques that involve the manipulation of DNA. D. Answer d is incorrect. The cells that produce these enzymes protect their own DNA because they do not work on modified DNA. 3. What is the basis of separation of DNA fragments by gel electrophoresis? a. The negative charge on DNA b. The size of the DNA fragments

2 c. The sequence of the fragments d. The presence of a dye A. Answer a is incorrect. The negative charge is necessary to cause the movement of the DNA fragments through the gel. The size of the DNA fragments B. Answer b is correct. Different size fragments move at different speeds with small fragments moving more quickly than larger fragments. C. Answer c is incorrect. The sequence of the fragments does not contribute to the separation. D. Answer d is incorrect. Dyes are used to visualize the fragments, but they do not contribute to the separation. 4. How is the gene for β-galactosidase used in the construction of a plasmid? a. The gene is a promoter that is sensitive to the presence of the sugar, galactose. b. It is an origin of replication. c. It is a cloning site. d. It is a marker for insertion of DNA. A. Answer a is incorrect. Promoters are binding sites for RNA polymerase. B. Answer b is incorrect. Beta-galactosidase is an enzyme not an origin of replication. C. Answer c is incorrect. The gene for β-galactosidase is located in cloning sites, but is not itself a cloning site. It is a marker for insertion of DNA. D. Answer d is correct. When DNA is inserted into the plasmid, the function of this enzyme is disrupted. The absence of color formation is a marker for recombination. 5. The basic logic of enzymatic DNA sequencing is to produce a. a nested set of DNA fragments produced by restriction enzymes b. a nested set of DNA fragments that each begin with different bases c. primers to allow PCR amplification of the region between the primers d. a nested set of DNA fragments that end with known bases A. Answer a is incorrect. Restriction enzymes cut specific sites but not frequently enough to allow sequencing.

3 B. Answer b is incorrect. It is important during enzymatic sequencing that the fragments all begin from the same primer, because they are then terminated at different bases. C. Answer c is incorrect. PCR amplification is now used for sequencing, but the logic still requires generating a nested set of fragments that end with the same base. a nested set of DNA fragments that end with known bases D. Answer d is correct. The logic still requires generating a nested set of fragments that end with the same base. This is accomplished using nucleotides that act as chain terminators. 6. A DNA library is a. an orderly array of all the genes within an organism b. a collection of vectors c. the collection of plasmids found within a single E. coli d. a collection of DNA fragments representing the entire genome of an organism A. Answer a is incorrect. Genomic libraries are made up of random fragments of DNA from the entire genome, not just genes. B. Answer b is incorrect. A library is a collection of vectors with DNA inserted into them. C. Answer c is incorrect. Multiple cells are required to hold all the vectors that carry the various DNA inserts that make up a genomic library. a collection of DNA fragments representing the entire genome of an organism D. Answer d is correct. A genomic library is a collection of random DNA fragments from the entire genome. 7. Molecular hybridization is used to a. generate cdna from mrna b. introduce a vector into a bacterial cell c. screen a DNA library d. introduce mutations into genes A. Answer a is incorrect. The enzyme reverse transcriptase is used to make cdna from mrna. B. Answer b is incorrect. Molecular hybridization refers to the ability of a labeled probe to bind to its complementary sequence in denatured DNA. screen a DNA library

4 C. Answer c is correct. Molecular hybridization refers to the ability of a labeled probe to bind to its complementary sequence in denatured DNA. This property is used to identify the presence of genes of interest within large DNA collections like libraries. D. Answer d is incorrect. Molecular hybridization refers to the ability of a labeled probe to bind to its complementary sequence in denatured DNA. 8. The enzyme used in the polymerase chain reaction is a. a restriction endonuclease b. heat-resistant RNA polymerase c. reverse transcriptase d. a heat-resistant DNA polymerase A. Answer a is incorrect. Restriction endonucleases are used to break up DNA. They are not part of the polymerase chain reaction. B. Answer b is incorrect. PCR generates DNA fragments, not RNA. C. Answer c is incorrect. Reverse transcriptase is used to convert RNA into DNA. PCR uses DNA as its template. a heat-resistant DNA polymerase D. Answer d is correct. PCR synthesizes DNA strands from a DNA template. The specific enzyme used, Taq polymerase, needs to be heat-resistant because the reaction cycles through high- and low-temperature phases. 9. How does the yeast two-hybrid system detect protein protein interactions? a. Binding of fusion partners triggers a signal cascade that alters gene expression. b. Fusion partners are detected using radioactive probes of Western blots. c. Protein protein binding of fusion partners triggers expression of a reporter gene. d. Protein protein binding of fusion partners triggers expression of the Gal4 gene. A. Answer a is incorrect. The fusion partners are two parts of the Gal4 transcriptional activator. No signal cascade is involved. B. Answer b is incorrect. The fusion partners are two parts of the transcriptional activator, Gal4. Gal4 activity results in a change in gene expression. Protein protein binding of fusion partners triggers expression of a reporter gene. C. Answer c is correct. If the fusion partners can bind, then Gal4 becomes functional and will trigger the expression of a reporter gene linked to the Gal4 promoter.

5 D. Answer d is incorrect. The fusion partners are hybrid proteins linking part of the Gal4 transcription activator with a protein of interest. Fusion partner binding triggers expression of a gene downstream from the Gal4 promoter. 10. In vitro mutagenesis is used to a. produce large quantities of mutant proteins b. create mutations at specific sites within a gene c. create random mutations within multiple genes d. create organisms that carry foreign genes A. Answer a is incorrect. The production of large quantities of proteins is made possible by expression vectors. create mutations at specific sites within a gene B. Answer b is correct. Scientists use in vitro mutagenesis to examine the effect of specific changes in gene sequence of the function of the protein product. C. Answer c is incorrect. In vitro mutagenesis allows for the generation of known mutations. D. Answer d is incorrect. Transgenic organisms carry foreign genes. 11. Insertion of a gene for a surface protein from a medically important virus such as herpes into a harmless virus is an example of a. a DNA vaccine b. reverse genetics c. gene therapy d. a subunit vaccine A. Answer a is incorrect. A DNA vaccine is based on splicing an internal protein into a plasmid to trigger a cellular immune response. B. Answer b is incorrect. Reverse genetics refers to the use of genetic knockouts to examine the role of a specific gene. C. Answer c is incorrect. Gene therapy involves the integration of a functional gene into an organism to replace or repair an existing genetic mutation. a subunit vaccine D. Answer d is correct. Subunit vaccines are based on packaging a gene for a unique marker protein for a pathogenic virus within a nonpathogenic virus which protects the individual by priming the immune response without exposing the person to the dangerous virus.

6 12. What is a Ti plasmid? a. A vector that can transfer recombinant genes into plant genomes b. A vector that can be used to produce recombinant proteins in yeast c. A vector that is specific to cereal plants like rice and corn d. A vector that is specific to embryonic stem cells The correct answer is a A vector that can transfer recombinant genes into plant genomes A. Answer a is correct. The Ti plasmid can trigger the integration of the genes within the plasmid into a plant cell s chromosome. In this way all the cells that arise from the transfected cell will carry the recombinant protein. The correct answer is a B. Answer b is incorrect. The Ti plasmid is specific to the plant bacteria Agrobacterium. The correct answer is a C. Answer c is incorrect. The Ti plasmid normally does not transfecting cereal plants. It has now been modified to allow use with cereals. The correct answer is a D. Answer d is incorrect. The Ti plasmid is specific for plants. 13. Which of the following is NOT a possible benefit of genetically modified crops? a. Increased nutritional value for people b. Enhanced resistance to insect pests c. Enhanced resistance to broad spectrum herbicides d. Enhanced resistance to insecticides A. Answer a is incorrect. Golden Rice is an example of a genetically modified crop with improved nutritional properties for humans. B. Answer b is incorrect. Bt crops are genetically modified to produce insecticidal proteins. C. Answer c is incorrect. Crops genetically engineered to overexpress EPSP synthase are an example of a plant that will resist herbicides. Enhanced resistance to insecticides D. Answer d is correct. Plants do not have to resist insecticides. Challenge Questions 1. Many human proteins, such as hemoglobin, are only functional as an assembly of multiple subunits. Assembly of these functional units occurs within the endoplasmic reticulum and Golgi apparatus of a eukaryotic cell. Discuss what limitations, if any, exist to the large-scale production of genetically engineered hemoglobin.

7 Answer Genetic engineering often means insertion of a gene of interest (and harvest of the protein of interest) from vast numbers of bacterial cells. Bacteria are prokaryotes, and as such do not have internal membrane systems like the endoplasmic reticulum and Golgi apparatus. An alternative would be to use a yeast system since yeast cells are eukaryotes. In either case, one of the major limiting factors is the ability to efficiently harvest the protein of interest and purify it away from all the other proteins normally produced by the host cell. 2. Enzymatic sequencing of a short strand of DNA was completed using dideoxynucleotides. Use the gel shown to determine the sequence of that DNA. G C A T Answer Read the gel starting from the smallest fragment (at the bottom) to the largest. The sequence is: CTGATAGTCAGCTG