Lodish 8e Ch06 Test Bank PDF

6-1 6 Molecular Genetic Techniques  6-2 Section 6.1 1. Organisms that are considered polyploid have: a. more than one genome. b. more than two copies of each chromosome. c. only on...

6-1 6 Molecular Genetic Techniques  6-2 Section 6.1 1. Organisms that are considered polyploid have: a. more than one genome. b. more than two copies of each chromosome. c. only one copy of each chromosome. d. multiple copies of a specific chromosome. Ans: b Question Type: Multiple Choice Chapter: 6 Blooms: Remembering, Understanding Difficulty: Easy 2. The chemical mutagen ethylmethane sulfonate (EMS) was used to treat cells, and after 24 hours of incubation, DNA was isolated from these cells and sent for sequencing. Based on the DNA sequence from untreated cells, those treated with EMS showed several of which of the following conversions? a. G–C to A–T b. G–A to C–T c. G–T to A–C d. G–G to T–G Ans: a Question Type: Multiple Choice Chapter: 6 Blooms: Applying, Analyzing Difficulty: Moderate 3. One common difference between meiosis and mitosis is that in meiosis, the end result is: a. four cells each having two chromosomes. b. two cells each having one chromosome. c. four cells each having one chromosome. d. two cells each having two chromosomes. Ans: c Question Type: Multiple Choice Chapter: 6 Blooms: Remembering, Understanding Difficulty: Easy 4. Which of the following tests is used to determine if different recessive mutations are in the same gene? a. centimorgan linkage test b. conditional test c. cohabitation test d. complementation test Ans: d Question Type: Multiple Choice Chapter: 6 Blooms: Applying, Analyzing Difficulty: Difficult  6-3 5. A mutation that changes a cysteine codon to a tryptophan codon is called a a. missense mutation. b. nonsense mutation. c. silent mutation. d. none of the above. Ans: a Question Type: Multiple Choice Chapter: 6 Blooms: Remembering, Understanding Difficulty: Easy 6. Crossing of a homozygous wild type with a mutant that is heterozygous for a dominant mutation will result in F1 progeny of which a. all show the mutant phenotype. b. half show the wild-type phenotype and half show the mutant phenotype. c. three-fourths show the wild-type phenotype and one-fourth show the mutant phenotype. d. all show the wild-type phenotype. Ans: b Question Type: Multiple Choice Chapter: 6 Blooms: Applying, Analyzing Difficulty: Moderate 7. A mutation in one gene that counteracts the effects of a mutation in another gene is known as a a. temperature-sensitive mutation. b. recessive mutation. c. conditional mutation. d. suppressor mutation. Ans: d Question Type: Multiple Choice Chapter: 6 Blooms: Remembering, Understanding Difficulty: Easy 8. Describe how genetic complementation can be used in yeast to determine whether two different recessive mutations are in the same or different genes. Ans: Yeasts are normally a haploid organism but can be made to form a diploid organism by mating. Genetic complementation is a phenomenon whereby the wild-type phenotype can be restored after mating two recessive mutants. If two recessive mutations are in the same gene then a diploid organism containing both mutations will show a mutant phenotype because neither allele provides a functional copy of the gene. If two recessive mutations are in different genes then the diploid yeast will show a wild-type phenotype because a wild-type allele of each gene will be present. Question Type: Essay Chapter: 6 Blooms: Applying, Analyzing Difficulty: Difficult 9. Describe the properties and utility of temperature-sensitive mutations.  6-4 Ans: A temperature-sensitive mutation is a mutation that expresses a wild-type phenotype at one temperature and a mutant phenotype at another temperature. For example, the protein may exist in a wild-type conformation at a permissive temperature of 23C; however, at a nonpermissive temperature of 36C, the protein undergoes a conformational change to a mutant form. Temperature-sensitive or conditional mutations are especially useful for isolation of mutations in essential genes. In this case, cells with the mutation can be propagated normally at the permissive temperature, and the effect of the mutation can be studied at the nonpermissive temperature. Question Type: Essay Chapter: 6 Blooms: Remembering, Understanding Difficulty: Difficult Section 6.2 10. A recombinant DNA vector: a. cannot replicate in a host. b. does not contain bacterial DNA sequences. c. contains DNA sequences from only one source. d. contains DNA sequences that can be cleaved by restriction enzymes. Ans: d Question Type: Multiple Choice Chapter: 6 Blooms: Remembering, Understanding Difficulty: Moderate 11. A polylinker increases the versatility of a DNA plasmid because it contains: a. a DNA recognition sequences for several different restriction enzymes. b. a DNA sequence to allow the vector to replicate. c. a DNA sequence that confers resistance to ampicillin. d. b and c. Ans: a Question Type: Multiple Choice Chapter: 6 Blooms: Remembering, Understanding Difficulty: Moderate 12. You have an E. coli plasmid containing a mammalian cDNA and want to isolate the cDNA to use in another experiment. The plasmid map reveals that you can isolate the complete cDNA by digesting the plasmid with EcoRI. How many bands would there be after a successful EcoRI restriction enzyme digest? a. one b. two c. three d. four Ans: b Question Type: Multiple Choice Chapter: 6 Blooms: Applying, Analyzing Difficulty: Moderate 13. Construction of a cDNA library involves several steps, the first usually involves:  6-5 a. digesting noncoding regions of DNA. b. cloning complementary double-strand DNA into a plasmid. c. isolating total RNA from the organism or cell type of interest. d. reverse transcribing RNA into cDNA. Ans: c Question Type: Multiple Choice Chapter: 6 Blooms: Remembering, Understanding Difficulty: Easy 14. Sequencing methods were used to test the integrity of the cDNA library and when this was done you found many of the sequences were those found in introns. After troubleshooting you determined that the reason these intronic sequences were present was because during the construction of this cDNA library: a. genomic DNA was present. b. retroviral DNA was a contaminant. c. endonucleases should have been added. d. RNA polymerase was present. Ans: a Question Type: Multiple Choice Chapter: 6 Blooms: Applying, Analyzing Difficulty: Difficult 15. In addition to sequencing, intronic sequences can be identified in a genomic DNA library using which of the following methods? a. Southern blotting b. in situ hybridization c. PCR d. Southern blotting and PCR Ans: d Question Type: Multiple Choice Chapter: 6 Blooms: Remembering, Understanding Difficulty: Easy 16. Which of the following series of steps does the polymerase chain reaction follow in order to amplify DNA in a test tube? a. DNA denaturation, primer elongation, primer annealing b. primer annealing, DNA ligation, primer elongation c. primer elongation, primer annealing, DNA ligation d. DNA denaturation, primer annealing, primer elongation Ans: d Question Type: Multiple Choice Chapter: 6 Blooms: Remembering, Understanding Difficulty: Moderate 17. You want to amplify a region of yeast DNA using PCR so that this fragment can be cloned into a plasmid. Which of the following is needed for the PCR? a. RNA polymerase b. DNA ligase  6-6 c. DNA helicase d. Taq polymerase Ans: d Question Type: Multiple Choice Chapter: 6 Blooms: Remembering, Understanding Difficulty: Moderate 18. To see if your polymerase chain reaction was successful and it amplified the right sequence of interest, you electrophorese the products from the reaction on an agarose gel. After staining the gel you find there are two bands of different sizes. Which one of the following is a possible explanation for these results? a. The PCR induced a frame shift mutation into the DNA sequence. b. RNA polymerase copied the DNA into two copies of RNA. c. The primers annealed to two different templates. d. a and b. Ans: c Question Type: Multiple Choice Chapter: 6 Blooms: Applying, Analyzing Difficulty: Difficult 19. Which of the following polymerase chain reaction techniques is widely used to study the amount of a specific mRNA within a cell or tissue? a. qualitative regional transcribed-PCR b. quantitative reverse targeted-PCR c. quantitative reverse transcriptase-PCR d. qualitative regional transposase-PCR Ans: c Question Type: Multiple Choice Chapter: 6 Blooms: Remembering, Understanding Difficulty: Easy 20. Which of the following enzymes will produce a blunt end (the cut site is indicated by the * in the recognition sequence)? a. TaqI (T*CGA) b. EagI (C*GGCCG) c. EcoRV (GAT*ATC) d. NsiI (ATGCA*T) Ans: c Question Type: Multiple Choice Chapter: 6 Blooms: Remembering, Understanding Difficulty: Easy 21. DNA ligase a. synthesizes DNA from an RNA template. b. forms a phosphodiester bond. c. joins Okazaki fragments. d. b and c  6-7 Ans: d Question Type: Multiple Choice Chapter: 6 Blooms: Remembering, Understanding Difficulty: Moderate 22. Which of the following is a functional element of a plasmid? a. origin of replication b. drug-resistance gene c. polylinker sequence d. all of the above Ans: d Question Type: Multiple Choice Chapter: 6 Blooms: Remembering, Understanding Difficulty: Moderate 23. Next generation sequencing is much more efficient than the Sanger method because a. it uses an RNA template instead of DNA. b. it uses gel electrophoresis to resolve end-labeled strands of DNA. c. it uses PCR amplification. d. it uses gel electrophoresis to resolve end-labeled strands of DNA, and it uses PCR amplification. Ans. c Question Type: Multiple Choice Chapter: 6 Blooms: Remembering, Understanding Difficulty: Difficult 45. Describe some typical features of a restriction enzyme recognition sequence. Ans: Restriction enzyme recognition sequences are typically 4–8 base pairs in length and are often palindromic, which means that the sequence read in the 5´ to 3´ direction is the same on each DNA strand. For example, the recognition sequence for EcoRI is 5´ GAATTC 3´. The complementary sequence is 3´ CTTAAG 5´, which is the same sequence when read in the 5´ to 3´ direction. Question Type: Essay Chapter: 6 Blooms: Remembering, Understanding Difficulty: Moderate 24. Describe the essential features of a yeast shuttle vector. Ans: A yeast shuttle vector is a plasmid that can replicate in both bacteria (E. coli) and yeast. For replication in bacteria, the plasmid needs a bacterial origin of replication and a gene for selection in transformed E. coli (e.g., for resistance to ampicillin). For replication in yeast, the plasmid needs a yeast origin of replication (autonomously replicating sequence, ARS), a gene for selection in transformed yeast (e.g., ura3), and a centromere. In addition, a polylinker sequence for the efficient cloning of foreign DNA is necessary. Question Type: Essay Chapter: 6 Blooms: Remembering, Understanding Difficulty: Difficult  6-8 Section 6.3 25. A specific nucleic acid fragment is used in all but one of the following techniques? a. Southern blotting b. Western blotting c. Northern blotting d. in situ hybridization Ans: b Question Type: Multiple Choice Chapter: 6 Blooms: Remembering, Understanding Difficulty: Easy 26. Which of the following is used to compare gene expression in cells under different conditions? a. Southern blotting b. P-element insertion c. stable transfection d. microarray analysis Ans: d Question Type: Multiple Choice Chapter: 6 Blooms: Remembering, Understanding Difficulty: Easy 27.The process by which genes cloned into specialized eukaryotic vectors are introduced into cultured animal cells for transient expression analysis is called: a. ligation. b. hybridization. c. transfection. d. transformation. Ans: c Question Type: Multiple Choice Chapter: 6 Blooms: Remembering, Understanding Difficulty: Difficult 28. You are growing a population of cells that appear to be resistant to G418, a chemical that is normally toxic to cells. Upon further analysis of the cells’ genome you find that it contains a genomic sequence that encodes the protein neomycin phosphotransferase. What would you do to test if this protein confers resistance to the G418 chemical? a. determine if the cells are also making β-galactosidase b. tag the protein with green fluorescent protein to see if it is degraded by G418 c. label a fragment of the gene to see where it is expressed d. clone the gene into a G418-sensitive cell line, then treat these cells with G418 Ans: d Question Type: Multiple Choice Chapter: 6 Blooms: Applying, Analyzing Difficulty: Difficult  6-9 29. Modifying the DNA sequence of a gene of interest by appending to it a DNA sequence that encodes a stretch of amino acids recognized by a known monoclonal antibody is called: a. epitope tagging. b. in situ hybridization. c. polymerase chain reaction. d. next-generation sequencing. Ans: a Question Type: Multiple Choice Chapter: 6 Blooms: Remembering, Understanding Difficulty: Moderate 30. Northern blotting is a technique used to detect which one of the following in cells and tissues? a. DNA. b. RNA. c. protein. d. carbohydrate. Ans: b Question Type: Multiple Choice Chapter: 6 Blooms: Remembering, Understanding Difficulty: Easy 31. In the large-scale production of a particular human protein in E. coli cells, the cDNA corresponding to the protein was modified so that the expressed protein would have six histidine residues at the C-terminus. The purpose of this modification was to a. facilitate transfer of the cDNA into the E. coli cells. b. provide a promoter for the transcription of the cDNA in E. coli. c. facilitate purification of the expressed protein though binding to an affinity column containing chelated nickel atoms. d. prevent degradation of the expressed protein by E. coli proteases. Ans: c Question Type: Multiple Choice Chapter: 6 Blooms: Applying, Analyzing Difficulty: Difficult 32. In situ hybridization is a powerful tool used in gene expression studies because it provides the investigator with information pertaining to a. the cellular and tissue-specific localization of the mRNA encoded by a particular gene. b. the activity of the protein translated from a particular mRNA. c. the size of the mRNA transcript. d. all of the above Ans: a Question Type: Multiple Choice Chapter: 6 Blooms: Applying, Analyzing Difficulty: Moderate  6 - 10 33. What is epitope tagging? What is its application? Ans: Epitope tagging is a method by which a cloned cDNA sequence is modified by the addition of a short DNA sequence that encodes a peptide recognized by a known monoclonal antibody. The short encoded peptide is called the epitope. This modified cDNA can be introduced and expressed in cells, and the location of the modified protein can be determined immunologically. Because the modified protein now expresses the epitope, the monoclonal antibody to the epitope can be used to detect the presence of the epitope-tagged protein. Epitope tagging allows the use of a single antibody for detection of any epitope-modified protein and eliminates the need to generate an antibody to each protein of interest. Question Type: Essay Chapter: 6 Blooms: Applying, Creating Difficulty: Easy 34. Compare the advantages and limitations of microarrays and Northern blots for analyzing gene expression. Ans: Microarrays allow a more global analysis of gene expression by analyzing thousands of genes simultaneously. Using the cluster-analysis program, groups of known and unknown genes that are regulated in a coordinated fashion can be revealed. Northern blots allow the analysis of only a few genes at a time; however, a Northern blot can reveal more information about the RNAs than a microarray. For example, a Northern blot can reveal the presence of multiple mRNAs that may be differentially expressed. The presence of multiple mRNAs could be missed by microarray analysis. Question Type: Essay Chapter: 6 Blooms: Analyzing, Evaluating Difficulty: Moderate Section 6.4 35. Which human disease is the result of an inherited X-liked recessive mutation? a. Tay-Sachs disease b. Duchenne muscular dystrophy c. cystic fibrosis d. none of the above Ans: b Question Type: Multiple Choice Chapter: 6 Blooms: Remembering, Understanding Difficulty: Difficult 36. Microsatellites, which often vary in DNA sequence between individuals, are also called: a. single-nucleotide polymorphisms. b. nucleotide tandem replacements. c. short tandem repeats. d. autosomal dominant polymorphism. Ans: c Question Type: Multiple Choice Chapter: 6 Blooms: Remembering, Understanding Difficulty: Difficult 37. The identification of disease genes can be made more complicated because of a. genetic heterogeneity.  6 - 11 b. location of the gene on the X chromosome. c. polygenic traits. d. genetic heterogeneity and polygenic traits. Ans: d Question Type: Multiple Choice Chapter: 6 Blooms: Remembering, Understanding Difficulty: Difficult 38. Linkage studies can map disease genes with a resolution of about one centimorgan. Typically, the physical distance of a DNA region this size is approximately: a. 7.5 × 102 base pairs b. 7.5 × 103 base pairs c. 7.5 × 104 base pairs d. 7.5 × 105 base pairs Ans: d Question Type: Multiple Choice Chapter: 6 Blooms: Remembering, Understanding Difficulty: Moderate 39. A haplotype is a set of closely linked genetic markers on a particular chromosome that tend to be inherited together. The genetic technique that looks at inheritance patterns and uses haplotypes in determining gene locations is: a. linkage mapping. b. linkage disequilibrium mapping. c. candidate gene approach. d. all of the above Ans: b Question Type: Multiple Choice Chapter: 6 Blooms: Remembering, Understanding Difficulty: Difficult 40. How can linkage analysis determine the position of genes on a chromosome? Ans: Linkage analysis examines the frequency of genetic recombination between two genes or markers on a chromosome. A genetic map or linkage map contains many markers mapped to a chromosome. Linkage analysis between the unknown gene and one of the known markers allows the rough localization of the gene on the chromosome. The basis for recombinational analysis is that two genes that are far apart on a chromosome will have a higher frequency of recombination than two genes that are close together. Thus, if recombination between the gene of interest and a marker is very low, then the gene is likely located near that marker gene. Question Type: Essay Chapter: 6 Blooms: Remembering, Understanding Difficulty: Difficult 41. How does genetic heterogeneity or polygenic traits make the identification of a disease gene more difficult? Ans: Some inherited diseases result from a mutation in a single gene; for example, a mutation in the -globin chain of hemoglobin causes sickle-cell anemia. However, some diseases result from mutations in not just one but multiple genes in an individual (a polygenic trait), whereas others result from mutations in any one of multiple different genes in different  6 - 12 individuals, a phenomenon known as genetic heterogeneity. When more than one gene is involved in a single individual or among individuals, then mapping of the disease gene to a marker gene on a linkage map becomes much more complex. Question Type: Essay Chapter: 6 Blooms: Remembering, Understanding Difficulty: Difficult Section 6.5 42. Although small hairpin RNAs are used in RNAi-mediated destruction of a target mRNA, in order for them to be effective, the double-stranded RNA must be cleaved by which of one of the following? a. RISC b. RNA polymerase c. exonuclease d. dicer Ans: d Question Type: Multiple Choice Chapter: 6 Blooms: Remembering, Understanding Difficulty: Moderate 43. CRISPR, one of the latest technologies to allow investigators to edit DNA sequences within a genome, can inactivate gene function if it produces a frameshift mutation leading to which one of the following sequences? a. TAG b. TTT c. TGT d. TTA Ans: a Question Type: Multiple Choice Chapter: 6 Blooms: Remembering, Understanding Difficulty: Difficult 44. A step in the generation of knockout mutations in mice includes selection of embryonic stem (ES) cells that are a. resistant to G418 and resistant to ganciclovir. b. sensitive to G418 and resistant to ganciclovir. c. resistant to G418 and sensitive to ganciclovir. d. sensitive to G418 and sensitive to ganciclovir. Ans: a Question Type: Multiple Choice Chapter: 6 Blooms: Remembering, Understanding Difficulty: Difficult 45. As an investigator you want to functionally inactivate a gene without altering its sequence. Which of the following would you use to accomplish this? a. gene knockout b. RNA interference c. dominant negative mutation d. RNA interference and dominant negative mutation  6 - 13 Ans: d Question Type: Multiple Choice Chapter: 6 Blooms: Applying, Analyzing Difficulty: Difficult 46. In RNA interference studies, the double-stranded RNA a. disrupts the target DNA sequence. b. results in the destruction of the target mRNA. c. destroys the target protein. d. all of the above Ans: b Question Type: Multiple Choice Chapter: 6 Blooms: Remembering, Understanding Difficulty: Easy 47. To study the function of the essential cytosolic Hsc70 genes in yeast, researchers constructed a shuttle vector in which a copy of the Hsc70 gene was ligated to the GAL1 promoter. The vector was then introduced into haploid yeast cells in which all four copies of the Hsc70 genes had been disrupted. Following introduction of the vector, you would expect that a. the yeast cells would grow on both glucose and galactose media. b. the yeast cells would grow on glucose but not galactose medium. c. the yeast cells would grow on galactose but not glucose medium. d. on transfer to either glucose or galactose medium, the vector-carrying cells would eventually stop growing because of insufficient Hsc70 activity. Ans: c Question Type: Multiple Choice Chapter: 6 Blooms: Applying, Analyzing Difficulty: Difficult 48. To inactivate the function of a wild-type small GTPase one could introduce a. a dominant negative allele of a GTPase gene that binds to and inactivates a guanine nucleotide exchange factor. b. an RNAi to silence the expression of the guanine nucleotide exchange factor. c. Cre recombinase. d. a and b. Ans. d Question Type: Multiple Choice Chapter: 6 Blooms: Applying, Analyzing Difficulty: Difficult 49. RNAi is used to functionally inactivate genes in cells and whole organisms like C. elegans. Describe the basics of how you would knock down the expression of a gene required for muscle formation in C. elegans and what method could you use to confirm that your results were specifically attributed to the RNAi? Ans: The cDNA coding sequence of the muscle target gene is cloned, in both the sense and antisense orientations, adjacent to a strong promoter into plasmids. In vitro transcription of both constructs using RNA polymerase and rNTPs yields RNA copies complementary to the coding and antisense sequences. These copies are allowed to anneal and are then microinjected into the gonad of C. elegans. In the developing embryos, Dicer cleaves the double-stranded RNA into siRNAs, which bind to the corresponding endogenous mRNA target. Binding leads either to the degradation of the message or it serves to prevent  6 - 14 the translation of the mRNA into a functional protein. In situ hybridization, Northern blot, or RT-PCR analyses are used to detect changes in the expression of the mRNA transcript, whereas immunocytochemistry or Western blot analysis are used to detect changes at the protein level. Question Type: Essay Chapter: 6 Blooms: Applying, Analyzing, Evaluating, Creating Difficulty: Difficult 50. How can the function of an essential gene required for embryonic development be studied in an adult knockout mouse? Ans: A standard knockout mouse cannot be created to study an essential embryonic gene because the loss of both copies of the gene would lead to embryonic death. The loxP-Cre system, however, can be employed to generate a conditional knockout mouse. In this case, the essential embryonic gene would be flanked by loxP sites. A transgenic mouse homozygous for the loxP modified embryonic gene would be mated with a mouse heterozygous for the loxP modified embryonic gene, which also expresses the Cre protein. The Cre protein would be expressed from a development-specific promoter that turns on Cre expression in a neonatal or adult mouse, leading to loss of the target gene only in post-embryonic mice. Question Type: Essay Chapter: 6 Blooms: Remembering, Understanding Difficulty: Difficult