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SC BIOL 101 - REEC7295_09_TB_chapter16

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Campbell's Biology, 9e (Reece et al.)Chapter 16 The Molecular Basis of InheritanceBecause most of the new material in textbook Chapter 16 is found in Concept 16.2, most of the Test Bank questions are found in this concept section as well. Multiple-Choice Questions1) In his transformation experiments, what did Griffith observe?A) Mutant mice were resistant to bacterial infections.B) Mixing a heat-killed pathogenic strain of bacteria with a living nonpathogenic strain can convert some of the living cells into the pathogenic form.C) Mixing a heat-killed nonpathogenic strain of bacteria with a living pathogenic strain makes the pathogenic strain nonpathogenic.D) Infecting mice with nonpathogenic strains of bacteria makes them resistant to pathogenic strains.E) Mice infected with a pathogenic strain of bacteria can spread the infection to other mice.Answer: BTopic: Concept 16.1Skill: Knowledge/Comprehension2) How do we describe transformation in bacteria?A) the creation of a strand of DNA from an RNA moleculeB) the creation of a strand of RNA from a DNA moleculeC) the infection of cells by a phage DNA moleculeD) the type of semiconservative replication shown by DNAE) assimilation of external DNA into a cellAnswer: ETopic: Concept 16.1Skill: Knowledge/Comprehension3) After mixing a heat-killed, phosphorescent strain of bacteria with a living nonphosphorescent strain, you discover that some of the living cells are now phosphorescent. Which observations would provide the best evidence that the ability to fluoresce is a heritable trait?A) DNA passed from the heat-killed strain to the living strain.B) Protein passed from the heat-killed strain to the living strain.C) The phosphorescence in the living strain is especially bright.D) Descendants of the living cells are also phosphorescent.E) Both DNA and protein passed from the heat-killed strain to the living strain.Answer: DTopic: Concept 16.1Skill: Synthesis/Evaluation1Copyright © 2011 Pearson Education, Inc.4) In trying to determine whether DNA or protein is the genetic material, Hershey and Chase made use of which of the following facts?A) DNA contains sulfur, whereas protein does not.B) DNA contains phosphorus, whereas protein does not.C) DNA contains nitrogen, whereas protein does not.D) DNA contains purines, whereas protein includes pyrimidines.E) RNA includes ribose, whereas DNA includes deoxyribose sugars.Answer: BTopic: Concept 16.1Skill: Knowledge/Comprehension5) Which of the following investigators was/were responsible for the following discovery?In DNA from any species, the amount of adenine equals the amount of thymine, and the amount of guanine equals the amount of cytosine.A) Frederick GriffithB) Alfred Hershey and Martha ChaseC) Oswald Avery, Maclyn McCarty, and Colin MacLeodD) Erwin ChargaffE) Matthew Meselson and Franklin StahlAnswer: DTopic: Concept 16.1Skill: Knowledge/Comprehension6) Cytosine makes up 42% of the nucleotides in a sample of DNA from an organism. Approximately what percentage of the nucleotides in this sample will be thymine?A) 8%B) 16%C) 31%D) 42%E) It cannot be determined from the information provided.Answer: ATopic: Concept 16.1Skill: Application/Analysis7) Which of the following can be determined directly from X-ray diffraction photographs of crystallized DNA?A) the diameter of the helixB) the rate of replicationC) the sequence of nucleotidesD) the bond angles of the subunitsE) the frequency of A vs. T nucleotidesAnswer: ATopic: Concept 16.1Skill: Knowledge/Comprehension2Copyright © 2011 Pearson Education, Inc.8) It became apparent to Watson and Crick after completion of their model that the DNA molecule couldcarry a vast amount of hereditary information in which of the following?A) sequence of basesB) phosphate-sugar backbonesC) complementary pairing of basesD) side groups of nitrogenous basesE) different five-carbon sugarsAnswer: ATopic: Concept 16.1Skill: Knowledge/Comprehension9) In an analysis of the nucleotide composition of DNA, which of the following will be found?A) A = CB) A = G and C = TC) A + C = G + TD) G + C = T + AAnswer: CTopic: Concept 16.1Skill: Application/Analysis10) Replication in prokaryotes differs from replication in eukaryotes for which of the following reasons?A) Prokaryotic chromosomes have histones, whereas eukaryotic chromosomes do not.B) Prokaryotic chromosomes have a single origin of replication, whereas eukaryotic chromosomes have many.C) The rate of elongation during DNA replication is slower in prokaryotes than in eukaryotes.D) Prokaryotes produce Okazaki fragments during DNA replication, but eukaryotes do not.E) Prokaryotes have telomeres, and eukaryotes do not.Answer: BTopic: Concept 16.2Skill: Knowledge/Comprehension11) What is meant by the description "antiparallel" regarding the strands that make up DNA?A) The twisting nature of DNA creates nonparallel strands.B) The 5' to 3' direction of one strand runs counter to the 5' to 3' direction of the other strand.C) Base pairings create unequal spacing between the two DNA strands.D) One strand is positively charged and the other is negatively charged.E) One strand contains only purines and the other contains only pyrimidines.Answer: BTopic: Concept 16.2Skill: Knowledge/Comprehension3Copyright © 2011 Pearson Education, Inc.12) Suppose you are provided with an actively dividing culture of E. coli bacteria to which radioactive thymine has been added. What would happen if a cell replicates once in the presence of this radioactive base?A) One of the daughter cells, but not the other, would have radioactive DNA.B) Neither of the two daughter cells would be radioactive.C) All four bases of the DNA would be radioactive.D) Radioactive thymine would pair with nonradioactive guanine.E) DNA in both daughter cells would be radioactive.Answer: ETopic: Concept 16.2Skill: Application/Analysis13) An Okazaki fragment has which of the following arrangements?A) primase, polymerase, ligaseB) 3' RNA nucleotides, DNA nucleotides 5'C) 5' RNA nucleotides, DNA nucleotides 3'D) DNA polymerase I, DNA polymerase IIIE) 5' DNA to 3'Answer: CTopic: Concept 16.2Skill: Knowledge/Comprehension14) In E. coli, there is a mutation in a gene called dnaB that alters the helicase that normally acts at the origin. Which of the following would you expect as a result of this mutation?A) No proofreading will occur.B) No replication fork will be formed.C) The DNA will supercoil.D) Replication will occur via RNA


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SC BIOL 101 - REEC7295_09_TB_chapter16

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