Chapter 16 Flashcards
1) 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.
B
2) How do we describe transformation in bacteria?
A) the creation of a strand of DNA from an RNA molecule
B) the creation of a strand of RNA from a DNA molecule
C) the infection of cells by a phage DNA molecule
D) the type of semiconservative replication shown by DNA
E) assimilation of external DNA into a cell
E
3) 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.
D
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.
B
5) 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 Griffith
B) Alfred Hershey and Martha Chase
C) Oswald Avery, Maclyn McCarty, and Colin MacLeod
D) Erwin Chargaff
E) Matthew Meselson and Franklin Stahl
D
6) 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.
A
7) Which of the following can be determined directly from X-ray diffraction photographs of crystallized DNA? A) the diameter of the helix B) the rate of replication C) the sequence of nucleotides D) the bond angles of the subunits E) the frequency of A vs. T nucleotides
A
8) It became apparent to Watson and Crick after completion of their model that the DNA molecule could carry a vast amount of hereditary information in which of the following? A) sequence of bases B) phosphate-sugar backbones C) complementary pairing of bases D) side groups of nitrogenous bases E) different five-carbon sugars
A
9) In an analysis of the nucleotide composition of DNA, which of the following will be found? A) A = C B) A = G and C = T C) A + C = G + T D) G + C = T + A
C
10) 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.
B
11) 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.
B
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.
E
13) An Okazaki fragment has which of the following arrangements?
A) primase, polymerase, ligase
B) 3’ RNA nucleotides, DNA nucleotides 5’
C) 5’ RNA nucleotides, DNA nucleotides 3’
D) DNA polymerase I, DNA polymerase III
E) 5’ DNA to 3’
C
14) 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 polymerase alone.
E) Replication will require a DNA template from another source.
B
15) Which enzyme catalyzes the elongation of a DNA strand in the 5' → 3' direction? A) primase B) DNA ligase C) DNA polymerase III D) topoisomerase E) helicase
C
16) Eukaryotic telomeres replicate differently than the rest of the chromosome. This is a consequence of which of the following?
A) the evolution of telomerase enzyme
B) DNA polymerase that cannot replicate the leading strand template to its 5’ end
C) gaps left at the 5’ end of the lagging strand
D) gaps left at the 3’ end of the lagging strand because of the need for a primer
E) the “no ends” of a circular chromosome
C
17) The enzyme telomerase solves the problem of replication at the ends of linear chromosomes by which method?
A) adding a single 5’ cap structure that resists degradation by nucleases
B) causing specific double-strand DNA breaks that result in blunt ends on both strands
C) causing linear ends of the newly replicated DNA to circularize
D) adding numerous short DNA sequences such as TTAGGG, which form a hairpin turn
E) adding numerous GC pairs which resist hydrolysis and maintain chromosome integrity
D
18) The DNA of telomeres has been found to be highly conserved throughout the evolution of eukaryotes. What does this most probably reflect?
A) the inactivity of this DNA
B) the low frequency of mutations occurring in this DNA
C) that new evolution of telomeres continues
D) that mutations in telomeres are relatively advantageous
E) that the critical function of telomeres must be maintained
E
19) At a specific area of a chromosome, the sequence of nucleotides below is present where the chain opens to form a replication fork: 3' C C T A G G C T G C A A T C C 5' An RNA primer is formed starting at the underlined T (T) of the template. Which of the following represents the primer sequence? A) 5' G C C T A G G 3' B) 3' G C C T A G G 5' C) 5' A C G T T A G G 3' D) 5' A C G U U A G G 3' E) 5' G C C U A G G 3'
D
20) Polytene chromosomes of Drosophila salivary glands each consist of multiple identical DNA strands that are aligned in parallel arrays. How could these arise?
A) replication followed by mitosis
B) replication without separation
C) meiosis followed by mitosis
D) fertilization by multiple sperm
E) special association with histone proteins
B
21) To repair a thymine dimer by nucleotide excision repair, in which order do the necessary enzymes act?
A) exonuclease, DNA polymerase III, RNA primase
B) helicase, DNA polymerase I, DNA ligase
C) DNA ligase, nuclease, helicase
D) DNA polymerase I, DNA polymerase III, DNA ligase
E) endonuclease, DNA polymerase I, DNA ligase
E
