Chapter 14—The Prokaryotic Chromosome: Genetic Analysis In Bacteria
Fill in the Blank
1. The three major evolutionary lineages of living organisms are __________,
__________, and __________.
Ans: bacteria, archea, eukaryotes
Difficulty: 1
2. Of the three major evolutionary lineages of living organisms, __________ and
__________ are much more similar to each other than either of them are to
__________.
Ans: bacteria, archea, eukaryotes
Difficulty: 1
3. Unchecked growth of E. coli cells would generate a mass of bacteria equal to the mass
of the earth in __________ days.
Ans: two
Difficulty: 2
4. A(n) __________ is unable to grow on __________ medium without at least one added
nutrient.
Ans: auxotroph, minimal
Difficulty: 2
5. The process of spontaneous uptake of DNA from the surrounding medium is known as
__________ __________, while the process of __________ __________ requires
laboratory procedures that make cell walls and membranes permeable to DNA.
Ans: natural transformation, artificial transformation
Difficulty: 2
6. The process by which N. gonorrhoeae obtained penicillin resistance is an example of
__________ __________.
Ans: natural transformation
Difficulty: 2
7. IS elements on F plasmids allow homologous recombination between an F plasmid and
the E. coli chromosome to create a(n) __________ bacterium.
Ans: Hfr
Difficulty: 1
8. Conjugation can produce recombinant cells only if an __________ number of
crossovers occurs.
Ans: even
Difficulty: 3
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9. Bacteria can exchange genetic information between different strains through
__________, __________, and __________.
Ans: transformation, conjugation, transduction
Difficulty: 1
10. The endosymbiont theory proposes that __________ and __________ in eukaryotes
originated when free-living bacteria were engulfed by primitive nucleated cells.
Ans: mitochondria, chloroplasts
Difficulty: 1
11. DNA replication in E. coli begins at __________ and proceeds __________ around the
circular genome to the terminator region.
Ans: oriC, bidirectionally
Difficulty: 2
12. In __________ transduction, phages can package any part of the donor genome.
__________ transduction is a property of lysogenic bacteriophages, which can package
only host genes adjacent to the integrated prophage genome.
Ans: generalized, specialized
Difficulty: 2
13. Although bacterial cells are haploid, F' plasmids carrying bacterial genes can create
specific regions of __________ __________.
Ans: partial diploidy
Difficulty: 2
14. The behavior of seeking out or avoiding chemicals is known as __________.
Ans: chemotaxis
Difficulty: 1
15. During chemotaxis, the one-second, one-way movement is called a __________
__________; the abrupt reorienting movement is a __________.
Ans: straight run, tumble
Difficulty: 2
16. Bacteria move up or down a concentration gradient by executing a __________
__________ __________ in which the time spent in a __________ __________ is
longer after the addition of __________ and shorter after the addition of __________.
Ans: biased random walk, straight run, attractant, repellent
Difficulty: 3
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Multiple Choice
17. Which of the following is a bacterial disease common among sexually active
individuals.
A) gonorrhea
B) Haemopilus infections
C) pneumonia
D) pancreatis
E) jaundice
Ans: A
Difficulty: 1
18. Which of the following is an infection in newborn infants when born to mother with
gonorrhea.
A) pneumonia
B) brain lesions
C) eye infection
D) heart disease
E) tendon inflammation
Ans: C
Difficulty: 1
19.
A)
B)
C)
D)
E)
________ are small circular pieces of DNA frequently found in bacteria.
Transposons
Insertion sequences
Plasmids
Bacteriophage
Centromere
Ans: C
Difficulty: 1
20. Penicillin works by inhibiting an enzyme involved in synthesis of cell walls in bacteria.
Mutations result in changes in the amino acid sequence of the enzyme so that it no
longer binds penicillin and cells are no longer inhibited. What will happen to a cell with
this mutation if no penicillin is present in the growth environment?
A) Nothing, it will not be selected and will remain a very minor component of the
microflora.
B) Mutant bacterium will come to dominant the population.
C) Mutant bacterium will not grow normally.
D) Mutant bacterium will not form a normal cell wall.
E) None of these will happen.
Ans: B
Difficulty: 2
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21. Another mutation for penicillin resistance is in an enzyme called a -lactamase.
Penicillin is a -lactam. Changes in the amino acid composition of the enzyme result in
it being able to degrade penicillin so it can't inhibit cell wall synthesis. What will
happen to a cell with this mutation if penicillin is present in the environment?
A) Nothing, it will not be selected and will remain a very minor component of the
microflora.
B) Mutant bacterium will come to dominant the population.
C) Mutant bacterium will not grow normally.
D) Mutant bacterium will not form a normal cell wall.
E) None of these will happen.
Ans: B
Difficulty: 2
22. We isolate nine different mutants unable to make histidine. They map in different
locations, by genetics means we can show there are nine different genes in the pathway.
How many enzymes can we anticipate finding in this pathway?
A) 1
B) 3
C) 6
D) 9
E) 12
Ans: D
Difficulty: 1
23. A bacterium is found that is resistant to the antibiotic gentimycin. The bacterium was
isolated in a hospital where patients were routinely given gentimycin for a variety of
infections. What was the selective pressure driving development of this resistant
population?
A) Presence of gentimycin in the environment.
B) High mutation rate for the bacterium.
C) Growth situation for the bacterium.
D) Patients not receiving antibiotic provided source of bacteria.
E) None of the above
Ans: A
Difficulty: 1
24.
A)
B)
C)
D)
E)
The bacterial chromosome is:
circular.
covalently closed.
1 strand DNA.
contains mobile sequences.
all of the above.
Ans: E
Difficulty: 1
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25. Organisms that lack a nuclear membrane and have no membrane-enclosed organelles
are termed:
A) bacteria.
B) archaea.
C) prokaryotes.
D) eukaryotes.
E) none of the above
Ans: C
Difficulty: 1
26. Bacterial numbers in the large intestine can exceed 1012 /gm in feces. There are perhaps
a half dozen species that predominate this flora and hundreds that play a very small role.
The bacteria grow and die, often lyse releasing DNA into the environment. What
protects viable bacteria from this DNA?
A) Largely impermeable cytoplasmic membrane.
B) Capsule composed of polysaccharide surrounding the cell.
C) Restriction endonucleases in the cells that degrade foreign DNA.
D) Nucleases excreted by bacteria to degrade DNA into sugar and the bases.
E) None of the above
Ans: C
Difficulty: 2
27. We spread 109 bacteria on a plate with 25 µg streptomycin/ml media. We get four
colonies to grow. These colonies include bacteria that are:
A) wild type.
B) resistant to streptomycin.
C) resistant to all antibiotics.
D) able to grow in unusual circumstance.
E) none of the above
Ans: B
Difficulty: 1
28. We grow up the streptomycin resistant (Strr) cells to 109 and plate them on medium with
25 µg/ml gentimycin. We get six colonies to grow. These cells should be:
A) Strr.
B) Gen r.
C) Strr Gen r.
D) wild type.
E) none of the above.
Ans: C
Difficulty: 3
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29. We plate bacteria out on a plate covered with a normally virulent bacteriophage. Two
colonies form. These bacteria in the colonies are now:
A) phage resistant.
B) phage sensitive.
C) wild type.
D) all of the above
E) none of the above
Ans: A
Difficulty: 1
30. How many plantings would it take to make a bacterium resistant to streptomycin,
gentimycin and a bacteriophage?
A) 1
B) 2
C) 3
D) 4
E) 5
Ans: C
Difficulty: 2
31. If resistance to streptomycin occurs about one in 109 cells (one amino acid gets changed
in one ribosomal protein), resistance to gentimycin occurs about one in 109 cells, how
often could you get double mutants, one that was mutated in both functions?
A) 109
B) 1018 (109  109)
C) 1
D) cannot estimate
E) none of the above
Ans: B
Difficulty: 2
32. The Ames test for carcinogens uses his- mutants of S. typhimurium. Carcinogens are
mutagens, and cause mutations. These mutants can be reverted, made to be his+ with
common mutagens. We spread 108 bacteria on a plate and then add the putative
mutagen/carcinogen on a piece of filter paper and count the colonies around the filter
paper. Lots of colonies mean lots of mutations, an effective mutagen. We test two
chemicals, A and B. One µg/ml A causes 14 colonies to appear, one µg B causes 96
colonies to form. Which is the most potent mutagen?
A) A
B) B
C) A or B
D) none of the above
E) both A and B
Ans: B
Difficulty: 3
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33.
A)
B)
C)
D)
E)
A strain of E. coli is trp- his- lac-. Which medium would this bacterium grow on?
lactose + histidine + tryptophan
glucose + histidine + tryptophan
maltose + histidine + tryptophan
minimal glucose, no histidine, no tryptophan
b and c
Ans: B
Difficulty: 2
34. What must an ORF contain to indicate that it codes for a protein?
A) Shine Delgarno sequence immediately upstream from the first codon to bind the
ribosomes to start translation.
B) Met codon, first amino acid.
C) A stop codon, to indicate the end of the protein.
D) A sequence indicative of a real protein, not simply poly phenylalanine.
E) All of the above
Ans: E
Difficulty: 2
35. Transposons are like insertion sequences but include a gene that confers selective
advantage to the bacterium, resistance to an antibiotic or to a heavy metal. Tn10 confers
resistance to tetracycline. We can isolate a Trp- mutant of E. coli by introducing a vector
for Tn10 (a plasmid that won't replicate in E. coli, a bacteriophage with the transposon),
growing for many generations and plating out on medium with tetracycline. This will
select the Tetr mutants, presumably from the transposon. Then the Tetr mutants can be
tested for their ability to grow with and without tryptophan. We have introduced Tn10
into wild type E. coli and want to select mutants unable to use lactose. We can do this
by plating the cells on:
A) tetracycline.
B) tetracycline + lactose.
C) tetracycline + glycerol + X-gal.
D) tetracycline + glucose.
E) tetracycline + maltose.
Ans: C
Difficulty: 3
36. E. coli DNA introduced into S. typhimurium occasionally undergoes recombination, but
this is very rare. Why is this so rare?
A) Restriction endonuclease in S. typhimurium degrade the E. coli DNA.
B) The DNA is dissimilar enough to make recombination less likely.
C) Some genes in E. coli are not found in S. typhimurium.
D) A region found in E. coli is simply absent from S. typhimurium.
E) All of these are good explanations.
Ans: E
Difficulty: 2
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37. We run a cross Hfr A+B+C+D+E+  F-A-B-C-D-E-. After 5 minutes we get F-B+, after 10
minutes F-B+D+, after 15 minutes F-B+D+E+, after 20 minutes, F-B+D+E+A+, after 25
minutes F-B+D+E+A+C+. The gene order is:
A) ABCDE.
B) BDEAC.
C) BDACE.
D) EDCAB.
E) EAEDB.
Ans: B
Difficulty: 3
38. We run a cross Hfr A+B+C+D+E+  F-A-B-C-D-E-. We count numbers of colonies with
each trait in the F- recombinants. We get 122 B+ 102 C+ 88 E+, 65 D+, 45 A+, 14 C+.
What is the gene order (this is called mapping by frequency)?
A) ABCDE
B) EDCBA
C) AEBCD
D) BCEDAC
E) AEDCB
Ans: D
Difficulty: 3
39. In the A region, we also find genes M and K. With transduction we run a cross: P1 (wild
type)  A-M-K-. We get 63 A+K+, 11 M+K+, and 33 M+K+. What is the gene order for
these three markers?
A) AMK
B) AKM
C) AMK
D) MKA
E) cannot determine
Ans: D
Difficulty: 3
40. In artificial transformation in E. coli, we add lots of calcium to the DNA. Why is this
done?
A) To inactivate any nucleases secreted by the cell.
B) To act as a counter ion for the - charges on DNA.
C) To keep RNA from hybridizing with DNA.
D) To keep the DNA single stranded.
E) None of the above
Ans: B
Difficulty: 2
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41. We have a mutant strain, A-B-. If these mutants revert, return to wild type (happens lots
with missense mutations, wrong amino acid) 1 10-7, what is the probability of having
both genes revert?
A) 1  10-7
B) 1  10-7 1  10-7 = 10-14
C) 1  10-3
D) 1  10-6
E) cannot determine
Ans: B
Difficulty: 2
42. We grow two bacteria together in a rich, undefined medium. One is A-B-, one is C-D-.
We grow to 109 cells/ml. We find about 1,000 cells will grow on minimal medium, that
is are A+B+C+D+. What must have happened?
A) A-B- reverted
B) C+D+ reverted
C) genetic exchange occurred
D) any of these could have occurred
E) cannot determine
Ans: C
Difficulty: 1
43. We grow the cells together with DNAase present. We still find the A+B+C+D+. Which
mechanism do we know is not occurring?
A) transduction
B) transformation
C) conjugation
D) mitosis
E) meiosis
Ans: B
Difficulty: 1
44. We grow the cells together with a 0.44 µm filter separating the cells. We don't find the
A+B+C+D+. Which mechanism is probably involved in this genetic exchange?
A) transduction
B) transformation
C) conjugation
D) mitosis
E) meiosis
Ans: C
Difficulty: 1
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45. We've got an Hfr strain with the plasmid inserted near the trp (tryptophan) locus. We
run a cross of this Hfr strain with a F- trp- strain. We isolate a trp+ strain that can readily
donate this trait to other trp- strains. No other traits are transferred. What has happened?
A) generated a F'trp plasmid
B) typical Hfr mating
C) transduction occurs
D) transformation occurs
E) cannot determine from this information
Ans: A
Difficulty: 2
46. By transduction we run a cross between A+B+C+ and A- B-C-. We run the cross and get
179 A+B+C+, 173 A-B-C-, 52 A-B+C+, 46 A+B-C-, 22 A-B+C-, 4 A+B+C- and 2 A-B-C+.
What is the gene order for these three genes (this is a classical three point cross)?
A) ABC
B) ACB
C) BAC
D) CBA
E) none of the above
Ans: B
Difficulty: 4
47. Bacteriophage  normally integrates in a region called int. It is possible to isolate
mutants that lack this site. In these mutants the phage will integrate at random at any
point in the chromosome. And like the wild type situation, occasionally when it excises
from the genome, it carries adjacent genes. Thus a  with nearly any gene can be
isolated. We have made a phage lysate in a int mutant. We take the lysate and infect a
trp- mutant with the lysate. We pick one colony that grows on minimal medium. What
has happened?
A) A  with trp attached has infected the cell.
B) We have a revertant.
C) Conjugation with wild type bacteria has occurred.
D) Transformation with wild type bacteria has occurred.
E) None of the above
Ans: A
Difficulty: 2
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48. We are working with wild type . We prepare a lysate on wild type cells and infect a
gal- strain and get a few colonies that grow on minimal galactose. What has happened?
A) generate a  dgal phage
B) isolated a revertant
C) conjugation
D) transformation
E) none of the above
Ans: A
Difficulty: 1
49. Transposon mutagenesis generates what act like deletion mutants. The function is
completely lost. The transposon has “jumped” into the middle of the gene and it is no
longer functional. A student has a mutant bacterium that is resistant to penicillin, it
makes a -lactamase that recognizes penicillin and cuts it. He infects this cell with Tn5.
He finds a cell that is now resistant to kanamycin (the antibiotic carried by Tn5) but is
no longer resistant to penicillin. What has happened?
A) Transposon alter bacterial metabolism.
B) Transposon alters how bacterium makes cell walls.
C) Transposon hopped into the gene for the -lactamase.
D) Transposon carries information for penicillin resistance in another form.
E) cannot determine
Ans: C
Difficulty: 2
50. We have isolated 15 mutant bacteria unable to use maltose. All 15 mutations map in the
same region by conjugation. We have tested the mutant for two enzymes in maltose
metabolism and find at least one of the mutants lacks one of the enzymes, other mutants
lack the other. What does this suggest?
A) Lots of enzymes in maltose metabolism.
B) All the genes for maltose metabolism are in a cluster, could be one operon.
C) There are genes scattered all over, we've simply found one set of mutants.
D) Maltose transport is very complex, these are transport mutants.
E) None of the above
Ans: B
Difficulty: 2
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51. We have determined that two genes are two minutes apart by mating and co-transduce
about 10% of the time with phage P1. We introduce a plasmid with an insertion
sequence in it. After many generations, we find the two genes are now three minutes
apart and no longer co-transduce at any frequency. What can we conclude?
A) Insertion sequence has inserted between the genes.
B) DNA has migrated to location between the genes and recombined.
C) An unidentified gene has mutated preventing co-transduction.
D) DNA has a loop between the genes that alters its ability to recombine.
E) Cannot determine
Ans: A
Difficulty: 2
52. We are interested in how a bacterium handles osmotic stress. We mutagenize with a
transposon, isolate the antibiotic resistant mutants and replica plate looking for bacteria
no longer able to grow in medium with 0.4 M NaCl (can't handle osmotic stress). How
can we clone this mutation?
A) Use a hybridization probe using a gene in E. coli known to be involved in
osmoregulation.
B) Use an immunological probe using a protein from E. coli† known to be made when cells
are osmotically stressed.
C) Use a hybridization probe using Tn5 DNA.
D) Use a genetic complementation probe, make a gene library form the mutant and test the
genes in wild type bacteria.
E) None of the above
Ans: C
Difficulty: 3
53. The bacteriophage Mu is a transposon, 44 kb long with inverted repeats at either end, a
transposase gene. It replicates by “jumping” from one gene to another in the genome of
the bacteria. With each jump it induces a mutation. We introduce Mu into E. coli and
get a Lac- mutant. How can we determine if this was induced by Mu?
A) Clone the gene, sequence, and look for Mu DNA sequences.
B) Clone the gene, introduce the plasmid with the cloned gene to another host and see if
any mutants.
C) Make a gene library and probe the library with Mu DNA and with lac DNA and see if
the same fragment binds both pieces of DNA.
D) Run an Hfr conjugal cross and see if genes to either side of lacT are a minute further
apart.
E) All of these approaches would work.
Ans: E
Difficulty: 3
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54. There are “promiscuous plasmids,” plasmids that can be interchanged between very
different bacteria. These often pick up transposons carrying antibiotic resistance. We are
in a hospital that has an outbreak of a variety of bacteria carrying resistance to
streptomycin, gentamycin, and penicillin. How could we determine how this was
occurring?
A) Test all the resistant bacteria for a plasmid.
B) See if the same plasmid could be isolated from all resistant bacteria.
C) See if the plasmid isolated from the bacteria conferred restance to a test bacterium.
D) With restriction mapping, see if the plasmid seemed to contain transposons known to
carry resistance to these antibiotics.
E) All of these would be good approaches.
Ans: E
Difficulty: 3
55. Resistance to streptomycin occurs about once in 109 cells in E. coli. Resistance to
penicillin occurs about once in 108. How often would a spontaneous double mutant
occur?
A) Once in 109
B) Once in 108
C) Once in 1017
D) Once in 1072
E) Cannot determine
Ans: C
Difficulty: 2
56. We want to isolate mutants temperature sensitive for utilization of glucose. What kind
of mutagen should we use?
A) transposon
B) uv to cause deletions
C) nitrous acid to cause missense mutations
D) frame shift mutagens
E) none of the above
Ans: C
Difficulty: 2
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57. There are protozoa that engulf cyanobacteria (photosynthetic bacteria) and maintain a
given number of the cyanobacteria even when they divide. The protozoa divide only
after the cyanobacteria have doubled their number. The cyanobacteria excrete glucose
that the protozoa consume. This is a good example of:
A) parasitism by protozoa.
B) example of how protozoa consume particulate food.
C) establishment of an endosymbiotic relationship.
D) symbiosis between two microorganisms.
E) none of the above
Ans: C
Difficulty: 1
58. What would happen if a transposon were to “jump” into the origin of the bacteria
genome?
A) Transposon would be replicated like the genome.
B) Transposon would “jump” to another location.
C) Transposon would inactivate the origin, no DNA replication possible.
D) Transposon would express its genes constituitively.
E) None of the above
Ans: C
Difficulty: 1
59. Bacteria like E. coli do not normally take up foreign DNA, they do not have the proteins
required to bring the DNA into the cell. We get around this with special washing
procedures, with electroporation. These treatments probably affect which aspect of the
cell?
A) The cell wall, open it to permit DNA to pass through.
B) The cytoplasmic membrane, alter it to permit DNA to pass through.
C) The nucleases in the cell, inactivate them.
D) Inactive RNA polymerase so genes don't get expressed as it happens.
E) None of the above
Ans: B
Difficulty: 2
60. Two bacteria grow together. One appears to donate DNA to the other. One of the
bacteria contains a bacteriophage that is normally quiescent but occasionally forms a
plaque. What sort of genetic exchange is occurring?
A) transformation
B) conjugation
C) transduction
D) electroporation
E) none of the above
Ans: C
Difficulty: 1
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61. We think we might have an insertion sequence in a gene causing a mutation. We know
the sequence of insertion sequence. How could we determine if the mutation is caused
by an insertion sequence?
A) Clone the gene and sequence.
B) Clone the gene and probe with insertion sequence DNA.
C) Make PCR mers and see if a ds DNA product is made by the cloned gene.
D) Carry out transduction of adjacent genes and see if they are further apart.
E) All of the above
Ans: E
Difficulty: 3
62. We want to isolate a temperature sensitive DNA polymerase, at 30o the enzyme works,
at 40o it doesn't. What sort of mutagen should be used?
A) Screen for spontaneous missense mutations.
B) Use a missense mutagen like nitrous acid.
C) Introduce a transposon.
D) Clone the gene and introduce an insertion sequence.
E) None of the above
Ans: B
Difficulty: 1
63. Bacteriophage proliferate in the host cell and then lyse the cell when they have perhaps
100 new phage. How many phage could be produced from a culture with 109 bacteria?
A) 109
B) 1010
C) 1011
D) 1012
E) 1013
Ans: C
Difficulty: 2
64. We have isolated a mutation in E. coli that makes the bacterium unable to grow with
high osmolarity. How can we map this gene, the first step?
A) conjugation
B) transformation
C) transduction
D) transposon mutagenesis
E) none of the above
Ans: A
Difficulty: 1
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65. Which technique is used to make a copy of the colonies on a petri plate? One plate can
be analyzed by a destructive method and the other retained to have the cells for future
use.
A) penicillin selection
B) replica plating
C) bifurcated plating
D) none of the above
Ans: B
Difficulty: 1
66. Early in the history of bacterial genetics, it was found that mutations in a pathway
usually mapped in the same region (certainly the same region by conjugation). These
clusters of genes were called “operons.” How would individual genes in each operon be
mapped to determine the gene order?
A) transduction
B) transformation
C) high resolution conjugation
D) a and b
E) b and c
Ans: D
Difficulty: 1
67.
A)
B)
C)
D)
How do you isolate a bacterial mutant resistant to streptomycin?
Plate bacteria on agar medium containing streptomycin.
Replicate plate on regular medium and medium with streptomycin.
Grow cells to mid log phase, add streptomycin and penicillin.
Grow cells to late stationary phase, add streptomycin, measure respiration in
representative colonies.
E) none of the above
Ans: A
Difficulty: 1
68. Rhizobia are bacteria that live symbiotically in the roots of leguminous plants. It has
been proposed that some enzymes in the bacteria were derived from the plant. What
mechanism would most likely have been involved in this?
A) transduction
B) transformation
C) conjugation
D) cannot determine
Ans: B
Difficulty: 2
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69. Many years ago Frederick Griffith isolated a mutant of Streptococcus pneumoniae that
was not pathogenic. It did not make a polysaccharide capsule surrounding the cell.
Griffith showed that he could mix dead pathogenic cells with live non-pathogenic cells
and isolate pathogenic cells. Oswald Avery went on to determine which chemical in the
dead cells conferred pathogenicity back to the live mutant cells?
A) polysaccharide
B) phospholipids
C) protein
D) RNA
E) DNA
Ans: E
Difficulty: 1
70. We have three genes A, B and C. By transformation we find that A and C cotransform
about 5% of the time. B and C cotransform about 5% of the time. A and B never
cotransform. How would this be diagrammed?
A) A........C......B
B) ACB
C) ABC
D) A..........B........C
E) none of the above
Ans: A
Difficulty: 1
Matching
71. Match the following taxonomic groups with their appropriate characteristics.
a. bacteria
1. circular genome
b. archaea
2. lack membrane-bound organelles
c. eukaryotes
3. contain nuclear membrane
4. haploid
5. most live in extreme environments
Ans: a. 1, 2, 4; b. 1, 2, 4, 5; c. 3
Difficulty: 2
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72. Match each item in the left column with the single most appropriate item in the right
column. Each number should be used only once.
a. auxotroph
1. requires cell-to-cell contact
b. conjugation
2. capable of taking up DNA
c. Hfr
3. uptake of DNA from external medium
d. transduction
4. requires at least one nutrient for growth
e. F'
5. F factor that has picked up a piece of the bacterial genome
f. transformation 6. E. coli strain with F factor integrated into bacterial genome
g. competent
7. requires virus
Ans: a. 4; b. 1; c. 6; d. 7; e. 5; f. 3; g. 2
Difficulty: 2
73. Match each item in the left column with the single most appropriate item in the right
column. Each number should be used only once.
a. Hfr strain
1. created through an error in excision
b. merozygote
2. partial diploid in which the two gene copies are identical
c. F' plasmid
3. partial diploid carrying different alleles of the same gene
d. heterogenote
4. created through an integration event
Ans: a. 4; b. 2; c. 1; d. 3
Difficulty: 2
74. Match the experiment to the appropriate medium onto which you would spread cells
from a lac-strs E. coli culture to:
a. select for Lac+ cells 1. rich medium + X-Gal
b. screen for Lac+ cells 2. rich medium + streptomycin
c. select for Strr cells
3. minimal medium + glucose
+
r
d. select for Lac Str
4. minimal medium + lactose
5. minimal medium + streptomycin + lactose
Ans: a. 4; b. 1; c. 2; d. 5
Difficulty: 3
75. Match each item in the left column with the single most appropriate item in the right
column. Each number should be used only once.
a. virulent
1. bacterial cell that contains an integrated phage genome
b. prophage
2. phage that always enters lytic cycle after infecting host
c. lysogenic
3. cycle in which phage DNA integrates into host chromosome
d. lysogen
4. cycle in which phage multiply rapidly and kill the cell
e. temperate
5. the integration of viral DNA into the host chromosome
f. lysogeny
6. integrated phage genome
g. lytic
7. phage that can enter either lytic or lysogenic cycle
Ans: a. 2; b. 6; c. 3; d. 1; e. 7; f. 5; g. 4
Difficulty: 3
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76. Match each item in the left column with the single most appropriate item in the right
column. Each number should be used only once.
a. specialized transduction
1. the simultaneous transduction of two or more
b. generalized transduction
bacterial marker genes
c. cotransformation
2. property of lysogenic phages, which can package only
d. cotransduction
host genes adjacent to the integrated prophage genome
3. the simultaneous transformation of two or more
bacterial marker genes
4. the ability of certain phages to transduce any gene in the
bacterial genome
Ans: a. 2; b. 4; c. 3; d. 1
Difficulty: 2
77. Match each item in the left column with the single most appropriate item in the right
column. Each number should be used once and only once.
a. pilus
1. contains bacterial chromosome
b. insertion sequence 2. group of 107-108 bacterial cells that are genetically identical
c. nucleoid body
3. appendage by which a donor cell contacts a recipient cell
d. capsule
4. small, circular, double-stranded, piece of
e. episome
extrachromosomal DNA
f. plasmid
5. thick, mucus-like coating around some bacterial cells
g. colony
6. small, transposable elements that allow integration of F
plasmid into host chromosome through homologous
recombination
7. plasmid that can integrate into host chromosome
Ans: a. 3; b. 6; c. 1; d. 5; e. 7; f. 4; g. 2
Difficulty: 2
78. Match the type of mutant in the left column with the appropriate isolation method or
methods in the right column.
a. auxotroph
1. screen
r
b. streptomycin
2. selection
c. mutant bacteria that form shiny colonies
3. screen or selection
d. temp-sensitive lethal
e. lacf. streptomycins
g. lac+
h. resistance to infection by T1 phage
Ans: a. 1; b. 2; c. 1; d. 1; e. 1; f. 1; g. 3; h. 2
Difficulty: 3
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True or False
79. Bacteria have a single circular chromosome of dsDNA.
Ans: True
Difficulty: 1
80. An open reading frame is a long stretch of DNA that codes for amino acids
uninterrupted by a stop codon.
Ans: True
Difficulty: 1
81. IS elements are plasmids that “insert” new phenotypes to bacteria.
Ans: False
Difficulty: 2
82. Small circular dsDNA molecules carrying non-essential genes, called plasmids, may
exist as multiple copies in some bacteria.
Ans: True
Difficulty: 1
83. A transductant is a bacteria that has been the recipient of donor DNA from a
bacteriophage.
Ans: True
Difficulty: 1
84. Electroporation is the use of high voltage and high salt to force foreign DNA into
bacteria.
Ans: False
Difficulty: 2
85. A plasmid that can integrate into the host bacterial genome is called an episome.
Ans: True
Difficulty: 1
86. Mating between bacteria can be disrupted using a kitchen blender.
Ans: True
Difficulty: 2
87. The lysogenic cycle of phage production results in bacterial cell lysis.
Ans: False
Difficulty: 1
88. An integrated copy of a temperate bacteriophage is called a prophage.
Ans: True
Difficulty: 1
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Short Answer
89. Describe five types of mutants that can be observed in bacteria.
Ans: Mutations affecting colony morphology; mutations conferring resistance to
antibiotics or bacteriophages; auxotrophic mutations; metabolic mutations
(affecting ability to break down and utilize complicated chemicals in the
environment); and mutations in essential genes.
Difficulty: 3
90. Since a mutation in an essential gene prevents a colony from growing, how would a
bacteriologist study such a strain?
Ans: Use a conditional lethal mutation such as a temperature-sensitive mutation.
Difficulty: 4
91. Explain why different Hfr cells can have F plasmids integrated at different sites.
Ans: Hfr cells are created when F plasmids integrate into the bacterial genome. F
plasmids contain insertion sequences (IS sequences). The bacterial genome also
contains IS sites, found at various positions along the bacterial chromosome. F
plasmids can integrate at any of these bacterial IS sites through homologous
recombination, resulting in Hfr strains with F plasmids integrated at different
sites.
Difficulty: 3
92. Explain the difference between the F-, F+, Hfr, and F' designations.
Ans: An F- cell has no F plasmid, an F+ cell does have an F plasmid, an Hfr cell has an
integrated F plasmid, and an F' cell has an F plasmid that has picked up a piece of
the bacterial genome.
Difficulty: 2
93. Plasmids are small, circular pieces of DNA that some bacteria carry in addition to their
own chromosome. Plasmids may include genes that benefit the bacterial host under
certain conditions. Describe what some of these genes may be.
Ans: One important group of plasmids (the F plasmids) carry genes that promote
conjugative gene transfer between two bacteria. Other beneficial genes include
genes that protect their hosts against toxic metals, genes that allow bacteria to
metabolize petroleum products, genes that contribute to pathogenicity, and genes
encoding resistance to antibiotics to name just a few.
Difficulty: 3
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94. Describe the difference between generalized and specialized transduction.
Ans: Transduction is a form of gene transfer in bacteria that depends on the packaging
of bacterial donor DNA in the protein coat of a bacteriophage. In generalized
transduction, phages can package any part of the donor genome. Specialized
transduction is a property of lysogenic bacteriophages, which can package only
host genes adjacent to the integrated prophage genome.
Difficulty: 3
95. Describe the difference between “selection” and “screen” with respect to mutant
isolation methods.
Ans: A “selection” uses conditions in which only the desired mutant will grow (e.g.
selection for resistance to an antibiotic by growing the cells on plates containing
the antibiotic). A “screen” requires the examination of each colony in a
population for its phenotype (e.g. to screen for colonies with different
morphological characteristics).
Difficulty: 3
96. Describe how to screen for bacteria that have lost their ability to carry out chemotaxis.
Ans: Inoculate bacteria onto center of a wet nutrient plate. Bacteria will multiply and
motile bacteria will swarm out to edges of plate in concentric circles, with the
leading edge moving toward the unused nutrient supply. After motile cells have
swarmed, pick cells from the center of the plate. These are the cells that could not
chemotax to unused nutrient supply.
Difficulty: 3
97. What type of processes could be defective in a mutant unable to carry out chemotaxis?
Ans: Some might be defective in making flagella (flagellum mutants); some in turning
the flagella (motor mutants); others in coordinating the behavior of flagella (signal
transduction mutants); and still others in detecting gradients of chemicals
(receptor mutants).
Difficulty: 3
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Experimental Design and Interpretation of Data
98. In the original interrupted mating experiment performed by Wollman and Jacob, they
mated HfrH strs thr+ azir tonr lac+ gal+ cells to F- strr thr- azis tons lac- gal- cells,
disrupted mating pairs at 1 minute intervals and plated the exconjugants on plates
containing streptomycin and lacking threonine. a. Why did they choose that type of
medium to plate on? b. In order for the above experiment to work, what must be true
about the location of the thr gene?
Ans: a. They used plates containing streptomycin and lacking threonine to select for
cells that had undergone conjugation. Donor cells are killed by streptomycin and
recipient cells cannot grow without threonine. b. The thr gene must be close to
the F plasmid integration site in that particular strain and thus is one of the first
genes transferred during the interrupted mating experiment.
Difficulty: 4
99. A P1 generalized transducing lysate from a wild-type strain is used to infect a leu-thrara- recipient strain. When ara+ transductants are selected, 47% of the ara+ also carry
leu+ and 2% of the ara+ transductants also carry thr+. When leu+ transductants are
selected, 50% also carry ara+, but none also carry thr+. What is the order of the three
genes?
Ans: The order is leu ara thr (ara closer to leu than to thr). The leu and thr genes were
not co-transduced but ara and thr were co-transduced at low frequency, so they
must be closer together than leu and thr.
Difficulty: 3
100. Describe how you would screen for Arg- cells arising from a wild-type culture of E.
coli.
Ans: To increase frequency of mutation, first expose wild-type culture to a mutagen.
Then plate mutagenized cells onto minimal plates containing arginine. After
colonies have grown up, replica plate onto minimal plates without arginine. Note
position of cells that grow on first set of plates, but not the second set. Those are
the Arg- mutants.
Difficulty: 3
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101. Through the use of different Hfr strains (H, 1, 2, and 3) that have the F factor inserted
into the chromosome at different points and in different directions, interrupted-mating
experiments can provide the order of genes in the E. coli genome. The linear order of
transfer of markers is shown below for four Hfr strains. Using this data, construct a
physical map of the E. coli chromosome.
HfrH
thi
gly
his
gal
pur
lac
Hfr1
pro
lac
pur
gal
his
gly
Hfr2
lac
pur
gal
his
gly
thi
Hfr3
gal
his
gly
thi
thr
pro
Ans: The E. coli chromosome is circular, and the order of markers is: thi gly his gal pur
lac pro thr.
Difficulty: 2
102. In an Hfr X F- cross, azir enters as the first marker, but the order of the other markers is
unknown. If the Hfr is wild-type and the F- is auxotrophic for each marker in the
experiment, what is the order of the markers in a cross where azir recombinants are
selected if 61% are lac+, 27% are gal+, 89% are tonr, and 1% are trp+?
Ans: The order of the markers used in this experiment is: azir, tonr, lac+, gal+, trp+.
Difficulty: 3
Page 278
103. A cross is made between an Hfr strain that is met+arg+gly+ and an F- strain that is metarg-gly-. Interrupted-mating experiments show that gly+ enters the recipient last, so the
gly+ recombinants are selected on minimal plates supplemented with methionine and
arginine. These recombinants are then tested for the presence of the met+ and arg+
alleles. The following numbers of colonies are found for each genotype:
gly+met+arg+
420
+
+
gly met arg
11
gly+met-arg+
0
+
gly met arg
49
a. Why was glycine (gly) left out of the selection medium?
b. What is the gene order?
c. What are the map distances in recombination units?
Ans: a. Glycine was left out to allow for selection of gly+ recombinants because gly is
the last marker to enter the recipient in this mapping experiment. This ensures
that all loci being mapped in the cross have already entered each recombinant that
is analyzed. b. Because we know that gly enters the cell last, there are only two
possible orders to consider: arg-met-gly and met-arg-gly. Because the gly+metarg+ recombinant did not show up in this experiment, we can conclude that it
required a quadruple crossover event, indicating that the order of genes must be
arg-met-gly. c. To determine map distance between two genes, count the number
of crossover events that occur between them. Recombination events between arg
and met will result in a Gly+Met+Arg- phenotype (remember, all cells were
selected first as gly+ recombinants). Thus, the distance between arg and met is
11/480 = 2.3 map units. Similarly, the map distance between met and gly is
49/480 = 10.2 map units.
Difficulty: 4
Page 279
104. A generalized transducing phage is used to transduce an a-b-c-d-e- recipient strain of E.
coli with an a+b+c+d+e+ donor. The recipient culture is plated on various media
indicated in the second column of the table below, with the results shown in the third
column. (Note that "a-" indicates a requirement for A as a nutrient, etc.)
Experiment Compounds added
Presence (+) or
Number
to minimal medium absence (-) of colonies
1
CDE
2
BDE
+
3
BCE
4
BCD
5
ADE
6
ACE
+
7
ACD
+
8
ABE
9
ABD
10
ABC
a. What genotype is being selected for in each experiment above?
b. Which of the above markers can be cotransduced?
c. What can you conclude about the linkage and order of the five genes?
Ans: a. Experiment 1 is selecting for a+b+ recombinants, experiment 2 for a+c+, 3 for
a+d+, then a+e+, b+c+, b+d+, b+e+, c+d+, c+e+, and d+e+ in experiment 10. b. The
markers that can be cotransduced are a and c, b and d, and b and e. c. b, d, and e
are relatively close to each other, with b in the middle. a and c are close to each
other, but far from b, d, and e.
Difficulty: 3
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