Chapter 14 key

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Use the same bubble sheet as for Chapter 9 homework part 2.
Chapter 14 homework. Use the Chapter 8 part 2 bubble sheet for this one.
15. The F episome is _____________.
A. a bacteriophage
B. a fertility factor*
C. a restriction modification system
16. The F episome __________.
A. can replicate as a plasmid*
B. can replicate as a phage
C. makes E. coli pathogenic.
17. The F episome can ____________.
A. only replicate as a plasmid
B. integrate into the bacterial genome*
18. Integration of a plasmid into the genome occurs _____________.
A. when crossing over occurs between a homologous region of the plasmid and
the genome*
B. when the cell is starved for essential nutrients.
C. when the DNA replicates
19. The F episome gives E. coli the ability to transfer DNA to ____________.
A. E. coli that contain an F episome
B. E. coli that do not contain an F episome*
C. other species of bacteria
20. Integration of the F into the genome occurs ____________________.
A. because of a cross over between an IS element on the episome and an IS
element in the genome*
B. because of a cross over between the F episome's pilus gene and the pilus gene
in the genome.
21. Horizontal or lateral transfer of genetic material refers to ___________.
A. transfer between contemporaries rather than transfer from parent to offspring*
B. transfer between the parent to ONLY one of the daughter cells
22. An Hfr cell is derived from a ____________.
A. a hfr minus cell
B. F- cell
C. F+ cell*
23. In an interrupted mating experiment between a Hfr and a F- cell, the purX
gene is transferred at 1 minute and the purY gene is transferred at 10
minutes. Which statement is true?
A. The F- gene is located at zero minutes.
B. The purX gene is closer to the origin of replicaiton than the purY gene.
C. The purX gene is closer to the the integrated F episome.*
D. The purY gene is closer to the the integrated F episome.
24. In a P1 bacteriophage transduction experiment you observe the following
cotransduction percentages for the ind+, pyrE+,and ile+ genes.
gene
cotransduction percentage
pyrE+ ile+
10%
pyrE+ ind+
30%
ind+ ile+
20%
Which genetic map is most correct?
ind
ile
pyrE
A)
*
ile
pyrE
ind
B)
pyrE
ind
ile
C)
pyr
ile
0 10
30
ind
D)
2
25.
In a P1 bacteriophage transduction experiment you generate the following
map. Choose the cotransduction percentages for the pyrB+, purA+,and malB+
genes that best match this map.
purA
malB
pyrB
A)
gene
C)
gene
cotransduction
percentage
*
cotransduction
percentage
pyrB+ malB+ 10%
pyrB+ malB+ 15%
pyrB+ purA+
pyrB+ purA+
50%
30%
purA+ malB+ 5%
purA+ malB+ 30%
B)
gene
D)
gene
cotransduction
percentage
cotransduction
percentage
pyrB+ malB+ 30%
pyrB+ malB+ 50%
pyrB+ purA+
pyrB+ purA+
15%
purA+ malB+ 15%
50%
purA+ malB+ 40%
26. Which of the following describes an Hfr E. coli strain?
A) A "male" E. coli.
B) E. coli that contains an F+ episome that is maintained as a plasmid.
C) E. coli in which the F+ episome has integrated into the genome. *
D) E. coli that does not contain an F+ episome
E) The E. coli cell immediately after mating.
F) The recombinant progeny of an E. coli mating.
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27. We have a Hfr E. coli strain and will use it to map the position of a number of
genes.
The cross is as follows
HfrH, cys+, gal+, lac+, phe+ X F-, cysA- ,galA- , lacZ- , pheAStrepR thrStrepS thr+
First all streptomycin resistant threonine prototrophs are selected and then genotyped.
Phenotype of colonies
Time at which sample was taken and
blended
0
none
5
cys-, gal-, lac-, phe-
10
cys-, gal-, lac+, phe-
15
cys-, gal-, lac+, phe-
20
cys-, gal+, lac+, phe-
25
cys-, gal+, lac+, phe-
30
cys-, gal+, lac+, phe-
35
cys-, gal+, lac+, phe-
40
cys-, gal+, lac+, phe-
45
cys-, gal+, lac+, phe-
50
cys-, gal+, lac+, phe+
55
cys-, gal+, lac+, phe+
60
cys-, gal+, lac+, phe+
65
cys-, gal+, lac+, phe+
70
cys-, gal+, lac+, phe+
75
cys+, gal+, lac+, phe+
80
cys+, gal+, lac+, phe+
85
cys+, gal+, lac+, phe+
90
cys+, gal+, lac+, phe+
95
cys+, gal+, lac+, phe+
100
cys+, gal+, lac+, phe+
4
Please draw the genetic map on the next page. Answer A on the bubblesheet
Correct bubblesheet answer is A*
DRAW THE GENETIC MAP.
USE AN ARROW TO INDICATE AN EXACT POSITION.
USE A BRACKET, { , TO IDENTIFY A RANGE OF POSSIBLE POSITIONS
**Answer**
Best answer
lacZ
0
}
95
5
90
10
85
15
20
80
}
cysA
}
galA
25
75
30
70
35
65
60
40
55
50
45
}
phenA
They might have drawn this as a linear map which begins and ends at time zero. That is
fine and would also receive full credit.
The second best answer is the next map.
5
90
lacZ
0
95
5
10
85
15
cysA
galA
20
80
25
75
30
70
35
65
60
40
55
50
45
phenA
28. In the previous question, at 75 minutes will all of the cys+, also be gal+.
A. They will.
B. They will not.*
29.
You are mapping E. coli by interrupted mating.
Here is the cross:
HfrH, met+, arg+, met+
X
StrepS thr+
F–, met– ,his–, arg–
StrepR thr–
met– = methionine auxotroph
his– = histidine auxotroph
arg– = arginine auxtroph
You mix the strains and at 1-minute intervals take a sample, blend it, and plate.
Plate on media containing Streptomycin and lacking threonine.
Then you replicate plate the StrepR thr+ colonies to
methionine dropout plates (has everything but methionine)
histidine dropout plates (has everything but histidine)
arginine dropout plates (has everything but arginine)
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Time at which the sample was taken.
Phenotype of colonies
4 minutes
met–, his–, arg-
5 minutes
met–, his–, arg+
6 minutes
met–, his–, arg+
37 minutes
met–, his–, arg+
38 minutes
met–, his+, arg+
39 minutes
met–, his+, arg+
79 minutes
met–, his+, arg+
80 minutes
met+, his+, arg+
Please draw the correct map. Answer B to get credit for this question.
**Answer**
B*
arg
met
80
5
38
his
30. In the question above, why are the bacteria plated on media containing
streptomycin and lacking threonine?
A. This is a "red-herring"–it is not done.
B. To eliminate the presence of the F+.
C. To prevent a second round of mating (F1 X F1).
D. To kill both parental strains.*
E. To prevent contaminants from growing.
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31. The FIRST part of these question will not be graded on the homework. It is a
map. However, it is important to do it. The questions that follow are going to
be graded.
You are mapping a large number of bacteria genes.
First you use an interrupted mating with mutants and a Hfr wild type.
You have these mutant lines
line 33
M- R- T- StrepR, thrline 52
B- O- StrepR, thrline 15
D- N- W- StrepR, thrline 21
T- X- Y- StrepR, thrline 47
N - O- R- StrepR, thrline 62
J- Z- StrepR, thrYou have this strain that is wild type for B, D, M, N, O, R, T, W, X, Y, and Z
Hfr B+ D+ M+ N+ O+ R+ T+ W+ X+ Y+ Z+ StrepS, thr+
You use interrupted mating to determine the position of the single letter genes. In each
cross shown below, the time of transfer of the wild type gene is given.
Hfr
Hfr
Hfr
Hfr
Hfr
Hfr
X
X
X
X
X
X
line 15 --> D+ at 9 min, N+ at 30 min, W+ at 5 min,
line 21 ---> T+ at 9 min, , X+ at 9 min, Y at 39 min
line 33 ---> M at 10 min, R at 20 min, T+ at 9 min
line 47 --> N+ at 30 min, O at 30 min, R+ at 20 min
line 52 --> B+ at 30 min, O at 30 min
line 62 --> J at 28.5 min, Z at 35 min
Now you do co-transduction mapping using a P1 transducing phage.
The ONLY cotransduction that occurs is shown below.
Selected marker
Cotransduction of unselected marker
Colonies were
selected for T
70% X
90% D
2% M
Colonies were
selected for X
90%D
70%T
0%M
Colonies were
selected for M
2%T
0.5%D
0%X
Colonies were
selected for J
1% O
10% B
30% N
Please draw the map. Answer C on the bubblesheet.
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Answer C*
5
9
W
X
9
D
9
T
10
M
20
R
J
N
O
Y
39
28.5
30
B
Z
35
This should be a circle because the E. coli genome is circular. However, for the XDT
and NBO regions I have drawn a straight line under them just to indicate that by
interrupted mating each group maps to one position--XDT at 9 min and NBO at 30 min.
This occurs because they are very close to each other. By transduction mapping we can
see that they are actually at distinct locations.
How should this be graded?
Interrupted mating map is worth 5 points
W @5, [email protected] (any order), [email protected], [email protected], [email protected], [email protected] (any order because
order cannot be determined by interrupted mating), [email protected], [email protected]
5 points for the P1 transduction portion.
Order of XDT and NBO determined by P1 transduction mapping. Not only must the
order of these three be correct but it must be correct with respect to the outside markers
(WXDTM and JNBOZ).
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32. What the StrepR, StrepS, thr+, and thr- are for? Choose the best answer.
A. So that each F- can be distinguished from the other F-'s.
B. So that recombinants can be distinguished from non-recombinants.*
C. So that the Hfr parent can be distinguished from the F- parent.
D. To kill the F- parent.
33. Based on these data tell me, tell me how many minutes of the genome fit into
a P1 phage head. Choose from this list.
A. 1 B. 1.5* C. 2
D. 2.5
E. 3
F. 3.5
G. 9 H. 30
34. Why don't you see co-transduction for any of the other pairs of genes?
A. Because they are in different complementation groups.
B. Because they are in the same complementation group.
C. There are too many.
D. They are on different chromosomes
E. They are too close.
F. They are too far apart. *
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35. You are mapping genes in E. coli. You are performing co-transduction using a
P1 transducing phage.
You have a wild type E. coli strain and a strain that carries the following mutations
a-, b-, c-, d-, eYou infect a wild type strain with the P1 bacteriophage and it lyses its host.
The offspring phage are used to infect the mutant strain described above.
The infected bacteria are plated on media that selected for different wild type versions
of the genes above. The colonies that grow are then tested to determine if they carry
wild type or mutant copies of the other genes.
Blank entries represent 0% cotransduction.
Cotransduction frequency of
the unselected marker
Selected marker
A+
B+
Colonies were selected for A+
-
2%
Colonies were selected for B+
2%
-
C+
D+
E+
50%
Colonies were selected for C+
-
1%
Colonies were selected for D+
1%
-
2%
2%
-
Colonies were selected for E+
The correct order for these genes is:
A. ABDCE
B. ABEDC*
C. BACDE
D. BADEC
E. CDABE
F. CDBEA
50%
G.
H.
I.
J.
DCAEB
DCBAE
EABAC
EDBAC
***Part of the key A--------B-E--------D--------------C ***
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36. Which statement is true? Answer this question with respect to the previous
question.
A. B is closer to A than it is to E
B. B is closer to E than it is to A*
C. B is equidistant from A and E.
37. Genetic material is transferred between from a Vibrio cholera bacteria to an
Escherichia coli bacteria. This is referred to as a ________________ transfer
of genetic material.
A.
B.
C.
D.
E.
F.
G.
galeic
lateral *
pangeic
pathogenic
trans
transpecial
vertical
38. Conjugation is ____________.
A. picking up of free DNA from the environment
B. the introduction transfer of one cell's DNA to another cell by a bacteriophage
C. transfer of genetic material between living bacterial cells - a form of mating *
39. Transformation is _____________.
A. picking up of free DNA from the environment*
B. the introduction transfer of one cell's DNA to another cell by a bacteriophage
C. transfer of genetic material between living bacterial cells - a form of mating
40. Transduction is _________________.
A. picking up of free DNA from the environment
B. the introduction transfer of one cell's DNA to another cell by a bacteriophage*
C. transfer of genetic material between living bacterial cells - a form of mating
41. What actually occurs when the F+ episome transfers itself the F- cell?
A.
B.
C.
D.
E.
Only one strand of the F+ DNA is transferred.*
The DNA is transferred as double stranded DNA.
The F+ cell is converted to an F- cell.
The F+ cell loses its F episome.
C and D
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42. Natural transformation can transfer genetic material between
A. completely different types of bacteria, for instance between Streptococcus and
Neisseria*
B. different species of Bacillus only
C. different types of bacteria as long as they are both either gram positive or gram
negative
43. Natural transformation is _____________
A. used by scientists to introduce DNA into E. coli.
B. used by scientists to introduce DNA into Bacillus subtillus*
44. The process by which Neisseria gonorrhoeae obtained penicillin resistance is
an example of __________ __________.
A. artificial transduction
B. conjugation
C. genetic engineering
D. natural transformation*
E. spontaneous mutation
F. transduction
45. How does an F+ episome integrate into the bacterial genome to produce an
Hfr strain?This occurs via site-specific crossing over between __________ in
the F episome and ______________ in the genome.
A. an IS element,
the Hfr site
B. an IS element,
an IS element*
C. an IS element,
the Origin
D. the Hfr site, an IS element
E. the Origin, an IS element
46. Which of the following is the best description/definition of an episome?
A. A naturally-occurring plasmid that confers on its host antibiotic resistance.
B. A naturally-occurring plasmid that has the ability to integrate into the bacterial
host's genome.*
C. A scientist-built plasmid that can be used to produce F+ male bacteria.
47. What does MRSA stand for and what causes it?
A. It stands for massively-resilient super antigens whose unique properties are
caused by a mutation in in the pra gene.
B. It stands for massively-resilient super antigens whose unique properties are
caused by a mutation in the receptor binding region.
C. It stands for methicillin-resistant Staphylococcus aureus whose unique
properties are caused by a mutation in the pra gene.
D. It stands for methicillin-resistant Staphylococcus aureus whose unique
properties are caused by a transposable element that encodes a penicillinbinding protein.*
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48. Why are transposable elements that carry antibiotic resistance genes OR
disease causing genes a major problem? Choose the single best answer.
A.
Because they facilitate the transfer of these undesirable traits to other bacteria
of the same species or even to bacteria of different species.*
B. Because they facilitate the transfer of these undesirable traits to other bacteria
of the same species only.
C. Because this type of bacteriophage can infect bacteria of different genera and
species.
49. Horizontal or lateral transfer of genetic material refers to ___________.
A. transfer between contemporaries rather than transfer from parent to offspring*
B. transfer between the parent to ONLY one of the daughter cells
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