MCB 421 HOMEWORK #6
Page 1 of 6
FALL 2011
Due October 13
Turn in 1-3
1. Given
a putA- point mutant and a P22 generalized transducing lysate grown on a
random pool of Tn10dCam (chloramphenicol resistant) transposon insertions:
(Tn10dcam is a derivative of Tn10 that has the cat gene encoding resistance to
chloramphenicol.)
a. How could you isolate a Tn10dCam near (but not in) the putA- gene? [Hint: this will
require several steps.] Draw a diagram showing how you would do the experiment and
indicate the medium you would use for each selection or screen. The phenotypes of the
WT and putA mutant are shown in the table below, where “+” indicates growth and “-“
indicates no growth.
putA+
putA-
Min + Glucose + NH4
+
+
Min + Glucose + Proline
+
-
ANSWER:
1. Isolate a Tn10dCam linked to the putA point mutation:
Random Pool
Tn10
putA+
putA-
X
Select CamR put+ on Min + Glucose + Cam + Proline medium.
putA+
Tn10
2. Make P22 HT lysate of above strain and transduce into CamS putA-. Select on
Min + Glucose + Cam + NH4 putA- will grow
Replica plate onto Min + Glucose + Cam + Proline putA- will not grow. This camR
putA- should be the desired strain.
MCB 421 HOMEWORK #6
Page 2 of 6
FALL 2011
Due October 13
b. How would you show that the putA allele is the same as the original putA?
Backcross to recipient strain and select as described above. All should be putA-.
putA-X
putA-X
All cam
R
c. How could you use a Tn10dCam insertion linked to the putA+ gene to isolate point
mutations in the putA gene? [Draw a figure indicating the donor and recipient and any
selection or screen you would use.]
ANSWER: Some generalized transduction particles carry camR and putA+.
Mutagenize with HA (specific for GC to AT). Some putA- mutants occur.
Infect putA+ recipient, select camR on min+ glu+NH4 + cam. Screen for
inability to use pro as a N source.
putA-X
putA-X
Min glu cam _ NH4
putA-X
2. Several E. coli mutations were isolated that produce a Lac- phenotype when present
on the chromosome. The lac operon has three gene products that can be assayed
independently. Complementation analysis was done with each of the new lac
mutations on multicopy plasmids and known lac mutations on the chromosome. The
results are shown in the following table. Based upon the results, indicate which gene
the mutation affects and explain your conclusion.\
MCB 421 HOMEWORK #6
Page 3 of 6
Plasmid
Chromosome
FALL 2011
Due October 13
Mutant
gene
Explanation
lacZ lacY lacA or genes
lac9001
+
-
+
lacY
Cannot complement a lacY mutant but can
complement lacZ and lacA
lac9002
+
+
-
lacA
Cannot complement a lacA mutant but can
complement lacZ and lacY
lac9003
-
+
+
lacZ
Cannot complement a lacZ mutant but can
complement lacY and lacA
lac9004
-
-
-
lacZ,Y,A
Cannot complement a lacY, lacZ, or lacA
mutants
lac9005
+
+
+
unknown
Complements lacZ, lacY, and lacA mutants -- the
three known genes
a.
Assuming that all the results described are valid and that no mutations outside of
the lac operon are required for lactose utilization, what could cause the results observed
for lac-9005 in row 5?
ANSWER: The lac-9005 mutation is known to have a Lac- phenotype when present
on the chromosome. The simplest explanation for this complementation result is
that the lac-9005 mutation results in overproduction suppression when present on a
multicopy plasmid. [An alternative possibility is that the mutation affects some
unknown gene that affects some new gene that results in a Lac- phenotype -- the
statement "the lac operon has three genes ..." was the clue that there really are only
three specific, structural genes that are required for lactose catabolism.
b.
How could you test your hypothesis?
ANSWER: A simple test for overproduction suppression is to do the
complementation analysis with a single copy of each lac mutant.
3. Salmonella Enteritidis expresses a specific fimbriae called Sef which facilitate
attachment of the bacteria to macrophages during infections. A large number of
mutations were isolated that prevent expression of sef genes, and complementation
analysis was done to determine the number of genes disrupted by the mutations. For each
mutation, one copy of the genes was present on the chromosome and a second copy of
the genes was present on a plasmid integrated into the chromosome at a different site.
The results are shown in the table below.
( indicates no Sef fimbriae produced and  indicates that Sef fimbriae are produced.)
MCB 421 HOMEWORK #6
Page 4 of 6
Copy
#1
+
sef
sef-51
sef-52
sef-53
sef-54
sef-55
sef-56
sef-57
sef-58
sef-59
sef-60
FALL 2011
Due October 13
Copy #2
sef-51
sef-52
sef-53
sef-54
sef-55
sef-56
sef-57
sef-58
sef-59
sef-60














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


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


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



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




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





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






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








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



















a. Based upon these results, how many complementation groups were detected?
ANSWER: 4 complementation groups.
b. Indicate which mutations fall into each complementation group:
ANSWER: Group #1 = sef-52, sef-55, sef-57; group #2 = sef-53 and sef-59;
group #3 = sef-54; group #4 = sef-58, sef-60; Note that sef-51 and sef-56
cannot be placed into a complementation group.
c. Suggest a likely explanation for the complementation behavior of the sef-51 mutation.
Would you expect this mutant to produce CRM? [Briefly explain your answer.]
ANSWER: sef-51 is recessive to the wild-type in trans, indicating that this
phenotype is NOT likely to be due to a trans-dominant negative (negativedominant) mutant. Thus, this phenotype is likely to be due to a cis-acting
mutation that prevents expression of all of the sef genes. Not likely to
produce CRM because mutations in a gene resulting in cis-dominance are
usually not made or truncated and thus rapidly degraded.
d. Suggest a likely explanation for the complementation behavior of the sef-56 mutation.
Would you expect this mutant to produce CRM? [Briefly explain your answer.]
ANSWER: sef-56 is dominant to the wild-type in trans, indicating that this
phenotype is likely to be due to a trans-dominant negative (negativedominant) mutant gene product. Likely to produce CRM because such
mutations are typically due to missense mutations.
4. A series of five mutants of phage T4 were isolated. Complementation tests were
MCB 421 HOMEWORK #6
Page 5 of 6
FALL 2011
Due October 13
performed by spotting a mixture of the two phage mutants to be tested (about 106 of
each) onto a lawn of bacteria at either 25°C or 40°C. The results are summarized below
(+ = complete lysis; − = no lysis or only a few plaques in the spot).
25°C
40°C
1
2
3
4
5
1
2
3
4
5
1
+
+
+
+
+
1
+
+
2
+
+
2
+
+
3
+
+
3
+
+
4
4
+
+
5
5
a. Describe the nature of the mutation in each mutant (for example, nonconditional,
temperature sensitive, cold sensitive)?
ANSWER:
Mutant 1 = TS (temperature sensitive)
Mutant 2,3,5 = nonconditional (or null would be acceptable)
Mutant 4 = CS (cold sensitive)
b. How many complementation groups do these mutants affect and which mutants are in
the same complementation groups?
ANSWER: Two complementation groups: group I = mutants 1,2,3 and group II =
mutants 4,5
5. A series of five mutants of phage T4 were isolated. Complementation tests were
performed by spotting a mixture of the two phage mutants to be tested (about 106 of
each) onto a lawn of bacteria at either 25°C or 40°C. The results are summarized below
(+ = complete lysis;  = no lysis or only a few plaques in the spot).
25°C




































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








40°C















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



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











(a) How many complementation groups do these mutants affect and which mutants are in
the same complementation groups?
ANSWER:
Two complementation groups: group I = mutants 1,2,3 and group II =
mutants 4,5
MCB 421 HOMEWORK #6
Page 6 of 6
FALL 2011
Due October 13
(b) Describe the nature of the mutation in each mutant (for example, nonconditional,
temperature sensitive, cold sensitive)?
ANSWER:
Mutant 1 = TS (temperature sensitive)
Mutant 2,3,5 = nonconditional (or null would be acceptable)
Mutant 4 = CS (cold sensitive)
5. Complementation analysis was done on six mutants that lack theonine
synthetase activity. The order of the mutational sites is not known. The results are
shown below:
thr-1 thr-2 thr-3
thr-1 +
+
thr-2
thr-3
thr-4
thr-5
thr-6
thr-4
+
+
-
thr-5
+
+
-
thr-6
-
1. How many complementation groups are represented?
ANSWER: At least two complementation groups: Group 1 = thr-1, thr-4,
thr-5; Group 2 = thr-2, thr-3; Ungrouped = thr-6
2. Suggest an explanation for the results for thr-6.
ANSWER: thr-6 fails to complement all of the other mutants. This could
be either due to a trans-dominant negative phenotype of this mutant (e.g. due
to a missense mutation that poisons threonine synthetase) or a cis-dominant
negative phenotype caused by the mutation (e.g. due to an amber mutation
that prevents expression of downstream genes). In either case, it is impossible
to determine whether the thr-6 mutation is within one of the two
complementation groups described by the other mutations or whether it is in
a different complementation group.