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ANSWER KEY
Name_______________________________
Bacterial Genetics,
BIO 4443/6443
Fall Semester 2003
Exam III
StudentpID#__________________________
1.) You have three compatible plasmids and you want to know if they are conjugative,
mobilizable, or nonmobilizable.
Plasmid A has an ampR gene (ampicillin resistance)
Plasmid B has a kanR gene (kanamycin resistance)
Plasmid C has a tetR gene (tetracycline resistance)
You use a recipient E.coli that is resistant to rifampicin and perform the following
experiments.
No
i.) A culture of cells that contains plasmid A is mixed with the recipient,
Colonies
allowed to incubate for several hours, and then the mixture is plated on
a plate containing ampicillin and rifampicin.
ii.) A culture of cells that contains plasmid B is mixed with the recipient,
allowed to incubate for several hours, and then the mixture is plated on
a plate containing kanamycin and rifampicin.
No
Colonies
iii.) A culture of cells that contains plasmid C is mixed with the recipient,
allowed to incubate for several hours, and then the mixture is plated on
a plate containing tetracycline and rifampicin.
iv.) A culture of cells that contains all three plasmids, A, B and C, is mixed with the recipient,
incubated for several hours, and then the mixture is spread on each of the plates shown below.
No
Colonies
ampicillin and rifampicin
kanamycin and rifampicin
tetracycline and rifampicin
1.) Is plasmid A self-transmissible, mobilizable, or nonmobilizable? Why (4pts)
Mobilizable. It is able to transfer to another cell, but only in the presence of the selftransmissable plasmid.
2.) Is plasmid B self-transmissible, mobilizable, or nonmobilizable? Why (4pts)
Nonmobilizable. It is unable to transfer to another cell.
3.) Is plasmid C self-transmissible, mobilizable, or nonmobilizable? Why (4pts)
Self-transmissible. It is able to transfer to another cell
.
4.) Which of the plasmids are likely to have the genes that encode for pilus formation? (4pts)
Self-transmissible plasmids generally encode the genes for pilus formation, plasmid C.
5.) Which plasmids are likely to have the genes that encode for a relaxase or nickase? (4pts)
Both self-transmissible and mobilizable plasmids would need a relaxase-like gene to initiate
transfer, plasmid A and plasmid C.
6.) Which antibiotic(s) were used to counterselect against the donor cells? (4pts)
rifampicin
Name_____________________________
page 2
7.)
Describe (or draw) the events involved in the conjugative transfer of a plasmid. Include
the following terms if appropriate (not all the terms are appropriate). Donor, Recipient,
Transconjugate, Transformant, oriV, oriT, RNAI, Leading strand replication, Lagging Strand
Replication, pilus, Dnase, Relaxase, Primase, Transformasome, Tra pore complex. (10 pts)
1. The donor cell forms a pilus and draws the recipient into contact and other transfer (Tra)
proteins form a pore complex so that transfer can occur.
2. A relaxase, or nicking enzyme, makes a nick at the origin of transfer, oriT, on the plasmid.
3. The 5’ end of the nicked template is held by the pore complex and the lagging strand template
is transferred into the recipient cell.
4. As the lagging strand template is transferred, the leading strand template is re-replicated.
5. The lagging strand template is simultaneously replicated in the recipient cell.
6. Once the entire template has been transferred, replication completes and the ends of the
molecule are joined, and the pore complex breaks down and both cells have the plasmid.
8.) Strain A is a his- trp+ Hfr strain of E. coli. Strain B is a his+ trp- recipient. You mix the two
strains and grow them under conditions where conjugation normally occurs, before spreading the
mixture on minimal plates that contain glucose plus the supplements indicated below. Following
an overnight incubation, draw how each plate would be expected to appear. BELOW EACH
PLATE, EXPLAIN WHAT GROWS AND WHY. (8 pts)
Plate 1:
Plate 2:
Plate 3:
Plate4:
Histidine
Tryptophan
Histidine and
No additions
Tryptophan
Rationale: Strain A grows. Strain B grows. Everything grows. Only transcojugates grow.
No counterselection No selection
No selection
Without supplements, the
against donor
against recipient against either strain. his+ recipient requires the
transfer of trp+.
You repeat the experiment above, this time using a his- trp+ strain of E. coli that carries an F
plasmid, Strain A, and the same his+ trp- recipient as a recipient, Strain B. After spreading and
incubation, draw how each plate would be expected to appear. BELOW EACH PLATE,
EXPLAIN WHAT GROWS AND WHY. (4 pts)
Plate 1:
Plate 2:
Plate 3:
Plate4:
Histidine
Tryptophan
Histidine and
No additions
Tryptophan
No
growth
Rationale: Strain A grows. Strain B grows. Everything grows. Only transcojugates grow.
No counterselection No selection
No selection
The his+ recipient still
against donor
against recipient against either strain. requires trp+ to grow, but
no transfer of
chromosomal genes occurs
using an F plasmid.
Name_____________________________
page 3
You have isolated a new mutant that confers resistance to the antibiotic naladixic acid (nalR) in
an Hfr strain and you would like to determine its position on the chromosome. You know that
trpA is located at 27 minutes on the chromosome, hisG is located at 45 minutes on the
chromosome, and argR is located at 74 minutes on the chromosome and your nalR strain is wild
type for all of these markers. You mate the Hfr strain to an E. coli recipient that is rifR, trpA-,
hisG-, argR- for 90 minutes and then spread the cells on a minimal plate that contains
i) rifampicin, histidine, and arginine. 100 colonies grow up on these plates.
When you replica plate these 100 colonies onto plates that contain
ii.) rifampicin and arginine, 24 colonies grow.
or
iii.) rifampicin and histidine, 3 colonies grow.
or
iv.) rifampicin, arginine, histidine, and naladixic acid, 19 colonies grow.
9.)
What gene(s) is selected for in plate (i), (ii), (iii), and (iv)? (5pts)
i.) trpA (27 min, 100%) ii.) hisG (45 min, 24%) iii) argR (74, min, 3%) iv.) nalR (?, 19%)
10.) Based on your results above, fill in the graph below to determine where the nal gene
maps on the E. coli chromsome AND indicate the minute position for nal next to the graph (9pts)
% recombinants
100
trpA
hisG
nalR
The nal gene is
located at approximately
10
___51___ minutes on the
E. coli chromosome
argR
1
20 30 40 50 60 70 80
Minutes on the chromosome
Name_____________________________
page 4
You’ve identified an operon that controls the expression of the genes that are required to
synthesize the amino acid tryptophan. You mutagenize the bacteria and isolate 8 trp- mutants
that map to this region. Hoping to understand more, you decide to determine how many different
genes are represented in your 8 mutants using a complementation analysis. You isolate F’
factors that each carry one of the trp- mutations.
One by one, you transfer each F’ into each mutant to see if the different trp- mutants can
complement each other. The ability of each transconjugate to grow in the absence of tryptophan
is shown below.
Recipient mutant #1 #2 #3 #4 #5 #6 #7 #8
which carries an F’ with mutant
#1
#2
#3
#4
#5
#6
#7
#8
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
-
+
+
-
-
11.) Which mutants have mutations that are in the same trp gene. How many different genes
have you isolated mutations in? (10pts)
Group A
Group B
Group C
Group D
mut #1, mut #2, mut #6
mut #3
mut #4
mut #5, mut #7, mut #8
The mutations fall into four different genes
You’ve isolated an additional mutant #9 that appears different from the other mutants in
your complementation tests. Its results are shown below.
Recipient mutant #1 #2 #3 #4 #5 #6 #7 #8 #9 #10
which carries an F’ with mutant
#9 -
-
-
-
-
-
-
-
-
12.) Describe a type of mutation that could produce the above result for mutant #9 and
identify where in the operon the mutation occurs? (5pts)
It could be a large deletion that encompasses all four genes or, if all the genes are in the same
operon, it could be a mutation in that inactivates the promoter region and prevents the
transcription of this operon.
Name_____________________________
page 5
The last mutant you isolate, mutant #10, maps to a different portion of the chromosome
from all the other mutations and you suspect that it has landed in a regulatory protein for the
operon. You isolate an F’ with mutation #10 on it and perform the complementation analysis as
before. In addition, you also try complementing the mutant with the wild type strain and obtain
the results shown below.
Recipient mutant #1 #2 #3 #4 #5 #6 #7 #8 #9 #10 WT
cell
which carries an F’ with mutant
#10 +
13.) Assuming that mutation #10 does occur in the regulatory gene for this operon, is the
mutation dominant or recessive to the wild type gene? (2pts)
Since the wild type pheontype is expressed when the two mutations are both present, the
mutation is recessive to the wild type gene.
14.) Assuming that this mutation inactivates the regulatory gene, is it more likely that the trp
operon is regulated by positively or negatively? Why? (3pts)
It is more likely to be a positively regulated operon. Since inactivation of the regulatory
gene renders the cell unable to express (or activate transcription) from the operon.
If it were a negatively regulated operon, inactivation of the regulatory protein would be
expected to result in constitutive expression of the genes since it would not be present to bind
and repress the operon.
15.) You have obtained the following four mutant strains of the naturally competent bacteria
B. subtilis. For each strain, describe what effect the mutation would most likely have on the
competence of the bacteria and WHY. (8 pts)
A strain deleted for comX:
ComX is the pheromone precursor peptide that is processed and secreted by the cell. In
the absence of the pheromone, the cell would not be able to sense and initiate the upregulation of
the compence genes that are needed for DNA uptake.
A strain with a point mutation in comA that renders the ComA protein constitutively active:
ComA is the transcription factor that, when activated by phosphorylation, acts as a positive
transcriptional regulator that turns on the genes necessary for DNA uptake and competence. If
ComA is constitutively active, cells would probably be competent most of the time.
A strain deleted for comP:
ComP is the transmembrane receptor protein that binds to the competence pheromone to
initiate the induction of the competence genes. If it is deleted, it would prevent the ability of the
cell to sense and turn on the competence genes.
Name_____________________________
page 6
You are trying to knockout the lacZ gene in the naturally competent bacterium
Acinetobacter calcoaceticus. You have purified a plasmid that contains the lacZ gene with the
gene for kanamycin resistance inserted into the middle of the gene to inactivate it. You mix
naturally competent culture of bacteria with a solution of plasmid DNA that has been treated in
the following way:
i.) The plasmid was linearized and denatured into single stranded.
ii.) You linearize the plasmid DNA with a restriction endonuclease before you mix it with the
cells
iii.) You leave the circular plasmid intact without doing anything
16.) Which procedure(s) will most likely yield transformants? Why or why not? (8pts)
Only procedure ii would have a chance to yield transformants because naturally
competent cells bind to double stranded DNA and take it up in linear form.
17.) Which of the above substrates would yield transformants if you were to induce competence
by using electroporation to transform the DNA? (4pts)
All of the procedures would allow the DNA to be taken up by the cell since the
electroporation technique generally makes the cells porous to almost any small molecule.
Obtaining a stable transformant would then depend on recombination occurring between the
transformed DNA and the cellular copy.