Gene Transfer in Bacteria and
Bacteriophage
Using Gene Transfer Between
Bacteria As a Means for Studying
Bacterial Genes
Types of
Traits Studied
• For bacteria
-need for nutrients
prototropic: can grow on minimal medium
auxotropic: must have specific nutrients added
to medium
-morphology of colonies
-resistance/sensitivity to antibiotics
• For bacteriophage
-host range (ability to infect specific bacteria)
-appearance of plaques (shows growth)
Testing for Nutritional Requirements
Replica plating transfers the pattern of
bacterial colonies to test plates.
DNA of Prokaryotic Cells
• Bacterial cells have a single, circular
chromosome and therefore have one
copy of each gene.
• Partial diploids (merozygotes) can be
formed by the introduction of genetic
material from another cell.
Gene Transfer Processes for
Bacteria and Their Viruses
1.
2.
3.
4.
Conjugation
Transformation
Transduction
Infection with bacteriophage
Gene Transfer Processes for
Bacteria and Their Viruses
1. Conjugation
Transfer of DNA from
one bacterial cell
to another
Donor cell (F+ or Hfr) transfers DNA to
recipient cell (F-)
Conjugation
Genetic Analyses Using Conjugation
A. Determining linkage from interrupted
mating experiments
B. Determining gene order from gradient
of transfer
C. Higher-resolution mapping by
recombination frequency
Genetic Analyses Using Conjugation
A. Determining linkage from
interrupted mating experiments
Combine Hfr strain (Strs) and F- strain.
Remove samples at specific time intervals.
Use blender to disrupt mating.
Plate on streptomycin to kill donor cells.
Test recipient cells for genes from Hfr
strain.
Genetic Analyses Using Conjugation
A. Determining linkage from interrupted
mating experiments
Problem 1, page 2-4
Three Hfr strains for E. coli are mated
individually with an auxotrophic F- strain using
an interrupted mating procedure. Approximate
times of entry of each gene are listed below.
Determine the map of the E. coli chromosome
and show the orientation of the F plasmid in
each Hfr strain.
Genetic Analyses Using Conjugation
A. Determining linkage from interrupted
mating experiments
Problem 1, page 2-4
Strain 1
lac+
3 min
Strain 2
Strain 3
argG+
19 min ilv+
5 min
gal+
his+
argG+
xyl+
12 min
39 min
63 min
74 min
xyl+
ilv+
thr+
lac+
ilv+
thr+
78 min gal+
95 min his+
30 min
34 min
51 min
59 min
xyl+
argG+
his+
gal+
68 min lac+
95 min thr+
9 min
20 min
44 min
71 min
80 min
88 min
Genetic Analyses Using Conjugation
A. Determining linkage from interrupted
mating experiments
lac
Problem 1, page 2-4
thr
17
ilv
8
9
1
3
gal
27
4
xyl
11
arg
2
his
24
Genetic Analyses Using Conjugation
B. Determining gene order from gradient
of transfer
Combine Hfr and F- strains.
Allow for natural disruption of conjugated
pairs.
Select for earliest transferred marker.
Test for markers transferred later in
conjugation.
Genetic Analyses Using Conjugation
B. Determining gene order from gradient of
transfer
Problem 2, page 2-4
An Hfr strain donates the genes xyl+ pro+
lac+ and gal+ to an F- strain. Recombinants
are selected for gal+. Tests are done to
determine the presence of the other three
genes in the gal+ recombinants. What is
the gene order?
Genetic Analyses Using Conjugation
2. Determining gene order from gradient of transfer
Problem 2, page 2-4
gal+
100% of strains
lac+
pro+
70% of strains
30% of strains
xyl+
10% of strains
Select for gal+
Test for lac+, pro+, xyl+
Gene order: Gal---Lac---Pro---Xyl
Recombination to Integrate
Transferred Genes
a+
a
b+ c+
b
c
a
b
c
a+
b+ c+
Genetic Analyses Using Conjugation
C. Higher-resolution mapping by
recombination frequency
Combine Hfr and F- strains.
Allow for natural disruption of conjugated
pairs.
Select for marker that enters LAST.
Test for unselected markers.
Genetic Analyses Using Conjugation
C. Higher-resolution mapping by
recombination frequency
Problem 3, page 2-4
An Hfr strain that is met+ arg+ leu+ strs is
conjugated with an F- strain that is met- argleu- strr. Interrupted mating studies show that
leu+ enters last. Recombinants that are leu+
strr are selected and then tested for the
presence of met+ and arg+. The following
numbers of bacteria are found for each of the
genotypes listed below. Determine the gene
order and the distances between the genes in
map units.
Genetic Analyses Using Conjugation
C. Higher-resolution mapping by
recombination frequency
Problem 3, page 2-4
leu+ met- argleu+ met+ argleu+ met+ arg+
leu+ met- arg+
50
80
Select for leu+
370
Test for met+, arg+
0
Genetic Analyses Using Conjugation
C. Higher-resolution mapping by
recombination frequency
Problem 3, page 2-4
leu+
met+
arg+
Hfr
met-
F-
Genetic Analyses Using Conjugation
Problem 3, page 2-4
Smallest number of offspring represents 4
crossovers, identifies middle gene.
Genotype will be leu+ met- arg+.
leu+
met+
arg+
Hfr
met-
F-
Genetic Analyses Using Conjugation
Problem 3, page 2-4
Recombination between leu and met gives
leu+ met- arg- offspring.
leu+
met+
arg+
Hfr
met-
F-
Genetic Analyses Using Conjugation
Problem 3, page 2-4
Recombination between met and arg gives leu+ met+
arg- offspring.
leu+
met+
arg+
Hfr
met-
F-
Genetic Analyses Using Conjugation
Problem 3, page 2-4
Leu  met
Met  arg
leu
50 = .1 = 10 map units
500
80 = .16 = 16 map units
500
met
10 map units
arg
16 map units
Gene Transfer Processes for
Bacteria and Their Viruses
2. Transformation
DNA taken up from external environment
Genetic Analysis Using
Transformation
Determining genetic distance with
transformation mapping
Transform bacteria with DNA containing two
markers (eg. his-, met-) in addition to penicillin
sensitivity.
Select transformants on minimal medium + penicillin
to kill non-transformants.
Plate survivors on complete medium to test for his-,
met-.
Genetic Analysis Using
Transformation
Determining genetic distance with
transformation mapping
Problem 4, page 2-5
DNA is isolated from E. coli strain A (his- met- pens)
and used to transform strain B (his+ met+ pens).
Transformants are selected on minimal medium +
penicillin to kill his+ met+ cells and survivors are
plated on complete medium. The classes and
numbers of cells obtained are listed below.
Determine the recombination frequency between the
his and met genes.
Genetic Analysis Using
Transformation
Determining genetic distance with
transformation mapping
Problem 4, page 2-5
Rf = number of single transformants
total number of transformants
Genetic Analysis Using
Transformation
Determining genetic distance with
transformation mapping
his-
met-
his- met+
his+ methis- met-
35
27
194
Genetic Analysis Using
Transformation
Determining genetic distance with
transformation mapping
Single transformants, his- met+ and his+ met-,
represent crossovers between the genes.
his-
met-
his-
met-
Genetic Analysis Using
Transformation
Determining genetic distance with
transformation mapping
Problem 4, page 2-5
Rf = 35 + 27 = 62 = .24 = 24 map units
256
256
Gene Transfer Processes for
Bacteria and Their Viruses
3. Transduction
Transfer of
bacterial genes
with a
bacteriophage
Transduction
Genetic Analysis Using Transduction
Determining cotransduction frequency
with three-factor transduction.
Cotransduction frequency = tendency
for genes to be transferred together on
same piece of transducing DNA
Genetic Analysis Using Transduction
Three-factor transduction:
Transducing bacteriophage are used to
transfer DNA with three markers to bacterial
cells.
Bacteria are selected for one of the markers
and tested for the presence of the other two
markers.
Gene order and cotransduction frequency
can be determined.
Gene Transfer Processes for
Bacteria and Their Viruses
Three-factor transduction
Problem 6, Page 2-5
Transducing phages that infected an A+B+C+
cell are used to infect an A-B-C- cell.
Transductants receiving the A+ marker were
tested for the presence of B+ and C+. The
classes and numbers of transductants
observed is shown below. Determine the gene
order and the cotransduction frequencies for
A+ with B+ and A+ with C+.
Gene Transfer Processes for
Bacteria and Their Viruses
Three-factor transduction
Problem 6, Page 2-5
A+ B+ C+
45
A+ B+ C-
80
A+ B-
C+
1
A+ B- C-
300
Select for A+
Test for B+ and C+
Genetic Analysis Using Transduction
Problem 6, page 2-5
Smallest number of offspring represents 4
crossovers, identifies middle gene. Genotype
will be A+ B- C+.
A+
A-
B+
B-
C+
C-
Genetic Analysis Using
Transduction
Problem 6, page 2-5
Cotransduction of A and B
A+B+C+
45
A+B+C-
80
125/426 = .29
Cotransduction of A and C
A+B+C+
45
A+B-C+
1
46/426 = .11
Genetic Analysis Using
Transduction
Problem 6, page 2-5
Cotransduction of A and B = .29
Cotransduction of A and C = .11
The higher the cotransduction frequency, the
closer the genes are to each other.
Therefore A and B are closer than A and C.
Gene Transfer Processes for
Bacteria and Their Viruses
4. Infection with
bacteriophage
In a mixed infection,
recombination can be
detected between
bacteriophage carrying
different genes.
Gene Transfer Processes for
Bacteria and Their Viruses
Infection with bacteriophage
Infect bacteria with bacteriophage of two
different genotypes.
Recombination can occur between
bacteriophage genes.
Determine genotypes of resulting
bacteriophage.
Rf = number of recombinant plaques
total number of plaques
Gene Transfer Processes for
Bacteria and Their Viruses
Infection with bacteriophage
lawn of bacterial cells
Plaque for
one genotype
Plaque for
alternate
genotype
Genetic Analysis for Infection
With Bacteriophage
r ara
Parental Types
ra r a+
h+
h-
h+
+
X
h-
Recombinant Types
r ar a+
hh+
Gene Transfer Processes for
Bacteria and Their Viruses
Infection with bacteriophage
Problem 5, Page 2-5
Three different bacteriophage T2 strains
carrying mutations in the r gene (ra, rb
and rc) were each involved in a cross r+ X r+ h-, where x=a, b or c. The
h
x
x
numbers of bacteriophage of each type
are listed below. Give any one of four
possible linkage maps for these genes.
Gene Transfer Processes for
Bacteria and Their Viruses
4. Infection with bacteriophage
Problem 5, Page 2-5
r -x h +
r+x h-
r+x h+
r -x h -
r-a h+ x r+a h-
340
420
120
120
r-b h+ x r+b h-
320
560
60
60
r-c h+ x r+c h-
390
590
8
12
Gene Transfer Processes for
Bacteria and Their Viruses
Infection with bacteriophage
Rf = number of recombinant plaques
total number of plaques
Rf = 120 + 120 = 240 = .24 = 24 map units
1000
1000
Gene Transfer Processes for
Bacteria and Their Viruses
Infection with bacteriophage
One possible map:
rc
h
2
rb
10
ra
12