Transcript BIOL 112 – Principles of Zoology

Recombination in bacteria
I. Bacterial Review
II. Conjugation
IV. Bacteriophage genetics
A.
B.
C.
V.
Phage cycle
Plaque assay
Phage cross
Transduction
A.
B.
Generalized transduction
Cotransduction
I. Bacterial Review
A. Can be grown both in liquid and agar medium,
colony = visible cluster of cells
B. Antibiotic resistant mutants = able to grow in the
presence of an antibiotic
C. Nutritional mutants
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
Prototrophs = wildtype cells can synthesize nutrients from
simple molecules present in the growth media
Auxotrophs = can’t synthesize an essential nutrient and
can’t grow unless that nutrient is supplied in the medium
Minimal medium = contains only inorganic salts, energy
source and carbon atoms
Nonselective medium = all wild type cells form colonies
Selective medium = medium that allows growth of only one
type of cell
Selective plating:
Allows the desired mutant to
reproduce but not wild-type
genotypes
 antibiotic resistance (mutants able
to grow in presence of antibiotic)
Strr (mutants that are resistant to
streptomycin)
 minimal medium supplemented
with specific nutrient
• Met- auxotroph is able to grow if
minimal medium has methionine
CNA (Columbia Naladixic Acid) Agar
selective for Gram-positive bacteria
II. Conjugation
A. Lederberg & Tatum’s
experiment illustrated that
DNA was transferred from
one bacterium to another.
Strain #1: B-M-P+T+
Strain #2: B+ M+ P- TThese colonies are due to the
transfer of genetic material
between two strains by
conjugation.
Bacterial sex – Sex Pili required for a good time!!!
B. F plasmid (F factor)
Ability to transfer based on presence of Fertility
factor, now known as F plasmid
~12% the size of the bacterial chromosome
 sex pili genes
 surface exclusion protein genes, preventing F+
conjugating with F+
 transfer origin
Episome – F plasmid can remain as a free
plasmid or be integrated into the bacterial
chromosome
1). Properties of F plasmid:
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Can be replicated inside the cell
Cells with F plasmid (F+) have sex pili
F+ and F- cells can conjugate
Transfer of information is one-way from donor (F+) to
recipient (F-)
Strain A
Strain B
F+
x
F(donor)
(recipient)
 F+ cannot conjugate with other F+ cells
 F+ can become integrated into the host chromosome
(rare event) – F+ carries one or more insertion sequence
elements (IS)

F may leave genome, taking copies of some genes
(F’)
host
F+ chromosome
F factor
F-
FF+
F pili promote
cell to cell contact
2) F plasmid replication: F replicates through rolling
circle replication and it is transferred to a recipient cell
via temporary cytoplasmic bridge between two cells.
A copy remains in the donor cell.
F+
3. Intigration of F (Hfr)
On rare occasions, the F plasmid is
integrated into the circular chromosome,
and there is genetic recombination…
this can then be transferred to a
recipient cell and incorporated into the
recipient's genome.
Hfr = high frequency
of recombination.
Still only partial
transfer, not
100%
i.e. prior to mating, the recipient was lac- and pro-, however after a
short time period of mating, the recipient is now lac+, but still proafter a longer mating, the cell is lac+ and pro+ (lac is always
transferred first, pro later)
Chromosome transferred in a linear
manner…
C. Mapping the E. coli chromosome using
interrupted mating
A. Interrupted mating – used a blender to separate
bacterial cells that were in the act of conjugation,
without killing them
Hypothesis: the time it takes for genes to enter a recipient cell is
directly related to their order along the bacterial chromosome
Interrupted matings would lead to various lengths of the Hfr
chromosome being transferred to the recipient.
To determine gene order,
colonies were picked from
previous plates and
restreaked on plates that
had azide or
bacteriophage
La T1, or on
minimal plates
c
Ga
l
λ
Whether or not bacteriaFcould grow
depended upon their genotypes
i.e. a cell that is Azs can’t grow on
azide plates…a cell that is Azr can
T1 A2
T
s
Rate of transfer
T
L
A2
%
100 100 90
Time 8
8
9
T1
70
11
Lac Gal
40 25
18 25
λ
15
26
S
90
Gradient of transfer:
Frequency of inheritance corresponds to the order of transfer
Genes are mapped according to time of appearance of
recombinants
Circular, low resolution map is made by combining maps from
different Hfr donors
Conclusions:
 there was a point of origin (O),
because transfer occurs from
a fixed point on the donor
chromosome
 O first, F last
 determined the gene order
based on the gradient of
transfer
 chromosome was circular
 the F is integrated at different
points and different directions
Last
first
ORDER OF TRANSFER
STRAIN
1
G
E
B
D
N
A
2
P
Y
L
G
E
B
3
X
T
J
F
P
Y
4
B
E
G
L
Y
P
Random reshuffling of genes???
Pattern: L next to G, G next to E, L next to B
ORDER OF GENES ON ORIGINAL F+: XTJFPYLGEBDNA
MARKERS
Hfr 1
Hfr 2
Hfr 3
R
10
20
-
I
40
-
5
U
25
5
-
E
-
60
-
C
-
45
-
S
55
-
20
gene order: CRUISE
Using the E. coli map (that is
based on 100 minutes),
identify the location of the
origin of both Hfr 1 & Hfr 3.
The Hfr2 origin has already
been identified.
MAP ORDER USING Hfr 2
FIRST, then you can map
the Hfr 1 & 3 relative to
that.
1
R
U
C
3
I
E
S
If leu+ str-r recombinants are desired from the cross:
Hfr leu+ str-s x F- leu- str-r, on what kind of medium should
the matings pairs be plated?
Plate on minimal medium that lacks leucine (select for leu+)
but contains streptomycin (selects for Str-r)
D. The F’ state and Merozygotes
Formation of a
“Partial diploid”
The F “pops out” often
taking a piece of the
chromosome with
it…becoming an F’ [F
prime], in the process
creating a stable
“partial diploid” =
MEROZYGOTE
recombination occur in asexual prokaryotes via:
Conjugation, Transduction, Transformation
IV. Bacteriophage genetics
A. Phage Cycle:
1) Virus binds to host cell
2) Tail sheath causes core
penetration of cell wall
3) DNA in head is extruded
Once inside cell:
a) All processes halted
b) Degradation of host
DNA initiated
c) Phage DNA replicated,
transcribed & translated
using host machinery
d) Virus parts assembled
e) Host cell ruptures
B. Plaque assay & the Phage cross
Plaques are clear zones formed in a lawn of cells due to
lysis by phage. Different phages produce distinct plaque
morphologies
1)
Two parental strains:
h - r + X h+ r 
h+ only infects strain#1

h- infects both strains

r+ small plaque

r- large plaque
Double infection
Phage lysate analyzed, looked at different plaque types
RF = (h+r+) + (h-r-)
total plaques
Double Infections:
Plaque phenotypes
produced by progeny of
the cross h-r+ x h+rFour plaque phenotypes
can be differentiated:
2 parental types,
(h-r+ and h+r-)
and 2 recombinant types
(h+r+ and h-r-)
Thus, phage genomes
can be mapped by
analysis of RF
V. Transduction
A. Generalized Transduction: Small fraction of DNA from the
donor bacterial strain are carried by the phage
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Only a few genes are carried
Any host marker can be transduced
Phage infects the recipient, transferring the DNA fragment,
creating a merozygote
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Remains in cytoplasm (abortive transduction)
Transduced bacterial genes may be incorporated into the bacterial
chromosome (complete transduction)
B. cotransduction
If 2 genes are close together along the
chromosome, a bacteriophage may
package a single piece of the
chromosome that carries both genes and
transfer that piece to another bacterium
The likelihood of this occurring depends
upon how close together they are
Can map genes using cotransduction…
Mapping genes using cotransduction
 Select for the transduction of one gene and then
monitor whether or not another gene is
cotransduced along with it
Donor: arg+ met+ strs (infected w/P1 and lysate
mixed w/recipient)
Recipient: arg- met- strr
1. Plated on minimal media w/arg and strep but no
met,
2. Pick each of the colonies and re-streak on plates
without met and without arg.
3. Calculate cotransduction frequency:
# growing on media (no arg)
total # colonies