Transcript powerpoint

Chapter 18
• Molecular Genetics of Bacteria and Viruses
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
Viruses
• Discovered in plants (TMV, 1883)
• TMV crystalized in 1935
• Used as model systems
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
Figure 18.3 Infection by tobacco mosaic virus (TMV)
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
Figure 18.1 T4 bacteriophage infecting an E. coli cell
0.5 m
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
Characteristics
• Genetic material; DNA or RNA
• Control metab. Of host
• Reproduce
• Structure includes protein
• Life ???
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
Structure
• Nucleic acid
• Protein
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
1 gagttttatc gcttccatga cgcagaagtt aacactttcg gatatttctg atgagtcgaa 61 aaattatctt gataaagcag gaattactac tgcttgttta cgaattaaat cgaagtggac 121
tgctggcgga aaatgagaaa attcgaccta tccttgcgca gctcgagaag ctcttacttt 181 gcgacctttc gccatcaact aacgattctg tcaaaaactg acgcgttgga tgaggagaag 241
tggcttaata tgcttggcac gttcgtcaag gactggttta gatatgagtc acattttgtt 301 catggtagag attctcttgt tgacatttta aaagagcgtg gattactatc tgagtccgat 361
gctgttcaac cactaatagg taagaaatca tgagtcaagt tactgaacaa tccgtacgtt 421 tccagaccgc tttggcctct attaagctca ttcaggcttc tgccgttttg gatttaaccg 481
aagatgattt cgattttctg acgagtaaca aagtttggat tgctactgac cgctctcgtg 541 ctcgtcgctg cgttgaggct tgcgtttatg gtacgctgga ctttgtggga taccctcgct 601
ttcctgctcc tgttgagttt attgctgccg tcattgctta ttatgttcat cccgtcaaca 661 ttcaaacggc ctgtctcatc atggaaggcg ctgaatttac ggaaaacatt attaatggcg 721
tcgagcgtcc ggttaaagcc gctgaattgt tcgcgtttac cttgcgtgta cgcgcaggaa 781 acactgacgt tcttactgac gcagaagaaa acgtgcgtca aaaattacgt gcggaaggag
841 tgatgtaatg tctaaaggta aaaaacgttc tggcgctcgc cctggtcgtc cgcagccgtt 901 gcgaggtact aaaggcaagc gtaaaggcgc tcgtctttgg tatgtaggtg gtcaacaatt
961 ttaattgcag gggcttcggc cccttacttg aggataaatt atgtctaata ttcaaactgg 1021 cgccgagcgt atgccgcatg acctttccca tcttggcttc cttgctggtc agattggtcg
1081 tcttattacc atttcaacta ctccggttat cgctggcgac tccttcgaga tggacgccgt 1141 tggcgctctc cgtctttctc cattgcgtcg tggccttgct attgactcta ctgtagacat 1201
ttttactttt tatgtccctc atcgtcacgt ttatggtgaa cagtggatta agttcatgaa 1261 ggatggtgtt aatgccactc ctctcccgac tgttaacact actggttata ttgaccatgc 1321
cgcttttctt ggcacgatta accctgatac caataaaatc cctaagcatt tgtttcaggg 1381 ttatttgaat atctataaca actattttaa agcgccgtgg atgcctgacc gtaccgaggc 1441
taaccctaat gagcttaatc aagatgatgc tcgttatggt ttccgttgct gccatctcaa 1501 aaacatttgg actgctccgc ttcctcctga gactgagctt tctcgccaaa tgacgacttc 1561
taccacatct attgacatta tgggtctgca agctgcttat gctaatttgc atactgacca 1621 agaacgtgat tacttcatgc agcgttacca tgatgttatt tcttcatttg gaggtaaaac 1681
ctcttatgac gctgacaacc gtcctttact tgtcatgcgc tctaatctct gggcatctgg 1741 ctatgatgtt gatggaactg accaaacgtc gttaggccag ttttctggtc gtgttcaaca 1801
gacctataaa cattctgtgc cgcgtttctt tgttcctgag catggcacta tgtttactct 1861 tgcgcttgtt cgttttccgc ctactgcgac taaagagatt cagtacctta acgctaaagg 1921
tgctttgact tataccgata ttgctggcga ccctgttttg tatggcaact tgccgccgcg 1981 tgaaatttct atgaaggatg ttttccgttc tggtgattcg tctaagaagt ttaagattgc 2041
tgagggtcag tggtatcgtt atgcgccttc gtatgtttct cctgcttatc accttcttga 2101 aggcttccca ttcattcagg aaccgccttc tggtgatttg caagaacgcg tacttattcg 2161
ccaccatgat tatgaccagt gtttccagtc cgttcagttg ttgcagtgga atagtcaggt 2221 taaatttaat gtgaccgttt atcgcaatct gccgaccact cgcgattcaa tcatgacttc 2281
gtgataaaag attgagtgtg aggttataac gccgaagcgg taaaaatttt aatttttgcc 2341 gctgaggggt tgaccaagcg aagcgcggta ggttttctgc ttaggagttt aatcatgttt 2401
cagactttta tttctcgcca taattcaaac tttttttctg ataagctggt tctcacttct 2461 gttactccag cttcttcggc acctgtttta cagacaccta aagctacatc gtcaacgtta 2521
tattttgata gtttgacggt taatgctggt aatggtggtt ttcttcattg cattcagatg 2581 gatacatctg tcaacgccgc taatcaggtt gtttctgttg gtgctgatat tgcttttgat 2641
gccgacccta aattttttgc ctgtttggtt cgctttgagt cttcttcggt tccgactacc 2701 ctcccgactg cctatgatgt ttatcctttg aatggtcgcc atgatggtgg ttattatacc 2761
gtcaaggact gtgtgactat tgacgtcctt ccccgtacgc cgggcaataa cgtttatgtt 2821 ggtttcatgg tttggtctaa ctttaccgct actaaatgcc gcggattggt ttcgctgaat 2881
caggttatta aagagattat ttgtctccag ccacttaagt gaggtgattt atgtttggtg 2941 ctattgctgg cggtattgct tctgctcttg ctggtggcgc catgtctaaa ttgtttggag 3001
gcggtcaaaa agccgcctcc ggtggcattc aaggtgatgt gcttgctacc gataacaata 3061 ctgtaggcat gggtgatgct ggtattaaat ctgccattca aggctctaat gttcctaacc
3121 ctgatgaggc cgcccctagt tttgtttctg gtgctatggc taaagctggt aaaggacttc 3181 ttgaaggtac gttgcaggct ggcacttctg ccgtttctga taagttgctt gatttggttg
3241 gacttggtgg caagtctgcc gctgataaag gaaaggatac tcgtgattat cttgctgctg 3301 catttcctga gcttaatgct tgggagcgtg ctggtgctga tgcttcctct gctggtatgg
3361 ttgacgccgg atttgagaat caaaaagagc ttactaaaat gcaactggac aatcagaaag 3421 agattgccga gatgcaaaat gagactcaaa aagagattgc tggcattcag
tcggcgactt 3481 cacgccagaa tacgaaagac caggtatatg cacaaaatga gatgcttgct tatcaacaga 3541 aggagtctac tgctcgcgtt gcgtctatta tggaaaacac
caatctttcc aagcaacagc 3601 aggtttccga gattatgcgc caaatgctta ctcaagctca aacggctggt cagtatttta 3661 ccaatgacca aatcaaagaa atgactcgca
aggttagtgc tgaggttgac ttagttcatc 3721 agcaaacgca gaatcagcgg tatggctctt ctcatattgg cgctactgca aaggatattt 3781 ctaatgtcgt cactgatgct gcttctggtg
tggttgatat ttttcatggt attgataaag 3841 ctgttgccga tacttggaac aatttctgga aagacggtaa agctgatggt attggctcta 3901 atttgtctag gaaataaccg tcaggattga
caccctccca attgtatgtt ttcatgcctc 3961 caaatcttgg aggctttttt atggttcgtt cttattaccc ttctgaatgt cacgctgatt 4021 attttgactt tgagcgtatc gaggctctta
aacctgctat tgaggcttgt ggcatttcta 4081 ctctttctca atccccaatg cttggcttcc ataagcagat ggataaccgc atcaagctct 4141 tggaagagat tctgtctttt cgtatgcagg
gcgttgagtt cgataatggt gatatgtatg 4201 ttgacggcca taaggctgct tctgacgttc gtgatgagtt tgtatctgtt actgagaagt 4261 taatggatga attggcacaa tgctacaatg
tgctccccca acttgatatt aataacacta 4321 tagaccaccg ccccgaaggg gacgaaaaat ggtttttaga gaacgagaag acggttacgc 4381 agttttgccg caagctggct
gctgaacgcc ctcttaagga tattcgcgat gagtataatt 4441 accccaaaaa gaaaggtatt aaggatgagt gttcaagatt gctggaggcc tccactatga 4501 aatcgcgtag
aggctttgct attcagcgtt tgatgaatgc aatgcgacag gctcatgctg 4561 atggttggtt tatcgttttt gacactctca cgttggctga cgaccgatta gaggcgtttt 4621 atgataatcc
caatgctttg cgtgactatt ttcgtgatat tggtcgtatg gttcttgctg 4681 ccgagggtcg caaggctaat gattcacacg ccgactgcta tcagtatttt tgtgtgcctg 4741 agtatggtac
agctaatggc cgtcttcatt tccatgcggt gcactttatg cggacacttc 4801 ctacaggtag cgttgaccct aattttggtc gtcgggtacg caatcgccgc cagttaaata 4861 gcttgcaaaa
tacgtggcct tatggttaca gtatgcccat cgcagttcgc tacacgcagg 4921 acgctttttc acgttctggt tggttgtggc ctgttgatgc taaaggtgag ccgcttaaag 4981 ctaccagtta
tatggctgtt ggtttctatg tggctaaata cgttaacaaa aagtcagata 5041 tggaccttgc tgctaaaggt ctaggagcta aagaatggaa caactcacta aaaaccaagc 5101
tgtcgctact tcccaagaag ctgttcagaa tcagaatgag ccgcaacttc gggatgaaaa 5161 tgctcacaat gacaaatctg tccacggagt gcttaatcca acttaccaag ctgggttacg
5221 acgcgacgcc gttcaaccag atattgaagc agaacgcaaa aagagagatg agattgaggc 5281 tgggaaaagt tactgtagcc gacgttttgg cggcgcaacc tgtgacgaca
aatctgctca 5341 aatttatgcg cgcttcgata aaaatgattg gcgtatccaa cctgca //
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
Figure 18.4 Viral structure
Capsomere
of capsid
RNA
Capsomere
Membranous
envelope
DNA
Head
Capsid Tail
sheath
RNA
DNA
Tail
fiber
Glycoprotein
18  250 mm
20 nm
(a) Tobacco mosaic virus
Glycoprotein
70–90 nm (diameter)
80–200 nm (diameter)
50 nm
50 nm
(b) Adenoviruses
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
(c) Influenza viruses
80  225 nm
50 nm
(d) Bacteriophage T4
HIV virus
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
Figure 18.2 Comparing the size of a virus,
a bacterium, and an animal cell
Virus
Bacterium
Animal
cell
Animal cell nucleus
0.25 m
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
Viral Reproduction
• Obligate intracellular parasites
• Limited host range
• Recognize host by cell surface receptor sites
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
Viral genome replication
• DNA  DNA (using host DNA polymerase)
• DNA  RNA (using viral RNA replicase)
• RNA  viral DNA (using reverse transcriptase)
viral genomic RNA or mRNA  protein
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
Figure 18.5 A simplified viral reproductive cycle
Entry into cell and
uncoating of DNA
DNA
VIRUS
Capsid
Transcription
Replication
HOST CELL
Viral DNA
mRNA
Viral DNA
Capsid
proteins
Self-assembly of new
virus particles and their
exit from cell
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
Reproductive cycles
• Lytic Cycles
– Results in death (lysis) of host cell
– Virulent viruses  lytic
– Restriction enzymes – natural bacterial
enzymes  protection f/ viruses  coevolution
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
Figure 18.6 The lytic cycle of phage T4, a virulent phage
1 Attachment. The T4 phage uses
its tail fibers to bind to specific
receptor sites on the outer
surface of an E. coli cell.
5 Release. The phage directs production
of an enzyme that damages the bacterial
cell wall, allowing fluid to enter. The cell
swells and finally bursts, releasing 100
to 200 phage particles.
2 Entry of phage DNA
and degradation of host DNA.
The sheath of the tail contracts,
injecting the phage DNA into
the cell and leaving an empty
capsid outside. The cell’s
DNA is hydrolyzed.
Phage assembly
4 Assembly. Three separate sets of proteins
self-assemble to form phage heads, tails,
and tail fibers. The phage genome is
packaged inside the capsid as the head forms.
Head
Tails
Tail fibers
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
3 Synthesis of viral genomes
and proteins. The phage DNA
directs production of phage
proteins and copies of the phage
genome by host enzymes, using
components within the cell.
Repro. Cycles (cont)
• Lysogenic cycle
– Incorp. Into host genome
– Temperate virus – integrate into host genome,
latent until lytic cycle initiated
– Prophage – phage genome incorp. Into
specific site, later  lytic
– Lysogenic cell – host w/ prophage
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
Figure 18.7 The lytic and lysogenic cycles of phage
, a temperate phage
Phage
DNA
The phage attaches to a
host cell and injects its DNA.
Many cell divisions
produce a large
population of bacteria
infected with the
prophage.
Phage DNA
circularizes
Phage
Occasionally, a prophage
exits the bacterial chromosome,
initiating a lytic cycle.
Bacterial
chromosome
Lytic cycle
Lysogenic cycle
Certain factors
determine whether
The cell lyses, releasing phages.
Lytic cycle
is induced
or
New phage DNA and
proteins are synthesized
and assembled into phages.
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
Lysogenic cycle
is entered
Prophage
Phage DNA integrates into
the bacterial chromosome,
becoming a prophage.
The bacterium reproduces
normally, copying the prophage
and transmitting it to daughter cells.
Table 18.1 Classes of Animal Viruses
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
Figure 18.8 The reproductive cycle of an enveloped RNA virus
1 Glycoproteins on the viral envelope
bind to specific receptor molecules
(not shown) on the host cell,
promoting viral entry into the cell.
Capsid
RNA
Envelope (with
glycoproteins)
2 Capsid and viral genome
enter cell
HOST CELL
Viral genome (RNA)
Template
5 Complementary RNA
strands also function as mRNA,
which is translated into both
capsid proteins (in the cytosol)
and glycoproteins for the viral
envelope (in the ER).
3 The viral genome (red)
functions as a template for
synthesis of complementary
RNA strands (pink) by a viral
enzyme.
mRNA
Capsid
proteins
ER
Glycoproteins
Copy of
genome (RNA)
4 New copies of viral
genome RNA are made
using complementary RNA
strands as templates.
6 Vesicles transport
envelope glycoproteins to
the plasma membrane.
8 New virus
7 A capsid assembles
around each viral
genome molecule.
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
Retroviruses e.g. HIV
• Use reverse transcriptase to transcribe DNA f/
RNA
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
Figure 18.9 The structure of HIV, the retrovirus that causes AIDS
Glycoprotein
Viral envelope
Capsid
Reverse
transcriptase
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
RNA
(two identical
strands)
Figure 18.10 The reproductive cycle of HIV, a retrovirus
HIV
Membrane of
white blood cell
1 The virus fuses with the
cell’s plasma membrane.
The capsid proteins are
removed, releasing the
viral proteins and RNA.
2 Reverse transcriptase
catalyzes the synthesis of a
DNA strand complementary
to the viral RNA.
HOST CELL
Reverse
transcriptase
Viral RNA
RNA-DNA
hybrid
4 The double-stranded
DNA is incorporated
as a provirus into the cell’s
DNA.
0.25 µm
HIV entering a cell
3 Reverse transcriptase
catalyzes the synthesis of
a second DNA strand
complementary to the first.
DNA
NUCLEUS
Chromosomal
DNA
Provirus
5 Proviral genes are
transcribed into RNA
molecules, which serve as
genomes for the next viral
generation and as mRNAs for
translation into viral proteins.
RNA genome
for the next
viral generation
mRNA
6 The viral proteins include capsid
proteins and reverse transcriptase
(made in the cytosol) and envelope
glycoproteins (made in the ER).
New HIV leaving a cell
9 New viruses bud
off from the host cell.
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
8 Capsids are
assembled around
viral genomes and
reverse transcriptase
molecules.
7 Vesicles transport the
glycoproteins from the ER to
the cell’s plasma membrane.
Figure 18.11 SARS (severe acute respiratory
syndrome), a recently emerging viral disease
(a) Young ballet students in Hong Kong
wear face masks to protect themselves
from the virus causing SARS.
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
(b) The SARS-causing agent is a coronavirus
like this one (colorized TEM), so named for the
“corona” of glycoprotein spikes protruding from
the envelope.
Figure 18.12 Viral infection of plants
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
Viruses and cancer
• Tumor viruses transform cells  cancerous
stste
• e.g. hepatitus B virus   liver cancer
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
Prions
Patogenic proteins
e.g. Mad Cow disease
Kuru
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
Figure 18.13 Model for how prions propagate
Prion
Original
prion
Many prions
Normal
protein
New
prion
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
Viruses
• May have evolved f/ other mobile genetic
elements, e.g plasmids, transposons
• Non living? – nature’s most complex
molecules?
• Living? – simplest forms of life?
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
Bacterial genetics
• Single circular chromosome (in nucleoid)
• Many also contain plasmids (sm. Extrachromo.
Ring of DNA)
• Reproduce by binary fission  clones
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
Figure 18.14 Replication of a bacterial chromosome
Replication
fork
Origin of
replication
Termination
of replication
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
Figure 18.15 Can a bacterial cell acquire genes
from another bacterial cell?
EXPERIMENT Researchers had two mutant strains, one that could make arginine but not tryptophan
(arg+ trp–) and one that could make tryptophan but not arginine (arg– trp+). Each mutant strain and a
mixture of both strains were grown in a liquid medium containing all the required amino acids. Samples
from each liquid culture were spread on plates containing a solution of glucose and inorganic salts (minimal
medium), solidified with agar.
Mixture
Mutant
strain
arg+ trp–
Mutant
strain
arg trp+
RESULTS Only the samples from the mixed culture, contained cells that gave rise to colonies on
minimal medium, which lacks amino acids.
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
Mixture
Mutant
strain
arg+ trp–
Mutant
strain
arg– trp+
No
colonies
(control)
Colonies
grew
No
colonies
(control)
CONCLUSION
Because only cells that can make both arginine and tryptophan (arg+ trp+ cells)
can grow into colonies on minimal medium, the lack of colonies on the two control plates showed that
no further mutations had occurred restoring this ability to cells of the mutant strains. Thus, each cell
from the mixture that formed a colony on the minimal medium must have acquired one or more genes
from a cell of the other strain by genetic recombination.
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
Genetic Recombination
1. Transformation
– gene transfer, bacteria get DNA from
surroundings (e.g. another bacterial cell)
 antibiotic resistance
–
assimilated DNA may be integrated into
bacterial chromo.
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
2. Transduction – gene transfer f/ one bacterial
cell to another by a phage (virus)
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
Figure 18.16 Generalized transduction
Phage DNA
1
Phage infects bacterial cell that has alleles A+ and B+
2
Host DNA (brown) is fragmented, and phage DNA
and proteins are made. This is the donor cell.
A+ B+
A+ B+
Donor
cell
3
A bacterial DNA fragment (in this case a fragment with
the A+ allele) may be packaged in a phage capsid.
A+
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
Figure 18.16 Generalized transduction
Phage DNA
1
Phage infects bacterial cell that has alleles A+ and B+
2
Host DNA (brown) is fragmented, and phage DNA
and proteins are made. This is the donor cell.
A+ B+
A+ B+
Donor
cell
3
A bacterial DNA fragment (in this case a fragment with
the A+ allele) may be packaged in a phage capsid.
A+
4
Phage with the A+ allele from the donor cell infects
a recipient A–B– cell, and crossing over (recombination)
between donor DNA (brown) and recipient DNA
(green) occurs at two places (dotted lines).
Crossing
over
A+
A– B–
Recipient
cell
5
The genotype of the resulting recombinant cell (A+B–)
differs from the genotypes of both the donor (A+B+) and
the recipient (A–B–).
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
A+ B–
Recombinant cell
3. Conjugation – direct gene transfer
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
Figure 18.17 Bacterial conjugation
Sex pilus
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
1 m
Figure 18.18 Conjugation and recombination in E. coli (layer 1)
F Plasmid
Bacterial chromosome
F+ cell
F+ cell
Mating
bridge
1
F+ cell
Bacterial
chromosome
F– cell
A cell carrying an F plasmid
(an F+ cell) can form a
mating bridge with an F– cell
and transfer its F plasmid.
2
A single strand of the F
plasmid breaks at a
specific point (tip of blue
arrowhead) and begins to
move into the recipient cell.
As transfer continues, the
donor plasmid rotates
(red arrow).
3
4
DNA replication occurs in
both donor and recipient
cells, using the single
parental strands of the
F plasmid as templates
to synthesize complementary
strands.
A+
F factor
The circular F plasmid in an F + cell
can be integrated into the circular
chromosome by a single crossover
event (dotted line).
B+
C+
(a) Conjugation and transfer of an
F plasmid from an F+ donor to
an F– recipient
Hfr cell
F+ cell
1
The plasmid in the
recipient cell
circularizes. Transfer
and replication result
in a compete F plasmid
in each cell. Thus, both
cells are now F+.
D+
A+
C+
B+
D+
2
The resulting cell is called an Hfr cell
(for High frequency of recombination).
D+ C+
B+
A+
D+ C+
B+
B+
A+
B–
A+
A+
F– cell
3
B–
B+
C– –
D
A–
B–
4
Since an Hfr cell has all
the F-factor genes, it can
form a mating bridge with
an F– cell and transfer DNA.
A+
–
B– C D–
A–
A single strand of the F factor
5 The location and orientation
of the F factor in the donor
breaks and begins to move
chromosome determine
through the bridge. DNA
the sequence of gene transfer
replication occurs in both donor
during conjugation. In this
and recipient cells, resulting in
example, the transfer sequence
double-stranded DNA
for four genes is A-B-C-D.
Temporary
partial
diploid
7
C– –
D
A–
B+
A+
B–
C– –
D
A–
Two crossovers can result
in the exchange of similar
(homologous) genes between
the transferred chromosome fragment
(brown) and the recipient cell’s
chromosome (green).
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
B–
A+
B+
C– –
D
A–
6
C–
A–
D–
The mating bridge
usually breaks well
before the entire
chromosome and
the rest of the
F factor are transferred.
Recombinant F–
bacterium
8 The piece of DNA ending up outside the
bacterial chromosome will eventually be
degraded by the cell’s enzymes. The recipient
cell now contains a new combination of genes
but no F factor; it is a recombinant F – cell.
(b) Conjugation and transfer of part
of the bacterial chromosome from
an Hfr donor to an F– recipient,
resulting in recombination
Figure 18.18 Conjugation and recombination in E. coli (layer 2)
F Plasmid
Bacterial chromosome
F+ cell
F+ cell
Mating
bridge
1
F+ cell
Bacterial
chromosome
F– cell
A cell carrying an F plasmid
(an F+ cell) can form a
mating bridge with an F– cell
and transfer its F plasmid.
2
A single strand of the F
plasmid breaks at a
specific point (tip of blue
arrowhead) and begins to
move into the recipient cell.
As transfer continues, the
donor plasmid rotates
(red arrow).
3
4
DNA replication occurs in
both donor and recipient
cells, using the single
parental strands of the
F plasmid as templates
to synthesize complementary
strands.
A+
F factor
The circular F plasmid in an F + cell
can be integrated into the circular
chromosome by a single crossover
event (dotted line).
B+
C+
(a) Conjugation and transfer of an
F plasmid from an F+ donor to
an F– recipient
Hfr cell
F+ cell
1
The plasmid in the
recipient cell
circularizes. Transfer
and replication result
in a compete F plasmid
in each cell. Thus, both
cells are now F+.
D+
A+
C+
B+
D+
2
The resulting cell is called an Hfr cell
(for High frequency of recombination).
D+ C+
B+
A+
D+ C+
B+
B+
A+
B–
A+
A+
F– cell
3
B–
B+
C– –
D
A–
B–
4
Since an Hfr cell has all
the F-factor genes, it can
form a mating bridge with
an F– cell and transfer DNA.
A+
–
B– C D–
A–
A single strand of the F factor
5 The location and orientation
of the F factor in the donor
breaks and begins to move
chromosome determine
through the bridge. DNA
the sequence of gene transfer
replication occurs in both donor
during conjugation. In this
and recipient cells, resulting in
example, the transfer sequence
double-stranded DNA
for four genes is A-B-C-D.
Temporary
partial
diploid
7
C– –
D
A–
B+
A+
B–
C– –
D
A–
Two crossovers can result
in the exchange of similar
(homologous) genes between
the transferred chromosome fragment
(brown) and the recipient cell’s
chromosome (green).
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
B–
A+
B+
C– –
D
A–
6
C–
A–
D–
The mating bridge
usually breaks well
before the entire
chromosome and
the rest of the
F factor are transferred.
Recombinant F–
bacterium
8 The piece of DNA ending up outside the
bacterial chromosome will eventually be
degraded by the cell’s enzymes. The recipient
cell now contains a new combination of genes
but no F factor; it is a recombinant F – cell.
(b) Conjugation and transfer of part
of the bacterial chromosome from
an Hfr donor to an F– recipient,
resulting in recombination
Figure 18.18 Conjugation and recombination in E. coli (layer 3)
F Plasmid
Bacterial chromosome
F+ cell
F+ cell
Mating
bridge
1
F+ cell
Bacterial
chromosome
F– cell
A cell carrying an F plasmid
(an F+ cell) can form a
mating bridge with an F– cell
and transfer its F plasmid.
2
A single strand of the F
plasmid breaks at a
specific point (tip of blue
arrowhead) and begins to
move into the recipient cell.
As transfer continues, the
donor plasmid rotates
(red arrow).
3
4
DNA replication occurs in
both donor and recipient
cells, using the single
parental strands of the
F plasmid as templates
to synthesize complementary
strands.
Hfr cell
A+
F factor
The circular F plasmid in an F + cell
can be integrated into the circular
chromosome by a single crossover
event (dotted line).
B+
C+
(a) Conjugation and transfer of an
F plasmid from an F+ donor to
an F– recipient
Hfr cell
F+ cell
1
The plasmid in the
recipient cell
circularizes. Transfer
and replication result
in a compete F plasmid
in each cell. Thus, both
cells are now F+.
D+
A+
C+
B+
D+
2
The resulting cell is called an Hfr cell
(for High frequency of recombination).
D+ C+
B+
A+
D+ C+
B+
B+
A+
B–
A+
A+
F– cell
3
B–
B+
C– –
D
A–
B–
4
Since an Hfr cell has all
the F-factor genes, it can
form a mating bridge with
an F– cell and transfer DNA.
A+
–
B– C D–
A–
A single strand of the F factor
5 The location and orientation
of the F factor in the donor
breaks and begins to move
chromosome determine
through the bridge. DNA
the sequence of gene transfer
replication occurs in both donor
during conjugation. In this
and recipient cells, resulting in
example, the transfer sequence
double-stranded DNA
for four genes is A-B-C-D.
Temporary
partial
diploid
7
C– –
D
A–
B+
A+
B–
C– –
D
A–
Two crossovers can result
in the exchange of similar
(homologous) genes between
the transferred chromosome fragment
(brown) and the recipient cell’s
chromosome (green).
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
B–
A+
B+
C– –
D
A–
6
C–
A–
D–
The mating bridge
usually breaks well
before the entire
chromosome and
the rest of the
F factor are transferred.
Recombinant F–
bacterium
8 The piece of DNA ending up outside the
bacterial chromosome will eventually be
degraded by the cell’s enzymes. The recipient
cell now contains a new combination of genes
but no F factor; it is a recombinant F – cell.
(b) Conjugation and transfer of part
of the bacterial chromosome from
an Hfr donor to an F– recipient,
resulting in recombination
Figure 18.18 Conjugation and recombination in E. coli (layer 4)
F Plasmid
Bacterial chromosome
F+ cell
F+ cell
Mating
bridge
1
F+ cell
Bacterial
chromosome
F– cell
A cell carrying an F plasmid
(an F+ cell) can form a
mating bridge with an F– cell
and transfer its F plasmid.
2
A single strand of the F
plasmid breaks at a
specific point (tip of blue
arrowhead) and begins to
move into the recipient cell.
As transfer continues, the
donor plasmid rotates
(red arrow).
3
4
DNA replication occurs in
both donor and recipient
cells, using the single
parental strands of the
F plasmid as templates
to synthesize complementary
strands.
Hfr cell
A+
F factor
The circular F plasmid in an F + cell
can be integrated into the circular
chromosome by a single crossover
event (dotted line).
B+
C+
(a) Conjugation and transfer of an
F plasmid from an F+ donor to
an F– recipient
Hfr cell
F+ cell
1
The plasmid in the
recipient cell
circularizes. Transfer
and replication result
in a compete F plasmid
in each cell. Thus, both
cells are now F+.
D+
A+
C+
B+
D+
2
The resulting cell is called an Hfr cell
(for High frequency of recombination).
D+ C+
B+
A+
D+ C+
B+
B+
A+
B–
A+
A+
F– cell
3
B–
B+
C– –
D
A–
B–
4
Since an Hfr cell has all
the F-factor genes, it can
form a mating bridge with
an F– cell and transfer DNA.
A+
–
B– C D–
A–
A single strand of the F factor
5 The location and orientation
of the F factor in the donor
breaks and begins to move
chromosome determine
through the bridge. DNA
the sequence of gene transfer
replication occurs in both donor
during conjugation. In this
and recipient cells, resulting in
example, the transfer sequence
double-stranded DNA
for four genes is A-B-C-D.
Temporary
partial
diploid
7
C– –
D
A–
B+
A+
B–
C– –
D
A–
Two crossovers can result
in the exchange of similar
(homologous) genes between
the transferred chromosome fragment
(brown) and the recipient cell’s
chromosome (green).
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
B–
A+
B+
C– –
D
A–
6
C–
A–
D–
The mating bridge
usually breaks well
before the entire
chromosome and
the rest of the
F factor are transferred.
Recombinant F–
bacterium
8 The piece of DNA ending up outside the
bacterial chromosome will eventually be
degraded by the cell’s enzymes. The recipient
cell now contains a new combination of genes
but no F factor; it is a recombinant F – cell.
(b) Conjugation and transfer of part
of the bacterial chromosome from
an Hfr donor to an F– recipient,
resulting in recombination
4. Transposons – DNA sequences that can
move f/ one chromo. site to another
“jumping genes”
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
Figure 18.19 Transposable genetic elements in bacteria
Insertion sequence
3
A T C C G G T…
A C C G G A T…
3
5
TAG G C CA…
TG G C CTA…
5
Transposase gene
Inverted
Inverted
repeat
repeat
(a) Insertion sequences, the simplest transposable elements in bacteria, contain a single gene that
encodes transposase, which catalyzes movement within the genome. The inverted repeats are
backward, upside-down versions of each other; only a portion is shown. The inverted repeat
sequence varies from one type of insertion sequence to another.
Transposon
Insertion
sequence
Antibiotic
resistance gene
Insertion
sequence
5
5
3
3
Transposase gene
Inverted repeats
(b) Transposons contain one or more genes in addition to the transposase gene. In the transposon
shown here, a gene for resistance to an antibiotic is located between twin insertion sequences.
The gene for antibiotic resistance is carried along as part of the transposon when the transposon
is inserted at a new site in the genome.
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
Genetic recombination  genetic variation
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
Control of gene expression
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
Figure 18.20 Regulation of a metabolic pathway
(a) Regulation of enzyme
activity
Precursor
Feedback
inhibition
Enzyme 1
Enzyme 2
Enzyme 3
(b) Regulation of enzyme
production
Gene 1
Gene 2
Regulation
of gene
expression
Gene 3
–
Enzyme 4
Gene 4
–
Enzyme 5
Tryptophan
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
Gene 5
The operon model (1961)
• Gene regulation in prokaryotes
Jacques Monod, François Jacob et André Lwoff
reçoivent le prix Nobel
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
Figure 18.21 The trp operon: regulated synthesis of
repressible enzymes
trp operon
Promoter
DNA
Promoter
Genes of operon
trpD
trpC
trpE
trpR
trpB
trpA
Operator
Regulatory
gene
mRNA
5
3
RNA
polymerase
Start codon
Stop codon
mRNA 5
E
Protein
Inactive
repressor
D
C
B
A
Polypeptides that make up
enzymes for tryptophan synthesis
(a) Tryptophan absent, repressor inactive, operon on. RNA polymerase attaches to the DNA at the
promoter and transcribes the operon’s genes.
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
DNA
No RNA made
mRNA
Protein
Active
repressor
Tryptophan
(corepressor)
(b) Tryptophan present, repressor active, operon off. As tryptophan
accumulates, it inhibits its own production by activating the repressor protein.
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
Figure 18.22 The lac operon: regulated synthesis of
inducible enzymes
Promoter
Regulatory
gene
DNA
Operator
lacl
lacZ
No
RNA
made
3
mRNA
Protein
RNA
polymerase
5
Active
repressor
(a) Lactose absent, repressor active, operon off. The lac repressor is innately active, and in
the absence of lactose it switches off the operon by binding to the operator.
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
lac operon
DNA
lacl
lacz
3
mRNA
5
lacA
RNA
polymerase
mRNA 5'
5
mRNA
-Galactosidase
Protein
Allolactose
(inducer)
lacY
Permease
Transacetylase
Inactive
repressor
(b) Lactose present, repressor inactive, operon on. Allolactose, an isomer of lactose, derepresses
the operon by inactivating the repressor. In this way, the enzymes for lactose utilization are induced.
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
Figure 18.23 Positive control of the lac operon by
catabolite activator protein (CAP)
Promoter
DNA
lacl
lacZ
CAP-binding site
Active
CAP
cAMP
Inactive
CAP
RNA
polymerase
can bind
and transcribe
Operator
Inactive lac
repressor
(a) Lactose present, glucose scarce (cAMP level high): abundant lac mRNA synthesized.
If glucose is scarce, the high level of cAMP activates CAP, and the lac operon produces
large amounts of mRNA for the lactose pathway.
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
Promoter
DNA
lacl
lacZ
CAP-binding site
Operator
RNA
polymerase
can’t bind
Inactive
CAP
Inactive lac
repressor
(b) Lactose present, glucose present (cAMP level low): little lac mRNA synthesized.
When glucose is present, cAMP is scarce, and CAP is unable to stimulate transcription.
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings