Download Microbial Genetics

Chapter 8
Microbial Genetics
© 2013 Pearson Education, Inc.
Lectures prepared by Christine L. Case
Copyright © 2013 Pearson Education, Inc.
Lectures prepared by Christine L. Case
Insert Fig CO 8
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Structure and Function of Genetic
Material
Learning Objectives
8-1 Define genetics, genome, chromosome, gene,
genetic code, genotype, phenotype, and
genomics.
8-2 Describe how DNA serves as genetic information.
8-3 Describe the process of DNA replication.
8-4 Describe protein synthesis, including
transcription, RNA processing, and translation.
8-5 Compare protein synthesis in prokaryotes and
eukaryotes.
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Terminology
 Genetics: the study of what genes are, how they
carry information, how information is expressed, and
how genes are replicated
 Gene: a segment of DNA that encodes a functional
product, usually a protein
 Chromosome: structure containing DNA that
physically carries hereditary information; the
chromosomes contain the genes
 Genome: all the genetic information in a cell
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Terminology
 Genomics: the molecular study of genomes
 Genotype: the genes of an organism
 Phenotype: expression of the genes
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Determine Relatedness
Insert table from Clinical Focus,
p. 220, bottom
If possible on this slide, include a title:
Determine Relatedness
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Determine Relatedness
 Which strain is more
closely related to the
Uganda strain?
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Strain
% Similar to
Uganda
Kenya
71%
U.S.
51%
Figure 8.1a A prokaryotic chromosome.
Chromosome
Insert Fig 8.1a
Disrupted E. coli cell
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Figure 8.1b A prokaryotic chromosome.
Insert Fig 8.1b
KEY
Amino acid metabolism
DNA replication and repair
Lipid metabolism
Carbohydrate metabolism
Membrane synthesis
Genetic map of the chromosome of E. coli
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Figure 8.2 The Flow of Genetic Information.
Parent cell
DNA
expression
recombination
Genetic information is used within a cell to
produce the proteins needed for the cell to
function.
Transcription
Genetic information can be
Insert
Fig 8.2
transferred between cells
of the same generation.
replication
Genetic information can
be transferred between
generations of cells.
New combinations
of genes
Translation
Cell metabolizes and grows
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Recombinant cell
Daughter cells
DNA
 Polymer of nucleotides: adenine, thymine, cytosine,
and guanine
 Double helix associated with proteins
 “Backbone” is deoxyribose-phosphate
 Strands are held together by hydrogen bonds
between AT and CG
 Strands are antiparallel
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Figure 8.3b DNA replication.
5′ end
3′ end
Insert Fig 8.3b
3′ end
5′ end
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The two strands of DNA are antiparallel. The sugar-phosphate
backbone of one strand is upside down relative to the backbone of
the other strand. Turn the book upside down to demonstrate this.
Figure 8.3a DNA replication.
Parental
strand
5′ end
3′ end
Parental
strand
1
1
2
2
1
The double helix of the
parental DNA separates as
weak hydrogen bonds
between the nucleotides on
opposite strands break in
response to the action of
replication enzymes.
Replication
fork
Insert Fig 8.3a
Hydrogen bonds form
between new complementary
nucleotides and each strand
of the parental template to
form new base pairs.
2
3
3
Enzymes catalyze the formation
of sugar-phosphate bonds
Daughter
strand
between sequential
nucleotides on each resulting
daughter strand.
The replication fork
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5′ end
Parental
strand
3′ end
Parental
strand
Daughter
strand
forming
New
Strand
Template
Strand
Sugar
Phosphate
1
Insert Fig 8.4
2
Hydrolysis of the phosphate
bonds provides the energy
for the reaction.
When a nucleoside
triphosphate bonds to
the sugar, it loses two
phosphates.
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DNA Synthesis
 DNA is copied by DNA polymerase






In the 5' → 3' direction
Initiated by an RNA primer
Leading strand is synthesized continuously
Lagging strand is synthesized discontinuously
Okazaki fragments
RNA primers are removed and Okazaki fragments joined
by a DNA polymerase and DNA ligase
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Table 8.1 Important Enzymes in DNA Replication, Expression, and Repair
Insert Table 8.1
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Figure 8.5 A summary of events at the DNA replication fork.
REPLICATION
Proteins stabilize the
unwound parental DNA.
The leading strand is
synthesized continuously
by DNA polymerase.
3'
DNA polymerase
5'
Enzymes
unwind the
1
parental double
helix.
Replication
fork
RNA primer
Insert Fig 8.5
Primase
5'
DNA polymerase
3'
Parental
strand
Okazaki fragment
DNA
polymerase
The lagging strand is
synthesized discontinuously.
Primase, an RNA polymerase,
synthesizes a short RNA
primer, which is then extended by
DNA polymerase.
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DNA polymerase
digests RNA primer
and replaces it with DNA.
DNA ligase
3'
5'
DNA ligase joins
the discontinuous
fragments of the
lagging strand.
Figure 8.6 Replication of bacterial DNA.
Replication forks
REPLICATION
An E. coli 1chromosome in the process of replicating
Origin of replication
Parental strand
Replication fork
Daughter strand
Replication fork
Termination of
replication
Bidirectional replication of a circular bacterial DNA molecule
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Replication of Bacterial DNA
ANIMATION DNA Replication: Overview
ANIMATION DNA Replication: Forming the Replication Fork
ANIMATION DNA Replication: Replication Proteins
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Check Your Understanding
 Give a clinical application of genomics. 8-1
 Why is the base pairing in DNA important? 8-2
 Describe DNA replication, including the functions of
DNA gyrase, DNA ligase, and DNA polymerase. 8-3
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Transcription
 DNA is transcribed to make RNA (mRNA, tRNA,
and rRNA)
 Transcription begins when RNA polymerase binds to
the promoter sequence
 Transcription proceeds in the 5' → 3' direction
 Transcription stops when it reaches the
terminator sequence
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Figure 8.7 The process of transcription.
RNA polymerase
DNA
TRANSCRIPTION
DNA
mRNA
1
Protein
RNA polymerase bound to DNA
Promoter (gene begins)
Template
strand of DNA
RNA polymerase
1
2
3
4
RNA polymerase binds to the
promoter, and DNA unwinds at
the beginning of a gene.
RNA is synthesized by complementary
base pairing of free nucleotides with
the nucleotide bases on the template
strand of DNA.
The site of synthesis moves
along DNA; DNA that has been
transcribed rewinds.
Insert Fig 8.7
RNA
RNA
nucleotides
RNA synthesis
Transcription reaches the
terminator.
5 RNA and RNA polymerase are
released, and the DNA helix re-forms.
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Terminator
(gene ends)
The Process of Transcription
ANIMATION Transcription: Overview
ANIMATION Transcription: The Process
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Figure 8.11 RNA processing in eukaryotic cells.
Exon
Intron
Exon RNA polymerase
Intron
Exon
DNA
Met
Met
1
In the nucleus, a gene
composed of exons and introns
Met
DNA
is transcribed to RNA by RNA polymerase.
2
Processing involves snRNPs in the nucleus to remove
the intron-derived RNA and splice together the exonderived RNA into mRNA.
RNA
transcript
1
Insert Fig 8.11
mRNA
Met
3
After further modification, the mature mRNA
Met
travels to the cytoplasm, where it
directs protein synthesis.
Nucleus
Cytoplasm
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Translation
 mRNA is translated in codons (three nucleotides)
 Translation of mRNA begins at the start codon:
AUG
 Translation ends at nonsense codons: UAA, UAG,
UGA
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Figure 8.2 The Flow of Genetic Information.
Parent cell
DNA
expression
recombination
Genetic information is used within a cell to
produce the proteins needed for the cell to
function.
Transcription
Genetic information can be
Insert
Fig 8.2
transferred between cells
of the same generation.
replication
Genetic information can
be transferred between
generations of cells.
New combinations
of genes
Translation
Cell metabolizes and grows
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Recombinant cell
Daughter cells
The Genetic Code
 64 sense codons on mRNA encode the 20 amino
acids
 The genetic code is degenerate
 tRNA carries the complementary anticodon
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Figure 8.8 The genetic code.
Insert Fig 8.8
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Figure 8.10 Simultaneous transcription and translation in bacteria.
TRANSLATION
RNA polymerase
DNA
Met
Met
Met
DNA
mRNA
Met
Protein
RNA
Ribosome
Peptide
1
Insert Fig 8.10
Met
Met
Met
Met
Direction of transcription
RNA
polymerase
DNA
5′
Peptide
Polyribosome
Ribosome
mRNA
Direction of translation
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Figure 8.9.1-2 The process of translation.
RNA polymerase
TRANSLATION
Ribosome
Met
DNA
mRNA
Ribosomal
subunit
Protein
P Site
Met
DNA
Leu
tRNA
Anticodon
1
Ribosomal
subunit
1
Start
Second
codon codon
mRNA
Components needed to begin
Insert Fig
translation come together.
2
mRNA
On the assembled ribosome, a tRNA carrying the first
8.9 (1)
and
(2)
amino
acid
is paired with the start codon on the mRNA.
The place where this first tRNA sits is called the P site.
A tRNA carrying the second amino acid approaches.
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Figure 8.9.3-4 The process of translation.
Peptide bond forms
Met
Met
Leu
Met
Met
DNA
A site
Leu
Phe
Gly
E site
1
mRNA
3 The second codon of the mRNA pairs with a tRNA carrying
the second amino acid at the A site. The first amino acid
joins to the second by a peptide bond. This attaches the
polypeptide to the tRNA in the P site.
Ribosome moves
along mRNA
4 The ribosome moves along the mRNA until the second tRNA is
in the P site. The next codon to be translated is brought into the
A site. The first tRNA now occupies the E site.
Insert Fig 8.9 (3) and (4)
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mRNA
Figure 8.9.5-6 The process of translation.
tRNA released
Met
Leu
Met
Gly
Leu
Growing
polypeptide chain
Phe
Gly
Met
Phe
mRNA
Insert Fig 8.9 (5) and (6)
5 The second amino acid joins to the third by another peptide
bond, and the first tRNA is released from the E site.
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mRNA
6 The ribosome continues to move along the mRNA,
and new amino acids are added to the polypeptide.
Figure 8.9.7-8 The process of translation.
Arg
Polypeptide
released
Gly Met
Met
Gly
Met
Met
Phe
Phe
Gly
Arg
mRNA
New protein
Insert Fig 8.9 (7) and (8)
mRNA
Stop codon
7 When the ribosome reaches a stop codon,
the polypeptide is released.
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8 Finally, the last tRNA is released, and the ribosome comes
apart. The released polypeptide forms a new protein.
The Genetic Code
ANIMATION Translation: Overview
ANIMATION Translation: The Genetic Code
ANIMATION Translation: The Process
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Check Your Understanding
 What is the role of the promoter, terminator, and
mRNA in transcription? 8-4
 How does mRNA production in eukaryotes differ
from the process in prokaryotes? 8-5
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The Regulation of Gene Expression
Learning Objectives
8-6 Define operon.
8-7 Explain pre-transcriptional regulation of gene
expression in bacteria.
8-8 Explain post-transcriptional regulation of gene
expression.
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Regulation
 Constitutive genes are expressed at a fixed rate
 Other genes are expressed only as needed
 Inducible genes
 Repressible genes
 Catabolite repression
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Operon
ANIMATION Operons: Overview
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Figure 8.12.1 An inducible operon.
Operon
Control region
I
P
Structural genes
Z
O
Y
A
DNA
Regulatory gene
1
Promoter
Operator
Structure of the operon. The operon consists of the promoter (P) and operator (O) sites and structural
genes that code for the protein. The operon is regulated by the product of the regulatory gene (I ).
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Figure 8.12.2 An inducible operon.
RNA polymerase
Transcription
Repressor
mRNA
Active
repressor
protein
Translation
Repressor active, operon off. The repressor protein binds with the
operator, preventing transcription from the operon.
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Figure 8.12.3 An inducible operon.
Transcription
Operon
mRNA
Translation
Allolactose
(inducer)
Inactive
repressor
protein
Permease
β-Galactosidase
Transacetylase
Repressor inactive, operon on. When the inducer allolactose binds to the repressor protein,
the inactivated repressor can no longer block transcription. The structural genes are transcribed,
ultimately resulting in the production of the enzymes needed for lactose catabolism.
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Figure 8.13.1 A repressible operon.
Operon
Control region
Structural genes
DNA
Regulatory
gene
Promoter
Operator
Structure of the operon. The operon consists of the promoter
(P) and operator (O) sites and structural genes that code for the
protein. The operon is regulated by the product of the regulatory
gene (I ).
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Figure 8.13.2 A repressible operon.
RNA polymerase
Transcription
Repressor
mRNA
Operon
mRNA
Translation
Inactive
repressor
protein
Polypeptides
comprising the
enzymes for
tryptophan
synthesis
Repressor inactive, operon on. The repressor is inactive,
and transcription and translation proceed, leading to the
synthesis of tryptophan.
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Figure 8.13.3 A repressible operon.
Active
repressor
protein
Tryptophan
(corepressor)
Repressor active, operon off. When the corepressor
tryptophan binds to the repressor protein, the activated
repressor binds with the operator, preventing transcription
from the operon.
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Repression
ANIMATION Operons: Induction
ANIMATION Operons: Repression
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Figure 8.14 The growth rate of E. coli on glucose and lactose.
Log10 of number of cells
Bacteria growing on glucose
as the sole carbon source
grow faster than on lactose.
Glucose
Lactose
Log10 of number of cells
Time
All glucose
consumed
Lactose used
Glucose
used
Time
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Lag
time
Bacteria growing in a medium
containing glucose and lactose
first consume the glucose and
then, after a short lag time, the
lactose. During the lag time,
intracellular cAMP increases,
the lac operon is transcribed,
more lactose is transported
into the cell, and
β-galactosidase is synthesized
to break down lactose.
Figure 8.15 Positive regulation of the lac operon.
Promoter
lacI
lacZ
DNA
CAP-binding site
Operator
RNA
polymerase
can bind
and transcribe
cAMP
Active
CAP
Inactive lac
repressor
Inactive
CAP
Lactose present, glucose scarce (cAMP level high). If glucose is scarce,
the high level of cAMP activates CAP, and the lac operon produces large amounts
of mRNA for lactose digestion.
Promoter
lacI
lacZ
DNA
CAP-binding site
Operator
RNA
polymerase
can't bind
Inactive lac
Inactive
repressor
CAP
Lactose present, glucose present (cAMP level low). When glucose is
present, cAMP is scarce, and CAP is unable to stimulate transcription.
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Epigenetic Control




Methylating nucleotides
Methylated (off) genes are passed to offspring cells
Not permanent
Biofilm behavior
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Applications of Microbiology 3.1 Paenibacillus.
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Applications of Microbiology 3.3 A fruiting body of a myxobacterium.
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Figure 8.16 MicroRNAs control a wide range of activities in cells.
DNA
Transcription of
miRNA occurs.
miRNA
miRNA binds to target
mRNA that has at least six
complementary bases.
mRNA
mRNA is degraded.
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Check Your Understanding
 Use the following metabolic pathway to answer the
questions that follow it. 8-6
enzyme a
Substrate A
enzyme b
Intermediate B
End-product C
a. If enzyme a is inducible and is not being synthesized at
present, a (1) __________ protein must be bound tightly to the
(2) __________ site. When the inducer is present, it will bind to
the (3) __________ so that (4) __________ can occur.
b. If enzyme a is repressible, end-product C, called a
(1) __________, causes the (2) __________ to bind to
the (3) __________. What causes derepression?
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Check Your Understanding
 What is the role of cAMP in catabolite repression?
8-7
 How does miRNA stop protein synthesis? 8-8
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Mutation: Change in the Genetic
Material
Learning Objectives
8-9 Classify mutations by type.
8-10 Describe two ways mutations can be repaired.
8-11 Describe the effect of mutagens on the mutation
rate.
8-12 Outline the methods of direct and indirect
selection of mutants.
8-13 Identify the purpose of and outline the procedure
for the Ames test.
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Mutation




A change in the genetic material
Mutations may be neutral, beneficial, or harmful
Mutagen: agent that causes mutations
Spontaneous mutations: occur in the absence of
a mutagen
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Mutation
 Base substitution (point mutation)
 Change in one base
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Figure 8.17 Base substitutions.
Parental DNA
Replication
During DNA replication,
a thymine is incorporated
opposite guanine by
mistake.
Daughter DNA
Daughter DNA
Daughter DNA
In the next round of replication,
adenine pairs with the new
thymine, yielding an AT pair
in place of the original
GC pair.
Replication
Insert Fig 8.17
Granddaughter DNA
mRNA
Amino acids Cysteine
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Transcription
When mRNA is transcribed
from the DNA containing
this substitution, a codon is
produced that, during
translation, encodes a
different amino acid:
tyrosine instead of
cysteine.
Tyrosine
Translation
Cysteine
Cysteine
Mutation
 Missense mutation
 Base substitution results in change in amino acid
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Figure 8.18a-b Types of mutations and their effects on the amino acid sequences of proteins.
DNA (template strand)
Transcription
mRNA
Translation
Amino acid sequence
Met
Lys
Phe
Gly
Stop
Normal DNA molecule
DNA (template strand)
mRNA
Amino acid sequence
Met
Missense mutation
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Lys
Phe
Ser
Stop
Mutation
 Nonsense mutation
 Base substitution results in a nonsense codon
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Figure 8.18ac Types of mutations and their effects on the amino acid sequences of proteins.
DNA (template strand)
Transcription
mRNA
Translation
Amino acid sequence
Met
Lys
Phe
Normal DNA molecule
Met
Stop
Nonsense mutation
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Gly
Stop
Mutation
 Frameshift mutation
 Insertion or deletion of one or more nucleotide pairs
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Figure 8.18ad Types of mutations and their effects on the amino acid sequences of proteins.
DNA (template strand)
Transcription
mRNA
Translation
Amino acid sequence
Met
Lys
Phe
Normal DNA molecule
Met
Lys
Leu
Ala
Frameshift mutation
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Gly
Stop
The Frequency of Mutation
 Spontaneous mutation rate = 1 in 109 replicated
base pairs or 1 in 106 replicated genes
 Mutagens increase to 10–5 or 10–3 per replicated
gene
ANIMATION Mutations: Types
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Figure 8.19a Oxidation of nucleotides makes a mutagen.
Stop
Met
Stop
Met
Insert Fig 8.19a
Adenosine nucleoside normally base-pairs by
hydrogen bonds with an oxygen and a hydrogen
of a thymine or uracil nucleotide.
Met
Stop
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Altered adenine will hydrogen bond with a
hydrogen and a nitrogen of a cytosine nucleotide.
Stop
Stop
Figure 8.19b Oxidation of nucleotides makes a mutagen.
Normal daughter DNA
Mutated granddaughter DNA
Stop
Met
HNO
Met 2
Normal parent DNA
Replication
Stop
Insert Fig 8.19b
Altered parent DNA
The altered adenine pairs with cytosine
instead of thymine.
Stop
Stop
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Replication
Altered daughter DNA
Altered granddaughter DNA
Stop
Chemical Mutagens
ANIMATION Mutagens
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Figure 8.20 Nucleoside analogs and the nitrogenous bases they replace.
Normal nitrogenous base
Analog
Stop
Met
Adenine nucleoside
2-Aminopurine nucleoside
Met
Stop
The 2-aminopurine is
incorporated into DNA in place of adenine but can pair with
Fig 8.20
cytosine, so an AT pair becomes a Insert
CG pair.
Stop
Stop
Stop
Thymine nucleoside
5-Bromouracil nucleoside
The 5-bromouracil is used as an anticancer drug because it is mistaken for
thymine by cellular enzymes but pairs with cytosine. In the next DNA replication, an
AT pair becomes a GC pair.
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Radiation
 Ionizing radiation (X rays and gamma rays) causes
the formation of ions that can react with nucleotides
and the deoxyribose-phosphate backbone
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Radiation
 UV radiation causes thymine dimers
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Figure 8.21 The creation and repair of a thymine dimer caused by ultraviolet light.
Ultraviolet light
Exposure to ultraviolet light causes adjacent thymines to
become cross-linked, forming a thymine dimer and disrupting
their normal base pairing.
Thymine dimer
An endonuclease cuts the DNA, and an exonuclease removes
the damaged DNA.
New DNA
DNA polymerase fills the gap by synthesizing new DNA, using
the intact strand as a template.
DNA ligase seals the remaining gap by joining the old and
new DNA.
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Repair
 Photolyases separate thymine dimers
 Nucleotide excision repair
ANIMATION Mutations: Repair
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Selection
 Positive (direct) selection detects mutant cells
because they grow or appear different
 Negative (indirect) selection detects mutant cells
because they do not grow
 Replica plating
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Figure 8.22 Replica plating.
Handle
Sterile velvet is
pressed on the
grown colonies
on the master
plate.
Velvet
surface
(sterilized)
Master plate with
medium
containing
histidine
Cells from each colony
are transferred from the
velvet to new plates.
Petri plate with
medium containing histidine
Auxotrophic mutant
Plates are incubated.
Growth on plates is compared.
A colony that grows on the
medium with histidine but could
not grow on the medium without
histidine is auxotrophic
(histidine-requiring mutant).
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Petri plate with
medium lacking
histidine
Colony missing
Figure 8.23 The Ames reverse gene mutation test.
Experimental
sample
Suspected
mutagen
Rat liver
extract
Experimental plate
Incubation
Cultures of
histidine-dependent
Salmonella
Media lacking histidine
Control
(no suspected
mutagen added)
Rat liver
extract
Colonies of
revertant
bacteria
Incubation
Control plate
Two cultures are
prepared of
Salmonella bacteria
that have lost the
ability to synthesize
histidine
(histidinedependent).
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The suspected mutagen
is added to the
experimental sample
only; rat liver extract
(an activator) is added
to both samples.
Each sample is poured onto a
plate of medium lacking
histidine. The plates are then
incubated at 37°C for two days.
Only bacteria whose histidinedependent phenotype has
mutated back (reverted) to
histidine synthesizing will grow
into colonies.
The numbers of colonies on the experimental
and control plates are compared. The control
plate may show a few spontaneous histidinesynthesizing revertants. The test plates will
show an increase in the number of histidine
synthesizing revertants if the test chemical is
indeed a mutagen and potential carcinogen.
The higher the concentration of mutagen
used, the more revertant colonies will result.
Check Your Understanding




How can a mutation be beneficial? 8-9
How can mutations be repaired? 8-10
How do mutagens affect the mutation rate? 8-11
How would you isolate an antibiotic-resistant
bacterium? An antibiotic-sensitive bacterium? 8-12
 What is the principle behind the Ames test? 8-13
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Genetic Transfer and Recombination
Learning Objectives
8-14 Differentiate horizontal and vertical gene
transfer.
8-15 Compare the mechanisms of genetic
recombination in bacteria.
8-16 Describe the functions of plasmids and
transposons.
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Genetic Recombination
 Vertical gene transfer: occurs during reproduction
between generations of cells
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Figure 8.2 The Flow of Genetic Information.
Parent cell
DNA
expression
recombination
Genetic information is used within a cell to
produce the proteins needed for the cell to
function.
Transcription
Genetic information can be
Insert
Fig 8.2
transferred between cells
of the same generation.
replication
Genetic information can
be transferred between
generations of cells.
New combinations
of genes
Translation
Cell metabolizes and grows
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Recombinant cell
Daughter cells
Genetic Recombination
 Horizontal gene transfer: the transfer of genes
between cells of the same generation
ANIMATION Horizontal Gene Transfer: Overview
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Genetic Recombination
 Exchange of genes between two DNA molecules
 Crossing over occurs when two chromosomes break
and rejoin
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Figure 8.24 Genetic recombination by crossing over.
DNA from one cell aligns with
DNA in the recipient cell.
Notice that there is a nick in
the donor DNA.
Donor DNA
Recipient
chromosome
DNA from the donor aligns with complementary
base pairs in the recipient’s chromosome. This
can involve thousands of base pairs.
Insert Fig 8.24
RecA protein catalyzes the
joining of the two strands.
RecA protein
The result is that the recipient’s chromosome
contains new DNA. Complementary base pairs
between the two strands will be resolved by DNA
polymerase and ligase. The donor DNA will be
destroyed. The recipient may now have one or
more new genes.
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Genetic Transformation
ANIMATION Transformation
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Figure 8.25 Griffith’s experiment demonstrating genetic transformation.
RECOMBINATION
Living encapsulated
bacteria injected into
mouse.
Living nonencapsulated
bacteria injected into mouse.
Heat-killed encapsulated
bacteria injected into
mouse.
Living nonencapsulated and
heat-killed encapsulated bacteria
injected into mouse.
Mouse died.
Mouse remained healthy.
Mouse remained healthy.
Mouse died.
No colonies were isolated
from mouse.
Colonies of encapsulated
bacteria were isolated
from dead mouse.
Colonies of encapsulated
bacteria were isolated
from dead mouse.
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A few colonies of nonencapsulated bacteria were
isolated from mouse;
phagocytes destroyed
nonencapsulated bacteria.
Figure 8.26 The mechanism of genetic transformation in bacteria.
Recipient cell
Chromosomal DNA
DNA fragments
from donor cells
Recipient cell takes
up donor DNA.
Donor DNA aligns
with complementary
bases.
Recombination occurs
between donor DNA
and recipient DNA.
Degraded
unrecombined
DNA
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Genetically transformed cell
Figure 8.27 Bacterial conjugation.
Mating
bridge
Sex pilus
F– cell
Insert Fig 8.27
F+ cell
Sex pilus
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Mating bridge
Figure 8.28a Conjugation in E. coli.
RECOMBINATION
Bacterial
chromosome
Mating bridge
Replication
and transfer
of F factor
Insert Fig 8.28a
F factor
F+ cell
F– cell
F+ cell
F+ cell
When an F factor (a plasmid) is transferred from a donor (F+)
to a recipient (F–), the F– cell is converted to an F+ cell.
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Figure 8.28b Conjugation in E. coli.
Recombination between
F factor and chromosome,
occurring at a specific site
on each
Insertion of F factor
into chromosome
Integrated F factor
Insert Fig 8.28b
F+ cell
Hfr cell
When an F factor becomes integrated into the chromosome of an
F+ cell, it makes the cell a high frequency of recombination (Hfr) cell.
© 2013 Pearson Education, Inc.
Figure 8.28c Conjugation in E. coli.
Replication
and transfer
of part of the
chromosome
In the recipient,
recombination
between the
Hfr chromosome
fragment and the
F– chromosome
Insert Fig 8.28c
Hfr cell
F– cell
Hfr cell
Recombinant
F– cell
When an Hfr donor passes a portion of its chromosome into
an F– recipient, a recombinant F– cell results.
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Conjugation in E. coli
ANIMATION Conjugation: F Factor
ANIMATION Conjugation: Overview
ANIMATION Conjugation: Hfr Conjugation
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Genetic Map of the Chromosome of
E. coli
ANIMATION Conjugation: Chromosome Mapping
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Figure 8.1b A prokaryotic chromosome.
Insert Fig 8.1b
KEY
Amino acid metabolism
DNA replication and repair
Lipid metabolism
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Carbohydrate metabolism
Membrane synthesis
Transduction by a Bacteriophage
ANIMATION Transduction: Generalized Transduction
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Figure 8.29 Transduction by a bacteriophage.
RECOMBINATION
Phage protein coat
Phage DNA
Bacterial
chromosome
A phage infects the donor bacterial cell.
Donor
cell
Phage DNA and proteins are made, and the
bacterial chromosome is broken into pieces.
Insert
Phage
DNA
Bacterial
DNA
Occasionally during phage assembly, pieces of
bacterial DNA are packaged in a phage capsid.
Then
the donor cell lyses and releases phage
Fig
8.29
particles containing bacterial DNA.
A phage carrying bacterial DNA infects
a new host cell, the recipient cell.
Recipient
cell
Donor
bacterial
DNA
Recipient
bacterial
DNA
5
Recombinant
cell
reproduces
normally
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Many cell
divisions
Recombination can occur, producing a
recombinant cell with a genotype
different from both the donor and
recipient cells.
Transduction by a Bacteriophage
ANIMATION Transduction: Specialized Transduction
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Figure 13.13 Specialized transduction.
Prophage
gal gene
Bacterial
DNA
Galactose-positive
donor cell
gal gene
Insert Fig 13.13
Galactosenegative
recipient cell
Galactose-positive
recombinant cell
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1 Prophage exists in
galactose-using host
(containing the gal gene).
2 Phage genome excises, carrying
with it the adjacent gal gene from the
host.
3 Phage matures and cell lyses,
releasing phage carrying gal
gene.
4 Phage infects a cell that cannot
utilize galactose (lacking gal
gene).
5 Along with the prophage, the
bacterial gal gene becomes
integrated into the new host's
DNA.
6 Lysogenic cell can now metabolize
galactose.
Plasmids
 Conjugative plasmid: carries genes for sex pili and
transfer of the plasmid
 Dissimilation plasmids: encode enzymes for
catabolism of unusual compounds
 R factors: encode antibiotic resistance
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Figure 8.30 R factor, a type of plasmid.
Origin of replication
mer
sul
str
Pilus and
conjugation
proteins
cml
Insert Fig 8.30
Origin of transfer
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tet
Transposons
 Segments of DNA that can move from one region of
DNA to another
 Contain insertion sequences for cutting and resealing
DNA (transposase)
 Complex transposons carry other genes
ANIMATION Transposons: Overview
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Figure 8.31a-b Transposons and insertion.
IS1
Transposase gene
Inverted repeat
Inverted repeat
(a) An insertion sequence (IS), the simplest transposon, contains a gene
for transposase, the enzyme that catalyzes transposition. The tranposase
gene is bounded at each end by inverted repeat sequences that function
as recognition sites for the transposon. IS1 is one example of an insertion
sequence, shown here with simplified IR sequences.
Tn5
Kanamycin resistance
IS1
IS1
(b) Complex transposons carry other genetic material in addition to
transposase genes. The example shown here, Tn5, carries the gene for
kanamycin resistance and has complete copies of the insertion sequence
IS1 at each end.
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Transposons
ANIMATION Transposons: Insertion Sequences
ANIMATION Transposons: Complex Transposons
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Figure 8.31c Transposons and insertion.
Transposase gene
Transposase cuts DNA, leaving sticky ends.
IS1
IS1
Sticky ends of transposon and target DNA anneal.
(c) Insertion of the transposon Tn5 into R100 plasmid
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Check Your Understanding
 Differentiate horizontal and vertical gene transfer.
8-14
 Compare conjugation between the following pairs:
F+ × F–, Hfr × F–. 8-15
 What types of genes do plasmids carry? 8-16
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Genes and Evolution
Learning Objective
8-17 Discuss how genetic mutation and
recombination provide material for natural
selection to act upon.
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Genes and Evolution
 Mutations and recombination provide diversity
 Fittest organisms for an environment are selected by
natural selection
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Clinical Focus, unnumbered figure, page 220.
Insert West Nile Virus figure
from Clinical Focus, p. 220
West Nile virus
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Clinical Focus, unnumbered table, page 220
Insert Table from Clinical Focus,
p. 220, bottom
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Evolution
 Which strain is more
closely related to the
Uganda strain?
 How did the virus change?
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Strain
% Similar to
Uganda
Kenya
71%
U.S.
51%
Check Your Understanding
 Natural selection means that the environment favors
survival of some genotypes. From where does
diversity in genotypes come? 8-17
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