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Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings PowerPoint Lectures for Biolog y: Concepts and Connections, Fifth Editi on    Campbell , Re e ce , T ayl or, an d S i mon  Lectures by Chris Romero Chapter 10 Molecular Biology of the Gene

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Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings

PowerPoint Lectures for 

Biolog y: Concepts and Connect ions, Fi f th Edi t ion 

 – Campbell , Reece, Taylor, and Simon  

Lectures by Chris Romero

Chapter 10

Molecular Biology of the

Gene

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10.2 DNA and RNA are polymers of nucleotides

• Nucleic acids are polynucleotides made of longchains of nucleotide monomers

 –  Nitrogenous bases

• Single-ring pyrimidines: thymine (T),

cytosine ( C)

• Double-ring purines: adenine (A), guanine

(G)

 –  Sugar-phosphate backbone

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• DNA and RNA are identical except for two

things

 –  Nitrogenous bases

• DNA: A, C, G, T

• RNA: A, G, C, U

 –  Sugars

•DNA: deoxyribose

• RNA: ribose

Animation: DNA and RNA Structure 

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LE 10-2a

Sugar-phosphate backbone Phosphate group Nitrogenous base 

Sugar  

DNA nucleotide 

DNA polynucleotide DNA nucleotide 

Sugar 

(deoxyribose)

Thymine (T) 

Nitrogenous base

(A, G, C, or T)Phosphate

group

T  T 

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LE 10-2b

Thymine (T)  Cytosine (C) Pyrimidines 

Adenine (A) Purines 

Guanine (G) 

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LE 10-2c

Phosphate

group

Nitrogenous base

(A, G, C, or U)

Sugar 

(ribose)

Uracil (U)

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LE 10-2dKey 

Hydrogen atom 

Phosphorus atom 

Carbon atom Nitrogen atom Oxygen atom 

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10.3 DNA is a double-stranded helix

• James Watson and Francis Crick worked outthe three-dimensional structure of DNA, based

on X-ray crystallography by Rosalind Franklin

• DNA consists of two polynucleotide strandswrapped around each other in a double helix

 –  Sugar-phosphate backbones are on the

outside and nitrogenous bases on the inside

Animation: DNA Double Helix 

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 –  Each base pairs with a complementary

partner 

•  A with T, and G with C

 –  Hydrogen bonds between the bases hold

the strands together 

• The Watson-Crick model of DNA suggested a

molecular explanation for genetic inheritance

LE 10 3

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LE 10-3c

Twist 

LE 10 3d

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LE 10-3d

Hydrogen bond Base

pair  

Ribbon model  Partial chemical structure  Computer model 

G  C T  A 

A  T 

T A 

C C 

G G 

G C 

T T 

T T 

A A A 

G  C A  T 

C G 

A T 

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DNA REPLICATION

10.4 DNA replication depends on specific base pairing

The Watson-Crick model of DNA structure suggesteda mechanism for its replication

 –  DNA strands separate

 – Enzymes use each strand as a template toassemble new nucleotides into complementary

strands

• The mechanism of DNA replication is

semiconservative

 –  Each new double helix consists of one old and one

new strand

LE 10 4

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LE 10-4a

Parental

molecule

of DNA

Both parental

strands serve

as templates

Two identical

daughter molecules

of DNA

Nucleotides 

C  G A 

G  C A 

A T 

A C 

C C 

G G 

A C 

A A  A 

C  C C 

C G 

T G 

T T 

T G 

T T 

G G 

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• DNA replication is a complex process

 – Some of the helical DNA molecule mustuntwist

Animation: DNA Replication Overview 

LE 10 4b

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LE 10-4b

G  C T A 

A  T G 

G C C 

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10.5 DNA replication: A closer look

• DNA replication begins at specific sites (originsof replication) on the double helix

 –  Proteins attach and separate the strands

 –  Replication proceeds in both directions,

creating replication bubbles

• Parent strands open, daughter strands

elongate

 –  Replication occurs simultaneously at many

sites

LE 10 5a

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LE 10-5a

Origin of replication 

Bubble 

Parental strand Daughter strand 

Two daughter DNA molecules 

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• DNA's sugar-phosphate backbones are

oriented in opposite directions

 –  The enzyme DNA polymerase adds

nucleotides at only the 3’ end 

•One daughter strand is synthesized as acontinuous piece

• The other strand is synthesized as a series

of short pieces

• The two strands are connected by the

enzyme DNA ligase

LE 10-5b

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LE 10-5b

3 end  5 end 

5 end  3 end 

4 A 

HO 

OH 

P 1  3 

5 P 

4 3 

1 2 

LE 10-5c

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LE 10-5cDNA polymerase

molecule 

Parental DNA Daughter strand

synthesized

continuously

Daughter 

strand

synthesized

In pieces

3

5

3

5

5

3

3

5

DNA ligase Overall direction of replication 

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THE FLOW OF GENETIC INFORMATIONFROM DNA TO RNA TO PROTEIN

10.6 The DNA genotype is expressed as

proteins, which provide the molecular basis for 

phenotypic traits

• The information constituting an organism's

genotype is carried in its sequence of DNA

bases

•  A particular gene—a linear sequence of many

nucleotides—specifies a particular polypeptide

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• The flow of genetic information

1. Transcription of the genetic information in DNAinto RNA

2. Translation of RNA into the polypeptide

Beadle-Tatum one gene-one enzyme hypothesis –  Studies of inherited metabolic disorders in mold

suggested that phenotype is expressed through

proteins

 –   A gene dictates production of a specific enzyme

 –  The hypothesis has been restated to one gene-

one polypeptide

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10.7 Genetic information written in codons is

translated into amino acid sequences

• Genetic information flows from DNA to RNA to

protein

• Nucleotide monomers represent letters in analphabet that can form words in a language

 –  Triplet code

• Three-letter words (codons)

• Each word codes for one amino acid in a

polypeptide

LE 10-7a

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DNA molecule 

Gene 1 

Gene 2 

Gene 3 

A A A A A  C  C  G G  C 

C C G G  G  U  U  U  U U U U 

A A DNA strand 

Transcription 

RNA 

Translation 

Polypeptide Amino acid 

Codon 

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10.8 The genetic code is the Rosetta stone of life

• The genetic code specifies thecorrespondence between RNA codons and

amino acids in proteins

 – Includes start and stop codons

 –  Redundant but not ambiguous

• Nearly all organisms use exactly the same

genetic code

LE 10-8a

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Second base 

UUU U  C  A  G 

G A A 

A C 

G U C A G 

UUC UUA UUG CUU CUC CUA CUG AUU AUC AUA 

Leu 

lle 

AUG  Met or 

start GUU GUC GUA GUG 

Val  Ala 

Thr  

GCG GCA GCC GCU ACG ACA ACC ACU 

Pro 

CCG CCA CCC CCU UCG UCA UCC UCU 

Ser  Tyr   Cys UAU 

UAC UAA UAG CAU CAC CAA CAG 

His 

Gln 

Stop Stop 

UGU UGC UGA UGG  Trp 

Stop 

Arg 

CGU CGU CGA CGG 

Asn 

Lys 

Asp 

Glu Gly 

Arg 

Ser  AAU  AGU AAC AAA AAG GAU 

GAG 

AGC AGA AGG GGU GGC 

GGG 

Phe 

Leu 

GAC GAA GAG 

GGA 

LE 10-8b

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Strand to be transcribed 

Transcription 

DNA T 

A  G T 

A  C  T  T 

T  T 

A  A 

A A  G 

T  T 

T A  A 

A  G U RNA 

Translation Start

codon 

A  A  U  U G  U  U  A  G 

Stop

codon 

Phe Lys Met Polypeptide 

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10.9 Transcription produces genetic messages in

the form of RNA

• One DNA strand serves as a template for the

new RNA strand

• RNA polymerase constructs the RNA strand ina multistep process

 –  Initiation

• RNA polymerase attaches to the promotor 

• Synthesis starts

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• Elongation:

 – RNA synthesis continues

 –  RNA strand peels away from DNA template

 –  DNA strands come back together in

transcribed region

• Termination

 – RNA polymerase reaches a terminator sequence at the end of the gene

 –  Polymerase detaches

LE 10-9a

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RNA

polymerase RNA nucleotides 

Template

strand of DNA Direction of 

transcription 

Newly made RNA 

A  A C  C 

C  C  A 

G  T  T 

U A 

G T  A 

LE 10-9b

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RNA polymerase 

Initiation 

Promoter 

DNA  Terminator 

DNA

Area shown

In Figure 10.9A

Initiation 

Elongation 

Termination  Growing

RNA

Completed RNA RNA

polymerase

DNA of gene 

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Animation: Transcription 

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10.10 Eukaryotic RNA is processed before

leaving the nucleus

• The RNA that encodes an amino acid

sequence is messenger RNA (mRNA)

• In prokaryotes, transcription and translationboth occur in the cytoplasm

• In eukaryotes, RNA transcribed in the nucleus

is processed before moving to the cytoplasmfor translation

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• RNA Splicing

 – Noncoding segments called introns are cutout

 –  Remaining exons are joined to form a

continuous coding sequence

 –  A cap and a tail are added to the ends

LE 10-10

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Exon Exon  Exon Intron Intron 

Cap DNA 

RNA

transcript

with cap

and tail

mRNA 

Coding sequence Nucleus 

Cytoplasm 

Exons spliced together  

Tail Introns removed 

Transcription

Addition of cap and tail 

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10.11 Transfer RNA molecules serve as interpreters

during translation

• Transfer RNA (tRNA) molecules match the right amino

acid to the correct codon

• tRNA is a twisted and folded single strand of RNA

 –   Anticodon loop at one end recognizes a particular 

mRNA codon by base pairing

 –   Amino acid attachment site is at the other end

• Each amino acid is joined to the correct tRNA by a

specific enzyme

LE 10-11a

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Amino acid attachment site 

Hydrogen bond 

RNA polynucleotide chain 

Anticodon 

LE 10-11bAmino acid

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Anticodon 

Amino acid

attachment site 

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10.12 Ribosomes build polypeptides

•  A ribosome consists of two subunits

 –  Each is made up of proteins and ribosomal

RNA (rRNA)

• The subunits of a ribosome

 –  Hold the tRNA and mRNA close together in

binding sites during translation

 –  Allow amino acids to be connected into a

polypeptide chain

LE 10-12a

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tRNA

molecules Growing

polypeptide 

Largesubunit 

Small

subunit mRNA 

LE 10-12btRNA-binding sites

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tRNA-binding sites 

Largesubunit 

Small

subunit 

mRNA

binding

site

LE 10-12c

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Growing

polypeptide 

mRNA 

Codons 

tRNA 

Next amino acid

to be added to

polypeptide 

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10.13 An initiation codon marks the start of an

mRNA message

• The initiation phase of translation

 –  Brings together mRNA, a specific tRNA,

and the two subunits of a ribosome

 –  Establishes exactly where translation will

begin

• Ensures that mRNA codes are translated in

the correct sequence

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• Initiation is a two-step process

 – Step 1

• mRNA binds to a small ribosomal subunit

• Initiator tRNA, carrying the amino acid Met,

binds to the start codon

 –  Step 2

 A large ribosomal subunit binds to the smallone, forming a functional ribosome

• Initiator tRNA fits into one binding site; the

other is vacant for the next tRNA

LE 10-13a

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Start of genetic message 

End 

LE 10-13b

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Initiator tRNA P site

 

mRNA Start codon 

Small ribosomal

subunit 

A site 

Large

Ribosomal

subunit

C A U  A A  U 

U G  A  G U 

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10.14 Elongation adds amino acids to the

polypeptide chain until a stop codon terminates

translation

• Once initiation is complete, amino acids are

added one by one in a three-step elongation

process

1. Codon recognition

2. Peptide bond formation

3. Translocation

• Elongation continues until a stop codon reaches

the ribosome's A site, terminating translation

LE 10-14Amino

id

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Polypeptide P site 

mRNA 

A site 

Codons Anticodon 

acid 

Condon recognition

Peptide bond

formation New

peptide

bond 

Translocation 

Stop

codon 

mRNA

movement 

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Animation: Translation 

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10.15 Review: The flow of genetic information in

the cell is DNA RNA protein

• The sequence of codons in DNA, via the

sequence of codons in RNA, spells out the

primary structure of a polypeptide

1. Transcription of mRNA from a DNA

template

2. Attachment of amino acid to tRNA

3. Initiation of polypeptide synthesis

4. Elongation

5. Termination

LE 10-15DNA  Transcription 

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mRNA RNA

polymerase mRNA is

transcribed from a

DNA template.

Each amino acid

attaches to its proper tRNA with the help of a

specific enzyme and ATP.

Translation Amino acid 

mRNA 

Enzyme 

ATP 

Anticodon Large

ribosomal

subunit Initiator 

tRNA 

Start Codon  Small

ribosomalsubunit 

U  A A U 

C G 

Initiation of 

polypeptide synthesis

The mRNA, the first

tRNA, and the ribosomal

Sub units come together.

Stop codon 

A succession of tRNAs

add their amino acids to

the polypeptide chain

as the mRNA is moved

through the ribosome,

one codon at a time.

Growingpolypeptide 

New peptidebond forming 

mRNA 

Polypeptide 

Codons 

Elongation 

Termination The ribosome recognizes

a stop codon. The poly-

peptide is terminated

and released.

tRNA 

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10.16 Mutations can change the meaning of 

genes

• Mutation: any change in the nucleotide

sequence of DNA

 – Caused by errors in DNA replication or recombination, or by mutagens

 –  Can involve large regions of a chromosome

or a single base pair 

 –  Can cause many genetic diseases, such as

sickle-cell disease

LE 10-16a

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Normal hemoglobin DNA 

mRNA 

C  T  T 

A A G 

Normal hemoglobin Glu 

mRNA 

Sickle-cell hemoglobin Val 

Mutant hemoglobin DNA 

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• Two general categories of genetic mutations

 – 

Base substitutions replace one base withanother 

• Most are harmful but may occasionally have

no effect or be beneficial

 –  Base insertions or deletions alter the

reading frame

• Result is most likely a nonfunctioning

polypeptide

• Mutagenesis caused by spontaneous error or 

a physical or chemical mutagen

LE 10-16bNormal gene 

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Base substitution 

Protein mRNA 

Base deletion 

Missing 

Met 

Met 

Met  Lys 

Lys 

Lys  Phe  Gly  Ala 

Ala Phe 

Ala 

Ser  

Leu  His 

A  A  A  A 

A  A  A 

A A A A A  U  U  U  U 

U U U U 

U  U  U  U  G  G  G G G 

G  G  G  G 

C  C 

C  C 

G  G  G  G  G  C C 

MICROBIAL GENETICS

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10.17 Viral DNA may become part of the host

chromosome

• Viruses are infectious particles consisting of 

nucleic acid enclosed in a protein capsid

• Viruses depend on their host cells for thereplication, transcription, and translation of 

their nucleic acid

 – DNA enters host bacterium, circularizes,and enters one of two pathways

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 –  Lytic cycle

Host produces more viruses

• Host cell lyses (breaks open) to release new

viruses

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 –  Lysogenic cycle

Phage DNA inserted by recombination intothe host chromosome; is now a prophage

• Prophages replicated each time host cell

divides; passed on to generations of 

daughter cells

• Does not destroy host

Environmental signal may trigger switchfrom lysogenic to lytic cycle

LE 10-17

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Phage 

Phage DNA 

Attaches

to cell

Cell lyses,

releasing phagesPhage injects DNA

Lytic cycle

Phages assemblePhage DNA

circularizes

New phage DNA and

proteins are synthesized

Bacterial

chromosome

OR

Lysogenic cycle

Prophage

Many cell

divisions

Lysogenic bacterium repro-

duces normally, replicating the

prophage at each cell division

Phage DNA inserts into the bacterial

chromosome by recombination

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10.18 Many viruses cause disease in animals

• Structure of a virus that invades animal cells

 –  Genetic material may be RNA (examples:

flu, HIV) or DNA (examples: hepatitis,

herpes)

 –  Protein coat

 –  Sometimes a membranous envelope with

glycoprotein spikes

• The envelope helps the virus enter and leave

the host cell during its reproductive cycle

LE 10-18aMembranous

envelope

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envelope 

RNA 

Protein

coat 

Glycoprotein spike 

LE 10-18bVIRUS 

Viral RNA  Glycoprotein spike P t i t

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(genome)  Protein coat Envelope 

Entry Plasma membrane

of host cell 

Uncoating Viral RNA

(genome) RNA synthesis

by viral enzyme RNA synthesis

(other strand) 

Template mRNA Protein

synthesis 

New

viral proteins 

New viral

genome Assembly 

Exit 

CONNECTION

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10.19 Plant viruses are serious agricultural pests

• Most plant viruses

 –  Have RNA genomes

 –  Enter their hosts via wounds in the plant's

outer layers

 –  May spread throughout the plant through

plasmodesmata

• There is no cure for most plant viruses

LE 10-19

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RNA 

Protein 

CONNECTION

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10.20 Emerging viruses threaten human health

•Emerging viruses have appeared suddenly or have recently come to the attention of 

scientists

 – Examples: HIV, SARS, Ebola, West Nile

• Processes contributing to emergence

 –  Mutation

 –  Contact between species

 –  Spread from isolated populations

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10.21 The AIDS virus makes DNA on an RNA

template

• HIV, the AIDS virus, is a retrovirus

 –  Flow of genetic information is RNA _ DNA

 –  Inside a cell, HIV uses its RNA as a

template for making DNA

 –  The enzyme reverse transcriptase catalyzes

reverse transcription

Animation: HIV Reproductive Cycle 

LE 10-21a

Envelope 

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p

Glycoprotein 

Proteincoat 

RNA

(two identicalstrands) 

Reverse

transcriptase 

LE 10-21b

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Viral RNA 

DNA

strand Double-

stranded

DNA

Viral

RNA

and

proteins

C YTOPLASM NUCLEUS 

Chromosomal

DNA Provirus

DNA 

RNA 

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10.22 Bacteria can transfer DNA in three ways

•Bacteria can transfer genes from cell to cell byone of three processes

 –  Transformation: the uptake of foreign DNA

from the surrounding environment

 –  Transduction: transfer of bacterial genes by

a phage

 – Conjugation: union of two bacterial cells andthe transfer of DNA between them

LE 10-22a

DNA enters

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cell 

Fragment of 

DNA from

another 

bacterial cell 

Bacterial chromosome

(DNA) 

LE 10-22bPhage 

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Fragment of DNA from

another bacterial cell

(former phage host)

LE 10-22cMating bridge 

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Recipient cell

(“female”) 

Sex pili 

Donor cell

(“male”) 

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• Once new DNA is in a bacterial cell, part of it

may integrate into the recipient's chromosome

 –  Occurs by crossing over between the two

molecules

 – 

Leaves the recipient with a recombinantchromosome

LE 10-22d

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Donated DNA  Crossovers  Degraded DNA 

Recombinant

chromosome Recipient cell’s 

chromosome 

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10.23 Bacterial plasmids can serve as carriers for 

gene transfer 

• The F factor is a piece of bacterial DNA

 –  Carries genes for things needed for 

conjugation

 –  Contains an origin of replication

 –  Can transfer chromosomal DNA by

integrating into the donor bacterium's

chromosome or entering the cell as a

plasmid

LE 10-23aF factor (integrated) 

Male (donor)

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Male (donor)

cell

Origin of F

replication

Bacterial

chromosome

F factor starts replication

and transfer of chromosome

Recipient cell 

Only part of the

chromosome transfers

Recombination can occur  

LE 10-23bF factor (plasmid) 

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Cell now male 

Plasmid completes

transfer and circularizes

F factor starts replication

and transfer 

Bacterial

chromosome

Male (donor)

cell

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• Plasmids

 – Small circular DNA molecules separatefrom the bacterial chromosome

 –  Can serve as carriers for the transfer of 

genes

LE 10-23c

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Plasmids 

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