viruses called bacteriophages can infect and set in motion a genetic takeover of bacteria, such as...

• Viruses called bacteriophages can infect and set in motion a genetic takeover of bacteria, such as Escherichia coli

• Bacteria are prokaryotes with cells much smaller and more simply organized than those of eukaryotes

• Viruses are smaller and simpler than bacteria

• Tobacco mosaic disease stunts growth of tobacco plants and gives their leaves a mosaic coloration

• In the late 1800s, researchers hypothesized that a particle smaller than bacteria caused the disease

• In 1935, Wendell Stanley confirmed this hypothesis by crystallizing the infectious particle, now known as tobacco mosaic virus (TMV)

• Viruses are not cells• Viruses are very small infectious particles

consisting of nucleic acid enclosed in a protein coat and, in some cases, a membranous envelope

• Viral genomes may consist of– Double- or single-stranded DNA– Double- or single-stranded RNA

• Depending on its type of nucleic acid, a virus is called a DNA virus or an RNA virus

Capsomereof capsid

RNA

18 250 mm

Tobacco mosaic virus20 nm

A capsid is the protein shell that encloses the viral genome and can have

various structures

Capsomere

Glycoprotein

70–90 nm (diameter)

DNA

Adenoviruses50 nm

• Some viruses have structures that have membranous envelopes that help them infect hosts

• These viral envelopes surround the capsids of influenza viruses and many other viruses found in animals

• Viral envelopes, which are derived from the host cell’s membrane, contain a combination of viral and host cell molecules

Glycoprotein

80–200 nm (diameter)

RNA

Capsid

Influenza viruses50 nm

Membranousenvelope

• Bacteriophages, also called phages, are viruses that infect bacteria

• Phages have an elongated capsid head that encloses their DNA

• A protein tailpiece attaches the phage to the host and injects the phage DNA inside

• Viruses use enzymes, ribosomes, and small host molecules to synthesize progeny viruses

80 225 nm

DNAHead

Tailsheath

Tailfiber

Bacteriophage T450 nm

DNAVIRUS

Capsid

HOST CELL

Viral DNA

Replication

Entry into cell anduncoating of DNA

Transcription

Viral DNA

mRNA

Capsidproteins

Self-assembly ofnew virus particlesand their exit from cell

The Lytic Cycle• The lytic cycle is a phage reproductive cycle

that culminates in the death of the host cell• The lytic cycle produces new phages and digests

the host’s cell wall, releasing the progeny viruses

• A phage that reproduces only by the lytic cycle is called a virulent phage

• Bacteria have defenses against phages, including restriction enzymes that recognize and cut up certain phage DNA

Attachment

Entry of phage DNAand degradation of host DNA

Synthesis of viralgenomes and proteins

Assembly

ReleasePhage assembly

Head Tails Tail fibers

The Lysogenic Cycle• The lysogenic cycle replicates the phage genome

without destroying the host• The viral DNA molecule is incorporated by genetic

recombination into the host cell’s chromosome• This integrated viral DNA is known as a prophage• Every time the host divides, it copies the phage

DNA and passes the copies to daughter cells• Phages that use both the lytic and lysogenic cycles

are called temperate phages

Phage

Phage DNA

The phage attaches to ahost cell and injects its DNA.

Phage DNAcircularizes

Bacterial chromosome

Lytic cycle

The cell lyses, releasing phages.Lytic cycleis induced

or Lysogenic cycleis entered

Certain factorsdetermine whether

Lysogenic cycle

Occasionally, a prophageexits the bacterial chromosome,initiating a lytic cycle.

The bacterium reproducesnormally, copying the prophageand transmitting it to daughter cells.

Prophage

Many cell divisionsproduce a large population of bacteria infected withthe prophage.

Daughter cellwith prophage

Phage DNA integrates into thebacterial chromosomes, becoming aprophage.

New phage DNA and proteins aresynthesized and assembled into phages.

Reproductive Cycles of Animal Viruses• Two key variables in classifying viruses that

infect animals:– DNA or RNA?– Single-stranded or double-stranded?

• Many viruses that infect animals have a membranous envelope

• Viral glycoproteins on the envelope bind to specific receptor molecules on the surface of a host cell

Viral Envelopes

RNA

ER

Capsid

HOST CELL

Viral genome (RNA)

mRNA

Capsidproteins

Envelope (withglycoproteins)

Glyco-proteins Copy of

genome (RNA)

Capsid and viral genomeenter cell

New virus

Template

RNA as Viral Genetic Material• The broadest variety of RNA genomes is found

in viruses that infect animals• Retroviruses use reverse transcriptase to copy

their RNA genome into DNA• HIV is the retrovirus that causes AIDS

Capsid

Viral envelopeGlycoprotein

Reversetranscriptase

RNA(two identicalstrands)

• The viral DNA that is integrated into the host genome is called a provirus

• Unlike a prophage, a provirus remains a permanent resident of the host cell

• The host’s RNA polymerase transcribes the proviral DNA into RNA molecules

• The RNA molecules function both as mRNA for synthesis of viral proteins and as genomes for new virus particles released from the cell

HOST CELL

ReversetranscriptionViral RNA

RNA-DNAhybrid

DNA

NUCLEUS

ChromosomalDNA

Provirus

RNA genomefor thenext viralgeneration

mRNA

New HIV leaving a cell

HIV entering a cell

0.25 µm

HIVMembrane ofwhite blood cell

Viroids and Prions: The Simplest Infectious Agents

• Viroids are circular RNA molecules that infect plants and disrupt their growth

• Prions are slow-acting, virtually indestructible infectious proteins that cause brain diseases in mammals

• Prions propagate by converting normal proteins into the prion version

Normalprotein

New prion

Prion Original prion

Many prions

The Bacterial Genome and Its Replication

• The bacterial chromosome is usually a circular DNA molecule with few associated proteins

• Many bacteria also have plasmids, smaller circular DNA molecules that can replicate independently of the chromosome

• Bacterial cells divide by binary fission, which is preceded by replication of the chromosome

Mutation and Genetic Recombination as Sources of Genetic Variation

• Since bacteria can reproduce rapidly, new mutations quickly increase genetic diversity

• More genetic diversity arises by recombination of DNA from two different bacterial cells

• Three processes bring bacterial DNA from different individuals together:– Transformation– Transduction– Conjugation

Transformation and Transduction• Transformation is the alteration of a bacterial

cell’s genotype and phenotype by the uptake of naked, foreign DNA from the surrounding environment

• For example, harmless Streptococcus pneumoniae bacteria can be transformed to pneumonia-causing cells

• In the process known as transduction, phages carry bacterial genes from one host cell to another

A+

Phage DNA

A+

Donorcell

B+

A+

B+

Crossingover

A+

A– B–

Recipientcell

A+ B–

Recombinant cell

Conjugation and Plasmids• Conjugation is the direct transfer of genetic

material between bacterial cells that are temporarily joined

• The transfer is one-way: One cell (“male”) donates DNA, and its “mate” (“female”) receives the genes

Sex pilus 5 µm

Mutantstrain

arg+ trp–

Mutantstrain

arg+ trp–

Mixture

Mixture

Nocolonies(control)

Nocolonies(control)

Coloniesgrew

Mutantstrain

arg– trp+

Mutantstrain

arg– trp+

The F Plasmid and Conjugation• Cells containing the F plasmid, designated F+

cells (fertile), function as DNA donors during conjugation

• F+ cells transfer DNA to an F recipient cell• Chromosomal genes can be transferred during

conjugation when the donor cell’s F factor is integrated into the chromosome

• A cell with a built-in F factor is called an Hfr cell• The F factor of an Hfr (high frequency of

recombination) cell brings some chromosomal DNA along when transferred to an F– cell

F plasmid Bacterial chromosome

F+ cellMatingbridge

F+ cell

F+ cellBacterial chromosome

F– cell

Conjunction and transfer of an F plasmid from and F+ donor to an F– recipient

F+ cell Hfr cell

F factor

Hfr cell

F– cell

Temporarypartialdiploid

Recombinant F–

bacterium

Conjugation and transfer of part of the bacterial chromosome from anHfr donor to an F– recipient, resulting in recombination

R plasmids and Antibiotic Resistance• R plasmids confer resistance to various antibiotics• When a bacterial population is exposed to an

antibiotic, individuals with the R plasmid will survive and increase in the overall population

• The DNA of a cell can also undergo recombination due to movement of transposable elements within the cell’s genome

• Transposable elements, often called “jumping genes,” contribute to genetic shuffling in bacteria

Transposition of Genetic Elements

Insertion Sequences• The simplest transposable elements, called

insertion sequences, exist only in bacteria• An insertion sequence has a single gene for

transposase, an enzyme catalyzing movement of the insertion sequence from one site to another within the genome

Insertion sequence

Transposase gene

53

Invertedrepeat

35

Invertedrepeat

Transposons• Transposable elements called transposons are

longer and more complex than insertion sequences• In addition to DNA required for transposition,

transposons have extra genes that “go along for the ride,” such as genes for antibiotic resistance

53

35

Transposon

Insertion sequence

Insertion sequence

Antibioticresistance gene

Transposase geneInverted repeat

Operons: The Basic Concept• In bacteria, genes are often clustered into

operons, composed of– An operator, an “on-off” switch– A promoter– Genes for metabolic enzymes

• An operon can be switched off by a protein called a repressor

• A corepressor is a small molecule that cooperates with a repressor to switch an operon off

Promoter Promoter

DNA trpR

Regulatorygene

RNApolymerase

mRNA

3

5

Protein Inactiverepressor

Tryptophan absent, repressor inactive, operon on

mRNA 5

trpE trpD trpC trpB trpA

OperatorStart codon Stop codon

trp operon

Genes of operon

E

Polypeptides that make upenzymes for tryptophan synthesis

D C B A

DNA

Protein

Tryptophan(corepressor)

Tryptophan present, repressor active, operon off

mRNA

Activerepressor

DNA

Protein

Tryptophan(corepressor)

Tryptophan present, repressor active, operon off

mRNA

Activerepressor

No RNA made

Repressible and Inducible Operons: Two Types of Negative Gene

Regulation• A repressible operon is one that is usually on;

binding of a repressor to the operator shuts off transcription

• The trp operon is a repressible operon• An inducible operon is one that is usually off; a

molecule called an inducer inactivates the repressor and turns on transcription

• The classic example of an inducible operon is the lac operon, which contains genes coding for enzymes in hydrolysis and metabolism of lactose

DNA lacl

Regulatorygene

mRNA

5

3

RNApolymerase

ProteinActiverepressor

NoRNAmade

lacZ

Promoter

Operator

Lactose absent, repressor active, operon off

DNA lacl

mRNA5

3

lac operon

Lactose present, repressor inactive, operon on

lacZ lacY lacA

RNApolymerase

mRNA 5

Protein

Allolactose(inducer)

Inactiverepressor

-Galactosidase Permease Transacetylase

• Inducible enzymes usually function in catabolic pathways

• Repressible enzymes usually function in anabolic pathways

• Regulation of the trp and lac operons involves negative control of genes because operons are switched off by the active form of the repressor

Positive Gene Regulation• Some operons are also subject to positive

control through a stimulatory activator protein, such as catabolite activator protein (CAP)

• When glucose (a preferred food source of E. coli ) is scarce, the lac operon is activated by the

binding of CAP• When glucose levels increase, CAP detaches

from the lac operon, turning it off

DNA

cAMP

lacl

CAP-binding site

Promoter

ActiveCAP

InactiveCAP

RNApolymerasecan bindand transcribe

Operator

lacZ

Inactive lacrepressor

Lactose present, glucose scarce (cAMP level high): abundant lacmRNA synthesized

DNA lacl

CAP-binding site

Promoter

RNApolymerasecan’t bind

Operator

lacZ

Inactive lacrepressor

InactiveCAP

Lactose present, glucose present (cAMP level low): little lacmRNA synthesized

• Two features of eukaryotic genomes are a major information-processing challenge:– First, the typical eukaryotic genome is much larger

than that of a prokaryotic cell– Second, cell specialization limits the expression of

many genes to specific cells

• The DNA-protein complex, called chromatin, is ordered into higher structural levels than the DNA-protein complex in prokaryotes

Nucleosomes, or “Beads on a String”• Proteins called histones are responsible for the

first level of DNA packing in chromatin• The association of DNA and histones seems to

remain intact throughout the cell cycle• In electron micrographs, unfolded chromatin

has the appearance of beads on a string• Each “bead” is a nucleosome, the basic unit of

DNA packing

DNA double helix

Histonetails

His-tones

Linker DNA(“string”)

Nucleosome(“bead”)

10 nm

2 nm

Histone H1

Nucleosomes (10-nm fiber)

30 nm

Nucleosome

30-nm fiber

300 nm

Loops

Scaffold

Protein scaffold

Looped domains (300-nm fiber)

Metaphase chromosome

700 nm

1,400 nm

Differential Gene Expression• Differences between cell types result from

differential gene expression, the expression of different genes by cells within the same genome

• In each type of differentiated cell, a unique subset of genes is expressed

• Many key stages of gene expression can be regulated in eukaryotic cells

Signal

NUCLEUS

DNA

RNA

Chromatin

Gene availablefor transcription

Gene

Exon

Intro

Transcription

Primary transcript

RNA processing

Cap

Tail

mRNA in nucleus

Transport to cytoplasm

CYTOPLASM

mRNA in cytoplasm

Translation

Degradationof mRNA

Polypeptide

CleavageChemical modificationTransport to cellular

destination

Degradation of protein

Active protein

Degraded protein

Regulation of Chromatin Structure• Genes within highly packed heterochromatin

are usually not expressed• Chemical modifications to histones and DNA of

chromatin influence both chromatin structure and gene expression

Histone Modification

• In histone acetylation, acetyl groups are attached to positively charged lysines in histone tails

• This process seems to loosen chromatin structure, thereby promoting the initiation of transcription

Histonetails

Amino acidsavailablefor chemicalmodification

DNAdouble helix

Histone tails protrude outward from a nucleosome

Acetylation of histone tails promotes loose chromatinstructure that permits transcription

Unacetylated histones Acetylated histones

DNA Methylation• DNA methylation, the addition of methyl groups to

certain bases in DNA, is associated with reduced transcription in some species

• In some species, DNA methylation causes long-term inactivation of genes in cellular differentiation

• In genomic imprinting, methylation turns off either the maternal or paternal alleles of certain genes at the start of development … lions, tigers, ligers?

The Roles of Transcription Factors• To initiate transcription, eukaryotic RNA

polymerase requires the assistance of proteins called transcription factors

• General transcription factors are essential for the transcription of all protein-coding genes

• In eukaryotes, high levels of transcription of particular genes depend on control elements interacting with specific transcription factors

Enhancers and Specific Transcription Factors

• Proximal control elements are located close to the promoter

• Distal control elements, groups of which are called enhancers, may be far away from a gene or even in an intron

• An activator is a protein that binds to an enhancer and stimulates transcription of a gene

Enhancer(distal control elements)

Proximal control elements

Upstream

DNA

Promoter

Exon Intron Exon Intron Exon

DownstreamTranscription

Poly-A signalsequence

Terminationregion

Intron Exon Intron Exon

RNA processing:Cap and tail added;introns excised andexons spliced together

Poly-A signal

Cleaved 3 endof primarytranscript

3

Poly-Atail

3 UTR(untranslated

region)

5 UTR(untranslated

region)

Startcodon

Stopcodon

Coding segment

Intron RNA

5 Cap

mRNA

Primary RNAtranscript(pre-mRNA)

5Exon

Distal controlelement Activators

Enhancer

DNA

DNA-bendingprotein

TATAbox

PromoterGene

Generaltranscriptionfactors

Group ofmediator proteins

RNApolymerase II

RNApolymerase II

RNA synthesisTranscriptionInitiation complex

Controlelements

Enhancer Promoter

Albumingene

Crystallingene

Availableactivators

Availableactivators

Albumingene notexpressed

Albumingeneexpressed

Liver cell Lens cell

Crystallin genenot expressed Crystallin gene

expressed

Liver cellnucleus

Lens cellnucleus

Some transcription factors function as repressors, inhibiting expression of a particular geneSome activators and repressors act indirectly by influencing chromatin structure

PrimaryRNAtranscript

DNA

or

Exons

RNA splicing

mRNA

In alternative RNA splicing, different mRNA molecules are produced from the same primary transcript, depending on which RNA segments are treated as exons and which as introns

mRNA Degradation• The life span of mRNA molecules in the cytoplasm

is a key to determining the protein synthesis• The mRNA life span is determined in part by

sequences in the leader and trailer regions

• RNA interference by single-stranded microRNAs (miRNAs) can lead to degradation of an mRNA or block its translation

• The phenomenon of inhibition of gene expression by RNA molecules is called RNA interference (RNAi)

Dicer

Hydrogenbond

Proteincomplex

miRNATarget mRNA

Degradation of mRNA

OR

Blockage of translation

Protein Processing and Degradation

• After translation, various types of protein processing, including cleavage and the addition of chemical groups, are subject to control

• Proteasomes are giant protein complexes that bind protein molecules and degrade them

Protein tobe degraded

Ubiquitinatedprotein

Proteasome

Protein entering aproteasome

Protein fragments(peptides)

Proteasomeand ubiquitinto be recycled

Ubiquitin

Types of Genes Associated with Cancer

• Genes that normally regulate cell growth and division during the cell cycle include:– Genes for growth factors– Their receptors– Intracellular molecules of signaling pathways

• Mutations altering any of these genes in somatic cells can lead to cancer

Oncogenes and Proto-Oncogenes• Oncogenes are cancer-causing genes• Proto-oncogenes are normal cellular genes that

code for proteins that stimulate normal cell growth and division

• A DNA change that makes a proto-oncogene excessively active converts it to an oncogene, which may promote excessive cell division and cancer

Proto-oncogene

DNA

Translocation or transposition:gene moved to new locus, under new controls

Newpromoter

Gene amplification:multiple copies of the gene

Point mutationwithin a controlelement

Oncogene Oncogene

Point mutationwithin the gene

Normal growth-stimulatingprotein in excess

Normal growth-stimulatingprotein in excess Normal growth-stimulating

protein in excessHyperactive ordegradation-resistant protein

Tumor-Suppressor Genes

• Tumor-suppressor genes encode proteins that inhibit abnormal cell division

• Any decrease in the normal activity of a tumor-suppressor protein may contribute to cancer

Interference with Normal Cell-Signaling Pathways

• Many proto-oncogenes and tumor suppressor genes encode components of growth-stimulating and growth-inhibiting pathways, respectively

• The Ras protein, encoded by the ras gene, is a G protein that relays a signal from a growth factor receptor to a cascade of protein kinases

• Many ras oncogenes have a mutation that leads to a hyperactive Ras protein that issues signals on its own, resulting in excessive cell division

• The p53 gene,“guardian angel of the genome” encodes a tumor-suppressor protein that is a specific transcription factor that promotes synthesis of cell cycle–inhibiting proteins

• Mutations that knock out the p53 gene can lead to excessive cell growth and cancer

• Increased cell division, possibly leading to cancer, can result if the cell cycle is over stimulated or not inhibited when it normally would be

Protein overexpressed

EFFECTS OF MUTATIONS

Protein absent

Cell cycle notinhibited

Increased celldivision

Cell cycle overstimulate

Effects ofmutations

Activeformof p53

DNADNA damagein genome

UVlight

Protein kinasesMUTATION

Defective ormissingtranscriptionfactor, such as p53, cannotactivatetranscription

Protein kinases(phosphorylationcascade)

Cell cycle-inhibitingpathway

Cell cycle-stimulatingpathway

Protein thatinhibitsthe cell cycle

NUCLEUS

DNA

Gene expression

Transcriptionfactor (activator)

Receptor

G protein

Growthfactor

MUTATIONHyperactiveRas protein(product ofoncogene)issues signalson its own

Protein thatstimulatesthe cell cycle

Colon

Colon wall

Loss oftumor-suppressorgene APC (orother)

Normal colonepithelial cells

Small benigngrowth (polyp)

Larger benigngrowth (adenoma)

Activation ofras oncogene

Loss oftumor-suppressorgene DCC

Loss oftumor-suppressorgene p53

Additionalmutations

Malignant tumor(carcinoma)

Individuals who inherit a mutant oncogene or tumor-suppressor allele have an increased risk of developing certain types of cancer (Inherited Predisposition to Cancer)

Transposable Elements and Related Sequences

• The first evidence for wandering DNA segments came from geneticist Barbara McClintock’s breeding experiments with Indian corn

• McClintock identified changes in the color of corn kernels that made sense only by postulating that some genetic elements move from other genome locations into the genes for kernel color

Movement of Transposons and Retrotransposons

• Eukaryotic transposable elements are of two types:– Transposons, which move within a genome by

means of a DNA intermediate– Retrotransposons, which move by means of an RNA

intermediate

DNA of genomeTransposon

is copied

Mobile transposon

Transposon

Insertion

New copy oftransposon

Transposon movement (“copy-and-paste” mechanism)

Retrotransposon movement

DNA of genome

Insertion

RNA

Reversetranscriptase

RetrotransposonNew copy of

retrotransposon

Genes and Multigene Families• Most eukaryotic genes are present in one copy

per haploid set of chromosomes• The rest of the genome occurs in multigene

families, collections of identical or very similar genes

• Globin gene family clusters also include pseudogenes, nonfunctional nucleotide sequences that are similar to the functional genes

DNA

Non-transcribedspacer

RNA transcripts

Transcription unit

DNA18S 5.8S 28S

rRNA

18S

5.8S28S

Part of the ribosomal RNA gene family

Heme

Hemoglobin

-Globin

-Globin

-Globin gene family -Globin gene family

Chromosome 11Chromosome 16

1 12 A

Embryo Embryo Fetus AdultFetus

and adult

The human -globin and -globin gene families

• DNA technology has revolutionized biotechnology, the manipulation of organisms or their genetic components to make useful products

• An example of DNA technology is the microarray, a measurement of gene expression of thousands of different genes

DNA Cloning and Its Applications• To work directly with specific genes, scientists

prepare gene-sized pieces of DNA in identical copies, a process called gene cloning

• Most methods for cloning pieces of DNA in the laboratory share general features, such as the use of bacteria and their plasmids

• Cloned genes are useful for making copies of a particular gene and producing a gene product

Bacterium

Bacterialchromosome

Plasmid

Gene inserted intoplasmid

Cell containing geneof interest

Gene ofinterest DNA of

chromosome

RecombinantDNA (plasmid)

Plasmid put intobacterial cell

Recombinantbacterium

Host cell grown in cultureto form a clone of cellscontaining the “cloned”gene of interest

Protein expressedby gene of interest

Protein harvested

Gene ofinterest

Copies of gene

Basicresearchon gene

Basicresearchon protein

Basic research andvarious applications

Gene for pestresistance insertedinto plants

Gene used to alterbacteria for cleaningup toxic waste

Protein dissolvesblood clots in heartattack therapy

Human growth hor-mone treats stuntedgrowth

Using Restriction Enzymes to Make Recombinant DNA

• Bacterial restriction enzymes cut DNA molecules at DNA sequences called restriction sites usually makes many cuts, yielding restriction fragments

• The most useful restriction enzymes cut DNA in a staggered way, producing fragments with “sticky ends” that bond with complementary “sticky ends” of other fragments

• DNA ligase is an enzyme that seals the bonds between the “sticky ends” restriction fragments

Restriction site

DNA 53

35

Restriction enzyme cutsthe sugar-phosphatebackbones at each arrow.

One possible combination

DNA fragment from anothersource is added. Base pairingof sticky ends producesvarious combinations.

Fragment from differentDNA molecule cut by thesame restriction enzyme

DNA ligaseseals the strands.

Recombinant DNA molecule

Sticky end

Isolate plasmid DNAand human DNA.

Cut both DNA samples withthe same restriction enzyme.

Mix the DNAs; they join by base pairing.The products are recombinant plasmidsand many nonrecombinant plasmids.

Bacterial cell lacZ gene(lactosebreakdown)

Humancell

Restrictionsite

ampR gene(ampicillinresistance)

Bacterialplasmid Gene of

interest

Stickyends

Human DNAfragments

Recombinant DNA plasmids

Introduce the DNA into bacterial cellsthat have a mutation in their own lacZgene.

Recombinantbacteria

Plate the bacteria on agarcontaining ampicillin and X-gal.Incubate until colonies grow.

Colony carrying non-recombinant plasmidwith intact lacZ gene

Colony carryingrecombinantplasmid withdisrupted lacZ gene

Bacterialclone

Amplifying DNA in Vitro: The Polymerase Chain Reaction (PCR)

• The polymerase chain reaction, PCR, can produce many copies of a specific target segment of DNA

• A three-step cycle—heating, cooling, and replication—brings about a chain reaction that produces an exponentially growing population of identical DNA molecules

Genomic DNA

Targetsequence

5

3

3

5

5

3

3

5

Primers

Denaturation:Heat brieflyto separate DNAstrands

Annealing:Cool to allowprimers to formhydrogen bondswith ends oftarget sequence

Extension:DNA polymeraseadds nucleotides tothe 3 end of eachprimer

Cycle 1yields

2molecules

Newnucleo-

tides

Cycle 2yields

4molecules

Cycle 3yields 8

molecules;2 molecules

(in white boxes)match target

sequence

Gel Electrophoresis

• One indirect method of rapidly analyzing and comparing genomes is gel electrophoresis

• This technique uses a gel as a molecular sieve to separate nuclei acids or proteins by size

• In restriction fragment analysis, DNA fragments produced by restriction enzyme digestion of a DNA molecule are sorted by gel electrophoresis

• Restriction fragment analysis is useful for comparing two different DNA molecules, such as two alleles for a gene

Cathode

Powersource

Anode

Mixtureof DNAmoleculesof differ-ent sizes

Gel

Glassplates

Longermolecules

Shortermolecules

Normal -globin allele

175 bp 201 bp Large fragment

Sickle-cell mutant -globin allele

376 bp Large fragment

Ddel Ddel Ddel Ddel

Ddel Ddel Ddel

Ddel restriction sites in normal and sickle-cell alleles of-globin gene

Normalallele

Sickle-cellallele

Largefragment

376 bp201 bp175 bp

Electrophoresis of restriction fragments from normaland sickle-cell alleles

Medical Applications• One benefit of DNA technology is identification

of human genes in which mutation plays a role in genetic diseases

• Scientists can diagnose many human genetic disorders by using PCR and primers corresponding to cloned disease genes, then sequencing the amplified product to look for the disease-causing mutation

• Even when a disease gene has not been cloned, presence of an abnormal allele can be diagnosed if a closely linked RFLP marker has been found

DNARFLP marker

Disease-causingallele

Normal allele

Restrictionsites

Human Gene Therapy• Gene therapy is the alteration of an afflicted

individual’s genes• Gene therapy holds great potential for treating

disorders traceable to a single defective gene• Vectors are used for delivery of genes into cells• Gene therapy raises ethical questions, such as

whether human germ-line cells should be treated to correct the defect in future generations

Cloned gene

Retroviruscapsid

Bonemarrowcell frompatient

Inject engineeredcells into patient.

Insert RNA version of normal alleleinto retrovirus.

Viral RNA

Let retrovirus infect bone marrow cellsthat have been removed from thepatient and cultured.

Viral DNA carrying the normalallele inserts into chromosome.

Bonemarrow

Pharmaceutical Products• Some pharmaceutical applications of DNA

technology:– Large-scale production of human hormones and

other proteins with therapeutic uses– Production of safer vaccines

Forensic Evidence

• DNA “fingerprints” obtained by analysis of tissue or body fluids can provide evidence in criminal and paternity cases

• A DNA fingerprint is a specific pattern of bands of RFLP markers on a gel

• The probability that two people who are not identical twins have the same DNA fingerprint is very small

• Exact probability depends on the number of markers and their frequency in the population

Defendant’sblood (D)

Blood from defendant’sclothes

Victim’sblood (V)

Animations and Videos

• Processing of Gene Information• Control of Gene Expression in Eukaryotes• RNA Interface• Regulatory Proteins Regulation by Repression• Tryptophan Repressor• Lac Operon• Bozeman - Lac Operon• The Lac Operon in E. coli• Bozeman - Gene Regulation

http://highered.mcgraw-hill.com/olc/dl/120077/bio25.swf

http://highered.mcgraw-hill.com/olc/dl/120080/bio31.swf


• Bozeman - Restriction Enzyme• DNA Transformation – 1• DNA Transformation – 2• Conjugation: Transfer of Chromosomal DNA• Conjugation - Transfer of F Plasmid• Integration and Excision of a Plasmid• Transduction (Generalized)• Bacterial Transformation• Lamda Phage Replication Cycle


• Early Genetic Engineering Experiment• Genetic Engineering• Cloning• Steps in Cloning a Gene• Tutorial - Human Cloning• Human Cloning - Reproductive Cloning• Human Cloning - Therapeutic Cloning• Genetic Engineering to Produce Insulin• Polymerase Chain Reaction


• PCR – 1• PCR – 2• Gel Electrophoresis – 1• Gel Electrophoresis – 2• DNA Fingerprinting• Construction of a DNA Library• DNA Restriction Enzymes• Restriction Enzyme Digestion of DNA• Restriction Endonucleases


• Principles of Biotechnology• Applications of Biotechnology• Constructing Vaccines• DNA Probe (DNA Hybridization)• Restriction Length Ploymorphisms• cDNA• Southern Blot• Entry of Virus into Host Cell• Replication Cycle of a Retrovirus


• HIV Replication• Mechanism for Releasing Enveloped Viruses• Treatment of HIV• Prions Disease• How Prions Arise• Stem Cells• Embryonic Stem Cells• Human Embryonic Stem Cells• Human Stem Cells


• Microarray• Sanger Sequencing• DNA Fingerprinting• RNA Interface• miRNA• Dicer• DNA Microarrays• Bozeman - DNA Fingerprinting• Bozeman - Viral Replication


• Genes into Plants Using the Ti-plasmid• Highput Through Sequencing• Sequencing the Genome• Cycle Sequencing• Bozeman - Effects of Change in Pathways• Chapter Quiz Questions – 1• Chapter Quiz Questions – 2• Chapter Quiz Questions – 3• Chapter Quiz Questions – 4

What does the operon model attempt to explain?• the coordinated control of gene expression

in bacteria• bacterial resistance to antibiotics• how genes move between homologous

regions of DNA• the mechanism of viral attachment to a host

cell• horizontal transmission of plant viruses

When tryptophan (an amino acid) is present in the external medium, the bacterium brings in the tryptophan and does not need to make this amino acid. Which of the following is true when there is no tryptophan in the medium?

• The repressor is active and binds to the operator.

• The repressor is inactive, and RNA polymerase moves through the operator.

• The operator is bound, and mRNA is made.• Genes are inactive.• The corepressor binds to the repressor.

Each of a group of bacterial cells has a mutation in its lac operon. Which of the following will make it impossible for the cell to metabolize lactose?• mutation in lac (-galactosidase gene)• mutation in lac (cannot bind to operator)• mutation in operator (cannot bind to

repressor)• mutation in lac (cannot bind to inducer)

Each of a group of bacterial cells has a mutation in its lac operon. Which of the following will make it impossible for the cell to metabolize lactose?

• mutation in lac (-galactosidase gene)• mutation in lac (cannot bind to operator)• mutation in operator (cannot bind to

repressor)• mutation in lac (cannot bind to inducer)

Which element(s) from the following list constitute(s) a bacterial operon?

• repressor gene• promoter• inducer• repressor protein• all of the above

Which of the following statements about specific transcription factors is false?

• The binding of specific transcription factors to the control elements of enhancers influences the rate of gene expression.

• Specific transcription factors include activators and repressors.

• MyoD is one.• Some act indirectly by affecting chromatin structure.• Interaction of specific transcription factors and RNA

polymerase II with a promoter leads to a low rate of initiation and production of a few RNA transcripts.

Approximately what proportion of the DNA in the human genome codes for proteins or functional RNA?

• 83%• 46%• 32%• 13%• 1.5%

A specific gene is known to code for three different but related proteins. This could be due to which of the following?

• premature mRNA degradation• alternative RNA splicing• use of different enhancers• protein degradation• differential transport

RNA is cut up into small 22-nucleotide fragments to regulate another “target” mRNA. Which of the following is/are true?• The target mRNA is degraded, and its protein is

not made.• The RNA fragments enhance protein synthesis by

the mRNA.• The RNA fragments bind the ribosome to enhance

use of the mRNA and protein synthesis.• The target mRNA is blocked from being used in

translation.• The RNA fragments act on the ribosome to shut

down translation of all mRNAs.

RNA is cut up into small 22-nucleotide fragments to regulate another “target” mRNA. Which of the following is/are true?

• The target mRNA is degraded, and its protein is not made.

• The RNA fragments enhance protein synthesis by the mRNA.

• The RNA fragments bind the ribosome to enhance use of the mRNA and protein synthesis.

• The target mRNA is blocked from being used in translation.

• The RNA fragments act on the ribosome to shut down translation of all mRNAs.

Even though the two cells have numerous transcription factors and many are present in both cells, the lens cell makes the crystallin protein (not albumin), whereas the liver cell makes albumin (not crystallin). Which of the following explains this cell specificity?

• Specific transcription factors made in the cell determine which genes are expressed.

• At fertilization, specific cells are destined for certain functions.

• The activators needed for expression of the crystallin gene are present in all cells.

• The promoters are different for the different genes.

Differential gene expression (different genes turned on in different cells) leads to different tissues developing in the embryo. Which of the following is not a cause of differential gene expression?

• cytoplasmic determinants• induction• the environment around a particular cell• corepressor proteins

Initially, cytoplasmic determinants are localized in one part of a zygote and could be which of the following? (Choose more than one answer.)

• gene• mRNA• transcription factor• ribosome• myoblast

Initially, cytoplasmic determinants are localized in one part of a zygote and could be which of the following? (Choose more than one answer.)• gene• mRNA• transcription factor• ribosome• myoblast

Scientists showed that bicoid mRNA, and then its Bicoid protein, is normally found in highest concentrations in the fly’s anterior. What would happen if Bicoid were injected at the posterior end?

• Anterior structures would form at both ends.• Posterior structures would form at both ends.• The embryo would have no dorsal-ventral axis.• Bicoid mRNA wouldn’t be translated into protein.

Mutations in _______ genes caused the development of legs in the place of antennae.

• homeotic• embryonic lethal• myoD• Ras• wild-type

Wild type

Eye

Mutant

The shape of an organ, the number of brain cells in an embryonic brain, the removal of mutated cells, and the webbing cells between the toes of a human embryo are all regulated by which of the following?

• certain cells becoming much larger• certain cells shrinking• certain cells dying• formation of embryonic cells• concentration of Bicoid protein

Which of the following would not typically cause a proto-oncogene to become an oncogene?• gene suppression• translocation• amplification• point mutation• retroviral activation

Which of the following would not typically cause a proto-oncogene to become an oncogene?

• gene suppression• translocation• amplification• point mutation• retroviral activation

Which of the following statements about the APC gene is false?

• It is a tumor-suppressor gene.• It is mutated in 60% of colorectal cancers.• It regulates cell migration and adhesion.• It may be deleted in colon cancer.• Mutations in one allele are enough to lose the

gene’s function.

Which of the following statements about the APC gene is false?

• It is a tumor-suppressor gene.• It is mutated in 60% of colorectal cancers.• It regulates cell migration and adhesion.• It may be deleted in colon cancer.• Mutations in one allele are enough to lose

the gene’s function.

The diagrams on the next slide show an intact DNA sequence (top) and three experimental DNA sequences. A red X indicates the possible control element (1, 2, or 3) that was deleted in each experimental DNA sequence. The area between the slashes represents the approximately 8 kilobases of DNA located between the promoter and the enhancer region. The horizontal bar graph shows the amount of reporter gene mRNA that was present in each cell culture after 48 hours relative to the amount that was in the culture containing the intact enhancer region (top bar = 100%).

Scientific Skills Exercise

What was the independent variable in this experiment?

• the length of time that the cells were incubated

• the relative level of reporter gene mRNA• the distance between the promoter and the

enhancer• the possible control element that was deleted

What was the independent variable in this experiment?



enhancer• the possible control element that was

deleted

What was the dependent variable in this experiment?


• how many of the artificial DNA molecules were taken up by the cells


enhancer

What was the control treatment in this experiment?• the reporter gene• the construct that had no DNA deleted from

the enhancer• the temperature, pH, and salt concentration

of the incubation medium• the construct that resulted in the lowest

amount of reporter mRNA

Do the data suggest that any of these possible control elements are actual control elements?

• Only control elements 1 and 2 appear to be control elements.

• Only control element 3 appears to be a control element.

• All three appear to be control elements.• None of the possible control elements appear

to be actual control elements.

Do the data suggest that any of these possible control elements are actual control elements?• Only control elements 1 and 2 appear to be

control elements.• Only control element 3 appears to be a

control element.• All three appear to be control elements.• None of the possible control elements appear

to be actual control elements.

Did deletion of any of the possible control elements cause a reduction in reporter gene expression? How can you tell?• Deletion of element 3 caused a reduction in

reporter gene expression; that construct resulted in less than 50% of the control level of mRNA.

• Deletion of elements 2 and 3 caused a reduction in reporter gene expression; those constructs resulted in less than the highest level of mRNA.

• None of the deletions caused a reduction in reporter gene expression; all of them still resulted in reporter mRNA being made.

Did deletion of any of the possible control elements cause a reduction in reporter gene expression? How can you tell?

• Deletion of element 3 caused a reduction in reporter gene expression; that construct resulted in less than 50% of the control level of mRNA.

• Deletion of elements 2 and 3 caused a reduction in reporter gene expression; those constructs resulted in less than the highest level of mRNA.

• None of the deletions caused a reduction in reporter gene expression; all of them still resulted in reporter mRNA being made.

If deletion of a control element causes a reduction in gene expression, what must be the normal role of that control element?• To repress gene expression; without the control element,

repressors are not able to bind to the enhancer, and the level of gene expression decreases.

• To activate gene expression; without the control element, activators are not able to bind to the enhancer, and the level of gene expression decreases.

• To repress gene expression; without the control element, repressors are not able to bind to the enhancer, and the level of gene expression increases.

• To activate gene expression; without the control element, repressors are not able to bind to the enhancer, and the level of gene expression increases.

If deletion of a control element causes a reduction in gene expression, what must be the normal role of that control element?

• To repress gene expression; without the control element, repressors are not able to bind to the enhancer, and the level of gene expression decreases.



• To activate gene expression; without the control element, repressors are not able to bind to the enhancer, and the level of gene expression increases.

Did deletion of any of the possible control elements cause an increase in reporter gene expression? How can you tell?• Deletion of control element 1 or 2 caused an increase

in reporter gene expression; both constructs resulted in over 100% of the control level of mRNA.

• Deletion of control element 1 caused an increase in reporter gene expression; that construct resulted in the highest level of mRNA.

• Deletion of control element 3 caused an increase in reporter gene expression; that construct resulted in less reporter mRNA than the control.

• All of the deletions caused an increase in reporter gene expression; all of them still resulted in reporter mRNA being made.

If deletion of a control element causes an increase in gene expression, what must be the normal role of that control element?• To activate gene expression; without the control element,

repressors are not able to bind to the enhancer, and the level of gene expression increases.

• To repress gene expression; without the control element, activators are not able to bind to the enhancer, and the level of gene expression decreases.



Which of the following is a property of life shared by prokaryotic cells and eukaryotic cells, but not viruses?

• nucleic acids used to store hereditary information

• order and complexity in arrangement of biological molecules

• the ability to process energy through metabolic reactions

• the capacity to evolve

Which of the following is a property of life shared by prokaryotic cells and eukaryotic cells, but not viruses?

a) nucleic acids used to store hereditary information

b) order and complexity in arrangement of biological molecules

c) the ability to process energy through metabolic reactions

d) the capacity to evolve

Which of the following is characteristic of the lytic cycle?

• Viral DNA is incorporated into the host genome.

• The virus-host relationship usually lasts for generations.

• A large number of phages are released at a time.

• Many bacterial cells containing viral DNA are produced.

• The viral genome replicates without destroying the host.

Which of the following is characteristic of the lytic cycle?

• Viral DNA is incorporated into the host genome. • The virus-host relationship usually lasts for

generations.• A large number of phages are released at a

time. • Many bacterial cells containing viral DNA are

produced. • The viral genome replicates without destroying

the host.

What is the function of reverse transcriptase in retroviruses?

• It converts host cell RNA into viral DNA.

• It hydrolyzes the host cell's DNA.

• It uses viral RNA as a template for making complementary RNA strands.

• It translates viral RNA into proteins.

• It uses viral RNA as a template for DNA synthesis.

Why are viruses referred to as obligate parasites?

• They use the host cell to reproduce.

• Viral DNA always inserts itself into host DNA.

• They invariably kill any cell they infect.

• They can incorporate nucleic acids from other viruses.

• They must use enzymes encoded by the virus itself.

Which of the following molecules make up the viral envelope?

• viral glycoproteins

• capsid

• phospholipids from human host cell membrane

• membrane proteins from human host cell

• viral DNA

You have isolated viral particles from a patient, but you are not sure whether they are adenoviruses or influenza viruses. The presence of which class of biological molecules would allow you to distinguish between the two types of virus?

• RNA

• phospholipids

• proteins

• glycoproteins

• DNA

The HIV virus attacks only a certain type of white blood cells, and not other cell types. Why?

• HIV receptors are not found on the other cell types.

• Reverse transcriptase cannot transcribe RNA to DNA.

• Viral mRNA cannot be transcribed from the integrated provirus.

• Viruses cannot bud from the host cell.

Which is not an accepted theory about the evolution of viruses:

a) Viruses originated from naked bits of cellular nucleic acids.

b) Genes coding for capsid proteins allowed viruses to bind cell membranes.

c) Plasmids and transposons may have been the original sources of viral genomes.

d) Viruses are the descendents of precellular life forms.

AZT is a nucleoside analog used to treat HIV infections. It is a modified nucleoside. Which step does AZT hamper in the reproductive cycle of the HIV virus?

• entry into the cell

• synthesis of DNA from RNA catalyzed by reverse transcription

• transcription of RNA from proviral DNA

• viral assembly within the cell

Which of the following most likely describes the vertical transmission of a plant virus?• The plant shows symptoms of disease after being

grazed on by herbivores.

• Sap from one plant is rubbed on the leaves of a second plant; both plants eventually show disease symptoms.

• Seeds are planted and reared under protected conditions, but mature plants show disease symptoms.

• After a gardener prunes several plants with the same shears, they all show disease symptoms.

The photograph shows Rainbow and CC (CC is Rainbow’s clone). Why is CC’s coat pattern different from Rainbow’s given that CC and Rainbow are genetically identical?

• random X chromosome inactivation

• heterozygous at coat color gene locus

• environmental effects on gene expression

• all of the above

Which is an incorrect statement about STRs (Short Tandem repeats)?

• They are tandemly repeated units of 5- to 10 nucleotide sequences

• The number of repeats is polymorphic from person to person

• Two alleles of an STR may differ in an individual• They occur in specific regions of the genome• PCR is used to amplify particular STRs.

Which is an incorrect statement about STRs (Short Tandem repeats)?

• They are tandemly repeated units of 5- to 10 nucleotide sequences

• The number of repeats is polymorphic from person to person

• Two alleles of an STR may differ in an individual

• They occur in specific regions of the genome• PCR is used to amplify particular STRs.

Which of the following beneficial traits have not resulted from DNA technology and genetic engineering of crop plants?

• Delayed ripening • Resistance to drought• Resistance to herbicides• Resistance to salinity• Superweeds

• Delayed ripening • Resistance to drought• Resistance to herbicides• Resistance to salinity• Superweeds

Which of the following beneficial traits have not resulted from DNA technology and genetic engineering of crop plants?

Which of the following is not a correct statement about third generation sequencing?• A single DNA molecule is sequenced on its

own• Different bases interrupt an electric current

for a particular length of time a compound and an isotope; a molecule

• DNA moves through a small nanopore a molecule and a compound; a molecule

• DNA must be cut into fragments or amplified

• A single DNA molecule is sequenced on its own

• Different bases interrupt an electric current for a particular length of time a compound and an isotope; a molecule

• DNA moves through a small nanopore a molecule and a compound; a molecule

• DNA must be cut into fragments or amplified

Which of the following is not a correct statement about third generation sequencing?

Place the steps in a cycle of PCR (Polymerase Chain Reaction) in the correct order:

• 3-1-2• 3-2-1• 1-2-3• 2-3-1• 1-3-2• 2-1-3

1. Annealing—Cool to allow primers to form hydrogen bonds with ends of target sequence

2. Extension—DNA polymerase adds nucleotides to the 3 end of each primer

3. Denaturation—Heat briefly to separate DNA strands

Place the steps in a cycle of PCR (Polymerase Chain Reaction) in the correct order:

1. Annealing—Cool to allow primers to form hydrogen bonds with ends of target sequence

2. Extension—DNA polymerase adds nucleotides to the 3 end of each primer

3. Denaturation—Heat briefly to separate DNA strands

• 3-1-2• 3-2-1• 1-2-3• 2-3-1• 1-3-2• 2-1-3

Which of the following is an example of “recombinant DNA”?• combining alternate alleles of a gene in a

single cell • manipulating a meiotic crossing-over event• cloning genes from homologous pairs of

chromosomes• introducing a human gene into a bacterial

plasmid • alternate alleles assorting independently

• combining alternate alleles of a gene in a single cell

• manipulating a meiotic crossing-over event• cloning genes from homologous pairs of

chromosomes• introducing a human gene into a

bacterial plasmid • alternate alleles assorting independently

Which of the following is an example of “recombinant DNA”?

This segment of DNA is cut at restriction sites 1 and 2, which creates restriction fragments A, B, and C. Which of the following electrophoretic gels represents the separation of these fragments?

a)

b)

c)

d)

1) The top diagram depicts the very large regulatory region upstream of the Hoxd13 gene. The area between the slashes represents the DNA located between the promoter and the regulatory region.

2) The diagrams to the left of the bar graph show, first, the intact DNA and, next, the three altered DNA sequences. A red X indicates the segment (A, B, and/or C) that was deleted in each line of transgenic mice.

3) The horizontal bar graph shows the amount of Hoxd13 mRNA that was present in the digit-formation zone of each transgenic 12.5-day-old embryo paw relative to the amount that was in the digit-formation zone of a wild-type mouse that had the intact regulatory region (top bar = 100%). The paw images have blue stain visible where the Hoxd13 mRNA is located.

Scientific Skills Exercise

Which of the four treatments was the control for the experiment?

• the wild-type mouse C• the transgenic mouse with all three segments

deleted N• the transgenic mouse with segments B and C

deleted• the transgenic mouse with only segment C

deleted

• the wild-type mouse C• the transgenic mouse with all three segments

deleted N• the transgenic mouse with segments B and C

deleted• the transgenic mouse with only segment C

deleted

Which of the four treatments was the control for the experiment?

The hypothesis was that all three segments of the regulatory region are required for highest expression of the Hoxd13 gene. Is this hypothesis supported by the results?

• Yes; when any of the segments were deleted, the expression level dropped to less than 100% of the control.

• No; they did not delete the promoter, so the gene could still be expressed even without the segments.

• Yes; when all three segments were present, the expression level was at 100%.

• No; even when segments were deleted, the Hoxd13 gene was still being expressed.

What was the effect on the amount of Hoxd13 mRNA when segments B and C were both deleted? • Only about 60% of the control amount of

Hoxd13 mRNA was produced.• The deletion of segments B and C had no effect

on the amount of Hoxd13 mRNA produced.• Only about 35% of the control amount of

Hoxd13 mRNA was produced.• Only about 5% of the control amount of Hoxd13

mRNA was produced.

What was the effect on the amount of Hoxd13 mRNA when segments B and C were both deleted?• Only about 60% of the control amount of

Hoxd13 mRNA was produced.• The deletion of segments B and C had no effect

on the amount of Hoxd13 mRNA produced.• Only about 35% of the control amount of

Hoxd13 mRNA was produced.• Only about 5% of the control amount of Hoxd13

mRNA was produced.

Look at the blue stain in the in situ hybridization for the transgenic mouse lacking segments B and C. How would you describe the spatial pattern of gene expression in the embryo paw as compared to the control?

• There is very light blue stain in the center of each digit zone as compared to the control.

• The blue stain is generally lighter than in the control, but all four digit zones are still visible.

• There is almost no blue stain anywhere in the paw as compared to the control.

• There is no blue stain at the base of the paw as compared to the control.

What was the effect on the amount of Hoxd13 mRNA when just segment C was deleted?

• The deletion of segment C had no effect on the

amount of Hoxd13 mRNA produced.• Only about 60% of the control amount of

Hoxd13 mRNA was produced.• Only about 35% of the control amount of

Hoxd13 mRNA was produced.• Only about 5% of the control amount of

Hoxd13 mRNA was produced.

How would you describe the spatial pattern of gene expression in the embryo paw lacking segment C as compared to the control and to the paw lacking segments B and C?

• The digit zones are not visibly stained as they are in the control and the paw lacking B and C.

• The top of the paw is stained darker than both the control and the paw lacking B and C.

• The base of the paw is stained darker than both the control and the paw lacking B and C.

• The digit zones are defined with darker stain than both the control and the paw lacking B and C.

Suppose the researchers had only measured the amount of Hoxd13 mRNA and not done the in situ hybridizations. What important information about the role of the regulatory segments would have been missed?

• The interaction of the regulatory region with the promoter would have been missed.

• The interaction among the different segments of the regulatory region would have been missed.

• The mRNA would not have been blue; therefore it could not have been measured for the results shown in the bar graph.

• The spatial patterns of Hoxd13 gene expression in the paws would have been missed.

Suppose the researchers had only done the in situ hybridizations and not measured the amount of Hoxd13 mRNA. What important information would have been missed?

• The information about which regulatory segments were deleted would have been missed.

• The spatial patterns of Hoxd13 gene expression in the paws would have been missed.

• Qualitative data about Hoxd13 mRNA levels would have been missed.

• Quantitative data about Hoxd13 mRNA levels would have been missed.

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