Viruses called bacteriophages can infect and set in motion a genetic takeover of bacteria, such as Escherichia coli Bacteria are prokaryotes with cells.

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)

Viral genomes may consist of
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

Capsomere of capsid RNA 18  250 mm Tobacco mosaic virus 20 nm A capsid is the protein shell that encloses the viral genome and can have various structures

Capsomere DNA Glycoprotein 70–90 nm (diameter) 50 nm Adenoviruses

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

Membranous envelope Capsid RNA Glycoprotein 80–200 nm (diameter) 50 nm Influenza viruses

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

Head DNA Tail sheath Tail fiber 80  225 nm 50 nm Bacteriophage T4

Entry into cell and uncoating of DNA VIRUS DNA Capsid Replication Transcription HOST CELL Viral DNA mRNA Viral DNA Capsid proteins Self-assembly of new virus particles and 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 DNA and degradation of host DNA Phage assembly Release Head Tails Tail fibers Assembly Synthesis of viral genomes and proteins

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 DNA The phage attaches to a host cell and injects its DNA. Daughter cell with prophage Many cell divisions produce a large population of bacteria infected with the prophage. Phage DNA circularizes Phage Bacterial chromosome Occasionally, a prophage exits the bacterial chromosome, initiating a lytic cycle. Lytic cycle Lysogenic cycle Certain factors determine whether The bacterium reproduces normally, copying the prophage and transmitting it to daughter cells. The cell lyses, releasing phages. Lytic cycle is induced or Lysogenic cycle is entered Prophage New phage DNA and proteins are synthesized and assembled into phages. Phage DNA integrates into the bacterial chromosomes, becoming a prophage.

Reproductive Cycles of Animal Viruses
Two key variables in classifying viruses that infect animals: DNA or RNA? Single-stranded or double-stranded? Viral Envelopes 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

Capsid Capsid and viral genome enter cell RNA HOST CELL Envelope (with glycoproteins) Viral genome (RNA) Template mRNA ER Capsid proteins Glyco- proteins Copy of genome (RNA) New virus

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

Glycoprotein Viral envelope Capsid RNA (two identical strands) Reverse transcriptase

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 Reverse transcription Viral RNA RNA-DNA hybrid
Membrane of white blood cell HIV HOST CELL Reverse transcription Viral RNA RNA-DNA hybrid 0.25 µm HIV entering a cell DNA NUCLEUS Provirus Chromosomal DNA RNA genome for the next viral generation mRNA New HIV leaving a 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

Original prion Prion Many prions New prion Normal protein

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

Phage DNA A+ B+ A+ B+ Donor cell A+ Crossing over A+ A– B– Recipient cell 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

Mixture Mutant strain arg+ trp– Mutant strain arg– trp+ Mixture Mutant strain arg+ trp– Mutant strain arg– trp+ No colonies (control) No colonies (control) Colonies grew

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+ cell F+ cell Mating bridge F– cell F+ cell Bacterial chromosome 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 Temporary partial diploid Recombinant F– bacterium Conjugation and transfer of part of the bacterial chromosome from an Hfr 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 Transposition of Genetic Elements 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

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 5¢ 3¢ 3¢ 5¢ Inverted repeat Transposase gene Inverted repeat

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

Transposon Insertion sequence Antibiotic resistance gene Insertion sequence 5¢ 3¢ 3¢ 5¢ Inverted repeat Transposase gene

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

Polypeptides that make up enzymes for tryptophan synthesis
trp operon Promoter Promoter Genes of operon DNA trpR trpE trpD trpC trpB trpA Operator Regulatory gene RNA polymerase Start codon Stop codon 3¢ mRNA 5¢ mRNA 5¢ E D C B A Protein Inactive repressor Polypeptides that make up enzymes for tryptophan synthesis Tryptophan absent, repressor inactive, operon on

DNA mRNA Protein Active repressor Tryptophan (corepressor) Tryptophan present, repressor active, operon off

DNA No RNA made mRNA Protein Active repressor Tryptophan (corepressor) Tryptophan present, repressor active, operon off

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

Regulatory gene Promoter Operator DNA lacl lacZ No RNA made 3¢ mRNA RNA polymerase 5¢ Active repressor Protein Lactose absent, repressor active, operon off

Lactose present, repressor inactive, operon on
lac operon DNA lacl lacZ lacY lacA RNA polymerase 3¢ mRNA mRNA 5¢ 5¢ Permease Transacetylase Protein -Galactosidase Inactive repressor Allolactose (inducer) Lactose present, repressor inactive, operon on

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

Promoter DNA lacl lacZ RNA polymerase can bind and transcribe CAP-binding site Operator Active CAP cAMP Inactive lac repressor Inactive CAP Lactose present, glucose scarce (cAMP level high): abundant lac mRNA synthesized

Promoter DNA lacl lacZ CAP-binding site Operator RNA polymerase can’t bind Inactive CAP Inactive lac repressor Lactose present, glucose present (cAMP level low): little lac mRNA 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

2 nm DNA double helix His- tones Histone tails Histone H1 10 nm Linker DNA (“string”) Nucleosome (“bead”) Nucleosomes (10-nm fiber)

30 nm Nucleosome 30-nm fiber

Protein scaffold Loops 300 nm Scaffold Looped domains (300-nm fiber)

700 nm 1,400 nm Metaphase chromosome

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

Chemical modification Degradation of protein
Signal NUCLEUS Chromatin DNA Gene available for transcription Gene Transcription RNA Exon Primary transcript Intro RNA processing Tail Cap mRNA in nucleus Transport to cytoplasm CYTOPLASM mRNA in cytoplasm Degradation of mRNA Translation Polypeptide Cleavage Chemical modification Transport to cellular destination Active protein Degradation of 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

Histone tails DNA double helix Amino acids available for chemical modification Histone tails protrude outward from a nucleosome Unacetylated histones Acetylated histones Acetylation of histone tails promotes loose chromatin structure that permits transcription

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

(distal control elements)
Enhancer (distal control elements) Proximal control elements Poly-A signal sequence Termination region Exon Intron Exon Intron Exon DNA Upstream Downstream Promoter Transcription Poly-A signal Primary RNA transcript (pre-mRNA) Exon Intron Exon Intron Exon Cleaved 3¢ end of primary transcript 5¢ RNA processing: Cap and tail added; introns excised and exons spliced together Intron RNA Coding segment mRNA 3¢ Start codon Stop codon 5¢ Cap 5¢ UTR (untranslated region) 3¢ UTR (untranslated region) Poly-A tail

Distal control element Activators Promoter Gene DNA Enhancer TATA box General transcription factors DNA-bending protein Group of mediator proteins RNA polymerase II RNA polymerase II Transcription Initiation complex RNA synthesis

Some transcription factors function as repressors, inhibiting expression of a particular gene
Some activators and repressors act indirectly by influencing chromatin structure Liver cell nucleus Lens cell nucleus Available activators Available activators Enhancer Promoter Control elements Albumin gene Albumin gene not expressed Crystallin gene Albumin gene expressed Crystallin gene not expressed Crystallin gene expressed Liver cell Lens cell

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 Exons DNA Primary RNA transcript RNA splicing or mRNA

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)

Protein complex Degradation of mRNA Dicer OR miRNA Target mRNA Hydrogen bond 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

Proteasome and ubiquitin to be recycled Ubiquitin Proteasome Protein to be degraded Ubiquitinated protein Protein fragments (peptides) Protein entering a proteasome

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 Point mutation within a control element Translocation or transposition: gene moved to new locus, under new controls Point mutation within the gene Gene amplification: multiple copies of the gene New promoter Oncogene Oncogene Normal growth-stimulating protein in excess Normal growth-stimulating protein in excess Normal growth-stimulating protein in excess Hyperactive or degradation- 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

Cell cycle-stimulating pathway
Growth factor MUTATION Hyperactive Ras protein (product of oncogene) issues signals on its own G protein Cell cycle-stimulating pathway Receptor Protein kinases (phosphorylation cascade) NUCLEUS Transcription factor (activator) DNA Gene expression Protein that stimulates the cell cycle Cell cycle-inhibiting pathway Protein kinases MUTATION Defective or missing transcription factor, such as p53, cannot activate Active form of p53 UV light DNA damage in genome DNA Protein that inhibits the cell cycle Effects of mutations EFFECTS OF MUTATIONS Protein overexpressed Protein absent Cell cycle overstimulate Increased cell division Cell cycle not inhibited

Individuals who inherit a mutant oncogene or tumor-suppressor allele have an increased risk of developing certain types of cancer (Inherited Predisposition to Cancer) Colon Loss of tumor- suppressor gene APC (or other) Activation of ras oncogene Loss of tumor- suppressor gene p53 Colon wall Loss of tumor- suppressor gene DCC Additional mutations Normal colon epithelial cells Small benign growth (polyp) Larger benign growth (adenoma) Malignant tumor (carcinoma)

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

New copy of transposon Transposon DNA of genome Transposon is copied Insertion Mobile transposon Transposon movement (“copy-and-paste” mechanism) New copy of retrotransposon Retrotransposon DNA of genome RNA Insertion Reverse transcriptase Retrotransposon movement

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 RNA transcripts Non-transcribed spacer Transcription unit DNA 18S 5.8S 28S rRNA 5.8S 28S 18S Part of the ribosomal RNA gene family

Heme -Globin Hemoglobin a-Globin a-Globin gene family -Globin gene family Chromosome 16 Chromosome 11    1 2 1  A     Fetus and adult Embryo Embryo Fetus Adult The human a-globin and b-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 Cell containing gene of interest Gene inserted into plasmid Bacterial chromosome Plasmid Gene of interest Recombinant DNA (plasmid) DNA of chromosome Plasmid put into bacterial cell Recombinant bacterium Host cell grown in culture to form a clone of cells containing the “cloned” gene of interest Gene of interest Protein expressed by gene of interest Copies of gene Protein harvested Basic research and various applications Basic research on gene Basic research on protein Gene for pest resistance inserted into plants Gene used to alter bacteria for cleaning up toxic waste Protein dissolves blood clots in heart attack therapy Human growth hormone treats stunted growth

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

One possible combination Recombinant DNA molecule
Restriction site DNA 5¢ 3¢ 3¢ 5¢ Restriction enzyme cuts the sugar-phosphate backbones at each arrow. Sticky end DNA fragment from another source is added. Base pairing of sticky ends produces various combinations. Fragment from different DNA molecule cut by the same restriction enzyme One possible combination DNA ligase seals the strands. Recombinant DNA molecule

Human DNA fragments Bacterial cell lacZ gene (lactose breakdown) Human
Isolate plasmid DNA and human DNA. Restriction site ampR gene (ampicillin resistance) Bacterial plasmid Gene of interest Sticky ends Cut both DNA samples with the same restriction enzyme. Human DNA fragments Mix the DNAs; they join by base pairing. The products are recombinant plasmids and many nonrecombinant plasmids. Recombinant DNA plasmids Introduce the DNA into bacterial cells that have a mutation in their own lacZ gene. Recombinant bacteria Plate the bacteria on agar containing ampicillin and X-gal. Incubate until colonies grow. Colony carrying nonrecombinant plasmid with intact lacZ gene Colony carrying recombinant plasmid with disrupted lacZ gene Bacterial clone

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

5¢ 3¢ Target sequence Genomic DNA 3¢ 5¢ Denaturation: Heat briefly to separate DNA strands 5¢ 3¢ 3¢ 5¢ Annealing: Cool to allow primers to form hydrogen bonds with ends of target sequence Cycle 1 yields 2 molecules Primers Extension: DNA polymerase adds nucleotides to the 3¢ end of each primer New nucleotides Cycle 2 yields 4 molecules Cycle 3 yields 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

Mixture of DNA molecules of different sizes Longer molecules Cathode Shorter molecules Power source Gel Glass plates Anode

Normal b-globin allele
175 bp 201 bp Large fragment Ddel Ddel Ddel Ddel Sickle-cell mutant b-globin allele 376 bp Large fragment Ddel Ddel Ddel Ddel restriction sites in normal and sickle-cell alleles of -globin gene Normal allele Sickle-cell allele Large fragment 376 bp 201 bp 175 bp Electrophoresis of restriction fragments from normal and 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

RFLP marker DNA Disease-causing allele Restriction sites Normal allele

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 Insert RNA version of normal allele into retrovirus. Viral RNA Let retrovirus infect bone marrow cells that have been removed from the patient and cultured. Retrovirus capsid Viral DNA carrying the normal allele inserts into chromosome. Bone marrow cell from patient Bone marrow Inject engineered cells into patient.

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

Blood from defendant’s
blood (D) Blood from defendant’s clothes Victim’s blood (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

Animations and Videos 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

Animations and Videos 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

Animations and Videos 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

Animations and Videos 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

Animations and Videos 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

Animations and Videos Microarray Sanger Sequencing DNA Fingerprinting RNA Interface miRNA Dicer DNA Microarrays Bozeman - DNA Fingerprinting Bozeman - Viral Replication

Animations and Videos 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 Answer: A

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 112

The repressor is active and binds to the operator.
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. Answer: B 113

The repressor is active and binds to the operator.
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. 114

mutation in lac (-galactosidase gene)
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) Answer: A Figure 18.4 115

mutation in lac (-galactosidase gene)
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) 116

Which element(s) from the following list constitute(s) a bacterial operon?
repressor gene promoter inducer repressor protein all of the above Answer: B 117

Which element(s) from the following list constitute(s) a bacterial operon?
repressor gene promoter inducer repressor protein all of the above 118

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. Answer: E 119

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. 120

Approximately what proportion of the DNA in the human genome codes for proteins or functional RNA?
83% 46% 32% 13% 1.5% Answer: E Concept 18.3 121

Approximately what proportion of the DNA in the human genome codes for proteins or functional RNA?
83% 46% 32% 13% 1.5% 122

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 Answer: B 123

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 124

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. Answer: A or D Figure 18.14 125

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. 126

At fertilization, specific cells are destined for certain functions.
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. Answer: A See Figure 18.11, although different combinations of transcription factors are present. 127

At fertilization, specific cells are destined for certain functions.
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. Answer: A See Figure 18.11, although different combinations of transcription factors are present. 128

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 Answer: (1) Cytoplasmic determinants. Molecules that are localized within the zygote get put into different cells during cleavage of the embryo and regulate transcription, mRNA usage, or protein activity to cause differentiation of the embryonic cell. (2) Induction. A hormone (or inducer, signaling molecule) will be released by one cell of the embryo and bind to another cell to induce changes in the other cell.

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 Answer: B, C 131

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 132

Anterior structures would form at both ends.
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. Answer: A. Figure Anterior structures would form at both ends. 133

Anterior structures would form at both ends.
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. Answer: A. Figure Anterior structures would form at both ends. 134

Mutations in _______ genes caused the development of legs in the place of antennae.
homeotic embryonic lethal myoD Ras wild-type Wild type Eye Mutant Answer: A. homeotic Figure 18.20 135

Mutations in _______ genes caused the development of legs in the place of antennae.
homeotic embryonic lethal myoD Ras wild-type Wild type Eye Mutant Answer: A. homeotic Figure 18.20 136

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 Answer: C 137

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 Answer: C 138

Which of the following would not typically cause a proto-oncogene to become an oncogene?
gene suppression translocation amplification point mutation retroviral activation Answer: A 139

Which of the following would not typically cause a proto-oncogene to become an oncogene?
gene suppression translocation amplification point mutation retroviral activation 140

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. Answer: E. Figure 18.26 141

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. Answer: E. Figure 18.26 142

Scientific Skills Exercise
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%).

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 Answer: D

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 dependent variable in this experiment?
the length of time that the cells were incubated how many of the artificial DNA molecules were taken up by the cells the relative level of reporter gene mRNA the distance between the promoter and the enhancer Answer: C

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 Answer: B

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. Answer: C

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. Answer: A

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. Answer: B

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.

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. Answer: A

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. 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, activators are not able to bind to the enhancer, and the level of gene expression decreases. Answer: C

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

nucleic acids used to store hereditary information
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 Answer: C

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 162

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. Answer: C Concept 19.2 163

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. 164

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. Answer: E Concept 19.2 165

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. 166

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. Answer: A 167

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. 168

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 Answer: A, C, D 169

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 170

RNA phospholipids proteins glycoproteins 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 Answer: B Students will need to recognize that influenza viruses have an envelope, but adenoviruses do not. 171

RNA phospholipids proteins glycoproteins 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 172

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. Answer: A Concept This question examines the concept that viruses are able to recognize and attach to host cells via receptor proteins and recognition molecules on the plasma membrane. 173

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. 174

Which is not an accepted theory about the evolution of viruses:
Viruses originated from naked bits of cellular nucleic acids. Genes coding for capsid proteins allowed viruses to bind cell membranes. Plasmids and transposons may have been the original sources of viral genomes. Viruses are the descendents of precellular life forms. Answer: D

Which is not an accepted theory about the evolution of viruses:
Viruses originated from naked bits of cellular nucleic acids. Genes coding for capsid proteins allowed viruses to bind cell membranes. Plasmids and transposons may have been the original sources of viral genomes. Viruses are the descendents of precellular life forms.

AZT is a nucleoside analog used to treat HIV infections
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 Answer: B Concept Because AZT is a thymine nucleoside, it will specifically block reverse transcription from RNA to DNA. 177

AZT is a nucleoside analog used to treat HIV infections
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 178

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. Answer: C Concept Students will need to distinguish between horizontal transmission of disease and vertical transmission. 179

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. 180

random X chromosome inactivation heterozygous at coat color gene locus
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 Answer: D

random X chromosome inactivation heterozygous at coat color gene locus
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 182

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. Answer: A 183

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. 184

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 Answer: E

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

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 Answer: D 187

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 188

Place the steps in a cycle of PCR (Polymerase Chain Reaction) in the correct order:
Annealing—Cool to allow primers to form hydrogen bonds with ends of target sequence Extension—DNA polymerase adds nucleotides to the 3 end of each primer 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 Answer: A Figure 20.8 189

Place the steps in a cycle of PCR (Polymerase Chain Reaction) in the correct order:
Annealing—Cool to allow primers to form hydrogen bonds with ends of target sequence Extension—DNA polymerase adds nucleotides to the 3 end of each primer Denaturation—Heat briefly to separate DNA strands 3-1-2 3-2-1 1-2-3 2-3-1 1-3-2 2-1-3 190

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 Answer: D Concept Recombinant DNA is a DNA molecule formed when segments of DNA from two different sources—often different species—are combined in vitro (in a test tube). 191

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 192

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) Answer: A See Figure Negative DNA moves toward the positive pole; smaller DNA fragments move faster. 193

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) 194

Scientific Skills Exercise
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. Answer: B 195

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 Answer: A 197

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 198

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. Answer: A 199

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. 200

Only about 60% 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. Answer: C 201

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. 202

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. Answer: B 203

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. 205

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. Answer: B 206

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. 207

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. Answer: A 208

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. 209

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. Answer: D 210

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. 211

Qualitative data about Hoxd13 mRNA levels 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. Answer: D 212

Qualitative data about Hoxd13 mRNA levels 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. 213

Viruses called bacteriophages can infect and set in motion a genetic takeover of bacteria, such as Escherichia coli Bacteria are prokaryotes with cells.

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Viruses called bacteriophages can infect and set in motion a genetic takeover of bacteria, such as Escherichia coli Bacteria are prokaryotes with cells.

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Presentation on theme: "Viruses called bacteriophages can infect and set in motion a genetic takeover of bacteria, such as Escherichia coli Bacteria are prokaryotes with cells."— Presentation transcript:

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