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Molecular Exam 3 Study Guide

by: Kiara Lynch

Molecular Exam 3 Study Guide Bio 413

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Kiara Lynch
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Chapter 7: Control of Gene Expression Chapter 8: Manipulating Proteins, DNA, and RNA
Dr. Stefan Samulewicz
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This 25 page Study Guide was uploaded by Kiara Lynch on Wednesday April 6, 2016. The Study Guide belongs to Bio 413 at La Salle University taught by Dr. Stefan Samulewicz in Spring 2016. Since its upload, it has received 63 views. For similar materials see Molecular in Biology at La Salle University.


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Date Created: 04/06/16
Molecular Exam 3 Notes Molecular Week 7 Notes Chapter 7: Control of Gene Expression  Neuron vs. lymphocyte o Same genome but express different RNAs and proteins o Both have housekeeping genes o Neuron- nervous system; sends information; long branches, electrical signals o Lymphocyte- immune system; WBC, fights invaders, moves freely throughout body  Differentiated cells contain all of the genetic info necessary to direct formation of a complete organism o Nucleus of a skin cell from an adult frog transferred into an enucleated egg leads to the growth of a tadpole o In many plants, differentiated cells retain the ability to differentiate  Single cell  clone of progeny cells  entire plant o Able to grow a calf from an epithelial cell  Differentiated cell nucleus from adult cow introduced into an enucleated egg calf  Different calves produced from the same differentiated cell donor are genetically identical (clones)  Remove genetic info of a cell and replace it with an epithelial cell  Epithelial cells from adult cow placed next to unfertilized egg cell which has had its chromosomes removed  Electric pulse causes donor cell to fuse with enucleated egg cell  The reconstructed zygote undergoes cell division and forms and embryo  The embryo is placed in a “foster” mother cow  Calf is produced  Shortened lifespan and susceptible to disease  Telomere shortening  loss of info/proofreading  Aging process has already started- mature cells from an adult cow  Microarray o Shows that there is a difference in mRNA expression patterns among different human cancer cells  Expression varies among tumors o Tissue samples down, genes across o mRNA shows which genes are activated o mRNA  cDNA  label o Microchip- grid; each has a sample of a different gene (single stranded DNA) o Hybridization- probe binds to DNA on chip  2-D gels (electrophoresis) o Molecular weight vs isolectric points o Different proteins are expressed by human tissues o Isoelectric point- pH at which protein has no net charge  How to control transcription for proper functioning of a cell o In nucleus  Transcriptional control (DNA), RNA processing control (RNA transcript) o In cytosol  RNA transport and localization control (mRNA)  mRNa degradation and translation controls (mRNA)  Protein activity control (protein) – active or inactive  Post-translational modifications (phosphorylation, acetylation, ubiquitylation, etc.)  Proteins and DNA interact at the promoter region o Double helical structure of DNA produces two grooves o Weak interactions- need about 20 to bind protein to DNA o Transcription factors prefer to bind in major groove  GC (AT) is different from CG (TA) basepairs o Transcribe proper strand- able to tell the difference between GC and CG base pairs o Different base pairs can be recognized by their edges without need to open the double helix o 4 possible configurations  H bond acceptors/donors, methyl groups form pattern  DNA recognition code o If substrate approaches at the minor groove- can’t tell the difference between GC and CG (specific pattern in major groove) o Edge of each base pair at the major or minor groove has a distinctive pattern of H bond donors, acceptors, and methyl groups o Major groove- 4 bp configuration, unique pattern o Minor groove- similar pattern  Gene regulatory proteins recognize specific DNA sequences o Highly specific interactions o Binding of a gene regulatory protein to the major groove of DNA  Takes about 10-20 contacts to bind  Contacts involve different amino acids, each contributing to the strength of the protein-DNA interaction  AA side chains are projected out  bind with elements in the nucleotide bases  Helix turn helix motif- DNA binding motif/domain o C terminal alpha helix is the recognition helix o Sequence specific recognition of DNA o Fits into major groove- contacts edges of base pairs o N terminal  Structural component; helps position recognition helix o Bind as dimers  2 copies of the recognition helix are separated by 1 turn of the DNA helix  Lambda repressor and Cro proteins control bacteriophage lambda gene expression  Tryptophan repressor and CAP control expression of sets of E. Coli genes o Specific sequences are recognized by the bacteriophage lambda cro protein  Sequences are arranged symmetrically  Allows each half of the DNA site to be recognized in the same way by each protein monomer  Homeodomain motif o Bound to specific DNA sequence o Regulates development in most eukaryotes o Homeodomain- folded into 3 alpha helices  packed tightly together by hydrophobic interactions  Helix turn helix motif  Recognition helix forms important contacts with the major groove  Flexible arm- forms contacts with the minor groove  Zinc finger motif o Zn binds to His from alpha helix and Cys from beta sheets and maintains the structure that fits in the major groove o Antiparallel beta sheets o Alpha helix o 4 amino acids bind to zinc and hold 1 end of the helix to 1 end of the beta sheet  DNA binding by a Zn-finger protein o Recognizes DNA by using 3 zinc fingers arranged as direct repeats o 3 fingers have similar amino acid sequences and contact DNA in similar ways o Multiple domains linked in 1 polypeptide  Zinc finger motif dimer o Binds to specific DNA sequence o Beta sheet and alpha helix; also has other helices o Zn holds together the structure o Each Zn finger domain has 2 zinc atoms  One stabilizes the DNA recognition helix  One stabilizes a loop involved in dimer formation o Each zinc is coordinated by 4 appropriately spaced cysteine residues o 2 recognition helices are separated by 1 turn of DNA  Prokaryotic Met repressor protein of the Methionine gene o Met repressor regulates genes that encode enzymes that catalyze methionine synthesis o If methionine is abundant, it binds to the repressor which leads to conformational changes which causes it to bind to DNA and turn off synthesis o Met binds to repressor  dimerize  subunits come together  fit in major groove  2 strand beta sheet of repressor binds to major groove o Met repressor needs 5-adenosyl Met to bind to DNA  Leucine Zipper o Dimer bound to DNA; homeodimer o 2 alpha helical binding domains dimerize through alpha helical leucine zipper region  Inverted Y shape  Arms mediate DNA binding to major groove  Each arm binds to ½ of the symmetrical DNA o String of leucine- single alpha helix o 2 alpha helices wrap around each other and leave recognition portion open to fit into major groove o Gcn4 protein- regulates transcription in response to available amino acids in the environment o Heterodimerization of leucine zipper proteins can alter their binding specificity  Homodimers bind to symmetrical DNA sequences  2 monomers can bind to form a heterodimer which recognizes a hybrid DNA sequence (recognizes 2 sequences)  Helix loop helix (HLH) o Dimer bound to DNA o 2 alpha helices held together by loop o Two monomers held together in a 4 helix bundle o Each monomer contributes 2 alpha helices and is connected by a flox loop of protein o Specific DNA sequence binds to the projecting alpha helices in bundle  Inhibiting regulated by truncated HLH proteins o HLH motif- responsible for dimerization and DNA binding o Active HLH homodimer- recognizes symmetrical DNA sequence o Inactive HLH heterodimer- binds full-length HLH protein to truncated HLH protein that is unable to bind DNA tightly  Dimerization but missing recognition part of motif  If present in excess, the truncated protein molecule blocks the homodimerization of the full length HLH protein and prevents it from binding to DNA  Binds bull length subunits but cannot bind to DNA  Zn fingers o Bind to different DNA sequences o 6 zinc fingers  Same overall structure, but each binds a different sequence o Arg-Gua contacts are common but guanine can also be recognized by ser, his, and lys  Same amino acid can recognize more than one base o Sequence specific interactions between 6 different Zn fingers and rec. sequences  Adds complexity to control over genes  Gel mobility shift assay o Radioactively label promotor; mix with nuclear extract; TF bind to DNA at the right places; proportion it so every fragment of DNA only gets 1 TF bound; DNA with little protein o Run on gel  DNA migrate  radioactive band o Shifted mobility of DNA- a big transcription factor shifts mobility a lot o Antibody produced cell line mixed with radioactive DNA fragment (160 nuc of regulating DNA sequence from a gene encoding light chain of antibody) o Polyacrylamide gel electrophoresis then autoradiography o Free DNA fragments migrate rapidly to bottom of gel o Proteins bound to fragments are stopped – 6 bands  6 different sequence specific DNA bound proteins for one sequence  Column chromatography o Fractionate extract o Each fraction is mixed with radioactive DNA fragments and applied to 1 lane of a polyacrylamide gel o Affinity exclusion by salt concentration to isolate proteins o Isolate one protein  Proteins that bind DNA are separated (nonspecific/weak affinity for bulk DNA; ionic attractions)  Proteins washed off DNA with salt solution  DNA of a particular sequence is eluted with salt  End with protein that tightly binds to 1 specific sequence  DNA affinity chromatography o Beads with sequence of DNA that are potential binding sequences for protein  First put in a variety of sequences then salt wash  elutes many different DNA binding proteins  Then do a second column with 1 specific DNA sequence then salt wash  elutes specific DNA binding protein Molecular Week 8 Notes  Methods to determine DNA sequence recognized by a gene regulating protein o Mixed with short fragments of DNA o Double stranded DNA fragments that bind to regulating proteins are separated through gel mobility shifts o Separation of DNA-protein complexes from free DNA o DNA fragments removed from protein  more rounds o Nucleotide sequence can be determined and a consensus DNA recognition sequence can be made  DNA footprinting o Protein bound DNA radioactively labeled o DNA cleaved (random single stranded cuts) o Protein removed o DNA denatured and separated into two strands o Resultant fragments from labeled strand  Alignments o Look for highly conserved regions in a promoter (where transcription factors bind)  Chip assay to identify transcription factors o Chromatin immunoprecipitation o Crosslink between gene regulating protein and DNA  chop DNA  isolate fragments with antibody against the protein (precipitate)  reverse crosslinks and remove protein left with DNA sequence that protein binds to (amplify) o Allows identification of all the sites in a genome that a regulatory protein occupies o The identities for the precipitated, amplified DNA fragments can be determined by hybridizing the mixture of fragments to DNA microarrays  Gene regulatory circuit o The complete set of genes controlled by 3 key regulatory proteins in budding yeast, as deduced from the DNA sites where the regulatory proteins bind o Reg. proteins- Mata 1, Matα1, and Matα2  Specify the two different haploid mating types (analogous to male and female gametes) of this unicellular organism o 16 chromosomes in the yeast genome- show where various combinations of the 3 regulatory proteins bind o Proteins act in complexes to regulate genes o Determinations of complete transcriptional circuits show that transcriptional networks are not infinitely complex o Chip analysis and phylogenetic footprinting  Operons- prokaryotes o Operon- transcriptional unit where all of the genes responsible for 1 process are together regulated by 1 promoter o TATA box and consensus sequence- sigma factor binds with polymerase o Operator- binds factor that gets in the way and prevents sigma factor and polymerase from binding which turns off the genes o Clustered genes in E. coli that code for enzymes that manufacture tryptophan  5 genes of the Trp operon are transcribed as a single mRNA molecule which allows their expression to be controlled coordinately  Clusters of genes transcribed as a single mRNA molecule are common in bacteria; each cluster is called an operon  Activators and repressors o If the level of tryptophan inside the cell is low, RNA polymerase binds to the promoter and transcribes the 5 genes of the Trp operon o If the level of tryptophan is high, the trp repressor is activated to bind to the operator where it blocks the binding of RNa polymerase to the promoter o Whenever the level of intracellular Trp drops, the repressor releases its tryptophan and becomes inactive which allows the polymerase to begin transcribing the genes  Tryptophan repressor o The binding of Trp to the tryptophan repressor protein changes its conformation o This structural change enables the gene regulatory protein to bind tightly to a specific DNA sequence (the operator), which blocks transcription of the genes encoding the enzymes required to produce tryptophan (the Trp operon) o Helix turn helix protein structure o Tryptophan binding increases the distance between the two recognition helices in the homodimer, allowing the repressor to fit snugly on the operator o Noncovalent bonds b/w trp and repressor o Trp binds  conformational change  recognition helix moves and can bind to major groove  Mechanism of gene regulatory proteins controlling gene transcription (prokaryotes) o Negative regulation o Positive regulation o Addition of a ligand can turn on a gene either by removing a gene repressor protein from the DNA or by causing a gene activator protein to bind o The addition of an inhibitory ligand can turn off a gene either by removing a gene activator protein from the DNA or by causing a gene repressor protein to bind o Activator protein helps attract RNA polymerase  Lambda repressor acting as an activator or repressor o Some bacterial gene regulatory proteins can act as either a transcriptional activator or a repressor, depending on the precise placement of their DNA-binding sites o Operator location- need correct distance from other sequence to let RNA polymerase bind o Ligand independent o Transcription is activated or repressed by lambda repressor (depends on conformation)  Dual control of the Lac operon o No cAMP then CAP can’t bind  Repressor not bound  No lactose o Glucose present  Prevents cAMP and CAP binding  Repressor binds in absence of ligand o No lactose  repressor bound No glucose  increase cAMP and CAP o No glucose  no repressor Lactose  CAP bound, increase cAMP o Glucose and lactose levels control the initiation of transcription of the Lac operon through their effects on CAP and the Lac repressor protein o LacZ, the first gene of the Lac operon, encodes the enzyme beta- galactosidase, which breaks down the disaccharide lactose to galactose and glucose o Lactose addition increases the concentration of allolactose, an isomer of lactose, which binds to the repressor protein and removes it from the DNA o Glucose addition decreases the concentration of cyclic AMP  Because cyclic AMP no longer binds to CAP, this gene activator protein dissociates from the DNA, turning off the operon o Several Lac repressor binding sites located at different positions along the DNA o The expression of the Lac operon never completely shuts down  A small amount of the enzyme beta-galactosidase is required to convert lactose to allolactose, thereby permitting the Lac repressor to be inactivated when lactose is added to the growth medium  DNA looping can stabilize protein-DNA interactions o Lac repressor is a tetramer- has multiple binding sites- can bind to 2 operators simultaneously  enhances repressor ability o The Lac operon has a total of three operators o At the concentrations of Lac repressor in the cell, and in the absence of lactose, the state in which 2 operators are bound is the most stable  To dissociate completely from the DNA, the Lac repressor must first pass through an intermediate where it is bound to only a single operator  The local concentration of the repressor is then very high in relation to the free operator, and the reaction to the double- bound form is favored over the dissociation reaction  Then even a low-affinity site can increase the occupancy of a high-affinity site and give higher levels of gene repression in the cell  Binding of two proteins to separate sites on the DNA double helix can greatly increase their probability of interaction o There is an optimal distance for DNA looping (about 500 bp) o Increases frequency of collision  Gene activation at a distance o Enhancers present hundreds of bp away- can loop around and encourage transcription o NtrC is a bacterial gene regulatory protein that activates transcription by directly contacting RNA polymerase and causing a transition between the initial DNA-bound form of the polymerase and the transcriptionally competent form o The transition stimulated by NtrC requires energy from ATP hydrolysis o Interaction of NtrC and RNA polymerase with the intervening DNA looped out  Interchangeable RNA polymerase subunits as a strategy to control gene expression in a bacterial virus o SPO1 uses a bacterial polymerase to transcribe its early genes immediately after the viral DNA enters the cell o 28 early gene encodes a sigmalike factor that binds to RNA polymerase and displaces the bacterial sigma factor  Initiates transcription of the SPO1 middle genes o Middle gene 34 displaces the 28 and directs RNA polymerase to transcribe the late genes  Produces the proteins that package the virus chromosome into a virus coat and lyse the cell  Sets of virus genes are expressed in the order in which they are needed  rapid and efficient viral replication  Gene control region- eukaryotes o Gene transcription factors and other regulating proteins can be adjacent to promoter or at a great distance upstream (can also be downstream in introns) o Hetero and dimeric proteins o The promoter- DNA sequence where TF and polymerase assemble o Regulatory sequences serve as binding sites for gene regulatory proteins, whose presence on the DNA affects the rate of transcription initiation o DNA looping allows gene regulatory proteins bound at any of these positions to interact with the proteins that assemble at the promoter o Many gene regulatory proteins act through a mediator while others influence the general transcription factors and RNA polymerase directly o Many gene regulatory proteins also influence the chromatin structure of the DNA control region which affects transcription initiation indirectly o The mediator and general transcription factors are the same for all polymerase II transcribed genes but the gene regulatory proteins and the locations of their binding sites relative to the promoter differ for each gene  The modular structure of a gene activator protein o Independent DNA-binding and transcription-activating domains in the yeast gene activator protein Gal4 o A functional activator can be reconstituted from the C-terminal portion of the yeast Gal4 protein if it is attached to the DNA-binding domain of a bacterial gene regulatory protein (LexA) by genetic engineering techniques o Resulting bacterial-yeast hybrid produced in yeast cells  activates transcription from yeast genes provided that the specific DNA-binding site for the bacterial protein has been inserted next to them o Gal4- responsible for activating the transcription of yeast genes that code for the enzymes that convert galactose to glucose o Chimeric gene regulatory protein requires a LexA recognition sequence to activate transcription o The control gene for a gene controlled by LexA was fused to the E. coli LacZ gene which codes for the enzyme beta-galactosidase (monitor expression level specified by a gene control region) o LacZ is a reporter gene- reports the activity of a gene control region o Phosphorylated galactose  activate gene with Gal4 TF regulating protein  2 domains- binds DNA consensus sequence and transcription activator domain o LacZ- gene construct under control of promoter that initially contained Gal 4  took off Gal4 DNA binding domain and put on LexA o Correct TAD but incorrect binding domain  no transcription  Fits into major groove but can’t H bond or interact o Need TAD and DNA binding domain for transcription  4 ways eukaryotic activator proteins can direct local alterations in chromatin structure to stimulate transcription initiation o Compact DNA- genes are tucked away o Activator protein with access to DNA close to gene  Can bind chromatin remodeling complexes, unravel, remove histones, or rearrange  expose gene  modify histone tails  unravel o Acetylation of histones makes it easier for histone chaperones to remove them from nucleosomes o Nucleosome remodeling and histone removal favor transcription initiation by increasing the accessibility of DNA and thereby facilitating the binding of Mediator, RNA polymerase, and the general TF and activator proteins o Transcription initiation and the formation of a compact chromatin structure can be regarded as competing biochemical assembly reactions and enzymes that increase the accessibility of DNA in chromatin will tend to favor transcription initiation  Writing and reading histone code during transcription initiation o Gene regulating protein binds to exposed DNA o Specific pattern triggered by 1 activator protein  3 modifications  exposes gene o Gene activator protein binds to DNA packaged into chromatin  attracts a histone acetyl transferase to acetylate lysine  histone kinase phosphorylates a serine  signals histone acetyl transferase rxn  histone code for transcription initiation is written o Writing is sequential- each histone modification depends on a prior modification o Final reading- general TF TFIID and chromatin remodeling complex bind  recognize acetylated histone tails through a bromodomain  Transcriptional synergy- activator proteins o Synergy- additive effect of multiple activators; multiple proteins cause huge increases in transcription o Observed b/w different gene activator proteins from the same organism and b/w activator proteins from different eukaryotic species when they are experimentally introduced into the same cell o High degree of conservation responsible for eukaryotic transcription initiation  6 ways in which eukaryotic gene repressor proteins can operate o Get in the way  Whichever substrate gets there first is expressed at a higher level  Competitive DNA binding o Eliminate activity of TAD  Both activator and repressor bind to DNA then bind to each other  Activator can’t do its job and bind to other proteins (chromatin remodeling complex/histone modification complex)  Masking the activation surface o Interfere with general TF  Both activator and repressor bind  Repressor prevents TF from coming together  Direct interaction with the general TF o Bind chromatin remodeling complex  Becomes more densely packed  not accessible  Growth arrest o Modify histone tails  Keep chromatin dense  Histone deacetylases o Modifications package and hide genes  Histone methyl transferase  histone methylation  proteins bind to methylated histones and hide genes  Coactivators/repressors o Bind to proteins that bind DNA- assemble on other DNA bound gene regulatory proteins o Given TF can fall into either category (on/off) depending on context it is used o Some have no other function than to attract others that do have activity  Fruit flies o Fertilized egg o Syncytium- 1 cell with many nuclei; nucleus divides o Nuclei migrate to sides of cell and cell boundaries start to form  Membrane engulfs individual cells around cytoplasm in middle o Cellular blastoderm  Cells around membrane begin to differentiate based on patterning o Anterior- nervous system and hoad o Segmented body  Extraembryonic membrane  Dorsal epidermis  Nervous system and ventral epidermis  Posterior part of digestive tract  Mesoderm  Anterior part of digestive tract o Follicle- female  Many signals and gradients  Single cell- 4 divisions  16 cells  1 oocyte and 15 nurse cells (prepare oocyte for journey)  When dividing  no time for transcription  Nurse cells synthesize mRNAs and proteins and pump into cell so when blastoderm forms, nuclei can tell where they are relative to each other  Follicle cells- inserting into receptors and synthesizing molecules to bind to receptors to send signals into oocytes o Provide terminal signal or dorsal/ventral signal  Morphogens- substance who’s concentration is read by cells to discover their position relative to a certain landmark or beacon  Dorsal and ventral sides send different signals  Nuclei are essentially equal but still differentiated o Dorsal and Ventral  Once dorsal is phosphorylated it releases cactus and can’t fit into nucleus to act as TF  Toll- receptor that accepts signal  2 proteins  Cactus- inhibitor of dorsal by binding to it  Dorsal- morphogen transcription factor (activator and repressor)  Twist- gene, TF; in activation of ventral side  Decapentaplegic (DPF)- dorsal represses this o Front and back  Gradient formed of signaling molecules o Anterior and posterior patterning  3 levels of delineation  1 - GAP- cut embryo in half  2 - pair rule- segments rd  3 - segment polarity (front and back of genes) o Focus on even-skipped gene (Eve gene)  Nonuniform distribution of four gene regulatory proteins in an early drosophila embryo- bicold, giant, hunchback, Kruppel (4 morphogens in oocyte)  Embryo is a syncytium with multiple nuclei in a common cytoplasm  Proteins are concentrated on the nuclei  Heterogeneous distributions of morphogens  Two and a half hours after fertilization, the egg was fixed and stained with antibodies that recognize the Eve protein and antibodies that recognize the giant protein  Yellow- Eve and Giant proteins both present  Egg contains about 4000 nuclei  Proteins in nuclei- stripes are about 4 nuclei wide o Isolated even-skipped genes  Module- gene construct of Eve gene regulatory region  Piece of Eve regulatory region removed and inserted upstream of a test promoter that directs the synthesis of the enzyme beta- galactosidase  When this artificial construct was reintroduced into the genome of Drosophila embryos, the embryos expressed beta- galactosidase in the position of the second Eve stripe o Close up view of the Eve stripe 2 unit  Regulatory sequences that bind one or another of four gene regulatory proteins  4 regulatory sequences are responsible for the proper expression of Eve in stripe 2  Flies that are deficient in the two gene activators fail to express efficiently Eve in stripe 2  DNA binding sites for these gene regulatory proteins were determined by cloning the genes encoding the proteins, overexpressing the proteins in E. coli, purifying them, and performing DNA-footprinting experiments  In some cases, the binding sites for the gene regulatory proteins overlap and the proteins can compete for binding to the DNA o Distribution of the gene regulatory proteins responsible for ensuring that Eve is expressed in stripe 2  Antibodies against the 4 proteins  The expression of Eve in stripe 2 occurs only at the position where the two activators (Bicoid and Hunchback) are present and two repressors are absent (Giant and Kruppel)  Lack of Kruppel expands stripe 2 posteriorly  If binding sites for Kruppel in stripe two module are inactivated by mutation then stripe 2 also expands posteriorly  Eve gene encodes a gene regulatory protein which after its pattern of expression is set up in seven stripes, regulates the expression of other Drosophila genes  Nuclei are exposed directly to positional cues in the form of concentrations of gene regulatory proteins  Integration of multiple inputs at a promoter o Transcription factor  Activator or repressor  Act by themselves or with other subunits  Bind in different ways  Activate strongly or weakly o Multiple sets of gene regulatory proteins can work together to influence transcription initiation at a promoter as they do in the Eve stripe 2 module o Final transcriptional activity of the gene most likely results from a competition b/w activators and repressors Molecular Week 9 Notes  Map of a plasmid vector o 1 plasmid per bacterial cell, 1 insert per plasmid o ColE1 origin  Origin of replication  Afl III 1153 o Pvu II 977 o MCS  Multiple cloning site; in the middle of LacZ  Sac I 759 and Kpn I 657 o lacZ  Reporter gene- makes blue product  Blue- no insert; white- insert  Selection that is inserted  Pvu I 503 and Pvu II 532 o f1 (+) origin f1 (-) origin  Nae I 333, Ssp I 22 Nae I 134, Ssp I 445 o Ssp I 2859 o Amp  Antibiotic resistance gene- selection method  Xmn I 2645, Sca I 2526, Pvu I 2416 Chapter 8: Manipulating Proteins, DNA, and RNA  C57 black mouse, MRL mouse o Expressed lupus- autoimmune disease  Light micrographs of cells in culture o Mouse fibroblasts o Chick myoblasts- fuse to form multinucleate muscle cells o + cheaper to grow cells; faster; no IAUCUC; 1 cell type shows consistency o – not a good in vivo model b/c it doesn’t show interaction of full system  Examination with light microscope o Movement of microtome arm o Specimen embedded in wax or resin o Steel blade slice into ribbon of sections  put on glass slide, stained and mounted under coverslip  viewed under microscope o Formaldehyde- preserves tissue architecture by forming chemical bonds (crosslinks)  H&E stain o Hematoxylin  Dark blue  Binds to DNA/RNA in nucleus o Eosin  Binds to cytoplasm  Denser cytoplasm is more red  Preparation of hybidomas that secrete monoclonal antibodies against a particular antigen o The selective growth medium used after the cell fusion step contains an inhibitor that blocks the normal biosynthetic pathways by which nucleotides are made o The cells then use a bypass pathway to synthesize their nucleic acids o This pathway is defective in the mutant cell line derived from the B cell tumor, but is intact in the normal cells obtained from the immunized mouse o Because neither cell type used for the initial fusion can survive and proliferate on its own, only the hybridoma cells do so o Mouse immunized with antigen X  makes anti-X antibody  B lymphocytes die after a few days in culture o Mutant cell line derived from a tumor of B lymphocytes  Cells grow indefinitely in normal medium but die in selective medium o Fusion  Resulting hybridoma cells cultured in multiple wells  Only hybidoma cells survive and peripherate in the selective medium  Supernatant tested for anti-X antibody and cells from positive well cultured at -1 cell per well  Cells allowed to multiply and individual supernatants tested for anti-X antibodies  Positive clones provide a continuing source of anti-X antibody  Immunohistochemistry o Using antibody to specifically identify protein within tissue section o Dark spots are positive for antibody  Fluorescently tag antibody o First barrier filter- lets through only blue light o Beam splitting mirror- reflects light below 510 nm and transmits light above 510 o Second barrier filter- cuts out unwanted fluorescent signals, passing the specific green light  Bromodexyuridine (BrdU) o Acts like a T residue o Can show which cells recently divided  Typical mouse ear o No wound, control  Can see recently divided skin cells o Mouse ear that was cut  In site hybridization o Want to look for RNA molecules o Single stranded DNA probes for gene A o Mixture of single stranded DNA molecules o Hybridization in formamide  Only A forms stable double helix  A, C, and E all form stable double helices; imperfect base pairing  Laser capture microdissection o Want to look for proteins and RNA within this tissue o Stand, slide, tissue o Can shoot laser that will melt part of cap and will infuse into tissue section o Ependorph tube cap with special coating  X-ray crystallography o A narrow parallel beam of x-rays is directed at a well-ordered crystal o Crystal of ribulose bisphosphate carboxylase (enz with role in CO2 fixation during photosynthesis) o The atoms in the crystal scatter some of the beam and the scattered waves reinforce one another at certain points and appear as a pattern of diffraction spots o Diffraction pattern and amino acid sequence can be used to produce an atomic model (structural features of protein)  NMR spectroscopy o 2D NMR spectrum derived from C-terminal domain of cellulose o Spots represent interactions between hydrogen atoms that are near neighbors in the protein and hence reflect the distance that separates them o Enables possible compatible structures to be derived o Multiple structures of the enzyme that satisfy the distance constraints equally well  gives a good indication of the probable 3D structure  Restriction enzymes o Blunt end cut- result is completely double stranded o Cutting palindromic sequences  Symmetry axis o Sticky ends- overhang of single strands  Allows to create recombinant  Forming recombinant molecule o Annealing o Giant chromosome o Chop up with restriction enzyme o Ligate into plasmid vectors using sticky ends to join fragments o Fragments with the same cohesive ends can readily join by complementary base-pairing between their cohesive ends  Cloning o Circular double stranded plasmid DNA (cloning vector) o Cleavage with restriction nuclease o DNA fragment to be cloned  covalent linkage by DNA ligase o Recombinant DNA o Insertion of a DNA fragment into a bacterial plasmid with the enzyme DNA ligase  Plasmid cut open with restriction nuclease, mixed with DNA fragment, DNA ligase and ATP added, cohesive ends base-pair, and DNA ligase seals the nicks in the DNA backbone  recombinant DNA molecule  Amplification of the DNA fragments inserted into a plasmid o Small bit of DNA  clone  amplify  extract plasmid  many copies o Bacterial cell  double stranded recombinant plasmid DNA introduced  cell culture produces new bacteria  many copies of purified plasmid isolated from lysed bacterial cells o Grown in culture, makes copy every time it divides- amplifies that fragment  Library of genetic info stored for future use o Two types- source of material and first couple of steps differ but the end result is similar (1 clone)  Genomic o cDNA Library is usually stored as a set of bacteria, each bacterium carrying a different fragment of human DNA  Genomic o Collection of entire genomic DNA o Chop up insert fragment into plasmid  1 fragment per plasmid o Leftover plasmids close on themselves- most do not have an insert  Need blue/white selection o Use DNA ligase to attach linkers o Restriction site cut in random places with restriction enzymes  Creates sticky ends  Put in plasmid transfection  Grow  Clones isolated  put in library o Genomic library  Chromosomal DNA  restriction nuclease digestion  DNA cloning  Introns and untranslated regions still present  2 genes- 3 clones contain each gene- 1 complete copy of each  add together  Transfection o Get 1 plasmid per bacterial cell o Separate clones  Agar plate with Amp  Incubate O/N  Colonies form  Bacterial cells that took up plasmid live  Plaque of different clones  Take each plaque that contains a unique fragment of genome  Clones into library o Recombinant DNA molecules  intro of plasmids into bacteria  genomic DNA library  cDNA library o Genes for 1 specific tissue that are expressed at that moment- not whole genome o Into tissue type o Extract mRNA (has poly A tail) o Reverse transcriptase  mRNA  DNA  need primer with free 3’ OH  oiligo-dt = T’s linked together  hybridize to poly A tail o copy mRNA template cDNA o RNaseH- digests hybridized double stranded DNA, degrades original template o cDNA library  Chromosomal DNA  transcription  RNA splicing  reverse transcription and DNA cloning  clones into library  Only looking at exons – spliced before mRNA formation  Multiple copies of a gene – different expression levels (activity)  Visualizing DNA o Acrylamide gel  Higher resolution, can tell a 1 bp difference in length o Agarose gel  For RNA and DNA  Can only tell hundreds to thousands of base pair differences  If you don’t need the exact length difference use this gel  Thousands of DNA molecules in each band  Ethidium bromide stain  Labeling DNA and RNA molecules o Create radioactive or nonradioactive probes o Heat- melt and separate two strands o Mix in random hexanucleotides o Hybridize to sequence in template o Add DNA polymerase and building blocks o Free 3’ OHs create another strand  Incorporates a labeled nucleotide  Normal nucleotide  attach something that can be detected by an antibody  acts like a normal nucleotide but is visible with an antibody o 1 double strand cold unlabeled fragment o 2 strands- 1 hot, 1 cold  melt  probes to search for a given sequence o Take plasmids cut with res. Enzymes  2 bands on gel for plasmid and insert  Detection of specific RNA or DNA molecules by gel transfer o RNA- is this gene being expressed? If so, at what level? o DNA- are there any easily recognizable mutations within this DNA? o Agarose gel, electrophoresis  Nucleic acids separated by size  Extract RNA and run  Extract DNA and cut with restriction enzyme and then run o Separated nucleic acids blotted onto nitrocellulose paper by suction of buffer through gel and paper  Layers- sponge in buffer, gel, membrane, paper towels, weight  Moisture wicks up into paper towels  As buffer moves up it draws DNA out of gel and sticks to membrane o Remove nitrocellular paper with tightly bound nucleic acids o Labeled probe hybridized to separated DNA o Labeled probe hybridized to complementary DNA bands visualized by autoradiography  Antibody/probe wash over membrane  looks for base pairs; single stranded o RNA  Is it there? (Is a band present or absent?)  The bigger the band the more expression o DNA  Extract with normal brain and tumor brain and cut with same restriction enzyme  Use probe  Can see the sequence mutation that causes the tumor o Types of blots  Blot with protein  western blot  Blot with RNA  northern blot  Blot with DNA  southern blot Molecular Week 10 Notes  Process of isolating colonies- Plating o Spread out colonies of bacteria containing recombinant plasmids o Nitrocellulase paper on top- treated paper o O/N incubation o Remove paper- some bacteria will stick o Lyse bacteria, denature DNA o Plaque formation- each plaque is a colony of 1 clone o Incubate with probe and wash  Radioactively labeled DNA probe  Labeled nucleotide- attach to complementary sequence  Or antibody- bind to protein  Gives colonies containing plasmid of interest o Expose paper to photographic film o Colonies with at least part of genome o Expression vector- produces protein associated with the protein of interest o Libraries contain many clones in vials in a freezer; plate clones to access them  Dideoxy chain termination sequencing technique - find overall sequence of gene o Codons for amino acids o Look for regions that show a low level of degeneracy o Synthesize all 16 possibilities labeled probe  screen library and isolate the complete clone and use it to get complete, correct nucleotide sequence o Dideoxynucleotide  Missing 3’ OH  New nucleotide can no longer be added- chain termination o Melt  single strand  add primer  allow polymerase to start synthesizing new strand  randomly introduce different dideoxynucleotides  chain termination  different length sequences  run 4 tubes of DNA on gel  separate by size  radioactive C’s are most active o Acrylamide with radioactive samples  Bands separated by 1 nucleotide in length o Label each dideoxynucleotide individually  Different color fluorescent tag  seen differently by spec  1 tube and run in 1 lane  Spec at bottom of gel  light up bands as they pass by  able to detect which nucleotide passed by  Steps to sequence a genome o Extract DNA o Restriction enzyme to cut  make library o Pick clones o Grow o Sequence o Repeat with different restriction enzyme to find overlap  put in order  Genomic library o Only about 1.5% of human genome is coding info o 3 stop codons out of 64 codons  1 is randomly introduced about every 20 codons o Open reading frame- no stop codon  sequence codes for protein  PCR- polymerase chain reaction o Amplify particular segments of DNA o 1. Melting o 2. Annealing – bind to sequence and create free 3’ end o 3. Elongation – Taq polymerase adds nucleotides in correct order creating second sequence o Primers  Clone directly- TA cloning  Anneal to plasmid st o 1 cycle  Double stranded chromosomal DNA  separate DNA strands and add primer  Produces two double stranded DNA molecules nd o 2 cycle  Separate DNA strands and anneal primer  DNA synthesis  Produces 4 double stranded DNA molecules o 3 cycle  Produces 8 double stranded DNA molecules Week 11 Notes  Recombinant proteins o Amplify DNA o Take domain 1 and link to domain 2 o Embed palindrome  Cut with restriction enzyme- sticky ends o Mix  hybridize o How you add an epitope tag  Designing genes o Cells  isolate DNA  DNA to be cloned  separate strands and add primers  PCR for amplification  genomic clones st o Cells  isolate mRNA  mRNA sequence to be cloned  add 1 primer, reverse transcriptase and deoxyribonucleoside triphosphates  DNA/RNA double strand  separate strands and add second primer  PCR amplification  cDNA clones  PCR for forensic analysis o VNTRs- variable nucleotide tandem repeats  In introns or spaces between genes  Dinucleotide  Quantify? Real time PCR o Instead of Eppendorf tubes  thin wall capillary tube – holds 5 uL sample o Instead of metal heating block  air temperature in chamer – small sample changes temperature quickly o PCR cycles and spectrophotometer  Every PCR cycle, ring spins and each sample is exposed to spec  Certain wavelength of light shines through o PCR mix contains SYBR green  Fluorescent dye- unique dye  Only binds to double stranded DNA  Only fluoresces when bound to DNA whereas ethidium bromide always fluoresces  SYBR indicates how much double stranded DNA there is o Results  Same amount of cDNA – amount of product determines time  Housekeeping gene – 2 lines overlap, same level in pre- adipocytes and adipocytes o Promoter  Temperature sensitive, pH sensitive, etc.  Express gene  mRNA to protein  extract protein from bacterial cells o DNA helicase  Temperature sensitive promoter  Increase temperature, increase amount of DNA helicase  Good way to make abundance of proteins for other experiments but might not always work  Limit of effectiveness – won’t know to fold or add post- translational modifications, etc. o Protein  determine partial amino acid sequence  synthesize DNA probe  screen cDNA or genomic DNA library  gene or cDNA  insert into expression vector  introduce into E. coli or other host cell  extract protein  repeat  Epitope tagging o Gene for protein of interest (expression vector) o Insert DNA encoding epitope tag o Introduce into cell o Epitope tagged protein use antibodies to locate or purify  Immunolocalization using antibodies to protein tag  Rapid purification of tagged protein and any associated proteins  Pulldown experiment o What proteins interact with your protein? o Recombinant DNA techniques are used to make fusion between protein X and GST (glutathione S-transferase)  GST binds to glutathione which binds to protein o Fusion protein bound to glutathione coated beads o When cell extract is added, interacting proteins bind to protein X o Glutathione solution elutes fusion protein together with proteins that interact with protein X  FRET o Under what conditions do the 2 proteins interact? o Unique epitope tags on each protein  Blue fluorescent protein – shine violet light  comes out blue  Green fluorescent protein – shine in blue light  comes out green o Interacting proteins  Shine violet light  comes out green o Non-interacting proteins  Shine violet light  blue light detected  Yeast to hybrid system o Find as many binding partners for this protein as you can o Reporter gene that under certain condition o Epitope tag on DNA binding domain – BAIT o Create cDNA library with all proteins produced by cell line and binding factors  Attach epitope tag – TAD – find potential binding partner with TAD o BAIT and target protein + binding partner and transcriptional activation domain  Recombinant genes encoding BAIT and prey introduced into yeast cell  Transcription of reporter gene  blue o Reporter gene on only if binding partner binds to BAIT  TAD turns on reporter gene  transcription  blue cell o Used to find an alcoholism gene  Wild type- feed together  Mutant- eat alone  Identify genes o Mutagenized – wild type cells exposed to something (ex: radiation, heat) o Mutagenized cells plated out in petri dish grow into colonies  colonies replicated onto two identical plates and incubated at two different temperatures  mutant cell that divides at the permissive temperature but fails to divide at the restrictive temperature o At elevated temp, some don’t replicate o Pathway of cell replication  Find step by step processes  ER  Golgi apparatus  secretory vesicles  Normal cell – protein secreted  Secretory mutant A – protein accumulates in ER  Secretory mutant B – protein accumulates in Golgi apparatus  Double mutant AB – protein accumulates in ER  Differentiated Display o Need to know PCR and how to run a big gel o Annealing temperature is critical  Only want a specific sequence to be amplified  Too high- breaks H bonds and primer won’t bind  Too low- don’t need as much energy to bind primer – can bind in other places o Cheap, quick, easy o 4 samples in each experiment- compare o Take cDNA from each sample  PCR with low annealing temp  a lot of bands appear  Find conserved bands – look for differences o Don’t know what genes the bands represent  Sequence  find gene  Find cause of disease/phenotype  Antisense RNA o Eliminating expression of your gene o Antisense RNA- double stranded mRNA – won’t work o Ex: Cell division interrupted because protein responsible for division is inhibited  Antisense RNA bound to normal RNA  Create specific mutation o Clone enzyme o Alter 1 amino acid o Melt double stranded plasmid into single strands o Add primer that hybridizes almost perfectly except for 1 nucleotide o Point mutation o Add DNA polymerase o Plasmids- 1 strand of wild type and 1 strand with point mutation  Microarray o More expensive and challenging- need technology for creating chips and reading them (spec and quantify) o Look for hundreds and thousands of genes at 1 time o Know what’s on blot, don’t know what the probe is o Take single stranded DNA from many different clones and spot onto an array on a microchip o cDNA library  put each in different spot on chip o Liver cells  treatment and control groups  expression of genes  extract mRNA  convert to cDNA (reverse transcriptase- single stranded)  incorporate label – fluorescently tagged  take cDNA and mix and float over chip – hybridization, complementary base pairs, gene was expressed if binding occurs  Housekeeping genes- yellow – expressed equally in both types of cells  Colors indicate activity of genes in cells o Gene expression pattern for nontoxic and toxic substances  Liver reacting to drugs as if it was a toxin o Limiting factor- what if gene responsible for your disease is not on chip  won’t come up  SAGE- serial analysis of gene expression o Population of mRNA that might get extracted o Looking for gene expression o Tags  Within every mRNA is a short stretch of nucleotides totally unique to that mRNA  Search database  determine genome it came from o Don’t need to create cDNA and sequence the whole thing o Single stranded mRNA  double stranded cDNA  Find and cut out tags  Link tags in longer stretches of DNA  sequence multiple at a time  Analyze expressed genes- qualitative and quantitative o mRNA  cDNA (oligo dt primer tagged with iron) o Anchoring enzyme binds to specific palindromic sequence and chop cDNA into pieces o Want piece closest to poly A tail o Iron sticks to magnet o Tag is at 5’ end of 3’ most fragment of cDNA o Synthesize linker – binds to sticky ends  Contains another palindromic sequence for the tagging restriction enzyme  Binds to sequence within linker but cuts about 12 nucleotides downstream o Take tags, link together (50-100 tags) o Clone and sequence o List of genes – tally o Can discover new genes o Time consuming, have to manipulate DNA and RNA, technical steps, sequencing, expensive o If you don’t know what the gene is, it will still show up  Creating a knock out organism- eliminating gene completely from organism o Clone gene you want to knock out  Myostatin- limits size your muscles grow to o Cut and insert another gene  Neo- codes for protein that gives cell resistance to a certain drug X  Neo in  cells live o Attach another gene to end  TK- makes cells sensitive to a drug Y o Insert gene into nucleus of cell o Allow recombination to occur  Gene could be randomly inserted into host – nonhomologous recombination  Myostatin gene randomly inserted with neo and TK  Drug Y in  cells die  Gene could find myostatin gene and undergo homologous recombination  Gene inserted with Neo but without TK  Drug Y in  cells live b/c no TK o Knockout mouse  Altered genes into cells, grow into colony, test for the rare colony in which the DNA fragment has replaced one copy of the normal gene  Take embryonic stem cell (homologous recombination)  Inject into embryo of mouse  Embryo into mother  Reproduce  Gamete producing cells – ¼ of progeny will be knockouts  Myostatin knockout mouse- big muscles


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