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GENETICS 2581 - FINAL EXAM NOTES (2014) LECTURE 13 - 24.docx

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Biology 2581B
Susanne Kohalmi

Biology 2581: Genetics Final Exam Lectures Lecture 13 – Chromosomal Rearrangements • Mouse VS human genome o At the sequence level – many similarities (same types of genes with similar sequences) o However when comparing chromosomes, staining patterns suggested no conservation • Same kinds of gene should have similar types of chromosomes…but this was not the case • 20 mouse VS 23 human • After both genomes were sequenced, each mouse chromosome could be “pieced” together from different human chromosomes • Syntenic segments – identity of genes, order of genes, orientation of transcription (almost the same) • Break the mouse genome in 100s of fragments but they can be aligned with human genome • Package size and content is different • Over time, the genome for human and mouse diverged (organized into different kinds of chromosomes) • You can line up the genomes all together, might have to cut it up • Are rearrangements always deleterious? No – generation of antibodies • How do only 2-3 x 10 genes encode 10 different antibodies? • How is the immune system able to respond to new challenges? • Chromosome 14 and B cell development • There are groups of segments • Recombination events can happen between different groups of segments • Each segment can combine with each other in different combinations, enormous of variation that can be made (fewer genes are needed) • If there is a lack of diversity in making antibodies, it can lead to a shift in formation of T and B cell clones o Observed in lymphomas • Deletions: loss of sequences • Small deletions affect a single gene while large deletions leads to loss of 10s / 100s of genes • Can be caused by X-ray or other chromosome damaging agents that break DNA backbone • Use PCR primers outside of deleted area and amplify it (piece missing, visibly smaller) • Can be detected through karyotyping • Most homozygous deletions are lethal and even most heterozygous • But there are exceptions: o Wolf-Hirschhorn Syndrome: deletion from short arm of chr 4 o Cri Du Chat Syndrome: deletion from short arm of chr 5 o Leads to severe phenotypes • Humans cannot survive if more than 3% of their genome is deleted • Deletions have an effect with gene dosage • A lot of gene regulation is dependent of gene dosage (how many copies do you have) • Ex: Drosophila – Notch gene o When you are heterozygous for deletion, wing has different appearance (not lethal) + o Only one copy of Notch • Deletion loop is formed when lined with homolog with a deletion (cannot be aligned)  • Recombination event can only happen between identical sequences, not happening in deletion loop • Genes in loop cannot be separated • Genetic distance between loci on either side will be underestimated • Deletion homologue is shorter because it’s missing deletion loop • Can visualize large scale deletions from the banding patterns • Use deletions to map mutations  look at banding patterns  look at phenotype and make correlations • Duplications: additions of sequences o Tandem duplications has direct repeats or inverted repeats in the same sequence o Nontandem duplications are dispersed (further away) • Caused by chromosome breaks o X-ray breaks one chromosome in two places and break homologous in another place • Can see different outcomes, duplicated matches other places in your chromosome (pair with each other, first loop pairs, second loop pairs) • Detected through karyotyping and fluorescence markers in situ hybridizations • Often there are no obvious phenotypic consequences, not losing something • But possible due to increase in copy number of particular gene, it can alter expression by placing gene in different chromosomal environment • Heterozygosity for duplications covering more than 5% of haploid genome is often lethal • Three copies of Notch gene – extra veins in aberrant (visible change in phenotype) • You need identical sequences for recombination • Duplication can cause unequal crossing overs in recombination • Inversions: half circle rotations of sequences, invert a certain segment in chromosome o In pericentric, part of inversion includes centromere but in paracentric, it is in one arm only (does not include centromere) • Intrachromosomal inversions are caused by recombination event (bend) between repeats in chromosomes • Most inversions do not result in an abnormal phenotype, they rearrange the order of genes (not adding or deleting = not effect on function) • Unless the breakpoint is in the middle of a gene, it will disrupt gene function • Or the changed chromosomal environment impacts regulation of gene expression • Inversion loops occur individuals which are heterozygous for inversion when homologous align during meiosis • Pericentric inversion: o Included centromere o Chromosome can line up again, recombination can happen • Paracentric inversion: o Only one arm o Can line up, outcome has broken / deleted products of dicentric chromosome • Translocation: part of chromosome relocates • Do not result in abnormal phenotype (no addition / deletion), only rearrange order of genes • Unless breakpoint is in the middle of a gene or change chromosomal environment impact on regulation of gene expression Lecture 14 – Prokaryotic Gene Regulation • E.Coli biphasic growth curve: o B-galactosidase (lactose metabolism enzyme) is produced in phase II but not in phase I o Only uses lactose once glucose depletes o Simple monosaccharide glucose is preferred over disaccharide lactose as a C and energy source • Two genes are critical for the breakdown of lactose into simpler sugars o LacY gene: encodes lactose permease (import lactose) o LacZ gene: encodes B-galactosidase (breaks apart lactose)  • Bacteria is unicellular cell, must be capable of extreme adaptation to changing environment • Can’t waste energy making products it doesn’t need, must be capable of switching genes on or off as needed • Gene regulation (on / off and volume) is very important • Catabolic genes involved with breakdown of complex products into simpler ones (using lactose) are only needed when complex compound is present • Lactose is a not a desirable sugar, E.coli uses it if no other choice is available • Most advantageous for E.coli cell to express LacZ and Y when: o i) lactose is present and ii) glucose is absent • Lac genes are normally off and are actively turned on (induced) when grown in presence of lactose (levels increase 1000 fold) • Derivative of lactose – allolactose is the inducer in this system • LacZ and Y localize to a tightly linked cluster of genes on the bacterial chromosome • LacA gene: encodes lactose transacetylase • Product of LacA gene is not required for catabolism of lactose • Operon: coordinates control, prokaryotes only • LacZ, Y and A genes are transcribed in unison as part of transcriptional unit referred to as an operon • All genes involved in the same biological process can be regulated together • Promoter: site from which RNA polymerase initiates transcription • Proteins are used to control ability of RNA polymerase to transcribe o Repressor: negative regulation, diminishes transcription when bound to cis-acting element o Activator: positive regulation, increases transcription when bound to cis-acting element + • Lac Cells: able to utilize lactose as a C and energy source (wild type E.coli show this phenotype) • Lac Mutants: unable to utilize lactose o Cells carrying loss of function mutations in LacZ and Y show Lac - phenotype • Mutations did not provide any insight into how lac operon was regulated • LacY mutants can not import lactose and LacZ mutants can not catabolize lactose • Models of induction: o Inducer positively regulates an activator o Inducer acts to relieve inhibition of transcription (inducer inhibits a repressor) • Constitutive Mutants: cells produced B-gal all the time, abnormally on in absence of lactose o Defined a new gene at a distant chromosomal locus (LacI) o Suggested a negative regulatory role for LacI gene product • PaJaMo Experiment: - - o Grow recipient cell (LacI , LacZ) in the absence of lactose (cytoplasm devoid of functional LacI and LacZ gene products) + + o Introduce copies of the LacI and LacZ genes into the recipient (produces merodiploid – cell with two copies of some genes) o Put into a plasmid and insert into cell o Monitor level of B-galactosidase synthesis o Induction is possible even in the absence of the inducer (inconsistent with 1 model) o Synthesis of B-galactosidase begins shortly after LacI and LacZ + genes were introduced o When inducer is added, synthesis continues o With no inducer, B-galactosidase synthesis stops, LacI is being expressed, producing a repressor o LacI must encode a repressor • First 2.5 hours of experiment: insufficient repressor protein is present in cytoplasm to prevent LacZ transcription, due to time needed for LacI transcription, B-gal levels rise • After 2.5 hours (no inducer): sufficient levels of LacI repressor build up, LacZ transcription stops and levels of B-gal fall • After 2.5 hours (inducer added): addition of lactose relieves inhibition of LacZ transcription by LacI repressor, B-gal levels rise • Lac Operon Model: o Binding of LacI repressor protein to a hypothetical site near promoter (operator) prevents RNA polymerase from recognizing promo-er o LacI mutants are unable to bind to operator, RNA polymerase can recognize promoter and transcribe operon o LacI repressor bound to the inducer (allolactose) can not bind to DNA of operator o LacI repressor not bound to allolactose can bind to DNA of operator c - • LacO mutants (similar to LacI ) produce B-gal constitutively, defined a regulatory sequence in the promoter (just regulating, not coding) c • Jacob and Monod showed that LacO mutations did not affect structure / function by cis / trans test with merodiploids o DNA sequence elements would influence only expression of adjacent genes on same DNA molecule (cis) o Proteins would be able to diffuse through the cytoplasm and act on target sites on any DNA molecule in cell (trans) o Cell is phenotypically normal = LacI is trans-dominant to LacI - o Normal LacI protein from alternate copy can act in trans and bind operator on chromosomal copy o Protein diffuses throughout cell and is able to bind to O o LacO is cis-dominant to LacO since normal LacO alternate copy can only act in cis o LacO defines a sequence if mutated LacI can’t bind and block • Non-Inducible Mutants: cells carrying mutations in LacI gene (LacI ) that s abolish ability of repressor to bind with inducer o Cell is phenotypically normal = LacI is trans-dominant to LacI + o Mutant LacI protein from alternate copy can act in trans and bind operator on chromosomal copy • Repression – no lactose in cell, repressor protein is bound to operator and Z, Y, A genes can’t be expressed • Induction – repressor, with inducer bound, can no longer bind to operator and Z, Y, A genes can be expressed  RNA polymerase can begin transcription because promoter is accessible Lecture 15 – Prokaryotic Gene Regulation II • Based on their genetic analysis, Jacob and Monod developed a simple, yet highly detailed model of induction • This hypothetical model gave rise to several testable predictions: • Repressor protein must physically bind the operator sequence o DNA footprinting experiments establish the region of DNA to which a proteins binds o Goal is to try to assay whether a given protein is able to bind a given segment of DNA o Start off with stretch of DNA that contains binding site, then the 5’ end of that fragment you include a radioactive marker (allows you to track where the fragments are going) o You have a control, fragments on their own o Experimental has fragments together with the binding protein o Take the pool of fragments and pool them together with DNAase1, doesn’t cut like a restriction enzyme, it cuts at random o You get a whole spectrum of sizes after treatment with the DNAase o Take the segments and run it through an agarose gel, after you take a piece of X-ray film and lay it overtop of the gel. Wherever that’s radioactivity, there will be a black band on the film o In contrast, if the proteins do bind to the DNA, then it will protect the DNA fragment. That will not allow the DNAase to bind and cut, it will not allow the full array of fragments o Presence of the footprint means that the protein actually does bind the fragment o When the repressor bind the operator, you will not get any cuts in that fragments which will give the footprint • Repressor protein must physically bind the wild-type operator sequence, c not the LacO mutant o Testing whether or not the repressor can bind the mutant operator or not o You start off with a labeled repressor, track where the protein is going in the experiment o You have 2 sequences: one with type copy and a mutant copy of the operator o Mix the protein with the LacO sequence, and then place it in the test tube, which contains a glycerol gradient; low at top high at bottom o You mix the protein and DNA together and lay it on top of the gradient. Put in a centrifuge and it’ll spin out to the gradient which is the same density as the sample o When you find the wild type one, it’ll be in the gradient because of radioactivity o When you do it with the mutant, you won't find radioactivity, so it does not bind • Repressor protein must have 2 distinct domains, one that binds DNA and one that bi-ds inducer o LacI mutations that affect DNA binding all cluster in the N-terminus of protein o LacI mutations affecting inducer binding all cluster in main body of protein o Indicates that there is two different domains o N-terminal part actually codes for a helix-turn-helix motif o Fits very nicely into the major groove of DNA o Specificity of binding determined by H bonds between helix and nucleotide bases o LacI repressor is a tetramer of 4 identical LacI-encoded subunits • Shape (conformation) of the LacI repressor must be altered by inducer binding (repressor is allosteric) o Allosteric: protein that undergoes reversible changes in conformation when bound to another molecule o When it’s not bound to an inducer, it’s able to bind to the operator sequence o When it is bound to the inducer, there's a 3.5 Angstram displacement which does not allow it to bind to DNA o Binding of allolactose to the LacI repressor alters the conformation of the DNA binding domain • However, Jacob and Monod failed to address the role of glucose in the control of transcription • Catabolite Repression: presence of the preferred catabolite (glucose) represses transcription of the LacZ gene • There is a small amount of induction when there is glucose and lactose, but nothing compared to just lactose (2500x more) • Why do E. Coli grown with both lactose and glucose not show full induction of LacZ transcription? • LacZ transcription is also dependent on a positive regulatory protein, cAMP receptor protein, CRP (catabolite activator protein, CAP) • CRP promotes the ability of RNA polymerase to initiate transcription • Level of cAMP is regulated by Adenyl Cyclase enzyme, which is dependent on the level of glucose present • Glucose indirectly controls the activity of CRP (aka CAP) through modulating cAMP levels • Lac operon is regulated by two superimposed layers of control: negative layer (LacI repressor) and a positive layer (CRP-cAMP)  • Both layers mediate their affect through controlling RNA polymerase • Operator site overlaps with the site that binds the cAMP • This addition to the model gives rise to more testable predictions o The binding of the repressor to the operator should prevent transcription by RNA polymerase o CRP-cAMP should bind near the promoter and aid in transcription by RNA polymerase • Different genes need different strategies of control • Anabolic genes are involved with the synthesis of simple products into more complex ones – ability to make tryptophan (one of the 20 amino- acids used in protein synthesis) • Needed all the time unless enough complex product is present • Tryptophan is constantly needed for protein synthesis • Trp genes are normally on and are actively turned off when: o i) Cell has made enough for present needs o ii) Abundant supply in the environment • Lac operon is an example of an inducible system • Trp operon is an example of a repressible system • TrpA, B, C, D, and E genes are requires for the biosynthesis of trp • TrpR: allosteric repressor protein whose conformation is modulated by tryptophan itself  • TrpR can only bind when there is lots of trp in the media o Will bind and keep trp concentrations low • Two different sized transcripts are formed in the absence of tryptophan o Full Transcript: runs through all structural genes o Short Transcript: ends right before first structural gene • In a trpR- mutant with trp, there should be full transduction • When you do experiment, it is only 33% • Attenuation: indicates there is another independent mechanism of repression • Within leader sequence, there are 4 regions that are able to form complementary base pairs o 1, 2, 3, 4 o Can fold back on itself to form double stranded region o Regions 1 with 2 together, regions 3 and 4 together, regions 2 and 3 together o There are 2 trp codons in there o Once you start translating if there are low levels of trp, ribosome will come along and if it hits stem and loop, it has to insert trp charged tRNA’s, but if there aren’t any, then it will just stop and wait o Transcription still occurs though o In higher levels of trp, when the ribosome comes to the stem and loop, it won't stall very long and it’ll insert the trp charged tRNA’s and continue running through the transcript  o Region 2 and 3 won't pair again, but 3 and 4 can pair and allow another regulation point o When 3 and 4 bind, it stops translation Lecture 16 – Quantitating Gene Regulation • So far we have examined gene regulation in a qualitative manner • Is it possible to examine transcription of genes in a quantitative manner? • If so…what could this tell us about transcriptional regulatory networks? • Goal: create mathematical models of two simple genetic networks • Activation o X  Y o X: transcriptional activator o Y: target gene o  : (+) regulation • Repression o X –| Y o X: transcriptional repressor o Y: target gene o –| : (-) regulation • dY/dt = b – aY o Y: level of gene Y mRNA within a cell (in units of nmoles) o b: rate of production (nmoles / min) o a: rate of degradation (1/min) o Source – production through transcription o Sink – degradation through nucleases • If you have a high rate for b and low rate for a, level of Y will start accumulating in the box • If you have a high a and low b, eventually amount of message will start decreasing until it hits 0 • Why is the rate of degradation not also in nmoles / min? • Rate of degradation depends on how much Y there is in the box at any given instant of time • Degradation Rate: o If a = 0.1 / min, this means 10% of Y degrades every minute o If at an instant in time, there is 1000 nmoles – degradation rate would be 100 nmoles / min o At another instant in time, there’s only 100 nmoles – degradation rate is only going to be 10 nmoles / min o a = 10 nmoles / (min * 100 moles) = 0.1 / min • Transcription of a gene can be simulated with the help of specialized programs • Steady State: o dY / dt = b – aY o 0 = b – aY st o Y =st / a o Rate of production divided by rate of degradation • Levels of Y mRNA will reach a steady state level within a cell based solely on production and degradation rates • Initial levels of Y mRNA will not affect the final steady state levels of the message • To quantitate amount of message at steady state, you need to know: i) production rate and ii) degradation rate • Describing activation mathematically: X  Y o X = activator  o S =Xsignal (converts X to active form X*) o Once in X*, it can bind and now we’ll get increased transcription • Example: o S =XcAMP o X = CRP, X* = CRP-cAMP o Rate of production (Y promoter activity) is a function of X* o Y promoter activity is proportional to X* o What is the function? • Hill Function for an activator o Y promoter activity = B(X*) / ( K + (X*) ) n o B: maximum expression level of Y, when cell is grown under perfect optimal conditions (upper limit) o K: concentration of X* at which gene is at half its maximal expression level o n: “order” of equation, determines how “switch-like” the gene is • n = 1 – as X* increases, you’ll see a slow steady increase in transcription • n = 4 – as X* increases, you’ll get little response at first and a very strong response around K • We can use this equation to model simple activation of the gene • Describing repression mathematically: X –| Y o X = repressor o S =Xsignal (converts X to active form X*) o Once in X*, it can bind and stop transcription • Example: o S = co-repressor X o X = repressor, X* = co.rep-rep complex o Y promoter activity is proportional to 1 / X o What is the function? • Hill Function for a repressor o Y promoter activity = B / ( 1 + (X*/K) ) o Inverse of Hill Function for activator o High concentration of repressor = low levels of expression o Low concentration of repressor = high levels of expression • Logic approximation is a simplification – it considers genes to be ON / OFF • Activation: o If X* < K, Yprom act0 o If X* > K, Y = B prom act • Repression: o If X* < K, Yprom actB o If X* > K, Y = 0 prom act • Example: o X = activator o S =Xmaltose o Genes F and G = hypothetical genes that produce a mRNA encoding a protein required for maltose metabolism o Gene G: b = 50 nmoles / min, a = 0.5 / min, Y = st0 nmoles o Gene F: b = 5 nmoles / min, a = 0.05 / min, Y = 100 nmoles st • Why would gene G have a high production rate only to degrade the product at a high rate anyways? • Production and degradation rates affect a gene’s response time • Advantage: lets you respond to the presence of the maltose much faster • Gene F can’t access the maltose right away • The same is true in reverse • Maltose in the media and removed – gene G is able to respond much more quickly to the environment (turns off faster compared to gene F) • Gene F has all this gene product there even when there is no maltose in the environment • Response Time: T 1/2 ln(2) / a o Gene G = 0.693 / 0.5 = 1.39 min o Gene F = 0.693 / 0.05 = 13.9 min o OFF  ON: time needed to see a 50% increase in message levels o ON  OFF: time needed to see a 50% decrease in message levels • Response time for gene G is much faster compared to gene F (more slow) Lecture 17 – Eukaryotic Gene Regulation • Different cell types display vastly different morphologies, functions, behaviours • Yet all the information required to give rise to a fully developed adult is present in a single, differentiated somatic cell • How is this possible? • Genes are differentially regulated across tissues – expression levels vary widely across tissue and time • Genes can be differentially regulated during development • Eukaryotes are more developmentally complex, multicellular systems • Need to coordinate gene expression: o In various tissues o At various times in cell cycle / development o In response to external signals • Genome size is greater than in prokaryotes • There are 3 RNA polymerases: o I transcribes rRNA genes o II transcribes mRNA genes (dealt with here) o III transcribes tRNA and 5S rRNA genes • DNA is in the form of chromatin • Eukaryotes can have up to 700x more DNA and 10x more genes than typical prokaryote o Human Genome  3 x 10 bp 9 6 o E.Coli Genome  4 x 10 bp • Regulated protein  12 bp sequence o How unique is the cis-acting sequence? 12 o 4 ~ 16000000 6 o E.Coli: (4 x 10 * 2) / 16000000 = 0.5 (not probably for E.coli to have cis-acting sequence) o Human: (3 x 10 * 2) / 16000000 = 375 (highly probable to have many cis-acting elements by chance) • Harder for eukaryotes to achieve specificity • Solution: o Higher affinity binding of proteins to DNA  lesser chance a protein to bind randomly o Combination of proteins required for transcriptional control • Combinatorial Control: control of gene expression requiring presence / absence of a particular combination of regulatory proteins • E.Coli has single LacI repressor binding the operator (one of the few interactions that controlled expression of that operon) • With eukaryotes, there are many different cis-acting elements and trans- acting factors • Proper combination of regulators assembled at the regulatory sequences at the gene in order to get proper transcriptional control • There are 3 polymerases (in prokaryotes, there’s only one) o RNA polymerase I transcribes ribosomal RNA genes o RNA polymerase III transcribes transfer RNA genes and 1 ribosomal RNA gene o RNA polymerase II is responsible for transcribing all protein-coding genes • Each protein coding gene has a complex and unique set of cis-regulatory regions that influence ability of RNA polymerase II to initiate transcription • There are actually many different layers to controlling gene expression • Eukaryotes organize their DNA in the form of chromatin (wrapped around nucleosome) • Tight packaging of DNA and associated histones act as a barrier to transcription • DNA must unwrap and histones must be moved to allow transcription • In order to get full levels, these nucleosomes need to be removed to get access to RNA polymerase and other transcription factors • In vitro experiment  take naked DNA with RNA polymerase II, basal transcription factors = high levels of transcription • If you do the same experiment with chromatin- form, you’ll get extremely low levels of transcription • Chromatin makes promoter sequences less accessible to RNA polymerase II  • Remodeling Proteins: in charge of removing nucleosomes or sliding them away for transcription to occur • Mechanisms affecting chromatin structure are critical to gene regulation in eukaryotes but not prokaryotes • In prokaryotes o Cis-Acting Factors: positive regulatory sequence, promoter, operator o Trans-Acting Factors: RNA polymerase, activator, repressor • In eukaryotes o Cis-Acting Factors: promoter, enhancer, silencer, insulator o Trans-Acting Factors: RNA polymerase II, basal factor, activator (pos), repressor (neg) • In order to get proper expression of the gene, not only do you need the core promoter you also have other elements: enhancer, silencers and insulator that can be located extremely far away from the actual gene (thousands of bp away) • How can these elements be defined? By mutation • Reporter Gene: o Take coding sequence for a protein and place under the regulatory region that’s controlling gene of interest o Now regulatory region is controlling reporter gene o Determine which bases are important for expression • Trans-acting factors can also be defined by mutation o Reporter gene under control of regulatory region of gene of interest o Conduct mutant screens and look for mutants in different genes that affect expression of reporter and such mutants would define transcription factors • Core Promoter: minimal DNA sequence needed to assemble RNA polymerase II • Enhancers: DNA elements that attract transcription factors, located at great distances from promoter • Transcription Factors: any protein other than RNA polymerase that is required for transcription o Basal and specific TF serve as activator o Bind to core promoter, proximal elements or to enhancers • Promoter alone doesn’t support high levels of expression • Promoter can recruit polymerase and basal factors and will support low levels of transcription but only a small basal level • Basal Transcription Apparatus: complex of basal factors and RNA polymerase II that assemble on promoter and are capable of initiating minimum levels of transcription • Full expression requires the help of enhancer elements • Core promoter has many different sequences • TATA Box: found in most genes, located upstream of transcriptional start o Named after nucleotide sequence (rich in T and A’s) • TBP: TATA box binding protein, binds to TBP associated factors (TAFs) o Helps position RNA polymerase II near transcriptional initiation site o TBP puts a kink in the DNA (only two H bonds between T and A) o Doesn’t take as much energy to pry apart the strands o Already prone to coming apart, TBP helps along the way o Binds DNA at the TATA box and puts physical stress (bends) to promote separation of the strands • On the x axis, looks at different positions on the promoter • If you start mutation the bases defining the TATA box, you will see a decrease in expression • Point mutations in these crucial regions greatly reduces the ability to transcribe • TATA box must have an important regulatory role in controlling expression of the gene • Binding of the polymerase and the basal apparatus is enough to get you low levels of expression • You need enhancers for full level – eukaryotes require activator bound to enhancer and promoter to fully activate transcription  • Where are the enhancers? o Anywhere within 200 kb of the gene (upstream / downstream) o Fully functional even when inverted • DNA can loop around (flexible enough that you can get trans-acting factor binding the enhancer sequence) • Bringing that trans-acting factor to the core promoter where the polymerase has bound • They can affect association of histone and DNA and help recruit / stabilize basal apparatus • DNA is too rigid for it to bend if the two components are too close together • If it’s further away, it’s easier to bend the DNA so those two proteins can interact • Protein binding to distant sites on DNA can increase probability of interaction • Deadspot: when two proteins are too close together, they won’t interact • Sweetspot: interaction of the two proteins, after that interaction decreases • Enhancers are primarily responsible for controlling tissue specific gene expression • Albumin Gene: o Expressed in the liver, secreted into the bloodstream o Liver specific enhancers o In the liver, the trans-acting elements responsible for binding to those elements will be expressed and that will promote transcription in the liver o In the brain, the same enhancer elements are there but those trans- acting factors aren’t expressed (not same expression of albumin gene) • Yellow gene in drosophila produces black pigment (made in wing, body, bristles) o Loss of function causes a yellow pigment o Take body enhancer and placed it so that it controlled expression of GFP o That body enhancer produces flies that were able to express GFP in their body o You can use these elements to control tissue specific expression • If you want to identify all the elements responsible for controlling expression of that gene • Use a reporter and replace coding sequence of AP1 with coding sequence of GFP  o You’d see green fluorescence in the eye and lung but not in the tail • You can start making deletions in the regulatory sequences to see the effects on transcription • Delete first portion – expression in the eye but not the lung anymore o That sequence must contain an enhancer for the lung expression • If you do another deletion – no effect • If you do another deletion – no expression anywhere o Must define enhancer for the eye • If you delete the last portion – expression in eye, lung and tail • Abnormal expression in tissue where it’s usually not expressed o That sequence must define a negative regulator (deleted = abnormal expression) o This defines a silencer • If you delete more – confirms position of eye enhancer (abolished expression of the eye) • Insulator: block activation by enhancers • Existence of insulator elements (cis-acting that recruit trans-acting that can affect efficiency of the enhancer elements) • Insulators are interacting, creating sub-domain, isolating enhancer and promoter so that it can affect transcription of any other gene in the vicinity • Insulator can sequester the enhancer, binding enhancer element and preventing for interacting with either of the promoter • Or it can sequester the promoter itself, leaving the enhancer free to modulate expression of the first promoter Lecture 18 – Eukaryotic Gene Regulation • Silencer: recruits a repressor protein • Full expression in eukaryotic gene, however if you have a repressor bound to a silencer it will override activation (next to no transcription) • Repressors: affect stability of polymerase, block movement / assembly, act more indirectly by affecting activators • Competition: case of mutually exclusive binding • Enhancer can recruit an activator or repressor  • If a repressor gets there first, blocks binding of the activator and can’t stimulate transcription • Quenching I: repressor binds to and blocks DNA binding region of an activator o DNA-binding domain is blocked o Activator cannot bind to enhancer • Quenching II: repressor binds to and blocks activation domain of an activator o Activator can still bind to enhancer elements but repressor is blocking activation domain, can’t stimulate basal machinery • For eukaryotic gene regulation, model system = budding yeast • Ability to engage in genetic analysis is crucial advantage of use of budding yeast • Expression of GAL7, 10, and 1 genes is required for yeast cells to use galactose as a C source • Inducible genes –turned on when galactose is present in media, codes for enzymes responsible for metabolism of galactose • All 3 genes are tightly linked on the same chromosome, each has their own promoter but all share the same enhancer (not an operon) • Coordinate control is achieved in this case through the use of a common enhancer that affects all 3 genes • If the enhancer is deleted, you will only get basal levels of transcription • There are two trans-acting factors that work through this enhancer: GAL4 and GAL80 • Loss of function GAL4 = basal levels of transcription, regulatory mutant • Loss of function GAL80 = constitutive expression (always transcribed no matter of presence of galactose) • Gal4 (activator)  Gal7, 10, 1 transcription • Gal80 (repressor) –| Gal7, 10, 1 transcription • Gal4 and Gal 80 work through enhancer • How do Gal4 and Gal80 interact with one another in this pathway controlling Gal7, 10, 1 transcription? • Alternative #1: Galactose –| Gal80 –| Gal4  Gal7, 10, 1 transcription • If you remove Gal4 = low levels of transcription • Alternative #2: Galactose  Gal4 –| Gal80 –| Gal7, 10, 1 transcription • If you remove Gal80 = nothing to inhibit Gal4 = high levels of transcription • Gal4’s positive effect can be inhibition of the inhibitor • If you lose Gal4, there’s nothing there to inhibit Gal80 = low levels of transcription • How to distinguish between the two alternatives? - - • If alternative #1 is true, then…predict that Gal4 Gal80 cells would display basal levels of transcription o A cell that has two mutations (one in Gal4 and one in Gal80) o When Gal4 is gone, you are going to get low levels of transcription but then Gal80 works through inhibiting Gal4 o Gal80 won’t have an effect on the system because there is no Gal4 to act on • In contrast, if alternative #2 is true, then…predict that Gal4 Gal80 would displace constitutive expression o Gal80 is gone, nothing inhibiting transcription = high levels but Gal4 functions through inhibiting Gal80 (no effect) o Gal80 is epistatic to Gal4, cells would show Gal80 phenotype • If we can make double mutant, we could distinguish between the two alternatives and come up with simple model of that transcription network • In budding yeast, very easy to do experiments o Budding yeast can either exist as a haploid or diploid o Haploid – exist in one or two mating types (α and a) o α can’t mate with α, a can’t mate with a o Cell from α and a will come together to fuse forming a diploid a/α o Diploid will go through meiosis and generate 4 haploid products o Some spores will contain both mutations, take some of those spores and provide rich-growth medium and grow vegetatively as a haploid and assay what’s happening in terms of regulation of Gal genes - - • Induce mating  generate Gal4 Gal80 mutant  assay  see only basal levels • This confirms alternative #1 and rules out alternative #2 o Gal4 is acting more downstream in the pathway • Type II quenching – Gal4 is an activator, binds enhancer element and in the absence of galactose, the Gal80 is bound and blocks access to activation domain (Gal4 can’t stimulate transcription)  • Induction: result of binding of galactose to Gal1 o
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