Brandl Lecture Notes

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Western University
Biochemistry 2280A
Derek Mc Lachlin

Brandl Lecture 1 Notes DNA – The Blueprint of Life – Fun Facts  Information is stored in DNA in the form of genes  The human genome encodes about 25,000 different protein encoding genes  In any one cell type, about 10,000 are expressed at various levels  The expression of the correct genes is essential for growth and differentiation of all cells Significance of Gene Expression  Many diseases are due to altered expression of one or more genes  Simple diseases involve often one gene, complex diseases often involve many genes  By manipulating gene expression, we have the potential to prevent or reverse disease The Central Dogma  DNA → RNA → PROTEIN  Two key steps: transcription (making RNA) and translation (making protein)  Why is there an RNA step in gene expression? Three main reasons include: 1. The RNA step provides an amplification which allows genes to be expressed at different levels 2. Because RNA can be easily degraded, expression of a gene can be stopped quickly 3. RNA provides additional opportunities to regulate the expression of genes Q: There are 2 steps at which genes can be regulated. True/False? ANS: False – there are many more than 2 steps Regulation of Gene Expression Steps at which gene expression can be regulated in a eucaryotic cell  Transcription o Initiation o Elongation o Termination  RNA processing o RNA editing o 5’ capping o Splicing o 3’ polyadenylation  mRNA export (movement of RNA from nucleus to cytoplasm for it to be expressed, occurs through nuclear pores – very important in coordinating gene expression)  Translation o Initiation o Elongation o Termination  mRNA degradation (amount of RNA is dependent on two factors – rate of synthesis and rate of decaying – can control amount of RNA in the cell by how rapidly it is degraded)  Protein modification (some proteins are synthesized in inactive form and require modification to become functional)  Protein degradation (level of protein also depends on rate of synthesis and rate of decaying) Q: Which step in a biological process is often the most highly regulated? ANS: The first step in a biological process is often highly regulated as it saves the energy required in the subsequent steps. If you regulate the last step, the cell would consume 2 ATP unnecessarily, whereas if you regulate the first step, you may not consume any ATP at all. Transcription – Significance  Because transcription is the first step in gene expression, it is often highly regulated  Many genes are principally controlled at the level of transcription  Many diseases result from defects in transcription factors (cancer)  The specificity of many transcription factors makes them logical drug targets First Focus on Prokaryotic Transcription  Mechanistically similar to eucaryotic transcription, but somewhat similar Terminology  Promoter – DNA sequences required to initiate transcription of a gene (or operon) o Bacterial promoter – promoter is almost always at the 5’ end  Operon – a set of genes transcribed from a single promoter and thus expressed from a common RNA o Multiple protein products from a common RNA  Terminator – the DNA sequence required to stop translation Key Features of a Bacterial Promoter  +1 site – start site for transcription  -10 site  -35 site  Structure of bacterial promoters – sequences were lined up at start site, and noticed there were common bases at -10 region (Pribnow box) and common bases at -35 region o Spacing between them can differ slightly but there is always some sort of commonality at -10 and -35  What is the most frequently occurring base at each position of the -10 and -35 element? o What is the “consensus sequence” for the -10 and -35 elements? o Most frequently TATAAT at -10 region o Most frequently TTGACA at -35 region  Consensus sequence is a general term for any specific type of DNA element  Consensus sequence – the most frequently occurring base in a group of functionally related DNA elements  Consensus is a statistical entity – thus, does not have to be exact same as specific elements Bacterial RNA Polymerase  The enzyme that makes specific RNA transcripts using DNA as a template and nucleoside triphosphates (NTPs) as substrates  It is a multi-subunit enzyme  Found in two forms: o Core enzyme – catalytic activity  In the test tube, the core enzyme will synthesize RNA from ends and nicks in the DNA, but it does not recognize promoters  Has no specificity o Holoenzyme – recognizes promoters  Core + sigma subunit = RNA polymerase holoenzyme  Has specificity  What does a sigma subunit do to give RNA polymerase specificity? o The promoter specificity is determined by the sigma subunit o Sigma enhances the binding of RNA polymerase to the promoter DNA by contacting the -10 and -35 sequences o Recognition between protein and DNA – this recruits RNA polymerase to the promoter  The RNA polymerase – promoter interaction is one of many examples demonstrating the importance of protein- nucleic acid interactions Steps in Initiation of Transcription 1. RNA polymerase holoenzyme binds the promoter o Closed complex – complex between RNA polymerase holoenzyme and promoter 2. RNA polymerase unwinds the DNA strands around the start site o Open complex – DNA strands are separated 3. The first NTP is brought to the template o No primer is required 4. Using NTPs as substrate (ATP, GTP, CTP and UTP) chain elongation proceeds in a 5’ – 3’ direction, following base pairing rules o RNA made is complementary to the template DNA o Phoshpodiester bonds are formed in this process and pyrophosphate is released 5. After addition of 5-10 nucleotides, sigma falls off the holoenzyme o Core enzyme goes on and transcribes the rest of the strand o Sigma is crucial for initiation but not required for elongation 6. The transcription bubble moves downstream (5’ – 3’) with the template DNA re-annealing behind 7. Chain elongation proceeds until a terminator is reached and RNA polymerase falls off 8. Sigma rebinds RNA polymerase and the cycle is repeated Brandl Lecture 2 Notes Methods of Regulation 1. Some genes have better -10 and -35 sequences  These genes will be expressed at higher levels  Will recruit RNA polymerase at higher efficiency  Not dynamic  Sequences cannot be regulated; are “stuck” with that sequences throughout time 2. There is more than one sigma factor  Each recognizes different promoters 70  The most prevalent is the “housekeeping” σ, σ  Others include σ nitrogen metabolism, σ HS heat shock  Heat shock responds to high temperature, is induced and reacts with RNA core enzyme 3. Gene specific regulatory proteins  Dynamic!  Can change genes in response to different environmental signals through these gene specific regulatory proteins  Largely used by eukaryotic cells  Negative regulation – factors repress transcription  Positive regulation – factors activate transcription Negative Regulation Example: trp operon  In E. coli the 5 genes for tryptophan biosynthesis are transcribed from a common promoter  The trp operon is regulated by the concentration of tryptophan in the environment  Trp operon is expressed when there is little tryptophan in the cellular environment  How is the trp operon regulated by tryptophan? Trp Promoter  σ regulated promoter  Between the -10 and -35 sequences is an element called the trp operator that binds a protein called the trp repressor  At high [tryptophan], trp repressor binds tryptophan  The trp repressor-tryptophan complex binds the operator DNA  The trp-repressor-tryptophan complex blocks RNA polymerase from the promoter  RNA polymerase cannot bind to the promoter; it is sterically inhibited from getting there  Polymerase can’t park at the promoter  When [tryptophan] is low, tryptophan dissociates from the trp repressor  The trp repressor no longer binds the trp operator  The trp repressor will only bind DNA when it itself is bound to tryptophan Result  RNA polymerase can access the trp operator, transcription occurs  Low [tryptophan] – genes are on  High [tryptophan] – genes are off  What is the structural basis for the regulation by tryptophan? Trp Repressor – Monomer  107 amino acid residues  Alpha helices 4 and 5 make up the helix-turn-helix motif  Helices 4 and 5 are critical  Helix-turn-helix motif is crucial for binding DNA  Many DNA-binding proteins act as dimers  Can be true dimers like (trp repressor) or pseudo-dimers Trp Repressor – Dimer  Protein dimer has 2-fold symmetry  Helix 3s contact to allow dimerization  Helix 5 of each monomer recognizes adjacent major grooves in the operator DNA  Major groove is wide enough to fit an alpha helix  Protein can read the bases in the major groove  Most proteins interact with major groove, not minor groove  Minor groove is too small, doesn’t have much discriminatory power  Repressor only binds to DNA when it has tryptophan bound to it  Tryptophan binding induces a conformational change in the Trp repressor which allows DNA binding  Binding of tryptophan is crucial to remaining conformation of protein that will bind to DNA  Tryptophan binds to protein, alters conformation of protein (from tilted helix 5s to new conformation)  Process has been selected for through evolution General Themes to take from the Trp Operon 1. Trp repressor is a site specific DNA binding protein  Binding to specific sequences very tightly  Recognize individual specific sequences 2. There is a binding site for trp repressor within the Trp promoter  Genes that don’t have an operator won’t be regulated by the trp repressor; binding site is needed 3. Trp repressor inhibits transcription by blocking access of RNAP to the promoter 4. Trp repressor is responsive to an environmental signal  For Trp repressor, this would be tryptophan (environmental signal)  All gene specific regulatory proteins are regulated by some signal  Q: Genetics has been critical in understanding the regulation of the trp operon. Different mutations in the trp repressor can result in? A: The trp operon can always be transcribed or can never be transcribed. Always transcribed – never bind DNA Never transcribed – always bind DNA Six Differences in Transcription between Procaryotes and Eucaryotes 1. Three RNA polymerases in eucaryotes  Pol I most rRNA genes  Pol II mRNAs (makes proteins)  Pol III tRNAs, 5S rRNA 2. No operons in eucaryotes  Each gene is transcribed as a single unit. (monocistronic; in contrast to polycistronic in bacteria) 3. Promoter structure:  Eucaryotes do not have -10 and -35 sequences – there is no sigma factor  Promoter recognition is determined by a set of proteins, one of which recognizes a TATA-element  TFIID binds to TATA box, TFIID contains TATA-Binding Protein (TBP) 4. In eucaryotes regulatory proteins often bind DNA several thousand base pairs from the start site  TATA box usually within 100 bases of start site 5. Combinational Control: groups of proteins work together to determine the expression of a gene  Eucaryotic promoter – usually many regulatory factors acting together to regulate any one gene  A range of expression through as much as a 1000-fold range can be achieved  A bacterial gene is either on or off; a eukaryotic gene can be on or off, but can be any range from any low ON to any high ON 6. Nucleosomes and higher order chromatin structure regulate transcription  Often nucleosomes repress transcription by locking access of transcription factors  Shield DNA from access to transcription factors  In general, chromatin acts to repress transcription  Need to reduce chromatic structure in order to allow transcription to occur Nucleosome position is influenced by many proteins  Some factors bind nicely to chromatin, either because sites are in right positions or they don’t mind presence of histones  Lot of regulatin in eukaryotic cell is to mediate chromatin structure  Two methods of regulating: o Gene-specific regulators recruit chromatin remodeling complex (shift nucleosomes around) o Gene specific regulats recruit histone modifiying proteins (post-translation)alters histones, may loosen them up a bit to allow transcription to occur Regulation of the Galactose inducible genes in yeast as a model for positive control  Q: When would you expect the genes required for galactose metabolism to be expressed? a) In the presence of galactose b) In the presence of glucose c) In the presence of galactose and absence of glucose d) All the time The GAL10 gene of yeast is regulated by galactose  UAS GAL(upstream activating sequence) is required for galactose induction  The transcriptional activator Gal4 binds specifically to binding sites in UAS gal  Gal4 is a site-specific binding protein  Binds to each of 4 boxes (binding sites)  Is a dimer  Lot of similarity with trp repressor  Regions of the Gal4 protein required for function have been identified through deletion analyses  Two regions required for activation  C-terminus – this region is required to activate gene expression  In the presence of galactose, Gal4 activates expression of the Gal genes  Through a mechanism involving interaction with transcription factors, it recruits RNA polymerase  get a lot of transcription  Gal80 inhibits Gal4 function in the absence of galactose  Gal80 blocks activation function of Gal4  Induction of Gal4 by galactose involves a conformational change induced by Gal3  The Gal genes are subject to catabolite repression (they are repressed by glucose)  In the presence of glucose, a protein called Mig1 binds to promoter and blocks RNA polymerase from coming in  This promoter also regulated through negative repression  Negatively regulated by Mig1  By a mechanism similar to what we saw in trp repression  In the absence of glucose Mig1 is phosphorylated and does not enter the nucleus  Glucose – Gal10 not expressed, Mig1 is dephosphorylated, can go into nucleus and block transcription  NO glucose – Gal10 expressed, Mig1 is phosphorylated, found in cytoplasm, basically invisible Brandl Lecture 3 Notes Formation of mRNAs: RNA Processing in Eucaryotic Cells  Primary transcript becomes a mRNA for export out of the nucleus and translation  Three main steps: o 5’ capping o 3’ polyadenylation o Splicing  Each requires a specific enzyme(s) 5’ Capping  When RNA is made, first nucleotide added is a triphosphate, and a 7-methylguanosine is added to it to the 5’ end through a unique 5’-5’ linkage  Marks the 5’ end of the mRNA as being intact  Required for mRNA export from the nucleus  Required for translation of the mRNA into protein (translational initiation) 3’ Polyadenylation  Happens while transcription is ongoing  Sequence that signals polyadenylation, endonuclease cleavage and addition of polyA tail  Helps protect the 3’ end of the mRNA from degradation  Indicates that the 3’ end of the mRNA is intact and therefore is important for: o Export out of the nucleus o Translation RNA Splicing  In eucaryotic cells, protein encoding sequences are interrupted by one or more noncoding sequences called introns  Coding sequences are called exons  Exons = expressed, introns = interruptions  Introns are spliced out of the primary transcript to give the mature RNA  The splicing pattern is often tissue specific Why is Splicing Important?  Differential splicing – a single RNA can be spliced in different ways to create related but distinct proteins  Splicing enhances the coding capacity of the genome – each gene can make more than one protein Mechanism of Splicing  Splicing requires specific sequences in the RNA – it is RNA that triggers splicing  Three key sequences: o 5’ splice junction – between exon and intron o 3’ splice junction – between exon and intron o Branch point – adenine must be present in branch point – essential for splicing  Introns are removed in two consecutive transesterification reactions 1. Attack by the 2’-OH of the adenine at the branch point with the 5’ splice junction – releases 5’ exon and creates lariat structure 2. Attack by the 3’-OH of the free 5’ splice junction with the 3’ splice junction – releases the lariat and joins the two exons Spliceosome  The enzymatic machinery for splicing  Composed of snRNPS, U1, U2, U4, U5, and U6  snRNPS contain both RNA and protein – RNA plays a key role in the splicesome as it holds thing together and is fundamental for molecular recognition  Through RNA-RNA and protein-protein interactions, the splicesome positions the RNA for splicing  Q: In evolutionary terms, the first enzymes were probably composed of? ANS: RNA Self-Splicing RNAs  RNAs can have catalytic activity  Some RNAs can self splice Splicing and Human Disease  Abnormal splicing of the β-globin RNA can result in defective β-globin and hemoglobin deficiency  Very minor changes in the intron can result in defects in splicing RNA Processing  RNA processing occurs when transcription is ongoing  They are concerted processes  Some of the co-ordination is mediated by RNA polymerase – RNA polymerase brings in a number of the processing factors mRNA Export  Processed RNA in the nucleus, must get it out into the cytoplasm of the cell  mRNA export from the nucleus requires interaction of the mRNA with protein carriers  Transport receptors and poly-A-binding protein bind the RNA, allowing it to be exported o Transport receptors act in coordinated manner  Transport occurs through nuclear pores  Nuclear pores are selective gates that control transport in and out of the nucleus o Span the double bilayer Translation: mRNA to Protein  Second half of central dogma Genetic Code  The genetic code “spells” out the amino acid sequence in 3 letter words called codons  Why is each codon 3 bases?  Q: Given that any nucleotide can be A, U, G, or C, how many different 3 base codons are there A: 64 o A one letter code with A, G, U, C would provide: 4 codons o A two letter code would provide 4 x 4 = 16 codons o A three letter code would provide 4 x 4 x 4 = 64 codons o As there are 20 amino acids, a 3 letter code (64 codons is sufficient)  Key feature in the genetic code o Universal o Non-overlapping – gives greatest flexibility in protein structure  Overlapping would allow more coding in the genome but it would place significant restrictions on what amino acid residues could follow each other o No gaps – continuous o Redundancy – some codons specify the same amino acid  Redundancy often occurs at the third position of the codon o Generally amino acids found less frequently in proteins have fewer codons  For example: Met – AUG, Trp – UGG o Functionally related amino acids have similar codons  Only different at the third position (e.g. Asp = GAC, GAU)  Question: Functional related amino acids have similar codons. Why is this advantageous? Answer: Increases the change of a functional protein in the case of a single base mutation  3 stop codons – UAA, UAG, UGA  1 start codon – AUG Brandl Lecture 4 Notes Types of Mutations  Missense mutation – results in a single amino acid change o May or may not have affect on the protein  Frameshift mutation – insertion or removal of bases o Insertion or deletion of a base results in a change in reading frame of the protein from the point of mutation onward  Nonsense mutation – results in premature termination of the protein o Usually have functional consequence, truncates protein from that point onwards tRNA  Transfer RNA  Bring the amino acid to the growing polypeptide chain  All tRNAs have a similar structure  All approximately 80 bases in length  Nonconventional bases o Put in post-transcriptionally  2D cloverleaf structure due to internal base pairing (stems and loops) o Three loops – T loop, D loop, anticodon loop o All RNAs have complex structures as they are single stranded  2 key single stranded regions o 3’ acceptor site – is at the 3’ end which is the region that attaches to the amino acid  All tRNAs end with CCA, that is added post-transcriptionally o Anticodon – base pairs with the codon  In 3D, the tRNAs look like an L Wobble  Most organisms have fewer than 45 different tRNAs  How can all 61 amino acid encoding codons be used?  Some tRNA species must pair with more than one codon  Molecular basis for Wobble: o Base pairing between the anticodon of some tRNAs and the codon, only requires matching at two positions  Example – Phe has two codons (UUC and UUU) o Base pairing interactions for the two Phe codons, with the Phe-tRNA  Accurate base pairing at all 3 positions of the codon  Wobble – accurate base pairing at only the first two positions of the codon Aminoacyl tRNA Synthetases  Couple the 3’ end of a specific tRNA to a specific amino acid  There are 20 aminoacyl tRNA synthetases – one for each amino acid  Specificity is crucial  Charging of tRNAs o Must exquisitely recognize the amino acid and tRNA o Store energy from ATP in a high energy ester linkage  High energy bond between amino acid and tRNA o Energy is used later on for translation o Provide specificity by coupling an amino acid with its correct tRNA – proofreading function  No further check after this coupling Ribosomes  Catalyze protein synthesis  There are two subunits – large subunit and small subunit o Large subunit – 49 proteins and 3 RNA molecules o Small subunit – 33 proteins and 1 RNA molecule  RNA plays a key role structurally and in catalysis o RNA has catalytic activity, not the protein  There are three sites for the binding of tRNAs on the ribosome o A site, P site, and E site o A site – the site that binds aminoacyl-tRNA o P site – the site that binds peptidyl-tRNA o E site – the site from which tRNAs exit  mRNA is bound close to the A and P sites Steps in Translation  Figure 7-33 assumes that translation has already started  A peptidyl-tRNA is in the P site of the ribosome  mRNA is decoded in a 5’ to 3’ direction, one codon at a time  Step 1 o An aminoacyl-tRNA binds to the A site of the ribosome o Requires basepairing between the tRNA and the codon  Step 2 o Enzymatic step o The energy from the ester bond of the peptidyl-tRNA in the P site is used to form a new peptide bond between the amino acids in the A and P sites  Step 3 o Peptide bond formation is coupled to a conformation change in the ribosome that shifts the large subunit “forward” (downstream towards the 3’ end of the RNA) o This conformational change repositions the tRNAs  P site (3) into E site and A site (4) into P site  Step 4 o Small subunit moves forward (towards the 3’ end) exactly 3 bases from the codon o The tRNA leaves the E site o Step 1 is repeated  Question: Specificity of translation comes from what step? Answer: Correct charging of the aa-tRNA synthetase and base-pairing of t-RNA and codon in the A-site of the ribosome Key Points for Translation  In eukaryotes, translation occurs in the cytoplasm (recently some reports of low levels of nuclear translation – still being debated)  Energy for peptide bind synthesis comes from the high energy aa-tRNA ester bond and indirectly comes from ATP  Specificity comes from o Specificity of the aminoacyl tRNA synthetase o The requirement for base pairing in the A site of the ribosome  Complex reaction involving o Both RNA and protein molecules (RNA plays a key structural and catalytic role) o Conformational (shape) changes in the ribosome are required  Occurs in a 5’ to 3’ direction along the RNA making a protein from the N to C terminus o First part of RNA exposed is 5 ‘end, which initiates translation  mRNA is decoded one codon at a time, 3 bases at a time – step by step process Initiation of Translation  Starting a protein out-of-frame could result in a nonfunctional protein  How is the start site of translation determined? o Eukaryotes  Initiator met-tRNA binds to small ribosomal subunits with translation initiation which binds to 5’ cap  Complex scans RNA downstream 5’ to 3’, looking for AUG  Once bound to AUG, large subunit comes in and initiator met-tRNA is in the P site o Prokaryotes  Ribosomes recognize internal ribosome binding sites found just upstream from each functional AUG  Question: Why must translational initiation in bacteria use a mechanism other than 5’ scanning? Answer: There are multiple proteins encoded from common mRNAs in bacteria Brandl Lecture 5 Notes Termination of Translation  Requires o One of the three stop codons  No tRNAs for stop codons o Specific termination factors  Binding of release factor to the A site occurs during termination  Release factor adds water to the chain rather than another amino acid, which releases the peptide from the tRNA and terminates translation o Catalysis reaction Antibiotics and Translation  Many antibiotics block bacterial translation  Its intricate molecular interactions and importance make it a prime therapeutic target  See table 7.3 – do not memorize mechanisms but understand how they work  It is important to recognize that most tools of recombinant DNA technology use chemical quantities of the DNA o While I will present the DNA as a single line, there are in fact many identical molecules in the reaction o Q: In a DNA sequencing reaction, you generally add about 100 ng of DNA. Assuming the DNA is 3000 base pairs, how many molecules of template are added? o One base pair has a mass of about 600 Daltons (g/mole) A: 3 x 100 5 6  1 mole of a DN6 of 3000 bp has a mass -7: 3000 bp x 600 g/bp = 18 x 10 (approx. 2 x 10 )  1 mole – 2 x 10 , X mole – 100 ng (1 x 10 )  X = 1 x 10 / 2 x 10 = 5 x 104 -14 23 10  How many molecules? 5 x 10 x 6.02 x 10 (Avogadro’s number) = 3 x 10 Recombinant DNA Technology Genetic Engineering  The techniques which allow DNA fragments from different sources to be recombined to make new molecules with unique features Synthetic Biology  The construction of new biological parts and systems  The re-design of existing, natural systems for useful purposes  Plant/crop and dog breeding are examples of humans manipulated the DNA of other organisms  What is novel about recombinant DNA technology? Comparison of Classical Genetic Techniques and Recombinant DNA Technology  Classical genetics o Slow: limited by the breeding time of the organisms and change genetic events o Limited to breeding species  Recombinant DNA technology o Rapid: as quick as a few days in some organisms o No limitations Significance of Recombinant DNA Technology  Research o Recombinant DNA technology is a powerful tool in understanding cellular and molecular structure and function o Recombinant DNA technology was critical in genome sequencing projects  Biotechnology o The use of organisms to do work for man o Modern biotechnology impacts many aspects of society and relies heavily on recombinant DNA technology o Remains one of the fastest growing segments of the North American economy Medicine  Drug production and design o Human insulin, human growth hormone, vaccines, Artemisinin (malaria), interleukin  Diagnosis of disease – detect pathogens and disease causing genes through their DNA signatures Example: o Blood sample from infected person o Remove cells by centrifugation o Rare HIV particle in serum of infected person o Extract viral RNA genome o Reverse transcriptase/PCR amplification o Control, using blood from non-infected person  Genetic counseling – whether someone is a carrier of a disease related allele  Potential for gene therapy – many diseases result from having a defective gene o We can add back functional genes to cure disease Agriculture  Production of crops with unique features  Examples include: o Vitamin A enhanced rice (golden rice) o Cold and drought resistant crops o Pest resistant corn o Non-spoiling tomatoes o Non-allergenic peanuts  Production of novel molecules in plants o For example: large scale production of therapeutic drugs  Production of organic fuels Manufacturing  Novel products (such as spider silk, valuable fibre)  More efficiently (such as detergents)  Environmentally friendly (such as biofuels) Environment  Bacteria that “eat” oil, styrofoam, etc. Forensics and Law  DNA fingerprinting Recombinant DNA Technology Case Study  Cloning and expression of your favorite gene (YFG) in E. Coli o You have identified a human protein that will be of tremendous value in treating disease o Your goal is to mass produce this protein (YFP)  Expression of YFP (your favorite protein) using recombinant DNA technology o Why is this important?  Many proteins are difficult to obtain from their native source because of their level of expression and/or difficulties in purification  Recombinant DNA technology can dramatically increase expression and facilitate purification of YFP  The process of expressing YFP will require: o Cloning YFG o Introducing YFG into E. coli o Purifying the protein from E. coli  E. coli are the most commonly used organisms for expressing YFP o They grow quickly and inexpensively o Protein extracts are made easily o Genetic engineering is simple o Multicopy plasmids and strong promoters can drive expression  How do we get enough of YFG to be able to clone it? o Synthetic oligonucleotides o Hybridization o PCR Brandl Lecture 6 Notes Synthetic Oligonucleotides  Fragments of single stranded DNA with defined sequence that are made synthetically  When they are short (20-30 bases), they are often referred to as primers or oligos Nucleic Acid Hybridization  Denaturation/melting  Renaturation/annealing/hybridization  Crucial in PCR and other processes DNA Denaturation  If you have a duplex DNA, and you heat it up, it will denature into two individual strands o Can be done by heating or hydroxide Melting temperature (Tm)  Given a specific DNA at defined concentration, the temperature at which 50% of the molecules are single stranded o Characteristic of every DNA molecule o At low temperatures, all DNA molecules are double stranded and at high temperatures, all DNA molecules are single stranded o Referring to chemical quantities of molecules, not one molecule  Factors that determine the Tm – intrinsic features of the molecule itself o G:C content  The more G:C content, the higher the Tm  This is because G-C base pair has three hydrogen bonds, whereas A-T only has two hydrogen bonds so it is easier to melt A-T bonds  G-C bonds have more base stacking interaction as well, thus they have a higher melting temperature o Length  The longer the DNA molecule, the higher its melting temperature  Melting temperature is approximately 4 degrees for every G:C and 2 degrees for every A:T o Degree of complementarity  Degree of complementarity between the two nucleic acid strands  DNAs that are not perfectly complementary (not perfect matches) will anneal or hybridize  If there are too many mismatches, they will not hybridize at all, h
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