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Biochemistry 2280A
Eric Ball

How to do well: READ THE TEXT BOOK + COURSE PACKAGE  Assignment: use of data bases, looking at gene structure as well as give you a practical on recombinant DNA technology and genomic section. GENE EXPRESSION (Topic 22-23) DNA- the blue print of life  Info stored in the form of genes  About 25000 genes in the human genome  Out of the ones from above only one cell type of about 10000 of the genes are expressed  Correct expression of the gene is essential for growth and differentiation of cells  Also critical in the growth and development of all organisms and their ability to respond to their environments. Significance of gene expression  Diseases are due to altered expression of one or more genes.  Trying to manipulate the gene expression; by manipulating the gene potential we have the ability to prevent or reverse disease states. Why is there RNA step in Gene Expression (DNA→RNA→Protein) 1. Since the genome has one copy of each gene, RNA provides an amplification step that allows genes to be expressed at different levels; probably one of the major reasons, why RNA is in gene expression. Can’t use DNA b/c can’t be directly amplified. 2. It can be rapidly degraded, make RNA to turn the gene on, and degrade RNA to turn the gene off really fast. (RNA can be easily degraded; Quickly stopping the expression of the gene.) 3. RNA provides many additional opportunities to regulate gene expression; so RNA allows many more steps to regulate the gene expression. (RNA provides many additional opportunities to regulate the expression of genes) Are there two steps in gene expression or more?  Despite the fact that the central dogma makes it appear that there are two steps, there are in fact many many more steps in gene expression. Gene Expression steps in EUCARIOTIC cell (DNA—transcription—→RNA—translation—→ protein) 1. Transcription a. Initiation b. Elongation c. Termination 2. RNA processing —Breaks the central dogma from DNA to RNA to protein, unchanged doesn’t always happen. Some RNA can be modified to actually change the sequence that is in the genetic code. (all of this is happening in the nucleus) a. RNA editing b. 5’capping c. Splicing d. 3’polyadenylation 3. mRNA export —Getting the mRNA into the cytoplasm, it is a highly regulated process. If the protein encoded gene (mRNA) doesn’t get to the cytoplasm this isn’t doing very much of the for the gene expression. It is a key step in regulation. 4. Translation —once the mRNA is in the cytoplasm this step will instantly be initiated. a. Initiation b. Elongation c. Termination 5. mRNA Degradation — The amount of mRNA is found in the cell depends on two factors: 1) rate of synthesis 2) rate of decay So if the RNA id decayed really fast, you would not have much of it around meaning that we would have less of the protein. The amount RNA in the cell depends both on the amount of transcription in the nucleus and the amount of degradation going on (outside the nucleus) and this is often controlled so it doesn't just occur randomly. 6. Protein Modification — b/c many proteins are inactive in their natural form they have to be modified in order to become active. Some of the modification is through phosphorylation and some through ubiquination, acetylation etc. all of these can affect protein function. 7. Protein Degradation — Similar to regulation of RNA levels; we can control the amount of proteins available by controlling the rate at which we make proteins through translation + rate of their degradation. Genes are regulated at many different levels it’s the balance if its synthesis and degradation that determines how much protein we have. Which step in a biological process is most highly regulated?  The first step because it saves energy required in subsequent steps. The first step in a biological process is often the most critical for regulation. Because transcription is the first step in gene regulation, it is often most highly regulated. A—ATP→ADP —→B—→C—→D—→E o Many disease results from defects in transcription factors, and the cures for these diseases are targeted to transcription. Focus on Prokaryotic Transcription  It involves fundamental mechanisms and molecules that are similar on both prokaryotic and eukaryotic. But we will focus on PROKARYOTIC transcription. Terminology o Promoter= the DNA sequence required for the specific initiation of transcription of a gene or operon. (The DNA sequence that is required to initiate transcription.) o Operon= set of genes transcribed from a single promoter and thus expressed from a common RNA. (Polysesranic- from many genes expression) NOT present in Eukaryotes. o Terminator= the DNA sequence required to stop transcription, it has no zero. Key features of the a Bacterial promoter  This is a double-helix DNA, and RNA is transcribed from it.  The RNA transcription site is the spot at which the RNA synthesis begins (+1).  The +1 is the starting point (initiation site) of transcription, next base is +2 and the base upstream to that is -1 (notice: there is NO zero base).  They were discovered by the alignment of sequences from the E.coli promoters… identified that the regions had some similar sequences.  Three key sequences are required for transcriptional initiation. One is a start site, and the other 2 are the -35 sequence and the -10 sequence.  The -35 has a consensus sequence of TTGACA and -10 have TATAAT. These sequences are required by RNA polymerase to recognize the promoter. They are positioned about 17 base pairs apart, and the -10 site is about -10 base pairs from the transcription start site.  Consensus sequence = most frequently occurring base in a group of functionally related DNA elements, not necessary and actual thing that happens  The consensus for image= C  Start sight +1, -10,-30,-35 BACTERIAL RNA POLYMERASE  The enzyme that will synthesize specific RNA transcripts using DNA as the template, and nucleoside triphosphates (NTPs) as substrates.  There are other enzymes that will make RNA but only this one is used when the bacteria makes RNA using DNA as a template.  Multiple-subunit enzyme o CORE enzyme o WHOLE enzyme  Core enzyme= synthesizes RNA from ends and nicks in the DNA, it has little specificity thus unable to recognize promoters; the promoter is recognized by an additional subunit, the sigma subunit.  The sigma is what recognizes the -35 and -10 sequences through protein-DNA interactions (contact) between the -35 DNA sequence and some DNA sequence and same with -10. This is what gives rise to the specificity.  The whole enzyme is the core + the sigma subunit= the whole enzyme that will recognize the promoters allowing for the initiation of specific transcription.  Sigma subunit makes a specific contact with the -10 and -35, the recognition site for RNA polymerase: Sigma decreases the binding of RNAP to nonspecific DNA  RNA polymerase-promoter interactions are one of many examples demonstrating the importance of protein-nucleic acid interactions. STEPS OF INITIATION 1. The whole enzyme must recognise/interact with the promoter; closed complex 2. RNA polymerase unwinds the DNA strands around the start site of transcription forming an open complex 3. The first NTP is brought to the template DNA—unlike replication no primer is required, comes to the +1 site. 4. Using the NTPs as substrates the elongation proceeds in a 5’—3’ direction is started and it follows the base rule paring a. Phosphodiester bonds are broken 5. After adding about 5—10 nucleotides, sigma falls off the holoenzyme 6. The transcription bubble moves down stream with the re-annealing template DNA remaining behind 7. Chain of elongation continues until a terminator is reached, and RNAP falls off 8. Sigma rebinds the RNAP again and the process starts again (also known as the sigma cycle) GENE EXPRESSION (Lecture 2) Regulation of Transcription (Three steps) 1. Strength of basic promoter elements  Some genes have better -10 and -35 sequence, some interact more tightly with sigma binders, some expressed at higher levels because they have better prefer RNA polymerase. Evolutionary change on the time scale of the cell, static thing. Can’t be regulated. NOT DYNAMIC 2. More than one sigma factor  Each recognizes a different promoter. Most predominant is the “housekeeping ” sigma,  , Others include… (nitrogen metabolism) and  (heat shock) 3. Gene specific regulatory proteins bind specific DNA sequences found in one or more promoters and serve to activate or repress transcription; used largely by eukaryotic cells • Dynamic! • Positive regulation – factors activate transcription – an activator protein (CAP) regulates transcription of the lac operon and other genes involved in carbon metabolism • Negative regulation – factors repress transcription - tryptophan (Trp) repressor Tryp Operon  In E. coli, the 5 genes (E, D, C, B, A) for tryptophan biosynthesis are transcribed as a single unit from a common promoter; they are transcribed, and regulated by the tryptophan in the environment.  The Trp operon is expressed when there is a little tryptophan in the cellular environment  When tryptophan concentration is low… o Tryptophan dissociates from the Trp promoter – Trp repressor no longer binds the Trp operator o RNA polymerase can access the Trp promoter – initiates transcription o RNA polymerase is recruited to the Trp promoter by the -10 and -35 sequences allowing transcription o Cell wants to turn on gene biosynthesis when little Trp around – if it can’t use Trp from environment, it will make its own gene o At low concentrations of tryptophan, the tryptophan comes off and now there is a structural change in the protein - the alhpa-5 helices tilt inwards and this form of the protein can no longer make good contacts to the DNA  When tryptophan concentration is high… o The Trp repressor binds tryptophan operon and blocks expression o A conformational change occurs in the repressor’s structure such that it can bind the operator site within the promoter o Binding of the repressor sterically inhibits access of RNA polymerase to the promoter Trp promoter:  Between the -35 and -10 sequences is an element called the Trp operator that binds a protein called the Trp repressor  At high concentrations of tryptophan, Trp repressor binds the amino acid tryptophan What is the structural basis for this regulation?  Trp repressor – monomer o 107 amino acid residues o Has a helix turn helix motif that is required for DNA binding o Alpha helices 4 and 5 make up the helix turn helix motif – make base specific interaction with DNA o Binds DNA as a dimer  Trp repressor – dimer o Protein dimer has 2 fold symmetry o Amino acid side chains on helix 5 make base specific contacts with the major groove of its operator sequence The Trp repressor dimer with the core of the protein held together via protein-protein interactions →  (a) Trp repressor in the absence of bound tryptophan – it cannot bind the operator DNA since the recognition helices are tilted inward  (b) Active Trp repressor (capable of binding its operator) when tryptophan is bound – recognition helix 5 on each of the 2 monomers makes base specific contacts in consecutive major grooves of the DNA General themes to take from the Trp operon: 1. Trp repressor is a site specific DNA binding protein • Protein makes base specific contact based on the DNA sequence • Only want binding in certain spots of the genome 2. There is a binding site for Trp repressor within the Trp promoter 3. Trp repressor inhibits transcription by blocking access of RNAP to the promoter 4. Trp repressor is responsive to an environmental signal • In this case, it is responsive to tryptophan • Can turn genes on and off in response to tryptophan Genetics question→ C; mutations that allow the operon to be always on…. All the mutations defined how the protein work… genetics crucial on determining how all this works. Q- When is the trp operon expressed? a) When there is a lot of tryptophan in the cellular environment. b) When there is a little tryptophan in the cellular environment. c) All the time*** d) Unsure Six difference in transcription between prokaryotic and eukaryotes: 1. Prokaryotes have one RNA polymerase and eukaryotes have 3:  Pol I – most rRNA genes  Pol II – mRNAs  Pol III – tRNAs and snRNA *Similar in structure but they differ enough that they recognize different promoters 2. No operons in eucaryotes  Each gene is transcribed as a single unit (monocistronic)  In prokaryotes, the contrast is polycistronic – bacteria 3. Promoter structure  Promoter recognition is through a distinct set of proteins  Eukaryotes do not have -10 and -35 sequences – there is no sigma factor  The role of these transcription factors is similar to bacterial sigma but more complex  One of these factors, TATA-binding protein, is a component of the transcription factor TFIID - recruits factors that results in RNA polymerase being recruited to the promoter  It binds a basal promoter element, the TATA-box, found ~20 base-pairs upstream of the transcriptional start site +1 4. In eukaryotes, regulatory elements are often bind the DNA many thousands of base pairs distant from +1  They may be brought into proximity of the promoter by DNA looping  The proteins bind DNA several thousands of base pairs away from the start site  In prokaryotes, the regulatory proteins bind very close to the start site 5. Nucleosomes and higher order chromatin structure have a profound effect on determining the access of transcription factors to DNA  Often nucleosomes repress transcription by blocking access of transcription factors o modification of nucleosome positioning is influenced by many proteins  Another set of factors modify the histones - that can help recruit chromatin remodeling complex or in turn allow chromatin remodeling and chromatin re-modification 6. Combined control: groups of proteins work together to determine the expression of a gene  In bacteria, promoters are often on or off and not in between  In eukaryotes, it is often like a dimmer switch, showing a wide range of expression  … ranges over approximately 1000 fold range; because many factors act on the promoter.  Lot of regulation to mediate the structure as well as allow for the transcription. ***Eukaryotic cells control gene expression through chromatin*** Positive Control:  Gene specific regulatory proteins can also act to enhance transcription  Increased transcription is generally accomplished by increasing the rate of recruitment of RNA polymerase or the activity of RNA polymerase  Both enhanced recruitment and/ or enhanced activity of RNA polymerase are usually achieved by protein-protein interactions between the activator protein and RNA polymerase  GAL10 require an up-stream activating sequences… Each of the boxes bind to a specific GAL14 as a dimer, the regions of the GAL14 that are required for function have been identified from the deletion analysis; one requires DNA other capable of transcription activation.  In the presence of glactose the GAL4 binds to the sequences, recruiting the RNA polymerase and achieves a lot of transcription… expression by direct and indirect interactions… much more complex process.  Without the presence of glactose the GLA80 that binds to it in order to block the activation of 80 inhibitors. There is another GAL3 that activates the GAL14… Book notes page 275-279… ***Positive regulation of the Lac operon is sensitive to the presence/absence of glucose in the cellular environment*** Negative Regulation at the Lac Operon:  Cells only require expression of the lac operon if lactose is in the media  In the absence of lactose… o The lac repressor binds the promoter, inhibiting the action of RNA polymerase o Inhibits transcription  In the presence of lactose… o The lac repressor binds lactose - results in a conformational change in the protein so that it no longer binds the operator sequence in the promoter (dissociates from the operator) o Thus allowing RNA polymerase to come in and transcription to occur  -10 and 35 are not very strong here **STUFF AT THE SIDE** Glucose, No Lactose  High glucose, low cAMP and any CAP around can’t bind to DNA  RNAP can’t interact with the promoter and can’t initiate transcription Glucose and Lactose  High glucose, low cAMP  Lac repressor is bound to the lactose and can’t bind to DNA  Polymerase can come in here so some transcription can occur…BUT…  Low levels of transcription because -10 and -35 not very strong Lactose, No Glucose  Low glucose, high cAMP  The -10 and -35 open, high cAMP, so CAP is bound to cAMP so it can bind to sequence which helps recruit RNA polymerase and you get high levels of transcription No Glucose, No Lactose  Lac repressor would be bound to the operator  Transcription would be inhibited almost completely FORMATION OF mRNAs: RNA PROCESSING (Lecture 3) The GAL genes in yeast are regulated by GLACTOSE and GLUCOSE. They are subject to positive and negative regulation, and the DNA binding proteins are important for both the negative and positive regulation. The protein interactions are important for regulation; the regulation occurs through the transcriptional machinery and changes in chromatin, structure. Whole together, the system is sensitive to the environmental signals; the signals result in change in gene expression. Primary RNA transcript ↓ Processed to become a translatable mRNA ↓ Exported out of the nucleus ↓ Translation  5’ capping  3’ polyadenylation  Splicing concerted processes (they happen together) **Each step requires specific enzymes** 5’Capping  5’end of the message is capped with 7 methyl guanosine (7-mG)  Capping is required for RNA export from the nucleus (marks the end of the mRNA as being intact), aids in stabilizing the mRNA from being degraded and acts as a translational signal 3’ Polyadenylation:  The addition of a long A-tract to the 3’ end of the RNA by enzymes downstream  The RNA is first cleaved ~30 bases following an AAUAAA sequence which is 3’ to the coding region – a string of A residues is then added (Downstream =3’ and Upstream= 5’)  Enzymes recognize AAUAAA sequence  clips  then polyA tail is added  The polyA tail helps protect the 3’end of the mRNA from degradation o Provides additional stability to the mRNA by reducing effects of 3’ endonucleases  Indicates that the 3’ end of the mRNA is intact and therefore is important for export out of the nucleus and for translation o If you have the cap to the polyA tail, the cell wants to export that RNA into the cytoplasm and ultimately translate it RNA splicing:  In eukaryotic cells, protein coding sequences are interrupted by one or more non-coding sequences called introns o Genes are not continuous, they are interrupted SPLICING QUESTION→D o Introns = interruptions  The coding sequences are called exons o Exons = expressed  Introns are spliced out of the primary transcript to form the mature mRNA  DIFFERENTIAL SPLICING= a single RNA (primary transcript) can be spliced in different ways to create related but distinct (mRNAs) proteins. o Often tissue specific… due to the location of the specific gene.  Splicing enhances the coding capacity of the genome.  It is unclear if prokaryotes lost introns or if eukaryotes gained introns SPLICOSOME— snRNPs  Introns are removed as the result of the catalytic activity found in small nuclear ribonucleoprotein particles (snRNPs = snurps) o The snRNPs as the name suggests are complexes of RNA (small nuclear RNA) and protein  The RNA component has a recognitions function, acting through base pairing with sequences on the precursor mRNA o The specific sequences that are required in RNA:  5’ splice junction  3’ splice junction  Branch point (adenine is essential for splicing)  The snRNPs are also critical in arranging the ends into position o Through RNA-RNA and protein-protein interactions, the spliceosome positions the RNA for splicing  The assembly of snRNPs that catalyze splicing is called a splicesome Splicing (intron removal) occurs by 2 consecutive transesterification reactions:  Cleavage at the exon 1-intron boundary results from the attack of the 5’ splice junction by adenine in the branch point  This generates a lariat structure as the result of the unique 2’-phosphodiester bond  In the second reaction, the 3’ OH of exon 1 reacts with the 3’ splice junction, cutting out the intron and joining exons 1 through 2  The enzymatic machinery for splicing - snRNPs, U1, U2, U4, U5, and U6 *The first enzymes were likely composed of RNA* Self-Splicing RNAs:  In pre mRNA in protozoa, mitochondria & chloroplasts, snRNPs are not required  Through RNA-RNA and protein-protein interactions, the spliceosome positions the RNA for splicing  The RNA is capable of catalyzing its own conversion to mRNA (self-splice) o This is one example of RNAs having catalytic potential o Another example is ribozymes, which catalyze the cleavage of other RNA molecules in a sequence specific fashion Splicing and Human Disease:  Abnormal splicing of the human -globin RNA can result in defective -globin and hemoglobin deficiency (problem due to splicing)  Normal adult -globin primary RNA transcript The mRNA export from the nucleus:  In eukaryotic cells, RNA is synthesized in the nucleus and translated in the cytoplasm; highly REGULATED  Export requires interaction of the mRNA with protein carriers  Export can be regulated by the control of export factors.  Mature mRNAs are transported from the nucleus to cytoplasm through nuclear pores - highly organized set of proteins in the nuclear membrane o Nuclear pores are selective gates that control transport in and out of the nucleus o Contain basket like structure on the inside, fibrils on the outside  The “waste” products from splicing (introns) are not transported  Transport through nuclear pores is a regulated process that requires the recognition of proteins bound to the poly-A-tail, the 5’ cap, and often internally o Somewhat schematic – extremely complicated structure so must be a fair bit of regulation going on Translation – mRNA to protein  The decoding of the mRNA to produce a protein occurs on the ribosome  This process of translation is complex, requiring many RNA and protein factors TRANSOLATION OF mRNA TO PROTEIN (Lecture 4) Genetic Code:  The genetic code “spells” out the amino acid sequence in 3 letter words called codons  Each protein has a specific reading frame that is determined by where the decoding process begins How many amino acids are used for protein synthesis in the cell (different codons does the cell needs?) – 20  Generally amino acids are found less frequently in proteins and have less codons.  Functionally related amino acids have similar codons. The code is:  As there are 20 amino acids, a 3-letter code (64 codons) is sufficient  Universal (found in all organisms)  Non-overlapping  Commaless, no gaps o (gaps were removed by splicing)  61 codons for 20 amino acids o therefore redundant  Redundancy – some codons specify the same amino acid o Occurs at the 3 position of the codon (wobble position)  3 stop codons: o UAA, UAG, UGA  1 start codon: o AUG (used in 99% of time by proteins) QUESTION 1_ D — the advantage of the functionally related amino acids that have similar codons is that it increases the chance of a functional protein in the case of a single base mutation. Types of Mutations  Missense Mutation – single base mutation which results in a single amino acid change  Frameshift mutation – could be the insertion of one or two or taking away one or two – insertion or deletion of a base results in a change in reading frame of the protein from the point of the mutation onward _Have serious consequences on the protein.  Nonsense mutation – a nonsense mutation results in premature termination of the protein tRNA (transfer):  Vehicle that bring amino acids to the growing polypeptide chain  Function in a codon specific fashion, relying on base pairing rules  All roughly ~80 nucleotides in length and have a cloverleaf secondary structure due to internal base pairing (stems and loops)  The anticodon of the tRNA hybridizes with the codon  TWO key single stranded regions in RNA which play crucial role in function: 1. 3’ acceptor site... 3’ amino acid receptor site is attached to amino acid, posttranscriptional, added after RNA is made 2. Other key single stranded region: anti codon. In anti-codon, base pair with codon using normal base pair rules except for wobble.  The correct amino acid is covalently linked to the 3’ end of the tRNA (3 acceptor site)  Wobble results from the fact that accurate base pairing for some tRNAs only requires matching at the first 2 positions o Anti-codon base pairs with the codon using normal base-pair rules  Non-conventional bases  Modified bases put in afterward  Don’t have clover leaf structure in real life – in 3D looks more like an L structure  Most organisms have fewer than 45 different tRNAs, some tRNA species pair with more than one codon. Molecular Basis for Wobble: 1. Some amino acids have more than one tRNA 2. Accurate base pairing for some tRNAs only requires matching at the first two positions of the codon  Base-pairing interactions for the 2 Phe codons, with the Phe-tRNA  Accurate base pairing at the 3 positions of the codon  Accurate base pairing at only the first two positions of the codon – WOBBLE Amino Acid Activation – Aminoacyl tRNA synthetases:  Using ATP as the energy source, the carboxyl group of a specific amino acid is coupled to the 3’ end of a specific tRNA in a high-energy bond  20 of these enzymes: one for each amino acid  Amino acid activation by aminoacyl t-RNA synthetases (otherwise known as charging of tRNAs) is important for: 1. Providing an energy source for later peptide bond formation – store energy from ATP in a high energy ester linkage 2. Providing specificity by coupling the correct amino acid to the specific tRNA Ribosomes: (molecular assembly that catalyzes protein synthesis)  Protein synthesis occurs on large multimeric protein RNA complexes called ribosomes  Ribosomes have 2 subunits, large and small
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