×
Log in to StudySoup
Get Full Access to UGA - BCMB 3100 - Study Guide - Final
Join StudySoup for FREE
Get Full Access to UGA - BCMB 3100 - Study Guide - Final

Already have an account? Login here
×
Reset your password

transcribeo

transcribeo

Description

School: University of Georgia
Department: Biology
Course: Intro to Biochem and Molecular Bio
Professor: Wood sabatini
Term: Fall 2016
Tags: DNA, translation, elongation, termination, and RNA
Cost: 50
Name: Final Quiz Study Guide
Description: Chapters 36, 37, 39, and 40
Uploaded: 12/02/2016
11 Pages 168 Views 6 Unlocks
Reviews


Quiz 16: CH 36,37,39,40 The quiz is about 25 multiple choice questions in 50 min ∙ RNA synthesis  o Gene- DNA sequence that is transcribed (including genes that don’t encode  proteins) o Housekeeping genes- encode proteins or RNA essential for cell function  ▪ Ex: tRNAs, rRNAs, enzymes in metabolic pathways o Biological information flows from nucleic acids to proteins and is irreversible, but  there are exceptions  ∙ Central Dogma revised  o Reverse transcriptase makes DNA from RNA in retroviruses  ∙ Genome- RNA or DNA can carry genetic info ∙ You can regulate transcription, processing (which is precursor RNA ???? mature RNA), or  translation  o Primary mechanism of gene regulation is transcription (to transcribe or not  transcribe)  o Precursor transcripts are the primary product of transcription of genes and are  often modified by splicing, editing, cleavage or modification ∙ Transcription o Initiation ???? elongation ???? termination o Initiation: Unwinding of DNA helix, synthesis of a short length of RNA. No helicase  or primase present. This is the rate-limiting step because it takes time for the  polymerase to bind at certain site and begin initiation  o Elongation: Addition of ribonucleotide bases matching the “coding” strand and  complementary to the “template” strand.  ▪ Polymerization is in the 5’→3’ direction. o Termination: RNA synthesis stops at “pause” sites—GC-rich sites on the DNA  (RNA). o RNA processing adds poly-A tail to precursor RNA to form mature RNA, which  regulates the amount of mRNA for protein production and it stabilizes the RNA ∙ Template strand provides template to make RNA by reading and providing complementary  base pairs o Convention for double-stranded DNA: Coding strand (top) is written: 5’→3’ Template strand (bottom) is written: 3’→5’ o Gene is transcribed from 5’ end to the 3’ end: Template strand of DNA is copied from the 3’ end to the 5’ end o Growth of RNA chain proceeds 5’→3’  ∙ Bacterial RNA polymerase catalyzes DNA- directed RNA synthesis  o RNA Pol is part of a larger transcription complex  o DNA is continuously unwound as RNA Pol catalyzes a processive elongation of the  RNA chain o RNA Pol “holoenzyme” recognizes and binds to duplex DNA and catalyzes the  synthesis of RNA in a 5’ to 3’ direction and is “processive” ∙ Subunits of RNA polymerase  o Core components are used for elongation, sigma factor is important for initiation of  transcription  o Core + sigma factor forms the holoenzyme  ∙ Promoters are DNA sequences that are upstream from transcription start sites and recruit  RNA polymerase and tell it to start initiation here  ∙ Bacterial Promoter  o DNA binding proteins bind to consensus sequences and allows transcription to  begin about 10 bases down from the sequence  o The sigma subunit is required for promoter recognition ∙ When the sigma unit is bound to DNA the core complex, it o decreases the affinity of the polymerase for DNA by 10,000 fold, allows the  enzymes to rapidly scan the DNA for a “promoter” o increases the affinity of the core polymerase for specific promoter sequences o Once the promoter is located and RNA synthesis begins, the σ subunit dissociates  from the enzyme to assist another polymerase in initiation ∙ Termination: transcription complexes disassemble at the 3’ end of genes at specific  termination sequences 1. Unstable elongation (rho-independent): G-C rich region in the RNA forms  hydrogen bonds to make an unpaired hairpin loop structure. This greatly pauses the  polymerase. The instability of the remaining U-rich region (oligU tail) allows for the  DNA-RNA complex to dissociate 2. Rho-dependent termination: When the polymerase pauses, the rho protein uses  ATP to act as a helicase as it pulls the RNA strand away from RNA polymerase and  the DNA template, disrupting transcription o Pause sites - regions of the gene where the rate of elongation slows down (10 to  100-fold) or stops temporarily  ∙ Regulation of Gene Expression  o Expression of housekeeping genes is constitutive ▪ These genes usually have strong promoters and are efficiently and  continuously transcribed  ▪ Genes whose products are required at low levels have weak promoters and  are infrequently transcribed ▪ Strong promoter=has sequence close to consensus sequence. These are  DNA sequences where the promoters bind. A weak promoter are regions of  DNA with a mutation that changes the consensus sequence.  o Regulated genes are expressed at different levels under different conditions Nutrients/stress ▪ Regulation can occur at any point in the flow of biological information  (transcription initiation is most frequent) ▪ Primary way to regulate a gene is transcription  ▪ Ex: lac operon in bacteria  ∙ Lac operon o An operator is a DNA sequence where a repressor binds, inhibiting transcription. The  gene that makes the repressor is lac I. ▪ When the repressor is bound to the operator, it forms a DNA loop so polymerase  is prevented from getting through to transcribe the genes  ▪ The genes are repressed when there is no lactose present or high glucose o When glucose is scarce or lactose is present, the lac operon genes are expressed and  transcribe in order to metabolize lactose ▪ Lac Z, Y, and A are genes involved in breaking down lactose o b-Galactosidase (lacZ gene product) – hydrolyzes lactose o Lactose permease (lacY gene product) - transporter for lactose o Thiogalactoside transacetylase (lacA gene product) – acetylates nonmetabolizable b galactosides o How does a cell sense lactose to kick the repressor off? Lactose is converted to  allolactose by B-galactosidase. Allolactose acts as an inducer and binds to repressor,  therefore preventing the repressor from binding to operator.  ▪ “escape synthesis” allows small level of permease and galactosidase which  serves to detect lactose when present o Further activation occurs when cyclic AMP binds to CRP: you get 50 fold increase of  transcription ▪ cAMP is produced when glucose levels are low   Glucose only Lac operon repressed Glucose + lactose Partially induced Lactose only Fully induced  (cAMP/cRP too)


What’s the difference in RNA polymerases in bacteria and eukaryotes?




How is it recognized by RNA Pol?




o How does a cell sense lactose to kick the repressor off?



Don't forget about the age old question of 1/2mv^2 + 1/2iw^2
If you want to learn more check out non cumulative exam
We also discuss several other topics like chem 101 exam 3
Don't forget about the age old question of the hollow space in the center of the gray commissure of the spinal cord that helps to circulate csf is called the
If you want to learn more check out marketers are particularly interested in postpurchase behavior because it
Don't forget about the age old question of gsu music

So, lactose +/- glucose affects gene expression  ∙ Gene expression in prokaryote vs Eukaryote  o Eukaryotes are more complex because… ▪ More polymerases  ∙ Pol I transcribes ribosomal RNAs ∙ Pol II transcribes mRNA  ∙ Pol III transcribes tRNA ▪ Multiple promoter elements ▪ Transcription and translation occur in different compartments ▪ Transcription initiation is influenced by elements outside the immediate  promoter region ∙ (enhancers and silencers)  ▪ Many accessory proteins ▪ RNA polymerase do not bind DNA directly ∙ Eukaryotic Pol II promoter components o Cis-acting elements: DNA sequence that regulates gene on the same molecule of DNA o TATA binding protein binds the TATA box region  ▪ TATA binding proteins bends and unwinds DNA and provides platform for other  components  o Transcription factor II is a factor for Pol II  o Initiator elements o DPE: downstream promoter element o Transcription Factor binding  a. TFIID binds TATA box then recruits other factors, including A, B, and H b. TFIIA/TFIIB/TFIIH bind c. Pol II binds d. TFIIH has a helicase activity allows separation of strands and kinase activity that  phosphorylates amino acids on the tail of Pol II  i. Phosphorylation promotes transition to the elongation complex to begin  transcription: “promoter clearance”. The closed complex becomes an open  complex so transcription can begin  o Basel transcription complex initiates transcription at a low frequency  o Additional factors that bind to other sites are needed to achieve high rate of mRNA  synthesis and allow gene regulation  ∙ Other cis-acting elements on eukaryotic transcription  o Enhancer mechanism: DNA sequence that can be far removed (thousand of base pairs)  from promoter, but it enhances transcription  ▪ Activator binds enhancer ▪ Activator binds basal transcription ▪ factors ▪ TBP binds TATA and captures ▪ RNA pol II ▪ DNA loops ▪ RNA pol II transcribes o There are proteins that help bridge gap between enhancer bound transcription factor and pol II: mediators  ▪ a multi-subunit co-activator complex that facilitate the binding and/or function  of Pol II at the promoter ▪ recruited to the DNA template via interactions with sequence-specific  transcriptional activators ▪ helps recruit and stabilize RNA polymerase II near specific genes that are then  transcribed. ∙ *Be able to differentiate between promoter and enhancer: pg 680* o Enhancer is far removed from the promoter o Promotor actually binds DNA  ∙ Application examples: Eukaryotic organisms have systems to respond to environmental stimuli  and regulate transcription o Estrogen stimulates cell growth by turning on transcription o Lipid soluble hormones control binding of a “nuclear receptor” (transcription factor) to  control gene elements  1. Activate transcription factor (binds to DNA sequence and ligand binds) 2. Recruitment of a coactivator  3. Coactivator recruits HAT, acetylation of lysine residues in the histone tails  4. Binding of a remodeling-engine complex to the acetylated lysine residues  5. ATP-dependent remodeling of the chromatin structure to expose DNA 6. Recruitment of RNA polymerase II ∙ Breast cancer: rapidly growing cells. Cancer needs estrogen to promote cell growth. To attack  the cancer, you want a drug that will block binding of estrogen  o When estrogen binds to certain proteins with receptors, leads to conformational change  that exposes binding site of a coactivator, which then binds and recruits HAT, a protein  that acetylates histones.  o Tamoxifen is a competitive inhibitor. It binds to the receptor but “antagonizes” its  activation. ▪ Part of a group of drugs called selective estrogen receptors modulators (SERMS) o The histone tail acts as a platform for proteins to bind to. At histone H3, there are 2  lysines at positions 8 and 9 that get acetylated by HAT. This provides a binding site for  an acetyl-lysine binding protein (bromodomain protein). This then recruits other  components for ATP-dependent chromatin remodeling. DNA around nucleosomes is  broken up to expose certain sequences. This leads to recruitment of TATA binding  protein and transcription occurs.  ∙ Points to remember about transcription 1. What is a promoter? 2. How is it recognized by RNA Pol? 3. What’s the difference in RNA polymerases in bacteria and eukaryotes? 4. What is the role of the sigma subunit of RNA pol? 5. How does transcription terminate? 7. How does the lac repressor repress and how is it derepressed?8. What is an enhancer and what binds to it? 9. How estrogen receptor regulates gene expression? 10. What role is the immunoglobulin enhancer playing in Burkitt’s lymphoma? ∙ Burkitt’s lymphoma is an uncommon type of Non-Hodgkin Lymphoma (commonly affects  children). It is a highly aggressive type of B-cell lymphoma that often starts and involves  body parts other than lymph nodes.  ∙ Analysis of the patient’s karyotype showed a chromosomal translocation. ∙ Because you moved a piece of DNA from one chromosome to another, there is a disruption  of a gene and enhancer element.  Ch 39 & 40: Protein Synthesis ∙ Genetic Code: relation (set of rules) between the sequence of bases in RNA (DNA) and the  sequence of amino acids in protein.  o It’s a 3-base code that is sequential, non-overlapping, and degenerate ∙ Translation: tRNA forms specific interactions with codon in the p-site of the ribosome. The a site is where the next tRNA comes in with its amino acid. The proximity of the 2 amino acids  results in peptide bonds. The ribosome shifts and opens up the next a-site for the next amino  acid  ∙ Reading frames of mRNA  o Translation of the correct message require selection of the correct reading frame: This  is why there is a start codon (typically AUG) o The concept of non-overlapping: message is read in a nonoverlapping triplet code. This  is important because a mistake of a nucleotide would only effect that particular codon  and have minimal effect  o Stop codons- UGA is an example we should know, but you don’t need to memorize any  other codons.  o The code is degenerate: there are multiple codons for a single amino acid o Code is unambiguous: each codon only codes for one amino acid  o The first two nucleotides of a codon are often enough to specify a given amino acid  (Gly = GG__ ) o Codons with similar sequences specify similar amino acids Ex: Glu and Asp = GA__o Only 61 of the 64 codons specify amino acids ▪ Termination (stop codons): UAA, UGA, UAG ▪ Initiation codon - Methionine codon (AUG) specifies initiation site ∙ The RNA Tie Club was a group of scientists who wanted to solve the riddle of RNA structure and  understand how it builds proteins  o They didn’t do any experiments, just discussed ideas  ∙ Richard Feynman  o Discovered the corner-cutting and bad science that NASA did that contributed to the  explosion of the space shuttle Challenger due to the defective o-ring  ∙ How we knew it was a 3-letter genetic code  o 3-letter code proposed by George Gamow  o RNA tie club did math to determine that using the 4 bases, there could only be 3-letter  combos to get an appropriate amount of proteins (43 = 64 combinations) o Crick discovered that t-RNA is the adapter between the codon and the amino acid and  that each tRNA must recognize at least 1 codon and there are at least 1 tRNA per amino  acid  ∙ Cloverleaf structure of t-RNA o There are Watson-crick base pairs, looped regions with unpaired bases, and the 5 bases  that are not paired in the anti-codon loop  ▪ Anti-codons are complementary to the codon. This is what specifies the unique  tRNA ▪ The unpaired base loops have specific base modifications (ex: 5-methylcytidine) o Anticodon base pairs with codon in mRNA o 3’ end always ends in CCA (added post transcription) o Amino acid is added to A residue at the 3’  end ∙ The 1st position of the anticodon is the “wobble position” This allows the anticodon on  the tRNA to interact with multiple codons for the same amino acids.  o Inosine is formed when A is deaminated  o Inosine maximizes the # of codons that can be read by a tRNAo This is one way the code is generate. More than one codon can code for the same  amino acid since the tRNA can recognize them when there’s an inosine at the  wobble position  o Allows bacteria to have less tRNAs by allowing them to interact with more  codons  o Minimizes effects of deleterious mutations ∙ Aminoacylation of tRNA  o The tRNA must have the correctly charged amino acids  o Charging is done by the enzyme Aminoacyl-tRNA Synthetase. The enzyme has an  active site for the tRNA and the amino acid, so it selectively binds tRNA based  on: ▪ Structure of tRNA  ▪ Anticodon  ▪ Acceptor stem  o Linkage of amino acid to tRNA requires ATP: the tRNA is now “activated”  ▪ High-energy linkage  o Energy stored in aminoacyl-tRNA used in formation of peptide bond ∙ The genetic code is translated by 2 adaptors  1. aminoacyl-tRNA synthetase. The linkage is through a high energy bond  created with ATP.  2. tRNA itself whose anticodon forms base pairs with the appropriate mRNA  codon.  An error an either adaptor leads to synthesis failure. The wrong amino acid is  incorporated into the protein  ∙ Aminoacyl- tRNA Synthetases- see slides 26-29 on CH 39/40 ppt if you want to see the steps,  but he didn’t say they were important  ∙ Specificity of Aminoacyl-tRNA Synthetase  o Binds the correct amino acid based on size, charge, hydrophobicity o Binds specific tRNA molecule based on anticodon, structural features, and  acceptor stem ∙ The variable sequence of each tRNA is at the acceptor stem  ∙ There is a specific binding pocket on tRNA where the amino acid can bind.  ∙ Serine can get into the active site and is charged to tRNA even though it needed valine. But there  is a proofreading mechanism that cleaves the serine off since it lacks the methyl group that  threonine and valine have o Threonyl-tRNA synthetase prevents coupling of incorrect amino acids to threonyl-tRNA ∙ The 70s ribosome is bound to the mRNA and has 2 sites for 2 charged tRNAS: the a-site (where  new tRNA binds) and p-site (growing peptide chain)o The ribosome moves 5’????3’ catalyzing peptide bond formation from the n terminus to  the c terminus  ∙ Ribosomes between prokaryotes and eukaryotes are still similar  o The ribosomal RNA component of the ribosome catalyzes peptide bond formation, so  there can’t be much deviation between ribosomal RNA sequence  o Ribosomes consist of rRNA and protein ∙ 50S, 30S, 70S etc. ribosomes are named based on their densities  ∙ The A and P sites are really close together in the acceptor stem to catalyze peptide bond  formation  o The e site is the exit site- where the tRNA leaves the ribosome  ∙ Mechanism of Translation o How do the 50s and 30s come together correctly over the AUG start codon?  ∙ Initiation a. translation complex assembles at the beginning of the mRNA coding sequence b. Complex consists of:  Ribosomal subunits mRNA template to be translated Initiator tRNA molecule Protein initiation factors o 30s ribosomal subunit binds to factors 1 and 3. Factor 2 is bound to the charged tRNA  (fMet-tRNA, methionine with a formyl group). These 3 factors assemble at the start  codon. Factors 1 and 3 are then kicked out, allowing for the 50s subunit to come in, GTP is hydrolyzed, and then factor 2 is kicked out. Then the translation initiation complex is  complete.  o Now, does this system know WHICH AUG to start from? There is a sequence upstream  that is recognized by ribosomal RNA  i. There is a sequence on the 3’ end of 16s rRNA (AGGA) that is the upstream  signal of the first AUG. ii. 30S ribosome binds to a region of the mRNA upstream of the initiation  sequence (Shine-Dalgarno sequence) via complementary base sequence at  the 3’ end of the 16S rRNA ∙ Initiator tRNA a. First codon translated is usually AUG b. Each cell contains at least two methionyl-tRNAMet molecules which recognize AUG 1. The initiator tRNA recognizes initiation codons i. Factor 2 is recognized by the specific fMet tRNA  2. Second tRNAMet recognizes only internal AUG∙ EF-Tu binds to all the charged tRNAS. Only the EF-Tu form that has GTP bound can bind to the  charged tRNAs. 2 important functions of this protein:  1. Protects ester linkage from hydrolysis 2. GTP hydrolysis inly when correct interactions can occur. Making sure the right  codon is paired with the right anticodon o After GTP hydrolysis, you need more GTP to continue elongation o So you need constantly supply of EF-Tu and EF-Ts enzyme that replenishes the GTP  o Translocation  o EF-G uses the energy of GTP hydrolysis to force itself into A site of 30S o Allowing movement of peptidyl-tRNA from A site to P site, Thus forcing the tRNA and  mRNA to move thru the ribosome a distance of 1 codon o Ribosomal RNA component, specifically 23S rRNA that catalyzes the formation of  the peptide bond. Adenine assists by extracting and adding a proton  ∙ Termination- there is no tRNA that can bind to a stop codon. SO, an RF binds and cleaves the  peptide chain∙ Protein Synthesis is energetically expensive. It takes energy to make order (low entropy).  o Four phosphoanhydride bonds are cleaved for each amino acid added to a  polypeptide chain ▪ Amino acid activation: 2 ~P bonds ATP ???? AMP + 2 Pi ▪ Chain elongation: 2 ~P bonds 2 GTP ???? 2 GDP + 2 Pi o It takes so much energy because you need to compensate for the loss of  entropy that results from having very specific order of amino acid in peptide,  amino acid linked to tRNA, and tRNA/codon interaction  Review of important points for quiz: ∙ Stronger promoter= more closely matches consensus sequence  ∙ Be able to recognize a promoter sequence: TATA box ∙ The sigma factor o decreases the affinity of the polymerase for DNA by 10,000 fold, allows the enzymes  to rapidly scan the DNA for a “promoter” o increases the affinity of the core polymerase for specific promoter sequences o It is NOT involved in elongation. Sigma factor only allows polymerase to bind  ∙ Mechanisms of termination  ∙ Add that the repressor forms a loop on the DNA so that the polymerase cant even get  through  ∙ Trans-acting factors are proteins that bind the promoter DNA elements (TATA binding  protein) o Cis acting elements are DNA elements that are on the same molecule (TATA box, and  enhancers) ∙ Enhancers  ∙ Estrogen mechanism o When HAT is recruited, it modifies nucleosomes and activates promoter ∙ Burkitt’s lymphoma- translocation between chromosome 8 and 14. Your b-cells are always  expressing immunoglobulin genes, so now you’ll be transcribing c-myc where you weren’t  previously
Page Expired
5off
It looks like your free minutes have expired! Lucky for you we have all the content you need, just sign up here