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Biochemistry Exam #3: Lectures #16-#23

by: Denise Croote

Biochemistry Exam #3: Lectures #16-#23 0280

Marketplace > Brown University > Biology > 0280 > Biochemistry Exam 3 Lectures 16 23
Denise Croote
Brown U
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These notes are very comprehensive! They include step by step descriptions of each mechanism and plenty of diagrams. The topics covered include: - Biological Amine and Nucleotide Metabolism - DNA...
Introductory Biochemistry
Arthur Salomon
Study Guide
DNA, gene, molecular cloning, transcription, translation, metabolism, biochemistry
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This 25 page Study Guide was uploaded by Denise Croote on Saturday January 23, 2016. The Study Guide belongs to 0280 at Brown University taught by Arthur Salomon in Spring 2016. Since its upload, it has received 23 views. For similar materials see Introductory Biochemistry in Biology at Brown University.


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Date Created: 01/23/16
Lecture 16 – Biological Amine and Nucleotide Metabolism  Receptor Guaylyl Cylases – convert GTP to the second messenger GMP when activated  NO is an unusual signaling molecule because it is a gas. Because NO is generated and acts within the same cell it can be classified as a second messenger, but it can also be considered a hormone because it can diffuse into neighboring cells  Ca2+ enter the cells via the acetylcholine receptor, a ligand gated ion channel o Acetylcholine receptor changes whether bulky hydrophobic side chains inhabit the channel (forcing it into a closed state) or whether smaller polar residues inhabit the channel (forcing it into a open state) o Ca2+ targets calmodulin, once activated calmodulin undergoes a large conformational change to expose a hydrophilic helix that interacts by wrapping around and activating domains of proteins.  Steroid hormone receptors regulate gene expression, steroids can travel across the membrane and then bind to receptor proteins in the nucleus  BIOLOGICAL AMINES – function as neurotransmitters and hormones o Epinephrine, Norepinephrine, Dopamine, GABA, Serotonin, Histamine o Generated by the decarboxylation of amino acids (with cofactor PLP) o Tyrosine  Dopa  Dopamine  norepinephrine  epinephrine o Glutamate  GABA o Histidine  Histamine o Tyrptophan  Serotonin o Decarboxylation is also the first step in polyamine synthesis (organic compounds having two or more primary amino groups), which is important for DNA compaction o Polyamines have lots of amino groups and a positive charge o Ornithine decarboxylase is a drug treatment for African Sleeping Sickness  DFMO: a suicide inhibitor of orthinine decarboxylase forms a covalent bond an enzyme, rendering it incapable of further action  THE TETRAPYRROLE FAMILY o Heme, cytochromes, oxidases, chlorophylls o Derived from glycine and succinyl-CoA in animals and from glutamate in plants o Formed when aminolevulinate is dimerized to porphobilinogen, which is tetramerized to form tetrapyrroles o Can take a aminolevulinate and create a heme with seven steps, add charged Fe2+ to the center of the molecule o When you degrade the heme you get biliverdin which is the reason bruises look green NUCLEOTIDE METABOLISM Base Nucleoside Nucleotide Nucleic Acid Purines Adenine Adenosine Adenylate RNA Deoxyadenosine Deoxyadenylate DNA Guanine Guanosine Guanylate RNA Deoxyguanosine Deoxyguanylate DNA Pyramidines Cytosine Cytidine Cytidylate RNA Deoxycytidint Deoxycytidylate DNA Thymine Thymidine Thymidylate DNA Deoxythymidine Deoxythymidylate Uracil Uridine Uridylate RNA  Nucleoside – sugar + base  Nucleotide – sugar + phosphate + base Structure:  Memorize the structures of the purine and pyrimidine bases  Abbreviations of ribonucleosides are GMP, GDP, GTP ect and the dexoyribonucleosides are dGMP, dGDP, dGTP  PRPP is the donor of the ribose-phosphate unit of nucleotides  PRPP is an intermediate in the biosynthesis of Trp, His, and nucleotides  PRPP: Free bases (A,C,G,T,U) are not synthesized directly and added to PRPP o For purines bases are built stepwise onto PRPP  First you add NH3 from glutamine to the sugar ring, and then you build off the NH2 by adding glycine, followed by rxn with formyl H4 folate, addition of another glutamine, and a bunch of enzymes to construct the two rings of the purine ring.  When you have created IMP in the final steps you reach a branching point. AMP and GMP can be formed from IMP in two different reactions.  Feedback regulation of purine biosynthesis: AMP, IMP, and GMP can inhibit the committed step of the common pathway. AMP and GMP inhibit the entry point for their specific synthesis from IMP o For pyrimidines a precursor base is synthesized and added to PRPP  Derived from aspartate and carbamoyl phosphate  The carbamoyl phosphate comes from a cytosolic pool, a separate pool from the urea cycle  Enzyme carbamoyl phosphate synthetase II has three different active sites connected by a channel  Orotate is the base precursor  ATCase (enzyme that adds aspartate to carbamoyl phosphate in the first step of the reaction) is regulated by CTP. ATP inhibits CTP (so it inhibits the inhibitor)  ATP is a purine, purines regulate the pyrimidine synthesis, if excess purines are present they ramp up the speed of the pyrimidine reaction to create equal amounts o Can interconvert between ATP + AMP  2ADP o Ribonucleotide reductase converts ribonucleotides to deoxyribonucleotides  rNDP + NADH + H  dNDP + NADP + H20 ------------------ Ribonucleotide reductase NADPH  NADP  Ribonucleotide reductase has a regulatory site and an allosteric site o It takes OH and makes H by protonating the OH to make water, which is a good leaving group, the waters leaves and the carbocation is stabilized in the reaction pocket, and H is put on o H next to the just added H gets taken off and put back on o Positively regulated by ATP (increases when energy charge is high) o Negatively regulated by dATP (feedback regulation) o Pur[pse: to provide a balanced pool of nucleotides  Take CDP and EDP and make dCDP and dUDP with ribonucloetide reductase. Can then make dTMP from these with thymidylate synthase  thymidylate synthase carries out a one carbon transfer reaction  Overall it goes dUMP to dTMP  This is an important target for chemotherapeutic agents – the reason why people get sick from chemo is because these agents often affect the healthy DNA as well as the cancer DNA NUCLEOTIDE DEGRADATION  General strategy is to remove the 5’ phosphate  nucleoside o Remove the ribose  free base  Purines are broken down and turned into uric acid and excreted o Uric acid is not a major nitrogenous catabolite in mammals but we do produce some in purine catabolism, overproduction causes gout  Prymidines can be recycled  Free bases (like adenine and guanine) can be directly added to PRPP to make GMP, AMP REVIEW: LECTURE 17 – DNA  Central Dogma: replication of DNA  transcription to RNA  translation to protein  DNA is a polymer of deoxyribonucleotides  DNA is a sequence of bases that carries the genetic information  Within the genes, nucleotides specify the amino acid sequences of proteins  Outside the genes, regulatory sequences direct DNA replication and mRNA synthesis  Deoxyribonucleotides include: DNA STRUCTURE  Nucleotides are linked by phosphodiester bonds (phosphate on 3’ carbon bonds to the 5’ carbon on the next nucleotide), their bases remain available for hydrogen bonding. They form with 5’ to 3’ polarity.  Linear hydrogen bonds are the strongest hydrogen bonds. A T have 2 hydrogen bonds. GC have 3 hydrogen bonds.  This base pairing holds the two strands of DNA together. The complementarity allows the information to be copied.  The DNA strands run antiparallel. The sugar phosphate backbone is on the outside.  The DNA is a double helix. It has a wide and shallow groove and a narrow and deep groove.  The structure is stabilized by hydrogen bonding and stacking interactions o B-DNA – most stable, in solution, 10.5 base pairs per turn, right handed, predominates in the cells o A-DNA – DNA,RNA and RNA, RNA helix, in solution, 11 base pairs per turn, also right handed o Z-DNA – left handed, found in cells, thought to be the contorted form of DNA as DNA is being transcribed  The B-DNA cannot bind with RNA because of the steric clash of its contortion with RNA. The A-DNA can incorporate RNA because it structure allows no steric clash. RNA STRUCTURE  Single stranded helix  The base pairs in RNA are A-U and G-C but G-U can also occur (but is a weaker non- linear base pairing)  Can have hairpin loops and segments that are not paired STABILITY OF DNA  The double stranded DNA can be denatured with high temperature or pH  Once the strands of the DNA molecule are separated it becomes very difficult for one half to find the other again  Stability is judged on the melting point, more GC binding in the DNA means it is more stable  As DNA denatures you see increased UV absorbance due to the disruption of base stacking interactions  The major groove can discriminate between AT, TA, GC, and CG  The minor groove can only discriminate AT/TA vs. GC/CG  Amino acid side chains for hydrogen bonds with bases in double helical DNA o Asn, Gln, Glu, Lys, Arg DNA Binding Proteins:  Generally bind DNA in the major groove –in B-DNA the a-helix fits nicely into the wide major groove and leads to the closes interaction of amino acids  Certain DNA binding motifs are common… o Helix-Turn-Helix: DNA recognition helix is positioned in the major groove o Zinc Finger: helix fits in the major groove, zinc helps hold the tertiary structure together o Homeodomain: widely conserved, a-helix positioned in the major groove o Leucine Zipper: DNA binding in the major groove, leu dimerizes in an external zipper region  o Basic Helix – Loop – Helix: two helices inserting into the DNA and two other helices binding to each other COMPACTION of DNA:  DNA molecules are very long, and must be compact to fit into cells  Supercoiling compacts the DNA – (like when you wind up a telephone cord), innate to each cell and highly regulated  Most DNA is underwound (strained) which is 7 turns, relaxed DNA is 8 turns  Underwinding facilitates strand separation  DNA underwinding is defined by the topological linking number, Lk: defined by the number of times the second stand breaks the plane of the first strand  You can think of underwound by two turns as a 1% unwindedness  Positive supercoils are right handed (Lk=200  Lk=202)  Negative supercoils are left handed (Lk=200  Lk=198)  These are referred to as topoisomers because they only differ in a topological property (the linking number) Type I Topisomerases:  Changes Lk in increments of 1  Mainly relaxes supercoils and works by catalyzing single or double stranded breaks, crossing one strand through the other, and resealing the breaks.  Mechanism: Binds DNA, cleaves on strand of the duplex DNA, passes a single strand through the break, DNA is religated  Toposiomerase I – when you unwind the DNA you get a larger area of DNA, which means the new segment is not as capable of traveling through a gel as well  Highly supercoiled DNA travels farther than unwound DNA Type II Topisomerases:  Changes Lk in increments of 2  Mechanism: binds two segments of the duplex DNA, passes other segment of DNA through break, cleaves both strands on one segment of DNA  Plectoemic supercoiling is less compact, solenoid coiling in more compact and found in most cells  Histone proteins act to package the DNA into nucleosomes  146bp of DNA are wrapped around the histone core  Packaging of DNA into nucleosomes facilitates supercoiling Lecture 18- DNA Replication and Repair  Semiconservative Model – hybrid duplex of old and new strand (right) o Proved by the Meselson-Stahl experiment, used 15N DNA and then found that after one replication cycle half of the DNA was 15N and the other half was 14N  Conservative Model – duplex of only old or only new synthesized DNA (wrong )  DNA synthesis is performed by DNA polymerases o Need a template strand to copy o Primer strand with 3’ OH o dNTP substrates o catalyzes 5’ to 3’ synthesis and phosphodiester bond formation  DNA synthesis is always 5’ to 3’ Accuracy and Fidelity:  DNA Polymerase I, DNA Polymerase II, DNA Polymerase III  All have 3’  5’ proofreading ability, but only DNA polymerase I has 5’ to 3’ exonucleases  Exonucleases cleave proteins one at a time from the end  Proofreading allows for only one net 8 error for every 10 base pairs added STAGE ONE: Initiation  Replication is bidirectional – both strands are replicated at the same time  Replication starts at the origin and the DNA is held open by a replication fork  Initiation requires specific sequences and proteins – it starts at a DUE – a DNA unwinding element. This element contains a high amount of ATP.  There are mostly A-T bonds at the start of the fork because these are weaker and easier to break apart  Single stranded binding proteins bind to the single strand of DNA to prevent it from being digested by nucleosomes before it is ready to be replicated  Helicases separate the strands of the DNA  DnaA protein – recognizes the origin sequence, opens the duplex at a specific site  DNA gyrases relieve the strain caused by unwinding STAGE TWO: Elongation  DNA primase creates an RNA primer that DNA polymerase uses to copy the DNA  DNA polymerase III is the main enzyme involved in the replication of DNA in E-coli o Nucleotides are added to the 3’ side  Leading Strand are synthesized continuously  Lagging Strand is synthesized discontinuously, add primer  make DNA  add primer  make DNA o DNA polymerase I comes along and does exonuclease activity in the 5’ to 3’ direction. It removes the RNA and 5’  3’ polymerase activity fills in with DNA o DNA Ligase uses ATP or NAD to seal the nicks  DNA ligase gets adenylated  Transfer of AMP to the 5’ phosphate of the donor  Formation of a phosphodiester bond between the 5’ phosphate donor and the 3’ hydroxyl of the acceptor  Basic Steps in Elongation for leading and lagging o Primase (DnaG protein) synthesizes the RNA primer o dNTPs (deoxyribose nucleotides) are added by DNA polymerase III o Replication for moves and DnaB helicase unwinds the DNA o Single stranded proteins stabilize the strands o DNA gyrases relieve the strain caused by unwinding  Structue of DNA pol III enzyme : o Has two cores for polymerase activity o Has DnaB helicase attached and a B clamp o The B Clamps holds onto the DNA as the polymerase works o For the lagging strand when an okazaki fragment is near completion, primase binds to DnaB, synthesizes a new primer, and then dissociates. A new B clamp is loaded onto the template by the primer and the process continues, each time leaving behind a B clamp and attaching a new one. STAGE THREE: Termination  Occurs when the replication forks meet and terminate  In E.coli because they have a circular DNA plasmid, they termination occurs when the forks meet each other at the opposite end of the plasmid  In eukaryotes the replication rate is slower and the chromosomes are longer, they have many origins of replication  Telomeres are repeated sequences at the end of the linear chromosomes  Telomeres are added by telomerase – it synthesizes DNA from an RNA template (a reverse transcriptase)  Every time somatic cell replicate they lose a bit of their DNA, they eventually die when they lose too much DNA REPAIR:  A mutation is a permanent change in the DNA sequence  Mutations can be … o Silent  no effect on gene function o Deleterious  impairs gene function o Advantageous  enhances gene function  Mutations lead to genetic diversity, cancer in somatic cells, and birth defects in germ cells  Transitions – changes a purine to another purine or a pyrimidine to another prymidine  Transversions – changes a pyrimidine to a purine or a purine to a pyrimidine  Insertions and deletions lead to frame shift mutations  Mutations re caused by mistakes in replication or DNA damage  Deaminating Agents are chemical mutagens that can remove an amino group from a nitrogenous base causing a point mutation o Examples: nitrous acid (HNO2) , NaNO2, NaNO3  UV irradiation is another source of DNA damage  De-purination can also occur – removal of the purine group from the by alkylating agents  Alkylation changes the base pairing properties  DNA damage on one strand can be repaired using the information from the other strand Mechanisms to Repair DNA:  Mismatch Repair: corrects for replication errors o Parental strand is marked and methylated  MutL-MutS + MutH travels down the strand and looks for a bulge that signifies that a base pair is incorrect. Then signals other molecules to come fix it. o Causes a nick, recruits an endonuclease, goes a little beyond, and polymerizes backwards o End up with DNA containing mis-match re-synthesized  Base Excision Repair: o The N-glycosyl bond in cleaved near the damaged base  DNA glycosylase does the cleavage – there is a different glycosylase for each base lesion o Sugar is removed by a endonuclease o DNA polymerase comes alone o DNA ligase seals the nick  Nucleotide – Excision Repair: o Used for the removal of large bulky lesions (like a pyrimidine dimer)  Excinucleases make two cuts and then remove the damaged DNA  Straddle the damaged area  Direct Repair: o Does not remove the base or the nucleotide o Repairs the defect but it costs one enzyme (suicide enzyme – it is inactivated) Classes on RNA Lecture 19 – Transcription  Messenger RNA – mRNA  Ribosomal RNA – rRNA  Transfer RNA – tRNA  Replication copies the entire genome  Small nuclear RNA – snRNA  Transcription copies only particular genes or groups of genes at a  Small interfering RNA –sIRNA time, some portions of the DNA genome are never transcribed  Only ONE DNA strand serves as a template for transcription  Micro RNA – mIRNA  RNA polymerases do not need a primer (de novo synthesis), they use a DNA template, synthesize in the 53 direction, need NTPs as substrates, but DO NOT have 35 exonuclease activity  They do not have proofreading ability because the mRNA transcripts are short lived, have a large turnover rate, and are not propagating enough genetic information  The nontemplate coding strand is the 53 strand, the DNA template strand is the 35 strand  Instead of incorporating T’s we incorporate U’s Structure of RNA polymerase:  Bound to the DNA and the action of pushing the DNA forward separates the strands  Incoming NTP is added to the 3’ end and 5’ end is squeezed out the backside INITIATION (prokaryotes):  RNA polymerase recognizes and binds to specific DNA sequence called promoters  TATAAT – is a sequence that helps to pull the two strands apart  Prokaryotes: o RNA poly of ecoli (with 5 subunits and a sigma protein) bind to the DNA promoter forming a closed complex. Binding occurs ONLY after all of the proteins have associated.  The consensus sequence is the sequence that would make the tightest bond to sigma 70 (advantageous) o Helix is unwound and the open complex is created. Transcription is then initiated and sigma protein leaves. o Can transcribe to the right and to the left of the promoter o Promoter is cleared and ELONGATION proceeds  TERMINATION: o Roe independent: RNA synthesis encounters a termination sequence, RNA folds on itself, forming a hairpin structure, occurs at the location of an A rich sequence (weak H bonds). o Roe dependent: requires a roe protein, roe protein pries the RNA-DNA helicase apart, requires energy, does not happen at an A rich site  Prokaryotic genes are arranged in operons (clusters of genes under the control of a single promoter) A single mRNA strand could code for more than one protein (polycistronic). Eukaryotic genes are monocistronic (an mRNA transcript codes for only one protein). INITIATION (eukaryotes):  We have three polymerases, promoters have little intrinsic affinity for RNA polymerase  RNA poly II finds and binds the promoter that may be very upstream, it binds to a TATA box  Requires a lot of protein factors, the closed complex is more than 30 proteins  Proteins are assembled onto the closed complex one by one AFTER RNA poly has bound to the DNA  TFIIHH unwinds the DNA, when the DNA is unwound the TFIHH dissociates and the RNA transcribes  Phosohpdiester bond formation occurs and the promoter is cleared  Serines are hyper-phoshorylated on the polypeptide, this leads to the recruitment of other machinery (like splicing machinery)  Reverse Transcriptase: used to go from RNA  DNA  Reverse transcriptase is required for retroviral integration (HIV)  Steps: o Use DNA polymerase to take and RNA template and turn it into a DNA-RNA hybrid o Chew away the RNA to leave a singles stranded DNA o Synthesize another strand of DNA to make double stranded DNA Eukaryotic Post Transcriptional Processing  Add a 5’ cap – enhances the stability since the (DNA polymerase) bond is a 5’ to 5’ triphosphate linkage. This is not recognized by endonucleases o Only done in eukaryotes  Add Poly A Tail – further increases the stability, (RNA polymerase) synthesized by polyA polymerase  Splice out the introns (non-coding sequences) – (Reverse Transcriptase) involved introns themselves that have an A nucleotide that excises itself out  Leave in exons ( each usually corresponds to a domain) (Replicases) 3 classes on introns:  Group I splicing – splicing involves a transesterification reaction o OH attacks 5’ exon, intermediate, then OH on 5’ becomes the new nucleophile completing the reaction o Found in bacteria, lower eukaryotes o Self Splicing ** o Group I introns are ribozymes (RNA molecules with catalytic activities)  They can do phosphodiester bond cleavage and transesterification  Group II splicing- form a lariat intermediate, adenosine in the lariat structure has three phosphodiester bonds  Nuclear splicing – forms a lariat intermediate and requires other proteins, squeezes out an adenosine  snRNA’s are required for mRNA splicing in NUCLEAR SPLICING ONLY o recognize splice sites that are specified by sequences at the ends of introns o important to double check that you have bound to the base pair before splicing o U1- the part of the RNA complementary to the 5’ splice site o U2 – the part of the mRNA complementary to the branch point which has the imperfect base pairing o U2 and U1 recognize their complementary sites, ATP hydrolysis drives conformational changes, and splice site and branch point are brought closer together leading to release of the intron o mRNA splicing is carried out on a spliceosome complex o different exons are spliced together, we trade the wasteful excision of introns (75% removed) for increased variability of proteins o Alternative RNA processing can produce multiple mRNAs from a single transcript REVIEW: Group 1 Group 2 Nuclear Transesterifications yes yes Yes Conservation of Phosphate Bond yes yes Yes 1 step: attack by OH Yes Yes Yes - OH from solvent Yes No No - OH from Intron A No Yes Yes Larlat intermediate No yes Yes 2 step: 3’OH of 5’ exon attacks 3’ splice site yes yes Yes RNA catalysis yes yes Yes Requirement for Proteins no No yes  Prokaryotics rRNA is also processed – they are methylated and then cleaved  Eukaryotic rRNA are processed in the nucleus using snoRNA direct modifications  tRNAs are processed with the following mechanism: o endonucleoytic cleavage (taking out intron) o activation of 5’OH o nucleophillic attack o formation of the 3-5’ phosphodiester, (phosphate comes from ATP)  this is distinct from group I,II, and Nuclear in that it requires ATP and endonucleases LECTURE 20 – TRANSLATION  A tRNA molecule has an amino acid attached, an amino acid binding site, an adaptor, and a anticodon that binds to the codon on the mRNA  Theories debated whether it was an overlapping or non overlapping code o Experiment: made a point mutation and it only affected one amino acid, proved that the code in non-overlapping  To have different sequences for all 20 amino acids we need 3 nucleotides per codon  A deletion or an insertion or two insertions or deletions will cause a frame shift mutation. Three insertions or deletions will cause the sequence to shift back into frame. This proved that codons are made up of 3 nucleotides.  Determined which codons code for which amino acids by using chemically synthesized homopolymers: ex.) Poly(U)  poly(Phe)  Each of the amino acids can be coded by several different codons (termed degeneracy) o For ex.) Alanine is coded by 4 codons o This is because the of “wobble” = multiple rd binding of the 3 base (3’ base of codon and 5’base of anticodon) o The 1 two pair bind strongly and the 3 more loosely  A-U bind and G-C bind, but G-U is also feasible  I pairs with A,U, or C  When the 5’base of the anticodon (wobble) is a C or an A only one codon is recognized, when the base is a U or G two different codons may be recognized, and when the base is an I three codons can be recognized  Amino acids recognize the entire surface of the tRNA they are supposed to bind to, not just the 3 amino acids in the binding arm.  Incoming amino acids form and ester linkage with the adenosine on the 3’ arm. Ribosomes:  Bacterial: (70S) made up of a 50S and a 30S subunit  Eukaryotic: (80S) made up of a 60S and a 40S subunit  rRNAs are key for function. Ribosomes depleted of protein can still catalyze peptide bond formation o cleaved one bond in the 16S rRNA and it abolished protein synthesis o Structural analysis of ribosomes show no protein near the catalytic center, only RNA o 3’ end of tRNA interacts with rRNA not proteins  The antibiotics that interfere with protein synthesis actually target rRNA  Prokaryotic Translation: the 16S rRNA interacts with the Shine –Dalgarno (S-D) sequence on mRNA and positions it on the ribosome. This SD sequence determines where the ribosome binds initially (necessary due to polycistronic nature)  Eukaryotic Translation: Scanning from the cap locates the start codon 1.) Prokaryotic 2.) Eukaryotic  The mRNA is read 5’ to 3’ while the protein is synthesized NC o Determined this through the experiment, added labeled Leu-tRNA to the C terminus and found that the C terminus was synthesized last STAGES of PROTEIN SYNTHESIS 1.) Stage One: Activation of Amino Acids: a. Amino acid + ATP  Aminoacyl-AMP + PPi b. Aminoacyl – AMP + tRNA  Aminoacyl-tRNA + AMP i. Ser-tRNA ser= charged and tRNA se= uncharged c. Must choose the correct amino acid for the proper tRNA d. Each amino acid has a specific aminoacyl-tRNA synthetase to carry out the reaction (requires so much energy because it must be done correctly). e. The AA-tRNA synthetase recognizes structural features on the tRNA and can carry put both steps in aminoacylation. It has two classes. Class I is a monomer and Class II is a dimer. 2.) Stage Two: Initiation a. In bacteria  the small 30S subunit binds initiation factor 3 and initiation factor 1. The 16rRNA on the 30S subunit binds the SD sequence on the mRNA at the P site. The IF-2 GTP binds fMET-tRNA and brings it to the P site. b. 50S binds and GTP is hydrolyzed, the initiation factors are released and the 70S complex is formed. GTP will only hydrolyze if the binding is tight and the tRNA is charged with an amino acid. (checking system) c. In Eukaryotes  more initiation factors that bind to the 5’ cap and the poly A tail, scan for start codon (no SD sequence) 3.) Stage Three: Elongation a. Binding of aminoacyl-tRNA to A-site – the molecule EF-TU-GTP hydrolyzed and Ef-TU-GDP dissociates. The hydrolysis will only occur if the correct base pairing stabilizes the association. b. Peptide Bond Formation – formation between the P site and the A site using peptidyl transferase. A-site nitrogen attacks the carboxylic acid to form a peptide bond. c. Translocation – EF-G-GTP binds A site to displace peptidyl-tRNA. Hydrolysis of GTP causes a conformation change. Causes the shift of A-site  P site  E site 4.) Stage Four: Termination – release factor (RF) binds termination codon in A-site, polypeptide is transferred to H2O causing dissociation of the peptide from the tRNA 5.) Stage Five: Post-Translational Processing a. N and C terminus modification b. Direct proteins to destination and then remove the signal sequences c. Glycosylation – attach carbohydrate side chains, occurs in the ER, important in protein targeting d. Add prosthetic groups e. Disulfide bond formation f. Add isoprenyl groups to anchor otherwise soluble proteins to the membrane  Protein synthesis requires 5 high energy phosphate bonds per peptide bond ** expensive  Antibiotics: they attack prokaryotic protein synthesis by blocking rRNA. o Ex.) Puromycin has a similar structure to adenosine, it binds to the A site and can be incorporated into the protein with the peptide bond. This leads to dissociation instead of translocation. Causes a nonfunctional protein to be formed. (suicide inhibitor) o Certain antibiotics target eukaryotes and certain target prokaryotes  Proteins targeting – N terminal signal sequence targets to membranes or mitochondria and chloroplasts. Nuclear localization sequences (NLS) target the nucleus.  SRP (signal receptor sequence) causes free ribosomes to stall until peptide is brought to the ER. SRP binds to a receptor on the ER and the protein is translated.  Proteins are degraded by modification by ubiquitinin. Multiple Ubiquitins are added to the protein for degradation. o The lifetime of the protein depends on the stability of the N terminus LECTURE 21 – GENE REGULATION  Eukaryotes have a 5’ cap that is not present in prokaryotes, allows for ribosomes to find the mRNA  In prokaryotes the moment the 5’ end of the mRNA is synthesized it is fed into the ribosome  Constitutive: gene is always on, constant transcription, unregulated  Induction: transcript activity is increased  Repression: transcript activity is decreased  In E.coli all genes under the control of sigma 70 have -10 and -35 regions. Sigma 70 functions in recognizing most of the E.coli promoters.  There are other sigma proteins (ex. Sigma 32 recognizes heat shock promoters) but sigma 70 is the most popular  Prokaryotic genes with UP elements are expressed at bigger levels.  Repressors impede RNA polymerase access and activators enhance the access of RNA polymerase. Activators bind at another location to increase the level of transcription. Signaling molecules can bind to the activator and increase the activator’s affinity for the activator binding site or can decrease the affinity.  An operator is a DNA sequence (NOT a protein) that binds to the repressors and blocks RNA polymerase to decrease transcription  Signaling molecules can affect the shape of the repressor to either increase or decrease the affinity for the operator The Lac Operon:  The operon is used to digest lactose when it is present and needs to be used as a source of fuel  In the absence of lactose, the lac repressor prevents the transcription of the enzyme needed to digest the lactose by binding to the operator  In the presence of lactose, allolactose binds to the repressors and decreases its affinity for the operator. The repressor does not bind the DNA, allowing transcription to occur.  The lac promoter is sub-optimal, you need high levels of activation for transcription to occur  CRP (cAMP receptor protein) is a glucose sensor, when glucose high, cAMP is low, and when glucose is low cAMP is high.  So when glucose is low and lactose is present, cAMP binds and increases the affinity for the activation site  high levels of transcription  If glucose is high (leading cAMP to be low), then low levels of transcription will result even if lactose is present  *** you need both low glucose (high cAMP) to activate (positively regulate) and high lactose to decrease repressor affinity (negative regulate) in order to have high transcription The trp Operon:  Codes for the genes responsible for tryptophan synthesis  When tryptophan is present in the environment, the genes for tryptophan synthesis are not expressed because the operator is blocked by the repressor  When tryptophan levels are low, transcription proceeds  When trp levels are high the ribosome blasts through the chain and covers the base pairs in sequence 2. As a result sequence 3 and 4 base pair.  When trp levels are low, ribosome gets to trp and waits for the charged trp to arrive, sequence 2 base pairs with sequence 3 instead.  Sequence 2,3 has no polyU, so the transcript cannot be terminated. Sequence 3,4 has a polyU and forms an attenuator structure and stop transcription ** SUMMARY: when trp is high there are a lot of charged trp molecules present, the ribose covers sequence 2, sequence 3 and 4 bind to form the attenuator structure that halts transcription. Eukaryotic Gene Regulation:  Initiation proteins are arranged in a stepwise fashion onto the DNA transcript  Yeast have upstream activating sequences for regulation and eukaryotes have enhancers  In Eukaryotes promoter access is restricted because… o DNA is compacted into chromatin. Condensed chromatin is inherently repressed. o Condensed chromatin is inaccessible and the chromatin must be remodeled for the DNA polymerase to gain access Activation of Transcription:  Basal machinery will assemble at promoters, but if unregulated will proceed at a low transcription rate. It cannot work until an activator comes around.  Basal transcription factors: recruit RNA polymerase to the promoters  HAT – histone acetyltransferases: transfer acetyl groups to DNA to increase gene expression  Transactivators: bind to the DNA at enhancers, they act as distant enhancers  Co-activators: non-DNA binding but mediate the interaction between RNA polymerase and transactivators o Co-activators can recruit TATA binding proteins (that make up the pre- initiation complex)  *** these help to shape the DNA to bring it into 3D proximity to transcription factors Repression of Transcription:  The DNA is too long for the eukaryotic genome to synthesize a repressor for every gene so repression inherently occurs in the tightness of the chromatin  Repressors block the interaction between the transactivators and the co- activators, do not touch the DNA Galactose Metabolism  Gal4 is a transactivator, has two domains, one to binds to the DNA and another binds to the co-activator  Gal 80p is a repressor  When Gal 80p is bound to Gal4, Gal4 can’t bind to the coactivator  When galactose binds to Gal3p the galactose-gal3p complex binds to the Gal80p-Gal4 complex and activates Gal4  Gal4 recuits SAGA, Mediator, and TFIID to the galactose promoters, leading to DNA polymerase II recruitment Transcription Regulation by Steroid Hormones: 1. Hormones are carried to the target tissue, diffuse across the plasma membrane, and bind to specific receptor proteins in the nucleus 2. Hormone binding changes the conformation of the receptor, forms a homo or heterodimers with another hormone receptor complex a. Hormone receptor complex has a transactivating domain, a DNA binding domain, and a ligand binding domain 3. This complex binds to specific regulatory regions called hormone response elements (HREs) on the DNA adjacent to specific genes 4. Receptor attracts coactivators or corepressor protein to regulate the adjacent gene 5. Altered levels of the hormone regulated gene product produce the cellular response of the hormone  External signals, such as phosphorylation, can regulate transcription factor activity by causing the importation of a protein that binds to the enhancers of DNA  Can also regulate posttranscriptional processing, you can produce multiple mRNAs from a single primary transcript o Can cleave at different locations and polyadenylate (add polyA tail) at different locations *** Regulation can occur at the stages of transcription initiation and post – transcriptional processing  Allows for global control, coordinate control, and gene-specific control LECTURE 22 – TECHNIQUES IN MOLECULAR CLONING  In order to clone DNA 5 basic procedures are necessary… 1. cut the DNA at precise locations  Using restriction endonucleases 2. covalently join two DNA molecules  Using DNA LIgase 3. move recombinant DNA from test tube to host  using transformation, infection, or electroporation 4. replicate DNA segments within the host  DNA vector 5. identify which host cells harbor the DNA Overview:  first you obtain the DNA fragment of interested by cleaving the chromosome with a restriction endonuclease  next the cloning vector is cleaved with restriction endonucleases to make room for the incoming DNA fragment  The fragments are ligated together to form a recombinant vector  The recombinant vector is introduced into the host cell  The transformed cell is cloned to make many copies of the recombinant DNA Restriction Endonucleases:  Biologically designed to cleave DNA that is not methylated, methylation blocks the enzyme  We have engineered these enzymes to ignore methylation and cut any DNA  Can make a staggering cut or a guillotine cut  The straight cut is less efficient for litigation  The overhangs of the staggered cuts must be compatible in order to be litigated together  If you want to introduce foreign DNA in a certain orientation, you must use two different restriction enzymes (introducing directionality)  Poly linker site – short segments of DNA that contains many restriction sites (sites where the DNA is cut). These are placed in the circular plasmid by DNA Ligase and are recognized by restriction endonucleases. Moving the DNA from the test tube to a host  Vector Type Method of Introduction Typical size of DNA Fragments plasmid transformation Up to 15,000 bp bacteriophage Phage infection Up to 23,000bp BACs electroporation Up to 400,000 bp  Plasmids  are stuck on the surface of E.coli, heat shocked, and taken up into the bacteria (transformation)  Plasmids have an origin of replication and unique restriction sites  They have tetracycline resistance genes and ampicillin resistance genes, these genes cleave and block destruction by antibiotics  Bacteriophages  certain parts of the bacterial genome are disposable, we take these parts out and put in our foreign DNA.  The viral load lands on the E.coli cell and injects the DNA into the cell (Phage infection)  We only want one viral particle to land on the virus so only one set of foreign DNA is injected (we want all the bacterial to be uniformly cloned)  Bacterial Artificial Chromosomes  these are used when additional machinery is needed for copying large DNA segments  (electroporation) is the process by which we apply a large electric field to increase the permeability of the cells membrane  The lac Z gene is cut in the BAC vector and the large DNA fragment with the appropriate sticky ends is added. Since the lacZ gene has been disrupted no B- galactosidase is produced  Blue White Selection: Sometimes the BAC vector is opened and resealed before the foreign DNA has been integrated. Some cells can take up a BAC vector void of foreign DNA and some can take up a BAC vector with foreign DNA. o If cells take up the BAC vector they will be able to live on the agar gel containing the chloramphenicol o Disruption of BAC with foreign DNA leaves a white colony. Those who don’t have the foreign DNA will turn blue when fed a substance. Methods to select host cells harboring recombinant DNA molecules   Plasmid starts with both ampicillin resistance and tetracycline resistance  Pst I cuts in the amp gene and adds the foreign DNA  Cells that do NOT take up a plasmid with be susceptible to tetracycline and ampicillin  Cells with plasmids but NO foreign DNA insert will be resistant to tetracycline and ampicillin  Cells with plasmids and foreign DNA inserts will be resistant to tetracycline but susceptible to ampicillin  Therefore: what you want is the bacteria that survive on tetracycline (meaning they have the plasmid) but die on tetracycline and ampicillin (meaning the foreign DNA has interrupted the ampicilling resistance) Libraries: collections of DNA fragments each carried in vectors 1. Genomic Library - contains genomic DNA (chromosomal DNA) a. organism specific and cell type independent b. genomic DNA segments are individually broken up and packaged into bacteriophages 2. cDNA Library – contains DNA copies of mRNA, organism specific, cell type specific a. the DNA is made from mRNA by reverse transcriptase b. first isolate the mRNA and then add a primer for reverse transcriptase c. reverse transcriptase makes the first strand of DNA and then the RNA is degraded by base. (base goes after the RNA because the 2’OH is better than the 2’H) d. DNA polymerase synthesizes the other strand of the double stranded DNA e. placed into individual bacteriophages, don’t want two bacteriophages with different DNA to infect the same cell  Denaturation and Annealing separates the strands into SSDNA, these can be used as probes  Nitrocellulose paper can be used to pick up bacteria from a culture, alkali solution can disrupt the cell and expose the DNA, DNA probes can be put on the paper and will anneal to the genes of interest Types of Probes:  We can generate a probe based on a protein sequence, if we have a protein we like, we work backwards and calculate a possible DNA sequence that could have made the protein.  It is difficult because there will be numerous triplets that could code for each amino acid  We take a protein sequence that will require us to make the least number of probes Amplification by PCR  Found a heat insensitive DNA polymerase, so you can heat the DNA and cause the two strands to separate without destroying the DNA  Once separated add the primer and the thermostable DNA polymerase to synthesize the new strands  Repeat this to receive millions of copies, disadvantage being that there is a high error rate  The genome has been sequenced, the DNA was digested into fragments and inserted into BACs, they then matched up all of the sequence overlaps to reveal the final sequence  Found that the DNA codes for much more than just proteins, proteins only made up 1% of the DNA  From a protein you can take the aa sequence and screen the library for which gene it could be. From a gene you can insert an expression vector to express whatever protein it codes for.  Cloned genes can be expressed if they have a origin of replication, promoters and operator sequences  Mutagens can be inserted into cells or you can insert a corrected DNA into a plasmid to be inserted into a cell Yeast 2-Hybrid Analysis  You can test whether domains interact with each other by mating gal4p DNA binding domains and gal4P DNA activation domains  plant them on a medium that requires they interact for survival  use sequence fusion proteins to identify which proteins are interacting LECTURE 23 – POST-TRANSCRIPTIONAL REGULATION  Can regulate at the translation stage o RNA is fed through the ribosome in the presence of tRNA until the start codon is found, you can regulate translation by blocking access to the start codon or you can block the productive formation of the ribosome complex (decrease translation) o you could shield the mRNA from exonucleases (increase translation)  Official Ways: o Inhibit eLF (eukaryotic initiation factor) activity through phosphorylation o Repressor proteins bind to the RNA o 4EBPs disrupts the interaction of eLF4E and eLF4G (kinases can turn off this disruption)  Can regulate the RNA Stability RNA interference – (RNAi) – RNA molecules inhibit gene expression, typically by destroying specific mRNA molecules o In the cell there are non-protein coding RNA, these RNA’s target mRNA for degradation  Dicer is a human enzyme that cleaves double stranded RNA into short fragments called siRNA and microRNA. Dicer leaves a two base overhang on the 3’end.  Each siRNA is unwound into two single stranded RNA (ssRNAs), one serves as the guide stand and is incorporated into the RISC complex. The other passenger strand is degraded. o Argonaute selects which strand is the guide strand; the guide strand is less thermodynamically stable at the 5’ end.  The guide strand binds with a complementary mRNA strand, this binding induces cleavage by Argonaute, the catalytic component of the RISC complex  Argonaute activity is capable of degrading mRNA (silencing it)  Main Function of RISC: regulate abundance of mRNA and accessibility to the mRNA  By binding to the mRNA the mRNA is no longer single stranded and it cannot be translated by the ribosome Cellular Iron Homeostasis  When iron is low your cells increase uptake and decrease storage  When iron is high your cells decrease uptake and increase storage  Ferritin is an iron core – it hold the iron that has entered the cell  Transferrin – binds to iron and carries it around the blood stream, it can bind to a transferrin receptor on the surface of the cell, causing the iron to be endocytosed. o When Fe in the cell decreases, increase uptake (increase TfR) and decrease storage (decrease ferritin) o When Fe in the cell increases, decrease uptake (decrease TfR) and increase storage (increase ferritin)  Regions of the 3’ UTR of the TfR transcript sense levels of iron  Regions of the 5’ UTR (stemloop) of ferreting sense levels of iron  Each of these regions are filled with IRE motifs (iron responsive elements) Paper: “Iron regulated ferritin mRNA translation through a segment of its 5’ UTR” 1. Found out which sequences regulate the translation of ferritin  Made artificial constructs with the 5’ UTR and 3’ UTR, just 5’ UTR, and just 3’UTR  Instead of the ferritin gene, used a reporter construct  Ran a TLC plate to see if an acetylated product resulted  Found that when the 5’UTR was missing, no regulation so they determined 5’UTR was important 2. How does the 5’ UTR regulate translation?  Found a protein IRP that binds to the IRE (iron responsive elements) of the 5’ UTR  Used SDS – Gel electrophoresis  Did a cross linking assay to see how many components of the cell bind to IRP (looked at the protein complexes) 3. What is the IRP?  Before they did affinity chromatography they decreased the presence of the other background compounds so that there was not noise (compounds to interact and bind to the important compound)  Purified the IRP through affinity chromatography = separating proteins base on the affinity for some molecule  Stuck the 5’ UTR on a bead to selectively capture the IRP  Saw a single band that represented the binding protein  Sequenced the protein by edman degradation i. Edman degradation is a process by which a person cycles the pH to remove amino acids one by one  Also sequenced the protein by tandem mass spectrometry  Discovered that IRP was cytopplasmic aconitase Aconitase:  Found that cytoplasmic and mitochondrial aconitase are isoforms  Aconitase has an iron sulfur complex in its active site  When iron is high, IRP (aconitase) closes onto itself making a narrow cleft and uselessly produces isocitrate. This allows translation of ferritin (which we want because we want to store the high levels of iron in the cell)  When iron is low, IRP (aconitase) opens up and binds to the IRE element of the ferritin RNA. This binding decreases the translation of ferritin, resulting in less storage (which is good because iron is low so we do not want to store it) o When the cleft if wide (open) IRE Binding Protein 1, also called IRP1, binds and inhibits translation ***** IRE is RNA and IRP is a protein


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