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BCHM 307 Class Notes Week 7

by: Sean Anderson

BCHM 307 Class Notes Week 7 BCHM 307

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Sean Anderson
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These notes cover the lecture material for exam 2 up to week 7.
Dr. Stefan Paula
Class Notes
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This 24 page Class Notes was uploaded by Sean Anderson on Sunday October 16, 2016. The Class Notes belongs to BCHM 307 at Purdue University taught by Dr. Stefan Paula in Fall 2016. Since its upload, it has received 4 views.


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Date Created: 10/16/16
Sean Anderson Professor Paula BCHM 307 10/31/16 BCHM 307 Exam 2 Study Guide Lipids  Lipids—biological molecules that are soluble in nonpolar solvents and poorly soluble in water o They are predominantly hydrophobic molecules o They do not form polymers and aggregate to form bilayers and micelles o Glycerophospholipids and sphingolipids  Lipid classes o Fatty acids o Triacylglycerol o Glycerophospholipids o Sphingolipids o Cholesterol  Fatty acids o Amphiphilies o Polar head group: Carboxylic acids o Nonpolar tail: Hydrocarbon tail o Fatty acids can be saturated, unsaturated, or polysaturated  Unsaturated fatty acids o Double bonds are usually in the cis-conformation  Triacylglycerol “triglycerides” o Function: energy storage in fat cells (adipocytes)  Glycerophospholipids o Function: major component of biological membranes  Lipid bilayers o Three common head groups on Glycerophospholipids  -Choline (no charge)  -Serine (- charge)  -Ethanolamine (no charge)  Phospholipases o They cleave phospholipids at specific sites o  Sphingolipids o Instead of glycerol, sphingolipids have sphingosine as their backbone o  Sphingomyelins o Function: sphingomyelin is found in the myelin sheath around nerves o Head group is a phophocholine or phosphoethanol group  Cerebrosides o Similar to sphingolipids, but carry a monosaccharide in their head groups (instead of phophocholine or phosphoethanolamine)  Gangliosides o A ganglioside is a sphingolipid with an oligosaccharide head group o Function: gangliosides are found on cell surfaces are involved in cell/cell recognition  Steroids: cholesterol o Cholesterol is derived from isoprene o Has a weak amphiphilic character o Cholesterol is found in the biological membrane o Cholesterol is a metabolic precursor of steroid hormones  Estrogen, testosterone, cortisol, aldosterone  Characteristics of a lipid bilayer o Fluid: no clearly defined geometry, head groups move up and down o Hydrocarbon tails wave o Asymmetric: different lipids are found in each leaflet  Only certain lipids can from bilayers o Fatty acids 1 hydrocarbon tail o Glycerophospholipids  2 hydrocarbon tails o Triacylglycerol’s  3 hydrocarbon tails o  Melting points of lipids o Double bonds put kinks into the acyl chain o The melting point of an acyl chain decreases as the degree of unsaturation increases  More double bonds = lower melting point –highly saturated  Less double bonds = higher melting point –highly unsaturated  Lipid motions in a bilayer o Transverse diffusion  Polar head group has to travel through the hydrophobic tails o Lateral diffusion  Polar head group travels along the hydrophilic membrane o  Three types of membrane proteins o Integral membrane protein o Peripheral membrane protein o Lipid-linked protein o  Lipids can be covalently linked to proteins in a variety of ways o Myristoylation o Palmitoylation o Prenylation o  Transmembrane proteins span the bilayer o Bacteriorhodopsin: 20 AA/helix o Beta Barrel: 8 strand minimum o  The Fluid Mosaic Model o Membrane proteins float in a sea of lipids and do not transverse the bilayer (= lateral diffusion) o Carbohydrates   Carbohydrates o They are formed from CO a2d H O 2 o Carbohydrates have the generic formula: (CH 2) n  Where n > 3 o Their roles:  Energy source in metabolism  Mediation of intercellular communication  Structural support (cell walls)  Carbohydrate classification o Monosaccharides  Glucose, fructose, ribose o Small polymers  Disaccharides – two sugars bound  Sucrose, lactose, maltose  Trisaccharides—three sugars bound  Oligosaccharides—several sugars bound o Polysaccharides—large polymer of sugars  Starch, glycogen, cellulose  Carbohydrate naming o Pent—5 o Hex—6 o Aldoses: monosaccharides made from aldehydes  Glyceraldehyde is the simplest aldose o Ketoses: monosaccharides made from ketones  Dihydroxyacetone is the simplest ketone  Isomers o Most carbohydrates are chiral o (picture)  Rules for designating D and L notation 1. Draw aldehyde or ketone group at the top—Fischer Projection 2. Identify lowest chiral center in molecules 3. Classify configuration based on position of the OH group  If OH is on the right, D form  If OH is on the left, L form o Most sugars in nature are in D form o Biological systems can distinguish D and L sugars  Plenty of isomers possible –be able to sketch them o Aldopentoses  D-Ribose o Aldohexoses  D-Glucose  D-Galactose  Ring formation o Same sugar can be sketched in different ways  Fisher projection  Haworth projection o  Cyclization of glucose o Since the aldehyde group has rotational freedom about the C1-C2 bond, two anomers are possible  α-anomer (-OH down)  β-anomer (-OH up) o α-D-Glucose is most prevalent in biochemistry  Bond formation between monosaccharides and other molecules o Glucose methylated at the hydroxyl group at C1 o The C1O bond is called a glycosidic bond (keeps monosaccharides together) o  Some more monosaccharide derivatives o Glucosamine—an amino group has replaces the OH substituent o Glucuronic acid—a carboxyl group has replaces the terminal CH OH 2 o Glyceraldehyde-3-phosphate—a phosphate group has replaced an OH substituent  Disaccharides: o Sucrose  Sucrose is table sugar  Sucrose has an α(12)β glycosidic bond  Sucrose is the most abundant sugar in nature  o Lactose  Lactose is milk sugar  Lactose has a β(14) glycosidic bond  Breakdown enzyme: lactose  (lactose intolerance)  o Cellulose  Structural support of plants  β-1,4 glycosidic bond  o Starch  α-amylose (100-1000 glucose units)  α-1,4 glycosidic bond   Amylopectin  α-1,6 glycosidic bond  branching   Polysaccharides o Starch  Long chain polymer with many glucose units is the principle store of carbohydrates in plants  Used as food by young plants until leaves grow  Animals use starch as an energy source by ingesting plants  Mixture of α-amylose and amylopectin o Glycogen  Found in animals, not plants  Similar to amylopectin (but higher degree of branching)  Glucose is needed by the body  quick removal of glucose monomers from multiple branch ends  Straight chains like amylose provide only one glucose molecule at a time (slower)   Glycoproteins o Glycoproteins are N-linked via—Asn o Glycoproteins are C-linked via—Ser or Thr o o ABO blood types:  Oligosaccharides on surface of red blood cells  Type Aantibodies against B  Type B antibodies against A  Type AB  no antibodies  Type O  antibodies against A and B o  Artificial sweeteners o Very sweet compounds o Compounds that cannot be metabolized  Aspartame DNA  DNA: general o Carrier of genetic information o Chromosomes: DNA and proteins o Virus: encapsulated DNA  DNA structure o DNA and RNA contain nitrogenous bases –learn to draw purines and pyrimidines  Purines – Adenine and Guanine  2 ring structures  Pyrimidines—Thymine and Cytosine  1 ring structures  RNA: uracil o RNA: uracil instead of thymine  Absence of methyl group  DNA/RNA: ribose o Bases are linked to a sugar via a N-glycosidic bond o Ribose: 2’ –OH group o Deoxyribose: 2’ –H group o Ring numbering on the sugars contains primes  Nucleotides o Add 1-3 phosphate groups at 2’ carbon atoms o Nucleoside = base + sugar o Nucleotide = base + sugar + phosphate  Some nucleotides have functions other than being part of DNA/RNA o Coenzyme A (CoA)  -SH at top of structure is the only active part of the molecule o Nicotinamide adenine dinucleotide (NAD)  nicotinamide structure at top is the only active part of the molecule  DNA: polymer formation o Nucleotides are connected by phosphodiester bonds o Direction: read sequence of bases 5’ 3’  DNA: 3D structure o Pauling model for DNA (incorrect) o Triple helix: phosphates and ribose in core with bases periphery o Watson and Crick model for DNA (correct) o Double helix—strands are antiparallel  DNA: the double helix o Base pairs come from the core of the double helix o Phosphate backbone forms the periphery o Structural features:  2 antiparallel strands  Bases in center, sugars and phosphates in periphery  Bases stacked on top of each other  Places of bases roughly perpendicular to axis  H-bonding between base pairs in a complementary fashion  2 grooves (major and minor) o Stabilizing forces:  Hydrophobic interactions  bases buried inside  H-bonds  between base pairs  Van-der-Waals interactions  core densely packed  Ionic interactions  cations around the phosphate groups reduce electrostatic repulsion  Base pairings in a double helix o Pyrimidines form 2 H-bonds o Purines form 3 H-bonds o Base pair width is similar o Chargaff’s rule provided initial evidence for base pairing  A + G = T + C or A = T and C = G  Complementary  DNA denaturing o DNA (like proteins) can denature (unfold, melt) o Monitor denaturing by measuring the absorbance at 260nm o T m melting point o  DNA renaturing o DNA can renature (refold, annealing) o Central Dogma of Central Biology   Central dogma o DNA  RNA  protein o Genes: sequences of DNA o Replication: copying DNA o Transcription: making RNA and DNA template o Reverse transcription: making DNA from RNA template o Translation: making proteins from an RNA template  Replication o Replication is semi-conservative  DNA is supercoiled o Topoisomerases: enzymes that alter supercoiling  Replication fork o A replication for is the site where DNA strands separate and 2 new strands are synthesized  Several proteins participate in DNA replication (“Replisome”) o Helicase o Single-stranded binding proteins o Primase o DNA polymerase o Sliding clamp o RNase o DNA ligase  Helicase o Helicases convert double-stranded DNA (dsDNA) to single-stranded DNA (ssDNA)  Single-stranded binding proteins (ssBP) o Single-stranded DNA is bound by single-stranded binding proteins  Protection from nucleases  Prevention of annealing  DNA polymerase o DNA polymerase faces two problems:  DNA polymerase can only extend a pre-existing chain  DNA polymerase cannot initiate polynucleotide synthesis  RNA polymerase (primase) can  Primase makes an RNA primer (10-60 nucleotides)  The RNA primer will be replaced by DNA later  DNA synthesis occurs in the 5’3’ direction o Antiparallel strands of DNA need to be replicated simultaneously  A DNA polymerase binds to each strand  The two polymerases work side-by-side  One strand must loop out  Lagging strand is synthesized in a discontinuous fashion o Okazaki fragments (100-2000 nucleotides long) o  Sliding clamp o Most DNA polymerases are processive (hold on to substrate for several catalytic cycles) because the sliding clamps holds the DNA and the DNA polymerase in place  3D structure of DNA polymerase o “Klenow” fragment of DNA polymerase I o  DNA polymerase: proofreading o DNA polymerase proofreads newly synthesized DNA:  Base pair mismatch  Local distortion of in helix sense by DNA polymerase  Hydrolysis of wrong nucleotides o 3’5’ exonuclease activity o  RNase and ligase o RNase: removal of primer o Ligase: sealing gaps  Replication: a potential problem at the 3’end o 2 replication forks moving in opposite directions o 3’ overhang of parent strand  Daughter strand shorter o Implication of problem:  Each round of DNA replication would lead to chromosome shortening  Telomeres o Solution:  Telomerase repeatedly adds nucleotide segments (“telomeres”) to the 3’ end of parent DNA strand  DNA polymerase can then complete the other DNA strand  DNA packing o Eukaryotic DNA wraps around histones (proteins) o The fundamental unit of DNA packing is the nucleosome  Levels of chromatin structure o Fibers loop to form packed chromosomes o Nucleosomes aggregate to form a 30-nm fiber (coil, solenoid) o DNA wraps around histones to form a nucleosome  Covalent modification of histones o Histone modification can regulate gene expression o Positively charged Lys residues normally strong interact with negatively charged DNA o Low expression levels, DNA not accessible to other proteins o Acetylation removes positive charge (neutral)  Weaker interaction with DNA (loosening)  Gene expression levels because of better access to DNA for transcription  Transcription   Transcription  Transcription is carried out by immobile protein complexes that reel in the DNA  RNA transcript is complementary to the anti-sense (non-coding) strand  o Chromatin remodeling  Histone Lys residues can be acetylated to activate transcription  DNA is loosened from the histone so that different DNA segments are exposed o Promoters  Transcription starts at promoters which are close to the protein- coding sequence  Promoters are recognized by RNA polymerase directly or by proteins that direct RNA polymerase to the area  RNA polymerase subunits α β,2β’, ω, σ (prokaryotes)  σ factor recognizes promoter  positions polymerase  leaves   Eukaryotic promoters have a variety of consensus sequences  Note:  Promoter position and length can vary  Promoters can be “upstream” (5’ side) and “downstream” (3’ side)  More than one promoter can be present  o Sigma factor  o Transcription factors  No sigma factor  Numerous transcription factors instead  Some roles of transcription factors:  Prepare DNA for transcription (helicase activity)  Direct and position RNA polymerase and DNA o Enhancers and activators  Enhancers are located further away from the transcription site than promoters are  Activators are proteins that bind to enhancers  Enhancers/activators regulate gene expression patterns in response to signals (internal or external)   Mediators are proteins that interact with both the activator and the transcription machinery  o Silencers and repressors  Interference with activators function  repressor binds to silencer sequence close to enhancer sequence, prevents binding to activators  Gene expression is no longer activated but suppressed  Counterparts: o Enhancer  silencer o Activator  repressor o  Prokaryotic operons: The lac operon o Operon:  A set of functionally related genes  Its expression is coordinated in response to external signals o Lac operon:  Set of genes for 3 proteins enabling E. coli to metabolize lactose  o Absence of lactose  no expression of Z, Y, and A o Why?  Lac suppressor firmly binds to “operator” sequence  (O – a silencer) by blocking the promoter site  o Lactose is present:  Repressor binds to lactose, then changes conformation, loosens its “grip” on operator  RNA polymerase can start its job (=transcription)   RNA polymerase   Bacteria: just one type of RNA polymerase  Eukaryotes: several types of RNA polymerase  RNA polymerase II: responsible for protein-coding genes  During transcription, the RNA forms a short double helix with the DNA template (hybrid helix)  No primer is needed (unlike DNA polymerase)  o Mechanism  Very similar to mechanism used by DNA polymerase  Nucleophilic attack by the 3’ OH group onto the P of the α- phosphate of dNTP  Polymerization in the 5’  3’ direction   Two Mg 2+ions are required for catalysis  Shielding of negative charge on phosphates  Otherwise, nucleophilic attack would be too difficult due to repulsion of negative charges o Shielding of negative charge/charge repulsion o o Proofreading  Sequence of events:  Incorrect nucleotide gets incorporated  Hybrid helix becomes distorted  RNA leaves active site  Hydrolysis takes place (3’  5’ exonuclease activity)  Hydrolysis is stimulated by a transcription factor and performed by RNA polymerase II  o Termination  How does RNA polymerase “know” where to stop?  Prokaryotes: o Option 1: (Rho—independent) o Symmetrical GC-rich sequences that favor a hairpin structure o Followed by sequences that are rich in U o Remember: AU pairs are weaker than GC pairs  Destabilization of RNA/DNA helix in “bubble” (RNA falls off)   RNA hairpins:  “stem loop” structures: RNA folds back on itself  High GC content enhances hairpin stability  o Option 2: (Rho—dependent)  Helicase (Rho)  Binds to mRNA at a recognition sequence  Destabilization of RNA/DNA helix in “bubble”  A:U-rich pairs easy to separate; RNA falls off   Fate of mRNA o Prokaryotic mRNA: immediate translation o Eukaryotic mRNA: needs to travel from nucleus to cytosol (location of ribosomes)  mRNA can be modified (processed)  Variety of gene products can be generated genes  Expression can be regulated  Capping of eukaryotic mRNA o 1.) Eukaryotic mRNAs receive a 5’ cap  Protection from exonucleases that would attack the 5’ end o 2.) The 3’ end receives a 3’ Poly A tail  Attachment of Poly A binding proteins  Protection from exonucleases that would attack the 3’ end o  Splicing of mRNA o Sections of DNA are missing their mRNA counterparts o These sections (introns) have been removed during mRNA processing o Only the exons remain  About 90% of the human genome is composed on introns o Splicing commences before transcription ends  Spliceosome proteins and small nuclear RNAs (snRNAs)   Splicing does not occur in prokaryotes  Two transesterification steps:   Alternative splicing of mRNA o Alternative splicing not all exons are utilized simultaneously  Variations in proteins easily accomplished often tissue specific  Assemble proteins in “building blocks”  Fewer genes needed   mRNA degradation o mRNA life span: 1-24 hr o Life span regulation of gene expression o  RNA interference (RNAi) o Purpose: regulation of gene expression o Average length: 20-25 nucleotides (= small) o Two types:  Small interfering RNAs (siRNAs)  MicroRNAs (miRNAs) o Both bind to complementary mRNA and lead to its destruction o Therapeutic value: antiviral, anticancer o  Using RNAi in research o How can RNAi be used in research?  Reverse genetics—disrupt or silence a gene and determine any phenotypic changes in the cell/organism  RNAi can be used to target a mRNA for degradation to prevent expression  dsRNA that matches the target gene will be designed, synthesized, ad delivered into the cell o How is the siRNA delivered? 1. Transfection—lipid vesicles 2. Electroporation—apply electric pulse to cells that temporarily weakens membrane integrity 3. Virus  RNAi as a potential therapeutic strategy o Current obstacle: Effective delivery of RNA to target sites o Problem:  RNA subject to nuclease degradation  RNA is unable to cross cell membranes o Potential solutions:  Lipid of polymer based nanocarriers  Need to optimize for delivery to target site, uptake of nanocarriers, and release of RNA once in the cells  Chemical modifications o Also need to identify good targets and minimize toxicity to patient  tRNA and rRNA processing o RNA polymerase I  rRNA (ribosomal RNA) o Polymerase II  mRNA o RNA polymerase II  tRNA (transfer RNA) o rRNA:  rRNA transcripts combine with ribosomal proteins  Splicing to remove introns  Modification of nucleotides  rRNA processing o Splicing:  o Example of chemical modification:   tRNA processing o Splicing o Addition of CCA at 3’ end (eukaryotes) o Covalent modifications (see below)   Versatility of RNA o RNA has properties similar to those of proteins:  RNA (like proteins) can adopt complex tertiary structures  RNA (like enzymes) can catalyze reactions o RNA nucleotides can base pair in a variety of ways:  Watson-Crick base paring  Nonstandard base pairing  Many more possibilities than in DNA  o RNase P: enzyme composed of RNA and proteins o Was there an early RNA world?  RNA could assume function of proteins (as catalysts) and DNA (store information)


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