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by: Gayatri


GPA 3.91
CHM 116
Dr. Nash

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CHM 116
Dr. Nash
Class Notes
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This 9 page Class Notes was uploaded by Gayatri on Thursday February 5, 2015. The Class Notes belongs to CHM 116 at Purdue University taught by Dr. Nash in Winter2015. Since its upload, it has received 210 views. For similar materials see CHM 116 in Chemistry at Purdue University.


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Date Created: 02/05/15
BIOL 23100 Exam 1 Study Guide Cells Organelles and Molecules Relative dimensions of the parts of cellscells 0 H2O molecule 4 Angstroms o Epithelial cell 30 um 0 DNA 2 nm 0 Paramecium 1 mm visible to o Lipid bilayer 4 nm human eyes 0 Mitochondrium 2 um o Frog egg 255 mm o Chloroplast 8 um 0 De ne o Prokaryote single celled basic but ef cient living organism 10A30 on earth possess no nucleus or organelles o Eukaryote cells with DNA nucleus organelles perform complex tasks I Protist single celled no division of labor within organism I Metazoan multicellular organisms diff cells have specialized diff functions Topology of cells membranes amp organelles Signi cance of each of the 7 boundary properties in the cell 1 Continuous keeps things in and out 2 Selectively permeable prevents free diffusion in and out movement is managed and regulated 3 Asymmetric different inside and outside faces lipids proteins carbohydrates kept in their respective places Disequilibrium creates electric voltage and chemical concentration gradients Locus for biochemical activity a plane area vs a space Respond to external signals outside receptors 9 transduction of signals inside the cell 7 Mediate cellcell interactions communication and response Derivation of surfacetovolume ratios 0 Sphere 3r o Cube 6 s 0 Any solid lr Explain how SV varies with size the linear dimension and the signi cance of this relationship for the life of the cell 0 As size linear dimension r increases SN decreases o It is harder to transport materials across a cell to support the volume when the radius is larger 0 Smaller cells have a better SN ratio more ef cient Explain the 3 compartments or spaces we de ned and the experiments you could use to demonstrate them 1 Cytosol inside area of the cell but not in organelles includes nuclear area 0 Experiment injecting a dye in the cytosol reveals that is it topologically separate from any other part of the cell because the dye does not diffuse into any organelles or to the outside of the cell except for inside the nucleus This is because the nuclear membrane has pores which allow the dye to diffuse in 2 Organelle lumens and 3 Outer membrane internal to cell 0 Both are part of the endomembrane system 0 Experiment placing dye in a dish with the cell allowing the cell to soak it up by endocytosis This reveals that the extracellular space is topologically equivalent to the inside of organelles the lumens All but the plastids mitochondria and chloroplast are penetrated Membrane fusion budding endo and exocytosis o Organelle interaction when two organelles fuse membrane lea ets and proteins keep their orientation while contents of organelle mix but remain in the lumen area reverse process budding PM o Membrane interaction when an organelle fuses with a plasma membrane the lipid lea ets and cytosolic parts of transmembrane proteins remain cytosolic while luminal lea ets luminal end of transmembrane proteins as well as lumen area become part of the extracellular area removal of contents exocytosis reverse endocytosis Biological Chemistry General properties of water 0 Water is a polar molecule its charge is asymmetrically distributed 0 This uneven charge distribution allows water all 3 atoms to form noncovalent bonds with other H20 molecules making it cohesive constant interaction 0 Dissociation of water H20 9 H OH 39 Pure water H OH lOA7 M I Dissociation constant Kw HOH lOA7 M De ne pH log H measure of protons 0 High pH less H basic low pH more H acidic CC single vs double bonds and their rotation o CC single bonds have free rotation of carbon and other atoms saturated o CC double bonds have no rotation carbon bond is locked in unsaturated Draw the 7 functional groups and note which are ionized under cellular conditions draw those in their ionized form noting the group that they release and the resulting charge H CH l 1 Methyl H not ionized under normal conditions 2 Carbonyl C not ionized under normal conditions 3 Hydroxyl O H as pH increases can lose H 9 O H s 4 Sulihydryl H as pH increases can lose H 9 S H O O I Ho P o 0 f o 5 Phosphate OH as pH increases can lose 2H 9 039 H O O lti OH lt5 6 Carboxyl acidic as pH increases can lose H 9 1 0 H T H N H 7 Amine basic N H as pH decreases can gain H 9 H De ne pK describe carboxylic acid and an amine above and below their pK o pK transition between neutral molecules and their charged state occurs at speci c pH value 0 At low pH carboxylic acid is a neutral molecule COOH and amine is present in the NH3 form 0 At high pH carboxylic acid is in the C00 form and amine is a neutral molecule NH2 Four important linkages 0 ol l Ester alcohol acid 9 l a Present in phospholipids O I II 2 Phosphoester alcohol phosphate 9 9 39 quot a Present in DNA backbones O O OgO 3 Phosphoanhydride phosphate phosphate 9 69 6 a Present in sugars ATP the negatively charged oxygens are forced to close even though they repel each other bc of the phosphate linkage causes energy release when the bond is broken Cu 4 Amide amine carboxylic acid 9 CN a Present in peptide bonds in protein important for structure b C N bond does NOT rotate double bond character Note All are covalent bonds and all reactions to form linkages involve release of H20 Explain compare and contrast the four noncovalent interactions 0 Hbonds I H is shared by two electronegative atoms 0 and N usually I Weakened by H20 which forms Hbonds very well with functional groups on macromolecules I Direction and orientation matter I Involved in protein structure alpha helices DNA base pairing I The straighter the path of the bond the stronger it is I The bonds are relatively weak but strong when combined bonds lining along the entire surfaces of macromolecules o Ionic bonds I and charge attraction I pH dependent I Weaker in presence of H20 bc charged groups are shielded by their interactions with H20 molecules BUT they can come close and exclude overcome water eg Enzyme substrate interactions 0 Hydrophobic forces I Groups that are forced together by water even though they have no attraction toward each other so that they have an overall smaller disruption effect on the solution I Driven by entropy working towards increasing disorder 0 van der Waals interactions I Nonspecific interaction can be between any two atoms I Small quick asymmetries in electron distribution when they come within a certain distance of each other I Very weak bonds can be strong when combined Strengths of bonds covalent gt ionic gt hydrogen bonds gt van der waals Structures Of Biological Macromolecules Four major classes of macromolecules l Lipids steroids phospholipids a Found in plasma membrane and organelle membranes 2 Sugars amp Polysaccharides a In glucose OH on the 1 carbon can be in alpha bottom or beta top position b Can be branched or unbranched i Cellulose beta unbranched ii Starch alpha unbranched iii Glycogen alpha branched c Monosaccharides can build the following i Oligosaccharides found on membrane proteins ii Polysaccharides cellulosestarch made of glucose 3 Nucleic Acids a Subunits are nucleotides composed on base 5carbon sugar phosphate group b Found in DNA and RNA building blocks 4 Amino Acids and Proteins a Structure consists of alpha carbon COOH carbonyl NH2 amine hydrogen R group b Properties amphoteric acidbase zwitterionic can be acidic basic uncharged polar nonpolar c Joined together with loss of H20 and formation of peptide bond has double bond properties does not rotate 9 forms polypeptides d Building blocks in proteins Two major kinds of lipids Steroids Phospholipids Ring backbones Acyl chains glycerol backbone ester linkages w fatty chains Shorter spread out structure Long chain made of polar head group glycerol backbone and fatty chains Cholesterol rigid planar structure with one OH Phosphatidylcholine exible structure has an amide group giving it some polarity on one side of the containing polar head group and a hydrophobic tail molecule m co o my RZ CO O CH 0 H30 Hzc o Fgt o CH2 CH2 Nquot CH3 CH3 P110 sphatidylcho line Carbohydrates macromolecules made up of sugars formed into polymers called polysaccharides which are linked by alpha and beta glycosidic bonds they work to promote structure and provide energy and storage 0 Glucose simple most common form of sugar 4 different methods of drawing 0 Numbering mechanism start with first OH carbon on right in structure 2 end with C in CHZOH H C OH CH20H OH F OH O lt30 Hem HO F H HO H l3 0H OH OHOH OH HO OH H C OH OH Polar OH groups on glucose make it water soluble and give it the ability to make hydrogen bonds with water molecules Explain starch glycogen and cellulose can have such different properties 0 Starch glycogen and cellulose are all polymers made from glucose have the same subunits But what gives them their properties are the bonds they are connected with giving them their own unique structures I Cellulose OH on 1 carbon in beta position unbranched chain I Starch OH on 1 carbon in alpha position unbranched chain I Glycogen OH on 1 carbon in alpha position branched chain dAMP and its structure 0 The functional groups on different bases are responsible for hydrogen bonds that make the base pairs 0 Bases are covalently linked to sugars Phospoester links are present between 5 carbon and phosphate group o The 1 carbon is the one that connects the H on the sugar to the N on N N the base I K N Ho e o N OH H2N O Basic structure for an amino acid in the charged and uncharged forms H 39I H2N C COOH H3N Cl COOquot R R Linkage of amino acids 0 Linked together with the loss of H20 H from one amino acid and OH from another 0 H and formation of a peptide bond 0 This peptide bond is a single bond between the C and N groups but has properties of a I double bond does NOT rotate c N R groups side chains of amino acids give the amino acid its properties polar nonpolar acidic basic Properties of the four different classes of macromolecules major differences 0 Phospholipids I Made of glycerol fatty acids phosphate have a polar head group and hydrophobic tail I Building blocks are linked by ester and phosphoesterphosphodiester bonds I Function is to form membranes and store energy 0 Carbohydrates I Made of sugars I Polymer is polysaccharides can be linear or branced I Polymers are linked by glycosidic alpha and beta bonds I Function is to provide structure and energy storage 0 Nucleic Acids I Made of nucleotides made of bases ribose or deoxyribose sugar and phosphate I Bonds are phosphoester between C and N I Polymer is nucleic acids DNA and RNA all unbranched I Polymer is linked by phosphodiester bonds I Function is to store info and catalyze protein formation rRNA 0 Proteins I Made of amino acids I Polymer is proteins I Polymer is linked by peptide amide bond I Function is to provide structure catalyze reactions and regulate processes 0 Maj or differences Carbohydrates vs Proteins and Nucleic Acids I Carbohydrates use identical building blocks connected in different ways to make different structures while proteins and nucleic acids use different building blocks connected in identical ways to do the same 0 Maj or differences Phospholipids vs Carbohydrates Proteins and Nucleic Acids I Phospholipids are not connected by polymerization into a large structure they are simply held close by hydrophobic interactions while carbohydrates proteins and nucleic acids are all products of polymerization held together by covalent bonds Energetics amp Thermodynamics In The Cell Keq for a chemical reaction ratio of the concentrations of products over reactants at equilibrium Laws of Thermodynamics 0 First Law energy is conserved and converted in all processes and is never created or destroyed 0 Second Law any system tends to move toward greater entropy disorder the lowest energy state Gibbs Free Energy free energy during a reaction at constant pressure and temperature 0 AG AH TAS The maximum energy available to do work AG is equal to the total energy change in a reaction AH minus the energy lost to entropy TAS o It allows us to predict the direction of a reaction Relationship between K and AG 0 K is the ratio of PR which is determined by the relative G of P and R 0 As long as the free energy of reactants is greater than free energy of products we expect the reaction to proceed spontaneously 0 At equilibrium K 1 and AG 0 so there is no further net reaction For a highly favorable reaction typical K is greater than 1 and AG is less than 0 For an unfavorable reaction typical K is less than 1 and AG is greater than 0 Coupled Reaction when a highly favorable reaction very negative AG is paired up with an unfavorable reaction high positive AG to make it occur 0 These are vital because most essential reactions in our bodies are unfavorable and nonspontaneous but they need energy and help from other reactions to proceed AG of a coupled reaction can be calculated by adding the AGs of the two reactions together K can be calculated if standard AG is given using AG 57kJmole gtlt logKeq Enzyme Kinetics Catalytic mechanism of enzymes 0 Enzymes catalyze reactions by lowering the activation energy E of a reaction by stabilizing the transition state and leaving AG unchanged enzymes are not consumed in the reaction itself Active site a domain of an enzyme where reactants bind and catalysis is carried out creating products Three ways in which enzymes catalyze reactions 1 By aligning with substrate molecules favorably 2 By changing the chemical environment and eXposing substrates to reactive R groups which are often charged or polar 3 By putting strain on covalent bonds leading to bent substrates changing the structure MichaelisMenten equation V VmaX X SS Km 0 Two important constants describing enzyme behavior 1 VmaX maximum reaction rate V for an enzyme 2 Km substrate concentration 8 that produces a reaction rate value at l2VmaX 035 7 7 7 7 g p 7 7 p 7 7 p 7 g 7 7 g 7 7 g g 77 030 X 025 1 020 015 Reaction rate 010 Km 005 000 v u v t 0 1000 2000 3000 4000 Substrate concentration LineweaverBurke equation lV KmVmaXlS lVmaX 0 Useful because it is in the simple line equation and can give us a fairly good estimate of Km and VmaX with the T plotting and best fit line equation of a few points because the y intercept is lVmaX and the X intercept is lKm K W Enzyme activity with regards to temperature enzymes have an optimum temperature and their activity is decreased below and above this point Enzyme activity with regards to pH enzymes have an optimum pH and their activity is decreased below and above this point Irreversible enzyme inhibition when inhibitors bind to an enzyme very tightly using covalent bonds usually on a critical R group necessary for function stopping the enzyme s activity permanently 0 Eg The antibiotic penicillin which inhibits bacterial trasnpeptidase 0 Eg The nerve agent sarin which inhibits acetylcholineesterase which breaks down Ach a neurotransmitter which stops muscle contraction Inhibition of AChase auses muscle spasms seizures paralysis etc Enzyme Inhibition And Allostery Competitive inhibitors compete with substrate for the active site on an enzyme have similar molecular structure as the substrate 0 Effect on catalysis the enzyme is still able to reach its normal VmaX but only if S is increased I VmaX is unchanged but Km increases Noncompetitive inhibitors bind to enzyme at a site different than the active site and have no direct effect on substrate binding to the enzyme 0 Effect on catalysis since S does not affect degree of inhibition the VmaX is affected by I but this value does not interfere with Km I VmaX decreases Km is unchanged NO inhibitor Noncompetitive Competitive t yquot393x Inhibition Inhibition vi Competitive iiiliitJititJti v Vmax U2 quot39ax Noncompetitwe inhibition 12 v5 K K S Slope of a LineweaverBurk plot KmVmaX o For both inhibitors the slope is greater than it is for the normal enzyme Allosteric regulation of an enzyme occurs when specific molecules bind the regulatory binding sites on enzymes which are located away from the active site and this leads to alteration of the structure of the enzyme affecting its activity and making the active site less accessible to substrates Allosteric regulator molecule vs noncompetitive inhibitor 0 Although the two seem similar and bind to sites other than the active site noncompetitive inhibitors do not bring about conformational change in the enzyme and do not affect substrate binding in any way like allosteric regulators do Protein Structure Generic amino acid structure 0 Two are linked together by a peptide bond The peptide bond is between the C and lil N atoms and has double bond character so it does not rotate The H and O are HzN 39 I3 39 COOH always in the trans opposite side position R Important amino acid structures HNZCo0 H2NCH OH o Glycine H W O H EH OH H2 4 N C C l CH2 1 o Aspartic ac1d H L oH HOCoo H2 H c lt ii o Lys1ne 013 Levels of protein structure 0 Primary linear sequence of amino acids provides info necessary to determine 3D structure of the protein 0 Secondary shape or conformation of portions of the peptide chain can be alpha helices or beta sheets both involve hydrogen bonds between atoms 0 Tertiary conformation of the entire protein or polypeptide chain formed based on interactions of functional groups that are on the different R groups of amino acids o Quaternary interaction of multiple polypeptides to form a protein or functional complex via noncovalent attractions Alpha helix coiled conformation secondary structure of DNA that forms a twisted backbone from the backbone atoms and has R groups sticking out along the outside 0 It is stabilized by the hydrogen bonds between H and N atoms which line up because of the way it coils and also by the peptide bonds between the C and O atoms H bonds are formed between atoms of backbone not the R groups Beta sheet made of strands of polypeptides that run along each other and interact because of the hydrogen bonds between the backbone atoms and the ones next to them forming a sheet 0 The R groups point up and down alternatively 0 Sheets run antiparallel to each other Coiledcoil structure an example of an quaternary structure of proteins where two alpha helices wrap around each other to make a combined helix 0 To form this structure the alpha helices must have small nonpolar amino acid residues along them so that they can line them up through hydrophobic interactions creating a stripe when the two helices coil with each other Protein Structure and Function Structural Basis of Protein Ligand Binding cAMP binding site many proteins bind cAMP they all have conserved sites Seven separate interactions are formed 0 Serine and threonine 3 hydrogen bonds with the base 0 Glutamic acid and backbone Nitrogen 2 hydrogen bonds with ribose o Arginine one ionic bond with phosphate 0 Serine one hydrogen bond with phosphate Ways to disrupt any of the above mentioned binding sites 0 Changing an important residue R group in site 0 Changing a residue that is important for proper folding and 3D structure Protease enzymes that carry out the reaction of breaking peptide bonds in other proteins leading to their degradation Serine proteases have serine residue in their active sites 0 Three residues that work to hydrolyze the peptide bond of substrate protein 1 Serine 2 Histidine 3 Aspartic Acid 0 Process I Electronegative O on SERl95 attacks OC carbonyl of the protein splitting the peptide bond and forming a transient covalent bond with N terminal fragment of the substrate protein I SERl95 is made more reactive by HST57 bc of the proton attraction I ASP plays a role in in uencing the ionic state of HISS 7 DNA and Chromatin Structure Important experiments 0 Griffiths 1920s studied streptococcus pnemonaie bacteria respiratory I S strain lethal and R strain benign I A certain factor transformed R 9 S I Experiment 1 S strain injected in mouse 9 it died I Experiment 2 R strain injected in mouse 9 it lived I Experiment 3 Heatkilled S strain injected in mouse 9 it lived I Experiment 4 Heatkilled S strain and R strain mixed and injected in mouse 9 it died I Found evidence of some kind of hereditary principle 0 Avery MacLeod and McCarty 1940s tried to purify and separate macromolecules using same R and S strain mixture I RNA mixed 9 gave R strain O I Protein mixed 9 gave R strain I Lipid mixed 9 gave R strain I Carbohydrate mixed 9 gave R strain I DNA mixed 9 gave S strain I Found that DNA carries heritable information Hershey and Chase 1950s worked with bacteriophages viruses that infect bacteria I Used two different viruses one with labeled DNA and one with labeled protein and used it to infect Ecoli bacteria I Found that the successfully infected bacteria had the presence of the virus with labeled DNA confirming DNA is the genetic material Basic structure of DNA double helix 0 O O 2 nm diameter 034 rise per base pair Paired up strands are antiparallel Sugar phosphate backbone is linked by phosphodiester bonds between the 3 C of one ribose and 5 C of the next Backbone structure gives rise to major and minor grooves proteins interact with DNA in major groove Each ribose is connected to a base which points inward into the helix perpendicular to the ladder rungs Bases next to each other are connected by hydrogen bonds Base pairs A and G are purines and C and T are pyrimidines so when A pairs with T and G pairs with C the distance between the two ribosephosphate backbones in the helix is nearly constant 339 and 539 ends of a nucleotide strand refer to the carbon atoms where the strands of DNA begin and end The DNA double helix is antiparallel meaning that as we move along the helix one strand is oriented 3 to 5 and the other is 5 to 3 Chromatin the mass of genetic material made of DNA and proteins that condenses into coils to form a chromosome which is a threadlike structure that contains the genetic material necessary for replication


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