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Exam 1 Study Guide

by: Luke Holden

Exam 1 Study Guide 3050

Marketplace > Clemson University > Microbiology > 3050 > Exam 1 Study Guide
Luke Holden
General Microbiology
Dr. Rudolph

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This is just a study guide that was helpful for me when preparing for my first biochem exam. Extremely thorough!
General Microbiology
Dr. Rudolph
Study Guide
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This 0 page Study Guide was uploaded by Luke Holden on Friday February 5, 2016. The Study Guide belongs to 3050 at Clemson University taught by Dr. Rudolph in Winter 2016. Since its upload, it has received 57 views. For similar materials see General Microbiology in Microbiology at Clemson University.

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Date Created: 02/05/16
Study Guide for BCHM EXAM 1 1 Understand different types of bonds 8 d Covelant Bonds These bonds share electrons with other atoms i Polar Covelant Very Different EN Unequal Sharing ii Nonpolar Covelant Same EN Equal Sharing Ionic bonds These are attracted ions that remain close to each other in order to have their octet rule satisfied These typically occur between nonmetals and metals Hydrogen Bonding This occurs with the atoms FON They are a special type of covalent bonding and they give water its many properties Metallic bonding occurs between two metal atoms 2 Understand the properties of water and how hydrogen bonding in uences them 8 Water can form bonds at 1 time Two on the oxygen and two on the hydrogen So water can bind with other water molecules because of hydrogen bonding Makes H20 the universal solvent This makes water form solvation spheres spheres that form around individual solute atoms 399 Cohesion This is the ability for water to stick together and form a relatively high surface tension in the liquid state This is due to the hydrogen bonding being able to occur among other water molecules Adhesion This is the ability of water to stick to other things such as the xylem and phloem of a plant stem This capability allows water to travel up the root system and nurture the plant Surface Tension This is the ability of water to remain taught Ice on water Ice oats on water because as the phase of water changes from liquid to solid lattices begin to from amongst the water molecules These lattices in essence open up the space in between each molecule which make the solid denser than the liquid High Melting point This is because H Bonds are the strongest Intramolecular force besides ionic bonding i Increase the solutes in water Increase in boiling point ii Increase the solutes in water Decrease the melting pointNaCl on roads 3 What is osmotic pressure Be able to calculate osmotic pressure a Osmotic pressure is the amount of pressure required to stop net ow of water across a membrane Equation Tr MR T i Pi osmotic pressure ii Ilfor this class iii MMolarity iv RGas Constant 008206 atmLmolK V TTemperature in Kelvins celcius to kelvin c273 4 Explain isotonic hypotonic and hypertonic solutions a Water will always go towards the most solute b Isotonic equal amount of solute entering and leaving the cell c Hypotonic In uX of water into the cell d Hypertonic release of water out of the cell 5 Be able to calculate pH or H ion concentration of a strong acid a Since strong acids dissociate fully then the pH is calculated using the logH concentration b To calculate the H concentration given the pH then do the inverse log 10quot pH c Equations to know i 14 pHpOH ii pH pKa log A lHA iii p log 6 Understand the significance of buffers and know the Henderson Hassalbalch equation a Buffers are so important because they keep certain systems of the body at a relative pH The Hernderson Hassalbauch equation is defined as AL HA b g pHpKalog 3 c So your dream buffer is when the pH pKa therefore A HA 7 Define or describe the Central Dogma in relation to the storage amp retrieval of genetic information the creation of proteins and the eXpression of traits a It can be described as Replication I Transcription Translation Expression of genetic material through proteins 8 Illustrate andor describe the general structural physical andor chemical characteristics of amino acids a The amino acid has 3 things i Carboxylic Acid ii Amine Group iii And an R group b The R group defines the amino acid and will determine whether or not it will fall into the one of the four categories Nonpolar Polar Acidic and Basic c Amino Acids have special characters such as the ability to amphoteric which is tha lity to act as an acid or base carry both a positive and negative charge on the same molecule NonPolar Methionine Met Cystine Cys C Glycine Gly G Ac1d1c Bas1c O Proline Pro P O O O Alanine Ala A Aspart1c Ac1d Lys1ne Lys K Polar A D Valine Val V Sp Arginine Arg R S 39 S S Gl t 39 A 39d Leucine Leu L eme er H mm Cl Histidine His H Glu E T T Y Isoleucine Ile I ymsme yr Th 39 Th T Phenylalanine reomne r Phe F Aspargine Asn A Tryptophan Trp W Glutamine Gln Q 9 Describe andor illustrate the titration of an amino acid including the pI s or pK s of ionizable groups a The pI of I oniable groups is the average of the two pKa s and depending on whether or not it is acidic or basic determines what pKas to use 140 120 100 00 11H 00 40 20 140 120 100 00 60 40 20 glutamic atin 43 22 00 139n I ma 15 2 D 140 120 100 30 60 40 20 30 25 lysine 22 00 10 20 III 5 1 5 equivalents DH ddecl 140 120 100 30 60 40 20 30 25 c For glutamic acid you will use the pKa values of pKl and pK2 Average these two together This is because at a pH of 1 all of the functional groups are protonated When it reaches a pKa of 22 this when 50 of the protons on the carboxylic acid have let go of there protns and thus will hang the charge of th amino acid in the balance When it reaches a pH of 43 then 50 of the protons on the R group have let go and thus pulls the chrge down another step Finally when it reaches a pKa of 97 all of the protons have dissociated and the amino acid is now basic You can create a table like this H Char e 1 1 22 1O 43 O 1 97 1 2 d For the basic amino acid lysine you will average the two pKa values at the top end pKa 2 and pKa 3 You start with a charge of 2 at a pH of l and then create a table just like acidic amino acid H l 22 W mn 90 10 Determine the charge of the amino acid at a certain pH based on their pK values a To determine the charge you just simply make a charge and pH chart like the previous question Once you find the pH of the amino acid you just go to the spot on the chart that has your pH value ie pH 7 on the basic amino acid 1 charge 11 Detrmine at what pH an aminoacid would have a particular charge Based on pK values a Do the same thing as before except work backwards l2 Identify andor describe the structures amp properties of the 20 standard amino acids a Refer to quizlet l3 Distinguish between essential vs non essential amino acids vs non standard amino acids including examples of the physiological roles a Essential amino acids are the amino acids that needed to be required through food i Ile Leu Lys Met Phy Try Thr Val b Non essential amino acids are the amino acids that can be made within the cell c Non standard amino acids These are mostly modified standard amino acids that have special biological roles Some but few are involved with the protein i GABA inhibitory NT of the brain involved in muscle relaxation sleep diminished emotional reaction and sedation ii Serotonin NT of the brain modulates mood appetite sexual activity aggression body temperature sleep smooth muscle constriction iii Melatonin secreted by the pineal gland during darkness linked to circadian rhythms and sleep wake cycles iv Thyroxin secreted by the thyroid increases rates of chemical reactions and metabolism in almost all cells of the body V Indole 3 acetic acid 1AA major plant hormone stimulates cell growth amp elongation rooting inhibits aXillary bud development 14 Understand primary structure of proteins a The primary structure of the protein is the simple order of the amino acids in a protein 15 Definedescribe different secondary structures and super secondary structures a A HeliX This is the structure of a protein such that it forms a heliX due to intramolecular forces i This structure is rigid and that is due to resonance and hydrogen bonding ii Every 4 th amino acid there is going to be a hydrogen bond between the Nitrogen and hydrogen 4 amino acids away This is then iii This loss of electrons is stabilized by resonance iv Proteins such as Gly and Pro due not do a heliX 1 Gly R group is a simple hydrogen and theus the structure will be to felixable 2 Pro Too rigid The big pentagon at the bond would not allow the bond to rotate to form a turn in the heliX In addition there is no N H possible 3 MALEK b B Pleated Sheet i This when a single polypeptide chain loops around and forms side by side each polypeptide chain This causes hydrogen bonds to form between the amine and carbonyl groups 1 Parallel Sheets This is when the R groups are on the same side and the single segment This is when C terminus and C Terminus Line up side by side HCHI HICH RCH FFCHI C CO HINI N C CEO 00 0 A 1 1 ll HN CEO HIN HIN HCFl PCH HCFl HCR 7 00 NH 00 0C 140A 7 7 7 130A N H oc NH NH IR CHl HC R R CH R CHI CoH N 00 00 quot N HEN 00 C N HN 39Hl N N HlCR H CH HlC R HCR Antiparallel Parallel N a 3 Antiparallel N terminus and C terminus line up Also the hydrogen bonds are straight across This makes antiparallel more stable than the parallel c Motif This is a mixture of a heliX and b pleated sheets i BaB motif This is two beta pleat sheets outline and connected by and alpha heliX They are stabilized by hydrophobic interactions by the nonpolar side chains of both the sheets and heliX ii B mender This looks like a solar panel which has multiple beta pleated sheets all connected by loops iii Aa unit this is to alpha helices connected by a turn iV B barrel this is a corkscrew looking barrel of beta pleated sheets V Greek Key This beta pleated sheets however the 1st goes to the 3 which loops back to the second which goes to the fourth m Vi a b c d e d Domain This is a big motif i Leucine zipper when it is open it has one function and when it is closed it has another 16 Understand how to classify them in a Ramachandran plot What are turns and loops in the context of secondary structures a Loops and turns are the connector portions of a protein These little guys are about 4 amino acids long and are quite huge in the structure of a protein i B turn is a very short loop that is created by the hydrogen bond formed between the 1st amino acid in the turn glycine and its bro proline which is 4 amino acids away Remember how proline can t do A helices Same thing bro Proline causes a big old fat turn to happen which is perfect for a loop And glycine is so passive it is cool with just about anything b Ramachandran Plots This is a logarithmic graph that allows you to determine whether it is an a heliX or a b pleated sheet Antiparrailleil Collagen triple 3 Slim Parallel 3391 55quot ighttwisted She 39 39 sheets 13 Leftbandied rrhEH 123 m ETD El IL in 3 H ivghtahanded if EHEME 12D quot E1 riiii iim m m 13 39 1B 35 degrees l7 Define tertiary protein structure and describe how hydrophobic amp electrostatic interactions and hydrogen amp covalent bonds can all be involved in stabilizing tertiary protein structure a Unique 3 d structure of one long polypeptide This is typically the end point for most proteins This is when it is originally very unorganized and it undergoes protein folding This makes it a highly organized molecule b It has several distinctive features i The amino acids that were far away are now close to each other ii This tight packing causes water to be eXpelled and the polar and nonpolar amino acids have ability to interact forming electrostatic bonds iii iv This contains domains So specifically it is called a domain however when in tertiary structure meaning it has taken on a 3d shape it is called a fold Most proteins are considered mosaic It is like the janitor protein carries around many keys Each key fits a lock in the school specific system but some keys go to his car or to his house which is not apart of the specific system He is still able to interact with other things but most specifically his system he is assigned to c This relies on a crap ton of interactions in order to be folded correctly i ii iii iv Hydrophobic interactions This when the water insoluble molecules are driven into the center of the molecule eXpelling the water and thus creating solvation spheres in the center of the molecule This increases the entropy of water However some water stays in the center and forms 4 different bonds stabilizing the backbone This frees up the molecule to move and be exible Ionic electrostatic interactions These are interactions where water ahs been eXpelled thus reveling the and charges When they bond it is called a SALT BRIDGE These are good for adjacent subunits adjacent polypeptide chains in a compleX proteins Hydrogen bonds ALL OVER THE PLACE These form literally everywhere there IS NOT WATER Covalent Bonds These are created whenever you have the modifications after translation But most prominent are the disulfide bonds that form as a result of covalent bonds These are the big foundation proteins because they allow the protein to remain folded to an extent adverse conditions Hydration water causes the protein to remain folding This is kind of like a lot of helpers holding up the protein which allows it to be sturdy This is also important because whenever a molecule comes to the active site of enzyme and the protein binds to the enzyme Water is moved thus increasing the entropy of water and ultimately the bind of the enzyme to the substrate Vi Salt 1 g bridge C Hydrophobic Interactions D sul de ii Bridge I 3 x U Hydrogen Fig 522 Bond 18 Describe the nature of the Anfinsen experiments and how or why they were important in protein folding a These experiments proved that when folding a protein you had to take a specific pathway I primary secondary tertiary and quaternary b His steps i ii iii iv He denatured the protein using BME and urea which BME destroyed the disulfide bonds and the urea disrupts the hydrophobic interactions and thus causes the protein to be denatured The first time through he removed the BME first and then the urea and only 1 activity returned The second time through he removed urea first and then BME and almost ALL function returned to the protein The point This experiment proved that the protein had to be folded a certain way and in a certain order When the BME was removed first the protein was able to take several conformations different than those of its original function Removing Urea allowed the protein to be folded correctly 19 What are heat shock proteins andor molecular chaperones and what is their role in protein folding and turnover a Molecular chaperons These guys assist in protein folding i They keep the unfolded protein from interacting during transition stages to prevent mis folding l Hsp 70s These are proteins that stabilize the early protein They have to binding sites one on for ATP and the other for the protein ii They actually aid with the process of folding this is the shake and bake l Hsp 60s These guys are the big can for folding proteins They have two big rings composed of 7 different subunits and will form on top of each other This creates a kind of machine that is boss at folding proteins First the folded protein enters the top of the Hsp 6O compleX into the GROEL rings Then the GROES cap comes and tops of the can 7ATP are hydrolyzed in the process which takes about 20 seconds Then the protein is released along with 7 ADP molecules Denatured Flilbosome unfolded protein k hsp70 Sig Folded protein TADP GFOES I l quotin V i 1 Unfolded protein 20 Define protein denaturation and describe the main reagents or treatments that denature proteins a Protein Denaturation This is when the application of some external factor cause the protein to un fold or misfold thus losing its function b Things that make it happen i Strong Acid or Base ii Organic Solvents iii Detergents iv Reducing Agents V Salt Concentration vi Heavy metal Ions vii Temperature Changes viii Mechanical Stress 21 Define quaternary protein structure and describe how or why it can be advantageous over tertiary structure in terms of protein function a This is when two or more polypeptides subunits join together to form a larger more difficult task b Why is this better than tertiary structure i It may be easier to synthesize 4 short polypeptide units rather than 1 long 0116 ii It is easier to repair one small subunit instead of a big one iii This increases biological function 22 How is protein degradation facilitated a Protein degradation in eukaryotes is facilitated by the ubiquitin U and protesasome pathway i U is activated and attached to the protein on lysine by the enzyme El which uses ATPDAMP ii Then E2 the conjugation enzyme and E3 the target enzyme comes and specifically binds U to the target protein iii Polyubiquination occurs and a crap ton of U gets attached to the U iv The US ag down the proteasome and then it destroys protein Een scmeel Humdegenaralinn magnesiumadaptive immunity f 171131 rh quot H111 u h 35quot 4391 F 1 111 Jr l 1 u Hi i Iquot 1 I Gail mam Repair Tira i i im 533555 335an want miml 5mm immune TEEFEE f ii EI Ti 553i E Ei Ubiqu t in lPmt asnme System 1 Protein Degradalinn V 23 Know structures of collagen and its function a Part of connective tissue synethesized 24 Special protein to the bone and skin C Lots of Gly and Pro d Each collegen fiber for the part it is put in has a different structure and thus a different function e General Structure i You have three Left hand a helicies all wrapped together in a Right hand formation This is one segment You take other segments and wrap it the same way around the first This creates a rope like structure f General Function i This is specific for each connective tissue matrix but the general function is structure ii Long repeated structures of Gly X Y Describe the structural and functional similarities amp differences between hemoglobin and myoglobin including the mechanisms of action of any physiological factors that can affect O2 binding Hemoglobin Myoglobin Lets go of O2 quickly Stores 02 for muscles Found in blood Found in muscle tissue 2A and 2B 8A Less Affinity for 02 High Affinity for O2 Cooperative Oxygen binding the binding of 1 O2 molecule will enhance the binding of another Gives Muscle its Red color Has 4 heme groups which contain 4 iorn atoms which provide the positive charge to oxygen s negative charge Has only 1 Heme group Factors that effect 1 Amount of O2 2 Amount of CO 3 Binding of first 02 When oxygenated it is in its relaxed state When deoxygenated it is in its taut state Extra Notes a Enzymes biological catalyst that amplify the rate of the reaction with out being used in the reaction i It does not change free energy but rather activation energy ii A catalyst eliminates randomness b Vocab i Substrate reactant ii Active site where substrate binds iii Transtion state intermediate C Two types of models for enzyme activity Lock and Key Induced Fit Hand in glove Insert the substrate and the enzyme works The substrate causes the enzyme to change structure to accommodate the substrate anlt thus will create the product d Classification of Enzymes i 6 Classes 1 OXioreductase Addition and removal of Protons and electrons Transferase Transfer one group to another Hydrolase cleaves bond in the presence of water Lyase removal of a functional group without water Isomerase converts it into the isomer 6 Ligase Mends something together e Cofactors loosely bound non protein components that aid in catalytic reactions 25 Describe why it is important for enzymes to be regulated a Enzyme regulation is huge because if you under regulate it to where too much 9599 product is made then the reaction will over work and thus to much product could be detrimental b If over regulated where there was too little product made then that would be determential 26 List and describe the various ways that enzymes can be regulated 27 Distinguish between competitive non competitive and uncompetitive inhibition and mixed inhibition including their mechanisms effects on enzyme kinetics and changes in Michaeilis Menton amp and Lineweaver Burke plots Factors Competitive Non Competitive Uncompetitive Mixed Information Binds to active site Binds to both the Binds to the enzyme and prevents Enzyme and the substrate compleX substrate from Enzyme substrate only binding compleX Km Increases No Change Decreases Increase or Decrease Reversible Yes No No Vmax No Change Decrease Decrease 28 29 Describe the various ways that enzymes can be regulated by covalent modification a This is the addition of a functional group to a protein with a covalent bond b Examples i Phosphorylation Dephosphorylation ii OxidationReduction iii Proenzyme An inactive enzyme precursor polypeptide that must be proteolytically cleaved or hydrolyzed in order to become active 30 Describe and exemplify what is meant by rational drug design and its applications 31 Describe the general characteristics of allosteric enzymes a An allosteric enzyme is an enzyme that has two active binding sites one for the substrate and the other for the activator or the inhibitor The enzyme can either be activated or inhibited Enzyme activity gt s F E R E E 2 2 The Rate of an EnzymeCatalyzed Reaction as a Function of Substrate igpmsepzaaon 32 Explain how and why cofactors pH and temperature affect or regulate enzyme activities a Changes in pH can effect the enzyme by i Changing the ionization of the protein molecules ii Changing the ionization of the substrate iii Precipitate the protein b Changes in temperature i Can effect the rate of reaction ii Extreme heat or cold can cause the protein to denature c Availability of the cofactors will also regulate the rate of the reaction 33 How is sickle cell hemoglobin different from normal adult hemoglobin and how do these differences contribute to hemoglobin function 8 Normal hemoglobin is spherical b For the sickle cell it has a Valine in the position for glycine at position 6 for the subunits Since Valine is hydrophobic it pushes the cell in creating a pocket and thus forming the famous sickle cell Because of this the sickle cell bumps through the capillaries and cause clots and decrease the oxygen transport This gives them immunity from malaria however which is cause by infecting healthy red blood cells 34 Describe amp exemplify the various types of enzyme cofactors 35 Describe some of the main objectives of enzyme kinetics studies 36 What is the meaning and significance of Vmax amp Km 8 b c d Vmax max velocity of the reaction Km 12 the concentration of 12 the maximum velocity Both are constants that help us identify enzymes Km enzyme affinity for a substrate i Decrease the Km Increase Substrate affinity 37 Distinguish between the Michaelis Menton and Lineweaver Burke equations and plots and their applications 8 The equations to remember i The MampM equation Vmax S I t39 W 39t 1 mm eoczy KmS 1 Km 1 l x v Vmax S Vmax 38 Distinguish between the following terms enzyme unit specific activity turnover number catalytic efficiency 8 Enzyme Unit the amount of enzyme that produces l micromole of product per minute Katal Specific Activity micromoles of product formed per minute per mg protein ie micromol minmg Turnover number amount of molecules that can be produced per unit of time for l enzyme molecule i KcatVmaxEnz Catalytic efficiency how good the enzyme is in the presence of low s


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