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Elementary Biochem Exam 1 Study Guide

by: Juliana Bernard

Elementary Biochem Exam 1 Study Guide Biochemistry 420

Marketplace > University of Massachusetts > Biochemistry > Biochemistry 420 > Elementary Biochem Exam 1 Study Guide
Juliana Bernard
GPA 4.0
Elementary Biochemistry
Dr. Becky Miller

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Exam 1 Guide covering: molecular interactions and nucleic acids, protein structure and function, and energetics and enzymes. Study guide questions and sample exam questions answered.
Elementary Biochemistry
Dr. Becky Miller
Study Guide
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This 6 page Study Guide was uploaded by Juliana Bernard on Monday October 12, 2015. The Study Guide belongs to Biochemistry 420 at University of Massachusetts taught by Dr. Becky Miller in Fall 2015. Since its upload, it has received 396 views. For similar materials see Elementary Biochemistry in Biochemistry at University of Massachusetts.


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Date Created: 10/12/15
Exam One Study Guide Biochem 420 Lecture 3 1 Compare and contrast the types of noncovalent interactions we discussed Chargecharge interactions simplest of the noncovalent bonds interaction between 2 fully charged atomsgroups permanent charges on atomgroup examples are ionic bonds or salt bridges Chargedipole interactions interaction between 1 atomgroup with full charge and a polar atomgroup with a partial charge permanent charges on atomgroup Dipoledipole interactions interaction between 2 polar atomgroups with partial charges permanent charges on atomgroup Van der Waals dispersion interactions intermolecular forces between neighboring molecules that either attract or repulse each other temporary interaction between 2 atomsmolecules Hydrogen bond interaction between an Hatom covalently bonded to another atom NFO and a pair of nonbonded electrons on a separate atom permanent interaction between 2 atomsmolecules high energy compared to other noncovalent bonds highly Rf Oquot39quotHOgtR directional cliugig mm 0ng 2 What types of noncovalent interactions isare present in liquid water Solid water ice Sodium chloride dissolved in water Olive oil A mixture of olive oil and water Liquid water hydrogen bonds between water molecules Soid water same as above Sodium chloride in water chargedipole interactions between Na and Cl and water and hydrogen bonds between water molecules Olive oil and water hydrogen bonds form between water molecules and wan der waals interactions tails of oil triglycerides 3 Compare and contrast the structures of the five nucleotides What features do they all share How do they differ The 2 types of nucleotides are pyrimidines and purines Pyrimidine nucleotides are aromatic cyclic rings including uracil thymine and cytosine Purines are composed of two aromatic cyclic rings and include adenine and guanine Adenine forms 2 hydrogen bonds with either uracil in RNA or with thymine in DNA Guanine forms 3 hydrogen bonds to cytosine in both DNA and RNA 4 Is a phosphodiester bond between nucleotides with A and T different than one between C and G What about A and T versus A and U Explain your answers Phosphodiester bonds exist between the phosphate of one nucleotide and the sugar 339 carbon of the next nucleotide forming the backbone of DNA or RNA regardless of the nucleotide The differences between nucleotides are present in the specific base attached whether it be adenine guanine cytosine thymine or uracil however these are not involved directly in the phosphodiester bonds that make up the backbone 5 Interactions with ions play a role in stabilizing the structures of nucleic acids Based on what you know about nucleic acids do you think that M92 ions or C ions or both ions interact with and stabilize nucleic acids Why Due to the phosphates in nucleic acids they hold a negative charge If added the positively charged Mg2 atoms would be attracted to and ultimately interact and stabilize nucleic acids while Cl ions would be repulsed 6 Compare and contrast base stacking and base pairing interactions in DNA Both base pairing and base stacking are used to stabilize the structure of DNA Base pairing in DNA refers to the secondary structure of the nucleic acid and is due to the hydrogen bonding that takes place between two opposite nucleotides on complementary strands of DNA or RNA Base stacking is responsible for much of the stability of DNA and is due to van der Waals dispersion forces between the bases above and below each other Lecture 4 1 Each amino acid is unique however all peptide bonds are essentially identical Why Peptide bonds are formed between the carboxyl end of one amino acid and the amino end of the other The sole unique part of amino acids is the R chain Because the R chain is uninvolved in peptide bonding of amino acids all peptide bonds are essentially identical H O H O H 0 H O H3N CII Clt H3N Clli Clt fH3N 13 g ITI c l Cf H20 till 0 39 R2 0 R1 H R2 0 Peptide bond 1 2 Given the following pKa values rank the acids from strongest to weakest Fumaric acid 303 Carbonic acid 6 37 Formic acid 3 75 Sulfurous 1 85 Lower pKa stronger acid Sulfurous Fumaric acid Formic acid Carbonic acid strongest weakest 3 Is an acid with a pKa of 50 mosty protonated or deprotonated at pH 70 Sketch a titration curve to explain your answer An acid with a pKa of 50 is mostly deprotonated at pH 7 When the pKapH which is known as the inflection point we know that the ionizable group is half protonated and half deprotonated At pH 5 the acid would be 50 protonated and 50 deprotonated As the pH increases beyond the inflection point the acid will continue to deprotonate until the equivalence point is met acid is completely deprotonated now all conjugate base ti I39D 511 3390 4396 5399 39ul39nllLum of Nal JIH ml 4 A region of a protein has a signi cant number of Arg Cys Thr Ser and His Where in the protein structure would that region most likely be found Arg His basic 1 charge at neutral pH Ser Cys Thr polar no charge at neutral pH hydrophilic Due to the significant number of polar amino acids this region of protein would most likely be found on the outside of the protein so they can move fluids into the the body 5 What would the overall charge of the peptide ValTyrGnGlu be at pH 125 Use the following table of amino acid pKas to answer this question Hint Think carefully about which functional groups are in peptide bonds and which ones are not NH3 Val Tyr Gln Glu COOH At neutral pH 0 O O O 1 0 At pH 125 0 O 1 O 1 1 Overall net charge 3 Lecture 5 1 Describe the relationship between the terms polypeptide and protein A polypeptide is a linear polymer of amino acids linked together by peptide bonds Protein39s are functional structures composed primarily of one or more folded polypeptide 2 You are studying the function of a leucine in the insulin receptor Is it possible for a naturallyoccurring point mutation to result in replacement of leucine with glycine With valine What possible amino acids would you predict to have the least effect on protein structurefunction the greatest effect Explain each of your answers Leucine codons UUA UUG CUU CUC CUG Glycine codons GGU GGC GGA GGG No naturallyoccurring point mutation could result in the replacement of leucine with glycine Valine codons GUU GUC GUA GUG Yes a point mutation of either CUU gt GUU CUC gt GUC CUG gt GUG or UUA gt GUA could result in the replacement of leucine with valine Because leucine is nonpolar a mutation resulting in another nonpolar amino acid would likely have the least effect on protein structurefunction Replacing leucine with a polar or fully charged amino acid would likely have a greater effect as it might disrupt noncovalent interactions that are important for protein structurefunction The greatest effect would be a mutation that causes a STOP codon which is possile for some codons that specify leucine 3 Look at the image of an a heix What kinds of noncovalent bonds provide the structure of an a heix Are the R groups involved in these bonds Only hydrogen bonds provide the structure of the alpha helix R groups are not involved with 56 these bonds 4 E curve for asparagine Label all inflection points equivalence points and pKas Indicate where the ioni 100 protonated and 100 deprotonated Am t39n pill 39i ll krjulm furnmgml Hitter Higgins gt If OHl added a In the image above asparagine is the curve in yellow The inflection points are indicated by the red arrows and the equivalence points are indicated by blue arrows The orangeyellow arrow represents where the carboxyl and amino groups are 100 protonated The first blue arrow indicates the place where the carboxyl group is 100 deprotonated but the amino group is 100 protonated The second blue arrow indicates where the carboxyl and amino groups are both 100 deprotonated 5 Tertiary structure of the insulin receptor is dependent on many intramoecuar interactions One of these is an interaction between glutamic acid and arginine What kind of interaction is this In what pH range would you expect this interaction to occur Would a decrease in pH to 3 effect the interaction An increase in pH to 9 Explain your answers An interaction between the acidic amino acid glutamic acid and the basic amino acid arginine would be a charge charge or ionic interaction likely to occur between pH 55 to 115 Arg R group is 1 and Glu R group is 1 At pH 3 the interaction would disappear because the glutamic acid R group is now protonated At pH 9 the interaction would remain because you have not yet reached the pKa of the arginine R group Lecture 7 1 Draw a concept map using the terms free energy enthalpy entropy standard free energy exergonic endergonic K and Q 2 What is the difference between AG and the standard free energy change A6 Which one is always the same for a speci c reaction Why The standard free energy change is always the same for a specific reaction measured at standard conditions Free energy is the distance in free energy from the equilibrium state of a given reaction and depends on the concentrations of the reactants and products in a cell 3 If the activation energy of Reaction A is smaller than the activation energy of Reaction B what does that tell you about the reactions If the activation energy if reaction A is smaller than that of reaction B then we can assume that reaction A will occur faster than reaction B 4 Divide the phrases below into two groups those that describe exergonic reactions and those that describe endergonic reactions NO TE You can put phrases in BOTH groups if they apply to both exergonic and endergonic reactions positive AG free energy of reactants is greater than free energy of products releases energy requires an enzyme requires activation energy to occur spontaneous Exergonic reactions Endergonic Reactions Releases energy Positive AG Requires activation energy to occur Requires activation energy to occur Spontaneous Requires an enzyme AG of reactants gt AG of products Requires and enzyme 5 Why DON T enzymes change the AG or Keq of a reaction Can an enzyme make a nonspontaneous reaction favorable Why or why not Enzymes make reactions faster by lowering the activation energy of the reaction they don39t change the AG or the Keq of a reaction because they don39t change the structure of the reactants or products or change the equilibrium state Thus an enzyme cannot make a nonspontaneous reaction favorable because the AG doesn39t change Lecture 8 1 Under certain conditions the AG of a reaction is 50 kJmol Under these conditions which statements isare TRUE The reaction will proceed more rapidly than a reaction with a AG of 10 kJmol CANNOT DETERMINE The AG for the reverse reaction is 50 kJmol TRUE The AG will decrease become more negative if an enzyme is added FALSE 2 If a potential enzyme is discovered that binds the substrates as tighty as it binds the transition state would it show an increase enhancement in the rate of the reaction Should it be classi ed as an enzyme Why or why not This would not be classified as an enzyme because it is unlikely to speed up the rate of reaction Binding the substrate as tightly as the transition state makes it less likely that the reaction will proceed since there is no change in the strength of interaction when forming the transition state There is no motivation for the reaction to proceed 3 Sketch a graph of enzyme activity versus amount of substrate keep the amount of enzyme constant Draw another line to represent what would happen if the amount of enzyme used at all the substrate concentrations was higher I Lli ileiiil e ll g While this graph shows three different enzyme concentrations the WW key point is that the Vmax increases while the Km value stays constant and does not change when enzyme concentration is lfl39ilno x39 doubled for each enzyme concentration Constant Km value 4 Explain why enzyme activity anayses most commonly measure initial velocities of reactions Early in the reaction the reaction rate velocity is fastest and reflects the actual speed of the enzyme As the reaction proceeds the rate slows down and then stops therefore scientists measure the initial rate to show know how quickly the enzyme is working at a given substrate concentration Lecture 9 1 Glucokinase and hexokinase are two enzymes involved in carbohydrate metabolism that catalyze the same reaction they add a phosphate group to glucose forming glucose6phosphate The Km of glucokinase for glucose is 20 mM and the Km of hexokinase is 015 mM Which enzyme has the higher af nity for glucose Which one would most likely be saturated at normal blood glucose levels around 50 mM If the turnover numbers for the two enzymes are the same which one is the most ef cient Note Both enzymes are categorized as MichaeisMenten enzymes Hexokinase has the higher affinity for glucose smaller Km higher affinity At normal blood glucose levels glucokinase has a larger Km value than 50mM which means the enzyme will not be saturated If the turnover number Kcat is the same for both enzymes the Km will be used to determine the efficiency Because the Km for hexokinase is smaller it will make the specificity constant higher kcatKm making hexokinase the more efficient enzyme higher kcatKm more efficient 2 Create a table to compare and contrast MichaelisMenten and allosteric enzymes or use the one provided in your post lecture slides Fill in as much of the table as possible without referring to your notes Then go back and add in more details MichaelisMenten Allosteric ncrease rate of a reaction stabilize transition state lnteract more favorably with transition state than substrates or products Can catalyze forwards and reverse reactions Specific for specific substrates Substrates interact in active site Do NOT alter AG or Keq Maximum activity at infinite substrate concentration enzyme saturated Graph of V0 vs S is sigmoidal Must have multiple active sites Show positive cooperativity of substrate binding Exist in two forms shapes with different substrate affinities Graph of V0 vs S is hyperbolic One active site or multiple 39independent39 active sites 3 Feedback control is a common strategy for turning off or on a biochemical pathway Why is this such a useful strategy Feedback control is a useful strategy because it helps the cell to determine when to stop or start a biochemical pathway and conserve energy by turning on a pathway when it is needed and turning of a pathway when the product is not needed A product of the pathway can go back and turn off an enzyme from earlier in the pathway This is called negative feedback mechanism A product can also activate an enzyme further in the biochemical pathway by turning it on This is a positive feedback mechanism 4 What makes irreversible inhibitors much more potent inhibitors than reversible inhibitors Irreversible inhibitors as their name implies make irreversible changes to enzymes They generally do this through covalent modifications which can result in complete loss of enzymatic activity With reversible inhibitors their modifications to the enzyme are only temporary When the reversible inhibitor is no longer present the enzyme can resume its function 5 Addition of molecule X decreases the amount of the T form of an allosteric enzyme Can you conclusivey determine whether molecule X is a substrate activator or inhibitor If so which one If not what could it be Explain your answer Decreasing the T form of an allosteric enzyme means that it is going to its more active R form Molecule X is therefore causing the allosteric enzyme to become active making it either a substrate or an activator The only way to tell a difference is if we knew whether molecule X bound to the active site or not A substrate would bind the active site and an activator would bind somewhere else other than the active site An inhibitor would cause an allosteric enzyme to be in its less active T form Practice Exam Questions 1 When an allosteric enzyme interacts with a molecule called MOLY the relative amount of Rform increases Based on that information MOLY must be an inhibitor must be an activator could be an inhibitor or a substrate could be an activator or a substrate could be an inhibitor or an activator WPQWFD 2 At low substrate concentrations addition of more substrate will the rate of the reaction At saturating substrate concentrations more substrate will the rate of the reaction A increase increase B increase not affect C not affect increase D not affect not affect 3 Erythromycin and penicillin are drugs used to treat bacterial infections Both drugs are weak acids with one ionizable hydrogen Penicillin is a stronger acid compared to erythromycin therefore penicillin has a Ka and a pKa compared to erythromycin A bigger bigger B bigger smaller C smaller bigger D smaller smaller 4 You have a solution containing one of the 20 standard amino acids and are trying to identify which amino acid is in the solution At a pH of 30 its overall charge is 1 At a pH of 120 its overall charge is 1 Which of the following amino acids could it be A Asp B Gln C Phe D Lys 5 What piece of information would allow you to conclusively determine a bond is nonpolar covalent A the atoms are similar in size B the atoms are located in different molecules C one atom holds onto the electrons more strongly than the other D the atoms share electrons equally 6 In your cells citrate is converted to isocitrate and the standard free energy change for the reaction is 670 kJmol Use this information to answer the questions below a Under standard conditions the concentration of citrate is the concentration of isocitrate at equilibrium Circle your answer less than greater than 7 In a cell the concentration of citrate is 150 mM the concentration of isocitrate is 005 mM and the temperature is 298 K What is the AG for the conversion of citrate to isocitrate under these conditions Please write the equation CIRCLE your answer and include units AG AG RT lnQ AG 670 kJmol 8315x10393 kJmolK298 Kln005mM15mM AG 670 kJmol 284 kJmol340 AG 670 kJmol 843 kJmol AG 173 kJmol


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