QUIZ 2 mega quizlet

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A variety of factors influence enzyme activity. Substances that bind to the enzyme and interfere with substrate binding or catalysis are inhibitors. Identify the type of inhibition associated with each of the descriptions and examples by classifying each statement as irreversible, competitive, or mixed inhibition.
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A variety of factors influence enzyme activity. Substances that bind to the enzyme and interfere with substrate binding or catalysis are inhibitors. Identify the type of inhibition associated with each of the descriptions and examples by classifying each statement as irreversible, competitive, or mixed inhibition.
Irreversible inhibition
inhibitor may permanently modify an enzyme
DIPFDIPF permanently modifies the hydroxyl group of a Ser residue at the active site

Competitive inhibition
inhibitor binds reversibly to an enzyme's active site
malonate, which resembles succinate, binds to the succinate dehydrogenase active site

Mixed inhibition
inhibitor binds to an enzyme at a site other than the active site
the Al+3Al3+ ion binds to acetylcholinesterase or to the acetylcholinesterase‑substrate complex
Competitive inhibitor
structurally similar to substrate
when present, 𝐾mKm of enzyme will increase
prevents substrate from binding enzyme

Uncompetitive inhibitor
when present, 𝐾mKm of enzyme will decrease
binds enzyme-substrate complex only

Mixed inhibitor
binds either enzyme or enzyme-substrate complex
when present, 𝐾mKm of enzyme will either increase or decrease
Irreversible inhibition
inhibitor may permanently modify an enzyme
DIPF permanently modifies the hydroxyl group of a SerSer residue at the active site

Competitive inhibition
inhibitor binds reversibly to an enzyme's active site
a transition state analog binds reversibly to isomerase

Noncompetitive inhibition
inhibitor binds to an enzyme at a site other than the active site
the Al3+Al3+ ion binds to acetylcholinesterase or to the acetylcholinesterase‑substrate complex
The oxygen dissociation curve is sigmoidal in shape ("S"‑shaped).
As oxygen binds to this molecule the shape of the molecule changes, enhancing further oxygen binding.
The binding pattern for this molecule is considered cooperative.
This molecule delivers oxygen more efficiently to tissues.

The oxygen dissociation curve is hyperbolic in shape.
This molecule has a greater affinity for oxygen.

Oxygen binds irreversibly to this molecule.
Carbon monoxide binds at an allosteric site, lowering oxygen binding affinity.
The graph shows the oxygen‑binding curves for myoglobin and hemoglobin. Label the graph and answer the questions.
Use the curves to determine the partial pressure of oxygen at 50% saturation for hemoglobin and myoglobin.
myoglobin 𝑃50=P50=
hemoglobin 𝑃50=P50=
Which protein has a higher affinity for oxygen?
The prosthetic group of hemoglobin and myoglobin is heme.
The organic ring component of heme is porphyrin.
Under normal conditions, the central atom of heme is Fe2+.
In deoxyhemeoglobin the central iron atom is displaced 0.4 Å out of the plane of the porphyrin ring system.
The central atom has 6 bonds: to 4 nitrogen atoms in the porphyrin, one to a histadine residue, and one to oxygen.
Both hemoglobin and myoglobin contain a prosthetic group called heme, which contains a central iron (Fe)(Fe) atom.
By itself, heme is not a good oxygen carrier. It must be part of a larger protein to prevent oxidation of the iron..
Hemoglobin is a heterotetramer, whereas myoglobin is a monomer.Molecular oxygen binds reversibly to the Fe(II)Fe(II) atom in heme.
What is the best method to precisely determine the mass of a protein?mass spectrometryThe first step in protein purification is called _______________________.differential centrifigationWhat can SDS polyacrylamide electrophoresis be used to do?determine the molecular weights of subunits of an oligomeric proteinAntibodies used as regents to quantify proteins is the basis of a lab method called _____________.2-D gel electrophoresisWhat is the method for ensuring that proteins are separated by size and not be charge during gel electrophoresis?a negatively charged detergent binds to amino acids and denatures the proteinAn assay is useful for what during protein purification?detecting a protein in a sampleEnzyme‑linked immunosorbent assays (ELISA) are useful in biochemistry because:they can be used to detect very small amounts of a specific material.Which statement is true of the proteome?- It includes the interactions of peptides that yield a functional unit. - It varies with cell type. - It varies with environmental conditions ^^^^All three statements are true.To answer this question, please reference the Problem Solving Video: Oxygen Dissociation Curves. The change in hemoglobin's oxygen affinity due to a change in pH is known as the Bohr effect. Physiological pH levels in tissues typically vary little, but these changes have a noticeable effect on the oxygen affinity of hemoglobin. The graph contains three oxygen dissociation curves for hemoglobin, each curve representing a different pH level. Identify the pH for each oxygen dissociation curve in the graph.Curve 1: pH 7.6 Curve 2: pH 7.4 Curve 3: pH 7.2 What effect will hemoglobin have on the bicarbonate reaction in blood in the lungs? It will decrease HCO−3HCO3− productionTo answer this question, please reference the Problem Solving Video: Oxygen Dissociation Curves. Match the features to either the sequential or concerted model of allosteric regulation.Sequential model ligand affinity of protein varies with number of bound ligands hybrid conformations between subunits allowed ligand affinity of empty subunits affected by ligand‑bound subunits negative homotropic effects allowed Concerted model conformational states always reach equilibrium between proteins symmetry rule dictates subunits all in same conformation ligand affinity dependent only on quaternary structureA researcher has observed that the oxygen affinity of adult hemoglobin (HbA) is higher than expected when using a purified sample rather than red blood cell lysate. She plots the oxygen affinity curves for the purified HbA, cell lysate HbA, and purified HbA in the presence of 2.5 mM2.5 mM 2,3‑bisphosphoglycerate (2,3‑BPG). 2,3‑BPG is known to be an allosteric regulator of HbA in red blood cells.Given the oxygen dissociation curves, which of the following statements are correct? Purified HbA has a higher oxygen affinity than purified HbF. HbF loads oxygen at lower 𝑝O2pO2 than does HbA in the presence of 2,3‑BPG. 2,3‑BPG only weakly interacts with HbF. Which residue in HbA β do you think contributes the most to the increased interaction between HbA aand 2,3‑BPG? His‑143The graph represents the adult hemoglobin binding curve (in green) at pH 7.4 in the presence of 2,3‑bisphosphoglycerate. The hemoglobin binding curve has a sigmoidal shape, due to four interacting oxygen‑bound sites. For comparison, the myglobin binding curve has only one oxygen‑bound site and has a hyperbolic curve. For each of the six scenarios, determine whether the hemoglobin binding curve would shift left or shift right.Shifts left The adult hemoglobin (HbA) is replaced by an infant's fetal hemoglobin (HbF). Hemoglobin is isolated from red blood cells and stripped of 2,3‑bisphosphoglycerate. Tetrameric hemoglobin is dissociated into its subunits. Shifts right The blood pH drops from 7.4 to 7.2. The CO2CO2 concentration in the blood increases. The concentration of 2,3‑bisphosphoglycerate increases during acclimation to high altitude. Answer BankAll of the cells in the body need oxygen. Hemoglobin molecules in red blood cells transport oxygen through the bloodstream. Oxygen is loaded onto hemoglobin molecules in the lungs and unloaded from the hemoglobin molecules in the tissues. What drives the unloading of oxygen from hemoglobin molecules in the tissues?the low partial pressure of oxygen in the tissuesAggregate of HbS molecules The mutated form of hemoglobin (hemoglobin S, or HbS) in sickle‑cell anemia results from the replacement of a glutamate residue by a valine residue at position 66 in the β chain of the protein. Normal hemoglobin is designated HbA. Under conditions of low [O2][O2] , HbS aggregates and distorts the red blood cell into a sickle shape. See image of eight aggregated HbS molecules. Sickled red blood cells are relatively inflexible and may clog capillary beds, causing pain and tissue damage. The sickled red blood cells also have a shorter life span, leading to anemia. Sickling occurs in deoxyhemoglobin S but not in oxyhemoglobin S. Oxyhemoglobin has a small, hydrophobic pocket in a β chain region located in the interior of the protein. In deoxyhemoglobin, however, this pocket is located on the surface of the protein. In deoxyhemoglobin S, Val 66 interacts with this surface pocket, leading to aggregation of HbS. How does HbS aggregation occur in sickle‑cell anemia? Place the steps in the correct order. Note that deoxyhemoglobin is in the T state; oxyhemoglobin is in the R state.Which amino acids would be expected to produce a similar sickling effect if substituted for Val at position 6?6? leucine, alanine Choose two amino acids that would be reasonable candidates for the pocket-Val 66 interaction. phenylalanine, leucine No aggregation [O2][O2] decreases due to vigorous exercise or high altitude. R state Hb shifts to T state Hb. Val interacts with the pocket of a β chain on another HbS. Additional T state HbS interact with the growing aggregate to form an insoluble fiber. Sickled red blood cellWhich three statements accurately describe the blood buffering system in humans?The blood buffering system maintains the pH of blood near 7.4. The blood buffering system is facilitated by the enzyme carbonic anhydrase, which interconverts carbon dioxide and water to carbonic acid, ionizing into bicarbonate and H+H+ . The blood buffering system utilizes the H2CO3H2CO3/HCO-3HCO3- conjugate acid/base pair..How does hemoglobin function as a pH buffer?Hemoglobin binds hydrogen ions after carbon dioxide enters the red blood cell.Select the scenarios which, according to the Bohr effect, result in hemoglobin's release of bound oxygen.The concentration of hydrogen ions in the blood increases. Carbon dioxide levels in the blood increase.What must be true of the free‑energy change (Δ𝐺)(ΔG) for a reaction to be spontaneous?it must be negativeA reaction has an equilibrium constant (𝐾eq)(Keq) of 50.50. When performed in the presence of an appropriate enzyme, the forward rate constant is increased 20‑20‑fold. What will happen to the reverse rate constant?it will increase 20-foldWhat role does an enzyme play in catalysis? An enzyme increases the energy of the transition state so that it breaks down more rapidly. An enzyme increases the equilibrium constant. An enzyme increases the rate of the forward reaction. An enzyme decreases the activation energy.An enzyme decreases the activation energy.The average volume of a red blood cell is 87μm3.87μm3. The mean concentration of hemoglobin in red blood cells is 0.34 g⋅ml−1.0.34 g⋅ml−1. What is the weight of the hemoglobin contained in an average red blood cell? How many hemoglobin molecules are there in an average red blood cell? Assume that the molecular weight of the human hemoglobin tetramer is 65 kDa.65 kDa.hemoglobin weight: 2.96x10^-11 number of hemoglobin molecules: 2.17x10^8Classify the overall structure of hemoglobin in its two conformational states, based on images depicting the conformational changes in heme.T State stabilized by increased ion pairings at the α1β2α1β2 and α2β1α2β1 interfaces iron ion protrudes from heme towards His F88 tense state of hemoglobin R State narrowed pocket between β subunits iron ion assumes a planar conformation relaxed state of hemoglobinWhat is the function of the hormone erythropoietin (EPO)?stimulates red blood cell production in the bone marrowWhich of the statements about enzymes are true?Catalysis occurs at the active site, which usually consists of a crevice on the surface of the enzyme. Generally, an enzyme is specific for a particular substrate. For example, thrombin catalyzes the hydrolysis of the peptide bond between Arg and Gly. A substrate must bind to the active site before catalysis can occur. An enzyme yields a specific product, whereas a nonbiological catalyst may produce more than one product with the occurance of side reactions.Hydrogen bonds are noncovalent interactions between a hydrogen atom with a partial positive charge, called the hydrogen bond donor, and an atom with a partial negative charge, called the hydrogen bond acceptor. Progesterone is a hormone that contains two ketone groups. The oxygen in the ketone group can function as a hydrogen bond acceptor.Select the amino acids that have side chains that can form a hydrogen bond with progesterone at pH 7. histidine asparagine serineSuppose that a glutamine residue in the active site of an enzyme was mutated to alanine. As expected, the alanine mutant was inactive, suggesting that the glutamine residue was critical to the catalytic mechanism. Which mutation is most likely to restore wild-type level of activity to the alanine mutant?A to NMatch each enzyme class to the type of reactions catalyzed.transferase - reactions involving the transfer of a functional group from one molecule to another ligase - energetically unfavorable reactions that require ATP to form new bonds oxidoreductase - oxidation-reduction reactions lyase - reactions that eliminate or form a double bond hydrolase - hydrolysis reactions isomerase - isomerization reactionsThe maximum velocity (𝑉max)(Vmax) of an enzyme‑catalyzed reaction isthe rate observed when all enzyme active sites are saturated with substrate.Each of the reactions is catalyzed by an enzyme. For each reaction, classify the enzyme according to one of the six main classes suggested by the International Union of Biochemistry.pictures on Lecture 10 question 5 lyase ligase transferaseWhen the rate constant for dissociation of the enzyme-substrate complex (𝑘−1)(k−1) is greater than the rate constant for conversion to product (𝑘2),(k2), the 𝐾MKM is most analogous tothe 𝐾d.Match the terms with the correct definitions.cofactor - a nonprotein molecule or ion that is needed for an enzyme to carry out catalysis apoenzyme - an enzyme unable to act as a catalyst because its cofactor is removed coenzyme - an organic molecule, such as a vitamin, needed by an enzyme to carry out catalysisWhat does an apoenzyme require to become a holoenzyme?cofactorSelect the definition that best describes a coenzyme.an organic molecule that assists with a protein's catalytic functionComplete the diagram of the mechanism of a general enzyme-catalyzed reaction. The abbreviations E, P, and S indicate the enzyme, product, and substrate, respectively.E+S↽−−⇀ES↽−−⇀E+PEnzymes are important molecules in biochemistry that catalyze reactions. The energy diagram illustrates the difference between a catalyzed reaction and an uncatalyzed reaction. Label the energy diagram.Lecture 10 Question 11 for labeled diagarm What ways do enzymatic catalysts increase the rates of reactions? They promote the formation of a transition state. They lower the activation energy of the reaction.Given the reaction‑progress curves for two different reactions, A and B, complete the passage.In reaction A, the stability of the substrate is less than the stability of the product. The free energy change, ΔG, of the reaction is negative, so the reaction is considered exergonic and spontaneous. In reaction B, the stability of the substrate is greater than the stability of the product. The free energy change, ΔG, of the reaction is positive, so the reaction is considered endergonic and not spontaneous.Which of the reactions are spontaneous (favorable)?C6H13O9P+ATP⟶C6H14O12P2+ADPΔ𝐺=−14.2 kJ/mol C2H4+H2→Rh(I)C2H6ΔG=−150.97 kJ/mol both have Negative delta GFor the reaction A+B↽−−⇀C+DA+B↽−−⇀C+D, assume that the standard change in free energy has a positive value. Changing the conditions of the reaction can alter the value of the change in free energy (Δ𝐺)(ΔG). Classify the conditions as to whether each would decrease the value of Δ𝐺ΔG, increase the value of Δ𝐺ΔG, or not change the value of Δ𝐺ΔG for the reaction. For each change, assume that the other variables are kept constant.Decrease Δ𝐺 coupling with ATP hydrolysis increasing [A] and [B] Increase Δ𝐺 increasing [C] and [D] No effect on ΔG adding a catalystThe value of Δ𝐺°′ for the conversion of 3-phosphoglycerate to 2-phosphoglycerate (2PG) is +4.40 kJ/mol. If the concentration of 3-phosphoglycerate at equilibrium is 2.95 mM, what is the concentration of 2-phosphoglycerate? Assume a temperature of 25.0°C. [2PG]= mMmM.499The change in Gibbs free energy under biochemical standard-state conditions is represented by ΔG°′. The equilibrium constant under biochemical standard-state conditions is represented by K′eq. For a given reaction, K′eq is directly related to ΔG°′. Complete the ΔG°′ values for the K′eq values given.K′eq 1 10^-5 10^4 10^2 10^-2 ΔG°′ 0 28.55 -28.84 -11.42 5.71 (kJ∕mol)Consider the reaction. S⥫𝑘2𝑘1⥬P What effects are produced by an enzyme on the general reaction?The activation energy for the reaction is lowered. The rate constant for the reverse reaction (𝑘2)(k2) increases. The formation of the transition state is promoted.The diagram shows the catalytic hydrolysis of sucrose into the molecules glucose and fructose by the enzyme sucrase. Identify the role of each component in the diagram.Lecture 10 question 18Which interactions can contribute to the intrinsic binding energy during enzymatic catalysis?electrostatic interactions van der Waals interactions hydrogen bondingFerrochelatase catalyzes the insertion of Fe2+ into protoporphyrin IX. Why can the antibody generated against N‑methylmesoporphyrin catalyze the same reaction as ferrochelatase?The antibody makes more favorable interactions with the transition state than the substrate.On what basis are enzymes and proteins with allosteric properties different from those without allosteric properties? allosteric enzymes havedifferent dependence on the substrate concentration.The effects of molecules other than substrate on allosteric enzymes are calledheterotropic effects.What effect will a heterotropic activator have on a sigmoidal kinetic plot of 𝑉V versus [S]?[S]?The curve will shift to the left.Which statement is true of studying the rates of enzyme‑catalyzed reactions? It is called enzyme kinetics. It can involve determining how fast the substrate disappears as it is converted to product. It can involve following the appearance of product formed over time. All three statements are true.all three statements are trueHow do you find the 𝐾MKM from a double‑reciprocal, or Lineweaver-Burk, plot?𝑥-intercept=-1/𝐾Mx-intercept=-1/KMTo obtain the turnover number of an enzyme (𝑘2),(k2), one must divide 𝑉V by 𝑉max.Vmax. divide 𝑉maxVmax by 𝑘cat.kcat. divide 𝑉maxVmax by 2.2. divide 𝑉maxVmax by the total enzyme concentration.divide 𝑉maxVmax by the total enzyme concentration.Each of the given reactions is catalyzed by an enzyme. For each reaction, classify the enzyme according to one of the six categories suggested by the International Union of Biochemistry. Consider the reaction. Lecture 10 question 6Which category does the catalyzing enzyme belong to?hydrolase Which category does the catalyzing enzyme belong to?oxidoreductase Which category does the catalyzing enzyme belong to?isomeraseThe alteration of enzyme structure on binding of a substrate to an active site is referred to asinduced fit.Which statement is true of carbonic anhydrase? It catalyzes a reaction involving water. It catalyzes a reaction involving CO2.CO2. It is a very fast enzyme. All of these answer choices are correct.all of these answer choices are correctWhich statement is true of a holoenzyme, but not of an apoenzyme?A holoenzyme contains its necessary cofactor.TPCK is a molecule with large hydrophobic groups. Why does TPCK inactivate chymotrypsin but not trypsin? .TPCK looks like the substrate for chymotrypsin, but not trypsin, and thus can bind in its active site and modify His‑57.Which statement is false of a competitive inhibitor?It irreversibly inhibits the enzyme by chemically modifying a group at the active site.An enzyme inhibitor that decreases the apparent 𝐾MKM and reduces the 𝑉maxVmax can be classified asan uncompetitive inhibitorDiisopropylphosphofluoridate (DIPF) inactivates chymotrypsin by covalently modifying serine 195. Which statement is true of DIPF's inhibitory mechanism?Serine 195 is in an environment that gives it a higher than normal reactivity with respect to DIPF.The mechanism of chymotrypsin can be viewed as a two‑step process, acylation of the enzyme active site followed by a deacylation reaction. What accounts for the observed "burst" in rapid kinetic studies of the hydrolysis of N‑acetyl‑L‑phenylalanine p‑nitrophenyl ester by chymotrypsin?The rate of the acylation reaction is faster than the rate of the deacylation reaction.Which mechanism does NOT regulate the activity of an enzyme? a competitive inhibitor binding to the active site a competitive inhibitor binding to the ES complex an uncompetitive inhibitor binding to the ES complex All three mechanisms regulate enzyme activity.a competitive inhibitor binding to the ES complexMany drugs are competitive inhibitors of specific enzymes. For drugs that act as competitive inhibitors,the amount of drug needed to achieve the same effect is increased as the substrate concentration increases.You have isolated a new protease that cleaves peptide bonds on the carboxyl side of Asp and Glu. Based on the enzyme's inactivation by DIPF, you suspect that it may utilize a mechanism similar to chymotrypsin. the differencepresence of a positively charged residue in the S1 binding pocket.Which statement regarding a system at equilibrium is false?ΔG′ is zero.What does a proteolytic enzyme cleave?peptide bondsIdentify the factors that directly favor the unloading of oxygen from hemoglobin in the blood near metabolically active tissues.an increase in blood acidity near the tissues the presence of a pressure gradient for oxygen an increase in blood temperature near the tissuesRefer to the Biochemistry in Focus section of your text for this chapter to answer this question. Which of the following statements about Ngb-H64Q is true?Ngb-H64Q has a higher affinity than hemoglobin for both carbon monoxide and oxygen. However, Ngb-H64Q binds to carbon monoxide with a higher affinity than to oxygen.The illustration shows several oxygen‑dissociation curves. Assume that curve 3 corresponds to hemoglobin with physiological concentrations of CO2CO2 and 2,3-bisphosphoglycerate (2,3-BPG) at pH 7.7. Assign each perturbation or the lack of a perturbation to the curve it represents.Curve 1 a loss of quaternary structure Curve 2 a decrease in CO2CO2 an increase in pH Curve 3 no perturbation Curve 4 an increase in 2,3-BPGThe binding of oxygen to myoglobin and hemoglobin has what effect on the heme iron?It causes the iron to move closer to the plane of the porphyrin ring.Which statements correctly describe how 2,3‑bisphosphoglycerate (2,3‑BPG) reduces hemoglobin's affinity for oxygen? 2,3‑BPG binds to the amino termini, stabilizing the R (high‑affinity) state. 2,3‑BPG binds to the interface between α1β1 and α2β2, stabilizing the R (high‑affinity) state. 2,3‑BPG binds to the heme iron and prevents oxygen from binding. 2,3‑BPG binds to positively charged Lys and His residues in the center of the hemoglobin, stabilizing the T (low‑affinity) state.2,3‑BPG binds to positively charged Lys and His residues in the center of the hemoglobin, stabilizing the T (low‑affinity) state.My Attempt Rapidly metabolizing tissues generate large amounts of protons and carbon dioxide. The result of this increase in protons and carbon dioxide is that the oxygen‑binding curve of hemoglobin changes shape from sigmoidal to hyperbolic. remains the same. shifts to the left, meaning more binding of O2O2 at lower O2O2 levels. shifts to the right, meaning lower saturation at higher O2O2 levels.shifts to the right, meaning lower saturation at higher O2 levels.In addition to transporting oxygen from the lungs to the tissues, hemoglobin is also involved in transporting carbon dioxide from the tissues to the lungs. How does hemoglobin transport carbon dioxide from the tissues to the lungs? Carbon dioxide competes for the 2,3‑BPG‑binding site. Carbon dioxide binds to the carboxy terminus of hemoglobin. Carbon dioxide reacts with an amino group, stabilizing the R state. Carbon dioxide competes for the oxygen‑binding site on the heme.Carbon dioxide reacts with an amino group, stabilizing the R state.Each chain of hemoglobin can be viewed as existing in either the R state or the T state. What is the relationship between these two hemoglobin states and oxygen binding?Oxygen binds with greater affinity to the R state, and oxygen binding converts hemoglobin to the R state.In fetal hemoglobin (HbF), the two α subunits are replaced with two γ subunits. As a result, fetal hemoglobin has a higher affinity for oxygen than does the mother's adult hemoglobin. The higher oxygen affinity of HbF is due to a decreased amount of cooperativity between the α and β subunits. decreased binding of 2,3‑BPG. decreased binding of CO2CO2 to the γ subunits. a different mode of binding between the heme and the γ subunits.decreased binding of 2,3‑BPG.What is the molecular consequence of the hemoglobin S mutation? Hemoglobin S has a lower solubility and tends to precipitate in the lungs. Hemoglobin S has a lower affinity for oxygen. Hemoglobin S forms aggregates and fibrous precipitates when oxygen is released. Hemoglobin S forms aggregates and fibrous precipitates when oxygen is bound.Hemoglobin S forms aggregates and fibrous precipitates when oxygen is released.Hemoglobin S, the abnormal form of hemoglobin responsible for sickle‑cell anemia, is the result of a mutation in the gene for the β subunit. The hemoglobin S mutation changes a hydrophobic amino acid R group to a positively charged amino acid R group. positively charged amino acid R group to a negatively charged amino acid R group. negatively charged amino acid R group to a hydrophobic amino acid R group. negatively charged amino acid R group to a positively charged amino acid R group.negatively charged amino acid R group to a hydrophobic amino acid R group.Protein function is lost or reduced when a protein is denatured. Which of the environmental factors listed can cause protein denaturation? excessive heatprotein‑digesting enzymesextreme pHexposure to waterexcessive heat extreme pHMatch the descriptions and compounds with the terms competitive, uncompetitive, and noncompetitive inhibition.lecture 13 question 4Determine which of the graphs represents the relationship of reaction rate (velocity) and substrate concentration when the enzyme concentration of the non‑allosteric enzyme is constant. Answer Bank Determine how reaction rate (velocity) varies with substrate concentration.looks like square root x graph Rate increases Additional substrate is added when substrate concentration is low. Rate decreases - nothing Rate is unchanged Additional substrate is added when substrate concentration is very high. Substrate is added when enzyme is saturated with substrate.The table below lists experimental conditions that can be applied to a reaction catalyzed by a hypothetical Michaelis-Menten enzyme. For each experimental condition described, complete the table to indicate as precisely as possible the effect on the maximal velocity, 𝑉max,Vmax, and Michaelis constant, 𝐾M ,KM , of the hypothetical enzyme.question 7 lecture 13The Michaelis‑Menten equation models the hyperbolic relationship between [S] and the initial reaction rate 𝑉0V0 for an enzyme‑catalyzed, single‑substrate reaction E+S↽−−⇀ES⟶E+PE+S↽−−⇀ES⟶E+P. The model can be more readily understood when comparing three conditions: [S]<<𝐾m[S]<<Km, [S]=𝐾m[S]=Km, and [S]>>𝐾m[S]>>Km. Match each statement with the condition that it describes. Note that "rate" refers to initial velocity 𝑉0V0 where steady state conditions are assumed. [Etotal][Etotal] refers to the total enzyme concentration and [Efree][Efree] refers to the concentration of free enzyme.[S]<<𝐾m Almost all active sites are empty. The rate is directly proportional to [S]. [S]=𝐾m [Efree] is equal to [ES] [S]>>𝐾m [ES] is much higher than [Efree] Reaction rate is independent of [S] Not true for any of these conditions Increasing [Etotal]will lower 𝐾mThe amino acid asparagine can promote cancer cell proliferation. Treating patients with the enzyme asparaginase is sometimes used as a chemotherapy treatment. Asparaginase hydrolyzes asparagine to aspartate and ammonia. Considering the provided Michaelis-Menten curves for two different asparaginase enzymes, complete the passage. The arrow indicates the concentration of asparagine in the human body.The 𝑉maxVmax of asparaginase 1 is faster than the 𝑉maxVmax of asparaginase 2. At the substrate concentration indicated by the arrow, asparaginase 1 reaction velocity is slower than asparaginase 2 reaction velocity. The 𝐾MKM of asparaginase 1 is greater than the 𝐾MKM of asparaginase 2. Considering the performance of the enzymes, asparaginase 2 would make a more effective chemotherapeutic agent.The plots show the effect of pH, temperature, and substrate concentration on enzyme activity and reaction rate. Answer the questions based on the plots. Figure 1 shows the activity of two enzymes, enzyme A and enzyme B, at different pH values. At what pH does enzyme B have maximum activity?lecture 13 question 12An enzyme catalyzes a reaction with a 𝐾m of 8.50 mM and a 𝑉maxof 2.30 mM⋅s−1.. Calculate the reaction velocity, 𝑣0, for each substrate concentration. [S]=2.50 mM 𝑣0: [S]=8.50 mM 𝑣0: [S]=14.0 mM 𝑣0: eq: for first - 2.3*2.5/(8.5+2.3).523 1.15 1.43One way of expressing the rate at which an enzyme can catalyze a reaction is to state its turnover number. The turnover number is the maximum number of substrate molecules that can be acted on by one molecule of enzyme per unit of time. The table gives the turnover number of four representative enzymes. EnzymeSubstrateTurnover number (per second)RibonucleaseRNA100Fumarasefumarate800Lactate dehydrogenaselactate1000Ureaseurea10,000 How many molecules of RNA can one molecule of ribonuclease act on in 18.7 min? RNA molecules:18.7*60*100 112200Determine the value of the turnover number of the enzyme catalase, given that 𝑅max (Vmax) for catalase is 41 mmol⋅L−1⋅s−1 and [E]t equals 3.2 nmol⋅L−1 . Catalase has a single active site. turnover number:12812500What are characteristics of allosteric enzymes?.They undergo conformational changes as a result of modulator binding. They may have binding sites for regulatory molecules that are separate from active sites. They generally have more than one subunitThe Lineweaver-Burk plot, which illustrates the reciprocal of the reaction rate (1/𝑣)(1/v) versus the reciprocal of the substrate concentration (1/[S])(1/[S]) , is a graphical representation of enzyme kinetics. This plot is typically used to determine the maximum rate, 𝑉maxVmax , and the Michaelis constant, 𝐾mKm , which can be gleaned from the intercepts and slope. Identify each intercept and the slope in terms of the constants 𝑉maxVmax and 𝐾mKm .xintercept: −1/𝐾m yintercept: 1/𝑉max slope: 𝐾m/𝑉maxSelect the graph that correctly illustrates the effect of a positive modifier (effector) on the velocity curve of an allosteric enzyme. Place the correct graph in the set of axes. The solid blue curve represents the unmodified enzyme. The dashed green curve represents the enzyme in the presence of the effector. R is the highly active form of the enzyme and T is the less active form of the enzyme. Assume that this is a positively cooperative enzyme, meaning that the affinity for substrate increases with increasing substrate concentration.green line above follows relatively same shape but not exact What would happen to [T]/[R] if substrate concentration is decreased? [T]/[R] would increase.To answer this question, please reference the Problem Solving Video: Michaelis-Menten Equations and Lineweaver-Burk Plots. Double‑reciprocal, or Lineweaver-Burk, plots can reveal the type of enzyme inhibition exhibited by an inhibitor. Competitive inhibitors increase the 𝐾MKM without affecting the 𝑉max.Vmax. Noncompetitive inhibitors decrease the 𝑉maxVmax without affecting the 𝐾MKM. Uncompetitive inhibitors decrease both the 𝐾MKM and 𝑉max.Vmax. Based on the Lineweaver-Burk plot, what type of enzyme inhibition is exhibited by the inhibitor?competitiveTo answer this question, please reference the Problem Solving Video: Michaelis-Menten Equations and Lineweaver-Burk Plots. Penicillinase, also known as β‑lactamase, is a bacterial enzyme that hydrolyzes and inactivates the antibiotic penicillin. Penicillinase follows simple Michaelis-Menten kinetics and reaches half its maximal rate of 6.8×10−10 µmol⋅min−16.8×10−10 µmol⋅min−1 when the penicillin concentration is 5.2×10−6 M.5.2×10−6 M. What would happen to the initial reaction velocity, 𝑉0,V0, of penicillinase if the penicillin concentration were 10.4×10−6 M?10.4×10−6 M?𝑉0would approach 𝑉max..The results of kinetic experiments of an unidentified enzyme were used to create a plot of the initial reaction velocity (𝑉0)(V0) versus the substrate concentration ([S]).([S]). The enzyme kinetics data were also used to create a double‑reciprocal plot of 1𝑉01V0 versus 1[S].1[S]. Does it appear that the enzyme follows Michaelis-Menten kinetics? Now use the double‑reciprocal plot to determine the maximum velocity, 𝑉maxVmax, and the Michaelis constant, 𝐾MKM, of the enzyme. 𝑉max = 𝐾M=yes .5 200An enzyme that follows simple Michaelis-Menten kinetics has an initial reaction velocity of 10 µmol⋅min−110 µmol⋅min−1 when the substrate concentration is five times greater than the 𝐾M.KM. What is the 𝑉maxVmax of this enzyme? 𝑉max=12Enzymes are regulated in many ways, including allosterically. Allosteric regulation involves compounds binding to an enzyme at sites other than the active site. The terms heterotropic and homotropic describe the relationship between an allosteric regulator and an enzyme. The schematic of a general enzymatic pathway is shown, in which numbered enzymes catalyze the conversion of lettered compounds. Categorize regulatory interactions as heterotropic or homotropic based on the schematic.Heterotropic regulation inhibition of 1 by C activation of 3 by G Homotropic regulation inhibition of 1 by A inhibition of 5 by C activation of 2 by BIncreasing substrate concentration restores enzyme activity that has been blocked bya competitive inhibitor only.The graph shows three plots of velocity (𝑣0)(v0) versus substrate concentration ([S])([S]) . Determine which curve represents an enzyme's reaction velocity without any inhibitor present, which curve represents the velocity in the presence of a noncompetitive inhibitor, and which curve represents the velocity in the presence of a competitive inhibitor.no inhibitor- top black line competitive inhibitor - line goes above vmax noncompetitive inhibitor - line does not go above vmax Select all of the true statements. When substrate is present in excess, the maximum reaction rate is unchanged in the presence of a competitive inhibitor. A higher substrate concentration is associated with a faster reaction rate if the enzyme is not saturated.What kind of inhibitors are transition state analogs usually classified as?competitive inhibitorsMy Attempt What is the catalytic triad of chymotrypsin, a type of serine protease?the amino acids cysteine, histidine, and aspartatethe amino acids serine, histidine, and glutamatethe enzyme−cofactor−substrate complexthe enzyme−cofactor−intermediate complexthe amino acids serine, histidine, and aspartatethe amino acids serine, histidine, and aspartateChymotrypsin, trypsin, and elastase are digestive enzymes called serine proteases. The serine proteases differ in substrate specificity. Chymotrypsin cleaves peptide bonds after aromatic or bulky hydrophobic side chains, trypsin requires basic amino acid residues, and elastase cleaves bonds following small uncharged side chains. A chart of amino acids is available for your reference. Determine which specificity pocket is a part of each enzyme. ChymotrypsinTrypsinElastase Answer Bank Which amino acids have side chains that fit into the specificity pocket of trypsin?lecture 13 question 8 lysine and arginineThe Michaelis-Menten equation is an expression of the relationship between the initial velocity 𝑉0V0 of an enzymatic reaction and substrate concentration [S][S]. There are three conditions that are useful for simplifying the Michaelis-Menten equation to an expression from which the effect of [S][S] on the rate can be more readily determined. Match the condition (e.g., [S]=𝐾m[S]=Km) with the statement or statements that describe it.[S]=0.1 K𝑚 Doubling [S][S] will almost double the rate. [S]= K𝑚 Half of the active sites are occupied by substrate. [S]=10 K𝑚 About 90% of the active sites are occupied by substrate. [S]=10 K𝑚 Doubling [S][S] will have little effect on the rate. [S]=0.1 K𝑚 Less than 10% of the active sites are occupied by substrate. [S]=10 K𝑚 This condition will result in the highest rate.My Attempt What kind of inhibitors are transition state analogs usually classified as? uncompetitive inhibitorscompetitive inhibitorsnoncompetitive inhibitorscompetitive inhibitorsConsider the reaction. 2X⟶Y+Z2X⟶Y+Z When the concentration of X is doubled, the reaction rate increases by a factor of 4. What is the order of the reaction? What is the rate equation? Calculate 𝑘k when [X]=0.11 and the rate of the reaction is 0.0040 M/s. 𝑘=second-order rate=k[X]^2 0.33The lock and key model and the induced fit model are two models of enzyme action explaining both the specificity and the catalytic activity of enzymes. Indicate whether each statement is part of the lock and key model, the induced fit model, or is common to both models.Lock and key model The enzyme active site has a rigid structure complementary to that of the substrate. Induced fit model The enzyme conformation changes when it binds the substrate so that the active site fits the substrate. Common to both models The substrate binds to the enzyme at the active site, forming an enzyme‑substrate complex. The substrate binds to the enzyme through noncovalent interactions.Choose the correct answer from the list below. Not all of the answers will be used.A reaction that is directly proportional to the concentration of reactant is a - first order Enzymes that do not obey Michaelis-Menten kinetics. Incorrect match - allosteric ___ substrates do not bind in a random order, a tertiary complex forms between enzyme and substrates. - ordered sequential The value Vo is called the ___. - initial reaction velocity A reaction with two substrates is considered a ___ reaction. - :bimolecular ___ is the study of rates of chemical reactions. - kinetics At ___ there will be no net change in the concentration of substrate or product. - equilibriumWhich of the following describes positive heterotropic regulationBinding of an effector molecule shifts the T-R equilibrium in favor of the R stateonsider the reaction mechanism below: Which of the following is true about this reaction? (pick all that apply)During the course of the reaction, the enzyme is covalently modified This is an example of a 'ping pong' or double displacement reaction19. The formula V0 = Vmax x [S]/([S] + KM), indicates the relationship betweenenzyme activity as a function of substrate concentration.Allosteric proteins: A. contain distinct regulatory sites and have multiple functional sites. B. display cooperativity.Both A and BHow is the negative charge of the tetrahedral transition state stabilized in chymotrypsin?Interaction with backbone amide residues in the oxyanion holeWhen chymotrypsin activity is monitored with a chromogenic substrate N-Acetyl-l-phenylalanine p-nitrophenyl, the kinetics shows a burst phase and a steady-state phase. What does this tell us about chymotrypsin's mechanism of catalysis?The burst phase is due to release of the p-nitrophenolate and formation of an enzyme-acyl intermediate.Choose the correct answer from the list below. Not all of the answers will be used. - A type of catalysis where the proton donor is not water. - substrate analog which reacts with the enzyme - An enzyme that temporarily undergoes covalent catalysis as part of its mechanism. - A type of enzyme inhibitor where KM is unaltered. - The type of reaction catalyzed by proteases. - An enzyme that is part of a pigment formation pathway and has a low-temperature optimum. - A protease enzyme with a low pH optimum. - The inhibitor that binds only to the ES complex and lowers the Vmax and KM. - The enzyme inhibition that can be overcome by increasing concentration of substrate.acid-base affinity label chymotrypsin noncompetitive hydrolysis tyrosinase pepsin uncompetitive competitiveThe Km is:equal to the substrate concentration when the reaction rate is half its maximal value.Allosteric effectors alter the equilibria between:the R and T forms of a proteinA critical feature of the Michaelis-Menten model of enzyme catalysis isformation of an ES complex.Choose the correct answer from the list below. Not all answers will be used. Some may be used more than once - Sigmoidal-shaped binding curve indicates this type of binding - This is the chemical form in which most of the carbon dioxide is transported in the blood. - The shape of the myoglobin binding curve that shows that it is not regulated allosterically. - Amino acid side chain coordinating with heme iron. - This is the molecule whose function is to facilitate diffusion of oxygen in muscle cells. - This is the chemical produced by the action of carbonic anhydrase - This condition is a result of a single-point mutation in the β chain of hemoglobin. - This type of hemoglobin has a lower affinity for 2,3 bis-phosphoglycerate - Stabilizes the T-state of hemoglobincooperative carbonate ion hyperbolic histidine myoglobin carbonate ion Sickle-cell anemia fetal carbamateThe model describing allosteric regulation that requires all subunits to be in the same state is called theconcerted modelThe conversion of glucose-6-phosphate to fructose-6-phosphate is catalyzed by an isomerase enzyme. Glucose-6-phosphate was mixed with the enzyme under standard conditions and the reaction was allowed to come to equilibrium. If the Keq′ is 0.50, what is the ∆G°′ in kJ/mol?+1.71Carbon dioxide forms carbamate groups in proteins such as hemoglobin by reaction with:N-terminal amino groups.Which of the following is correct concerning the oxygenation plot of proteins X and Y shown in the graph below?Protein X corresponds to fetal hemoglobin, and protein Y corresponds to normal adult hemoglobin.Choose the correct answer from the list below. Not all of the answers will be used. - Enzymes will decrease the energy of activation but do not change the ____ of a chemical reaction. - The site on the enzyme where the reaction occurs. - A tightly bound cofactor might be called a ___. - Enzymes that cleave molecules by addition of water are called ___. - The substance that the enzyme binds and converts to product. - An endergonic reaction requires an input of ___ to proceed. - A reaction that is exergonic will be ___. - Enzymes that do not have the required cofactor bound are called - Which model is more appropriate to describe an enzyme binding to its substrate? - Enzymes that transfer electrons are called ___.equilibria active site prosthetic group hydrolases substrate energy spontaneous apoenzymes Induced Fit oxidoreductasesFor the two reactions a) A→B ∆Go′ = 2 kJmol-1 and b) X→Y ∆Go′ = -3.5 kJmol-1, which of the following statements is correct?Reaction a is not spontaneous.What is the Bohr effect?the regulation of hemoglobin binding by pH and carbon dioxide