BIOCHEM 420
Be able to calculate the net charge of a polypeptide.
-determine whether the functional group of the N-terminal α-amino, C-terminal α-carboxyl, and each R group is protonated / deprotonated / unaffected by comparing pH to pKa -determine the charge of each functional group -add all the charges together for the net charge
Why is it advantageous for biochemical reactions to be organized into catabolism and anabolic pathways? • They increase the efficiency of energy transfer because you can capture more energy when it's broken down into several reactions More points of regulation allows you to conserve energy
1. They increase the efficiency of energy transfer because you can capture more energy when it's broken down into several reactions 2. More points of regulation allows you to conserve energy
For an unfavorable reaction, the free energy change is favored. A. positive; reactants B. positive; products C. negative; products D. negative; reactants
A
Which of the following characteristics, if any, is a unique property of a Michaelis-Menten enzyme as compared to an allosteric enzyme? A. The graph of Vο vs. [S] for the enzyme is hyperbolic. B. The enzyme can catalyze both the forward and reverse reaction. C. There are two forms of the enzyme each with a different activity and a different substrate affinity. D. None of the above is a unique property of a Michaelis-Menten enzyme.
A
9. In what way is phosphorylation similar to feedback control? A. Both regulatory mechanisms lead to changes in enzyme conformation that alter activity. B. Both regulatory mechanisms use covalent modification to alter enzyme activity C. Both regulatory mechanisms always decrease the Vmax of the target enzyme. D. Both regulatory mechanisms are irreversible.
A B and D are saying the same thing can't guarantee that C is correct
Which of the following about chymotrypsin's active site are true? I. Histidine participates in general acid/base catalysis II. Aspartic acid participates in electrostatic interaction with histidine III. Serine engages in covalent catalysis with histidine A. IandII B. II and III C. I and III D. I, II, and III
A aspartic acid stabilizing histidine's +1 charge is electrostatic catalysis histidine accepting a proton from serine is base catalysis serine interacting with the substrate is covalent catalysis
iClicker: Why does this graph level off? [S] versus initial velocity A. The substrate is used up B. The enzyme active sites are all filled C. The reaction is reaching equilibrium D. The rate of the reverse reaction becomes significant
A-->--> no because we're at the beginning of the reaction. No substrate is used up yet B. yes. Levels off because the enzyme is saturated. Therefore, the initial speed has reached its fastest speed c.--> no because you're measuring the a. Allow reaction to proceed long enough, then you'll see the equilibrium d. --> haven't reached equilibrium, so very little of the reverse reaction is happening at this moment
1. At low substrate concentrations, addition of more substrate to a Michaelis-Menten enzyme 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
B
5. Which of the following factors is/are altered when an enzyme is added to a chemical reaction? A. Km B. rate C. change in Gibbs free energy D. All of the above are altered.
B
If K < Q, what direction will a chemical reaction proceed in and why? A. Forward, there are more products than reactants B. Reverse, there are more products than reactants C. Forward, there are more reactants than products D. Reverse, there are more reactants than products
B
Select the TRUE statement(s). I. In a cell, a reaction is always spontaneous if the ΔG° is negative. II. In a cell, a reaction is always spontaneous if the ΔG is negative. III. In a cell, a reaction is always spontaneous if an enzyme is present that can catalyze the reaction. A. I only B. II only C. III only D. I and II only E. I, II and III are TRUE
B
Similarities and Differences Between MM and Allosteric Enzymes
BOTH: - increase the rate of a reaction by stabilizing the transition state - interact more favorably with the transition state than the substrates or products -can catalyze forwards and reverse reaction -specific for specific substrates -do NOT change ΔG of Keq - maximum activity at infinite substrate concentration - enzyme saturated MM: - graph of V0 versus [substrate] is hyperbolic - have one active site or many "independent" active sites for different substrates - Km is relevant Allosteric: - graph of V0 versus [substrate] is sigmoidal -multiple "dependent" active sites for substrates - binds of one substrate alters the shape of the enzyme and influences binding of additional substrate modules (POSITIVE COOPERATIVITY) - exist in two forms (shapes) with different substrate affinities
What are the most important factors for determining protein charge? A. pH of environment and α-carboxyl pKa B. hydrophobicity and R group pKa C. pH of environment and R group pKa D. hydrophobicity and α- carboxyl pKa
C
Addition of molecule X decreases the T/R ratio of an allosteric enzyme. Can you conclusively determine whether molecule X is a substrate, activator, or inhibitor? If so, which one? If not, what could it be? Explain your answer.
COULD BE AN ACTIVATOR OR SUBSTRATE. Decreasing the T form of an allosteric enzyme means that more enzyme subunits are transitioning to the 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 (not the active site). An inhibitor would cause an allosteric enzyme to be in its less active T form.
Why are changes in pH or temperature NOT effective regulatory mechanisms in a cell?
Changing the pH or temperature in a cell is not selective for a specific reaction and will affect all reactions. Regulatory mechanisms need to be specific in order to allow chemical reactions to respond differently to the environmental conditions.
For each of the four inhibitors list: enzyme interaction effects on substrate interaction effects on Vmax Effects on Km
Competitive: E only - active site blocks substrate binding decreases Vmax no change in Km uncompetitive: ES only decreases substrate release decreases Vmax decreases Km noncompetitive: ES or E - no preference doesn't change substrate binding because it doesn't bind to the active site no change in Km decreases VMax mixed: ES or E - usually prefers one over the other alters substrate binding decreases Km increases or decreases VMax
What piece(s) of information would allow you to conclusively determine that an inhibitor was a competitive inhibitor? Uncompetitive inhibitor? Noncompetitive inhibitor? Mixed inhibitor?
Competitive: inhibitor binds to the active site, prevents binding of the substrate, and Vmax is unchanged (ANY OF THESE ALONE IS ENOUGH TO SUPPORT THAT YOU HAVE A COMPETITIVE INHIBITOR BECAUSE IT'S ORIGINAL) - Uncompetitive: inhibitor binds ONLY to the ES complex and decreases substrate release AND both Vmax AND Km decrease (must know the effect on both Vmax AND Km or that the inhibitor binds to ES only) - Noncompetitive: inhibitor does not affect substrate interaction, Vmax decreases but Km is unchanged (change in Vmax is not enough to conclusively determine) - Mixed: each inhibitor binds to E and ES AND decreases Vmax while Km increases or decreases (must have all of these together)
When an allosteric enzyme interacts with a molecule called MOLEY the T/R ratio decreases. Based on this information, MOLEY A. must be an inhibitor. B. must be an activator. C. must be a substrate. D. could be an inhibitor or a substrate. E. could be an activator or a substrate. F. could be an inhibitor or an activator.
E
Divide the phrases below into two groups - those that describe exergonic reactions and those that describe endergonic reactions. NOTE - You can put a phrase in BOTH groups if it applies to both exergonic and endergonic reactions. - positive ΔG - free energy of reactants is greater than free energy of products - releases energy - requires an enzyme to occur in a cell - requires activation energy to occur - favorable (spontaneous)
Endergonic: - positive ΔG - requires energy -free energy of products > free energy of reactants - not favorable (not spontaneous) BOTH: - requires an enzyme to occur in a cell - requires activation energy to occur Exergonic: - Negative ΔG - releases energy - free energy of reactants is greater than free energy of products - favorable (spontaneous)
Why DON'T enzymes change the ΔG or Keq of a reaction? Can an enzyme make a nonspontaneous reaction favorable? Why or why not?
Enzymes don't change the ΔG or the Keq because they only affect the rate of a reaction. This means that they allow us to get to ΔG and Keq faster, but they do not directly affect each of these things. They do not affect products and reactant concentrations ΔG Keq They do affect Ea Since ΔG affects whether a reaction is spontaneous (favorable) or not spontaneous (unfavorable), and enzymes have no affect on ΔG, they cannot affect whether a reaction is spontaneous or not spontaneous.
Define free energy and describe the two components.
Free energy is a quantitative measure of useful work and is composed of two parts: enthalpy (measure of the energy in bonds and types of bonds) and entropy (measure of disorder in the environment/system containing the molecules of the chemical reaction).
Draw a concept map that relates the terms: free energy, enthalpy, entropy, standard free energy, exergonic, endergonic, K, and Q. A concept map is a diagram that helps describe relationships between terms.
Free energy is the difference between enthalpy (the heat of breaking bonds and forming new bonds) and the entropy (the amount of disorder/chaos in a reaction). The standard free energy change is constant for a specific reaction under standard conditions • Standard Free Energy - "Artificial free energy" - not what would happen in a cell Includes: - temperature = 298 K - pressure = 1 atm - [reactants] and [products] = 1M - pH = 7.0 In a cell NOT at equilibrium, use ΔG° and Q Free Energy - what would happen in a cell exergonic reactions have a -ΔG and endergonic reactions have a +ΔG K is the ratio of products to reactants when ΔG = 0, or it is at its equilibrium. This can be determined experimentally from doing the equation of the standard free energy. and G is the ratio of products to reactants when ΔG is anything except 0.
Sketch a Lineweaver-Burk plot for an enzyme with a Km of 0.25 mM and a Vmax of 20 mM/sec. Don't worry about exact scales, but be sure to indicate the values at which the line hits both the x-axis and y-axis and label your axes.
Km - X intercept is 5/2 Max - y intercept is .05
You are investigating the activity of a recently discovered enzyme called Loonyase. Loonyase has a single active site and a Vmax of 0.10 mM/sec. From this information, can you determine the Km of Loonyase's substrate? If so, what is the Km? If not, explain why not.
Km= substrate concentration at 1⁄2 Vmax. However, the information provided DOES NOT GIVE THE RELATIONSHIP BETWEEN THE SUBSTRATE CONCENTRATION and Vmax so there is not enough information to know the value of Km. INCREASE IN [SUBSTRATE] INCREASES VMAX BUT KM DOESN'T GIVE US THE EXACT RELATIONSHIP BETWEEN SUBSTRATE CONCENTRATION INCREASE WITH VMAX INCREASE.
How could you conclusively determine (know for certain) whether a newly discovered enzyme should be classified as a Michaelis-Menten or allosteric enzyme?
MM - single unit allosteric - multiple units Performing the following experiments would help distinguish between MM or allosteric enzymes: (1.) if the substrate(s) is/are known, an enzyme KINETIC ASSAY would reveal whether Vo as a function of increasing [S] is hyperbolic vs. sigmoidal Sigmoidal shows positive cooperativity of substrate binding (2.) solving its structure to determine whether it has multiple subunits/active sites or not. Most multi-subunit enzymes are allosteric enzymes.
Describe how phosphorylation could increase or decrease the activity of an enzyme.
Phosphorylation is a covalent modification that changes the local microenvironment of the phosphorylated residue, which often affects some level of conformational change on the protein, either activating or inactivating it.
• Explain how positive cooperativity leads to the sigmoidal shape of an allosteric enzyme activity graph. Be thorough and discuss what happens at the beginning, middle, and end of the graph. Also, include 'T form' and 'R form' in your answer.
Positive cooperativity occurs when the binding of a substrate to one enzyme subunit induces a conformational change in the neighboring subunits that makes them more likely to bind substrate. Sigmoidal shape because low substrate concentrations means that most of the enzyme subunits are in the T form and will be not be bound to substrate so the activity of those enzymes is very low. With increasing [substrate] the T/R ratio decreases and more and more enzymes have at least one subunit in the R form and bound to substrate, this results in a rapid increase in activity because the subunits all cooperate together. These R form subunits cause the neighboring subunits to shift into the R shape even if that are not binding substrate. One all of the subunits are in the R form and repeatedly binding substrate and creating product at a maximal rate the activity curve levels off.
• describe the general mechanisms used by enzymes to stabilize the transition state. • describe one common mechanism used by some enzymes to enhance the rate of a reaction.
Rate enhancing mechanisms depend on the induced fit model, which involves a conformation change of the substrate. Enzymes change reaction kinetics because they optimize the number of interactions made between the substrate and the transition state. Every covalent bond they make releases energy, so the more interactons they make with the enzyme, the more binding energy that is released. This ultimately lowers the free energy of the transition state. Enzymes change reaction kinetics (rates).
Review the calculations from class and the homework questions making sure you know when to use each free energy equation.
Remember that you should be using the ΔG°' (biochemical standard conditions) ΔG° (standard conditions) if you are calculating standard free energy ΔG equation for cellular conditions.
Explain what is meant by the terms 'reversible' and 'irreversible'. Does reversible mean the same thing as 'favorable'? Explain your answers.
Reversible means that a reaction is BIDIRECTIONAL AND AT EQUILIBRIUM OR CLOSE TO EQUILIBRIUM (ΔG = 0 or +/_ 3) can be exergonic or endergonic Irreversible reaction is one that proceeds in ONE DIRECTION because the reaction is so far from equilibrium. This means that the conditions cannot be altered to make it go in the opposite direction Favorability - the likelihood of SPONTANEOUSLY TURNING REACTANTS INTO PRODUCTS. -ΔG = favorable +ΔG = unfavorable
The Km of glucokinase for glucose is 20 mM, and the Km of hexokinase for glucose is 0.15 mM. However, hexokinase and glucokinase can also interact with fructose and catalyze the addition of a phosphate group. Based on the provided information, what do you know, if anything, about the affinity of each enzyme for fructose? Explain your answer.
The Km is substrate specific so the exact value cannot be known without measuring it experimentally. However, a prediction can be made that hexokinase will have a greater affinity for fructose than glucokinase because fructose is similar in structure to glucose and hexokinase has a greater affinity (lower Km) for glucose. (NOTE: Fructose Km is 12 mM for glucokinase in liver and 1 mM for hexokinase in brain.)
Summarize the assumptions that Michaelis and Menten made when modeling the hyperbolic curve and generating a mathematical equation to fit the curve. Explain why they made each of these assumptions.
The amount of EP (product) is negligible so this term does not occur in the reaction. Also, the following reverse reaction is not occurring: E+P --> ES and this means that you are not going backwards from the enzyme and product to form ES. Therefore, only k2 matters (you can ignore k-2) 2. [ES] is constant when measuring initial velocity 3. Vmax can be measured. Aka, you can always reach a maximal velocity. altogether, these assumptions theoretically simplify the real situation such that: reaction velocity can be related to the [substrate] for a system when S reversibly binds to E to form ES, and then ES reacts irreversibly to generate product P and regenerate free enzyme E
The reason why we can't generalize that the whole outside of an enzyme is hydrophilic or hydrophobic can be credited to enzyme rate enhancement mechanisms and the induced fit model. Describe the rate enhancing mechanisms / strategies that enzymes use to catalyze reactions. Then, describe which mechanisms chymotrypsin uses. Be sure to include the specific active site amino acids that are involved.
The induced fit model describes the formation of the E-S complex as a result of the interaction between the substrate and a flexible active site. The substrate produces changes in the conformation on the enzyme aligning properly the groups in the enzyme. It allows better binding and catalytic effects. microenvironment (electrostatic catalysis and acid/base catalysis)- what is happening to the active site of an enzyme to speed up the rate of a reaction. R groups of active site are polar and form electrostatic (salt bridges and noncovalent) interactions with the substrate to catalyze reactions faster. distortion - change in the shape of the enzyme which alters the bond angles and facilitates ES binding and release direct participation (covalent catalysis) - covalent bonds formed with substrate to facilitate binding to active site orientation/proximity- enzyme will help the substrate get into the right position (which takes less time than substrate on its own) Chymotrypsin: uses electrostatic catalysis, acid/base catalysis, and covalent catalysis to catalyze the cleavage of a peptide bond. The stabilization of histidine's + charge by aspartate is an example of electrostatic catalysis, the acceptance of a proton by histidine from serine is general base catalysis, and the formation of a covalent bond between serine and the substrate is an example of covalent catalysis.
Based on the K equilibrium at 5.4 x 10^-2, which of the following is true? DHAP ---> G3P <----- A. The concentration of DHAP is greater than the concentration of G3P at equilibrium B. The concentration of G3P is greater than the concentration of DHAP at equilibrium C. The concentrations of DHAP and G3P are roughly equivalent at equilibrium
The rates of forward and reverse reactions are zero at equilibrium Since ka= [products]/[reactants] this means that given .054 or a fraction, more products than reactants If Ka = 1, equal ratio If Ka = (integer) x 10 ^ (integer) If Ka = >1 then more products than reactants Equilibrium is about the RATE of forward and reverse reactions, not about the CONCENTRATIONS of the products and reactants
What is the difference between ΔG and the standard free energy change (ΔG°' or ΔG°)? Which one is always the same for a specific reaction? Why is one always the same and why does the other one change?
The standard free energy change is always the same for a specific reaction when measured at standard conditions. This is because it represents the free energy change at equilibrium. ΔG is the distance from the equilibrium state of a given reaction and VARIES DEPENDING ON the concentrations of the reactants and products in a cell.
Explain why an enzyme interacts most strongly with the transition state and not the substrate (reactant).
The transition state, although it lasts for a short period of time, is necessary for substrates to pass through in order to be converted to products. Enzymes lower the activation energy of the transition state, allowing more substrates to be converted to products in a shorter period of time. The transition state is also higher in energy and more unstable than the reactants. By BINDING MOST STRONGLY TO THE TRANSITION STATE, this delayed period of optimization allows the substrate to maximize the amount of interactions WITH HIGH ENERGY CONFORMATION WHICH WILL REDUCE THE activation energy of the transition state and ALLOW MORE MOLECULES TO ADOPT THIS SHAPE. DEMONSTRATION/MAINPOINT- breaking the stick on her knee To increase the rate of the reaction, the enzyme optimizes interactions with the ___________ which stabilizes the structure and allows more product to be formed in a given amount of time. • Beginning --> substrate • Middle --> transition state ○ The most favorable interactions with the transition state which stabilizes the structure and lowers the Ea needed for formation ○ R groups of the enzyme make the most interactions with the substrate during the transition state End --> product state
What's the difference between turnover number and the specificity constant?
Turnover number, or kcat, the catalytic rate constant, measures the number of substrate molecules turned over per enzyme molecule per unit time when the enzyme is saturated with substrate. The specificity constant, described as kcat / Km, is distinct because it also factors in the affinity or Km of the enzyme for any given substrate(s).
Explain why enzyme activity analyses most commonly measure INITIAL velocity of a reaction.
Under Michaelis Menden conditions, analysis assumes that 1) substrate concentration does not significantly change 2) the amount of product is negligible 3) the reverse reaction rate does not contribute. Also, initial velocity depends on known enzyme and substrate concentrations and is the beginning/linear region of the curve where velocity does not change with time.
Enzyme A and Enzyme B both catalyze the same reaction. If Enzyme A has a bigger turnover number than Enzyme B, what, if anything, does that tell you about their specificity constants? Explain your answer.
We don't have all of the parameters to determine the specificity constant - we need Kcat AND Km to determine the specificity constant. The turnover rate is the the number o of substrate molecules that can be turned into product per enzyme molecule in a certain period of time. Simply put, it is the rate of [ES] --> E+P. Since the specificity constant depends on both the turnover rate and enzyme affinity for substrate (Km AND Kcat), we cannot compare their specificity constants unless we have all the numbers.
You perform an enzyme activity assay (experiment) using 1.0 mM of enzyme and determine that the Vmax is 0.5 mM/sec. If you repeat the assay using 2.0 mM of enzyme, would the increase in enzyme concentration affect Vmax? Explain your answer.
Yes, Vmax would increase because there are more active sites available at every substrate concentration so the substrate can be used up more quickly leading to a rate increase.
If a glutamic acid in the active site of an enzyme has an arginine nearby, will the pKa of glutamic acid be affected? If so, explain how. If not, explain why not. arginine has a pKa of 12.5
Yes. Around physiological pH, arginine will remain protonated because it has a pKa of 12.5. This +1 charge will favor the formation of a -1 charge on glutamic acid, which is possible because glutamic acid has an ionizable COOH in it's R group. Therefore, the interaction between a +1 and -1 charge will encourage glutamic acid to ionizer sooner and will decrease the pKa of glutamic acid. Ionizable R groups involve: glutamate/glutamic acid aspartate/aspartic acid histidine lysine arganine cysteine tyrosine
iClicker: Which statement about free energy is true? A. Free energy is wasted energy and cannot be used by a cell B. Chaos in a system decreases the available useful energy C. Change in free energy can be directly measured in a cell D. All of the above are true statements
a. False--> it is useful energy for a cell b. True c. Change in STANDARD free energy can be measured in a cell
• Use the enzyme activity graph to the right to describe what you know about the enzyme. When is it active? When is it inactive? What happens at pH = 12? What happens at pH = 3? Where is 50% activity? Can you tell how many ionizable groups are in the active site? If so, how many?
active between pH of (5 or) 6 - 8 below pH of 5, the enzymes are inactive at pH of 6.5, enzymes are 50% active we can't tell how many ionizable groups are present in the active site here.
On a single free energy diagram, illustrate an endergonic reaction in the presence and absence of an enzyme.
enzymes lower the activation energy of an endergonic reaction ΔG is unaffected by this process
Explain the relationship between a graph of [product] vs. time and initial velocity (V0) vs. [substrate].
to find the velocity for a given enzyme and substrate concentration, a graph containing [product] versus time is created. These data fit with a hyperbolic curve. The initial slope is the the initial rate to be determined. An initial velocity and [substrate] fit a single point on a graph of initial velocity versus [substrate].