Ch. 12 Enzyme Kinetics

What does x-intercept represent on a Lineweaver-Burk plot?

-1/Km

Given the initial rate of vo = 2x10-8 s-1, Vmax = 1x10-7 M s-1 and [S] = 10 mM, calculate KM.

0.04 Vo = (Vmax[S]) / (KM + [S]) 2x10-8 s-1 = (1x10-7 M s-1)(10 mM) / (KM + 10 mM) KM = 0.4

What are the 2 definitions of Vmax?

1. A constant for each enzyme, equal to kcat[E]T 2. The theoretical maximal rate of the reaction (but it is never achieved in reality)

What are the three definitions of Km?

1. A constant, but it is also the substrate concentration is ½ Vmax → [S] 2. A constant derived from rate constants, (k-1 + k2) / k1 3. An estimate of the dissociation constant of E from S

Michaelis-Menten Assumptions

1. Assumes the formation of an enzyme-substrate complex: ES 2. Assumes that the [ES] is constant, also known as the "steady-state assumption" 3. Assumes that a high [S], the breakdown of ES (k2) to form product P, is the rate limiting step, equal to the catalytic rate constant, kcat

How do you obtain the Lineweaver-Burk Equation for a line?

1. Begin with ν = Vmax [S]/(Km + [S]) and take the reciprocal of both sides. 2. Rearrange to obtain the Lineweaver-Burk Equation for a line

The three hallmarks of uncompetitive inhibition are:

1. Both the KM and Vmax are changing. 2. KM / Vmax are not changing, so they have the same slope in the Lineweave plots of their kinetics. 3. Only observed in enzymes have two or more substrates.

What are the 2 ways to lose ES complex?

1. Can go forward and form product and enzyme back in original state P: ES → E + P (k2) 2. Can dissociate back into enzyme + substrate: ES ⇋ E + S (k-1)

Ping-pong reaction in serine protease:

1. Chymotrypsin (E) binds with first substrate, peptide (A). Peptide bound to substrate is (AE) and PE' forms right away which is tetrahedral oxyanion 2. R' with new N terminus is released (P) which forms acyl enzyme (E'). And then second substrate binds, water (B) to active site and form oxyanion tetrahedral intermediate (E'B) supported by the tetrahedral oxyanion hole 3. Water (B) helps resolve acyl enzyme (E') and releases the R with new C terminus (Q) and gets back chymotrypsin (E)

What are the 4 types of reversible enzyme inhibitors and how do you distinguish between them using linear plots?

1. Competitive: raises Km with no change in Vmax 2. Noncompetitive: decreases Vmax with no change in Km 3. Mixed noncompetitive: alters both Km and Vmax 4. Uncompetitive: alters both Km and Vmax but with same slope, Km / Vmax

Schematic for Ping-Pong reaction:

1. Enzyme first binds to substrate (A) and then it forms a covalent intermediate PE' 2. P is released but E' is still modified and bind to second substrate B forming E'B 3. Then forms product EQ and then release product Q and get enzyme back to beginning.

What are the two effects of temperature on enzymatic activity?

1. Enzyme rate typically doubles in rate for every 10oC, as long as the enzyme is stable and active. 2. At higher temperatures, the protein becomes unstable and denaturation occurs (loss of structure and function)

What are the three assumptions of enzyme kinetics?

1. Solutions are behaving ideally 2. Our constant are indeed constant 3. For our reactions, substrate isn't forming product without the help of enzyme at a big enough rate for us to consider and that it's negligible.

What are the 3 equations that need to memorize?

1. v = (Vmax [S]) / (Km + [S]) 2. Vmax =k2[ET] 3. Km = (k-1 + k2) / k1

What does y-intercept represent on a Lineweaver-Burk plot?

1/Vmax

What does the x-axis represent on a Lineweaver-Burk plot?

1/[S]

What is the Lineweaver-Burk equation?

1/v = (Km/Vmax)(1/[S]) + 1/Vmax

What does the y-axis represent on a Lineweaver-Burk plot?

1/vo

Given the following kinetic parameters, determine the value of KM. k1 = 700 M-1 sec-1, k-1 = 400 sec-1, k2 = 1000 sec-1, [E]T = 700 nM.

2 KM = (k-1 + k2) / k1 KM = (400 sec-1 + 1000 sec-1) / 700 M-1 sec-1) KM = 2

On the Michaelis-Menten plot, where does zero-order occur?

Above KM, at higher [S] no longer sensitive to [S] concentration and begins to plateau, called zero-order because saturating enzyme with substrate

What is the saturation effect?

At low substrate concentrations, you have a pseudo first order reaction where the initial velocity appears to be directly proportional to the substrate concentration → linear part of MM plot As approach higher substrate concentration, begin to approach zero-order kinetics where there is no more change in the rate. Which means initial velocity is equal to maximal velocity → flat (zero change) No change because at very high substrate concentration the enzyme is saturated → enzyme is already working at full tilt

On the Michaelis-Menten plot, where does pseudo first-order occur?

Below KM, at low [S] the relationship between [P] formation and [S] concentration looks linear, called pseudo first-order

Is the binding step reversible or irriversible? Explain.

Binding is reversible so rate constant for ES binding is k1 and rate constant for dissociation is k-1

Enzymes that catalyze reactions involving two (or more) substrates

Bisubstrate reactions

can occur by Ordered- or Random-Sequential mechanisms or by a Ping-Pong mechanism

Bisubstrate reactions

type of reversible inhibitor that binds to the same site as your substrate (binds to the active site)

Competitive inhibition

Why are enzyme-catalyzed reactions are more complex than chemically catalyzed reactions?

Complex have multiple steps: 1. Binding step: enzyme and substrate bind each other: E + S ⇋ ES 2. Once ES complex is formed → catalysis si what's rate limiting

proceed via formation of a covalently modified enzyme intermediate, E'. Product (P) of the reaction with A is released prior to reaction of the enzyme with the second substrate, B, release of product (Q) and regeneration of E.

Double-Displacement (Ping-Pong) reaction

seeks to determine the initial and maximal reaction velocity (or rate) that enzyme can attain and the binding affinities for substrates and inhibitors

Enzyme kinetics

T/F: Non Covalent modifications of an enzyme cannot significantly influence its activity.

False

T/F: The KI of a competitive inhibitor is equal to [ES][I] / [ESI]

False

Initial Velocity of Product Formation Increases with

Increasing [S]

affecting an enzyme under its normal conditions by adding a inhibitor

Inhibition

a substance or a therapeutic that interacts with the enzyme and slows down its activity.

Inhibitor

Describe the Michaelis-Menten plot.

Initial velocity (νO, Y axis) is plotted as a function of substrate concentration (X axis)

The effects of pH may be due to effects on

KM (binding) or Vmax (catalysis) or both

What is the symbol for Michaelis-Menten constant? And what does it equal?

Km = [S] when VO = ½ Vmax

Why does Km change in competitive inhibition?

Km is changing because enzyme inhibitor is interfering with the ability of the enzyme to bind with its substrate.

Explain why Km does not change in pure noncompetitive inhibtion.

Km is not changing because enzyme inhibitor is not interfering with the ability of the enzyme to bind with its substrate.

An enzyme has a KM of 8 μM in the absence of a competitive inhibitor and a KMapp of 12 μM in the presence of 3 μM of the inhibitor. Calculate KI.

Kmapp = ɑKm → ɑ = Kmapp / Km → ɑ = 12 μM / 8 μM ⇒ ɑ = 1.5 ɑ = 1 + [I] /KI → ɑ - 1 = [I] /KI → KI = [I] / ɑ - 1 → KI = 3 μM / 1.5 - 1 ⇒ KI = 6 μM

Is a Lineweaver-Burk plot linear or nonlinear? Explain.

Linear A plot 1/ν versus 1/[S] should yield a straight line: y = mx + b

What distinguishes between types of Bisubstrate Reactions?

Lineweaver-Burk Plot

offer another useful form of the rate law that described enzyme catalyzes reactions

Lineweaver-Burk Plots

can be used to present kinetic data and to calculate values for KM and Vmax, Useful for analysis of inhibitors.

Lineweaver-Burk plot

The rate law for enzyme-catalyzed reactions is described by what equation?

Michaelis-Menten equation

relates the initial velocity of a reaction to the maximal reaction velocity and the Michaelis constant for a particular enzyme and substrate.

Michaelis-Menten equation

Which reversible inhibitor alters both Km and Vmax (decreases)?

Mixed noncompetitive

Why is Vmax a theoretical maximal velocity?

Never going to reach Vmax because in reality, encounters between E and S are asynchronous, not simultaneous.

Steady-State Assumption

No change in the concentration of the ES complex However, rapid flux means the ES complex either dissociates, or results in product formation, as rapidly as new ES complexes are forming

Is the Michaelis-Menten equation linear or nonlinear? Explain.

Nonlinear because the enzyme has to first encounter substrates and bind them and then in the second step perform the catalysis

The leading substrate (A) must bind first, followed by B. Reactions between A and B occurs in the tertiary complex and is usually followed by an ordered release of the products, P and Q (in either order).

Ordered single-displacement reaction

an irreversible (so-called "suicide") inhibitor of the enzyme, glycoprotein peptidease, which catalyzes an essential step in bacterial cell all synthesis

Penicillin

What is an example of an irreversible inhibitor?

Penicillin is an irreversible, suicide inhibitor

What is the hallmark of single-displacement reaction?

Presence of ternary complex (three-part complex)

In this type of sequential reaction, all possible binary enzyme-substrate and enzyme-product complexes are formed rapidly and reversibly when enzyme is added to a reaction mixture containing A, B, P, and Q.

Random, single-displacement reaction

Define rate and affinity in enzyme kinetics.

Rate - velocity or initial velocity of the reaction Affinity - formation of ES complex

How is enzymatic activity influenced by temperature?

Rates of enzyme-catalyzed reactions generally increase with increasing Temperature (T)

points where reaction can be turned up or turned down are catalyzed by specific enzymes that we may want to target for therapeutics

Regulatory points

What are the two classes of inhibitors?

Reversible & Irreversible

display a linear plot of the substrate (A) or product (P) concentration, as a function of time

Simple first-order reactions

How can single-displacement be distinguished from double-displacement?

Single-displacement reactions have intercepting lines. Double-displacement reactions have parallel lines.

What does small Km and large Km mean?

Small Km means little dissociation (tight binding), e.g. 10-6 M Large Km means lots of dissociation (weak binding), e.g. 10-2 M

double-displacement

Substrate → product → substrate → product = "ping-pong"

The decay of a hypothetical radioisotope has a rate constant of 0.01 s-1. How much time is required for half of a 1-g sample of isotope to decay?

The half-life for a first-order process is: t1/2 = ln2 / k → = 0.693 / 0.01s-1 ⇒ t1/2 = 69.3 s

T/F: The catalytic activity of an enzyme can be regulated by controlling the amount of enzyme available and by structural alteration of the enzyme that influences substrate binding.

True

T/F: The effects of uncompetitive inhibition on Vmax are not reversed by increasing substrate concentration.

True

T/F: The effects of uncompetitive inhibition on Vmax are not reversed by increasing the substrate concentration.

True

T/F: The vo versus [S] plot for ATCase is sigmoidal, rather than hyperbolic as it is in the enzyme that follow the Michaelis-Menten model.

True

Which reversible inhibitor alters both Km and Vmax but with same slope, Km / Vmax?

Uncompetitive

If the M-M model fits, then: k2 = kcat = ?

Vmax / ET

At temperature about 50℃ to 60℃, enzymes typically show

a decline in activity

What describes the Michaelis-Menten equation?

a rectangular hyperbolic dependence of ν on [S]

Which of the following equations is used in Lineweaver-Burk plots? a. 1/vo = (Km/Vmax)(1/[S])+(1/Vmax) b. 1/vo = (Km/Vmax)([S])+(Vmax) c. vo = (Km/Vmax)(1/[S])+(1/Vmax) d. vo = (Km/Vmax)([S])+(Vmax) e. 1/vo = (Km/Vmax)([S])+(1/Vmax)

a. 1/vo = (Km/Vmax)(1/[S])+(1/Vmax)

For a given enzyme, k1 = 300 sec-1, k2 = 600 sec-1, k-1 = 100 sec-1, and Vmax = 30 μM sec-1, what is [E]T? a. 50 nM b. 50 nM sec-1 c. 20 μM sec-1 d. 20 μM e. 30 μM

a. 50 nM Vmax = k2[E]T 30 μM sec-1 = (600 sec-1) [E]T [E]T = 0.05 μM [E]T = 50 nM

Which of the following statements about inhibitors of enzyme-catalyzed reactions is TRUE? a. A competitive inhibitor does not affect Vmax. b. A pure non-competitive inhibitor typically affects KM but not kcat. c. A mixed non-competitive inhibitor will always bind at the active site and affect KM. d. A competitive inhibitor binds irreversibly to the enzyme at the active site. e. An uncompetitive inhibitor bind to either free enzyme, or the enzyme-substrate complex, irreversibly.

a. A competitive inhibitor does not affect Vmax.

Which of the following statements is INCORRECT with respect to allosteric enzymes? a. Allosteric enzymes always have hyperbolic reaction kinetics. b. Allosteric enzymes must have quaternary structures. c. Allosteric enzymes bind molecules that cause a change in the tertiary structure of the protein. d. Allosteric enzymes have two states, one that has low activity, and one that has high activity.

a. Allosteric enzymes always have hyperbolic reaction kinetics. False. Allosteric enzymes exhibit sigmoidal reaction kinetics, not hyperbolic.

For an enzyme that displays Michaelis-Menten kinetics, what is the effect on KM of doubling the concentration of substrate? a. No effect c. Decrease

a. No effect The KM of an enzyme is a concentration of substrate, the concentration of at which the initial velocity (Vo) is half maximal velocity (Vmax). As such, it cannot be affected by change sin substrate concentration.

Which enzyme has the highest catalytic efficiency? Enzyme X: kcat = 10^6 s-1 & Km 10^-3 M Enzyme Y: kcat = 10^7 s-1 & Km 10^-1 M Enzyme Z: kcat = 10^8 s-1 & Km 10 M a. X b. Not enough information to determine catalytic efficiency. c. Y and Z have the same catalytic efficiency. d. Z e. Y

a. X

A covalent enzyme intermediate is a modification found in a _______ reaction. a. double-displacement (ping-pong) b. random sequential c. ordered sequential d. allosteric e. concerted displacement

a. double-displacement (ping-pong)

A rate law describes how the rate of the reaction is related to _______. a. substrate concentration b. product concentration c. free energy of the reaction d. efficiency of the reaction e. rate constants

a. substrate concentration

Which possesses the highest free energy? a. transitions state b. reactant c. initial state d. final state e. product

a. transitions state

When is the only time you see Vmax?

at high [S] concentrations much greater than KM

An enzyme has a Vmax of 150 units of product formed sec-1, per unit of enzyme. The KM for its substrate is 5.5 mM. Calculate vo when [S] is 0.5 mM. a. 75 micromol sec-1 b. 12.5 micromol sec-1 c. 37.5 micromol sec-1 d. 100 micromol sec-1

b. 12.5 micromol sec-1 Vo = (Vmax[S]) / (KM + [S]) Vo = (150 s-1)(0.5 mM) / (5.5 mM + 0.5 mM) Vo = 12.5 mM sec-1

Which of the following of inhibition can be overcome by large enough concentration of substrate? a. All of these types of inhibition can be overcome given a large enough substrate concentration. b. Competitive inhibition c. Uncompetitive inhibition d. Mixed competitive inhibition e. Pure noncompetitive inhibition

b. Competitive inhibition

Which of the following describes the effect of an enzyme on the initial velocity of a given reaction? Use image above. a. Decrease b. Increase c. No change

b. Increase Enzymes are catalysts which "speed up" the reactions that occur in living things. That is, they increase the initial velocity of these reactions.

Which of the following features in a Lineweaver-Burk plot is altered by a competitive inhibitor in an enzyme-catalyzed reaction? a. Intercept on the y-axis b. Intercept on the x-axis c. Interceptor on the y-axis which is the 1/Vo axis d. Intercept on the 1/Vo axis e. Shape of the plot.

b. Intercept on the x-axis A Lineweaver-Burk double-reciprocal plot is used to investigate the relationship between concentration of [S], and initial rate of the reaction, VO. The x-axis is used to plot 1/[S] and the intercept is equal to 1/KM. Since KM is increased by competitive inhibitors, the intercept will change.

Which of the following describes the effect of an enzyme on the equilibrium constant of a given reaction? a. Decrease b. No change c. Increase

b. No change The equilibrium position of a reaction is determined by free energy considerations and is not affected by a catalyst.

Which of the following statements is TRUE concerning the "plateau" region of the curve? a. All of the available substrate has been converted to product. b. The enzymes active site is saturated. c. The enzyme is in the tense state. d. A and C. e. A and B.

b. The enzymes active site is saturated. True. The plateau indicated that the enzyme is working close to he maximal activity under the experimental conditions. That is, the active site is saturated.

Which enzyme has the highest turnover number? Enzyme X: kcat = 10^6 s-1 & Km 10^-3 M Enzyme Y: kcat = 10^7 s-1 & Km 10^-1 M Enzyme Z: kcat = 10^8 s-1 & Km 10 M a. Not enough information to determine turnover number. b. Z c. X d. Y e. Both Y and Z

b. Z

A Lineweaver-Burk plot is also referred to as _________. Select all that apply. a. a sigmoidal plot. b. a linear plot. c. a Michaelis-Menten plot. d. a double reciprocal plot.

b. a linear plot. d. a double reciprocal plot.

Parallel lines on a Lineweaver-Burk plot indicate _______. Select all that apply. a. an increase in KM. b. decrease in KM. c. decrease in Vmax d. uncompetitive inhibition.

b. decrease in KM. c. decrease in Vmax d. uncompetitive inhibition.

When [S] = KM, vo = ( ___ ) x (Vmax). a. [S] b. 0.75 c. 0.5 d. Km e. kcat

c. 0.5

Which of the following statements is INCORRECT with respect to covalent modification of an enzyme via phosphorylation? a. Ser, Thr or Tyr residues can be phosphorylated. b. Phosphorylation modifies the size, shape and charge of the phosphorylated amino acid. c. Covalent modification via phosphorylation is irreversible. d. Phosphorylation may either decrease or increase enzyme activity.

c. Covalent modification via phosphorylation is irreversible. False. Covalent modification via phosphorylation is reversible, not irreversible.

Which of the following is true for the maximal velocity of an enzyme-catalyzed reaction? a. Maximal velocity is not affected by the presence of a noncompetitive inhibitor. b. Maximal velocity increases when pH increases. c. Maximal velocity can be calculated from the initial rate of the reaction at a concentration of substrate that is equal to the KM. d. Maximal velocity is reduced in the presence of a transition state analog competitive inhibitor.

c. Maximal velocity can be calculated from the initial rate of the reaction at a concentration of substrate that is equal to the KM. Vmax is equal to twice the initial rate at the concentration of substrate equal to the KM.

Which of the following statements about phosphorylation of enzymes is FALSE? a. The phosphorylation of enzymes is catalyzed by other enzymes called kinases. b. Phosphorylation is a reversible covalent modification. c. The phosphorylation of an enzymes results in change in its conformation, thereby increasing its activity. d. The phosphorylation of enzymes is one mechanism via which hormone elicit an intracellular response.

c. The phosphorylation of an enzymes results in change in its conformation, thereby increasing its activity. False. The phosphorylation of an enzyme does alter its conformation, but this may cause either an increase OR a decrease in activity.

Which of the following statements is false for an enzyme that follows Michaelis-Menten kinetics? a. The initial velocity of the reaction is dependent on substrate concentration. b. Maximal velocity occurs when the enzyme is entirely in ES form. c. The relationship between substrate concentration and reaction rate is sigmoidal. d. The Michaelis-Menten equatio assumes that ES is maintained at a steady state.

c. The relationship between substrate concentration and reaction rate is sigmoidal. False. Enzymes which follow Michaelis-Menten kinetics have a rectangular hyperbolic relationship between substrate concentration and reaction rate.

Which of the following is TRUE about competitive inhibitors? a. Competitive inhibitors lower the KM and the Vmax of the enzyme. b. Competitive inhibitors structurally resemble the substrate and so they bind to the active site and become covalently attached the enzyme. c. Transition state analogs often make better competitive inhibitors than do substrate analog. d. A and C are both true. e. A, B, and C are all true.

c. Transition state analogs often make better competitive inhibitors than do substrate analog. Transition state analogs often make excellent competitive inhibitors because enzyme are designed to bind the transition state intermediate even better then they bind their substrates. Statements A and B are both false. Competitive inhibitors increase the KM of an enzyme and have no effect on the Vmax. Competitive inhibitors bind reversibly in the active site, not irreversibly.

In order for an enzymatic reaction obeying the Michaelis-Menten equation to reach ¾ of its maximum velocity, a. [S] would need to be equal to KM. b. [S] would need to be ½ KM. c. [S] would need to be 3KM. d. [S] would need to be ¾ KM. e. not enough information is given to make this calculation.

c. [S] would need to be 3KM.

Protein kinases are involved in a. the digestion of drugs to potentially toxic byproducts. b. the degradation of enzymes to the component amino acids. c. the phosphorylation of a wide variety of proteins. d. the metabolism fo drugs to water soluble, extractable compounds. e. all of the above.

c. the phosphorylation of a wide variety of proteins.

Kinetics _______ prove a reaction mechanism

cannot

Which reversible inhibitor raises Km with no change in Vmax?

competitive

Which of the following statements about inhibitors of enzyme-catalyzed reactions is TRUE? a. A competitive inhibitor bind irreversibly to the enzyme at the active site. b. A mixed non-competitive inhibitor will always bind at the active site and affect KM. c. A pure non-competitive inhibitor typically affects KM but not kcat. d. A competitive inhibitor does not affect Vmax. e. An uncompetitive inhibitor bind to either free enzyme, or the enzyme-substrate complex, irreversibly.

d. A competitive inhibitor does not affect Vmax.

What is the DIFFERENCE between an allosteric inhibitor and a competitive inhibitor? a. Allosteric inhibitors act only on multisubunit enzymes; competitive inhibitors act only on monomeric enzymes. b. Competitive inhibitors are substrate analogs; allosteric inhibitors are transition state analogs. c. A competitive inhibitor reduces Vmax; an allosteric inhibitor does not affect Vmax. d. An allosteric inhibitor bound to one subunit alters substrate binding to other subunits; a competitive inhibitor bound at one active site alters binding at only that active site.

d. An allosteric inhibitor bound to one subunit alters substrate binding to other subunits; a competitive inhibitor bound at one active site alters binding at only that active site. Although an allosteric inhibitor molecule may bind at only one subunit of an enzyme, through cooperativity it alters the activity of all catalytic subunits in the enzyme.

A molecular decreases the activity of an enzyme by binding to a site other than the active site. What is the name of this molecule? a. A covalent catalyst. b. A cofactor. c. A competitive inhibitor. d. An allosteric inhibitor. e. A transition state analog.

d. An allosteric inhibitor.

Enzyme A has KM value of 3.5 mM, and enzyme B has KM value of 2.1x10-2 mM. Which of the following is TRUE? a. Enzyme B binds the transition state of its substrate better than enzyme A. b. Enzyme A binds its substrate with induced fit while B does not. c. Enzyme A is more catalytically efficient than enzyme B. d. Enzyme B has tighter substrate binding than A. e. Enzyme B has higher Vmax than enzyme A.

d. Enzyme B has tighter substrate binding than A.

All of the following statements concerning uncompetitive inhibition are true, EXCEPT: a. The y-intercept on the Lineweaver-Burk plot is altered b. Vmax is altered c. Km is altered d. No exceptions. All of these are true. e. Km/Vmax remains unaltered.

d. No exceptions. All of these are true.

Which of the following statements about allosteric enzyme si FALSE? a. Allosteric enzymes have tertiary structure. b. Allosteric effectors elicit a change in shape of the enzyme via a process termed cooperativity. c. The substrate of an allosteric enzyme may also be an allosteric effector that for enzyme. d. Regulation of enzymes by allosteric activators is irreversible.

d. Regulation of enzymes by allosteric activators is irreversible. False. The interaction between allosteric effectors nad allosteric proteins is reversible, not irreversible.

The presence of an uncompetitive inhibitor in an enzyme-catalyzed reaction will alteer which of the following in a Lineweaver-Burk plot? a. Intercept on the x-axis b. Slope of the plot c. Intercept on the y-axis d. The intercept on both axes e. Shape of the plot

d. The intercept on both axes On a Lineweaver-Burk double reciprocal plot the intercept on the x-axis is equal to 1/KM and the intercepto on the y-axis is equal to 1/Vmax. Both KM and Vmax are affected by uncompetitive inhibitors and so the intercept on both axes will change.

Which of the following is true regarding the effect of inhibitors on Michaelis-Menten reactions? a. Competitive inhibition have no effect on Vo. b. Competitive inhibitors decrease the apparent KM. c. Competitive inhibitors make it impossible to calculate the Vmax of an enzyme for its substrate. d. Uncompetitive inhibitors decrease the apparent KM and decrease the apparent Vmax, e. Noncompetitive inhibitors have no effect on apparent KM,

d. Uncompetitive inhibitors decrease the apparent KM and decrease the apparent Vmax,

KM is a. a measure of the catalytic efficiency of the enzyme. b. equal to half of Vmax. c. the rate constant for the reaction ES → E + P. d. the [S] that half-saturated the enzyme. e. a ratio of substrate concentration relative to catalytic power.

d. the [S] that half-saturated the enzyme.

Kinetic analysis is a __________ reasoning tool.

deductive

What is the initial velocity at [S] = 15 mM, if Vmax = 150 sec-1 and KM = 1.5 mM. a. 100 sec-1 b. 75 sec-1 c. 150 sec-1 d. 37.5 sec-1 e. 136 sec-1

e. 136 sec-1 Vo = (Vmax[S]) / (KM + [S]) Vo =(150 sec-1)(15 mM) / (1.5 mM + 15 mM) Vo = 136 sec -1

Which enzyme has the highest affinity for its substrate? Enzyme X: kcat = 10^6 s-1 & Km 10^-3 M Enzyme Y: kcat = 10^7 s-1 & Km 10^-1 M Enzyme Z: kcat = 10^8 s-1 & Km 10 M a. Y and Z have the same substrate affinity. b. Y c. Z d. X and Z have te same substrate affinity. e. X

e. X

What does simple rate equations describe the progress of?

first-order and second-order reactions

A → P is a _______-order reaction, with units of ______.

first; s-1

What is the hallmark of double-displacement reactions?

involves covalent interaction

Inhibitors that bind covalently at the active site that can't be dialyzed away and destroys enzyme activity

irreversible inhibitors

What is the order for a reactant given by?

its exponent in the rate equation

Rate of formation of ES is ____, while rate of dissociation is _____.

k1 ; k-1

Rate of product formation is _____.

k2

Catalytic efficiency

kcat / Km a way to score how well an enzyme catalyzes a reaction

What is the turnover number?

kcat. = k2 the number of substrate molecules converted to product per enzyme molecule, per unit of time (sec-1), when E is saturated with substrate.

An enzyme's overall catalytic efficiency is expressed as

kcat/KM

a branch of science concerned with the rates of reactions

kinetics

Are first-order rates linear or nonlinear?

linear

type of reversible inhibitor that interferes with the ability of the enzyme to bind to its normal substrate, but the inhibitor still binds to the enzyme at a different place on the substrate.

mixed noncompetitive

Which reversible inhibitor decreases Vmax with no change in Km?

noncompetitive

What is enzyme-substrate recognition and catalysis are greatly dependent on?

pH

In enzyme-catalyzed reactions at low [S], the rate is proportional to [S], as in a

pseudo first-order reaction: ν = k[S]

type of reversible inhibitor that binds to free enzyme or enzyme bound to its substrate because not competing with substrate for active site

pure noncompetitive

Single-displacement (aka sequential) reactions can be of two distinct classes:

random or ordered

the proportionality constant in the rate equation.

rate constant k

Going from ES complex → E + P is described by

rate constant k2 (aka kcat)

V = k[A]1 = k[A] is the ________.

rate law

the mathematical relationship between the reaction rate, or velocity, "ν", and concentration of reactants, for ex., [A]

rate law

the amount of A consumed (or P formed) per unit time (t).

rate, or velocity

What does m represent on a Lineweaver-Burk plot?

ratio of KM/Vmax

Inhibitors that bind non-covalently at the active site or at some other site that can be dialyzed away and restore enzyme activity

reversible inhibitors

Kinetics can only ________ various alternative hypotheses, because they don't fit the data.

rule out

What is an example of double-displacement reaction?

serine protease mechanism involves two substrates, the polypeptide substrates and water, and can't bind water until the first product leaves

Bisubstrate reactions can fall into 2 classes:

single-displacement (sequential) double-displacement (ping-pong)

What is the upper limit of kcat / Km?

the diffusion limit, approaching 1x109, the rate at which E and S encounter each other by diffusion

type of reversible inhibitor interferes with ES complex so substrate now binds to enzyme differently and shows parallel slopes on Lineweaver-Burk plot

uncompetitive

Michaelis-Menten equation

v = (Vmax [S]) / (Km + [S])

Determine the velocity of the elementary reaction X + Y → Z when the sample contains 3 μM X and 5 μM Y and k for the reaction is 400 M-1•s-1.

v = k[X][Y] → (400 M-1•s-1)(3 μM)(5 μM) → (400 M-1•s-1)(3x10-6 M)(5x10-6 M) → 6x10-9 M•s-1 ⇒ 6 nM•s-1

An enzyme-catalyzed reaction has a KM of 1 mM and a Vmax of 5 nM•s-1. What is the reaction velocity when the substrate concentration is (a) 0.25 mM, (b) 1.5 mM, or (c) 10 mM?

vo = Vmax[S] / Km+ [S] When [S] = 0.025 mM → vo = (5 nM•s-1)(0.25 mM) / (1 mM + 0.25 mM) → vo = 1.25 nM•s-1 / 1.25 ⇒ vo = 1 nM•s-1 When [S] = 1.5 mM → vo = (5 nM•s-1)(1.5 mM) / (1 mM + 1.5 mM) → vo = 7.5 nM•s-1 / 2.5 ⇒ vo = 3 nM•s-1 When [S] = 10 mM → vo = (5 nM•s-1)(10 mM) / (1 mM + 10 mM) → vo = 50 nM•s-1 / 11 ⇒ vo = 4.5 nM•s-1

random single-displacement

where either substrate may bind first, followed by the other substrate

ordered single-displacement

where leading substrate bind first, followed by the other substrate

In enzyme-catalyzed reactions at high [S], the enzyme reaction approaches

zero-order kinetics: ν = Vmax

a number equal to the slope at the beginning of each time course

νO (initial velocity)