Ch12

What is Km defined as?

1/2 of Vmax

in line-weaver-burke plot, what is the vertical intercept?

1/Vmax

Compounds that function as "mixed inhibitors" A Option A : bind to the enzyme reversibly. B Option B : can bind to the enzyme/substrate complex. C Option C : interfere with substrate binding to the enzyme.

All of the above

Enzyme activity in cells is controlled by which of the following? A Option A : feedback inhibition B Option B : modulation of expression levels C Option C : covalent modifications D Option D : allosteric effectors

All of the above

In uncompetitive inhibition, the inhibitor binds only to the ______.

ES complex

When [S] = KM, ν0 = (_____)× (Vmax). A Option A: 0.5 B Option B: KM C Option C: 0.75 D Option D: kcat E Option E: [S]

Option A: 0.5

For the reaction, the steady state assumption A Option A: ES breakdown occurs at the same rate as ES formation B Option B: [S] = [P] C Option C: implies that k−1 and k2 are such that the [ES] = k1[ES] D Option D: [P]>>[E] E Option E: implies that k1=k−1

Option A: ES breakdown occurs at the same rate as ES formation

Which of the following is correct in regards to the diagram below? Option A: X=B, Y=A, Z=Q B Option B: X=A, Y=B, Z=P C Option C: X=E, Y=A, Z=E D Option D: X=E, Y=B, Z=P E Option E: X=E, Y=B, Z=Q

Option A: X=B, Y=A, Z=Q

KM is A Option A: the [S] that half-saturates the enzyme. B Option B: a ratio of substrate concentration relative to catalytic power. C Option C: equal to half of Vmax. D Option D: a measure of the catalytic efficiency of the enzyme. E Option E: the rate constant for the reaction ES → E + P.

Option A: the [S] that half-saturates the enzyme.

A Lineweaver-Burk plot is also referred to as: choose all possible(s) A Option A : a sigmoidal plot. B Option B : a linear plot. C Option C : a Michaelis-Menten plot. D Option D : a double reciprocal plot.

Option B : a linear plot. Option D : a double reciprocal plot.

Parallel lines on a Lineweaver-Burk plot indicate A Option A : an increase in KM. B Option B : decrease in Vmax. C Option C : uncompetitive inhibition. D Option D : decrease in KM.

Option B : decrease in Vmax. C Option C : uncompetitive inhibition. D Option D : decrease in KM.

Find the initial velocity for an enzymatic reaction when Vmax = 6.5 × 10-5 mol·sec-1, [S] = 3.0 × 10-3 M, KM = 4.5 × 10-3 M and the enzyme concentration at time zero is 1.5 × 10-2 μM. A Option A: 1.4 × 10-2 mol·sec-1 B Option B: 2.6 × 10-5 mol·sec-1 C Option C: 3.9 × 10-5 mol·sec-1 D Option D: Not enough information is given to make this calculation. E Option E: 8.7 × 10-3 mol·sec-1

Option B: 2.6 × 10-5 mol·sec-1

________ clinical trials are focused on evaluating the efficacy of new drug candidates, and usually use _____ test. A Option A: Phase 2; double blind B Option B: Phase 2; single blind C Option C: Phase 1; double blind D Option D: Phase 3; double blind E Option E: Phase 1; single blind

Option B: Phase 2; single blind

Reaction that is first order with respect to A and B A Option A: is independent of reactant concentration. B Option B: is dependent on the concentration of A and B. C Option C: is dependent on the concentration of A. D Option D: is always faster than first-order reactions due to loss of concentration dependence. E Option E: has smaller rate constants than first-order reactions regardless of reactant concentration.

Option B: is dependent on the concentration of A and B.

KM A Option A: All of the above are correct. B Option B: is the concentration of substrate where the enzyme achieves ½Vmax. C Option C: is equal to Ks. D Option D: measures the stability of the product E Option E: is high if the enzyme has high affinity for the substrate.

Option B: is the concentration of substrate where the enzyme achieves ½Vmax.

For a reaction A + B → C, if the concentration of B is much larger than A so that [B] remains constant during the reaction while [A] is varied, the kinetics will be A Option A: hyperbolic. B Option B: pseudo-first-order. C Option C: unimolecular. D Option D: sigmoidal. E Option E: zero-order.

Option B: pseudo-first-order.

he Michaelis constant KM is defined as A Option A : [S] = [ES] B Option B : [ES]/2 C Option C : (k-1 + k2)/k1 D Option D : ½ Vmax

Option C : (k-1 + k2)/k1

I propose to design a new drug which will act as an inhibitor for an enzyme. If I have used all current information about the mechanism of this enzyme to design this inhibitor and I carefully engineer it with similar chemical properties of the transition state, what type of inhibitor am I attempting to engineer and how will I know if I have succeeded? A Option A: A uncompetitive inhibitor, collect kinetic data both in the presence and absence of inhibitor and watch for a change in Vmax. B Option B: None of the above. C Option C: A competitive inhibitor, collect kinetic data both in the presence and absence of inhibitor and watch for a change in KM. D Option D: A uncompetitive inhibitor, collect kinetic data both in the presence and absence of inhibitor and watch for a change in KM. E Option E: A competitive inhibitor, collect kinetic data both in the presence and absence of inhibitor and watch for a change in Vmax.

Option C: A competitive inhibitor, collect kinetic data both in the presence and absence of inhibitor and watch for a change in KM.

A compound that distorts the active site, rendering the enzyme catalytically inactive is called A Option A: an allosteric effector B Option B: none of the above C Option C: a uncompetitive inhibitor D Option D: a competitive inhibitor E Option E: an inactivator

Option C: a uncompetitive inhibitor

What is the velocity of a first-order reaction at 37oC when the reactant concentration is 6 × 10-2 M and the rate constant is 8 × 103 sec-1? A Option A: 1.33 × 105 M-1·sec-1 B Option B: Not enough data are given to make this calculation C Option C: 7.5 × 10-2 M·sec D Option D: 4.8 × 102 M·sec-1 E Option E: 1.33 × 105 M·sec

Option D: 4.8 × 102 M·sec-1

The catalytic efficiency of an enzyme can never exceed A Option A: k-1. B Option B: k-1 + k2. C Option C: k2. D Option D: k1. E Option E: (k-1 + k2)/k1.

Option D: k1.

This diagram refers to a (an) A Option A: X, Y, and Z must be provided in order to answer correctly. B Option B: ordered bisubstrate reaction. C Option C: Ping Pong reaction. D Option D: random bisubstrate reaction. E Option E: double order ping pong reaction.

Option D: random bisubstrate reaction.

Enzyme E is responsible for conversion of substrate X to product U. As a result of this conversion electrons are transported to a coenzyme (FAD) within Enzyme E. In order for the reaction to be completed, a second substrate NAD+ must also bind Enzyme E and collect stored electrons (which converts it to product, NADH). The graph below shows the data while varying X, with fixed concentrations of NAD+. What type of multi-substrate mechanism does enzyme E utilize? Substrates: X NAD+ ↓ ↓ Products: U NADH A Option A: Sequential but the data cannot differentiate between ordered and random. B Option B: simultaneous addition C Option C: Ping Pong D Option D: sequential - Random E Option E: sequential - Ordered

Option D: sequential - Random

At substrate concentrations much lower than the enzyme concentration, A Option A: first order enzyme kinetics are not observed. B Option B: the KM is lower. C Option C: the rate of reaction is independent of substrate concentration. D Option D: the rate of reaction is expected to be directly proportional to substrate concentration. E Option E: the rate of reaction is expected to be inversely proportional to substrate concentration.

Option D: the rate of reaction is expected to be directly proportional to substrate concentration.

Determine the KM and Vmax from the following graph. (Note: On the x-axis the minor tick mark spacing is 0.005; on the y-axis the minor tick mark spacing is 0.002) A Option A: KM = [165]; Vmax = 33/s B Option B: KM = [270]; Vmax = 68/s C Option C: KM = [0.006]; Vmax = 0.0075/s D Option D: KM = [0.196]; Vmax = 0.0075/s E Option E: KM = [33]; Vmax = 167/s Mark for Review Submit Answer

Option E: KM = [33]; Vmax = 167/s

Pseudo-first-order reaction kinetics would be observed for the reaction A + B → C A Option A: if [C]>[A] and [C]>[B]. B Option B: if [A] or [B] = 0. C Option C: if [C] = 0. D Option D: if [A] or [B] > [C]. E Option E: None of the above.

Option E: None of the above.

Which of the following is (are) true? A Option A: It describes a double displacement reaction. B Option B: All of the above are true. C Option C: The [ES] will remain constant if k2 > k1 and k−1 < k2. D Option D: None of the above is true. E Option E: The reaction is zero order with respect to [S] if [S] >> [E].

Option E: The reaction is zero order with respect to [S] if [S] >> [E].

In order for an enzymatic reaction obeying the Michaelis-Menten equation to reach 3/4 of its maximum velocity, A Option A: not enough information is given to make this calculation B Option B: [S] would need to be equal to KM C Option C: [S] would need to be ½ KM D Option D: [S] would need to be ¾ KM E Option E: [S] would need to be 3KM

Option E: [S] would need to be 3KM

Irreversible enzyme inhibitors A Option A: inhibit competitively. B Option B: function via Ping Pong mechanism. C Option C: maximize product by minimizing ES → E+S. D Option D: behave allosterically. E Option E: inactivate the enzyme.

Option E: inactivate the enzyme.

The KM can be considered to be the same as the dissociation constant KS for E + S binding if A Option A: ES → E + P is fast compared to ES → E + S. B Option B: this statement cannot be completed because KM can never approximate KS. C Option C: the concentration of [ES] is unchanged. D Option D: k1 >> k2. E Option E: k2 << k-1.

Option E: k2 << k-1.

Fourth-order reactions. A Option A: have three or more sequential rate determining steps. B Option B: require a 'Ping Pong' mechanism. C Option C: are best analyzed using Lineweaver-Burk plots. D Option D: exist only when enzymatically catalyzed. E Option E: none of the above.

Option E: none of the above.

Allosteric activators A Option A: stabilize conformations with higher Ks. B Option B: All of the above. C Option C: None of the above. D Option D: bind via covalent attachment. E Option E: stabilize conformations with higher substrate affinity.

Option E: stabilize conformations with higher substrate affinity.

A two-substrate enzymatic reaction in which one product is produced before the second substrate binds to the enzyme has a ______ mechanism.

Ping Pong

Assume a first order reaction, the rate of the reaction 2A → B is dependent on ______.

[A]

The E+S → E+P reaction is ______.

bimolecular

In ______, the inhibitor binds to a site involved in both substrate binding and catalysis.

competitive inhibition

The type of enzyme inhibition in which Vmax is unaffected is ______.

competitive inhibition

A new drug has been discovered which inhibits the reaction catalyzed by enzyme A. Based on the information shown below, what is this drug? Km Changed

competitive inhibitor

Different enzymes that catalyze the same reaction, although may be found in different tissues, are known as ______.

isozymes

how can you find Km on a curve?

look for where the linearness ends..

Following several experiments, the data presented on the graph below was obtained. What can you determine from this graph? only 1/Vmax is changed

non competitive

A common type of covalent modification of regulatory enzymes involves ______ of serine residues.

phosphorylation

If A → B is a zero-order reaction, the rate is dependent on ______.

the rate constant

A lab recently developed a new drug which is hypothesized to inhibit the enzyme cyclooxygenase-2 (COX-2) and reduce inflammation. In their first test they monitored the reaction of substrate as it is converted to product in the presence of the new drug (data shown below). If the hypothesis is correct the observed initial rate will be at least 2 times slower than the normal reaction without the drug. If the normal initial rate is 30 mM/s, does the data below indicate that the team has designed a successful inhibitor? curved downwards

yes