CH 12 biochem

Pataasin ang iyong marka sa homework at exams ngayon gamit ang Quizwiz!

From the graph below plotting data that was collected under steady state conditions, velocity on the y-axis in units of μM/s and substrate concentration of the x-axis in units of μM, what is the Vmax? A) 0.24 μM/s B) 18 μM C) 0.2 μM D) 0.24 μM E) 0.12 μM/s

A) 0.24 μM/s

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) A competitive inhibitor, collect kinetic data both in the presence and absence of inhibitor and watch for a change in Vmax. B) A competitive inhibitor, collect kinetic data both in the presence and absence of inhibitor and watch for a change in KM. C) A uncompetitive inhibitor, collect kinetic data both in the presence and absence of inhibitor and watch for a change in KM. D) A uncompetitive inhibitor, collect kinetic data both in the presence and absence of inhibitor and watch for a change in Vmax. E) None of the above.

A) A competitive inhibitor, collect kinetic data both in the presence and absence of inhibitor and watch for a change in Vmax.

The Michaelis constant KM is defined as I. (k-1 + k2)/k1 II. ½ Vmax III. [S] = [ES] IV. [ES]/2 A) I B) I, II C) II D) I, IV E) II, IV

A) I

The following data were collected under conditions indicated in the graph below during the time period of 0-5 seconds. Upon plotting the Lineweaver-Burk plot, the information given in the table below was determined. Based on this available information which of the following is FALSE? x-intercept - 0.002 (Units on the x-axis are 1/M) y-intercept 0.005 (Units on the y-axis are 1/s) slope 2.50 A) The Vmax equals 200 M/s B) The Ks equals 500 M C) The kapp equals 200 per second D) The data was collected prior to reaching steady state. E) The kcat cannot be determined for this information.

A) The Vmax equals 200 M/s

In the plot below, can the KM be determined? If so, what is its value? A) Yes, it is 30 mM. B) Yes, it is 30 mM/sec. C) Yes, it is 60 mM/sec D) Yes, it is 60 mM E) No this data does not follow Michaelis-Menten kinetics

A) Yes, it is 30 mM.

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? A) Yes. B) No. C) This cannot be determined with the information given. D) The data is dependent on the maximal velocity. E) The answer is dependent on the substrate concentration.

A) Yes.

A new drug has been discovered which inhibits the reaction catalyzed by enzyme A. Based on the information shown below, what is this drug? A) competitive inhibitor B) uncompetitive inhibitor C) mixed inhibitor D) allosteric activator E) More information is required to answer the question.

A) competitive inhibitor

Irreversible enzyme inhibitors A) inactivate the enzyme B) inhibit competitively C) maximize product by minimizing ESE+S D) behave allosterically E) function via Ping Pong mechanism

A) inactivate the enzyme

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

A) is dependent on the concentration of A and B.

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

A) is the concentration of substrate where the enzyme achieves ½ Vmax.

The breakdown of dopamine is catalyzed by the enzyme monoamine oxidase (MAO). What is the final concentration of product if the starting dopamine concentration is 0.050 M and the reaction runs for 5 seconds. (Assume the rate constant for the reaction is 0.249 s−1.) A) 0.050 M B) 0.014 M C) 0.018 M D) 1.2 M E) 0.025 M

B) 0.014 M

From the graph below plotting data that was collected under steady state conditions, velocity on the y-axis in units of μM/s and substrate concentration of the x-axis in units of μM, what is the KM? A) 0.24 μM/s B) 18 μM C) 0.2 μM D) 0.24 μM E) 0.12 μM/s

B) 18 μM

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) 3.9 × 10-5 mol•sec-1 B) 2.6 × 10-5 mol•sec-1 C) 1.4 × 10-2 mol•sec-1 D) 8.7 × 10-3 mol•sec-1 E) Not enough information is given to make this calculation

B) 2.6 × 10-5 mol•sec-1

Find kcat for a reaction in which Vmax is 4 × 10-4 mol•min-1 and the reaction mixture contains one microgram of enzyme (the molecular weight of the enzyme is 200,000 D). A) 2 × 10-11 min-1 B) 8 × 107 min-1 C) 8 × 109 min-1 D) 2 × 10-14 min-1 E) 4 × 108 min-1

B) 8 × 107 min-1

The following questions (29 and 30) refer to the overall transformation shown in the following reaction: 28. Which of the following is (are) true? A) The [ES] will remain constant if k2>k1 and k−1< k2. B) The reaction is zero order with respect to [S] if [S]>>[E] C) It describes a double displacement reaction D) All of the above are true. E) None of the above is true.

B) The reaction is zero order with respect to [S] if [S]>>[E]

Following several experiments, the data presented on the graph below was obtained. What can you determine from this graph? A) This data may have been collected both in the absence (solid line) and presence (dashed line) of a competitive inhibitor. B) This data may have been collected both in the absence (solid line) and presence (dashed line) of a mixed (noncompetitive) inhibitor. C) This data may have been collected both in the absence (solid line) and presence (dashed line) of mechanism based inhibitor. D) This data may have been collected both in the absence (solid line) and presence (dashed line) of an inhibitor which binds the active site. E) More than one of the above are correct.

B) This data may have been collected both in the absence (solid line) and presence (dashed line) of a mixed (noncompetitive) inhibitor.

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

B) X=B, Y=A, Z=Q

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

B) k1.

An enzyme is near maximum efficiency when A) its turnover number is near Vmax. B) kcat/KM is near 108 M-1s-1. C) k1 << k-1. D) kcat/KM is equal to kcat. E) KM is large when k2 exceeds k1.

B) kcat/KM is near 108 M-1s-1. `

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) sigmoidal. B) pseudo-first-order. C) unimolecular. D) zero-order. E) hyperbolic.

B) pseudo-first-order.

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) sequential - Ordered B) sequential - Random C) simultaneous addition D) Ping Pong E) Sequential but the data cannot differentiate between ordered and random.

B) sequential - Random

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

B) the rate of reaction is expected to be directly proportional to substrate concentration.

[S] = KM for a simple enzymatic reaction. When [S] is doubled the initial velocity is A) 2 Vmax B) equal to Vmax C) (1/3) Vmax D) 0.5 Vmax E) 2 KM/[S]

C) (1/3) Vmax

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

C) 0.5

Based on the figure in the question above (question 54), which of the following expressions would correctly define KM? A) A= KM B) KM = A/2 C) B = KM D) C = - KM E) D= 1/ KM

C) B = KM

Parallel lines on a Lineweaver-Burk plot indicate I. an increase in KM. II. decrease in KM. III. decrease in Vmax. IV. uncompetitive inhibition. A) I, IV B) II, III, IV C) I or II, III D) I or III, II E) I, III, IV

C) I or II, III

A lead compound would be most promising if it had: A) KI = 4.7 × 105 M. B) KI = 1.5 × 108 M. C) KI = 1.5 × 10-8 M. D) KI = 4.7 × 10-5 M. E) KM = 4.7 × 105 M.

C) KI = 1.5 × 10-8 M

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

C) Phase 2; single blind

In order for an enzymatic reaction obeying the Michaelis-Menten equation to reach 3/4 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

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

C) k1 >> k2

The following questions (33 and 34) refer to the diagram (with boxes where it has been left incomplete): 32. This diagram refers to a (an) A) Ping Pong reaction. B) ordered bisubstrate reaction. C) random bisubstrate reaction. D) double order ping pong reaction E) X, Y, and Z must be provided in order to answer correctly

C) random bisubstrate reaction.

Allosteric activators A) bind via covalent attachment. B) stabilize conformations with higher Ks. C) stabilize conformations with higher substrate affinity. D) all of the above E) none of the above.

C) stabilize conformations with higher substrate affinity.

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 of drugs to water soluble, excretable compounds. E) all of the above

C) the phosphorylation of a wide variety of proteins.

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) 1.33 × 105 M-1•sec-1 B) 1.33 × 105 M•sec C) 7.5 × 10-2 M•sec D) 4.8 × 102 M•sec-1 E) Not enough data are given to make this calculation

D) 4.8 × 102 M•sec-1

Based on the figures below, which of the following expressions would be correct? A) Vmax = 1/B B) C = 1/ Vmax C) D= Vmax D) D = 1/ Vmax E) A = 1/ Vmax

D) D = 1/ Vmax

A Lineweaver-Burk plot is also referred to as I. a sigmoidal plot. II. a linear plot. III. a Michaelis-Menten plot. IV. a double reciprocal plot. A) II B) II, III C) IV D) II, IV E) III, IV

D) II, IV

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) KM = [0.006]; Vmax = 0.0075/s B) KM = [0.196]; Vmax = 0.0075/s C) KM = [165]; Vmax = 33/s D) KM = [33]; Vmax = 167/s E) KM = [270]; Vmax x = 68/s

D) KM = [33]; Vmax = 167/s

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

D) a competitive inhibitor

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-saturates the enzyme. E) a ratio of substrate concentration relative to catalytic power.

D) the [S] that half-saturates the enzyme.

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

E) ES breakdown occurs at the same rate as ES formation

Compounds that function as "mixed inhibitors" I. interfere with substrate binding to the enzyme. II. bind to the enzyme reversibly. III. can bind to the enzyme/substrate complex. A) I B) II C) III D) II, III E) I, II, III

E) I, II, III

Enzyme activity in cells is controlled by which of the following? I. covalent modifications II. modulation of expression levels III. feedback inhibition IV. allosteric effectors A) I B) II C) III D) III, IV E) I, II, III, IV

E) I, II, III, IV

An extremely efficient enzyme called "efficase" catalyzes the conversion of "A" to "B." A researcher decides to mutate the enzyme in order to try to improve its performance. Following active site mutations, a significant reduction in the value of KM and Vmax was observed. Which of the following may have occurred? A) The affinity of the enzyme for the substrate was increased to a point which did not favor propagation (continuation) of the reaction. B) The decrease in Vmax was not related to the decrease in KM. C) If the reaction was first-order, the change in KM cannot have affected Vmax. D) The stability of E+S (E+A as written above) was increased, thereby increasing the KM. E) The reverse reaction (breakdown of EA to E+A) was favored, slowing the Vmax.

E) The reverse reaction (breakdown of EA to E+A) was favored, slowing the Vmax.

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

E) none of the above

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

E) none of the above.


Kaugnay na mga set ng pag-aaral

MANAGEMENT EXAM - MANAGEMENT BY OBJECTIVES (MOB)

View Set

1 - Basic Principles of Life and Health Insurance and Annuities

View Set

MOET - Grade 8 - Unit 3 - Grammar 1 - Must vs. have to vs. ought to

View Set

Chapters 7-9 Supply Chain Questions

View Set

Team Response Scenario: Bill Goodman

View Set

Biology Chapter 12 - Understand and Apply Questions, Chapter 15 Quiz, Biology Test 2 Test Questions:, BIO Test #1 Answer key, Bio test #3 !, Biology 1 Exam 4

View Set