Protein Techniques exam 2

1. What does thermodynamics primarily tell us about a chemical reaction?

A. How quickly the reaction occurs

B. Whether the reaction is energetically favorable

C. Which enzyme catalyzes the reaction

D. How much enzyme is present

B. Whether the reaction is energetically favorable

What does thermodynamics primarily tell us about a chemical reaction?

Whether the reaction is energetically favorable

1. A competitive inhibitor of an enzyme is usually:

a. a highly reactive compound.

b. a metal ion such as Hg2+ or Pb2+.

c. structurally similar to the substrate.

d. water insoluble.

e. a poison.

c. structurally similar to the substrate.

catalytic triad composed of three amino acids:

serine, histidine, asparatic acid

Serine of catalytic triad

Acts as the nucleophile that attacks the peptide bond.

Acts as the nucleophile that attacks the peptide bond.

Serine of catalytic triad

Histidine of catalytic triad

Acts as a general base and acid to facilitate proton transfers.

Acts as a general base and acid to facilitate proton transfers.

Histidine of catalytic triad

Aspartate of catalytic triad .

Stabilizes the histidine residue through hydrogen bonding

Stabilizes the histidine residue through hydrogen bonding

Aspartate of catalytic triad .

6. The transition state of a catalyzed reaction is lower in energy than that of the uncatalyzed reaction

a) because of enthalpic interactions between the enzyme and the transition state.

b) because of favorable interactions with the substrate.

c) because of a smaller delta S between the [ES] and [EX] complex.

d) because enzymes only work at low temperatures.

e) both a and c are correct

e) both a and c are correct

6. The transition state of a catalyzed reaction is lower in energy than that of the uncatalyzed reaction

because of enthalpic interactions between the enzyme and the transition state.

because of a smaller delta S between the [ES] and [EX] complex.

enthalpic interactions between the enzyme and the transition state. is why the

transition state of a catalyzed reaction is lower in energy

Rate of formation of ES

k1 [E][S]

k1 [E][S] is what

Rate of formation of ES

k-1 is what

dissociation of enzyme from substrate

dissociation of enzyme from substrate

k-1 is what

Rate of dissociation of ES =

k2 [ES] + k-1 [ES]

k2 [ES] + k-1 [ES] is what

Rate of dissociation of ES

Factors affecting Km

pH

Temperature

Ionic strengths

Nature of substrate

Which substrate has the highest affinity to elastase?

PAPA|A

If a mixture of the three substrates was presented to elastase with the concentration of each peptide equal to 0.5mM, which would be digested most rapidly? Which most slowly? (Assume enzyme is present in excess.)

With enzyme in excess, the initial digestion rate for each substrate is proportional to (kcat}}[S]/(Km+[S]). At equal ([S]=0.5\) mM:

Fastest: PAPA — (37(0.5)/(1.5+0.5)=9.25\)

Slowest: PAPI — (12(0.5)/(5.2+0.5)=1.05\)

Elastase is closely related to chymotrypsin. Suggest two types/features of residues you might find at or near the active site.

Like chymotrypsin, elastase likely has the conserved serine-protease catalytic triad—Ser, His, and Asp—and an oxyanion hole with backbone NH groups that stabilize the tetrahedral intermediate. Elastase’s specificity pocket is relatively shallow/narrow, which favors small side chains.

Based on these data, suggest what features of amino acid sequence dictate the specificity of proteolytic cleavage by elastase.

The residues flanking the cleaved bond matter, especially the residue immediately N-terminal to it (the P1 residue). Elastase favors small, neutral residues there, such as alanine; a bulkier hydrophobic residue such as isoleucine is less favorable. Neighboring residues also influence cleavage efficiency.

KM,app​:

measured km when inhibitor is present

equation for apparent \(K_M\) in competitive inhibition

Kmapp: Km(1+i/ki)

for an competitive inhibitor ic50 equals

ki(1+s/km)

measured km when inhibitor is present

KM,app​

The catalytic efficiency of many enzymes depends on pH. Chymotrypsin (active site shown below) shows a maximum value of kcat/Km at pH 8. Detailed analysis shows that kcat increases rapidly between pH 6 and 7 and remains constant at higher pH. Km increases rapidly between pH 8 and 10. Suggest explanations for these observations.

The change in kcat between pH 6 and 7 must involve the loss of a proton in the active site that affects catalytic rate. (The best candidate is His57; pKa 6.04.) The increase in Km at higher pH must involve a change in binding at the active site. (The best candidate may be deprotonation of the amide groups of Gly193 & Ser195 that hold the substrates carbonyl group in place during catalysis. There may also be other R-groups not shown that would deprotonate under these conditions that help to hold the substrate in place.)

The table below shows the result of an experiment designed to determine the apparent Michaelis

constant (KM) and the maximum velocity (Vmax) for a particular enzyme. The kinetic data shows how

the initial rate of reaction, measured in a spectrophotometer as the change in absorption (A) per

minute, varies as the substrate concentration is increased. Use this information to determine the value

of KM and Vmax for the enzyme by plotting the data using a suitable graphical method.

The kinetic parameters can be calculated from the properties of the linear (Lineweaver-Burk) graph

as follows. The x-intercept is equal to -1/KM while the y-intercept is 1/Vmax. Since -1/KM = -0.192, KM =

5.217 mM (the reciprocal of -1/KM multiplied by –1). And 1/Vmax = 1.268 so Vmax = 0.789 mM min-1.

NAD, coA, vitamins are all

Coenzymes

Coenzymes examples

NAD, coA, vitamins

coA is a

conenzyme

Zn++ is a

cofactor

The higher the Vmax the

more substrate an enzyme can process per min

the more substrate an enzyme can process per min when

The higher the Vmax

Most enzymes operating in the human body work best at a temperature of

37°C.

allopurinol is an example what

suicide inhibitor

aspirin

penicillin

DIPF (nerve gas)

allopurinol

are examples of what

suicide inhibitor

examples of suicide inhibitor

aspirin

penicillin

DIPF (nerve gas)

allopurinol

An enzymatic reaction has a baseline maximum velocity of 120 umol/min. When a non-competitive inhibitor is added at a concentration of 15uM, the apparent maximum velocity (Vmax-app) drops to 30umol/min. Calculate the inhibition constant (Ki) of this inhibitor

0.168uM

1. A competitive inhibitor of an enzyme is usually:

structurally similar to the substrate.

which inhibitor is structurally similar to the substrate.

competitive inhibitor

2. A non-competitive inhibitor of an enzyme catalyzed reaction

a. binds to the Michaelis complex (ES).

b. decreases Vmax.

c. is without effect at saturating substrate concentrations.

d. can actually increase reaction velocity in rare cases.

e. The first and second choices are both correct.

e. The first and second choices are both correct.

2. A non-competitive inhibitor of an enzyme catalyzed reaction

a. binds to the Michaelis complex (ES).

b. decreases Vmax.

which inhibitor binds to the Michaelis complex (ES).

non-competitive inhibitor

3. A competitive inhibitor has the following effect on a Lineweaver-Burke (double reciprocal) Plot.

a. It moves the entire curve to the right.

b. It moves the entire curve to the left.

c. It changes the y-intercept.

d. It changes the x-intercept.

e. It has no effect on the slope.

d. It changes the x-intercept.

3. A competitive inhibitor has the following effect on a Lineweaver-Burke (double reciprocal) Plot.

It changes the x-intercept.

Which inhibitor has this effect: It changes the x-intercept.

competitive inhibitor

4. The degree of inhibition α by a competitive inhibitor is obtained from

a. measurement of Vmax.

b. measurement of the y-intercept on a Limeweaver-Burke Plot.

c. measurement of KM.

d. crystallographic studies.

e. is unrelated to the binding affinity of the inhibitor to the enzyme.

c. measurement of KM.

4. The degree of inhibition α by a competitive inhibitor is obtained from

measurement of KM.

5. An allosteric inhibitor of an enzyme usually

a. binds to the active site.

b. participates in feedback regulation.

c. denatures the enzyme.

d. causes the enzyme to work faster.

e. is a hydrophobic compound.

b. participates in feedback regulation.

5. An allosteric inhibitor of an enzyme usually

b. participates in feedback regulation.

participates in feedback regulation, which inhibitor

allosteric inhibitor

1. Which of the following best describes the assumption made in steady state kinetic analysis?

a) The concentration of [S] is decreasing.

b) The concentration of [ES] is constant.

c) Both a and b are correct.

d) Neither a nor b are correct.

e) The total amount of enzyme decreases.

b) The concentration of [ES] is constant.

1. Which of the following best describes the assumption made in steady state kinetic analysis?

The concentration of [ES] is constant.

2. When [S]=KM, the velocity of an enzyme catalyzed reaction is about:

a) 0.1*VMAX.

b) 0.2*VMAX.

c) 0.3*VMAX.

d) 0.5*VMAX.

e) 0.9*VMAX.

d) 0.5*VMAX.

2. When [S]=KM, the velocity of an enzyme catalyzed reaction is about:

0.5*VMAX.

3. The substrate KM in an enzyme-catalyzed reaction

a) is usually less than Kd, the dissociation constant.

b) is never less than Kd.

c) cannot be equal to Kd.

d) is estimated from the Y-intercept of a Lineweaver-Burk plot.

e) cannot be derived from simple rate equations.

b) is never less than Kd.

3. The substrate KM in an enzyme-catalyzed reaction is never

less than Kd.

The what? in an enzyme-catalyzed reaction is never less than Kd.

substrate KM

4. In an enzyme catalyzed reaction, provides information on and provides information on .

a) KM, chemical step, VMAX, substrate binding.

b) VMAX, chemical step, kCAT, substrate binding.

c) KD, substrate binding, kCAT, chemical step.

d) KM, substrate binding, VMAX, chemical step.

e) kCAT, substrate binding, VMAX, chemical step.

d) KM, substrate binding, VMAX, chemical step.

4. In an enzyme catalyzed reaction, provides information on and provides information on .

KM, substrate binding, VMAX, chemical step.

In an enzyme catalyzed reaction, KM provides information on

substrate binding

In an enzyme catalyzed reaction, VMAX provides information on

chemical step

5. The initial velocity, vo, of an enzyme catalyzed reaction reaches Vmax

a) at [S] = KM.

b) at [S] = 10*KM.

c) at 1/[S] = 1/KM.

d) only as 1/[S] --> 0.

e) The third and fourth choices are both correct.

d) only as 1/[S] --> 0.

5. The initial velocity, vo, of an enzyme catalyzed reaction reaches Vmax

only as 1/[S] --> 0.

6. The of a double reciprocal plot equals and the equals .

a) slope, VMAX, y-intercept, KM.

b) slope, KM, y-intercept, 1/VMAX.

c) slope, KM/VMAX, y-intercept, 1/VMAX.

d) slope, KM/VMAX, y-intercept, VMAX.

e) slope, KM/VMAX, x-intercept, VMAX.

c) slope, KM/VMAX, y-intercept, 1/VMAX.

6. The of a double reciprocal plot equals and the equals .

slope, KM/VMAX, y-intercept, 1/VMAX.

The what of a double reciprocal plot equals KM/VMAX and the y-intercept equals 1/VMAX.

slope

The slope of a double reciprocal plot equals what and the y-intercept equals 1/VMAX.

KM/VMAX

The slope of a double reciprocal plot equals KM/VMAX and the what equals 1/VMAX.

y-intercept

The slope of a double reciprocal plot equals KM/VMAX and the y-intercept equals what

1/VMAX

The slope of a double reciprocal plot equals

KM/VMAX

The what of a double reciprocal plot equals KM/VMAX

slope

7. When [S] = 0.1*KM, the velocity of an enzyme catalyzed reaction is about:

a) 0.1*Vmax.

b) 0.3*Vmax.

c) 0.5*Vmax.

d) 0.7*Vmax.

e) 0.9*Vmax.

a) 0.1*Vmax.

7. When [S] = 0.1*KM, the velocity of an enzyme catalyzed reaction is about:

0.1*Vmax.

8. When [S] = 10*KM, the velocity of an enzyme catalyzed reaction is about:

a) 0.1*Vmax.

b) 0.3*Vmax.

c) 0.5*Vmax.

d) 0.7*Vmax.

e) 0.9*Vmax.

e) 0.9*Vmax.

8. When [S] = 10*KM, the velocity of an enzyme catalyzed reaction is about:

0.9*Vmax.

2. What does kinetics primarily describe?

A. Whether a reaction is thermodynamically favorable

B. The equilibrium constant only

C. How fast a reaction occurs

D. The structure of DNA

C. How fast a reaction occurs

2. What does kinetics primarily describe?

How fast a reaction occurs

3. A reaction has ΔG < 0. What does this mean?

A. The reaction cannot occur

B. The reaction is thermodynamically favorable

C. The reaction must occur immediately

D. The reaction requires an enzyme

B. The reaction is thermodynamically favorable

3. A reaction has ΔG < 0. What does this mean?

The reaction is thermodynamically favorable

4. Why might a thermodynamically favorable reaction occur very slowly?

A. Its products are always unstable

B. It has a high activation-energy barrier

C. ΔG is always positive

D. Enzymes prevent favorable reactions

B. It has a high activation-energy barrier

4. Why might a thermodynamically favorable reaction occur very slowly?

It has a high activation-energy barrier

5. What is activation energy?

A. Energy released when products form

B. Energy stored permanently in products

C. Energy required to destabilize the existing state and reach the transition state

D. Energy required to destroy an enzyme

C. Energy required to destabilize the existing state and reach the transition state

5. What is activation energy?

Energy required to destabilize the existing state and reach the transition state

6. What happens when an enzyme catalyzes a reaction?

A. ΔG becomes more negative

B. ΔG becomes zero

C. Activation energy decreases

D. The equilibrium changes permanently

C. Activation energy decreases

6. What happens when an enzyme catalyzes a reaction?

Activation energy decreases

7. Which statement about enzymes is correct?

A. Enzymes change the thermodynamic favorability of a reaction

B. Enzymes increase activation energy

C. Enzymes lower activation energy

D. Enzymes are permanently consumed

C. Enzymes lower activation energy

7. Which statement about enzymes is correct?

Enzymes lower activation energy

8. Why do enzymes allow many reactions to occur at reasonable temperatures?

A. They eliminate all molecular collisions

B. They lower the activation-energy barrier

C. They increase ΔG

D. They prevent substrate binding

B. They lower the activation-energy barrier

8. Why do enzymes allow many reactions to occur at reasonable temperatures?

They lower the activation-energy barrier

9. For a chemical reaction to occur, molecules generally need to:

A. Collide randomly at any angle

B. Avoid each other

C. Meet, have appropriate orientation, and have sufficient energy

D. Always be at 0°C

C. Meet, have appropriate orientation, and have sufficient energy

9. For a chemical reaction to occur, molecules generally need to:

Meet, have appropriate orientation, and have sufficient energy

10. Which is an example of a productive collision?

A. Molecules collide with insufficient energy

B. Molecules collide in an orientation that allows bonds to form or break

C. Molecules never interact

D. Molecules collide but are permanently separated

B. Molecules collide in an orientation that allows bonds to form or break

10. Which is an example of a productive collision?

Molecules collide in an orientation that allows bonds to form or break

11. What is one major advantage of an enzyme's active site?

A. It prevents substrates from interacting

B. It brings substrates together in a favorable orientation

C. It increases substrate randomness

D. It destroys substrate before binding

B. It brings substrates together in a favorable orientation

11. What is one major advantage of an enzyme's active site?

It brings substrates together in a favorable orientation

12. The reactants that bind to an enzyme are called:

A. Products

B. Cofactors

C. Substrates

D. Coenzymes

C. Substrates