Question 1

Give the Michaelis-Menten equation and define each term in it. Does this equation apply to all enzymes? If not, to which kind does it not apply?

Accepted Answer

The Michaelis-Menten equation is given by:

V = (Vmax * [S]) / (Km + [S])

Where:
- V is the initial velocity of the reaction
- Vmax is the maximum velocity of the reaction
- [S] is the substrate concentration
- Km is the Michaelis constant, which is the substrate concentration at which the reaction velocity is half of Vmax

This equation applies to enzymes that follow Michaelis-Menten kinetics, which are enzymes that exhibit saturation kinetics and have a characteristic hyperbolic curve when plotting reaction velocity against substrate concentration. Not all enzymes follow this kinetics, particularly enzymes that exhibit cooperative kinetics or allosteric regulation. These enzymes do not follow the Michaelis-Menten equation and require different kinetic models to describe their behavior.

Question 2

Why is a transition-state analog not necessarily the same as a competitive inhibitor?

Accepted Answer

A transition-state analog is a molecule that mimics the structure of a substrate in its transition state during an enzymatic reaction. The transition state is a high-energy, unstable state that occurs during the conversion of substrates into products. Transition-state analogs are designed to bind to the enzyme's active site with high affinity because enzymes are often most complementary to the transition state of the reaction they catalyze. By resembling the transition state, these analogs can act as potent inhibitors of the enzyme.

On the other hand, a competitive inhibitor is a molecule that competes with the substrate for binding to the active site of an enzyme. It resembles the substrate's structure (not the transition state) and thus can occupy the active site, preventing the actual substrate from binding. Competitive inhibitors typically reduce the rate of the reaction by decreasing the amount of enzyme available to bind to the substrate.

The key difference between a transition-state analog and a competitive inhibitor lies in what they mimic and how they interact with the enzyme:

1. Structural Mimicry:
   - Transition-state analog: Mimics the high-energy transition state of the substrate.
   - Competitive inhibitor: Mimics the substrate itself in its ground state.

2. Binding Affinity:
   - Transition-state analogs often bind with much higher affinity to the enzyme than either the substrate or competitive inhibitors because enzymes are highly optimized to stabilize the transition state.
   - Competitive inhibitors generally have a lower affinity compared to transition-state analogs because they are competing with the substrate for the active site rather than exploiting the enzyme's highest affinity state.

3. Inhibition Mechanism:
   - Transition-state analogs inhibit the enzyme by stabilizing the enzyme in a form that is complementary to the transition state, effectively "trapping" the enzyme in a non-productive complex.
   - Competitive inhibitors simply block the active site, preventing substrate binding without necessarily stabilizing a transition-state-like conformation of the enzyme.

4. Overcoming Inhibition:
   - The inhibition caused by transition-state analogs is often very difficult to overcome because of their high affinity for the enzyme.
   - Competitive inhibition can often be overcome by increasing the concentration of the substrate, which can outcompete the inhibitor for the active site.

In summary, while both transition-state analogs and competitive inhibitors can prevent an enzyme from catalyzing a reaction, they do so by different mechanisms and with different degrees of affinity and specificity. Transition-state analogs are specialized inhibitors that exploit the unique high-affinity state of the enzyme for the transition state, while competitive inhibitors simply block substrate access to the active site.

Question 3

Enzymes with a kcat /Km ratio of about 108 M-1s-1 are considered to show optimal catalytic efficiency. Fumarase, which catalyzes the reversible-dehydration reaction   has a ratio of turnover number to the Michaelis-Menten constant, (kcat/Km) of 1.6 × 108 for the substrate fumarate and 3.6 × 107 for the substrate malate. Because the turnover number for both substrates is nearly identical, what factors might be involved that explain the different ratio for the two substrates?

Accepted Answer

not answered

Question 4

The Mg2+ metal ions involved in the enolase reaction mechanism are required for which reason?

Accepted Answer

A)  They form covalent bonds with a porphoryin ring to coordinate the substrate. 
B)  They enhance the electron-withdrawing potential of the carbonyl group of 2-phosphoglycerate. 
C)  They stabilize the phosphate group of 2-phosphoglycerate. 
D)  They allow a Lys in the active site to donate a proton to the 2-phosphoglycerate substrate. 
E)  They prevent the reverse catalysis of 3-phosphoglycerate to 2-phosphoglycerate from taking place. 
A)  They form covalent bonds with a porphoryin ring to coordinate the substrate. 
B)  They enhance the electron-withdrawing potential of the carbonyl group of 2-phosphoglycerate. 
C)  They stabilize the phosphate group of 2-phosphoglycerate. 
D)  They allow a Lys in the active site to donate a proton to the 2-phosphoglycerate substrate. 
E)  They prevent the reverse catalysis of 3-phosphoglycerate to 2-phosphoglycerate from taking place.

Question 5

Michaelis-Menten kinetics is sometimes referred to as "saturation" kinetics. Why?

Accepted Answer

Michaelis-Menten kinetics is often refer

Question 6

Vmax for an enzyme-catalyzed reaction:

Accepted Answer

A)  generally increases when pH increases. 
B)  increases in the presence of a competitive inhibitor. 
C)  is limited only by the amount of substrate supplied. 
D)  is twice the rate observed when the concentration of substrate is equal to the Km. 
E)  is unchanged in the presence of a uncompetitive inhibitor. 
A)  generally increases when pH increases. 
B)  increases in the presence of a competitive inhibitor. 
C)  is limited only by the amount of substrate supplied. 
D)  is twice the rate observed when the concentration of substrate is equal to the Km. 
E)  is unchanged in the presence of a uncompetitive inhibitor.

Question 7

In the following diagram of the first step in the reaction catalyzed by the protease chymotrypsin, the process of general base catalysis is illustrated by the number _____, and the process of covalent catalysis is illustrated by the number _____.

Accepted Answer

A)  1; 2 
B)  1; 3 
C)  2; 3 
D)  2; 1 
E)  3; 2 
A)  1; 2 
B)  1; 3 
C)  2; 3 
D)  2; 1 
E)  3; 2

Question 8

A good transition-state analog:

Accepted Answer

A)  binds covalently to the enzyme. 
B)  binds to the enzyme more tightly than the substrate. 
C)  binds very weakly to the enzyme. 
D)  is too unstable to isolate. 
E)  must be almost identical to the substrate. 
A)  binds covalently to the enzyme. 
B)  binds to the enzyme more tightly than the substrate. 
C)  binds very weakly to the enzyme. 
D)  is too unstable to isolate. 
E)  must be almost identical to the substrate.

Question 9

An enzyme follows Michaelis-Menten kinetics. Indicate (with an "x") which of the kinetic parameters at the left would be altered by the following factors. Give only one answer for each.

Accepted Answer

The answer of An enzyme follows Michaelis-Menten kinetics. Indicate (with...

Question 10

Write an equilibrium expression for the reaction S $\rightarrow$  P and briefly explain the relationship between the value of the equilibrium constant and free energy.

Accepted Answer

The equilibrium expression for a chemica

Question 11

An enzyme can catalyze a reaction with either of two substrates, S1 or S2. The Km for S1 was found to be 2.0 mM, and the Km, for S2 was found to be 20 mM. A student determined that the Vmax was the same for the two substrates. Unfortunately, he lost the page of his notebook and needed to know the value of Vmax. He carried out two reactions: one with 0.1 mM S1, the other with 0.1 mM S2. Unfortunately, he forgot to label which reaction tube contained which substrate. Determine the value of Vmax from the results he obtained:
Tube number
Rate of formation of product
1
0.5
2
4.8
3
4
5

Accepted Answer

To determine the value of V_max from the

Question 12

Both water and glucose share an -OH that can serve as a substrate for a reaction with the terminal phosphate of ATP catalyzed by hexokinase. Glucose, however, is about a million times more reactive as a substrate than water. The BEST explanation is that:

Accepted Answer

A)  glucose has more -OH groups per molecule than does water. 
B)  the larger glucose binds better to the enzyme; it induces a conformational change in hexokinase that brings active-site amino acids into position for catalysis. 
C)  the -OH group of water is attached to an inhibitory H atom, while the glucose -OH group is attached to C. 
D)  water and the second substrate, ATP, compete for the active site resulting in a competitive inhibition of the enzyme. 
E)  water normally will not reach the active site because it is hydrophobic. 
A)  glucose has more -OH groups per molecule than does water. 
B)  the larger glucose binds better to the enzyme; it induces a conformational change in hexokinase that brings active-site amino acids into position for catalysis. 
C)  the -OH group of water is attached to an inhibitory H atom, while the glucose -OH group is attached to C. 
D)  water and the second substrate, ATP, compete for the active site resulting in a competitive inhibition of the enzyme. 
E)  water normally will not reach the active site because it is hydrophobic.

Question 13

Which statement regarding enzyme activity is CORRECT?

Accepted Answer

A)  Enzymes bind their substrates better than the transition state. 
B)  Enzymes bind the transition state better than the reaction products. 
C)  Enzymes reduce the activation energy required for the reaction to take place. 
D)  Enzymes bind their substrates better than the transition state and the transition state better than the reaction products. 
E)  Enzymes bind the transition state better than the reaction products and reduce the activation energy required for the reaction to take place. 
A)  Enzymes bind their substrates better than the transition state. 
B)  Enzymes bind the transition state better than the reaction products. 
C)  Enzymes reduce the activation energy required for the reaction to take place. 
D)  Enzymes bind their substrates better than the transition state and the transition state better than the reaction products. 
E)  Enzymes bind the transition state better than the reaction products and reduce the activation energy required for the reaction to take place.

Question 14

One of the enzymes involved in glycolysis, aldolase, requires Zn2+ for catalysis. Under conditions of zinc deficiency, when the enzyme may lack zinc, it would be referred to as the:

Accepted Answer

A)  apoenzyme. 
B)  coenzyme. 
C)  holoenzyme. 
D)  prosthetic group. 
E)  substrate. 
A)  apoenzyme. 
B)  coenzyme. 
C)  holoenzyme. 
D)  prosthetic group. 
E)  substrate.

Question 15

For a reaction that can take place with or without catalysis by an enzyme, what would be the effect of the enzyme on the:
(a) standard free energy change of the reaction?
(b) activation energy of the reaction?
(c) initial velocity of the reaction?
(d) equilibrium constant of the reaction?

Accepted Answer

(a) The standard free energy change of t

Question 16

If chemical reactions will eventually reach an equilibrium state, what is the purpose of enzymes in a biological system?

Accepted Answer

A)  Enzymes are consumed to speed up chemical reactions. 
B)  Enzymes speed up chemical reactions without being used up in the process. 
C)  Enzymes slow down chemical reactions. 
D)  Enzymes alter the equilibrium state between reactants and products. 
E)  Enzymes prevent the formation of unstable reaction intermediates. 
A)  Enzymes are consumed to speed up chemical reactions. 
B)  Enzymes speed up chemical reactions without being used up in the process. 
C)  Enzymes slow down chemical reactions. 
D)  Enzymes alter the equilibrium state between reactants and products. 
E)  Enzymes prevent the formation of unstable reaction intermediates.

Question 17

Which amino acid is NOT capable using its side chain (R group) to participate in general acid-base catalysis?

Accepted Answer

A)  Asp 
B)  His 
C)  Ser 
D)  Val 
E)  Lys 
A)  Asp 
B)  His 
C)  Ser 
D)  Val 
E)  Lys

Question 18

An enzyme-catalyzed reaction was carried out with the substrate concentration initially a thousand times greater than the Km for that substrate. After 9 minutes, 1% of the substrate had been converted to product, and the amount of product formed in the reaction mixture was 12 $\mu$mol. If, in a separate experiment, one-third as much enzyme and twice as much substrate had been combined, how long would it take for the same amount (12$\mu$mol) of product to be formed?

Accepted Answer

A)  1.5 min 
B)  13.5 min 
C)  27 min 
D)  3 min 
E)  6 min 
A)  1.5 min 
B)  13.5 min 
C)  27 min 
D)  3 min 
E)  6 min

Question 19

The following data were obtained in a study of an enzyme known to follow Michaelis-Menten kinetics: V0
Substrate added
($\mu$mol/min)
(mmol/L)
217
0)8
325
2
433
4
488
6
647
1,000
The Km for this enzyme is approximately:

Accepted Answer

A)  1 mM. 
B)  1000 mM. 
C)  2 mM. 
D)  4 mM. 
E)  6 mM. 
A)  1 mM. 
B)  1000 mM. 
C)  2 mM. 
D)  4 mM. 
E)  6 mM.

Question 20

The combination of amoxicillin and clavulanic acid into the drug known as Augmentin is widely used as an antibacterial agent. The purpose of amoxicillin in this drug is to target _____, which catalyzes crosslinking of peptidoglycan, and the purpose of clavulanic acid in this drug is to target _____, which normally would inactivate the amoxicillin. Both of these compounds act as _____ inhibitors of enzyme function.

Accepted Answer

A)  transpeptidase; hexokinase; competitive 
B)  peptidoglycanase; $\beta$ lactamase; uncompetitive 
C)  transpeptidase; $\beta$ lactamase; suicide 
D)  glycanprotease; penicillinase; competitive 
E)  $\beta$ lactamase; lysozyme; suicide 
A)  transpeptidase; hexokinase; competitive 
B)  peptidoglycanase; $\beta$ lactamase; uncompetitive 
C)  transpeptidase; $\beta$ lactamase; suicide 
D)  glycanprotease; penicillinase; competitive 
E)  $\beta$ lactamase; lysozyme; suicide

Exam 6: Enzymes

Give the Michaelis-Menten equation and define each term in it. Does this equation apply to all enzymes? If not, to which kind does it not apply?

Why is a transition-state analog not necessarily the same as a competitive inhibitor?

The Mg2+ metal ions involved in the enolase reaction mechanism are required for which reason?

Michaelis-Menten kinetics is sometimes referred to as "saturation" kinetics. Why?

Vmax for an enzyme-catalyzed reaction:

In the following diagram of the first step in the reaction catalyzed by the protease chymotrypsin, the process of general base catalysis is illustrated by the number _____, and the process of covalent catalysis is illustrated by the number _____.

A good transition-state analog:

An enzyme follows Michaelis-Menten kinetics. Indicate (with an "x") which of the kinetic parameters at the left would be altered by the following factors. Give only one answer for each.

Write an equilibrium expression for the reaction S →\rightarrow→ P and briefly explain the relationship between the value of the equilibrium constant and free energy.

Both water and glucose share an -OH that can serve as a substrate for a reaction with the terminal phosphate of ATP catalyzed by hexokinase. Glucose, however, is about a million times more reactive as a substrate than water. The BEST explanation is that:

Which statement regarding enzyme activity is CORRECT?

One of the enzymes involved in glycolysis, aldolase, requires Zn2+ for catalysis. Under conditions of zinc deficiency, when the enzyme may lack zinc, it would be referred to as the:

For a reaction that can take place with or without catalysis by an enzyme, what would be the effect of the enzyme on the: (a) standard free energy change of the reaction? (b) activation energy of the reaction? (c) initial velocity of the reaction? (d) equilibrium constant of the reaction?

If chemical reactions will eventually reach an equilibrium state, what is the purpose of enzymes in a biological system?

Which amino acid is NOT capable using its side chain (R group) to participate in general acid-base catalysis?

The following data were obtained in a study of an enzyme known to follow Michaelis-Menten kinetics: V0 Substrate added ( μ\muμ mol/min) (mmol/L) 217 0)8 325 2 433 4 488 6 647 1,000 The Km for this enzyme is approximately:

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Amino Acid Oxidation and the Production of Urea

Oxidative Phosphorylation and Photophosphorylation

Carbohydrate Biosynthesis in Plants and Bacteria

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Hormonal Regulation and Integration of Mammalian Metabolism

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The Mg²⁺ metal ions involved in the enolase reaction mechanism are required for which reason?

V_max for an enzyme-catalyzed reaction:

In the following diagram of the first step in the reaction catalyzed by the protease chymotrypsin, the process of general base catalysis is illustrated by the number _, and the process of covalent catalysis is illustrated by the number _.

Write an equilibrium expression for the reaction S $\rightarrow$ P and briefly explain the relationship between the value of the equilibrium constant and free energy.

One of the enzymes involved in glycolysis, aldolase, requires Zn²⁺ for catalysis. Under conditions of zinc deficiency, when the enzyme may lack zinc, it would be referred to as the:

The following data were obtained in a study of an enzyme known to follow Michaelis-Menten kinetics: V₀ Substrate added ( $\mu$ mol/min) (mmol/L) 217 0)8 325 2 433 4 488 6 647 1,000 The K_m for this enzyme is approximately: