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7.9Problems

 

 

Outline two ways in which the characteristics of enzymes differ from those of inorganic catalysts, and explain these characteristics in relation to the structure of enzymes.

Enzymes catalyze biochemical reactions.

1)

Explain how enzymes facilitate biochemical reactions.

2)

Describe the structure of the active site of enzymes.

3)

Explain the meanings of Km and Vmax in the Michaelis-Menten equation.

4)

If Km of substrate A for an enzyme is 0.2 mM and that of substrate B is 0.02 mM, which has a higher affinity for the enzyme?

It is assumed that protein A and protein B form the complex AB. Consider a case in which the concentrations of A, B and AB in a cell are each 1 μM and they exist in a state of equilibrium.

1)

Calculate the equilibrium constant for the reaction A + B ⇄ AB. Care should be taken with the units used.

2)

If the concentrations of A, B and AB are all one thousandth (1 nM) and they are in a state of equilibrium, indicate how the equilibrium constant changes compared with 1).

 

 

Glycolysis and the citric acid cycle can be understood as mechanisms that release the entire carbon backbone of glucose as CO2 and store the free energy contained in glucose as ATP and NADH. Glycolysis is regulated by ATP (final product), acetyl-CoA (intermediate), citric acid, etc. Identify two of the main enzymes involved in this regulation and briefly explain their regulation mechanisms.

 

 

Briefly explain the principles of allosteric enzymes and allosteric regulation.

 

 

Outline how the activity of trypsin changes if the serine residue in its active center is replaced with alanine (assuming that the overall structure of the protein does not significantly change). Also, if this variant enzyme and normal trypsin are mixed at a ratio of 1:1, explain how the apparent Vmax and Km values change provided that a fixed amount of enzyme is used.

 

 

Outline two ways in which the characteristics of enzymes differ from those of inorganic catalysts, and explain these characteristics in relation to the structure of enzymes.

Each enzyme has a unique tertiary structure, which limits the substrates that the active center of the enzyme can come into contact with. This is one of the characteristics of enzymes, and is known as substrate specificity. The amino acids in the active center of an enzyme exhibit activity unique to that enzyme, and enzymes normally catalyze only one reaction type. This is known as reaction specificity.

Enzymes catalyze biochemical reactions.

1)

Explain how enzymes facilitate biochemical reactions.

Enzymes facilitate chemical reactions by reducing the amount of activation energy they require.

2)

Describe the structure of the active site of enzymes.

The tertiary structure made of peptide bonds forms an active site (a keyhole) that is entered by a specific substrate (a key). Low-molecular substances, such as metal ions, may be needed at active sites.

3)

Explain the meanings of Km and Vmax in the Michaelis-Menten equation.

Km: a scale of affinity between an enzyme and a substrate. Vmax: since Vmax = kcat [E0], higher values of Vmax mean a higher reaction rate; in other words, the catalytic effect of the enzyme at a particular unit concentration is higher.

4)

If Km of substrate A for an enzyme is 0.2 mM and that of substrate B is 0.02 mM, which has a higher affinity for the enzyme?

The affinity of substrate B is higher.

It is assumed that protein A and protein B form the complex AB. Consider a case in which the concentrations of A, B and AB in a cell are each 1 μM and they exist in a state of equilibrium.

1)

Calculate the equilibrium constant for the reaction A + B ? AB. Care should be taken with the units used.

Equilibrium constant K = [AB] / [A][B] = 10-6 / 10-6 x 10-6 = 106M-1

2)

If the concentrations of A, B and AB are all one thousandth (1 nM) and they are in a state of equilibrium, indicate how the equilibrium constant changes compared with 1).

The equilibrium constant is calculated in accordance with 1), producing K = 109M-1. The constant is therefore 1,000 times higher.

 

 

Glycolysis and the citric acid cycle can be understood as mechanisms that release the entire carbon backbone of glucose as CO2 and store the free energy contained in glucose as ATP and NADH. Glycolysis is regulated by ATP (final product), acetyl-CoA (intermediate), citric acid, etc. Identify two of the main enzymes involved in this regulation and briefly explain their regulation mechanisms.

Phosphofructokinase (PFK: see Fig. 7-4): allosterically inhibited by ATP and citric acid and allosterically activated by AMP.
Pyruvate kinase (PK: see Fig. 7-4): allosterically inhibited by ATP and acetyl-CoA and activated by fructose biphosphate.
Pyruvate dehydrogenase (PDH: see the text in 7.5): allosterically inhibited by CoA and NADH. In addition, it is phosphorylated by pyruvate dehydrogenase kinase (which is activated by CoA, NADH and ATP), thereby lowering its activity.

 

 

Briefly explain the principles of allosteric enzymes and allosteric regulation.

Allosteric control is a mechanism in which an effector binds to an enzyme site other than the active center, which changes the conformation of the enzyme, thus altering its affinity to the effector and the catalytic activity of the enzyme; enzymes regulated by this mechanism are called allosteric enzymes. See in 7.7.

 

 

Outline how the activity of trypsin changes if the serine residue in its active center is replaced with alanine (assuming that the overall structure of the protein does not significantly change). Also, if this variant enzyme and normal trypsin are mixed at a ratio of 1:1, explain how the apparent Vmax and Km values change provided that a fixed amount of enzyme is used.

The amino acids in the active center play an important role in terms of enzymatic activity, and cannot be substituted with other amino acids in many cases. In the case of trypsin, it loses activity if the serine in the active center is replaced with alanine. However, this amino acid replacement does not change the tertiary structure of trypsin, and it is considered that the enzyme can still bind to substrates. The Vmax value of the enzyme mixture therefore becomes half of the original, but its Km level does not change.

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