Chapter 6 Basic Concepts of Enzyme Action

Matching Questions

Use the following to answer questions 1-10: apoenzymes hydrolyases active site transition state spontaneous induced fit energy prosthetic group lock and key substrate(s) oxidoreductases equilibria

Choose the correct answer from the list below. Not all of the answers will be used.

The site on the enzyme where the reaction occurs.

The substance that the enzyme binds and converts to product.

Enzymes that do not have the required cofactor bound are called

A tightly bound cofactor might be called a(n) .

Enzymes will decrease the energy of activation but do not change the of a chemical reaction.

A reaction that is exergonic will be .

An endergonic reaction requires an input of to proceed.

Enzymes that transfer electrons are called _.

Enzymes that cleave molecules by addition of water are called _.

Which model is more appropriate to explain an enzyme binding to its substrate?

Fill-in-the-Blank Questions

194 Enzymes accelerate the rate of a chemical reaction by the free energyof activation of the reaction.

The difference between the standard-state free energy, ΔGº, and the biochemical standard-state free energy is that ΔGº refers to the standard free-energy change at .

196 An enzyme that loosely binds substrate will have a level of specificity.

197 Organic cofactors are referred to as

198 A reaction can occur spontaneously only if ΔG is

199 When ΔG for a system is zero, the system is at

200 An enzyme that has been stripped of small molecules needed for activity is called

201 The total change of free energy in a reaction depends on and

202 The difference in values for G and Go′ is in the .

Competitive inhibitors that mimic the substrate while in the transition state are called inhibitors.

Multiple-Choice Questions

What is the common strategy by which catalysis occurs? increasing the probability of product formation shifting the reaction equilibrium stabilization the transition state

All of the above.

None of the above

An enzyme will specifically bind its substrate because of a tight lock and key binding mechanism. a high number of hydrophobic amino acids in the center of the protein. a large number of weak interactions at the active site. additional nonprotein cofactors

None of the above.

Examples of cofactors include: biotin and thiamine pyrophosphate. pyridoxal phosphate and coenzyme A. B and C.

Zn+2, Mg+2, and Ni+2.

All of the above.

A cofactor is best defined as another protein a covalently bound inorganic molecule a small molecule that holds the substrate in the active site a molecule responsible for most of the catalytic activity of the enzyme

None of the above

Which of the following is true?

Enzymes force reactions to proceed in only one direction. Enzymes alter the equilibrium of the reaction. Enzymes alter the standard free energy of the reaction. All of the above.

None of the above

The Gibbs free energy of activation is: the difference between the substrate and the transition state. the difference between the substrate and the product the difference between the product and the transition state. All of the above.

None of the above.

At equilibrium, the Gibb’s free energy is _.

The rate of a reaction, or how fast a reaction will proceed, is best determined by

None of the above.

The relationship between Go′ and G is best described as . determined by the temperature described by changes in Keq differ from standard state to physiological or actual concentrations of reactants and products dependent on the reaction mechanism of the reaction differ only in terms of the types of reactions used for each value

For the two reactions a) A→B Go′ = 2 kJmol-1 and b) X→Y Go′ = –3.5 kJmol-1, which of the following statements is correct?

Reaction a is not spontaneous at cellular concentrations.

Reaction b will react very quickly.

Reaction a is a more thermodynamically favorable reaction than b. Neither reaction is reversible.

None of the above

A graph of product versus time (as in Fig. 6.2 in your textbook) for an enzyme is determined to be hyperbolic. Why does the amount of product level off as time increases? The reaction has reached equilibrium, that is, the forward and reverse reactions are occurring at a fixed rate. There is a product inhibition of the enzyme. The reaction runs out of reaction materials. The enzyme has finished accelerating the reaction. None of the above.

The free energy of activation is _. the amount of chemical energy available in the transition state the difference in free energy between the substrate and product the free energy gained by adding a catalyst the difference in free energy between the transition state and the substrate All of the above.

The molecular structure that is short-lived and neither substrate nor product is known as substrate analog transition state free energy stabilization state catalysis state equilibrium intermediate

Riboflavin is a water-soluble organic substance that is not synthesized by humans. Metabolically, it is chemically converted into a substance called flavin adenine dinucleotide, which is required by succinate dehydrogenase. Which of the following statements is most correct?

Riboflavin is a coenzyme. Flavin adenine dinucleotide is a vitamin. Succiniate dehydrogenase is a coenzyme. Flavin adenine dinucleotide is a coenzyme.

The active site of an enzyme . is a series of amino acids that bind the enzyme is a linear sequence of amino acids that react with each other binds covalently to the substrate allows water to enter into the solvate the substrate None of the above.

The conversion of glucose-6-phosphate to fructose-6-phosphate is catalyzed by an isomerase enzyme. Glucose-6-phosphate was mixed with the enzyme under standard conditions and the reaction was allowed to come to equilibrium. If the Keq′ is 0.50 and the equilibrium [glucose6-phosphate] is 1.43 M, what is the equilibrium [fructose-6-phosphate]?

The conversion of glucose-6-phosphate to fructose-6-phosphate is catalyzed by an isomerase enzyme. Glucose-6-phosphate was mixed with the enzyme under standard conditions and the reaction was allowed to come to equilibrium If the Keq′ is 0.50, what is the G°′ in kJ/mol?

The conversion of glucose-6-phosphate to fructose-6-phosphate is catalyzed by an isomerase enzyme. Under cellular conditions (37oC), the glucose-6-phosphate is 6.6 μM and the fructose-6-phosphate is 1.3 μM. If the Keq′ is 0.50, what is the ΔG in kJ/mol? (Hint: Use the the previous question.)

That many transition-state analogs bind more tightly than the native substrate reinforces the concept that: transition-state analogs are planar structures. transition-state analogs are highly charged at physiological pH. binding to the transition state is through a lock-and-key-mechanism. transition-state analogs are hydrophobic. binding to the transition state is through an induced-fit mechanism.

Short-AnswerQuestions

What is the relation between an enzyme-catalyzed reaction and the transition state of a reaction?

What is the difference between prosthetic groups and coenzymes?

How do enzymes facilitate the formation of the transition state?

How is the substrate bound to the active site?

You believe a substrate fits into a cleft like a key into a lock, but your roommate does not. Who is right?

In an enzymatic reaction in a test tube, the reaction will eventually reach equilibrium. Why does this not happen in living organisms?

How is free energy useful for understanding enzyme function?

While some enzymes have very specific substrates, others are more promiscuous. What would you suspect is the reason for this?

Multiple dilution and dialysis of a purified protein results in a loss of enzymatic activity. What might be the cause for this? Assume the structure of the protein is retained.

If Keq = 1, what is the G°′? If Keq >1, what is the G°′? If Keq <1, what is the G°′?

The free energy change (ΔG′) for the oxidation of the sugar molecules in a sheet of paper into CO2 and H2O is large and negative (the = G°′ – 2833 kJ/mol) Explain why paper is stable at room temperature in the presence of oxygen (O2).

The G°′ for the hydrolysis of ATP to ADP + Pi is approximately –31kJ/mole. Calculate the equilibrium constant for this reaction (R = 8.314J/°mole) at the cellular temperature of 37°C If the cellular concentrations of ATP, ADP, and Pi are 8, 1, and 8mM, respectively, is the above reaction at equilibrium in the cell?

How does a rigid, lock and key model for substrate binding not fit with the formation of the transition state?

A mutation of a proteolytic enzyme described in Section 6.1 results in a stable covalent bond between one of the catalytic amino acids of the protease with its protein substrate. What would be the most likely outcome of enzyme function?

What are transition state analogs?

Chapter 7 Kinetics and Regulation

Matching Questions

Use the following to answer questions 1-10: first-order reaction second-order reaction metabolism ensemble biomolecular turnover number

Choose the correct answer from the list below. Not all of the answers will be used.

Michaelis equilibrium sequential kinetics initial reaction velocity allosteric ping-pong