Chapter 6 Basic Concepts of Enzyme Action
Matching Questions
Use the following to answer questions 1-10:
Choose the correct answer from the list below. Not all of the answers will be used.
a) apoenzymes
b) hydrolyases
c) active site
d) transition state
e) spontaneous
f) induced fit
g) energy
h) prosthetic group
i) lock and key
j) substrate(s)
k) oxidoreductases
l) equilibria
1.
____________ The site on the enzyme where the reaction occurs.
2.
____________ The substance that the enzyme binds and converts to product.
Ans:
Section: 6.1
3.
Enzymes that do not have the required cofactor bound are called ____________.
Ans:
Section: 6.2
4.
A tightly bound cofactor might be called a(n) ____________.
Ans:
h
Section: 6.2
5.
Enzymes will decrease the energy of activation but do not change the ____________ of a
chemical reaction.
Ans:
Section: 6.3
Ans:
Section: 6.4
Chapter 6 Basic Concepts of Enzyme Action
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6.
A reaction that is exergonic will be ____________.
7.
An endergonic reaction requires an input of ____________ to proceed.
Ans:
g
Section: 6.3
8.
Enzymes that transfer electrons are called ___________.
Ans:
k
Section: 6.1
9.
Enzymes that cleave molecules by addition of water are called ____________.
Ans:
b
Section: 6.4
10.
Which model is more appropriate to explain an enzyme binding to its substrate?
Ans:
Section: 6.4
Fill–in-the-Blank Questions
11.
Enzymes accelerate the rate of a chemical reaction by the free energy of activation of the
reaction.
Ans: lowering Section: 6.3
12.
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 .
Ans: pH 7 Section: 6.3
13.
An enzyme that loosely binds substrate will have a level of specificity.
Ans: low Section: 6.1
14.
Organic cofactors are referred to as .
Ans: coenzymes Section: 6.2
15.
A reaction can occur spontaneously only if ΔG is .
Ans: negative Section: 6.3
16.
When ΔG for a system is zero, the system is at .
Ans: equilibrium Section: 6.3
17.
An enzyme that has been stripped of small molecules needed for activity is called .
Ans:
Section: 6.3
Chapter 6 Basic Concepts of Enzyme Action
3
18.
The total change of free energy in a reaction depends on and .
19.
The difference in values for G and Go′ is in the .
Ans: concentration of reactants and products Section: 6.3
20.
Competitive inhibitors that mimic the substrate while in the transition state are called
inhibitors.
Ans: transition-state analog Section: 6.4
Multiple-Choice Questions
21.
What is the common strategy by which catalysis occurs?
A)
increasing the probability of product formation
B)
shifting the reaction equilibrium
C)
stabilization the transition state
D)
All of the above.
E)
None of the above.
Ans: C Section: 6.4
22.
An enzyme will specifically bind its substrate because of____________
A)
a tight lock and key binding mechanism.
B)
a high number of hydrophobic amino acids in the center of the protein.
C)
a large number of weak interactions at the active site.
D)
additional nonprotein cofactors.
E)
None of the above.
Ans: C Section: 6.4
23.
Examples of cofactors include:
A)
Zn+2, Mg+2, and Ni+2.
B)
biotin and thiamine pyrophosphate.
C)
pyridoxal phosphate and coenzyme A.
D)
B and C.
E)
All of the above.
Ans: E Section: 6.2
24.
A cofactor is best defined as ______________.
A)
another protein
B)
a covalently bound inorganic molecule
C)
a small molecule that holds the substrate in the active site
D)
a molecule responsible for most of the catalytic activity of the enzyme
E)
None of the above.
Ans: E Section: 6.2
Ans: the substrate G; the G of the product Section: 6.3
Chapter 6 Basic Concepts of Enzyme Action
4
25.
Which of the following is true?
A)
Enzymes force reactions to proceed in only one direction.
B)
Enzymes alter the equilibrium of the reaction.
C)
Enzymes alter the standard free energy of the reaction.
D)
All of the above.
E)
None of the above.
26.
The Gibbs free energy of activation is:
A)
the difference between the substrate and the transition state.
B)
the difference between the substrate and the product.
C)
the difference between the product and the transition state.
D)
All of the above.
E)
None of the above.
Ans: A Section: 6.4
27.
At equilibrium, the Gibb’s free energy is ___________.
A)
a positive value
B)
neutral
C)
a negative value
D)
zero
E)
one
Ans: D Section: 6.3
28.
The rate of a reaction, or how fast a reaction will proceed, is best determined by __________.
A)
R
B)
G‡
C)
Gº′
D)
H
E)
None of the above.
Ans: B Section: 6.3
29.
The relationship between Go′ and G is best described as ______________.
A)
determined by the temperature
B)
described by changes in Keq
C)
differ from standard state to physiological or actual concentrations of reactants and
products
D)
dependent on the reaction mechanism of the reaction
E)
differ only in terms of the types of reactions used for each value
Ans: C Section: 6.3
30.
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?
A)
Reaction a is not spontaneous at cellular concentrations.
B)
Reaction b will react very quickly.
C)
Reaction a is a more thermodynamically favorable reaction than b.
D)
Neither reaction is reversible.
E)
None of the above.
Ans: E Section: 6.3
Ans: E Section: 6.3
Chapter 6 Basic Concepts of Enzyme Action
5
31.
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?
A)
The reaction has reached equilibrium, that is, the forward and reverse reactions are
occurring at a fixed rate.
B)
There is a product inhibition of the enzyme.
C)
The reaction runs out of reaction materials.
D)
The enzyme has finished accelerating the reaction.
E)
None of the above.
32.
The free energy of activation is _______________.
A)
the amount of chemical energy available in the transition state
B)
the difference in free energy between the substrate and product
C)
the free energy gained by adding a catalyst
D)
the difference in free energy between the transition state and the substrate
E)
All of the above.
Ans: D Section: 6.4
33.
The molecular structure that is short-lived and neither substrate nor product is known
as_______.
A)
substrate analog
B)
transition state
C)
free energy stabilization state
D)
catalysis state
E)
equilibrium intermediate
Ans: B Section: 6.4
34.
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?
A)
Riboflavin is a coenzyme.
B)
Flavin adenine dinucleotide is a vitamin.
C)
Succiniate dehydrogenase is a coenzyme.
D)
Flavin adenine dinucleotide is a coenzyme.
Ans: D Section: 6.2
35.
The active site of an enzyme_____________.
A)
is a series of amino acids that bind the enzyme
B)
is a linear sequence of amino acids that react with each other
C)
binds covalently to the substrate
D)
allows water to enter into the solvate the substrate
E)
None of the above.
Ans: E Section: 6.4
Ans: A Section: 6.3
Chapter 6 Basic Concepts of Enzyme Action
6
36.
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 [glucose-6-
phosphate] is 1.43 M, what is the equilibrium [fructose-6-phosphate]?
A)
1.00 M
B)
1.33 M
C)
0.667 M
D)
0.250 M
E)
0.150 M
37.
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?
A)
+0.99
B)
+1.71
C)
0, as defined by equilibrium conditions
D)
–0.99
E)
–2.27
Ans: B Section: 6.3
38.
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 G°′ from
the previous question.)
A)
+4.19
B)
–1.81
C)
–4.03
D)
–2.50
E)
–1.75
Ans: D Section: 6.4
39.
That many transition-state analogs bind more tightly than the native substrate reinforces the
concept that:
A)
transition-state analogs are planar structures.
B)
transition-state analogs are highly charged at physiological pH.
C)
binding to the transition state is through a lock-and-key-mechanism.
D)
transition-state analogs are hydrophobic.
E)
binding to the transition state is through an induced-fit mechanism.
Ans: E Section: 6.4
Ans: C Section: 6.3
Chapter 6 Basic Concepts of Enzyme Action
7
Short-Answer Questions
40.
What is the relation between an enzyme-catalyzed reaction and the transition state of a reaction?
41.
What is the difference between prosthetic groups and coenzymes?
groups are either covalently bound to the enzyme or nearly irreversibly associated with
the protein.
Section: 6.2
42.
How do enzymes facilitate the formation of the transition state?
interactions with the enzyme. The full complement of interactions occurs when the
transition state is achieved. This causes maximal release of free energy.
Section: 6.4
43.
How is the substrate bound to the active site?
various functional groups in the binding site.
Section: 6.4
44.
You believe a substrate fits into a cleft like a key into a lock, but your roommate does not. Who
is right?
substrate to actually modify the shape of the site a bit, a hypothesis known as induced fit.
See textbook Figs. 6.5 and 6.6 for further detail.
Section: 6.4
45.
In an enzymatic reaction in a test tube, the reaction will eventually reach equilibrium. Why does
this not happen in living organisms?
Ans:
In a cell, the product may be utilized for a subsequent reaction, thus the reaction may not
reach equilibrium.
Section: 6.3
46.
How is free energy useful for understanding enzyme function?
catalyzed-reaction will occur. Understanding if a reaction is thermodynamically
favorable is the first step in our knowledge the basic function of an enzyme.
Section: 6.3
Enzymes do this by binding the substrate with several weak interactions and forming a
temporary transition state intermediate.
Sections: 6.3 and 6.4
Chapter 6 Basic Concepts of Enzyme Action
8
47.
While some enzymes have very specific substrates, others are more promiscuous. What would
you suspect is the reason for this?
48.
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.
Ans:
The dilution and dialysis must have separated small molecules from the enzyme. These
are likely cofactors (cosubstrates) required for the activity. One could test this hypothesis
by adding back potential cofactors and observing a reconstituted activity.
Section: 6.2
49.
If Keq = 1, what is the G°′? If Keq >1, what is the G°′? If Keq <1, what is the G°′?
Section: 6.3
50.
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).
Ans:
Activation energy! While the reaction is highly thermodynamically favorable, it is not
going to happen without a catalyst or sufficient added energy due to the energy barrier.
Section: 6.4
51.
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?
Section: 6.3
52.
How does a rigid, lock and key model for substrate binding not fit with the formation of the
transition state?
flexible and dynamic binding and active site.
Section: 6.4
substrate is due to many weak interactions between the substrate and the amino acids of
interactions for binding.
Section: 6.4
Chapter 6 Basic Concepts of Enzyme Action
9
53.
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?
54.
What are transition state analogs?
process. They bind very tightly to the catalytic site and are useful in determining the
structure and catalytic mechanism of the enzyme.
and would not drive forward the rest of the reaction. This is much like the transition state
analog discussed in Section 6.4.
Section: 6.4