©FYNDLAY EXAM SOLUTIONS 2024/2025
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BIOC 3021 Exam 2 Questions And Answers
(Guaranteed A+)
In addition to the basic ES intermediate in an enzyme reaction, what other types of
intermediates are formed and why are these intermediates so important? - answer✔Transition
states occur in enzyme reactions that are important in product stabilization and changes in
energy.
Why would researchers try to determine the initial rate of enzyme activity (V0) at a very early
time point in the reaction? What happens later in the reaction? Why does this happen? -
answer✔The initial velocity should be determined early in the reaction because the velocity will
change later due to changes in substrate concentration.
Why does a plot of V vs S taper off and eventually reach a plateau at higher S levels? -
answer✔Substrate has filled all enzyme active sites and adding more will not increase the rate
of reaction.
What is the Michaelis-Menten Equation? The M-M Equation uses the terms (V), ( Vmax), (S) and
(Km). What do each of these terms mean? - answer✔The velocity of the reaction equals
maximum velocity times substrate concentration divided by the sum of the substrate
concentration and the Michaelis constant, which has units of concentration.
Why can the rate constant K-2 be ignored in the derivation of the M-M Equation? -
answer✔There is little product formed early in the reaction, so the reverse reaction can be
ignored.
Why is a V vs S plot linear at low S concentrations? Why does the plot curve off at intermediate
S levels, and why does it plateau at high S levels? - answer✔The V vs S plot is linear at low S
concentrations because the M-M equation reduces to V = Vmax * [S] / Km, which is a linear
equation. The plot then curves at intermediate levels with increasing substrate concentration,
and plateaus at high S levels due to the M-M equation approaching Vmax.
What does the Km term tell us about an enzyme? What does it signify if an enzyme has a low
Km or a high Km? What is the enzyme rate when the concentration of S = Km? - answer✔Km
tells us about an enzyme's efficiency; a low Km indicates an efficient enzyme, while a high Km
1|Page
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ALL RIGHTS RESERVED.
indicates a less efficient enzyme. When the Km and substrate concentration are equal, the
velocity of the reaction is equal to half the maximum possible velocity of the reaction.
What is the definition of turnover number and how does that constant relate to the Vmax of an
enzyme? What is the absolute value of the turnover number for the enzyme catalase and what
does that mean in terms of the amount of substrate used by a single molecule of the enzyme
per second? Why doesn't the turnover number of an enzyme change as the enzyme is purified?
- answer✔The turnover number is the number of molecules of substrate that can be converted
per second per molecule of enzyme. It is equal to Vmax divided by the concentration of the
enzyme. The enzyme catalase the absolute value of the turnover number is forty million, so it
can still function efficiently even with a poor Km. Turnover number doesn't change as the
enzyme is purified because it's an intrinsic property of enzymes.
What is the Lineweaver-Burk equation? How does it relate to the M-M Equation and how is it
used to determine the Km and Vmax of an enzyme? - answer✔1/v = 1/vmax + Km/(vmax*[S]). It
is essentially the M-M equation flipped. -1/Km = 1/[S] intercept, 1/vmax = 1/v intercept.
What is the difference between a reversible and an irreversible inhibitor? What is the mode of
action for the two most common reversible inhibitors, competitive and noncompetitive? When
examined using a Lineweaver-Burk plot, how do each of these inhibitors change the plots?
What information can be gained by examination of the intersection of these plots with the X
and Y intercepts? - answer✔Reversible inhibitors cause covalently-modified changes to
enzymes that cannot be undone, while reversible inhibitors do not. Competitive inhibitors
reversibly bind at an enzyme's active site. This increases the Km and thus shifts the LB plot to
the right. Noncompetitive inhibitors reversibly bind at a non-active site of an enzyme and exert
their effects allosterically. This decreases the Vmax of the reaction and shifts the LB plot up the
y-axis (1/v axis).
What is the mode of action of irreversible inhibitors? Why are they irreversible? What is the
specificity of action of TPCK, DIFP, and iodoacetamide? How can these inhibitors help
determine the character of an enzyme's active site? - answer✔Irreversible inhibitors cause
covalently-modified changes to enzymes, usually in the active sites. TPCK binds at the active site
of chymotrypsin with a histidine residue. DIFP reacts with active sites of serine proteases like
chymotrypsin and modifies the active site serine residue of acetylcholinesterase.
Iodoacetamide reacts with the active site of cysteine proteases by covalently bonding with the
sulfur atom. These inhibitors can help determine the character of an enzyme's active site, since
they will only inhibit if certain residues exist.
What are the common mechanisms used by enzymes to achieve catalysis? How do these
mechanisms reduce or increase the energy level of an enzyme-substrate or enzyme-transition
state intermediate? Which of these mechanisms specifically reduce the energy of the transition
2|Page
, ©FYNDLAY EXAM SOLUTIONS 2024/2025
ALL RIGHTS RESERVED.
state of a reaction and lower the energy of activation? - answer✔Proximity and orientation
effects: reduce entropy of reactants and overall entropy. General acid and general base
catalysis: acts as proton donor or acceptor to raise reaction rate, reduces energy of transition
state. Electrostatic effects: relieving electrostatic forces between substrate and enzyme
accelerates the reaction. Nucleophilic and electrophilic catalysis: catalysis by electron removal.
Structural flexibility: enzyme flexibility reduces energy of activation, tightest bonding is with
intermediates and enzyme. Entropy loss: binding to enzyme reduces tumbling and lowers the
entropy of the substrate. Desolvation: releases water and makes the substrate more reactive.
Covalent catalysis: formation of covalent intermediate can substitute for nucleophilic attack
and reduce the energy of later transition states.
What is involved in proximity and orientation effects, and how might these mechanisms speed
the rate of a reaction? - answer✔Close proximity and correct orientation raise reaction rate by
increasing the number of collisions between reacting groups.
What is acid-base catalysis? How can an enzyme function in acid-base catalysis in a way that
cannot be accomplished in a non-enzymatic chemical reaction? How does the ability of an
enzyme to donate hydrogen ions or hydroxyl ions enhance the rate of a reaction? What amino
acid R-group(s) are often involved in acid-base catalysis? - answer✔An enzyme acts a a proton
donor or acceptor. Substitution of enzyme functional groups for hydrogen ions and hydroxyl
groups permits rapid enzyme catalysis at neutral pH. Often involves histidine imidazole group,
since it can act as an acid or base at neutral p. May lower the energy of activation for
intermediates.
What is covalent catalysis, and how is this process used by enzymes to speed the rate of a
reaction? What examples of covalent catalysis involving enzymes or cofactors are cited in your
lectures? - answer✔Covalent catalysis involves the substrate forming a transient covalent bond
with the amino acid R-group in the active site of an enzyme. The formation of a covalent
intermediate can reduce energy of later transition states and substitute for nucleophilic attack
by a hydroxyl group. Examples from lecture include serine proteases forming covalent acyl-
enzyme intermediates, covalent Schiff base formation in acetoacetate decarboxylase, covalent
catalysis with cofactors like PLP and TPP, and covalent catalysis with pyridoxal phosphate.
What is entropy loss and how might it change the energy level during the course of an enzyme
reaction? - answer✔Entropy loss is a reduction in the amount of "randomness", or possible
configurations, for the molecules in a system Entropy decrease reduces energy level and favors
formation fo the ES complex.
What is desolvation and how does it relate to enzyme catalysis? - answer✔Desolvation is the
process of removing water from a substrate. When an enzyme does that is can make the
substrate more reactive.
3|Page
ALL RIGHTS RESERVED.
BIOC 3021 Exam 2 Questions And Answers
(Guaranteed A+)
In addition to the basic ES intermediate in an enzyme reaction, what other types of
intermediates are formed and why are these intermediates so important? - answer✔Transition
states occur in enzyme reactions that are important in product stabilization and changes in
energy.
Why would researchers try to determine the initial rate of enzyme activity (V0) at a very early
time point in the reaction? What happens later in the reaction? Why does this happen? -
answer✔The initial velocity should be determined early in the reaction because the velocity will
change later due to changes in substrate concentration.
Why does a plot of V vs S taper off and eventually reach a plateau at higher S levels? -
answer✔Substrate has filled all enzyme active sites and adding more will not increase the rate
of reaction.
What is the Michaelis-Menten Equation? The M-M Equation uses the terms (V), ( Vmax), (S) and
(Km). What do each of these terms mean? - answer✔The velocity of the reaction equals
maximum velocity times substrate concentration divided by the sum of the substrate
concentration and the Michaelis constant, which has units of concentration.
Why can the rate constant K-2 be ignored in the derivation of the M-M Equation? -
answer✔There is little product formed early in the reaction, so the reverse reaction can be
ignored.
Why is a V vs S plot linear at low S concentrations? Why does the plot curve off at intermediate
S levels, and why does it plateau at high S levels? - answer✔The V vs S plot is linear at low S
concentrations because the M-M equation reduces to V = Vmax * [S] / Km, which is a linear
equation. The plot then curves at intermediate levels with increasing substrate concentration,
and plateaus at high S levels due to the M-M equation approaching Vmax.
What does the Km term tell us about an enzyme? What does it signify if an enzyme has a low
Km or a high Km? What is the enzyme rate when the concentration of S = Km? - answer✔Km
tells us about an enzyme's efficiency; a low Km indicates an efficient enzyme, while a high Km
1|Page
, ©FYNDLAY EXAM SOLUTIONS 2024/2025
ALL RIGHTS RESERVED.
indicates a less efficient enzyme. When the Km and substrate concentration are equal, the
velocity of the reaction is equal to half the maximum possible velocity of the reaction.
What is the definition of turnover number and how does that constant relate to the Vmax of an
enzyme? What is the absolute value of the turnover number for the enzyme catalase and what
does that mean in terms of the amount of substrate used by a single molecule of the enzyme
per second? Why doesn't the turnover number of an enzyme change as the enzyme is purified?
- answer✔The turnover number is the number of molecules of substrate that can be converted
per second per molecule of enzyme. It is equal to Vmax divided by the concentration of the
enzyme. The enzyme catalase the absolute value of the turnover number is forty million, so it
can still function efficiently even with a poor Km. Turnover number doesn't change as the
enzyme is purified because it's an intrinsic property of enzymes.
What is the Lineweaver-Burk equation? How does it relate to the M-M Equation and how is it
used to determine the Km and Vmax of an enzyme? - answer✔1/v = 1/vmax + Km/(vmax*[S]). It
is essentially the M-M equation flipped. -1/Km = 1/[S] intercept, 1/vmax = 1/v intercept.
What is the difference between a reversible and an irreversible inhibitor? What is the mode of
action for the two most common reversible inhibitors, competitive and noncompetitive? When
examined using a Lineweaver-Burk plot, how do each of these inhibitors change the plots?
What information can be gained by examination of the intersection of these plots with the X
and Y intercepts? - answer✔Reversible inhibitors cause covalently-modified changes to
enzymes that cannot be undone, while reversible inhibitors do not. Competitive inhibitors
reversibly bind at an enzyme's active site. This increases the Km and thus shifts the LB plot to
the right. Noncompetitive inhibitors reversibly bind at a non-active site of an enzyme and exert
their effects allosterically. This decreases the Vmax of the reaction and shifts the LB plot up the
y-axis (1/v axis).
What is the mode of action of irreversible inhibitors? Why are they irreversible? What is the
specificity of action of TPCK, DIFP, and iodoacetamide? How can these inhibitors help
determine the character of an enzyme's active site? - answer✔Irreversible inhibitors cause
covalently-modified changes to enzymes, usually in the active sites. TPCK binds at the active site
of chymotrypsin with a histidine residue. DIFP reacts with active sites of serine proteases like
chymotrypsin and modifies the active site serine residue of acetylcholinesterase.
Iodoacetamide reacts with the active site of cysteine proteases by covalently bonding with the
sulfur atom. These inhibitors can help determine the character of an enzyme's active site, since
they will only inhibit if certain residues exist.
What are the common mechanisms used by enzymes to achieve catalysis? How do these
mechanisms reduce or increase the energy level of an enzyme-substrate or enzyme-transition
state intermediate? Which of these mechanisms specifically reduce the energy of the transition
2|Page
, ©FYNDLAY EXAM SOLUTIONS 2024/2025
ALL RIGHTS RESERVED.
state of a reaction and lower the energy of activation? - answer✔Proximity and orientation
effects: reduce entropy of reactants and overall entropy. General acid and general base
catalysis: acts as proton donor or acceptor to raise reaction rate, reduces energy of transition
state. Electrostatic effects: relieving electrostatic forces between substrate and enzyme
accelerates the reaction. Nucleophilic and electrophilic catalysis: catalysis by electron removal.
Structural flexibility: enzyme flexibility reduces energy of activation, tightest bonding is with
intermediates and enzyme. Entropy loss: binding to enzyme reduces tumbling and lowers the
entropy of the substrate. Desolvation: releases water and makes the substrate more reactive.
Covalent catalysis: formation of covalent intermediate can substitute for nucleophilic attack
and reduce the energy of later transition states.
What is involved in proximity and orientation effects, and how might these mechanisms speed
the rate of a reaction? - answer✔Close proximity and correct orientation raise reaction rate by
increasing the number of collisions between reacting groups.
What is acid-base catalysis? How can an enzyme function in acid-base catalysis in a way that
cannot be accomplished in a non-enzymatic chemical reaction? How does the ability of an
enzyme to donate hydrogen ions or hydroxyl ions enhance the rate of a reaction? What amino
acid R-group(s) are often involved in acid-base catalysis? - answer✔An enzyme acts a a proton
donor or acceptor. Substitution of enzyme functional groups for hydrogen ions and hydroxyl
groups permits rapid enzyme catalysis at neutral pH. Often involves histidine imidazole group,
since it can act as an acid or base at neutral p. May lower the energy of activation for
intermediates.
What is covalent catalysis, and how is this process used by enzymes to speed the rate of a
reaction? What examples of covalent catalysis involving enzymes or cofactors are cited in your
lectures? - answer✔Covalent catalysis involves the substrate forming a transient covalent bond
with the amino acid R-group in the active site of an enzyme. The formation of a covalent
intermediate can reduce energy of later transition states and substitute for nucleophilic attack
by a hydroxyl group. Examples from lecture include serine proteases forming covalent acyl-
enzyme intermediates, covalent Schiff base formation in acetoacetate decarboxylase, covalent
catalysis with cofactors like PLP and TPP, and covalent catalysis with pyridoxal phosphate.
What is entropy loss and how might it change the energy level during the course of an enzyme
reaction? - answer✔Entropy loss is a reduction in the amount of "randomness", or possible
configurations, for the molecules in a system Entropy decrease reduces energy level and favors
formation fo the ES complex.
What is desolvation and how does it relate to enzyme catalysis? - answer✔Desolvation is the
process of removing water from a substrate. When an enzyme does that is can make the
substrate more reactive.
3|Page
