BIOC 3021 EXAM 2 QUESTIONS AND ANSWERS 2024/2025 | VERIFIED
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? - 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? - 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? - 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? - 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? - 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? - 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? - Km tells us about an
enzyme's efficiency; a low Km indicates an efficient enzyme, while a high Km 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? - 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? - 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? - 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? - 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 state of a reaction and lower
the energy of activation? - 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? - 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? - 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? -
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?
- 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? - Desolvation is the process of
removing water from a substrate. When an enzyme does that is can make the substrate more reactive.
What is electrostatic destabilization and how could such effects contribute to catalysis? -
Electrostatic destabilization is the presence of repulsive forces. When an enzyme and substrate's
repulsive forces can be relieved, the reaction rate can be accelerated.
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? - 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? - 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? - 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? - 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? - 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? - 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? - Km tells us about an
enzyme's efficiency; a low Km indicates an efficient enzyme, while a high Km 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? - 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? - 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? - 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? - 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 state of a reaction and lower
the energy of activation? - 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? - 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? - 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? -
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?
- 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? - Desolvation is the process of
removing water from a substrate. When an enzyme does that is can make the substrate more reactive.
What is electrostatic destabilization and how could such effects contribute to catalysis? -
Electrostatic destabilization is the presence of repulsive forces. When an enzyme and substrate's
repulsive forces can be relieved, the reaction rate can be accelerated.
