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 energy changes.
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 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 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
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 energy changes.
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 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 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
