Question 1

The double-reciprocal transformation of the Michaelis-Menten equation, also called the Lineweaver-Burk plot, is given by 1/V₀ = K_m /(V_max[S]) + 1/V_max To determine K_m from a double-reciprocal plot, you would: A) multiply the reciprocal of the x-axis intercept by -1. B) multiply the reciprocal of the y-axis intercept by -1. C) take the reciprocal of the x-axis intercept. D) take the reciprocal of the y-axis intercept. E) take the x-axis intercept, where V₀ = 1/2 V_max.

Accepted Answer

The x-axis intercept of a Lineweaver-Burk plot represents -1/Km, so to find Km, you multiply the reciprocal of the x-axis intercept by -1.

Question 2

A small molecule that DECREASES the activity of an enzyme by binding to a site other than the catalytic site is termed a(n):
A) allosteric inhibitor.
B) alternative inhibitor.
C) competitive inhibitor.
D) stereospecific agent.
E) transition-state analog.

Accepted Answer

An allosteric inhibitor is a small molecule that binds to a site on the enzyme other than the catalytic site and diminishes its activity by causing a conformational change that affects the enzyme's ability to bind to its substrate. A competitive inhibitor binds directly to the enzyme's active site, whereas an alternative inhibitor and stereospecific agent are not terms commonly used to describe enzyme inhibitors. A transition-state analog is a molecule that resembles the transition state of a reaction and can bind to the enzyme's active site to inhibit its activity.

Question 3

In a plot of l/V against 1/[S] for an enzyme-catalyzed reaction, the presence of a competitive inhibitor will alter the: A) curvature of the plot. B) intercept on the l/[S] axis. C) intercept on the l/V axis. D) pK of the plot. E) V_max.

Accepted Answer

A competitive inhibitor decreases the effective concentration of the enzyme by binding to the active site, competing with the substrate. This results in an increase in the apparent Km value for the substrate. As a result, the intercept on the l/[S] axis increases in the presence of a competitive inhibitor. The curvature of the plot remains the same, and the intercept on the l/V axis and pK of the plot are not affected by the presence of a competitive inhibitor. V_max is not a relevant parameter in this context.

Question 4

Which factor has NOT been shown to play a role in determining the specificity of protein kinases?
A) disulfide bonds near the phosphorylation site
B) primary sequence at phosphorylation site
C) protein quaternary structure
D) protein tertiary structure
E) residues near the phosphorylation site

Accepted Answer

Disulfide bonds near the phosphorylation site have not been shown to play a role in determining the specificity of protein kinases, unlike the other factors listed. The primary sequence at the phosphorylation site, protein quaternary and tertiary structure, and residues near the phosphorylation site have all been shown to influence kinase specificity.

Question 5

How is trypsinogen converted to trypsin? A) A protein kinase-catalyzed phosphorylation converts trypsinogen to trypsin. B) An increase in Ca²⁺ concentration promotes the conversion. C) Proteolysis of trypsinogen forms trypsin. D) Trypsinogen dimers bind an allosteric modulator, cAMP, causing dissociation into active trypsin monomers. E) Two inactive trypsinogen dimers pair to form an active trypsin tetramer.

Accepted Answer

Trypsinogen is converted to trypsin through proteolysis, which involves the cleavage of a peptide bond within trypsinogen. This cleavage exposes the active site of the enzyme, allowing it to function as a protease. This conversion can be catalyzed by a variety of enzymes, including trypsin itself, as well as other proteases such as enteropeptidase in the small intestine.

Question 6

In competitive inhibition, an inhibitor: A) binds at several different sites on an enzyme. B) binds covalently to the enzyme. C) binds only to the ES complex. D) binds reversibly at the active site. E) lowers the characteristic V_max of the enzyme.

Accepted Answer

In competitive inhibition, the inhibitor binds reversibly at the active site of the enzyme, competing with the substrate for binding. This results in an increase in the apparent Km (affinity of the enzyme for the substrate) without affecting the Vmax (maximum velocity of the reaction). As a result, higher substrate concentrations are required to achieve the same reaction rate, which can be overcome by increasing the substrate concentration.

Question 7

Both water and glucose share an -OH that can serve as a substrate for a reaction with the terminal phosphate of ATP catalyzed by hexokinase. Glucose, however, is about a million times more reactive as a substrate than water. The BEST explanation is that:
A) glucose has more -OH groups per molecule than does water.
B) the larger glucose binds better to the enzyme; it induces a conformational change in hexokinase that brings active-site amino acids into position for catalysis.
C) the -OH group of water is attached to an inhibitory H atom, while the glucose -OH group is attached to C.
D) water and the second substrate, ATP, compete for the active site resulting in a competitive inhibition of the enzyme.
E) water normally will not reach the active site because it is hydrophobic.

Accepted Answer

The larger size and shape of glucose allows it to bind better to hexokinase and induce a conformational change that brings active-site amino acids into position for catalysis. This increases the reactivity of glucose as a substrate compared to water. The number of -OH groups per molecule (A) and the attachment of -OH groups to C vs. H (C) are not relevant factors in the difference in reactivity between glucose and water. There is no evidence of competitive inhibition (D) or exclusion by hydrophobicity (E) in this scenario.

Question 8

A transition-state analog:
A) is less stable when binding to an enzyme than the normal substrate.
B) resembles the active site of general acid-base enzymes.
C) resembles the transition-state structure of the normal enzyme-substrate complex.
D) stabilizes the transition state for the normal enzyme-substrate complex.
E) typically reacts more rapidly with an enzyme than the normal substrate.

Accepted Answer

Transition-state analogs are designed to resemble the transition-state structure of the normal enzyme-substrate complex, not the substrate itself, which allows them to bind more tightly to the enzyme than the substrate does.

Question 9

The number of substrate molecules converted to product in a given unit of time by a single enzyme molecule at saturation is referred to as the:
A) dissociation constant.
B) half-saturation constant.
C) maximum velocity.
D) Michaelis-Menten number.
E) turnover number.

Accepted Answer

The turnover number, also known as kcat, represents the number of substrate molecules that can be converted to product by a single enzyme molecule per unit time when the enzyme is fully saturated with substrate. It is a measure of the catalytic efficiency of the enzyme.

Question 10

A good transition-state analog:
A) binds covalently to the enzyme.
B) binds to the enzyme more tightly than the substrate.
C) binds very weakly to the enzyme.
D) is too unstable to isolate.
E) must be almost identical to the substrate.

Accepted Answer

A good transition-state analog should bind to the enzyme more tightly than the substrate in order to effectively mimic the transition state and inhibit the enzyme's activity. Covalent binding (A) may not accurately mimic the transition state and can also permanently inactivate the enzyme. Weak binding (C) would not effectively inhibit the enzyme. Being too unstable to isolate (D) would make the analog ineffective for use. While being almost identical to the substrate (E) is important for specificity, the most critical aspect for a good transition-state analog is tight binding.

Question 11

The Lineweaver-Burk plot is used to:
A) determine the equilibrium constant for an enzymatic reaction.
B) extrapolate for the value of reaction rate at infinite enzyme concentration.
C) illustrate the effect of temperature on an enzymatic reaction.
D) solve, graphically, for the rate of an enzymatic reaction at infinite substrate concentration.
E) solve, graphically, for the ratio of products to reactants for any starting substrate concentration.

Accepted Answer

The Lineweaver-Burk plot is a graphical representation used to determine the kinetic parameters of an enzyme-catalyzed reaction. It involves plotting the inverse of the initial reaction rate against the inverse of substrate concentration. The resulting plot is a straight line with a y-intercept equal to 1/Vmax and an x-intercept equal to -1/Km. By extrapolating the line to the y-axis, the value for 1/Vmax can be obtained, which represents the reaction rate at infinite enzyme concentration. Therefore, option D, "solve, graphically, for the rate of an enzymatic reaction at infinite substrate concentration," is the best choice.

Question 12

V_max for an enzyme-catalyzed reaction: A) generally increases when pH increases. B) increases in the presence of a competitive inhibitor. C) is limited only by the amount of substrate supplied. D) is twice the rate observed when the concentration of substrate is equal to the K_m. E) is unchanged in the presence of a uncompetitive inhibitor.

Accepted Answer

Vmax is the maximum rate of an enzyme-catalyzed reaction and is achieved when the enzyme is saturated with substrate. At this point, increasing substrate concentration does not increase the rate. When substrate concentration equals Km, the reaction rate is half of Vmax, so Vmax is twice the rate at Km.

Question 13

Penicillin and related drugs inhibit the enzyme _____; this enzyme is produced by _____.
A) $\beta$-lacamase; bacteria
B) transpeptidase; human cells
C) transpeptidase; bacteria
D) lysozyme; human cells
E) aldolase; bacteria

Accepted Answer

Penicillin and related drugs inhibit the bacterial enzyme transpeptidase, which is involved in cell wall synthesis in bacteria. This inhibition ultimately leads to bacterial cell death.

Question 14

Allosteric enzymes:
A) are regulated primarily by covalent modification.
B) usually catalyze several different reactions within a metabolic pathway.
C) usually have more than one polypeptide chain.
D) usually have only one active site.
E) usually show strict Michaelis-Menten kinetics.

Accepted Answer

Allosteric enzymes usually have more than one polypeptide chain, which allows for the binding of regulatory molecules to sites separate from the active site.

Question 15

Enzyme X exhibits maximum activity at pH = 6.9. X shows a fairly sharp decrease in its activity when the pH goes much lower than 6.4. One likely interpretation of this pH activity is that:
A) a Glu residue on the enzyme is involved in the reaction.
B) a His residue on the enzyme is involved in the reaction.
C) the enzyme has a metallic cofactor.
D) the enzyme is found in gastric secretions.
E) the reaction relies on specific acid-base catalysis.

Accepted Answer

The sharp decrease in activity below pH 6.4 suggests involvement of a histidine (His) residue, as its side chain has a pKa around 6, making it sensitive to changes in pH around this value. Histidine can act as a proton donor or acceptor near its pKa, playing a crucial role in enzyme catalysis at this pH range.

Question 16

The role of the metal ion (Mg²⁺) in catalysis by enolase is to: A) act as a general acid catalyst. B) act as a general base catalyst. C) facilitate general acid catalysis. D) facilitate general base catalysis. E) stabilize protein conformation.

Accepted Answer

The metal ion in enolase acts as a facilitator of general base catalysis by coordinating with the substrate and helping to abstract a proton from the substrate's hydroxyl group. This lowers the activation energy for the reaction and increases the rate of the enzyme-catalyzed reaction.

Question 17

Which statement about allosteric control of enzymatic activity is FALSE? A) Allosteric effectors give rise to sigmoidal V₀ versus [S] kinetic plots. B) Allosteric proteins are generally composed of several subunits. C) An effector may either inhibit or activate an enzyme. D) Binding of the effector changes the conformation of the enzyme molecule. E) Heterotropic allosteric effectors compete with substrate for binding sites.

Accepted Answer

Heterotropic allosteric effectors do not compete with substrates for binding sites since they bind to different sites on the enzyme.

Question 18

Phenyl-methane-sulfonyl-fluoride (PMSF) inactivates serine proteases by binding covalently to the catalytic serine residue at the active site; this enzyme-inhibitor bond is not cleaved by the enzyme. This is an example of what kind of inhibition?
A) irreversible
B) competitive
C) noncompetitive
D) mixed
E) pH inhibition

Accepted Answer

PMSF binds covalently to the catalytic serine residue at the active site, irreversibly inhibiting the enzyme. This inactivation is not reversible by simply removing the inhibitor, and the enzyme-inhibitor bond is not cleaved by the enzyme. Therefore, this is an example of irreversible inhibition.

Question 19

For the simplified representation of an enzyme-catalyzed reaction shown below, the statement "ES is in steady-state" means that: A) k₂ is very slow. B) k₁ = k₂. C) k₁ = k_-₁. D) k₁[E][S] = k_-₁[ES] + k₂[ES]. E) k₁[E][S] = k_-₁[ES].

Accepted Answer

The statement "ES is in steady-state" means that the concentration of the intermediate complex ES remains constant over time. The equation given in choice D is the mathematical representation of steady-state conditions, where the rate of formation of ES (k₁[E][S]) is equal to the sum of the rates of its consumption by the forward reaction (k_-₁[ES]) and the reverse reaction (k₂[ES]).

Question 20

To calculate the turnover number of an enzyme, you need to know: A) the enzyme concentration. B) the initial velocity of the catalyzed reaction at [S] >> K_m. C) the initial velocity of the catalyzed reaction at low [S]. D) the K_m for the substrate. E) both the enzyme concentration and the initial velocity of the catalyzed reaction at [S] >> K_m.

Accepted Answer

The turnover number of an enzyme is calculated by dividing the initial velocity of the catalyzed reaction at saturating substrate concentration ([S] >> Km) by the enzyme concentration. This requires knowledge of both the enzyme concentration and the initial velocity under these conditions.

Quiz 6: Enzymes

The double-reciprocal transformation of the Michaelis-Menten equation, also called the Lineweaver-Burk plot, is given by 1/V₀ = K_m /(V_max[S]) + 1/V_max To determine K_m from a double-reciprocal plot, you would:

A small molecule that DECREASES the activity of an enzyme by binding to a site other than the catalytic site is termed a(n) :

In a plot of l/V against 1/[S] for an enzyme-catalyzed reaction, the presence of a competitive inhibitor will alter the:

Which factor has NOT been shown to play a role in determining the specificity of protein kinases?

How is trypsinogen converted to trypsin?

In competitive inhibition, an inhibitor:

Both water and glucose share an -OH that can serve as a substrate for a reaction with the terminal phosphate of ATP catalyzed by hexokinase. Glucose, however, is about a million times more reactive as a substrate than water. The BEST explanation is that:

A transition-state analog:

The number of substrate molecules converted to product in a given unit of time by a single enzyme molecule at saturation is referred to as the:

A good transition-state analog:

The Lineweaver-Burk plot is used to:

V_max for an enzyme-catalyzed reaction:

Penicillin and related drugs inhibit the enzyme _; this enzyme is produced by _.

Allosteric enzymes:

Enzyme X exhibits maximum activity at pH = 6.9. X shows a fairly sharp decrease in its activity when the pH goes much lower than 6.4. One likely interpretation of this pH activity is that:

The role of the metal ion (Mg²⁺) in catalysis by enolase is to:

Which statement about allosteric control of enzymatic activity is FALSE?

Phenyl-methane-sulfonyl-fluoride (PMSF) inactivates serine proteases by binding covalently to the catalytic serine residue at the active site; this enzyme-inhibitor bond is not cleaved by the enzyme. This is an example of what kind of inhibition?

For the simplified representation of an enzyme-catalyzed reaction shown below, the statement "ES is in steady-state" means that:

To calculate the turnover number of an enzyme, you need to know: