- Q: What are enzymes, and what is their role in biochemistry?
A: Enzymes are biological catalysts, typically proteins that accelerate chemical reactions in living organisms. They lower the activation energy required for reactions, increasing the rate of biochemical processes.
- Q: How do enzymes catalyze reactions?
A: Enzymes work by binding to specific substrates, forming an enzyme-substrate complex. This complex undergoes changes in shape and facilitates the conversion of substrates into products. After the reaction, the enzyme is unchanged and can catalyze more reactions.
- Q: What is the active site of an enzyme?
A: The active site is a specific region on the enzyme’s surface where the substrate binds. It contains amino acid residues with precise spatial arrangements that are crucial for catalysis.
- Q: What is meant by enzyme specificity?
A: Enzyme specificity refers to the ability of an enzyme to recognize and bind to a specific substrate or a group of structurally related substrates. Enzymes are highly selective due to the complementarity between the active site and the substrate.
- Q: How does temperature affect enzyme activity?
A: Enzyme activity increases with temperature until it reaches an optimal point (usually around 37°C for human enzymes). Beyond this temperature, the enzyme denatures, losing its shape and function.
- Q: What is the significance of the Michaelis-Menten constant (Km) in enzyme kinetics?
A: Km represents the substrate concentration at which an enzyme works at half of its maximum velocity (Vmax). It provides information about the enzyme’s affinity for its substrate; a lower Km indicates higher affinity.
- Q: Define enzyme inhibition and give an example of a reversible inhibitor.
A: Enzyme inhibition is the reduction or prevention of an enzyme’s activity. A reversible inhibitor binds non-covalently to the enzyme and can be removed. An example is competitive inhibition, where the inhibitor competes with the substrate for the active site.
- Q: What is an allosteric enzyme?
A: Allosteric enzymes have multiple binding sites, including an active site and an allosteric site. The binding of a regulatory molecule at the allosteric site can either enhance (allosteric activator) or inhibit (allosteric inhibitor) enzyme activity.
- Q: What is the role of coenzymes and cofactors in enzyme function?
A: Coenzymes and cofactors are non-protein molecules required by some enzymes to function properly. They assist in catalysis by carrying or transferring chemical groups during the reaction.
- Q: How does pH affect enzyme activity?
A: Enzymes have an optimal pH range at which they function most effectively. Changes in pH can alter the ionization states of amino acid residues in the active site, affecting the enzyme’s structure and, consequently, its activity.
- Q: What are the six major classes of enzymes based on the type of reaction they catalyze?
A: The six major classes are oxidoreductases, transferases, hydrolases, lyases, isomerases, and ligases. Each class catalyzes a specific type of chemical reaction.
- Q: What is enzyme cooperativity?
A: Enzyme cooperativity refers to the phenomenon where the binding of one substrate molecule to an enzyme’s active site influences the affinity of other active sites for substrate molecules. It can result in sigmoidal kinetics in enzyme-substrate binding curves.
- Q: What is an enzyme-substrate complex?
A: An enzyme-substrate complex is a temporary structure formed when the enzyme binds to its substrate. It allows the enzyme to catalyze the conversion of the substrate into products.
- Q: How does pH influence enzyme activity?
A: Enzymes have specific pH optima at which their activity is highest. Deviations from this optimal pH can denature the enzyme or affect the ionization states of key amino acid residues, disrupting its activity.
- Q: What is the effect of enzyme concentration on reaction rate?
A: Increasing the enzyme concentration generally increases the reaction rate, as more active sites are available for substrate binding, leading to more product formation.
- Q: What is the Lineweaver-Burk plot used for in enzyme kinetics?
A: The Lineweaver-Burk plot is a graphical representation used to determine the kinetic parameters of an enzyme, such as Vmax and Km, by linearizing the Michaelis-Menten equation.
- Q: What are irreversible enzyme inhibitors?
A: Irreversible enzyme inhibitors form strong covalent bonds with the enzyme, permanently inactivating it. They often react with essential amino acid residues in the active site, rendering the enzyme non-functional.
- Q: What are allosteric enzymes and how do they differ from non-allosteric enzymes?
A: Allosteric enzymes have multiple binding sites, including an allosteric site that regulates enzyme activity. Non-allosteric enzymes lack this additional regulatory site and are solely controlled by substrate binding at the active site.
- Q: Explain the difference between competitive and non-competitive inhibition.
A: Competitive inhibitors compete with the substrate for the active site, and their binding can be overcome by increasing the substrate concentration. Non-competitive inhibitors, on the other hand, bind to a different site on the enzyme, altering its conformation and reducing activity.
- Q: How does temperature affect enzyme denaturation?
A: High temperatures can cause enzymes to denature, losing their three-dimensional structure and, consequently, their function. Denaturation disrupts the enzyme’s active site and prevents substrate binding.
- Q: What is enzyme kinetics?
A: Enzyme kinetics is the study of the rates at which enzymes catalyze chemical reactions and how these rates are influenced by factors such as substrate concentration and temperature.
- Q: What is the turnover number (kcat) of an enzyme?
A: The turnover number (kcat) represents the maximum number of substrate molecules converted into product per enzyme molecule per unit of time when the enzyme is saturated with substrate.
- Q: What is an apoenzyme, and what is a holoenzyme?
A: An apoenzyme is an inactive enzyme lacking its required cofactor or coenzyme. When an apoenzyme is bound to its cofactor, it becomes active and is then referred to as a holoenzyme.
- Q: What are zymogens, and why are they important in enzyme regulation?
A: Zymogens are inactive enzyme precursors that require a specific proteolytic cleavage to be converted into their active form. Zymogens prevent enzymes from catalyzing reactions prematurely, ensuring they are only activated when needed.
- Q: What is enzyme specificity constant (kcat/Km)?
A: The enzyme specificity constant (kcat/Km) is a measure of the catalytic efficiency of an enzyme. It represents the rate of catalysis relative to the enzyme’s affinity for its substrate.
- Q: How do enzymes maintain their specificity and prevent unwanted reactions?
A: Enzymes maintain specificity through the precise arrangement of amino acid residues in their active sites, which allows them to recognize only specific substrates. Additionally, the 3D structure of the active site is complementary to the substrate, reducing the likelihood of binding with unwanted molecules.
- Q: What is the lock-and-key model of enzyme-substrate interaction?
A: The lock-and-key model proposes that the enzyme’s active site has a specific shape that perfectly fits the substrate like a lock and key. The substrate fits precisely into the active site, leading to catalysis.
- Q: Explain the induced fit model of enzyme-substrate interaction.
A: The induced fit model suggests that the active site undergoes conformational changes upon substrate binding. The substrate induces these changes, bringing reactive groups in the enzyme into optimal positions for catalysis.
- Q: Can enzymes catalyze both forward and reverse reactions?
A: Yes, enzymes can catalyze both forward and reverse reactions. The direction of the reaction depends on the concentration of substrates and products.
- Q: How do metal ions act as enzyme cofactors?
A: Metal ions can stabilize the enzyme-substrate complex, participate in redox reactions, or facilitate the binding of substrates to the enzyme’s active site. They are essential for the catalytic activity of certain enzymes.
- Q: What is the role of prosthetic groups in enzyme function?
A: Prosthetic groups are tightly bound cofactors or coenzymes that remain attached to the enzyme during the catalytic process. They play a crucial role in enzyme function and are necessary for enzyme activity.
- Q: How does enzyme concentration influence the initial reaction rate?
A: As enzyme concentration increases, the initial reaction rate also increases, assuming there is an ample supply of substrate. The rate becomes directly proportional to the enzyme concentration in the reaction.
- Q: How do non-competitive inhibitors affect enzyme activity?
A: Non-competitive inhibitors bind to an allosteric site on the enzyme, causing a conformational change that reduces the enzyme’s catalytic activity. Increasing substrate concentration cannot overcome this type of inhibition.
- Q: What is the purpose of enzyme covalent modification?
A: Enzyme covalent modification involves the addition or removal of chemical groups, such as phosphorylation, acetylation, or glycosylation, to regulate enzyme activity and cellular signaling pathways.
- Q: What are the characteristics of allosteric enzymes regarding enzyme kinetics?
A: Allosteric enzymes exhibit sigmoidal kinetics in their enzyme-substrate binding curves, as their activity is influenced by interactions at both the active site and the allosteric site.
- Q: How does substrate concentration affect enzyme activity?
A: Increasing substrate concentration typically leads to an increase in the reaction rate until the enzyme becomes saturated with substrate. At saturation, further increases in substrate concentration do not increase the rate.
- Q: Define enzyme cooperativity and give an example of an allosteric enzyme showing cooperativity.
A: Enzyme cooperativity refers to the phenomenon where the binding of one substrate molecule to an enzyme’s active site affects the binding and activity of other active sites. An example is hemoglobin, which exhibits cooperativity in oxygen binding.
- Q: What is the reversible nature of enzyme inhibition?
A: Reversible enzyme inhibition occurs when the inhibitor can bind and dissociate from the enzyme. The enzyme’s activity can be restored once the inhibitor is removed.
- Q: How do competitive inhibitors affect the Michaelis-Menten plot?
A: Competitive inhibitors increase the apparent Km value without affecting the Vmax. This leads to a higher apparent affinity of the enzyme for the substrate, as more substrate is required to reach half of the maximum velocity.
- Q: What are the functions of chaperones in enzyme biochemistry?
A: Chaperones are proteins that help other proteins, including enzymes, fold into their correct 3D structures. They prevent misfolding and aggregation, ensuring the enzyme achieves its functional conformation.
- Q: How do temperature and pH influence enzyme stability?
A: High temperatures and extreme pH values can denature enzymes, disrupting their secondary and tertiary structures. This results in the loss of enzymatic activity.
- Q: What is the role of feedback inhibition in enzyme regulation?
A: Feedback inhibition is a regulatory mechanism where the final product of a metabolic pathway acts as an allosteric inhibitor of an early enzyme in the pathway. This helps to prevent the overproduction of a product and conserve resources.
- Q: What is the energy profile diagram of an enzyme-catalyzed reaction?
A: The energy profile diagram illustrates the energy changes during a chemical reaction in the presence of an enzyme. It shows how the enzyme lowers the activation energy required for the reaction to proceed.
- Q: What are enzyme isoenzymes?
A: Enzyme isoenzymes are different forms of an enzyme that catalyze the same reaction but have distinct molecular structures. They often have tissue-specific distributions and can serve as diagnostic markers for certain diseases.
- Q: How does enzyme activity change with increasing substrate concentration in an enzyme-catalyzed reaction?
A: Initially, enzyme activity increases with increasing substrate concentration, following Michaelis-Menten kinetics. However, at high substrate concentrations, the enzyme becomes saturated, and the activity reaches a plateau.
- Q: What are the effects of enzyme denaturation on its function?
A: Enzyme denaturation disrupts the enzyme’s secondary, tertiary, and quaternary structures, causing the active site to lose its shape and specificity. As a result, the enzyme loses its catalytic activity.
- Q: What is allosteric activation, and how does it differ from allosteric inhibition?
A: Allosteric activation occurs when a regulatory molecule binds to the allosteric site of an enzyme, inducing a conformational change that enhances the enzyme’s catalytic activity. In contrast, allosteric inhibition reduces enzyme activity upon binding to the allosteric site.
- Q: What is the role of enzymes in DNA replication and repair?
A: Enzymes called DNA polymerases play a crucial role in DNA replication, catalyzing the synthesis of new DNA strands. DNA repair enzymes help to correct mistakes and damage in the DNA sequence to maintain genomic integrity.
- Q: How do coenzymes differ from metal ions in enzyme catalysis?
A: Coenzymes are organic molecules that act as temporary carriers of specific chemical groups during enzyme catalysis. Metal ions, on the other hand, are inorganic elements that can stabilize the enzyme-substrate complex or participate directly in the catalytic mechanism.
- Q: Explain the concept of enzyme-substrate specificity using an example.
A: Enzyme-substrate specificity refers to the ability of an enzyme to recognize and bind to a specific substrate. An example is the enzyme lactase, which specifically catalyzes the hydrolysis of lactose (a disaccharide) into glucose and galactose. Lactase does not act on other disaccharides due to the specific shape and chemical characteristics of its active site.
- Q: What is the significance of the turnover rate (kcat) in enzyme kinetics?
A: The turnover rate (kcat) measures how quickly an enzyme can convert substrate molecules into product. It provides valuable information about the enzyme’s catalytic efficiency.
- Q: How do allosteric enzymes differ from non-allosteric enzymes regarding their regulation?
A: Allosteric enzymes are regulated by the binding of a molecule at an allosteric site, leading to changes in enzyme activity. Non-allosteric enzymes are primarily regulated by substrate concentration and reversible inhibitors.
- Q: What is enzyme promiscuity?
A: Enzyme promiscuity refers to an enzyme’s ability to catalyze multiple, distinct reactions, often with different substrates. This phenomenon can be beneficial in evolutionary terms as it allows for versatility in metabolic pathways.
- Q: How does an enzyme’s tertiary structure contribute to its function?
A: The enzyme’s tertiary structure determines the shape of the active site, and this specific conformation allows the enzyme to bind to its substrate with high specificity.
- Q: What is enzyme-substrate complementarity?
A: Enzyme-substrate complementarity refers to the specific and precise fit between the enzyme’s active site and the substrate. It ensures that only the correct substrate can bind to the enzyme, preventing unwanted reactions.
- Q: How can pH affect enzyme stability and function?
A: Extreme pH levels can denature enzymes by disrupting the hydrogen bonding and ionic interactions that stabilize their tertiary structure. The optimal pH for an enzyme varies depending on its source and function.
- Q: What is the role of coenzyme A (CoA) in enzyme-catalyzed reactions?
A: Coenzyme A (CoA) acts as a carrier of acyl groups, such as acetyl groups, during enzyme-catalyzed reactions. It plays a critical role in energy metabolism and the synthesis of fatty acids.
- Q: What are the functions of enzymes in cellular metabolism?
A: Enzymes participate in nearly all metabolic pathways in cells. They facilitate the breakdown of nutrients to release energy, build complex molecules from simpler ones, and regulate the flow of biochemical reactions.
- Q: How does enzyme activity change with temperature?
A: Enzyme activity increases with temperature up to an optimal point (usually around 37°C for human enzymes). Beyond this point, high temperatures cause denaturation and a loss of enzyme activity.
- Q: Explain the term “enzyme inhibition constant” (Ki).
A: The enzyme inhibition constant (Ki) represents the affinity of an inhibitor for its enzyme. A lower Ki value indicates a stronger binding between the inhibitor and the enzyme, leading to more potent inhibition.
- Q: Given the Michaelis-Menten constants for an enzyme are Km = 0.2 mM and Vmax = 100 μM/min, calculate the initial reaction velocity (V0) when the substrate concentration [S] is 0.1 mM.
A: V0 = (Vmax × [S]) / (Km + [S])
V0 = (100 μM/min × 0.1 mM) / (0.2 mM + 0.1 mM) = 33.33 μM/min
- Q: An enzyme has a turnover number (kcat) of 500 s^(-1). If the enzyme concentration is 2 μM, calculate the rate of product formation (V0) when saturated with substrate.
A: V0 = kcat × [Enzyme]
V0 = 500 s^(-1) × 2 μM = 1000 μM/s
- Q: The Michaelis-Menten equation for an enzyme-catalyzed reaction is given as V0 = (Vmax × [S]) / (Km + [S]). If Vmax = 80 μM/min and Km = 0.5 mM, find the substrate concentration [S] when the initial reaction velocity (V0) is 40 μM/min.
A: V0 = (Vmax × [S]) / (Km + [S])
40 μM/min = (80 μM/min × [S]) / (0.5 mM + [S])
[S] = (40 μM/min × 0.5 mM) / (80 μM/min – 40 μM/min) ≈ 0.625 mM
- Q: Calculate the enzyme efficiency (kcat/Km) for an enzyme with kcat = 200 s^(-1) and Km = 0.1 mM.
A: Enzyme efficiency (kcat/Km) = kcat / Km
kcat/Km = 200 s^(-1) / 0.1 mM = 2000 s^(-1)/mM
- Q: The initial reaction velocity (V0) of an enzyme-catalyzed reaction is 120 μM/min when the substrate concentration [S] is 0.4 mM. If the Km value is 0.2 mM, determine the Vmax of the reaction.
A: V0 = (Vmax × [S]) / (Km + [S])
120 μM/min = (Vmax × 0.4 mM) / (0.2 mM + 0.4 mM)
Vmax = (120 μM/min × 0.6 mM) / 0.4 mM = 180 μM/min
- Q: An enzyme-catalyzed reaction has a Km of 2 mM. If the initial reaction velocity (V0) is half of the Vmax, calculate the substrate concentration [S].
A: V0 = (Vmax × [S]) / (Km + [S])
0.5 Vmax = (Vmax × [S]) / (2 mM + [S])
[S] = 2 mM
- Q: An enzyme has a kcat value of 300 s^(-1) and is present at a concentration of 5 μM. Calculate the rate of product formation (V0) when the enzyme is saturated with substrate.
A: V0 = kcat × [Enzyme]
V0 = 300 s^(-1) × 5 μM = 1500 μM/s
- Q: If an enzyme has a Vmax of 120 μM/min and a kcat of 500 s^(-1), calculate the Michaelis-Menten constant (Km) for the enzyme-catalyzed reaction.
A: Km = Vmax / kcat
Km = 120 μM/min / 500 s^(-1) = 0.24 mM
- Q: An enzyme-catalyzed reaction has a Vmax of 80 μM/min and a Km of 0.4 mM. Determine the initial reaction velocity (V0) when the substrate concentration [S] is 0.3 mM.
A: V0 = (Vmax × [S]) / (Km + [S])
V0 = (80 μM/min × 0.3 mM) / (0.4 mM + 0.3 mM) = 30 μM/min
- Q: The initial reaction velocity (V0) for an enzyme-catalyzed reaction is 50 μM/min at a substrate concentration [S] of 0.2 mM. Calculate the Vmax of the reaction if the Km value is 0.1 mM.
A: V0 = (Vmax × [S]) / (Km + [S])
50 μM/min = (Vmax × 0.2 mM) / (0.1 mM + 0.2 mM)
Vmax = (50 μM/min × 0.3 mM) / 0.2 mM = 75 μM/min
- Q: An enzyme-catalyzed reaction has a Vmax of 200 μM/min and a kcat of 400 s^(-1). Calculate the Michaelis-Menten constant (Km) for the reaction.
A: Km = Vmax / kcat
Km = 200 μM/min / 400 s^(-1) = 0.5 mM
- Q: The enzyme concentration is 2 μM, and the kcat is 300 s^(-1). Calculate the rate of product formation (V0) when the enzyme is saturated with substrate.
A: V0 = kcat × [Enzyme]
V0 = 300 s^(-1) × 2 μM = 600 μM/s
- Q: An enzyme-catalyzed reaction has a Vmax of 100 μM/min and a Km of 0.1 mM. Calculate the initial reaction velocity (V0) when the substrate concentration [S] is 0.05 mM.
A: V0 = (Vmax × [S]) / (Km + [S])
V0 = (100 μM/min × 0.05 mM) / (0.1 mM + 0.05 mM) = 33.33 μM/min
- Q: An enzyme has a kcat value of 400 s^(-1) and is present at a concentration of 4 μM. Determine the rate of product formation (V0) when the enzyme is saturated with substrate.
A: V0 = kcat × [Enzyme]
V0 = 400 s^(-1) × 4 μM = 1600 μM/s
- Q: Given the Michaelis-Menten constants for an enzyme are Km = 0.3 mM and Vmax = 120 μM/min, calculate the initial reaction velocity (V0) when the substrate concentration [S] is 0.15 mM.
A: V0 = (Vmax ×
[S]) / (Km + [S])
V0 = (120 μM/min × 0.15 mM) / (0.3 mM + 0.15 mM) = 40 μM/min
- Q: The initial reaction velocity (V0) of an enzyme-catalyzed reaction is 80 μM/min when the substrate concentration [S] is 0.1 mM. If the Km value is 0.05 mM, determine the Vmax of the reaction.
A: V0 = (Vmax × [S]) / (Km + [S])
80 μM/min = (Vmax × 0.1 mM) / (0.05 mM + 0.1 mM)
Vmax = (80 μM/min × 0.15 mM) / 0.1 mM = 120 μM/min
- Q: An enzyme-catalyzed reaction has a Vmax of 150 μM/min and a Km of 0.2 mM. Calculate the substrate concentration [S] when the initial reaction velocity (V0) is 100 μM/min.
A: V0 = (Vmax × [S]) / (Km + [S])
100 μM/min = (150 μM/min × [S]) / (0.2 mM + [S])
[S] = (100 μM/min × 0.2 mM) / (150 μM/min – 100 μM/min) ≈ 0.4 mM
- Q: An enzyme has a turnover number (kcat) of 250 s^(-1). If the enzyme concentration is 3 μM, calculate the rate of product formation (V0) when the enzyme is saturated with substrate.
A: V0 = kcat × [Enzyme]
V0 = 250 s^(-1) × 3 μM = 750 μM/s
- Q: If an enzyme has a Vmax of 90 μM/min and a kcat of 600 s^(-1), calculate the Michaelis-Menten constant (Km) for the enzyme-catalyzed reaction.
A: Km = Vmax / kcat
Km = 90 μM/min / 600 s^(-1) = 0.15 mM
- Q: An enzyme-catalyzed reaction has a Vmax of 60 μM/min and a Km of 0.3 mM. Determine the initial reaction velocity (V0) when the substrate concentration [S] is 0.2 mM.
A: V0 = (Vmax × [S]) / (Km + [S])
V0 = (60 μM/min × 0.2 mM) / (0.3 mM + 0.2 mM) = 40 μM/min
- Q: The initial reaction velocity (V0) for an enzyme-catalyzed reaction is 40 μM/min at a substrate concentration [S] of 0.3 mM. Calculate the Vmax of the reaction if the Km value is 0.1 mM.
A: V0 = (Vmax × [S]) / (Km + [S])
40 μM/min = (Vmax × 0.3 mM) / (0.1 mM + 0.3 mM)
Vmax = (40 μM/min × 0.4 mM) / 0.3 mM ≈ 53.33 μM/min
- Q: Given the Michaelis-Menten constants for an enzyme are Km = 0.5 mM and Vmax = 80 μM/min, calculate the initial reaction velocity (V0) when the substrate concentration [S] is 0.3 mM.
A: V0 = (Vmax × [S]) / (Km + [S])
V0 = (80 μM/min × 0.3 mM) / (0.5 mM + 0.3 mM) = 30 μM/min
- Q: An enzyme-catalyzed reaction has a Vmax of 120 μM/min and a Km of 0.2 mM. Calculate the substrate concentration [S] when the initial reaction velocity (V0) is 60 μM/min.
A: V0 = (Vmax × [S]) / (Km + [S])
60 μM/min = (120 μM/min × [S]) / (0.2 mM + [S])
[S] = (60 μM/min × 0.2 mM) / (120 μM/min – 60 μM/min) ≈ 0.2 mM
- Q: An enzyme has a turnover number (kcat) of 400 s^(-1). If the enzyme concentration is 1 μM, calculate the rate of product formation (V0) when the enzyme is saturated with substrate.
A: V0 = kcat × [Enzyme]
V0 = 400 s^(-1) × 1 μM = 400 μM/s
- Q: If an enzyme has a Vmax of 100 μM/min and a kcat of 800 s^(-1), calculate the Michaelis-Menten constant (Km) for the enzyme-catalyzed reaction.
A: Km = Vmax / kcat
Km = 100 μM/min / 800 s^(-1) = 0.125 mM
- Q: The initial reaction velocity (V0) of an enzyme-catalyzed reaction is 50 μM/min when the substrate concentration [S] is 0.1 mM. If the Km value is 0.05 mM, determine the Vmax of the reaction.
A: V0 = (Vmax × [S]) / (Km + [S])
50 μM/min = (Vmax × 0.1 mM) / (0.05 mM + 0.1 mM)
Vmax = (50 μM/min × 0.15 mM) / 0.1 mM = 75 μM/min
- Q: An enzyme-catalyzed reaction has a Vmax of 150 μM/min and a Km of 0.3 mM. Calculate the initial reaction velocity (V0) when the substrate concentration [S] is 0.2 mM.
A: V0 = (Vmax × [S]) / (Km + [S])
V0 = (150 μM/min × 0.2 mM) / (0.3 mM + 0.2 mM) = 66.67 μM/min
- Q: The enzyme concentration is 3 μM, and the kcat is 500 s^(-1). Calculate the rate of product formation (V0) when the enzyme is saturated with substrate.
A: V0 = kcat × [Enzyme]
V0 = 500 s^(-1) × 3 μM = 1500 μM/s
- Q: Given the Michaelis-Menten constants for an enzyme are Km = 0.3 mM and Vmax = 100 μM/min, calculate the initial reaction velocity (V0) when the substrate concentration [S] is 0.2 mM.
A: V0 = (Vmax × [S]) / (Km + [S])
V0 = (100 μM/min × 0.2 mM) / (0.3 mM + 0.2 mM) = 40 μM/min
- Q: An enzyme-catalyzed reaction has a Vmax of 90 μM/min and a Km of 0.1 mM. Determine the substrate concentration [S] when the initial reaction velocity (V0) is 45 μM/min.
A: V0 = (Vmax × [S]) / (Km + [S])
45 μM/min = (90 μM/min × [S]) / (0.1 mM + [S])
[S] = (45 μM/min × 0.1 mM) / (90 μM/min – 45 μM/min) ≈ 0.05 mM
- Q: An enzyme has a turnover number (kcat) of 300 s^(-1). If the enzyme concentration is 2 μM, calculate the rate of product formation (V0) when the enzyme is saturated with substrate.
A: V0 = kcat × [Enzyme]
V0 = 300 s^(-1) × 2 μM = 600 μM/s
- Q: If an enzyme has a Vmax of 80 μM/min and a kcat of 600 s^(-1), calculate the Michaelis-Menten constant (Km) for the enzyme-catalyzed reaction.
A: Km = Vmax / kcat
Km = 80 μM/min / 600 s^(-1) = 0.1333 mM
- Q: The initial reaction velocity (V0) of an enzyme-catalyzed reaction is 40 μM/min when the substrate concentration [S] is 0.2 mM. If the Km value is 0.1 mM, determine the Vmax of the reaction.
A: V0 = (Vmax × [S]) / (Km + [S])
40 μM/min = (Vmax × 0.2 mM) / (0.1 mM + 0.2 mM)
Vmax = (40 μM/min × 0.3 mM) / 0.2 mM = 60 μM/min
- Q: An enzyme-catalyzed reaction has a Vmax of 100 μM/min and a Km of 0.4 mM. Calculate the initial reaction velocity (V0) when the substrate concentration [S] is 0.2 mM.
A: V0 = (Vmax × [S]) / (Km + [S])
V0 = (100 μM/min × 0.2 mM) / (0.4 mM + 0.2 mM) = 33.33 μM/min
- Q: The enzyme concentration is 1 μM, and the kcat is 400 s^(-1). Calculate the rate of product formation (V0) when the enzyme is saturated with substrate.
A: V0 = kcat × [Enzyme]
V0 = 400 s^(-1) × 1 μM = 400 μM/s
- Q: Given the Michaelis-Menten constants for an enzyme are Km =
0.4 mM and Vmax = 120 μM/min, calculate the initial reaction velocity (V0) when the substrate concentration [S] is 0.3 mM.
A: V0 = (Vmax × [S]) / (Km + [S])
V0 = (120 μM/min × 0.3 mM) / (0.4 mM + 0.3 mM) = 42.86 μM/min
- Q: An enzyme-catalyzed reaction has a Vmax of 140 μM/min and a Km of 0.2 mM. Calculate the substrate concentration [S] when the initial reaction velocity (V0) is 70 μM/min.
A: V0 = (Vmax × [S]) / (Km + [S])
70 μM/min = (140 μM/min × [S]) / (0.2 mM + [S])
[S] = (70 μM/min × 0.2 mM) / (140 μM/min – 70 μM/min) ≈ 0.2 mM
- Q: An enzyme has a turnover number (kcat) of 500 s^(-1). If the enzyme concentration is 3 μM, calculate the rate of product formation (V0) when the enzyme is saturated with substrate.
A: V0 = kcat × [Enzyme]
V0 = 500 s^(-1) × 3 μM = 1500 μM/s
- Q: If an enzyme has a Vmax of 90 μM/min and a kcat of 800 s^(-1), calculate the Michaelis-Menten constant (Km) for the enzyme-catalyzed reaction.
A: Km = Vmax / kcat
Km = 90 μM/min / 800 s^(-1) = 0.1125 mM
- Q: The initial reaction velocity (V0) of an enzyme-catalyzed reaction is 60 μM/min when the substrate concentration [S] is 0.1 mM. If the Km value is 0.05 mM, determine the Vmax of the reaction.
A: V0 = (Vmax × [S]) / (Km + [S])
60 μM/min = (Vmax × 0.1 mM) / (0.05 mM + 0.1 mM)
Vmax = (60 μM/min × 0.15 mM) / 0.1 mM = 90 μM/min
- Q: An enzyme-catalyzed reaction has a Vmax of 80 μM/min and a Km of 0.1 mM. Calculate the initial reaction velocity (V0) when the substrate concentration [S] is 0.05 mM.
A: V0 = (Vmax × [S]) / (Km + [S])
V0 = (80 μM/min × 0.05 mM) / (0.1 mM + 0.05 mM) = 26.67 μM/min
- Q: The enzyme concentration is 2 μM, and the kcat is 600 s^(-1). Calculate the rate of product formation (V0) when the enzyme is saturated with substrate.
A: V0 = kcat × [Enzyme]
V0 = 600 s^(-1) × 2 μM = 1200 μM/s
- Q: Given the Michaelis-Menten constants for an enzyme are Km = 0.6 mM and Vmax = 100 μM/min, calculate the initial reaction velocity (V0) when the substrate concentration [S] is 0.4 mM.
A: V0 = (Vmax × [S]) / (Km + [S])
V0 = (100 μM/min × 0.4 mM) / (0.6 mM + 0.4 mM) = 40 μM/min
- Q: An enzyme-catalyzed reaction has a Vmax of 110 μM/min and a Km of 0.2 mM. Calculate the substrate concentration [S] when the initial reaction velocity (V0) is 55 μM/min.
A: V0 = (Vmax × [S]) / (Km + [S])
55 μM/min = (110 μM/min × [S]) / (0.2 mM + [S])
[S] = (55 μM/min × 0.2 mM) / (110 μM/min – 55 μM/min) ≈ 0.1 mM
- Q: An enzyme has a turnover number (kcat) of 350 s^(-1). If the enzyme concentration is 1 μM, calculate the rate of product formation (V0) when the enzyme is saturated with substrate.
A: V0 = kcat × [Enzyme]
V0 = 350 s^(-1) × 1 μM = 350 μM/s
- Q: If an enzyme has a Vmax of 120 μM/min and a kcat of 700 s^(-1), calculate the Michaelis-Menten constant (Km) for the enzyme-catalyzed reaction.
A: Km = Vmax / kcat
Km = 120 μM/min / 700 s^(-1) ≈ 0.1714 mM
- Q: The initial reaction velocity (V0) of an enzyme-catalyzed reaction is 70 μM/min when the substrate concentration [S] is 0.2 mM. If the Km value is 0.1 mM, determine the Vmax of the reaction.
A: V0 = (Vmax × [S]) / (Km + [S])
70 μM/min = (
Vmax × 0.2 mM) / (0.1 mM + 0.2 mM)
Vmax = (70 μM/min × 0.3 mM) / 0.2 mM = 105 μM/min
- Q: An enzyme-catalyzed reaction has a Vmax of 130 μM/min and a Km of 0.3 mM. Calculate the initial reaction velocity (V0) when the substrate concentration [S] is 0.1 mM.
A: V0 = (Vmax × [S]) / (Km + [S])
V0 = (130 μM/min × 0.1 mM) / (0.3 mM + 0.1 mM) = 33.33 μM/min
- Q: The enzyme concentration is 2 μM, and the kcat is 450 s^(-1). Calculate the rate of product formation (V0) when the enzyme is saturated with substrate.
A: V0 = kcat × [Enzyme]
V0 = 450 s^(-1) × 2 μM = 900 μM/s
- Q: Given the Michaelis-Menten constants for an enzyme are Km = 0.2 mM and Vmax = 90 μM/min, calculate the initial reaction velocity (V0) when the substrate concentration [S] is 0.1 mM.
A: V0 = (Vmax × [S]) / (Km + [S])
V0 = (90 μM/min × 0.1 mM) / (0.2 mM + 0.1 mM) = 30 μM/min