Enzymes and Enzyme kinetics MCQ Quiz in मराठी - Objective Question with Answer for Enzymes and Enzyme kinetics - मोफत PDF डाउनलोड करा
Last updated on Mar 23, 2025
Latest Enzymes and Enzyme kinetics MCQ Objective Questions
Top Enzymes and Enzyme kinetics MCQ Objective Questions
Enzymes and Enzyme kinetics Question 1:
In a Lineweaver-Burk plot, the slope of the line is equal to
Answer (Detailed Solution Below)
Enzymes and Enzyme kinetics Question 1 Detailed Solution
The correct option is:1
Explanation:
- A Lineweaver-Burk plot, or double reciprocal plot, is a graphical representation of enzyme kinetics. It is derived by taking the reciprocal of both sides of the Michaelis-Menten equation:
- \([ \frac{1}{v} = \frac{K_M}{V_{max}} \cdot \frac{1}{[S]} + \frac{1}{V_{max}} ]\)
Where:
- ( v ) is the reaction velocity
- ( [S] ) is the substrate concentration
- ( Km) is the Michaelis constant
- ( Vmax ) is the maximum reaction velocity
In this double reciprocal plot:
- The y-axis represents \(( \frac{1}{v} )\)
- The x-axis represents \(( \frac{1}{[S]} )\)
The equation of a straight line is ( y = mx + b ), where:
- ( m ) is the slope of the line
- ( b ) is the y-intercept
Comparing the Lineweaver-Burk equation to the equation of a straight line:
- The slope (m) is \(( \frac{K_M}{V_{max}} )\)
- The y-intercept (b ) is \(( \frac{1}{V_{max}} )\)
Thus, the slope of the Lineweaver-Burk plot is equal to \(( \frac{K_M}{V_{max}} )\).
Enzymes and Enzyme kinetics Question 2:
The turnover number (kcat) of an enzyme is defined as:
Answer (Detailed Solution Below)
Enzymes and Enzyme kinetics Question 2 Detailed Solution
The correct option is: 1
Explanation:
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The turnover number (kcat) of an enzyme is a measure of its catalytic efficiency. It represents the maximum number of substrate molecules that a single enzyme molecule can convert to product per second when the enzyme is fully saturated with substrate. This value is often referred to as the enzyme's "rate constant" for its catalytic reaction.
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The substrate concentration at Vmax refers to the Michaelis constant (KM) and is not related to the turnover number, which is about enzyme efficiency, not substrate concentration.
-
The enzyme concentration at saturation is incorrect because turnover number is a measure of enzyme activity, not its concentration.
-
The velocity at KM is related to enzyme kinetics but does not define the turnover number, which is specifically the number of substrate molecules converted per enzyme molecule per second at maximum efficiency.
- The turnover number is a useful parameter for comparing different enzymes or understanding an enzyme's efficiency. A higher turnover number means the enzyme is more efficient at converting substrate to product. This value is often used alongside the Michaelis constant (KM) in calculating catalytic efficiency (kcat/KM), which is another important measure of enzyme performance.
Enzymes and Enzyme kinetics Question 3:
The enzyme hexokinase catalyzes the phosphorylation of glucose to glucose-6-phosphate. This reaction is an example of:
Answer (Detailed Solution Below)
Enzymes and Enzyme kinetics Question 3 Detailed Solution
The correct option is:3
Explanation:
- The enzyme hexokinase catalyzes the phosphorylation of glucose to form glucose-6-phosphate. This reaction involves the transfer of a phosphate group from ATP to glucose. This process is known as a group transfer reaction because it involves the transfer of a functional group (in this case, a phosphate group) from one molecule (ATP) to another molecule (glucose).
The reaction catalyzed by hexokinase: \([ \text{Glucose} + \text{ATP} \rightarrow \text{Glucose-6-phosphate} + \text{ADP} ]\)
- Group transfer reactions involve the transfer of a functional group, such as phosphate, from one molecule to another.
- Oxidation reactions involve the loss of electrons from a molecule.
- Reduction reactions involve the gain of electrons by a molecule.
- Hydrolysis reactions involve the breaking down of a molecule using water.
| Type of Reaction | Bond Involved | Enzyme Involved | Other Properties |
|---|---|---|---|
| Oxidation | C-H, C-C, C-N (electron transfer) | Dehydrogenase, oxidase | Involves the loss of electrons, often accompanied by the addition of oxygen or removal of hydrogen. |
| Reduction | C-H, C-C, C-N (electron transfer) | Reductase, hydrogenase | Involves the gain of electrons, typically accompanied by the addition of hydrogen or removal of oxygen. |
| Group Transfer | Phospho-, methyl, acetyl, etc. | Kinase, methyltransferase | Transfer of functional groups (e.g., phosphate) between molecules. Crucial in metabolic processes like glycolysis. |
| Hydrolysis | C-O, C-N (breaking bonds) | Hydrolase | Involves the breaking of a bond with the addition of water, such as in the breakdown of polymers into monomers. |
| Condensation | C-O, C-N (bond formation) | Ligase, synthetase | Formation of a bond between two molecules with the release of water. Important in polymer synthesis. |
| Isomerization | C-C, C-H (molecular rearrangement) | Isomerase | Rearrangement of atoms within a molecule to form isomers. Often seen in metabolic pathways. |
| Ligase Reactions | C-C, C-O, C-N (bond formation) | Ligase | Joins two molecules, often using energy from ATP hydrolysis. Plays a role in DNA repair and replication. |
Enzymes and Enzyme kinetics Question 4:
A propionate kinase enzyme utilizes two substrates, propionate and acetate, with Km for propionate being half that of acetate. Which one of the following options about the rate of the reaction at very low substrate concentrations is correct?
Answer (Detailed Solution Below)
Enzymes and Enzyme kinetics Question 4 Detailed Solution
The correct answer is The rate of propionate utilization is double that of acetate.
Explanation:
Michaelis-Menten kinetics, which describes the rate of enzyme-catalyzed reactions at different substrate concentrations.
The Michaelis-Menten equation is \(v = \frac{V_{max} [S]}{K_m + [S]}\)
where:
- v is the initial velocity of the reaction,
- Vmax is the maximum velocity,
- S is the substrate concentration,
- Km is the Michaelis constant.
Given that the for propionate is half that of acetate, it implies that the enzyme has a higher affinity for propionate than for acetate. Let's denote the
\(K_m^{\text{propionate}} = \frac{K_m^{\text{acetate}}}{2} \)
- At very low substrate concentration [S} <
For propionate, the reaction rate is \(v_{\text{propionate}} \propto \frac{[S_{\text{propionate}}]}{K_m^{\text{propionate}}} \)
- Since \(K_m^{\text{propionate}} = \frac{K_m^{\text{acetate}}}{2} \)
- \(v_{\text{propionate}} \propto \frac{[S_{\text{propionate}}]}{\frac{K_m^{\text{acetate}}}{2}} = 2 \times \frac{[S_{\text{propionate}}]}{K_m^{\text{acetate}}} \)
For acetate, the reaction rate is \(v_{\text{acetate}} \propto \frac{[S_{\text{acetate}}]}{K_m^{\text{acetate}}} \)
Thus, for the same substrate concentration, the rate of propionate utilization will be double that of acetate because the enzyme has a higher affinity (lower
Thus, the correct answer is The rate of propionate utilization is double that of acetate.
Enzymes and Enzyme kinetics Question 5:
For an enzyme following Michaelis-Menten kinetics, when [S]=KM then, the velocity v is
[S] is substrate concentration, KM is Michaelis constant, Vmax is maximal velocity
Answer (Detailed Solution Below)
Enzymes and Enzyme kinetics Question 5 Detailed Solution
The correct answer is 0.5 ×Vmax
Explanation:
The basic form of the Michaelis-Menten equation is: \(v = \frac{V_{max} [S]}{K_m + [S]}\)
where:
- v is the initial velocity of the reaction,
- Vmax is the maximum velocity,
- S is the substrate concentration,
- Km is the Michaelis constant.
If [S] = Km, we can substitute Km in place of [S] in the Michaelis-Menten equation:
- \(v = \frac{V_{max} [S]}{K_m + [S]}\)
- \( v = \frac{V_{max} K_m}{K_m + K_m}\)
- \(v = \frac{V_{max} K_m}{2 K_m}\)
- \( v = \frac{V_{max}}{2}\)
So, when the substrate concentration [S] is equal to the Michaelis constant Km , the velocity v is half of the maximum velocity Vmax. This is because the Michaelis-Menten equation simplifies to show that the enzyme is operating at half of its maximum capacity under these conditions.
Therefore, the correct answer is 0.5 ×Vmax
Enzymes and Enzyme kinetics Question 6:
For a reversible non competitive inhibition of an enzyme, choose the plot that you would use to determine Km:
Answer (Detailed Solution Below)
Enzymes and Enzyme kinetics Question 6 Detailed Solution
The correct answer is Option 3.
Explanation:
In Option 3, the given plot is Lineweaver Burk plot. The Lineweaver Burk plot is a graphical tool that simplifies the analysis of enzyme kinetics data, allowing researchers to determine Km and Vmax from linear regression analysis of reciprocal transformed data points.
Key PointsFor reversible non competitive inhibition of an enzyme, you would use a Lineweaver Burk plot also known as a double reciprocal plot to determine the apparent Km.
In a Lineweaver Burk plot, the x axis represents the reciprocal of substrate concentration 1/[S], and the y axis represents the reciprocal of reaction velocity 1/V.
The equation for the Lineweaver Burk plot is:
1/V= Km/Vmax . 1/[S] + 1/Vmax
Where:
- V is the reaction velocity.
- Km is the Michaelis Menten constant.
- Vmax is the maximum reaction velocity.
- S is the substrate concentration.
In the presence of non competitive inhibition, both Km and Vmax are affected. However, the apparent Km the apparent Michaelis constant can be determined from the Lineweaver Burk plot as the x intercept of the line. The y intercept of the line gives 1/Vmax.
By plotting the reciprocal of reaction velocity against the reciprocal of substrate concentration for various substrate concentrations, you can determine the apparent Km and Vmax from the slope and the intercepts, respectively, of the resulting line.
Additional Information
The Lineweaver Burk plot, also known as the double reciprocal plot, is a graphical method used to analyze enzyme kinetics data. It provides a linear representation of the Michaelis Menten equation, which describes the relationship between substrate concentration and reaction velocity in enzyme catalyzed reactions. The Lineweaver Burk plot is particularly useful for determining kinetic parameters such as the Michaelis constant Km and the maximum reaction velocity Vmax.
Michaelis Menten Equation:
The Michaelis Menten equation describes the relationship between substrate concentration [S] and reaction velocity V in enzyme catalyzed reactions:
Where:
- V is the reaction velocity.
- Vmax is the maximum reaction velocity the rate of the reaction when the enzyme is fully saturated with substrate.
- Km is the Michaelis constant the substrate concentration at which the reaction velocity is half of Vmax.
- [S] is the substrate concentration.
Linearization:
Taking the reciprocal of both sides of the Michaelis Menten equation yields:
This equation has the form of a straight line equation y=mx+b, where y is 1/V, m is Km/Vmax, x is 1/[S], and b is 1/Vmax.
Determining Kinetic Parameters:
- From the Lineweaver Burk plot, the slope of the line is Km/Vmax, the y intercept is 1/Vmax, and the x intercept is −1/Km.
- Km can be determined by taking the negative reciprocal of the x intercept −1/Km.
- Vmax can be determined from the y intercept 1/Vmax.
- These parameters provide valuable insights into the enzyme's affinity for its substrate Km and its catalytic efficiency Vmax.
Assumptions:The Lineweaver Burk plot assumes that the enzyme catalyzed reaction follows Michaelis Menten kinetics, which may not always be the case in complex systems or under certain conditions.
Conclusion: In summary, the Lineweaver Burk plot is a graphical tool that simplifies the analysis of enzyme kinetics data, allowing researchers to determine Km and Vmax from linear regression analysis of reciprocal transformed data points.
Enzymes and Enzyme kinetics Question 7:
Which of the following Statement is correct about enzymes?
Answer (Detailed Solution Below)
Enzymes and Enzyme kinetics Question 7 Detailed Solution
Key PointsEnzyme
- An enzyme is a biocatalyst that increases the rate of chemical reaction without itself being changed in the overall process.
- Virtually all cellular reactions or processes are mediated by enzymes.
- Enzymes have several properties that make them unique.
- Most (but not all) enzymes are proteins.
- With the exception of a small group of catalytic RNA molecules, all Enzymes are highly specific.
- They are specialized proteins and have a high degree of specificity for their substrates.
- Enzymes exhibit enormous catalytic power.
- It increases the rate of a reaction by lowering the activation energy.
- Enzymes do not change the equilibrium state of a biochemical reaction.
- It changes only the rate at which equilibrium is achieved.
- Proteinaceous enzymes can be divided into two general classes: simple enzymes, consist entirely of amino acids and conjugated enzymes, consist of proteins as well as non-protein components.
- The non-protein component is called a cofactor, which is required for catalytic activity.
- Removal of cofactor from conjugated enzyme leaves only protein component, called an apoenzyme, which generally is biologically inactive.
- The complete, biologically active conjugated enzyme (simple enzyme plùs cofactor) is called a holoenzyme.
- A cofactor can be linked to the protein portion of the enzyme either covalently or non-covalently.
- Some cofactors are simple metal ions and other cofactors are complex organic compounds, which are also called coenzymes.
- Many coenzymes are vitamins or contain vitamins as part of their structure.
- Some coenzymes are only transiently associated with a given enzyme molecule, so that they function as cosubstrates.
- Coenzymes which are tightly associated with the protein covalently or non-covalently are called prosthetic group.
Explanation:
- Enzyme increase reaction rates without altering the chemical equilibrium between reactants and products.
- Note that the enzyme (E) is not altered by the reaction, so the chemical equilibrium remains unchanged, determined solely by the thermodynamic properties of S and P.
Hence the correct answer is option 1
Enzymes and Enzyme kinetics Question 8:
In non-competitive inhibition:
Answer (Detailed Solution Below)
Enzymes and Enzyme kinetics Question 8 Detailed Solution
- Competitive Inhibition.
- Non-competitive Inhibition.
- Uncompetitive Inhibition.
Non-Competitive Competition -
- In non-competitive inhibition, the inhibitor bind to the enzyme at a site other than the active site or binding alters enzyme's three-dimensions configuration and blocks the reaction.
- There are two types of non competitive inhibition pure and mixed.
- In pure non-competitie inhibition, substrate and inhibitor bind at different sites on enzyme and binding of inhibitor does not affect binding of substrat.
- The inhibitior binds to eiher free enzyme or the ES complex.
- Hence, inhibition is not reversed by increasing the concentration of substrates.
- Pure non-competitive inhibition occurs if KI=KI'. In this type of inhibition, Vmax decreases and Km stays constant.
- In mixed non-competitive inhibition, the binding of inhibitor with enzyme influences the binding of substrate with enzyme.
- Either the binding sites for inhibitor and substrate are near one another or conformational changes in enzyme caused by inhibitor affect substrate binding.
- In this case, KI, and KI' are not equal and both KM and Vmax are altered by the presence of inhibitor.
- Pure non-competitive inhibition occurs if KI=KI'. In this type of inhibition, Vmax decreases and Km stays constant.
Hence option 1 is correct.
Enzymes and Enzyme kinetics Question 9:
Which of the following is CORRECT regarding aminotransferases?
Answer (Detailed Solution Below)
Enzymes and Enzyme kinetics Question 9 Detailed Solution
Concept:
- The first step in the catabolism of most L-amino acids, once they have reached the liver, is removal of the α-amino groups, promoted by enzymes called aminotransferases or transaminases.
- In these transamination reactions the α-amino group is transferred to the α-carbon atom of α-ketoglutarate, leaving behind the corresponding α-keto acid analog of the amino acid.
- There is no net deamination (loss of amino groups) in these reactions, because the α-ketoglutarate becomes aminated as the α-amino acid is deaminated.
- The effect of transamination reactions is to collect the amino groups from many different amino acids in the form of L-glutamate.
- Glutamate then functions as an amino group donor for biosynthetic pathways or for excretion pathways that lead to the elimination of nitrogenous waste product Cells contain different types of aminotransferases.
- Many are specific for α-ketoglutarate as the amino group acceptor but differ in their specificity for the L-amino acid. The enzymes are named for the amino group donor (alanine aminotransferase and aspartate aminotransferase, for example). The reactions catalyzed by aminotransferases are freely reversible, having an equilibrium constant of about 1.0 (ΔG′ ≈ 0 kJ/mol).
- All aminotransferases have the same prosthetic group and the same reaction mechanism. The prosthetic group is pyridoxal phosphate (PLP), the coenzyme form of pyridoxine, or vitamin B6.
- As a coenzyme in the glycogen phosphorylase reaction, but its role in that reaction is not representative of its usual coenzyme function. Its primary role in cells is in the metabolism of molecules with amino groups.
- Pyridoxal phosphate functions as an intermediate carrier of amino groups at the active site of aminotransferases.
- It undergoes reversible transformations between its aldehyde form, pyridoxal phosphate, which can accept an amino group, and its aminated form, pyridoxamine phosphate, which can donate its amino group to an α-keto acid.
- Pyridoxal phosphate is generally covalently bound to the enzyme’s active site through an aldimine (Schiff base) linkage to the ε-amino group of a Lys residue.
- Pyridoxal phosphate participates in a variety of reactions at the α, β, and γ carbons (C-2 to C-4) of amino acids.
- The highly conjugated structure of PLP (an electron sink) permits delocalization of the negative charge.
- Aminotransferases are classic examples of enzymes catalyzing bimolecular Ping-Pong reactions, in which the first substrate reacts and the product must leave the active site before the second substrate can bind.
- Thus the incoming amino acid binds to the active site, donates its amino group to pyridoxal phosphate, and departs in the form of an α-keto acid.
- The incoming α-keto acid then binds, accepts the amino group from pyridoxamine phosphate, and departs in the form of an amino acid.
Explanation:
Option A- Incorrect : Aminotransferases usually require α-ketoglutarate and glutamate as a reacting pair (please note: glutamate and not glutamine). Transamination reactions are reversible reactions. Aminotransferases can catalyze reactions involving essential amino acids as well.
Option B - Incorrect:
In these transamination reactions the α-amino group is transferred to the α-carbon atom of α-ketoglutarate, leaving behind the corresponding α-keto acid analog of the amino acid. There is no net deamination (loss of amino groups) in these reactions, because the α-ketoglutarate becomes aminated as the α-amino acid is deaminated. The effect of transamination reactions is to collect the amino groups from many different amino acids in the form of L-glutamate.
Option C- Incorrect : The reactions catalyzed by aminotransferases are freely reversible, having an equilibrium constant of about 1.0 (ΔG′ ≈ 0 kJ/mol).
Option D- Correct: Aminotransferases have the same prosthetic group and the same reaction mechanism. The prosthetic group is pyridoxal phosphate (PLP), the coenzyme form of pyridoxine, or vitamin.
Hence option d is correct
Enzymes and Enzyme kinetics Question 10:
Blocking enzyme action by blocking its active site is known as
Answer (Detailed Solution Below)
Enzymes and Enzyme kinetics Question 10 Detailed Solution
Key Points
- Reversible and irreversible inhibitors are chemicals which bind to an enzyme to suppress its activity.
- One method to accomplish this is to almost permanently bind to an enzyme. These types of inhibitors are called irreversible.
- However, other chemicals can transiently bind to an enzyme, these are called reversible.
- Reversible inhibitors either bind to an active site (competitive inhibitors), or to another site on the enzyme (non-competitive inhibitors).
Competitive Inhibitors
- Competitive inhibitors compete with the substrate at the active site and therefore increase Km (the Michaelis-Menten constant).
- However, Vmax is unchanged because, with enough substrate concentration, the reaction can still complete.
- The graph plot of enzyme activity against substrate concentration would be shifted to the right due to the increase of the Km, whilst the Lineweaver-Burke plot would be steeper when compared with no inhibitor.
Non-Competitive Inhibitors
- Non-competitive inhibitors bind to another location on the enzyme and as such decrease VMAX.
- However, KM is unchanged.
- This is demonstrated by a lower maximum on a graph plotting enzyme activity against substrate concentration and a higher y-intercept on a Lineweaver-Burke plot when compared with no inhibitor.
Allosteric inhibition:
- When an inhibitor binds to the enzyme, all the active sites of the protein complex of the enzyme undergo conformational changes so that the activity of the enzyme decreases.
- In other words, an allosteric inhibitor is a type of molecule which binds to the enzyme specifically at an allosteric site.
Feedback inhibition:
- This is extremely useful to limit the amount of an enzyme’s product, as the product can then go on to inhibit the same type of enzyme to ensure the amount of product is not excessive. This is known as feedback inhibition.
- For example, ATP allosterically inhibits pyruvate kinase to prevent increased formation of pyruvate, so less ATP is eventually formed.
- Additionally, phosphofructokinase is allosterically inhibited by citrate, an intermediate of the TCA cycle. This means that glycolysis will be limited when there is high ATP generation from the TCA cycle.
Hence, the correct answer is option 3.