Category: CSIR NET Life Science Previous Year Questions and Solution on Enzymes Kinetics

CSIR NET Life Science Previous Year Questions and Solutions on Enzyme Kinetics

Enzyme kinetics is one of the most high-scoring topics in CSIR NET Life Science. Every year, at least a few direct or conceptual questions are asked from Michaelis-Menten equation, Lineweaver-Burk plots, enzyme inhibition, and allosteric regulation. If you’re aiming for a top rank, mastering this section is non-negotiable.

In this article, we’ll go through:

Why enzyme kinetics is important for CSIR NET.
Some previous year questions on enzyme kinetics.
Step-by-step solutions and explanations.
Pro tips to crack enzyme-based questions quickly in the exam.

Why Enzyme Kinetics is Important in CSIR NET Life Science?

Enzyme kinetics doesn’t just test memory—it tests understanding. The examiners want to see if you can:

Interpret graphs like Michaelis-Menten and Lineweaver-Burk.
Differentiate between competitive, non-competitive, and uncompetitive inhibition.
Apply logic to enzyme efficiency and catalytic turnover problems.

Since enzyme kinetics is a foundation for biochemistry, it overlaps with multiple units such as metabolism, molecular biology, and biotechnology—making it a high-return topic.

Graph-Based Question Example

Lineweaver-Burk Plot Analysis:

Competitive inhibition → slope increases, intercept same.
Non-competitive inhibition → slope increases, intercept changes.
Uncompetitive inhibition → parallel lines with same slope.

These are frequently asked in graph-based MCQs.

Pro Tips to Crack Enzyme Kinetics Questions

Memorize equations (Michaelis-Menten, Lineweaver-Burk).
Understand inhibitor effects with a small chart—this saves time in the exam.
Practice previous year questions regularly.
Don’t just solve numericals—analyze patterns of questions asked.
Revise enzyme concepts from trusted sources like Lehninger Biochemistry and Let’s Talk Academy CSIR NET notes.

Final Words

Enzyme kinetics may look tricky at first, but once you understand the logic, it becomes one of the easiest scoring topics in CSIR NET Life Science. Practicing previous year questions ensures you won’t be surprised in the exam hall.

If you’re serious about cracking CSIR NET, make enzyme kinetics a regular part of your study routine. Start with basic numericals, move to inhibition-based problems, and finally master graph interpretations.

👉 Keep revising, keep practicing, and you’ll see your accuracy in enzyme questions go up.

(MODEL PAPER) 79. A biochemist purifies a new enzyme, generating the following purification table. The most effective purification step is (1) iv. (2) iii. (3) v (4) ii.

Affinity Chromatography: The Most Effective Step in Enzyme Purification

$(DEC 2013 GU) 78. You have isolated spleen cells and purified your desired enzyme from the spleen cell extract by a sequence of steps involving ammonium sulfate fractionation and gel chromatography. You found an increase in the specific activity of the enzyme in the purified fraction compared to cell free extract and made the following speculations: A. Cell free extract contained inhibitor bound enzyme B. Structure of the enzyme was modified during the course of purification C. Purification steps gradually removed the inhibitors D. The final fraction contained concentrated enzyme preparation Which one of the following combination is appropriate? (1) A, B and D (2) A, C and D (3) A, B and C (4) B, C and D$

Understanding Enzyme Purification: Role of Inhibitors and Enzyme Concentration on Specific Activity

(NOV 2020-11) 77. In regulating the quantity of enzyme, its degradation plays a pivotal role. Following statements are made to represent the degradation of enzymes in the 26S proteasome. A. The active sites of proteolytic subunits face exterior of the proteasome cylinder B. The active sites of proteolytic subunits face interior of the proteasome cylinder C. Degrading enzymes are targeted to exterior of proteasome by covalent attachment of one or more molecules of ubiquitin D. Degrading enzymes are targeted to interior of proteasome by covalent attachment of one or more molecules of ubiquitin Which one of the following combinations of statements represent correct mode of enzyme degradation? (1) A and B (2) B and C (3) B and D (4) A and C

Mechanism of Enzyme Degradation by the 26S Proteasome: Orientation of Active Sites and Ubiquitin Targeting

(DEC 2004) 76. The best example of regulation of enzyme by reversible covalent modification is- (1) Phosphorylation (2) Acetylation (3) Glycosylation (4) proteolytic cleavage

Phosphorylation: The Premier Example of Enzyme Regulation by Reversible Covalent Modification

(SEPT 2022-1) 75. Dixon plot is used to study the enzyme inhibition by plotting various expressions of velocity (v) and inhibitor concentration [l] on the X-axis (column A) and Y-axis (column B) as given below: Which one of the following options is the correct combination from columns A and B to draw the Dixon plot? 1. Column A -iv, Column B- i 2. Column A -i, Column B- ii 3. Column A- ii, Column B- iii 4. Column A- iii, Column B- iv

Using Dixon Plot for Enzyme Inhibition: Correct Axes Selection Explained

(NOV 2020-11) 74. A plot with which one of the following axes is drawn to exhibit enzyme inhibition kinetics applying Dixon's plot?

How to Use Dixon’s Plot for Enzyme Inhibition Kinetics: The 1/Vi1/Vi vs. [I][I] Approach

The correct answer is (3) Non-competitive inhibitor. Title Interpreting Double Reciprocal Equations: Identifying Non-Competitive Inhibition in Enzyme Kinetics Meta Description Learn how to distinguish non-competitive inhibition using changes in Lineweaver-Burk plot equations. See why only Vmax changes—classic for non-competitive inhibitors—in detailed biochemical analysis. Slug noncompetitive-inhibition-lineweaver-burk-double-reciprocal-equation Introduction Enzyme kinetics is foundational for understanding metabolic regulation and drug mechanisms. Identifying different inhibition types is pivotal when analyzing altered Lineweaver-Burk plots or double reciprocal equations. Non-competitive inhibition is characterized by a change in the y-intercept (reflecting Vmax) while Km remains unchanged. This article shows how to recognize non-competitive inhibition from the mathematical forms and the biochemical impact it has on enzyme activity. Double Reciprocal Equations Overview Given: Without inhibitor (X): 1 v 0 = K m V m a x ( 1 [ S ] ) + 1 V m a x v 0 1 = V max K m ( [S] 1 )+ V max 1 With inhibitor (X): 1 V 0 = K m V m a x ( 1 [ S ] ) + 1 V m a x ( 1 + [ X ] K X ) V 0 1 = V max K m ( [S] 1 )+ V max 1 (1+ K X [X] ) Analysis: Identifying Inhibition Type In the presence of inhibitor, the slope K m V m a x V max K m (linked to Km) is unchanged. The y-intercept is increased by a factor ( 1 + [ X ] K X ) (1+ K X [X] ), decreasing apparent Vmax but leaving Km unaltered. This matches the action of a non-competitive inhibitor, which binds at a site distinct from the substrate binding site and reduces enzyme activity regardless of substrate concentration. In competitive inhibition, the slope (dependent on Km) would increase; in uncompetitive inhibition, both Km and Vmax are affected proportionally. Mechanistic Understanding Non-competitive inhibitors bind to either free enzyme or the ES complex at an allosteric site. This reduces the overall number of catalytically competent enzyme molecules, decreasing Vmax. Importantly, Km (substrate affinity) remains unchanged—substrate can bind just as well, but some enzyme can’t convert it to product. Summary Table Inhibitor Type Double Reciprocal Equation Change Km Vmax Effect on Plot/Activity Competitive Slope increases, y-intercept unchanged ↑ — Can be overcome by substrate Non-competitive y-intercept increases, slope unchanged — ↓ Can’t be overcome by substrate Uncompetitive Both slope and y-intercept change ↓ ↓ Lower Km and Vmax Visual Summary Non-competitive inhibition: Lines (with and without inhibitor) have the same x-intercept (Km unchanged) but different y-intercepts (Vmax decreased). Conclusion The molecule X is a non-competitive inhibitor, as indicated by the unchanged Km and decreased Vmax represented in the altered double reciprocal equation. This is a classic kinetic pattern, crucial for enzyme pharmacology and molecular biology.

Interpreting Double Reciprocal Equations: Identifying Non-Competitive Inhibition in Enzyme Kinetics

(SEPT 2022-11) 72. An enzyme has a Km of 5 x 10-5 M and a Vmax of 100 µmoles.lit-1. min-1(Km is the Michaelis constant and Vmax is the maximal velocity). What is the velocity in the presence of 1 x 10-4 M substrate and 2 x 10-4 M competitive inhibitor, given that the Ki for the inhibitor is 2 x 10-4 M? (1) 0.005 µmoles. Lit-1. Min-1 (2) 50 µmoles. Lit-1. Min-1, (3) 5 µmoles. Lit-1. Min-1, (4) 500 µmoles. Lit-1. Min-1,

Calculating Enzyme Velocity with Competitive Inhibition: A Practical Example

(FEB 2022-1) 71. Following statements are made about uncompetitive inhibition of an enzyme: A. Uncompetitive inhibitor binds to both free enzyme as well as an enzyme- substrate complex. B. Addition of uncompetitive inhibitor lowers the Vmax of the reaction. C. Apparent KM of the enzyme is lowered. D. Apparent KM of the enzyme remains unchanged. Which one of the following option represents the correct combination of the statements? (1) B and C (2) A and C (3) A and B (4) A and D

Uncompetitive Enzyme Inhibition: Binding Mechanism and Effects on KmKm and VmaxVmax

(DEC 2012) 70. An enzyme catalyzed reaction was measured in the presence and absence of an inhibitor for an uncompetitive inhibition, (1) only Km is increased (2) only is Vmax decreased (3) both Km and Vmax are decreased (4) both Km and Vmax are not affected

Uncompetitive Inhibition: Effects on Enzyme Kinetics and Determination of Km and Vmax

(DEC 2013 GU) 69. If the product of an enzyme binds to the enzyme- substrate complex to exhibit its activity through a decrease in both Km and Vmax, this type of inhibition is called (1) competitive inhibition (2) non-competitive inhibition. (3) uncompetitive inhibition. (4) partially-competitive inhibition.

Demystifying Competitive Enzyme Inhibition: Substrate Competition and Reversibility

(DEC 2016) 68. A researcher was investigating the substrate specificity of two different enzymes, X and Y,. on the same substrate. Both the enzymes were subjected to treatment with either heat or an inhibitor which inhibits the enzyme activity. Following are the results heat obtained where, a= inhibitor treatment, b = heat treatment, c= control Which of the following statements is correct? (1) Only protein X is specific for the substrate, S (2) Only protein Y is specific for the substrate, S (3) Both X and Y are specific for the substrate, S (4) Both X and Y are non-specific for the substrate, S

Interpreting Enzyme Specificity Using Activity Profiles After Inhibitor and Heat Treatments

(DEC 2018) 67. For a reversible non-competitive inhibition of an enzyme, choose the plot that you would use to determine Km: _______ No inhibitor - - - - - - + Inhibitor

Lineweaver-Burk Plot Analysis for Km Determination in Reversible Non-Competitive Enzyme Inhibition

(JUNE 2002) 66. Which of following is true for non-competitive inhibition? (1) Increases Vmax (2) Decreases Vmax (3) Increases Km (4) Decreases Km

Non-Competitive Enzyme Inhibition: Effects on Kinetic Parameters VmaxVmax and KmKm

(MODEL PAPER) 65. Which one of the following enzyme reaction represents noncompetitive initiation?

Noncompetitive Enzyme Inhibition: Reaction Schemes, Mechanism, and Impact on Kinetics

(DEC 2001) 64. False statement of competitive inhibition is- (1) Structure same as substrate (2) Inhibits substrate binding (3) Binds to active site (4) Reaction cannot be favorably biased by increasing substrate concentration

Understanding Competitive Inhibition: How Substrate Concentration Overcomes Inhibition

(JUNE 2004) 63. During inhibition of enzyme action Vmax remains unchanged while Km is increased. Inhibition is- (1) Competitive (2) Non-competitive (3) Un-competitive (4) Allosteric

Understanding Enzyme Inhibition: Why Competitive Inhibition Increases KmKm but Not VmaxVmax

(JUNE 2016) 62. Penicillin acts as a suicide substrate. Which one of the following steps of catalysis does a suicide inhibitor affect? (1) K1 (2) K2 (3) K3 (4) K4

Mechanism-Based (Suicide) Inhibition in Enzyme Catalysis: Penicillin’s Site of Action Explained

(DEC 2019 ASSAM) 61. Choose the correct statement from the following: (1) Iodoacetamide inactivates an enzyme by reaction with a critical serine residue at neutral PH. (2) Proline racemase causes isomerisation of L- proline to D-proline. Ribose will be an appropriate transition state analog. (3) Tosyl-1-phenylalanine chloromethyl ketone binds at the active site of chymotrypsin and modifies an essential arginine residue. (4) binds to triose phosphate isomerase at the active site and covalently modifies a glutamic acid residue required for enzyme activity.

Covalent Enzyme Modification and Transition State Analogs: Key Mechanisms in Enzyme Inhibition

(DEC 2017) 60. Indicate the INCORRECT statement from the following: (1) Allosteric enzymes function through reversible noncovalent binding of allosteric modulators or effectors. (2) Monoclonal antibodies that catalyze hydrolysis of esters or carbonates can be produced. (3) Enzymes are not inhibited irreversibly by heavy metals such as Hg2+, Ag+ (4) Acid phosphatases hydrolyze biological phosphate esters at ̴ pH 5.0.

Enzyme Inhibition by Heavy Metals and Catalytic Activity of Monoclonal Antibodies: Clarifying Common Misconceptions

(DEC 2014) 59. Reaction products inhibit catalysis in enzymes by (1) covalently binding to the enzyme. (2) altering the enzyme structure (3) occupying the active site. (4) form a complex with the substrate.

How Reaction Products Inhibit Enzyme Catalysis: Mechanisms and Impact

(DEC 2014) 58. Enzyme parameters of four isozymes is given below. These isozymes are localized in different tissues. In liver the substrate concentration is 0.2 micromolar. The liver isozyme is likely to be (1) A (2) B (3) C (4) D

Selecting the Optimal Isozyme for Low Substrate Concentrations: Liver Enzyme Kinetics Explained

(JUNE 2019) 57. Choose the INCORRECT statement from the following statements made for an enzyme- catalyzed reaction (1) The kinetic properties of allosteric enzyme do not diverge from Michaelis-Menten behaviour. (2) In feedback inhibition, the product of a pathway inhibits an enzyme of the pathway (3) An antibody that binds tightly to the analog of the transition state intermediate of the reaction S → P, would promote formation of P when the analog is added to the reaction. (4) An enzyme with Kcat = 1.4 x 104 s-1 and Km = 9 x 10-5 M has activity close to the diffusion controlled limit.

Correcting Misconceptions in Enzyme Kinetics: Allosteric Behavior and Enzymatic Efficiency

(FEB 2022-1) 56. A schematic of a metabolic pathway is shown below. Under which of the following conditions would stoichiometric amounts of end products K and L be obtained if a concerted feedback inhibition mechanism were in operation? (1) K inhibits F→G and L inhibits F→H; D→E is inhibited at equal amounts of K and L (2) D→E is inhibited at equal amounts of K and L; K inhibits F→H and L inhibits F→G (3) D→E is inhibited at equal amount of G and H; K inhibits F→H and L inhibits F→G (4) K inhibits F→H and L inhibits F→G

Concerted Feedback Inhibition in Metabolic Pathways: Achieving Stoichiometric End Product Balance

(DEC 2019 ASSAM) 55. An allosteric enzyme has two heterotropic effectors, X and Y. The allosteric constant, L for the enzyme in the absence of effector molecules is 180. For the X- saturated form, the value of L increases from 180 to 1200, while for Y-saturated form it decreases to 60. What kind of effector molecules are X and Y? (1) X and Y both are positive regulators. (2) X is a negative regulator while Y is a positive regulator. (3) X is a positive regulator while Y is a negative regulator. (4) X and Y are not allosteric regulators.

Understanding Allosteric Regulation: Identifying Positive and Negative Effectors Through the Allosteric Constant LL

(DEC 2006) 54. If an enzyme obeying Hills reaction shows negative cooperativity. It means (1) Binding of substrate to any one site of multi-subunit enzyme decreases affinity for other substrate to other sub-units (2) Binding of substrate to any one site of single subunit enzyme decreases affinity for other substrate (3) Binding of substrate to any one site of multi-subunit enzyme increases affinity for other substrate to other sub-units (4) Binding of substrate to any one site of multi-subunit enzyme makes enzyme non-functional

Understanding Negative Cooperativity in Multi-Subunit Enzymes: Mechanism and Implications

(NOV 2020-11) 53. The Hill equation and its plot describe the following enzyme kinetic behaviours A. Saturation Kinetics B. Cooperative Kinetics C. Log Vi/(Vmax — Vi) versus Log [s] D. Log (Vmax — Vi)/Vi versus Log [s]-1 Which one of the following combination represent correct descriptions? (1) A and C (2) B and C (3) B and D (4) A and D

Understanding the Hill Equation and Hill Plot: Describing Cooperative Enzyme Kinetics

$(JUNE 2023-11) 52. The plot below has two curves (A, B) that show the fractional occupancy of hemoglobin and myoglobin by oxygen as a function of the amount of oxygen. The two reactions are i. E + S ⇄ ES and ii. E + nS ⇄ ESn. where S is O2 and E is myoglobin or haemoglobin The equations that could be used to fit the two curves are: Yo2 is the fraction of oxygen-binding sites occupied by oxygen. pO2 is partial pressure of oxygen. From the options given below, select the option with the right curve (A, B), reaction (i, ii) and equation/s (l, ll, lll, lV) that describe oxygen binding to haemoglobin and myoglobin. (1) Myoglobin: curve A, reaction i, equations lll and lV. Hemoglobin: curve B, reaction ii, equations l and ll. (2) Myoglobin: curve B, reaction i, equations ll and lV. Hemoglobin: curve A, reaction ii, equations l and lll. (3) Myoglobin: curve A, reaction ii, equations lll and lV. Hemoglobin: curve B, reaction i, equations l and ll. (4) Myoglobin: curve A, reaction ii, equations l and ll. Hemoglobin: curve B, reaction i, equations lll and lV.$

Common Misconceptions in Enzyme Kinetics: Understanding Allosteric Behavior, Energy Change, Inhibition, and Catalytic Efficiency

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Understanding Michaelis Constant (Km) and Its Relationship with Substrate Concentration and Initial Velocity in Enzyme Kinetics

(FEB 2022-1) 25. The following statements are being made to define the Michaelis constant (KM). It is: A. Independent of enzyme concentration [E] and substrate concentration [S] B. Equal to the dissociation constant when the [ES] complex dissociates more rapidly than product formation C. Equal to the dissociation constant when product formation is more rapid than [ES] complex dissociation D. An intrinsic property of an enzyme and does not depend on pH, temperature and ionic strength Which one of the following combination of statements is correct? (1) A and B only (2) A, B and D only (3) C and D only (4) A and D only