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Enzymes and Enzyme Kinetics Questions and Answers

11.

Which one of the following acts as a catalyst in an animal cell?

Exam Year(s)

RSMSSB Livestock Assistant - 2025, RSMSSB - 2025,

Answer: D

The catalysts of an animal cell are proteins.

Enzymes, which speed up the chemical reactions of the cell without being used up, are almost all proteins. Digestive enzymes, the enzymes of respiration and DNA polymerase are examples.

DNA stores genetic information, ATP is the energy currency that many reactions consume, and the cytoplasm is the medium in which reactions occur. None of them is itself the catalyst.

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12.

Who proposed the lock and key model of Enzyme?

Exam Year(s)

RSMSSB - 2025, NHM Rajasthan Pharmacist - 2025, Rajasthan NHM Pharma Assistant - 2025,

Answer: C

The lock and key model was proposed by Emil Fischer.

In 1894 the German chemist Emil Fischer suggested that an enzyme's active site has a rigid shape that fits only its specific substrate, as a lock fits only its key. This explained enzyme specificity.

Daniel Koshland later (1958) proposed the induced fit model, in which the active site changes shape as the substrate binds. Alfred Nobel invented dynamite and founded the Nobel Prizes. Alice Ball developed an injectable treatment for leprosy from chaulmoogra oil.

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13.

What is the value of \(V_{max}\) for an enzyme which follows simple Michaelis-Menten Kinetics, if \(V_{o}\) = 2 μ mol min-1 at 10 km ?

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: B

\(V_{max}\) is 2.2 μmol min-1.

Given \(v_{0} = 2\) μmol min-1 when [S] = 10 Km (the paper's '10 km'). The Michaelis-Menten equation is:

\[ v_{0} = \frac{V_{max}[S]}{K_{m} + [S]} \]

\[ \begin{aligned} 2 &= \frac{V_{max} \times 10K_{m}}{11K_{m}} \\ &= \frac{10}{11}V_{max} \\ V_{max} &= \frac{2 \times 11}{10} = 2.2 \end{aligned} \]

As a check, at a substrate concentration ten times Km the enzyme runs at 10/11 of its maximum, so Vmax must be a little above 2. The value 1.1 is half of 2.2, 3.3 has no basis in the equation, and 11 comes from multiplying by 11 and forgetting to divide by 10.

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Answer: A

Isoenzymes are oligomeric proteins which have different physico-chemical properties and catalyse the same reaction.

Isoenzymes catalyse the same reaction but differ in structure, and therefore in charge, electrophoretic mobility, kinetics and regulation. The classic example is lactate dehydrogenase, a tetramer built from two kinds of subunit, H and M, combined in five ways (H4, H3M, H2M2, HM3, M4); creatine kinase, a dimer, has three isoenzymes. These oligomeric examples are the basis of the textbook definition.

Options C and D are wrong because isoenzymes catalyse the same reaction, and option B ignores the oligomeric make-up of the standard examples.

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15.

Which one of the following is not a characteristic of competitive inhibition of an enzyme ?

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: B

The statement that is not a characteristic is There is no change in Km.

A competitive inhibitor resembles the substrate and binds the active site, so it competes with substrate. More substrate is needed to reach half the maximum rate: the apparent Km rises. Vmax is unchanged, because enough substrate outcompetes the inhibitor completely.

An increase in Km, binding at the active site, and relief by raising the substrate concentration are all features of competitive inhibition. An unchanged Km is the mark of pure non-competitive inhibition, where Vmax falls instead.

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16.

This type of enzyme regulation is irreversible in nature :

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: C

The irreversible type is Zymogen activation.

A zymogen (proenzyme) such as trypsinogen or chymotrypsinogen is activated by proteolytic cleavage of one or more peptide bonds. The cut cannot be rejoined, so the active enzyme cannot return to the inactive form; it can only be destroyed or blocked by an inhibitor.

Allosteric regulation depends on reversible binding of effectors. Covalent modification, such as phosphorylation, is reversed by enzymes such as phosphatases, and cAMP-mediated regulation ends when phosphodiesterase breaks down the cAMP.

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17.

The enzyme acetyl CoA carboxylase belongs to which class of enzyme ?

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: D

Acetyl CoA carboxylase is a ligase (EC class 6).

Ligases join two molecules using the energy of ATP hydrolysis. Acetyl CoA carboxylase couples bicarbonate to acetyl CoA to form malonyl CoA, the committed step of fatty acid synthesis, and the reaction is driven by ATP through a biotin carrier:

\[ \begin{aligned} &\text{acetyl CoA} + \mathrm{HCO_3^-} + \text{ATP} \\ &\rightarrow \text{malonyl CoA} + \text{ADP} + \mathrm{P_i} \end{aligned} \]

Carboxylases that use ATP like this are classed as carbon-carbon ligases (EC 6.4.1.2). Transferases move a group between molecules, hydrolases split bonds with water, and lyases add or remove groups without ATP and without hydrolysis.

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18.

Both the substrates A and B must be present at the enzyme active site simultaneously is a characteristic feature of ...................

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: C

The requirement is a feature of single displacement reactions.

In single displacement (sequential) bisubstrate reactions, ordered or random, both substrates bind to form a ternary complex E-A-B before any product leaves, so A and B must be at the active site together.

Double displacement (ping-pong) reactions work differently: substrate A binds, transfers a group to the enzyme and leaves as product P before B binds and receives the group. A and B are never on the enzyme together, as in aminotransferases. So the feature is not true of all bisubstrate reactions, nor of double displacement ones, and 'Both (B) and (C)' fails with B.

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19.

Which of the following is not a method of enzyme immobilization ?

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: C

Absorption is not a method of enzyme immobilization.

Enzymes are immobilised by four standard methods: adsorption onto a surface such as alumina or ion-exchange resin, through weak bonds; covalent binding to a support; entrapment in a gel lattice such as calcium alginate or polyacrylamide, or encapsulation in membranes; and cross-linking of enzyme molecules to each other with reagents such as glutaraldehyde.

Absorption, taking a substance into the bulk of another as a sponge takes up water, is not one of them. The question turns on the one-letter difference between adsorption, which is a method, and absorption.

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20.

When the velocity of enzyme activity is plotted against substrate concentration, which of the following is obtained ?

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: A

The plot of velocity against substrate concentration is a hyperbolic curve.

For an enzyme obeying Michaelis-Menten kinetics,

\[ v = \frac{V_{max}[S]}{K_m + [S]} \]

At low [S] velocity rises almost in proportion to [S]; at high [S] the enzyme is saturated and \(v\) levels off towards \(V_{max}\). That shape is a rectangular hyperbola.

A straight line with positive slope is what the Lineweaver-Burk double reciprocal plot gives, and a negative slope is the Eadie-Hofstee plot; neither is the direct v against [S] plot. Allosteric enzymes give a sigmoid curve, not a parabola.

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