MCQ on Enzymes

Enzymes refers to proteins that take part in our bodies’ metabolism, or chemical reactions, go more quickly. They construct some substances and deconstruct others. Enzymes are found in all living organisms. Enzymes are produced naturally in our bodies. Although, enzymes can be found in both synthetic and natural foods. Digestive assistance is one of the most significant functions of enzymes. Digestion represents the process of converting food into energy. Enzymes can be found in our saliva, pancreas, intestines, and stomach, for example. Fats, proteins, and carbohydrates are all broken down by them. Further, these nutrients are used by enzymes for cell development and repair.

Q1.A _________ is a biocatalyst that boosts the pace of a reaction without changing the reaction itself.

1. Silicon dioxide

2. Aluminum oxide

3. Enzyme

4. Hydrogen peroxide

Ans. The correct answer is option ‘c’ enzyme. The single choice is an enzyme, which is a biocatalyst that catalyses the chemical reaction without changing, whereas the other options are catalysts that increase or decrease the rate of reaction depending on their concentration.

Q2. The nature of an enzyme is 

1. Carbohydrate 

2. Lipid

3. Vitamin

4. Protein

Ans. The correct answer is option ‘d’ protein. Except for the catalytic RNA molecule, all enzymes are proteins. An enzyme’s catalytic activity is determined by its original protein structure. The catalytic activity of an enzyme is lost when it is denatured.

Q3. What information does a Lineweaver-Burk plot provide that a typical Michaelis-Menten plot does not?

1. Vi

2. Km

3. Vmax

4. None of these answers

Ans. The correct option is ‘d’ in none of these answers. The information in both graphs is identical. A Michaelis-Menten plot depicts the connection between substrate concentration and initial reaction rate (Vi versus [S]). A Lineweaver-Burk plot depicts the relationship between the inverses of the same two variables, but it is much easier to see crucial data on one. There are points of interest in the x-intercept, y-intercept, and slope. The Lineweaver-Burk plot, on the other hand, has the disadvantage of being more prone to inaccuracy if the accumulated data is flawed.

Q4. Which of the following modifications to an enzyme-catalysed process will change the Vmax?

1. Addition of a non-competitive inhibitor

2. Addition of a competitive inhibitor

3. Increasing substrate to supraphysiological concentrations

4. None of these options

Ans. The correct answer is option ‘a’ addition of a non-competitive inhibitor. The addition of a non-competitive inhibitor is the sole way to change the Vmax . This inhibitor reduces the quantity of free enzyme available to catalyse the reaction, lowering the Vmax by lowering the effective enzyme concentration. The Km is affected by the addition of a competitive inhibitor, but not the Vmax. Once saturation is attained, increasing the substrate concentration has little impact.

Q5. An enzyme that facilitates a process is known as a catalyst. What does a catalyst do to the reaction in terms of free energy to encourage it to proceed?

1. Catalysts increase the rate of the reaction by reducing the free energy of the transition state, which lowers the activation energy. 

2. Catalysts increase the amount of energy contributed to the reaction through heat from the reaction environment, thus increasing the rate of the reaction. 

3. Catalysts induce a global increase of entropy to the product state of the reaction, making it more favorable and thus occurring at a faster rate. 

4. Catalysts increase the rate of reaction by decreasing the free energy of the transition state, which increases the activation energy. 

Ans. The correct answer is option ‘a’ catalysts increase the rate of the reaction by reducing the free energy of the transition state, which lowers the activation energy. The reactants must travel through high-energy transition states before evolving into the products during a reaction. Because the activation energy of the chemical process has been lowered, the catalyst reduces the free energy of this transition state, making it ‘easier’ for the reactant to undergo the chemical reaction.

Q6. When an inhibitor binds to an enzyme at a place other than the active site, but only when the enzyme and substrate are already bound in complex, which of the following best represents the situation?

1. Competitive inhibition 

2. Allostery 

3. Uncompetitive inhibition

4. Non-competitive inhibition 

Ans. The correct answer is option ‘c’ uncompetitive inhibition. The creation of an enzyme-substrate complex provides an alternate site for an inhibitor to bind to the enzyme. Because the inhibitor and substrate are not competing for the same binding site on the enzyme, this method is considered uncompetitive.

Q7. What is the major mechanism through which enzymes boost reaction rates?

1. They decrease the stability of the transition state.

2. They lower the activation energy needed in the reaction.

3. They decrease the internal energy of the final product.

4. They decrease the reverse reaction rate and increase the forward reaction rate.

Ans. The correct answer is option ‘b’ they lower the activation energy needed in the reaction. Enzymes affect reaction rates by lowering the amount of energy required for the reaction to continue. As a result, the activation energy of the enzyme will drop. It is important to note that an enzyme increases both the forward and reverse reaction rates. If this were not the case, the uncatalysed reaction would produce more product, and enzymes have no effect on reaction equilibrium constants.

Q8. In a chemical process, which of the following modifications cannot be achieved by an enzyme?

1. Change in enthalpy

2. Change in reverse reaction rate

3. Change in activation energy

4. Change in forward reaction rate

Ans. The correct answer is option ‘a’ change in enthalpy. An enzyme is a type of biological catalyst that speeds up a reaction. This is done by decreasing the activation energy required to initiate the reaction. The reaction’s equilibrium, on the other hand, remains unaffected. This means that the presence of an enzyme has no effect on enthalpy or entropy.

Q9. Which of the following best describes the class of enzymes that break bonds by generating a new double bond or ring structure rather than by hydrolysis or oxidation?

1. Ligases

2. İsomerases

3. Transferases

4. Lyases

Ans. The correct answer is option ‘d’ lyases. Lyases is the correct answer. The forward reaction of this class of enzymes only requires one substrate.Ligases create bonds between molecules, isomerases rearrange atoms, kinases phosphorylate molecules, and transferases transfer or connect functional groups from one molecule to another.