Most Expected Questions of Life Science: Enzymes and Enzyme Kinetics! (Download PDF)

Most Important Questions of Enzymes and Enzyme Kinetics (Download PDF)

1. Enzyme Activity can be calculated by which unit?

1. Katal. 
2. IU. 
3. Mol per minute per ml. 
4. All 

2. What is known as Michaelis Menten Equation?

1. V_Max= V0. (s)/ Km+(S) 
2. S= V_max. (Km)/ Vo+(S) 
3. Vo=V_max. (s)/ Km+(S) 
4. Km=V_max. (s)/ Vo+(S) 

3. In a Lineweaver burk plot, enzyme initial rate and substrate concentration is given, graph is shown in figure, concentration of enzyme is maintained at constant level that is 1µM (10-6).

Calculate the Km Value and Vmax of the reaction.

1. Km=10µM, Vmax=5µM/min 
2. Km=5µM, Vmax=5µM/min 
3. Km=5µM, Vmax=4µM/min 
4. Km=4µM, Vmax=5µM/min 

4. Which statement is true for allosteric enzyme?

1. It regulates the binding of molecule at one site that directly affect binding of another molecule in distinct site. 
2. Generally, this enzyme is multi subunit protein in which one sub unit perform catalytic function and another performs regulatory function. 
3. Both
4. None of the above

5. What is true about Michaelis constant (Km)?

1. It indicates the substrate concentration at which rate of reaction is half with reference to its maximum value. 
2. It indicates the substrate concentration at which rate of reaction is double with reference to its maximum value. 
3. It indicates that the substrate concentration is free from rate of reaction is double with reference to its maximum value. 
4. None 

6. In a catalytic reaction why does Vmax not get increase while adding substrate continuously?

1. In an enzyme, Vmax is saturated with enzyme 
2. In an enzyme Vmax is saturated with Substrate 
3. Both 
4. None of the above

7. In the picture graph shown below, between substrate concentration and rate of reaction upper line shows rate of reaction without inhibitor and lower shows rate of reaction with inhibitors, what type of inhibition is this?

1. Competitive inhibition 
2. Non-competitive inhibition 
3. Mixed inhibition 
4. None 

8. In a process of glycolysis where the concentration of glucose continuously increases, what will be the state of rate of reaction of glucokinase?

1. Increase. 
2. Decrease. 
3. Can’t Change. 
4. None

9. Four statements are given about enzyme inhibition. Choose the correct one:

1. In pure competitive inhibition, the inhibitor binds to free enzyme and enzyme-substrate complex with equal affinity. 
2. In pure non-competitive inhibition or mix inhibition, the inhibitor binds to free enzyme and enzyme-substrate complex with equal affinity. 
3. Both 
4. None

10. Four statements are given about mechanism of enzyme :

I) Enzyme acts through the reduction of activation energy.

II) Enzyme acts through decrease in the pH.

III) Enzyme acts through increase in pH.

IV) Enzyme acts through enhancement of activation energy.

Choose the correct option

1. I 
2. II & III 
3. III 
4. IV 

Solutions

Solution-1

Enzyme activity can be represented by all the above three units.

Solution-2

In enzyme kinetics, Vo=V_max. (s)/ Km+(S), is known as Michaelis Menten Equation, where Vo =initial rate of reaction

V_max=maximal rate of reaction

Km= Michaelis Menten constant

S=concentration of substrate

Solution-3

Calculation of Km = Km is calculated by X intercept, that is -1/Km (in Lineweaver burk plot)

In given plot X intercept is -2

Thus -1/-2=1/2=0.5

According to plot 0.5*10^-5=5*10^-6M

Km=5µM

Km=1/2Vmax

Vmax=2Km

Vmax=5*2=10

Vmax=10µM/Min

Solution-4

Allosteric is a type of enzyme that regulate the binding of molecules which affect distinct site and it is a multiunit enzyme which regulate the catalytic functions.

Solution-5

Michaelis constant represented as Km value, and defined as concentration of substrate at which half of active site is occupied, and indicates the higher affinity to enzyme for substrate, lower the Km value.

Solution-6

Vmax is known as maximum velocity in an enzyme-catalysed reaction. A reaction reaches at this point when enzyme’s active site is occupied by substrate. So, at this point enzyme is saturated by substrate and no change is observed in Vmax.

Solution-7

In this graph, observe that Vmax is same for both type of reactions; this indicates that inhibitor has no effect on maximal growth and thus  this is a type of competitive inhibition

Solution-8

In an enzyme catalysed reaction, the relation is usually hyperbolic; it indicates as the concentration of substrate increases the rate of reaction  also increases.

Solution-9

In pure none competitive inhibition or also known as mixed inhibition, Inhibitors are able to bind enzyme-substrate complex and free enzyme both end equally.

Solution-10

Enzyme makes reaction rate faster and for faster reaction rate activation energy should be less.

Answer keys

1. D
2. C
3. B
4. C
5. A
6. B
7. A
8. A
9. B
10. A

Download Important Questions on Enzyme and Enzyme Kinetics  - Download PDF Here

Check Out: 

• Study Notes for CSIR NET Life Science - (Download PDF)
• Study Notes for CSIR NET Chemical Science - Download PDF Here!

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