Enzyme catalysis and Michaelis Menton Equation | Chemical Kinetics | MCQs for CSIR NET & GATE - Download PDF

Vijay Ishwar
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Enzyme catalysis and Michaelis Menton Equation
MCQs for CSIR NET & GATE

Chemical Kinetics
Topic wise Asignment-7

QUIZ        PDF

Q1. The expression for the equilibrium constant (Keq) for the enzyme catalyzed reaction given below, is

            (a) k1k3 / k2k4

            (b) k1k2 / k3k4

            (c) k2k3 / k1k4

            (d) k1k4 / k2k3

Q2. In an enzyme catalyzed reaction

k2= 3.42 x 104 s-1. If [E]o= 1 x 10-2 mol dm-3, the magnitude of maximum velocity and turn over number using Michaelis-Menten kinetics are

a)     3.42 x 102 mol dm-3 s-1; 3.42 x 104 s-1

b)    3.42 x 106 mol dm-3 s-1; 3.42 x 104 s-1

c)     3.42 x 104 mol dm-3 s-1; 3.42 x 106 s-1

d)    3.42 x 104 mol dm-3 s-1; 3.42 x 102 s-1

Q3. The slope and intercept obtained from 1/r against 1/[S] of an enzyme catalyzed reaction are 300 and 2 x 105, respectively. The Michaelis-Menten constants of the enzyme in this reaction is

            (a) 5 x 106 M

            (b) 5 x 10-6 M

            (c) 1.5 x 103 M

            (d) 1.5 x 10-3 M

Q4. For an enzyme-substrate reaction

The slope and the intercept of the plot between 1/r and 1/[S] are 10-2 s and 102  M-1 s, respectively. If Eo= 10-6 M and k-1/k2= 1000, the value of k1 will be close to (in units of M-1 s-1) [r is the rate of reaction and Eo is the initial concentration of the enzyme]

            (a) 1 x 1011

            (b) 1 x 104

            (c) 1 x 108

            (d) 1 x 106

Q5. The Lineweaver-Burk Plot of (initial rate)-1 vs (initial substrate concentration)-1 for an enzyme catalyzed reaction following Michaelis-Menten mechanism, the y-intercept is 5000 M-1s. If the initial enzyme concentration is 1 x 10-9 M, the turnover number is

            (a) 2.5 x 103

            (b) 1.0 x 104

            (c) 2.5 x 104

            (d) 2.0 x 105

Q6. For an enzyme catalyzed reaction, a Lineweaver-Burk plot gave the following data:

Slope= 40s, intercept= 4 (mmol dm-3 s-1)-1

If the initial concentration of enzyme is 2.5 x 10-9 mol dm-3, what is the catalytic efficiency (dm3 mol-1 s-1) of the reaction?

            (a) 105

            (b) 106

            (c) 107

            (d) 104

Q7. For an enzyme-substrate reaction the Michaelis constant is 0.042 mol dm-3. The rate of this reaction is 2.45 x 10-5 mol dm-3 s-1 when the substrate concentration is 0.89 mol dm-3. The maximum velocity (mol dm-3 s-1) of this enzyme catalysis is

            (a) 2.57 x 10-5

            (b) 2.45 x 10-5

            (c) infinite

            (d) 1.31 x 10-5

Q8. Vmax and Km for an enzyme catalyzed reaction are 2.0 x 10-3 Ms-1 and 1.0 x 10-6 M, respectively. The rate (s-1) of the reaction when the substrate concentration is 1.0 x 10-6 is

            (a) 3.0 x 10-3

            (b) 1.0 x 10-3

            (c) 2.0 x 10-3

            (d) 0.5

Q9. For an enzyme substrate reaction, a plot between 1/v and 1/[S] yields a slope of 40 s. If the enzyme concentration is 2.5 μM, then the catalytic efficiency (L mol-1 s-1) of the enzyme is

            (a) 40

            (b) 10-4

            (c) 107

            (d) 104

Ans Key

Q1.b Q2.a Q3.d Q4.a Q5.d Q6.a Q7.a Q8.b Q9.d 

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