Carbonic anhydrase (2.5 × 10-9mol dm-3) catalyses hydration of CO2 in red blood cells at pH 7.1 and 274 K. The rate of the reaction, v (in mol dm-3s-1) reaches its maximum value when varied with the substrate (S) concentration (in mmol dm-3) according to the following equation

\(\frac{1}{v}=4\left\{1+\frac{10}{[S]_0}\right\}\)

The catalytic efficiency of the enzyme (in dm3mol-1s-1) is

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CSIR-UGC (NET) Chemical Science: Held on (18 Sept 2022)
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  1. 4 × 105
  2. 106
  3. 107
  4. 104

Answer (Detailed Solution Below)

Option 3 : 107
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Detailed Solution

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Concept:

  • Enzymes are biological catalysts that accelerate chemical reactions manyfold, without getting consumed in the catalytic process.
    Several biological processes involving the conversion of substrates to products, thus, rely crucially on the catalytic activity of enzymes.
  • For enzyme catalyst reaction always follows the Michaelis-Menten mechanism. The relation between the rate of the reaction and enzyme-substrate is given by, 

\(\frac{1}{v} = \frac{1}{v_{max}} + \frac{k_M}{v_{max}}\frac{1}{[S]}\)......(1) 

where v = velocity of the reaction,

Vmax = maximum rate achieved by the system

[s] = substrate concentration

kM = Michaelis Constant

  • The catalytic efficiency (CE) can be calculated as the ratio of Vmax/Km

Explanation:

  • Given, the rate of the reaction, v (in mol dm-3s-1) reaches its maximum value when varied with the substrate (S) concentration (in mmol dm-3) according to the following equation

\(\frac{1}{v}=4\left\{1+\frac{10}{[S]_0}\right\}\)

or, \(\frac{1}{v}=\left\{4+\frac{40}{[S]_0}\right\}\).......(2)

  • Comparing equation (1) and (2) we got,

Vmax/Km = 0.025 s-1

  • Now, Carbonic anhydrase (2.5 × 10-9mol dm-3) catalyses the hydration of CO2 in red blood cells at pH 7.1 and 274 K.
  • Thus, catalytic efficiency

= 0.025 s-1/2.5 × 10-9mol dm-3

=107 dm3mol-1s-1

Conclusion:-

Hence, the catalytic efficiency of the enzyme (in dm3mol-1s-1) is 107 dm3mol-1s-1.

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