Q19 / 200 · NEET 2005 · Botany

The catalytic efficiency of two different enzymes can be compared by the:

  • A
    formation of the product
  • B
    the $\mathrm{pH}$ of optimum value
  • Correct answer
    C
    the $\mathrm{K}_{\mathrm{m}}$ value
  • D
    molecular size of the enzyme.

Solution

The catalytic efficiency of two different enzymes can be compared by comparing their $\mathrm{K}_{\mathrm{m}}$ value or Michaelis Menten constant. The Michaelis constant is the substrate concentration at which the reaction rate is at half-maximum. The $\mathrm{K}_{\mathrm{m}}$ describes the affinity of enzyme for a substrate molecule. Greater the affinity lower is the $\mathrm{K}_{\mathrm{m}}$ value and sooner the $\mathrm{V}_{\max }$ can be attained and vice versa. Related Theory $K_m$ is the concentration of substrate which permits the enzyme to achieve half Vmax. An enzyme with a high $K_m$ has a low affinity for its substrate, and requires a greater concentration of substrate to achieve Vmax. The relationship is defined by the MichaelisMenten equation: $$ v=V_{\max } /\left(I+\left(K_m /[S]\right)\right) $$ The Lineweaver-Burk double reciprocal plot rearranges the Michaelis-Menten equation as: $$ 1 / v=1 / V_{\max }+K_m / V_{\max } \times 1 /[S] $$ Plotting 1/v against 1/[S] give a straight line: where: $y$ intercept $=1 / V_{\max }$ gradient $=K_m / V_{\max }$ $x$ intercept $=-1 / K_m$ Figure

Hence, the correct answer is option: 'C'.

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