The catalytic efficiency of two different enzymes can be compared by the:
The catalytic efficiency of two different enzymes can be compared by the:
formation of the product
the $\mathrm{pH}$ of optimum value
the $\mathrm{K}_{\mathrm{m}}$ value
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 $=-1 / K_m$
