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Question

The catalytic efficiency for an enzyme is definedas

The correct answer is
$\frac{k_{cat}}{K_m}$

The catalytic efficiency of an enzyme is a measure of how efficiently an enzyme converts a substrate into a product. It is defined by the ratio of two key kinetic parameters: the catalytic constant (\( k_{cat} \)) and the Michaelis constant (\( K_m \)). Mathematically, it is expressed as:

\(\frac{k_{cat}}{K_m}\)

Let's understand why this particular ratio is used to define the catalytic efficiency:

  1. Catalytic Constant (\( k_{cat} \)): This represents the turnover number of an enzyme, indicating the number of substrate molecules converted to product per enzyme molecule per unit time when the enzyme is fully saturated with substrate. A higher \( k_{cat} \) signifies a faster turnover.
  2. Michaelis Constant (\( K_m \)): This constant reflects the affinity of the enzyme for its substrate. A lower \( K_m \) indicates a higher affinity, as the enzyme can achieve its half-maximal velocity at a lower substrate concentration.

The ratio \(\frac{k_{cat}}{K_m}\) effectively combines these two properties to determine how well an enzyme functions under conditions of low substrate concentration, thereby providing a measure of catalytic efficiency.

Now, let's examine the options to justify the correct answer:

  • \( k_{cat} \): This only describes the turnover rate and not the efficiency considering substrate affinity.
  • \( \frac{V_{max}}{k_{cat}} \): This expression does not relate to catalytic efficiency but could imply the amount of enzyme active sites contributing to \( V_{max} \).
  • \( \frac{k_{cat}}{K_m} \): This is the correct definition of catalytic efficiency as it considers both the turnover rate and substrate affinity.
  • \( \frac{k_{cat}}{V_{max}} \): This ratio does not meaningfully describe catalytic efficiency. \( V_{max} \) is impacted by enzyme concentration, not by inherent enzyme properties.

Thus, the expression that defines catalytic efficiency is \(\frac{k_{cat}}{K_m}\), which is why the correct answer is this option.

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Important Questions from Enzyme Kinetics and Michaelis Menten Equation

  1. An enzyme (E) catalyzes the biochemical reaction $A \rightarrow B$ with $k_{cat}$ equal to $500 s^{-1}$. If the initial reaction velocity ($V_0$) is $10 \mu M.s^{-1}$ at the total enzyme concentration $[E_t]$ of 30 nM and substrate concentration $[A]$ of $40 \mu M$, the value of $K_m$ (in $\mu M$) is ________
  2. Within the Michaelis-Menten framework, the ratio of $v_0/V_{max}$ 

    when $[S] = 20 \times K_m$ is _________. 

    (Round off to two decimal places)

  3. The enzyme $\alpha$-amylase used in starch hydrolysis has an affinity constant ($K_m$) value of $0.005$ M. To achieve one-fourth of the maximum rate of hydrolysis, the required starch concentration in mM (rounded off to two decimal places) is____.

  4. An enzymatic reaction exhibits Michaelis-Menten kinetics. For this reaction, on doubling the concentration of enzyme while maintaining [S] >> [$E_o$],

  5. In an assay of the type II dehydroquinase of molecular mass 18 kDa, it is found that the $V_{max}$ of the enzyme is $0.0134 \ \mu mol.min^{-1}$ when $1.8 \ \mu g$ enzyme is added to the assay mixture. If the $K_m$ for the substrate is $25 \ \mu M$, the $k_{cat}/K_m$ ratio will be ____________________ $\times 10^4 \ M^{-1}.s^{-1}$.
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