Enzyme Kinetics: Calculating $k_{cat}/K_m$ Ratio
This solution details the calculation of the $k_{cat}/K_m$ ratio for Type II dehydroquinase, using given kinetic parameters.
Dehydroquinase Catalytic Efficiency ($k_{cat}/K_m$) Calculation
The $k_{cat}/K_m$ ratio, also known as the specificity constant, measures an enzyme's catalytic efficiency. It is calculated using the maximum reaction velocity ($V_{max}$), the enzyme's molecular mass, the amount of enzyme used, and the Michaelis constant ($K_m$).
Enzyme Moles Determination
- Molecular mass of enzyme = 18 kDa = $18 \times 10^3 \ g.mol^{-1}$
- Mass of enzyme added = $1.8 \ \mu g = 1.8 \times 10^{-6} \ g$
- Moles of enzyme ($[E]_T$) = $\frac{\text{Mass of enzyme}}{\text{Molecular mass}}$
- $[E]_T = \frac{1.8 \times 10^{-6} \ g}{18 \times 10^3 \ g.mol^{-1}} = 0.1 \times 10^{-9} \ mol = 1.0 \times 10^{-10} \ mol$
Turnover Number ($k_{cat}$) Calculation
- Given $V_{max} = 0.0134 \ \mu mol.min^{-1}$.
- Convert $V_{max}$ to $mol.s^{-1}$: $V_{max} = 0.0134 \times 10^{-6} \ mol.min^{-1} \times \frac{1 \ min}{60 \ s} = \frac{0.0134}{60} \times 10^{-6} \ mol.s^{-1}$
- The turnover number ($k_{cat}$) represents the number of substrate molecules converted to product per enzyme molecule per unit time. It is calculated as $V_{max}$ divided by the molar concentration of the enzyme.
- $k_{cat} = \frac{V_{max}}{[E]_T}$
- $k_{cat} = \frac{\frac{0.0134}{60} \times 10^{-6} \ mol.s^{-1}}{1.0 \times 10^{-10} \ mol} = \frac{0.0134}{60} \times 10^4 \ s^{-1}$
- $k_{cat} \approx 2.233 \ s^{-1}$
$k_{cat}/K_m$ Ratio Calculation
- Given $K_m = 25 \ \mu M = 25 \times 10^{-6} \ M$.
- The ratio $k_{cat}/K_m$ quantifies the enzyme's efficiency at low substrate concentrations.
- $\frac{k_{cat}}{K_m} = \frac{2.233 \ s^{-1}}{25 \times 10^{-6} \ M}$
- $\frac{k_{cat}}{K_m} = \frac{2.233}{25} \times 10^6 \ M^{-1}.s^{-1}$
- $\frac{k_{cat}}{K_m} = 0.08932 \times 10^6 \ M^{-1}.s^{-1}$
- $\frac{k_{cat}}{K_m} = 8.932 \times 10^4 \ M^{-1}.s^{-1}$
The calculated ratio is approximately $8.932 \times 10^4 \ M^{-1}.s^{-1}$.
Important Questions from Enzyme Kinetics and Michaelis Menten Equation
- 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 ________
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)
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____.
An enzymatic reaction exhibits Michaelis-Menten kinetics. For this reaction, on doubling the concentration of enzyme while maintaining [S] >> [$E_o$],
- A single subunit enzyme converts 420 µmoles of substrate to product in one minute. The activity of the enzyme is __________ $ \times 10^{-6} $ Katal.
