You are characterizing a new enzyme isolated and purified in the laboratory. If the maximum velocity of the enzyme is $1800 \text{ } \mu moles \text{ L}^{-1} \text{min}^{-1}$ and the total concentration of the enzyme in the reaction mixture is $1.5 \mu \text{M}$, then the turnover number of the enzyme is _______ $\text{s}^{-1}$. (answer in integer)
Enzyme Turnover Number Calculation
The turnover number ($k_{cat}$) signifies the maximum rate at which a single enzyme molecule converts substrate into product per unit time. It is determined when the enzyme is operating at its maximum velocity ($V_{max}$) relative to the total enzyme concentration ($[E]_T$).
Enzyme Turnover Number Formula
The turnover number is calculated using the following formula:
$k_{cat} = \frac{V_{max}}{[E]_T}$
Provided Data
- Maximum velocity, $V_{max} = 1800 \text{ } \mu \text{moles } \text{L}^{-1} \text{ min}^{-1}$
- Total enzyme concentration, $[E]_T = 1.5 \mu \text{M}$
Step 1: Convert Units
The question requires the turnover number in $\text{s}^{-1}$. First, convert the maximum velocity ($V_{max}$) from $\text{min}^{-1}$ to $\text{s}^{-1}$.
Given that $1 \text{ min} = 60 \text{ s}$:
$V_{max} = \frac{1800 \text{ } \mu \text{moles } \text{L}^{-1}}{1 \text{ min}} \times \frac{1 \text{ min}}{60 \text{ s}} = 30 \text{ } \mu \text{moles } \text{L}^{-1} \text{ s}^{-1}$
The enzyme concentration $[E]_T$ is $1.5 \mu \text{M}$, which is equivalent to $1.5 \mu \text{moles } \text{L}^{-1}$.
Step 2: Calculate Turnover Number
Substitute the converted $V_{max}$ and the enzyme concentration $[E]_T$ into the formula:
$k_{cat} = \frac{30 \text{ } \mu \text{moles } \text{L}^{-1} \text{ s}^{-1}}{1.5 \text{ } \mu \text{moles } \text{L}^{-1}}$
$k_{cat} = 20 \text{ s}^{-1}$
Result
The calculated turnover number is 20 $\text{s}^{-1}$. The question asks for the answer as an integer.
Therefore, the turnover number is 20 $\text{s}^{-1}$.
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] >> [$E_o$],
- 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 ____________________ $\times 10^4 \ M^{-1}.s^{-1}$.
